Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB 111976677X, 9781119766773

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Introduction to Convective Heat Transfer

Introduction to Convective Heat Transfer A Software-Based Approach Using Maple and MATLAB®

Nevzat Onur Emeritus Professor, Department of Mechanical Engineering Gazi University, Turkey

This edition first published 2023 © 2023 John Wiley & Sons, Inc All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, except as permitted by law. Advice on how to obtain permission to reuse material from this title is available at http://www.wiley.com/go/permissions. The right of Nevzat Onur to be identified as the author of this work has been asserted in accordance with law. Registered Office John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, USA For details of our global editorial offices, customer services, and more information about Wiley products visit us at www.wiley.com. Wiley also publishes its books in a variety of electronic formats and by print-on-demand. Some content that appears in standard print versions of this book may not be available in other formats. Limit of Liability/Disclaimer of Warranty MATLAB® is a trademark of The MathWorks, Inc. and is used with permission. The MathWorks does not warrant the accuracy of the text or exercises in this book. This work’s use or discussion of MATLAB® software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLAB® software. While the publisher and authors have used their best efforts in preparing this work, they make no representations or warranties with respect to the accuracy or completeness of the contents of this work and specifically disclaim all warranties, including without limitation any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives, written sales materials or promotional statements for this work. This work is sold with the understanding that the publisher is not engaged in rendering professional services. The advice and strategies contained herein may not be suitable for your situation. You should consult with a specialist where appropriate. The fact that an organization, website, or product is referred to in this work as a citation and/or potential source of further information does not mean that the publisher and authors endorse the information or services the organization, website, or product may provide or recommendations it may make. Further, readers should be aware that websites listed in this work may have changed or disappeared between when this work was written and when it is read. Neither the publisher nor authors shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. Library of Congress Cataloging-in-Publication Data applied for Hardback ISBN: 9781119766766 [LCCN 2022058932] Cover Design: Wiley Cover Image: © engel.ac/Shutterstock Set in 9.5/12.5pt STIXTwoText by Straive, Chennai, India

To my wife: Ayfer To our children: Hakan and Funda Erdem and Gökcen To our grandchildren: Ada, Aslan, and Mila

Science is the most genuine guide for spiritual and intellectual enlightenment in life Mustafa Kemal Atatürk

vii

Contents Preface xv About the Author xvii About the Companion Website

xviii

1 1.1 1.2 1.3 1.4 1.5 1.5.1 1.5.2 1.6 1.7 1.8 1.9 1.9.1 1.9.2 1.10 1.10.1 1.10.2 1.11 1.12

Foundations of Convective Heat Transfer 1 Fundamental Concepts 1 Coordinate Systems 1 The Continuum and Thermodynamic Equilibrium Concepts 2 Velocity and Acceleration 3 Description of a Fluid Motion: Eulerian and Lagrangian Coordinates and Substantial Derivative 4 Lagrangian Approach 4 Eulerian Approach 5 Substantial Derivative 7 Conduction Heat Transfer 10 Fluid Flow and Heat Transfer 11 External Flow 11 Velocity Boundary Layer and Newton’s Viscosity Relation 11 Thermal Boundary Layer 12 Internal Flow 19 Mean Velocity 19 Mean Temperature 20 Thermal Radiation Heat Transfer 22 The Reynolds Transport Theorem: Time Rate of Change of an Extensive Property of a System Expressed in Terms of a Fixed Finite Control Volume 22 Problems 28 References 31

2 2.1 2.2 2.2.1 2.2.2 2.2.3 2.3 2.3.1 2.3.1.1 2.3.1.2 2.3.2 2.3.2.1 2.3.2.2 2.3.2.3

Fundamental Equations of Laminar Convective Heat Transfer 33 Introduction 33 Integral Formulation 33 Conservation of Mass in Integral Form 33 Conservation of Linear Momentum in Integral Form 34 Conservation of Energy in Integral Form 36 Differential Formulation of Conservation Equations 38 Conservation of Mass in Differential Form 38 Cylindrical Coordinates 41 Spherical Coordinates 41 Conservation of Linear Momentum in Differential Form 42 Equation of Motion for a Newtonian Fluid with Constant Dynamic Viscosity μ and Density ρ 45 Cartesian Coordinates (x, y, z) 45 Cylindrical Coordinates (r, θ, z) 46

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Contents

2.3.2.4 2.3.3 2.3.3.1 2.3.3.2 2.3.3.3 2.3.3.4 2.3.3.5 2.3.3.6 2.3.3.7 2.3.3.8 2.3.3.9 2.3.3.10 2.3.3.11 2.3.3.12 2.3.3.13

Spherical Coordinates (r, θ, ϕ) 46 Conservation of Energy in Differential Form 47 Mechanical Energy Equation 53 Thermal Energy Equation 53 Thermal Energy Equation in Terms of Internal Energy 54 Thermal Energy Equation in Terms of Enthalpy 55 Temperature T and Constant Volume Specific Heat cv 55 Temperature and Constant Pressure Specific Heat cp 56 Special Cases of the Differential Energy Equation 58 Perfect Gas and the Thermal Energy Equation Involving T and cp Perfect Gas and the Thermal Energy Equation Involving T and cv An Incompressible Pure Substance 58 Rectangular Coordinates 59 Cylindrical Coordinates (r, θ, z) 59 Spherical Coordinates (r, θ, ϕ) 59 Problems 64 References 67

3 3.1 3.2 3.3 3.3.1 3.4 3.4.1 3.5

Equations of Incompressible External Laminar Boundary Layers 69 Introduction 69 Laminar Momentum Transfer 69 The Momentum Boundary Layer Concept 70 Scaling of Momentum Equation 71 The Thermal Boundary Layer Concept 76 Scaling of Energy Equation 77 Summary of Boundary Layer Equations of Steady Laminar Flow 82 Problems 82 References 83

4 4.1 4.2 4.3 4.3.1 4.3.2 4.4 4.5 4.6 4.6.1 4.6.2 4.6.3 4.7 4.8

Integral Methods in Convective Heat Transfer 85 Introduction 85 Conservation of Mass 85 The Momentum Integral Equation 87 The Displacement Thickness δ1 88 Momentum Thickness δ2 89 Alternative Form of the Momentum Integral Equation 90 Momentum Integral Equation for Two-Dimensional Flow 90 Energy Integral Equation 91 Enthalpy Thickness 93 Conduction Thickness 93 Convection Conductance or Heat Transfer Coefficient 93 Alternative Form of the Energy Integral Equation 94 Energy Integral Equation for Two-Dimensional Flow 94 Problems 94 References 96

5 5.1 5.2 5.2.1 5.2.2 5.2.3

Dimensional Analysis 97 Introduction 97 Dimensional Analysis 101 Dimensional Homogeneity 102 Buckingham π Theorem 102 Determination of π Terms 103

58 58

Contents

5.3 5.4 5.5 5.5.1 5.5.2 5.5.3 5.5.4 5.5.5 5.5.6 5.5.7 5.5.8 5.5.9 5.5.10 5.5.11 5.6

Nondimensionalization of Basic Differential Equations 116 Discussion 125 Dimensionless Numbers 125 Reynolds Number 125 Peclet Number 126 Prandtl Number 126 Nusselt Number 126 Stanton Number 126 Skin Friction Coefficient 126 Graetz Number 127 Eckert Number 127 Grashof Number 127 Rayleigh Number 127 Brinkman Number 127 Correlations of Experimental Data 128 Problems 136 References 147

6 6.1 6.2 6.3 6.4

One-Dimensional Solutions in Convective Heat Transfer 149 Introduction 149 Couette Flow 151 Poiseuille Flow 156 Rotating Flows 171 Problems 175 References 180

7 7.1 7.2 7.2.0.1 7.2.0.2 7.2.0.3 7.2.0.4 7.2.0.5 7.2.0.6 7.2.0.7 7.2.0.8 7.3 7.4 7.4.1 7.4.2 7.4.2.1 7.4.2.2 7.5

Laminar External Boundary Layers: Momentum and Heat Transfer 183 Introduction 183 Velocity Boundary Layer over a Semi-Infinite Flat Plate: Similarity Solution 183 x-Component of Velocity – Uu 190 ∞ Boundary Layer Thickness δ(x) 190 Wall Shear Stress τw 191 Local Skin Friction Coefficient cf (x) 191 Drag Force D 192 Average Skin Friction Coefficient cf 192 Displacement Thickness δ1 (x) 192 Momentum Thickness δ2 (x) 192 Momentum Transfer over a Wedge (Falkner–Skan Wedge Flow): Similarity Solution 195 Application of Integral Methods to Momentum Transfer Problems 201 Laminar Forced Flow over a Flat Plate with Uniform Velocity 203 Two-Dimensional Laminar Flow over a Surface with Pressure Gradient (Variable Free Stream Velocity) 204 The Correlation Method of Thwaites 207 A Thwaites Type Correlation for Axisymmetric Body 212 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate: Similarity Solution for Uniform Wall Temperature Boundary Condition 212 Low-Prandtl-Number Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate: Similarity Solutions for Uniform Wall Temperature Boundary Condition 225 High-Prandtl-Number Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate: Similarity Solutions for Uniform Wall Temperature Boundary Condition 228 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate: Similarity Solution for Uniform Heat Flux Boundary Condition 230

7.6 7.7 7.8

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7.9 7.9.1 7.10 7.11 7.12 7.12.1 7.12.1.1 7.12.2 7.12.2.1 7.13 7.13.1 7.13.1.1 7.13.1.2 7.13.1.3 7.13.2 7.13.2.1 7.13.2.2 7.13.2.3 7.13.3 7.13.3.1 7.13.3.2 7.13.3.3 7.13.3.4 7.13.4 7.13.4.1 7.13.4.2 7.14 7.15 7.16 7.17

8 8.1 8.2 8.2.1 8.2.2 8.2.3 8.3 8.4 8.4.1 8.4.1.1 8.4.1.2

Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate: Similarity Solutions for Variable Wall Temperature Boundary Condition 237 Superposition Principle 245 Viscous Incompressible Constant Property Flow over a Wedge (Falkner–Skan Wedge Flow): Similarity Solution for Uniform Wall Temperature Boundary Condition 249 Effect of Property Variation 252 Application of Integral Methods to Heat Transfer Problems 253 Viscous Flow with Constant Free Stream Velocity Along a Semi-Infinite Plate Under Uniform Wall Temperature: With Unheated Starting Length or Adiabatic Segment 256 The Plate Without Unheated Starting Length 262 Viscous Flow with Constant Free Stream Velocity Along a Semi-Infinite Plate with Uniform Wall Heat Flux: With Unheated Starting Length (Adiabatic Segment) 262 The Plate with No Unheated Starting Length 265 Superposition Principle 265 Superposition Principle Applied to Slug Flow over a Flat Plate: Arbitrary Variation in Wall Temperature 266 Boundary Condition: Single Step at x = 0 266 Boundary Condition: Two Steps at x = 0 and x = ξ1 268 Boundary Condition: Three Steps at x = 0, x = ξ1 , and x = ξ2 268 Superposition Principle Applied to Slug Flow over a Flat Plate: Arbitrary Variation in Wall Heat Flux 272 Boundary Condition: Single Step at x = 0 273 Boundary Condition: Two Steps at x = 0 and x = ξ1 274 Boundary Condition: Triple Steps at x = 0, x = ξ1 , and x = ξ2 275 Superposition Principle Applied to Viscous Flow over a Flat Plate: Stepwise Variation in Wall Temperature 278 First Problem 278 Second Problem 279 Heat Flux for 0 < x < ξ 279 The Heat Flux for x > ξ1 280 Superposition Principle Applied to Viscus Flow over a Flat Plate: Stepwise Variation in Surface Heat Flux 282 First Problem 282 Second Problem 283 Viscous Flow over a Flat Plate with Arbitrary Surface Temperature Distribution 284 Viscous Flow over a Flat Plate with Arbitrarily Specified Heat Flux 289 One-Parameter Integral Method for Incompressible Two-Dimensional Laminar Flow Heat Transfer: Variable U∞ (x) and Constant Tw − T∞ = const 293 One-Parameter Integral Method for Incompressible Laminar Flow Heat Transfer over a Constant Temperature of a Body of Revolution 295 Problems 299 References 310 Laminar Momentum and Heat Transfer in Channels 313 Introduction 313 Momentum Transfer 313 Hydrodynamic Considerations in Ducts 313 Fully Developed Laminar Flow in Circular Tube 318 Fully Developed Flow Between Two Infinite Parallel Plates 323 Thermal Considerations in Ducts 326 Heat Transfer in the Entrance Region of Ducts 335 Circular Pipe: Slug Flow Heat Transfer in the Entrance Region 337 Heat Transfer for Low-Prandtl-Number Fluid Flow (Slug Flow) in the Entrance Region of Circular Tube Subjected to Constant Wall Temperature 337 Heat Transfer to Low-Prandtl-Number Fluid Flow (Slug Flow) in the Entrance Region of the Circular Tube Subjected to Constant Heat Flux 345

Contents

8.4.1.3 8.4.2 8.4.2.1 8.4.2.2 8.4.2.3 8.4.2.4 8.4.2.5 8.5 8.5.1 8.5.1.1 8.5.1.2 8.5.1.3 8.5.2 8.5.2.1 8.6 8.6.1 8.6.1.1 8.6.1.2 8.6.1.3 8.6.1.4 8.6.2 8.6.2.1 8.6.2.2 8.6.2.3 8.7 8.8 8.9 8.10 8.11

9 9.1 9.2 9.3 9.4 9.5 9.6

Empirical and Theoretical Correlations for Viscous Flow Heat Transfer in the Entrance Region of the Circular Tube 350 Parallel Plates: Slug Flow Heat Transfer in the Entrance Region 355 Heat Transfer to a Low-Prandtl-Number Fluid (Slug Flow) in the Entrance Region of Parallel Plates: Both Plates Are Subjected to Constant Wall Temperatures 355 Heat Transfer for Low-Prandtl-Number Fluid Flow (Slug Flow) in the Entrance Region of Parallel Plates: Both Plates Are Subjected to UHF 358 Heat Transfer for Low-Prandtl-Number Fluid Flow (Slug Flow) in the Entrance Region of Parallel Plates: Upper Plate Is Insulated While the Lower Plate Is Subjected to Constant Wall Temperature 363 Heat Transfer for Low-Prandtl-Number Fluid Flow (Slug Flow) in the Entrance Region of Parallel Plates: Upper Plate Is Insulated While the Lower Plate Is Subjected to Constant Heat Flux 367 Empirical and Theoretical Correlations for Viscous Flow Heat Transfer in the Entrance Region of Parallel Plates 370 Fully Developed Heat Transfer 372 Circular Tube 372 HFD and TFD Laminar Forced Convection Heat Transfer for Slug Flow in a Circular Pipe Subjected to Constant Wall Heat Flux 372 HFD and TFD Laminar Forced Convection Heat Transfer for Viscous Flow in a Circular Tube Subjected to Constant Wall Heat Flux 375 HFD and TFD Laminar Forced Convection Heat Transfer for Viscous Flow in a Circular Tube Subjected to Constant Wall Temperature 378 Infinite Parallel Plates 382 HFD and TFD Laminar Forced Convection Heat Transfer for Viscous Flow Between a Parallel Plate Channel. Both Plates Are Subjected to Constant Wall Heat Flux Boundary Condition 383 Heat Transfer in the Thermal Entrance Region 387 Circular Tube 388 Graetz Problem: HFD and Thermally Developing Flow in a Circular Tube under Constant Wall Temperature Boundary Condition 388 The Leveque Solution: UWT Boundary Condition 401 Graetz Problem: HFD and Thermally Developing Flow for Viscous Flow in Circular Tube Under Uniform Wall Heat Flux Boundary Condition 406 Empirical and Theoretical Correlations for Viscous Flow in the Thermal Entrance Region of the Pipe 415 Two Infinite Parallel Plates 419 Graetz Problem: HFD and Thermally Developing Flow Between Parallel Plates Subjected to Constant Wall Temperature 419 Graetz Problem: HFD and Thermally Developing Flow Between Parallel Plates Subjected to Constant Wall Heat Flux 428 Empirical and Theoretical Correlations for Viscous Flow in Thermal Entrance Region of Parallel Plates 436 Circular Pipe with Variable Surface Temperature Distribution in the Axial Direction 438 Circular Pipe with Variable Surface Heat Flux Distribution in the Axial Direction 443 Short Tubes 446 Effect of Property Variation 448 Regular Sturm-Liouville Systems 449 Problems 450 References 463 Foundations of Turbulent Flow 465 Introduction 465 The Reynolds Experiment 465 Nature of Turbulence 466 Time Averaging and Fluctuations 467 Isotropic Homogeneous Turbulence 470 Reynolds Averaging 470

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9.7 9.8 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.15.1 9.15.2 9.16

Governing Equations of Incompressible Steady Mean Turbulent Flow 474 Turbulent Momentum Boundary Layer Equation 477 Turbulent Energy Equation 478 Turbulent Boundary Layer Energy Equation 479 Closure Problem of Turbulence 480 Eddy Diffusivity of Momentum 481 Eddy Diffusivity of Heat 482 Transport Equations in the Cylindrical Coordinate System 483 Experimental Work on the Turbulent Mean Flow 484 Turbulent Flow in Pipe: Velocity Profiles 485 Turbulent Flow over a Flat Plate: Velocity Profiles 491 Transition to Turbulent Flow 496 Problems 498 References 504

10 10.1 10.2 10.3 10.3.1 10.3.1.1 10.3.1.2 10.3.1.3 10.4 10.4.1 10.5 10.5.1 10.5.2 10.6

Turbulent External Boundary Layers: Momentum and Heat Transfer 507 Introduction 507 Turbulent Momentum Boundary Layer 507 Turbulence Models 508 Zero-Equation Models 508 Boussinesq Model 508 Prandtl’s Mixing-Length Model 508 Van Driest Model 509 Turbulent Flow over a Flat Plate with Constant Free-Stream Velocity: Couette Flow Approximation 510 Inner Region 510 The Universal Velocity Profile 511 Three-Layer (von Karman) Model for the Velocity Profile 511 Other Velocity Models 514 Approximate Solution by the Integral Method for the Turbulent Momentum Boundary Layer over a Flat Plate 514 Laminar and Turbulent Boundary Layer 519 Other Eddy Diffusivity Momentum Models 521 Turbulent Heat Transfer 522 Analogy Between Momentum and Heat Transfer 529 Reynold’s Analogy 529 Chilton–Colburn Analogy 531 Prandtl–Taylor Analogy 532 Von Karman Analogy 535 Some Other Correlations for Turbulent Flow over a Flat Plate 539 Turbulent Flow Along a Semi-infinite Plate with Unheated Starting Length: Constant Temperature Solution 542 Flat Plate with Arbitrarily Specified Surface Temperature 550 Constant Free-Stream Velocity Flow Along a Flat Plate with Uniform Heat Flux 553 Turbulent Flow Along a Semi-Infinite Plate with Arbitrary Heat Flux Distribution 554 Turbulent Transition and Overall Heat Transfer 558 Property Variation 564 Problems 564 References 569

10.7 10.8 10.9 10.10 10.10.1 10.10.2 10.10.3 10.10.4 10.11 10.12 10.13 10.14 10.15 10.16 10.17

11 11.1 11.2

Turbulent Internal Flow: Momentum and Heat Transfer 573 Introduction 573 Momentum Transfer 573

Contents

11.2.1 11.2.1.1 11.2.1.2 11.2.1.3 11.2.1.4 11.2.1.5 11.2.1.6 11.2.2 11.2.2.1 11.2.2.2 11.2.2.3 11.2.2.4 11.2.2.5 11.2.2.6 11.2.2.7 11.2.2.8 11.2.2.9 11.2.2.10 11.2.2.11 11.2.2.12 11.3 11.3.1 11.3.1.1 11.3.2 11.3.2.1 11.3.2.2 11.3.2.3 11.4 11.4.1 11.4.2 11.4.2.1 11.4.2.2 11.5 11.5.1 11.5.2 11.5.3 11.5.3.1 11.5.3.2 11.5.4 11.5.4.1 11.5.4.2 11.5.4.3 11.5.5 11.5.6 11.5.7 11.6 11.7 11.7.1.1 11.7.1.2 11.7.1.3 11.7.1.4 11.7.1.5

Momentum Transfer in Infinite Two Parallel Plates 573 The Entrance Region 574 The HFD Region 575 Prandtl’s Mixing-Length Model 578 Buffer Region 579 The Mean Velocity 582 Skin Friction Coefficient or Fanning Friction Factor cf 582 Momentum Transfer in Circular Pipe Flow 585 Entrance Region 585 The HFD Region 586 Average Velocity V 589 Skin Friction Factor cf 589 Moody Friction Factor f 589 Prandtl Mixing-Length Model 590 Laminar Sublayer 591 Buffer Region 591 Turbulent Region 591 Moody Friction Factor 592 Fanning Friction Factor 593 The Power Law Velocity Distribution 596 Fully Developed Turbulent Heat Transfer 597 TFD and HFD Turbulent Flow Between Parallel Plates Subjected to UHF 598 Mean Stream Temperature 602 TFD and HFD Turbulent Flow in a Pipe Subjected to UHF 605 Laminar Viscous Sublayer: 0 < y+ < 5 609 Buffer Layer: 5 < y+ < 30 610 Turbulent Region: y+ > 30 610 HFD Thermally Developing Turbulent Heat Transfer 618 Circular Duct with UWT 618 Circular Duct with Uniform Wall Heat Flux 625 Solution for Fully Developed Temperature Distribution θ1 626 Solution for the Entry Region Temperature Distribution θ2 627 Analogies for Internal Flow 629 Reynolds Analogy 629 Colburn Analogy 631 Prandtl–Taylor Analogy 631 Laminar Sublayer 632 Turbulent Core 632 von Karman Analogy 633 Laminar Sublayer: 0 ≤ y+ ≪ 5 634 Buffer Layer: 5 ≤ y+ ≪ 30 635 Turbulent Core: y+ ≥ 30 635 The Analogy of Kadar and Yaglom 636 The Analogy of Yu et al. 637 Martinelli Analogy 639 Combined Entrance Region 641 Empirical and Theoretical Correlations for Turbulent Flow in Channels 642 Colburn Correlation 645 Dittus and Boelter Correlation 646 Sieder–Tate Correlation 646 Hausen Correlations 647 Petukhov Correlation 647

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Contents

11.7.1.6 11.7.1.7 11.7.1.8 11.7.1.9 11.8 11.8.1 11.8.2 11.8.2.1 11.8.2.2 11.8.2.3 11.8.2.4 11.9

Gnielinski Correlation 649 Gnielinski Correlation with Modification 650 Sleicher and Rouse Correlation 650 Nusselt Correlation 651 Heat Transfer in Transitional Flow 652 Friction Factor in the Transitional Flow 653 Heat Transfer in the Transition Region 654 Tam and Ghajar Approach 654 Churchill Approach 655 Gnielinski Approach 656 Abraham et al. Approach 657 Effect of Property Variation 660 Problems 660 References 670

12 12.1 12.2 12.3 12.4 12.4.1 12.4.2 12.5

Free Convection Heat Transfer 675 Introduction 675 Fundamental Equations and Dimensionless Parameters of Free Convection 675 Scaling in Natural Convection 679 Similarity Solution for Laminar Boundary Layer over a Semi-Infinite Vertical Flat Plate 681 Constant Wall Temperature 681 Uniform Heat Flux 688 Integral Method (von Karman–Pohlhausen Method): An Approximate Analysis of Laminar Free Convection on a Vertical Plate 695 Constant Wall Temperature 697 Uniform Heat Flux 700 Turbulent Free Convection Heat Transfer on a Vertical Plate 702 Empirical Correlations for Free Convection 704 Vertical Plate 704 Horizontal Plate 712 Inclined Plates 715 Vertical Cylinders 719 Horizontal Cylinder 722 Inclined Cylinder 723 Free Convection from Vertical Cylinders of Small Diameter 724 Free Convection Within Parallel Plate Channels 725 Vertical Parallel Plate Channel 725 Horizontal Parallel Plate Channel 731 Inclined Parallel Plate Channel 732 Rectangular Enclosures 735 Horizontal Rectangular Enclosure (θ = 0) 735 Vertical Rectangular Enclosure 737 Inclined Rectangular Enclosure 740 Horizontal Concentric Cylinders 743 Concentric Spheres 744 Spheres 744 Problems 745 References 752

12.5.1 12.5.2 12.6 12.7 12.7.1 12.7.2 12.7.3 12.7.4 12.7.5 12.7.6 12.7.7 12.8 12.8.1 12.8.2 12.8.3 12.9 12.9.1 12.9.2 12.9.3 12.10 12.11 12.12

Index

755

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Preface This textbook covers sufficient material to support a one-semester graduate-level course in convective heat transfer. Topics such as boiling and condensation are not included, since these are beyond the scope of this book. This textbook represents the teaching methodology I developed over 30 years of my experience in teaching engineering students. The textbook focuses on teaching the fundamental concepts of convective heat transfer. In this textbook, students will see a careful balance of theory and engineering applications. Special effort has been made to provide a physical interpretation of topics covered in the textbook. Careful attention has been given to the derivation of dimensionless parameters and theoretical as well as semiempirical equations to establish the students’ confidence. This process will help students to develop their judgment concerning fundamental concepts. The main goal of this textbook is to provide students with the capability, tools, and confidence to solve engineering problems in convective heat transfer, starting from basic principles. I believe that an understanding of the physics and development of mathematical models is essential to the analysis and design of engineering processes. In classroom, emphasis should be placed on teaching the students how to develop mathematical models of physical phenomena. Students should learn how to formulate the convection problems and gain physical insight. It is evident that the mechanisms of heat transfer and their operation in a given engineering system are very complex. A general representation for convective heat transfer is revealed through the development of basic equations in differential forms. Many students have difficulties in understanding and developing solutions to these differential equations. Solution methods, such as Bessel functions, Laplace transforms, separation of variables, Duhamel’s theorem, and eigenfunction expansion, are required to solve these differential equations. Most students face difficulties in using such methods in convective heat transfer, and solutions can be extremely difficult. Because of this, there is a tendency toward a descriptive approach in most textbooks on convective heat transfer. Some textbooks present a survey of literature, skipping the details of formulations and solutions. In the author’s opinion, this is not desirable from a teaching standpoint, and this approach does not provide engineering students with adequate fundamental knowledge to understand and interpret the solutions. The book is devoted to a comprehensible exposition of the principles of convective heat transfer as well as the mathematical formulation and solution of processes encountered in convection heat transfer. The treatment of subjects is based on a background in basic heat transfer, thermodynamics, fluid mechanics, and mathematics. The problems provided at the end of each chapter require the students to think critically, but are not difficult to comprehend. The textbook differs from existing convective heat transfer textbooks in many ways. First, to improve readability, the derivations of continuity, momentum, and energy equations are presented completely without skipping any steps. Sentences such as “it can be done,” “it can be shown,” etc., are eliminated as much as possible. To reduce students’ frustration, solutions of governing equations of the problems considered in the textbook are presented completely without skipping any steps, and this will improve readability. An important advantage of the textbook as compared to existing convective heat transfer textbooks is the integration of modern computational tools such as Maple and MATLAB. The specific commands associated with these software packages are used in the solution of examples. It is easy for students to use the extensive symbolic and numerical capabilities of Maple and MATLAB. Students using the computational software tools will find it easier to overcome mathematical difficulties and obtain analytical and/or numerical solutions. I believe that, using Maple and MATLAB, students will be able to explore convective heat transfer without getting bogged down in the complicated numerical and analytical methods required to solve convective heat transfer problems.

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Preface

Almost none of the problems encountered in convective heat transfer can be solved with a hand calculator. For this reason, engineering students should be proficient with computational tools, and should make full use of their capabilities. The equations developed in the analysis can be solved easily by use of computational tools and, in fact, this process is a motivator to many students. The computational software tools used in this book have existed for more than three decades and are commonly used in academy and industry. It is not likely that they will disappear. The use of these tools should not present an economic difficulty for any engineering college or student, since academic versions of these software packages are available. I believe that, in time, engineering education will evolve in this way. Both simple as well complex problems can be solved by using Maple and MATLAB. This approach will provide students with ample time to concentrate on understanding all the steps involved in the physics of convection heat transfer. This textbook is organized into 12 chapters. Chapter 1 reviews fundamental concepts of heat transfer and presents a cursory development of the Reynolds transport theorem. Chapters 2 and 3 cover fundamental equations of laminar convective heat transfer. Chapter 4 is devoted to the development of integral equations for boundary layer flows. Dimensional analysis, nondimensionalization of differential equations, and experimental heat transfer are covered in Chapter 5. Fundamental concepts are presented in terms of one-dimensional solutions in Chapter 6. The conclusions drawn from one-dimensional solutions are broad and valuable. Laminar external boundary layers of momentum and heat transfer are discussed in Chapter 7. The superposition principle is explained in terms of slug flow. Variable surface temperature and heat flux are introduced. Similarity solutions are also introduced in this chapter. Solutions are obtained by numerical and integral methods. Chapter 8 includes laminar momentum and heat transfer in ducts. An introduction to turbulent flows is presented in Chapter 9. Some available experimental data for internal and external turbulent flows are introduced. Turbulent external boundary layers for momentum and heat transfer are discussed in Chapter 10. Turbulent internal flows are discussed in Chapter 11, where analogies between momentum and heat transfer are also presented. Finally, Chapter 12 is devoted to free convection heat transfer, and concepts of internal and external natural convection are discussed in this chapter. I gratefully acknowledge the help of my teachers, students, and colleagues. Their contributions and criticisms are reflected in the pages of this book. The inconsistencies, obscurities, and errors that remain in the textbook are all mine. I have taught convective heat transfer from several textbooks over the years and have incorporated material from these books in my course notes. Over the past many years, the origin of the material has been forgotten. I hope that readers will bring to my attention the material in the textbook to warrant acknowledgment to the original sources.

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About the Author Nevzat Onur is Emeritus Professor in the Department of Mechanical Engineering at Gazi University. He pursued his undergraduate studies in mechanical engineering at the University of California, Davis (CA, USA), where he received his BS degree in 1974. He then attended the Tennessee Technological University, Cookeville (TN, USA), where he completed his MS and PhD degrees in 1976 and 1980, respectively. His 40 years of academic experience includes appointments at different universities in Turkey. All of his university appointments have been in teaching and research in the thermal sciences (thermodynamics, fluid mechanics, and heat transfer), along with administrative duties. These administrative duties include positions such as the Chairman of the Department of Mechanical Engineering and the Dean of Engineering Faculty. He has authored or coauthored several refereed research papers, and is well-qualified to write on the topic of convective heat transfer. He lives in Ankara, Turkey.

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About the Companion Website This book is accompanied by a companion website: www.wiley.com/go/introtoconvectiveheattransfer The instructor website includes ● ●

The solutions manual Maple and MATLAB® files

The student website includes ●

Maple and MATLAB® files

1

1 Foundations of Convective Heat Transfer 1.1 Fundamental Concepts Heat transfer is an energy transfer process because of a temperature gradient or difference. This temperature difference is called a driving force that causes heat to flow from a high-temperature region to a low-temperature region. There are basic mechanisms or modes for heat transfer: conduction, convection, and radiation. In this chapter, we will present their fundamental equations.

1.2 Coordinate Systems In the solution of heat transfer problems, we need to be able to provide the geometric identification of various points in the system under study. This is done by the use of a coordinate system. If the coordinate system is not accelerating and rotating, it is called an inertial system. The most common right-hand coordinate systems are illustrated in Figure 1.1. The transformation equations for Cartesian (rectangular) and cylindrical coordinate systems are given as x = r cos(θ)

(1.1a)

y = r sin(θ)

(1.1b)

z=z

(1.1c)

The spherical coordinate system may be transformed into the rectangular coordinate system. Transformation equations for Cartesian and spherical coordinate systems are given as x = r sin(θ) cos(ϕ)

(1.2a)

y = r sin(θ) sin(ϕ)

(1.2b)

z = rcos(θ).

(1.2c)

z

z

r P(z, r, θ)

P(x, y, z)

x

x

y (a)

Figure 1.1

P(r, θ, ϕ) θ

z

z

0

z

y

x

θ

r

ϕ

y

y

x

(b)

(c)

(a) Cartesian coordinate system. (b) Cylindrical coordinate system. (c) Spherical coordinate system.

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

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1 Foundations of Convective Heat Transfer

1.3

The Continuum and Thermodynamic Equilibrium Concepts

The convection heat transfer depends on the properties of material, such as density, thermal conductivity, pressure, and specific heat. These properties are assumed to be well defined at infinitely small points, and these properties are assumed to vary continuously from one point to another. In convective heat transfer, we will use the continuum model. In the continuum model, we deal with the macroscopic or average effect of molecules. The continuum assumption is very useful since it erases the molecular discontinuities by averaging the microscopic quantities on a small sampling volume. Macroscopic quantities such as density ρ, pressure p, temperature T, etc. are assumed to vary continuously and smoothly from point to point within the flow. The state of continuum may be described by continuous functions such as density ρ = ρ(x, y, z, t), → − → − temperature T = T(x, y, z, t), and velocity V = V(x, y, z, t). The continuum and thermodynamic equilibrium concepts are discussed in [1–3]. Continuum assumption breaks down under certain conditions. The criterion for the validity of continuum assumption depends on the Knudsen number Kn, and the Knudsen number is defined as λ Kn = (1.3) Le where λ is the molecular mean free path length, which is the average distance traveled by molecules before they collide. The characteristic length Le can be the diameter of a pipe or equivalent diameter of a channel. In other words, the length scale, Le , represents the overall dimension of the flow. In general, the physics of liquid flow in micro devices is not well known. Suppose we restrict our discussion to gases. The kinetic theory of gases provides an expression for this mean free path as follows: 1 λ= √ 2 n π σ2

(1.4)

where n is the number density (the number of molecules per unit volume) and σ is the molecular diameter. This number represents an effective diameter of collusions for gas molecules. The values of σ for different gases are presented in [4]. Table 1.1 gives the values of σ for some gases. The number density n is given as p n= (1.5) kT where p is the absolute pressure, T is the absolute temperature, and k = 1.38065 × 10−23 J/K is the Boltzmann constant. When we combine these two equations, we get the equation for mean free path length λ kT λ= √ 2 π p σ2

(1.6a)

A different form of Eq. (1.6a) for mean free path λ is √ μ π λ= RT p 2

(1.6b)

where R is the specific gas constant and μ is the dynamic viscosity of the gas. In this model, as discussed in [3], ideal gas is modeled as rigid spheres. The continuum model is valid as long as λ ≪ Le . If this condition is violated, flow is not in equilibrium; the relation between stress and rate of strain and the no-slip boundary condition are not valid anymore. In other words, the continuum Navier–Stokes model is valid when λ is much smaller than a characteristic dimension Le of Table 1.1 Molecular diameter values for different gases. Gas

𝛔 × 1010 (m)

Air

3.66

N2

3.70

CO2

4.53

O2

3.55

1.4 Velocity and Acceleration

the flow. In a similar way, the linear relation between heat flux and temperature gradient and the no-jump temperature boundary condition at the solid–fluid interface are also not valid. Note that the continuum concept forms the basis of mass, momentum, and energy equations. This means that fluid and flow properties are distributed in space, and point properties can be defined; derivatives for these properties may be determined. We will now present the classification of flow regimes. a) The continuum regime: In this regime, the Knudsen number is Kn < 10−3 . Flows in the continuum regime can be modeled by the conservation of mass, momentum, and energy with no-slip and no-jump boundary conditions. b) The slip flow regime: In this regime, the Knudsen number is in 10−3 < Kn < 10−1 range; the conservation of mass, momentum, and energy is valid with slip and temperature jump boundary conditions. The no-slip and no-jump assumptions are not valid anymore. c) The transition regime: The Knudsen number is in 10−3 < Kn < 10−1 range. In this regime, the conservation of mass, momentum, and energy is not valid. Flow must be solved using molecular-based models such as the Boltzmann equation. d) Free molecule regime: In this regime, Kn > 10. The collusions between molecules are neglected, and collisionless Boltzmann transport equations are required. The continuum approach also requires that the sampling volume be in thermodynamic equilibrium. Under thermodynamic equilibrium, we have → − → − V solid = V fluid no slip (1.7) Tsolid = Tfluid

no temperature jump

(1.8)

If the fluid is gas, Eqs. (1.7) and (1.8) will not be valid if the mean free path length λ is not much less than the flow length scale Le . If Kn ≪ 1, slip is negligible. If Kn = O(10−1 ), there is slip. If Kn = O(1) or greater, the slip concept is not valid anymore. These points are discussed in [3, 5]. Example 1.1 This problem involves the mean free path and the Knudsen number. Compute the Knudsen number for a small probe in air at a temperature of 288.15 K for a characteristic length of 0.001 m. The air pressure is 101 325 Pa. Solution Using Eq. (1.6a), we obtain: (1.38065 × 10−23 )(288.15 ) = √ = 6.59 × 10−8 m −10 2 2 2πpσ 2 π (101325) (3.66 × 10 ) 6.59 × 10−8 m λ = 6.59 × 10−5 = Kn = Le 0.001 λ= √

kT

The analysis in this book is based on continuum and thermodynamic equilibrium. The conservation of mass, momentum, and energy equations is valid as long as continuum assumption is valid. The no-velocity slip and no-temperature jump at the solid boundary are valid if thermodynamic equilibrium is justified.

1.4 Velocity and Acceleration It is important to describe the time rate of change of position of a particular element of fluid. This time rate of change of − −s ∕dt, is called velocity, and it is denoted by the symbol → position, d→ V, i.e. → − → − ds (1.9) V= dt → − In a rectangular coordinate system, the velocity field V is given as → − → − V = V(x, y, z, t). (1.10) → − → − The components of the vector V in rectangular coordinate system consist of the projections of V on the x-, y-, and z-axes, → − as shown in Figure 1.2. The velocity vector V point P is shown as an arrow. The magnitudes of the components of velocity

3

4

1 Foundations of Convective Heat Transfer

y

Figure 1.2

Velocity components in the rectangular coordinate system.

v V P w

u x

z

→ − → − → − → − vector V in the x-, y-, and z-directions are denoted by u, v, and w, respectively. If i , j , and k are unit vectors, then the → − velocity vector V can be expressed as → − → − → − → − V = u i + v j + wk. (1.11) The magnitude of the velocity at each point at any instant can be obtained as √ → − V = | V| = u2 + v2 + w2 . The acceleration of particle is given by → − − dv → − dw → − dV du → → − a = = i + j + k. dt dt dt dt

(1.12)

(1.13)

1.5 Description of a Fluid Motion: Eulerian and Lagrangian Coordinates and Substantial Derivative In fluid mechanics, the term field describes a quantity defined as a function of both position and time for a given region. There are two approaches to describe the motion of fluid and its associated properties in fluid mechanics: 1) Lagrangian approach 2) Eulerian approach. Ultimately, they are equivalent, and they describe the same thing, but they serve different purposes. In order to use Newton’s equations, we need to describe the flow for a moving particle.

1.5.1

Lagrangian Approach

We can observe the flow from a reference frame that is moving with the fluid. This is the Lagrangian approach. In the Lagrangian approach, the physical variables are described for a particular element of fluid as it moves in the flow. A fluid particle is selected and pursued on its onward course observing the changes in path, velocity, density, pressure, temperature, or any property associated with the particle at any instant. It is assumed that the observer is moving with the fluid particle. Each fluid particle is identified by its initial position (a, b, c) relative to the origin of the coordinate system at some arbitrarily chosen initial time t0 . We follow the path of fluid particle of fixed identity. Then, at any instant of time t, the position (x, y, z) and continuum properties, such as temperature, density associated, with fluid particle, are function of a, b, c and time t. The coordinate of this selected fluid particle is written as x = x(a, b, c, t)

(1.14a)

y = y(a, b, c, t)

(1.14b)

z = z(a, b, c, t)

(1.14c)

and the velocity and acceleration components are u=

𝜕x 𝜕t

ax =

𝜕u 𝜕t

(1.15a)

1.5 Description of a Fluid Motion: Eulerian and Lagrangian Coordinates and Substantial Derivative

v=

𝜕y 𝜕t

ay =

𝜕v 𝜕t

(1.15b)

w=

𝜕z 𝜕t

az =

𝜕w 𝜕t

(1.15c)

where u, v and w, and ax , ay , and az are, respectively, the x, y, and z components of velocity and acceleration of fluid particle, which occupies the position (x, y, z) and time t. Note that observer moves with the fluid. Example 1.2 A weather balloon is launched into the atmosphere by a meteorologist, as shown in Figure 1.E2. Figure 1.E2

Illustration of Lagrangian approach. Weather balloon

Temperature transmitting instrument

Suppose that the balloon reaches an altitude where it is neutrally buoyant, and it sends information about weather conditions to the local monitoring station on the ground. Suppose that we place a temperature sensor attached to this balloon, having negligible mass, floating in the atmosphere, and recording the atmosphere temperature or the temperature of the flow field. The following temperature data are recorded by the sensor: Location (x1 , y1 , z1 ) (x2 , y2 , z2 ) ⋮ (xn , yn , zn )

Time t1 t2 ⋮ tn

Temperature T1 T2 ⋮ Tn

This is a Lagrangian approach of temperature measurement. The time rate of temperature in such a measurement is denoted by DT/Dt, and it is called the material derivative or substantial derivative. We will talk about this derivative later. It reflects the time rate of temperature of the labeled particle by an observer (temperature sensor) moving with the particle. The Lagrangian method is difficult to use in fluid mechanics and convective heat transfer because of the relative motion of too many fluid particles with respect to one another.

1.5.2 Eulerian Approach In the Eulerian approach, we observe the flow from a reference frame that is fixed in space. Flow sweeps fluid properties past around this stationary observer. The Eulerian approach gives the value of a fluid variable at a given fixed point and at a given time without regard to the identity of the particles, which occupy the point at a location. This means that the individual particles are not identified. At a later time, different particles will pass through the same point. This is what the stationary observer sees, and this is the way continuum equations are usually formulated. Example 1.3 A stationary probe is placed in fluid flow and measures pressure as a function of time at a fixed position, as shown in Figure 1.E3. Since we are measuring a field variable at a particular fixed location in space, this is called an Eulerian approach of pressure measurement. In the Eulerian method, we compute the pressure field p(x, y, z, t) of the flow pattern, and we do not compute the pressure changes p(t) that a particle experiences as the particle moves through the field. Note that different particles pass through the location and the pressure is measured by the stationary probe. Suppose

5

6

1 Foundations of Convective Heat Transfer

we take the following data at a fixed point: t1 t2 t3 ⋮ tn

p1 p2 p3 ⋮ pn

Figure 1.E3

Illustration of Eulerian approach.

Flow direction Pressure probe

The time rate of change of pressure in such a measurement is denoted as (𝜕p/𝜕t)x, y, z and notice that suffix (x, y, z) implies that the observer records the change in pressure at a fixed location (x, y, z) and, the (𝜕p/𝜕t)x, y, z is also called the local rate of change of pressure. In the Eulerian reference frame, the velocity field is written as → − → − → − → − → − V = V(x, y, z, t) = u i + vj + w k

(1.16)

where the respective velocities are u = f(x, y, z, t)

(1.17a)

v = g(x, y, z, t)

(1.17b)

w = h(x, y, z, t).

(1.17c)

The variables x, y, z, and t are independent variables. In the vast majority of fluid mechanics problems, the Eulerian velocity field is used. It is important to understand the transformation between Eulerian and Lagrangian methods. Example 1.4 Fluid motion in Lagrangian description is given as x(t) = α(1 + λ β t)2 y(t) = 2β∕(1 + λ β t) where λ, α and β are constant parameters. a) Determine the Lagrangian particle velocity components. b) Determine the Eulerian velocity components. Solution a) We find the velocity components using Maple 2020. x(t) = α(1 + λ β t)3 dx u= = 3α(1 + λ β t)2 λ β dt y(t) = 2β∕(1 + λ β t)

(1) (2) (3)

2

v=

dy 2β λ =− dt (1 + λ β t)2

(4)

1.6 Substantial Derivative

b) In order to determine the Eulerian velocity components, we eliminate a and b among Eqs. (1)–(4). First, we solve Eqs. (1) and (3) for a and b. This will give us x(t) (1 + λ β t)3

(5)

β = y(t)∕(2 − λy(t)t)

(6)

α=

Next, we substitute Eqs. (5) and (6) into Eqs. (2) and (4). Then, we obtain the Eulerian velocity components. u=− v=

λy(λyt − 2)x 2

λy2 4

1.6 Substantial Derivative The fundamental laws of nature (conservation of mass, momentum, and energy) are Lagrangian in nature and that means that they are formulated for a particle of fixed identity. The laws of conservation of mass, momentum, and energy are related to the time rate of change of some property of a particle of fixed identity. We need a relationship that bridges the Lagrangian and Eulerian approaches. Let us explain this relationship with an example. Le B be the mass-dependent extensive property whose specific value b is b=

B . m

(1.18)

We need a relationship that connects the Eulerian time derivative to the Lagrangian time derivative. Now, assume an − → → − observer on a motorboat moving through water at absolute velocity W, which may be different than the fluid velocity V. The observer wishes to study the temperature of river. In this case, b = T. The observer may encounter three cases: 1) In the first case, the boat is tied to the riverbank. The observer records the time rate of change of temperature at a fixed location. The partial derivative of T with respect to time t at constant location x, y, z is measured. The result is ( ) dT 𝜕T = . (1.19) dt 𝜕 t x,y,z Here, we observe how temperature changes with time at this fixed position. − → 2) In the second case, the power of the boat is turned on. The powered boat moves with absolute velocity W relative to riverbank. All the time, the observer on the boat is measuring temperature. At any instant of time, the time rate of change of temperature is ( ) ( ) ( ) ( ) dT 𝜕T 𝜕T 𝜕T 𝜕T = + wx + wy + wz (1.20a) dt 𝜕t x,y,z 𝜕x x,y,z 𝜕y x,y,z 𝜕z x,y,z dy

, wy = dt , wz = dz are the components of boat velocity. The boat velocity is different than the river velocity. where wx = dx dt dt This total time derivative can also be written as →→ − dT 𝜕T − = + W• ∇ T (1.20b) dt 𝜕t 3) In the third case, the power of the boat is turned off. The powerless boat drifts downstream with river velocity. The − → → − velocity W of the boat (observer) is the same as the velocity V of the river which has components u, v, and w. At any instant of time, the time rate of change of temperature is given as ( ) ( ) ( ) ( ) 𝜕T 𝜕T 𝜕T DT 𝜕T = +u +v +w . (1.21a) Dt 𝜕t 𝜕x 𝜕y 𝜕z

7

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1 Foundations of Convective Heat Transfer

The first term on the right-hand side of Eq. (1.21a) is the local change of temperature and the last three terms represent the rate of change of temperature due to fluid motion. We may express Eq. (1.21a) as −→ − DT 𝜕T → (1.21b) = + V • ∇ T. Dt 𝜕t In general, the material derivative in the Cartesian coordinate system may be represented as 𝜕 𝜕 𝜕 𝜕 D = +u +v +w . (1.21c) Dt 𝜕t 𝜕x 𝜕y 𝜕z The material derivative in cylindrical coordinates v 𝜕 𝜕 𝜕 𝜕 D = + vr + θ + vz . (1.21d) Dt 𝜕t 𝜕r r 𝜕𝜃 𝜕z The material derivative in spherical coordinates vϕ v 𝜕 𝜕 D 𝜕 𝜕 = + vr + θ + . (1.21e) Dt 𝜕t 𝜕r r 𝜕𝜃 r sin(θ) 𝜕ϕ We used the symbol DT/Dt instead of dT/dt to stress that we are reporting the rate of change temperature as we follow the water (fluid particle). This total derivative is called as substantial derivative, material derivative, particle derivative, or the derivative following the motion. Recall that the concept of an observer moving with the substance observed is called the Lagrange viewpoint. The substantial derivative represents the observed time rate of change of a quantity caused by two factors: a) The change in that quantity with time at each point in the flow field b) The change in that quantity with time due to fluid motion. This change requires an elapse of time to move from one position to another position where the quantity has a different value. Example 1.5 Clarification of Terms in Substantial Derivative Dρ 𝜕ρ Explain the different interpretations for Dt = 0, 𝜕t = 0, and ρ = const Dρ Dt 𝜕ρ 𝜕t

= 0 means that the density of fluid element moving with the flow field does not change with time.

= 0 means that the density of fluid element at a fixed point in the flow field does not change with time ρ = const means that the density everywhere in the flow field is constant and does not change with time. Example 1.6 The velocity field and pressure distribution in a two-dimensional channel is given by u = (1 + 2x) m∕s v = −2y m∕s p = 101 − 2.5(x + x2 + y2 ) kPa where x (m) is along the channel centerline and y (m) is the vertical direction normal to the channel centerline. Determine the rate of change of pressure along the channel. Solution We apply Eq. (1.21c). Then, we obtain Dp 𝜕p 𝜕p 𝜕p = +u +v Dt 𝜕t 𝜕x 𝜕y 𝜕p =0 𝜕t 𝜕p = −5x − 2.5 𝜕x 𝜕p = −5y 𝜕y Dp = (1 + 2x)(−5x − 2.5) + (−2y)(−5y) = −10x − 2.5 − 10x2 + 10y2 Dt

1.6 Substantial Derivative

Example 1.7 Suppose that a swimmer is swimming in a large circular channel with flowing water, as shown in Figure 1.E7. Swimmer is attached a pressure-sensing device, such as a pressure transducer of the resistance type and is measuring the pressure. Swimmer will be called an observer. The position of the observer x, y, z is time dependent. This is because it takes for the observer a definite finite time to change its position from one location to another. Swimmer’s velocity is wx = dx/dt and velocity of water is u. Assume that flow is in the positive x-direction, and pressure drop dp/dx in the positive x-direction is negative (there is a pressure drop in the direction of the flow). The rate of change pressure p measured with respect to time by the swimmer is ( ) ( ) dp ( dx ) dp dp = wx = dt dt dx dx

r x 0

Figure 1.E7

swimmer

Illustration of substantial derivative of pressure.

This equation expresses the total rate of change of pressure with time as the point of observation of pressure p is moved an infinitesimal distance in the fluid. We may have the following cases: a) If the swimmer is moving downstream, the rate of change pressure with time will be negative. b) If the swimmer is moving upstream, the rate of change pressure with time will be positive. Suppose that the observer moves along with fluid. This means that the swimmer stopped swimming and is moving with the fluid. The swimmer’s velocity is the same as the fluid velocity, and this tells us that wx = u; then, the pressure change for the swimmer would be ( ) Dp dp =u Dt dx Suppose that the pressure at all points within the fluid is changing with time, and this change of pressure at the location of the swimmer was given by 𝜕 p/𝜕 t; the rate of change pressure observed by the swimmer moving with the fluid will be ( ) dp Dp 𝜕p = +u dt 𝜕t dx We used again the symbol Dp/Dt instead of dp/dt to stress that we are reporting the rate of change temperature as we follow the fluid particle. Example 1.8 A boat is instrumented with a temperature measuring probe. The boat is floating with the fluid, and it is measuring the fluid temperature. It is assumed that both flow and temperature are under unsteady-state conditions. The temperature and velocity distribution in the flow are given as T = x2 + 2yz + t u = 3x v = 4yt2 w=1 where T(o C), x(m), y(m), z(m). Determine the rate of change of the temperature recorded by the temperature probe at time t = 1 s when the boat is at the point (1,1,1).

9

10

1 Foundations of Convective Heat Transfer

Solution The material derivative of temperature is 𝜕T 𝜕T 𝜕T DT 𝜕T = +u +v +w Dt 𝜕t 𝜕x 𝜕y 𝜕z 𝜕T =1 𝜕t 𝜕T 𝜕T 𝜕T = 2x, = 2z, = 2y 𝜕x 𝜕y 𝜕z DT = 1 + (3x)(2x) + (4yt2 )(2z) + (1)(2y) Dt DT = 8yt2 z + 6x2 + 2y + 1 Dt t = 1, x = 1, y = 1, z = 1 DT = 17 (o C∕m) Dt Let us generalize the result. Let b be any specific fluid property such as density, temperature, and state of stress. It is defined as B b= m where B is an extensive flow property. The derivative flowing a fluid particle is ( ) ( ) ( ) ( ) 𝜕b 𝜕b 𝜕b Db 𝜕b = +u +v +w . (1.22) Dt 𝜕t 𝜕x 𝜕y 𝜕z The material derivative can be expressed in index notation as follows: Db 𝜕b 𝜕b = + Vi i = 1, 2, 3. Dt 𝜕t 𝜕xi

(1.23)

Here, let us explain the meaning of each term. Db Dt 𝜕b 𝜕t

= Lagrangian change property b. = local change of property b.

𝜕b Vi 𝜕x = convective change of property b. i

1.7

Conduction Heat Transfer

If there exists a temperature gradient in a body, it is experimentally observed that there is an energy transfer from the high-temperature region to the low-temperature region. This energy transfer is called conduction heat transfer. Heat can be conducted through gases, liquids, and solids. In an isotropic medium, the vectorial form of Fourier’s law of heat conduction is given in the Cartesian coordinate system as − → → − (1.24) q′′ = −k ∇T. The components of this equation are given as 𝜕T 𝜕x 𝜕T ′′ qy = −k 𝜕y 𝜕T q′′z = −k 𝜕z where k is the thermal conductivity (W/m.K). The conduction heat transfer is a vector quantity. q′′x = −k

(1.25a) (1.25b) (1.25c)

1.9 External Flow

T

T

T(x, t)

T(x, t)

𝜕T >0 𝜕x

𝜕T π/2 over the control surface common to region 1 and CS. To clarify some concepts, we will digress for the moment and consider fluid flow through a cylinder, as shown in Figure 1.18.

dA α

V dA

System boundary at time t ΔL

Figure 1.17

Subregion 1 at time t.

1.12 The Reynolds Transport Theorem: Time Rate of Change of an Extensive Property of a System Expressed in Terms of a Fixed Finite Control Volume

Figure 1.18

Cylinder containing fluid.

A

V

n Δx Figure 1.19

Slanted cylinder containing fluid.

Δx cos θ 90°

θ

A

V

θ

n Δx

Fluid enters the cylinder at the left face and reaches the right face in time Δt, and it covers a distance of Δx. The velocity does not vary across the cross-sectional area A of the cylinder. The volume of the fluid in the cylinder is ∀ = A Δx.

(1.60)

The volume flow rate of fluid is A Δx = A V (m3 ∕s) ∀̇ = Δt where the magnitude of the velocity is given as Δx Δt and the mass flow rate of fluid is

(1.61)

V=

(1.62)

→ − − ṁ = ρ A V = ρ V ⋅ → n A (m3 ∕s)

(1.63)

Next, consider the fluid in the slanted cylinder. See Figure 1.19. Fluid passes through the area A and velocity of fluid − makes angle θ with the unit normal vector → n to area A. The velocity does not vary across the cross-sectional area A. Fluid enters from the left face and reaches right face during time interval Δt and covers a distance Δx. The height of the slanted cylinder is Δx cos θ. The volume of fluid contained in the slanted cylinder is ∀ = (Δx cos θ) A. The volume flow rate of fluid is Δx cos θ A = VA cos θ. ∀̇ = Δt The mass flow rate of fluid is → − − ṁ = ρ V A cos θ = ρ V ⋅ → n A.

(1.64)

(1.65)

(1.66)

We now return to our problem. We let the mean length of the subregion be ΔL. The volume of the subregion is d∀1 = −(ΔL cos α dA)1 where cos α will be negative since α > π/2.

(1.67)

25

26

1 Foundations of Convective Heat Transfer

The mass contained in subregion 1 is dm1 = −ρ(ΔL cos α dA)1 . The property B contained in subregion I is dBI (t) = (ρ b d∀)t = {ρ b (−ΔL cos α dA)}t

(1.68)

where density ρ is the average value, which, in the limit, is the value of ρ at the control surface. Integrating over the control surface common to region 1, B1 (t) = −

∫

{𝜌 b (ΔL cos α dA)}t

(1.69)

CS1

where B1 (t) is the total property B contained in region 1 at time t. Notice that ΔL is the distance traveled by a particle on the system surface (along a streamline that existed at time t) in time interval Δt. Evaluation of B3 (t + Δt). To evaluate B3 (t + Δt), consider the enlarged view of region III shown in Figure 1.20. For region III, we have dBIII (t + Δt) = (ρ b d∀)t+Δt = {ρ b ΔL cos α dA}t+Δt

(1.70)

and integrating over region 3, we have B3 (t + Δt) =

∫

{ρ b ΔL cos α dA}t+Δt

(1.71)

CS3

−→ where CS3 is the surface area common to region 3 and control surface, and the angle between the area vector dA and the → − velocity vector V is α < π/2. We now substitute Eqs. (1.69) and (1.71) into Eq. (1.59), and we obtain DBsys Dt

=

∫ {ρ b ΔL cos α dA}t+Δt − ∫ {ρ b (ΔL cos α dA)}t CS3 CS1 d − lim ρ b d∀ + lim Δt→0 Δt→0 dt ∭ Δt Δt

(1.72)

CV

Let us now evaluate the third term on the right-hand-side of Eq. (1.72) as − ∫ {ρ b (ΔL cos α dA)}t lim

CS1

Δt

Δt→0

⎫ ⎧ ) ( ⎪ ⎪ ΔL = − lim ⎨ ρ b cos α dA⎬ Δt→0 ∫ Δt ⎪ ⎪CS1 ⎭t ⎩ = − ρ b V cos α dA ∬ CSI

ΔL

System boundary at time t

dA α

III

V

dA

Figure 1.20

Subregion III at time t + Δt.

(1.73)

1.12 The Reynolds Transport Theorem: Time Rate of Change of an Extensive Property of a System Expressed in Terms of a Fixed Finite Control Volume

where lim (ΔL∕Δt) = V is the velocity. Recall that the area of the control surface is independent of time interval Δt, and it Δt→0 is taken into the integral. For our problem, the limit of the integral is equal to the integral of the limit by Leibniz’ rule as discussed in [7]. Now, we evaluate the second term on the right-hand side of Eq. (1.72) as ∫ {ρ b ΔL cos α dA}t+Δt lim

CS3

Δt→0∫ CS3

Δt

Δt→0

{ = lim =

∫

( ρb

ΔL Δt

)

} cos α dA t+Δt

{ρ b (V) cos α dA}

(1.74)

CS3

Substituting Eqs. (1.73) and (1.74) into Eq. (1.72), we obtain DBsys Dt

=

d ρ b d∀ + ρ b (ΔL cos α dA) + {ρ b (V) cos α dA} ∬ ∬ dt ∭ CV

CSI

(1.75)

CS3

Noting that the remaining portion of control surface, which is neither CS1 nor CS3, has the property that either α = π/2 → − or V = 0. Thus, ∬

{ρ b (V) cos α dA} =

CS

∬

{ρ b (V) cos α dA} +

CS3

∬

ρ b (ΔL cos α dA)

CSI

Then, we have the following expression for (DB/dt)sys : DBsys Dt

=

d ρ b d∀ + {ρ b (V) cos α dA} ∬ dt ∭ CV

(1.76)

CS

This can be written as DBsys Dt

=

→ − −→ d ρ b d∀ + b(ρ V ⋅ dA) ∬ dt ∭ CV

(1.77)

CS

−→ → − where the dot product of dA and V is given as → − −→ V ⋅ dA = V cos α dA. Equation (1.77) is the Lagrangian-to-Eulerian transformation of the rate of change of extensive property B, and it is the general Reynolds transport theorem. For control volume fixed in space and ∀cv = const, dtd = 𝜕t𝜕 , and the order of differentiation and integration is interchangeable. The limits on the volume integral are independent of time. Under these conditions, Eq. (1.77) becomes DBsys Dt

=

→ − −→ 𝜕 (ρ b)d∀ + b(ρ V ⋅ dA). ∭ 𝜕t ∬ CV

(1.78)

CS

Let us review the physical interpretation of each term: DBsys represents the rate of change of an extensive property B of the system of fluid particles, defined from a Lagrangian Dt description at time t. d ∭ b ρ d∀ represents the time rate of change of extensive property B within the control volume which coincides with dt CV

system at time t. This is a local change. → − −→ ∬ b(ρ V ⋅ dA) represents the net flow of extensive B through the control surfaces at time t. This is a convective change. In CS

other words, this integral represents the flux of extensive property B across control surface. In summary, we may state that this equation relates the rate of change of an extensive property B of the system of fluid particles (defined from a Lagrangian description at time t) to the changes of the same property within control volume (defined → − → − → − from Eulerian description). Notice that ρ = ρ(x, y, z, t), b = (x, y, z, t), V = V(x, y, z, t), i.e. ρ, b, V are Eulerian functions. A more rigorous derivation is also available in [8, 10].

27

28

1 Foundations of Convective Heat Transfer

Problems 1.1

A m-kg cube with sides L-cm-long slides down an oil-coated inclined plane. If the inclined plane makes an α degree angle with the horizontal plane and the oil layer is b mm thick, determine the constant speed as the block slides down the inclined plane. The dynamic viscosity of the oil is μ (kg/m.s) see Figure 1.P1.

m

Figure 1.P1

oil

A block sliding down the inclined plane.

b α

g

1.2

Figure 1.P2 shows a liquid film flown down an inclined plane. The liquid is a Newtonian fluid having viscosity μ (kg/m.s).The dimensionless velocity profile is given by u(y) = 2η − η2 V where 𝜂 = (y/H) is the dimensionless distance and V is the average velocity. Determine the shear stress at the fluid–solid interface (y = 0), and at the free surface (at the liquid–air interface) (y = H ). Air

y

H

Liquid x g

Figure 1.P2

Liquid film flowing down the inclined plane.

1.3

Water flows with an average velocity of 6 m/s down a pipe. The pressure drop in the direction of the flow is 1.1 Pa per meter of length. a) An observer is moving with the fluid. What is the rate of change of pressure with respect to time for the observer? b) An observer is moving upstream at a velocity of 3 m/s. What is the rate of change of pressure with respect to time for the observer?

1.4

A two-dimensional velocity field is described by → − V = 3x3 yi + 2xy2 j. Find the total acceleration at x = 2 and y = 3

1.5

Consider laminar flow in a pipe. Velocity distribution in the pipe is given by ) [ ( ( )2 ] p1 − p2 r vz = R2 1 − 4μL R and μ is the dynamic viscosity and R is the pipe radius. Assume a temperature distribution across the pipe cross section as given below ) ( r T = Tw + (Tc − Tw ) 1 − R where Tc is the centerline temperature and Tw is the pipe wall temperature. Obtain an expression for the bulk temperature.

Problems

1.6

The velocity profile for a turbulent flow in a circular pipe is given by ( ) r 1∕7 u = 1− Vc R where R is the pipe radius. Average velocity for flow in a pipe is defined as R

2 u(r) r dr R2 ∫0 Derive an expression for the average velocity. V=

1.7

The local heat transfer coefficient h(x) for a fluid over an L-m-long and W-m-wide flat plate varies with distance along the plate starting from the leading edge of the plate as given below h(x) = C x−1∕4 where C is a constant. Determine the average heat transfer coefficient.

1.8

Mean fluid temperature for flow in a pipe is defined by ∫ uTdA Tm =

A

∫ u dA A

where V is the average fluid velocity in the pipe. The pipe radius is denoted by R. Velocity and temperature profiles for flow in a pipe are given by [ ( )2 ] r u = 2V 1 − R [( ) ( )] r 2 1 r 4 − T=B+C R 4 R where A, B, and C are constants. Obtain the mean fluid temperature. 1.9

Consider flow over a triangular plate as shown. Fluid free stream temperature is T∞ and plate surface temperature is Tw . Experimental studies indicate that the local Nusselt number Nux is given by Nux = CRem x where Nux = h(x) x/k and local Reynolds number is Rex = U∞ x/ν. Here, x is measured from the apex of the triangle, as shown in Figure 1.P9. Obtain an expression for the average Nusselt number. Tw

Flow T∞

0

x

U∞

L

Figure 1.P9

Flow over a triangular plate.

B

29

30

1 Foundations of Convective Heat Transfer

1.10

The velocity and temperature profiles for a viscous fluid flowing in a circular tube are given by [ ( )2 ] r vz = A 1 − R ( )3 r T(r) = F + G R where A, F, and G are constants and R is the tube radius. Determine the fluid bulk temperature in the tube.

1.11

Consider a viscous fluid flow over a flat plate. Flow is steady, two-dimensional, and laminar. Temperature distribution in the fluid may be expressed in the following form: T(x, y) = T∞ + B exp(−β x y) where B and β are constants. Here, T∞ is the free stream temperature. Determine the local heat transfer coefficient.

1.12

The Lagrangian description of fluid motion is given as x = a exp(λt) y = b exp(−λt) z = c. Obtain the Eulerian description of the fluid motion.

1.13

Temperature distribution in a 4-m-wide, 5-m-long, and 3-m-high room is given as T = 21 + 3x + 5y − 3z∘C where x (m) and y (m) are the coordinates in horizontal plane and z (m) is the vertical coordinate. A fly is at 2 m high from the floor and is moving with a velocity of 3 m/s in the x-direction. What is the rate of change temperature at this position.

1.14

Experiments are performed on a 30-mm diameter and 1-m-long cylinder in air. Free stream temperature of air temperature is 27 ∘C and surface temperature of cylinder is kept at 400 K. Radiation effects are neglected. Data are given in the following table:

q(W)

V(m/s)

540.0

1.2000

789.6

2.4000

1179

3.6000

1808.4

4.8000

2355.6

6.0000

Calculate the average heat transfer coefficient for each air speed. 1.15

Engine oil flows over a flat plate. At a fixed position, the temperature of oil is measured as function of distance y from the surface of plate. Measurements are plotted in Figure 1.P15. Estimate the local heat transfer coefficient at this fixed position.

References

60

T in degree celcius

55 50 45 40 35 30 25 20

0

0.5

Figure 1.P15

1.16

1

1.5 y (in mm)

2

2.5

Variation oil temperature as a function of distance y.

Consider hydrodynamically and thermally fully developed laminar flow of fluid in a pipe. Dimensionless temperature distribution is given as [ ( ) ( )] T(r) − Tw 1 r 4 1 r 2 96 3 + = − θ= Tm − Tw 11 16 16 R 4 R where R is the pipe radius, Tm is the mean fluid temperature, and Tw is the pipe surface temperature. a) Plot θ as function of (r/R) b) Obtain the heat transfer coefficient h ( ) k 𝜕T q′′w 𝜕r r=R h= = . Tw − Tm Tw − Tm Assume that energy flows from pipe surface to fluid.

1.17

The local heat transfer coefficient h(x) for a laminar flow over a flat plate is given by √ hx = 0.332 Rex Pr 1∕3 k where Rex =

U∞ x ν

is the local Reynolds number and x is measured from the leading edge of the plate. Develop an

expression for the average heat transfer coefficient h over the distance x = L measured from the leading edge of the plate.

References 1 Jiji, L.M. (2009). Heat Convection, 2e. Springer. 2 Agrawal, A., Kushwaha, H.M., and Jadhav, R.S. (2020). Microscale Flow and Heat Transfer. Springer. 3 Gad-el-Hak, M. (2005). Flow physics. In: The MEMS Handbook: Introduction and Fundamentals (ed. M. Gad-el-Hak). CRS Press. 4 Faghri, A., Zhang, Y., and Howell, J. (2010). Advanced Heat and Mass Transfer. Global Digital Press. 5 White, F.M. and Majdalani, J. (2021). Viscous Fluid Flow. McGraw Hill. 6 Arpaci, V.S. and Larsen, P.S. (1984). Convection Heat Transfer. Prentice Hall. 7 Sokolnikoff, I.S. and Redheffer, R.M. (1966). Mathematics of Physics and Modern Engineering, 2e. McGraw Hill.

31

32

1 Foundations of Convective Heat Transfer

8 9 10 11 12 13 14 15 16

Whitaker, S. (1977). Fundamental Principles of Heat Transfer. New York: Pergamon Press. Whitaker, S. (1968). Introduction to Fluid Mechanics. Prentice Hall. Slattery, J. (1968). Momentum, Energy, and Mass Transfer in Continua. McGraw Hill. Papanastasiou, T.C., Georgiou, G.C., and Alexandrou, A.N. (2000). Viscous Fluid Flow. CRC Press. Sultanian, B.K. (2016). Fluid Mechanics. CRC Press. Cengel, Y. and Cimbala, J.M. (2014). Fluid Mechanics, 3e. McGraw-Hill. Boas, M.L. (2006). Mathematical Methods in the Physical Sciences, 3e. Wiley. Panton, R.L. (2013). Incompressible Flow. Wiley. Bird, R.B., Steward, W.E., and Lightfoot, E.N. (2002). Transport Phenomena. New York: Wiley.

33

2 Fundamental Equations of Laminar Convective Heat Transfer 2.1 Introduction Since we are considering moving media, in this chapter, we will present the fundamental equations applicable to the open systems or control volume. The Eulerian approach is preferred in the study of convective heat transfer in determining temperature, pressure, forces, etc. at a particular location in space. Motivation for the use of the control volume technique is that measurement sensors, such as thermometers and pressure transducers, are located at a fixed position. Measurements give Eulerian fluid properties, for example, temperature T (x, y, z, t). We described fluid properties and flow characteristics in terms of field (Eulerian) concept. We will develop the integral form of the mass, momentum, and energy. In each development, the finite system approach will first be undertaken, and afterward, this finite system formulation will be converted to the case of finite control volume by means of the Reynolds transport theorem. The Reynolds transport theorem is repeated here for convenience. → d D ⃗ ⋅− b ρ d∀ = ρb d∀ + b(ρV dA). ∫ Dt ∫ dt ∫ sys

CV

CS

The Reynolds transport theorem gives a relation between the time rate of change within the system volume and that within the control volume. Finite integral formulations of the laws of mass, momentum, and energy provide us average values of quantities but cannot provide us the details of the flow. Using these integral formulations, we will develop differential formulations of the laws of mass, momentum, and energy valid at a point. To accomplish this task, we will consider infinitesimally differential control volume, and the infinitesimal control volume approaches to a point in the limit. There is a need for the differential formulation if the value of a properties at each point in a fluid varies or if the properties are known to vary with time. We will consider three natural laws: 1. Conservation of mass or continuity equation 2. Linear momentum equation 3. First law of thermodynamics or energy equation

2.2 Integral Formulation We are interested in moving media, and we wish to present the fundamental equations valid for the control volume. The Eulerian approach is very useful, and certainly, it is advantageous to study convective heat transfer. It is especially useful in determining forces, pressure, temperature, etc., at a particular location. We need only to place the measurement sensors (thermocouples, pressure transducers, etc.) on the surface of control volume. Quantities required in engineering can be expressed in terms of integrals. For example, heat transfer is the integral of heat flux over an area, and force is the integral of stress over a surface. Integral formulation is discussed in [1–3] in detail, and a cursory presentation will be given here.

2.2.1 Conservation of Mass in Integral Form One of the physical laws of nature useful in predicting fluid behavior is the law of conservation of mass. We start with the development of integral form of the continuity equation for a finite system.

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

34

2 Fundamental Equations of Laminar Convective Heat Transfer

System analysis: The conservation of mass states that mass can be neither created nor destroyed. An exception occurs when mass is converted into energy or energy into mass. We will not consider such changes in this text since such changes in mass are not important in convective heat transfer. The mass of a closed system (system or control mass) is constant, and it remains invariant with respect to time, and knowing that the density can change from point to point in the system, the conservation of mass can be expressed in integral form as ( ) Dm D = ρd∀ = 0 (2.1) Dt system Dt ∭ system

where D/Dt is used since we are following a specified group of material particles in a system. Control volume analysis: Consider a control volume fixed in space. Using the Reynolds transport theorem, we obtain the mathematical equation for the conservation of mass flow through control volume at any time t as follows: a) Extensive property B is m, the mass of fluid system. b) The quantity b is specific property of extensive property B, and it is b = B/m = m/m = 1. Thus, the conservation of mass for a control volume referring to Eq. (1.77) is d ⃗ ⋅ dA ⃗ =0 ρd∀ + ρV ∬ dt ∭

(2.2a)

CS

∀

or Eq. (2.2a) takes the equivalent form for a fixed nondeformable control volume 𝜕ρ ⃗ ⋅ dA ⃗ = 0. d∀ + ρV ∬ ∭ 𝜕t

(2.2b)

CS

∀

Note that we can choose a system of any shape at time t; then, we say that Eq. (2.2a) or Eq. (2.2b) is valid for any control ⃗ and time derivative d/dt is measured with respect to control volume. volume at time t. Velocity field V

2.2.2

Conservation of Linear Momentum in Integral Form

Newton’s second law is also called the linear momentum equation, and it states that resultant force acting on a system is equal to the rate of change of linear momentum. Force is measured in an inertial reference frame. System analysis: Let us now consider a finite system moving in a flow, as depicted in Figure 2.1. Since we are following the system, we use the notation D/Dt, and Newton’s law can be expressed in an inertial reference frame as ∑ ⃗| D ⃗=F ⃗S + F ⃗ B = DP || ⃗ F = ρVd∀ (2.3) Dt ||system Dt ∭ system ∑⃗ ⃗ is the velocity vector, P ⃗ = mV ⃗ is the linear momentum, where F is the sum of all the external forces acting on the system, V and the time derivative is taken for an inertial reference. z

Figure 2.1

System y x V

ρ

System at time t for Newton’s second law.

2.2 Integral Formulation

Equation (2.3) states that the rate of change of the linear momentum of the system is equal to the sum of external forces acting on the system. Let us consider each term in Eq. (2.3). ⃗ Linear momentum is defined as Linear momentum P: ⃗= P

∭

⃗ dm = V

⃗ d∀. ρV

∭

m

(2.4)

CV

⃗ B : The body forces (volume forces) are long-range forces and act on the material inside the system boundary. Body forces F Most common examples are gravity, electromagnetic, and centrifugal forces. Body force acting on a differential system d∀ is ⃗ B = ⃗fρ d∀ dF

(2.5a)

and integrating ⃗B = F

∭

⃗f ρd∀

(2.5b)

CV

where ρ is the density of fluid and d∀ is the differential volume element. Body forces penetrate the body and act on all the parts of the body. In convective heat transfer, the body force is usually gravity, so we may write ⃗B = F

∭

ρ⃗g d∀ = m⃗g.

(2.5c)

CV

⃗ S : Surface forces are short-range forces of molecular origin. Examples include tangential forces (shear) Surface forces F and normal forces (pressure and normal stress). These external forces act on the boundary of the system, and sometimes, it is called surface tractions. We are talking about the force per unit area on the boundary surfaces. We have expressed Newton’s law for a finite system, and again, D/Dt is used to provide the rate of change since Newton’s second law is applied to a system, which we are following in a flow field. Notice that both velocity and density may change from point to point in the system. Control volume analysis fixed in inertial space: The velocities and time derivates are measured with respect to control volume. We assume that the control volume is stationary in inertial space. Note that linear momentum is an extensive system property. Therefore: ⃗ = mV ⃗ a) Extensive property is linear momentum and B = P ⃗ ⃗ b) Specific intensive property b = B∕m = mV∕m = V. ⃗ = mV ⃗ and noting that the intensive property is Thus, applying the Reynolds transport theorem to extensive property P ⃗ we have b = V, ⃗ || d DP ⃗ V ⃗ ⋅ dA). ⃗ ⃗ V(ρ (2.6) = (ρV)d∀ + | ∬ Dt ||system dt ∭ CV CS Therefore, the derivative on the left-hand side of Eq. (2.6) is taken for an inertial reference frame. We can then use ⃗ Newton’s law, Eq. (2.3), to replace the (DP∕Dt) system term on the left-hand side of this last equation, and the desired momentum equation for a control volume can be written as ⃗s + F

∭

⃗fρd∀ = d ⃗ ⃗ V ⃗ ⋅ dA). ⃗ (ρV)d∀ + V(ρ ∬ dt ∭

CV

CV

(2.7)

CS

At time t, the system and control volume have the same shape. The linear momentum equation is a vector equation. The scalar components of Eq. (2.7) in terms of the Cartesian coordinate system can be expressed as Fsx +

∭

fx ρd∀ =

CV

Fsy +

∭ CV

∬

⃗ ⋅ dA) ⃗ + d u(ρV (ρu)d∀ dt ∭

CS

fy ρd∀ =

∬ CS

⃗ ⋅ dA) ⃗ + v(ρV

d (ρv)d∀ dt ∭

CS

(2.8b)

CV

⃗ ⋅ dA) ⃗ + d Fsz + f ρd∀ = w(ρV (ρw)d∀. ∭ z ∬ dt ∭ CV

(2.8a)

CV

CV

(2.8c)

35

36

2 Fundamental Equations of Laminar Convective Heat Transfer

2.2.3

Conservation of Energy in Integral Form

Closed system analysis: The concept of conservation of energy is used extensively in engineering problems, and all the engineering systems obey this law, and these concepts are considered in any engineering analysis. When a system undergoes a cyclic process, the first law of thermodynamics may be expressed as ∮

δQ =

1 δW J∮

(2.9a)

where the symbol ∮ refers to a “cyclic integral.” The quantity J is called the mechanical equivalent of heat and numerically is equal to J = 778.17 ft lbf /BTU in American engineering units. In the SI system of units, J = 1 N. m/J. Now, note that ∮ δQ represents net amounts of energy transfer to the system and ∮ δW represents the net amount of work done by the system. In the SI unit system, work and heat have the same units. Both heat and work are path functions; the differentials of heat and work are inexact differentials. For this reason, they are denoted by the symbols δQ and δW, respectively. Consider the system depicted in Figure 2.2. The first law of thermodynamics relates the heat transfer, work, and energy change of the system. The energy balance in differential form is dE = δQ − δW

(2.9b)

where dE is the differential of energy and energy is a property. The instantaneous time rate form of energy balance is δQ || δW || DE || D − = = eρd∀ (2.9c) | | dt |system dt |system Dt ||system Dt ∭ sys

where E = ∭ eρd∀ is the total energy of the system and the term e is the specific energy, and it is given as sys

1 E ̃ + V2 + φ =u e= m 2

(2.10)

δQ∕dt = Q̇ = heat transfer rate to the system ̇ = rate of work done by the system δW∕dt = W φ = gz = specific potential energy e = specific total energy ̃ = specific internal energy u V2 /2 = specific kinetic energy In its basic form stated here, the first law of thermodynamics applies only to a system. Notice that both density and specific energy may change from point to point in the system. The value of a state variable is defined to be that which would be seen by an observer moving with the fluid. Since we are following the system, we used the substantial derivative notation, D/Dt. Control volume analysis: To develop a control volume approach, we take E to be the extensive property to be used in Reynold’s transport equation. a) Extensive property, B = E b) Intensive property, b = E/m = e We can then use Reynold’s transport theorem to obtain the equation of the first law of thermodynamics for a finite control volume DE || ⃗ ⋅ dA) ⃗ + d = e(ρV (eρ)d∀. (2.11) Dt ||system ∬ dt ∭ CS

∀

Figure 2.2

ρ, e

System for first law of thermodynamics.

2.2 Integral Formulation

As time Δt → 0, the system and control volume coincide, and for this reason, we can write the following form of the equation of conservation of energy for control volume in a flow field: δQ δW ⃗ ⋅ dA) ⃗ + d − = e(ρV (eρ)d∀ ∬ dt dt dt ∭ CS

where ⋅

δQ | | δt |sys

=

δQ | | δt |cv

(2.12)

∀ ⋅

= Q and

δW | | δt |sys

=

δW | | δt |cv

⋅

= W. Thus, we have replaced the left-hand side of Eq. (2.11) by Eq. (2.9c). The ⋅

term Q represents the rate of heat transfer into control volume across control surface due to temperature difference and W represents the rate of work done (power) by the fluid (i.e. the system) in control volume on its surroundings. This equation needs to be further transformed into a useful form so that it will be of any value in solving heat transfer problems. This transformation is presented next. In this development, δW/dt term does not include the work done by the body force since this effect is included in the specific energy term as potential energy. We need to have a clear understanding of the work term, and this work term needs elaboration. This work rate term will be split into two parts: ⋅

a) Shaft work rate Wshaft

⋅

b) Surface force work rate Wsurface Thus, we have δW δWshaft δWsuface = + . dt dt dt Now, let us examine each work term.

(2.13) ⋅

⋅

a) Consider a rotating shaft in a control volume. The shaft work Ws = δWs ∕dt is the work done by the system on the surroundings (outside the control volume) as a result of work being transferred through the control surface by a rotating shaft. Rotating shaft transmits a torque, and this torque results from the fluid forces acting on the rotating shaft within the control volume. b) The surface force work is the work required in deforming the system boundary to make it coincident with the control surface. This is really the work done against surface forces (i.e. normal and shear forces). The surface work is really a mechanical work term, and the mechanical work requires a force to act through a distance. Forces acting on the boundary can be broken down into shear forces and a force normal to the area. To evaluate the surface work rate, ⃗ and tangential consider the subregion, as illustrated in Figure 2.3. The normal to surface is designated by unit vector n direction by unit vector ⃗s. There are two stresses acting on the control surface: 1. Normal stresses 2. Tangential stresses The stress vector τ⃗ can be expressed as ⃗ + τns ⃗s τ⃗ = τn n n

(2.14)

where τn n is the normal stress component, and it has a positive sign for tension. The τn s is the shear (tangential) stress component. The external force acting on the differential element area dA of the system is given as ⃗ = τ⃗ dA dF

(2.15) s

τ ns

n V ΔL τn nn dA dA

Figure 2.3

Subregion at time t.

System boundary at time t

Control surface

37

38

2 Fundamental Equations of Laminar Convective Heat Transfer

where τ⃗ is the stress vector. Assume that the distance ΔL is the mean length of the subregion. The work done by the system in displacing the system’s boundary element to coincide with control surface element in time interval Δt is given as ⃗ ⋅ (ΔL). ⃗ (ΔWsuface ) = −dF

(2.16)

There is a negative sign in Eq. (2.16) since work is done on the system by surface forces. Upon substituting Eq. (2.15) into Eq. (2.16), dividing by Δt, and taking the limit as Δt → 0, the average rate of work is ) ) ) ( ( ( ⃗ ⋅ ΔL ⃗ ⃗ ΔWsurface dF ΔL ⃗ ⃗ = − lim = −(⃗τdA) ⋅ lim == −(⃗τ ⋅ V)dA (2.17) lim Δt→0 Δt→0 Δt→0 Δt Δt Δt ⃗ = lim (ΔL∕Δt) ⃗ where V is the velocity vector. In order to obtain the contributions from all the parts of system surface Δt→0 boundary, which coincides with the control surface as Δt → 0, we integrate over the entire control surface δWsurface ⃗ ⃗ ⃗ + τns ⃗s) ⋅ VdA = − (⃗τ ⋅ V)dA = − (τn n n ∬ ∬ dt CS

(2.18)

CS

⃗ where subscript CS represents control surface. Thus, ∬ (⃗τ ⋅ V)dA represents the rate of work done by the system of fluid CS

within the control volume along its surface against viscous forces. This work consists of a friction work due to viscous shear and an expansion work since volume changes opposed by normal stresses. Next, we will treat the heat transfer term δQ/dt. Heat enters the control volume by diffusion relative to bulk fluid motion. It is possible that some energy, such as chemical reactions, electrical heating, atomic, or electromagnetic form, may enter control volume. This energy is converted into thermal energy leading to an internally distributed heat source. Hence, we may split the heat term into two parts: − → δQ || ⃗ + u′′′ d∀ = − q′′ ⋅ dA | ∬ ∫ dt |in CS

(2.19)

CV

− → ′′′ where q′′ is the heat flux vector to be discussed later and u is the energy generated by per unit volume. Thus, using Eqs. (2.13), (2.18), and (2.19) with Eq. (2.12), we obtain general form of the energy equation: ⋅ − → d ⃗ ⋅ dA) ⃗ = − q′′ ⋅ dA ⃗ − Wshaft + ⃗ (eρ)d∀ + e(ρV (⃗τ ⋅ V)dA + u′′′ d∀. ∬ ∬ ∬ ∫ dt ∭ ∀

2.3

CS

CS

CS

(2.20)

CV

Differential Formulation of Conservation Equations

If we need the knowledge of the value of properties at a specific location in the fluid itself, conservation equations are more useful when expressed in differential form. The conservation of mass equation will be expressed in differential form using an infinitesimal control volume by the use of a step-by-step accounting of mass flowing into and out of this infinitesimal control volume. We will follow this approach to give the reader better insight.

2.3.1

Conservation of Mass in Differential Form

The continuity equation for a finite control volume is given by ∬ CS

⃗ ⋅ dA ⃗+ 𝜕 ρV ρd∀ = 0. 𝜕t ∭ ∀

We will now apply the continuity equation to an infinitesimal control volume. We are now seeking a differential form of the continuity equation. Consider the infinitesimal rectangular parallel piped as the three-dimensional differential control volume element with the volume center at the point (x, y, z), as shown in Figure 2.4. Each face of the differential control volume element is numbered as shown in Figure 2.4. This control volume element is fixed in coordinate space (x, y, z). Since there is no size limitation on the integral form of the continuity equation, it can be applied to this infinitesimal small differential control volume element. Imagine that this is a small element of fluid removed from a boundary

2.3 Differential Formulation of Conservation Equations

Figure 2.4

y

Three-dimensional differential control volume element.

z

x

(x, y, z)

layer such as viscous fluid flow over a flat plate. Notice that density and velocity are Eulerian variables, and we are using Cartesian coordinates to express them, we may write ρ = ρ(x, y, z, t) u = u(x, y, z, t)

x-component of velocity

v = v(x, y, z, t)

y-component of velocity

w = w(x, y, z, t)

z-component of velocity

Note that u, v, and w are scaler functions. We wish to evaluate the integrals assuming that mass flux to each face is uniform. ⃗ ⋅ dA: ⃗ To evaluate this integral for our differential control volume, we will first note that the control surface is a) ∬ ρV CS

composed of six plane surfaces so that ∬

=

CS

∬ CS1

+

∬ CS2

+

∬ CS3

+

∬

+

CS4

∬

+

CS5

∬ CS6

=

6 ∑ i=1

∬

⃗ ⋅ dA. ⃗ ρV

(2.21)

CSi

The integral over each of these plane surfaces will be approximated by the value of the integrand at the area center of the surface. Note that CS in Eq. (2.21) represents control surfaces. As an example, we now wish to consider the integral for surface 1. Surface 1 is the surface at x + Δx/2. This integral is approximated as ∬

⃗ ⋅ dA ⃗ ≅ [(ρV) ⃗ (x+Δx∕2,y,z,t) ] ⋅ dA ⃗1 ρV

CS 1

⃗ ⃗ ≈ {[ρ(u ⃗i + v ⃗j + wk)] (x+Δx∕2,y,z,t) } ⋅ i ΔyΔz ≈ (ρu)x+Δx∕2 ΔyΔz. This is the x-component of mass flow rate at x + Δx/2 through control surface 1 CS1 . See Figure 2.4. We now approximate the other surface integrals, and summing over all faces and rearranging, we get { ⃗ ⋅ dA ⃗ ≅ [(ρu)x+Δx∕2 − (ρu)x−Δx∕2 ]ΔyΔz + [(ρv)y+Δy∕2 − (ρv)y−Δy∕2 ] ΔxΔz ρV ∬ Cs } + [(ρw)z+Δz∕2 − (ρw)z−∇z∕2 ]ΔxΔy . (2.22) b)

𝜕 ∭ 𝜕t ∀

ρ d∀: We will use an average value of density over the control volume element. The volume integral becomes 𝜕̃ ρ 𝜕 ΔxΔyΔz ρ d∀ ≈ 𝜕t ∭ 𝜕t

(2.23)

∀

where ̃ ρ is the average value of density. Substituting Eqs. (2.22) and (2.23) into Eq. (2.2a) and dividing by Δx Δy Δz, we get } { (ρu)x+Δx∕2 − (ρu)x−Δx∕2 (ρv)y+Δy∕2 − (ρv)y−Δy∕2 (ρw)z+Δz∕2 − (ρw)z−Δz∕2 ̃ 𝜕ρ + + + = 0. (2.24) Δx Δy Δz 𝜕t

39

40

2 Fundamental Equations of Laminar Convective Heat Transfer

Taking the limit of Eq. (2.24) as Δx → 0, Δy → 0, Δz → 0, we get the differential form of conservation of mass 𝜕ρ 𝜕 𝜕 𝜕 (ρu) + (ρv) + (ρw) + = 0. 𝜕x 𝜕y 𝜕z 𝜕t

(2.25)

⃗ The divergence of ρ V ⃗ is The first three terms of Eq. (2.25) comprise the divergence of vector ρ V. ⃗ =∇ ⃗ ⃗ ⋅V div(ρV)

(2.26a)

where divergence is defined as ⃗ = ⃗i 𝜕 + ⃗j 𝜕 + k⃗ 𝜕 ∇ 𝜕x 𝜕y 𝜕z and velocity vector is ⃗ = i⃗u + ⃗j v + k⃗ w. V Thus, the divergence of velocity vector [ ] ⃗ = ⃗i 𝜕 + ⃗j 𝜕 + k⃗ 𝜕 ⋅ [i⃗u + ⃗j v + k⃗ w] = 𝜕 u + 𝜕 v + 𝜕 w . ⃗ ⋅V ∇ 𝜕x 𝜕y 𝜕z 𝜕x 𝜕y 𝜕z

(2.26b)

Equation (2.25) satisfies the requirements of the conservation of mass at any arbitrary position in a fluid. This equation is referred to as the differential form of the conservation of mass or the continuity equation. It is satisfied at all points (x, y, z) within the fluid at all times. Equation (2.25) is equivalent to 𝜕ρ ⃗ → − + ∇ ⋅ (ρ V) = 0. (2.27) 𝜕t The continuity equation can be expressed in a number of different ways. For example, an alternate form is obtained by differentiating the product terms of Eq. (2.25) ] [ ] [ 𝜕ρ 𝜕ρ 𝜕ρ 𝜕ρ 𝜕u 𝜕v 𝜕w + + + +u +v +w = 0. (2.28) ρ 𝜕x 𝜕y 𝜕z 𝜕t 𝜕x 𝜕y 𝜕z Note that last four terms represent the substantial derivative of ρ. Recall that the substantial derivative was defined as D 𝜕 𝜕 𝜕 𝜕 = +u +v +w . D t 𝜕t 𝜕x 𝜕y 𝜕z Then, Eq. (2.28) can be expressed as { } 𝜕u 𝜕v 𝜕w 1 Dρ + + + = 0. ρ Dt 𝜕x 𝜕y 𝜕z

(2.29)

⃗ Thus, Eq. (2.29), Last three terms on the right-hand side of Eq. (2.29) represents the divergence of the velocity vector V. can be written as Dρ ⃗ = 0. ⃗ V) + ρ(∇. (2.30) Dt This form of continuity equation does not refer to any particular coordinate system. Equation (2.30) describes the variation of density with time as seen by an observer moving with a particular fluid element. In many convective heat transfer problems, the continuity equation may be required in other coordinate systems. Cylindrical coordinate system and spherical coordinate system are shown in Figures 2.5 and 2.6, respectively. z

Figure 2.5

(r, θ, z)

x = r cos (θ) y = r sin (θ) z=z y

θ x

Cylindrical coordinates.

2.3 Differential Formulation of Conservation Equations

z (r, θ, ϕ)

x = r sin (θ) cos (ϕ) y = r sin (θ) sin (ϕ)

θ y

z = r cos (θ)

ϕ x Figure 2.6

Spherical coordinates.

2.3.1.1 Cylindrical Coordinates

In cylindrical coordinate system, the continuity equation is given in the following form. 𝜕ρ 1 𝜕 1 𝜕 𝜕 + (ρ rvr ) + (ρv ) + (ρvz ) = 0. 𝜕t r 𝜕r r 𝜕θ θ 𝜕z Here, the vr , vθ , and vz are the velocity components.

(2.31)

2.3.1.2 Spherical Coordinates

In spherical coordinate system, the continuity equation is given in the following form. 𝜕ρ 1 𝜕 1 1 𝜕 𝜕 + (ρ r2 vr ) + (ρv sin(θ)) + (ρv ) = 0. 𝜕t r2 𝜕r r sin(θ) 𝜕θ θ r sin(θ) 𝜕φ φ

(2.32)

Here, the vr , vθ , and vϕ are the velocity components. Example 2.1 Consider that incompressible constant property fluid is flowing between two infinite parallel plates. Far from the entrance of the channel, entrance effects are not important and fluid particles move parallel to plates (streamlines are parallel to plates) and flow is called fully developed. We wish to write the continuity equation in a fully developed region of the channel. Solution Assumptions 1) Incompressible flow (ρ = const) 2) Streamlines are parallel to plates. Flow is fully developed (𝜕/𝜕 x = 0) Continuity equation 𝜕ρ 𝜕 𝜕 𝜕 (ρu) + (ρv) + (ρw) + = 0. 𝜕x 𝜕y 𝜕z 𝜕t For incompressible flow, density ρ = const, and we have 𝜕ρ = 0. 𝜕t Since the channel formed by two infinite parallel plates is very wide, changes in the flow properties in the z-direction are negligible. There are no edge effects. Thus, 𝜕 = 0. 𝜕z Since flow is in axial direction, there is no motion in the z-direction. Thus, we state that w = 0. For fully developed flow, we have 𝜕u = 0. 𝜕x

41

42

2 Fundamental Equations of Laminar Convective Heat Transfer

Continuity becomes 𝜕v = 0. 𝜕y

2.3.2

Conservation of Linear Momentum in Differential Form

If the value of properties at each point in a fluid is required or if the properties vary with time, differential form of the linear momentum equation becomes very useful. To provide a good understanding of physical process, we will proceed with usual approach for the derivation of differential form of linear momentum equation. The infinitesimal stationary cubical differential control volume is shown in Figure 2.7, and the center of the cubical element is at the point (x, y, z). For clarity, we have shown stresses acting on the faces. We imagine that this infinitesimal differential control volume element of fluid is removed from a viscous boundary layer. We removed the surfaces perpendicular to the z-axis for clarity. Forces on the fluid in the control volume are due to both surface stresses and body forces. These forces are listed below: 1) Surface forces a) pressure forces b) viscous forces 2) Body forces We will assume that the any one of the following body forces may act on this fluid element. a) Centrifugal force b) Coriolis force c) Electromagnetic force d) Gravity force

(x, y, z)

y v x z

u w

Figure 2.7

Convention for stress components on a differential control volume element for a viscous fluid.

2.3 Differential Formulation of Conservation Equations

For simplicity, let us work with only the x-component of the momentum equation, i.e. Fsx +

∭

fx ρd∀ =

CV

⃗ ⋅ dA) ⃗ + u(ρV

∬

d (ρu)d∀. dt ∭

CS

CV

To determine the x-component of the surface force, Fsx , consider Figure 2.7. The forces and stresses acting on the fluid element are the following: 1) Normal stress, for example, τx x , is the normal stress on the surface normal to the x-axis and acting in the x-direction, and this stress includes static pressure (thermodynamic pressure) force. 2) Shear stress, for example, τx y is the shear stress on the surface normal to the x-axis and acting in the y-direction. 3) Body forces. We will consider only the gravitational body force. We will use fx , fy , fz for the gravitational force per unit mass in the x, y, and z coordinate directions, respectively. Note that for simplicity, we will evaluate the forces acting on the infinitesimal control volume acting in the x-direction. We shall apply the x-component of the integral form of the momentum equation to infinitesimal control volume to obtain the x-component of the Navier–Stokes equation in Cartesian coordinates. We now evaluate each term of the x-component of the integral form of the momentum equation as follows. a) External surface forces: These are external forces acting on the surface of control volume. The stresses in the x-direction give rise to these forces. These forces are | | ̃ τxx ||x+ Δx ΔyΔz − τxx ||x− Δx ΔyΔz + τyx | Δy ΔxΔz − τyx | Δy ΔxΔz Fsx = |y+ 2 |y− 2 2 2 + τzx ||z+ Δz ΔxΔy − τzx ||z− Δz ΔxΔy. 2 2 Dividing by Δx Δy Δz, we obtain Fsx ΔxΔyΔz

= ̃

τxx ||x+ Δx − τxx ||x− Δx 2

2

Δx

+

| | τyx | Δy − τyx | Δy |y+ 2 |y− 2 Δy

+

τzx ||z+ Δz − τzx ||z− Δz 2

2

Δz

.

Then, taking the limit as Δx → 0, Δy → 0, Δz → 0, we obtain ] 𝜕 τxx 𝜕 τyx 𝜕 τzx x-component of surface force F′′′ = ̃ = + + . sx per unit volume 𝜕x 𝜕y 𝜕z

(2.33)

b) Body forces: Body forces act on the distributed mass inside the control volume. The x-direction body force acting on the infinitesimal control volume is ∭

fx ρd∀ ≈ fx ρ ΔxΔyΔz.

∀

Upon dividing by the volume Δx Δy Δz and taking the limit as Δx Δy Δz → 0, we obtain ] Body force per F′′′ = ≈ ρfx . Bx unit volume

(2.34)

c) Convective momentum ⃗ ⋅ dA) ⃗ into six parts, and we assume that Since our control volume has six faces, we will break up the integral ∬ u(ρV CS

momentum flux is uniform on each face. For example, momentum flux on control surface 1 (CS1) is ∬

⃗ ⋅ dA) ⃗ ≈ [u(ρ u)]x+Δx∕2 ΔyΔz. u(ρV

CS 1

Evaluating momentum flux from each face and adding them up, we obtain ∬

⃗ ⋅ dA) ⃗ = u(ρV ̃ (uρu)x+ Δx ΔyΔz − (uρu)x− Δx ΔyΔz + (uρv)y+ Δy ΔxΔz − (uρv)y− Δy ΔxΔz 2

2

2

CS

+ (uρw)z+ Δz ΔxΔy − (uρw)z− Δz ΔxΔy. 2

2

2

43

44

2 Fundamental Equations of Laminar Convective Heat Transfer

Upon dividing by ΔxΔyΔz and letting Δx → 0, Δy → 0, Δz → 0, we get M′′′ x =

𝜕 𝜕 𝜕 (uρu) + (uρv) + (uρw). 𝜕x 𝜕y 𝜕z

(2.35)

d) Rate of change of momentum of fluid in control volume (CV): For the x-component CV linear momentum of fluid, we have 𝜕 d (ρu)d∀ = ̃ (ρu)ΔxΔyΔz dt ∭ 𝜕t ∀

where we have used an average value of the density–velocity product. Dividing by the volume Δx Δy Δz and taking the limit as Δx Δy Δz → 0, we obtain ⋅

𝜕 (ρ u). (2.36) 𝜕t Substituting Eqs. (2.33)–(2.36) into Eq. (2.8a), we get the differential form of the linear momentum equation for the x-direction as 𝜕τyx 𝜕τzx 𝜕τ 𝜕 𝜕 𝜕 𝜕 (2.37a) (uρu) + (uρv) + (uρw) + (ρu) = xx + + + ρfx . 𝜕x 𝜕y 𝜕z 𝜕t 𝜕x 𝜕y 𝜕z M′′′ x =

In a similar fashion, we can write equations for the y and z directions 𝜕τxy 𝜕τyy 𝜕τzy 𝜕 𝜕 𝜕 𝜕 (vρu) + (vρv) + (vρw) + (ρv) = + + + ρfy 𝜕x 𝜕y 𝜕z 𝜕t 𝜕x 𝜕y 𝜕z 𝜕τyz 𝜕τzz 𝜕τ 𝜕 𝜕 𝜕 𝜕 (wρu) + (wρv) + (wρw) + (ρw) = xz + + + ρfz . 𝜕x 𝜕y 𝜕z 𝜕t 𝜕x 𝜕y 𝜕z

(2.37b) (2.37c)

We can put these equations in a more familiar form. Consider the terms on the left-hand side of the x-component of the momentum equation, Eq. (2.37a), as { } 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕 𝜕 (uρu) + (uρv) + (uρw) + (ρu) = u (ρu) + (ρv) + (ρw) + (ρ) 𝜕x 𝜕y 𝜕z 𝜕t 𝜕x 𝜕y 𝜕z 𝜕t { } 𝜕u 𝜕u 𝜕u 𝜕u +ρ u +v +w + . 𝜕x 𝜕y 𝜕z 𝜕t Recall that the substantial derivative is 𝜕 𝜕 𝜕 𝜕 D =u +v +w + . Dt 𝜕x 𝜕y 𝜕z 𝜕t The term on the left-hand side now becomes 𝜕 𝜕 𝜕 Du 𝜕 (uρu) + (uρv) + (uρw) + (ρu) = ρ +ρ 𝜕x 𝜕y 𝜕z 𝜕t Dt

{

𝜕u 𝜕u 𝜕u 𝜕u u +v +w + 𝜕x 𝜕y 𝜕z 𝜕t

} .

(2.38)

The second term on the right-hand side is zero because of the continuity equation. Thus, the x-component of the momentum equation becomes ρ

D u 𝜕 τxx 𝜕 τyx 𝜕 τzx = + + + ρfx . Dt 𝜕x 𝜕y 𝜕z

(2.39a)

We can write the other components of the momentum equation in a similar way, and they are given as D v 𝜕 τxy 𝜕 τyy 𝜕 τzy = + + + ρfy Dt 𝜕x 𝜕y 𝜕z D w 𝜕 τxz 𝜕 τyz 𝜕 τzz ρ = + + + ρfz . Dt 𝜕x 𝜕y 𝜕z ρ

(2.39b) (2.39c)

The problem is reduced to obtaining useful expressions for normal and shear stresses. We have obtained the viscous stress equations of motion in the Cartesian coordinate system. These equations relate the acceleration of a fluid element to the net body and surface forces. We need to relate normal and shearing stresses to velocity field. This is accomplished through the introduction of phenomenological relations known as constitutive equation. A constitutive equation is an empirical relationship that attempts to relate the stress in a continuum to the manner in which the material is deformed. Stress is a

2.3 Differential Formulation of Conservation Equations

linear function of rate of strain for Newtonian fluid. See [2–4]. For a Newtonian fluid, the shearing stresses are symmetric and are written in rectangular coordinates as follows: ) ( 𝜕u 𝜕v + (2.40a) τxy = τyx = μ 𝜕y 𝜕x ) ( 𝜕u 𝜕w + (2.40b) τxz = τzx = μ 𝜕z 𝜕x ) ( 𝜕v 𝜕w + . (2.40c) τyz = τzy = μ 𝜕z 𝜕y The normal stresses are

( ) [ ] 𝜕u 2 𝜕u 2 𝜕u 𝜕v 𝜕w ⃗ − μ + + = −p + μ 2 − div(V) 𝜕x 3 𝜕x 𝜕y 𝜕z 𝜕x 3 ( ) [ ] 𝜕v 2 ⃗ ⃗ 𝜕v 2 𝜕u 𝜕v 𝜕w τyy = −p + 2μ − μ + + = −p + μ 2 − ∇ ⋅V 𝜕y 3 𝜕x 𝜕y 𝜕z 𝜕y 3 ( ) [ ] 𝜕w 2 ⃗ ⃗ 𝜕w 2 𝜕u 𝜕v 𝜕w τz z = −p + 2μ − μ + + = −p + μ 2 − ∇⋅V . 𝜕z 3 𝜕x 𝜕y 𝜕z 𝜕z 3 τx x = −p + 2μ

(2.41a) (2.41b) (2.41c)

The term μ = ρ ν is called the coefficient of viscosity, dynamic viscosity, or absolute viscosity and ν is the kinematic viscosity of fluid and ρ is the density of the fluid. Fluid viscosity μ is a fluid property and depends on temperature and pressure. The ⃗ is called the divergence of V ⃗ and is defined in rectangular coordinates by ⃗ ⋅V term ∇ ⃗ = div(V) ⃗ = 𝜕u + 𝜕v + 𝜕w . ⃗ ⋅V ∇ 𝜕x 𝜕y 𝜕z With these equations for normal and shearing stresses, we obtain the differential form of the equations of motion [ ( )] [ ( )] 𝜕p Du 𝜕 𝜕u 2 ⃗ ⃗ 𝜕 𝜕u 𝜕v ρ = ρfx − + μ 2 − ∇⋅V + μ + Dt 𝜕x 𝜕x 𝜕x 3 𝜕y 𝜕y 𝜕x [ ( )] 𝜕 𝜕u 𝜕w + μ + (2.42a) 𝜕z 𝜕z 𝜕x [ ( )] [ ( )] 𝜕p 𝜕 𝜕v 2 ⃗ ⃗ 𝜕 Dv 𝜕u 𝜕v = ρfy − + μ 2 − ∇ ⋅V + μ + ρ Dt 𝜕y 𝜕y 𝜕y 3 𝜕x 𝜕y 𝜕x [ ( )] 𝜕v 𝜕w 𝜕 μ (2.42b) + + 𝜕z 𝜕z 𝜕y [ ( )] [ ( )] 𝜕p 𝜕 𝜕w 2 ⃗ ⃗ 𝜕 𝜕u 𝜕w Dw = ρfz − + μ 2 − ∇⋅V + μ + ρ Dt 𝜕z 3 𝜕x 𝜕z 𝜕x [𝜕z ( 𝜕z )] 𝜕 𝜕v 𝜕w + μ + . (2.42c) 𝜕y 𝜕z 𝜕y These are the equations of motion for viscous compressible fluid and usually are called the celebrated Navier–Stokes equations. 2.3.2.1 Equation of Motion for a Newtonian Fluid with Constant Dynamic Viscosity 𝛍 and Density 𝛒

When the dynamic viscosity and density are constant, the fluid is called incompressible, and effect temperature variations are small, Navier–Stokes equations can be simplified. These simplified equations are given below in different coordinate systems. Presenting these equations in different coordinates will be very useful. For example, the use of cylindrical coordinates in pipe flow or for external flow over bodies of revolution is the most convenient coordinate system. On the other hand, one uses the spherical coordinate system for flow over a sphere. 2.3.2.2 Cartesian Coordinates (x, y, z)

x-direction: ) ( 2 ) ( 𝜕p 𝜕u 𝜕u 𝜕u 𝜕 u 𝜕2u 𝜕2u 𝜕u +u +v +w = ρfx − +μ + 2 + 2 ρ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x 𝜕x2 𝜕y 𝜕z

(2.43a)

45

46

2 Fundamental Equations of Laminar Convective Heat Transfer

y-direction: ) ( 2 ) ( 𝜕p 𝜕v 𝜕v 𝜕v 𝜕 v 𝜕2 v 𝜕2 v 𝜕v +u +v +w = ρfy − +μ + + ρ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕y 𝜕x2 𝜕y2 𝜕z2 z-direction: ) ) ( ( 2 𝜕p 𝜕w 𝜕w 𝜕w 𝜕w 𝜕 w 𝜕2 w 𝜕2 w = ρfz − . + + ρ +u +v +w +μ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕z 𝜕x2 𝜕y2 𝜕z2

(2.43b)

(2.43c)

This set of equations may be more compactly written in vector notation as ρ

⃗ DV ⃗ ⃗ + μ∇2 V. = ρ⃗f − ∇p Dt

(2.43d)

2.3.2.3 Cylindrical Coordinates (r, 𝛉, z)

In many cases, it is required to formulate a flow problem in cylindrical coordinate systems. r-direction: ] [ v 𝜕v v 2 𝜕vr 𝜕v 𝜕v 𝜕p + vr r + θ r − θ + vz r = ρfr − ρ 𝜕t 𝜕r r 𝜕θ r 𝜕z 𝜕r [ 2 2 2 ] 𝜕v 𝜕 𝜕 vr 1 𝜕vr vr 𝜕 v v 1 2 − 2 + 2 2r − 2 θ + 2r +μ + r 𝜕r 𝜕r2 r r 𝜕θ r 𝜕θ 𝜕z θ-direction: ] [ 𝜕v 𝜕v v 𝜕v vv 𝜕vθ 1 𝜕p + vr θ + θ θ + r θ + vz θ = ρfθ − ρ 𝜕t 𝜕r r 𝜕θ r 𝜕z r 𝜕θ [ ] 2 2 2 𝜕 vθ 1 𝜕vθ vθ 1 𝜕 vθ 2 𝜕vr 𝜕 vθ +μ + + 2 − 2 + 2 + r 𝜕r 𝜕r2 r r 𝜕θ2 r 𝜕θ 𝜕z2 z-direction: ] [ v 𝜕v 𝜕v 𝜕v 𝜕vz 𝜕p + vr z + θ z + vz z = ρfz − ρ 𝜕t 𝜕r r 𝜕θ 𝜕z 𝜕z [ 2 2 2 ] 𝜕 𝜕 v v 𝜕 vz 1 𝜕vz 1 z + 2 2z + +μ + r 𝜕r 𝜕r2 r 𝜕θ 𝜕z2

(2.44a)

(2.44b)

(2.44c)

where vr , vθ , and vz are the velocity components in the r, θ, and z directions, and fr , fθ , and fz are components of body force per unit volume. 2.3.2.4 Spherical Coordinates (r, 𝛉, 𝛟)

r-direction: ( ) vϕ 𝜕vr v2θ + v2ϕ 𝜕vr vθ 𝜕vr 𝜕vr 𝜕p ρ + vr + + − = ρfr − 𝜕t 𝜕r r 𝜕θ r sin(θ) 𝜕ϕ r 𝜕r [ ( ) 2 2 ( ) 𝜕 vr 𝜕v 1 1 𝜕 1 𝜕 + μ 2 2 r2 vr + 2 sin(θ) r + 2 2 𝜕θ r 𝜕r r sin(θ) 𝜕θ r sin (θ) 𝜕ϕ2 ] 𝜕vϕ 2v 2 2 𝜕v 2 − 2 r − 2 θ − 2 vθ cot(θ) − 2 r r 𝜕θ r r sin(θ) 𝜕ϕ θ-direction ) ( vϕ 𝜕vθ vr vθ − v2ϕ cot(θ) 𝜕vθ 𝜕vθ vθ 𝜕vθ 1 𝜕p + vr + + + = ρfθ − ρ 𝜕t 𝜕r r 𝜕θ r sin(θ) 𝜕ϕ r r 𝜕θ [ ( ) ( ) 2 𝜕v 𝜕 vθ 1 𝜕 1 𝜕 1 1 𝜕 r2 θ + 2 vθ sin(θ) + +μ 2 2 𝜕r r 𝜕r r 𝜕θ sin(θ) 𝜕θ r2 sin (θ) 𝜕ϕ2 ] 2 cot(θ) 𝜕vϕ 2 𝜕v + 2 r − 2 r 𝜕θ r sin(θ) 𝜕ϕ

(2.45a)

(2.45b)

2.3 Differential Formulation of Conservation Equations

ϕ-direction ( ) vϕ 𝜕vϕ vr vϕ + vθ vr cot(θ) 𝜕vϕ vθ 𝜕vϕ 𝜕vϕ 𝜕p 1 ρ + vr + + + = ρfϕ − 𝜕t 𝜕r r 𝜕θ r sin(θ) 𝜕ϕ r r sin(θ) 𝜕ϕ [ ( ) ( ) 2 𝜕 𝜕v v ϕ ϕ 1 𝜕 1 1 𝜕 1 𝜕 r2 + 2 v sin(θ) + +μ 2 2 2 2 𝜕r r 𝜕r r 𝜕θ sin(θ) 𝜕θ ϕ r sin (θ) 𝜕ϕ ] 𝜕vr 2 cot(θ) 𝜕vθ 2 + + 2 r sin(θ) 𝜕ϕ r2 sin(θ) 𝜕ϕ

(2.45c)

where vr , vθ , and vϕ are the velocity components in the r, θ, and ϕ directions, and fr , fθ , and fϕ are components of body force per unit volume.

2.3.3 Conservation of Energy in Differential Form Whenever a fluid element is distorted, energy conversions will take place. Work done on fluid element may be converted into other forms of energy. For example, it may result in an increase in the internal energy of fluid element. Stated in integral form for a finite control volume, the first law is ⋅ − → d ⃗ ⋅ dA) ⃗ = − q′′ ⋅ dA ⃗ − Wshaft + ⃗ (eρ)d∀ + e(ρV (⃗τ ⋅ V)dA + u′′′ d∀. ∬ ∬ ∬ ∫ ∭ dt CS

∀

CS

CS

CV

We will now apply the integral form of energy equation to a control volume for which no mechanical or shaft work crosses the control surface to derive the differential form of the conservation of energy principle. This form of energy equation is very useful to study heat transfer problems in engineering. The derivation will be carried out in a step-by-step manner. We need a development of equations, which represents the energy conversion so that we will have a good understanding of the processes taking place. For this purpose, we choose a cubical control volume, containing fluid element, as shown in Figure 2.8. The control volume (cubical fluid element) has the dimensions ΔxΔyΔz. Its volume center is the point, (x, y, z). Each term in the energy integral equation will now be evaluated by using suitable approximations and meaning of each term will be discussed. 1)

d ∭ dt ∀

(eρ)d∀: The time rate of total energy storage will be evaluated. Using the average value of product of density ρ and

specific energy e over the volume element, our differential control volume element becomes [ ] 𝜕 d (eρ)d∀ ≈ (̃ ρe) ΔxΔyΔz dt ∭ 𝜕t

(2.46)

∀

or after dividing by the volume and letting Δx → 0, Δy → 0, Δz → 0 go to zero, we get ] Rate of change of 𝜕 = (ρe). stored energy per unit volume 𝜕t

(2.47)

⃗ ⋅ dA): ⃗ The rate of energy efflux from the control volume will now evaluated. This will be done by evaluating the 2) ∬ e(ρV CS

surface integral for each of the six surfaces of the control volume. For our six-sided control volume, we have ⃗ ⋅ dA) ⃗ ≅ e(ρV

∬

6 ∑

⃗ ⋅ dA) ⃗ k (eρV

(2.48)

k=1

CS

⃗ for each surface by evaluating the properties at the area center of each upon approximating the mean value of e ρV −→ −→ surface. For example, for the control surface CS 1 of Figure 2.8, energy convected at surface dA1 (dA1 is the surface at x + Δx/2) is ⃗ ⋅ dA) ⃗ = (e ρ u)x+Δx∕2 ΔyΔz. e(ρV

∬ CS1

Thus, forming the similar terms for the remaining surfaces, adding them up, and dividing by volume, we have ⃗ ⋅ dA) ⃗ ∬ e(ρV CS

ΔxΔyΔz

{ ≅

eρu|x+ Δx − eρu|x− Δx 2

2

Δx

+

eρv|y+ Δy − eρv|y− Δy 2

2

Δy

+

eρw|z+ Δz − eρw|z− Δz 2

2

Δz

} .

(2.49a)

47

48

2 Fundamental Equations of Laminar Convective Heat Transfer

After taking the limit as Δx → 0, Δy → 0, Δz → 0, control volume dimensions approach zero, and we get ] Net energy efflux 𝜕 𝜕 𝜕 = (e ρu) + (e ρv) + (e ρw). per unit volume 𝜕x 𝜕y 𝜕z

(2.49b)

| − → ⃗ || is the rate of heat transfer into control volume by conduction. That is, we assume that heat enters to 3) ∬ q′′ ⋅ dA | CS |diffusion the control volume by diffusion due to temperature gradient in the fluid relative to the bulk flow of fluid. Since this energy transfer process occurs at the control surface, and since this surface is made up of six plane surfaces, we have 6 ∑ − → − → ⃗ = ⃗ k. ̃ (q′′ ⋅ dA) q′′ ⋅ dA

∬ CS

(2.50)

k=1

See Figure 2.8. The integral over each of these plane surfaces will be approximated by the value of the integrand at the area center of the surface. The result of evaluation is − → ⃗ q′′ ⋅ dA | | ⎧ ′′ | ⎫ ′′ | ∬ q′′y | Δy − q′′y | Δy q′′z ||z+ Δz − q′′z ||z− Δz ⎪ |y+ 2 |y− 2 ⎪ qx |x+ Δx2 − qx |x− Δx2 CS 2 2 (2.51) =⎨ + + ⎬. Δx Δy Δz Δx Δy Δz ⎪ ⎪ ⎩ ⎭ Taking the limit as Δx → 0, Δy → 0, Δz → 0, we get the heat transfer rate per unit volume − → ⃗ q′′ ⋅ dA { } ∬ 𝜕q′′y − → 𝜕q′′x 𝜕q′′z CS ⃗ ⋅ q′′ = + + =∇ Δx Δy Δz 𝜕x 𝜕y 𝜕z

(2.52)

⃗ is where ∇ ⃗ ⃗ = 𝜕 ⃗i + 𝜕 ⃗j + 𝜕 k. ∇ 𝜕x 𝜕y 𝜕z The component of heat flux vector in each direction is given as 𝜕T q′′x = −k 𝜕x 𝜕T q′′y = −k 𝜕y 𝜕T q′′z = −k . 𝜕z The heat transfer rate per unit volume can also be expressed as ( ) ( )} ] { ( ) Heat conduction 𝜕T 𝜕 𝜕T 𝜕 𝜕T 𝜕 k + k + k . =− per unit volume 𝜕x 𝜕x 𝜕y 𝜕y 𝜕z 𝜕y

(x, y, z)

y x z Figure 2.8

Heat conduction terms on a differential control volume element.

(2.53)

2.3 Differential Formulation of Conservation Equations

We can add the rate of energy addition by radiation per unit volume; fortunately, the temperatures in convection problems are low enough so that we neglect the radiation term. δW 4) dt s : The rate of doing shaft work is assumed to be zero. ⃗ dA is the flow work term, and it is the work done by surface forces. This work rate occurs at the control volume 5) ∬ (⃗τ ⋅ V) CS

surface and is formally evaluated by carrying out the integration. But this is tedious and difficult task, and we will estimate the work term as follows. Again, recall that we are considering a stationary control volume in rectangular coordinates. We are interested in the rate at which work is done on the control volume by surface forces. The flow work term is evaluated at each face of the cubical control volume. We will again break the control surface into six plane surface elements and evaluate the stress vector τ⃗ and the velocity ⃗ at the area center of each surface element. To clarify this point, work terms on the left face as well as right face vector V of the cubical fluid element are shown in Figure 2.9. For our six-sided control volume, we have ⃗ (⃗τ ⋅ V)dA ≅

∬

6 ∑

⃗ [(⃗τ ⋅ V)dA] k.

(2.54)

k=1

CS

This means that we will evaluate the work term for each of the six faces of control volume. The work done per unit time on a face of the control volume (cubical element) by the environment is the force on that face multiplied by the velocity component in the direction of that force. We are talking about the rate at which work is done by the environment on the fluid mass inside the control volume. Note that the rate of work is negative when force and velocity are oppositely directed. For clarity, we have shown only the terms on the two faces of the cubical element. See Figure 2.9. Before we proceed, let us clarify this point with an example. We wish to evaluate the rate of work done on element on x faces due to normal stresses. See Figure 2.10. Net rate of work on element on x faces due to normal stress = (force on the left face) (velocity on the left face) + (force on the right face) (velocity on the right face) ] net rate of work on element = (ux+Δx∕2 )(τx x )x+Δx∕2 (ΔyΔz) − (ux−Δx∕2 )(τx x )x−Δx∕2 (ΔyΔz) on x face due to normal stress = (uτxx )x+Δx∕2 (ΔyΔz) − (uτxx )x−Δx∕2 (ΔyΔz).

z y x Figure 2.9

Differential control volume element for a viscous fluid.

ux–Δx/2 τxx–Δx/2

(x,y,z)

ux+Δx/2 τxx+Δx/2 Δz

Δx

Figure 2.10

Normal stress and x-velocity component on x faces.

49

50

2 Fundamental Equations of Laminar Convective Heat Transfer

Dividing by Δx Δy Δz and taking limit as Δx → 0, Δy → 0, Δz → 0, we obtain net rate of work on element ⎤ 𝜕(uτxx ) . on x face due to normal stress⎥ = ⎥ 𝜕x per Δx Δy Δz ⎦ In a similar manner, we can write the rate of work on element on x faces due to shear stress 𝜕 𝜕 (v τx y ) + (w τx z ). 𝜕x 𝜕x We now proceed to evaluate the work term for each of the six faces of the control volume. We will again break the control ⃗ at the area center volume surface into six plane surface elements and evaluate the stress vector τ⃗ and velocity vector V of each element. Consider the right face of the cubical element as =

⃗ ⃗ ⃗ ⃗ ⃗ τ⃗ = [τx x⃗i + τx y⃗j + τx z k] x+Δx∕2 and V = [ui + vj + wk]x+Δx∕2 ∬

⃗ ⃗ ⃗ ⃗ ⃗ (⃗τ ⋅ V)dA = [τx x⃗i + τx y⃗j + τx z k] x+Δx∕2 ⋅ [ui + vj + wk]x+Δx∕2

CS1

= [uτx x + vτx y + wτx z ]x+Δx∕2 . We now consider the left face of the cubical element as ⃗ ⃗ ⃗ ⃗ ⃗ τ⃗ = −[τx x⃗i + τx y⃗j + τx z k] x−Δx∕2 and V = [ui + vj + wk]x−Δx∕2 ∬

⃗ ⃗ ⃗ ⃗ ⃗ (⃗τ ⋅ V)dA = −[τx x⃗i + τx y⃗j + τx z k] x+Δx∕2 ⋅ [ui + vj + wk]x+Δx∕2

CS3

= −[uτx x + vτx y + wτx z ]x−Δx∕2 . Consider now the top surface of the cubical element as ⃗ ⃗ ⃗ ⃗ ⃗ τ⃗ = [τy x⃗i + τy y⃗j + τy z k] y+Δy∕2 and V = [ui + vj + wk]y+Δy∕2 ∬

⃗ ⃗ ⃗ ⃗ ⃗ (⃗τ ⋅ V)dA = [τy x⃗i + τy y⃗j + τy z k] y+Δy∕2 ⋅ [ui + vj + wk]y+Δy∕2

CS2

= [uτy x + vτy y + wτy z ]y+Δy∕2 . Consider now the bottom surface of the cubical element as ⃗ ⃗ ⃗ ⃗ ⃗ τ⃗ = −[τy x⃗i + τy y⃗j + τy z k] y−Δy∕2 and V = [ui + vj + wk]y−Δy∕2 ∬

⃗ ⃗ ⃗ ⃗ ⃗ (⃗τ ⋅ V)dA = −[τy x⃗i + τy y⃗j + τy z k] y−Δy∕2 ⋅ [ui + vj + wk]y−Δy∕2

CS4

= −[uτy x + vτy y + wτy z ]y−Δy∕2 . Consider now the front face of the cubical element as ⃗ ⃗ ⃗ ⃗ ⃗ τ⃗ = [τz x⃗i + τz y⃗j + τz z k] z+Δz∕2 and V = [ui + vj + wk]z+Δz∕2 ∬

⃗ ⃗ ⃗ ⃗ ⃗ (⃗τ ⋅ V)dA = [τz x⃗i + τz y⃗j + τz z k] z+Δz∕2 ⋅ [ui + vj + wk]z+Δz∕2

CS5

= [uτz x + vτz y + wτz z ]z+Δz∕2. Finally, we consider the back face of the cubical element as ⃗ ⃗ ⃗ ⃗ ⃗ τ⃗ = −[τz x⃗i + τz y⃗j + τz z k] z−Δz∕2 and V = [ui + vj + wk]z−Δz∕2 ∬

⃗ ⃗ ⃗ ⃗ ⃗ (⃗τ ⋅ V)dA = −[τz x⃗i + τz y⃗j + τz z k] z−Δz∕2 ⋅ [ui + vj + wk]z−Δz∕2

CS5

= −[uτz x + vτz y + wτz z ]z−Δz∕2 .

2.3 Differential Formulation of Conservation Equations

After evaluating work term for each surface, we add them up, and after rearranging, we get ⃗ ∬ (⃗τ ⋅ V)dA CS

≅

ΔxΔyΔz

[uτxx + vτxy + wτxz ]x+ Δx − [uτxx + vτxy + wτxz ]x− Δx 2

+

2

Δx [uτyx + vτyy + wτyz ]y+ Δy − [uτyx + vτyy + wτyz ]y− Δy 2

2

Δy [uτzx + vτzy + wτzz ]z+ Δz − [uτzx + vτzy + wτzz ]z− Δz

2 . (2.55a) Δz Upon taking the limit as Δx → 0, Δy → 0, Δz → 0, we get ] Flow work per 𝜕 𝜕 = {uτxx + vτxy + wτxz } + {uτyx + vτyy + wτyz } unit volume 𝜕x 𝜕y 𝜕 (2.55b) + {uτzx + vτzy + wτzz }. 𝜕z 6) ∫ u′′′ d∀ is the energy generation term. This is the net release of thermal energy within the control volume due to

+

2

CV

volumetric effects. It is possible for some energy to be generated within control volume due to chemical reaction, electromagnetic effects, nuclear effects, induction heating, and the dissipation of electrical energy. This form of energy is converted into thermal energy and leads to internally distributed heat source. These are the rates of energy input from external sources per unit volume. This is called volumetric thermal energy generation rate. These energy generation ′′′ effects are included in a single term, u (W/m3 ). Then, we have ∫ u′′′ d∀ CV

Δx Δy Δz

=

] Energy generation = u′′′ . per unit volume

(2.56)

Substituting Eqs. (2.47), (2.49b), (2.53), (2.55b), and (2.56) into Eq. (2.20), the integral form of the first law of thermodynamics, we obtain the first law of thermodynamics in differential form { ( ( ) ( )} ) 𝜕 𝜕 𝜕 𝜕 𝜕T 𝜕 𝜕T 𝜕 𝜕T 𝜕 (ρe) + (e uρ) + (e vρ) + (e wρ) = k + k + k + u′′′ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x 𝜕x 𝜕y 𝜕y 𝜕z 𝜕y 𝜕 + {uτxx + vτxy + wτxz } 𝜕x 𝜕 + {uτyx + vτyy + wτyz } 𝜕y 𝜕 (2.57) + {uτzx + vτzy + wτzz }. 𝜕z This form of energy equation is not in a useful form since it contains shear and normal stresses. Note that stresses are not easy to measure. Considering the terms on the left-hand side of Eq. (2.57) and rearranging these terms by performing the differentiations, we obtain 𝜕 𝜕 𝜕 𝜕 𝜕e 𝜕e 𝜕e 𝜕e (ρue) + (ρve) + (ρwe) + (ρe) = ρu + ρv + ρw + ρ 𝜕x 𝜕y 𝜕z 𝜕t 𝜕x 𝜕y 𝜕z 𝜕t } { 𝜕ρu 𝜕ρv 𝜕ρw 𝜕ρ + + + (2.58) +e 𝜕x 𝜕y 𝜕z 𝜕t ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ 0

where the first term on the right-hand side of Eq. (2.58) is the substantial derivative of e, continuity. Then, we have 𝜕 𝜕 𝜕 De 𝜕 (uρe) + (vρe) + (wρe) + (ρe) = ρ . 𝜕x 𝜕y 𝜕z 𝜕t Dt We can now write Eq. (2.57) in the following form: { ( ( ) ( )} ) 𝜕T 𝜕 𝜕T 𝜕 𝜕T De 𝜕 = k + k + k + u′′′ ρ Dt 𝜕x 𝜕x 𝜕y 𝜕y 𝜕z 𝜕y 𝜕 + {uτxx + vτxy + wτxz } 𝜕x

De , Dt

and the second term is zero by (2.59)

51

52

2 Fundamental Equations of Laminar Convective Heat Transfer

𝜕 {uτyx + vτyy + wτyz } 𝜕y 𝜕 + {uτzx + vτzy + wτzz }. 𝜕z Also, in our development of the first law, we have taken the total energy per unit mass to be . 1 ̃ ̃ + (u2 + v2 + w2 ) + ϕ e=u 2 where +

(2.60)

e = specific total energy ̃ = specific internal energy u 1 2 V = 12 (u2 + v2 + w2 ) = specific kinetic energy 2 ̃ = specific potential energy ϕ

̃ = ϕ(x, ̃ y, z), i.e. that ϕ ̃ is independent of time. To obtain the first law as we now have it, we implicitly assumed that ϕ Thus, the first term, De/Dt, on the left-hand side of the first law, Eq. (2.60), becomes [ ] Dϕ ̃ De D̃ u D 1 2 = + (u + v2 + w2 ) + (2.61a) Dt Dt Dt 2 Dt where Dϕ/Dt is ̃ ̃ ̃ ̃ 𝜕ϕ 𝜕ϕ 𝜕ϕ Dϕ =u +v +w + Dt 𝜕x 𝜕y 𝜕z

̃ 𝜕ϕ . 𝜕t ⏟⏟⏟

(2.61b)

0

Noting that ̃ ̃ ̃ ̃ =. 𝜕 ϕ ⃗i + 𝜕 ϕ ⃗j + 𝜕 ϕ k⃗ = −⃗f = −{f ⃗i + f ⃗j + f k} ⃗ ⃗ϕ ∇ x y z 𝜕x 𝜕y 𝜕z

(2.61c)

we have ̃ ̃ ̃ 𝜕ϕ 𝜕ϕ 𝜕ϕ = −fx , = −fy , = −fz . 𝜕x 𝜕y 𝜕z

(2.61d)

Therefore, ̃ Dϕ = −{ufx + vfy + wfz }. Dt This is the work done by the body forces; then, the De/Dt term takes the following form: [ ] u D 1 2 De D̃ = + (u + v2 + w2 ) − {ufx + vfy + wfz }. Dt Dt Dt 2 Finally, after rearrangement, the first law becomes [ ] 1 D ̃ + (u2 + v2 + w2 ) = ρ{ufx + vfy + wfz } u ρ Dt 2 ( ) ( )} { ( ) 𝜕T 𝜕 𝜕T 𝜕 𝜕T 𝜕 k + k + k + u′′′ 𝜕x 𝜕x 𝜕y 𝜕y 𝜕z 𝜕y 𝜕 𝜕 𝜕 + {uτxx + vτxy + wτxz } + {uτyx + vτyy + wτyz } + {uτzx + vτzy + wτzz } 𝜕x 𝜕y 𝜕z or

[ ] D 1 2 D̃ u +ρ V = ρ{ufx + vfy + wfz }+ ρ {Dt ( Dt ) 2 ( ) ( )} 𝜕T 𝜕 𝜕T 𝜕 𝜕T 𝜕 k + k + k + u′′′ 𝜕x 𝜕x 𝜕y 𝜕y 𝜕z 𝜕y 𝜕 𝜕 𝜕 + {uτxx + vτxy + wτxz } + {uτyx + vτyy + wτyz } + {uτzx + vτzy + wτzz } 𝜕x 𝜕y 𝜕z

(2.62)

(2.63)

(2.64)

(2.65)

2.3 Differential Formulation of Conservation Equations

where V2 = u2 + v2 + w2 . This energy equation is known as the total energy equation since it contains both thermal and mechanical energy terms. It contains shear and normal stresses. It is not an easy task to measure these stresses. We need to eliminate these unknowns. This form of the energy equation is not the most convenient one to solve most convection heat transfer problems. We need a conservation equation that isolates thermal phenomena, which will be convenient one in convection heat transfer problems. First, let us develop the conservation equation for mechanical energy. 2.3.3.1 Mechanical Energy Equation

Starting with the general form of the momentum equation, multiplying the x-component of momentum equation by u, the y-component of momentum equation by v, and the z-component of momentum equation by w, we get ( ) } { 𝜕τxx 𝜕τyx 𝜕τzx D u2 Du =ρ = ρufx + u + + (2.66a) ρu Dt Dt 2 𝜕x 𝜕y 𝜕z ( ) } { 𝜕τyx 𝜕τyy 𝜕τyz Dv D v2 ρv =ρ = ρvfy + v + + (2.66b) Dt Dt 2 𝜕x 𝜕y 𝜕z ( ) } { 𝜕τzx 𝜕τzy 𝜕τzz Dw D w2 ρw =ρ = ρwfz + w + + . (2.66c) Dt Dt 2 𝜕x 𝜕y 𝜕z Adding up these equations and rearranging, we obtain [ ] [ ] D u2 + v2 + w2 D V2 ρ =ρ = ρ{ufx + vfy + wfz } Dt 2 Dt 2 } { } { } { 𝜕τyx 𝜕τyy 𝜕τyz 𝜕τzx 𝜕τzy 𝜕τzz 𝜕τxx 𝜕τyx 𝜕τzx +v +w . +u + + + + + + 𝜕x 𝜕y 𝜕z 𝜕x 𝜕y 𝜕z 𝜕x 𝜕y 𝜕z (2.67) Notice that at this point, we obtained the energy equation from laws of mechanics and this equation is called mechanical energy equation. 2.3.3.2 Thermal Energy Equation

To obtain the thermal energy equation, we take the expression for ρD( V2 /2)/Dt term in the mechanical energy equation, Eq. (2.67), and substitute into the total energy equation, Eq. (2.65). This substitution yields } { } { 𝜕τyx 𝜕τyy 𝜕τyz 𝜕τxx 𝜕τyx 𝜕τzx D̃ u ρ + ρ{ufx + vfy + wfz } + u + + +v + + Dt 𝜕x 𝜕y 𝜕z 𝜕x 𝜕y 𝜕z } ( ) ( )} { ( { ) 𝜕τzx 𝜕τzy 𝜕τzz 𝜕T 𝜕 𝜕T 𝜕 𝜕T 𝜕 + + = ρ{ufx + vfy + wfz } + k + k + k + u′′′ +w 𝜕x 𝜕y 𝜕z 𝜕x 𝜕x 𝜕y 𝜕y 𝜕z 𝜕y 𝜕 𝜕 𝜕 + {uτxx + vτxy + wτxz } + {uτyx + vτyy + wτyz } + {uτzx + vτzy + wτzz }. (2.68a) 𝜕x 𝜕y 𝜕z Carrying out the indicated differentiations and after some algebra, we obtain { ( ( ) ( )} ) 𝜕T 𝜕 𝜕T 𝜕 𝜕T D̃ u 𝜕u 𝜕v 𝜕w 𝜕 = k + k + k + u′′′ + τxx + τyy + τzz ρ Dt 𝜕x 𝜕x 𝜕y 𝜕y 𝜕z 𝜕y 𝜕x 𝜕y 𝜕z ) ) ( ( ) ( 𝜕w 𝜕u 𝜕v 𝜕u 𝜕w 𝜕v + + τyz + + τxz + . + τxy 𝜕x 𝜕y 𝜕y 𝜕z 𝜕x 𝜕z

(2.68b)

Equation (2.68b) is called the thermal energy equation. Up to this point, we have made no assumption about the fluid. We assume that a Newtonian fluid and the second coefficient of viscosity are negligible. Note that we are now working with a Newtonian fluid. Notice that for Newtonian fluid shear stresses are symmetric, i.e. τxy = τyx , τyz = τzy , and τzz = τzx , as discussed in Kays et al. [2] and Schlichting and Gersten [4]. The normal stress terms can be modified as follows: ( )2 𝜕u 2 ⃗ ⃗ 𝜕u 𝜕u 𝜕u = −p − μ∇ ⋅ V + 2μ (2.69a) τxx 𝜕x 𝜕x 3 𝜕x 𝜕x ( )2 𝜕v 𝜕v 2 ⃗ ⃗ 𝜕v 𝜕v τyy (2.69b) = −p − μ∇ ⋅ V + 2μ 𝜕y 𝜕y 3 𝜕y 𝜕y ( ) 𝜕w 2 ⃗ ⃗ 𝜕w 𝜕w 2 𝜕w τzz = −p − μ∇ ⋅ V + 2μ . (2.69c) 𝜕z 𝜕z 3 𝜕z 𝜕z

53

54

2 Fundamental Equations of Laminar Convective Heat Transfer

We add up these equations to obtain 𝜕u 𝜕v 𝜕w ⃗ − 2 μ(∇ ⃗ 2 + 2μ ⃗ ⋅ V) ⃗ ⋅ V) τxx + τyy + τzz = −p(∇ 𝜕x 𝜕y 𝜕z 3

{ (

𝜕u 𝜕x

(

)2 +

𝜕v 𝜕y

)2

(

𝜕w + 𝜕z

)2

} .

Again, we modify the shear stress relations to obtain the following equations: ( ) ( )2 𝜕u 𝜕v 𝜕u 𝜕v + =μ + τxy 𝜕y 𝜕x 𝜕y 𝜕x ) ( ) ( 𝜕u 𝜕w 2 𝜕u 𝜕w τxz + =μ + 𝜕z 𝜕x 𝜕z 𝜕x ) ( )2 ( 𝜕v 𝜕w 𝜕v 𝜕w + =μ + . τyz 𝜕z 𝜕y 𝜕z 𝜕y

(2.70)

(2.71a) (2.71b) (2.71c)

When we substitute Eqs. (2.70), (2.71a), (2.71b), and (2.71c) into the thermal energy equation, Eq. (2.68b), we now get the thermal energy equation for a differential control volume. 2.3.3.3 Thermal Energy Equation in Terms of Internal Energy

ρ

( ) ( )} ( ) 𝜕T 𝜕 𝜕T 𝜕 𝜕T 𝜕 ⃗ 2 ⃗ − 2 μ(∇ ⃗ ⋅ V) ⃗ ⋅ V) k + k + k + u′′′ − p(∇ 𝜕x 𝜕x 𝜕y 𝜕y 𝜕z 𝜕y 3 } { )2 )2 ( ( ) ) ( )2 ( )2 ( ( 𝜕u 𝜕v 𝜕u 𝜕v 𝜕w 2 𝜕u 𝜕w 2 𝜕v 𝜕w + + + +μ + 2μ + + +μ +μ . 𝜕x 𝜕y 𝜕z 𝜕y 𝜕x 𝜕z 𝜕x 𝜕z 𝜕y

D̃ u = Dt

{

(2.72)

This is the thermal energy equation in terms of internal energy for a Newtonian fluid and thermal energy equation can be more compactly written in the following form: D̃ u ⃗ + μΦ ⃗ ⋅ (k∇T) ⃗ + u′′′ − p(∇ ⃗ ⋅ V) =∇ Dt where the viscous dissipation function Φ is { } { }2 { }2 { ( )2 ( )2 ( ) } 𝜕u 𝜕w 2 𝜕u 𝜕w 2 𝜕v 𝜕v 𝜕w 𝜕u 𝜕v Φ=2 + + + + + + + + 𝜕x 𝜕y 𝜕z 𝜕y 𝜕x 𝜕z 𝜕x 𝜕z 𝜕y { }2 2 𝜕u 𝜕v 𝜕w + + − . 3 𝜕x 𝜕y 𝜕z ρ

(2.73)

(2.74)

The viscous dissipation function Φ is always positive. At this point, let us look at each term in Eq. (2.72): D̃ u = rate of change of stored internal energy Dt ⃗ ⃗ ∇ ⋅ k∇T = energy transfer by conduction ′′′ u = internal(energy generation ) ⃗ = p 𝜕u + 𝜕v + 𝜕w = the reversible rate of compression work per unit volume. ⃗ ⋅ V) p(∇ 𝜕x 𝜕y 𝜕z μΦ = the irreversible rate of conversion of work into internal energy per unit volume. This term is the rate of work per unit volume being done by the stresses in shearing and distorting the fluid element. It is dissipated within the fluid as friction heat. If the shear rates are small, the viscous dissipation function may be neglected. If the fluid flows through a porous media, viscous dissipation can become large even for very low velocities. ρ

The work done on the element changes the potential and kinetic energy of the element. The work done on the element goes into compressing the element or changing the internal energy of the element. We can write an alternative form of this equation by making use of continuity equation as follows: ρ

D̃ u ⃗ ⋅ k∇T ⃗ + u′′′ + p Dρ + μΦ. =∇ Dt ρ Dt

(2.75)

⃗ = − 1 . The quantity μΦ represents the portion of the mechanical work, ⃗ ⋅ V) In deriving this equation, we have used (∇ ρ Dt which is irreversibly converted into heat by the viscous stresses, and it is called viscous dissipation function. Dρ

2.3 Differential Formulation of Conservation Equations

2.3.3.4 Thermal Energy Equation in Terms of Enthalpy

̃. We can write this equation in terms of the We often prefer to work with in terms of enthalpy h instead of internal energy u ̃ + p∕ρ. Let us take the derivative of h since specific enthalpy, h = u p Dρ u D(p∕ρ) D̃ u 1Dp Dh D̃ = + = + − 2 . Dt Dt Dt Dt ρ Dt ρ Dt From continuity, − 1ρ

Dρ Dt

(2.76)

⃗ Therefore, ⃗ ⋅V =∇

D̃ u Dp Dh ⃗ ⃗ ⋅ V). =ρ + + p( ∇ (2.77) Dt Dt Dt Combining Eq. (2.77) with the thermal energy equation, Eq. (2.73), expressed in terms of internal energy, we obtain the thermal energy equation in terms of enthalpy ρ

Dh ⃗ ⋅ k∇T ⃗ + u′′′ + Dp + μΦ. =∇ (2.78) Dt Dt It is common for a fluid to undergo a nearly constant pressure process for which Dp/Dt ≈ 0. This form is useful in phase change and thermal energy storage problems, as discussed in [5]. It appears that the thermal energy equation can be manipulated into different forms depending on the needs. We need to reformulate the thermal energy equation so that we can use the boundary conditions given in temperature and heat flux. ρ

2.3.3.5 Temperature T and Constant Volume Specific Heat cv

We can write thermal energy equation in terms constant volume specific heat cv and temperature T. Consider a pure substance in the absence of motion, surface tension, and electromagnetic effects, as discussed in [6]. Let us consider some thermodynamic relations. These will be used in expressing the thermal energy in terms of temperature. For a pure sub̃=u ̃(T, v), and differentiating this relation, we obtain stance, assume u 𝜕̃ u || 𝜕̃ u || dT + dv (2.79) 𝜕T ||v 𝜕v ||T where the coefficient of the first term on the right-hand side is the specific heat at constant volume, cv , and it is given as d̃ u=

cv ≅

𝜕̃ u || . 𝜕T ||v

(2.80)

Using the constant volume specific heat cv , the change in internal energy takes the following form: d̃ u = cv dT +

𝜕̃ u || dv. 𝜕v ||T

(2.81)

In order to evaluate the coefficient of the second term on the right-hand side, we first consider the first Tds equation as follows: d̃ u = Tds − pdv

(2.82)

where temperature T is the absolute temperature and ds is the specific entropy change. Now, let s = s(T, v), and let us find its total differential of entropy s as follows: ( ) ( ) 𝜕s 𝜕s ds = dT + dv. (2.83) 𝜕T v 𝜕v T Next, we substitute this equation into the first Tds equation, Eq. (2.82), and rearranging it yields [ ( ) ] ( ) 𝜕s 𝜕s dT + T − p dv. d̃ u=T 𝜕T v 𝜕v T Comparing Eqs. (2.81) and (2.84), we obtain ( ) ( ) 𝜕u 𝜕s =T − p. 𝜕v T 𝜕v T We now eliminate the entropy term by use of the following Maxwell relation: ( ) ( ) 𝜕p 𝜕s = 𝜕v T 𝜕T v

(2.84)

(2.85)

(2.86)

55

56

2 Fundamental Equations of Laminar Convective Heat Transfer

and the result is ( ) ( ) 𝜕p 𝜕u =T − p. 𝜕v T 𝜕T v Inserting Eq. (2.87) into Eq. (2.81), one obtains the desired relation [ ( ) ] 𝜕p − p dv. d̃ u = cv dT + T 𝜕T v

(2.87)

(2.88)

Noting that d̃ u, dT, and dv are the differential change in specific internal energy, temperature, and specific volume, respectively, of a fluid particle, and the time rate of change of the specific internal energy of a fluid particle is written as [ ( ) ] 𝜕p DT D̃ u Dv = cv + T . (2.89) −p Dt Dt 𝜕T v Dt Now, we can construct the Dv/Dt term as follows: 1 Dρ Dv D(1∕ρ) = =− 2 . Dt Dt ρ Dt

(2.90)

From continuity, we have −

1 Dρ ⃗ ⃗ ⋅V =∇ ρ Dt

so that Dv/Dt term becomes Dv 1 ⃗ ⃗ = ∇ ⋅ V. Dt ρ Then, the D̃ u∕Dt term takes the following form: [ ] 𝜕p | DT D̃ u 1⃗ ⃗ = cv + T || − p ∇ ⋅ V. Dt Dt 𝜕T |v ρ Let us multiply this equation by density ρ, and the result is [ ( ) ] 𝜕p D̃ u DT ⃗ ⃗ ⋅ V. ρ = ρcv + T −p ∇ Dt Dt 𝜕T v Substituting Eq. (2.94) into the internal energy form of the thermal energy equation, Eq. (2.73), we get ( ) 𝜕p DT ′′′ ⃗ + μΦ. ⃗ ⃗ ⃗ ⋅ V) = ∇ ⋅ k∇T + u − T (∇ ρcv Dt 𝜕T v

(2.91)

(2.92)

(2.93)

(2.94)

(2.95)

Assuming that k, μ, and ρ are constant, the kinematic viscosity is ν = μ/ρ, and the thermal diffusivity is α = k/ρcp . Many problems in convection heat transfer can be treated as incompressible constant property flows. For incompressible sub⃗ = 0. Hence, the third term on the right-hand side of Eq. (2.95) ⃗ ⋅V stance, the divergence of velocity vector is zero, i.e. ∇ drops out. In the Cartesian coordinate system, under these conditions, the thermal internal energy equation reduces to ) ( 2 ( ) 𝜕T 𝜕T 𝜕T 𝜕 T 𝜕2 T 𝜕2 T 𝜕T +u +v +w =k + u′′′ + μΦ. + + (2.96) ρcv 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x2 𝜕y2 𝜕z2 2.3.3.6 Temperature and Constant Pressure Specific Heat cp

We can also write the thermal energy equation in terms constant pressure specific heat cp and temperature T. Starting with h = h(T, p), we get 𝜕 h || 𝜕 h || dh = dT + dp (2.97) | 𝜕T |p 𝜕p ||T where the coefficient of the first term on the right-hand side is the specific heat at constant pressure, cp , and it is given as cp ≅

𝜕 h || . 𝜕T ||p

The change in specific enthalpy takes the following form: 𝜕 h || dh = cp dT + dp. 𝜕p ||T

(2.98)

(2.99)

2.3 Differential Formulation of Conservation Equations

The second T ds equation is given as dh = T ds + v dp.

(2.100)

We now choose entropy to be function of temperature T and pressure p, that is we take s = s(T, p) and take its total differential ( ) ( ) 𝜕s 𝜕s dT + dp. (2.101) ds = 𝜕T p 𝜕p T We substitute Eq. (2.101) into the second T ds equation, Eq. (2.100), and rearrange it as [ ( ) ] ( ) 𝜕s 𝜕s dp. dh = T dT + v + T 𝜕T p 𝜕p T Comparing Eqs. (2.99) and (2.102), we obtain the desired result ( ) ( ) 𝜕h 𝜕s =T + v. 𝜕p T 𝜕p T We will eliminate the entropy term in this equation using the following Maxwell relation: ( ) ( ) 𝜕v 𝜕s − = . 𝜕p T 𝜕T p Then, Eq. (2.103) becomes ( ) ( ) 𝜕v 𝜕h = −T + v. 𝜕p T 𝜕T p Inserting Eq. (2.105) into Eq. (2.99), one obtains the desired result [ ( ) ] 𝜕v 1 −T dp dh = cp dT + ρ 𝜕T p where v = 1/ρ. From this relation, we can write following equation: [ ( ) ] Dp Dh DT 1 𝜕v = cp + −T . Dt Dt ρ 𝜕T p Dt Let us multiply this relation by density ρ to get [ ( ) ] Dp Dh DT 𝜕v = ρcp + 1 − Tρ . ρ Dt Dt 𝜕T p Dt Next, we need to determine the term ρ(𝜕v/𝜕T)p . This is done as follows: [ [ ]) ( ) ( ) ] ( ( ) 𝜕v 1 𝜕ρ 1 𝜕ρ 𝜕 1 ρ =ρ =ρ − 2 . =− 𝜕T p 𝜕T ρ p ρ 𝜕T p ρ 𝜕T p Then, the ρ(Dh/Dt) term takes the following form: [ ( ) ] Dp Dh T 𝜕ρ DT ρ = ρcp + 1− . Dt Dt ρ 𝜕T p Dt We insert Eq. (2.110) into the enthalpy form of the thermal energy equation, Eq. (2.78), and the result is ( ) Dh ⃗ ⋅ k∇T ⃗ + q′′′ − T 𝜕ρ Dp + μΦ. ρ =∇ Dt ρ 𝜕T p Dt

(2.102)

(2.103)

(2.104)

(2.105)

(2.106)

(2.107)

(2.108)

(2.109)

(2.110)

(2.111)

This equation can be expressed in terms of the coefficient of thermal expansion β, and the coefficient of thermal expansion β is defined as ( ) 1 𝜕ρ β=− . (2.112) ρ 𝜕T p Next, we insert Eq. (2.112) into Eq. (2.111), and the result is ρcp

DT ⃗ ⋅ (k∇T) ⃗ + q′′′ + βT Dp + μΦ =∇ Dt DT

(2.113)

57

58

2 Fundamental Equations of Laminar Convective Heat Transfer Dp

where the βT DT term represents the compressibility effect. This is one of the useful general forms of the thermal energy equation. In the Cartesian coordinate system, Eq. (2.113) can be written in the following form: ) ) ( ( ) ( 𝜕p 𝜕p 𝜕p 𝜕p 𝜕v 𝜕T 𝜕T 𝜕T 𝜕T = ρT + ρcp +u +v +w +u +v +w 𝜕t 𝜕x 𝜕y 𝜕z 𝜕T p 𝜕t 𝜕x 𝜕y 𝜕z ⃗ ⋅ k∇T ⃗ + μΦ + u′′′ ∇ (2.114) where Φ is the viscous dissipation term. Several problems in convection heat transfer can be treated as incompressible constant property flows. Assuming that β ≈ 0, density ρ, and thermal conductivity k are constants and introducing the kinematic viscosity ν = μ/ρ and thermal diffusivity α = k/ρcp , the thermal enthalpy energy equation reduces to ) ( 2 ( ) 𝜕T 𝜕T 𝜕T 𝜕 T 𝜕2 T 𝜕2 T 𝜕T +u +v +w =k + μΦ + u′′′ + + (2.115) ρcp 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x2 𝜕y2 𝜕z2 The first term on the right-hand side is the rate of energy storage per unit volume for the control volume. The second term on the right-hand side is the rate at which energy is carried out by fluid motion per unit volume or the convective terms. 2.3.3.7 Special Cases of the Differential Energy Equation

We will present two special cases of the energy equation. Let us consider each case. 2.3.3.8 Perfect Gas and the Thermal Energy Equation Involving T and cp

The ideal gas law gives ρ=

p . RT

Substituting into the coefficient of thermal expansion β, Eq. (2.112), gives ( ) 1 𝜕ρ 1 p 1 β=− = = . ρ 𝜕T p ρ R T2 T

(2.116)

(2.117)

Substituting Eq. (2.117) into the thermal energy equation involving T and cp , Eq. (2.113), gives ρcp

DT ⃗ ⋅ (k∇T) ⃗ + q′′′ + Dp + μφ. =∇ Dt DT

(2.118)

2.3.3.9 Perfect Gas and the Thermal Energy Equation Involving T and cv

We now consider the perfect gas law as p 𝜕p = ρR = . 𝜕T T

(2.119)

Substituting Eq. (2.119) into Eq. (2.95) yields ρcv

DT ⃗ + μΦ. ⃗ ⋅ k∇T ⃗ + q′′′ − p(∇ ⃗ ⋅ V) =∇ Dt

(2.120)

2.3.3.10 An Incompressible Pure Substance

In this case, for an incompressible pure substance, since density ρ is constant, we have β = 0. From thermodynamics, we have also cv = cp = c. Then, the thermal energy equation takes the following form: ρc

DT ⃗ ⋅ k∇T ⃗ + q′′′ + μΦ. =∇ Dt

(2.121)

This equation can be used for low-speed flows with constant thermal conductivity. Thermal energy equation and viscous dissipation function for Newtonian fluid with constant ρ, μ, and k in different coordinate systems.

2.3 Differential Formulation of Conservation Equations

2.3.3.11 Rectangular Coordinates

[ ρcp

] [ 2 ] 𝜕T 𝜕T 𝜕T 𝜕 T 𝜕2 T 𝜕2 T 𝜕T +u +v +w =k + q′′′ + μΦ + + 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x2 𝜕y2 𝜕z2

(2.122)

where viscous dissipation is { } ( )2 ( )2 ( ) ) ( )2 ( )2 ( 𝜕u 𝜕w 2 𝜕u 𝜕w 2 𝜕v 𝜕w 𝜕v 𝜕v 𝜕u + + + Φ=2 + + + + + 𝜕x 𝜕y 𝜕z 𝜕x 𝜕y 𝜕y 𝜕z 𝜕z 𝜕x ( )2 2 𝜕u 𝜕v 𝜕w + + − . 3 𝜕x 𝜕y 𝜕z 2.3.3.12 Cylindrical Coordinates (r, 𝛉, z)

[ ρcp

] 𝜕T vθ 𝜕T 𝜕T 𝜕T + vr + + vz =k 𝜕t 𝜕r r 𝜕θ 𝜕z

{

} ( ) 1 𝜕 𝜕T 1 𝜕2 T 𝜕2 T r + 2 2 + 2 + q′′′ + μΦ r 𝜕r 𝜕r r 𝜕θ 𝜕z

(2.123)

where viscous dissipation is ]2 [ ] ]2 [ [ ( ) 𝜕vr 𝜕vz 2 1 𝜕vr 𝜕 vθ 1 𝜕vz 𝜕vθ + + + + + r Φ= r 𝜕θ 𝜕z 𝜕z 𝜕r 𝜕r r r 𝜕θ {( } )2 [ ]2 ( )2 𝜕vr 𝜕vz 1 𝜕vθ vr + +2 + + . 𝜕r r 𝜕θ r 𝜕z 2.3.3.13 Spherical Coordinates (r, 𝛉, 𝛟)

[ ρcp

] [ ( ) ( ) vϕ 𝜕T 1 𝜕T 𝜕 𝜕T vθ 𝜕T 1 𝜕 2 𝜕T 𝜕T + vr + + =k 2 r + 2 sin(θ) 𝜕t 𝜕r r 𝜕θ r sin(θ) 𝜕ϕ 𝜕r 𝜕θ r 𝜕r r sin(θ) 𝜕θ ] 2 1 𝜕 T + + q′′′ + μΦ r2 sin2 (θ) 𝜕ϕ2

(2.124)

where viscous dissipation Φ is ]2 [ ( )]2 [ ( ) 1 𝜕vr 1 𝜕vr 𝜕 vϕ 𝜕 vθ + + +r Φ= r 𝜕r r r 𝜕θ r sin(θ) 𝜕ϕ 𝜕r r ( ) ]2 [ v 𝜕v sin(θ) 𝜕 ϕ 1 θ + + r 𝜕θ sin(θ) r sin(θ) 𝜕ϕ {( ) )2 ( )2 } ( 𝜕vr 2 1 𝜕vθ vr 1 𝜕vϕ vr + vθ cot(θ) + + . 2 + + 𝜕r r 𝜕θ r r sin(θ) 𝜕ϕ r Example 2.2 Consider the steady incompressible laminar flow of fluid down the inclined plane. Using the Navier–Stokes equation of motion, obtain the velocity distribution. Evaluate the shear stress on the plane. See Figure 2.E2. Solution 1) Two-dimensional flow 2) Steady axial flow. No motion in the z-direction (w = 0) 3) Flow is along the x-axis and streamlines are parallel 4) Fully developed flow 5) Incompressible flow (ρ = const) 6) Newtonian fluid 7) Constant properties (ρ, μ, cp ) 8) Atmospheric pressure is uniform 9) No edge effects 𝜕/𝜕 z = 0

59

60

2 Fundamental Equations of Laminar Convective Heat Transfer

y p0

H

0

g

x θ

Figure 2.E2

Schematic of viscous fluid flow over an inclined plane.

Continuity Equation 𝜕ρ 𝜕 𝜕 𝜕 (ρu) + (ρv) + (ρw) + = 0. 𝜕x 𝜕y 𝜕z 𝜕t Constant density ρ 𝜕ρ = 0. 𝜕t Streamlines are parallel v = 0. Thus, no motion in the y-direction. No edge effects and 𝜕 w∕𝜕z = 0, w = 0. Continuity becomes 𝜕u = 0. 𝜕x This means that flow is fully developed in the x-direction and u is independent of x. We now state u = u(y). Navier–Stokes equations: x-direction: ) ( ) ( 2 𝜕p 𝜕u 𝜕u 𝜕u 𝜕u 𝜕 u 𝜕2 u 𝜕2 u ρ + + +u +v +w = ρfx − +μ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x 𝜕x2 𝜕y2 𝜕z2 1) 2) 3) 4) 5) 6)

𝜕u = 0 steady flow 𝜕t v=0 w=0 𝜕u = 0 fully developed flow, continuity. 𝜕x fx = g sin θ gravitational component 2 2 In most unidirectional flows 𝜕𝜕xu2 ≪ 𝜕𝜕yu2

7) No edge effects and

𝜕 𝜕z

= 0, and we may write

𝜕2 u 𝜕z2

=0

Under these assumptions, the equation of motion in the x-direction reduces to ( 2 ) 𝜕p 𝜕 u 0 = ρg sin (θ) − . +μ 𝜕x 𝜕y2 Since u = u(y), then ( 2 ) 𝜕p du μ =− + ρg sin (θ). 𝜕x dy2

2.3 Differential Formulation of Conservation Equations

y-direction: ( ) ( 2 ) 𝜕p 𝜕v 𝜕v 𝜕v 𝜕v 𝜕 v 𝜕2 v 𝜕2 v ρ +u +v +w = ρfy − +μ + + 𝜕t 𝜕x 𝜕y 𝜕z 𝜕y 𝜕x2 𝜕y2 𝜕z2 𝜕v = 0 steady flow 1) 𝜕t 2) v = 0 no motion in y-direction. Parallel flow 3) fy = − g cos(θ) The equation of motion in the y-direction reduces to 0 = −ρg cos(θ) −

𝜕p 𝜕y

or 𝜕p = −ρg cos(θ). 𝜕y We now integrate this equation to get p = ρg cos(θ)y + f(x) where f(x) is an integration constant. At y = H, p = p0 Using this condition, we have p0 = ρg cos(θ)H + f(x) ⇒ f(x) = p0 − ρg cos(θ)H. Thus, p = ρg cos(θ)y + p0 − ρg cos(θ)H. We now differentiate this equation with respect to x 𝜕p = 0. 𝜕x The momentum equation becomes ( 2 ) du = ρg sin (θ). μ dy2 The boundary conditions are y = 0, u = 0 ) ( 𝜕v y = H, τxy = μ 𝜕u = 0 since shear stress is zero at the free surface of the liquid. + 𝜕y 𝜕x Using Maple 2020, we obtain > de ≔ 𝛍 ⋅ diff(u(y), y, y) = 𝛒 ⋅ g ⋅ sig(𝛉); ( de ≔ 𝛍

) d𝟐 u(y) = 𝛒 g sin(𝛉) dy𝟐

> sol ≔ dsolve({de, u(𝟎) = 𝟎, u′ (H) = 𝟎}, u(y)); sol ≔ u(y) =

𝝆 gsin(𝛉)y𝟐 y H 𝛒 g sin(𝛉) − . 𝟐𝛍 𝛍

The solution becomes u=

[ ( ) ( )] ρ g sin(θ) H2 y y 2 2 − . 2μ H H

61

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2 Fundamental Equations of Laminar Convective Heat Transfer

Shear stress at the plane wall is ( ) du = ρ gH sin(θ). τw = μ dy Example 2.3 High-speed Newtonian incompressible constant property fluid is moving between parallel plates. See Figure 2.E3. Fluid motion is generated by constant pressure gradient. Assume that flow is steady and two-dimensional, and fluid particles move parallel to planes (streamlines are parallel to plates). Flow is hydrodynamically fully developed (HFD). Body forces are neglected. Taking viscous dissipation into account, we wish to write the energy equation for this flow. y

0

Figure 2.E3

x

Pressure driven flow between parallel plates.

2H

Solution Assumptions: 1) 2) 3) 4) 5) 6) 7) 8)

Steady axial flow. No motion in the z-direction (w = 0) Parallel streamlines No internal energy generation Constant properties (ρ, μ, cp ) Incompressible flow ρ = const No edge effects (𝜕/𝜕 z = 0) No end effects; fully developed flow with respect to the x-direction Body forces are neglected Governing equations are listed below. Continuity equation 𝜕ρ 𝜕 𝜕 𝜕 (ρu) + (ρv) + (ρw) + = 0. 𝜕x 𝜕y 𝜕z 𝜕t Momentum Equation x-direction: ) ( 2 ) ( 𝜕p 𝜕u 𝜕u 𝜕u 𝜕 u 𝜕2 u 𝜕2 u 𝜕u +u +v +w = ρfx − +μ + + ρ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x 𝜕x2 𝜕y2 𝜕z2 y-direction: ) ( 2 ) ( 𝜕p 𝜕v 𝜕v 𝜕v 𝜕 v 𝜕2 v 𝜕2 v 𝜕v +u +v +w = ρfy − +μ . + + ρ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕y 𝜕x2 𝜕y2 𝜕z2 Energy equation ) ( 2 ( ) 𝜕T 𝜕T 𝜕T 𝜕 T 𝜕2 T 𝜕2 T 𝜕T +u +v +w =k ρcp + 2 + 2 + μΦ + u′′′ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x2 𝜕y 𝜕z { } { ) } }2 ( { 2 2 ( )2 ( { ) } 𝜕u 𝜕w 2 𝜕u 𝜕w 2 𝜕v 𝜕v 𝜕w 𝜕u 𝜕v Φ=2 + + + + + + + + 𝜕x 𝜕y 𝜕z 𝜕y 𝜕x 𝜕z 𝜕x 𝜕z 𝜕y { }2 2 𝜕u 𝜕v 𝜕w + + − . 3 𝜕x 𝜕y 𝜕z

2.3 Differential Formulation of Conservation Equations

First, let us consider the continuity equation 𝜕ρ 𝜕 𝜕 𝜕 (ρu) + (ρv) + (ρw) + = 0. 𝜕x 𝜕y 𝜕z 𝜕t Constant density 𝜕ρ = 0. 𝜕t Flow is two-dimensional. There are no edge effects. Recall that streamlines are parallel to plates. For this configuration, the flow is independent of the z-direction. Thus, we have 𝜕 =0 𝜕z w = 0. For fully developed flow with no end effects, we have 𝜕u = 0. 𝜕x The continuity equation reduces to 𝜕v = 0. 𝜕y The integration of this equation yields v = C = const. To determine the integration constant C, we apply the no-slip boundary condition at the upper plate. At y = H, v = 0. Notice that the wall is impervious, and we find that C = 0. We now conclude that v = 0 everywhere in the flow field. Since the vertical component of velocity vanishes everywhere, we may state that the streamlines are parallel. The x-component of velocity u is independent of x and fully developed with respect to x. The velocity profile will appear the same for all the values of x. Flows in tubes or between two infinite parallel plates of constant cross section can be treated as fully developed if the tube is sufficiently long such that inlet and outlet effects can be neglected. The momentum equation simplifies to x-direction: Notice 0=−

𝜕p +μ 𝜕x

(

𝜕2 u 𝜕2 u + 2 𝜕x2 𝜕y

) .

For most unidirectional flows, we have 𝜕2u 𝜕2 u ≪ . 𝜕x2 𝜕y2 Then, the momentum equation reduces to ( 2 ) dp du μ = . 2 dx dy y-direction: 𝜕p = 0. 𝜕y Energy equation Steady flow 𝜕T = 0. 𝜕t

63

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2 Fundamental Equations of Laminar Convective Heat Transfer

Energy equation becomes ( 2 ) 𝜕T 𝜕 T 𝜕2T ρcp u =k + μΦ + 𝜕x 𝜕y2 𝜕y2 ( )2 𝜕u Φ= . 𝜕y

Problems 2.1

The steady two-dimensional x-component of the momentum equation for incompressible, constant property flow is given in the rectangular coordinate system ) ( 2 ) ( 𝜕p 𝜕u 𝜕u 𝜕 u 𝜕2u +v = ρfx − +μ + 2 . ρ u 𝜕x 𝜕y 𝜕x 𝜕x2 𝜕y Discuss each term in the equation.

2.2

Consider two-dimensional flow between two infinite parallel plates. Flow is steady, incompressible, and HFD. The fluid has constant property. The continuity equation is given as 𝜕u 𝜕v + = 0. 𝜕x 𝜕y Discuss the assumptions to simplify the continuity equation.

2.3

Consider steady, incompressible, two-dimensional x-component of the momentum equation for flow between two infinite parallel plates. The flow is called fully developed if v = 0 everywhere in the flow. Neglect the body forces. Simplify the continuity and momentum equations.

2.4

Consider steady, incompressible, two-dimensional flow between two infinite parallel plates. The energy equation is given as ) ( 2 ( ) 𝜕T 𝜕T 𝜕 T 𝜕2T +v =k + 2 + μΦ. ρcp u 𝜕x 𝜕y 𝜕x2 𝜕y Discuss each term in the energy equation.

2.5

Consider steady, incompressible, two-dimensional flow between two infinite parallel plates. The energy equation including viscous dissipation is given as ) ( 2 ( ) 𝜕T 𝜕T 𝜕 T 𝜕2T +v =k + μΦ. ρcp u + 𝜕x 𝜕y 𝜕x2 𝜕y2 Assume that flow is fully developed and heat conduction in the x-direction is negligible. There is no energy generation. Simplify the energy equation.

2.6

Far away from the inlet of a parallel plate channel, the entrance effects are not important, and streamlines are parallel to the channel centerline. An incompressible constant property viscous fluid flows steadily between two infinite parallel plates, as shown in Figure 2.P6. Flow is HFD, and flow is driven by the steady motion of the upper plate. The pressure gradient is zero everywhere and gravity term is absent. Such a flow is also called shear driven flow. The upper plate is moving with a uniform velocity V. The distance between the plates is H. The upper plate is kept at constant temperature T0 . A uniform heat flux q′′0 is applied to lower plate. Neglect body forces (gravity term). Energy generation and axial conduction are negligible. We wish to write the governing equations along with their boundary conditions.

Problems

T0

y Flow

u(y)

0

Figure 2.P6

2.7

u=V

x

H

q″0

Geometry and coordinate system for Couette flow.

Consider steady, two-dimensional, fully developed, laminar flow between infinite parallel plates, assuming constant fluid properties. See Figure 2.P7. The fluid has a uniform velocity V at the channel inlet. As the fluid moves in the parallel plate channel, a velocity boundary layer begins to develop along the parallel plate channel. Eventually, boundary layers meet at the channel centerline, and velocity profile becomes fully developed. The x and y coordinates are chosen along and normal to flow direction. The flow is driven by constant pressure gradient. Pressure gradient may be created by a pushing device such as a pump or fan. Body forces are neglected. The distance between the plates is 2H. Body forces are neglected. Far away from the inlet of a parallel plate channel, the entrance effects are not important, and streamlines are parallel to the channel centerline. Both plates are subjected to uniform temperature T0 . Energy generation and axial conduction are negligible. Including the effect viscous dissipation, we wish to write the continuity, momentum, and energy equation for HFD laminar flow region between parallel plates. T0

y u(x, y)

V 0 Ti

x

2H

Hydrodynamic entry region

Figure 2.P7

2.8

u(y)

Hydrodynamically fully developed region

T0

Geometry and coordinate system for laminar flow between infinite parallel plates.

Consider steady laminar flow in a tube assuming constant fluid properties. See Figure 2.P8. The fluid has a uniform velocity at the tube inlet. As the fluid moves in the tube, a velocity boundary layer begins to develop along the tube. Eventually, boundary layers meet at the tube centerline, and the velocity profile becomes fully developed. Flow is driven by a constant pressure gradient. Far away from the inlet of the tube, the entrance effects are not important, and streamlines are parallel to tube centerline. Heating begins after the velocity profile is fully developed, and a uniform heat flux q′′0 is applied to pipe surface. Energy generation and axial conduction are negligible. Taking the viscous dissipation into account, we wish to write the continuity, momentum, and energy equation for the HFD laminar flow region in a tube. q″0

r vz(r, z) 0

z

2R

Hydrodynamic entry region

Figure 2.P8

vz(r)

Hydrodynamically fully developed region

Geometry and coordinate system for laminar flow in a circular pipe.

65

66

2 Fundamental Equations of Laminar Convective Heat Transfer

Consider steady two-dimensional incompressible laminar flow between infinite parallel plates, assuming constant properties. The x and y components are shown in Figure 2.P9. A constant pressure gradient creates the fluid motion. Both the plates are subjected to uniform temperature Tw . The distance between the plates is 2H. The fully developed velocity profile is given by [ ( y )2 ] 3 . u= V 1− 2 H

2.9

Body forces are neglected. Neglect the energy generation and axial conduction. Including the effect of viscous dissipation, we wish to write the energy equation for the HFD region. TW

y

0

Figure 2.P9 Geometry and coordinate system for laminar flow between infinite parallel plates.

x

2H TW

Consider liquid metal flow past a semi-infinite flat plate as shown in Figure 2.P10. The plate surface temperature Tw is maintained at constant temperature. The temperature of the fluid far from the plate is T∞ . Write the energy equation and its boundary conditions for the flow.

2.10

y

Figure 2.P10 plate.

Thermal boundary layer for low Prandtl number flow over a flat

Flow T∞ U∞

Δ(x) 0

x

TW

A liquid metal flows steadily between two infinite parallel plates, as shown in Figure 2.P11. The separation distance between the parallel plates is H. The upper plate is subjected to uniform heat flux q′′0 , while the lower plate is kept at uniform temperature Tw . The inlet temperature of the liquid metal is Ti . Neglect energy generation, axial conduction, and viscous dissipation. We wish to write the energy equation along with boundary conditions for HFD flow.

2.11

q″0

y Flow

H

Ti 0

2.12

Figure 2.P11 Geometry and coordinate system for liquid metal flow between infinite parallel plates.

x

Tw

Consider viscous fluid flow between two infinite parallel plates (See Figure 2.P12). Flow is steady, two-dimensional, incompressible, and laminar. Flow enters the parallel plate channel with velocity V and temperature Ti . Heating begins when the flow is HFD. The fully developed velocity profile is given by [ ( y )2 ] 3 . u= V 1− 2 H Fluid motion is generated by a constant pressure gradient. Both the plates are subjected to uniform temperature Tw . The distance between the plates is 2H. Body forces are neglected. Neglect the viscous dissipation, energy generation, and axial conduction. We wish to write down the energy equation and its boundary conditions.

References

y

Tw

u(y) Ti

0

2H

x

Tw

Figure 2.P12

2.13

Geometry and coordinate system for laminar viscous flow between two infinite parallel plates.

Consider steady two-dimensional laminar flow between infinite parallel plates (See Figure 2.P13). The fluid properties are constant. Flow enters the parallel plate channel with a temperature Ti . The upper plate is insulated and is moving with uniform velocity V, while the lower plate is subjected to uniform heat flux q′′0 . Fluid particles move in the direction parallel to plates. The lower plate is stationary. The distance between the plates is H. The fully developed velocity profile is given by (y) u = . V H Body forces are neglected. Neglect the viscous dissipation, energy generation, and axial conduction. We wish to write the energy equation for HFD laminar flow between parallel plates under steady conditions. insulated

y

V Ti 0

H

x

q″0 Figure 2.P13

Geometry and coordinate system for laminar Couette flow.

References 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Sissom, L.E. and Pitts, D.R. (1972). Elements of Transport Phenomena. McGraw-Hill. Kays, W., Crawford, M., and Weigand, B. (2005). Convective Heat and Mass transfer, 4e. McGraw Hill Companies, Inc. White, F.M. and Majdalani, J. (2021). Viscous Fluid Flow. McGraw-Hill. Schlichting, H. and Gersten, K. (2000). Boundary Layer Theory, 8th revised ed. Springer. Eckert, E.R.G. and Drake, R.M. (1972). Analysis of Heat and Mass Transfer. McGraw-Hill. Moran, M.J., Shapiro, H.N., Boetner, D.D., and Bailey, M.B. (2018). Principles of Engineering Thermodynamics. Wiley. Arpaci, V.S. and Larsen, P.S. (1984). Convection Heat Transfer. Prentice Hall. Jiji, L.M. (2009). Heat Convection, 2e. Springer. Kakac, S., Yener, Y., and Pramuanjaroenkij, A. (2014). Convective Heat Transfer, 3e. CRC Press. Oosthuizen, P.H. and Naylor, D. (1999). An Introduction to Convective Heat Transfer Analysis. McGraw-Hill. Ghiaasiaan, S.M. (2018). Convective Heat and Mass Transfer, 2e. CRC Press. Burmeister, L.C. (1993). Convective Heat Transfer, 2e. Wiley. Cebeci, T. and Bradshaw, P. (1988). Physical and Computational Aspects of Convective Heat Transfer. Springer-Verlag. Bejan, A. (2013). Convective Heat Transfer, 4e. Wiley. Pai, S.I. (1956). Viscous Flow Theory. D. Von Nostrand Company.

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3 Equations of Incompressible External Laminar Boundary Layers 3.1 Introduction The concept of boundary layer theory was proposed by Prandtl in 1904. The boundary layer theory is discussed in depth by several people, such as Schlichting and Gersten [1], Rosenhead [2], and Schetz [3]. Consider heat transfer from a flat plate to a fluid stream in external flow. Assume that flat plate temperature is Tw , and free stream fluid velocity and temperature are U∞ and T∞ , respectively. We are interested in the following: a) The heat transfer from the plate to fluid b) The net force exerted on the flat plate by the fluid. We will look for answers to these questions.

3.2 Laminar Momentum Transfer Many engineering problems in convective heat transfer can be approximated, in the limit, by the flow over a thin flat plate. Consider a thin semi-infinite flat plate in a steady two-dimensional stream of flow having a free stream velocity of U∞ (x). The plate is aligned with the streamlines so that there will be no disturbance in the inviscid flow. The coordinate system is located at the leading edge of the plate. At this point, it should be noted that a mirror image exists underneath our hypothetically semi-infinite thin plate. For some distance from the leading edge of the plate, the boundary layer is laminar. See Figure 3.1. After the laminar region, the flow field has characteristics of both the laminar and turbulent flows. This region is called the transition region. After the transition region, the flow is fully turbulent. Experiments indicate that the occurrence of laminar–turbulent boundary layer transition depends on the Reynolds number defined as Rex =

ρU∞ x μ

(3.1)

where U∞ (m/s) is the free stream velocity, μ (N s/m2 ) is the dynamic viscosity of fluid, and ρ (kg/m3 ) is the fluid density. The value of Reynolds number at which transition takes place depends on surface roughness and the flow disturbances in the fluid outside the boundary layer. Based on experiments, the transition might take place over the range 2 × 104 ≤ Re ≤ 2 × 106 .

(3.2)

The transition region is partially incorporated into the laminar region and partially into the turbulent region, and the critical Reynolds number Rexc may be taken as Rexc = 500 000.

(3.3)

Suppose that we are interested in finding the net force exerted by the fluid stream on the plate. How do we calculate this net force acting on the plate?

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

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3 Equations of Incompressible External Laminar Boundary Layers

y

Inviscid region

U∞ Viscous region

X Transition

Laminar Figure 3.1

3.3

Turbulent

Two-dimensional boundary layer.

The Momentum Boundary Layer Concept

We wish to discuss the boundary layer concept. Let us consider the two-dimensional flow of an incompressible fluid over a two-dimensional thin flat plate. See the flow shown in Figure 3.2. We will discuss each region shown in Figure 3.2. Region I (x > 0, y > δ(x)): This region is outside the boundary layer. The velocity gradients are small, and the fluid behaves as if it were frictionless. Viscous forces are not important. Thus, the flow is essentially treated as potential flow. In this region, the velocity field is essentially equal to u = U∞ , v = 0, and p = p∞ . The flow outside the boundary layer is not affected by the presence of the flat plate. Region II (x > 0, 0 ≤ y ≤ δ(x)): In this region, the velocity gradients are appreciable, and thus, viscous forces are important. The region between 0 < x < L and 0 ≤ y ≤ δ(x) is called a boundary layer. The velocity changes from u = 0 on the solid wall to u = U∞ , where U∞ is the free stream velocity. The velocity gradients across the boundary layer are much greater than those in the direction parallel to the flat plate. For this reason, certain terms in the governing equations are negligibly small compared with the remaining terms. These small terms can be neglected without much of loss. We can now state that the effect of viscosity is confined to a region near the wall, and this region is the velocity boundary layer. The following conditions are required for the velocity boundary layer to exist: a. A slender body without flow separation. b. High Reynolds number (Rex > 100). The boundary layer concept was developed by the German engineer–mathematician Ludwig Prandtl, in a series of publications starting in 1904. Prandtl carved out of the entire flow field over the solid surface, a region in which the effects of viscosity are important. Outside of this region, flow is not affected by the solid surface. It is one of the most significant discoveries in the development of fluid mechanics and permits an understanding of many seeming paradoxes in real fluid behavior. It reveals ways to analyze problems too complicated to be solved by the direct integration of the full set y

Inviscid region

U∞ I

U∞

δ(x) II

u

Viscous region

x Flat plate L Figure 3.2

Velocity boundary layer flow over a flat plate.

3.3 The Momentum Boundary Layer Concept

of equations of motion and continuity. Creeping motion and boundary layer flow are extreme manifestations of viscosity effects. Broadly speaking, the former occurs for very viscous fluids and the latter for slightly viscous fluids. On the other hand, while creeping flow is uniquely laminar, boundary layers may be laminar or turbulent. Consider the flow over the flat plate, as shown in Figure 3.2. Flow is steady, laminar, two dimensional, and incompressible. The Newtonian fluid properties are uniform. The boundary layer flow over a surface has a free stream velocity U∞ (x), which may vary with x in an arbitrary manner. Based on experimental studies, we assumed that the boundary layer region δ × L is slender such that δ ≪ L. Assume that y is of the same order as boundary layer thickness δ, in which the velocity component u changes from u = 0 at the surface, i.e. at y = 0 to roughly u = U∞ at the edge of the boundary layer. In the δ × L region, the longitudinal momentum equation accounts for the competition among three types of forces. These forces are inertia, pressure, and viscous forces. In order to study the boundary layer flow, we must solve the conservation equations for mass and momentum as given below Continuity: 𝜕u 𝜕v + = 0. 𝜕x 𝜕y

(3.4)

x-direction momentum: u

𝜕u 1 𝜕p 𝜕u +v =− +ν 𝜕x 𝜕y ρ 𝜕x

y-direction momentum: u

𝜕v 𝜕v 1 𝜕p +v =− +ν 𝜕x 𝜕y ρ 𝜕y

(

(

𝜕2 u 𝜕2 u + 2 𝜕x2 𝜕y

𝜕2 v 𝜕2 v + 𝜕x2 𝜕y2

) .

(3.5)

) .

(3.6)

3.3.1 Scaling of Momentum Equation Suppose that flow over a surface has a main direction, and geometry has a gradual variation. Under these conditions, upstream conditions influence the flow. Examples of such flows are flows in channels and flows over plane surfaces. The equation of continuity and momentum may be simplified. In simplification processes, we take the following items into consideration: a. The diffusion of momentum in the main flow direction is much smaller than convective transport. b. The axial velocity component in the main flow direction is much larger than the velocity components in other directions. c. The pressure gradient in the main flow direction is much larger than the pressure gradient in the vertical direction of the flow. We will now present boundary layer approximations to Eqs. (3.4)–(3.6) by scaling. A scaling procedure is presented in detail in [4, 5]. Prandtl imagined a free stream outside the boundary layer. This free stream (inviscid flow) is not affected by the presence of the solid wall, and it is characterized by u = U∞ , v = 0, and p = p∞ . The order of magnitude of each term is estimated by scaling, and higher-order terms are dropped. The free stream velocity is U∞ , the characteristic length is L, and the boundary layer thickness is δ(x). In essence, we are considering a region defined by the space of height δ and length L along the flow direction. The order of magnitude of the distance in which u changes from 0 at the wall to U∞ in the free stream is denoted by the boundary layer thickness δ. We assume that the ratio of the boundary layer thickness δ(x) to characteristic length L is defined as δ(x)∕L ≪ 1

(3.7)

since the region under study is slender. Let us assume that Eq. (3.7) is valid for the problem under consideration. The dependent variable u and the independent variables x and y are assigned the following scales: u ∼ U∞

(3.8)

x∼L

(3.9)

y ∼ δ.

(3.10)

71

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3 Equations of Incompressible External Laminar Boundary Layers

These equations, Eqs. (3.8)–(3.10), will be applied to continuity equation to obtain a scale for the y-component of velocity v. The continuity equation is expressed as 𝜕u 𝜕v =− . 𝜕y 𝜕x

(3.11)

Using Eqs. (3.8)–(3.10), in the δ × L region, we obtain the following equation: U v ∼ ∞. δ L

(3.12)

Solving Eq. (3.12) for v, we get δ v ∼ U∞ . L

(3.13)

Based on Eq. (3.13), we see that v ≪ U∞ . Next, we will determine the order of magnitude of the inertia and viscous terms of the x-component of momentum equation. First inertia term u

U2 U 𝜕u ∼ U∞ ∞ = ∞ . 𝜕x L L

(3.14)

Second inertia term v

U 𝜕u ∼ v ∞. 𝜕y δ

(3.15)

We now eliminate v in Eq. (3.15) using Eq. (3.13), and the result is v

U2 𝜕u δ U ∼ U∞ . ∞ = ∞ . 𝜕y L δ L

(3.16)

We see that the both the inertia terms have the same order of magnitude. Next, we will determine the order of magnitude of two viscous terms in the x-component of momentum equation. First viscous term 𝜕 2 u U∞ ∼ 2 . (3.17) 𝜕x2 L Second viscous term 𝜕 2 u U∞ ∼ 2 . 𝜕y2 δ

(3.18)

Comparing Eq. (3.17) with Eq. (3.18) yields U∞ 𝜕2 u ( )2 2 𝜕x = L2 ∼ δ . 2 U∞ L 𝜕 u 𝜕y2 δ2

(3.19a)

Since δ ≪ L, we obtain the following relation: 𝜕2 u 𝜕2 u ≪ 2. 2 𝜕x 𝜕y

(3.19b)

Thus, 𝜕 2 u/𝜕x2 term in the x-component of the momentum equation can be neglected. Next, we will scale the two viscous terms in the y-components of the momentum equation, Eq. (3.6). First viscous term 𝜕2 v v ∼ 2. 𝜕x2 L

(3.20)

Second viscous term 𝜕2 v v ∼ 2. 𝜕y2 δ

(3.21)

3.3 The Momentum Boundary Layer Concept

Comparing Eq. (3.20) with Eq. (3.21) yields v 𝜕2 v ( )2 𝜕x2 ∼ L2 = δ . v L 𝜕2 v 2 2 δ 𝜕y

(3.22a)

Keeping in mind that δ ≪ L, we obtain the following relation: 𝜕2 v 𝜕2 v ≪ 2. 2 𝜕x 𝜕y

(3.22b)

Using Eqs. (3.19b) and (3.22b), we can now express the x- and y-components of the momentum equation in the following form: 𝜕u 1 𝜕p 𝜕2 u 𝜕u +v =− +ν 2 (3.23) u 𝜕x 𝜕y ρ 𝜕x 𝜕y 𝜕v 𝜕v 1 𝜕p 𝜕2 v u +v =− +ν 2. (3.24) 𝜕x 𝜕y ρ 𝜕y 𝜕y We dropped the higher-order terms in the momentum equation. Next, the order of magnitude of 𝜕p/𝜕x and 𝜕p/𝜕y will be determined using scaling. A balance between axial pressure and first inertia term in Eq. (3.23) yields 𝜕p 𝜕u ∼ 𝜌u . 𝜕x 𝜕x Equation (3.25) is scaled using Eqs. (3.8) and (3.9), and the result is given as U2 𝜕p ∼ ρ ∞. 𝜕x L Again, a balance between the pressure term and the second inertia term of Eq. (3.24) 𝜕p 𝜕v ≈ 𝜌v 𝜕y 𝜕y

𝜕p 𝜕x

∼

ρU2∞ Lδ2 ρ

U2∞ L

(3.26)

(3.27)

and the scaling of Eq. (3.27) by using Eqs. (3.8)–(3.10) and (3.13) yields )( ) ( )( )( ) 𝜕p ( v 1 δ δ δ δ ∼ U∞ = U∞ U∞ ≈ ρU2∞ 2 . 𝜕y L δ L δ L L Comparing Eq. (3.26) with Eq. (3.28) yields 𝜕p 𝜕y

(3.25)

∼

ρU2∞ Lδ2 ρ

U2∞

∼

( ) δ . L

(3.28)

(3.29a)

L

Using the assumption (δ/L) ≪ 1 yields the following result: 𝜕p 𝜕p ≪ . 𝜕y 𝜕x

(3.29b)

This equation tells us that the pressure gradient in the vertical direction is negligible compared pressure gradient in the axial flow direction. For two-dimensional flow, pressure depends on both the x and y variables. This can be expressed mathematically as follows: p = p(x, y) 𝜕p 𝜕p dx + dy. dp = 𝜕x 𝜕y Dividing Eq. (3.31) by dx and rearranging, we obtain [ ] dp 𝜕p (𝜕p∕𝜕y)dy = 1+ . dx 𝜕x (𝜕p∕𝜕x)dx The gradient dy δ ∼ . dx L

dy dx

(3.30) (3.31)

(3.32)

is scaled next, and the scaling result is given as (3.33)

73

74

3 Equations of Incompressible External Laminar Boundary Layers

Now, let us combine Eqs. (3.26), (3.28), and (3.33) with Eq. (3.32) to get [ ( )2 ] dp 𝜕p δ . = 1+ dx 𝜕x L

(3.34)

Since δ ≪ L, Eq. (3.34) becomes dp 𝜕p ≈ . (3.35) dx 𝜕x This result tells us that the boundary layer pressure distribution depends on the axial position x, and the variation of pressure in the y-direction is negligible. This means that at any position x, the pressure p(x) inside the boundary layer is equal to free stream pressure p∞ (x) at the edge of the boundary layer. Thus, p(x, y) = p∞ (x)

(3.36)

or 𝜕p dp∞ ≈ . (3.37) 𝜕x dx Thus, we can conclude that pressure within the boundary layer can be determined using the potential flow outside the boundary layer. The x-momentum equation becomes u

𝜕u 1 dp∞ 𝜕2 u 𝜕u +v =− +ν 2. 𝜕x 𝜕y ρ dx 𝜕y

(3.38)

Recall that the flow outside the boundary layer is assumed to be inviscid. Using the Bernoulli’s equation, we can write dp − 1ρ dx∞ term as follows: −

dU 1 dp∞ = U∞ ∞ . ρ dx dx

(3.39)

This equation relates the pressure gradient term in the boundary layer to the free stream velocity distribution U∞ (x). The free stream velocity distribution U∞ (x) is assumed to be available from the solution of the potential flow problem. The continuity equation will remain unchanged. We can now combine Eq. (3.39) with the x-component of momentum equation, and the result is u

dU 𝜕u 𝜕2 u 𝜕u +v = U∞ ∞ + ν 2 . 𝜕x 𝜕y dx 𝜕y

(3.40)

Let us reconsider the x-component of the momentum equation. The first two terms in this equation represent the inertia force, and these two terms are of the same order. The last term in this equation represents the viscous forces. The viscous term can be scaled to U 𝜕2 u . (3.41) ν 2 ∼ν ∞ 𝜕y δ2 The second inertia term on the left-hand-side of Eq. (3.40) is scaled to ( ) ( U ) U2 𝜕u δ ∞ v = U∞ = ∞. 𝜕y L δ L

(3.42)

Now, a balance between inertia and viscous forces yields U U2∞ ∼ν ∞ . L δ2 Rearranging Eq. (4.43) yields √ δ υ ∼ . L LU∞

(3.43)

(3.44)

This last result can be expressed as 1 δ ∼ √ L ReL

(3.45)

3.3 The Momentum Boundary Layer Concept

where ReL is the Reynolds number, and it is defined as ρU∞ L U∞ L = . μ ν

ReL =

(3.46)

√ We conclude that δ/L ≪ 1 when ReL ≫ 1. We also notice that the boundary layer solution will fail at the leading edge of the plate. The leading edge of the plate is a singularity point. We can now generalize Eq. (3.45) by setting L = x, and the result is δ(x) 1 ∼ √ (3.47) x Rex where Rex = 𝜌U∞ x/μ = U∞ x/ν is the local Reynolds number, ν is the kinematic viscosity, and δ(x) is the momentum (boundary) layer thickness. This equation gives us information about the variation of boundary layer thickness with the Reynolds number. We now return to the problem of determining the force exerted by the fluid stream on the plate. The wall shear stress τw is ( ) 𝜕u τw = μ . (3.48) 𝜕y wall Force exerted by the fluid on the plate is obtained by L

FD =

∫0

τw (x) W dx

(3.49)

where W is the plate width. The local skin friction coefficient cf is defined as cf =

τw ρ U2∞ ∕2

.

(3.50)

The wall shear stress τw may be scaled as τw ∼ μ

( ) U U∞ 1 −1∕2 = ρU2∞ ReL =μ ∞ √ δ 1 (L∕ ReL )

and it is found that the dimensionless skin friction coefficient cf is a function of Reynolds number −1∕2

cf ∼ ReL

.

(3.51)

We will now look in the y-component of the momentum equation. Each term is of order δ. Recall that the streamlines are nearly parallel over slender body, and the vertical velocity components are very small. Thus, the y-component momentum equation reduces to 0≈

1 𝜕p → p ≠ p(y). ρ 𝜕y

(3.52)

Since pressure gradient in the vertical direction is negligible, we can safely neglect the y-component of momentum equation. We can summarize our findings as follows. The two-dimensional form of continuity and momentum equations can be replaced by the following Prandtl boundary layer equations. Continuity: 𝜕u 𝜕v + = 0. 𝜕x 𝜕y x-direction momentum: 𝜕u 𝜕u 1 dp 𝜕2 u u +v =− +ν 2. 𝜕x 𝜕y ρ dx 𝜕y

(3.53)

(3.54)

y-direction momentum 0≈−

1 𝜕p ρ 𝜕y

(3.55)

75

76

3 Equations of Incompressible External Laminar Boundary Layers

subject to the following boundary conditions:

3.4

1. u(0, y) = U∞

(3.56)

2. v(0, y) = 0

(3.57)

3. u(x, 0) = 0 on the solid wall (no slip)

(3.58)

4. v(x, 0) = 0 impermeability on the solid wall

(3.59)

5. Uniform flow u(x, ∞) = U∞ for far from the solid wall in y-directions

(3.60)

6. Uniform flow v(x, ∞) = 0 for far from the solid wall in the y-directions

(3.61)

The Thermal Boundary Layer Concept

We again consider flow over a flat plate having a characteristic length L. The plate surface temperature is Tw . The free stream temperature and the velocity are T∞ and U∞ , respectively. If the plate surface temperature Tw is larger than the free stream fluid temperature T∞ , a thermal boundary layer develops along the plate. See Figure 3.3. Energy diffuses from the hot wall into moving fluid, and it is transported downstream. Temperature distribution is confined to a thin region, and this thin region is called thermal boundary layer, and its thickness is denoted by Δ(x). We again assume that (Δ/L) ≪ 1. We are interested in the thermal energy equation for flow over the thin flat plate. For this purpose, we assume: (1) (2) (3) (4) (5)

Steady state Two-dimensional flow Laminar flow Constant fluid properties No viscous dissipation.

Energy equation and its boundary conditions are given as ) ( 2 ( ) 𝜕T 𝜕T 𝜕 T 𝜕2 T ρcp u +v =k + 𝜕x 𝜕y 𝜕x2 𝜕y2 −k

𝜕T(x, 0) = q′′w 𝜕y

(3.62) (3.63)

or T(x, 0) = Tw

(3.64)

T(0, y) = T∞

(3.65)

T(x, ∞) = T∞

(3.66)

The Eckert number may appear in some forced convection problems and is defined as Ec =

U2∞ cp (Tw − T∞ )

(3.67)

where cp is the specific heat, U∞ is the freestream velocity, Tw is the surface temperature, and T∞ is the free stream velocity. In the present analysis, the Eckert number will be neglected.

3.4 The Thermal Boundary Layer Concept

T∞

y

U∞

T∞

Δ

Δ

U∞

U∞ δ

T∞

δ

Tw Figure 3.3

x

Tw

Thermal boundary layer for Δ > δ or Pr ≪ 1.

3.4.1 Scaling of Energy Equation The scaling procedure for the thermal boundary layer is discussed in [4, 5]. We will again use the scaling in order to examine the order of magnitude of each term of the energy equation. We will consider flow over a flat plate of having characteristic length L. The free stream velocity is U∞ , and the free stream temperature is T∞ . The velocity boundary layer thickness is δ, and the thermal boundary layer thickness is Δ as shown in Figure 3.3. The plate surface temperature is Tw . Again, we will assume that the velocity boundary layer region is characterized by δ ≪ L, and the thermal boundary layer is characterized by Δ ≪ L. (It is assumed here that the thermal boundary layer and the velocity boundary layer start at the leading edge of the plate.) We assume that Δ∕L ≪ 1. The scale for x: x ≈ L.

(3.68)

The scale for y: y ≈ Δ.

(3.69)

The scale for temperature difference is (Tw − T∞ ). Scales for u and v depend on whether thermal boundary layer thickness Δ is larger or smaller than velocity boundary layer thickness 𝛿. Thus, we will study two cases as given below: a) Δ(x) > δ(x) b) Δ(x) < δ(x). Case I: Δ(x) > δ(x) This is the thick thermal boundary layer case. The axial velocity u is within the thermal boundary layer. Thus, the scale for u is given as u ∼ U∞ .

(3.70)

The scaling of the continuity equation gives U v ∼ ∞ Δ L or

( v∼

U∞ L

(3.71) ) Δ.

(3.72)

The two convection terms of the energy equation can be scaled as (T − T∞ ) 𝜕T ∼ U∞ w 𝜕x L [( ) ] (Tw − T∞ ) (T − T∞ ) U∞ 𝜕T ∼ Δ ∼ U∞ w . v 𝜕y L Δ L u

(3.73) (3.74)

77

78

3 Equations of Incompressible External Laminar Boundary Layers

Thus, we see that both the convection terms are of the same order. Next, we will scale the two conduction terms on the right-hand side of the energy equation: 𝜕 2 T (Tw − T∞ ) ∼ 𝜕x2 L2 𝜕 2 T (Tw − T∞ ) ∼ . 𝜕y2 Δ2

(3.75) (3.76)

Taking the ratio of Eqs. (3.75) and (3.76), we get (Tw − T∞ ) 𝜕2 T ( )2 Δ 𝜕x2 ∼ L2 ∼ . 2 (Tw − T∞ ) L 𝜕 T 𝜕y2 Δ2 Since (Δ/L) ≪ 1, Eq. (3.77) gives 𝜕2 T 𝜕2 T ≪ . 𝜕x2 𝜕y2

(3.77)

(3.78)

This tells us that the axial conduction term is negligible compared with the conduction term in the y-direction. Thus, the boundary layer energy equation simplifies to u

𝜕T 𝜕T 𝜕2 T +v =α 2. 𝜕x 𝜕y 𝜕y

(3.79)

Consider now the energy Eq. (3.79), and let us look at the balance between the convection term in the x-direction and the conduction term in the y-direction. We see that u

𝜕2 T 𝜕T ∼α 2. 𝜕x 𝜕y

(3.80)

The scaling of the convection term yields (T − T∞ ) 𝜕T ∼ U∞ w . 𝜕x L The scaling of the conduction term in the y-direction yields u

α

(T − T ) 𝜕2 T ∼α w 2 ∞ . 𝜕y2 Δ

(3.81)

(3.82)

From Eqs. (3.81) and (3.82), we can write U∞

(T − T ) (Tw − T∞ ) ∼α w 2 ∞ . L Δ

Rearranging Eq. (3.83), we obtain √ α Δ ∼ . L U∞ L Using the definition of thermal diffusivity α = k/ρcp , Eq. (3.84) gives √ k Δ ∼ . L ρcp U∞ L

(3.83)

(3.84)

(3.85)

Let us introduce the definition of Prandtl number Pr and Reynolds number ReL as μcp Pr = k ρU∞ L . ReL = μ Using the definitions of Prandtl and Reynolds numbers, Eq. (3.85) can be written as 1 Δ ∼ √ . L Pr ReL

(3.86)

3.4 The Thermal Boundary Layer Concept

Thus, we conclude that √ Δ ≪ 1 when Pr ReL ≫ 1. L The product of the Prandtl and Reynolds numbers is called Peclet number, Pe, and it is defined as Pe = Pr ReL .

(3.87)

(3.88)

Let us take the ratio of thermal boundary layer thickness Δ to the velocity boundary layer thickness δ as given below Δ∕L Δ = = δ∕L δ

1 √ ReL Pr 1 √ ReL

1 ∼ √ . Pr

(3.89)

√ Thus, Δ > δ when Pr ≪ 1, and this represents the range covered by liquid metals. The heat transfer coefficient h is defined as k(𝜕T∕𝜕y)wall h= . Tw − T∞ The heat transfer coefficient is scaled as k[(Tw − T∞ )∕Δ] k h∼ ∼ . (Tw − T∞ ) Δ

(3.91)

The heat transfer coefficient for low Prandtl number limit is obtained as k l k√ h∼ ∼ ∼ Pr ReL L Δ L √ Pr

(3.90)

(3.92)

ReL

or this result can be casted in terms of the Nusselt number NuL = h L/k hL √ ∼ Pr ReL Pr ≪ 1 (3.93) NuL = k This case represents the liquid metals. Case II: Δ(x) < δ(x) This is the thin thermal boundary layer case as shown in Figure 3.4. This case represents a class of fluids, such as air (i.e. Pr ∼ 1) or water or oil (i.e. Pr > 1). The axial velocity u within the thermal boundary layer is smaller than the free stream velocity U∞ . The scale of u in Δ layer is not U∞ but we assume that the velocity profile within the thermal boundary layer is linear, as shown in Figure 3.5. From the similar triangles, we can obtain a scale for u as Δ . δ Performing a scale analysis on continuity equation gives ( ) v u u ∼ ⇒v∼Δ . L Δ L Combining Eq. (3.94) with Eq. (3.95) yields

(3.94)

u ∼ U∞

v ∼ U∞

Δ2 . L𝛿

(3.96) U∞

y

U∞

T∞

δ

U∞ T∞

Tw

T∞

δ

Δ

Figure 3.4

(3.95)

Δ x

Thermal boundary layer for Δ < δ or Pr ≫ 1.

Tw

79

80

3 Equations of Incompressible External Laminar Boundary Layers

U∞

Figure 3.5

Similar triangles.

u

δ

Δ

Next, we scale the convection terms of energy equation. First convection term 𝜕T Δ Tw − T∞ ∼ U∞ . u 𝜕x δ L Second convection term 𝜕T Δ2 Tw − T∞ Δ Tw − T∞ ∼ U∞ ∼ U∞ . v 𝜕y δL Δ δ L

(3.97)

(3.98)

A comparison of Eqs. (3.97) and (3.98) yields that the two convection terms are of the same order. Next, we will scale the conduction terms in the energy equation. Conduction in the axial direction is scaled as 𝜕 2 T (Tw − T∞ ) ∼ . (3.99) 𝜕x2 L2 Conduction in the normal direction to flow is scaled as 𝜕 2 T (Tw − T∞ ) ∼ . (3.100) 𝜕y2 Δ2 A comparison of Eqs. (3.99) and (3.100) reveals ]/ [ ( )2 (Tw − T∞ ) (Tw − T∞ ) (Tw − T∞ ) 𝜕 2 T∕𝜕x2 Δ Δ2 ∼ ≃ = . 2 2 2 (T − T ) L 𝜕 2 T∕𝜕y2 Δ L L w ∞

(3.101)

Since we assumed that Δ/L ≪ 1, Eq. (3.101) gives 𝜕2 T 𝜕2 T ≪ 2. 𝜕x2 𝜕y

(3.102)

Equation (3.102) indicates that axial conduction is negligible compared with conduction in the normal direction of the flow. The energy equation takes the following form: u

𝜕T 𝜕T 𝜕2 T +v =α 2. 𝜕x 𝜕y 𝜕y

(3.103)

We will perform a balance between convection and conduction terms to determine the condition for which Δ ≪ δ u

𝜕T 𝜕2 T ∼α 2. 𝜕x 𝜕y

(3.104)

Scaling each term in the above equation gives: U∞

(T − T ) Δ (Tw − T∞ ) ∼α w 2 ∞ . δ L Δ

(3.105)

Substituting the definition of thermal diffusivity α = k/ρcp , multiplying and dividing the left-hand side by L2 , and rearranging this last equation yield ( )3 Δ k δ . (3.106) ∼ L ρcp U∞ L L √ Next, we use δ∕L ≃ 1∕ ReL to eliminate δ/L in the above equation. This will give us ( )3 Δ k 1 ∼ . (3.107) √ L ρcp U∞ L Re L

3.4 The Thermal Boundary Layer Concept

Multiplying and dividing the right-hand side by dynamic viscosity μ, we obtain ( )3 μ k Δ 1 ∼ √ L μcp ρU∞ L ReL or

( )3 Δ 1 1 1 ∼ . √ L Pr ReL ReL

(3.108)

(3.109)

After rearrangement, one obtains 1 Δ ∼ . √ L Pr 1∕3 ReL

(3.110)

Thus, we conclude that √ Δ∕L ≪ 1 when Pr 1∕3 ReL ≫ 1.

(3.111)

Let us now establish the condition under which Δ < δ is valid. This is done by taking the following ratio of thermal boundary layer thickness to velocity boundary layer thickness, and after simplification, we obtain √ √ (Δ∕L) = (1∕Pr 1∕3 Re)∕(1∕ Re) (3.112) (δ∕L) or 1 Δ ∼ 1∕3 . δ Pr

(3.113)

Therefore, we find that Δ < δ when Pr 1∕3 ≫ 1. Setting L = x, the heat transfer coefficient h and the Nusselt number vary as √ k k ⇒ Pr 1∕3 Rex h∼ Δ x

(3.114)

(3.115)

and Nux =

√ xh ∼ Pr 1∕3 Rex . k

(3.116)

This gives us an idea for the correlation of experimental data in forced convection heat transfer. The thermal energy equation is expressed as } { 𝜕T 𝜕2 T 𝜕T ρcp u +v =k 2 (3.117) 𝜕x 𝜕y 𝜕y along with the following boundary conditions. Two commonly used boundary conditions will be considered here, i.e. wall temperature or uniform wall heat flux. The boundary conditions are given as T(0, y) = T∞

(3.118)

T(x, 0) = Tw

(3.119a)

or −k

𝜕T(x, 0) = q′′w 𝜕y

T(x, ∞) = T∞ .

(3.119b) (3.120)

In this discussion, the viscous dissipation term is neglected. We now have three equations and three unknowns u, v, and T. The solutions of these equations will be discussed later [6–14]. The velocity field is independent of temperature field. Thermal boundary layer approximation indicates that the diffusion of thermal energy in the flow direction is negligible.

81

82

3 Equations of Incompressible External Laminar Boundary Layers

3.5

Summary of Boundary Layer Equations of Steady Laminar Flow

In developing the equations, several assumptions are made. These assumptions are listed below. 1. 2. 3. 4. 5. 6. 7. 8. 9.

Steady laminar flow Viscous dissipation is neglected There is no gravitational effect Energy generation is neglected Newtonian fluid Continuum Two-dimensional flow Constant fluid properties Slender body

The equations for the boundary layer are as follows: Continuity: 𝜕u 𝜕v + = 0. 𝜕x 𝜕y

(3.121)

x-direction momentum: u

𝜕u 𝜕u 1 𝜕p 𝜕2 u +v =− +ν 2. 𝜕x 𝜕y ρ 𝜕x 𝜕y

(3.122)

The boundary conditions are: u(x, 0) = 0

(3.123)

v(x, 0) = 0 solid wall

(3.124)

u(x, ∞) = U∞

(3.125)

u(x, 0) = U∞

(3.126)

Energy: { } 𝜕T 𝜕T 𝜕2 T +v =k 2 ρcp u 𝜕x 𝜕y 𝜕y

(3.127)

T(0, y) = T∞

(3.128)

T(x, 0) = Tw

(3.129a)

or −k

𝜕T(x, 0) = q′′w 𝜕T

T(x, ∞) = T∞ .

(3.129b) (3.130)

Problems 3.1

Using scale analysis, show that in a boundary layer, the pressure gradient in the cross-stream direction is negligible compared to the pressure gradient in the streamwise direction. See Figure 3.P1. Discuss the significance of this finding.

References

y

Figure 3.P1 plate.

U∞

U∞ 0

δ x

Typical velocity boundary layer for laminar flow over a flat

u (x, y)

L

References 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Schlichting, H. and Gersten, K. (2017). Boundary Layer Theory, Translated into English by K. Mayes, 9e. Springer. Rosenhead, L. (1963). Laminar Boundary Layers. Dover Publications, Inc. Schetz, J.A., Boundary Layer Analysis, AIAA Education Series, 2010. Bejan, A. (2013). Convection Heat Transfer, 4e. Wiley. Jiji, L.M. (2009). Heat Convection, 2e. Springer. Oosthuizen, P.H. and Naylor, D. (1999). Introduction to Convection Heat Transfer. McGraw-Hill. Kakac, S., Yener, Y., and Pramuanjaroenkij, A. (2014). Convective Heat Transfer. CRC Press. Kays, W.M., Crawford, M.E., and Weigand, B. (2005). Convective Heat and Mass Transfer. McGraw-Hill. Cebeci, T. (2002). Convective Heat Transfer. Springer. Burmeister, L.C. (1993). Convective Heat Transfer, 2e. Wiley. Arpaci, V.S. and Larsen, P.S. (1984). Convection Heat Transfer. Prentice-Hall. White, F.M. and Majdalani, J. (2021). Viscous Fluid Flow. McGraw-Hill. Marinett, M.F. and Tardu, S. (2009). Convective Heat Transfer Solved Problems. Wiley. Ghiaasiaan, S.M. (2018). Convective Heat and Mass Transfer, 2e. CRC Press.

83

85

4 Integral Methods in Convective Heat Transfer 4.1 Introduction When electronic computers were not available, the integral method of solving boundary layer problems was very popular. If the solution procedure becomes tedious, it is very useful to use approximate methods. The integral method is one such method. The integral method is capable of handling arbitrary variations of free stream velocity U∞ (x) and surface temperature Tw (x). If the exact solution is not known or difficult to obtain, then the integral method is very useful. Fortunately, an integral method due to von Karman is available, as discussed in [1]. The integral method is used very often in heat transfer because of its simplicity. It applies to both the laminar and turbulent flows. However, the basic laws are satisfied in average sense, and naturally, the integral method gives approximate solutions. The accuracy of the solution depends on the assumed velocity and temperature profiles. The assumed velocity and temperature profiles can range from linear to more complex forms. The sophisticated functions are better able to capture the characteristics of velocity and temperature profiles. The undetermined coefficients of these velocity and temperature profiles are selected so that the boundary conditions are satisfied. It is true that the integral method is considerably less accurate than its differential counterpart, but the success it has enjoyed in many engineering applications justifies the usage of the method. The derivation of the integral form of momentum and energy equations is discussed in textbooks, such as [2–6].

4.2 Conservation of Mass Consider steady, incompressible, constant-property fluid flow over a curved surface of the body of revolution. See Figure 4.1. There is an axisymmetric flow over this curved surface of the body of revolution. We assume that there is a thin boundary layer over this surface. Since the boundary layer is thin, the pressure across the boundary layer is assumed to be constant. The boundary layer thickness is much smaller than R(x), where R(x) is the transverse radius of curvature. The flow outside the boundary layer is inviscid. As R → ∞, a two-dimensional surface is obtained. The surface can also be inside of a circular-section nozzle, and in this case, R(x) is measured from the centerline of the fluid stream. The free stream velocity just outside the boundary layer is denoted by U∞ (x). Stream fluid density is denoted by ρ. The edge of the viscous boundary layer does not coincide with a streamline, and for this reason, mass can enter the boundary layer from external flow. We choose a stationary control volume in the boundary layer. There is no fluid flow through body surface by injection or suction. See Figure 4.2. The chosen control volume extends above the boundary layer to a distance Y. The thickness of the control volume in the x-direction is δx, and the width of the control volume is Rδϕ. The height of the control volume is finite in the y-direction and denoted by Y, and notice that Y ≪ R. Recall that the integral form of the conservation of mass equation is d ⃗ ⋅ dV ⃗ =0 ρd∀ + ρV ∬ dt ∭ cv

cs

which for steady flow reduces to ∬

⃗ ⋅ dV ⃗ =0 ρV

(4.1)

cs

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

86

4 Integral Methods in Convective Heat Transfer

Figure 4.1 equation.

Coordinate system and control volume in the boundary layer over a body of revolution for the integral momentum

Figure 4.2

Control volume for the development of the integral equation conservation of mass.

or for the control volume having multiple inlet and outlet, it may be expressed as ∑ ∑ ṁ in − ṁ out = 0 in

(4.2)

out

Applying the conservation of mass to the control volume indicated by ABCD, we obtain the following equation for the conservation of mass [ ] d [ṁ x + δṁ Y ] = ṁ x + (ṁ x )dx (4.3) dx The rate at which mass passes through the strip on AB, as indicated in Figure 4.2, is ρ u Rδϕ dy, and integrating from y = 0 to y = Y gives the mass flow rate across the Rδϕ × AB. Thus, ṁ x = R δϕ

Y

∫0

ρudy

(4.4)

The rate at which mass enters the control volume through the face Rδϕ × BC is ṁ Y = R δϕ δx vY ρ Substituting Eqs. (4.4) and (4.5) into Eq. (4.3) gives [ ( ] [ ) ] Y Y Y d R δϕ R δϕ ρ u dy + R δϕ δx vY ρ = R δϕ ρudy + ρudy δx ∫0 ∫0 ∫0 dx

(4.5)

(4.6)

4.3 The Momentum Integral Equation

Recall that R ≫ Y and after a simple algebra, for constant ρ, we obtain the following form of the conservation of mass: ( ) Y 1 d R ρudy (4.7) vY ρ = ∫0 R dx

4.3 The Momentum Integral Equation Neglecting body forces, the x-component of the conservation of linear momentum is ∑ d ⃗ ⋅ dA) ⃗ Fx = ρ u d∀ + u(ρV ∬ dt ∭ cv

(4.8)

cs

Under steady flow conditions, Eq. (4.8) reduces to ∑ ⃗ ⋅ dA) ⃗ u(ρV Fx = ∬

(4.9)

cs

This equation can be written as ∑ Fx = Ṁ out − Ṁ in

(4.10)

where Ṁ in and Ṁ out are the rates of momentum in and out, respectively. Consider the control volume shown in Figure 4.3. The rate at which the momentum enters through the strip dy on AB is equal to u(ρuRδϕdy) = ρ u2 Rδϕdy, and integrating this quantity from y = 0 to y = Y will give the rate at which momentum enters through the face Rδϕ × AB at x: Ṁ x = Rδϕ

Y

∫0

ρu2 dy

(4.11)

The rate at which momentum enters through the face Rδϕ × BC at y = Y Ṁ Y = (Rδϕδx ρ uY )vY

(4.12)

The total momentum entering into the control volume is Ṁ in = Ṁ x + Ṁ Y = Rδϕ

Y

∫0

ρu2 dy + Rδϕδx vY ρ uY

The rate at which momentum leaves through the face Rδϕ × CD at x + δx ( ) Y Y d 2 2 ̇Mout = Ṁ x + d (Ṁ x )δx = Rδϕ Rδϕ ρu dy + ρu dy δx ∫0 ∫0 dx dx Pressure and viscous forces acting on the control volume are ) ] [( ∑ dp δx Fx = τY (Rδϕδx) − τw (Rδϕδx) + p (Rδϕ Y) − Rδϕ Y p + dx

Figure 4.3

Control volume for the development of the integral equation conservation of momentum.

(4.13)

(4.14)

(4.15)

87

88

4 Integral Methods in Convective Heat Transfer

Substituting Eqs. (4.13)–(4.15) into Eq. (4.10) and simplifying, we obtain ( ( ) ) Y dp 1 d τY − τw − Y = R ρu2 dy − ρvY uY ∫0 dx R dx Combining Eq. (4.16) with the conservation of mass equation, Eq. (4.7), we obtain ( ( ( ) ) ) Y Y u d dp 1 d τY − τw − Y= R R ρu2 dy − Y ρudy ∫0 ∫0 dx R dx R dx Since the pressure is constant across the boundary layer, the Y(dp/dx) term can be written as ) ( ) Y( dp dp = dy Y ∫0 dx dx Then, the momentum equation, Eq. (4.17), becomes ( ( ) ) ) Y Y Y( uY d dp 1 d 2 R R dy ρu dy − ρudy + τY − τw = ∫0 ∫0 ∫0 R dx R dx dx

(4.16)

(4.17)

(4.18)

(4.19a)

Now, assume that Y is sufficiently large such that uY → U∞ and τY → 0. This means that Y > δ(x), where δ(x) is the velocity boundary layer thickness. Equation (4.19a) becomes ( ( ) ) ) Y Y Y( U∞ d dp 1 d 2 R R dy (4.19b) ρu dy − ρudy + −τw = ∫0 ∫0 ∫0 R dx R dx dx where τw is the wall shearing stress and U∞ is the local free stream fluid velocity at the edge of the boundary layer. Since Y is much larger than the velocity boundary layer thickness, we can now use the Bernoulli equation for inviscid flow along streamline to relate the pressure gradient to free stream velocity gradient. In the undisturbed outer flow, the Bernoulli equation applies and may be written as p + ρ U2∞ ∕2 = const Differentiating Eq. (4.20) with respect to x, we get ) ( dU∞ dp = −ρU∞ dx dx Introducing Eq. (4.21) into the momentum equation, Eq. (4.19b), we obtain ( ( ) ) ) Y Y Y( dU∞ U∞ d 1 d 2 −τw = −ρU∞ R R dy ρu dy − ρudy + ∫0 ∫0 ∫0 R dx R dx dx We can set the upper limit of integrals as δ(x) since the integrals are zero for y > δ(x) ( ( ) ) ) δ δ δ( U∞ d dU∞ 1 d 2 −ρU∞ −τw = R R dy ρu dy − ρudy + ∫0 ∫0 ∫0 R dx R dx dx

(4.20)

(4.21)

(4.22)

(4.23)

where δ is the velocity boundary layer thickness, and it is defined as the distance from the wall to the point where u = 0.99 U∞ . We notice that U∞ and dU∞ /dx may be moved inside or taken outside the integration since they depend on x. Equation (4.23) is the integral momentum equation of a steady, laminar, and incompressible boundary layer flow. This integral equation is valid for both the laminar and turbulent boundary layer flows. Since the surface is curved, the free stream velocity U∞ (x) and pressure p(x) vary along the surface. Notice that the effect of gravity is neglected. Equation (4.23) is developed based on the conservation of mass and linear momentum. If the velocity distribution is known, τw may be easily determined by ( ) du τw = μ (4.24) dy y=0 This equation is valid for flow inside as well as over an axisymmetric body. The boundary layer thickness is much less than the local radius of curvature. For a two-dimensional body, the radius of curvature R(x) drops out. Next, we will introduce the concept of displacement thickness δ1 and momentum thickness δ2 in addition to boundary layer thickness δ.

4.3.1

The Displacement Thickness 𝛅1

Consider a two-dimensional flow along a flat surface. See Figure 4.4. The boundary layer begins to grow starting from x = 0. Surface AD is a solid wall, and major velocity changes take place within the boundary layer. Let us apply the conservation of mass to this control volume.

4.3 The Momentum Integral Equation

Figure 4.4

Control volume for the development of displacement and momentum thickness.

The mass flow rate across AB is Y

ṁ AB =

ρU∞ dy

∫0

(4.25)

The mass flow rate across CD is ṁ D C =

Y

(4.26)

ρudy

∫0

The mass flow rate across AD is ṁ AD = 0

(4.27)

The mass flow rate across BC is ṁ BC = ṁ DC − ṁ AB ṁ BC =

(4.28)

Y

∫0

Y(

Y

ρU∞ dy −

∫0

ρudy = ρU∞

∫0

1−

u U∞

) dy

Let us define displacement thickness δ1 as ) Y( u 1− dy δ1 ρ U∞ = ρ U∞ ∫0 U∞ Now, let Y → ∞ so that boundary layer is included. Thus, ) ) ∞( δ( u u δ1 = 1− dy ≈ 1− dy ∫0 ∫0 U∞ U∞

(4.29)

(4.30)

(4.31)

We need to integrate up to velocity boundary layer thickness since integral is zero for y > δ(x). Notice that for y > δ(x), u = U∞ and the displacement thickness δ1 is a measure of the displacement of mainstream due to the presence of flat plate and boundary layer. In other words, the displacement thickness δ1 is the distance by which the external streamlines are shifted due to the presence of the boundary layer.

4.3.2 Momentum Thickness 𝛅2 Consider again Figure 4.4. The x-momentum entering across AB is Ṁ AB =

Y

ρU2∞ dy

∫0

(4.32)

x-momentum leaving across CD is Ṁ CD =

Y

∫0

ρu2 dy

(4.33)

If Y → ∞, then u → U∞ . Then, the x-momentum entering across BC is obtained by multiplying Eq. (4.29) by U∞ as [ ) ] Y( u 1− dy (4.34) Ṁ BC = U∞ U∞ ρ ∫0 U∞

89

90

4 Integral Methods in Convective Heat Transfer

We can now write the net rate of momentum loss in control volume as ) ] [ Y( Y Y Y u 1− dy = ρU2∞ dy − ρu2 dy − U2∞ ρ ρu(U∞ − u)dy ∫0 ∫0 ∫0 ∫0 U∞

(4.35)

Let Y → ∞ and define the momentum thickness δ2 as ∞

δ2 U2∞ ρ =

∫0

or ∞

δ2 =

∫0

u U∞

ρu(U∞ − u)dy ( 1−

u U∞

(4.36)

)

δ

dy =

∫0

u U∞

( 1−

u U∞

) dy

(4.37)

Again, we needed to integrate up to velocity boundary layer thickness δ(x) since integral is zero for y > δ(x). Notice that for y > δ(x), u = U∞ . The momentum thickness δ2 is a measure of the momentum flux decrement due to boundary layer.

4.4

Alternative Form of the Momentum Integral Equation

Equation (4.23) may be expressed in terms of displacement thickness δ1 and momentum thickness δ2 , as given in [2] ] [( ) τw δ1 dδ2 1 dR 1 dU∞ (4.38) 2 + = + δ + 2 dx δ2 U∞ dx R dx ρU2∞ We now have a general form of integral momentum equation. Equation (4.38) represents the relation between the overall rate of momentum flux across a section of the boundary layer and the surface forces. Equation (4.38) applies to laminar flow as well as turbulent flow since we derived it without a specific assumption about the flow regime. Partial differential equations of conservation of mass and linear momentum are reduced to ordinary differential equation for δ2 . The solution of this ordinary differential equation is not an easy task. For many engineering problems in convective heat transfer, it is often acceptable if an approximate solution may be obtained. Equation (4.38) is the basis of many approximate boundary layer solutions.

4.5

Momentum Integral Equation for Two-Dimensional Flow

For flow over a two-dimensional surface without suction or blowing, R(x) → ∞ and R(x) drops out, and the momentum integral equation becomes ] [( ) τw δ1 dδ2 1 dU∞ + δ2 2 + (4.39) = dx δ2 U∞ dx ρU2∞ Equation (4.39) may be used to study boundary layer flows with pressure gradient. This equation will be discussed later in detail. In principle, to solve Eq. (4.39), we need to find an expression for the free stream velocity U∞ (x), and U∞ (x) may be obtained from the potential flow solution or experimental study. dU dp For flow with constant free stream velocity U∞ over a two-dimensional impermeable surface, dx∞ = dx = 0, and the momentum integral equation reduces to τw ρU2∞

=

dδ2 dx

The local friction coefficient is defined as τw cf = ρU2∞ ∕2

(4.40)

(4.41)

Using Eq. (4.41), the momentum equation takes the following form: dδ cf = 2 2 dx

(4.42)

4.6 Energy Integral Equation

Note the following: 1) 2) 3) 4)

The integral momentum equation applies to laminar as well as turbulent flow. The effect of gravity is neglected in the above equation. The integral momentum equation represents the integral formulation of both conservation of momentum and mass. Although u is a function of x and y, once the integrals in the equation are evaluated, one obtains a first-order ordinary differential equation with x as the independent variable.

4.6 Energy Integral Equation We will now discuss the energy integral equation. The derivation of the integral form of the energy equation is discussed in textbooks such as [1–6]. Consider the steady flow of incompressible, constant-property fluid at temperature T∞ over a surface of a body of revolution. See Figure 4.5. Since we are interested in low-speed flow, the viscous dissipation effects are negligible, and there is no energy generation. The heat conduction is important only in the y-direction. The surface is maintained at a different temperature, and thus, heat exchange takes place between surface and fluid. The temperature and velocity boundary layers form over the surface and the thickness of boundary layers are much smaller than the radius R(x) of revolution. There is no variation of temperature and velocity in the ϕ direction, and it is also assumed that the free stream enthalpy is constant. Surface temperature varies in the flow direction. We will assume that velocity and temperature variations in the flow direction are not important compared to the corresponding variations in the y-direction. There is no work done on control volume by viscous stresses. Let us review the assumptions we made. 1) 2) 3) 4) 5) 6)

negligible changes in potential energy axial conduction is not important dissipation is neglected work done by normal viscous stresses and body forces are not important there is no volumetric heat source no shaft work is done

Consider control volume depicted in Figure 4.6. Our control volume is of infinitesimal extent δx in the main flow direction, but it has a finite height Y in the direction normal to the surface. Control volume is fixed to the solid surface. In a manner analogous to the development of the integral momentum equation, we will derive an integral energy equation. The conservation of energy for a control volume can be written as Ė in − Ė out + Ė g = Ė st The application of the conservation of energy to control volume under steady-state conditions yields [Ė x + Ė Y + q̇ w ] = [Ė x+δx ]

Figure 4.5

Coordinate system and control volume in the boundary layer over a body of revolution.

(4.43)

91

92

4 Integral Methods in Convective Heat Transfer

Figure 4.6

Control volume for the development of the integral equation conservation of energy.

The rate at which mass crosses the strip (Rδϕ)dy on AB, as shown in Figure 4.6, will be Rδϕρudy; then, the rate at which enthalpy entering into control volume by fluid motion at position x through this strip is given by (Rδϕ)(ρu h* )dy. The rate at which enthalpy enters the control volume through the face (Rδϕ)AB in the flow direction is given by Y

ρuh∗ dy Ė x = Rδϕ ∫0

(4.44)

where h* = h + V2 /2 is the stagnation enthalpy. The rate at which enthalpy leaves the control volume at x + δx through (Rδϕ)CD face is given by [ ] Y Y d Rδϕ Ė x+δx = Rδϕ ρuh∗ dy + ρuh∗ dy δx (4.45) ∫0 ∫0 dx Energy entering by conduction through the lower surface (Rδϕ δx) of the control volume at y = 0 is qw = R δϕ δx q′′w

(4.46)

q′′w

where is the conduction heat flux. Enthalpy entering through the upper surface (Rδϕ δx) of the control volume at y = Y by the free stream mass flow rate Ė Y = R δϕ δx (ρvY ) h∗Y We substitute Eqs. (4.44)–(4.47) into the energy balance equation, Eq. (4.43), and we obtain } { [ ] } { Y Y Y d Rδϕ ρuh∗ dy + R δϕ δx q′′w + R δϕ δx (ρvY ) h∗Y = Rδϕ ρuh∗ dy + ρuh∗ dy δx Rδϕ ∫0 ∫0 ∫0 dx

(4.47)

(4.48)

We now assume that Y is large enough so that all fluid properties at position Y are free stream properties. This means that Y > Δ(x) and Δ(x) is the thermal boundary layer thickness. Now, after algebra } { [ ] Y d Rδϕ (4.49) ρuh∗ dy δx − R δϕ δx (ρvY ) h∗Y R δϕ δx q′′w = ∫0 dx The expression for the mass flow rate through the upper surface was developed when deriving the momentum integral equation. This mass flow rate entering the control volume through the upper surface Rδϕδx at y = Y is given as ( ) Y 1 d R ρudy vY ρ = ∫0 R dx Substituting Eq. (4.7) into the energy equation, Eq. (4.49), we get [ ( ] ) Y Y d d Rδϕ R ρuh∗ dy δx − δϕ ρuh∗Y dy δx R δϕ δx q′′w = ∫0 ∫0 dx dx

(4.50)

4.6 Energy Integral Equation

After simplification of Eq. (4.50), we obtain ] [ Y ) ( 1 d R q′′w = ρu h∗ − h∗Y dy R dx ∫0 If we let Y → ∞, the above equation becomes ] [ ∞ ) ( ∗ 1 d ∗ ′′ R qw = ρu h − h∞ dy R dx ∫0

(4.51)

(4.52)

We can write Eq. (4.52) in terms of temperature for low-velocity, constant-property flow of perfect gas as follows: [ ] ∞ q′′w 1 d R = u(T − T∞ ) dy (4.53) ρcp R dx ∫0 This is the energy integral equation for the boundary layer we are seeking for. We will write this equation in a different form.

4.6.1 Enthalpy Thickness Let us define the enthalpy boundary layer thickness 𝛥2 as )( ∗ ) ∞( h − h∗∞ u dy Δ2 = ∫0 U∞ h∗w − h∗∞

(4.54)

We can write Eq. (4.54) in terms of temperature for low-velocity, constant-property flow of perfect gas as follows: )( ) ∞( T − T∞ u Δ2 = dy (4.55) ∫0 U∞ Tw − T∞

4.6.2 Conduction Thickness Conduction thickness is defined as k(Tw − T∞ ) Δ4 = q′′w where k is the thermal conductivity of the fluid.

(4.56)

4.6.3 Convection Conductance or Heat Transfer Coefficient Finally, we define convection conductance or heat transfer coefficient as q′′w h= Tw − T∞

(4.57)

Thus, we have conduction thickness in terms of heat transfer coefficient h and fluid thermal conductivity k k Δ4 = (4.58) h Let us write Eq. (4.53) in a different form. Assume that fluid properties are constant. We will expand the integral on the right-hand side of Eq. (4.53) [ [ ] ] ∞ ∞ ∞ 1 dR d 1 d R u(T − T∞ ) dy = u(T − T∞ ) dy + R u(T − T∞ ) dy R dx ∫0 R dx ∫0 dx ∫0 ] [ ∞ ∞ d 1 dR u(T − T∞ ) dy + u(T − T∞ ) dy = R dx ∫0 dx ∫0 We substitute this result into Eq. (4.53), and we get [ ] ∞ ∞ q′′w d 1 dR = u(T − T∞ ) dy + u(T − T∞ ) dy ρcp R dx ∫0 dx ∫0 We now let Y → Δ [ ] Δ Δ q′′w d 1 dR = u(T − T∞ ) dy + u(T − T∞ ) dy ρcp R dx ∫0 dx ∫0

(4.59)

(4.60)

where Δ(x) is the thermal boundary layer thickness. The first term on the right-hand side of Eq. (4.60) represents the effect of curvature.

93

94

4 Integral Methods in Convective Heat Transfer

4.7

Alternative Form of the Energy Integral Equation

An alternative form of the energy integral equation is given as [ ( )] q′′w dΔ2 1 dR 1 dU∞ 1 d = + Δ2 (Tw − T∞ ) + + ρcp U∞ (Tw − T∞ ) dx (Tw − T∞ ) dx U∞ dx R dx

4.8

(4.61)

Energy Integral Equation for Two-Dimensional Flow

Let us now consider the energy integral equation for flow over a flat plate. For a two-dimensional impermeable flat plate with constant ρ, U∞ , Tw , the energy integral equation, Eq. (4.61), reduces to q̇ ′′w dΔ2 = ρcp U∞ (Tw − T∞ ) dx

(4.62)

The local heat transfer coefficient is defined by Eq. (4.57). Using Eq. (4.57), the energy integral equation, Eq. (4.62), becomes dΔ2 h (4.63) = ρcp U∞ dx The group of variables on the left-hand side of Eq. (4.63) is called the local Stanton number St. Thus, St =

dΔ2 h ⇒ St = ρU∞ cp dx

(4.64)

A more general form of energy integral equation is presented by Kays et al. [2] and Ghiaasiaan [7].

Problems 4.1

Develop Eq. (4.38).

4.2

Develop Eq. (4.38).

4.3

The following polynomial profile may be used in a laminar boundary layer: u = 2η − 2η3 + η4 U∞ where η = y/δ. Using the momentum integral equation, we wish to obtain the shape factor H = δ1 /δ2 and an expression that relates δ/x and Rex .

4.4

Consider a steady forced convection of liquid metal (Pr ≪ 1) over a horizontal flat plate. The flat plate is subject to a uniform heat flux q′′0 . See Figure 4.P4. We wish to obtain the Nusselt number based on the linear temperature profile.

Figure 4.P4

Linear temperature profile for liquid metal flow over a flat plate.

Problems

4.5

Chanson [16] reports that at a certain location from the leading edge of the plate, the experimental velocity distribution in a laminar boundary layer is given as y(mm)

u(m/s)

y(mm)

u(m/s)

0.0

0.0

21

10.79

0.5

0.271

35

11.30

1

0.518

41

11.52

2

1.113

58

11.55

3.5

1.93

60.5

11.51

6

3.137

66

11.50

8

4.157

70

11.60

10.2

5.32

90

11.59

14

7.28

100

11.50

18

9.22

Estimate the boundary layer thickness, the displacement thickness and the momentum thickness. Assume that air at 20 ∘ C is flowing over a flat plate. 4.6

Chanson [16] reports that in a water tunnel, turbulent velocity distributions were conducted in a developing boundary layer along a flat plate in the absence of pressure gradient. y(mm) x(m) =

u (m∕s) 0.25

u (m∕s) 0.50

u (m∕s) 0.75

0

0

0

0

0.5000

5.5700

3.9400

4.1400

1.0000

6.2300

5.3000

4.8200

2.0000

6.6000

6.3800

5.8900

3.5000

7.3000

6.6100

6.1200

6.0000

7.7200

7.0200

6.3000

8.0000

7.7700

7.2000

6.8300

10.2000

7.7000

7.4200

7.0700

14.0000

7.6400

7.6100

7.1300

18.0000

7.6900

7.8700

7.3800

21.0000

7.6800

7.6600

7.8700

35.0000

7.8400

7.6700

7.6900

41.0000

7.6900

7.7200

7.7100

58.0000

7.7600

7.6900

7.6600

60.5000

7.6700

7.7200

7.7200

66.0000

7.6800

7.7000

7.6800

70.0000

7.7700

7.7300

7.7000

90.0000

7.8000

7.6500

7.6900

100.0000

7.6900

7.6800

7.6800

Assume that water is at 17 ∘ C. a) Plot the velocity distribution at each station. b) Calculate the boundary layer thickness δ, displacement thickness δ1 and momentum thickness δ2 at x = 0.25 m, 0.5 m and 0.75 m. c) Estimate the average wall shear stress.

95

96

4 Integral Methods in Convective Heat Transfer

4.7

Consider incompressible constant property turbulent flow over a smooth flat plate. Assume that Pr = Prt ≈ 1 and δ ≈ Δ. There is no pressure gradient in the flow and free stream velocity is U∞ and free stream temperature is T∞ . Prandtl suggested that the mean velocity and temperature profiles are given ( y )1∕7 u = U∞ δ ( y )1∕7 Tw − T = Tw − T∞ Δ Prandtl suggested that the turbulent velocity boundary layer thickness is estimated by 0.37 δ = 1∕5 x Rex Using integral method, develop an expression for the Stanton number St.

4.8

Consider turbulent flow over a flat plate. The velocity distribution is given by ( y )n u = U∞ δ Determine δ1 /δ, δ2 /δ, δ3 /δ for n = 1/7. Here, δ3 represents energy thickness and defined as )( )2 ) ( δ( u u δ3 = dy 1− ∫0 U∞ U∞

References 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Schlichting, H. and Gersten, K. (2017). Boundary Layer Theory, Translated into English by K. Mayes, 9e. Springer. Kays, W.M., Crawford, E., and Weigend, B. (2005). Convective Heat and Mass Transfer, 4e. McGraw Hill. Burmeister, L.C. (1993). Convective Heat Transfer, 2e. Wiley. Oosthuizen, P.H. and Naylor, D. (1999). Introduction to Convective Heat Transfer Analysis. McGraw Hill. White, F.M. and Majdalani, J. (2021). Viscous Fluid Flow, 4e. McGraw Hill. Cebeci, T. (2002). Convective Heat Transfer, 2e. Springer. Ghiaasiaan, M.S. (2018). Convective Heat and Mass Transfer. CRC Press. Rosenhead, L. (1963). Laminar Boundary Layers. Dover Publications. Walz, A. (1969). Boundary Layers of Flow and Temperature. Cambridge: The M.I.T. Press. Thwaites, B. (1948). Approximate calculation of the laminar boundary layer. VII, The International Congress of Applied Mechanics, Imperial College, London (September 1948). Thwaites, B. (1960). Incompressible Aerodynamics. Oxford: Clarendon. Yuan, S.W. (1967). Foundations of Fluid Mechanics. Prentice Hall. Schetz, J.A. (2010). Boundary Layer Analysis. AIAA. Hibbeler, H.C. (2015). Fluid Mechanics. Pearson Prentice Hall. Smith, A.G. and Spalding, D.B. (1958). Heat transfer in a laminar boundary layer with constant fluid properties and constant wall temperature. J. R. Aeronaut. Soc. 4: 137–142. Chanson, H. (2014). Applied Hydrodynamics, An Introduction, CRC Press.

97

5 Dimensional Analysis 5.1 Introduction We may use the following methods to solve convective heat transfer problems. a) Analytical or numerical solution is used to obtain the temperature distribution. Here, temperature gradient at the wall is used to determine the heat transfer coefficient. b) Analogy between mass, momentum, and heat transfer. Here, analogy coupled with wall friction or mass transfer measurements provides the heat transfer coefficient. We will study this topic later. c) Dimensional analysis. In the absence of an analytical solution or analogy between heat and momentum transfer, we then use experimental methods. The heat transfer coefficient is obtained from the correlation of experimental data. Convection heat transfer is heavily involved with experimental testing if the analytical or numerical solution does not provide accurate results. This is especially true in turbulent and separating flow heat transfer. Solutions obtained by computational heat transfer yield only fair approximations for turbulent heat transfer. If we have a good understanding of the problem and are able generate a mathematical model but we are having problem generating a solution, it is a good idea to use experimental analysis. Therefore, there is strong need for experimental study in convection heat transfer. If more than three parameters influence a convective heat transfer problem, it becomes very difficult to study the effect of each parameter on the problem. For this reason, if an engineer wishes to deal with the heat transfer problem empirically, dimensional analysis is the tool to determine the pertinent dimensionless parameters and the required relationship among these parameters. Dimensional analysis is used to present the relationship in a compact form since it combines several variables into dimensionless groups. These dimensionless groups facilitate the interpretation and extend the range of application of experimental data. In order to apply the dimensional analysis to the convection heat transfer problem, the dimensional parameters that influence the convection problem must be known. These parameters may be established on the basis of a through physical understanding of the problem at hand. If an equation is in dimensionless form, it is applicable in all dimensional systems. An engineer takes a great step forward when he or she learns both analytical and experimental methods in convective heat transfer. The nondimensionalizing technique is very useful. Although the dimensional analysis is usually attributed Rayleigh, more compact and complete discussion of theory may be found in the works of Buckingham [1, 2]. We will use the SI system, and this system considers that fundamental mechanical quantities are mass (M), length (L), and time (T). In the case of thermal phenomena, it is necessary to introduce the fundamental quantity temperature (θ). The dimensions of frequently used physical quantities in fluid mechanics and heat transfer are given in Table 5.1. Solutions of many convective heat transfer problems are obtained by the use of experimental, analytical, and numerical methods. Engineers working on convective heat transfer should have a sound understanding of both the experimental and analytical methods. They can interpret and make use of experimental data obtained by others as well as their own experimental data. In this chapter, we will consider some techniques that are important in experimental work. Suppose that we have a complete understanding of the physical problem under investigation and are able to formulate the problem, but we cannot solve the problem analytically. This means that we have mathematical difficulty in obtaining the solution. Thus, the experimental methods are the main approach for the solution of the problem under investigation. This is usually the case for turbulent flow problems since there is no analytical solution. The results obtained by experimental study are presented in the form of generalized correlations using the method of dimensional analysis. Such correlations are called empirical relations since they are based on the experimental work.

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

98

5 Dimensional Analysis

Table 5.1

Dimensions of common variables. MLT𝛉 Dimension

Quantity

Symbol

Mass

m

Force

F

Drag force

FD

Lift force

FL

Equivalent sand roughness

ks

[L]

Length

l

[L]

Time

t

[T]

[M] ] [ ML 2 T [ ] ML T2 [ ] ML T2

[L2 ]

Area ∀

[L3 ]

Angle (radian measure)

α

Velocity

V

Shear velocity

u*

Acceleration

a

Angular velocity

ω

Angular acceleration

α

Linear momentum

P = mV

Angular momentum

H

Mass flow rate

ṁ

Density

ρ

Specific volume

v

Specific weight

γ

Volume flow rate

∀̇ =

Mass velocity

G

Torque or moment

Γ

Pressure

p

Change in pressure

Δp

[1] [ ] L T [ ] L T [ ] L T2 [ ] 1 T [ ] 1 T2 [ ] ML T [ ] ML2 T [ ] M T [ ] M L3 [ 3] L M [ ] M L2 T2 [ 3] L T [ ] M L2 T [ ] ML2 T2 [ ] M L T2 [ ] M L T2

Volume

(

ṁ ρ

)

5.1 Introduction

Table 5.1

(Continued)

Quantity

Symbol

Total pressure

pt

Vapor pressure

pv

Piezometric pressure

pz

Stress (normal)

σ

Stress (shear)

τ

Strain

ε

Modulus of elasticity

E

Surface tension

σ

Frequency

f

Dynamic viscosity

μ

Kinematic viscosity

ν

MLT𝛉 Dimension

[ [ [

M L T2 M L T2

] ]

] M L T2 [ ] M LT2 [ ] M LT2 1 M LT2 [ ] M T2 [ ] 1 T [ ] M LT [ 2] L T [ML2 ]

Moment of inertia of mass Moment of inertia of area

I

Work (energy)

W

Work rate, power

̇ P=W

[L4 ] ] [ ML2 T2 [ ] ML2 T3

Temperature

ϑ or T

[θ]

Total temperature

Tt

Heat

Q

Heat flow per unit time

q

Heat flux (heat flow per unit time and area)

q′′w

[θ] ] [ ML2 T2 [ ] ML2 T3 [ ] M T3

Mechanical equivalent of heat (J)

— [

Buoyant force

B = g βΔθ

Thermal conductivity

k

Specific heat at constant pressure

cp

[ [

L T2

]

ML T3 θ L2 T2 θ

] ]

99

100

5 Dimensional Analysis

Table 5.1

(Continued)

Quantity

Symbol

Specific heat at constant volume

cv

Thermal diffusivity

α

Specific enthalpy

h

Specific entropy

s

Coefficient of thermal expansion

β

Dimensional constant

gc

Convective heat transfer coefficient h and overall transfer coefficient U

h or U

Gas constant

R

Enthalpy

h

MLT𝛉 Dimension

] L2 T2 θ [ 2] L T [ 2] L T2 [ 2 ] L T2 θ 1 θ [

1 [

[

M T3 θ

L2 T2 θ [ 2] L T2

]

]

The method used to describe the dimensionless description of a physical problem is called dimensional analysis. The application of dimensional analysis is seen in nearly all fields of engineering, and it is extensively used in fluid mechanics and heat transfer. It is an important tool for studying fluid mechanics and heat transfer problems. Because of size, it is not always feasible to carry out tests on a full-scale prototype. For this reason, engineers prefer to test a subscale model and take necessary measurements. Experimental study is usually costly and time consuming, and for this reason, it is important to minimize the amount of experimental data that need to be collected. Dimensional analysis results in fewer experiments and improved possibilities for the presentation of the experimental data. Example 5.1 In this example, the representation of experimental data having two variables will be illustrated. Suppose that we have the following relationship for a set of experimental data: H = f(x, y). This relationship H may be represented graphically at four values of x holding y constant, as shown in Figure 5.E1. This is a satisfactory representation of the experimental data.

•

H

• •

• • •

•

y = y2 •

•

•

•

x

Figure 5.E1

y = y1

•

•

•

•

0

•

y = y3 y = y4

A plot of H vs. x for fixed value of y.

5.2 Dimensional Analysis

Example 5.2 In this example, the representation of experimental data having three variables will be illustrated. Suppose that we have the following relationship for a set of experimental data: H = f(x, y, z). A plot of H as a function of x for fixed value of y and z is shown in Figure 5.E2. Since there are three variables, the presentation of data became more complex. Above two examples reveal that as the number of variables increases, the number of graphs will increase, and presentation will be very cumbersome. Hence, we conclude from this discussion that the problem of handling a greater number of variables will increase the difficulty. It is also very difficult to gain real insight into the physical phenomenon with large mass data.

H

y = y1 • • • y=y 2 • • y = y3 • • • y = y4 • • • • • • • •

z = z2

z = z1

H

y = y1 • • • y = y2 • • y = y3 • • • y = y4 • • • • • • • •

x

x

(a)

H

(b)

y = y1 • • • y = y2 • • y = y3 • • • y = y4 • • • • • • • • z = z3

z = z4 H

• • • • • • • •

• y = y1 • • y = y2 y = y3 • • y = y4 • • •

x (c) Figure 5.E2

x (d)

(a–d) A plot H vs. for fixed values of y and z.

5.2 Dimensional Analysis The best way to eliminate these difficulties is to perform a dimensional analysis of all the relevant physical variables and fluid properties. Variables pertinent to physical problem is organized into sets of dimensionless numbers. In other words, dimensional analysis is a way of compacting technique to reduce the number of variables to be studied in an experimental study. Once these dimensionless numbers are obtained, we can use them to obtain the maximum amount of information from a minimum number of experiments. Dimensional analysis is a technique to gain insight into fluid flow and heat transfer. Dimensional analysis deals with the dimensions of the variables to produce a relationship among these variables. Such a relationship consists of one or more dimensionless groups. These dimensionless groups include some constants and exponents. These unknown constants and exponents are determined experimentally. For example, using dimensional analysis, it is possible that the case of H = f(x, y, z) may be reduced to π1 = g(π2 )

(5.1)

and the result may be plotted as shown in Figure 5.1. An examination of Figure 5.1 shows that the experimenter reduced his job to performing four tests only instead of performing several tests. Based on the discussion above, we see that dimensional analysis combines the several variables into dimensionless groups that simplify the interpretation and extend the range of application of the experimental data. In order to apply the dimensional analysis, it is necessary to know the variables, which influence the physical problem under consideration. The success of the method depends on the selection of the variables.

101

102

5 Dimensional Analysis

Figure 5.1

A plot dimensionless π1 as a function of π2 .

π1

π2

Therefore, for the application of the dimensional analysis to convection problems, we must have a through physical understanding of the problem under investigation. Once the correlations among the dimensionless quantities (numbers) are established, it remains to identify a mathematical equation to connect these dimensionless numbers to each other.

5.2.1

Dimensional Homogeneity

The method of dimensional analysis is based on the principle of dimensional homogeneity. Dimensional homogeneity tells us that an analytically derived equation representing a physical phenomenon must be valid for all the systems of units. The principle of dimensional homogeneity states that all the equations that describe the physical system must be dimensionally homogeneous. This means that the fundamental dimensions of different terms on both sides of an equation must be the same. Consider the following equation: f1 (x1 , x2 , . … xn ) + f2 (x1 , x2 , . … xn ). … … … fn (x1 , x2 , . … xn ).

(5.2)

If each of f1 , f2 , . …fn has the same dimension, then the equation becomes meaningful. Example 5.3 In this example, the concept of dimensional homogeneity is explained. Consider now the Bernoulli equation. Suppose we are working with the MLTθ system: p V2 + + gz = const. ρ 2 The dimensions of each term become p M∕LT2 L2 = = ρ M∕L3 T2 2 2 (L∕T) L2 V = = 2 2 T ( ) L2 L (L) = 2 . gz = 2 T T We see that each term of the equation has the dimension of L2 /T2 . Thus, the equation is dimensionally homogeneous.

5.2.2

Buckingham 𝛑 Theorem

The Buckingham π theorem is discussed by Buckingham [1, 2] in detail. Consider a physical problem described by n number of independent variables like x1 , x2 , x3 ,………xn . Suppose that x1 is the dependent variable and other variables x2 , x3 ,………xn are independent variables. Then, the dependent variable x1 can be expressed in terms of independent variables as x1 = f(x2 , x3 , ……… xn )

(5.3)

where n is the number of variables. The Buckingham π theorem states that Eq. (5.3) can be arranged into (n − m) independent dimensionless numbers or groupings. Each of these groupings is called a π term. The Buckingham π theorem

5.2 Dimensional Analysis

gives us the number of independent dimensionless groups of variables or dimensionless parameters needed to correlate the variables in a given physical process. These dimensionless groups of variables can be related by π1 = f1 (π2 , π3 , ……… πn )

(5.4)

where m is the number of primary dimensions. When the functional relationship among the π terms is established, the next step is to investigate experimentally to see the effect of each π term on the problem under consideration. The π terms having the most influence are retained and the π terms having the least influence effect on the problem are rejected. Finally, this process will lead to an empirical equation. Any unknown coefficients and exponents are then determined by experiments. The proof of the Buckingham theorem can be found in [2], and we are interested in its application. Dimensional analysis can only give us the functional relationship between dimensionless groups, and it does not give the form of function relating these groups. Before we proceed to specific applications of dimensional analysis, it will be useful to list some of the forces encountered in fluid mechanics and convective heat transfer. Inertia forces: The magnitude of inertia force is proportional to ( ) dV ds dV dV V =m = mV ≈ (ρ L3 )V Fi = ma = m dt ds dt ds L

(5.5)

where ρ is the fluid density, L is the characteristic dimension, and t is the characteristic time. The inertia force now becomes Fi ∝ ρ L2 V2 .

(5.6)

Viscous forces: The viscous force arises from the shear stress in the flowing fluid: τ=μ

dV dy

Fv = τA ∝

(5.7) ) ( ( ) V 2 dV A∝μ L = μVL. μ dy L

(5.8)

Pressure forces: The pressure force arises due to pressure difference in the fluid. It can be written as Fp = Δp A ∝ Δp L2

(5.9)

where Δp is the pressure drop along the flow. Gravity force: The gravity force on a fluid element is its weight. Hence, Fg = mg ∝ ρL3 g

(5.10)

where g is the acceleration of gravity.

5.2.3 Determination of 𝛑 Terms There are various methods of dimensional analysis to determine the number of independent dimensionless groups for a physical problem under investigation. In this chapter, we will use one of the following methods: a) b) c) d)

The method of repeating variables. The Rayleigh method. The Ipsen method. The use of differential equations.

Example 5.4 In this example, the Rayleigh’s method is used, and this method is discussed in [3]. Suppose that we are interested in the drag force exerted on a sphere in a flowing fluid. See Figure 5.E4. The first step is to identify the parameters on which the drag force depends. Our experience tells us that this force is a function of fluid density ρ, fluid velocity V, the fluid dynamic viscosity μ, and the sphere diameter D. Find a relationship among these variables.

103

104

5 Dimensional Analysis

Figure 5.E4

V, 𝞺, 𝞵

Forced flow over a sphere.

D

F

Solution The functional equation is given by FD = f(V, ρ, μ, D) We chose the MLTθ system, and the dimensions of significant variables are listed as follows:

F

[

ML T2

]

V

𝛒

[ ] L T

[

𝛍

M L3

]

[

D

M LT

] [L]

We now construct the product FD = C(Va ρb μc Dd ) [ ]a [ ]b [ ]c L M ML M =C [L]d . 2 3 T LT T L Dimensional homogeneity requires that we equate the powers of dimension on each side L ∶ a − 3b − c + d = 1 M∶ b+c=1 T ∶ −a − c = −2. Solve for exponents a, b, and c in terms of d as follows: a = d, b = d − 1, c = 2 − d. Next, we write the dimensional equation in terms of exponents as FD = C[Vd ρd−1 μ2−d Dd ] or

] [ ρd μ2 d FD = C Vd D ρ μd )d ( ρFD ρVD = C . μ μ2 Notice that the form of any π group can be altered by multiplying or dividing by another π group. ρF ρVD Dividing μ2D by the square of μ yields FD

(

ρVD =C 2 2 μ ρV D

)d .

Thus, the functional form of the equation is ) ( FD ρVD . = f μ ρ V2 D2 If we plot FD /ρ V2 D2 as a function of the Reynolds number ReD = ρ V D/μ, we obtain a single curve. This curve is independent of the choice of sphere diameter, fluid type, and fluid velocity, which moves over the sphere.

5.2 Dimensional Analysis

Example 5.5 In this example, the method of repeating variables is used, and this method is discussed in [4]. Consider flow over a flat plate. The wall shear stress τw in a boundary layer depends on the distance from the leading edge of the plate x, the fluid density ρ, the dynamic viscosity μ of the fluid, and the approaching free stream velocity U∞ , and the functional relationship is τw = f(x, ρ, μ, U∞ ). We wish to determine the dimensionless groups. Solution We will use the MLT system. The variables and their dimensions are listed as follows: 𝛕w

[

M L T2

𝛒

x

]

[ [L]

𝛍

M L3

]

[

U∞

M LT

]

[ ] L T

We will choose ρ, x, U∞ as the repeating variables and set up the equations π1 = τw ρa xb Uc∞ [ M0 L0 T0 =

M L T2

][

M L3

]a [L]b

[ ]c L T

M: 1+a = 0 L : − 1 − 3a + b + c = 0 T: −2−c = 0 Solutions of equations are a = − 1, b = 0, c = − 2 The dimensionless group is τ π1 = w2 ρU∞ Following a similar procedure, we have π2 = μ ρa xb Uc∞ [

M MLT = LT 0 0 0

][

M L3

]a [L]b

Constructing equations, we have M∶ 1+a=0 L ∶ −1 − 3a + b + c = 0 T ∶ −1 − c = 0 The solution of the equations is a = −1, b = −1, c = −1 The dimensionless group is μ π2 = ρ x U∞ The functional relation is ) ( τw μ = f ρ x U∞ ρU2∞

[ ]c L T

105

106

5 Dimensional Analysis

Example 5.6 Consider a submarine moving in water. The drag force on the submarine is a function of the size of submarine, the velocity of the submarine in water, density, and the dynamic viscosity of water. We wish to study the drag force on submarine by the Ipsen method [5]. Solution The relevant variables are the drag force on submarine (F), the size of the submarine (S), the velocity of the submarine (V), the density of water (ρ), and the dynamic viscosity of water (μ). The variables can be expressed in the following functional form: F = f(V, S, ρ, μ)

(a)

The variables and their dimensions are listed as follows: 𝛒

F

[

ML T2

]

[

𝛍

M L3

]

[

M LT

]

V

S

[ ] L T

[L]

There are five variables and three primary dimensions. Therefore, n = 5 and m = 3. So, there are 5 − 3 = 2 dimensionless groups. We express Eq. (a) in terms of fundamental dimensions [ ] [ ]) ] ([ ] [ L M M ML , [L], = f , T LT T2 L3

(b)

We will rearrange Eq. (a) by making it independent of one fundamental dimension at a time. Let us eliminate [M]. We choose one of the mass-dependent terms on the right-hand side. Let us choose dynamic viscosity μ and combine it with force F and density ρ in such a way that mass dependence disappears. Thus, [ ] ML [ 2] T2 F L = [ ] = M μ T LT [ ] M [ ] ρ L3 T = [ ] = 2 M μ L LT

Thus, we now have ) ( ρ F = f1 V, S, μ μ

(c)

Equation (c) has the following fundamental dimensions: ([ ] [ 2] [ ]) L L T = f1 , [L], 2 T T L We now eliminate time dependence of Eq. (c). To eliminate [T], we will pick velocity V and combine it with F/μ and ρ/μ in such a way that time dependence disappears [ 2][ ] [ 2][ ] T L 1 L F 1 × = = = [L] μ V T L∕T T L [ ][ ] [ ] ρ 1 L T ×V= 2 = μ T L L So, we now have ) ( ρV F1 = f2 S, . μV μ

(d)

5.2 Dimensional Analysis

Equation (d) has the fundamental dimensions ) ( 1 [L] = f2 L, L Finally, we eliminate [L] by combining S with other terms of Eq. (d) in such a way that length dependence disappears ( )( ) [ ] F 1 1 L = = [0] μ V S L ( ) [ ] ρV 1 = [L] = [0] (S) μ L where [0] refers to a dimensionless quantity. We now get final form ( ) ρVS F = f3 μVS μ

(e)

where Re = ρVS/μ is the Reynolds number. Example 5.7 In this example, the Ipsen method [5] will be used. Consider steady constant property low-speed flow over a flat plate. The plate is at uniform temperature Tw . It is assumed that both the velocity and thermal boundary layers start from the leading edge of the plate. See Figure 5.E7. Using dimensional analysis, find the dimensionless groups pertinent to the growth of (i) hydrodynamic boundary layer thickness δ and (ii) thermal boundary layer thickness Δ along the plate.

U∞ Δ x Figure 5.E7

δ

Tw

Velocity and thermal boundary layers for flow over a flat plate.

Solution The growth of the boundary layer is a diffusion process characterized by the kinematic viscosity or momentum diffusivity. a) We assume that the velocity boundary layer thickness depends on δ = f(U∞ , x, ν)

(a)

We will use the MLTθ system. The dimensions of the variables may be written as follows: 𝛅

U∞

x

[L]

[ ] L T

[L]

𝛎

[

L2 T

]

In terms of fundamental dimensions, we have ([ ] [ 2 ]) L L [L] = f , [L], T T We begin the elimination process, and we eliminate fundamental dimension [L] using distance x. We combine x with free stream velocity U∞ , boundary layer thickness δ, and kinematic viscosity ν so that Eq. (a) becomes independent of length dimension: δ [L] = = [0] x [L]

107

108

5 Dimensional Analysis

where [0] refers to a dimensionless quantity U∞ [L∕T] [ 1 ] = = x [L] T ] [ 2 [L ∕T] 1 ν = = T x2 [L2 ] Thus, we now have ( ) U∞ ν δ = f1 , 2 x x x

(b)

and Eq. (b) has dimensions ([ ] [ ]) 1 1 [0] = f1 , T T To eliminate time dimension [T], we pick ν/x2 and combine with U∞ /x so that combination becomes independent of time dimension [T]. U∞ x2 [1∕T] = = [0] x ν [1∕T] Thus, δ = f2 x

(

U∞ x ν

) (c)

or δ = f2 (Rex ) x where Rex = U∞ x/ν is the Reynolds number. b) Heating is assumed to start from the leading edge of the plate, and for this reason, we choose x as the appropriate length. The thermal boundary layer thickness Δ is expected to depend on the free stream velocity U∞ , the distance x from the leading edge, the kinematic viscosity ν, and the thermal diffusivity α. Functional relation among the variables is Δ = f(U∞ , x, ν, α)

(d)

We will use the MLTθ system. The dimensions of the variables may be written as follows: 𝚫

U∞

x

[L]

[ ] L T

[L]

𝛎

[

𝛂

L2 T

]

[

L2 T

]

In terms of fundamental dimensions, Eq. (d) is equivalent to ([ ] [ 2 ] [ 2 ]) L L L [L] = f , [L], , T T T Next, we will rearrange Eq. (d) so that we can eliminate length dimension [L]. To eliminate length dimension [L], we combine distance x with thermal boundary layer thickness Δ, kinematic viscosity ν, and thermal diffusivity α in Eq. (d): Δ [L] = = [0] x [L] U∞ [L∕T] [ 1 ] = = x [L] T ] [ 2 [L ∕T] 1 ν = = 2 2 T x [L ] ] [ 2 [L ∕T] α 1 = = T x2 [L2 ]

5.2 Dimensional Analysis

Thus, we now have ( ) U∞ ν α Δ = f1 , 2, 2 x x x x

(e)

and this equation has the following dimension: ([ ] [ ] [ ]) 1 1 1 [0] = f1 , , T T T To eliminate time dimension [T], we pick ν/x2 , and we combine x2 /ν with U∞ /x and α/x2 so that the equation becomes independent of time dimension [T] ] [ U∞ x2 [ 1 ] L2 = = [0] x ν T L2 ∕T [L2 ∕T] [L2 ] α x2 = = [0] x2 ν [L2 ] [L2 ∕T] Thus, we get ( ) U∞ x2 α x2 Δ = f2 , 2 x x ν x ν or Δ = f2 x

(

U∞ x 1 , ν ν∕α

) (f)

We can express this equation as Δ = f2 (Rex , Pr ) x where Rex = U∞ x/ν is the Reynolds number and Pr = ν/α is the Prandtl number. Example 5.8 In this example, we will use the Ipsen method [5]. External steady forced flow across a cylinder is shown in Figure 5.E8. The velocity of the fluid and the fluid properties influence the heat transfer rate from the cylinder. The heat transfer is expected to depend on the upstream velocity V, diameter D, fluid density 𝜌, and dynamic viscosity, 𝜇. The wall heat wall flux q′′w is expected to depend on these variables as well as on the temperature difference, ΔT = Tw − T∞ , fluid specific heat cp , and thermal conductivity k. The functional relationship can be written as q′′w = f(V, D, ρ, μ, k, ΔT) We wish to obtain the pertinent parameters of the problem under consideration.

Figure 5.E8

Forced flow across a cylinder.

Tw T∞ V ρ

D

Solution Using the definition of the heat transfer coefficient h, this functional relationship can be expressed as follows: h = f(D, V, ρ, μ, cp , k)

(a)

Here, we will use the MLTθ system. The fourth dimension [θ] is the dimension of temperature. Dimensions of the variables can be written as

109

110

5 Dimensional Analysis

h

[

M T3 θ

D

V

𝛒

[

[L]

[ ] L T

]

𝛍

M L3

]

[

cp

M LT

]

[

L2 T2 θ

k

]

[

ML T3 θ

]

We will eliminate successively fundamental dimensions. Substituting these dimensions into Eq. (a) results in [ ] ( ]) [ ] [ ] [ ] [ 2 ] [ M L M M L ML , 3 , , 2 , 3 = f [L], T LT T3 θ L Tθ Tθ We will begin the elimination process by mass. We will eliminate [M]. Let us combine density 𝜌 with other mass-dependent quantities [ ] M [ 3 ] T3 θ L h = [ ] = M ρ T3 θ 3 [L ] ML [ 4 ] T3 θ k L = [ ] = M ρ T3 θ 3 [L] M [ 2] μ LT L = [ ] = M ρ T L3

We now have ( ) μ k h = f1 D, V, , cp , ρ ρ ρ or

) ( k h = f1 D, V, ν, cp , ρ ρ

(b)

where ν = μ/ρ is the kinematic viscosity of the fluid. In terms of fundamental dimensions, Eq. (b) has the following dimensions: [ 3 ] ( [ ] [ 2 ] [ 2 ] [ 4 ]) L L L L L , 2 , 3 = f1 [L], , 3 T T Tθ Tθ Tθ Next, let us eliminate temperature, [θ]. Following a similar procedure, we combine cp with other temperature-dependent quantities [ 4 ] [ 4 ][ 2 ] [ 2] k L L 1 Tθ L = = = [ ] L2 ρcp T T3 θ T3 θ L2 2 [ 3 ] [ T 2θ ] [ ] L h L Tθ = = ρcp T T3 θ L2 This will yield ( ) h k = f2 D, V, ν, ρcp ρcp

(c)

where α = k/ρcp is the thermal diffusivity. The dimensions of Eq. (c) now become ( [ ] [ 2 ] [ 2 ]) [ ] L L L L = f2 [L], , , T T T T We now eliminate time dependence of Eq. (c). We eliminate time dimension [T] by combining velocity V with other time-dependent quantities ] [ ][ h L 1 = = [0] ρcp V T L∕T

5.2 Dimensional Analysis

[L∕T] V = 2 ν [L ∕T] [L∕T] V = 2 α [L ∕T]

[ ] 1 = L [ ] 1 = L

Thus, we get

( ) V V h = f3 D, , ρcp V ν α

(d)

Dimensions of Eq. (d) are [ [ ] [ ]] 1 1 , [0] = f3 [L], L L where [0] represents a dimensionless quantity. Finally, we eliminate length dependency. In order to eliminate length dimension [L], combine D with other length-dependent quantities so that we will have dimensionless quantities, and this will give us the following relation: ) ( VD VD h = f4 , (e) ρcp V ν α or St = f4 (ReD , Pe) where St = h∕ρ cp V is the Stanton number and Pe = VD/α is the Peclet number. We may express the Peclet number as V D ρν cp V D μ cp V D V D ρcp = = = = ReD Pr α 1 k ν k ν k NuD NuD = St = Pe ReD Pr Pe =

where Pr = ν/α is the Prandtl number and NuD = h∕kD is the Nusselt number. We can now express this equation in the following form: NuD = f(ReD , Pr ) where ReD = VD/ν is the dimensionless Reynolds number. Experimental data associated with water, gases, and viscous oils can be correlated in this form. We have neglected the buoyancy force. The generalized correlation can be written as NuD = ϕ(ReD )ψ(Pr ) Example 5.9 In this example, we will use the Ipsen method [5]. Consider steady forced convection heat transfer through a circular tube having uniform wall temperature. See Figure 5.E9. Fluid is heated externally. The temperature of the pipe wall is higher than the average fluid temperature. The change in temperature of the fluid is due to the external heat transfer from the tube walls, and frictional heating is neglected. Assume that the heat transfer coefficient h is influenced by the tube diameter D, average fluid velocity V, and fluid properties such as density ρ, dynamic viscosity μ, thermal conductivity k, specific heat cp , and tube length z. We wish to find a functional expression for forced convection heat transfer between fluid flowing through tube and its wall.

V D z Figure 5.E9

Problem description for Example 5.9.

111

112

5 Dimensional Analysis

Solution The heat transfer coefficient h in the functional form is given as h = f(D, V, ρ, μ, cp , k, z)

(a)

The physical quantities and their dimensions are given as

h

[

M T3 θ

D

V

𝛒

[

[L]

[ ] L T

]

𝛍

M L3

]

[

cp

M LT

]

[

L2 T2 θ

k

]

[

Z

ML T3 θ

] [L]

The dimensions of the variables are [ ] ( ] ) [ ] [ ] [ ] [ 2 ] [ M L M M L ML , , , = f [L], , , [L] T LT T3 θ L3 T2 θ T3 θ First, we eliminate mass dimension [M] using density ρ [ ] M [ ] [ 3] [ 3 ] T3 θ L M L h = M = = × 3 3 ρ M T θ T θ 3 [L ] ML ] [ 3] [ 4 ] [ T3 θ k L L ML = [ ] = = × M ρ M T3 θ T3 θ L3

and we obtain ( ) k h = f1 D, V, ν, cp , , z ρ ρ where ν = μ/ρ is the kinematic viscosity. This equation has the dimension [ 3 ] ( ) [ ] [ 2] [ 2 ] [ 4 ] L L L L L [L], , , = f , , [L] 1 T T T3 θ T2 θ T3 θ Next, we eliminate temperature dimension [θ] by combining cp with other temperature-dependent quantities. [ 3 ] [ 2 ] [ ] h L L Tθ = = × 3 2 ρcp T Tθ L [ 4 ] [ 2 ] [ 2] k L L Tθ = = × 3 2 ρcp T Tθ L This elimination process yields h = f2 (D, V, ν, α, z) ρ cp where α = k/ρ cp is the thermal diffusivity of the fluid. The dimension of this equation is ) ( [ ] [ ] [ 2] [ 2] L L L L = f2 [L], , , , [L] T T T T Now, we eliminate time dimension [T] combining velocity V with other time-dependent quantities [ ] [ ] [ ] L 1 V T = × 2 = ν T L L [ ] [ ] [ ] V 1 L T = × 2 = α T L L [ ] [ ] h L T = × = [0] ρ cp V T L

5.2 Dimensional Analysis

This gives us ) ( V V h = f3 D, , , z ρ cp V ν α This equation has the following dimensions: ) ( 1 1 [0] = f3 L, , , L L L Finally, we eliminate length dimension [L] combining diameter D with other length-dependent quantities [ ] VD 1 = × [L] = [0] ν L [ ] 1 VD = × [L] = [0] α L [L] z = = [0] D [L] Then, we get ( ) h VD VD z = f4 , , ρ cp V ν α D where St = h/ρ cp V is the Stanton number and Pe = VD/α is the Peclet number. Recall that Pe = ReD Pr; then, the Stanton number may be expressed as St =

NuD NuD = Pe ReD Pr

where Pr = ν/α = ρ cp /k is the Prandtl number. Then, we can generalize the discussion as NuD = f(ReD , Pr , z∕D The actual relationship among NuD , ReD , Pr, and z/D is determined experimentally. This form of relationship is especially true for short tubes. For long ducts, this relationship may be expressed in the following form: NuD = f(ReD , Pr )

Example 5.10 Consider natural convection around a hot vertical plate, as shown in Figure 5.E10. In this example, the Ipsen method [5] is used to find pertinent parameters. The plate is at temperature Tw and surrounding cold fluid is at temperature T∞ (Tw > T∞ ). Heat transfer from the hot plate to the fluid creates density differences. The density ρ of the hot fluid is less than the density ρ∞ of the cold fluid far from the hot plate. Hydrodynamic and thermal boundary layers develop along the vertical hot plate. The cold fluid outside the boundary layer is assumed to be stationary. We wish to find the dimensionless parameters.

FB = ρ∞ g

Flow

ρ, T

ρ, T H Tw

T∞ ρ∞

x y

W= ρ g A

Free convection on a vertical plate.

A

Figure 5.E10

113

114

5 Dimensional Analysis

Solution First, we wish to study the underlying physics. Imagine, we take a hot fluid element as shown in Figure 5.E10. Let us look at the forces acting on the spherical fluid element. The buoyancy force acting on this imaginary fluid element is FB = ∀ρ∞ g The weight of the hot fluid element is W = mg = ρ∀g Therefore, the net upward force that pushes the light fluid upward is F = FB − W = g∀(ρ∞ − ρ) For small temperature difference, we assume that ( ) 𝜕ρ ΔT Δρ = 𝜕T and this equation can be expressed in terms of the coefficient of thermal expansion β as ( ) 1 ρ −ρ 1 𝜕ρ ≃− ∞ β=− ρ 𝜕T ρ T∞ − T Therefore, ρ∞ − ρ = βρ(T − T∞ ) Consequently, the net force acting on the fluid element becomes F = g∀βρ(T − T∞ ) The net force per unit mass is g β(T − T∞ ) = g βΔT The upwards motion of the fluid element is caused by this force. We now have the parameters that influence the free convection h = f(H, gβΔT, ρ, μ, cp , k)

(a)

where h is the heat transfer coefficient. Of these seven variables, the product g β ΔT is considered as one variable. The dimensions of the variables are given as h

[

H

M T3 θ

] [L]

(g 𝛃 𝚫T)

𝛒

[

[

L T2

]

𝛍

M L3

]

[

cp

M LT

]

[

L2 T2 θ

k

]

[

ML T3 θ

]

We will begin the elimination process with mass-dependent terms. We will eliminate mass dimension [M] by combining the thermal conductivity term k with other mass-dependent terms ] [ 3 ] [ ] [ 1 h Tθ M = × = 3 k ML L Tθ [ ][ ] [ 4 ] k ML L3 L = = ρ T3 θ M T3 θ [ 3 ] [ 2 ] μ [M] Tθ Tθ . = × = k LT ML L2 We now get the following relation: ( ) k μ h = f1 H, gβΔT, , , cp k ρ k

(b)

5.2 Dimensional Analysis

After simplifications, this equation has the following fundamental dimensions: ( [ ] [ 4 ] [ 2 ] [ 2 ]) [ ] 1 L L L Tθ , 2 . = f1 [L], 2 , 3 , 2 L T Tθ L Tθ Next, we will combine cp with other temperature-dependent quantities to eliminate temperature dimension [θ] [ 2 ][ 2 ] μcp L Tθ = = [0] 2 k L T2 θ [ 4 ][ 2 ] [ 2] k L L Tθ . = = 3 2 ρ cp T Tθ L This elimination process yields h = f2 (H, gβΔT, α, Pr ) k

(c)

where α = k/ρcp is the thermal diffusivity and Pr = μcp /k = ν/α is the Prandtl number. Equation (c) can be expressed in terms of fundamental dimensions as ( [ ] [ ]) [ ] 1 L T = f2 [L], 2 , 2 . L T L Combining α with gβΔT to eliminate the time-dependent term [T]. [ ] [ ] [ ] [ ] gβΔT T T 1 L × 2 × 2 = 3 . = 2 α2 T L L L We now get

( ) gβΔT h = f3 H, , Pr . k α2

(d)

This equation has the following fundamental dimensions: [ [ ]] [ ] 1 1 = f3 [L], 3 . L L Lastly, we eliminate length-dependent term [L] using H [ ] 1 hH = [L] = [0] k L [ ][ 3] gβΔTH3 L 1 = [0]. = 3 1 α2 L We finally get the following relation: ( ) gβΔθH3 hH = f4 , Pr k α2 or hH = f4 (π, Pr ) k

(e)

where π = gβΔθH3 /α2 is a dimensionless number. Equation (e) can be put in the following form by combining with Prandtl number Pr: g β ΔT H3 1 g β ΔT H3 α g β ΔT H3 1 . . = = . Pr ν α ν α2 α2 3 g β ΔT H ν = . = GrH Pr = RaH α ν2 where Ra is the Rayleigh number. Finally, we can express Eq. (e) as NuH = f(RaH ).

115

116

5 Dimensional Analysis

5.3

Nondimensionalization of Basic Differential Equations

A different technique of obtaining dimensionless groups is to nondimensionalize the governing differential equations of the phenomena under investigation. Suppose that we have a complete understanding of the physical problem under investigation and are able to formulate the problem. In many engineering problems, the number of dimensional variables and parameters is very large. Consequently, the presentation of the solutions becomes very cumbersome. Before obtaining a solution, the first step should be to place the equations and boundary conditions in a dimensionless form having as few parameters as possible. Nondimensionalization will simplify the differential equation and its boundary conditions by reducing the parameters and rescaling the individual variables. In this way, we put the differential equations and boundary conditions in their most efficient form so that the usefulness of the solutions can be increased. Notice that the nondimensionalization is an analytical tool. Suppose that there is a formulation of physical problem in terms of governing equations. The term-by-term nondimensionalization of these governing differential equations leads directly to the related dimensionless numbers. We nondimensionalize each variable with proper, constant reference quantities. This method is not useful if the problem cannot be formulated in terms of governing differential equations. The method is also discussed in [6] and will be explained by examples. We will consider only incompressible fluids. Example 5.11 We wish to study the two-dimensional boundary layer problem over a flat plate as shown in Figure 5.E11. We assume a steady, incompressible, constant property flow over the flat plate with a constant free stream velocity U∞ and temperature T∞ . The pressure gradient is absent since U∞ = constant. The flow at the leading edge of the plate has a uniform velocity U∞ and temperature T∞ . The plate length is L, and it is subject to constant wall temperature Tw . The fluid density is ρ and the dynamic viscosity is μ. The velocity components in the x- and y-directions are u and v, respectively. Gravity force and viscous dissipation are neglected. We are interested in determining any dimensionless parameters that arise.

y

Figure 5.E11

U∞

Problem description for Example 5.11.

U∞ T∞ x

L Tw

Solution For the present problem, the equations of momentum and energy are uncoupled, and they can be solved separately. Incompressible constant property flow: Velocity problem Invoking the boundary layer approximations, two-dimensional forms of the basic equations of mass and momentum are given below. The conservation of mass 𝜕u 𝜕v + = 0. 𝜕x 𝜕y The momentum equation ( ) 𝜕u 𝜕u 𝜕2u ρ u +v =μ 2. 𝜕x 𝜕y 𝜕y The pressure gradient dp/dx = 0 for constant free stream velocity U∞ . The boundary conditions are u(x, 0) = 0 v(x, 0) = 0

5.3 Nondimensionalization of Basic Differential Equations

u(x, ∞) = U∞ u(0, y) = U∞ . We will put the governing equations in dimensionless form. We wish to have as few parameters as possible. Now, we choose the following constant reference quantities appropriate for the flow: a) Length L of the plate is taken as the reference length b) Free stream velocity U∞ is the reference velocity The dimensionless variables are denoted by capital letters U=

y u v x ,V = ,X = ,Y = . U∞ U∞ L L

Substituting dimensionless variables into the continuity equation and rearranging the equation, one obtains the dimensionless form of continuity equation 𝜕u 𝜕U∞ U 𝜕U 𝜕U dX U∞ 𝜕U = = U∞ = U∞ = 𝜕x 𝜕x 𝜕x 𝜕X dx L 𝜕X 𝜕V 𝜕V dY U∞ 𝜕V 𝜕v 𝜕U∞ V = = U∞ = U∞ = . 𝜕y 𝜕y 𝜕y 𝜕Y dy L 𝜕Y Thus, the continuity equation becomes U∞ 𝜕U U∞ 𝜕V + =0 L 𝜕X L 𝜕Y or after simplification 𝜕U 𝜕V + = 0. 𝜕X 𝜕Y

(E.1)

Repeating the process for the equation of motion, one obtains u

2 𝜕U U 𝜕U 𝜕U dX U∞ 𝜕U 𝜕u = U∞ U ∞ = U2∞ U = U2∞ U = U 𝜕x 𝜕x 𝜕x 𝜕X dx L 𝜕X

𝜕UU∞ 𝜕u 𝜕U dY U∞ 𝜕U = U∞ V = U2∞ V = V 𝜕y 𝜕y 𝜕Y dy L 𝜕Y ( ) ( ) ( ) U∞ 𝜕 ( 𝜕U ) dY U∞ 𝜕 2 U 𝜕 U∞ 𝜕U 𝜕 𝜕u 𝜕2 u = = = 2 = 𝜕y 𝜕y 𝜕y L 𝜕Y L 𝜕Y 𝜕Y dy 𝜕y2 L 𝜕Y2 2

v

Then, we obtain the partial differential equation ( 2 ) [ ] U∞ 𝜕U U2∞ 𝜕U U∞ 𝜕 2 U ρ U + V =μ . L 𝜕X L 𝜕Y L2 𝜕Y2 After simplification, we obtain the dimensionless form of momentum equation ) ( 𝜕U 1 𝜕2U 𝜕U +V = U 𝜕X 𝜕Y ReL 𝜕Y2

(E.2)

with the boundary conditions U(X, 0) = 0, V(X, 0) = 0, U(X, ∞) = 1, U(0, Y) = 1. Notice that with this choice of dimensionless variables, the Reynolds number ReL = ρU∞ L/μ appeared explicitly in the momentum equation itself. An inspection of Eqs. (E.1) and (E.2) indicates that the solution will be of the following form: U = f1 (X, Y, ReL ) V = f2 (X, Y, ReL ).

117

118

5 Dimensional Analysis

Our purpose in solving momentum and continuity equations is to obtain the local skin friction coefficient. The stress exerted by the fluid on the plate surface is ) ) ( ( ) ( ) ( ) μU ( ) ( 𝜕U∞ U 𝜕U 𝜕U dY 𝜕u ∞ 𝜕U τw = μ = =μ = μU∞ = μU∞ . 𝜕y y=0 𝜕y 𝜕y y=0 𝜕Y Y=0 dy L 𝜕Y Y=0 y=0 The dimensionless local skin friction coefficient cf is defined as ) ( τ cf μ ( 𝜕U ) 1 𝜕U = w2 = = . 2 ρLU∞ 𝜕Y Y=0 ReL 𝜕Y Y=0 ρ U∞ Hence, for a flat plate, we have cf =

2 f (X, ReL ). ReL 1

Incompressible constant property flow: Temperature problem The energy equation ) ( 𝜕T 𝜕2 T 𝜕T +v =k 2. ρcp u 𝜕x 𝜕y 𝜕y The boundary conditions for temperature are T(x, 0) = Tw T(0, y) = T∞ T(x, ∞) = T∞ . The choice of dimensionless variables is U=

T − Tw y u v x , V= , X= , Y= , θ= U∞ U∞ L L T∞ − Tw

where θ is the dimensionless temperature. The next step is to nondimensionalize the energy equation T = Tw + θ(T∞ − Tw ) 𝜕 𝜕θ 𝜕T = (U∞ U) [Tw + θ(T∞ − Tw )] = (U∞ )(T∞ − Tw )U u 𝜕x 𝜕x 𝜕x 𝜕θ dX (U∞ )(T∞ − Tw ) 𝜕θ = U = (U∞ )(T∞ − Tw )U 𝜕X dx L 𝜕X 𝜕 𝜕θ dY 𝜕T = (VU∞ ) [Tw + θ(T∞ − Tw )] = (VU∞ )(T∞ − Tw ) v 𝜕y 𝜕y 𝜕Y dy (U )(T − Tw ) 𝜕θ V = ∞ ∞ L 𝜕Y 𝜕θ dY 𝜕 𝜕T = [T + θ(T∞ − Tw )] = (T∞ − Tw ) 𝜕y 𝜕y w 𝜕Y dy (T − Tw ) 𝜕θ = ∞ (L ) 𝜕Y [ ] [ ] 𝜕 𝜕T 𝜕2 T 𝜕 (T∞ − Tw ) 𝜕θ 𝜕 (T∞ − Tw ) 𝜕θ 𝜕Y = = = 𝜕y 𝜕y 𝜕y L 𝜕Y 𝜕Y L 𝜕Y 𝜕y 𝜕y2 (T∞ − Tw ) 𝜕 2 θ . = L2 𝜕Y2 We substitute these into energy equation, and this gives us [ ( ( ) ] ) ) ] [( T∞ − Tw T∞ − Tw T∞ − Tw 𝜕 2 T 𝜕θ 𝜕T ρcp U∞ . U + U∞ V =k L 𝜕X L 𝜕Y L2 𝜕Y2 After simplification in the energy equation, we get [ 2 ] [ ] 𝜕θ 𝜕T k 𝜕 T U +V = . 𝜕X 𝜕Y ρcp U∞ L 𝜕Y2

5.3 Nondimensionalization of Basic Differential Equations

Rearrange the coefficient on the right-hand side of the energy equation by multiplying and dividing with dynamic viscosity μ as follows: ( )( )( ) ( ) μ 1 1 k k k 1 = = = . ρcp U∞ L cp ρU∞ L μcp ρU∞ L Pr ReL The reciprocal of k/μcp is the Prandtl number, and it contains only thermophysical properties of the fluid. The value of Prandtl number is important in heat transfer computations, and the Prandtl number is defined as μcp Pr = k and the reciprocal of μ/ρU∞ L is the Reynolds number that also includes the thermophysical properties of the fluid, and it is defined as ρU∞ L ReL = μ The energy equation simplifies to the following form: 𝜕T 1 𝜕2 T 𝜕θ +V = . (E.3) 𝜕X 𝜕Y ReL Pr 𝜕Y2 Notice again that with this choice of dimensionless variables, the Reynolds number ReL = ρU∞ L/μ and the Prandtl number Pr = μcp /k appeared in the energy equation itself. This means that the dimensionless energy equation contains two dimensionless parameters. The dimensionless forms of boundary conditions for temperature are U

θ(0, Y) = 0, θ(X, 0) = 0, θ(X, ∞) = 1. The dimensionless temperature θ may be expressed as θ = f3 (X, Y, ReL , Pr ). The local heat flux q′′w on the plate wall is ( ) k(T∞ − Tw ) ( 𝜕θ ) 𝜕T ′′ qw = −k =− . 𝜕y y=0 L 𝜕Y Y=0 The local heat transfer coefficient h is ( ) k(T∞ −Tw ) 𝜕θ − ( ) ′′ qw L 𝜕Y Y=0 k 𝜕θ h= = = . Tw − T∞ Tw − T∞ L 𝜕Y Y=0 Knowing the local heat transfer coefficient, the local Nusselt number Nux is ( ) ( ) x 𝜕θ hx 𝜕θ = =X . Nux = k L 𝜕Y Y=0 𝜕Y Y=0 Let us compare Eqs. (E.2) and (E.3). This comparison tells us that if Pr = 1, then the solutions to momentum and energy equations are identical. The dimensionless temperature profile θ is identical to the dimensionless velocity profile U. This means that under certain conditions, a simple relation between the heat transfer coefficient and the skin friction coefficient can be derived. The solution of the energy equation may be expressed in the following form for Pr = 1: θ = U = f(X, Y). Let us look at the heat transfer coefficient again. We found that the heat transfer coefficient is expressed as ( ) k 𝜕θ . h= L 𝜕Y Y=0 Next, let us look at the skin friction coefficient cf . We found that the skin friction coefficient cf is expressed as ( ) 2 𝜕U . cf = ReL 𝜕Y Y=0 ( ) ( ) 𝜕θ = 𝜕U for Pr = 1, we can write the following relation: Since 𝜕Y 𝜕Y wall wall ) ( ) ( )( μcp (2∕ReL ) cf 2 L 1 Pr = = =2 =2 . h (k∕L) (ρU∞ L∕μ) k k ρcp U∞ ρcp U∞

119

120

5 Dimensional Analysis

Thus cf = 2

(

) h ρcp U∞

(Pr ) = St Pr .

But we assumed that Pr ≈ 1, and we have now a relation between cf and h: c St = f 2 where St =

h ρcp U∞

is the Stanton number. This is known as the Reynolds analogy, and it is valid across the boundary layer

under the following conditions: 1) Zero pressure gradient 2) Constant wall temperature 3) For fluids having Pr ≈ 1 We have obtained the relevant dimensionless parameters for low-speed, forced convection heat transfer over a flat plate using a different method, and notice that we also have obtained information about the problem even before attempting for a solution to the differential equations. We have done this by nondimensionalizing the governing equations of mass, momentum, and energy equations. Notice that the nondimensionalization of the differential equations and boundary conditions arranges the parameters into the minimum number of groups that may be used to describe the physical problem under consideration. For the given same dimensionless boundary conditions, all the physical processes described by the same differential equations will have the same solution. Solutions of these equations are sensitive to the values of the coefficients in the equations. Solutions of these dimensionless equations along with dimensionless boundary conditions provide information required to determine the temperature and velocity distributions. The dimensionless temperature gradient on a solid surface can be expressed in terms of the Nusselt number. Then, we can determine the heat transfer from the surface. Solutions obtained in this way can be used for any fluid. Suppose that we wish to study heat transfer over a flat plate in laboratory. Only Reynolds and Prandtl numbers need to be varied, greatly reducing the number of measurements required. Using the dimensionless form of governing equations, we have eliminated the guess work and obtained the dimensionless parameters immediately. One weakness of the Buckingham π theorem method is that in many cases, the important parameters are not apparent. Example 5.12 We will consider steady, constant property, laminar flow in a tube with a fully developed velocity profile as shown in Figure 5.E12. The pipe length is L. The velocity profile vz is given as [ ( )2 ] r . (a) vz = 2V 1 − R The energy equation is ( 2 ) 𝜕T 𝜕 T 1 𝜕T ρ cp vz + =k . 𝜕z r 𝜕r 𝜕r2

(b)

The boundary conditions for the problem are T(0, r) = T0 T(z, R) = Tw

for z > 0

𝜕 T(z, 0) = 0 symmetry. 𝜕r We notice that there two characteristic lengths for this problem: the tube length L and pipe radius R. We will introduce the following dimensionless variables: z+ =

T − Tw v z r , η = , U = z, θ = Z0 R V Ti − Tw

(c)

5.3 Nondimensionalization of Basic Differential Equations

where V is the average fluid velocity and Z0 is the unknown parameter to be determined. Here, we assume low-speed flow, and for this reason, we neglected the viscous dissipation. Axial conduction is also neglected. We wish to gain some insight into the probable form of the solution of the energy equation. Obtain dimensionless form of energy equation and its boundary conditions.

r

Ti

Tw

Ti 0 vz (r)

D = 2R

z

L Figure 5.E12

Coordinate system for laminar flow in a tube.

Solution We proceed to nondimensionalize the energy equation and its boundary conditions 𝜕 𝜕θ 𝜕T = (V U) [Tw + (Ti − Tw )θ] = [(Ti − Tw )V]U 𝜕z 𝜕z 𝜕z [(Ti − Tw )V] 𝜕θ 𝜕θ dz+ = U + = [(Ti − Tw )V]U + 𝜕z dz Z0 𝜕z ( ) ) ( (T − T 1 𝜕T 1 𝜕 1 𝜕θ dη i w) = [Tw + (Ti − Tw )θ] = r 𝜕r η R 𝜕r R η 𝜕η dr (Ti − Tw ) 1 𝜕θ = η 𝜕η R2 𝜕 𝜕T 𝜕θ = [Tw + (Ti − Tw )θ] = (Ti − Tw ) 𝜕r 𝜕r 𝜕r 𝜕θ dη (Ti − Tw ) 𝜕θ = = (Ti − Tw ) 𝜕η dr R 𝜕η [( ) ] [( ) ] ( ) 2 T Ti − Tw 𝜕θ dη Ti − Tw 𝜕 2 θ − T 𝜕 T 𝜕 𝜕θ 𝜕 i w = . = = 𝜕r R 𝜕η 𝜕η R 𝜕η dr 𝜕 r2 𝜕η2 R2

vz

Substituting these equations into the energy equation and simplifying, we get ] ) [( [ρ cp (Ti − Tw )V] 𝜕θ Ti − Tw 𝜕 2 θ (Ti − Tw ) 1 𝜕θ . U + =k + Z0 𝜕z η 𝜕η 𝜕η2 R2 R2 We may now simplify Eq. (d) to get [ ] ] [ 2 ρ cp V 2 R2 𝜕θ 𝜕 θ 1 𝜕θ . + (1 − η2 ) + = k Z0 𝜕z 𝜕η2 η 𝜕η

(d)

(e)

The coefficient appearing in the bracket on the left-hand side of the energy equation is a dimensionless quantity, and we set it equal to unity. It may be expressed as [ ] ] [ ρ cp V 2 R2 ρ cp V 2 R2 = 1 ⇒ Z0 = . k Z0 k The dimensionless form of the energy equation now becomes ] [ 2 𝜕θ 𝜕 θ 1 𝜕θ . + (1 − η2 ) + = 𝜕z 𝜕η2 η 𝜕η

(f)

121

122

5 Dimensional Analysis

We now determined the unknown parameter Z0 . Thus, the axial dimensionless distance z+ becomes (z∕R) (z∕R) 2(z∕D) z = = = 2 μc ρ V (2R) Re Pr Re p (ρ cp V 2 R ) D D Pr μ k k ( ) where D = 2R is the diameter and GzD = Pr ReD Dz is called the Graetz number. The dimensionless boundary conditions are z+ =

θ(0, η) = 1,

𝜕θ(z+ , 0) = 0, θ(z+ , 1) = 0. 𝜕η

(g)

The boundary conditions do not introduce further dimensionless parameters. The dimensionless temperature θ and, therefore, the dimensionless temperature gradient at the wall depend on the Graetz number Gz. The Nusselt number NuD is based on the temperature difference between wall and mean fluid temperature. First, let us introduce the local heat transfer coefficient h as follows: ( ) ( ) ( ) (T −T ) k 𝜕θ 𝜕T k 0 R w 𝜕𝜕 ηθ k ′′ + R 𝜕 η η=1 q (z ) 𝜕 r r=R η=1 = = =− . h= w Tw − Tm Tw − Tm Tw − Tm θm Thus, the local Nusselt number becomes ( ) 2 𝜕θ hD =− . NuD = k θm 𝜕 η η=1 We expect that the solution to various problems of this kind may be expressed in the following form: ) ( ReD Pr . NuD = f(Gz) = f L∕D This equation gives an idea about the general functional relationship that could be expected in a theoretical solution or in the correlation of experimental data. Experimental data on heat transfer coefficient is correlated using dimensionless groups. A different dimensionless number in the literature is the Stanton number St. The Stanton number is defined as St =

h . ρ cp V

The Stanton number is related to the Reynolds number ReD = ρ V D/μ and the Prandtl number Pr = ν/α = μcp /k as follows: NuD NuD = St = ReD Pr Pe where Pe is the Peclet number. We conclude this discussion by stating that NuD = f(ReD , Pr , geometric ratio).

Example 5.13 Consider a vertical plate with a constant, uniform surface temperature Tw in an incompressible constant property fluid. See Figure 5.E13. The temperature of the fluid far from the plate surface is T∞ and is assumed to be constant. We assume that Tw > T∞ , and this means that heat is added to fluid from the plate surface. Suppose that the plate is made of copper and copper has a high thermal conductivity. There is a heating element within the plate. The heating element essentially maintains the surface at constant temperature Tw . Local fluid temperature near the plate is denoted by T. We allow the existence of body forces due to buoyancy in the y-direction. Fluid motion with low velocity takes place in the vertical direction, and for this reason, there will be no viscous dissipation. We wish to obtain dimensionless parameters that may arise in the problem.

5.3 Nondimensionalization of Basic Differential Equations

Figure 5.E13

g

L x O

Tw

Coordinate system for free convection on a vertical plate.

T∞

y

Solution We first write the governing equations of mass, momentum, and energy. The equation of conservation of mass is 𝜕u 𝜕v + =0 𝜕x 𝜕y and the momentum equation is ( ) 𝜕u 𝜕2 u 𝜕u ρ u +v = ρgβ(T − T∞ ) + μ 2 𝜕x 𝜕y 𝜕y where the term β is the coefficient of thermal expansion of fluid and g is the x-component of the acceleration due to gravity. The boundary conditions for the velocity field are u(x, 0) = 0 u(x, ∞) = 0 v(x, 0) = 0 u(0, y) = 0. The energy equation is ) ( 𝜕T 𝜕2 T 𝜕T +v =k 2. ρcp u 𝜕x 𝜕y 𝜕y The boundary conditions for the energy equation are T(0, y) = T∞ T(x, 0) = Tw T(x, ∞) = T∞ . Now, we choose constant reference properties appropriate for the flow. a) Length L of the plate is taken as the reference length b) Unknown velocity u0 is the reference velocity We introduce the following dimensionless variables: u v U= , V= u0 u0 where u0 is the unknown reference velocity and will be determined later. y x X= , Y= L L where L is the height of the plate. Dimensionless temperature θ is defined as θ=

T − T∞ . Tw − T∞

123

124

5 Dimensional Analysis

First, we consider the continuity equation [ ] ] ] (u ) [ [ 𝜕(u0 U) 𝜕U 𝜕U 𝜕U 𝜕X 𝜕u 0 = = u0 = u0 = 𝜕x 𝜕x 𝜕x 𝜕X 𝜕x L 𝜕X [ ] ] ( ) [ ] [ 𝜕(u0 V) u0 𝜕V 𝜕v 𝜕V 𝜕V 𝜕Y = = u0 = u0 = 𝜕y 𝜕y 𝜕y 𝜕Y 𝜕y L 𝜕X The dimensionless form of the continuity equation becomes 𝜕U 𝜕V + = 0. 𝜕X 𝜕Y Next, consider the momentum equation ] [ ] ] [ [ 𝜕(u0 U) 𝜕u 𝜕U 𝜕U dX u = (Uu0 ) = u20 U = u20 U 𝜕x 𝜕x 𝜕x 𝜕X dx ( ) 2 [ ] u0 𝜕U 1 𝜕U = u20 U = U 𝜕X L L 𝜕X [ [ ] ] 𝜕(u0 U) 𝜕U 𝜕u = (u0 V) = u20 V v 𝜕y 𝜕y 𝜕y [ ] ( 2) u0 𝜕U dY 𝜕U = = u20 V V 𝜕Y dy L 𝜕Y [ ] [ ] [ ]( ) (u ) 2 𝜕 u0 𝜕U 𝜕 u0 𝜕U 𝜕 𝜕u 𝜕2 u dY 𝜕 U 0 = = = = . 𝜕y 𝜕y 𝜕y L 𝜕Y 𝜕Y L 𝜕Y dy 𝜕y2 L2 𝜕Y2 Using this scheme and rearranging the momentum equation, one obtains )[ 2 ] ( ] gβ(T − T )L [ 𝜕U 𝜕U 𝜕 U ν w ∞ +V = . θ + U 𝜕X 𝜕Y Lu0 u20 𝜕Y2 ( ) ν Next, we may choose Lu = 1 → u0 = νL . However, this velocity scale is too small for free convection heat transfer, and 0 we reject this parameter. Therefore, we choose √ gβ(Tw − T∞ )L = 1 ⇒ u0 = gβ(Tw − T∞ )L. 2 u0 This is the characteristic velocity scale to be used. Then, we look at the dimensionless quantity ν/Lu0 , and this dimensionless quantity of free convection is the counterpart of the Reynolds number in forced convection heat transfer. This quantity can be expressed as 1 ν 1 ν = √ = √ = √ 3 Lu0 gβ(Tw −T∞ )L GrL L gβ(Tw − T∞ )L ν2

)L3 /ν2

where GrL = gβ(Tw − T∞ is called the Grashof number. Finally, the dimensionless form of the momentum equation becomes ] [ 𝜕U 1 𝜕2U 𝜕U +V =θ+ √ . U 𝜕X 𝜕Y GrL 𝜕Y2 The boundary conditions of the momentum equation are U(X, 0) = 0, U(X, ∞) = 0, V(X, 0) = 0, U(0, Y) = 0. With this choice of characteristic velocity, the dimensionless variables become uL vL = √ , V= √ ν GrL gβ(Tw − T∞ )L ν GrL y x X= , Y= L L T − T∞ θ= . Tw − T∞

U= √

u

5.5 Dimensionless Numbers

We will now consider the energy equation. Substituting the same variables into the energy equation, one obtains [ ] u (T − T∞ ) 𝜕T 𝜕θ 𝜕 dX u = (u0 U) [T∞ + (Tw − T∞ )θ] = 0 w U 𝜕x 𝜕X dx L 𝜕X [ ] u0 (Tw − T∞ ) 𝜕θ 𝜕T 𝜕 dY = (u0 V) [T∞ + (Tw − T∞ )θ] = V v 𝜕y 𝜕Y dy L 𝜕Y [ ] (Tw − T∞ ) 𝜕θ 𝜕 dY 𝜕T = [T∞ + (Tw − T∞ )θ] = 𝜕y 𝜕Y dy L 𝜕Y [ ] [ ] 2 (T − T ) 𝜕 𝜕 𝜕T 𝜕 T w ∞ 𝜕θ = = 𝜕y 𝜕y 𝜕y L 𝜕Y 𝜕y2 [ ] 𝜕 (Tw − T∞ ) 𝜕θ dY (Tw − T∞ ) 𝜕 2 θ = = . 𝜕Y L 𝜕Y dy L2 𝜕Y2 The energy equation becomes ] ] } [ {[ ] [ (Tw − T∞ ) 𝜕 2 θ u0 (Tw − T∞ ) u0 (Tw − T∞ ) 𝜕θ 𝜕θ U V =k ρcp . + L 𝜕X L 𝜕Y L2 𝜕Y2 Now, dividing by the leading coefficient on the left-hand side and rearranging the equation yields the new form of the dimensionless energy equation ] [ 𝜕θ 1 𝜕2 θ 𝜕θ +V = U √ 𝜕X 𝜕Y Pr GrL 𝜕Y2 where Pr = μcp /k is the Prandtl number. The boundary conditions for the energy equation are θ(0, Y) = 0, θ(X, 0) = 1, θ(X, ∞) = 0.

5.4 Discussion Let us make some generalizations based on dimensional analysis by nondimensionalizing the governing differential equations and their boundary conditions. The boundary conditions may also contain important dimensionless parameters. Notice that the method explained in Section 5.3 is useful if the differential equations describing the physical situation can be written. All the dimensionless parameters can be obtained immediately. The dimensionless form of the equations is the most efficient form. This method arranges the parameters into the minimum number of groups, and these groups can be used to describe the physical problem under investigation. If the differential equations are not known, then other methods such as the π theorem can be used to obtain the dimensionless groups. One weakness of the π theorem is that it is not easy to see the important parameters. Notice now that (i) the governing equations are dimensionless, (ii) the variables vary between 0 and 1, and (iii) the solution obtained is valid for all the parameters regardless of the choice physical dimension, type of fluid, fluid properties, and fluid velocity. By identifying the dimensionless parameters in a given physical problem, experimental studies can be planned. This will reduce the number of experiments required. This is also true for numerical solutions to be generated. The presentation of results is done efficiently when presentation is carried out by dimensionless numbers.

5.5 Dimensionless Numbers We now present some dimensionless numbers in convective heat transfer.

5.5.1 Reynolds Number ρU∞ L U∞ L inertia force = = ReL = μ ν viscous force

(5.11)

where ReL is the Reynolds number, U∞ is the characteristic velocity, L is a characteristic length, and ν is the kinematic viscosity.

125

126

5 Dimensional Analysis

5.5.2

Peclet Number ρcp U∞ D U∞ D convection = = = Pe conduction k α

(5.12)

where Pe is the Peclet number, α = k/ρcp is the thermal diffusivity, k is the fluid thermal conductivity, ρ is the density, cp is the constant pressure specific heat, and D is the characteristic length.

5.5.3

Prandtl Number μcp ν Diffusion of momentum = . = Pr = k α Diffusion of heat

(5.13)

The Prandtl number may be interpreted as the ratio of viscous effects to conduction effects and has the following range of values: Fluid

Pr

Liquid metals

0.004–0.03

Gases

0.7–1.0

Water

1.7–13.7

Light organic liquids

5–50

Oils

50–10 000

5.5.4

Nusselt Number temperature gradient at wall hL = NuL = k overall temperature difference

(5.14)

where L is a characteristic length, k is the thermal conductivity of fluid, h is the heat transfer coefficient, and NuL is the Nusselt number.

5.5.5

Stanton Number

We may also nondimensionalize the heat transfer coefficient in a different way. A widely used parameter is Stanton number St Heat transfer at wall h = St = . energy transported by stream ρ cp k

(5.15)

The Stanton number may be expressed as St =

Nu Re Pr

(5.16)

where the Stanton number is a regrouping of Re, Pr, and Nu.

5.5.6

Skin Friction Coefficient cf =

2 τw ρ V2

(5.17)

where τw is the wall shear stress. The Stanton number St is related to the skin friction coefficient cf for certain geometries such as smooth pipes and flat plates. Then, the relationship, if it is valid, takes the following form: St = f(cf , Pr , generic shape).

5.5 Dimensionless Numbers

5.5.7 Graetz Number Heat transfer by convection in entrance region D ρcp VD . = × z k Heat transfer by conduction Heat transfer data can be expressed in terms of Graetz number. For forced convection heat transfer in a pipe, using dimensional analysis, we obtained the following relationship: NuD = f(ReD , Pr , z∕D). We now introduce the Graetz number GzD as ) ( ReD Pr D = ReD Pr GzD = z∕D z

(5.18)

(5.19)

and NuD = f(GzD ).

(5.20)

5.5.8 Eckert Number U2∞ Temperature rise due to energy conversion = Ec. ∼ cv ΔT Temperature difference

(5.21)

5.5.9 Grashof Number An examination of the energy equation for free convection along a vertical plate reveals that Buoyancy force gβΔT L3 = Viscous force ν3

(5.22)

where GrL is the Grashof number and ΔT is the temperature difference.

5.5.10 Rayleigh Number On the other hand, in free convection, we encounter the Rayleigh number RaL . The Rayleigh number RaL can be expressed in terms of Grashof GrL and Prandtl Pr numbers as RaL =

g β ΔTL3 ν g β ΔTL3 = = GrL Pr . να α ν2

(5.23)

The magnitude of the Rayleigh number determines whether the flow is laminar or turbulent.

5.5.11 Brinkman Number The product of Eckert number Ec, and the Prandtl number Pr is the Brinkman number, Br. It can be seen that Br =

V2 • μcp = Ec Pr cp ΔT k

(5.24)

The magnitude of EcPr is a measure of the importance of the viscous dissipation. For high-speed flow, Nu = f(Re, Pr , Br)

(5.25)

127

128

5 Dimensional Analysis

5.6

Correlations of Experimental Data

In general, the heat transfer coefficient h may depend on system parameters: a) b) c) d) e) f) g) h) i)

Stream velocity, V Body shape Body size, D Wall roughness height, ε Wall temperature, Tw Stream temperature, T∞ Fluid properties, μ, cp , k Fluid density, ρ If there is a significant density changes across the boundary layer, then buoyant specific weight g Δρ becomes important Consider flow over a sphere. We may now express heat transfer coefficient h as h = f(V, D, ε, Tw , T∞ , ρ, μ, cp , k, gΔρ, shape).

(5.26)

Each variable may be expressed in terms four primary dimensions: mass, length, time, and temperature. We may reduce these variables by dimensional analysis. Body shape has a strong effect in convection heat transfer problems. We cannot equate cylinder data to the flat plate. A circular pipe gives different results than a rectangular duct. The orientation of a shape is also important. For a given shape, several experiments may be required to establish a functional relationship. To simplify the problem, we may make the following assumptions: a) b) c) d)

The surface may be hydraulically smooth. Temperature differences may be small. The flow may be considered as forced convection. In this case buoyant effect is neglected. Flow velocities may be low, and fluid is not too viscous. Then, the viscous dissipation may be neglected. As we will see later, we may express the heat transfer coefficient by dimensional analysis in the following form: NuD = f(ReD , Pr, generic shape)

(5.27)

where NuD = h D/k is the Nusselt number, ReD = ρ V D/μ is the Reynolds number, and Pr = μ cp /k is the Prandtl number. If we plot the experimental data on a log–log paper, data will often fall on a straight line. The functional relationship may be represented as NuD = C Rem D Pr

n

(5.28)

where the unknown constants C, m, and n are determined experimentally, and they depend on generic shape. This power law relation is valid for both the laminar and turbulent flows. The correlation of experimental data is discussed in [7–9]. Example 5.14 Experiments were performed in a wind tunnel to study cross flow heat transfer from a circular cylinder. Working fluid was air and Pr = 0.7. The results of experiments were given below in terms of Reynolds and average Nusselt numbers in Table 5.14a. We wish to correlate the data in terms of appropriate dimensionless parameters. Solution We assume that properties of air is independent of temperature. The appropriate dimensionless parameters are the average Nusselt number (NuD ) = hD∕k, the Reynolds number (ReD = ρVD/μ) and the Prandtl number (Pr = μcp /k). Air tables indicate that the Prandtl number of air is constant over a wide range of temperature, equal to Pr ≈ 0.7. For this reason, we may present the experimental data with the general functional relationship of the form NuD = f(ReD )

(5.29)

Fluid properties are evaluated at film temperature Tf = (Tw + T∞ )/2. The temperature Tw is the cylinder surface temperature and the T∞ is the free stream temperature.

5.6 Correlations of Experimental Data

Table 5.14a Experimental data for Example 5.14. ReD

NuD

6 500

39.90

14 700

66.10

18 900

77.20

25 000

91.75

29 350

101.40

Next, we plot the data on a log–log plot (ln NuD vs. ln ReD ). Figure 5.E14 shows plot of the data. An examination of the Figure 5.E14 reveals that these data points are nearly on a straight line. Actually, Figure 5.E14 applies to any fluid whose Prandtl number is approximately 0.7. Fitting this data with a linear least squares regression will give us the desired equation. Dimensional analysis shows that a simple power law of the form NuD = CRem D

(5.30)

is a good correlation formula for the data. However, this power function does not represent linear relationship between the ReD and NuD variables. The power function is transformed into a linear function by doing a little algebra. To be able to use linear regression, the form of this nonlinear equation of two variables is changed so that the new form is linear with terms that contain the original variables. For this purpose, we take the log of both sides of Eq. (5.30). ln(NuD ) = ln C + m ln(ReD )

(5.31)

4.8 4.6

Nu

4.4 4.2 4 3.8 3.6 8.6

8.8

9

9.2

9.4

9.6

9.8

10

10.2 10.4

Re Figure 5.E14

Average Nusselt number versus Reynolds number for flow over a cylinder.

This equation is linear for ln(NuD ) in terms of ln(ReD ). We define new variables X and Y. X = ln(ReD ) Y = ln(NuD ) A = ln(C)

129

130

5 Dimensional Analysis

Table 5.14b

Reorganized experimental data for Example 5.14.

ReD

NuD

ln(ReD )

ln(NuD )

6 500

39.90

8.7796

3.6864

14 700

66.10

9.5956

4.1912

18 900

77.20

9.8469

4.3464

25 000

91.75

10.1266

4.5191

29 350

101.40

10.2870

4.6191

Equation (5.31) now becomes Y = mX + A

(5.32)

Since we will use the p = polyfit(x, y, n) command in MATLAB 2021a to obtain the desired correlation, we need to reorganize the data given in Table 5.14a. To reorganize the data, we take the natural logarithm of Reynolds number ReD , and natural logarithm of Nusselt number NuD . See the data generated in Table 5.14b. The p = polyfit (x, y, n) command in MATLAB 2021b, fits a polynomial of degree n to data described by the vectors x and y, where x is the independent variable. It returns a row vector p of length n + 1 that contains the coefficients of polynomial in order of descending powers. Application of polyfit command in MATLAB 2021a p = polyfit(X, Y, 𝟏) gives us the row vector p p = [0.6185 − 1.7436] Thus, m = 0.6185 and A = −1.7436. Then, we find that C = exp(−1.7436) = 0.1749. The desired correlation for the Nusselt number is NuD = 0.1749Re0.6185 D This equation applies to any combination of cylinder diameter and air velocity, provided that the Reynolds number is in the range of 6.5 × 103 ≤ ReD ≤ 2.9 × 105 . Example 5.15 Figure 5.E15a shows an experimental setup to study forced convection heat transfer in a copper tube. An electrical heater may be used on the tube to provide energy to fluid passing through the heating section of the tube. Thermocouples are bonded on to the surface of the copper tube using a heat sink compound to measure surface temperatures. Copper has a high thermal conductivity, and for this reason, the isothermal boundary condition could be obtained. Temperatures at the inlet and outlet of the test section were measured using calibrated thermocouples. Electric power supplied to the tube can be measured using a wattmeter. A turbine-type flow meter can be used to measure the flow rate to calculate the average velocity. After taking care of heat losses, the average heat transfer coefficient can be calculated using q = hπDL(Tw − Tm ) ) ( Tmo + Tmi Tm = 2 where D is the inner diameter of the tube and L is the length of the heated section of the tube. The external surface of the tube is insulated to ensure that all electric energy is dissipated in the working fluid passing through the heated section. Suppose that the experiments were performed for air, water, and ethylene glycol. Ethylene glycol has high boiling point and low freezing point. It has stability over a wide range of temperatures along with high specific heat and thermal conductivity. It also has a low viscosity and reduced pumping requirements. Assume that the results of experiments are given in tabular form in Table 5.E15a. We wish to correlate the data in terms of appropriate dimensionless parameters.

5.6 Correlations of Experimental Data

Table 5.E15a

Experimental data. Air

Water

Ethylene glycol

0.7

2.29

34.6

ReD

NuD

NuD

NuD

10 000

32.21

51.50

153.50

12 000

37.50

59.90

177.50

15 000

44.6

72.60

212.80

20 000

56.40

91.60

267.20

25 000

67.10

107.70

319.4

Pr

Thermocouples

Tmi

Tmo

Flow meter

Insulation Flow direction Insulation Insulation Insulation Powersource leads Figure 5.E15a

An experimental rig for investigating forced convection heat transfer in a tube.

Solution We assume that properties of each fluid are independent of temperature. If we plot each data set given in Table 5.E15a on a log–log axis, we see that the data points appear to line up along a straight line. This indicates that fitting the data with a linear least squares regression will give us the desired equation. We will follow the procedure discussed in Example 5.14 to obtain the desired correlation for the average Nusselt number for each fluid. Using polyfit command in MATLAB 2021a, we obtain the correlation of the form NuD = CRem D

(5.33)

for each fluid. The correlation developed for each fluid is given below. Air: ln(NuD ) = 0.8001 ln(ReD ) − 3.8971 ( ln

ln(NuD ) = ln (ReD )0.8001 − 3.8971 ) NuD = −3.8971 Re0.8001 D NuD = 0.0203Re0.8001 D

(5.34)

Water: ln(NuD ) = 0.8103(ln ReD ) − 3.5159 NuD = 0.029Re0.8103 D

(5.35)

131

132

5 Dimensional Analysis

Ethylene Glycol: ln(NuD ) = 0.7998 ln(ReD ) − 2.3322 NuD = 0.0971Re0.7998 D

(5.36)

The variation of NuD as a function of ReD for each fluid is shown in Figure 5.E15b. It may be seen that for each fluid, the data are well correlated with straight line.

300 Ethylene Glycol Pr = 34.6 0.7998 Nu D = 0.0971 Re D

200 ND Water Pr = 2.29 Nu D = 0.029 Re0.8103 D 100

Air Pr = 0.7 0.8001 Nu D = 0.0203 Re D 0 15 000

20 000

25 000

RD Figure 5.E15b

Average Nusselt number versus Reynolds number for flow in a tube with three different fluids.

We wish to collapse the straight lines corresponding to different Prandtl numbers shown in Figure 5.E15b into a single line. In other words, we wish to generate a single correlation that represents all the data. Based on our experience in convective heat transfer, the mathematical description of the physical process is represented by the general functional relationship of the form NuD = f(ReD , Pr )

(5.37)

Dimensional analysis shows that a simple power law of the form NuD = cRem D Pr

n

(5.38)

is a good correlation formula. This time, we need to determine the constants C, m, and n in Eq. (5.38). To obtain a correlation of this form, first, we correlate NuD and ReD irrespective of the Prandtl numbers. By plotting average Nusselt number NuD vs. Reynolds number ReD on a log–log paper, we can guess the nature of the correct equation, NuD = f1 (ReD ). Next step is to generate a plot of NuD ∕f1 (ReD ) vs. Pr. This plot will reveal the dependence upon Pr. Let us now proceed to develop a general correlation for all available experimental data. Following the procedure discussed in Example 5.14, we will determine the variation of average Nusselt number NuD as a function of Reynolds number ReD irrespective of the Prandtl numbers. In other words, at this stage, we do not take the effect of Prandtl number into consideration. To perform this task, available experimental data are reorganized and presented in Table 5.E15b. Again, we consider Eq. (5.33) and we take log of both sides of Eq. (5.33). ln(NuD ) = m ln(ReD ) + ln C

(5.39)

5.6 Correlations of Experimental Data

Table 5.E15b

Reorganized experimental data.

ReD

NuD

Pr

log(NuD )

log(ReD )

10 000

32.21

0.7

3.4723

9.2103

12 000

37.50

0.7

3.6243

9.3927

15 000

44.6

0.7

3.7977

9.6158

20 000

56.40

0.7

4.0325

9.9035

25 000

67.10

0.7

4.2062

10.1266

10 000

51.50

2.29

3.9416

9.2103

12 000

59.90

2.29

4.0927

9.3927

15 000

72.60

2.29

4.2850

9.6158

20 000

91.60

2.29

4.5174

9.9035

25 000

107.70

2.29

4.6793

10.1266

10 000

153.50

34.6

5.0337

9.2103

12 000

177.50

34.6

5.1790

9.3927

15 000

212.80

34.6

5.3604

9.6158

20 000

267.20

34.6

5.5880

9.9035

25 000

319.4

329.4

5.7664

10.1266

We next define the following variables: X = ln(ReD ) Y = ln(NuD ) A = ln(c) Substituting these variables in Eq. (5.39), we now have the following linear equation Y = mX+A

(5.40)

We now apply the polyfit command of MATLAB 2021b to determine unknown constants. >> polyfit (X, Y, 𝟏) ans = 0.8035 − 3.2484 Then, we have ln(NuD ) = 0.8035 ln(ReD ) − 3.2484 We find that m = 0.8035 and A = −3.2484. With this information we find that C = 0.0388. Form of our correlation is given by NuD = 0.0388Re0.8035 D

(5.41)

Next, we need to determine the unknown constant n. For this purpose, we need(to determine ) the variation of average Nusselt number NuD with Prandtl number Pr. This can be determined by plotting ln

NuD 0.0388Re0.8035 D

as a function of ln(Pr). In

order to perform this task, we now generate the data given in Table 5.E15c. Again, we use the polyfit command in MATLAB 2021b to determine unknown constant n. Consider now the following equation ) ( NuD = n ln(Pr ) + ln(D) (5.42) ln 0.0388Re0.8035 D

133

134

5 Dimensional Analysis

Table 5.E15c ( log

Reorganized experimental data. )

NuD

log(Pr)

0.0388Re0.8035 D

−0.6789

−0.3567

−0.6733

−0.3567

−0.6792

−0.3567

−0.6756

−0.3567

−0.6812

−0.3567

−0.2096

0.8286

−0.2050

0.8286

−0.1920

0.8286

−0.1907

0.8286

−0.2081

0.8286

0.8825

3.5439

0.8813

3.5439

0.8834

3.5439

0.8799

3.5439

0.8790

3.5439

However, this time, we define the following variables. X = ln(Pr ) ( Y = ln

NuD

)

0.0388Re0.8035 D

A = ln(D) Substituting these variables in Eq. (5.41), we now have the following linear equation Y = nX+A

(5.43)

We use the polyfit command polyfit (X, Y, 1) ans = 0.3995 − 0.5339 So, we now have ) ( NuD = 0.3995 log(Pr ) − 0.5339 ln 0.0388Re0.8035 D

(5.44)

We now combine Eq. (5.44) to get the final form of the correlation for the experimental data. NuD = 0.02274Re0.8035 Pr 0.03995 D

(5.45)

Examination of data in Table 5.E15c shows a scatter of Prandtl numbers but we have a good fit with a straight least square regression line. This correlation is valid for 1 × 104 ≤ ≤ 2.5 × 104 and 0.7 ≤ Pr ≤ 34.6. We may even make some modest extrapolation outside these Reynolds and Prandtl number ranges. Of course, we may use multivariable least squares fit of all the data points. Multiple linear regression is discussed in [10] and [11]. This is the preferred procedure to determine the constants C and exponents m and n. First, let us assume that the

5.6 Correlations of Experimental Data

linear relationship is given by y = a0 + a1 x1 + a2 x2 .

(5.46)

The above equation is useful when fitting experimental data having two variables. The sum of squares of error could be expressed as S=

N ∑ [yi − a0 + a1 x1i + a2 x2i ]2 .

(5.47)

n=1

The conditions for minimum of S is written as 𝜕S 𝜕S 𝜕S = 0, = 0, = 0. 𝜕a0 𝜕a1 𝜕a2

(5.48)

The coefficients yielding the minimum sum of squares of the residuals are given as ∑ ∑ ∑ x1i x2i ⎤ ⎧a0 ⎫ ⎧ yi ⎫ ⎡n ⎪ ⎪ ⎪ ⎪ ∑ ∑ ∑ ∑ ⎢ x x2 x x ⎥ a1 ⎬ = ⎨ x1i yi ⎬ . ∑ 1i2 2i ⎥ ⎨ ∑ ⎢∑ 1i ∑ 1i ⎪ ⎪ ⎪ x1i x2i x2i ⎦ ⎪ ⎣ x2i ⎩a2 ⎭ ⎩ x2i yi ⎭ We take log of both sides of Eq. (5.46) to obtain log(Nu) = log(C) + m log(ReD ) + n log(Pr ) y = log(Nu) a0 = log(C) a1 = m x1 = log(ReD ) a2 = n x2 = log(Pr ). We apply Eq. (5.49) to Example 5.15, and we get the following matrix: 144.7468 20.0787 ⎤ ⎧a0 ⎫ ⎧ 67.5765 ⎫ ⎡15 ⎪ ⎪ ⎪ ⎢144.7468 1.3984e + 03 193.7547⎥ ⎪ a1 ⎬ = ⎨ 653.4294⎬ . ⎢ ⎥⎨ ⎪ ⎪ ⎪ 66.8631 ⎦ ⎪ ⎣20.0787 193.7547 ⎩a2 ⎭ ⎩ 106.4299⎭ The solution obtained by MATLAB2021a is a0 = −3.9359 a1 = 0.8193 a2 = 0.3995 Thus, we get the following coefficients: m = 0.8193 n = 0.3995 C = 0.0195

(5.49)

135

136

5 Dimensional Analysis

The following is the correlation we seek for: NuD = 0.0195 Re0.8193 Pr D

0.3995

.

In order to predict the behavior of a proposed design in engineering, we may follow different approaches. One approach is to use empirical equations based on previous experience in similar situations. A different approach is to perform laboratory experiments to develop the needed information. This approach is very common in turbulent flows. In turbulent flows, most engineering correlations of convective heat transfer are obtained from experimental data. For some very simple flows (e.g. laminar flow over a flat plate and laminar flow in a tube), we have exact solutions. The reliability of this method is limited only by the number of simplifying assumptions made. Another method is the use of analogies. Analogy between momentum and heat transfer is very common. Analogy is very helpful for evaluating the heat transfer in internal forced flows. For example, we may use mass transfer data to obtain the heat transfer data. The use of numerical methods to solve the governing differential equations is rapidly increasing. This activity is commonly called computational fluid dynamics (CFD). The computer codes such as ANSYS CFD yield results that resemble the data obtained from physical experiments. The term numerical experiments are used quite often to describe this activity.

Problems 5.1

Consider the steady incompressible laminar flow of liquid down the inclined plane. The inclined plane makes an angle θ with horizontal. See Figure 5.P1. Using the Navier–Stokes equation of motion, the velocity distribution was obtained. The velocity component u in the flow direction was given as [ ( ) ( )] ρ g sin(θ) H2 y y 2 2 − u= 2μ H H where ρ is the fluid density, μ is the dynamic viscosity of the fluid, and g is the acceleration of gravity. Determine whether this equation is dimensionally homogeneous. y p0

g

0

H x θ

Figure 5.P1

5.2

Laminar flow of liquid over an inclined plane.

A large tank has a water that is maintained at a constant temperature of T∞ . A hot object of mass m is dropped into the tank. The equation governing the temperature T of the object can be obtained using the Newton’s cooling law as mcp

dT = hA(T − T∞ ) dt

where cp is the specific heat of object, h is the average heat transfer coefficient, A is the surface area of the object, and t is the time. Dimensions of the variables can be taken from Table 5.1. What is the dimension of the heat transfer coefficient? 5.3

A small spherical steel is heat treated by quenching at constant temperature bath. Liquid is at temperature T∞ , and it is assumed that the temperature of the sphere above the bath temperature (T − T∞ ) depends on the following parameters: a) The time t after immersion of small spherical steel into bath b) The temperature of the sphere above the bath temperature Ti − T∞

Problems

c) The thermal heat capacity of small spherical steel C = ρc∀ d) The thermal surface resistance of sphere, R = hA. Here, h is the average heat transfer coefficient and A is the surface area of the sphere. What is the dimensional form of the governing equation for this process. 5.4

In this example, the Ipsen method is used. Consider forced convection heat transfer from a very long hot horizontal cylinder rotating about its axis in still air. See Figure 5.P4. Free convection and viscous dissipation effects are negligible. The cylinder surface temperature Tw is higher than the air temperature T∞ . The diameter of the cylinder is D. Use dimensional analysis to obtain the dimensionless groups related to heat transfer.

Figure 5.P4

Rotating cylinder about its axis.

Tw

D

T∞

ω

5.5

In this problem, turbulent flow near a wall is illustrated by the Ipsen method. Consider turbulent flow near the smooth walls of an infinite parallel plate channel. Let Uc be the centerline velocity in the channel. The channel height is dented by H. See Figure 5.P5. Since the walls are assumed smooth, any surface roughness will not affect the flow. Prandtl concluded that the average velocity u depends on the distance y from the wall surface, fluid kinematic viscosity ν, fluid density ρ, and the wall shear stress τw that fluid exerts on the channel walls. We assume that turbulence is independent of the flow conditions further away from the wall. The velocity field is assumed to be restricted to y/δ < 0.2 where δ is the velocity boundary layer thickness. Obtain dimensionless groups among the variables.

Figure 5.P5

Coordinate system for Problem 5.5.

y

5.6

H

In this problem, turbulent heat transfer near a wall is studied by the Ipsen method. Consider turbulent flow near the smooth walls of an infinite parallel plate channel. Let Uc be the centerline velocity in the channel. The channel height is dented by H. See Figure 5.P6. Since the walls are assumed smooth, any surface roughness will not affect the flow. Again we expect that the distance y from the wall surface, fluid dynamic viscosity μ, fluid density ρ, and the wall shear stress τw that fluid exerts on the channel walls will influence the temperature profile. We expect that the temperature field depends on the rate of heat transfer q′′w from the surface to flow. The rate of heat transfer has the unit of W/m2 . The temperature field is assumed to be restricted to y/Δ < 0.2, where Δ is the thermal boundary layer thickness. The heating begins on the plane wall starting from the origin of the thermal boundary layer, and thus, entire fluid is heated. We assume that turbulence is independent of the flow conditions further away from the wall. Obtain dimensionless groups among the variables.

137

138

5 Dimensional Analysis

Figure 5.P6

Tw

Coordinate system for Problem 5.6.

H

y 0

5.7

Consider an inclined enclosure, as shown in Figure 5.P7. Inclined enclosures are found in flat plate solar collectors. Natural convection in an inclined enclosure will be studied by the Ipsen method. We wish to obtain dimensionless parameters for this problem.

Figure 5.P7

H

W

TC

γ

5.8

Inclined enclosure.

TH

Experimental studies indicate that wall shear stress in a pipe flow depends on distance x from the inlet, diameter D, average velocity V, fluid density ρ, fluid dynamic viscosity μ, and the wall surface roughness ε. We may write the following functional relationship: τw = f(x, D, V, ρ, μ, ε) or f(τw , x, D, V, ρ, μ, ε) = 0. We wish to obtain the dimensionless parameters for this problem.

5.9

Consider a sphere falling in a tank containing fluid as shown in Figure 5.P9. When the drag force FD is balanced by the gravitational force, sphere attains its terminal velocity. Variables pertinent to the problem are: a) Radius of sphere, R b) Velocity of sphere, V c) Viscosity of fluid, μ d) Density of fluid, ρ e) Drag force, FD Thus, we write an equation for the force to have the following form: FD = f(ρ, R, V, μ). We wish to obtain an expression for the drag force as a function of sphere velocity and fluid properties.

Problems

Figure 5.P9

5.10

Free fall of a sphere.

Consider a vertical cylinder as shown in Figure 5.P10. Diameter of the cylinder is D, and the cylinder height is H. Constant property fluid except density is flowing as shown in Figure 5.P10. Experimental data will be collected for the effect of free and forced convection over the vertical cylinder. The problem involves both free and forced convection. The average heat transfer coefficient is expected to have the following functional dependence: f(h, β gΔT, V, ρ, μ, k, cp , H, D) = 0. Notice that there are two characteristic lengths in the problem. These are height H and diameter D. We expect that diameter D and height H characterize the forced convection and free convection, respectively. Obtain the pertinent dimensionless group to correlate the experimental data.

Figure 5.P10

D

Description of Problem 5.10.

g V

H

Flow

5.11

Consider forced convection over an isothermal flat plate of length L. Assume that forced convection boundary layer flow starts as laminar and then becomes turbulent at some distance x = xc . Assume unit depth into the paper. The temperature difference (Tw − T∞ ) is assumed to be constant. The local Nusselt number is given by the following correlations: hx 1∕2 = 0.332Rex Pr 1∕3 Laminar flow region: Nux = k hx 1∕3 = 0.0288Re0.8 Turbulent flow region: Nux = . x Pr k The transition region is simply neglected for simplicity. The average Nusselt number is defined as NuL =

qw L hL = . k k(Tw − T∞ )

where qw is average wall heat flux. Obtain an expression for the average Nusselt number.

139

140

5 Dimensional Analysis

5.12

Consider steady incompressible flow in the entrance region of a circular tube. The problem coordinate system is shown in Figure 5.P12. Hornbeck [12] uses boundary layer model to study the problem. This model provides good accurate solution for engineering purposes. The axial component of momentum equation that describes the fluid motion in a pipe due to an applied pressure gradient under steady flow is ) ( ) ( 2 ) ( 𝜕 vz 1 𝜕vz 𝜕v 𝜕v 1 dp +ν + vz z + vr z = − 𝜕z 𝜕r ρ dz r 𝜕r 𝜕 r2 The boundary conditions are vz (r, 0) = U0 vz (R, z) = 0

No slip

𝜕vz (0, z) =0 𝜕r

Symmetry

vr (R, z) = 0 p(0) = p0 where U0 is the inlet velocity profile. Using the following dimensionless variables nondimensionalize this z-component of momentum equation and determine the dimensionless form the differential equation and its boundary conditions v U= z U0 Rvr V= 𝜈 p − p0 P= ρU20 r η= R (2R)U0 2(z∕R) νz Re = Z= 2 = Re ν R U0

r U0 p0

0

Figure 5.P12

5.13

z

D = 2R

Coordinate system of Problem 5.P12.

Consider the following boundary layer momentum and energy equations for laminar flow with constant fluid properties (ρ, μ, cp , k). Assume negligible gravity and pressure gradient effects 𝜕u 𝜕u 𝜕2 u +v =ν 2 𝜕x 𝜕y 𝜕y ) )2 ( ( 𝜕T 𝜕2 T 𝜕T 𝜕u +v =k 2 +μ ρcp u 𝜕x 𝜕y 𝜕y 𝜕y

u

Problems

Nondimensionalize these equations using the following dimensionless variables. Assume that flow has a characteristic velocity U∞ and a characteristic length L X=

x L

Y=

T − T∞ y u v U= V= θ= . L U∞ U∞ Tw − T∞

Comment on the parameters that appear. 5.14

Figure 5.P14 shows a heated cylinder forced convection experiment. The heated section of the cylinder has diameter D = 4 cm and length L = 10 cm. The temperature of the heated section is Tw = 350 K. The free stream temperature of the air is T∞ = 300 K and air moves in cross flow with speed V (m/s). Power is adjusted to hold Tw constant and possible the variation of fluid properties with temperature are neglected. The experimental data are given as follows: V(m/s)

8.3

11.9

15.9

19.8

23.8

27.8

q (W)

29.3

37.6

46

53.8

61.2

68.2

We wish to fit these data to the correlation given below hD = C Rem D. k Do not consider Prandtl effect since only one fluid is used in the experiment. Estimate the average heat transfer coefficient h in (W/m2 K) when V = 32 m/s. NuD =

Flow V, T∞ q Insulation

Test section

Insulation Thermocouples and heater leads

Tw L

Figure 5.P14

5.15

An experimental rig to study forced convection heat transfer from a cylinder.

Consider fully developed turbulent flow in a smooth pipe. a) Close to the wall, u depends on density ρ, dynamic viscosity μ, wall shear stress τw , and y. This can be expressed as u = f(ρ, μ, τw , y). Perform a dimensional analysis to show that u∕uτ = f

( √ y ν

τw ρ

)

b) Near the center, velocity may be represented by u − Vc = f(ρ, R, τw , y). Perform a dimensional analysis to obtain a functional form of velocity distribution. c) If u = f(ρ, μ, τw , y, R), use dimensional analysis to obtain a general functional relationship. d) There is a buffer region. Here, both (a) and (b) apply. What can you say about this region. 5.16

Consider viscous flow between two infinite parallel plates. The coordinate system is shown in Figure 5.P16. Assume steady, laminar, fully developed flow and heat transfer. A uniform heat flux is applied to each plate and inlet fluid temperature is Ti

141

142

5 Dimensional Analysis

Assume Newtonian, incompressible, constant property fluid. The axial component of momentum equation that describes the fluid motion in the parallel plate channel is due to an applied constant pressure gradient. Neglect viscous dissipation and body forces. a) Write down the governing equation of motion b) Write down the boundary conditions c) Solution of momentum equation gives the velocity distribution ( )[ ( y )2 ] H2 dp 1− u=− 2μ dx H Obtain the centerline or maximum velocity. The mean velocity in the channel defined as H

2 u dy H ∫0 What is the relation between average velocity V and centerline velocity uc ? Show that velocity profile can be written as [ ( y )2 ] 3 u= V 1− 2 H V=

d) Non-dimensionalize the governing equation and its boundary conditions using y u η= U= H uc where uc is the centerline velocity. e) Write down the energy equation f) Write down the boundary conditions g) Non-dimensionalize the governing energy equation and its boundary conditions using T−T y x ξ= , η= θ = ′′ i , x0 H (q0 H)∕k where x0 is unknown parameter to be determined. y

Ti

q0ʺ

0 u(y)

2H

x q0ʺ

Figure 5.P16

5.17

Coordinate system and steady fully developed laminar flow in a parallel plate channel.

Consider incompressible constant property flow over a flat plate. The basic equations of continuity and momentum and energy, making the usual boundary layer approximations, are Continuity 𝜕u 𝜕v + =0 𝜕x 𝜕y Momentum ( ) dp 𝜕u 𝜕u 𝜕2 u ρ u +v =− +μ 2 𝜕x 𝜕y dx 𝜕y The boundary conditions are u(x, 0) = 0 v(x, 0) = 0

Problems

u(x, ∞) = U∞ (x) u(0, y) = U∞ (0) Energy

) 𝜕2 T 𝜕T 𝜕T =k 2 ρcp u +v 𝜕x 𝜕y 𝜕y The boundary conditions are (

T(x, 0) = Tw

(Constant wall temperature)

or −k

𝜕T (x, 0) = q′′w 𝜕y

(Constant wall heat flux)

T(0, y) = T∞ T(x, ∞) = T∞ These equations can be solved numerically but it is better to place these equations in dimensionless form so that we will have as few parameters as possible. Employ the following dimensionless variables suggested by Hornbeck [13]. p − p0 ρvL u P= V= U= U∞ μ ρU2∞ y xμ X= 2 Y= L L ρU∞ T − Tw θ= T∞ − Tw k θ = ′′ (T − T∞ ) qw L where U∞ is the characteristic velocity and L is measured along the plate surface and the plate length is the characteristic length. Show that dimensionless form of differential equations is given as Continuity 𝜕U 𝜕V + =0 𝜕X 𝜕Y Momentum equation U

𝜕V dP 𝜕 2 U 𝜕U +V =− + 𝜕X 𝜕Y dX 𝜕Y2

U(X, 0) = 0 V(X, 0) = 0 U(X, ∞) =

u = U∞ (X) U∞

U(0, Y) = U∞ (0) Energy equation ) ( 𝜕θ 1 𝜕2 θ 𝜕θ +V = U 𝜕X 𝜕Y Pr 𝜕Y2 Constant wall temperature θ(X, 0) = 0 θ(0, Y) = 1 θ(X, ∞) = 1

Pr =

μcp k

143

144

5 Dimensional Analysis

Constant wall heat flux 𝜕θ (X, 0) = −1 𝜕Y θ(0, Y) = 0 θ(X, ∞) = 0 Compare your solution with Example 5.11. 5.18

Consider plug flow between two infinite parallel plates as shown in Figure 5.P18. Both plates are subjected to uniform heat flux q′′0 . Inviscid fluid has an initial temperature of Ti and flows between the parallel plates. The energy equation may be written as ρcp V

𝜕2 T 𝜕T =k 2 𝜕x 𝜕y

T(0, y) = Ti 𝜕T (x, 0) = 0 𝜕y 𝜕T k (x, H) = q′′0 𝜕y Using the following dimensionless temperature and spatial quantities, as discussed in Bennett [43], obtain the dimensionless form of the energy equation and its boundary conditions. θ=

Figure 5.P18

5.19

k(T − Ti ) Hq′′0

ξ=

x H

η=

y H

Coordinate system and temperature field for steady slug flow in a parallel plate channel of constant heat flux.

Consider plug flow between two infinite parallel plates as shown in Figure 5.P19. Both plates are subjected to uniform temperature Tw . Inviscid fluid has an initial temperature of Ti and flows between the parallel plates. The energy equation may be written as ρcp V

𝜕2 T 𝜕T =k 2 𝜕x 𝜕y

The boundary conditions are T(0, y) = Ti 𝜕T (x, 0) = 0 𝜕y T(x, H) = Tw Using the following dimensionless temperature and spatial quantities obtain the dimensionless form of energy equation and its boundary conditions. See Bennett [43]. θ=

(T − Ti ) Tw − Ti

ξ=

x H

η=

y H

Problems

y

Figure 5.P19 Coordinate system and temperature field for steady slug flow between parallel plates of constant surface temperature.

Ti V

0

Tw 2H

T(y)

x

Tw

5.20

Experiments were performed by Onur and Hewitt [15] to study forced convection heat transfer on a tilted flat plate. The experimental data were collected from the tilted flat plate exposed to five different flow velocities and sample data are tabulated below U∞ (m/s)

h(W∕m2 K)

2.51

18.50

4.36

23.94

6.50

28.93

8.89

33.62

11.00

37.25

Plate length L = 0.1679 m, kinematic viscosity of air ν = 1.6188 × 10−5 m2 /s, and thermal conductivity of air k = 0.02677 W/m K. Develop a correlation in the form h = f(U∞ ). 5.21

Experiments were performed by Turgut and Sari [16] to study flow and heat transfer characteristics in transition and turbulent regions. The working fluid was air. The tube used in experiments was smooth and had a hexagonal cross section. ReD

NuD

2 322

8.90

3 609

12.00

4 337

13.58

4 940

15.10

7 737

20.42

11 000

30.40

14 731

40.20

17 122

45.90

20 471

55.80

30 232

70.30

32 061

74.60

38 386

81.40

47 394

96.90

52 058

103.30

The turbulent flow is strongly influenced by the character of the tube entrance. The tube has a well-designed nozzle entrance to obtain a uniform velocity at the tube inlet. Develop a correlation of the form NuD = C RenD .

145

146

5 Dimensional Analysis

5.22

Consider Example 5.13. Hornbeck [13] recommends the use of the following dimensionless variables to put continuity, momentum, and the energy equations in dimensionless form. a) Use these dimensionless variables to put these equations, and boundary conditions in dimensionless form uL νu = L2 gβ(Tw − T∞ ) νGrL Lv V= ν T − T∞ θ= Tw − T∞ U=

x ν2 x = L gβ(Tw − T∞ ) LGrL y Y= L X=

4

gβ(Tw − T∞ )L3 is the Grashof number and L is a characteristic length in the x-direction. ν2 b) The local Nusselt number is defined by where

Nux =

q′′w x . k(Tw − T∞ )

Obtain the Nusselt number in terms of dimensionless variables. 5.23

Consider Example 5.13. The vertical flat plate is subjected to uniform heat flux q′′w . In this case, the boundary conditions become u(x, 0) = 0 u(x, ∞) = 0 v(x, 0) = 0 u(0, y) = 0 −k

𝜕T(x, 0) = q′′w 𝜕y

T(x, ∞) = T∞ T(0, y) = T∞ . Hornbeck [13] recommends the following dimensionless variables: νku L3 gβq′′w Lv V= ν (T − T∞ )k θ= q′′w L U=

x ν2 k L5 gβq′′w y Y= . L X=

Reformulate the problem using these dimensionless variables.

References

5.24

Consider Problem 5.12. Pipe is subjected to uniform heat flux q′′w . The energy equation becomes ( 2 ) ) ( 𝜕T 𝜕T 𝜕 T 1 𝜕T + vr =k . + ρcp vz 𝜕z 𝜕r r 𝜕r 𝜕r2 The boundary conditions are T(r, 0) = Ti 𝜕T(R, z) = q′′w 𝜕r 𝜕T(0, z) = 0. 𝜕r Hornbeck [13] recommends the following dimensionless variable for temperature: k

θ=

k (T − Ti ) q′′w R

Obtain the dimensionless form of the energy equation and its boundary conditions. 5.25

Consider constant property viscous incompressible fluid flow over the equilateral triangular area, as shown in Figure 5.P25. Free stream velocity and temperature of the fluid are U∞ and T∞ , respectively. Using the correlation of the general form for the local Nusselt number for a flat plate Nux = CRenx . obtain an expression for the average Nusselt number between for the heat transfer between triangular area and the fluid.

Figure 5.P25

Coordinate system for flow over a triangular plate.

y T∞ x

H

U∞

L

References 1 2 3 4 5 6 7 8 9 10 11 12

Buckingham, E. (1914). On physically similar systems: illustrations of the use of dimensional analysis. Phys. Rev. 4: 345. Buckingham, E. (1915). Model experiments and the form of empirical equations. Trans. ASME 37: 263–296. Karwa, R. (2020). Heat and Mass Transfer. Springer. Hibbeler, R.C. (2015). Fluid Mechanics. Pearson Prentice Hall. Ipsen, D.C. (1960). Units, Dimensions, and Dimensionless Numbers. McGraw-Hill. White, F.M. and Majdalani, J. (2021). Viscous Fluid Flow, 4e. McGraw-Hill. Mills, A.F. (1999). Heat Transfer, 2e. Prentice Hall. Incropera, F.P., Dewitt, D.P., Bergman, T.L., and Lavine, A.S. (2013). Foundations of Heat Transfer, 6e. Wiley. Kreith, F. and Manglik, R. (2018). Principles of Heat Transfer, 8e. CENGAGE Learning. Rao, S.S. (2002). Applied Numerical Methods for Engineers and Scientists. Prentice Hall. Chapra, S.C. (2017). Applied Numerical Methods with MATLAB for Engineers and Scientists, 4e. McGraw-Hill. Hornbeck, R.W. (1963). Laminar flow in the entrance region of a pipe. Appl. Sci. Res., Sec. A 13: 224–232.

147

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13 Hornbeck, R.W. (1973). Numerical marching techniques for fluid flows with heat transfer, Special Publication, NASA-SP-297. 14 Bennett, T.D. (2013). Transport by Advection and Diffusion. Wiley. 15 Onur, N. and Hewitt, J.C. (1980). A study of wind effects on collector performance. ASME Paper 80-C2/Sol.4. 16 Turgut, O. and Sari, M. (2013). Experimental and numerical study of turbulent flow and heat transfer inside hexagonal duct. Heat Mass Transfer 49: 543–554. 17 Tekir, A.M., Taskesen, E., Aksuc, B. et al. (2020). Comparison of bi-directional multi-wave alternating magnetic field effect on ferromagnetic nanofluid flow in a circular pipe under laminar flow conditions. Appl. Therm. Eng. 179: 1–11. 18 Klinkenberg, A. (1955). Dimensional systems and systems of units in physics with special reference to chemical engineering. Chem. Eng. Sci. 4: 67–177. 19 Langhaar, H.L. (1951). Dimensional Analysis and Theory of Models. Wiley. 20 Gibbings, J.C. (2001). Dimensional Analysis. Springer. 21 Horning, H.G. (2006). Dimensional Analysis. Dover Publications. 22 Simon, V., Weigand, B., and Gomma, H. (2017). Dimensional Analysis. Springer. 23 Hansen, A.G. (1964). Similarity Analysis of Boundary Value Problems in Engineering. Prentice-Hall. 24 Zohuri, B. (2015). Dimensional Analysis and Self Similarity Methods for Engineers and Scientists. Springer. 25 Lemons, D.S. (2017). A Student’s Guide to Dimensional Analysis. Cambridge University Press. 26 Yarin, L.P. (2012). The Pi-Theorem with Applications to Fluid Mechanics and Heat and Mass Transfer. Springer. 27 Taylor, E.S. (1974). Dimensional Analysis for Engineers. Oxford: Clarendon Press. 28 Arpaci, V.S. and Larsen, P.S. (1984). Convection Heat Transfer. Prentice-Hall. 29 Eckert, E.R.G. and Drake, R.M. (1972). Analysis of Heat and Mass Transfer. McGraw-Hill. 30 Cebeci, T. and Bradshaw, P. (1984). Physical and Computational Aspects of Convective Heat Transfer. Springer-Verlag. 31 Oosthuizen, P.H. and Naylor, D. (1999). Introduction to Convective Heat Transfer Analysis. McGraw-Hill. 32 Jiji, L.M. (2009). Heat Convection, 2e. Springer-Verlag. 33 Schlichting, H. (1979). Boundary Layer Theory, 7e. McGraw-Hill. 34 Bodia, J.R. and Osterle, J.F. (1961). Finite difference analysis of plane Poiseuille and Couette flow developments. Appl. Sci. Res. A10: 265–276. 35 Bridgeman, P.W. (1931). Dimensional Analysis. Yale University Press. 36 Benallou, A. (2018). Energy and Mas Transfer, vol. 1. Wiley. 37 Arpaci, V.S., Selamet, A., and Kao, S.H. (2000). Introduction to Heat Transfer. Prentice Hall. 38 Daniel, C. and Wood, F.S. (1980). Fitting Equations to Data. Wiley. 39 Long, A.C. (1999). Essentials of Heat Transfer. Pearson Education Limited. 40 Levenspiel, O. (2014). Engineering Flow and Heat Exchange, 3e. Springer. 41 Brown, S.H. (2009). Multiple linear regression analysis: A matrix approach with MATLAB, Alabama Journal of Mathematics, Spring/Fall: 1–3. 42 Sheikh, A.J., Beck, J.V. and Amos, D.E. (2008). Axial heat conduction effects in the entrance region of parallel plate ducts. Int. J. Heat and Mass Transfer, 51: 5811–5822. 43 Magrab, E.B., Azarm, S., Balachandran, B., Duncan, J.H., Herold, K.E. and Walsh, G.C. (2011). An Engineer’s Guide to MATLAB, Pearson Education, Inc. 44 Meade, D.B., Michael May, S.J., Cheung, C.K. and Keough, G.E. (2009). Getting Started with MAPLE, John Wiley and Sons, Inc. 45 Gilbert, R.P., Hsiao, G.C. and Ronkese, R.J. (2021). Differential Equations-A Maple Supplement 2e, CRC Press.

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6 One-Dimensional Solutions in Convective Heat Transfer 6.1 Introduction In this chapter, one-dimensional flow and heat transfer problems will be considered. In one-dimensional flows, the streamlines are straight lines. In one-dimensional rectilinear flows, u = u(x) and v = w = 0. In axisymmetric rectilinear flows, vz = vz (r) and vr = vθ = 0. We will discuss these points shortly. One-dimensional solutions may be found in [1–3]. Although the applications of one-dimensional problems are restrictive, conclusions that we draw from the solutions of one-dimensional flow problems are very instructive to understand the fundamental concepts. The reader will develop an appreciation for the physical significance of each term in the governing equations of the problem and gain insight to identify the conditions under which the terms may be neglected. The meaningful simplification of the governing equations is critical to obtaining solutions. Energy, continuity, and momentum equations are not coupled in general. For this reason, velocity distribution is obtained first. With velocities known, temperature distribution may be obtained. We will make certain simplifying assumptions in the solution of the problems, and here are some of these assumptions: 1. 2. 3. 4.

Flow is laminar. Steady flow, 𝜕/𝜕t = 0. In general, the fluid properties are constant. In general, energy, momentum, and continuity equations are decoupled. This means that velocity field can be determined independently from the temperature field. 5. Streamlines are parallel. Consider the flow between two infinite parallel plates, as depicted in Figure 6.1. It is assumed that there is no fluid motion in the z-direction, i.e. w = 0. Since the streamlines are parallel, the velocity component v normal to streamline is zero; thus: v = 0. Knowing this result, continuity gives 𝜕u = 0. (6.1) 𝜕x In this case, the flow inside the duct is assumed to be fully developed. This means that the certain properties of flow do not change with the axial distance. This is true far from the duct entrance where entrance effects can be neglected. Fluid properties are constant, and forms of the velocity and temperature profiles do not change with the axial distance. Since Eq. (6.1) is valid in the flow, we can also state that 𝜕2 u = 0. (6.2) 𝜕x2 Consider flow in a pipe. We assume that the temperature and velocity profiles are expressed in the following form: (y) T − Tc (6.3) =g Tw − Tc R (y) u (6.4) =f Vc R

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

150

6 One-Dimensional Solutions in Convective Heat Transfer

y

Figure 6.1

Streamlines between infinite parallel plates.

x

0

where Tc is the centerline temperature, Tw is the surface temperature, Vc is the pipe centerline velocity, and R is the pipe radius. In this case, in the fully developed flow, the velocity and temperature profile functions f and g are independent of axial position. 6. Axial conduction will be neglected in channel flows for Pe = RePr > 100, as discussed in [4]. This will yield 𝜕2T = 0. (6.5) 𝜕x2 7. Consider a shaft rotating concentrically inside a sleeve, as depicted in Figure 6.2. Fluid is rotating under steady conditions with an angular velocity ω in the direction, as shown in Figure 6.2. We assume that there are no axial variations. The streamlines are concentric. For axisymmetric conditions along with negligible axial variation, we may then write 𝜕 vθ =0 𝜕θ 𝜕T = 0. 𝜕θ From Eqs. (6.6a) and (6.6b), we conclude that

(6.6a) (6.6b)

𝜕 2 vθ

=0 (6.7a) 𝜕θ2 𝜕2 T = 0. (6.7b) 𝜕θ2 8. Consider flow between two infinite parallel plates. We may state that the axial variation of temperature is constant 𝜕T = const. (6.8) 𝜕x Equation (6.8) is exact in certain flows and reasonable approximation for other cases. The following conditions may lead to the validity of Eq. (6.8). a) Parallel streamlines b) Uniform heat flux boundary condition c) Negligible entrance effects 9. We will see later in channel flows that, for uniform wall temperature boundary condition, the heat flux approaches zero and the bulk (mean) temperature Tm approaches wall temperature Tw as the axial distance x increases. This region in Figure 6.2

TW

θ

Concentric streamlines.

6.2 Couette Flow

which the mean temperature Tm no longer changes with the axial distance x is characterized by 𝜕T/𝜕x = 0. This called asymptotic thermally developed flow. Asymptotic thermally developed flows occur in annular as well as Couette flow systems. This point is discussed in Example 6.5.

6.2 Couette Flow These flows are first considered by Couette as discussed by White and Majdalani [5]. Couette flow provides a simple model to study heat transfer for flow between parallel plates. We will improve our understanding of the physical significance of parameters governing the velocity and temperature distribution in the flow by studying the simple geometries. A general case of Couette flow includes the effects of both the plate motion and the pressure gradient. The plates are assumed to be infinite in extend, and thus, there are no end effects. The streamlines are parallel to plates. Example 6.1 Consider two infinite parallel impermeable plates. Incompressible, constant property fluid fills the space between plates. While the lower plate is stationary, the upper plate is moving with a constant velocity U in the positive x-direction. Lower plate temperature is greater than the moving plate temperature. See Figure 6.E1a. Taking the viscous dissipation into account, determine (i) velocity distribution, (ii) temperature distribution, (iii) heat flux at both the surfaces, (iv) the location of maximum temperature, (v) a Nusselt number, and (vi) the Fanning friction factor for the lower surface. Neglect gravity effect and pressure variation in the fluid. Figure 6.E1a

y

Heat transfer in Couette flow.

U

T1

g

H

0

x

T0

Solution Notice that plates are infinite to the extent that there is no variation in the x- and z-directions. Also, we notice that there is no motion in the z-direction, i.e. w = 0. Energy is generated by viscous dissipation in the flow. Flow is fully developed since all the fluid particles move parallel to plates. We need velocity distribution to determine the temperature distribution, heat flux, and Nusselt number. We make the following assumptions. 1. 2. 3. 4. 5. 6. 7. 8.

Steady flow Laminar flow Constant fluid properties (ρ, cp , k, μ) No pressure gradients in the flow direction No edge effects 𝜕/𝜕z = 0 (Infinite plates) No end effects 𝜕/𝜕x = 0 (Infinite plates) Newtonian fluid Fully developed flow (Both the velocity and temperature profiles do not change in the flow direction. This means that 𝜕u/𝜕x = 0 and 𝜕T/𝜕x = 0.) 9. No body forces in the x-direction 10. No motion in the z-direction, w = 0 Continuity equation Dρ ⃗ =0 ⃗ •V) + ρ(∇ Dt

151

152

6 One-Dimensional Solutions in Convective Heat Transfer

or 𝜕ρ 𝜕ρ 𝜕ρ 𝜕ρ +u +v +w +ρ 𝜕t 𝜕x 𝜕y 𝜕z

(

𝜕u 𝜕v 𝜕w + + 𝜕x 𝜕y 𝜕z

) =0

For constant density (incompressible fluid), we have 𝜕ρ 𝜕ρ 𝜕ρ 𝜕ρ = = = = 0. 𝜕t 𝜕x 𝜕y 𝜕z The continuity equation becomes 𝜕u 𝜕v 𝜕w + + = 0. 𝜕x 𝜕y 𝜕z We have two infinite parallel plates; there are no edge effects as well as end effects, and since there is no motion in the z-direction, we see that w = 0. 𝜕w =0 (No edge effects) 𝜕z 𝜕u =0 (No end effects and hydrodynamically fully developed (HFD) flow) 𝜕x Then, using these simplifications, the continuity equation reduces to 𝜕v = 0. 𝜕y Integration yields v(x, y) = f(x). We know that at the wall, we can apply no slip boundary condition y = 0 v = 0 (Impermeable wall) This means that f(x) = 0. The y-component velocity becomes v(x, y) = 0. This means that streamlines are parallel in the flow field, and the y-component of velocity is zero everywhere. Consider now the y-component of the momentum equation ) ( 2 ) ( 𝜕p 𝜕v 𝜕v 𝜕v 𝜕 v 𝜕2 v 𝜕2 v 𝜕v +u +v +w = ρfy − +μ . + + ρ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕y 𝜕x2 𝜕y2 𝜕z2 Then, the y-component of the momentum equation reduces to 0 = ρfy −

𝜕p 𝜕y

fy = −g. Integration yields p = p0 − ρgy. Next, we now consider the x-component of the momentum equation ( ) ( 2 ) 𝜕p 𝜕u 𝜕u 𝜕u 𝜕u 𝜕 u 𝜕2 u 𝜕2 u ρ +u +v +w = ρfx − +μ + 2 + 2 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x 𝜕x2 𝜕y 𝜕z 𝜕u =0 𝜕t

(Steady flow)

fx = 0

(No body force in the x-direction)

𝜕u =0 𝜕x

(HFD flow)

6.2 Couette Flow

𝜕2 u =0 𝜕x2 𝜕u =0 𝜕z

since

𝜕u =0 𝜕x

(No edge effects)

𝜕2 u 𝜕z2 𝜕p =0 𝜕x

𝜕u =0 𝜕z

since

(No axial pressure variation in the flow)

The x-component of the momentum equation reduces to d2 u = 0. dy2 The boundary conditions are given as u(0) = 0 u(H) = U. Let us use the Maple 2020 to get the solution > restart; > interface (displayprecision = 𝟓); 𝟓 > de ≔ u′′ (y) = 𝟎; de ≔ D(𝟐) (u)(y) = 𝟎 > sol ≔ dsolve({de, u(𝟎) = 𝟎, u(H) = U}, u(y)); Uy sol ≔ u(y) = H The solution can be expressed as y u(y) = . U H With the known velocity distribution, we next consider the energy equation and the viscous dissipation function [ ] ] [ 2 𝜕T 𝜕T 𝜕T 𝜕T 𝜕 T 𝜕2 T 𝜕2 T + q′′′ + μΦ ρcp +u +v +w =k + + 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x2 𝜕y2 𝜕z2 { ( Φ=2

𝜕u 𝜕x

(

)2 +

𝜕v 𝜕y

)2

(

𝜕w + 𝜕z

𝜕T =0 𝜕t

(Steady state)

𝜕T =0 𝜕x

(No end effects)

v

𝜕T =0 𝜕y

)2

}

( +

𝜕v 𝜕u + 𝜕x 𝜕y

)2

( +

𝜕w 𝜕v + 𝜕y 𝜕z

)2

(

𝜕u 𝜕w + + 𝜕z 𝜕x

(Since v = 0 because of continuity equation)

𝜕T =0 𝜕z

(No edge effect and w = 0 since no motion in the z-direction)

𝜕2 T = 0. 𝜕x2

(Negligible axial conduction.)

)2

2 − 3

(

𝜕u 𝜕v 𝜕w + + 𝜕x 𝜕y 𝜕z

)2

153

154

6 One-Dimensional Solutions in Convective Heat Transfer

𝜕2T =0 𝜕z2

(No edge effects. Since there are no edge effects

q′′′ = 0

(No energy generation)

𝜕T 𝜕2 T 𝜕T = 0 and since = 0, we have 2 = 0.) 𝜕z 𝜕z 𝜕z

The energy equation becomes k

d2 T + μΦ = 0. dy2

With the assumptions we have made previously, the viscous dissipation term reduces to ( )2 𝜕u . Φ= 𝜕y The energy equation takes the following form: ( μ ) ( )2 U d2 T = − . 2 k H dy The boundary conditions are T(0) = T0 T(H) = T1 . The solution is obtained by Maple 2020. > restart; > > de ≔ T′′ (y) = −

(𝛍) (

U H

)𝟐

; k 𝛍 U𝟐 de ≔ D(𝟐) (T)(y) = − 𝟐 H k > sol ≔ dsolve({de, T(𝟎) = T𝟎 , T(H) = T𝟏 }, T(y)); sol ≔ T(y) = −

•

𝟐 𝟐 𝟏 𝛍 U y𝟐 𝟏 (U 𝛍 − 𝟐 k T𝟎 + 𝟐 k T𝟏 )y + 𝟐 H𝟐 k 𝟐 Hk

+ T𝟎 The solution is ( 2)( )[ (y) ( y )] y μU T − T0 = [(T1 − T0 )] + 1− . H 2k H H The first term in this equation is due to conduction, and the second term is due to viscous dissipation in the fluid. In dimensionless form, we have Ec Pr (η − η2 ) θ=η+ 2 where θ =

T−T0 , T1 −T0

η=

y , H

Pr =

μcp k

= Prandtl number, and Ec =

U2 cp (T1 −T0 )

= Eckert number. The Eckert number indicates the

effect of viscous dissipation. The product of Prandtl and Eckert numbers is called Brinkman number Br = EcPr. We will plot dimensionless temperature as a function dimensionless distance y/H for different Br numbers. See Figure 6.E1b. Assume that T1 > T0 . We wish to find the heat flux at both the plates. Heat flux at the lower plate (stationary plate) is ) k(T1 − T0 ) ( Br 1+ . q′′w = − H 2 If we consider heat transfer on the stationary plate, we see that heat flows from the fluid to the plate.

6.2 Couette Flow

1.8 1.6 Br = 10

1.4 1.2

Br = 4

1 θ 0.8 Br = 2

0.6

Br = 1 Br = 0

0.4 0.2 0 0

Figure 6.E1b

0.2

0.4

η

0.6

0.8

1

Dimensionless temperature distribution in Couette flow for different Brinkman numbers.

Heat flux at the upper plate (moving plate) is ) k(T1 − T0 ) ( Br q′′w = − 1− . H 2 Consider now heat transfer on the moving plate. An examination of dimensionless temperature distribution reveals that heat transfer is from the moving plate to fluid for Br < 2, and heat transfer is from fluid to the moving plate when Br > 2. If Br = 2, then there is no heat transfer from the moving plate. The case Br = Ec Pr = 0 corresponds to no flow condition, temperature distribution is a straight line, and this represents the pure conduction heat transfer. The location of maximum temperature will be determined next. First, we will determine the first derivative of temperature by Maple 2020 and express it as ( )] [ 2y Br 1 dT 1 = (T1 − T0 ) + − 2 =0 dy H 2 H H Solving this equation yields ] [ 1 1 ymax = H + . 2 Br Second, we determine the second derivative as (T − T ) d2 T = − 2 2 1 Ec Pr < 0 2 H dy and is always negative. Therefore, ymax is the location of maximum temperature, and the maximum temperature is Tmax − T0 1 Br (2 + Br)2 1 + = . = + T1 − T0 2 2Br 8 8Br A Brickman number of unity or greater implies a temperature increase in the fluid. At low-speed flows, viscous fluids have high Brinkman numbers. Except high viscous fluids, in low-speed flow, we normally neglect viscous dissipation effects. Let us clarify this point with an example.

155

156

6 One-Dimensional Solutions in Convective Heat Transfer

Suppose that U = 15 m/s and ΔT = T2 − T1 = 15 ∘ C. Calculate the Brinkman number for different fluids.

𝛍(kg/m. s)

Fluid

1.139 × 10−3

Water

−6

Air

17.93 × 10

Engine oil

123.4 × 10−2

k (W/m. K)

cp (kJ/kg/ ∘ C)

Pr

Brinkman number

0.5911

4.187

8.07

0.02891

0.02511

1.006

0.714

0.01044

0.145

1.9

2920

23.052

We can define the Nusselt number based on lower surface as follows. First, we define the heat transfer coefficient h ( )] [ ( ) 2y Br 1 1 dT + − 2 −k −k(T1 − T0 ) dy y=0 H 2 H H q′′w y=0 h= = = T0 − T1 T0 − T1 T0 − T1 ( )] [ Br 1 1 −k(T1 − T0 ) + ( )] [ Br 1 1 H 2 H + . h= =k T0 − T1 H 2 H Next, we introduce the Nusselt number [ ] hDH Br =2 1+ k 2 DH = 2H is the hydraulic diameter. Suppose that we wish to base the heat transfer coefficient and the Nusselt number on the moving plate ( ) ( )] [ 2y Br 1 dT 1 −k −k(T1 − T0 ) + − dy y=H H 2 H H2 y=H q′′w h= = = T0 − T1 T0 − T1 T0 − T1 ] [ Br 1 1 − h=k H 2 H NuDH =

. Next, we introduce the Nusselt number [ ] hDH Br =2 1− k 2 DH = 2H is the hydraulic diameter. The Fanning friction factor cf for the lower surface is ( ) du μ U μH dy y=0 (τ)y=0 4μ 4μ 2μ = = = = 1 = cf = 1 1 2 2 2 ρUH ρU(2H) ρUDH ρU ρU ρU NuDH =

2

2

2

4 cf = ReDH where ReDH =

6.3

ρUDH μ

is the Reynolds number and DH = 2H is the hydraulic diameter.

Poiseuille Flow

This class problem deals with axial fluid flow in infinitely long channels. There are no end effects. The driving mechanism for the fluid flow is the pressure gradient. The flow streamlines are parallel to channel walls. These flows are named after J.L.M. Poiseuille, as discussed in [5]. Both the surfaces of parallel plates are usually considered to be stationary.

6.3 Poiseuille Flow

Example 6.2 Consider two infinite stationary parallel plates. Incompressible, constant property fluid fills the space between plates, while the lower plate is at temperature T0 and the upper plate is at temperature T1 . We assume that the upper plate temperature is greater than the lower plate temperature (T1 > T0 ). Flow is created by the pressure gradient in the axial flow direction. This means that dp/dx = const pressure gradient exists in the flow. Flow is steady and fully developed. Taking the viscous dissipation into account, determine (i) the velocity distribution, (ii) temperature distribution, (iii) maximum temperature and location of maximum temperature, and (iv) heat flux at both the surfaces (Figure 6.E2a). y

(p1 > p2)

g

T1

p1

p2 0

x

2H T0

Figure 6.E2a

Coordinate system and problem description for Example 6.2.

Solution Observe that plates are infinite in extent and no motion in the z-direction; thus, w = 0. Energy is generated by viscous dissipation in the flow. We need temperature distribution to determine heat flux and the Nusselt number. Assumptions: 1. 2. 3. 4. 5. 6. 7. 8. 9.

Steady flow Laminar flow Constant fluid properties (ρ, cp , k, μ) No edge effects 𝜕/𝜕z = 0 No end effects 𝜕/𝜕x = 0; except for pressure, 𝜕p/𝜕x ≠ 0 Newtonian fluid Pressure gradient in the x-direction is constant Continuum No motion in the z-direction and w = 0

First, we need velocity distribution to determine the temperature distribution. We start with the steady form of continuity equation. Continuity: 𝜕u 𝜕v 𝜕w + + = 0. 𝜕x 𝜕y 𝜕z Since there is no motion in the z-direction, we have w = 0. We assume that flow is fully developed in the axial direction, and thus, we have 𝜕u = 0. 𝜕x Notice that fluid particles move in the direction parallel to plates. Continuity reduces to dv = 0. dy Integration yields v = f(x) To determine the integration constant f(x), we use no slip boundary condition on the lower plate. This means that v(x, 0) = 0. Both the plates are stationary solid walls and are impermeable. Therefore, we have f(x) = 0, and the y-component of velocity is v(x, y) = 0

157

158

6 One-Dimensional Solutions in Convective Heat Transfer

Again, notice that since v(x, y) = 0, streamlines are parallel everywhere. Consider the y-component of the momentum equation ) ( 2 ) ( 𝜕p 𝜕v 𝜕v 𝜕v 𝜕 v 𝜕2 v 𝜕2 v 𝜕v +u +v +w = ρfy − +μ ρ + + 𝜕t 𝜕x 𝜕y 𝜕z 𝜕y 𝜕x2 𝜕y2 𝜕z2 This equation reduces to dp = −ρg. dy Integration yields p = p0 − ρgy. This means that pressure is hydrostatic in the y-direction. Now, the x-component of the momentum equation is ) ( 2 ) ( 𝜕p 𝜕u 𝜕u 𝜕u 𝜕 u 𝜕2 u 𝜕2 u 𝜕u +u +v +w = ρfx − +μ + 2 + 2 ρ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x 𝜕x2 𝜕y 𝜕z Under the stated assumptions, we have 𝜕u =0 𝜕t

(Steady state)

fx = 0

(No body force in the x − direction)

𝜕p = constant 𝜕x 𝜕u =0 𝜕x 𝜕u =0 𝜕y

(No end effects. Fully developed flow in the x-direction) since v = 0

𝜕u =0 𝜕z

(No edge effects and recall that no motion in the z-direction, and we have w = 0)

𝜕2u =0 𝜕x2

since

𝜕u =0 𝜕x

𝜕u 𝜕2u = 0. =0 since 𝜕z 𝜕z2 Based on these facts, the x-momentum equation reduces to 1 dp d2 u = μ dx dy2 Boundary conditions for the momentum equation are At y = −H

u=0

(no slip)

At y = H

u=0

(no slip)

The solution will be obtained by Maple 2020 ( ) ( ) 𝟏 • dp ′′ ; > de ≔ u (y) = 𝝁 dx de ≔ D(𝟐) (u)(y) =

dp 𝝁 dx

> sol ≔ dsolve({de, u(0) = 0, u(H) = 𝟎}, u(y)); sol ≔ u(y) = This last equation can be put in the following form: )( 2)[ ( ( y )2 ] dp H 1− . u= − dx 2μ H

1 dp y𝟐 1 dp H y − . 2 𝝁 dx 2 dx 𝝁

6.3 Poiseuille Flow

Let us now define the average fluid velocity H

V=

1 u dy. 2H ∫−H

Integration yields ( 2)( ) dp H V= − . 3μ dx The maximum velocity occurs at the duct centerline where y = 0 ) ( 2)( dp H − . Vc = 2μ dx Then, we can express centerline velocity Vc in terms of average velocity V Vc =

3 V. 2

Thus, we can write velocity as [ [ ( y )2 ] ( y )2 ] 3V u = Vc 1 − 1− = . H 2 H With known velocity distribution, the energy equation and the viscous dissipation are considered next: ] [ 2 [ ] 𝜕T 𝜕T 𝜕T 𝜕 T 𝜕2 T 𝜕2 T 𝜕T ρcp +u +v +w =k + 2 + 2 + q′′′ + μΦ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x2 𝜕y 𝜕z { ( Φ=2

𝜕u 𝜕x

(

)2 +

𝜕v 𝜕y

)2

(

𝜕w + 𝜕z

)2

}

( +

𝜕v 𝜕u + 𝜕x 𝜕y

)2

( +

𝜕w 𝜕v + 𝜕y 𝜕z

𝜕T =0 𝜕x

(No end effects)

𝜕T =0 𝜕z

(No edge effects)

𝜕T =0 𝜕t

(Steady conditions)

𝜕2 T =0 𝜕x2

(Negligible axial conduction) Since

𝜕2 T =0 𝜕z2

Notice that

𝜕u =0 𝜕x

(No end effects. Fully developed velocity profile)

𝜕2T 𝜕T = 0 since no edge effects. We have 2 = 0. 𝜕z 𝜕z

d2 T + μΦ = 0 dy2 ( )2 du Φ= dy k

. Boundary conditions for the energy equation are

At y = H T = T1 .

( +

𝜕2 T 𝜕T = 0, we have 2 = 0. 𝜕x 𝜕x

With these simplifications, the energy equation becomes

At y = −H T = T0

)2

𝜕u 𝜕w + 𝜕z 𝜕x

)2 −

2 3

(

𝜕u 𝜕v 𝜕w + + 𝜕x 𝜕y 𝜕z

)2

159

160

6 One-Dimensional Solutions in Convective Heat Transfer

The solution of the energy equation is obtained by Maple 2020: > #Solution of energy equation > #Vc is the centerline velocity in the channel >

( 𝟏−

> u ≔ (Vc )

•

( u ≔ Vc 𝟏 −

( y )𝟐 )

)

y𝟐

H

;

H𝟐

> > 𝚽 ≔ (diff(u, y))𝟐 ; 𝚽≔

𝟒 V𝟐c y𝟐 H𝟒

> de ≔ T′′ (y) +

(𝛍)

de ≔ D(𝟐) (T)(y) +

•𝚽; k 𝟒 𝛍 V𝟐c y𝟐

k H𝟒 > sol ≔ dsolve({de, T(−H) = T𝟎 , T(H) = T𝟏 }, T(y));

sol ≔ T(y) = −

μ V𝟐c y𝟒 𝟒

−

(T𝟎 − T𝟏 )y 𝟐 μ V𝟐c + 𝟑 k T𝟎 + 𝟑 k T𝟏 + 𝟐H 𝟔k

𝟑kH Temperature distribution can be put into the following form: ) ) ( 2)( ( μVc T1 − T0 ( y) y4 1+ + 1− 4 T(y) = T0 + 2 H 3k H

or it can be casted into dimensionless form taking Eckert and Prandtl numbers into consideration [ ( y )] (Ec Pr ) [ ( y )4 ] T − T0 1 1+ + 1− = T1 − T0 2 H 3 H or θ=

(Ec Pr ) 1 (1 + η) + (1 − η4 ) 2 3

μcp T − T0 V2c y , and Pr = . We can plot this dimensionless temperature as a function and η = H , Ec = T1 − T0 cp (T1 − T0 ) k of dimensionless position for different EcPr numbers, as shown in Figure 6.E2b. To find the location of the maximum temperature, first, we set dT/dy = 0 and solve for y ( ) (T − T0 )Ec Pr 4y3max dT 1 T1 − T0 = − 1 = 0. dy 2 H 3 H4

where θ =

The second derivative of temperature is √ ymax = H 3

d2 T dy2

< 0; the solution of this equation gives

3 . 8(Ec)(Pr )

If we substitute this value into temperature distribution, we obtain the maximum temperature in the fluid: [ ( )1∕3 ] ( )[ ( )4∕3 ] Tmax − T0 ( 1 ) Ec. Pr 3 1 3 1 + . 1+ 1− = T1 − T0 2 8 Ec. Pr 3 8 Ec. Pr Assume that T1 > T0 . Let us find heat flux at each plate: [ ] Ec Pr 4y3 dT 1 = − (T1 − T0 ). dy 2H 3 H4

6.3 Poiseuille Flow

1.2 EcPr = 2

1

0.8

EcPr = 0.375

θ 0.6 EcPr = 0

0.4

0.2

0 –0.5

Figure 6.E2b

0 η

0.5

1

Dimensionless temperature distribution in flow between parallel plates including the effect of viscous dissipation.

Heat flux at the lower plate is ) ( ) ( ] T1 − T0 [ 8 dT ′′ 1 + Ec Pr . qw = −k = −k dy y=−H 2H 3 Heat flux at the upper surface is ) ( ) ( ] T1 − T0 [ 8 dT ′′ = −k qw = −k 1 − Ec Pr . dy y=−H 2H 3 Consider heat transfer at the lower plate. Heat flows from fluid to the plate. If we look at Figure 6.E2b, we see the following cases for heat transfer at the upper plate. a) If Ec Pr < 0.375, heat flows from the plate to the fluid. b) If Ec Pr > 0.375, heat flows from the fluid to the plate. c) If Ec Pr = 0.375, there is no heat transfer at the upper plate.

Example 6.3 Consider an incompressible fluid flowing in an infinitely long tube. The fluid motion is generated by the axial pressure gradient. Fluid properties are constant. Flow is laminar, fully developed, steady, and axisymmetric. The pipe surface is maintained at a uniform temperature Tw . See Figure 6.E3. Gravity, axial temperature variation, and end effects are r TW

2R

Figure 6.E3

O

z

Coordinate system and problem description for Example 6.3.

TW R 0

161

162

6 One-Dimensional Solutions in Convective Heat Transfer

neglected. Taking viscous dissipation into consideration, determine (i) temperature distribution, (ii) maximum temperature and its location, (iii) surface heat flux, (iv) Nusselt number based on temperature difference Tw − Tc , where T(0) = Tc is the temperature at r = 0, and (v) Nusselt number based on temperature Tw − Tm , where Tm is the mean temperature in the fluid. Solution Assumptions (1) Steady laminar flow, 𝜕/𝜕t = 0. (2) Since the tube is infinitely long, flow field does not vary in the flow direction, 𝜕/𝜕z = 0. This means HFD flow. No end effects. (3) Incompressible fluid. (4) Flow is axisymmetric, 𝜕/𝜕θ = 0. (5) Newtonian fluid. (6) No gravitational effects. (7) No energy generation. (8) Negligible axial temperature variation. (9) Continuum. (10) Constant fluid properties (ρ, cp , k, μ). Energy is generated by viscous dissipation in the flow, and it is removed from the fluid by conduction on the pipe surface. We need temperature distribution to determine heat flux and the Nusselt number. First, we determine the velocity distribution. We start with the continuity equation. Continuity: 𝜕ρ 1 𝜕 1 𝜕 𝜕 + (ρ rvr ) + (ρv ) + (ρvz ) = 0 𝜕t r 𝜕r r 𝜕θ θ 𝜕z For constant density 𝜕ρ 𝜕ρ 𝜕ρ 𝜕ρ = = = =0 𝜕r 𝜕θ 𝜕r 𝜕z 𝜕ρ =0 (Steady flow) 𝜕t vθ =

𝜕 =0 𝜕θ

(Axisymmetric flow)

𝜕vz =0 (Very long tube; no end effects; HFD flow) 𝜕z The continuity equation reduces to d ( rvr ) = 0. dr Integration yields rvr = f(z) No slip boundary condition is applied at the pipe surface to determine the integration constant: At r = R

vr = 0.

Then, we find that f(z) = 0, and therefore, vr (r, z) = 0. This tells us that streamlines are parallel to the pipe surface. The radial component of velocity vr is zero everywhere. Next, we consider the z-component of the Navier–Stokes equation.

6.3 Poiseuille Flow

The z-momentum equation ] [ v 𝜕v 𝜕v 𝜕v 𝜕vz 𝜕p + vr z + θ z + vz z = ρfz − ρ 𝜕t 𝜕r r 𝜕θ 𝜕z 𝜕z ( ) [ 2 2 ] 𝜕v 𝜕 v v 𝜕 1 1 𝜕 z r z + 2 2z + +μ r 𝜕r 𝜕r r 𝜕θ 𝜕z2 𝜕vz =0 𝜕t vr

𝜕vz = 0 since vr = 0 𝜕r

𝜕vz = 0, vθ = 0 𝜕θ vz

𝜕v 𝜕vz = 0 since z = 0 𝜕z 𝜕z

(Steady flow) (Continuity) (Axisymmetric flow) (Very long tube; no end effects; flow is fully developed)

fz = ρgz = 0

(No gravitational effect)

2 𝜕v 1 𝜕 vz = 0 since z = 0 2 2 𝜕θ r 𝜕θ

(Axisymmetric flow)

𝜕vz =0 𝜕z

(No end effects; fully developed flow)

𝜕 2 vz 𝜕v (Very long tube; no end effects) = 0 since z = 0 𝜕z 𝜕z2 The z-momentum equation reduces to ( ) dv dp μ d r z = = g(r) r dr dr dz First, we integrate with respect to z dp = g(r) dz p = g(r)z + C0 where C0 is just an integration constant. Let us look the r-component of the momentum equation: ] [ 𝜕vr vθ 𝜕vr v2θ 𝜕vr 𝜕vr 𝜕p + vr + − + vz = ρfr − ρ 𝜕t 𝜕r r 𝜕θ r 𝜕z 𝜕r [ ( ) ] 2 𝜕2 v 1𝜕 v 2 𝜕v 𝜕 1 𝜕 (r vr ) +μ + 2 2r − 2 r + 2r 𝜕r r 𝜕r r 𝜕θ r 𝜕θ 𝜕z . Since vr = 0 is everywhere, the r-momentum equation reduces to 𝜕p = 0. 𝜕r Integration of this equation gives us p = f(z) where f(z) is an integration constant. We equate the pressure distributions g(r)z + C0 = f(z).

163

164

6 One-Dimensional Solutions in Convective Heat Transfer

This tells us that one side depends on z and the other side depends on z and r. But this is not possible. The only admissible solution is g(r) = C where C is a constant. Note that the axial pressure gradient, dp/dz = C, is constant in the flow direction. Now, we can integrate the z-momentum equation along with the boundary conditions. Boundary conditions are dvz (0) =0 dr vz (R) = 0. The solution is obtained by Maple 2020:

( ) ( ) r • dp ; > de ≔ r•diff(vz (r), r, r) + diff(vz (r), r) = 𝛍 dz ( 𝟐 ) r dp d d vz (r) + r v (r) = de ≔ z 𝟐 dr 𝛍 dz dr } ) ({ > sol ≔ dsolve de, vz (R) = 𝟎, v′z (𝟎) = 𝟎 , vz (r) ;

sol ≔ vz (r) =

𝟐 𝟏 r𝟐 dp 𝟏 R dp − 𝟒 𝛍 dz 𝟒 𝛍 dz

This solution can be put in the following form: ( ) 1 dp (R2 − r2 ). vz = − 4μ dz Let us now define mean velocity V ( ) R 2 R2 dp . V= 2 vz r dr = − 8μ dz R ∫0 Velocity distribution can be expressed in terms of mean velocity as [ ( )2 ] r . vz = 2V 1 − R The energy equation is { } ] ( ) [ 𝜕T 1 𝜕2 T 𝜕2 T 𝜕T 𝜕T vθ 𝜕T 𝜕T 1 𝜕 + vr + + vz =k r + 2 2 + 2 + q′′′ + μΦ ρcp 𝜕t 𝜕r r 𝜕θ 𝜕z r 𝜕r 𝜕r r 𝜕θ 𝜕z {( } ) ]2 ( ) ]2 [ ]2 [ [ ( ) 𝜕vr 2 𝜕vz 2 1 𝜕vr 𝜕 vθ 1 𝜕vθ vr 1 𝜕vz 𝜕vθ + + + Φ=2 + + + + r 𝜕r r 𝜕θ r 𝜕z 𝜕r r r 𝜕θ r 𝜕θ 𝜕z ]2 [ 𝜕vr 𝜕vz + + 𝜕z 𝜕r 𝜕T =0 𝜕t

(Steady flow)

𝜕T =0 𝜕z

(Very long tube; no end effect; we neglect axial temperature variation)

𝜕2T =0 𝜕z2

(Negligible axial conduction)

𝜕T =0 𝜕θ

(Axisymmetric flow)

𝜕T 𝜕2T =0 = 0 since 2 𝜕θ 𝜕θ

(Axisymmetric flow)

q′′′ = 0

(No energy generation)

6.3 Poiseuille Flow

The energy equation and the viscous dissipation function reduce to ( ) k d dT r + μΦ = 0 r dr dr ) ( dvz 2 . Φ= dr Substituting velocity distribution into viscous dissipation term yields ) )2 ( ) ( ( dvz 2 4Vr 16V2 2 r. = − 2 = Φ= dr R R4 The energy equation now becomes ) ( ( ) μ 16V2 3 d dT r. r =− dr dr k R4 Boundary conditions are dT(0) = 0 or T(0) = finite dr T(R) = Tw . The solution of the energy equation is obtained by Maple 2020: > restart;

(

r𝟐 > vz ≔ 𝟐 V 𝟏 − 𝟐 R ) ( r𝟐 vz ≔ 𝟐 V 𝟏 − 𝟐 R > 𝚽 ≔ (diff(vz , r))𝟐 ; •

𝚽≔

)

•

;

𝟏𝟔 V𝟐 r𝟐 R𝟒

> 𝚽≔

𝟏𝟔 V𝟐 r𝟐 R𝟒

( 𝛍 •r ) •𝚽; > de𝟏 ≔ r•diff(T(r), r, r) + diff(T(r), r) = − k ) ( 𝟐 𝟏𝟔 𝛍 r𝟑 V𝟐 d d T(r) + T(r) = − de𝟏 ≔ r 𝟐 dr k R𝟒 dr > sol ≔ dsolve({de𝟏, T′ (𝟎) = 𝟎, T(R) = Tw }, T(r));

V𝟐 𝛍 + k Tw . k kR This solution can be put in the following form: [ ( )4 ] V2 μ r 1− . T = Tw + k R sol ≔ T(r) = −

𝛍 r𝟒 V𝟐 𝟒

+

Maximum temperature occurs at the tube center, and it is determined by setting r = 0 in the temperature distribution: Tmax = T(0) = Tc = Tw +

μV2 . k

The temperature at r = 0 is V2 μ . k Surface heat flux is obtained as [ ] ( ) V2 μ ( r )3 ( 1 ) 4μV2 dT 4 =k =− . q′′w = k dr r=R k R R r=R R Tc = Tw +

165

166

6 One-Dimensional Solutions in Convective Heat Transfer

The heat transfer coefficient based on temperature difference - Tw − Tc 4μV2 R = 4k h= = Tw − Tc R V2 μ − k and the Nusselt number becomes ( ) 2R h NuD = = 8. k Let us define the heat transfer coefficient based on the temperature difference, Tw − Tm , where Tm is the mean fluid temperature. First, we will define the mean fluid temperature −

q′′

vz TdAc

∫ Tm =

Ac

vz dAc

∫ Ac

R

Tm =

{

∫0

[ [ ( )4 ]]} ( )2 ] [ V2 μ r r 1− 2V 1 − Tw + 2πrdr R k R [ R{ ( )2 ]} r 2V 1 − 2πrdr ∫0 R

Integration is carried out by Maple 2020: 2

5 μV . 6 k The mean fluid temperature is also referred to as the bulk fluid temperature or mixing cup temperature. It is the temperature that characterizes the average convected thermal energy at an axial position in the tube. The heat transfer coefficient based on Tw − Tm is defined as Tm = Tw +

4μV2 R = 24k . h= = 2 Tw − Tm 5R 5 μV − 6 k The Nusselt number is −

q′′w

NuD =

h(2R) hD 48 = = = 9.6. k k 5

Example 6.4 Consider two infinite stationary parallel plates. See Figure 6.E4. Incompressible, constant property fluid fills the space between plates. While the upper plate is at temperature Tw , the lower plate is insulated. Flow is created by the pressure gradient in the axial flow direction. Flow is steady and fully developed. Taking the viscous dissipation into account, determine (i) the velocity distribution, (ii) temperature distribution, (iii) heat flux at upper surface, and (iv) Nusselt number. y

TW

g

H

0 Figure 6.E4

x

Insulated

Coordinate system and problem description for Example 6.4.

6.3 Poiseuille Flow

Solution Assumptions: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Steady flow, 𝜕/𝜕t = 0 Laminar flow Constant fluid properties No edge effects 𝜕/𝜕z = 0 No end effects 𝜕/𝜕x = 0; except for pressure, 𝜕p/𝜕x ≠ 0 Newtonian fluid Pressure gradient in the x-direction is constant No energy generation in the fluid No gravitational effect in the x-direction No fluid motion in the z-direction, w = 0

We start with the steady form of the continuity equation: 𝜕u 𝜕v 𝜕w + + = 0. 𝜕x 𝜕y 𝜕z Using the stated assumptions, we reduce the continuity equation to the following form: dv = 0. dy Integration yields v = f(x) constant. To determine the integration constant f(x), we use no slip boundary condition on the lower plate. Thus, we write v(x, 0) = 0. Solid walls are impermeable. We get f(x) = 0, and the y-component of velocity becomes v(x, y) = 0. This tells us that streamlines are parallel to plates. Consider the y-component of the momentum equation as ) ( 2 ) ( 𝜕p 𝜕v 𝜕v 𝜕v 𝜕 v 𝜕2 v 𝜕2 v 𝜕v +u +v +w = ρfy − +μ . + + ρ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕y 𝜕x2 𝜕y2 𝜕z2 Since v(x, y) = 0 everywhere and fy = − g, the momentum equation reduces to dp = −ρg. dy The integration of this equation yields p = p0 − ρgy. This means that pressure is hydrostatic in the y-direction. Next, we consider the x-component of the momentum equation: ) ( 2 ) ( 𝜕p 𝜕u 𝜕u 𝜕u 𝜕 u 𝜕2u 𝜕2u 𝜕u +u +v +w = ρfx − +μ . + + ρ 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x 𝜕x2 𝜕y2 𝜕z2 We now look at each term of the x-momentum equation: 𝜕u =0 𝜕t

(Steady flow)

u

𝜕u =0 𝜕x

v

𝜕u = 0 since v = 0 𝜕y

(Continuity equation)

𝜕u 𝜕u = 0 since =0 𝜕z 𝜕z

(No edge effects; there is no fluid motion, w = 0)

w

since

𝜕u =0 𝜕x

(No end effects)

167

168

6 One-Dimensional Solutions in Convective Heat Transfer

The momentum equation becomes 1 dp d2 u . = μ dx dy2 Boundary conditions for the momentum equation are given by At y = 0

u=0

(No slip)

At y = H

u=0

(No slip)

The solution will be obtained by Maple 2020: ( ) ( ) 𝟏 • dp > de ≔ u′′ (y) = ; 𝛍 dx dp de ≔ D(2) (u)(y) = μ dx > sol ≔ dsolve({de, u(𝟎) = 𝟎, u(H) = 𝟎}, u(y)); sol ≔ u(y) =

1 dp y2 1 dp H y − 2 μ dx 2 dx μ

This equation can be put in the following form: ) ( 2 ) [( ) ( ) ] ( y y 2 dp H − . u= − dx 2μ H H Let us now define the average fluid velocity H

V=

1 u dy. H ∫0

Integration yields ) ( 2 )( dp H − . V= 12μ dx Thus, we can write velocity as [( ) ( ) ] y y 2 u = 6V . − H H Knowing the velocity distribution, we now consider the energy equation and viscous dissipation: ] [ 2 [ ] 𝜕T 𝜕T 𝜕T 𝜕 T 𝜕2 T 𝜕2 T 𝜕T +u +v +w =k + 2 + 2 + q′′′ + μΦ ρcp 𝜕t 𝜕x 𝜕y 𝜕z 𝜕x2 𝜕y 𝜕z } { )2 ( )2 ( ( )2 ( ) ) ( )2 ( )2 ( 𝜕v 𝜕w 2 𝜕w 𝜕v 𝜕u 𝜕w 2 2 𝜕u 𝜕v 𝜕w 𝜕v 𝜕u 𝜕u + + + + − . + + + + + + Φ=2 𝜕x 𝜕y 𝜕z 𝜕x 𝜕y 𝜕y 𝜕z 𝜕z 𝜕x 3 𝜕x 𝜕y 𝜕z Let us look at each term of the energy equation and the viscous dissipation function: 𝜕T =0 𝜕t 𝜕T =0 𝜕x 𝜕T v = 0 since v = 0 𝜕y w

𝜕T =0 𝜕z

(Steady conditions) (No axial temperature variation; no end effects) (Continuity) since w = 0;

𝜕T = 0 since no edge effects. 𝜕z

𝜕2T =0 𝜕x2

(No axial conduction.)

𝜕T 𝜕2T = 0 since =0 𝜕z 𝜕z2

(No edge effects)

6.3 Poiseuille Flow

Under these assumptions, the energy equation and the dissipation function reduce to the following form: d2 T + μΦ = 0 dy2 ( )2 du Φ= dy k

. Boundary conditions for the energy equation are At y = 0

dT =0 dy

At y = H

T = Tw .

The solution is obtained by Maple 2020: > restart; >

( ) y ( y )𝟐 − ; > u ≔ (𝟔•v)• H H ( ) y𝟐 y u ≔ 𝟔V − 𝟐 H H > 𝚽 ≔ (diff(u, y))𝟐 ; ( )𝟐 𝟐y 𝟏 𝚽 ≔ 𝟑𝟔 V𝟐 − 𝟐 H H (𝛍) ′′ •𝚽; > de ≔ T (y) + k

)𝟐 𝟐y 𝟏 − 𝟑𝟔 𝛍 V H H𝟐 de ≔ D(𝟐) (T)(y) + k > sol ≔ dsolve({de, T′ (𝟎) = 𝟎, T(H) = Tw }, T(y)); 𝟐

(

𝟐 𝟐 𝟑 𝛍 V (H − 𝟐 y)𝟒 𝟔 𝛍 V y − 𝟒 𝟒 Hk kH 𝟐 𝟐𝟕 V 𝛍 + 𝟒 k T 𝟏 w + 𝟒 k The temperature distribution is { [ } ( y )]4 (y) μV2 3 27 − 1−2 + . −6 T − Tw = k 4 H H 4

sol ≔ T(y) = −

Temperature T0 on the insulated surface is obtained by setting y = 0, and it is given as 6 μV2 + Tw . k Temperature distribution can be put in the following dimensionless form: { [ } ( y )]4 (y) μcp T − Tw 3 27 V2 − 1−2 + = −6 Tw − T0 cp (Tw − T0 ) k 4 H H 4 T0 =

T − Tw = Ec Pr T0 − Tw

} { [ ( y )]4 (y) 27 3 + . − 1−2 −6 4 H H 4

On the other hand, Pr =

μcp k

and Ec =

V2 . Let us assume that the upper surface is hot and heat flows from the upper cp (Tw −T0 )

surface into the fluid. Heat flux at the upper surface is given by ( ) 12μV2 dT . =− q′′w = k dy y=H H

169

170

6 One-Dimensional Solutions in Convective Heat Transfer

The mean fluid temperature Tm is H

Tm =

1 uTdy. VH ∫0

The substitution of velocity and temperature distribution yields { [ }} [( ) ( ) ]} { H{ ( y )]4 (y) y 2 y μV2 3 27 1 Tm = − − 1−2 + dy. 6V Tw + −6 VH ∫0 H H k 4 H H 4 Integration gives the mean fluid temperature: 2

129 μV . 35 k The heat transfer coefficient is defined as Tm = Tw +

h=

12μV2 420 k H = = 2 129 H 129 μV − 35 k −

q′′w Tw − Tm

h(2H) hDH 2 × 420 = = = 6.51 k k 129 where DH = 2H is the hydraulic radius. NuDH =

Example 6.5 Consider laminar flow between two infinite parallel plates subjected to uniform wall temperature Tw , as shown in Figure 6.E5. Suppose that we wish to write the energy equation for the following cases assuming HFD flow: a) Thermally developing flow b) Thermally fully developed flow c) Asymptotic thermally developed flow y

TW

g

H

0 Figure 6.E5

x

TW

Coordinate system and problem description for Example 6.5.

Solution The energy equation for developing laminar flow between parallel plates can be simplified to the following form: ) ( 𝜕T 𝜕2 T 𝜕T +v =k 2. ρcp u 𝜕x 𝜕y 𝜕y This form of the energy equation is valid in the thermally developing region. Energy equation for hydrodynamically fully developed region between parallel plates can be simplified to the following form by setting v = 0. ρcp u

𝜕2 T 𝜕T =k 2 𝜕x 𝜕y

6.4 Rotating Flows

where u = 6V 𝜕 𝜕x

(

[( ) y H

T − Tw Tm − Tw

− )

( y )2 ] H

is the velocity distribution. For thermally fully developed region, we have

= 0.

This gives us for uniform wall temperature ( ) )( T − Tw dTm 𝜕T = . 𝜕x Tm − Tw dx The energy equation can be written as [( ) ( ) ] ( ) )( T − Tw dTm y y 2 𝜕2 T 6V − =α 2 H H Tm − Tw dx 𝜕y k is the thermal diffusivity. This form of the energy equation is valid in the thermally fully developed region. ρcp As the axial distance x increases, temperature T approaches wall temperature Tw and dTm /dx approaches zero. This means that we have 𝜕T/𝜕x → 0. Then, the energy equation reduces to

where α =

𝜕2 T = 0. 𝜕y2 This form of the energy equation is valid for asymptotic thermally developed flow associated with uniform wall temperature boundary condition and flow between parallel plates.

6.4 Rotating Flows It is possible to generate angular fluid motion between two concentric cylinders by moving one cylinder while fixing the other one. A typical example is the fluid motion in the clearance space between a journal and its bearing. The streamlines are concentric circles. Example 6.6 Consider incompressible constant property lubrication oil in the clearance between two infinitely long concentric cylinders. See Figure 6.E6. This is also called circular Couette flow and is the basis for Couette rotational-type viscometers. Flow between concentric cylinders is steady and laminar. The radius of the inner cylinder and the outer cylinder is ri and r0 , respectively. The angular rotational velocity of the insulated inner cylinder is ω. The outer cylinder is fixed, and its temperature is T0 . Assuming laminar flow and taking viscous dissipation into consideration, obtain the temperature distribution within lubrication oil. Figure 6.E6

Circular Couette flow.

r

T0

Insulation

θ ri

r0

Solution Assumptions: 1. The concentric cylinders are very long and, for this reason, axial variation in velocity and temperature is negligible, 𝜕/𝜕z = 0

171

172

6 One-Dimensional Solutions in Convective Heat Transfer

2. 3. 4. 5. 6. 7. 8. 9. 10.

Velocity and temperature do not vary with the angular direction, 𝜕/𝜕θ = 0 Lubrication oil is assumed to be Newtonian Steady flow, 𝜕/𝜕t = 0 Laminar flow Axisymmetric flow, 𝜕/𝜕θ = 0 Constant fluid properties No end effects No gravitational effects Continuum

Fluid motion is generated by the rotation of the inner cylinder. Heat is generated by viscous dissipation, and it is removed from the fluid at the outer cylinder. We start with the continuity equation. The continuity equation is 𝜕ρ 1 𝜕 1 𝜕 𝜕 + (ρr vr ) + (ρ vθ ) + (ρvz ) = 0. 𝜕t r 𝜕r r 𝜕θ 𝜕z For constant density 𝜕ρ 𝜕ρ 𝜕ρ 𝜕ρ = = = = 0. 𝜕t 𝜕r 𝜕θ 𝜕z For axisymmetric flow 𝜕vθ = 0. 𝜕θ Since the cylinders are very long, there are no end effects, and axial changes are negligible: 𝜕vz = 0. 𝜕z Under these conditions, the continuity equation reduces to d (r vr ) = 0. dr The integration of this equation yields r vr = C. The unknown constant C is determined using the no slip boundary condition At the surface of the outer cylinder: r = r0

vr = 0.

Thus, we find that C = 0. Then, we see that vr (r, z) = 0. This means that the streamlines are concentric. We now consider the θ-component of the momentum equation. θ-direction: ] [ 𝜕v 𝜕v v 𝜕v vv 𝜕vθ 1 𝜕p + vr θ + θ θ + r θ + vz θ = ρfθ − ρ 𝜕t 𝜕r r 𝜕θ r 𝜕z r 𝜕θ ] [ 2 2 ( ) 𝜕 vθ 1 𝜕 vθ 2 𝜕v 𝜕 1 𝜕 (rv ) + 2 + 2 r + +μ 𝜕r r 𝜕r θ r 𝜕θ2 r 𝜕θ 𝜕z2 . We now consider the terms in this equation: 𝜕vθ =0 (Steady state) 𝜕t fθ = 0

(No body force)

𝜕vθ =0 𝜕θ

(Axisymmetric flow)

𝜕vθ =0 𝜕z

(No end effects)

6.4 Rotating Flows

𝜕vθ = 0 since vr = 0 everywhere. (Continuity) 𝜕r vr v θ = 0 since vr = 0 everywhere (Continuity) r

vr

𝜕 2 vθ

= 0 since

𝜕vθ =0 𝜕θ

𝜕θ 2 𝜕vr = 0 since vr = 0 r2 𝜕θ 2

(Axisymmetric flow) (Continuity)

𝜕vθ 𝜕 2 vθ = 0 since =0 (No end effects) 2 𝜕z 𝜕z Applying the assumptions, we see that the momentum equation simplifies to ( ) d 1 d (rvθ ) = 0. dr r dr The boundary conditions are r = ri

vθ = ω r i .

r = r0

vθ = 0.

The integration of this differential equation yields C C1 r+ 2 2 r where C1 and C2 are constants, and these constants are obtained using boundary conditions. Using Maple 2020 vθ =

C1 = − C2 =

2 ω r2i r20 − r2i

ω r2i r20 r20 − r2i

The final solution is expressed as ] [ [ 2 2] 2 ω r2i r 1 ω ri r0 vθ = + − 2 2 r r20 − r2i r0 − r2i or vθ (r) (r0 ∕ri )2 (ri ∕r) − (r∕ri ) . = ω ri (r0 ∕ri )2 − 1 The next step is to examine the energy equation in cylindrical coordinates: } { ] ( ) [ 𝜕T 1 𝜕2 T 𝜕2 T 𝜕T 𝜕T vθ 𝜕T 𝜕T 1 𝜕 ρcp + vr + + vz =k r + 2 2 + 2 + q′′′ + μΦ 𝜕t 𝜕r r 𝜕θ 𝜕z r 𝜕r 𝜕r r 𝜕θ 𝜕z where viscous dissipation is {( ) ]2 ( )} [ ]2 [ ]2 [ ( ) 𝜕vz 2 𝜕vr 2 1 𝜕vr 𝜕 vθ 1 𝜕vθ vr 1 𝜕vz 𝜕vθ + + + + r + + + Φ=2 𝜕r r 𝜕θ r 𝜕z 𝜕r r r 𝜕θ r 𝜕θ 𝜕z ]2 [ 𝜕vr 𝜕vz + + 𝜕z 𝜕r . We now look at the energy equation and the viscous dissipation function: 𝜕T =0 𝜕t vr

𝜕T =0 𝜕r

(Steady state) vr = 0

(Continuity)

173

174

6 One-Dimensional Solutions in Convective Heat Transfer

vθ 𝜕T 𝜕T = 0 since =0 r 𝜕θ 𝜕θ

(Axisymmetric)

𝜕T 𝜕T = 0 since =0 𝜕z 𝜕z

(No end effects)

1 𝜕2T 𝜕T =0 = 0 since 𝜕θ r2 𝜕θ2

(Axisymmetric)

𝜕2T 𝜕T =0 = 0 since 𝜕z 𝜕z2

(No end effects; no axial conduction)

q′′′ = 0

(No energy generation)

vz

Based on the assumptions, the energy equation is reduced to ( ) dT k d r + μΦ = 0 r dr dr where the viscous dissipation Φ is simplified to [ ] [ ( )] dvθ vθ 2 𝜕 vθ 2 = . Φ= r − 𝜕r r dr r The substitution of velocity into viscous dissipation yields 4ω2 r4i r40 Φ= ( )2 r20 − r2i r4 or

[ Φ=

2 ω r2i 1 − (ri ∕r0 )2

]2 1 . r4

We now substitute viscous dissipation term into the energy equation to get [ ]2 ( ) 2 ω r2i μ dT 1 1 d r =− . r dr dr k 1 − (ri ∕r0 )2 r4 The boundary conditions of this equation are r = ri

dT =0 dr

r = r0

T = T0

Integrating twice yields ]2 [ 2 ω r2i μ 1 + C3 ln r + C4 . T=− 4 k 1 − (ri ∕r0 )2 r2 The unknown constants are obtained using the boundary conditions [ ]2 2 ω r2i μ 1 C3 = − 2 k 1 − (ri ∕r0 )2 r2i [ ]2 [ ] 2 ω r2i μ 2 1 C4 = T0 + + ln (r0 ) 4k 1 − (ri ∕r0 )2 r20 r2i See [1] for further discussion.

Problems

Problems 6.1

Consider laminar flow in the tube. The tube diameter is 4 cm. The velocity and temperature profiles are given as [ )] ( r 2 m∕s vz = 0.2 1 − 0.02 [( ] ) ( ) r 4 3 r 2 1 T = 400 + 140 K − − 0.02 4 0.02 4 where r is in meters. Determine the bulk temperature.

6.2

Consider incompressible fluid with constant properties between two infinite-parallel plates. The upper plate is moving with constant velocity V. The lower plate is stationary. The space between plates is H. The lower plate is at temperature T0 , while the upper plate exchanges heat with ambient by convection. The heat transfer coefficient between the upper plate and ambient is h. The ambient fluid temperature is T∞ . See Figure 6.P2. The velocity profile between the plates is given by (y) u = . V H We wish to determine: (a) The temperature distribution in the fluid by taking the viscous dissipation into consideration (b) Surface temperature of the moving plate (c) Surface heat flux at the moving plate y

h , T∞

V

H

0 Figure 6.P2

6.3

x

T0

Couette flow with convective boundary condition.

Consider incompressible fluid with constant properties between two infinite parallel plates. Space between plates is H. The lower plate is at temperature T0 , while the upper plate is at temperature T1 . See Figure 6.P3. The velocity profile between the plates is given by [( ) ( ) ] y 2 y . − u = 6V H H

Figure 6.P3

Coordinate system and problem description.

y

T1

H

0

x

T0

175

176

6 One-Dimensional Solutions in Convective Heat Transfer

We wish to determine: a) The temperature distribution in the fluid for asymptotic thermally developed flow. b) The heat flux at the lower plate. c) The mean temperature in the fluid. d) Nusselt number. 6.4

Consider the Couette flow problem in Example 6.1. Assume that both plates are kept at temperature T0 . Determine (i) the temperature distribution, (ii) the location of the maximum temperature and its value, and (iii) heat flux at the plates.

6.5

Consider incompressible fluid flow in a tube. Flow in the tube is generated by a constant pressure gradient in the flow direction. The velocity profile in the tube is given by [ ( )2 ] r vz = 2V 1 − . R The pipe wall surface is maintained at a constant uniform heat flux. See Figure 6.P5. The axial temperature variation is neglected. The viscous dissipation term is not negligible. Determine: (a) The temperature distribution. (b) The wall heat flux. r

q0ʺ

Figure 6.P5

6.6

D = 2R

z

0

Coordinate system and problem description.

Consider two infinite parallel plates. The spacing between the plates is H. Incompressible fluid fills the space between plates. The fluid properties are constant except the thermal conductivity k. The thermal conductivity k varies with temperature according to √ k=β T where β is a constant. While the lower plate is stationary, the upper plate is moving with a constant velocity U in the positive x-direction. Both the plates are at temperature T0 . See Figure 6.P6. The viscous dissipation is assumed to be important. Neglect gravity effect and pressure variation in the fluid. Determine temperature distribution. y

U

T0

g

H

O Figure 6.P6

6.7

x

T0

Heat transfer for laminar flow with variable thermal conductivity.

Consider incompressible fluid with constant properties between two infinite parallel plates. The upper plate is moving constant velocity V. The lower plate is stationary and insulated. The space between plates is H. See Figure 6.P7.

Problems

The velocity profile between the plates is given by (y) u = . V H At large distances from the inlet, the flow is asymptotically thermally fully developed. We wish to determine: (a) The temperature distribution in the fluid by taking the viscous dissipation into consideration. (b) Surface temperature at the stationary plate. y

T0

V

H

x

0 Figure 6.P7

6.8

Insulation

Coordinate system and problem description for Couette flow.

Consider steady incompressible constant property Newtonian viscous fluid between infinitely long concentric pipes. See Figure 6.P8. The flow is induced by rotation of pipes and there is no motion in the z-direction. Relevant parameters of the problem are shown in the figure. Viscous dissipation is important. There are no body forces in the flow. We wish to obtain the velocity and temperature distributions in the flow.

Figure 6.P8

T0, ω0

Heat transfer in flow between rotating concentric pipes.

r0 Ti, ω ri

6.9

vi v0

Consider incompressible fluid flow in a pipe. The pipe radius is R and the heat transfer coefficient between the external surface of the pipe and the ambient is h. The ambient fluid temperature is kept at constant temperature T∞ . See Figure 6.P9. The fluid motion in the pipe is created by an axial pressure gradient 𝜕p/𝜕z. Viscous dissipation is not negligible, and flow is axisymmetric and laminar. Axial temperature variation and end effects are neglected. a) Make the necessary assumptions to reduce the energy and momentum equations b) Write the differential equation of motion and energy c) Specify the boundary conditions d) Obtain the velocity distribution e) Obtain the volume flow rate r

2R

Figure 6.P9

O

h, T∞

z

Heat transfer in pipe flow with convective boundary condition.

R 0

h, T∞

177

178

6 One-Dimensional Solutions in Convective Heat Transfer

f) Obtain the temperature distribution g) Obtain the pipe surface temperature h) Obtain the heat flux at the pipe wall 6.10

Consider the flow of liquid film over an inclined plate. The thickness of the film is H and liquid film flows down the inclined plane due to gravity. The liquid film is subjected to uniform heat flux q′′0 . The inclined surface is kept at temperature Tw . See Figure 6.P10. Assume laminar incompressible flow. The end effects, internal energy generation, and axial variation of temperature are neglected. The velocity distribution is given by [ ] ρ g H2 sin θ y y2 − . u= μ H 2H2 The viscous dissipation term is not negligible. Based on the given information: a) State the assumptions. b) Determine the temperature distribution. y q0ʺ u(y) g

0

H

x θ

Figure 6.P10

6.11

Heat transfer in liquid flow over an inclined plane with uniform heat flux boundary condition.

Consider the flow of incompressible liquid film over an inclined plate. The thickness of the film is H and the liquid film flows down the inclined plane due to gravity. The free surface of the liquid film is at temperature T0 , and the inclined surface is at temperature Tw . See Figure 6.P11. Assume laminar incompressible flow. The end effects and axial variation of velocity and temperature are neglected. The velocity distribution is given by ] [ ρ g H2 sin θ y y2 . − u= μ H 2H2 The viscous dissipation term is not negligible, and based on given information: a) State the assumptions b) Determine the temperature distribution. y T0 u(y) g

0

x

H

Tw θ

Figure 6.P11

Heat transfer in liquid flow over an inclined plane with constant temperature boundary condition.

Problems

c) Mixing cup temperature at any point along x location is defined as the temperature obtained if all the fluid issuing at x was collected in a container and mixed completely. Thus, mixing cup temperature is defined as ∫ ρcp u T dAc Ac

Tm = ṁ =

̇ p mc

∫

ρ u dAc .

Ac

The mixing cup temperature for this problem becomes H

Tm =

∫H u T dy H

∫H u dy

.

Obtain an expression for the mixing cup temperature. 6.12

Consider Example 6.1. Nondimensionalize the momentum and energy equations along with their boundary conditions. Use the following parameters: θ=

6.13

T − T0 T1 − T0

u∗ =

u U

η=

y . L

Consider Couette flow between parallel plates. See Figure 6.P13. Lower plate at y = 0 is stationary and upper plate at y = H is moving with velocity U. y

U

H

O Figure 6.P13

x

Stationary

Couette flow with constant pressure gradient.

a) Define a dimensionless pressure gradient parameter β ( )( 2) 1 dp H β=− 2μ dx U Plot dimensionless velocity u/U as a function of dimensionless distance y/H for different values of pressure parameter β. b) Obtain an expression for the Fanning friction factor for lower plate as well as upper plate. 6.14

Consider Couette flow between parallel plates. See Figure 6.P14. Lower plate at y = 0 is stationary and upper plate at y = H is moving with velocity U = 110 m/s. Fluid between the plates is air. For air, k = 0.0257 W/m. K and μ = 1.914 × 10−5 N. s/m2 . Including viscous dissipation effects, determine (a) the maximum temperature rise in the fluid (b) heat flux at the plate surface.

179

180

6 One-Dimensional Solutions in Convective Heat Transfer

y

U

T0 = 17 °C

H = 1.0 mm

x

O Figure 6.P14

6.15

Stationary

T0 = 17 °C

Heat transfer in Couette flow.

Consider laminar flow between two infinite parallel plates subjected to uniform wall temperatures T1 and T2 . See Figure 6.P15. a) Write down the energy equation and its boundary conditions for the thermally fully developed (TFD) region. b) Write down the energy equation and its boundary conditions for asymptotically fully developed region. Develop a solution for temperature distribution for asymptotic thermally developed flow. Additionally, we wish to obtain wall heat flux at each plate and Nusselt number. y

T2

g

H x

0 Figure 6.P15

6.16

T1

Coordinate system and problem description.

Consider Couette flow between parallel plates. See Figure 6.P16. Lower plate at y = 0 is stationary and upper plate at y = H is moving with velocity V = 30 m/s. The viscous fluid between the plates is lubricating oil. For lubricating oil, k = 0.138 W/m.K and μ = 0.03195 Pa.s. Lower plate is kept at 80 ∘ C and the upper plate is insulated. Determine (a) the temperature of the upper plate and (b) heat flux at the stationary plate. Insulated

y

V = 30 m/s

H = 1.0 mm

O Figure 6.P16

x

T0 = 80 °C

Stationary

Heat transfer in Couette flow.

References 1 2 3 4

Jiji, L.M. (2009). Heat Convection. Springer. Arpaci, V.S. and Larsen, P.S. (1984). Convection Heat Transfer. Prentice Hall Inc. Burmeister, L.C. (1993). Convective Heat Transfer, 2e. Wiley. Kays, W.M., Crawford, M.E., and Weigend, B. (2005). Convective Heat and Mass Transfer, 4e. McGraw Hill.

References

White, F.M. and Majdalani, J. (2021). Viscous Fluid Flow, 4e. McGraw Hill. Arpaci, V.S., Selamet, A., and Kao, S.H. (2000). Introduction to Heat Transfer. Prentice Hall. Ghiaasiaan, S.M. (2018). Convective Heat and Mass Transfer, 2e. CRC Press. Bird, R.B., Steward, W.E., Lightfoot, E.N. et al. (2015). Introductory Transport Phenomena. Wiley. Pai, S.I. (1956). Viscous Flow Theory I-Laminar Flow. Van Nostrand. Bennett, T.D. (2013). Transport by Advection and Diffusion. Wiley. Middleman, S. (1998). An introduction to Mass and Heat Transfer. Wiley. Cebeci, T. (2002). Convective Heat Transfer, 2e. Springer. Magrab, E.B., Azarm, S., Balachandran, B., Duncan, J.H., Herold, K.E. and Walsh, G.C. (2011). An Engineer’s Guide to MATLAB, Pearson Education, Inc. 14 Meade, D.B., Michael May, S.J., Cheung, C.K. and Keough, G.E. (2009). Getting Started with MAPLE, John Wiley and Sons, Inc. 15 Gilbert, R.P., Hsiao, G.C. and Ronkese, R.J. (2021). Differential Equations-A Maple Supplement 2e, CRC Press. 5 6 7 8 9 10 11 12 13

181

183

7 Laminar External Boundary Layers: Momentum and Heat Transfer 7.1 Introduction In this section, the search for a similarity method is the main concern. With the similarity variable, the partial differential equation of the physical problem is expected to be transformed into an ordinary differential equation. After the original work of Blasius [1] on the boundary layer problem of flat plate, Goldstein [2] studied the problem in depth. Following a different approach, Sedov [3] developed a method to find a similarity variable using dimensional analysis. The similarity method is discussed in several books such as [4–8]. Later, Arpaci and Larsen [9] presented a method to search for a similarity variable. Their method is based on dimensional analysis, and this method will be used in this book as much as possible. Details of the method are discussed by Arpaci and Larsen, and a brief outline will be given here: (1) The dependent and independent variables are made dimensionless in terms of characteristic properties. If there is no characteristic property, arbitrarily selected reference quantities are used. (2) All the arbitrarily selected reference quantities are eliminated successively by employing the mathematical and physical principles. The mathematical principle states the invariance of the number of dependent and independent variables of a mathematical expression under any transformation. On the other hand, the physical principle concerns with the dimensional homogeneity of a physical expression. The idea for the similarity solutions is to transform a partial differential equation to an ordinary differential equation by combining the variables in a special way. Transformation by a similarity variable is successful if the similarity variable remains as the only independent variable. Transformation reduces the number of independent variables by one, and for this reason, two of the original initial boundary conditions reduce to one. When fluid flows over a surface, a boundary layer grows in a thin region near the wall shown in Figure 7.1, and the effect of fluid viscosity is confined to this thin region. This notion is developed by L. Prandtl in 1904. The momentum boundary layer thickness δ(x) is defined as the distance from the wall surface where the velocity u(x, y) reaches 99% of the free stream velocity U∞ . The only physically significant length scale is the boundary length thickness δ(x) in the direction normal to wall surface. On the other hand, the streamwise length scale L for advection is much larger than the characteristic length δ(x) in the direction normal to wall surface. Since δ ≪ L, the diffusion across the boundary layer is much larger than the axial diffusion. In forced convection, the boundary layer thickness affects the imposed external flow over a flat surface minimally.

7.2 Velocity Boundary Layer over a Semi-Infinite Flat Plate: Similarity Solution Consider viscous incompressible constant property parallel flow over a sharp-edged thin smooth semi-infinite flat plate. This is the classical problem of flow over a semi-infinite flat plate set at zero angle of incidence to a uniform stream of velocity U∞ . See Figure 7.1. The upstream velocity is uniform and parallel to the flat plate. We assume that the plate is thin enough not to disturb the uniform exterior flow so that fluid flows at a constant velocity of U∞ over a flat plate. For constant U∞ = constant, −(1/ρ)(𝜕p/𝜕x) = 0. Flow is assumed to be steady and two-dimensional along with constant fluid properties.

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

184

7 Laminar External Boundary Layers: Momentum and Heat Transfer

U∞

y U∞

δ(x) 0 L

Figure 7.1

X

Steady boundary layer flow over a flat plate.

At the edge of the boundary layer, the velocity is U∞ . The Reynolds number is large enough so that the boundary layer assumptions are valid. Based on these assumptions, the governing equations are given as follows: Continuity 𝜕u 𝜕v + =0 𝜕x 𝜕y

(7.1)

x-Momentum u

𝜕u 𝜕2 u 𝜕u +v =ν 2. 𝜕x 𝜕y 𝜕y

(7.2)

Boundary conditions for the momentum equation are u(x, 0) = 0 (No slip)

(7.3a)

v(x, 0) = 0 (Solid wall)

(7.3b)

u(x, ∞) → U∞ .

(7.3c)

The boundary condition at the leading edge of the plate is specified as follows: u(0, y) → U∞ .

(7.3d)

This problem was solved by the Blasius as discussed in [5] in 1908 by the similarity method. The problem under consideration has no characteristic length in any direction. Velocity profiles in the boundary layer are expected to be similar at all the values of x starting from the leading edge of the plate, as shown in Figure 7.2. That is, u(x1 , y) at location 1 is similar to u(x2 , y) at location 2, assuming that the y-coordinate is stretched by the change in boundary layer thickness δ(x). Thus, we can state that despite the growth of the boundary layer with distance x from the leading edge, the shapes of velocity profiles are geometrically similar at different x-locations along the plate differing only by a stretching factor in the y-direction. We assume that the x-component of velocity u has a similar profile at all x positions along the flat plate for any fluid and for any free stream velocity U∞ . It is very convenient to discuss the velocity profile within the boundary layer in terms of dimensionless variables, as shown in Figure 7.3. Distance y is normalized with local boundary layer thickness δ(x), while the local velocity component u is normalized with free stream velocity U∞ . We can conclude that the dimensionless velocity profile shown in Figure 7.3 is the same for all laminar flows along the flat plate for all positions on the flat plate. The similarity of profiles implies that a unique function exists such that u = f1 (y∕δ) (7.4) U∞ U∞ U∞

y

δ(x)

U∞

δ(x) x

Figure 7.2

(1)

Similar cubic velocity profiles.

(2)

7.2 Velocity Boundary Layer over a Semi-Infinite Flat Plate: Similarity Solution

Figure 7.3

Dimensionless velocity profile.

1 y δ

0

1 u U∞

where δ(x) is the local boundary layer thickness. Except very close to the wall, the velocity u is of the order of the free stream velocity U∞ and the distance y measured from the plate wall is of the order of boundary layer thickness δ(x). Thus, u ∼ U∞

(7.5)

y ∼ δ.

(7.6)

Using Eqs. (7.5) and (7.6), the continuity equation becomes U∞ v + ∼0 x δ

(7.7a)

or δU∞ . x Again, using these order of magnitude terms in the momentum equation, we get ( )( ) ( ) U∞ U∞ δ U∞ U∞ U∞ + ≈ν x x δ δ2 v∼

or

√ δ∼

νx . U∞

Equation (7.8b) can be expressed in dimensionless form √ ν 1 δ ∼ = √ x U∞ x Rex where Rex = U∞ x/ν is the local Reynolds number. Thus, the similarity principle becomes ) ( √ U∞ u = f2 y U∞ νx

(7.7b)

(7.8a)

(7.8b)

(7.8c)

(7.9)

where the term in parentheses is a dimensionless coordinate. This fact is consistent with the experimental measurements made with laminar boundary layer flows. Let us obtain the similarity variable for the problem under consideration by the method described in [9]. Choosing arbitrarily selected reference lengths x0 and y0 , the inherent characteristic velocity U∞ , and the arbitrarily selected reference velocity V0 , we define the following dimensionless quantities: x (7.10a) x∗ = x0 y y∗ = (7.10b) y0 u (7.10c) u∗ = U∞ v v∗ = . (7.10d) V0

185

186

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Using Eqs. (7.10a)–(7.10d), we can nondimensionalize continuity and momentum equations. With nondimensionalization, the continuity and momentum equations now become ( ) U∞ y0 𝜕u∗ 𝜕v∗ + =0 (7.11a) V0 x0 𝜕x∗ 𝜕y∗ ) ( ) ( ∗ ∗ U∞ y20 V 0 y0 𝜕 2 u∗ ∗ 𝜕u ∗ 𝜕v v + = . (7.11b) u ν x0 𝜕x∗ ν 𝜕y∗ 𝜕y∗2 This implies that ( ) 2 u x y V0 y0 U∞ y0 U∞ y0 , = f1 , , , U∞ x 0 y0 ν V0 x0 ν x0 ( ) 2 x y V0 y0 U∞ y0 U∞ y0 v , = g1 , , , . V0 x 0 y0 ν V0 x0 ν x0

(7.12a)

(7.12b)

The reference quantities x0 , y0 , and V0 are not characteristic properties. Thus, we can eliminate these quantities one by one. The mathematical principle permits us to transform the variables in any way we need, but we are not permitted to change the number of variables. Let us start with V0 . We transform Eqs. (7.12a) and (7.12b) such that only one term remains depending on V0 . We will introduce a new parameter in place of U∞ y0 /V0 x0 . This parameter is obtained multiplying U∞ y0 /V0 x0 by V0 y0 /ν; thus, ) ) ( )( ( U∞ y20 V0 y0 U ∞ y0 = V0 x0 ν ν x0 but Eqs. (7.12a) and (7.12b) contain this parameter. Therefore, we have the following mathematical expression: ( ) 2 u x y V0 y0 U∞ y0 , . = f2 , , U∞ x 0 y0 ν ν x0 Next, we multiply the left-hand side of Eq. (7.12b) with ( )( ) vy V0 y0 v = 0. V0 ν ν

V0 y0 . ν

(7.13a)

Thus, we get

Therefore, Eq. (7.12b) becomes ( ) 2 v y0 x y V0 y0 U∞ y0 = g2 , , , . ν x 0 y0 ν ν x0 Equations (7.13a) and (7.13b) have physical significance in u and v if they are independent of V0 . Thus, ( ) 2 u x y U ∞ y0 = f2 , , U∞ x 0 y0 ν x 0 ( ) 2 v y0 x y U ∞ y0 = g2 . , , ν x 0 y0 ν x 0

(7.13b)

(7.14a)

(7.14b)

We now eliminate x0 . This done by multiplying U∞ y20 ∕ν x0 by x0 /x. The result of multiplication is U∞ y20 U∞ y20 x0 = . ν x0 x νx Then, Eqs. (7.14a) and (7.14b) become ) ( 2 u x y U ∞ y0 = f3 , , U∞ x 0 y0 νx ) ( 2 v y0 x y U ∞ y0 = g3 . , , ν x 0 y0 νx

(7.14c)

(7.14d)

7.2 Velocity Boundary Layer over a Semi-Infinite Flat Plate: Similarity Solution

According to the physical principle, Eqs. (7.14c) and (7.14d) assume significance when the left- and right-hand sides of these equations are dimensionally homogeneous. There is no characteristic length in the x-direction, and consequently, the right-hand side of Eqs. (7.14c) and (7.14d) are independent of x0 . Then, we have ) ( y U∞ y20 u (7.15a) = f3 , U∞ y0 νx ) ( v y0 y U∞ y20 = g3 . (7.15b) , ν y0 νx Next, we eliminate y0 . This is carried out by multiplying U∞ y20 ∕νx by y2 ∕y20 in Eqs. (7.15a) and (7.15b) and multiplying v y0 /ν by y/y0 on the left-hand side of Eq. (7.15b). The result is ) ( y U ∞ y2 u (7.15c) = f4 , U∞ y0 νx ) ( y U ∞ y2 ν . (7.15d) v = g4 , y y0 νx According to the physical principle, Eqs. (7.15c) and (7.15d) assume significance only when the left- and right-hand sides are dimensionally homogeneous. There is no characteristic length in y-direction. Consequently, the right-hand side of Eqs. (7.15c) and (7.15d) must be independent of y0 . The result of the elimination process is ) ( U∞ y2 u (7.16a) = f4 U∞ νx ) ( U ∞ y2 ν . (7.16b) v = g4 y νx Equation (7.16a) can be rearranged as u = f5 (η) U∞

(7.17a)

where η is the similarity variable and is given below η = y∕(νx∕U∞ )1∕2 .

(7.17b)

Notice that the similarity parameter contains both x and y, and this variable may be used to transform the partial differential equation into an ordinary differential equation in one independent variable. Note that it is much simpler to solve an ordinary differential equation than a partial differential equation. Let us rearrange the similarity variable η as follows: ( )1∕2 1 ν U∞ ν = . (7.17c) y η x Then, the v component of velocity becomes ( )1∕2 1 ν U∞ g4 (η). v= η x Using g5 = g4 /η, we have ) ( ν U∞ 1∕2 g5 (η). v= x

(7.17d)

(7.17e)

We take 𝜕u/𝜕x and 𝜕v/𝜕y df 𝜕η 𝜕 𝜕u = [U∞ f5 (η)] = U∞ 5 𝜕x 𝜕x dη 𝜕x df5 [ η ] = U∞ − dη 2x

(7.18)

187

188

7 Laminar External Boundary Layers: Momentum and Heat Transfer

𝜕 𝜕v = 𝜕y 𝜕y =

[√

] √ ν U∞ ν U∞ dg5 𝜕η g5 (η) = x x dη 𝜕y

U∞ dg5 . x dη

(7.19)

Next, we substitute Eqs. (7.18) and (7.19) into the continuity equation df [ η ] U∞ dg5 + =0 U∞ 5 − dη 2x x dη and we obtain the following mathematical expression: dg5 η df5 = . dη 2 dη

(7.20)

This tells us that dependent variables f5 and g5 are related to each other according to this relation. The integration of the above differential equation yields within a constant g5 =

1 η df5 . 2∫

(7.21)

Using the integration by parts yields ( ) 1 η f5 − f dη . g5 = ∫ 5 2

(7.22)

Now, using a new dependent variable f = ∫ f5 dη and from Eq. (7.17a), we have df u . = U∞ dη

(7.23)

We also obtain the following relation using Eqs. (7.22) and (7.17e): ) ( √ df 1 v ν −f η = U∞ 2 x U∞ dη where the similarity variable is given as y . η= √ νx∕U∞

(7.24)

(7.25)

Equations (7.23) and (7.24) are in the form originally obtained by Blasius. We now introduce these equations into the momentum equation. First, we will determine the derivative 𝜕u du 𝜕η = . 𝜕x dη 𝜕x Now, using the expression for

u and the similarity variable η, we have U∞

) df d2 f U∞ = U∞ 2 dη dη ( √ ) √ U∞ y U∞ 1 η 𝜕η 𝜕 = y =− =− . 𝜕x 𝜕x νx 2 νx x 2x

d du = dη dη

(

(7.26)

(7.27)

(7.28)

Then, we obtain U d2 f d2 f η 𝜕u = −U∞ 2 = − ∞ η 2. 𝜕x 2x dη dη 2x 𝜕2 u 𝜕u and 2 in terms of η 𝜕y 𝜕y ( √ ) √ ( ) U∞ U∞ d2 f d 𝜕u du 𝜕η df 𝜕 = = U∞ y = U∞ 𝜕y dη 𝜕y dη dη 𝜕y νx νx dη2

(7.29)

Next, we will express

(7.30)

7.2 Velocity Boundary Layer over a Semi-Infinite Flat Plate: Similarity Solution

) ( √ U∞ d2 f 𝜕η d = U∞ U∞ dη νx dη2 𝜕y ) ( √ ( √ ) U 2 d3 f U∞ d2 f U∞ d 𝜕 U∞ y = ∞ = . 2 dη νx dη 𝜕y νx νx dη3

𝜕 𝜕2 u = 𝜕y 𝜕y2

(

√

U∞ d2 f νx dη2

)

After substituting Eqs. (7.23), (7.24), and (7.29)–(7.31) into the momentum equation, we obtain )] (√ ( ( )] [ ) [ √ U∞ d2 f U∞ df 1 ν U∞ df d2 f − η η −f × U∞ 2 + U∞ dη 2 x dη2 2 x dη xν dη [ ) 3 ] ( U∞ d f = ν U∞ . x ν dη3

(7.31)

(7.32)

Finally, after simplification, the celebrated Blasius equation is obtained along with boundary conditions d3 f 1 d2 f + f = 0. dη3 2 dη2

(7.33)

Equation (7.33) is a third-order nonlinear ordinary differential equation. The boundary conditions of Eq. (7.33) are η=0

df =0 dη

(7.34a)

η = 0,

f=0

(7.34b)

η = ∞,

df = 1. dη

(7.34c)

The leading edge is a mathematical discontinuity in u. There is no exact analytical solution of this equation. Blasius solved the problem in 1908 by using a power series for small η and an asymptotic expansion for large η. Oosthuizen and Naylor [10] presented numerical solutions. A rather exactly accurate tabulation of functions f, df/dη, and d2 f/dη2 is given by Howard [11] in 1938. We will use MATLAB 2021a and Maple 2020 to solve Eq. (7.33). MATLAB provides a convenient and easy to use routine. This routine is bvp4c and bvp5c. Both bvp4c and bvp5c can solve fairly complicated problems. We will now present a numerical solution of Eq. (7.33) using Maple 2020. The results of calculations are > restart; > #Laminar External Boundary Layer #Solution of Eq.(𝟕.𝟑𝟑).Velocity Boundary layer over a flat plate. Pohlhausen Similarity Solution. y > #𝜼 is the dimensionless similarity variable and 𝜼 = √ (v x) U∞ ) ( df u = ># U∞ d𝜼 >

> Digits ≔ 𝟏𝟔; Digits ≔ 𝟏𝟔 > interface (rtablesize = 𝟐𝟎𝟎); 𝟐𝟎𝟎 > interface (displayprecision = 𝟔); 𝟔 > 𝛃 ≔ seq(𝛈, 𝛈 = 𝟎..𝟓, 𝟎.𝟐) ∶ > de ≔ diff( f(𝛈), 𝛈, 𝛈, 𝛈) + 𝟎.𝟓 ⋅ f(𝛈) ⋅ diff( f(𝛈), 𝛈, 𝛈); ( 𝟐 ) d𝟑 d f(𝛈) + 𝟎.𝟓𝟎𝟎𝟎𝟎𝟎 f(𝛈) f(𝛈) de ≔ d𝛈𝟑 d𝛈𝟐

189

190

7 Laminar External Boundary Layers: Momentum and Heat Transfer

> #We take 𝜼 = 10 instead of 𝜼 = ∞; > G ∶= dsolve({de, f(𝟎) = 𝟎, f′ (𝟎) = 𝟎, f′ (𝟏𝟎) = 𝟏}, type = numeric, f(𝛈), output = Array([𝛃])) ∶ > F ≔ G(𝟐); 7.2.0.1

x-Component of Velocity –

u U∞

In terms data given in Table 7.1, the variation of longitudinal velocity (df/dη) = u/U∞ as a function of η is shown in Figure.7.4. The numerical solution is compared with data provided by Junkhan and Serovy [12, 13]. The close agreement between theory and experiment is apparent 7.2.0.2

Boundary Layer Thickness 𝛅(x)

Boundary layer thickness δ(x) is defined as the distance y from the plate where the velocity ratio u/U∞ = 0.99. In other words, the boundary layer thickness δ(x) is the location where √ the velocity u is 99% of free stream velocity U∞ . From Table 7.1, we see that u/U∞ = 0.99 when η ≅ 5.0, and since η ≡ y U∞ ∕νx, we have √ U∞ . 5.0 ≅ δ(x) νx Table 7.1

Velocity distribution in laminar boundary layer.

𝛈

f

df/d𝛈

d2 f/d𝛈2

0.000000

0.000000

0.000000

0.332057

0.200000

0.006641

0.066408

0.331984

0.400000

0.026560

0.132764

0.331470

0.600000

0.059735

0.198937

0.330079

0.800000

0.106108

0.264709

0.327389

1.000000

0.165572

0.329780

0.323007

1.200000

0.237949

0.393776

0.316589

1.400000

0.322982

0.456262

0.307865

1.600000

0.420321

0.516757

0.296663

1.800000

0.529518

0.574758

0.282931

2.000000

0.650025

0.629766

0.266752

2.200000

0.781193

0.681310

0.248351

2.400000

0.922290

0.728982

0.228092

2.600000

1.072506

0.772455

0.206455

2.800000

1.230977

0.811510

0.184007

3.000000

1.396808

0.846045

0.161360

3.200000

1.569095

0.876082

0.139128

3.400000

1.746950

0.901761

0.117876

3.600000

1.929525

0.923330

0.098086

3.800000

2.116030

0.941118

0.080126

4.000000

2.305747

0.955518

0.064234

4.200000

2.498040

0.966957

0.050520

4.400000

2.692361

0.975871

0.038973

4.600000

2.888248

0.982683

0.029484

4.800000

3.085321

0.987789

0.021871

5.000000

3.283274

0.991542

0.015907

7.2 Velocity Boundary Layer over a Semi-Infinite Flat Plate: Similarity Solution

1

0.8

circle Re = 48 000 cold

0.6

square Re = 48 000 hot

u/U∞ 0.4

0.2 Blasius profile 0 0

Figure 7.4

1

2

3

η

4

5

6

Comparison of longitudinal velocity profile with experimental data of Junkhan and Serovy. Source: [12, 13].

We can now express the boundary layer thickness δ(x) in the following form: δ(x) 5 = √ x Rex

(7.35)

where Rex is the local Reynolds number and is defined as Rex = U∞ x/ν. Boundary layer thickness δ(x) is useful in gauging the influence of viscous diffusion. 7.2.0.3 Wall Shear Stress 𝛕w

The Blasius solution may also be used to calculate the drag on a flat plate. We can evaluate the wall shear stress as follows: √ ( ) U∞ d2 f || 𝜕u = μ U∞ (7.36a) τw (x) = μ 𝜕y y=0 νx dη2 ||η=0 ′′

where the gradient of u is obtained from Eq. (7.30). From Table 7.1, we get f (0) = 0.332, and consequently, Eq. (7.36a) becomes √ U∞ τw (x) = 0.33206 μ U∞ (7.36b) νx or τw (x) =

0.33206 μU∞ √ Rex x

(7.36c)

where Rex = U∞ x/ν is the local Reynolds number. 7.2.0.4 Local Skin Friction Coefficient cf (x)

We can obtain the dimensionless local skin friction coefficient as given below [ ] 0.33206 μU∞ √ 2 Re x τw 2τw x 0.664 = = √ . = cf = 2 2 2 ρU∞ ∕2 ρU∞ ρU∞ Rex

(7.37)

191

192

7 Laminar External Boundary Layers: Momentum and Heat Transfer

7.2.0.5

Drag Force D

The drag force D on one side of the plate assuming that the solution is valid from the leading edge is obtained as follows. Assuming a depth of b units into paper, the drag force is obtained using Eq. (7.36a) √ L L U∞ d2 f || dx τw (x)(b dx) = μ b U∞ D= √ ∫x=0 ν dη2 ||η=0 ∫x=0 x √ √ U∞ d2 f || 2 L (7.38a) = μU∞ | 2 ν dη |η=0 or √ D = 0.664μ b U∞ ReL (7.38b) where ReL = ρU∞ L/μ is the mean Reynolds number and b is the width of the plate. 7.2.0.6

Average Skin Friction Coefficient cf

It is customary to express drag data in terms of a mean drag coefficient cf . The mean skin friction coefficient cf is defined as √ 0.664μ b U∞ ReL D cf ≡ = ρU∞ 2 A∕2 ρU2∞ bL∕2 or 1.328 cf = √ (7.39) ReL where ReL = U∞ L/ν is the mean Reynolds number based on the entire plate length L. In other words, we first find the total shear force experienced by the surface of axial length L L

Lτw =

∫0

τw (x) dx

where τw is the average shear stress. The average skin friction coefficient cf is cf = 7.2.0.7

τw ρU2∞ ∕2

.

Displacement Thickness 𝛅1 (x)

] ∞[ δ1 (x) df 1 1− = √ dη x dη Rex ∫0

(7.40)

and for the flow being analyzed using integration command trapz in MATLAB 2021a trapz(eta, 𝟏-fp) ans = 𝟏.𝟕𝟐𝟐𝟏 Therefore, the displacement thickness δ1 (x) is δ1 (x) 1.7221 = √ x Rex 7.2.0.8

(7.41)

Momentum Thickness 𝛅2 (x) ∞ δ2 (x) 1 = √ x Rex ∫0

(

df dη

)[

] df 1− dη dη

(7.42)

7.2 Velocity Boundary Layer over a Semi-Infinite Flat Plate: Similarity Solution

and for the flow being analyzed using integration command trapz in MATLAB 2021a >> F = fp.∗ (𝟏-fp); >> trapz(eta, F) ans = 𝟎.𝟔𝟔𝟐𝟎 Momentum thickness δ2 (x) can be given as δ2 (x) 0.6620 = √ . x Rex

(7.43)

Example 7.1 Air is flowing over a thin flat plate at atmospheric pressure at 20 ∘ C and at a free stream velocity of 3.5 m/s. The plate length in the flow direction is 30 cm and width of the plate is 100 cm. At 10 cm from the leading edge of the plate, determine: (a) (b) (c) (d) (e) (f)

boundary layer thickness displacement thickness momentum thickness local wall shear stress local friction coefficient plot the variation of x-component of velocity as a function y

Solution At 20 ∘ C, fluid properties are: ν = 15.26 × 10−6 m2 ∕s Reynolds number: U∞ L (3.5)(0.3) = ≈ 68807. ν 15.26 × 10−6 Flow is laminar over the entire plate. At x = 0.1 m U x (3.5)(0.1) Rex = ∞ = ≈ 22936. ν 15.26 × 10−6 (a) Boundary layer thickness δ ReL =

5 (0.1) 5x = 0.003301 m = √ δ= √ Rex 22 936 (b) Displacement thickness- δ1 (x) 1.7221x (1.7221)(0.1) = √ ≈ 0.001137 m δ1 (x) = √ Rex 22936 (c) Momentum thickness 0.6620 x (0.6620) (0.1) = √ = 0.0004371 m δ2 (x) = √ Rex 222936 (d) Local skin friction coefficient 0.664 0.664 = 0.004384 m cf = √ = √ Rex 22936

193

194

7 Laminar External Boundary Layers: Momentum and Heat Transfer ′

(e) Using Table 7.1, the variation of f (η) with η, a table will be constructed as follows: √

𝛎x U∞

u df = U∞ d𝛈

u = U∞

0

0

0

0.00033015

0.1659

0.5806

1.0

0.0006603

0.3297

1.154

1.5

0.00099045

0.4868

1.704

2.0

0.0013206

0.6297

2.204

2.5

0.0016508

0.7514

2.630

3.0

0.0019809

0.8460

2.961

3.5

0.0023111

0.9131

3.196

4.0

0.0026412

0.9554

3.344

4.5

0.0029714

0.9794

3.428

5.0

0.0033015

0.9914

3.470

𝛈

y=𝛈

0 0.5

df d𝛈

The y values are calculated using the relation √ νx = 0.0006603 η. y=η U∞ The variation of u as a function y is plotted as in Figure 7.E1. Figure 7.E1 distance y.

0.003

Variation of longitudinal velocity u as a function of vertical

0.002 y 0.001

0 0

1

2

3

u

Example 7.2 Air flows at a velocity of 6 m/s over a plate that has a length of 0.2 m in the direction of the flow. The air ahead of the plate has a temperature of 300 K. The surface of the plate is also at 300 K. Velocity within the hydrodynamic boundary layer is measured by hotwire anemometer. The x-component of velocity u = 3 m/s at x = 0.1 m measured from the leading edge of the plate. Calculate the y-component v of the velocity and calculate the position of the y-component of velocity with respect to the plate. Solution At 300 K, ν = 15.89 × 10−6 m2 ∕s, ReL =

ρ = 1.1614 kg∕m3

ρU∞ L U∞ L (6)(0.2) = = = 75519. μ ν 15.89 × 10−6

7.3 Momentum Transfer over a Wedge (Falkner–Skan Wedge Flow): Similarity Solution

Flow is laminar since ReL < 500 000. 3 u df u = = 0.5 and for = 0.5 from the following solution: = U∞ 6 dη U∞ f

df d𝛈

1.4000

0.3229

0.4562

1.600

0.4203

0.5167

𝛈

Using interpolation for

df = 0.5, dη

η ≈ 1.5447 and f ≈ 0.3934 √ √ (15.89 × 10−6 )(0.1) νx ⇒y=η ⇒ y = 1.5447 ⇒ y ≈ 0.000794 m η= √ U∞ (6) νx U∞ √ √ 1 (15.89 × 10−6 )(6) 1 νU∞ ′ (ηf − f) ⇒ v = [(1.5447)(0.5) − (0.3934)] v= 2 x 2 (0.2) y

⇒ v ≈ 0.00586

m . s

7.3 Momentum Transfer over a Wedge (Falkner–Skan Wedge Flow): Similarity Solution Consider symmetrical flow over a wedge, as shown in Figure 7.5a. Flow is assumed to be steady, laminar, and incompressible. This is an interesting problem in engineering, and a solution is developed by Falkner and Skan [14] for this problem. The boundary layer equations will have similarity solutions if at least one dimension extends to infinity. The wedge geometry has this characteristic. Fluid velocity and pressure upstream of the wedge are uniform. However, flow outside the boundary layer is inviscid, and velocity and pressure just outside the velocity boundary layer vary with distance x along the wedge. Continuity and momentum equations for the wedge flow are 𝜕u 𝜕v + =0 𝜕x 𝜕y u

(7.44)

𝜕u 𝜕u 1 dp∞ 𝜕2 u +v =− +ν 2. 𝜕x 𝜕y ρ dx 𝜕y

(7.45)

δ U∞(x) V

y β

x

x β=0

π

m=0 Wedge flow (a)

Flat plate

Stagnation point flow

(b)

V-Uniform free stream approach velocity Figure 7.5

(a–c) Wedge flow with various wedge angles.

(c)

= 90° 2 m=1

β=π

195

196

7 Laminar External Boundary Layers: Momentum and Heat Transfer

The pressure gradient is retained in the momentum equation since external flow is now accelerated over the wedge. The pressure distribution of the potential flow outside the velocity boundary layer is governed by the Bernoulli’s equation as given dU 1 dp = −U∞ ∞ . ρ dx dx

(7.46)

We now need a solution for the velocity distribution U∞ (x) in the inviscid flow region. Solution to inviscid flow outside the thin boundary layer is obtained by potential flow theory. Using potential flow theory as discussed in Evans [5], an expression for the free stream velocity at the edge of the boundary layer is U∞ = Cxm

(7.47)

where C is a constant, the exponent m is related to wedge angle β by m=

β β∕π x dU∞ = = 2 − β∕π 2π − β U∞ dx

(7.48)

where β is given in radians. Angle β may be expressed in terms of parameter m as 2πm m+1 and the free stream velocity U∞ (x) at the wedge surface varies with distance from the wedge tip. β=

(7.49)

(1) When β > 0 (measured in radians), the free stream velocity increases along the wedge surface. (2) When β < 0 (measured in radians), the free stream velocity decreases along the wedge surface. A wedge of negative opening angle is physically impossible, but this situation could be realized when boundary layer experiences suction, as discussed in [15]. (3) When β = 0, the flow is parallel to flat plate. See Figure 7.5b. (4) When β = π, the flow is normal to flat plate, i.e. m = 1, corresponds to flow over a flat plate set normal to the flow direction of free stream. This closely represents the flow near a stagnation point of a bluff body. See Figure 7.5c. At the edge of the boundary layer, free stream velocity distribution is dU dp∞ ρm 2 = −ρU∞ ∞ = −ρCxm (Cmxm−1 ) = − U . (7.50) dx dx x ∞ Upon combining the momentum equation with Eq. (7.50), we get the following form of momentum boundary layer equation: u

2 𝜕u 𝜕u mU∞ 𝜕2 u +v = +ν 2. 𝜕x 𝜕y x 𝜕y

(7.51)

The boundary conditions are u(x, 0) = 0

(7.52a)

v(x, 0) = 0

(7.52b)

u(x, ∞) = U∞ (x) = Cxm .

(7.52c)

Falkner and Skan [14] discovered the similarity transformation for the wedge flow and presented a numerical solution. A similarity variable η is defined as √ √ √ U∞ C m−1 C m−1 =y x x 2 η=y =y (7.53) νx ν ν and a dimensionless stream function ψ that satisfies the continuity equation is defined as √ √ ψ = ν x U∞ f(η) = ν Cxm+1 f(η).

(7.54)

7.3 Momentum Transfer over a Wedge (Falkner–Skan Wedge Flow): Similarity Solution

The x-component of velocity u is given as

( √ ) ( ) m+1 m−1 𝜕ψ 𝜕ψ 𝜕η 𝜕 √ 𝜕 C u= = = y x 2 νC x 2 f(η) 𝜕y 𝜕η 𝜕y 𝜕η 𝜕y ν ) ) (√ (√ m+1 df m−1 C df df x 2 = U∞ . = νC x 2 = C xm dη ν dη dη

The first partial derivative of u with respect to y is ( √ ) ( ) m−1 𝜕 𝜕u 𝜕u 𝜕η df C 𝜕 = = Cxm y x 2 𝜕y 𝜕η 𝜕y 𝜕η dη 𝜕y ν √ √ 2 U∞ d2 f Cxm m d f = U∞ . = Cx 2 νx νx dη2 dη

(7.55)

(7.56)

The second derivative of u with respect to y is ] ) [ √ ( √ ( ) U∞ d2 f U∞ d2 f 𝜕η 𝜕 𝜕 𝜕u d 𝜕2 u =ν =ν U∞ U∞ ν 2 =ν 𝜕y 𝜕y 𝜕y νx dη2 dη νx dη2 𝜕y 𝜕y ( √ ) ) (√ U∞ d2 f U∞ d =ν U∞ dη νx dη2 νx =

U2∞ d3 f . x dη3

(7.57)

The first partial derivative of u with respect to x is ( ) ( ) 𝜕 𝜕 𝜕u df m df m df m 𝜕 = Cx = (Cx ) + (Cx ) 𝜕x 𝜕x dη 𝜕x dη 𝜕x dη = C m xm−1

d2 f 𝜕η df + (Cxm ) 2 dη dη 𝜕x

[ √ ] 2 df Cxm m d f 𝜕 = Cmx + (Cx ) 2 y dη 𝜈x dη 𝜕x [ √ ] 2 C m−1 m−1 df m d f 𝜕 + (Cx ) 2 y x 2 = Cmx dη 𝜈 dη 𝜕x [ ] √ 2 y C m−3 m−1 df m d f + (Cx ) 2 (m − 1) x 2 = Cmx dη 𝜈 dη 2 [ ] √ ( m) 2 (m − 1) C m−3 df Cx m d f 2 + (Cx ) 2 y x =m x dη 2 𝜈 dη [ ] √ ( m) 2 (m − 1) df C m−3 Cx m d f + (Cx ) 2 y x =m x dη 2 𝜈 dη [ ] √ ( m) 2 (m − 1) d f C df Cx + (Cxm ) 2 y xm−1 x−2 =m x dη 2 𝜈 dη m−1

=m

) ( df U∞ U∞ d2 f m−1 + . η dη x x dη2 2

(7.58)

197

198

7 Laminar External Boundary Layers: Momentum and Heat Transfer

The y-component of velocity v is obtained as follows: ] [ m+1 𝜕ψ 𝜕 √ =− v=− 𝜈C x 2 f(η) 𝜕x 𝜕x [ (√ ] ) √ m+1 m+1 df 𝜕η 𝜕 =− 𝜈C x 2 f(η) + 𝜈C x 2 𝜕x dη 𝜕x ( √ )] [ ( ) √ m+1 m+1 df 𝜕 m−1 C 𝜕 √ y x 2 𝜈C x 2 f(η) + 𝜈C x 2 =− 𝜕x dη 𝜕x 𝜈 [ √ { ]} ) m−1 ( ) m−3 ( √ √ m+1 df m − 1 m+1 C x 2 f(η) + 𝜈C x 2 y x 2 − 𝜈C 2 dη 𝜈 2 [ {√ ]} √ ( ) ( ) √ m−1 df 𝜈 Cxm m + 1 Cxm−1 m f(η) + Cx 𝜈x y =− x 2 dη 2 𝜈x2 } {√ √ ( ) Cxm 𝜈x df [( m − 1 ) ] 𝜈 Cxm m + 1 f(η) + η =− x 2 x dη 2 } {√ √ ( ) [( ) ] 𝜈 Cxm m + 1 Cxm 𝜈x df m−1 f(η) + η =− x 2 dη 2 x2 } {√ √ 𝜈 U∞ ( m + 1 ) U∞ 𝜈 df [( m − 1 ) ] f(η) + η =− x 2 x dη 2 √ [ ] U∞ 𝜈 ( m + 1 ) 1 − m df f− η . =− x 2 1 + m dη

(7.59)

The pressure gradient with respect to x is dU 1 dp = −U∞ ∞ = −mC2 x2 m−1 ρ dx dx

(7.60)

where U∞ = Cxm , and substituting Eqs. (7.55)–(7.60) into the momentum boundary layer equation, one finds that constant C and variable x cancel out, and we obtain the celebrated Falkner–Skan nonlinear ordinary differential equation in terms of f(η) [ ( )2 ] d2 f d3 f 1 df + (m + 1)f 2 + m 1 − = 0. (7.61) dη dη3 2 dη The boundary conditions are f(0) = 0

(7.62a)

df(0) =0 dη

(7.62b)

df(∞) = 1. dη

(7.62c)

This is a nonlinear boundary value problem, and this equation is also solved by Hartree [16]. We could make it an initial ′′ ′ ′′ value problem by specifying f (0) and letting f (∞) “float”). Missing f (0) is approximated by Forbrich [17] by [ ]0.53394399 β∕π + 0.1988377 1 ′′ . (7.63) f (0) = √ 0.8160218 2 − β∕π See also [18] for small inaccuracies. The equation for the Falkner–Skan wedge flows can be solved numerically by using the dsolve command in Maple 2020 or bvp4c in MATLAB 2021a for the indicated boundary conditions. A summary of the results is given in Table 7.2.

7.3 Momentum Transfer over a Wedge (Falkner–Skan Wedge Flow): Similarity Solution

Table 7.2 𝛃

Falkner–Skan solutions with impermeable surface. m

d2 f(0)

d2 f(0)

d𝛈2

d𝛈2

Kays et al. [26]

Maple 2020

–0.625

–0.0904

0.0

0.0

–0.314

–0.0476

0.2203

0.2203

0

0

0.3321

0.3320

π/5

0.111

0.51182

0.5116

π/2

0.333

0.75713

0.7574

π

1

1.2325

1.2325

5π/3

5

2.6852

2.6852

2π

∞

∞

∞

Separation Flat plate

Stagnation flow

Assuming a unit depth into the paper, the local wall shear stress τw at the wedge surface is obtained as follows: [( ( )] )) (√ ( ) [( ) ( )] U∞ 𝜕η d df 𝜕u du τw = μ =μ U∞ =μ 𝜕y y=0 dη 𝜕y dη dη νx η=0 ( 2 ) μU∞ √ df = Rex x dη2 η=0

(7.64)

where Rex = U∞ x/ν is the Reynolds number. Wall shear stress is conventionally expressed in dimensionless form. For this purpose, a local friction coefficient is defined as ( 2 ) μU∞ √ df Rex ( 2 ) x dη2 η=0 τw 2 df cf = (7.65a) = = √ ρU2∞ ∕2 ρU2∞ ∕2 Rex dη2 η=0 where Rex = [U∞ (x)x]/ν = Cxm + 1 /ν is the Reynolds number. In a similar manner, we may write the displacement thickness δ1 and momentum thickness δ2 . Recall that the definitions of the displacement thickness δ1 and momentum thickness δ2 are given by ] ∞[ u 1− dy δ1 = ∫0 U∞ )[ ] ∞( u u δ2 = 1− dy. ∫0 U∞ U∞ √ Using the dimensionless distance η = y U∞ ∕νx, we have √ νx dy = dη U∞ df u = U∞ dη and we obtain the displacement thickness δ1 and momentum thickness δ2 for wedge flow ] ∞[ δ1 df 1 dη 1− = √ x dη Rex ∫0 )[ ] ∞( δ2 1 df df = √ 1− dη. x dη dη Rex ∫0

(7.65b)

(7.65c)

199

200

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Equation (7.47) may be applied near the leading edge of a blunt object such as cylinder and sphere. Free stream velocity on the stagnation point of cylinder and sphere can be predicted by potential flow theory, as discussed in [6] ( ) ( ) x x U∞ = 2V sin ≈ 4V for cylinder R D U∞ =

( ) ( ) 3 x x V sin ≈ 3V for sphere 2 R D

where V is the oncoming normal velocity, x is the distance along the surface measured from the stagnation point, and D is the cylinder diameter. U∞ is the velocity just outside the boundary layer. Example 7.3 Air at 1-atm pressure is flowing normally across a 6-cm cylinder. Free stream temperature and velocity of air is 300 K and 10 m/s, respectively. Very close to the forward stagnation point of the cylinder, the free stream velocity U∞ (velocity just outside the boundary layer) along the surface is given by U∞ =

4Vx D

where V is the oncoming normal velocity, x is the distance along the surface measured from the stagnation point, and D is the cylinder diameter. See Figure 7.E3. We wish to calculate the displacement thickness δ1 and momentum thickness δ2 at the stagnation point. Figure 7.E3

x

V = 10 m/s T∞ = 300 K

Flow across a cylinder.

D

Stagnation point

Solution For the stagnation point, flow belongs to the Falkner–Skan similarity flows, and free stream velocity is given by U∞ = Cxm . The parameter m is m=

(β∕π) . 2 − (β∕π)

For stagnation point flow, β = π and m = 1. Thus, we can write 4V . D The momentum thickness δ1 is ] ∞[ df x 1− δ1 = √ dη dη Rex ∫0 ] ∞[ df x dη 1− = √( ) ∫0 dη U x C=

∞

ν

x = √( ) ∫0 2 Cx ν

∞

] [ df dη 1− dη

7.4 Application of Integral Methods to Momentum Transfer Problems

= √ (

1

∞

[ ] df 1− dη dη

) 4V 1 ∫0 D ν √ ] ∞[ df Dν dη. 1− = 4 V ∫0 dη

At 300 K, the viscosity of air is ν = 15.89 × 10−6 m2 /s. Next, we solve numerically Eq. (7.61) for m = 1. This is done with MATLAB 2021a. We define y1 = f y2 = f ′ y3 = f′′ . Then, we reduce Eq. (7.61) to first-order differential equations. dy1 = y2 dη dy2 = y3 dη dy3 = (y2 )2 − y1 y3 − 1 dη along with y1 (0) = 0, y2 (0) = 0, and y3 (4.4) = 1. We chose 4.4 instead of ∞. Solution is given in Table 7.E3. df Using trapz commend in MATLAB 2021a, we evaluate the integral. Let fp represent and eta represent η. We get dη >> trapz(eta, 1-fp) ans = 0.6336. The displacement thickness δ1 is √ 0.06 × 15.89 × 10−6 (0.6336) ≈ 0.051 mm. δ1 = 4 × 10 The momentum thickness δ2 is √ )[ ] ∞( df Dν df 1− dη δ2 = 4 V ∫0 dη dη >> trapz(eta, fp.∗ (1-fp)) ans = 0.2763 √ δ2 =

0.06 × 15.89 × 10−6 (0.2763) ≈ 0.042 mm. 4 × 10

7.4 Application of Integral Methods to Momentum Transfer Problems If the free stream velocity varies in arbitrary manner along the surface, the solution of partial differential equations requires considerable amount of time and effort. An approximate method is widely used to get quick estimates. The approximate method fulfills the conservation of mass and momentum in an integral manner for control volume, which extends from surface to free stream. Control volume includes the entire boundary layer. The integral form of conservation of momentum was derived in Chapter 4. The integral method is unbelievably valuable when an exact solution is not available, or it is difficult to obtain.

201

202

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Table 7.E3

Solution of Eq. (7.61) for m = 1.

𝛈

f

df/d𝛈

d2 f/d𝛈2

0

0

0

1.2330

0.1000

0.0060

0.1183

1.1332

0.2000

0.0233

0.2267

1.0349

0.3000

0.0510

0.3254

0.9390

0.4000

0.0881

0.4146

0.8468

0.5000

0.1336

0.4949

0.7587

0.6000

0.1867

0.5666

0.6755

0.7000

0.2466

0.6302

0.5977

0.8000

0.3125

0.6863

0.5256

0.9000

0.3837

0.7355

0.4593

1.0000

0.4594

0.7783

0.3986

1.1000

0.5392

0.8154

0.3437

1.2000

0.6223

0.8473

0.2943

1.3000

0.7085

0.8745

0.2506

1.4000

0.7971

0.8975

0.2119

1.5000

0.8879

0.9170

0.1779

1.6000

0.9804

0.9333

0.1483

1.7000

1.0744

0.9468

0.1228

1.8000

1.1697

0.9580

0.1011

1.9000

1.2660

0.9671

0.0827

2.0000

1.3631

0.9746

0.0672

2.1000

1.4608

0.9806

0.0542

2.2000

1.5591

0.9855

0.0435

2.3000

1.6579

0.9894

0.0348

2.4000

1.7570

0.9925

0.0277

2.5000

1.8564

0.9950

0.0220

2.6000

1.9560

0.9969

0.0175

2.7000

2.0558

0.9985

0.0139

2.8000

2.1557

0.9997

0.0111

2.9000

2.2557

1.0007

0.0090

3.0000

2.3558

1.0016

0.0073

3.1000

2.4560

1.0022

0.0061

3.2000

2.5563

1.0028

0.0052

3.3000

2.6566

1.0033

0.0045

3.4000

2.7569

1.0037

0.0041

3.5000

2.8573

1.0041

0.0038

3.6000

2.9577

1.0045

0.0036

3.7000

3.0582

1.0048

0.0034

3.8000

3.1587

1.0051

0.0034

3.9000

3.2592

1.0055

0.0034

4.0000

3.3598

1.0058

0.0034

4.1000

3.4604

1.0062

0.0034

4.2000

3.5610

1.0065

0.0035

4.3000

3.6617

1.0068

0.0035

4.4000

3.7624

1.0072

0.0036

7.4 Application of Integral Methods to Momentum Transfer Problems

7.4.1

Laminar Forced Flow over a Flat Plate with Uniform Velocity

Consider steady incompressible constant property laminar flow over a semi-infinite solid plate. Free stream velocity U∞ over the plate is constant. Determine the boundary layer thickness δ(x) and skin friction coefficient cf (x). We assume a third-order polynomial for velocity profile u = a0 + a1 x + a2 x2 + a3 x3 .

(7.66)

The coefficients an , n = 0, 1, 2, 3 are determined using the following boundary conditions on the velocity. We now list all the known boundary conditions for the flow. u(x, 0) = 0 (No-slip boundary condition at the wall)

(7.67a)

u(x, δ) = U∞

(7.67b)

𝜕u (x, δ) = 0 𝜕y 𝜕2 u (x, δ) = 0. 𝜕y2

(7.67c) (7.67d)

Second and third boundary conditions are approximate since the edges of the boundary layer are not uniquely defined. On the other hand, the last boundary condition, Eq. (7.67d), is obtained from the momentum equation. The momentum equation is applicable on the plate surface, and for this reason, u = v = 0 at y = 0, thus, u

𝜕u 𝜕2 u 𝜕2 u 𝜕u +v = ν 2 ⇒ 2 = 0. 𝜕x 𝜕y 𝜕y 𝜕y

Using the polynomial velocity profile and boundary conditions, we evaluate the unknown constants a0 = 0 a1 =

(7.68a)

3 U∞ 2 δ

a2 = 0 a3 = −

(7.68b) (7.68c)

1 U∞ . 2 δ3

Substituting these coefficients into the polynomial velocity profile, we get ( ) ( ) 1 y 3 3 y u − = . U∞ 2 δ 2 δ

(7.68d)

(7.69)

Thus, the assumed velocity profile is expressed in terms of the unknown variable δ(x), and this unknown δ(x) variable is the boundary layer thickness. This unknown variable is determined using the integral form of the momentum equation. Under the present assumptions, the momentum integral equation reduces to τw ρU2∞

=

dδ2 dx

where δ2 is the momentum thickness and is given as ( ) δ u u 1− dy. δ2 = ∫0 U∞ U∞

(7.70)

(7.71)

The integral equation can also be written in the following form: ( ) δ τw u d u = 1 − dy dx ∫0 U∞ U∞ ρU2∞ or

[ δ ] ( ) d 𝜕u = u(U∞ − u)dy . ν 𝜕y y=0 dx ∫0

(7.72)

203

204

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Substituting the velocity profile, Eq. (7.69), into Eq. (7.72), we get 39 2 dδ 3 1 νU = U 2 ∞ δ 280 ∞ dx 140 ν dx δdδ = 13 U∞ at x = 0

δ=0

(7.74) x

δ

∫0

(7.73)

δdδ =

140 ν dx. 13 U∞ ∫0

(7.75)

Evaluating the integral and rearranging, we obtain √ 280∕13 4.64 δ δ = √ ⇒ = √ x x Rex Rex where Rex = U∞ x/ν is the local Reynolds number. The wall shear stress τw is given as } ( ) { [ ( ) ( )] 1 y 3 𝜕u 𝜕 3 y U∞ − τw = μ =μ 𝜕y y=0 𝜕y 2 δ 2 δ y=0 } {[ ( ) ] ( ) ( ) 3 y 2 1 3 1 − τw = μU∞ 2 δ 2 δ δ y=0 3 μU∞ . τw = 2 δ Substituting the value of velocity boundary layer thickness δ into the equation of wall shear stress τw , we get μU∞ 3 3 μU∞ = √ 2 δ 2 4.64x∕ Rex 1 0.6466 τw = ρU2∞ √ . 2 Rex

(7.76)

(7.77)

τw =

(7.78)

The local friction coefficient cf over the plate is obtained as τw 0.6466 = √ . cf = ρU2∞ ∕2 Rex

(7.79)

The average frictional drag over the plate wall τw is determined by integrating the local shear stress τw (x) over the length L. The average friction coefficient as we did for the Blasius solution yields cf =

τw

ρU2∞ ∕2 1.293 cf = √ ReL

=

1 L ∫ τ (x)dx L 0 w ρU2∞ ∕2

=

L L τw (x) 1 1 dx = c (x)dx L ∫0 ρU2∞ ∕2 L ∫0 f

(7.80)

where ReL = U∞ L/ν is the average Reynolds number.

7.4.2

Two-Dimensional Laminar Flow over a Surface with Pressure Gradient (Variable Free Stream Velocity)

Consider the following form of the integral momentum equation: ] [( ) τw δ1 dδ2 1 dU∞ + δ2 2 + . = dx δ2 U∞ dx ρU2∞

(7.81)

The boundary layer integral equation may be used to study boundary layer flows with pressure gradient. This equation is valid for incompressible flow without wall injection or suction. Equation (7.81) relates the displacement thickness δ1 , momentum thickness δ2 , and wall shear stress τw to unknown boundary layer thickness δ. Now, the momentum integral equation is used to determine the absolute value of these various quantities. Basically, there are two approaches to the solution of integral equations:

7.4 Application of Integral Methods to Momentum Transfer Problems

(a) Guessed velocity or temperature profiles (b) Empirical correlations among integral parameters The method proposed by von Karman [19] is one of the earliest methods. This technique was first introduced by von Karman and Pohlhausen in papers on boundary layer analysis, and it is known as the von Karman–Pohlhausen method. It is most widely used for the solution of the boundary layer equation. The Von Karman–Pohlhausen method is also discussed in [6, 20–22]. An essential part of the integral method is the selection of a velocity profile, and the selected velocity profile must satisfy the following requirements: (a) (b) (c) (d)

Must satisfy no slip at the wall Smooth curve at y = δ (u = U∞ , 𝜕u/𝜕y = 0 ) Must allow for inflection points for positive pressure gradient. No a priori assumption about the separation can be made.

For the solution of the integral momentum equation, Pohlhausen assumed a fourth-order polynomial for the velocity profile u = aη + bη2 + cη3 + dη4 (7.82) U∞ where 0 ≤ η ≤ 1 η = y/δ(x) and δ(x) is the boundary layer thickness. To determine the unknown coefficients, the following boundary conditions are used: y = 0,

u=0

(7.83a)

y = 0,

v=0

(7.83b)

y = 0,

ν

y = δ,

u = U∞

y = δ, y = δ,

dU 𝜕 2 u 1 dp = −U∞ ∞ = 2 ρ dx dx 𝜕y

𝜕u =0 𝜕y 𝜕2 u = 0. 𝜕y2

(7.83c) (7.83d) (7.83e) (7.83f)

The boundary condition, Eq. (7.83c), comes from the boundary layer form of the momentum equation and the no-slip boundary condition. Equation (7.82) already satisfies the boundary conditions at y = 0 and u = 0. The application of the remaining four boundary conditions gives us Λ Λ Λ Λ b=− c = −2 + d=1− 6 2 2 6 where we write the velocity profile in the following form: u = f(η) = G1 (η) + ΛG2 (η) U∞ a=2+

(7.84)

where G1 (η) and G2 (η) are given as G1 (η) = 2η − 2η3 + η4 η 1 (η − 3η2 + 3η3 − η4 ) = (1 − η)3 6 6 and the parameter Λ is called Pohlhausen pressure gradient parameter, and it is defined as G2 (η) =

δ2 (x) dU∞ . (7.85) ν dx This parameter is a measure of the pressure gradient in the outer flow, and it is the ratio of pressure force to viscous force. The dimensionless quantity Λ is introduced by Pohlhausen, and this dimensionless quantity plays the role of a form parameter for the velocity profile. Λ=

205

206

7 Laminar External Boundary Layers: Momentum and Heat Transfer

1.2 Λ = 30

1

Λ = 12

0.8 Λ = –12 u/U∞ 0.6 Λ=0 0.4

0.2

0 0

Figure 7.6

0.2

0.4

η

0.6

0.8

1

Velocity profiles for different values of Λ.

The Pohlhausen pressure gradient parameter Λ varies with the local pressure gradient, and it is a dimensionless variable that is a measure of the pressure gradient in the outer flow. In other words, the Pohlhausen parameter determines the effect of an external pressure gradient on the shape of the velocity profile. Figure 7.6 shows the velocity profiles corresponding to various values of the Pohlhausen pressure parameter Λ. The pressure parameter Λ is limited in the range −12 ≤ Λ ≤ 12, and Λ = 7.052 corresponds to the velocity profile at the stagnation point. We present important points without detail; see [6] for more information. Λ=0 Flow past a flat plate. Λ = −12;

τw = (𝜕u∕𝜕y)y=0 = 0

Separation occurs. This means at y = 0 𝜕u/𝜕y = 0. Λ > 12

(u∕U∞ ) > 1

Velocity boundary layer is not expected to exceed that of the outer flow locally. This is not correct physically. Λ < −12 Negative velocity. Reverse flow. Boundary layer theory is not valid. −12 ≤ Λ ≤ 12 Inspection of figure reveals that (u/U∞ ) < 1 and form parameter Λ must stay in this range. λ = 7.052 corresponds to the velocity profile at the stagnation point, as discussed in [6]. The various quantities can be evaluated using the approximate velocity profile. Schlichting and Gersten [6] give the following relations in terms of the Pohlhausen pressure gradient parameter Λ. The displacement thickness is ] 1[ ∞ δ1 u Λ 3 1− dη = = − . [1 − G1 − ΛG2 ]dη = ∫0 ∫0 δ U∞ 10 120 The momentum thickness δ2 is )[ ] 1( 1 δ2 Λ Λ2 u 37 u = − − . 1− dη = [G1 + ΛG2 ][1 − G1 − ΛG2 ]dη = ∫0 ∫0 δ U∞ U∞ 315 945 9072

7.4 Application of Integral Methods to Momentum Transfer Problems

The wall shear stress τw at the surface of the body is given by ( ) ) μU∞ ( 𝜕u Λ = 2+ . τw = μ 𝜕y y=0 δ 6 The shear correlation S is ) ) ( )( τw δ2 ( 37 Λ Λ2 Λ − − . 2+ S= μU∞ 6 315 945 9072 The shape parameter H is ) ( Λ 3 − δ 10 120 H= 1 = ( ). δ2 Λ Λ2 37 − − 315 945 9072 These expressions relate the displacement and momentum thicknesses and wall shear stress to unknown boundary layer thickness. The solution of Eq. (7.81) with the Pohlhausen method is very tedious and bothersome. Many calculations have been carried out and compared with experiments. The problem has been reformulated by Holstein and Bohlen. The method of Holstein and Bohlen is discussed in [6]. Based on work of Holstein and Bohlen, Walz [23] and Thwaites [24] reformulated the problem into a convenient form. Studies indicate that the separation criterion is rather crude, as discussed in [25]. 7.4.2.1 The Correlation Method of Thwaites

The method of Thwaites is found to be a very elegant, and it is an extension of the method due to Holstein and Bohlen. The method of Thwaites [24] does not assume a priori that the velocity profile may be approximately represented by a polynomial. In other words, Thwaites abandoned the universal velocity profile idea. Consider now that the integral momentum equation, Eq. (7.81), is multiplied by the parameter δ2 U∞ /ν and rearranged as δ2 dU∞ τw δ2 U δ dδ = ∞ 2 2 + (2 + H) 2 μ U∞ ν dx ν dx or τw δ2 U d = ∞ μ U∞ 2 dx

(

δ22 ν

) + (2 + H)

δ22 dU∞ ν dx

(7.86a)

(7.86b)

where H = δ1 /δ2 is the shape parameter and is a dimensionless function of the profile shape. Expressions for different quantities that appear in Eq. (7.86b) need to be established as functions of pressure parameter Λ(x). The pressure parameter Λ(x) is given by Λ(x) =

δ2 dU∞ . ν dx

Thwaites defined the following one-parameter correlations based on an analysis of the existing exact solutions boundary layer equations and available experimental data S=

τ δ δ2 = w 2 (shear correlation) δ4 μU∞

(7.87)

where δ4 is the shear thickness, and it is defined as δ4 =

μU∞ . τw

Shape factor correlation is H = δ1 ∕δ. Introducing Eqs. (7.87) and (7.88) into Eq. (7.86b), one gets ( ) δ2 dU∞ U∞ d δ22 S= + (2 + H) 2 . 2 dx ν ν dx

(7.88)

(7.89)

207

208

7 Laminar External Boundary Layers: Momentum and Heat Transfer

The Holstein and Bohlen parameter K is defined as ( )( ) δ22 dU∞ K= . ν dx This expression parameterizes the profile shape, and it may be called pressure gradient factor correlation. The Holstein and Bohlen parameter K is related to the Pohlhausen pressure gradient parameter Λ by ( )( ) ( 2)( 2)( ) ( 2) δ2 δ2 δ22 dU∞ dU∞ δ = = Λ K= 2 ν dx ν dx δ δ2 ( ) Introducing the Holstein and Bohlen parameter K = δ22 ∕ν (dU∞ ∕dx) into Eq. (7.89), the momentum thickness δ2 is eliminated, and the result is ( ) U d K + (2 + H)K (7.90a) S= ∞ 2 dx dU∞ ∕dx or ( ) d K ≈ F(K) (7.90b) U∞ dx dU∞ ∕dx where F(K) is a universal function of K and is given by U∞ dδ22 = 2S − 2(2 + H)K. (7.91) ν dx Notice that Eq. (7.90b) is a universal relationship and universal implies that it is applicable all types of variations of free stream velocity U∞ (x) including the Falkner-Skan velocity profile U∞ = Cxm . Notice also that the parameter S is related to the skin friction and K is related to the pressure gradient. If we take K as an independent variable, we see that S and H are functions of K. Notice that F is a combination of S and H. In the Holstein–Bohlen method, the function F(K) is given by the von Karman–Pohlhausen polynomial. On the other hand, Thwaites determines functions S and H from various existing numerical solutions and available experimental data of boundary layers. Each such solution or family of solutions gives a curve for (τw δ2 )/(μU∞ ) versus K. An average and approximate curve of S is then found from a curve that approximates all accurate curves as closely as possible. In a similar way, H is found from accurate computations of δ1 /δ2 . Thwaites plotted parameter F(K) as a function of K and reported that the points show very little scatter over the entire range. An improvement in accuracy is observed especially for regions of positive pressure gradient. Thwaites proposed that the function F(K) can be replaced by F(K) =

F(K) = a − bK = 0.45 − 6 K. With this linear fit, Eq. (7.90b) becomes ( ) d K U∞ = 0.45 − 6 K. dx dU∞ ∕dx Equation (7.93) is rewritten using the Holstein and Bohlen parameter as given below ( ) 2 2 6 dU∞ δ2 0.45 d δ2 + = . U∞ dx ν U∞ dx ν U∞

(7.92)

(7.93)

(7.94)

Equation (7.94) is a first-order linear homogeneous differential equation in δ22 ∕ν. Now, let us define a new parameter Z = δ22 ∕ν

(7.95)

and we transform Eq. (7.94) to the following form: 6 dU∞ 0.45 dZ + Z= . dx U∞ dx U∞

(7.96)

We now have first-order differential equation and Z is the dependent variable. An integrating factor for this first-order ordinary differential equation is [ ( ) ] 6 dU∞ I = exp dx = exp[6 ln U∞ ] = U6∞ . ∫ U∞ dx

7.4 Application of Integral Methods to Momentum Transfer Problems

Next, the first-order differential equation is multiplied by this integrating factor, and it can be written as d ( 6 ) U∞ Z = 0.45U5∞ dx

(7.97)

and the solution is x ( )| U6∞ (x)Z(x) − U6∞ Z | = 0.45 U5 dξ. |x=0 ∫0 ∞

(7.98)

( )| Taking x = 0 as the leading edge of the surface, U6∞ Z | = 0 since either U6∞ ||x=0 = 0 or δ2 |x = 0 = 0 at the leading edge |x=0 δ2 where x = 0 since Z = ν2 . Then, the solution is x

U6∞ Z = a

∫0

U5∞ dξ.

(7.99)

Using the relation Z = δ22 ∕ν, Eq. (7.99) can be expressed as x

δ22 ≈

0.45ν U5∞ (ξ)dξ. U6∞ ∫0

(7.100)

Note that x = 0 starts from the stagnation point. Thawaites indicates that Eq. (7.100) predicts momentum thickness δ2 very accurately. Once we find momentum thickness δ2 , we can evaluate K = δ22 (dU∞ ∕dx)∕ν along with local shear stress τw τw =

μU∞ S(K) δ2

and displacement thickness δ1 δ1 = δ2 H(K). We may also find the local friction coefficient as cf =

2μ S(K). ρ U∞ δ2

Schetz [25] presents the following curve fits for values of parameters S(K) and H(K): 0 ≤ K ≤ 0.1 S = 0.22 + 1.57 K − 1.8 K2

(7.101a)

H = 2.61 − 3.75 K + 5.24 K2 −0.1 ≤ K ≤ 0 S = 0.22 + 1.402 K + 0.018 K∕(0.107 + K)

(7.101b)

H = 2.088 + 0.0731∕(0.14 + K). Recall that solutions obtained from this method are based on a wide range of profile shapes as well as on experimental data, and therefore, these solutions appear to be better than those obtained from the Pohlhausen method. Equation (7.100) has proved to be an especially useful technique in the solution of a variety of problems. Separation occurs when K = −0.09. For a given boundary shape, an approximate solution may be estimated in the following simple way. For the given shape, the potential flow solution gives the outer velocity U∞ (x). Then, Eq. (7.100) is used to determine the momentum thickness δ2 (x) x

δ22 ≈

0.45ν U5∞ (ξ)dξ. U6∞ ∫0

209

210

7 Laminar External Boundary Layers: Momentum and Heat Transfer

The pressure parameter Λ(x) is evaluated from the relation )2 ( δ2 dU∞ Λ Λ2 37 − − . Λ= 2 315 945 9072 ν dx Now, since we know Λ(x), we can calculate the boundary layer thickness δ(x) from ( ) δ2 Λ Λ2 37 = − − δ 315 945 9072 and displacement thickness δ1 (x) from the following relation: δ1 3 Λ = − . δ 10 120 The wall shear stress τw at the surface of the body is calculated from the following relation: ( ) ) μU∞ ( Λ 𝜕u 2+ = τw = μ 𝜕y y=0 δ 6 where Eq. (7.84) may be used for the velocity profile. Example 7.4 Consider flow over a flat plate with constant velocity u = U∞ . Calculate the following: (a) (b) (c) (d) (e)

The momentum thickness δ2 The displacement thickness δ1 The wall shear stress τw The skin friction coefficient cf The boundary layer thickness δ

Solution a) The momentum thickness δ2 is x

δ22 =

0.45 ν U5∞ dξ U6∞ ∫0

For flow past a flat plate, the potential flow is u = U∞ and is constant; hence, the above equation gives δ22 =

0.45 νx U∞

or δ2 0.67 = √ . x Rex b) The displacement thickness δ1 For flow over a flat plate, K = 0 and shape factor H = 2.61. Thus, the displacement thickness δ1 is δ1 = Hδ2 ) ( √ √ νx νx = 1.75 δ1 = 2.61 × 0.671 . U∞ U∞ c) The wall shear stress For K = 0, shear correlation S = 0.22. See Table 7.3. The wall shear stress is τw δ2 μU ≈ S(K) ⇒ τw = S ∞ μU∞ δ2 √ μU∞ μU∞ U∞ τw = S = (0.220) = 0.3278ρ ν U∞ √ δ2 νx νx 0.671 U∞ )√ ( ρ U2∞ ν . τw = 0.6557 2 U∞ x

7.4 Application of Integral Methods to Momentum Transfer Problems

Table 7.3

Shear and shape functions correlated by Twaites [24].

K

H(K)

S(K)

K

H(K)

S(K)

0.25

2

0.5

–0.048

2.87

0.138

0.2

2.07

0.463

–0.052

2.90

0.130

0.14

2.18

0.404

–0.056

2.94

0.122

0.12

2.23

0.382

–0.060

2.99

0.113

0.1

2.28

0.359

–0.064

3.04

0.104

0.080

2.34

0.333

–0.068

3.09

0.095

0.064

2.39

0.313

–0.072

3.15

0.085

0.048

2.44

0.291

–0.076

3.22

0.072

0.032

2.49

0.268

–0.080

3.30

0.056

0.016

2.55

0.244

–0.084

3.39

0.038

0.0

2.61

0.220

–0.008

2.64

0.208

–0.016

2.67

0.195

–0.024

2.71

0.182

–0.032

2.75

0.168

–0.040

2.81

0.153

Stagnation point (approx.)

Flat plate Separation

–0.086

3.44

0.027

–0.088

3.49

0.015

–0.090

3.55

0.0

d) The skin friction coefficient is τw 0.6557 . = √ cf = ρ U2∞ ∕2 Rex e) The boundary layer thickness δ(x). Let us apply the Karman–Pohlhausen approximation to determine δ(x). For the flat plate, U∞ is constant and the momentum thickness δ2 is obtained as δ22 =

0.45 νx . U∞

We now need to know pressure parameter Λ(x), and it is given by )2 ( δ2 dU∞ Λ Λ2 37 − − . Λ= 2 315 945 9072 ν dx Since U∞ is constant, we have dU∞ /dx = 0, and this will give us ( )2 Λ Λ2 37 − − Λ=0 315 945 9072 )𝟐 ( 𝟏 𝚲𝟐 𝟑𝟕 ; − − > eq ≔ 𝚲 ⋅ 𝟑𝟏𝟓 𝟗𝟒𝟓 𝟗𝟎𝟕𝟐 ( )𝟐 𝟏 𝟏 𝟑𝟕 − 𝚲− 𝚲𝟐 eq ≔ 𝚲 𝟑𝟏𝟓 𝟗𝟒𝟓 𝟗𝟎𝟕𝟐 > > fsolve(eq, 𝚲); − 𝟑𝟕.𝟕𝟗𝟒𝟓𝟒𝟓𝟎𝟎, −𝟑𝟕.𝟕𝟗𝟒𝟓𝟒𝟓𝟎𝟎, 𝟎., 𝟐𝟖.𝟏𝟗𝟒𝟓𝟒𝟓𝟎𝟎, 𝟐𝟖.𝟏𝟗𝟒𝟓𝟒𝟓𝟎𝟎 Only acceptable solution is Λ = 0 δ2 37 Λ Λ2 = − − . δ 315 945 9072

211

212

7 Laminar External Boundary Layers: Momentum and Heat Transfer

For Λ = 0, we have δ2 37 315 = ⇒δ= δ δ 315 37 2 ) √ √ ( )( 315 νx νx = 5.83 δ= 0.686 37 U∞ U∞ or 5.83 δ . = √ x Rex We may also calculate displacement thickness δ1 and wall shear stress τw as ) ( ) ( Λ 3 3 − =δ δ1 = δ 10 120 10 δ1 1.75 = √ x Rex ) 2μU μU∞ ( Λ ∞ τw = 2+ = δ 6 δ 2τw 0.686 = √ . ρU2∞ Rex 7.4.2.2

A Thwaites Type Correlation for Axisymmetric Body

Kays et al. [26] consider an approximate laminar boundary layer solution for arbitrarily varying free stream velocity over a body of revolution. Kays et al. had chosen a different approximation for F(K) F(K) = a − bK = 0.45 − 5.68 K. With this choice for F(K), they give the following relation for the momentum thickness δ2 : √ √ x 0.664 ν U4.68 (ξ) R2 (ξ) dξ. δ2 = ∫0 ∞ R(x)U2.84 ∞ (x)

(7.102a)

Once the momentum thickness δ2 is determined, local shear stress can be evaluated from Table 7.3. White and Majdalani [27] report an equation in the following form: x

δ22 ≈

0.45ν R2 U5∞ dx R2 U6∞ ∫0

(7.102b)

where separation assumed to occur at K = − 0.09, and Rott and Crabtree [28] discuss this problem. If R is constant or much larger than x, then this equation reduces to Thwaites’ equation for two-dimensional plane flow. In these equations, R(x) takes care of curvature effects.

7.5 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate: Similarity Solution for Uniform Wall Temperature Boundary Condition We will consider the classical problem of flow over an isothermal semi-infinite flat plate set at zero angle of incidence to a uniform stream of velocity U∞ . The plate is at constant temperature Tw , as depicted in Figure 7.7. Free stream temperature T∞ is constant. Body forces are negligible. Since flow is assumed to be low velocity, viscous dissipation is neglected. Flow is assumed to be viscous incompressible, steady, and two-dimensional along with constant fluid properties. Since fluid properties are assumed to be constant, velocity distribution is independent of temperature distribution. This means energy and momentum equations are uncoupled. We also assume that the difference between the plate temperature Tw and free stream temperature T∞ is small. For this reason, the variation of fluid properties with temperature is negligible. The Reynolds number is large enough so that the boundary layer assumptions are valid. This is the companion thermal problem to the momentum transfer problem of Blasius. We are interested in determining the thermal boundary layer thickness,

7.5 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

δ

U∞

y

Ts (x)

T∞

U∞

Δ

T∞

T

u

T∞ Tw

0

Tw

x

Flat plate (a) Figure 7.7

x

0 (b)

Velocity and thermal boundary layer over a flat plate.

surface heat flux, heat transfer coefficient, and Nusselt number. Based on these assumptions, the governing equations of the problem under consideration are given as Continuity 𝜕u 𝜕v + =0 𝜕x 𝜕y x-momentum 𝜕u 𝜕2 u 𝜕u +v =ν 2. u 𝜕x 𝜕y 𝜕y

(7.103)

(7.104)

Boundary conditions for the momentum equation are u(x, 0) = 0

(No slip)

(7.105a)

v(x, 0) = 0

(Solid Wall)

(7.105b)

u(x, ∞) → U∞ .

(7.105c)

The initial condition at the leading edge of the plate is specified as follows: u(0, y) → U∞ .

(7.105d)

Notice that we know the solution of the momentum equation. We will simply focus on the solution of the energy equation. The energy equation is u

𝜕T 𝜕2 T 𝜕T +v =α 2 𝜕x 𝜕y 𝜕y

(7.106)

where α = k/ρcp . The initial and boundary conditions are T(0, y) = T∞

(7.107a)

T(x, 0) = Tw

(7.107b)

T(x, ∞) = T∞ .

(7.107c)

The energy equation is expressed in terms of dimensionless temperature θ defined as θ=

T − Tw . T∞ − Tw

(7.108)

The energy equation now becomes u

𝜕θ 𝜕2 θ 𝜕θ +v = α 2. 𝜕x 𝜕y 𝜕y

(7.109)

The boundary conditions become θ(0, y) = 1

(7.110a)

213

214

7 Laminar External Boundary Layers: Momentum and Heat Transfer

θ(x, 0) = 0

(7.110b)

θ(x, ∞) = 1.

(7.110c)

The correct form of the similarity variable is the same as the similarity variable used in momentum transfer √ U∞ . η=y νx

(7.111)

The dimensionless temperature θ is assumed to depend on η, i.e. θ = θ(η). The velocity components in the energy equation are borrowed from the Blasius solution df u = U∞ dη [ ] √ df v 1 ν η −f . = U∞ 2 U∞ x dη

(7.112) (7.113)

Next, we will determine the following derivatives: ( √ ) U∞ η dθ dθ 𝜕η dθ 𝜕 𝜕θ = = y =− 𝜕x dη 𝜕x dη 𝜕x νx 2x dη dθ 𝜕η dθ 𝜕 𝜕θ = = 𝜕y dη 𝜕y dη 𝜕y

( √ y

U∞ νx

)

√ =

(7.114)

U∞ dθ νx dη

(7.115)

) (√ ) U∞ dθ U∞ dθ 𝜕η d = νx dη dη νx dη 𝜕y (√ ) (√ ) U∞ dθ U∞ U d2 θ d = = ∞ 2. dη νx dη νx νx dη

𝜕 𝜕2 θ = 𝜕y 𝜕y2

(√

(7.116)

Substitution of these equations into the energy equation gives dθ d2 θ Pr =0 + f(η) 2 dη dη2

(7.117)

where θ = θ(η) and f = f(η). The function f(η) comes from the solution of the Blasius equation. The boundary conditions are η = 0, η → ∞,

θ=0

(7.118a)

θ = 1.

(7.118b)

We will now integrate Eq. (7.117). Let us define p=

dθ . dη

(7.119)

Then, we introduce Eq. (7.119) into Eq. (7.117), and the differential equation becomes dp Pr + f(η)p = 0. dη 2 Separating the variables and integrating and recalling that p = ( ) η Pr dθ = C1 exp − f(η) dη . dη 2 ∫0 Integrating Eq. (7.121), we get ( )] η[ η Pr exp − θ = C1 f(η)dη dη + C2 . ∫0 2 ∫0

(7.120) dθ , dη

one obtains (7.121)

(7.122)

7.5 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

Application of the boundary condition θ(0) = 0 yields C2 = 0. Next, we will use the boundary condition θ(∞) = 1. This will give us C1 =

∞ ∫0

1 [ ( )] . η Pr exp − f(η)dη dη 2 ∫0

(7.123)

Therefore, the temperature distribution θ(η) becomes [ ( )] η Pr η ∫0 exp − f(η)dη dη 2 ∫0 . θ(η) = [ ( )] η Pr ∞ ∫0 exp − f(η)dη dη 2 ∫0

(7.124)

This equation can be integrated numerically using cumtrapz command in MATLAB 2021a for any Prandtl number provided that the function f(η) is known. The solution of the energy equation can also be numerically obtained by Maple 2020. Let us present the solution for air at 1-atm pressure (Pr = 0.7). The result is presented in Table 7.4. > #Solution of Energy Equation, Eq.(𝟕.𝟏𝟏𝟕), for Pr = 𝟎.𝟕 > restart; > interface(rtablesize = 𝟓𝟎); 50 > Digits ≔ 𝟏𝟔; Digits ≔ 16 > interface(displayprecision = 𝟔); 6 > > > de𝟏 ≔ diff(f(𝛈), 𝛈, 𝛈, 𝛈) + 𝟎.𝟓 ⋅ f(𝛈) ⋅ diff(f(𝛈), 𝛈, 𝛈); d3 de1 ≔ 3 f (η) + 0.500000 f (η) dη

(

) d2 f (η) dη2

> Pr ≔ 𝟎.𝟕; Pr ≔ 0.700000 ) Pr ⋅ f(𝛈) ⋅ diff(𝛉(𝛈), 𝛈); > de𝟐 ≔ diff(𝛉(𝛈), 𝛈, 𝛈) + 𝟐 ( ) 2 d d de2 ≔ 2 θ(η) + 0.350000 f (η) θ(η) dη dη > (

> > #We take 𝛈 = 𝟔 instead of 𝛈 = ∞ > 𝛃 ≔ seq(𝛈, 𝛈 = 𝟎..𝟔, 𝟎.𝟐𝟓) ∶ > G ≔ dsolve({de𝟏, de𝟐, f(𝟎) = 𝟎, f′ (𝟎) = 𝟎, f′ (𝟔) = 𝟏, 𝛉(𝟎) = 𝟎, 𝛉(𝟔) = 𝟏}, {f(𝛈), 𝛉(𝛈)}, type = numeric, output = Array([𝛃])) ∶ > > > F ≔ G(𝟐) ∶ d𝛉 , 𝛉(𝛈) d𝛈 > DATA ≔ F[… , [𝟏, 𝟓, 𝟔]];

> #Value of 𝛈,

215

216

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Table 7.4 Solution of the energy equation for Pr = 0.7 under uniform wall boundary condition. 𝛈

𝛉

d𝛉 d𝛉

0.000000

0.000000

0.294442

0.250000

0.073605

0.29435

0.500000

0.147132

0.293729

0.750000

0.220381

0.292044

1.000000

0.293023

0.288793

1.250000

0.364609

0.283524

1.500000

0.434586

0.275865

1.750000

0.502321

0.265568

2.000000

0.567141

0.252537

2.250000

0.628369

0.236858

2.500000

0.685372

0.218804

2.750000

0.737610

0.198830

3.000000

0.784677

0.177535

3.250000

0.826325

0.155612

3.500000

0.862491

0.133785

3.750000

0.893284

0.112744

4.000000

0.918978

0.093083

4.250000

0.939979

0.075260

4.500000

0.956786

0.059572

4.750000

0.969954

0.046153

5.000000

0.980052

0.034993

5.250000

0.987629

0.025962

5.500000

0.993192

0.018847

5.750000

0.997190

0.013386

6.000000

1.000000

0.009302

The dimensionless temperature θ is plotted as a function of dimensionless distance η for different Prandtl numbers in Figure 7.8. When we review the momentum and thermal boundary layer equations, we notice that (1) temperature and velocity profiles are identical for Pr = 1 (2) an examination of Figure 7.8 reveals that temperature profiles depends on Prandtl number (3) for Pr ≪ 1, the thermal boundary layer thickness Δ(x) is greater than the momentum boundary layer thickness δ(x), i.e. Δ(x) > δ(x) (4) for Pr ≫ 1, the thermal boundary layer thickness Δ(x) is less than the momentum boundary layer thickness δ(x), i.e. Δ(x) < δ(x). ′′

Local heat flux qw (x) on the plate surface can be expressed as [ ] 𝜕T(x, 0) dθ(0) 𝜕η ′′ = −k (T∞ − Tw ) qw = −k 𝜕y dη 𝜕y ] [ √ [ ] dθ(0) U∞ dθ(0) 1 √ = k (Tw − T∞ ) Rex = −k (T∞ − Tw ) dη νx dη x

(7.125a)

7.5 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

1

0.8

Pr = 0.1

0.6 θ(η)

Pr = 1 0.4

Pr = 3 Pr = 10

0.2

0 0

Figure 7.8

2

4

η

6

8

10

Dimensionless temperature as a function dimensionless distance for different Prandtl numbers.

or q′′w x dθ(0) √ = Rex . k(Tw − T∞ ) dη

(7.125b)

Since the local heat transfer coefficient h(x) is defined as h=

q′′w (Tw − T∞ )

the local Nusselt number becomes hx dθ(0) √ = Rex Nux = k dη or dθ(0) −1∕2 Nux Rex = dη

(7.126)

(7.127a)

(7.127b)

where Rex = ρU∞ x/μ is the local Reynolds number. Since we know the temperature distribution, we can write dθ(0)/dη as dθ(0) = dη

∞

∫0

1 . [ ( )] η Pr exp − f(η)dη dη 2 ∫0

(7.128a)

Since we know f(η), we can evaluate the integral for any given Prandtl number. Such calculations were performed by Elzy and Sisson [30]. Integration can be carried out using cumtrapz and trapz commands in MATLAB 2021a since we have tabulated data for the f(η). The Nusselt number depends on the dimensionless temperature gradient, (dθ/dη)η = 0 at the surface for a given Prandtl number Pr. The values of (dθ/dη)η = 0 for various Prandtl numbers Pr are reproduced by Maple −1∕2 2020 and is plotted in Figure 7.9. The values of Nux Rex for various values of Pr are presented in Table 7.5. It is known that there is a strong dependence of the thermal boundary layer thickness on Prandtl numbers. For low values of Prandtl numbers, the thermal boundary layer thickness is greater than the momentum boundary layer thickness. On the other hand, for high Prandtl numbers, the thermal boundary layer thickness is smaller than the momentum boundary layer thickness. Low-Prandtl-number solutions are given in Table 7.6, as discussed in [26, 31]. An examination of Table 7.5 reveals that dθ(0)/dη for the Prandtl numbers between 0.6 and 15 may be represented by ( ) dθ = 0.3328Pr0.3401 . dη η=0

217

218

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Numerical solution 2 box symbol-Curve fit of

1 dθ dη

η=0

0.8 0.6 0.4

0.2

0.1

Figure 7.9

Table 7.5

Variation of

1

dθ(0) dη

10 Pr

100

1000

as function of Prandtl number.

−1∕2

The values of Nux Rex

for various Prandtl numbers.

Pr

0.7

1

−1∕2 Nux Rex

0.2944

0.3325

0.4852

0.5766

0.7281

0.8341

Pr

50

75

500

750

1000

1500

1.2472

1.4279

2.6882

3.077

3.3870

3.8722

−1∕2

Nux Rex

3

−1∕2

Table 7.6 The values of Nux Rex numbers.

5

10

15

for low Prandtl

Pr

0.001

0.01

0.1

0.5

−1∕2 Nux Rex

0.0173

0.0516

0.140

0.259

For simplicity, the exponent of Pr is taken as 1/3, and the last expression is written as ( ) dθ ≈ 0.332Pr1∕3 . dη η=0

(7.128b)

In engineering, we are interested in the total heat transfer rate from the entire plate surface. Consider a rectangle of length L in the x-direction and a constant unit width. Unit width is considered since flow is two-dimensional. The total heat transfer rate per unit width between surface and fluid along the surface having a length L is L

q=

∫0

q′′w (x) dx.

The local heat transfer rate is given by Eq. (7.125b) k(Tw − T∞ ) dθ(0) q′′w x dθ(0) √ Rex ⇒ q′′w = = k(Tw − T∞ ) dη x dη

√(

) U∞ √ x. ν

7.5 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

Thus, total heat transfer q is

√( ) k(Tw − T∞ ) dθ(0) U∞ √ x dx q= ∫0 x dη ν √( ) ⎡ L U∞ ⎤⎥ dθ(0) 1 ⎢ = k(Tw − T∞ ) √ dx ⎢ dη ν ⎥ ∫0 x ⎣ ⎦ √( ) ⎡ U∞ ⎤⎥ √ dθ(0) 2 L = ⎢k(Tw − T∞ ) ⎢ dη ν ⎥ ⎦ ⎣ √( ) ⎡ U∞ L ⎤⎥ dθ(0) ⎢ = 2 k(Tw − T∞ ) ⎢ ⎥ dη ν ⎣ ⎦ ] [ √ dθ(0) k(Tw − T∞ ) ReL . =2 dη L

The average heat transfer coefficient h for a flat plate is defined as ] [ √ dθ(0) k(Tw − T∞ ) ReL 2 dη q = h= (L × 1)(Tw − T∞ ) L × (Tw − T∞ ) ] √ [ dθ(0) k ReL h=2 dη L

(7.129)

(7.130) (7.131)

where ReL = ρU∞ L/μ is the average Reynolds number based on plate length. We can now write the average Nusselt number NuL as NuL =

√ dθ(0) hL = 2 ReL . k dη

(7.132)

Consider a plate of length L and width W. Total heat transfer q L

L

q=

h(Tw − T∞ )(Wdx) = (Tw − T∞ )(W) hdx ∫0 ∫0 ( ) L 1 hdx . = (Tw − T∞ )(WL) L ∫0

(7.133)

Finally, total heat transfer may be expressed as q = hA(Tw − T∞ )

(7.134)

where A = WL is the plate surface area, L is the plate length in the flow direction, and W is the plate width. The local Nusselt number Nux can be estimated with the following relation for the range of Prandtl numbers from 0.6 to 50: √ hx = 0.332 Rex Pr1∕3 (7.135a) Nux = k 0.6 < Pr < 15 Tw = const. The local Stanton number may be written as St =

Nux . Rex Pr

(7.135b)

Equation (7.135a) may also be written in terms of the local Stanton number −1∕2

Stx = 0.332Rex

Pr−2∕3

(7.135c)

219

220

7 Laminar External Boundary Layers: Momentum and Heat Transfer

0.6 < Pr < 15 Tw = const. The average Nusselt Number NuL is obtained by integrating local heat transfer coefficient NuL =

√ hL = 0.664 ReL Pr1∕3 k

(7.136)

0.6 < Pr < 15 Tw = const. Notice that the average Nusselt number for the entire plate is twice the local Nusselt number at the trailing edge, i.e. at x = L. One has to be careful in using Eqs. (7.135a), (7.135b), and (7.136). These equations are valid for the indicated Prandtl number range. In this indicated Prandtl number range, we have common gases such as air, oxygen, nitrogen, hydrogen, and carbon dioxide. In addition to gases, low-viscosity liquids, such as water and refrigerants, are also included in this Prandtl number range. These equations are not valid for high-viscosity fluids, such as heavy oils and glycerin, or for liquid metals that have very low Prandtl numbers. These equations are derived based on the assumption of constant fluid properties (ρ, cp , k), but, in practice, fluid properties vary with temperature. In convection heat transfer, fluid properties are evaluated at the film temperature Tf = (Tw + T∞ )/2. The ratio of thermal boundary layer thickness Δ(x) to the momentum boundary layer thickness δ(x) for flow over a flat plate is estimated by the following relation: Δ ≈ Pr−1∕3 δ

(7.137a)

Pr > 0.6 or 5 Δ ≈ 1∕2 1∕3 x Pr Rex

(7.137b)

Pr > 0.6. When Pr = 1, δ = Δ, and the shape of the velocity and temperature profiles are identical. Viscous and conduction effects are the same, resulting in similar velocity and temperature profiles. For a very wide range of Prandtl numbers, Churchill [32] has suggested the following empirical correlation for local Nusselt number in laminar flow over an isothermal flat plate for all the Prandtl numbers. The constants in the correlation are based on the best available theoretical and experimental results √ 0.3387 Rex Pr1∕3 Nux = [ (7.138) ) ]1∕4 ( 0.0468 2∕3 1+ Pr Pe = Rex Pr > 100 Tw = Const where Nux = h x/k is the local Nusselt number and Rex = x U∞ /ν is the local Reynolds number. This formula is for laminar boundary layer flow over a flat plate at constant wall temperature and is accurate to ±1% over the entire range of Prandtl number. Fluid properties are evaluated at the film temperature Tf = (Tw + T∞ )/2. We now wish to find the average Nusselt number for Eq. (7.138). The average Nusselt number NuL was defined as NuL =

hL k

(7.139)

where h was the average heat transfer coefficient and defined as L

h=

1 h(x) dx. L ∫0

(7.140)

7.5 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

Substituting Eq. (7.140) into Eq. (7.139), for average NuL , we get [ ] ] [ L L L Nux L 1 1hx dx = dx. NuL = h dx = ∫0 ∫0 x k k L ∫0 x

(7.141)

Nusselt number correlations are expressed in terms of Reynolds number Rex , and for this reason, it is convenient to change the integration coordinates from x to Rex . Thus, we use Rex =

μ Rex ρ U∞ x →x= . μ ρ U∞

(7.142)

From Eq. (7.142), we have μ dx = dRex . ρ U∞

(7.143)

Substituting Eq. (7.143) into Eq. (7.141), for the average Nusselt number, we get L

NuL =

∫0

ReL Nux dx = ∫0 x

Nux μ dRex . μ ρ U ∞ Rex ρ U∞

(7.144)

Notice that x = 0,

Rex = 0

x = L,

Rex = ReL .

Finally, we have the following expression for NuL after carrying out the integration: ReL

NuL =

∫0

Nux dRex Rex

0.3387 Pr1∕3 NuL = [ ( ) ]1∕4 ∫0 0.0468 2∕3 1+ Pr

ReL

dRex = [ √ Rex

√ 0.3387 Pr1∕3 Re [2 Rex ]0 L . ] ( )2∕3 1∕4 0.0468 1+ Pr

After simplification, the average Nusselt number becomes √ 0.6774 Pr1∕3 ReL NuL = [ for ReL < Rexc ) ]1∕4 ( 0.0468 2∕3 1+ Pr

(7.145)

(7.146)

Tw = Const. Example 7.5 Atmospheric air at 17 ∘ C flows at a velocity of 4 m/s over a 2 m by 1 m flat plate, as shown in Figure 7.E5. The plate surface temperature is 77 ∘ C. Determine the total convective heat transfer from the one surface of the plate. T∞ = 17 °C p = 1 atm U∞ = 4 m/s

Tw = 77 °C

L=2m

Figure 7.E5

Geometry and problem description for Example 7.5.

Solution First, we evaluate air properties at the film temperature Tf as Tf =

Tw + T∞ 77 + 17 = = 47 ∘ C = 320 K. 2 2

221

222

7 Laminar External Boundary Layers: Momentum and Heat Transfer

At the film temperature, the properties we need are given below ν = 17.9 × 10−6 m2 ∕s k = 0.0277 W∕m.K Pr = 0.703. The free stream velocity U∞ is U∞ = 4 m∕s. The Reynolds number based on the plate length is ReL =

U∞ L 4 m∕s × 2 m = 4.47 × 105 . = ν 17.9 × 10−6 m2 ∕s

Since the Reynolds number is less than the 500 000, flow is laminar for the entire plate. We will use Eq. (7.146) to evaluate the Nusselt number √ √ 0.6774 Pr1∕3 ReL 0.6774 (0.703)1∕3 4.47 × 105 NuL = [ ⇒ NuL = = 387.67. [ ) ]1∕4 ) ]1∕4 ( ( 0.0468 2∕3 0.0468 2∕3 1+ 1+ Pr 0.703 The average heat transfer coefficient h is ) ( ( ) 0.0277 W∕m.K W k (387.67) = 5.34 2 . h= NuL = L 2m m K The total convective heat transfer from the one surface of the plate is q = h A(Tw − T∞ ) = (5.34 W∕m2 K) × (2 m2 )(350 − 290) ≈ 640 W. Example 7.6 Water at 290 K is flowing over a 30 cm by 30 cm square plate at a velocity of 0.5 m/s. The plate is at a uniform temperature of 330 K. (a) (b) (c) (d) (e)

Plot the variation of local heat flux in W/m2 along the plate Obtain the mean heat flux Plot the variation of the u-velocity profile in the boundary layer at x = 20 cm from the leading edge of the plate Plot the variation of the temperature profile in the boundary layer at x = 20 cm from the leading edge of the plate Calculate the heat lost by the plate

Solution Let us determine the film temperature Tf =

290 + 330 = 310 K. 2

The properties of water are kg N.s , μ = 695 × 10−6 2 , m3 m W , Pr = 4.62. k = 0.628 m.K The Reynolds number is ρ = 993

ReL =

ν = 6.99 × 10−7

ρU∞ L (993)(0.5)(0.3) = = 214 316 < 500 000 μ 695 × 10−6

so the flow is laminar.

m2 , s

7.5 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

(a) Local heat flux We will solve the energy equation for Pr = 4.62 so that we can obtain the dimensionless temperature distribution. Using dθ(0) = 0.5614. Hence, dimensionless temperature, we will plot the temperature variation. From the solution, dη x q′′w dθ(0) √ = Rex k(Tw − T∞ ) dη √ k(T − T ) ρU∞ x dθ(0) w ∞ q′′w = x dη μ √ √ dθ(0) ρU∞ 1 = k(Tw − T∞ ) dη μ x =

11919.8 . √ x

Note that x is in meter. The plot of q′′w as a function of x is shown in Figure 7.E6a. (b) Mean heat flux is L

q′′w = Figure 7.E6a

1 1 q′′ (x) dx = L ∫0 w 0.3 ∫0

0.3

11919.8 dx ≈ 43525 W∕m2 . √ x

Variation of local heat flux along the plate.

140000 120000 100000 qw″

80000 60000 40000 20000 0

0.1

0.2 X (a)

The solution of the energy equation for Pr = 4.62 is given in the following table: 𝛈

df/d𝛈

𝛉

0.0

0.0

0.0

0.5000

0.1658

0.2796

1.0000

0.3297

0.5442

1.5000

0.4867

0.7617

2.0000

0.6297

0.9040

2.5000

0.7512

0.9722

3.0000

0.8460

0.9946

3.5000

0.9130

0.9993

0.3

223

224

7 Laminar External Boundary Layers: Momentum and Heat Transfer

𝛈

df/d𝛈

𝛉

4.0000

0.9555

0.9999

4.5000

0.9795

0.9999

5.0000

0.9915

0.9999

√ √ (6.99 × 10−7 )(0.2) νx y=η =η ≈ 0.0005287 η. U∞ 0.5 The fluid velocity is in terms of

df dη

df df = 0.5 . dη dη The variation of velocity u with position y is plotted in Figure 7.E6b. The variation of temperature with distance is obtained as follows. Using the dimensionless temperature θ, we have T − Tw θ= T∞ − Tw u = U∞

T = Tw + (T∞ − Tw )θ T = 330 + (290 − 330)θ or T = 330 − 40 θ. Figure 7.E6b distance y.

0.0025

Variation of longitudinal velocity u as a function of vertical

0.0020

0.0015 y 0.0010

0.0005

0 0

0.1

0.2

0.3

0.4

u (b)

The variation of temperature T with distance y is shown in Figure 7.E6c. Total heat transfer from the plate is calculated as √ NuL = 0.664 ReL Pr1∕3 NuL = 511.98 (0.628)(511.98) W k NuL = = 1071.74 2 . L 0.3 mK Heat transfer from one surface of the plate is h=

q = hA(Tw − T∞ ) = (1071.74 )(0.3)2 (330 − 290) = 3858 W.

7.6 Low-Prandtl-Number Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

Figure 7.E6c

Variation of temperature as a function of vertical distance y.

0.0025

0.0020

0.0015 y (m) 0.0010

0.0005

0 300

310 T (K)

320

330

(c)

7.6 Low-Prandtl-Number Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate: Similarity Solutions for Uniform Wall Temperature Boundary Condition For low-Prandtl-number fluids, the velocity boundary layer is much thinner than the thermal boundary layer (δ ≪ Δ). See Figure 7.10. Then, the longitudinal velocity, u(x, y), can be assumed to be uniform everywhere and equal to free stream velocity U∞ , that is: u ≈ U∞ ,

u df = ≈ 1. dη U∞

(7.147)

Consider the energy equation again dθ d2 θ Pr =0 + f(η) 2 dη dη2 where the dimensionless similarity variable is √ U∞ η=y νx and the dimensionless temperature is T − Tw θ= . T∞ − Tw

(7.148)

(7.149)

(7.150)

We know express Eq. (7.148) as Pr θ′′ =− f 2 θ′

(7.151) T∞ U∞

U∞,T∞

Tw

δ Tw (a)

Figure 7.10

Ts (x)

∆ (x)

y

Velocity and Temperature profile for Pr ≪ 1.

T∞ x

x

0 (b)

225

226

7 Laminar External Boundary Layers: Momentum and Heat Transfer

and differentiate this equation with respect to η. This will give us [ ] Pr df d θ′′ . =− dη θ′ 2 dη Since d f/d η = 1, the energy equation, Eq. (7.152), becomes [ ] Pr d θ′′ =− . dη θ′ 2

(7.152)

(7.153)

Integrating once with respect to η, we obtain θ′′ Pr = − η + C1 . 2 θ′ ′ We let p = dθ/dη = θ , and Eq. (7.154a) becomes p′ Pr = − η + C1 . p 2

(7.154b)

Integrating Eq. (7.154b) again with respect to η, we get ] [ Pr ln(p) = − η2 + C1 η + ln C2 4 or

(7.154a)

[ ] Pr p = C2 exp − η2 + C1 η . 4 Integrating once more with respect to η yields η [ ] Pr θ= C2 exp − η2 + C1 η dθ + C3 . ∫0 4

(7.155a)

(7.155b)

(7.156)

The boundary conditions for evaluating the integrating constants are θ(0) = 0

(7.157a)

θ(∞) = 1.

(7.157b)

We need a third boundary condition. At the plate surface u = 0 and v = 0, and from the energy equation, we obtain the third boundary condition as d2 θ || = 0. dη2 ||η=0

(7.157c)

The application of the boundary conditions yields 1 and C3 = C1 = C2 = 0 ( ) . Pr ∞ ∫0 exp − η2 dη 4 Using Maple 2020, we obtain that

(7.158)

> assume(Pr > 0); > ∞

>

∫0

( ) Pr exp − ⋅ η2 dη; 4

√ √ 1 4 π . √ 2 Pr ∼

Then, we express integral as √ ∞ ) ( Pr 2 π . exp − η dη = ∫0 4 Pr Thus, C3 =

√

Pr . π

(7.159)

(7.160)

7.6 Low-Prandtl-Number Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

Then, the temperature distribution becomes √ η ( ) Pr Pr exp − η2 dη. θ= π ∫0 4

(7.161)

Equation (7.161) can be integrated using Maple 2020 > > >

√

) ( 𝛈 Pr ⋅ 𝛈𝟐 Pr ⋅ d𝛈; exp − 𝝅 ∫𝟎 𝟒 ) (√ √ Pr 𝛈 Pr √ 𝛑 erf 𝛑 𝟐 √ Pr

> simplify(%);

(√ erf

Pr 𝛈 𝟐

)

and the final solution can be expressed as (η√ ) θ = erf Pr . 2 The error function is defined as

(7.162)

x

2 exp(−ξ2 ) dξ. erf(x) = √ ∫ π 0 We can plot the dimensionless temperature θ as a function of dimensionless distance η for different low Prandtl numbers. Let us determine the local Nusselt number for this case. The local heat transfer coefficient h is ( ) ( ) 𝜕θ 𝜕T −k −k(T∞ − Tw ) 𝜕y y=0 𝜕y y=0 q′′w = = h= Tw − T0 Tw − T∞ Tw − T∞ ) ( ) (√ ( ) ( ) U∞ 𝜕η dθ 𝜕θ dθ =k h=k =k 𝜕y y=0 dη η=0 𝜕y dη η=0 νx ( ) √ ( ) ( ) √ U∞ x k dθ k dθ Rex = h= x dη η=0 ν x dη η=0 and the local Nusselt number is ( ) √ hx dθ = Rex . Nux = k dη η=0

(7.163a)

Let us find (dθ/dη)η = 0 . Using Maple 2020 ( √ ) Pr ; > θ ≔ erf η ⋅ 2 θ ≔ erf

(

1 √ η Pr 2

> diff(θ, η); 1 2 − η Pr √ e 4 Pr √ π

)

227

228

7 Laminar External Boundary Layers: Momentum and Heat Transfer

> subs(η = 0, diff(θ, η)); √ e0 Pr √ π Then, we write √ ( ) dθ Pr = dη η=0 π √ Nux = 0.564 Pex

(7.163b)

Tw = Const where Pex = Rex Pr is the Peclet number. The average Nusselt number can be written as √ NuL = 1.128 ReL Pr1∕3

(7.164)

Tw = Const. Eckert and Drake [33] recommend the use of the following relation for a flat plate heated along its entire length for low-Prandtl-number fluids. The plate surface temperature is assumed to be constant √ Nux Pr = √ √ √ Rex 1.55 Pr + 3.09 0.372 − 0.15 Pr Tw = Const 0.005 < Pr < 0.05.

(7.165)

This relation is useful for calculating heat transfer in liquid metals.

7.7 High-Prandtl-Number Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate: Similarity Solutions for Uniform Wall Temperature Boundary Condition For high-Prandtl-number fluids such as oils, the velocity boundary layer is much thicker than the thermal boundary layer (δ ≫ Δ). See Figure 7.11. This means that the thermal boundary layer is within the momentum boundary layer. Consider again the energy equation dθ d2 θ Pr =0 + f(η) 2 dη dη2

(7.166)

d2 f = 0.332 dη2

(7.167)

√ where θ = (T − Tw )/(T∞ − Tw ) and η = y∕ U∞ ∕νx. A good approximation for the dimensionless velocity profile for high-Prandtl-number fluids is to assume that the velocity profile is linear within the thermal boundary layer, and slope is given by the value at the wall. At η = 0, we have

and let A = 0.332. Integrating Eq. (7.167) twice, we obtain the dimensionless velocity A 2 η + C1 η + C2 . (7.168) 2 Using the boundary conditions f(0) = 0 and (df/dη)η = 0 = 0, we find that C1 = 0 and C2 = 0. Thus, the dimensionless velocity becomes f=

f=

A 2 η. 2

(7.169)

7.7 High-Prandtl-Number Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

U∞ δ (x) Ts (x)

y U∞

∆ (x)

(a)

Tw

Tw T∞

u

T∞

x

x

0 (b)

Velocity and Temperature profile for Pr ≫ 1.

Figure 7.11

Substituting this dimensionless velocity into the energy equation, we obtain ( 2) η d2 θ dθ = 0. + (0.332 Pr) 2 4 dη dη

(7.170)

The boundary conditions are θ(0) = 0

(7.171a)

θ(∞) = 1.

(7.171b)

Let C = 0.3321 P r and p = dθ/dη; then, the differential equation becomes dp η2 + C p = 0. dη 4 Separating and integrating, we get ( ) C p = B exp − η3 . 12 Introducing p = dθ/dη and integrating with respect to η once more, the differential equation becomes η ( ) C exp − η3 dη + D. θ=B ∫0 12 Using the boundary condition θ(0) = 0 gives D = 0 η ) ( C exp − η3 dη. θ=B ∫0 12

(7.172)

(7.173)

(7.174)

(7.175)

Applying the boundary condition θ(∞) = 1 gives the unknown constant B as B=

∞

∫0

1 ( ) . C exp − η3 dη 12

Finally, the dimensionless temperature distribution becomes η ) ( C exp − η3 dη ∫ 12 θ = 0∞ ( ) . C exp − η3 dη ∫0 12 Let us evaluate the integral in the denominator of Eq. (7.177) using Maple 2020 > assume(C, positive); > ∞

>

∫𝟎

) ( C exp − ⋅ 𝛈𝟑 d𝛈; 𝟏𝟐

(7.176)

(7.177)

229

230

7 Laminar External Boundary Layers: Momentum and Heat Transfer

√ 𝟏∕𝟑 𝟐 𝟏𝟐 𝛑 𝟑 ( ) 𝟐 𝟗 C∼𝟏∕𝟑 𝚪 𝟑 Inserting C = 0.3321 Pr into this result and evaluating gives the value of integration ∞ ( ) C 2.9521 exp − η3 dη = . ∫0 12 Pr1∕3 Finally, we obtain the temperature distribution as η ( ) Pr η3 dη. θ = 0.3387Pr1∕3 exp − ∫0 36.1336

(7.178)

(7.179)

This integral can be evaluated numerically for any Prandtl number. Let us formulate the local Nusselt Number Nux Nux =

−x (𝜕T∕𝜕y)y=0 xq′′w xh = = k k(Tw − T∞ ) (Tw − T∞ )

T = Tw + (T∞ − Tw )θ ( ) 𝜕η 𝜕T dθ = (T∞ − Tw ) 𝜕y dη η=0 𝜕y ( ) 1√ dθ = (T∞ − Tw ) Rex . x dη η=0 Thus, we obtain ( ) √ dθ Nux = Rex dη η=0 ( ) dθ and we obtain from the temperature solution θ(η) as dη η=0 ( ) dθ = 0.3387Pr1∕3 . dη η=0

(7.180) (7.181)

(7.182)

(7.183)

(7.184)

The local Nusselt number becomes √ Nux = 0.3387Pr1∕3 Rex .

(7.185)

The average Nusselt number is √ NuL = 0.678 ReL Pr1∕3 .

(7.186)

7.8 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate: Similarity Solution for Uniform Heat Flux Boundary Condition In many practical engineering problems, the heat flux is essentially constant, and temperature varies along the plate. As an example, we may show the electrically heated plate. The convective heat transfer coefficient h as well as the temperature difference between the plate and the free stream varies continuously with the x-location along the plate. The main objective is to determine the plate surface temperature distribution. We will consider again the classical problem of flow over a semi-infinite flat plate set at zero angle of incidence to a uniform stream of velocity U∞ . The plate is subjected to uniform heat flux. Body forces are negligible. Since flow is assumed to be low velocity, viscous dissipation is neglected. Flow is assumed to be steady and two-dimensional along with constant fluid properties. The Reynolds number is large enough so that the boundary layer assumptions are valid. Based on these assumptions, the governing equations of the problem under consideration are given as

7.8 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

Continuity 𝜕u 𝜕v + =0 𝜕x 𝜕y

(7.187)

x-momentum u

𝜕u 𝜕2 u 𝜕u +v =ν 2. 𝜕x 𝜕y 𝜕y

(7.188)

Boundary conditions are At y = 0,

u=0

(No slip)

(7.189a)

v=0

(Solid wall)

(7.189b)

At y = ∞, u → U∞ .

(7.189c)

The boundary condition at the leading edge of the plate is specified as follows: x = 0, u = U∞ .

(7.189d)

The difference between plate surface temperature and free stream fluid temperature is small, and the variation of fluid properties with temperature is negligible. The momentum equation is independent of the energy equation. The similarity solution of the momentum equation is valid in the present problem, and the solution of the momentum equation is given as df u = U∞ dη √ ( ) df 1 ν U∞ η −f v= 2 x dη √ U∞ η=y . νx Consider now the energy equation and its boundary conditions u

𝜕T 𝜕T 𝜕2 T +v =α 2 𝜕x 𝜕y 𝜕y

(7.190) (7.191) (7.192)

(7.193)

where α = k/ρcp . The boundary conditions are T(0, y) = T∞ −k

𝜕T(x, 0) = q′′w 𝜕y

T(x, ∞) = T∞ .

(7.194a) (7.194b) (7.194c)

The energy equation is expressed in terms of temperature difference θ = T − T∞ as u

𝜕θ 𝜕θ 𝜕2 θ +v = α 2. 𝜕x 𝜕y 𝜕y

(7.195)

The boundary conditions become θ(0, y) = 0 −k

𝜕θ(x, 0) = q′′w 𝜕y

θ(x, ∞) = 0.

(7.196a) (7.196b) (7.196c)

231

232

7 Laminar External Boundary Layers: Momentum and Heat Transfer

We need to determine a similarity variable for the energy equation, and the method for obtaining a similarity variable is discussed in [8, 9]. We now define the following dimensionless temperature: F = θ∕θR

(7.197)

where θR is an unknown parameter, and we cannot define θR as wall temperature since the wall temperature is not spatially constant. Consider the boundary condition −k

𝜕θ(x, 0) = q′′w 𝜕y

𝜕θ 𝜕F dF 𝜕η dF = kθR = kθR = kθR k 𝜕y 𝜕y dη 𝜕y dη

√

U∞ . νx

Thus, we have √ √ q′′ dF dF U∞ νx = q′′w ⇒ − = w . −kθR dη νx dη kθR U∞ Next, we set √ √ q′′ q′′w νx νx = 1 ⇒ θR = w . kθR U∞ k U∞ Then, we have the following transformation as a similarity variable for temperature: √ q′′ νx F(η) θ= w k U∞

(7.198)

(7.199)

(7.200a)

or

√ q′′w νx F(η) (7.200b) T − T∞ = k U∞ √ where η = y∕ U∞ ∕νx. The parameter θR is determined such that the wall boundary condition produced the similarity variable. It appears that the boundary conditions have important role in determining the similarity variable. We now take the partial derivatives as shown below [ ′′ √ [√ ] ] q′′ 𝜕 𝜕θ 𝜕 qw νx νx = F(η) = w F(η) 𝜕x 𝜕x k U∞ k 𝜕x U∞ { [√ } ] √ ′′ 𝜕 νx νx 𝜕 𝜕θ qw = F(η) [F(η)] + 𝜕x k 𝜕x U∞ U∞ 𝜕x ] √ ′′ [√ η νx F νx dF 𝜕θ qw (7.201) = − 𝜕x k U∞ 2x 2x U∞ dη ) ′′ ( 𝜕θ qw dF = (7.202) 𝜕y k dη )−1∕2 ( 2 ) ′′ ( 𝜕 2 θ qw νx dF = . (7.203) k U∞ 𝜕y2 dη2 Substituting these into the energy equation, one obtains ] [ ′′ ] [√ ] [ √ qw η df νx F νx dF − U∞ dη k U∞ 2x 2x U∞ dη [ ( [ √ ) ( )] ( )] [ ′′ ( )] qw dF q′′w νx −1∕2 d2 F df 1 ν U∞ . η −f =α + 2 x dη k dη k U∞ dη2 After simplification, the energy equation can be reduced to the ordinary differential equation ( ) d2 F Pr dF df f − F = 0. + 2 dη dη dη2

(7.204)

(7.205)

7.8 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

The boundary conditions become dF(0) = −1 dη

(7.206a)

F(∞) = 0.

(7.206b)

The energy equation, Eq. (7.205), along with the momentum equation may be solved numerically for different Prandtl numbers using bvp4c in MATLAB 2021a or Maple 2020. As an example, we present a solution for Pr = 0.7 using Maple 2020 as given below. > restart; > Digits ≔ 𝟏𝟔; Digits ≔ 𝟏𝟔 > interface(rtablesize = 𝟓𝟎); 𝟓𝟎 > interface(displayprecision = 𝟔); 𝟔 > > #Momentum Equation > > de𝟏 ≔ diff(f(𝛈), 𝛈, 𝛈, 𝛈) + 𝟎.𝟓 ⋅ f(𝛈) ⋅ diff(f(𝛈), 𝛈, 𝛈); de1 ≔

d𝟑 f(𝛈) + 𝟎.𝟓𝟎𝟎𝟎𝟎𝟎 f(𝛈) d𝛈𝟑

(

) d𝟐 f(𝛈) d𝛈𝟐

> #Energy Equation > #F is the dimensionless temperature > #Prandtl number Pr > Pr ≔ 𝟎.𝟕; Pr ≔ 𝟎.𝟕𝟎𝟎𝟎𝟎𝟎 ) Pr ⋅ (f(𝛈) ⋅ diff(F(𝛈), 𝛈) de𝟐 ≔ diff(F(𝛈), 𝛈, 𝛈) + 𝟐 − diff( f(𝛈), 𝛈) ⋅ F(𝛈)); (

> d𝟐 F(𝛈) + 𝟎.𝟑𝟓𝟎𝟎𝟎𝟎f(𝛈) d𝛈𝟐 ) ( d f(𝛈) F(𝛈) − 𝟎.𝟑𝟓𝟎𝟎𝟎𝟎 d𝛈

de2 ≔

(

) d F(𝛈) d𝛈

> > #Solution is given below > > #We take 𝛈 = 𝟔 instead of 𝛈 = ∞ > 𝛃 ≔ seq(𝛈, 𝛈 = 𝟎..𝟔, 𝟎.𝟐𝟓) ∶ > G ≔ dsolve({de𝟏, de𝟐, f(𝟎) = 𝟎, f′ (𝟎) = 𝟎, f′ (𝟔) = 𝟏, F′ (𝟎) = −𝟏, F(𝟔) = 𝟎}, {f(𝛈), F(𝛈)}, type = numeric, output = Array([𝛃])) ∶ > >

233

234

7 Laminar External Boundary Layers: Momentum and Heat Transfer

> #Solution of Energy Equation > F ≔ G(𝟐) ∶ dF , F(𝛈) d𝛈 > DATA ≔ F[… , [𝟏, 𝟐, 𝟑]]; > #Values of 𝛈,

The solution of Eq. (7.205) is presented in Table 7.7 for Prandtl number Pr = 0.7. Let us evaluate the local Nusselt number. The local Nusselt number is Nu =

q′′w x . k(Tw − T∞ )

(7.207)

The Nusselt number is evaluated noting that Tw = T(x, 0) and √ q′′ νx F(0) Tw (x) − T∞ = w k U∞ and the result for the local Nusselt number is 1 √ Nux = Rex . F(0)

(7.208)

(7.209)

Table 7.7 Solution of the energy equation for uniform heat flux boundary condition. 𝛈

F(𝛈)

dF/d𝛈

0.000000

2.461215

−1.000000

0.250000

2.211942

−0.991351

0.500000

1.966878

−0.966641

0.750000

1.729799

−0.927801

1.000000

1.503982

−0.876907

1.250000

1.292164

−0.816198

1.500000

1.096503

−0.748056

1.750000

0.918550

−0.674951

2.000000

0.759235

−0.599354

2.250000

0.618882

−0.523635

2.500000

0.497247

−0.449954

2.750000

0.393580

−0.380161

3.000000

0.306719

−0.315722

3.250000

0.235185

−0.257676

3.500000

0.177297

−0.206627

3.750000

0.131272

−0.162771

4.000000

0.095324

−0.125953

4.250000

0.067744

−0.095738

4.500000

0.046958

−0.071491

4.750000

0.031564

−0.052464

5.000000

0.020357

−0.037859

5.250000

0.012331

−0.026893

5.500000

0.006668

−0.018838

5.750000

0.002723

−0.013050

6.000000

0.000000

−0.008981

7.8 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

Substituting the value of F(0) = 2.4612 for Pr = 0.7 into Eq. (7.209), we get √ Pr = 0.7. Nux = 0.4063 Rex ,

(7.210a)

We have developed a local Nusselt number for Pr = 0.7, and we can develop equations for other Prandtl numbers in a similar way. When heat flux q′′w is specified at the plate surface, we are interested in the plate surface temperature Tw (x). The average heat transfer coefficient for uniform heat flux is defined as ′′

h=

qw

′′

qw

=

L

1 ΔT(x) dx L ∫0

ΔT

.

(7.210b)

Kays et al. [26] give the following relation for the local Nusselt number for constant heat flux, q′′w = const: √ Nux = 0.453 Rex Pr1∕3 .

(7.210c)

This result is obtained by the integral method. Recall that the local heat transfer coefficient is h=

q′′w . Tw (x) − T∞

Then, the local Nusselt number is given as ′′

Nux =

qw x hx = . k k[Tw (x) − T∞ ]

(7.210d)

We now combine Eq. (7.210d) with Eq. (7.210c), and we get q′′w x

Tw (x) − T∞ =

1∕2

0.453 kRex Pr1∕3

.

(7.210e)

This equation tells us that the surface temperature increases in the x-direction (Tw ∼ x1/2 ) along the plate starting from the leading edge. We now must be careful in discussing average results. The average temperature difference along the plate may be obtained by performing the integration L

Tw − T∞ =

1 [Tw (x) − T∞ ]dx. L ∫0

(7.211a)

Substituting Eq. (7.210e) into Eq. (7.211a), we get Tw − T∞ =

1 L ∫0

L

q′′w x dx ) ( √ U∞ Pr1∕3 x 0.453 k ν

and performing the integration, we get Tw − T∞ =

q′′w L . √ k 0.6795 ReL Pr1∕3

(7.211b)

The average heat transfer coefficient h is defined as h=

q′′w TW − T∞

.

We combine this equation with Eq. (7.211b) with Eq. (7.211c) to get the average Nusselt number NuL h=

or h=

q′′w q′′w L √ k 0.6795 ReL Pr1∕3 ( ) √ k 0.6795 ReL Pr1∕3 L

(7.211c)

235

236

7 Laminar External Boundary Layers: Momentum and Heat Transfer

or √ hL = 0.6795 ReL Pr1∕3 . k Let us express Eqs. (7.210c) and (7.211d) as √ k h = 0.6795 ReL Pr1∕3 L and √ k h = 0.453 Rex Pr1∕3 . x Next, let us evaluate the local heat transfer coefficient h at the trailing edge of the plate √ k h|x=L = 0.453 ReL Pr1∕3 . L NuL =

(7.211d)

Now, we evaluate the ratio h∕ h|x=L 0.6795 h = = 1.5. h|x=L 0.453 For a wide range of Prandtl numbers, Churchill and Ozeo [32] give following correlation valid for uniform heat flux. In developing this correlation, they combined numerical solution with available experimental data √ 0.4637 Rex Pr1∕3 (7.212a) Nux = [ ) ]1∕4 ( 0.02052 2∕3 1+ Pr Pex > 100 q′′w = const. Integrating Eq. (7.212a), the average Nusselt number NuL for the entire range of Pr can be obtained as √ 0.696 ReL Pr1∕3 NuL = [ ( ) ]1∕4 0.02052 2∕3 1+ Pr

(7.212b)

PeL > 100. Fluid properties are evaluated at the film temperature. Note that for surfaces with uniform heat flux, the local convective heat transfer coefficient is used to determine the local surface temperature. Example 7.7 Engine oil at 330 K flows over a square flat plate with a velocity of 3 m/s. The plate length is 0.5 m. A uniform heat flux of 10 000 W/m2 is applied to the surface. Determine the average surface temperature of the plate. Solution Since the plate surface temperature is unknown, the fluid temperatures cannot be determined. We evaluate the fluid properties at the free stream temperature T∞ = 330 K ν = 96.6 × 10−6 m2 ∕s

k = 141 × 10−3 W∕∕m.K

Reynolds number U∞ L 3 × 0.5 = = 15 528. ν 96.6 × 10−6 For average Nusselt number, we choose Eq. (7.212b) √ 0.696 ReL Pr1∕3 NuL = [ ) ]1∕4 ( 0.02052 2∕3 1+ Pr ReL =

Pr = 1205.

7.9 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

√ 0.696 15528 (1205)1∕3 NuL = = 922.76. )] [ ( 0.02052 1+ 1205 Next, we evaluate the average temperature difference using Eq. (7.211c) Tw − T∞ =

′′ L qw k Nu

L

0.5 10 000 = 38.42 K. Tw − T∞ = 0.141 922.76 Average surface temperature is Tw = T∞ + 38.42 K = 368.42 K. Next, we calculate the film temperature Tw + T∞ 368.42 + 300 = = 349 K ≈ 350 K. 2 2 We now correct the oil properties based on the new film temperature Tf =

ν = 41.7 × 10−6 m2 ∕s

k = 138 × 10−3 W∕∕m.K

Pr = 546.

The corrected Reynolds number is U∞ L 3 × 0.5 = = 35 971. ν 41.7 × 10−6 The corrected average Nusselt number is ReL =

NuL = 1078.6. Average temperature difference is 0.5 10 000 = 33.59 K. 0.138 1078.6 The corrected average surface temperature is Tw − T∞ =

Tw = 330 + 33.59 K = 363.59 K. If needed, a new film temperature is determined and calculation is repeated.

7.9 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate: Similarity Solutions for Variable Wall Temperature Boundary Condition Similarity solutions for some cases where the plate surface temperature varies with x may be obtained by similarity variables. In each case, it is assumed that the velocity field and temperature field is uncoupled. This problem was discussed by researchers such as Schlichting and Gersten [6], Oosthuizen and Naylor [10], and Cebeci and Bradshaw [35]. In this section, a flat plate with variable surface temperature will be considered. Assume laminar flow along a semi-infinite flat plate with constant free stream velocity U∞ and constant free stream temperature T∞ . Physical properties of fluid are constant. Plate wall temperature varies with axial position x as follows: Tw − T∞ = Bxn

(7.213)

where B and n are constants and T∞ is the free stream velocity approaching to flat plate. The energy equation and its associated boundary conditions are given as u

𝜕T 𝜕2 T 𝜕T +v =α 2 𝜕x 𝜕y 𝜕y

(7.214)

where α = k/ρcp is the thermal diffusivity. T(x, 0) = Tw

where Tw = T∞ + B xn

(7.215a)

237

238

7 Laminar External Boundary Layers: Momentum and Heat Transfer

T(0, y) = T∞

(7.215b)

T(x, ∞) = T∞ .

(7.215c)

Velocity distribution is independent of the temperature distribution, and we have assumed constant property flow. We already have a solution for the momentum equation. The velocity components are as follows: df dη √ ( ) 1 νU∞ df v= η −f . 2 x dη

(7.216)

u = U∞

(7.217)

The dimensionless distance η is √ U∞ . η=y νx We will define a dimensionless temperature θ(x, y) = θ(η) as follows: θ=

Tw − T Tw − T∞

(7.218)

(7.219)

or T = Tw − (Tw − T∞ )θ or T = Tw − T∞ + (Tw − T∞ )θ + T∞ T = T∞ + (Tw − T∞ )(1 − θ) T = T∞ + Bxn − Bxn θ.

(7.220)

We need various derivatives of temperature T(x, y). The partial derivative of T with respect to x is 𝜕T 𝜕θ = Bnxn−1 − Bnxn−1 θ − Bxn 𝜕x 𝜕x ⎡ ⎢ U∞ 𝜕η 𝜕θ dθ dθ ⎢ y = = − √ 𝜕x dη 𝜕x dη ⎢ 2 U∞ ⎢ νx2 ⎣ νx

(7.221) ⎤ ⎥ ⎥ = − η dθ . ⎥ 2x dη ⎥ ⎦

(7.222)

Combining Eqs. (7.221) and(7.222), we obtain ) ( η dθ 𝜕T = Bnxn−1 − Bnxn−1 θ − Bxn − 𝜕x 2x dη η dθ 𝜕T = B n xn−1 − B n xn−1 θ + B xn−1 . 𝜕x 2 dη

(7.223)

The partial derivative of temperature T with respect to y is 𝜕θ dθ 𝜕η 𝜕T = −Bxn = −Bxn 𝜕y 𝜕y dη 𝜕y √ U∞ n dθ = −Bx . dη νx The second partial derivative of T with respect to y is [ ] [ ] √ √ U∞ U∞ 𝜕η d 𝜕 𝜕2 T n dθ n dθ −Bx = −Bx = 𝜕y dη νx dη dη νx 𝜕y 𝜕y2 [ ]√ √ U∞ d dθ U∞ −Bxn = dη dη νx νx

(7.224)

7.9 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

U∞ d2 θ νx dη2 U d2 θ = −Bxn−1 ∞ 2 . ν dη = −Bxn

Substituting these equations into the energy equation yields ][ ] [ η dθ df B n xn−1 − B n xn−1 θ + B xn−1 U∞ dη 2 dη ] [ √ √ ( )] [ [ ] 2 U df dθ 1 νU∞ ∞ n n−1 U∞ d θ η −f −Bx = α −Bx + 2 x dη dη νx ν dη2

(7.225)

(7.226)

and after some algebra, we obtain the following ordinary differential equation: df d2 θ Pr dθ + nPr (1 − θ) = 0. + f 2 dη dη dη2

(7.227)

The boundary conditions of Eq. (7.227) are θ(0) = 0

(7.228a)

θ(∞) = 1.

(7.228b)

The partial differential equation is reduced to an ordinary differential equation. We need to solve Eq. (7.227) for a given n value and Prandtl number Pr to determine the temperature distribution. The local heat transfer coefficient h (x) is defined as 𝜕T(x, 0) −k q′′w 𝜕y = h= (Tw − T∞ ) (Tw − T∞ ) 𝜕T(x, 0) 𝜕 𝜕θ = [T + (T∞ − Tw )θ] = (T∞ − Tw ) 𝜕y 𝜕y w 𝜕y ] [ dθ(0) 𝜕η = (T∞ − Tw ) dη 𝜕y [ ] √ dθ(0) U∞ = (T∞ − Tw ) dη νx [ ] √ dθ(0) 1 xU∞ = (T∞ − Tw ) dη x ν [ ] dθ(0) 1 √ = (T∞ − Tw ) Rex . dη x Substituting Eq. (7.230) into Eq. (7.229) gives us the local heat transfer coefficient √ dθ(0) U∞ h=k dη νx or h=k

dθ(0) 1 √ Rex . dη x

The local Nusselt number Nux becomes hx dθ(0) √ = Rex . Nux = k dη We need to evaluate

(7.229)

(7.230)

(7.231a)

(7.231b)

(7.232)

dθ(0) for a given temperature distribution on the plate. The average heat transfer coefficient h is dη

L

h=

1 h(x) dx. L ∫0

(7.233)

239

240

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Substituting the local heat transfer coefficient, Eq. (7.231b), into Eq. (7.233) and integrating, we obtain: √ dθ(0) U∞ 1 L dx h=k √ dη ν L ∫0 x √ dθ(0) U∞ 1 √ (2 L) h=k dη ν L or h=2

( )| ( )√ k dθ | ReL | . L dη ||η=0

Then, the average Nusselt number becomes ( )| √ hL dθ | = 2 ReL NuL = | . k dη ||η=0 We can also evaluate the local heat flux q′′w (x) at the plate surface as follows: [ ] 𝜕T(x, 0) dθ(0) 1 √ = −k (T∞ − Tw ) q′′w = −k Rex 𝜕y dη x

(7.234)

(7.235)

(7.236)

(7.237)

(7.238a)

or q′′w x dθ(0) √ = Rex . k(Tw − T∞ ) dη

(7.238b)

Example 7.8 Air flows at a velocity of 4 m/s over a flat plate. The plate length in the flow direction 50 cm. Free stream temperature of the air approaching the plate is 20 ∘ C. The plate is heated by the embedded heating element, and the surface temperature Tw of the plate is measured at several different locations along the length of the plate in the flow direction. A curve fit of the data gives the surface temperature of the plate given by Tw = 20 + 40(x∕L)∘ C x is in cm and measured from the leading edge of the plate. Using the similarity solution, plot the variation of local heat flux along the plate. Solution L = 0.5 m TW = 20 + 40

( ) x ∘ C L

x is in centimeters. The average surface temperature Tw is L 40 [ ( )] 1 1 x dx = 40 ∘ C. Tw = Tw dx = 20 + 40 L ∫0 40 ∫0 L Film temperature Tf is Tf = (Tw + T∞ )∕2 = (40 + 20)∕2 = 30 ∘ C = 303 K. Air properties at this film temperature are estimated as k ≈ 26.5 × 10−3 W∕m∘ C ν = 16.19 × 10−6 m2 ∕s Pr ≈ 0.706.

7.9 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

The Reynolds number ReL is ReL =

U∞ L (4m∕s)(0.5 m) = = 123 533 < 500 000 ν 16.19 × 10−6 m2 ∕s

and flow is laminar. We rearrange the plate temperature Tw as ( ) x ∘ C Tw − 20 = 40 0.5 where x is in meters. The plate temperature variation is in the form Tw − T∞ = B x (∘ C) For this case, we find that n = 1. We now solve the energy equation for n = 1 and Pr = 0.706. See the solution given below. Solution is obtained using Maple 2020 > restart; > > Digits ≔ 𝟏𝟔; interface(rtablesize = 𝟏𝟎𝟎); > 𝟏𝟎𝟎 > interface(displayprecision = 𝟒); 𝟒 n ≔ 𝟎.𝟓 > #Variable surface temperature > n ≔ 𝟏;

#Tw (x) = T∞ + Bxn n≔𝟏

> Digits ≔ 𝟏𝟔 > Pr ≔ 𝟎.𝟕𝟎𝟔;

#Fluid is air

Pr ≔ 𝟎.𝟕𝟎𝟔𝟎 𝟏 > de𝟏 ≔ diff(f(𝛈), 𝛈$𝟑) + ⋅ f(𝛈) ⋅ diff(f(𝛈), 𝛈$𝟐); 𝟐 ( 𝟐 ) d𝟑 𝟏 d f(𝛈) de1 ≔ f(𝛈) + f(𝛈) 𝟐 d𝛈𝟑 d𝛈𝟐 > de𝟐 ≔ diff(𝛉(𝛈), 𝛈$𝟐) + n ⋅ Pr ⋅ diff(f(𝛈), 𝛈) ⋅ (𝟏 − 𝛉(𝛈)) ( ) Pr + ⋅ f(𝛈) ⋅ diff(𝛉(𝛈), 𝛈); 𝟐 > ) ( d𝟐 d f(𝛈) (𝟏 − 𝛉(𝛈)) 𝛉(𝛈) + 𝟎.𝟕𝟎𝟔𝟎 d𝛈 d𝛈𝟐 ) ( d 𝛉(𝛈) + 𝟎.𝟑𝟓𝟑𝟎 f(𝛈) d𝛈

de2 ≔

> 𝛃 ≔ seq(𝛈, 𝛈 = 𝟎..𝟏𝟎, 𝟎.𝟓) ∶ sol ≔ dsolve({de𝟏, de𝟐, f(𝟎) = 𝟎, f′ (𝟎) = 𝟎, f′ (𝟏𝟎) = 𝟏, 𝛉(𝟎) = 𝟎, > 𝛉(𝟏𝟎) = 𝟏}, type = numeric, output = Array([𝛃])) ∶

241

7 Laminar External Boundary Layers: Momentum and Heat Transfer

The solution gives us that

dθ | | dη |η=0

= 0.4817 and heat flux

k(Tw − T∞ ) dθ | √ | Rex x dη ||η=0 √ [ ( )] U∞ x x dθ || k 40 = | x 0.5 dη |η=0 ν

q′′w =

q′′w = (26.5 × 10−3 )

[(

40 0.5

)]

√ (0.4817)

√ W 4x = 507.6 x 2 . −6 m 16.19 × 10

The variation of local heat flux q′′w (x) with axial position x is shown in Figure 7.E8. Figure 7.E8

Variation of local heat flux along the plate.

300

qʺw(X)

242

200

100

0 0

0.1

0.2

0.3

0.4

0.5

X

Here, we notice that n = 0 corresponds to a uniform wall temperature boundary condition. On the other hand, n = 1/2 represents the uniform heat flux boundary condition. Let us discuss this point. The surface heat flux can be expressed as q′′w = h(x) (Tw − T∞ ).

(7.239)

Substituting the surface temperature variation into the above equation, we obtain q′′w = h(x) C xn .

(7.240)

According to this equation, surface heat flux is constant if h(x) ∝ x1∕n . The local heat transfer coefficient h(x) is obtained from the local Nusselt number as ( ) k h(x) = Nux . x The local Nusselt number for a plate with surface temperature variation described earlier is ( ) √ U∞ x dθ . Nux = dη η=0 ν Therefore, the local heat transfer coefficient becomes √ ( )( ) U∞ x dθ k h(x) = x dη η=0 ν

(7.241)

(7.242)

(7.243)

(7.244a)

7.9 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

or

( ) √ U∞ −1∕2 dθ x . h(x) = k dη η=0 ν

(7.244b)

Therefore, heat flux can be expressed as ( ) √ U∞ −1∕2 dθ . x q′′w = C xn k dη η=0 ν

(7.245)

The examination of this equation shows that heat flux q′′w is constant and independent of x if we set n = 1/2. This means that for the constant wall heat flux boundary condition for flat plate, the surface temperature varies as 1

Tw = T∞ + Bx 2 .

(7.246)

We conclude that a similarity solution exists for flow over a flat plate with uniform heat flux. Example 7.9 Consider air flow over flat plate. The plate length in the flow direction is 15 cm. The plate width is 1 m. A uniform heat flux 2.5 kW/m2 is applied to plate surface. Air velocity approaching the plate is 10 m/s. The free stream air temperature is 20 ∘ C. Plot the local wall temperature of the plate along the plate. Solution Plate surface temperature along the flow direction will vary with position. It is not possible to evaluate the film temperature since the plate surface temperature is not known. For this reason, the air properties first will be evaluated at the free stream temperature, T∞ = 20 ∘ C. At free stream temperature T∞ = 20 ∘ C, the air properties are k = 0.0256 W∕m.K,

ν = 15.26 × 10−6 m2 ∕s,

Pr = 0.708.

The Reynolds number ReL is ReL =

U∞ L (10)(0.15) = = 98 296. ν 15.26 × 10−6

Since ReL < 5 × 105 , flow is laminar over the plate. Since a uniform heat flux is applied on the plate surface, plate surface temperature will vary as follows: Tw = T∞ + Bx0.5 . This means that for a uniform heat flux, n = 0.5. The Prandtl number for air is Pr = 0.708, and we will solve the following equations numerically using Maple 2020: Energy equation df d2 θ Pr dθ + f + nPr (1 − θ) = 0 2 dη dη dη2 where n = 0.5. The boundary conditions are θ(0) = 1 θ(∞) = 1. Momentum equation d3 f 1 d2 f + f = 0. dη3 2 dη2 The boundary conditions are f(0) = 0,

f′ (0) = 0,

f′ (∞) = 1.

243

244

7 Laminar External Boundary Layers: Momentum and Heat Transfer

The solution of these equations will give us > restart; > interface(rtablesize = 𝟏𝟎𝟎); > 𝟏𝟎𝟎 ># n ≔ 𝟎.𝟓 > #Flat plate with uniform heat flux since n = 𝟎.𝟓 > n ≔ 𝟎.𝟓;

#Tw (x) = T∞ + Bxn n ≔ 𝟎.𝟓

> > Pr ≔ 𝟎.𝟕;

#Fluid is air

Pr ≔ 𝟎.𝟕 𝟏 > de𝟏 ≔ diff(f(𝛈), 𝛈$𝟑) + ⋅ f(𝛈) ⋅ diff(f(𝛈), 𝛈$𝟐); 𝟐 ( 𝟐 ) d𝟑 𝟏 d de𝟏 ≔ f(𝛈) f(𝛈) + f(𝛈) 𝟐 d𝛈𝟑 d𝛈𝟐 > de𝟐 ≔ diff(𝛉(𝛈), 𝛈$𝟐) + n ⋅ Pr ⋅ diff(f(𝛈), 𝛈) ⋅ (𝟏 − 𝛉(𝛈)) ( ) Pr ⋅ f(𝛈) ⋅ diff(𝛉(𝛈), 𝛈); + 𝟐 de𝟐 ≔

) d f(𝛈) (𝟏 − 𝛉(𝛈)) d𝛈 ) ( d 𝛉(𝛈) + 𝟎.𝟑𝟓𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 f(𝛈) d𝛈

d𝟐 𝛉(𝛈) + 𝟎.𝟑𝟓 d𝛈𝟐

(

> 𝛃 ≔ seq(𝛈, 𝛈 = 𝟎..𝟏𝟎, 𝟎.𝟓) ∶ > sol ≔ dsolve({de𝟏, de𝟐, f(𝟎) = 𝟎, f′ (𝟎) = 𝟎, f′ (𝟏𝟎) = 𝟏, 𝛉(𝟎) = 𝟎, 𝛉(𝟏𝟎) = 𝟏}, type = numeric, output = Array([𝛃])) ∶ > > We obtain the value of (dθ/dη)η = 0 from the solution dθ(0) ≈ 0.4058. dη Since heat flux is given as q′′w x dθ(0) √ = Rex . k(Tw − T∞ ) dη Plate surface temperature can be written as √ q′′w x q′′w x = (Tw − T∞ ) = . √ dθ(0) √ dθ(0) U∞ k Rex k dη dη ν Heat flux q′′w = 2500 W∕m2 and

√ √ (2500)( x) (Tw − T∞ ) == = 297.27 x √ 10 (0.0256)(0.4058) 15.26 × 10−6 Tw = 20 + 297.27x0.5 .

7.9 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

Next, we will determine mean plate temperature 1 0.15 ∫0

Tw =

0.15

√ [20 + 297.27 x]dx = 96.75 ∘ C

We now calculate the fluid film temperature. The fluid film temperature Tf is T∞ + Tw 20 + 96.75 = ≈ 58.38 ∘ C ≈ 332 K. 2 2 Air properties at this temperature Tf =

k = 0.0286 W∕m2 K, ν = 19.1 × 10−6 m2 ∕s,

Pr = 0.702.

The Reynolds number ReL is ReL =

U∞ L (10)(0.15) = 78 534 = ν 19.10 × 10−6

ReL < 5 × 105 . Flow is laminar. We now calculate temperature difference (Tw − T∞ ) as √ √ (2500)( x) (Tw − T∞ ) = = 297.7 x √ 10 (0.0286)(0.4058) 19.1×10 −6 √ Tw = 20 + 297.7 x. The difference between the last two calculations√is small. No need to do any more iteration. We can now accept that the surface temperature variation is Tw (x) = 20 + 297 x. Wall temperature variation is depicted in Figure 7.E9. Figure 7.E9

Variation of surface temperature along the plate.

180 160 140 Tw(x)

120 100 80 60 40 20 0 0

0.1

0.2

0.3

X

7.9.1

Superposition Principle

In Section 7.9, we have obtained a similarity solution for the temperature distribution Tw (x) − T∞ = Bxn and the similarity solution was represented by Eq. (7.227). The energy equation of thermal boundary layer is linear, and this fact makes it possible to extend this solution. Thus, the results for varying surface temperature represented by Eq. (7.213) may be used to obtain the solution for the polynomial temperature variation on a plate as given below Tw (x) − T∞ = a0 + a1 x + a2 x2 + .. … =

∞ ∑

am xm .

m=0

(7.247)

245

246

7 Laminar External Boundary Layers: Momentum and Heat Transfer

The superposition principle is discussed by Oosthuizen and Naylor [10], White and Majdalani [27], Eckert and Drake [33], Cebeci and Bradshaw [35], and Chapman and Rubesin [36], and since the energy equation is linear, the sum of the solutions for different wall temperature variations is also a solution. Dimensionless temperature θ is expressed as T −T T − T∞ θ= w =1− (7.248) Tw − T∞ Tw − T∞ and we assume that dimensionless temperature θ depends on similarity variable η. Then, we may express this equation as T − T∞ = (1 − θ)(Tw − T∞ ).

(7.249)

Suppose that we have (Tw − T∞ )A and (Tw − T∞ )B wall temperature distributions. We wish to write the solution of the energy equation for the wall temperature distribution (Tw − T∞ )A + (Tw − T∞ )B . We assume that θA is the similarity solution for wall temperature distribution (Tw − T∞ )A and θB is the similarity solution for wall temperature distribution (Tw − T∞ )B . Then, we can write following relations. The solution of the energy equation for wall temperature distribution (Tw − T∞ )A is given by (T − T∞ )A = (1 − θA )(Tw − T∞ )A . The solution of the energy equation for wall temperature distribution (Tw − T∞ )B is given by (T − T∞ )B = (1 − θB )(Tw − T∞ )B . We may now write the solution of the energy equation for the wall temperature distribution (Tw − T∞ )A + (Tw − T∞ )B as T − T∞ = (1 − θA )(Tw − T∞ )A + (1 − θB )(Tw − T∞ )B . Next, we wish to calculate the rate of heat transfer at the plate surface as ( ) 𝜕T . q′′w = −k 𝜕y y=0 The derivative (𝜕T/𝜕y)y = 0 is evaluated and combined with the Fourier law [( ] ) ) ( ( ) 𝜕θA 𝜕θB 𝜕T =− (T − T∞ )A + (T − T∞ )B 𝜕y y=0 𝜕y y=0 w 𝜕y y=0 w ] [( ) ) ( dθB 𝜕η dθA 𝜕η (T − T∞ )A + (T − T∞ )B . =− dη 𝜕y y=0 w dη 𝜕y y=0 w √ U At this point, we use the similarity variable η = y νx∞ [( ]√ ) ) ( dθA dθB U∞ . =− (Tw − T∞ )A + (Tw − T∞ )B dη η=0 dη η=0 νx Heat flux becomes ]√ [( ) ) ( U∞ dθA dθB ′′ qw = k (Tw − T∞ )A + (Tw − T∞ )B dη η=0 dη η=0 νx or q′′w x = k(Tw − T∞ )A

[(

dθA dη

)

( + η=0

dθB dη

)

(Tw − T∞ )B (T w − T∞ )A η=0

]

√

Rex .

Example 7.10 Air flows at a velocity of 5 m/s over a flat plate aligned in the flow direction. Free stream temperature of the air is 17 ∘ C. The plate has a length of 30 cm in the flow direction. The plate is heated by the embedded heating element and the surface temperature Tw of the plate is measured at several different locations along the length of the plate in the flow direction. A curve fit of the data gives the surface temperature of the plate given by ( )0.6 x Tw = 43 + 30 30 where x is in centimeters. Using the similarity solution method, plot the variation of local heat flux along the plate.

7.9 Viscous Incompressible Constant Property Parallel Flow over a Semi-Infinite Flat Plate

Solution The plate surface temperature Tw is ( )0.6 x . Tw = 43 + 30 30 The plate temperature varies from 40 ∘ C to 73 ∘ C. Its average temperature is L 30 [ ( )] 1 1 x TW = Tw dx = 43 + 30 dx = 61.75 ∘ C L ∫0 30 ∫0 30 The film temperature Tf is Tw + T∞ 61.5 + 17 = = 39.37 ∘ C ≈ 312.37 K. 2 2 Air properties at this temperature are given as Tf =

ν = 17.13 × 10−6 m2 ∕s,

k = 0.0271 W/m.K,

Pr = 0.705.

The Reynolds number ReL is 5 × 0.3 UL = = 87 565. ν 17.13 × 10−6 Flow is laminar since ReL < 500 000. The surface temperature variation consists of a step jump at the leading edge of the plate plus an increase along the plate. Then, the surface temperature can be written in the following form: ReL =

Tw − T∞ = 26 + 30(x∕0.3)0.6 = 26 + 61.78x0.6 where T∞ = 17 ∘ C. Step jump at the leading edge is 26 ∘ C and the exponent of temperature variation along the plate is n = 0.6. Solutions are obtained by Maple 2020 ( ) dθ = 0.2934 for n = 0. dη η=0 ( ) dθ = 0.4238 for n = 0.6. dη η=0 Heat flux is [ ( )] √ q′′w x 5x 61.78 x0.6 = 0.2934 + (0.4238) 0.0271 × 26 26 17.13 × 10−6 q′′w =

111.68 0.1 √ + 383.34x . x

The variation of heat flux with axial position is shown in Figure 7.E10. We may now generalize this superposition idea. Variation of local heat flux along the plate.

1600

1400

1200 qʺw(x)

Figure 7.E10

1000

800

600 0.05

0.10

0.15 X

0.20

0.25

0.30

247

248

7 Laminar External Boundary Layers: Momentum and Heat Transfer

(a) θ0 is the similarity solution for the flat plate with uniform wall temperature Tw0 − T∞ = Tw0 = a0 . We can now write the solution of the energy equation as (T − T∞ )0 = (1 − θ0 )(Tw0 − T∞ ) or (T − T∞ )0 = (1 − θ0 )a0 .

(7.250a)

(b) θ1 is the similarity solution for the wall temperature distribution Tw1 − T∞ = a1 x. We can now write the solution of the energy equation as (T − T∞ )1 = (1 − θ1 )(Tw1 − T∞ ) or (T − T∞ )1 = (1 − θ1 ) a1 x. (c) θ2 is the similarity solution for the wall temperature distribution Tw2 − T∞ = a2 the energy equation

(7.250b) x2 .

We can now write the solution of

(T − T∞ )2 = (1 − θ2 )(Tw2 − T∞ ) or (T − T∞ )2 = (1 − θ2 )a2 x2 .

(7.250c)

Note θ0 , θ1 , and θ2 are the similarity solutions for the three different wall temperature distributions. Then, the solution for the three wall temperature distributions is (T − T∞ ) = (1 − θ0 )a0 + (1 − θ1 )a1 x + (1 − θ2 )a2 x2 . We now evaluate the wall heat flux ) ) ) ] [( ( ( 𝜕θ 𝜕θ 𝜕θ q′′w = −k − 0 a0 + − 1 a1 x + − 2 a2 x2 . 𝜕y 𝜕y 𝜕y √ Using the similarity variable η = (y∕x) Rex , we have )( ) )( ) )( ) ] ( [( ( dθ0 dθ2 dθ1 𝜕η 𝜕η 𝜕η a0 + a1 x + a2 x2 q′′w = k dη 𝜕y 0 dη 𝜕y 1 dη 𝜕y 3 [ ( ) ) ) ( ] ( dθ0 √ dθ2 √ dθ1 √ k 2 a = Rex + a1 x Rex + a2 x Rex . x 0 dη dη dη ( ) √ dθ Recall that Nu = dη Rex , and we can write the following relation: q′′w

(7.251)

(7.252)

(7.253)

(7.254)

wall

k q′′w = [a0 Nu0 + a1 xNu1 + a2 x2 Nu2 ] (7.255) x where Nu0 is the Nusselt number for constant wall temperature Tw − T∞ = Tw0 = a0 , Nu1 is the Nusselt number for a wall temperature variation Tw − T∞ = a1 x, and Nu2 is the Nusselt number for a wall temperature variation Tw − T∞ = a2 x2 . Eckert and Drake [33] give the following general relation for this case: q′′w =

∞ ∑ Nui k Nu0 ai xi x Nu0 i=0

where Nui is the Nusselt number for a wall temperature variation according to Tw − T∞ = B xn .

(7.256)

7.10 Viscous Incompressible Constant Property Flow over a Wedge

7.10 Viscous Incompressible Constant Property Flow over a Wedge (Falkner–Skan Wedge Flow): Similarity Solution for Uniform Wall Temperature Boundary Condition Heat transfer from a wedge surface can be obtained by solving the laminar boundary layer energy equation. Flow is steady and incompressible. Fluid properties are assumed to be constant. Viscous dissipation is neglected. The energy equation and its boundary conditions are given as u

𝜕T 𝜕T 𝜕2 T +v =α 2 𝜕x 𝜕y 𝜕y

(7.257)

T(0, y) = T∞

(7.258a)

T(x, 0) = Tw

(7.258b)

T(x, ∞) = T∞

(7.258c)

where α = k/ρcp is the thermal diffusivity. Wedge surface is at constant temperature Tw . Dimensionless temperature θ is defined as T − Tw θ= . (7.259) T∞ − Tw The dimensionless temperature θ is a function of the similarity parameter η, that is θ = θ(η), and the energy equation and its boundary conditions are expressed in terms of dimensionless temperature θ(η) u

𝜕θ 𝜕2 θ 𝜕θ +v =α 2 𝜕x 𝜕y 𝜕y

(7.260)

θ(0, y) = 1

(7.261a)

θ(x, 0) = 0

(7.261b)

θ(x, ∞) = 1.

(7.261c)

The energy equation will be transformed into an ordinary differential equation by the method of similarity transformation. Similarity transformation η and velocity components u and v are borrowed from laminar momentum boundary layer flow over a wedge. Recall that free stream velocity U∞ is U∞ = C xm .

(7.262)

Using Eq. (7.262), the similarity variable η is expressed as √ √ U∞ Cxm−1 =y . η=y νx ν Velocity components are u = U∞

df dη

( √ m+1 v = − C ν xm−1 2

(7.263)

(7.264) )[

( f−

]

) df 1−m η . 1+m dη

(7.265)

The partial derivative of temperature θ with respect to x is dθ 𝜕 η 𝜕θ = 𝜕x dη 𝜕 x √ ( ) m−3 dθ C m−1 𝜕θ = y x 2 . 𝜕x dη ν 2

(7.266)

249

250

7 Laminar External Boundary Layers: Momentum and Heat Transfer

The partial derivative of temperature θ with respect to y is [ √ ] dθ 𝜕η dθ 𝜕 𝜕θ Cxm−1 = = y 𝜕y dη 𝜕y dη 𝜕y ν √ dθ C m−1 𝜕θ = x 2 . 𝜕y dη ν

(7.267)

The second partial derivative of temperature θ with respect to y is [ ] √ ( ) 𝜕η d dθ 𝜕 𝜕θ 𝜕2 θ C m−1 2 = x = 𝜕y 𝜕y dη dη ν 𝜕y 𝜕y2 [ ] [√ ] √ d Cxm−1 Cxm−1 dθ = dη dη ν ν =

d2 θ C xm−1 . ν dη2

(7.268)

Substituting these equations along with u and v components of velocity into energy equation, we obtain the following form of the energy equation: ) ( dθ m+1 d2 θ Prf(η) = 0. (7.269) + 2 dη dη2 The boundary conditions are: θ(0) = 0

(7.270a)

θ(∞) = 1.

(7.270b)

Since we know the similarity solution of the momentum equation, we can obtain a similarity solution for the energy equation. Solutions may be obtained by bvp4c in MATLAB 2021a or Maple 2020. We will obtain the numerical solutions by Maple 2020. The local wall heat flux is obtained as ( ) ( ) ( ) 𝜕η 𝜕T dθ ′′ qw = −k = −k(T∞ − Tw ) 𝜕y y=0 dη η=0 𝜕y (√ ) ( ) U∞ dθ = −k(T∞ − Tw ) dθ η=0 νx ( ) √ k(Tw − T∞ ) dθ Rex (7.271) = x dθ η=0 where Re = U∞ x/ν is the Reynolds number. We can now define the local heat transfer coefficient as (√ ) ( ) U∞ (x) dθ k(Tw − T∞ ) ) (√ ( ) dθ η=0 νx q′′w U∞ (x) dθ = =k h= (Tw − T∞ ) (Tw − T∞ ) dθ η=0 νx (√ ) ( ) x U∞ (x) dθ k = dθ η=0 x ν or local Nusselt number is √ ( dθ ) Nux = Rex dθ η=0

(7.272)

(7.273)

√ xU where Rex = ν∞ is the local Reynolds number. Values of Nux ∕ Rex are presented in Table 7.8. Next, we are interested in the average Nusselt number for wedge of length L. The average heat transfer coefficient for a wedge of length L may be defined as L

h=

1 h(x) dx L ∫0

(7.274)

7.10 Viscous Incompressible Constant Property Flow over a Wedge

√ Table 7.8 Values of Nux ∕ Rex for uniform wall temperature boundary condition on a wedge, U∞ = C xm . 𝛃

m

Pr = 0.7

Pr = 1

Pr = 5

Pr = 10

Pr = 15

–0.624

–0.0904

0.19713

0.21702

0.32789

0.38937

0.43016

–0.314

–0.0476

0.2670

0.3014

0.5131

0.6436

0.7348

0

0

0.29268

0.33205

0.57668

0.72814

0.83411

π/5

0.1111

0.3331

0.37778

0.66917

0.85021

0.97692

π/2

0.333

0.38408

0.43999

0.79208

1.0113

1.16489

π

1

0.49586

0.57046

1.0434

1.33879

1.5457

8π/5

4

0.81355

0.93897

1.73889

2.23984

2.5909

where h(x) is the local heat transfer coefficient, and it is given by √ ] U∞ (x) dθ . h(x) = k νx dη η=0 Substituting this into the equation of the average heat transfer coefficient, we obtain √ √ ] ] L L U∞ (x) dθ U∞ (x) 1 k dθ h= k dx = dx. L ∫0 νx dη η=0 L dη η=0 ∫0 νx

(7.275)

(7.276)

The external free stream potential flow is given as U∞ = Cxm and this equation is substituted into the above equation to get √ ] ] √ L L√ k dθ k dθ C xm−1 C dx = h= xm−1 dx. L dη η=0 ∫0 ν L dη η=0 ν ∫0 Evaluating this integral, we obtain √ ] √ ] √ C 2 2 C Lm+1 k dθ k dθ m+1 h= L = L dη η=0 ν m + 1 L dη η=0 m + 1 ν √ √ ] ] U∞ (L)L (C Lm )L k dθ k dθ 2 2 = h= L dη η=0 m + 1 ν L dη η=0 m + 1 ν ] k dθ 2 √ h= ReL . L dη η=0 m + 1 Finally, the average Nusselt number is obtained as ] ( ) √ hL 2 dθ = NuL = ReL . k m + 1 dη η=0

(7.277)

(7.278)

(7.279)

(7.280)

Example 7.11 We wish to study heat transfer at the stagnation point line of an isothermal cylinder, as shown in Figure 7.E11. Notice that m = 1 corresponds to stagnation flow. Suppose that air with Pr = 0.7 is flowing over the cylinder. The air stream far from the cylinder has velocity V and temperature T∞ . We are interested in the local heat transfer coefficient at the stagnation point in symmetric two-dimensional flow about a long circular cylinder. Solution The two-dimensional stagnation point flow is a good approximation for flow near the stagnation line of a cylinder. The potential flow solution for a cylinder in the vicinity of stagnation point is x 2Vx U∞ = for small (a) R R

251

252

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Figure 7.E11

U∞ x

V

Heat transfer at the stagnation point.

Tw R

T∞

where x is measured along the circular perimeter away from the stagnation point. We will compare this equation with U∞ = Cxm , and we see that C = 2V/R = 4V/D and m = 1. It is more convenient use a Reynolds number based on the undisturbed free stream velocity ahead of the body, V, instead of the local free stream velocity, U∞ , just outside the boundary layer. Next, we define the Reynolds number ReD as ρ U∞ x ρ x ( 4Vx ) ρ VD (4 x2 ) 4 x2 = = = ReD 2 (b) Rex = 2 μ μ D μ D D and ( ) xh hD x x Nux = = = NuD . (c) k k D D √ We can now get the value of Nux ∕ Rex for the stagnation point from Table 7.8 for m = 1 and Pr = 0.7. Thus, √ Nux ≈ 0.496 Rex . (d) Substituting (b) and (c) into (d), we get √ ( ) 4 x2 x = 0.496 ReD 2 . NuD D D Thus, after simplification, we obtain the Nusselt number for the stagnation point of an infinitely long cylinder as √ NuD = 0.992 ReD . The effect of Prandtl numbers taken into account by curve fitting is discussed in [34] √ NuD = 1.14 ReD Pr0.4

(e)

(f)

(g)

VD hD where ReD = and NuD = . Expression (g) is reported to be very accurate for ν k 0.6 ≤ Pr ≤ 1.1.

7.11 Effect of Property Variation In general, if the fluid properties change with temperature, velocity and temperature profiles will change. This will yield different friction and heat transfer coefficients compared to constant property friction and heat transfer coefficients. All the solutions we discussed up to now are based on steady incompressible constant property flow. If the property values depend on temperature, we need a method to take care of the effect of temperature-dependent properties using semiempirical methods. For engineering calculations, constant property solutions obtained by analytic or experimental methods are corrected to account for property variation with temperature. Kays et al. [26], Ghiaasiaan [39], and Kakac et al. [40] discuss the property variation with temperature. For liquids: The viscosity variation is responsible for most of the effect, and the following method is used for correcting the constant property solution results for the effects of property variation ( )n μw Nu St = = (7.281a) Nucp Stcp μ∞ ( )m μw cf = (7.281b) cfcp μ∞

7.12 Application of Integral Methods to Heat Transfer Problems

where Tw is the surface temperature. T∞ is the free stream temperature. cp refers to constant property solutions or small temperature difference experimental results. ∞ refers to free stream. For gases: The viscosity, thermal conductivity, and density are functions of absolute temperature ) ( Tw n St Nu = = Nucp Stcp T∞ )m ( Tw cf = . cfcp T∞ (Tw > T∞ ) Heating Gas

m

U∞ = const Liquids

(7.282b)

(Tw < T∞ ) Cooling

n

−0.1

2D Stagnation point

(7.282a)

m

n

−0.01

−0.05

0.4

0.1

0.3

0.20

−0.25

0.09

0 0.07 −0.25

All properties are evaluated at free stream temperature for external flows. Kays et al. [26] and Kakac et al. [40] report that the exponents m and n are functions of geometry and type of flow and give detailed information about the exponents m and n. Information about m and n is too lengthy and is not included here. Interested reader may refer to these references.

7.12 Application of Integral Methods to Heat Transfer Problems In general, it is difficult to obtain exact solutions of boundary layer equations. The integral method is a powerful and useful technique to obtain an approximate solution of boundary layer problems encountered in engineering. The boundary layer conservation equations are integrated over the boundary layer thickness. It is based on the approximate velocity and temperature profiles in the boundary layer. This method was introduced by von Karman and developed by Pohlhausen. Detailed information is given in [6]. Example 7.12 Sodium, potassium, and mercury are called liquid metals. These have high thermal conductivities with α ≫ ν, and they have excellent heat transport properties and are used in nuclear reactors to remove heat. Liquid metal (Pr ≪ 1) flows over a horizontal flat plate subject to a uniform heat flux q′′0 . See Figure 7.E12. Heat transfer is taking place by forced convection. Assume a uniform velocity and linear temperature profile to determine the Nusselt number. y

T∞ Δ(X)

U∞

0 Figure 7.E12

x

qʺ0

qʺw (x) qʺ0 0

Geometry and problem description for Example 7.12.

x

253

254

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Solution We will use the integral energy equation −k

𝜕T || d = ρcp u(T − T∞ )dy 𝜕y ||y=0 dx ∫0

ρcp

d u(T − T∞ )dy = q′′w . dx ∫0

Δ

or Δ

Integrating from 0 to x, we get x

Δ

ρcp

∫0

u(T − T∞ )dy =

∫0

q′′w dx.

For our problem, q′′w = q′′0 is uniform heat flux. We assume a linear temperature profile T = a0 + a 1 y The boundary conditions are y = Δ,

T = T∞ 𝜕T −k = q′′0 . 𝜕y

y = 0,

Using boundary conditions, we evaluate the unknown coefficients. From the second boundary condition, we get −ka1 = q′′0 q′′0

. k The application of the first boundary condition yields ( ′′ ) q T ∞ = a0 + − 0 Δ k ( ′′ ) q0 Δ. a0 = T∞ + k a1 = −

Substituting expressions for a0 and a1 into the temperature profile, we get the temperature distribution in the fluid q′′0

(Δ − y). k We now substitute temperature and velocity profiles into the integral energy equation, and we obtain Δ ( q′′ ) x 0 ρcp U∞ (Δ − y)dy = q′′ dx. ∫0 0 ∫0 k T(x, y) − T∞ =

Carrying out the integration, we get √ √ αx . Δ= 2 U∞ Wall temperature distribution Tw is Tw = T(x, 0) = T∞ +

Δ q′′0

= T∞ +

k The local heat transfer coefficient h is

Δ q′′0 k

.

q′′0 q′′w k k k h= = = = = √ √ Tw − T∞ Δ √ Δ q′′0 αx 2x 2 U∞ k The local Nusselt number Nux is 1 √ hx = √ Pex Nux = k 2 where Pex is the Peclet number.

√

U∞ x . α

7.12 Application of Integral Methods to Heat Transfer Problems

Example 7.13 Liquid metal is flowing over a flat plate. Surface temperature of the plate varies by √ Tw − T∞ = x where x is measured from the leading edge of the plate. For liquid metals, we may assume u ≈ V since Pr ≪ 1. Assuming a third-order temperature profile, obtain the expression for the local Nusselt number. Solution We assume a third-degree polynomial for temperature T(x, y) = b0 + b1 y + b2 y2 + b3 y3 . The boundary conditions are given as y = 0,

T = Tw

y = Δ,

T = T∞

y = Δ,

𝜕T =0 𝜕y

y = 0,

𝜕2 T = 0. 𝜕y2

The fourth boundary condition is obtained from the energy equation since u = v = 0 on the solid wall. Constants of the temperature profile are determined using the boundary conditions, and these constants are given as b0 = Tw b2 = 0 3 1 (T − Tw ) 2 ∞ Δ 1 1 b3 = − (T∞ − Tw ) 3 . 2 Δ b1 =

Therefore, the assumed third-degree temperature distribution becomes [ ( ) ( )] 1 y 3 3 y − . T − T∞ = (T∞ − Tw ) 2 Δ 2 Δ The local heat transfer coefficient h is ( ) 𝜕T −k 𝜕y y=0 q′′w h= = . Tw − T∞ Tw − T∞ The local heat transfer coefficient is in terms of thermal boundary layer thickness Δ(x) is h=

3k . 2Δ

We find Δ(x) by the integral energy equation ( ) Δ d 𝜕T = u(T − T∞ )dy. −α 𝜕y y=0 dx ∫0 We substitute velocity and temperature profiles into the energy equation as ( ) [ ( ) Δ { ( )] } Tw − T∞ d 1 y 3 3 y 3 α = − dy V (T∞ − Tw ) 2 Δ dx ∫0 2 Δ 2 Δ

255

256

7 Laminar External Boundary Layers: Momentum and Heat Transfer

and we get the following expression after the evaluation of integral: ( ) (T − T ) 3 d 3 α w ∞ = {[Tw − T∞ ]Δ}. 2 V Δ 8 dx Surface tempeerature is given by √ Tw − T∞ = x and substitute into the expression given above (√ ) ( ) x 3 d √ 3 α = { x Δ} 2 V Δ 8 dx (√ ) ( ) x α d √ 4 = { x Δ}. V Δ dx √ Let z2 = x Δ. Then, we get ( ) α dz2 dz x = z2 = 2z3 4 V dx dx Δ(0) = 0 → z(0) = 0. The solution of this ODE is ( ) α 2 x = z4 . 4 V The elimination of z yields the thermal boundary layer thickness Δ(x) √( ) α x. Δ=2 V The Nusselt number Nux is x3k 3x hx = = . k k2Δ 2Δ Finally, the local Nusselt number can be expressed in the following form: √ 3 Vx 3 x . Nux = √( ) = 2 4 α αx 2 V Nux =

7.12.1 Viscous Flow with Constant Free Stream Velocity Along a Semi-Infinite Plate Under Uniform Wall Temperature: With Unheated Starting Length or Adiabatic Segment Laminar flow over a semi-infinite with an unheated starting length has practical applications in engineering. A practical application of this problem may be seen in electronic chip cooling. In chip cooling, heat source may be located at some distance from the leading edge. Thermal boundary layer begins to develop after velocity boundary layer. Consider the uniform, laminar, steady, two-dimensional, constant property flow over semi-infinite flat plate. Constant fluid free stream temperature and velocity are T∞ and U∞ , respectively. The temperature of the plate surface varies along the length x in a stepwise manner. The unheated length ξ of semi-infinite plate is insulated and is at temperature T∞ of the approaching fluid stream. On the other hand, for x > ξ, the heated section of the plate is at constant temperature Tw0 and Tw0 > T∞ . A thermal boundary layer begins stating at the distance x = ξ and the thickness of thermal boundary layer is denoted by Δ(x) and increases in the downstream direction of the flow, as sketched in Figure 7.12. We wish to determine the local Nusselt number. We assume profiles for both velocity u and temperature T in the boundary layer. The closer the assumed profiles to real profiles, the better the result of calculations will be. This implies the satisfaction of the boundary conditions. A third-order velocity profile is assumed as ( ) ( ) 1 y 3 3 y u − = . (7.283) U∞ 2 δ 2 δ

7.12 Application of Integral Methods to Heat Transfer Problems

Tw0 > T∞ Tw (x)

y δ (x) U∞, T∞

∆(x) for Pr>1

T∞ 0

Tw0

ξ

Figure 7.12

Tw0 T∞

x 0

x=ξ

Laminar flow over a flat plate with an unheated length.

For a two-dimensional impermeable flat plate with constant ρ, U∞ , Tw0 , the energy integral equation reduces to q̇ ′′w dΔ2 = ρcp U∞ (Tw0 − T∞ ) dx where Δ2 is ∞

Δ2 =

∫0

(

u U∞

)(

T − T∞ Tw0 − T∞

) dy.

The integral formulation of the energy equation may be expressed in the following form: ( ) Δ d 𝜕T −α = u(T − T∞ )dy. 𝜕y y=0 dx ∫0

(7.284)

The standard boundary conditions on temperature are y = 0,

T = Tw0

(7.285a)

y = Δ,

T = T∞

(7.285b)

y = Δ, y = 0,

𝜕T =0 𝜕y 𝜕2 T = 0. 𝜕y2

(7.285c) (7.285d)

The fourth boundary condition is obtained from energy equation since u = v = 0 on the solid wall. Assume a third-degree polynomial for temperature T(x, y) = b0 + b1 y + b2 y2 + b3 y3 .

(7.286)

Constants of the temperature profile are determined using the boundary conditions, and these constants are given as b0 = Tw0 b2 = 0 3 1 (T − Tw0 ) 2 ∞ Δ 1 1 b3 = − (T∞ − Tw0 ) 3 . 2 Δ b1 =

Therefore, the assumed third-degree temperature distribution becomes ( ) ( ) T − Tw0 3 y 1 y 3 = . − T∞ − Tw0 2 Δ 2 Δ

(7.287)

This equation can also be written in a different form. First, the left-hand side of Eq. (7.287) is expressed as T − Tw0 + T∞ − T∞ T − T∞ + T∞ − Tw0 T − Tw0 T − T∞ = = = + 1. T∞ − Tw0 T∞ − Tw0 T∞ − Tw0 T∞ − Tw0

(7.288)

257

258

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Then, the temperature distribution becomes [ ( ) ( )] T − T∞ 1 y 3 3 y + . = 1− Tw0 − T∞ 2 Δ 2 Δ

(7.289)

Knowing the dimensionless temperature distribution, the local heat transfer coefficient h at the wall is determined from ( ) 𝜕T −k ′′ 𝜕y y=0 qw h= = . (7.290) Tw0 − T∞ Tw0 − T∞ We evaluate the temperature gradient at the wall as ( ) 3 (Tw0 − T∞ ) 𝜕T . =− 𝜕y y=0 2Δ The local heat transfer coefficient h becomes 3 k . h= 2 Δ(x)

(7.291)

(7.292)

Thus, the heat transfer coefficient at the wall is determined by knowing the thermal boundary layer thickness Δ(x). The thermal boundary layer thickness Δ(x) is obtained using the integral form of the energy equation. Substituting these temperature and velocity profiles into the integral energy equation, we obtain { ( ) [ Δ ( ) }{ ( ) ( ) ]} 3α (Tw0 − T∞ ) d 1 y 3 1 y 3 3 y 3 y − = − + (Tw0 − T∞ ) 1 − dy. U∞ 2 Δ dx ∫0 2 δ 2 δ 2 Δ 2 Δ After integration, we get ( ) { [ ( )2 ( ) ]} T∞ − Tw0 d 3 Δ 4 3 3 Δ α = (T∞ − Tw0 )U∞ δ . − 2 Δ dx 20 δ 280 δ

(7.293)

Velocity boundary layer thickness δ(x) is known. Eliminating δ(x) in the above equation gives the first-order ordinary differential equation in terms of thermal boundary layer thickness Δ(x). However, Eq. (7.293) is simplified first. For Prandtl number greater than unity, the thermal boundary layer is smaller than the viscous boundary layer. That is: for Pr > 1,

Δ < δ.

Based on this condition, the last term in the integral energy equation, Eq. (7.293), can be neglected ( ) ( ) 3 Δ 2 3 Δ 4 ξ, the wall heat flux q′′w is ( ) ( ) 𝜕T 3 ′′ . (7.306) qw = −k = k(Tw0 − T∞ ) 𝜕y y=0 2Δ Substituting the expression for thermal boundary layer thickness Δ(x) into Eq. (7.306) yields [ ( )3∕4 ]−1∕3 ( ) ξ k 1∕2 ′′ 1∕3 qw = 0.3312 (Tw0 − T∞ )Pr Rex 1 − . x x The local heat transfer coefficient h(x) can be obtained as follows: [ ( )3∕4 ]−1∕3 ( ) q′′w ξ k 1∕2 1∕3 h(ξ, x) = h = Pr Rex 1 − = 0.3312 . Tw0 − T∞ x x The local Nusselt number Nux is obtained using the local heat transfer coefficient h(ξ, x) = h √ 0.3312Pr1∕3 Rex hx Nux = = [ ( )3∕4 ]1∕3 k ξ 1− x

(7.307)

(7.308)

(7.309)

x ≥ ξ. Here, notice that this equation is singular at x = ξ. This problem is similar to the singularity of wall stress in boundary layer theory as x → 0 and note that the boundary layer approximation is invalid near singularity. The results are valid for x − ξ ≫ Δ(x). However, in practice, we are usually interested in overall heat transferred over a finite region rather than on local values. The quantity we are interested in could be determined by numerical integration. Since the singularity is integrable, the result is acceptable with reasonable accuracy. We may also express Eq. (7.309) in terms local Stanton number St St =

−1∕2 Nux 0.332 Pr−2∕3 Rex h = [ = . ] ( )3∕4 1∕3 ρ cp U∞ Rex Pr ξ 1− x

(7.310)

Ameel [38] gives the following relation for the average Nusselt number NuL (averaged over heated portion of the plate) p )[ ( ( ) p+1 ] p+1 ξ p+2 L NuL = NuL,ξ=0 (7.311) 1− L−ξ L where NuL,ξ=0 is the average Nusselt number obtained from Eq. (7.146) and parameter p = 2 for laminar flow and p = 8 for turbulent flow. Remember that L is the total plate length and the quantity NuL,ξ = 0 is the average Nusselt number for a plate of length L when heating starts at the leading edge of the plate. Example 7.14 Air at 20 ∘ C and 1 atm flows at a velocity of 4 m/s over a flat plate. The length in the flow direction is 1 m and its width is 1 m. Assuming that heating starts at 0.25 m from the leading edge of the plate, compute the power required to maintain the plate surface at 100 ∘ C.

7.12 Application of Integral Methods to Heat Transfer Problems

Solution The mean fluid temperature Tf is Tf =

Tw0 + T∞ 100 + 20 = = 60 ∘ C 2 2

The fluid properties at this temperature are ρ = 1.0515 kg∕m3 ,

μ = 20.0 × 10−6 kg∕m.s,

k = 28.7 × 10−3 W∕m.∘ C

Pr = 0.702. Reynolds number is ρU∞ L 1.0515 × 4 × 1 = = 210 130 < 500 000. μ 20.0 × 10−6

ReL =

Thus, since ReL < 500 000, flow is laminar. Let us first compute the local heat transfer coefficient h [ ( )3∕4 ]−1∕3 ( ) ξ k 1∕2 . Pr1∕3 Rex 1 − h = 0.3312 x x The Reynolds number Rex is Rex =

ρU∞ x 1.0515 × 4 x = = 210 130 x. μ 20 × 10−6

Substituting property values into the equation for the local heat transfer coefficient h yields ( h = 0.3312 × h=

√

28.7 × 10−3 x

)

[ ) ]−1∕3 ( √ 0.25 3∕4 (0.7021∕3 )( 210130 x) 1 − x

3.87 . [ ( ) ]1∕3 0.25 0.75 x 1− x

The average heat transfer coefficient h is L

h=

1

1 1 3.87 h dx = dx ≈ 7.73 W∕m2 .K. ) ]1∕3 ( L − ξ ∫ξ 0.75 ∫0.25 √ [ 0.25 0.75 x 1− x

Total heat transfer q is q = hA(Tw − T∞ ) = h(L − ξ)W(Tw − T∞ ) = 7.73 × (1 × 0.75) × (100 − 20) ≈ 464 W.

Tw0 > T∞ y δ (x)

T∞

Tw (x)

∆(x)

U∞ 0

Figure 7.13

Tw0

Tw0 x

T∞ 0

Laminar flow over a flat plate with constant surface temperature.

261

262

7 Laminar External Boundary Layers: Momentum and Heat Transfer

7.12.1.1

The Plate Without Unheated Starting Length

Figure 7.13 shows a flat plate having a uniform surface temperature Tw . Both the velocity and temperature profiles start from the leading edge of the plate. The plate has no unheated starting section (i.e. ξ = 0). The temperature distribution is given as [ ( )] 1 y 3 3y − . T(x, y) = Tw0 + (T∞ − Tw0 ) 2Δ 2 Δ The earlier results we obtained reduce to [ ] 13 1 1∕3 0.975 Δ = = 1∕3 δ 14 Pr Pr 4.528 Δ = √ x Pr1∕3 Rex

(7.312a) (7.312b)

( ) k 1∕2 (Tw0 − T∞ )Pr1∕3 Rex x ( ) q′′w k 1∕2 = 0.3312 h= Pr1∕3 Rex Tw0 − T∞ x

q′′w = 0.3312

(7.313) (7.314)

√ xh = 0.3312Pr1∕3 Rex . k

Nux =

(7.315)

We may also define the average heat transfer coefficient h L

h=

1 1∕2 h dx = 0.6624ReL Pr1∕3 L ∫0

(7.316)

and average Nusselt number NuL NuL =

hL 1∕2 = 0.6624 ReL Pr1∕3 . k

(7.317)

7.12.2 Viscous Flow with Constant Free Stream Velocity Along a Semi-Infinite Plate with Uniform Wall Heat Flux: With Unheated Starting Length (Adiabatic Segment) Consider the uniform, laminar, steady, two-dimensional, constant property flow over semi-infinite flat plate. Fluid free stream temperature and velocity are T∞ and U∞ , respectively. The unheated length ξ of the semi-infinite plate is insulated and is at temperature T∞ . The heated section of the plate is subjected to a uniform heat flux q′′0 for x ≥ x0 . See Figure 7.14. We wish to determine the local Nusselt number and the temperature distribution. The unknown quantity is the wall-fluid temperature difference, Tw (x) − T∞ , that is the temperature variation along the plate. The application of Newton’s law of cooling gives the local heat flux q′′0 = h[Tw (x) − T∞ ]. y

(7.318) Tw(x) > T∞

″ (x) qw

Insulation T∞ U∞

∆(x)

T∞ 0 q″0 = 0 ξ Figure 7.14

Tw (x) q″0

δ (x)

q0″

x

0

x=ξ

Laminar flow over a flat plate with an unheated length subjected to uniform heat flux.

x

7.12 Application of Integral Methods to Heat Transfer Problems

From this relation, we can write the local heat transfer coefficient as h(x) =

q′′0 Tw (x) − T∞

.

(7.319)

The local Nusselt number can be written as x q′′0 . Nux = k[Tw (x) − T∞ ]

(7.320)

Therefore, once the surface temperature distribution is determined, the local Nusselt number Nux can be obtained. Surface temperature distribution Tw (x) is obtained using the integral form of the energy equation q′′w dΔ2 = ρcp (Tw − T∞ ) dx ( ) ( ) Δ T − T∞ u Δ2 = dy ∫0 U∞ Tw − T∞ or the integral equation can be expressed as ( ) Δ d 𝜕T −α = u(T − T∞ )dy. 𝜕y y=0 dx ∫0 A third-order polynomial for velocity profile is assumed: ( ) ( ) 1 y 3 3 y u − = u∞ 2 δ 2 δ and assume a third-order polynomial for the temperature profile T(x, y) T = b0 + b1 y + b2 y2 + b3 y3 .

(7.321)

The boundary conditions for the temperature are 𝜕T = q′′0 𝜕y

y = 0;

−k

y = Δ;

T = T∞

(7.322a) (7.322b)

𝜕T =0 𝜕y 𝜕2 T = 0. 𝜕y2

y = Δ; y = 0;

(7.322c) (7.322d)

The last boundary condition, Eq. (7.322d), is obtained from the energy equation since u = 0 and v = 0 on the plate surface. The four unknown coefficients are determined by the application of the boundary conditions ′′

2 Δq0 , 3 k

q′′0

q′′0

. 3kΔ2 Substituting these constants into Eq. (7.321), the temperature profile becomes ( )] ( ′′ ) [ (y) q0 1 y3 2 Δ − + . T − T∞ = k 3 Δ 3 Δ3 b0 = T∞ +

b1 = −

k

,

b2 = 0,

b3 =

(7.323)

The surface temperature is obtained by setting y = 0 in Eq. (7.323) given above, and the result is ′′

2 q0 Δ(x) (7.324) 3 k where T(x, 0) = Tw (x) is the surface temperature. The local heat transfer coefficient h is obtained using Eq. (7.324) as Tw (x) − T∞ =

h=

q′′0 Tw (x) − T∞

=

3 k . 2 Δ(x)

(7.325)

The local Nusselt number in terms of thermal boundary layer thickness Δ(x) is Nux =

3 x . 2 Δ(x)

(7.326)

263

264

7 Laminar External Boundary Layers: Momentum and Heat Transfer

It remains to determine the thermal boundary layer thickness Δ(x). Substituting the velocity and temperature profiles into the integral energy equation, we obtain the thermal boundary layer thickness Δ(x) ( ′′ ) { Δ[ ][ ] ( ′′ ) } q0 q0 1 y3 d 3 y 1 y3 2 α = U∞ − Δ − y + dy . (7.327) 3 2 k dx ∫0 2δ 2δ 3 3Δ k Evaluating integrals, we have { [ ( ) ]} α 1 Δ 3 d 1 Δ Δ2 − . (7.328) = U∞ dx 10 δ 140 δ For Pr ≫ 1, we have Δδ ≪ 1, and thus ( ) 1 Δ 1 Δ 3 . T∞

″ (x) qw

T∞ U∞

∆(x) 0

x

q0′′ Figure 7.15

q0″

δ (x)

0

x

Laminar flow over a flat plate subjected to uniform heat flux.

7.12.2.1 The Plate with No Unheated Starting Length

Figure 7.15 shows a flat plate having a uniform surface heat flux q′′0 . Both the velocity and temperature profiles start from the leading edge of the plate. Since the plate has no unheated starting length (i.e. ξ = 0), (Δ/x) becomes 3.594 Δ = . √ x Pr1∕3 Rex Wall temperature is obtained by setting ξ = 0 in Eq. (7.336) [ ] ( ) 1 x ′′ . Tw (x) − T∞ = q0 2.396 √ k Pr1∕3 Rex

(7.339)

The local heat transfer coefficient is ( ) √ k 3 k = 0.417 Pr1∕3 Rex . h= 2 Δ(x) x The local Nusselt number is √ Nux = 0.417Pr1∕3 Rex .

(7.340)

Since we know the uniform heat flux q′′0 and surface area A, total heat transfer is given by q = Aq′′0 , but we may wish to determine the average plate surface temperature. This average temperature may be evaluated from ( ) L q′′0 L 1 x dx (7.341) Tw (x) − T∞ = (Tw (x) − T∞ )dx = L ∫ξ L ∫ξ k Nux where Nux is an appropriate correlation. Using Eq. (7.342), we obtain the average surface temperature Tw (x) − T∞ =

q′′0 L

(7.342)

kNuL 1∕2

where NuL is given by NuL ≈ 0.680ReL Pr1∕3 .

7.13 Superposition Principle We have seen cases in which wall temperature or wall heat flux varies along the plate in a stepwise fashion. Other wall temperature or wall heat flux variations are also of interest to engineers. We have seen that the solution of problems involving both variable surface temperature and variable surface heat flux is difficult. The variation of the surface temperature or wall heat flux along the x-direction has two effects on the temperature boundary layer. It influences: (a) the shape of the temperature profile (b) the thickness Δ(x) of the thermal boundary layer

265

266

7 Laminar External Boundary Layers: Momentum and Heat Transfer

We will approach these problems with a superposition method. The superposition method is discussed by different researchers such as Kays et al. [26], White and Majdalani [27], Rubesin [42], and Hanna and Myers [46]. Superposition is an immensely powerful technique used to obtain the solutions of complex problems from solutions simpler problems. Consider constant property flows. In such flows, the solution of the momentum equation is independent of the solution of the energy equation. Since the boundary layer energy equation is linear in temperature, the superposition principle can be used to construct solutions to problems with variable wall temperatures and heat fluxes from simple step function solutions we have obtained in previous sections. Since the energy equation is linear, the sum of solutions is also a solution. Analytical solutions are difficult if the boundary condition is not a step change. Let us explain the fundamental concepts behind the superposition principle with simple examples involving slug flow. Slug flow is discussed in [43]. For this purpose, we study longitudinal inviscid flow over a flat plate with constant free stream velocity under uniform wall temperature. We assume that the fluid viscosity is vanishingly small. A fluid with negligible viscosity will not adhere to the plate wall. This solution will give us a feeling how heat transfer coefficient changes along the flow direction. This model also enables us to calculate the temperature variation and the heat transfer coefficient very easily.

7.13.1 Superposition Principle Applied to Slug Flow over a Flat Plate: Arbitrary Variation in Wall Temperature Consider steady incompressible inviscid flow over the semi-infinite plate. The flow is heated by a wall temperature distribution Tw (x). The wall temperature distribution is shown in Figure 7.16b. A thermal boundary layer of thickness Δ(x) is formed as shown in Figure 7.16a. Since flow is inviscid, there is no momentum boundary layer and flow is called slug flow. Axial conduction is neglected compared to conduction in the y-direction. 7.13.1.1

Boundary Condition: Single Step at x = 0

The boundary layer energy equation and its boundary conditions for this problem can be written as ρcp U∞

𝜕2 T 𝜕T =k 2 𝜕x 𝜕y

(7.343)

T(0, y) = T∞

(7.344a)

T(x, 0) = Tw0

(7.344b)

T(x, ∞) = T∞ .

(7.344c)

Notice that there is a step change in wall temperature above fluid temperature T∞ at x = 0 (The wall temperature has undergone a temperature step change from T∞ to Tw0 ). We now define a transformation θ = T − T∞ , and this transformation reduces two of the nonhomogeneous boundary conditions to homogeneous conditions without affecting the homogeneity of the differential equation. The formulation of the problem in θ becomes ρcp U∞

y

𝜕2 θ 𝜕θ =k 2 𝜕x 𝜕y U∞

(7.345)

T∞ Tw (x)

T∞

∆(x)

U∞

Tw0 T∞

0

Tw (a)

Figure 7.16

x

Tw0 – T∞ 0 x

x=0 (b)

(a) A constant temperature flat plate in inviscid flow. (b) Step change in surface temperature.

7.13 Superposition Principle

θ(0, y) = 0

(7.346a)

θ(x, 0) = Tw0 − T∞

(7.346b)

θ(x, ∞) = 0.

(7.346c)

We now define an auxiliary problem. The formulation of the auxiliary problem is U∞ 𝜕f 𝜕2 f = 2 α 𝜕x 𝜕y

(7.347)

f(0, y) = 0

(7.348a)

f(x, 0) = 1

(7.348b)

f(x, ∞) = 0

(7.348c)

where α = k/ρ cp is the thermal diffusivity. Here, we defined an auxiliary problem where we replaced the boundary condition (Tw0 − T∞ ) by a constant surface temperature of magnitude of unity. The solution of this problem is obtained by the Laplace transform method. Now, let us define the Laplace transform of f(x, y) with respect to x as Lx [f(x, y)] = f(s, y).

(7.349)

Taking the Laplace transform of the partial differential equation with respect to x, we obtain ] [ 2 ] [ U∞ 𝜕f 𝜕 f = L Lx x α 𝜕x 𝜕y2 U d2 f = ∞ [s f(s, y) − f(0, y)]. 2 α dy

(7.350) (7.351)

Using the boundary condition f(0, y) = 0, we obtain d2 f U∞ s f = 0. − α dy2

(7.352)

The Laplace transforms of the boundary conditions are 1 f(s, 0) = s f(s, ∞) = 0.

(7.353a) (7.353b)

The solution of this ordinary differential equation, Eq. (7.352), is obtained by Maple 2020 > ( > de ≔ diff(f(y), y, y) −

U∞ 𝛂

) ⋅ (s) ⋅ f(y) = 𝟎;

de ≔

U s f(y) d𝟐 =𝟎 f(y) − ∞ 𝟐 𝛂 dy

> sol ≔ dsolve(de, f(y)); √ sol ≔ f(y) = C𝟏 e

√ √ √ U∞ s y U∞ s y − √ √ 𝛂 𝛂 + C𝟐 e

1 Using the boundary condition f(s, ∞) = 0 yields that _C1 = 0, and using the boundary condition f(0, s) = gives us s 1 C2 = , and the solution becomes s ( √ ) U∞ √ 1 f = exp −y s . (7.354) s α

267

268

7 Laminar External Boundary Layers: Momentum and Heat Transfer

The inverse Laplace transform is obtained from [44] ( √ ) ( √ ) U∞ U∞ f(x, y) = erfc y = 1 − erf y 4αx 4αx

(7.355)

where f(x, y) is the solution for a unit step change in the wall temperature at x = 0. The solution in terms of θ is θ = (Tw0 − T∞ ) f(x, y)

(7.356a)

or the solution in terms of the temperature T(x, y) is T − T∞ = (Tw0 − T∞ ) f(x, y) or

(7.356b)

)] ) ( √ ( √ y U∞ y U∞ T − T∞ = (Tw0 − T∞ ) 1 − erf = (Tw0 − T∞ )erfc . 2 αx 2 αx [

7.13.1.2

(7.356c)

Boundary Condition: Two Steps at x = 0 and x = 𝛏1

Consider now the same problem with the following two steps. See Figure 7.17. This time, temperature has a step at x = 0 and x = ξ1 . For x ≥ ξ1 , the temperature distribution due to the effect of these two steps is ) ( √ y U∞ T(x, y) − T∞ = (Tw0 − T∞ ) erfc 2 αx ( √ ) U∞ y + (Tw1 − Tw0 ) erfc . (7.357) 2 α(x − ξ1 ) We may express the temperature in the following form: T(x, y) − T∞ = (Tw0 − T∞ )f(x, y) + (Tw1 − Tw0 )f(x − ξ1 , y)

(7.358)

where f(x − ξ1 , y) is

) ( √ U∞ y . f(x − ξ1 , y) = erfc 2 α(x − ξ1 )

7.13.1.3

Boundary Condition: Three Steps at x = 0, x = 𝛏1 , and x = 𝛏2

We again consider the same problem with the new boundary conditions. See Figure 7.18. The temperature distribution for x > ξ2 is ) ) ( √ ( √ U∞ y U∞ y + (Tw1 − Tw0 )erfc T − T∞ = (Tw0 − T∞ )erfc 2 αx 2 α(x − ξ1 ) ) ( √ U∞ y . + (Tw2 − Tw1 )erfc 2 α(x − ξ2 ) Tw (x) Tw1 Tw0

(Tw1 – Tw0)

(Tw0 – T∞) 0 T∞ Figure 7.17

x=0

x = ξ1

Wall temperature variation having two steps.

x

(7.359)

.

7.13 Superposition Principle

Tw (x) Tw2 (Tw2 – Tw1)

Tw1 (Tw1 – Tw0)

Tw0 (Tw0 – T∞)

T∞

0 x = ξ1

x=0 Figure 7.18

x = ξ2

x

Wall temperature variation with three steps.

Temperature distribution may be represented as T − T∞ = (Tw0 − T∞ ) f(x, y) + (Tw1 − Tw0 ) f(x − ξ1 , y) + (Tw2 − Tw1 ) f(x − ξ2 , y)

(7.360)

where f(x − ξ2 , y) is

) ( √ U∞ y . f(x − ξ2 , y) = erfc 2 α(x − ξ2 )

We wish to generalize these ideas. Consider now slug flow over a flat plate. The temperature steps from T∞ to Tw0 at x = ξ. See Figure 7.19. The response of fluid temperature due to the effect of a stepwise discontinuity in surface temperature at x = ξ is T(x, y) − T∞ = (Tw0 − T∞ ) f(x − ξ, y)

(7.361)

where f(x − ξ, y) is ⎛ y f(x − ξ, y) = erfc ⎜ ⎜2 ⎝

√

U∞ ⎞⎟ . α(x − ξ) ⎟ ⎠

It is known that heat transfer in a boundary layer flows involving arbitrarily specified wall temperature is an important engineering problem. The method used here is the superposition of step-wall temperature solutions. Rubesin [42] used this method to make heat transfer calculations. We are now in a position to generalize the discussion. We have examined sudden change in temperature along the boundary. Figure 7.20 shows an arbitrary temperature distribution on a surface. The wall temperature distribution is approximated by breaking up the surface temperature distribution into several constant temperature step functions. Thus, for a series of steps, the fluid temperature is the sum of the effects of the individual constant steps in wall temperature. Now, we generalize this relation as follows: T(x, y) − T∞ = (Tw0 − T∞ )f(x, y) + (Tw1 − Tw0 )f(x − ξ1 , y) + (Tw2 − Tw1 )f(x − ξ2 , y) + ..… (Tw,n − Tw,n−1 )f(x − ξn , y)

Tw (x) Tw0 0

Tw0 – T∞ T∞ x=ξ

Figure 7.19

x

Wall surface temperature variation with a step at x = ξ.

(7.362a)

269

270

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Tw (x)

Tw2 Tw1 Tw0 T∞ 0 Figure 7.20

ξ1

ξ2

x

Arbitrary temperature distribution on a surface.

G(x)

0

x

x=ξ

Figure 7.21

Function with a discontinuity.

or T − T∞ = (Tw0 − T∞ ) f(x, y) +

N ∑ f(x − ξn , y)[Tw (ξn ) − Tw (ξn−1 )] n=1 N

= (Tw0 − T∞ ) f(x, y) +

] [ ∑ T (ξ ) − Tw (ξn−1 ) (ξn − ξn−1 ) f(x − ξn , y) w n ξn − ξn−1 n=1

(7.362b)

where f(x, y) is the response of unit temperature input at the leading edge of the plate. Define ΔTw = Tw (ξn ) − Tw (ξn − 1 ) and Δξ = ξn − ξn − 1 . Letting N → ∞, Eq. (7.362b) can be expressed in terms of Stieltjes integral as given below ) | ( x dTw | dξ| T − T∞ = f(x − ξ, y) . (7.362c) | ∫0 dξ |Stieltjes The Stieltjes integral is discussed in [45], and we will very briefly discuss here. It is a shorthand notation for expressing a sum plus an integral. Consider now the integral, I, as x

I=

Φ(x)

∫0

dG(x) dx. dx

If Φ(x) and dΦ(x)/dx are well-behaved functions, there is no difficulty in finding the Riemann integral I. Suppose that G(x) is given as shown in Figure 7.21. A discontinuity in G(x) occurs at x = ξ. Everywhere except x = ξ, the integral is evaluated as Riemann integral. At x = ξ, we note that Φ(x) = Φ(ξ) and now x

I=

∫0

Φ(x)

ξ−ε dG(x) dG(x) dx = dx Φ(x) ∫0 dx dx

( Rieman)

7.13 Superposition Principle ξ+ε

+

∫ξ−ε

Φ(x)

x dG(x) dG(x) dx + dx. Φ(x) ∫ξ+ε dx dx

( Rieman)

(7.363)

Consider now the second integral on the right-hand side. We let ε → 0. ξ+ε

∫ξ−ε

Φ(x)

ξ+ε dG(x) dG(x) dx = Φ(ξ) dx ∫ dx dx ξ−ε

= Φ(ξ)[G(ξ + ε) − G(ξ − ε)] = Φ(ξ)ΔG(ξ).

(7.364)

We now consider a function with several jumps. See Figure 7.22. The integral may be written as x

I=

∫0

Ψ(x)

dG(x) dx + Ψ(ξ0 )ΔGξ0 + Ψ(ξ1 )ΔGξ1 + Ψ(ξ2 )ΔGξ2 … … dx

(7.365a)

Ψ(x)

N ∑ dG(x) Ψ(ξn )ΔG(ξn ). dx + dx n=0

(7.365b)

or x

I=

∫0

Stieltjes integral may be represented by an ordinary Riemann integral where G is continuous, plus a summation of the contribution of discontinuities, if any. We now return to Eq. (7.362c), and this equation may be expressed as ) | ) ( ( x x dTw dTw | dξ| dξ f(x − ξ, y) = f(x − ξ, y) T − T∞ = | ∫0 ∫0 dξ dξ |Stieltjes Riemann +

N ∑

f(x − ξn , y) ΔTw,n

n=0

all jumps in T (7.366a) ( +) ( −) where ΔTw,n = Tw ξn − Tw ξn . So, we have a Riemann integral plus a summation. The Riemann integral is evaluated without taking into consideration any discontinuities in the wall temperature distribution. In other words, the integral term accounts for the portions where continuous wall temperature variations take place. The dTw is the infinitesimal wall temperature variation at location ξ. We will talk about the summation term later. The summation accounts for any discontinuity in the wall temperature at ξn location. In other words, the second term on the right-hand side sums over N number of ( ) discontinuous jumps in wall temperature that may occur downstream of the leading edge. The temperature Tw ξ+n is the ( −) wall temperature immediately downstream of the temperature jump, and the temperature Tw ξn is the wall temperature immediately upstream of the temperature jump, which occurs at the position ξn . Each discontinuity gives an additional contribution to temperature distribution. We will use this concept to study the effect of axial variation of wall temperature variation. Figure 7.22

Function with several jumps.

G(x)

∆G2

∆G1 ∆G0

x = ξ1

x = ξ2

x

271

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7 Laminar External Boundary Layers: Momentum and Heat Transfer

Heat flux at the wall may be obtained as q′′w = −k

x 𝜕f(x − ξ, y) || 𝜕T || = −k | | ∫0 𝜕y |y=0 𝜕y |y=0

−k

(

dTw dξ

) dξ

N ∑ 𝜕f(x − ξn , y) || | ΔTw,n . 𝜕y |y=0 n=0

(7.366b)

Example 7.15 Consider slug flow over a flat plate. Suppose that surface temperature distribution is given by Tw (x) = T∞ + Bx. We wish to find temperature distribution T(x, y) in fluid and the wall heat flux. Solution Unit step solution at x = ξ is ( f(x − ξ, y) = 1 − erf

y

(

x

∫0

.

√ 2 α(x − ξ)∕U∞

Temperature distribution is { T(x, y) − T∞ =

)

1 − erf

)} (

y

√ 2 α(x − ξ)∕U∞

dTw dξ

) dξ +

∑

{

( 1 − erf

n=0

y

√ 2 α(x − ξn )∕U∞

)} ΔTw,n .

Since there is no step change in temperature distribution, the second term drops out. Thus, we have { ( )} x y T(x, y) − T∞ = 1 − erf (B)dξ. √ ∫0 2 α(x − ξ)∕U ∞

Heat flux q′′w

q′′w

at the wall is

{ ( )}| x | y 𝜕 T || 𝜕 | (B)dξ 1 − erf = −k = −k √ | | ∫ 𝜕 y |y=0 | 0 𝜕y 2 α(x − ξ)∕U∞ |y=0 [ ] √ x 2k B x kB 1 . = √ dξ = √ √ ∫ 0 π α∕U π α∕U x−ξ ∞

7.13.2

∞

Superposition Principle Applied to Slug Flow over a Flat Plate: Arbitrary Variation in Wall Heat Flux

The flow over the semi-infinite plate is heated by a uniform heat flux qw0 , and a thermal boundary layer of thickness Δ(x) is formed as shown in Figure 7.23a. Since flow is inviscid, there is no momentum boundary layer and flow is called slug flow (Figure 7.23b). Axial conduction is neglected compared to conduction in the y-direction. y

U∞

T∞

T∞

″ (x) qw

∆(x)

U∞ 0

x (a)

Figure 7.23

″ qw0

0 x=0

x (b)

(a) A constant heat flux over flat plate in inviscid flow. (b) Step change in surface heat flux variation.

7.13 Superposition Principle

7.13.2.1 Boundary Condition: Single Step at x = 0

The boundary layer energy equation and its boundary conditions for this problem can be written as ρcp U∞

𝜕2 T 𝜕T =k 2 𝜕x 𝜕y

T(0, y) = T∞ −k

𝜕T || = q′′w0 𝜕y ||y=0

T(x, ∞) = T∞ .

(7.367) (7.368a) (7.368b) (7.368c)

Notice that there is a step change in wall heat flux at x = 0. We define a function f(x, y) =

T(x, y) − T∞ q′′w0 ∕k

(7.369)

to reduce the nonhomogeneous boundary conditions to homogeneous conditions. The partial differential equation and its boundary conditions become U∞ 𝜕f 𝜕2 f = 2 α 𝜕x 𝜕y f(0, y) = 0 −

𝜕f || =1 𝜕y ||y=0

f(x, ∞) = 0.

(7.370) (7.371a) (7.371b) (7.371c)

The nonhomogeneous boundary condition is now a stepwise disturbance of unity. In other words, we now have a unit step change in wall heat flux. Solution of this problem is obtained by Laplace transform method. Now, let us define the Laplace transform of f(x, y) with respect to x as Lx [f(x, y)] = f(s, y). This will give us ] [ 2 ] [ U∞ 𝜕f 𝜕 f = L Lx x α 𝜕x 𝜕 y2

(7.372)

or U d2 f = ∞ [s f(s, y) − f(0, y)]. 2 α dy

(7.373)

Since f(0, y) = 0, the differential equation becomes d2 f U∞ s f = 0. − α dy2

(7.374)

The Laplace transform of the boundary conditions are df(s, 0) 1 =− dy s

(7.375a)

f(s, ∞) = 0.

(7.375b)

273

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7 Laminar External Boundary Layers: Momentum and Heat Transfer

The solution of this ordinary differential equation, Eq. (7.375a), is obtained by Maple 2020. The application of boundary conditions yields ( √ ) √ U∞ √ α 1 f= exp −y s . (7.376) U∞ s3∕2 α The inverse Laplace transform is obtained from [44] [ √ ( √ )] ( 2 ) y U∞ y U∞ xα exp − f(x, y) = 2 − y erfc . π U∞ 4αx 2 αx

(7.377)

Equation (7.377) is the solution of the problem resulting from a unit step change heat input. The solution of the original problem is ] [ f(x, y) (7.378a) T(x, y) − T∞ = q′′w0 k or ( √ [ √ )] ) ( 2 ( ′′ ) 2 y U∞ y y U∞ xα T(x, y) − T∞ = qw0 − erfc exp − . (7.378b) k π U∞ 4αx k 2 αx Wall temperature distribution may be obtained by setting y = 0 in Eq. (7.378b) ( √ ) xα 2 Tw (x) − T∞ = q′′w0 k π U∞

(7.379)

where T(x, 0) = Tw (x). The surface temperature distribution may be expressed as Tw (x) − T∞ = q′′w0 F(x)

(7.380)

where F(x) is

) ( √ αx 2 . F(x) = k π U∞

We have local Nusselt number Nux q′′w0 q′′w0 x hx x = = ( √ ) k Tw (x) − T∞ k k q′′w0 xα 2 k π U∞ √ √ x π U∞ = = 0.886 Rex Pr1∕2 . 2 αx Consider the same problem with a new boundary condition as defined below. Nux =

7.13.2.2

(7.381)

Boundary Condition: Two Steps at x = 0 and x = 𝛏1

Heat flux variation with two steps is shown in Figure 7.24. Wall temperature distribution for x > ξ1 may be written as √ ( √ ) ) ⎛ 2 α (x − ξ1 ) ⎞ ( ′′ αx 2 ′′ ′′ ⎟. (7.382) + qw1 − qw0 ⎜ Tw (x) − T∞ = qw0 ⎜k k π U∞ π U∞ ⎟ ⎠ ⎝ Temperature may be represented as ) ( Tw (x) − T∞ = q′′w0 F(x) + q′′w1 − q′′w0 F(x − ξ1 ) where F(x − ξ1 ) is F(x − ξ1 ) = A A=

2 k

√

√

x − ξ1

α . π U∞

Consider the same problem with new boundary conditions as given below.

(7.383)

7.13 Superposition Principle ″ (x) qw ″ qw1 ″ qw0

0

x

ξ1

Figure 7.24

Heat flux variation with two steps.

7.13.2.3 Boundary Condition: Triple Steps at x = 0, x = 𝛏1 , and x = 𝛏2

Heat flux variation with three steps is shown in Figure 7.25. Wall temperature distribution for x > ξ2 becomes √ √ ) 2 ( 2 α √ α √ Tw (x) − T∞ = q′′w0 x + q′′w1 − q′′w0 x − ξ1 k π U∞ k π U∞ √ ) 2 ( α √ + q′′w2 − q′′w1 x − ξ2 k π U∞

(7.384a)

or ) ) ( ( Tw (x) − T∞ = q′′w0 F(x) + q′′w1 − q′′w0 F(x − ξ1 ) + q′′w2 − q′′w1 F(x − ξ2 ) where F(x − ξ2 ) is F(x − ξ2 ) = A 2 k

A=

√

(7.384b)

√ x − ξ2

α . π U∞

Let us generalize these ideas. Consider now slug flow over a flat plate. Surface heat flux variation is shown in Figure 7.26. ″ (x) qw ″ qw2

″ – q″ qw2 w1

″ qw1 ″ – q″ qw1 w0

″ qw0

ξ1

0 Figure 7.25

ξ2

x

Heat flux variation with triple steps.

″ (x) qw

″ qw0

0 Figure 7.26

″ =0 qw0

ξ

x

Heat flux variation with single step along a flat plate.

275

276

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Temperature response of the fluid for step-in heat flux at x = ξ is √ √ ) ( 2 ⎛ ⎡ U∞ ⎞⎟⎤⎥ y U∞ y y α(x − ξ) ′′ ⎢ 2 ⎜ . T(x, y) − T∞ = qw0 − erfc exp − ⎜ 2 α(x − ξ) ⎟⎥ ⎢k π U∞ 4αx k ⎣ ⎝ ⎠⎦ Surface temperature is evaluated by setting y = 0 in Eq. (7.385) [ √ ] 2 α(x − ξ) ′′ Tw (x) − T∞ = qw0 . k π U∞

(7.385)

(7.386)

We may express surface temperature as Tw (x) − T∞ = q′′w0 F(x − ξ) where F(x − ξ) is

(7.387)

√

2 F(x − ξ) = k

α(x − ξ) . π U∞

It is known that heat transfer in a boundary layer flows involving arbitrarily specified heat flux is an important engineering problem. The method used here is the superposition of step-wall heat flux solutions. Hanna and Myers [46] and Hanna [47] used this method to make heat transfer calculations. We are now able to generalize the discussion. Consider surface heat flux distribution given in Figure 7.27. We may now write the surface temperature distribution ( ( ) ) Tw (x) − T∞ = q′′w0 F(x) + q′′w1 − q′′w0 F(x − ξ1 ) + q′′w2 − q′′w1 F(x − ξ2 ) ( ( ) ) (7.388) + q′′w3 − q′′w2 F(x − ξ3 ). … … + q′′w,n − q′′w,n−1 F(x − ξn ) where F(x) is the response of heat input at the leading edge of the plate, and we rearrange Eq. (7.388) as shown below ( ) N q′′w,n − q′′w,n−1 ∑ F(x − ξn )(ξn − ξn−1 ). (7.389) Tw (x) − T∞ = q′′w0 F(x) + (ξn − ξn−1 ) n=1 ( ) Introducing Δq′′w = q′′w,n − q′′w,n−1 and Δξ = (ξn − ξn − 1 ), and letting N → ∞, we may now replace summation by integral (

x

dq′′w dξ

)

N ∑

Δq′′w,n F(x − ξn ) (7.390) ∫0 n=0 ( ) ( ) where Δq′′w,n = q′′w ξ+n − q′′w ξ−n . Step-in heat flux at the leading edge is included in the summation. Equation (7.391) may be written in Stieltjes sense ( ′′ ) x dqw dξ|Stieltjes . Tw (x) − T∞ = F(x − ξ) (7.391) ∫0 dξ Tw (x) − T∞ =

F(x − ξ)

dξ +

Example 7.16 Consider slug flow over a flat plate. The flat plate has heat flux variation with a single step, as shown in Figure 7.E16. We wish to determine the wall temperature distribution. Figure 7.27

″ (x) qw ″ qw2 ″ qw1 ″ qw0

0

ξ1

ξ2

x

Arbitrary heat flux distribution on a surface.

7.13 Superposition Principle

Figure 7.E16

Single step heat flux distribution along the plate.

qʺw (x) qʺw 0 qʺw = 0 0

a

x

Solution We have a step at x = a. For x > a, the unit step solution is √ 2 α(x − ξ) . F(x − ξ) = k π U∞ We now apply Eq. (7.391). Then, we obtain ( ′′ ) N x ∑ dqw dξ + F(x − ξ) Δq′′w,n F(x − ξn ) Tw (x) − T∞ = ∫0 dξ n=0 ) ( √ √ α 2 q′′w0 x − a. Tw (x) − T∞ = 0 + k π U∞ We have only one step and do not have continuous function. For this reason, integral in Eq. (7.391) is zero. Example 7.17 Consider slug flow over a flat plate. Heat flux variation is shown in Figure 7.E17. The variation of heat flux is given as q′′w = C x. We use Eq. (7.391). Then, we obtain, We wish to determine the wall temperature distribution. ( ′′ ) N x ∑ dqw dξ + F(x − ξ) Δq′′w,n F(x − ξn ). Tw (x) − T∞ = ∫0 dξ n=0 We do not have a step, and for this reason, summation in Eq. (7.391) is zero. We have only a continuous function ) ( √ x√ α 2 Tw (x) − T∞ = C x − ξ dξ + 0. ∫0 k π U∞ Integration by Maple 2020 yields x

>

∫𝟎

√ x − 𝛏 d𝛏;

𝟐 x𝟑∕𝟐 𝟑 Wall temperature distribution becomes ) [ ( √ ] α 2 3∕2 2 x . Tw (x) − T∞ = C k π U∞ 3 Later, we will use the superposition principle to build a general formula for variable wall heat flux boundary condition on a surface. Figure 7.E17

Linear heat flux distribution along the plate.

qʺw (x)

0

x

277

278

7 Laminar External Boundary Layers: Momentum and Heat Transfer

7.13.3 Superposition Principle Applied to Viscous Flow over a Flat Plate: Stepwise Variation in Wall Temperature Now, consider steady flow of a viscous fluid with constant free stream velocity U∞ and temperature T∞ along a flat plate. The variation of plate temperature Tw (x) is given in Figure 7.28. The boundary layer flow over the flat plate is steady and laminar. Fluid properties are constant. Given the wall temperature Tw (x), we are interested in determining the fluid temperature T(x, y), local heat transfer coefficient, and local Nusselt number. Recall that we have obtained a solution for unheated starting length problem over a flat plate by the integral method. The boundary layer energy equation is ) ( 𝜕T 𝜕2 T 𝜕T +v =k 2. ρ cp u (7.392) 𝜕x 𝜕y 𝜕x With the following boundary conditions: { Tw0 x < ξ1 y = 0; T= Tw1 x > ξ1

(7.393a)

y → ∞;

T = T∞

(7.393b)

x = 0;

T = T∞

(7.393c)

The energy equation is linear, and the solution T(x, y) may be written as T(x, y) − T∞ = T0 (x, y) + T1 (x, y).

(7.394)

We can now break up the problem as follows: 7.13.3.1

First Problem

ρ cp

(

𝜕T 𝜕T u 0 +v 0 𝜕x 𝜕y

) =k

𝜕 2 T0 𝜕x2

(7.395)

with following boundary conditions x>0

T0 = Tw0 − T∞ ;

y = 0; y → ∞;

(7.396a)

T0 = 0

x = 0;

(7.396b)

T0 = 0.

(7.396c)

The solution of this problem may be written as [ ( ) ( )3 ] 1 y 3 y + T0 (x, y) = (Tw0 − T∞ ) 1 − 2 Δ0 2 Δ0

Figure 7.28

Tw (x) Tw1 Tw0 T∞ x=0

x > 0.

ξ1

x

(7.397)

Flat plate with stepwise variation in the wall temperature.

7.13 Superposition Principle

7.13.3.2 Second Problem

The energy equation in this second problem is ) ( 𝜕T 𝜕2 T 𝜕T ρ cp u 1 + v 1 = k 21 𝜕x 𝜕y 𝜕x with the following boundary conditions { 0, 0 < x < ξ1 y = 0; T1 = Tw1 − Tw0 , x > ξ1

(7.398)

x>0

(7.399a)

y → ∞;

T1 = 0

(7.399b)

x = ξ1 ;

T1 = 0.

(7.399c)

The solution to the second problem is [ ( ) ( )3 ] 1 y 3 y + T1 (x, y) = (Tw1 − Tw0 ) 1 − 2 Δ1 2 Δ1

(7.400)

We can now construct the temperature distribution of the original problem. The temperature distribution is given by T(x, y) − T∞ = T0 (x, y);

0 < x < ξ1

(7.401)

and T(x, y) − T∞ = T0 (x, y) + T1 (x, y);

x > ξ1

(7.402)

In Eqs. (7.401) and (7.402), Δ0 and Δ1 may be calculated from Δ0 = δr0 ; Δ1 = δr1 ;

0.976 r0 = √ 3 Pr

[ ( )3∕4 ]1∕3 ξ1 0.976 1− r1 = √ 3 x Pr

4.64 δ = √ . x Rex Next, we can write the local heat flux q′′w for the original problem. 7.13.3.3 Heat Flux for 0 < x < 𝛏

First, we write the temperature distribution [ ( ) ( )3 ] 1 y 3 y . + T(x, y) − T∞ = (Tw0 − T∞ ) 1 − 2 Δ0 2 Δ0 Using Fourier law, we have ( ) ( ( ) ) 𝜕T 3 1 3 1 = −k(Tw0 − T∞ ) − q′′w = −k = k(Tw0 − T∞ ) 𝜕y y=0 2 Δ0 2 δr0 ( ) √ √ 3 Rex Pr 3 q′′w = k(Tw0 − T∞ ) × × 2 4.64 x 0.976 ( )√ k q′′w = 0.331 Rex Pr1∕3 (Tw0 − T∞ ) x or q′′w = h(x)(Tw0 − T∞ )

(7.403)

279

280

7 Laminar External Boundary Layers: Momentum and Heat Transfer

where h(x) is h(x) = 0.331 7.13.3.4

( )√ k Rex Pr1∕3 . x

The Heat Flux for x > 𝛏1

First, we write the temperature distribution [ [ ( ) ( )3 ] ( ) ( )3 ] 1 y 1 y 3 y 3 y T(x, y) − T∞ = (Tw0 − T∞ ) 1 − + + + (Tw1 − Tw0 ) 1 − 2 Δ0 2 Δ0 2 Δ1 2 Δ1 ) ( ) ( 1 1 3 𝜕T 3 × + k(Tw1 − Tw0 ) × × = k(Tw0 − T∞ ) q′′w = −k 𝜕y y=0 2 δ × r0 2 δ × r1 and finally, we get q′′w

[ ⎧ ( )3∕4 ]−1∕3 ⎫ ( )√ ξ1 ⎪ k 1∕3 ⎪ = 0.331 Rex Pr ⎨(Tw0 − T∞ ) + (Tw1 − Tw0 ) 1 − ⎬ x x ⎪ ⎪ ⎩ ⎭

(7.404)

or q′′w = h(x)(Tw0 − T∞ ) + h(ξ1 , x)(Tw1 − Tw0 )

(7.405)

We can now generalize Eq. (7.405) as follows: q′′w = h(x)(Tw,0 − T∞ ) +

N ∑ h(ξn , x)(Tw,n − Tw,n−1 )

(7.406)

n=1

where h(x) and h(ξn , x) are ( ) √ k h(x) = 0.331 Pr1∕3 Rex x and

(7.407)

[ ( )3∕4 ]1∕3 ( ) √ ξn k Pr1∕3 Rex 1 − h(ξn , x) = 0.331 . x x

(7.408)

Example 7.18 Flat plate is cooled by air at 20 ∘ C and 1-atm pressure. Heat transfer is accomplished by forced convection. Air is flowing over the flat plate with a free stream velocity of 15 m/s. The length of the flat plate is 20 cm. The plate is heated by embedded heating elements, and power supply can be adjusted to obtain the desired temperature distribution. The surface temperature distribution across the plate is given as follows: 0 − 5 cm

T = 40 ∘ C

5 − 10 cm

T = 60 ∘ C

x > 10 cm

T = 80 ∘ C

Calculate the heat flux at 20 cm from the leading edge of the plate. Determine the local heat transfer coefficient. Solution Variation in surface temperature is shown in Figure 7.E18. We wish to develop an expression for the heat flux. In every region encountered when moving from the leading edge to x = 10 cm, we may write the heat flux as Tw0 = 40 ∘ C

Tw1 = 60 ∘ C

ξ1 = 5 cm

ξ2 = 10 cm

Tw2 = 80 ∘ C

7.13 Superposition Principle

Tw (x)

Tw 2

Tw 1 Tw 0 T∞ 0 Figure 7.E18

ξ1

ξ2

x

Stepwise temperature distribution along the plate.

Air properties are evaluated at the film temperature. At x = 10 cm, the air temperature is 80 ∘ C and film temperature. 80 + 20 = 50 ∘ C = 323 K Tf = 2 Air properties at this temperature are ν = 18.2 × 10−6 m2 ∕s k = 0.028 W∕m.K Pr = 0.703. The Reynolds number Rex is U x 15 × 0.20 = 164 800. Rex = ∞ = ν 18.2 × 10−6 Since Rex < 500 000, flow is laminar. Heat flux from the plate in 0 < x < 5 is given by ( ) k 1∕2 q′′w = 0.332 Pr1∕3 Rex (Tw0 − T∞ ). x Heat flux from the plate in 5 < x < 10 is given by ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ( ) (T − T ) k 1∕2 w1 w0 ⎥. Pr1∕3 Rex ⎢(Tw0 − T∞ ) + [ q′′w = 0.332 ( )3∕4 ]1∕3 ⎥ ⎢ x ξ 1 ⎢ ⎥ 1− ⎢ ⎥ x ⎣ ⎦ Heat flux from the plate for x > 10 cm is given by ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ( ) (Tw1 − Tw0 ) (Tw2 − Tw1 ) k 1∕2 ⎥. q′′w = 0.332 Pr1∕3 Rex ⎢(Tw0 − T∞ ) + [ +[ ] ] 1∕3 1∕3 ( )3∕4 ( )3∕4 ⎢ ⎥ x ξ ξ 1 2 ⎢ ⎥ 1 − 1 − ⎢ ⎥ x x ⎣ ⎦ We now substitute numerical values for to evaluate the heat flux at x = 20 cm ⎡ ⎤ ⎢ ⎥ ( ) √ (60 − 40) (80 − 60) 0.028 ⎥ q′′w = 0.332 + (0.703)1∕3 (164 800) ⎢(40 − 20) + [ ] [ ] ( )3∕4 1∕3 ( )3∕4 1∕3 ⎥ ⎢ 0.20 5 10 ⎢ ⎥ 1− 1− ⎣ ⎦ 20 20 q′′w = 1177 W∕m2 .

281

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7 Laminar External Boundary Layers: Momentum and Heat Transfer

If there are N number of steps present in the surface temperature distribution over a distance x, heat flux q′′w can be obtained from the relation given below q′′w = 0.332

N ( ) ∑ ΔTw,i k 1∕2 Pr1∕3 Rex [ ) ]1∕3 ( x i=0 ξi 3∕4 1− x

where ξi is the distance of the ith step-in surface temperature measured from the leading edge and ΔTw0 = Tw0 − T∞ .

7.13.4 Flux

Superposition Principle Applied to Viscus Flow over a Flat Plate: Stepwise Variation in Surface Heat

Viscous fluid is flowing over a flat plate. Consider now a two-dimensional laminar boundary layer problem over a flat plate. The flow is incompressible with constant properties. Free stream velocity U∞ and temperature T∞ are constant. Suppose that wall heat flux distribution is given, as in Figure 7.29. We wish to develop an expression by assuming third-order velocity and temperature profiles by the integral method to estimate the wall temperature variation with distance x measured from the leading edge of the plate. The energy equation and its boundary conditions are u

𝜕T 𝜕2 T 𝜕T +v =α 2 𝜕x 𝜕y 𝜕y

x=0 y=0 y=0 y=∞

(7.409)

T = T∞ ( ) 𝜕T = q′′w0 −k 𝜕y y=0 ( ) 𝜕T = q′′w1 −k 𝜕y y=0

(7.410a) 0 < x < ξ1

(7.410b)

x > ξ1

(7.410c) (7.410d)

T = T∞

The energy equation is linear, and the solution may be written as T − T∞ = θ1 (x, y) + θ2 (x, y)

(7.411)

We can now split the problem into two simpler problems. 7.13.4.1

First Problem

The first problem is valid in 0 < x < ξ1 u

𝜕θ 𝜕2 θ 𝜕θ1 + v 1 = α 21 𝜕x 𝜕y 𝜕y

x=0 y=0 y=∞

(7.412)

θ1 = 0 ) ( 𝜕θ1 −k = q′′w0 𝜕y y=0

(7.413a) 0 < x < ξ1

θ1 = 0

(7.413c) Figure 7.29

″ qw ″ qw1

″ qw0

0

ξ1

(7.413b)

x

Variation of heat flux.

7.13 Superposition Principle

We have already obtained the solution to this problem by the integral method [ ( ) ( )3 ] q′′w0 Δ1 2 1 y 1 y θ1 = − + k 3 2 Δ1 3 Δ1

(7.414)

Δ1 3.594 = where Rex = U∞ x/ν is the Reynolds number and Pr = μcp /k is the Prandtl number. The √ x Pr1∕3 Rex temperature distribution for 0 < x < ξ1 may be written as [ ( ) ( )3 ] q′′w0 Δ1 2 1 y 1 y + − T(x, y) − T∞ = (7.415) k 3 2 Δ1 3 Δ1 where Δ1 is

Wall temperature distribution in this region is obtained by setting y = 0 [ ( ) ] x −1∕2 Pr−1∕3 Rex Tw (x) − T∞ = q′′w0 2.396 k

(7.416)

7.13.4.2 Second Problem

u

𝜕θ 𝜕2 θ 𝜕θ2 + v 2 = α 22 𝜕x 𝜕y 𝜕y

x=0 y=0 y=∞

(7.417)

θ2 = 0 ) ( 𝜕θ2 = q′′w1 − q′′w0 −k 𝜕y y=0

(7.418a) x > ξ1

θ2 = 0

(7.418b) (7.418c)

The solution to this problem is [ ) ( ′′ ( ) ( )3 ] qw1 − q′′w0 Δ2 2 1 y 1 y − + θ2 = k 3 2 Δ2 3 Δ2

(7.419)

where Δ2 is

[ ( )]1∕3 ξ1 Δ2 3.594 = 1 − . √ 1∕3 x x Pr Rex

We now construct the temperature distribution for the original problem T − T∞ = θ1 (x, y) + θ2 (x, y); or

x > ξ1

[

( ) ( )3 ] 2 1 y 1 y − + T − T∞ = k 3 2 Δ1 3 Δ1 [ ) ( ′′ ( ) ( )3 ] ′′ qw1 − qw0 Δ2 2 1 y 1 y − + + k 3 2 Δ2 3 Δ2 q′′w0 Δ1

(7.420a)

Wall surface temperature distribution for the original problem is obtained by setting y = 0 in Eq. (7.420b) ) ( ′′ ′′ ′′ 2 qw0 Δ1 2 qw1 − qw0 Δ2 Tw (x) − T∞ = + 3 k 3 k or substituting expressions for Δ0 and Δ1 into Eq. (7.421), we get { [ ( )]1∕3 } ) ( ′′ ξ1 x −1∕3 −1∕2 ′′ ′′ qw0 + qw1 − qw0 1 − Rex Tw (x) − T∞ = 2.396 Pr k x

(7.420b)

(7.421)

(7.422)

Example 7.19 Consider viscous fluid flow over a flat plate. The variation of surface heat flux is shown in Figure 7.E19. Air at atmospheric pressure is flowing over the plate with free stream temperature T∞ = 17 ∘ C and free stream velocity

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7 Laminar External Boundary Layers: Momentum and Heat Transfer

Figure 7.E19

qʺw (x)

Double step heat flux distribution along the plate.

100W/m2 50 W/m2

0

5

8

x (cm)

U∞ = 4 m/s. Heat transfer is taking place by forced convection under laminar flow conditions. Develop an expression for the variation of wall temperature with axial position x and estimate the temperature at x = 8 cm. Temperature distribution is { [ ( )]1∕3 } ) ( ′′ ξ x −1∕3 −1∕2 ′′ ′′ Rex qw0 + qw1 − qw0 1 − . Tw (x) − T∞ = 2.396 Pr k x We do not know the average surface temperature, and for this reason, we cannot evaluate the film temperature. As first approximation, we evaluate the air properties at T∞ = 17 ∘ C = 290 K. ν = 15 × 10−6 m2 ∕s k = 0.0255 W∕m.K Pr = 0.704 The Reynolds number Rex is Ux 4 × 0.08 = 21 333. = ν 15 × 10−6 Since Rex < 500 000, flow is laminar. Surface temperature distribution is { ) )] } [ ( ( 0.05 1∕3 x 1 (0.0704)−1∕3 √ 50 + (100 − 50) 1 − . Tw (x) = 17 + 2.396 0.0255 x 4x 15 × 10−6 Estimate the temperature at x = 8 cm = 0.08 m Rex =

Tw ≈ 20 ∘ C Now, we can estimate the film temperature and repeat the calculation.

7.14 Viscous Flow over a Flat Plate with Arbitrary Surface Temperature Distribution The heat transfer problem in a boundary layer flow involving arbitrary specified wall temperature is especially important in engineering. The method used to carry out nonisothermal heat transfer calculations is based on the superposition of step-wall temperature solutions. Step-wall temperature solution is reviewed briefly. Rubesin [42] used the superposition method to make heat transfer calculations for different wall temperature distributions. Rubesin has shown that the heat transfer rate for an arbitrary temperature variation may be determined by superimposing several “step temperature distribution.” Klein and Tribus [45] presented solutions for nonisothermal surfaces. Baxter and Reynolds [48] have given basic results for both laminar and turbulent flow past a flat plate based on step-wall temperature solutions. Levy [49] and Harnett et al. [50] presented solutions for nonisothermal surfaces. An experimental study on convective heat transfer to air from a flat plate with a stepwise discontinuous surface temperature was performed by Scesa [51]. Harnett et al. [52] present a simplified procedure. We can now employ the superposition ideas that we discussed before to develop a general formula for varying surface temperature.

7.14 Viscous Flow over a Flat Plate with Arbitrary Surface Temperature Distribution

Tw (x)

∆Tw2 Tw (x) ∆Tw1 Tw0

Tw0 – T∞ T∞ ξ0 = 0 ξ1

T∞

ξ2

ξn–1

dξ

ξn

x

(a) Figure 7.30

0

x (b)

Semi-infinite plate with variable surface temperature distribution.

Consider now steady incompressible viscous flow over a flat plate. We will consider the two-dimensional laminar boundary layer problem for variable surface temperature Tw (x). Suppose that wall temperature varies as illustrated in Figure 7.30. Next, we write the energy equation and its boundary conditions as u

𝜕T 𝜕2 T 𝜕T +v =α 2 𝜕x 𝜕y 𝜕y

(7.423)

y = 0;

T = Tw (x)

(7.424a)

y → ∞;

T = T∞

(7.424b)

x = 0;

T = T∞

(7.424c)

Here viscous dissipation is neglected. Under these conditions, the energy and momentum equations are uncoupled. Governing the energy equation of the problem under consideration is linear. The principle of superposition can be applied to obtain the temperature distribution in the fluid corresponding to the arbitrary continuous boundary condition Tw (x). Surface temperature distribution can be estimated by constant temperature steps. Then, the constant temperature solutions for each step are superposed. To find the total effect of wall temperature variation on fluid, we can sum over a series of small constant temperature steps as we have done before. We will employ Eq. (7.366a), and it is repeated here for convenience: ) | ( N x ∑ dTw | dξ| f(x − ξ, y) + f(x − ξn , y)ΔTw,n (ξn ) (7.425) T(x, y) − T∞ = | ∫ξ=0 dξ |Stieltjes n=0 Equation (7.425) will be written in the following form: ) ( x dTw T(x, y) − T∞ = (Tw0 − T∞ )f(x, y) + dξ f(x − ξ, y) ∫ξ=0 dξ +

N ∑

f(x − ξn , y)ΔTw,n (ξn )

(7.426)

n=1

where Tw (0) = Tw0 at the leading edge. This equation takes care of finite steps as well as continuous variations occurring ( ) ( ) in the wall temperature. Here, ΔTw,n (ξn ) = Tw,n ξ+n − Tw,n ξ−n denotes the temperature rise across the nth discontinuity ( ) in the wall temperature. The temperature Tw,n ξ+n is the temperature immediately downstream of temperature jump and ( −) Tw,n ξn is the temperature immediately upstream of the jump. Each discontinuity gives an additional contribution to local

285

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7 Laminar External Boundary Layers: Momentum and Heat Transfer

heat transfer. The integral portion represents an ordinary Riemann integral. It is evaluated without taking into consideration any discontinuities in the wall temperature distribution. On the other hand, the summation term accounts for the any discontinuity in the wall temperature distribution. Here, the combination of Riemann integral and the summation is called Stieltjes integral. Heat flux from the surface is determined using Fourier’s law ( ) 𝜕T (7.427) q′′w = −k 𝜕y y=0 We differentiate Eq. (7.426) with respect y and set y = 0; we obtain an expression for local heat flux on the plate ] ( ) x [ dTw 𝜕f(x − ξ, y) 𝜕f(x, y) || ′′ dξ −k qw = −(Tw0 − T∞ )k ∫ξ=0 𝜕y ||y=0 𝜕y dξ y=0 N ∑ 𝜕f(x − ξn , y) || −k | ΔTw,n (ξn ). 𝜕y |y=0 n=1

(7.428)

Recall that f(x − ξ, y) is a known quantity to be determined later. We will now express the heat flux in terms of the heat transfer coefficient h(ξ, x). Temperature distribution for a step change in wall temperature at x = ξ is T − T∞ = (Tw − T∞ )f(x − ξ, y). The local heat transfer coefficient h(ξ, x) resulting from a step change in surface temperature at x = ξ is obtained as follows: 𝜕f(x − ξ, y) || 𝜕T | −k || −k(Tw − T∞ ) | ′′ 𝜕y |y=0 𝜕y qw |y=0 = = h(ξ, x) = Tw − T∞ Tw − T∞ Tw − T∞ 𝜕f(x − ξ, y) || = −k (7.429) | . 𝜕y |y=0 The heat transfer coefficient h(x) at the leading edge of the plate is 𝜕T | 𝜕f | −k || −k(Tw − T∞ ) || ′′ 𝜕y 𝜕y qw |y=0 |y=0 𝜕f | h(x) = = = = −k || Tw − T∞ Tw − T∞ Tw − T∞ 𝜕y |y=0

(7.430)

We can also write h(ξn , x) = −k

𝜕f(x − ξn , y) || | 𝜕y |y=0

(7.431)

Notice that h(ξ, x) is the heat transfer coefficient at resulting from a step change in wall temperature at location x = ξ, and it was found by the integral method for a laminar boundary layer flow heat transfer. Substituting Eqs. (7.429)–(7.431) into Eq. (7.428), we obtain an expression for the heat flux q′′w . ) ( N x ∑ dTw dξ + h(ξ, x) h(ξn , x) ΔTw,n (7.432) q′′w = (Tw0 − T∞ )h(x) + ∫0 dξ n=1 Recall that velocity distribution Eq. (7.432) is for a particular U∞ . How do we obtain the kernel h(ξ, x)? First, we find f(x − ξ, y). Recall that we have obtained the temperature distribution for a step change in wall temperature at x = ξ by the integral method. [ ( ) ( )] 1 y 3 3 y + (7.433) T − T∞ = (Tw0 − T∞ ) 1 − 2 Δ 2 Δ or T − T∞ = (Tw0 − T∞ ) f(x − ξ, y) where f(x − ξ, y) is given as [ ( ) ( )] 1 y 3 3 y + . f(x − ξ, y) = 1 − 2 Δ 2 Δ

(7.434)

7.14 Viscous Flow over a Flat Plate with Arbitrary Surface Temperature Distribution

Notice that we have expression for the thermal boundary layer thickness Δ, which is given here for convenience [ ( )3∕4 ]1∕3 ξ 4.52 x . Δ= 1− √ 1∕3 x Pr U x∕ν ∞

We now evaluate for 𝜕f(x − ξ, y)/𝜕y|y = 0 [ √ ( )3∕4 ]−1∕3 Pr1∕3 U∞ x∕ν 𝜕 f(x − ξ, y) || ξ 3 3 1 1− | = − 2 Δ = − 2 × 4.52 𝜕y x x |y=0 [ √ ( )3∕4 ]−1∕3 Pr1∕3 U∞ x∕ν ξ 1− = −0.332 x x and for 𝜕f(x, y)/𝜕y|y = 0 , we have Pr 𝜕 f(x, y) || = −0.332 𝜕 y ||y=0

1∕3

√

U∞ x∕ν

(7.435)

(7.436)

x

Now, we express the kernel h(ξ, x) and h(ξn , x) as

[ ( )3∕4 ]−1∕3 𝜕f(x − ξ, y) || ξ k 1∕3 1∕2 h(ξ, x) = −k = 0.332 Pr Rex 1 − | 𝜕y x x |y=0 [ ( )3∕4 ]−1∕3 𝜕f(x − ξn , y) || ξn k 1∕3 1∕2 h(ξn , x) = −k = 0.332 Pr Rex 1 − | 𝜕y x x |y=0

(7.437a) (7.437b)

and the heat transfer coefficient h(x) at the leading edge is h(x) = −k

𝜕f(x, y) || k 1∕2 = 0.332 Pr1∕3 Rex 𝜕y ||y=0 x

(7.437c)

Finally, substituting Eqs. (7.437a)–(7.437c) into Eq. (7.432), we may write Eq. (7.432) in the following form: [ ) ( )3∕4 ]−1∕3 ( x dTw ξ k 1∕3 1∕2 k 1∕3 1∕2 ′′ qw = 0.332 Pr Rex (Tw0 − T∞ ) + 0.332 Pr Rex dξ 1− ∫0 x x x dξ [ ( )3∕4 ]−1∕3 N ξn k 1∕3 1∕2 ∑ + 0.332 Pr Rex ΔTw,i 1− x x n=1

(7.438)

Knowing the local heat transfer coefficient, we can get local wall heat flux distribution. Since this equation is not exact, the heat transfer performance predicted by this expression is compared with available exact solutions. Such a comparison is possible for the case where the wall temperature varies as Tw − T∞ = B xn

(7.439)

Harnett et al. [50] report that over the range of Prandtl numbers from 0.7 to 20 and values of n from 0 to 10, the deviation of the approximate results from the exact solutions is less than 5.5%. We may conclude that this approximate solution adequately represents the variation of local heat transfer coefficient within the laminar boundary layer flow heat transfer. Example 7.20 Air is flowing over a flat plate with a speed of 10 m/s. Air has a temperature and pressure 20 ∘ C and 1 atm, respectively. Plate surface temperature distribution is given in Figure 7.20. Calculate the local heat flux at x = 8 cm. Solution We will use Eq. (7.438) q′′w

k k 1∕2 1∕2 = 0.332 Pr1∕3 Rex (Tw0 − T∞ ) + 0.332 Pr1∕3 Rex ∫0 x x

x

[

) ( )3∕4 ]−1∕3 ( dTw ξ dξ 1− x dξ

287

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7 Laminar External Boundary Layers: Momentum and Heat Transfer

[ ( )3∕4 ]−1∕3 N ξn k 1∕3 1∕2 ∑ 1− + 0.332 Pr Rex ΔTw,n . x x n=1 The second term in Eq. (7.438) is neglected since there is no continuous function, and we have a step at the leading edge and a single step at ξ = ξ1 [ ( )3∕4 ]−1∕3 N ξn k 1∕3 1∕2 k 1∕3 1∕2 ∑ ′′ qw = 0.332 Pr Rex (Tw0 − T∞ ) + 0.332 Pr Rex ΔTw,n 1− x x x n=1 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ √ (T − T ) k w1 w0 ⎥. q′′w = 0.332 Rex Pr1∕3 ⎢(Tw0 − T∞ ) + [ ] ( )3∕4 1∕3 ⎥ ⎢ x ξ ⎢ ⎥ 1− ⎢ ⎥ x ⎣ ⎦ At x = 8 cm, the film temperature is 20 + 80 = 50 ∘ C 2 From air tables, properties at Tf = 50 ∘ C and 1 atm are Tf =

Pr = 0.703; Rex =

k = 0.028 W∕m.K;

ν = 18.2 × 10−6 m2 ∕s

U∞ x 10 × 0.08 = = 43 956 ν 18.2 × 10−6

⎡ ⎤ ⎢ ⎥ √ (0.0280) (80 − 50) ⎥ 43 956(0.703)1∕3 ⎢(50 − 20) + [ q′′w = 0.332 ] )3∕4 1∕3 ⎥ ( ⎢ 0.08 0.05 ⎢ ⎥ 1− ⎣ ⎦ 0.08 q′′w ≈ 1623 W∕m2 . Example 7.21 Viscous fluid is flowing over the plate with uniform free stream velocity U∞ and a constant free stream temperature T∞ . Fluid properties are constant and viscous dissipation is neglected. Flow is laminar over the entire plate. The wall surface temperature distribution over the flat plate is given by Tw (x) − T∞ = a + Bx where a and b are constants and x is the distance measured from the leading edge of the plate. See Figure 7.21. Obtain the local heat flux, the local heat transfer coefficient, and the local Nusselt number. Solution We have a single step in wall temperature at the leading edge of the plate and a linear variation in wall temperature starting at the leading edge. So, we have dTw /dξ = b, and there is only one term in the summation. We will use the following relation: ) ( N x ∑ dTw dξ + h(ξ, x) h(ξn , x) ΔTw,n q′′w = [Tw (0) − T∞ ]h(0, x) + ∫0 dξ n=1 where h(ξ, x) for the single-step solution at x = ξ is [ ( )3∕4 ]−1∕3 ξ k 1∕3 1∕2 . h(ξ, x) = 0.332 Pr Rex 1 − x x Thus, q′′w

[ [ ( )3∕4 ]−1∕3 ( )3∕4 ]−1∕3 ( ) x ( ) √ √ ξ ξ 1∕3 k 1∕3 k 1− 1− = 0.331 Rex Pr (B)dξ + 0.331 Rex Pr [Tw (0) − T∞ ] x ∫0 x x x

7.15 Viscous Flow over a Flat Plate with Arbitrarily Specified Heat Flux

where Tw (0) = Tw0 and Tw (0) − T∞ = a. Finally, this equation can be rearranged as [ ( )3∕4 ]−1∕3 ⎤ x ( )⎡ √ ξ ′′ 1∕3 k ⎢ qw = 0.331 Rex Pr B 1− dξ + a⎥ . ⎥ x ⎢ ∫0 x ⎣ ⎦ Integration is carried out by Maple 2020, and the result is ( ) √ k [a + 1.6122 B x]. q′′w = 0.331 Rex Pr1∕3 x The local heat transfer coefficient h(x) can be obtained as ( ) √ 0.331 Rex Pr1∕3 kx [a + 1.6122 B x] q′′w . h= = Tw − T∞ a + Bx The local Nusselt number is √ 1∕3 h x 0.331 Rex Pr [a + 1.6122 B x] Nux = = . k a + Bx Example 7.22 Consider an arbitrary surface temperature variation, as shown in Figure 7.E22. There is no step or jump in the surface temperature distribution. Obtain an expression for the surface heat flux distribution. Figure 7.E22

Arbitrary surface temperature distribution along the plate.

Tw (x)

0

Solution We will use Eq. (7.438) q′′w

k k 1∕2 1∕2 = 0.332 Pr1∕3 Rex (Tw0 − T∞ ) + 0.332 Pr1∕3 Rex ∫0 x x [ ( )3∕4 ]−1∕3 N ξn k 1∕3 1∕2 ∑ + 0.332 Pr Rex ΔTw,n . 1− x x n=1

x

ξ

x

[

) ( )3∕4 ]−1∕3 ( dTw ξ dξ 1− x dξ

We do not have any steps and heat flux becomes x ( ) √ (dTw ∕dξ) k Pr1∕3 Rex dξ. q′′w = 0.332 ( )3∕4 ]1∕3 ∫0 [ x ξ 1− x

7.15 Viscous Flow over a Flat Plate with Arbitrarily Specified Heat Flux In some heat transfer problems, the arbitrarily varying heat flux distribution is known, and the surface temperature distribution is required. Klein and Tribus [45] extended the calculation method to take care of arbitrary wall heat flux problems. Hanna and Myers [46] discusses the problem of heat transfer from a flat plate having a step-in wall heat flux in boundary layer flow. Hanna [47] presents a technique for the superposition of step-wall heat flux solutions of the boundary layer energy equation. Smith and Shah [53] presented approximate equations for calculating heat transfer over a flat plate with arbitrary heat flux variation. Now, as we have done before, we will again use the superposition principle for the variable

289

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7 Laminar External Boundary Layers: Momentum and Heat Transfer

dq′′w qw (x)

″ – q″ Δq″w,1 = qw,1 w,0 ″ – q″ Δq″w,2 = qw,2 w,1

″ Δqw2

qw (x) ″ Δqw1

q″w

q0

=0 q w0 ξ1 0 ξ=0

Figure 7.31

ξ2

dξ

x

0

x=ξ

x

(b)

(a)

(a–b) Semi-infinite plate with variable surface heat flux distribution.

heat flux problem. Heat flux variation is shown in Figure 7.31a. Surface heat flux distribution can be estimated by small uniform heat flux steps. Since the energy equation is linear with respect to temperature, a superposition technique can be used to obtain temperature distribution for arbitrary wall heat flux over the surface. Consider the energy equation and heat flux distribution, as shown in Figure 7.31b. u

𝜕T 𝜕2 T 𝜕T +v =α 2 𝜕x 𝜕y 𝜕y

at x = ξ,

T = T∞

at y = 0,

−k

at y = ∞,

(7.440) (7.441a)

𝜕T = q′′0 ; 𝜕y

x>ξ

(7.441b) (7.441c)

T = T∞

To get a solution, we will employ Eq. (7.391). Equation (7.391) is repeated here for convenience ( ′′ ) | x dqw | dξ| F(x − ξ) Tw (x) − T∞ = | ∫0 dξ |Stieltjes

(7.442a)

We can write Eq. (7.442a) as (

x

Tw (x) − T∞ =

∫0

F(x − ξ)

dq′′w dξ

) dξ +

N ∑ F(x − ξn )Δq′′w,n

(7.442b)

n=0

F(x − ξ) may be obtained by the integral method. We recall that we obtained the wall temperature distribution for a step change in wall heat flux at x = ξ by the integral method. This solution was obtained by assuming third-order polynomial for temperature and velocity using the integral method, and it was given as ( ′′ ) [ (y) ( )] q0 1 y 3 2 Δ − + . T(x, y) − T∞ = k 3 Δ 3 Δ Surface temperature is obtained by setting y = 0, and surface temperature distribution becomes ( ) ( q′′ ) 2 0 Tw (x) − T∞ = Δ 3 k where Tw (x) = T(x, 0) and the thermal boundary layer thickness Δ(x) is given by ]1∕3 [ ξ 3.594 Δ = 1 − . 1∕2 x x Pr1∕3 Re x

(7.443a)

7.15 Viscous Flow over a Flat Plate with Arbitrarily Specified Heat Flux

For a step change in heat flux at x = ξ, surface temperature can now be expressed as ]1∕3 [ ( ′′ ) q0 x ξ 2.396 Tw (x) − T∞ = 1− √ k x Rex Pr1∕3 or

{ Tw (x) − T∞ =

q′′0

]1∕3 } [ ( ) ξ 2.396 x 1− √ k x Rex Pr1∕3

Here, F(x − ξ) is given by [ ]1∕3 ( ) ξ x 2.396 1− F(x − ξ) = √ k x Rex Pr1∕3

(7.443b)

(7.443c)

(7.443d)

We may substitute Eq. (7.443c) into Eq. (7.442a) or Eq. (7.442b). Let us substitute F(x − ξ) into Eq. (7.442a), and this substitution will give us ]1∕3 ( ′′ ) | x[ ( ) dqw ξ 2.396 x | dξ| 1 − (7.444a) Tw (x) − T∞ = √ | k x dξ Rex Pr1∕3 ∫0 |Stieltjes Integral in Eq. (7.444a) is in the Stieltjes sense. We may write Eq. (7.444a) in the following form: ]1∕3 ( ′′ ) x[ dqw ξ x x 2.396 −1∕2 dξ 1− Tw (x) − T∞ = 2.396 Pr−1∕3 Rex q′′w0 + √ k k Rex Pr1∕3 ∫0 x dξ ] [ N ξn 1∕3 x 2.396 ∑ ′′ + √ 1− Δq k Rex Pr1∕3 n=1 w,n x

(7.444b)

Suppose that we have continuous function of surface heat flux distribution without any steps or jumps. For this heat flux distribution, Eq. (7.444b) becomes ]1∕3 ( ′′ ) x[ dqw ξ x 2.396 1− Tw (x) − T∞ = √ dξ (7.444c) k Rex Pr1∕3 ∫0 x dξ Using Eq. (7.444c), we may obtain the local Nusselt number Nux Nux =

q′′w x = k(Tw − T∞ )

q′′w (x)x ]1∕3 ( ′′ ) x[ dqw ξ x 2.396 dξ 1− k √ k Rex Pr1∕3 ∫0 x dξ

√ 0.418 Rex Pr1∕3 q′′w (x) ≈ ]1∕3 ( ′′ ) x[ dqw ξ 1− dξ ∫0 x dξ

(7.445)

Again, suppose that we have a continuous function of heat flux without any steps or jumps; then, we integrate in [ ]1∕3 1 ξ 1 dξ and v = q′′w . and dv = dq′′w . Then, we have du = − Eq. (7.444c) by parts. Let u = 1 − x 2∕3 ⎛ ξ⎞ x 3⎜1− ⎟ ⎜ x ⎟⎠ ⎝ The result of the integration by parts is given as ]−2∕3 x[ ξ 1 0.797 1− q′′w (ξ)dξ (7.446) Tw (x) − T∞ = √ k Rex Pr1∕3 ∫0 x We may think of Eq. (7.446) in Stieltjes sense. Equation (7.446) may be utilized in computing the distribution of surface temperature for any prescribed surface heat flux distribution. Notice that q′′w (ξ) may include step jumps as well as continuous variations. Example 7.23 Consider a flat plate having a two steps heat flux distribution as given in Figure 7.E23. Air at 1 atm pressure and T∞ ∘ C is flowing over the plate with a velocity of U∞ m/s . Develop an expression for the variation of wall temperature with distance measured from the leading edge of the plate.

291

292

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Figure 7.E23

qʺw qʺw 1

Double step heat flux distribution along the plate.

W/m2

qʺw 0 W/m2 ξ

0

Solution We use Eq. (7.444a) Tw (x) − T∞ =

x 2.396 √ k Rex Pr1∕3 ∫0

x

x[

1−

ξ x

]1∕3 (

dq′′w dξ

) dξ

] [ N ξ 1∕3 x x 2.396 ∑ ′′ −1∕2 + 2.396 Pr−1∕3 Rex q′′w0 + √ Δqw,n 1 − n . k k Rex Pr1∕3 n=1 x The first part of the equation is zero. Thus, we have ] [ N ξ 1∕3 x x 2.396 ∑ ′′ −1∕2 Δqw,n 1 − n . Tw (x) − T∞ = 2.396 Pr−1∕3 Rex q′′w0 + √ k k Rex Pr1∕3 n=1 x We have the following temperature distribution: ′′ x 2.396 qw0 0≤x≤ξ √ k Rex Pr1∕3 [ ][ [ ] ] ) ( ′′ ξn 1∕3 x 2.396 ′′ ′′ qw0 + qw1 − qw0 1 − Tw (x) − T∞ = √ k Rex Pr1∕3 x

Tw (x) − T∞ =

x ≥ ξ.

On the other hand, Kays et al. [26] give the following relation for arbitrarily specified surface heat flux: x

Tw (x) − T∞ =

∫0

g(ξ, x)q′′w (ξ) dξ

(7.447)

where kernel g(ξ, x) is given as [ ( )3∕4 ]−2∕3 −1∕2 Pr−1∕3 Rex ξ g(ξ, x) = 1− ( ) ( ) x 6 Γ 43 Γ 53 (0.332) k

(7.448)

Substituting Eq. (7.448) into Eq. (7.447) gives an expression for the surface temperature of the plate due to arbitrary flux distribution. Here, we assume that viscous fluid flows over a flat plate [ ( )3∕4 ]−2∕3 )( ( ) x ξ 0.623 −1∕2 −1∕3 Pr Tw (x) − T∞ = Rex q′′w (ξ)dξ (7.449) 1− ∫0 k x Example 7.24 Consider a flat plate having a single-step heat flux distribution as given in Figure 7.E24. Viscous fluid flows over the plate with free stream temperature of T∞ ∘ C and velocity of U∞ m/s. Using Eq. (7.449), develop an expression for the variation of wall temperature with distance measured from the leading edge of the plate. Figure 7.E24

qʺw (x) qʺ0 (W/m2)

0

x

Single step heat flux distribution along the plate.

7.16 One-Parameter Integral Method for Incompressible Two-Dimensional Laminar Flow Heat Transfer

Solution

[ ( )3∕4 ]−2∕3 )( ) x ξ 0.623 −1∕2 Tw (x) − T∞ = Pr−1∕3 Rex q′′0 dξ 1− ∫0 k x [ ( )3∕4 ]−2∕3 x )( ) ( ξ 0.623 −1∕2 ( ′′ ) −1∕3 Pr q0 1− = Rex dξ. ∫0 k x Integral is evaluated by Maple 2020 𝟐 ( ( ) 𝟑 )− 𝟑 x 𝛏 𝟒 d𝛏; > 𝟏− ∫𝟎 x (

𝟖 × 𝛑𝟐 ( )𝟑 𝟐 𝟗𝚪 𝟑 > evalf(%); 𝟑.𝟓𝟑𝟑𝟐𝟕𝟕𝟓𝟎𝟓 x Then, the surface temperature distribution becomes ) )( ( 0.623 −1∕2 ( ′′ ) q0 (3.5332 x). Pr−1∕3 Rex Tw (x) − T∞ = k The local Nusselt number is q′′0 x q′′0 x Nux = = ( )( ) 0.623 k(Tw − T∞ ) −1∕2 ( ′′ ) q0 (3.5332x) Pr−1∕3 Rex k k 1∕2 = 0.454Pr1∕3 Rex .

7.16 One-Parameter Integral Method for Incompressible Two-Dimensional Laminar Flow Heat Transfer: Variable U∞ (x) and Constant Tw − T∞ = const Consider laminar flow over a surface. Fluid properties are assumed to be constant. Surface temperature of the body is constant and denoted by Tw . Free stream velocity U∞ (x) is variable. White and Majdalani [27] give an expression to calculate the Nusselt number based on a method proposed by Spalding and Smith [54]. In this method, they give an expression for conduction thickness Δ4 Δ2 dU∞ U∞ dΔ24 ≈ a(Pr) − b(Pr) 4 ν dx ν dx where Δ4 is defined by Δ4 =

k(Tw − T∞ ) k = h q′′w

(7.450)

(7.451)

and Eq. (7.450) is a general relation, and it is valid for all the Prandtl numbers. The constants a(Pr) and b(Pr) are given in Table 7.9. The solution to Eq. (7.450) is given by White and Majdalani as [ x ] Δ24 a b−1 = b U∞ dx + C (7.452) ∫x0 ν U ∞

where C = 0 if x0 is at the stagnation point. Here, velocity distribution U∞ (x) is variable, and temperature difference (Tw − T∞ ) is constant. The terms a(Pr) and b(Pr) can be estimated in the range 0.1 ≤ Pr ≤ 10 by the following relations: 1 0.35 √ ≈ 0.332Pr a

(7.453)

b ≈ 2.95Pr0.07

(7.454)

293

294

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Table 7.9 Values of a and b for different Prandtl numbers Pr. Pr

a−1/2

b

0

0

2.0

0.001

0.0173

2.06

0.01

0.0760

2.17

0.1

0.140

2.46

0.3

0.215

2.68

0.72

0.296

2.88

1

0.332

2.95

2

0.422

3.10

3

0.485

3.18

10

0.728

3.38

100

1.572

3.61

1 000

3.387

3.72

10 000

7.297

3.76

Source: White and Majdalani [27].

White and Majdalani give a dimensionless version of Eq. (7.452) NuL a−1∕2 = [ √ ]1∕2 ) ( ) ( ReL U∞ −b x U∞ b−1 dx ∫0 U0 U0 L

(7.455)

where dimensionless quantities are given as ReL =

U0 L ν

NuL =

hL . k

Here, L and U0 denote reference length and velocity, respectively. Let us go back to Eq. (7.450). For Pr = 0.7, Eq. (7.450) reduces to Δ2 dU∞ U∞ dΔ24 ≈ 11.68 − 2.87 4 ν dx ν dx

(7.456)

Equation (7.456) can be integrated to get an equation valid for air x

Δ24 =

11.68 ν∫0 U1.87 ∞ dx U2.87 ∞

(7.457)

if either U∞ or Δ4 is zero at x = 0. At the stagnation point where U∞ = x(dU∞ /dx)x = 0 , Δ24 =

4.07ν (dU∞ ∕dx)

(7.458)

Equation (7.457) is evaluated for a given velocity distribution keeping in mind that Δ4 = k/h and working fluid is air. Recall that the heat transfer coefficient h can be expressed in terms of Stanton number St St =

q′′w Nux k = = Rex Pr ρ cp U∞ (Tw − T∞ ) ρ cp U∞ Δ4

(7.459)

7.17 One-Parameter Integral Method for Incompressible Laminar Flow Heat Transfer over a Constant Temperature of a Body of Revolution

7.17 One-Parameter Integral Method for Incompressible Laminar Flow Heat Transfer over a Constant Temperature of a Body of Revolution Consider laminar flow over a body of revolution. See Figure 7.32. Fluid properties are assumed to be constant. With Pr = 0.7, Kays et al. [26] give the following relation for axisymmetric case: x

2

R

Δ24

=

2 11.68 ν ∫0 U1.87 ∞ R dx

(7.460)

U2.87 ∞

where R is the radius of revolution in these equations, and the fluid properties are constant. Note that Δ4 = k/h. Equation (7.460) is expressed in terms of the Stanton number for Pr = 0.7 √ 0.418 ν R U0.435 ∞ St = [ x (7.461) ]0.5 2 ∫0 U1.187 R dx ∞ For variable density and different Prandtl numbers, the following relation is given by Kays et al.: √ C1 μ GC2 R Nux = St = Rex Pr [∫ x GC3 R2 dx]1∕2

(7.462)

0

where G = ρU∞ and the constants C1 , C2 , and C3 are given in Table 7.10 Example 7.25 Air at 300 K flows over a cylinder with diameter D and a constant surface temperature of Tw . The temperature of the fluid ahead of the cylinder is at temperature T∞ and approaching fluid velocity is V. See Figure 7.E25a. The velocity distribution from the potential flow is given as U∞ = 2V sin(φ).

U∞ x

R

T∞ V

Figure 7.32 Table 7.10

Flow over an axisymmetric body. Values C1 , C2 , and C3 for different Prandtl numbers Pr.

Pr

C1

C2

C3

0.7

0.418

0.435

1.87

0.8

0.384

0.450

1.90

1

0.332

0.475

1.95

5

0.117

0.595

2.19

10

0.073

0.685

2.37

Source: Kays et al. [26].

295

296

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Figure 7.E20

Figure 7.E21

Double step temperature distribution along the plate.

A step-ramp surface temperature distribution.

Assume that this is a reasonable approximation for flow over the upper side of the cylinder. On the forward portion of the cylinder (0 < φ < 80), Kreith and Manglik [57] give the following empirical correlation for the local value of heat transfer coefficient h(φ) at any angular position φ: [ ( φ )3 ] √ h(φ)D NuD = . = 1.14 ReD Pr0.4 1 − k 90 On the other hand, experimental data provided by Giedt [56] for different Reynolds number are presented in Figure 7.E25b. Data from Giedt’s plot are obtained by WebPlotDigitizer-4.5-win32-x64. We wish to determine the Nusselt number as a function of φ for 0 ≤ φ ≤ π2 and compare the results with experimental data. Figure 7.E25a

U∞

Geometry and problem description for Example 7.25.

TW x

V

R

φ

T∞

(a)

Solution Prandtl number of the air at 300 K is given as Pr = 0.707. The general solution is x

Δ24 =

aν b−1 U∞ dx. Ub∞ ∫0

For Pr = 0.707, we estimate values of a and b from Eqs. (7.453) and (7.454) 1 0.35 √ ≈ 0.332Pr a b(Pr) ≈ 2.95Pr0.07 .

7.17 One-Parameter Integral Method for Incompressible Laminar Flow Heat Transfer over a Constant Temperature of a Body of Revolution

280 260 240

Nusselt number

220 200 180 160 140 120 100 80

0

20

40

60 80 100 120 140 𝜃 Degrees from stagnation point (b)

160

180

200

Figure 7.E25b Variation of local Nusselt number with angular position for air flow past a nearly isothermal circular cylinder. Source: Data from Giedt [56]. Courtesy of American Society of Mechanical Engineers.

1 theory

Nusselt/sqrt(Reynolds)

0.9

Giedt

0.8 0.7 0.6 0.5 0.4 0.3

0

Figure 7.E25c

20

40 Degrees (c)

60

80

Comparison of experimental and theoretical study.

We estimate a and b from these equations a ≈ 11.56

b ≈ 2.87 x

11.56 ν U1.87 ∞ dx ∫ U2.87 0 ∞ )2 φ ( )2 ( 11.56 ν∫0 [2V sin(φ)]1.87 Rdφ k D Δ24 = = = h hD∕k [2V sin(φ)]2.87 Δ24 =

297

298

7 Laminar External Boundary Layers: Momentum and Heat Transfer

( ( (

(

1 NuD 1 NuD 1 NuD 1 NuD

)2

φ

= )2

1.87 1.87 1 (11.56) × ν × (2V) × R ∫0 [sin(φ)] dφ (2V)2.87 × [sin(φ)]2.87 D2 φ

1.87 1 (11.56) × ν × D ∫0 [sin(φ)] dφ = 2 2 × (2V) × [sin(φ)]2.87 D

)2

φ

= )2

(11.56) × ∫0 [sin(φ)]1.87 dφ ( ) 4 × VD × [sin(φ)]2.87 ν φ

=

(11.56) × ∫0 [sin(φ)]1.87 dφ

4 × (ReD ) × [sin(φ)]2.87 √ √ φ ( ) ∫0 [sin(φ)]1.87 dφ (11.56) × 1 = √ √ NuD 2 (ReD ) × [sin(φ)]2.87 ( ) NuD [sin(φ)]1.435 2 = √ √ √ φ (ReD ) (11.56) ∫0 [sin(φ)]1.87 dφ NuD 0.5882[sin(φ)]1.435 = √ . √ φ (ReD ) ∫0 [sin(φ)]1.87 dφ At the stagnation point, we must do an approximation ( ) x . U∞ ≈ 4V D We substitute this approximation into integral x[ ( )]1.87 11.56 ν x dx Δ24 = [ ( )]2.87 4V ∫0 D x 4V D x [( )]1.87 x [( )]1.87 ( )2 11.56 ν(4V)1.87 11.56 ν k x x = dx = dx )] )] [( [( 2.87 2.87 ∫ ∫ h D D x x 0 0 (4V)2.87 (4V) D D )2 ( x [( )]1.87 x 1 1 1 x 11.56 ν 11.56 ν 1 = 2 dx = 2 x1.87 dx [( 2.87 )] 1.87 )] [( 2.87 ∫0 NuD D x D D D ∫0 x (4V) (4V) D D2.87 NuD 2 = 0.9965. = √ √ ReD 11.56 × 0.3484 Let us construct a table to compare the theoretical results with experimental results φ (degrees)

0

20

38

57

67

71

Nu √ D ReD

0.9845

0.9786

0.9325

0.8544

0.8029

0.7790

Empirical data from W.H. Giedt [56]

0.9885

0.9570

0.9198

0.7708

0.5285

0.3515

Empirical data from Kreith and Manglik [57]

0.9884

0.9775

0.9140

0.7373

0.5806

0.5031

Problems

Problems 7.1

Air at an average film temperature of 37 ∘ C flows over a flat plate with a constant free stream velocity of 4 m/s. The plate length along the flow direction is 2 m. The plate surface temperature is uniform. At a point located at x = 1 m and y = 0.493 cm on the plate, determine the velocity components u and v.

7.2

For flow over a surface, the local shear stress is given by √ ρ μ 3∕2 τw (x) = 0.4 U . x ∞ Determine the average skin friction coefficient.

7.3

Consider steady two-dimensional laminar boundary layer equations over a flat plate. Governing equations are given as 𝜕u 𝜕v + =0 𝜕x 𝜕y 𝜕u 𝜕2 u 𝜕u +v =ν . u 𝜕x 𝜕y 𝜕y Kays et al. [26] present a method to obtain a similarity variable. The velocity profiles at all the x positions are geometrically similar and differing only by a multiplying factor. This means that u = f[y g(x)]. The idea here is to combine the two independent variables x and y into single variable η(x, y) and assume that depends on η. Assume a similarity variable in the following form:

u U∞

η(x, y) = y g(x) u = F(η). U∞ Obtain Eq. (7.33). 7.4

We wish to estimate the variation of surface velocity along a wall if the wedge angle is β = 30∘

7.5

Viscous fluid is flowing over a flat plate. The velocity profile in a laminar boundary layer is described by (y) u = U∞ δ for y < δ(x) and for y > δ(x), u = U∞ . Using the integral method: (a) Derive an expression for the boundary layer thickness δ(x) (b) Derive an expression displacement thickness δ1 (c) Derive an expression momentum thickness δ2

7.6

Fluid is flowing over a flat plate with constant free stream velocity. Assume that the velocity profile in a laminar boundary layer is approximated by ( y )2 (y) ( y )3 ( y )4 u +c =a+b +d +e U∞ δ δ δ δ with the following boundary conditions: at y = 0,

u = 0,

y = δ,

u = U∞ ,

𝜕2 u =0 𝜕y2 𝜕2 u 𝜕u = 0, = 0. 𝜕y 𝜕y2

299

300

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Determine: (a) the constant coefficients a, b, c, and d (b) the boundary layer thickness (c) the local wall shear stress (d) the local skin friction coefficient 7.7

The velocity profile in a laminar boundary layer over a flat plate is given by ( y) u π . = sin U∞ 2δ Determine: (a) the boundary layer thickness (b) the wall shear stress (c) the skin friction coefficient (d) the momentum thickness δ2 (e) the displacement thickness δ1 (f) the shape factor H

7.8

7.9

7.10

For flow in the stagnation point of a body, the potential flow is given by ( ) x U∞ = V L Use Thwaites’ method to determine: (a) the displacement thickness δ1 (b) the momentum thickness δ2 (c) skin friction coefficient cf Consider laminar boundary layer flow over a flat plate. Dimensionless laminar boundary layer profile is given in the table below. Use the integral form of the momentum equation to determine boundary layer thickness y/𝛅

u/U∞

0.0

0.0

0.08

0.134

0.17

0.266

0.25

0.395

0.33

0.518

0.41

0.630

0.49

0.730

0.56

0.812

0.65

0.877

0.73

0.924

0.81

0.957

0.89

0.977

0.97

0.989

1.0

1.0

Howard [11] uses the following velocity profile for laminar boundary layer over as surface: U∞ = 1 − x. Predict the separation point by using the equation of Thwaites.

Problems

7.11

Consider flow over a cylinder as shown in Figure 7.P11. Potential flow gives the following velocity distribution on the circular cylinder: U∞ (x) = 2V sin(φ) x = Rφ where V is the oncoming normal velocity and x is the distance along the surface measured from the stagnation point. Determine the momentum thickness δ2 at the stagnation point.

Figure 7.P11

Flow over a cylinder.

U∞ (x) x V

7.12

r

R

φ

Consider flow over a cylinder, as shown in Figure 7.P12. (a) Potential flow gives the following velocity distribution the flow past the circular cylinder: U∞ (x) = 2V sin(φ) where φ = x/R is the local angle in radians. (b) Schlichting and Kersten [6] report the experimental work of Hiemenz for flow over a cylinder at ReD = 1.9 × 104 as given below U∞ (x) = 1.814(φ) − 0.271(φ)3 − 0.0471(φ)5 V

φ=

U∞ D ν

=

x . R

Apply the Thwaites method to predict the separation point. Figure 7.P12

Flow over a cylinder.

U∞ (x) x V

7.13

φ

r

R

Air at a temperature of 17 ∘ C flows along a flat plate under steady laminar flow conditions. Air free stream velocity is 7 m/s. The plate surface is maintained at a surface temperature of 57 ∘ C. At a position 0.5 m from the leading edge, determine: (a) the velocity in the x-direction, u, 1.32 mm from the wall (b) local boundary layer thickness δ(x) (c) the local coefficient of friction cf (x) (d) the local heat transfer coefficient h(x)

301

302

7 Laminar External Boundary Layers: Momentum and Heat Transfer

7.14

Air is flowing over a constant temperature flat plate. The free stream temperature of air is 27 ∘ C and air is at 1-atm pressure. The free stream velocity of the air is 10 m/s and the flat plate temperature is 47 ∘ C. We wish to calculate the local temperature in a laminar boundary layer along a flat plate at a point 10 cm from the leading edge and 2 mm from the plate wall.

7.15

Engine oil at 50 ∘ C flows over a 2-m-long flat plate. The free stream velocity of the engine oil is 1 m/s. The surface temperature of the plate is 85 ∘ C. The bottom surface of the plate is insulated. Determine the total heat transfer from the plate.

7.16

Air at 1-atm pressure and 17 ∘ C flows over a square flat plate. One side of the plate is 1 m, and the bottom surface of the plate is insulated. The plate surface temperature is 77 ∘ C. The free stream velocity of the air is 1 m/s. (a) Plot the variation of local heat flux along the plate. (b) Obtain the mean heat transfer rate.

7.17

Consider constant property viscous fluid flow over a flat plate having the shape of an equilateral triangle. See Figure 7.P17. Free stream velocity and temperature of the fluid are U∞ and T∞ , respectively. The surface temperature of the plate is T0 . Assume T0 > T∞ . Develop an expression for the heat transfer between triangular area and the fluid.

y

T∞

H 0

U∞

Figure 7.P17

x

Geometry and coordinate system for Problem 7.17.

7.18

Air at atmospheric pressure is flowing over a smooth 5-m-long plate. Free stream air velocity is 3 m/s. Width of the plate is 1 m. The plate surface temperature is 710 K and free stream air temperature is 290 K. Plate is divided into five equal segments. Determine the heat loss from each segment.

7.19

Air at atmospheric pressure is flowing over a smooth 60 cm by 60 cm square plate. Free stream air velocity is 6 m/s. The plate is heated electrically and generates 1100 W power. The bottom of the plate is perfectly insulated. The free stream air temperature is 300 K. Determine the average plate temperature.

7.20

Air at atmospheric pressure flows over a square flat plate that has a length of 1 m in the flow direction. Free stream velocity and temperature of air are 4 m/s and 27 ∘ C, respectively. A uniform heat flux of 2000 W/m2 is applied on the surface of the plate. We wish to plot the local wall temperature along the flow direction.

7.21

Air flows over a flat plate with a free stream velocity 10 m/s. The length of the plate in the flow direction is 20 cm. The free stream temperature of the air is 17 ∘ C. The surface temperature of the plate varies as ( ) x ∘ C Tw = 17 + 40 20 where x is measured from the leading edge of the plate, and it is in centimeters. We wish to plot the variation of local heat flux q′′w along the plate.

Problems

7.22

Air at 1 atm flows at a velocity of 5 m/s over a flat plate. The length of the plate in the flow direction is 30 cm. The free stream temperature of the air is 15 ∘ C, and the plate surface temperature variation is given by ( )0.6 x ∘C Tw = 30 + 25 30 where x is measured from the leading edge of the plate, and it is in centimeters. Plot the variation of local heat flux along the plate.

7.23

Consider the flow configuration shown in Figure 7.P23. Atmospheric air is flowing over the surface with free stream temperature T∞ = 25 ∘ C. The plate surface temperature is kept at 125 ∘ C. At the tail edge, the speed of flow external to the boundary layer is measured to be 15 cm/s. State the boundary layer equations for both the momentum and thermal energy equations. Estimate the momentum and thermal boundary layer thicknesses at the center of the plate. Obtain an expression for the average Nusselt number. T∞ u=V 20 cm 35°

Figure 7.P23

7.24

Geometry and coordinate system for Problem 7.23.

Consider two-dimensional fluid flow over a cylinder with diameter D and a surface temperature of Tw . The temperature of the fluid ahead of the cylinder is at temperature T∞ and fluid velocity is V. See Figure 7.P24. We wish to obtain an expression of the form: −1∕2

NuD ReD

= bPrm

to estimate the heat transfer in the vicinity of the stagnation point. The potential flow solution for flow normal to cylinder is given by U∞ = Figure 7.P24

4Vx for small (x∕R). D

Flow over a cylinder.

U∞ x

V

Tw R

T∞

7.25

Consider flow of fluid past a body of revolution. The fluid flows parallel to body axis as illustrated in Figure 7.P25a. Steady flow of a constant property fluid is assumed. Here r is the radial distance from the body centerline to the surface location. The relationship between the two-dimensional boundary layer equations and rotationally symmetric flows is discussed in Burmeister [15]. For rotationally symmetric flows, the Mangler’s coordinate transformation equations can be used. In Mangler’s transformation, the equivalent two-dimensional coordinates (x and y) are related to those for rotational symmetry by x

x = L−2

∫0

r2 (x)dx y = L−1 r(x)y

(1)

303

304

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Figure 7.P25a Coordinate system for flow parallel to the axis of body of revolution the surface.

y

U∞ x

r(x)

V

(a)

where L is constant having dimension of length. For rotationally stagnation flow, it is possible to take r ≈ x and U∞ = Cx. Eq. (1) gives x = x3 ∕3L

and y = xy∕L

(2)

For this coordinate system, the velocity outside the boundary layer becomes U∞ = C(3L2 x)1∕3

(3)

We compare this velocity distribution with Falkner-Wedge flows. This tells us that the equivalent 2D flow is U∞ = Cxm with m = 1/3. In Eq. (1), if we take x = x, then we find that arbitrary constant L becomes L = 31/3 . Burmeister shows that the rotationally symmetric boundary layer is thinner than the 2D boundary layer by a factor of 31/2 . (a) Show that the local Nusselt number for stagnation point is given as 1∕2

Nux ≈ 0.76 Pr0.4 Rex

(b) Consider fluid flow over a sphere with diameter D and a surface temperature of Tw . The temperature of the fluid ahead of the sphere is at temperature T∞ and fluid velocity is V. See Figure 7.P25b. Develop an expression of the form for the Nusselt number: −1∕2

NuD ReD

= bPrm

to estimate the heat transfer in the vicinity of the stagnation point. Potential flow solution for flow normal to sphere is given by U∞ =

3Vx 3Vx = 2R D

for small (x∕R).

Figure 7.P25b

U∞ x

Flow over a sphere.

Tw R

V T∞ (b)

7.26

Air (Pr = 0.72) flows over a sphere with diameter D and a constant surface temperature of Tw . The temperature of the fluid ahead of the sphere is at temperature T∞ and approaching fluid velocity is V. See Figure 7.P26. The velocity

Problems

distribution from the potential flow is given as: 3 U∞ = V sin(φ). 2 Assume that this is a reasonable approximation for flow over the upper side of the sphere. We wish to determine the Nusselt number as a function of φ for 0 ≤ φ ≤ π2 . Figure 7.P26

Flow over a sphere.

U∞ Tw V T∞

7.27

x

φ

R

Consider slug flow over a semi-infinite flat plate. The free stream temperature and velocity are T∞ and U∞ , respectively. The plate is maintained at uniform temperature T0 . The temperature distribution is given by: (πy) T − T0 . = sin T∞ − T0 2Δ Use the integral method to determine the local Nusselt number.

7.28

Consider slug flow over a semi-infinite flat plate. The free stream temperature and velocity are T∞ and U∞ , respectively. The plate surface temperature varies with distance from the leading edge according to: Tw (x) = T∞ + C x1∕2 . The temperature distribution is given by (πy) T − Tw (x) = sin . T∞ − Tw (x) 2Δ Use the integral method to determine the local Nusselt number.

7.29

Fluid is flowing over a flat plate with constant free stream velocity. There is an unheated starting length on the plate. See Figure 7.P29. Assume that the temperature profile in a laminar flow on a flat plate can be approximated by the following polynomial: (y) ( y )2 ( y )3 ( y )4 T=a+b +c +d +e . Δ Δ Δ Δ δ(x)

y

T∞ Δ(x)

U∞ T∞ 0 ξ Figure 7.P29

Tw

x

Heat transfer from a flat plate with unheated section.

305

306

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Tw > T∞ y u = U∞

T∞ 0

Figure 7.P31

Tw

Δ(x)

T∞

Tw

ξ

X

T∞ 0

x=ξ

Laminar flow over a flat plate with an unheated length.

(a) Evaluate the constants and obtain an expression for the dimensionless temperature in terms of dimensionless distance. Use the following boundary conditions: y=0

T = Tw

y=Δ

T = T∞ ,

𝜕2 T =0 𝜕y2 𝜕T = 0, 𝜕y

𝜕2 T = 0. 𝜕y2

(b) Derive an expression for the local Nusselt number along the plate as a function of Reynolds number. 7.30

Very low Prandtl (Pr ≪ 1) fluid is flowing over a semi-infinite flat plate with uniform velocity U∞ and a constant free stream temperature T∞ . Fluid properties are constant, and flow is laminar. Wall surface temperature distribution is given by Tw (x) = T∞ + b x where b is a constant and x is the distance measure from the leading edge of the plate. For low-Prandtl-number fluids, it is reasonable to assume that u ≈ U∞ . Assume a third-degree profile for temperature distribution. Using the integral method, obtain an expression for local Nusselt number.

7.31

Consider liquid metal flow over a flat plate. Flow is laminar, steady, two-dimensional, constant property, with velocity u ≈ U∞ and temperature T∞ along the flat plate shown in Figure 7.P31. Axial conduction and viscous dissipation are neglected. A certain portion of the leading edge of the plate is insulated, as shown in Figure 7.P31, and the remaining portion of the plate is at uniform temperature Tw . The plate length is L. Assuming a linear temperature distribution in the boundary layer, develop an expression for the Nusselt number.

7.32

Consider liquid metal flow over a flat plate. Flow is steady and laminar with velocity U∞ and temperature T∞ . The variation of surface temperature is given by [ xξ where m is constant, and x is measured from the leading edge of the plate. Assume a linear temperature variation. Using the integral method, obtain an expression for the local Nusselt number.

7.33

Liquid metal (Pr ≪ 1) is flowing over a flat plate. The flat plate is subjected to uniform heat flux q′′0 . The boundary layer is laminar. Neglecting viscous dissipation and assuming a linear temperature variation, determine the local Nusselt number.

7.34

Highly viscous oil (Pr ≫ 1) is flowing over a flat plate. The flat plate is subjected to uniform heat flux q′′0 . The boundary layer is laminar. The velocity and temperature profiles are given by (y) u = U∞ δ q′′ T = T∞ + w [Δ − y]. k

Problems

Neglecting viscous dissipation and assuming a linear temperature variation, determine the local Nusselt number. 7.35

Engine oil at 17 ∘ C flows over a flat plate. The free stream velocity is 1 m/s. The plate is heated to 30 ∘ C, starting at 10 cm from the leading edge of the plate. The total plate length is 100 cm. The bottom surface of the plate is insulated. Determine the total heat loss from the plate.

7.36

Air at 14 ∘ C and 1 atm flows over a flat plate at 10 m/s. The temperature distribution along the plate is shown in Figure 7.P36. What is the heat transfer rate 12 cm from the leading edge of the plate? Tw 140 °C

100 °C 50 °C T∞ = 14°C

0 Figure 7.P36

7.37

8 cm

4 cm

x(cm)

12 cm

Stepwise surface temperature distribution.

Viscous fluid flows over a flat plate. The plate surface temperature is given by √ Tw (x) − T∞ = B x. Determine an expression for heat flux variation with x along the plate and determine the local Nusselt number Nux .

7.38

Air at 20 ∘ C and 5 m/s flows over a flat plate. The plate surface temperature is given in Figure 7.P38. Determine expressions for heat flux variation with x along the plate. T

150 °C

70°C

50°C

T∞ = 20°C 0 Figure 7.P38

0.5 Temperature variation on the plate.

1

x(m)

307

308

7 Laminar External Boundary Layers: Momentum and Heat Transfer

7.39

Consider slug flow over a flat plate. Assume that the surface temperature over a flat plate varies as Tw − T∞ = D(x − 2) where D is a constant. Determine the heat flux over the plate.

7.40

Consider viscous fluid flow over a flat plate. Suppose that heat flux over a flat plate varies as q′′w = m(x − a) where m and a are constants. See Figure 7.P40. Determine the temperature distribution over the plate and develop an expression for the local Nusselt number.

″ qw ″ = m (x – a) qw

0

a

Figure 7.P40

7.41

x

Delayed ramp heat flux distribution along the plate.

Consider viscous fluid flow over a flat plate. Suppose that heat flux over a flat plate varies as q′′w = Dxα where D and α are constants. Determine the temperature distribution over the plate.

7.42

Consider constant property viscous flow over a semi-infinite plate. Flow is steady, two-dimensional, and laminar. Free stream velocity and temperature are U∞ and T∞ , respectively. The distribution of heat surface flux is given by q′′w = C x where C is a constant and x is measured from the leading edge of the plate. We wish to develop an expression for the local surface temperature variation along the plate. We also wish to develop also an expression for the local Nusselt number.

7.43

An incompressible fluid having a Prandtl number of 5 is flowing over a body. The potential flow at the stagnation point of the body is given by U∞ = Ax1∕3 where A is a constant. Calculate the local Stanton number at stagnation point for Pr = 5.

7.44

Consider viscous fluid with high Prandtl number over a flat plate. The energy equation and its boundary conditions may be written as u

𝜕θ 𝜕2 θ 𝜕θ +v =α 2 𝜕x 𝜕y 𝜕y

y=0

θ=0

y=∞

θ=1

where θ = (T − Tw )/(T∞ − Tw ) is the dimensionless temperature. For such fluids, Δ ≪ δ, and under these conditions, the velocity component u can be approximated by a linear function as given below ( ) 𝜕u u = (u)y=0 + (y − 0) + . … … 𝜕y y=0

Problems

Then, the velocity component u is given by u ≈ B(x) y. (a) Apply this equation to the two-dimensional continuity equation to get an expression for the y-component of velocity V. (b) Now, use the u and v velocity components in the energy equation to get:

B(x)y

𝜕θ 1 − 𝜕x 2

(

) dB 2 𝜕θ 𝜕2 θ y =α 2 dx 𝜕y 𝜕y

and state the boundary conditions. (c) Use the combination of variables η = y/g(x) and obtain the following form of function g(x): [ ]1∕3 x 1 g(x) = √ 9α B dx . ∫0 B (d) Transform the dimensionless form of energy equation into the following form: d2 θ dθ =0 + 3η2 dη dη2 where η = [ (e) Set p =

yB ]1∕3 , and this is known as the Lighthill transformation. √ x 9α∫0 Bdx

dθ and obtain the following expression for temperature distribution: dη η

θ=

∫0

exp(−ξ3 )dξ 0.892979

where ξ is a dummy variable of integration. (f) Obtain an expression for the local heat flux. 7.45

Eckert proposed a method to study flow over a constant-temperature body of arbitrary shape as discussed in Kays et al. [26]. In this method, it is assumed that the rate of growth of any of the thermal boundary layer thickness, Δ, is a function of local parameters alone, i.e., in non-dimensional form ( ) Δ24 dU∞ U∞ dΔ24 =f , Pr ν dx ν dx It is reported that this formulation is obtained dimensional analysis. Conduction thickness Δ4 is defined as Δ4 =

k h

The function f for arbitrary variation of free-stream velocity is assumed to be the same as for the family of wedge flows. The wedge flows are evaluated via similarity solutions. For wedge flow solutions, we have −1∕2

Nux Rex

= C1 = constant

where C1 depends on Prandtl number Pr and parameter m. For wedge flows, we have U∞ = Cxm where C is a dimensional constant. Using the wedge flow solutions, for Prandtl number, Pr = 0.7, obtain the following relation Δ2 dU∞ U∞ dΔ24 = 11.68 − 2.87 4 ν dx ν dx

309

310

7 Laminar External Boundary Layers: Momentum and Heat Transfer

Integrate this equation to obtain x

Δ24 =

11.68 ν ∫0 U1.87 ∞ dx U2.87 ∞

Finally develop an expression for the local Nusselt number Nux 7.46

Chanson [70] reports that at a certain location from the leading edge of the plate, the experimental velocity distribution in a laminar boundary layer is given as

y(mm)

u(m/s)

y(mm)

u(m/s)

0 0.5

0

21.0

10.79

0.271

35.0

11.30

1

0.518

41.0

11.52

2

1.113

58.0

11.55

3.5

1.93

60.5

11.51

6

3.137

66.0

11.50

8

4.157

70.0

11.60

10.2

5.32

90.0

11.59

14

7.28

100.0

11.50

18

9.22

Estimate the boundary layer thickness, the displacement thickness and the momentum thickness. Assume that air at 20 ∘ C is flowing over a flat plate. Courtesy of H. Chanson.

References 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Blasius, H. (1950). The boundary layers in fluids with little friction, NACA TM-256. Goldstein, S. (1965). Modern Developments in Fluid Dynamics. Dover Publication. Sedov, L.I. (1993). Similarity and Dimensional Methods in Mechanics, 10e. CRC Press. Hansen, A.G. (1976). Similarity Analysis of Boundary Value Problems in Engineering. Prentice-Hall. Evans, H.L. (1968). Laminar Boundary Layer Theory. Addison-Wesley Publishing Company. Schlichting, H. and Gersten, K. (2017). Boundary Layer Theory, Translated into English by K. Mayes, 9e. Springer. Yarin, L.P. (2012). The Pi-Theorem: Applications to Fluid Mechanics and Heat and Mass Transfer. Springer. Hellums, J.D. and Churchill, S.W. (1964). Simplification of the mathematical description of boundary and initial value problems. AIChE J. 10 (1). Arpaci, V.S. and Larsen, P.L. (1984). Convection Heat Transfer. Prentice-Hall. Oosthuizen, P.H. and Naylor, D. (1999). Introduction to Convection Heat Transfer. McGraw Hill. Howard, L. (1938). On the solution of the laminar boundary layer equations. Proc. R. Soc. Ser. A 164: 547–579. Junkhan, G.H. and Seroyvy, G.K. (1966). Effects of free stream turbulence and pressure gradient on flat plate boundary layer velocity profiles and on heat transfer, ASME Paper 66-WA/HT-4 (November). Junkhan, G.H. and Seroyvy, G.K. (1967). Effects of free stream turbulence and pressure gradient on flat plate boundary layer velocity profiles and on heat transfer. ASME J. Heat Transfer 89 (2): 169–176. Falkner, W.M. and Skan, S.W. (1931). Solution of boundary layer equations. Philos. Mag. 12: 865–896. Burmeister, L.C. (1993). Convective Heat Transfer, 2e. Wiley. Hartree, D.R. (1937). On an equation occurring in Falkner and Skan’s approximate treatment of the equations of the boundary layer. Proc. Cambridge Philos. Soc. 33 (Part II): 223–239.

References

17 Forbrich, S.A. (1982). Improved solutions to the Falkner-Skan boundary layer equation. AIAA J. 20: 1306–1307. 18 Chuang, H. (1985). Comment on improved solutions to the Falkner-Skan boundary-layer equation. AIAA J. 23 (12): 2004–2005. 19 Von Karman, T. (1946). On laminar and turbulent friction, NACA TM 1092. 20 Pai, S.H. (1956). Viscous Flow Theory: Laminar Flow. Van Nostrand Company Inc. 21 Rosenhead, L. (1962). Laminar Boundary Layers. Oxford University Press. 22 Curle, N. and Davies, H.J. (1968). Modern Fluid Dynamics. Van Nostrand Company LTD. 23 Walz, A. (1969). Boundary Layers of Flow and Temperature. M.I.T. Press. 24 Thwaites, B. (1949). Approximate calculation of laminar boundary layer. Aeronaut. Q. 1: 245–280. 25 Schetz, J.A. (2010). Boundary Layer Analysis. AIAA. 26 Kays, W.M., Crawford, M.E., and Weigand, B. (2005). Convective Heat and Mass Transfer. McGraw-Hill. 27 White, F.M. and Majdalani, J. (2021). Viscous Fluid Flow. McGraw Hill. 28 Rott, N. and Crabtree, L.F. (1952). Simplified laminar boundary layer calculation for bodies of revolution and yawed wings. J. Aeronaut. Sci. 19: 553–565. 29 Cebeci, T. (2002). Convective Heat Transfer, 2e. Springer. 30 Elzy, E. and Sisson, R.M. (1967). Tables of Similar Solutions to the Equations of Momentum, Heat and Mass Transfer in Laminar Boundary Layer Flow. Bulletin No.40, Engineering Experiment Station, Oregon State University, Corvalis, Oregon (February). 31 Sparrow, E.M. and Gregg, J.L. (1959). Details of exact low Prandtl number boundary layer solutions for forced and free convection, NASA MEMO 2-27-59E. 32 Churchill, S.W. (1976). A comprehensive correlating equation for forced convection from flat plates. AIChE J. 22 (2): 264–268. 33 Eckert, E.R.G. and Drake, R.M. (1972). Analysis of Heat and Mass Transfer. McGraw-Hill. 34 Churchill, S.W. and Ozeo, H. (1973). Correlations for laminar forced convection with uniform heating in flow over a plate and in developing and fully developed flow in a tube. ASME J. Heat Transfer 78–84. 35 Cebeci, T. and Bradshaw, P. (1984). Physical and Computational Aspects of Convective Heat Transfer. Springer. 36 Chapman, D.R. and Rubesin, W.M. (1949). Temperature and velocity profiles in the compressible laminar boundary layer with arbitrary distribution of surface temperature. J. Aeronaut. Sci. 16 (9): 547–565. 37 Bejan, A. (2013). Convective Heat Transfer. Wiley. 38 Ameel, T. Average effects of forced convection over a flat plate with an unheated starting length. Int. Commun. Heat Mass Transfer 24 (8): 1113–1120. 39 Ghiaasiaan, S.M. (2018). Convective Heat and Mass Transfer, 2e. CRC Press. 40 Kakac, S., Yener, Y., and Pramuanjaroenkij, A. (2014). Convective Heat Transfer. CRC Press. 41 Forsberg, C.H. (2021). Heat Transfer: Principles and Applications. Elsevier. 42 Rubesin, M.W. (1951). The effect of an arbitrary surface temperature variation along a flat plate on convective heat transfer in an incompressible turbulent boundary layer, NACA TN 2345. 43 Galante, S.R. and Churchill, S.W. (1990). Applicability of solutions for convection in potential flow. Adv. Heat Transfer 20, 353–388. 44 Ozisik, M.N. and Hahn, D.W. (2012). Heat Conduction, 3e. Wiley. 45 Klein, J. and Tribus, M. (1952). Forced convection from non-isothermal surfaces. Engineering Research Institute, University of Michigan, Ann Arbor, August (Contract AF 18 (AF)-51, E.O. No.462-BR-1). 46 Hanna, O.T. and Myers, J.F. (1962). Heat transfer in boundary layer flows past a flat plate with a step wall-heat flux. Chem. Eng. Sci. 17: 1053–1055. 47 Hanna, O.T. (1962). Step-wall heat flux superposition for heat transfer in the boundary layer flows. Chem. Eng. Sci. 17: 1041–1051. 48 Baxter, D.C. and Reynolds, W.C. (1958). Fundamental solutions for heat transfer from non-isothermal flat plates. J. Aeronaut. Sci. 25: 403–404. 49 Levy, S. (1952). Heat transfer to constant property laminar boundary layer flows with power-function free stream velocity and wall temperature variation. J. Aeronaut. Sci. 19 (5): 341–348. 50 Harnett, J.P., Eckert, E.R.G., Birkebak, R., and Sampson, R.L. (1956). Simplified Procedures for the Calculation of Heat Transfer to Surfaces with Non-uniform Temperatures. TR-56-373, WADC Tech. Rep. 56-373, ASTIA Document No: AD 110450, Department of Mechanical Engineering, University of Minnesota.

311

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51 Scesa, S. (1951). Experimental investigation of convective heat transfer to air from a flat plate with a stepwise discontinuous surface temperature. M.S. Thesis. University of California, Berkeley. 52 Harnett, J.P., Eckert, E.R.G., Birkebak, R., and Sampson, R.L. (1956). Simplified Procedures for the Calculation of Heat Transfer to Surfaces with Non-uniform Temperatures. WADC Tech. Rep. 56-373. 53 Smith, A.G. and Shah, V.L. (1962). Theoretical and Experimental Study of Heat and Mass Transfer from Non-isothermal Surfaces. Tech. Rep., Note No.135, ASTIA Document No: AD, Contract AF 61(0521)-267, College of Aeronautics, Cranfield, England. 54 Smith, A.G. and Spalding, D.B. (1958). Heat Transfer in a Laminar Boundary Layer with Constant Fluid Properties and Constant Wall Temperature. J. Roy. Aeronaut. Soc. 62: 60. 55 Nellis, G.F. and Klein, S.A. (2021). Introduction to Engineering Heat Transfer. Cambridge University Press. 56 Giedt, W.H. (1949). Investigation of variation of point unit-heat transfer coefficient around a cylinder normal to an air stream. Trans. ASME 71: 375–381. 57 Kreith, F. and Manglik, R. (2018). Principles of Heat Transfer, 8e. 58 Bennet, T.D. (2013). Transport by Advection and Diffusion: Momentum, Heat and Mass Transfer. Wiley. 59 Tribus, M. and Klein, J. (1955). Forced convection through a laminar boundary layer over an arbitrary surface with an arbitrary temperature variation. J. Aeronaut. Sci. 22 (1): 62–64. 60 Aziz, A. (2009). A similarity solution for laminar thermal boundary layer over a flat plate with convective surface boundary condition. Commun. Nonlinear Sci. Numer. Simul. 14 (4): 1064–1068. 61 Churchill, R.V. (1972). Operational Mathematics. McGraw-Hill. 62 Abramowitz, M. and Steigun, I. (1972). Handbook of Mathematical Functions, Graphs and Tables. Dover Publicatios. 63 Churchill, S.W. and Bernstein, M. (1977). A correlating equation for forced convection from gasses and liquids to a circular cylinder in cross flow. ASME J. Heat Transfer 99: 300–306. 64 Whitaker, S. (1972). Forced convection heat transfer correlations for flow in pipes, past flat plates, single cylinders, single spheres, and flow in packed beds and tube bundles. AIChE J. 18: 361–371. 65 Long, C.A. (1999). Essential Heat Transfer. Pearson Education. 66 Jacob, M. (1949). Heat Transfer. Wiley. 67 Whitaker, S. (1976). Elementary Heat Transfer Analysis. Pergamon Press Inc. 68 Nellis, G. and Klein, S. (2009). Heat Transfer. Cambridge University Press. 69 Sparrow, E.M. and Lin, S.H. (1965). Boundary layers with prescribed heat flux – application to simultaneous convection and radiation. Int. J. Heat Mass Transfer 8: 437–448. 70 Magrab, E.B., Azarm, S., Balachandran, B., Duncan, J.H., Herold, K.E. and Walsh, G.C. (2011). An Engineer’s Guide to MATLAB, Pearson Education, Inc. 71 Meade, D.B., Michael May, S.J., Cheung, C.K. and Keough, G.E. (2009). Getting Started with MAPLE, John Wiley and Sons, Inc. 72 Gilbert, R.P., Hsiao, G.C. and Ronkese, R.J. (2021). Differential Equations-A Maple Supplement 2e, CRC Press.

313

8 Laminar Momentum and Heat Transfer in Channels 8.1 Introduction Internal laminar flow through channels, such as pipes and parallel plates, will be considered in this chapter. Fluid is constrained on all sides by solid boundaries. In these channels, continuum assumption is assumed to be valid. In many engineering devices, such as minichannels and microchannels, flow is laminar. We will emphasize on the analytical solutions, and I believe that students will gain insight provided by the analytical solutions. There is a geometric length scale imposed on the flow in the cross-stream direction, and the channel diameter DH is the pertinent length scale. The Reynolds number for flow in a channel is defined as ReDH =

ρVDH μ

(8.1)

where V is the mean velocity, DH is the inside hydraulic diameter, ρ is the fluid density, and μ is dynamic viscosity of the fluid. Many engineering devices have noncircular cross sections; heat transfer and friction coefficients are needed for these configurations. For this reason, it is obvious that geometric length scale DH has significant physical importance in internal flows, and therefore, it is common to use hydraulic diameter DH for the length scale. Hydraulic diameter DH is defined as 4 × Ac P where Ac and P are the cross-sectional flow area and wetted flow perimeter, respectively. In a fully developed laminar flow, the critical Reynolds number to the onset of turbulence is DH =

(8.2)

ReD ≈ 2300

(8.3)

and for ReD < 2300, the flow is called laminar. For ReD ≥ 2300, the flow will be considered to be turbulent. Actually, flow becomes fully turbulent for ReD ≥ 10 000.

8.2 Momentum Transfer We will study momentum transfer inside smooth tubes and between two infinite parallel plates to develop pertinent parameters. Many applications of fluid flow occur in round smooth tubes and between parallel plates. Flow between two infinite parallel plates and flow in a smooth tube serve as a stepping-stone to study more complex engineering problems. Laminar tube flow is encountered generally in heat exchangers, heating or cooling of fluids such as oils and in many other engineering applications.

8.2.1 Hydrodynamic Considerations in Ducts First, we will consider momentum flow in a smooth round tube, and later, we will study momentum transfer between two infinite parallel plates.

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

314

8 Laminar Momentum and Heat Transfer in Channels

r

V

Edge of boundary layer

vr (1)

Core

Flow

vz

(2)

D = 2R

Vc z

δ(z)

Developing velocity profile

Fully developed velocity profile

LH Developing region

Figure 8.1

Fully developed region

Growth of velocity boundary layer along a circular duct.

Circular tubes At a large distance from the duct entrance, the flow forgets the nature of its initial velocity entrance configuration. Assuming constant fluid properties, there will be a characteristic axial velocity profile independent of the axial position. There will be no velocity component in the direction normal to flow. Axial pressure gradient required against the viscous forces will be constant along the channel. The flow is hydrodynamically fully developed (HFD). Both analytical and experimental correlations are available for the prediction of friction factors. Based on velocity distribution, there are two regions in the tubes, as shown in Figure 8.1.

(a) Hydrodynamic entrance region (b) Fully developed region Consider a viscous fluid moving in a circular duct. Here, we have chosen the z-axis along the pipe. Fluid enters the duct at reference Section 8.1 with uniform velocity V. Hydrodynamically developing region A velocity boundary layer develops along the circular duct, as illustrated in Figure 8.1. In the entrance region of a tube, there is a developing velocity boundary layer, and this region is called hydrodynamic entrance region. Eventually, the boundary layer fills the tube, the boundary layer thickness δ approaches the same magnitude as radius R = D/2, δ ≈ R, and flow becomes fully developed at reference Section 8.2. The distance between inlet and the point where flow becomes fully developed is called hydrodynamic entrance length and is denoted by LH . This region has the following characteristic futures:

(a) (b) (c) (d)

The radial velocity component vr does not vanish (vr ≠ 0), and for this reason, streamlines are not parallel to tube walls. Core velocity Vc increases with axial distance z. Pressure decreases with axial distance, dp/dz < 0. Velocity boundary layer thickness is less than tube radius δ < R.

Hydrodynamic entrance length The hydrodynamic entrance length LH is discussed in [1] and [2]. In the inlet region of a duct,

the core of the duct near the inlet is filled with fluid that is unaffected by the wall conditions. The fluid in the core region may be regarded as an external flow. The fluid very near the duct wall may be modeled as a boundary layer flow. Boundary layer approximation may be used for flow in the entrance region of tube since there is a boundary layer development. The boundary layer thickness for external flow is δ 1 . ∼ √ z Rez

(8.4)

Applying this equation in the entrance region, we obtain 1 D ∼ √ . LH ReLH

(8.5)

8.2 Momentum Transfer

Next, we express ReLH in terms of the Reynolds number ReD based on diameter D, L V LH V D LH = = ReD H . ν ν D D Thus, we have √ 1 1 D D ∼ √ = √ LH L LH Re H D ReD D ReLH =

(8.6a)

or LH ∼ ReD (8.6b) D where ReD = ρ V D/μ is the Reynolds number based on the channel diameter. The length required for the flow to be fully developed in laminar flow is important. Various authors, such as Kays et al. [1], Shah and London [3], and Shah [4], have given the hydrodynamic entrance lengths for laminar flow inside ducts. Chen [5] applied the integral momentum equation to laminar flow development in the hydrodynamic entrance region of the circular tube without the use of boundary layer assumptions. The entrance length is defined as the distance along the axis where the centerline velocity reaches 99% of its fully developed value. A correlation for hydrodynamic entrance length of circular tubes is reported by Shah and London [3] due to Chen LH 0.6 = + 0.056 ReD D 0.035 ReD + 1

(8.7)

where ReD = ρVD/μ is the Reynolds number based on the inside diameter of the tube. As discussed in [1], the hydrodynamic entrance length LH for any duct having an arbitrary cross-sectional area may be estimated by LH = 0.05 DH ReDH

(8.8)

where ReDH = ρ V DH ∕μ is the Reynolds number based on the hydraulic diameter DH . HFD region The region beyond the hydrodynamic entrance region is called fully developed region. This region has the

following characteristics: (a) For two-dimensional incompressible flow, the velocity profile vz does not change with the axial distance, i.e. 𝜕 vz /𝜕 z = 0. (b) Flow is essentially streamlined flow, and the streamlines are parallel to tube centerline (vr = 0). We call this region fully developed region. Friction factor In the design of engineering equipment, such as compact heat exchangers, a calculating equation for the friction factor is valuable. If we examine Figure 8.1, we see that the velocity profile in the hydrodynamic entrance region of a tube changes along the axial direction. Pressure variation in the axial direction is caused by frictional loss and change in momentum flux. Thus, pressure drop calculations in a duct must consider the effect of momentum flux variation due to flow acceleration and effect of surface shear forces. The shear stress at the tube wall must be related to pressure gradient in the flow direction. In a heat exchanger design, the pressure drop experienced by fluid must be overcome by an external device such as a fan or pump. Consider the control volume shown in Figure 8.2. The momentum balance on control volume yields

−

⎛ ⎞ d d ⎜ (pAc ) = τw P + ρ v2z dAc ⎟ ⎟ dz dz ⎜∫ ⎝ Ac ⎠

(8.9)

where Ac is the cross-sectional area, p is the pressure, P is the perimeter, and τw is the wall shear stress. Eq. (8.9) shows that pressure force balances the shear force at the wall as well as change in the momentum flux of the fluid. For a duct having a constant cross-sectional area, Eq. (8.9) reduces to −

⎞ ⎛ d p τw P d ⎜ 1 = + ρ v2z dAc ⎟ . ⎟ dz Ac dz ⎜ Ac ∫ ⎠ ⎝ Ac

(8.10)

315

316

8 Laminar Momentum and Heat Transfer in Channels

ρ v2z dAc

ρ v2z dAc +

Ac

Ac

p Ac

p Ac +

d dz

d dz

ρ v2z dAc dz Ac

(p Ac) dz

τw P dz dz

Figure 8.2

A differential momentum balance.

We see from Eq. (8.10) that the pressure gradient will be largest in the developing region. This is true since the core of the flow is accelerating, and wall shear stress is higher. The second term on the right-hand side of Eq. (8.10) is related to momentum flux. Even though the flow enters the tube at uniform velocity, due to shear forces, it becomes nonuniform. In the fully developed region, the velocity profile does not change along the flow direction, and the second term in Eq. (8.10) will be zero. Thus, we have −

d p τw P = . dz Ac

(8.11)

To avoid confusion, local Fanning or skin friction factor cf and Darcy or Moody friction factor f are defined next. Fanning friction factor:

cf =

2 τw

(8.12a)

ρ V2

where ρV2 /2 is the channel dynamic pressure. The friction factor defined by Eq. (8.12a) is a local friction factor, and it is based on the local wall shear stress. The mean Fanning friction factor is based on the mean shear stress τw , and the mean shear stress τw is based on the overall length L L

τw =

1 τ (x) dx. L ∫0 w

(8.12b)

Then, the mean Fanning friction factor is cf =

2 τw ρ V2

L

=

1 c (x) dx. L ∫0 f

(8.12c)

Moody friction factor: For internal flow, the pressure gradient is correlated using the Moody (Darcy) friction factor f,

defined as f=

−(dp∕dz)DH ρV2 ∕2

(8.12d)

where the term ρV2 /2 is the pipe dynamic pressure. The friction factor defined by Eq. (8.12d) is a local Moody friction factor, and it is based on the local pressure gradient. An average friction factor f based on the overall length L of the pipe is defined as (p − px=L )(D) f = x=0 . (8.12e) L(ρV2 ∕2) Equation (8.12e) can be expressed in terms of local friction factor f as ) L( L 2DH L ρV2 D dp 1 1 − dx = f dx = f= H 2 f dx. 2 L ρV ∕2 ∫0 dx L ∫0 LρV ∫0 2D

(8.12f)

8.2 Momentum Transfer

The apparent or average Fanning friction factor cf, app and the apparent or average Moody friction factor fapp are defined to include the entrance effects and will be discussed later. In the fully developed region, pressure gradient is constant and decreases linearly. Design engineers are interested in determining the pressure drop in the entrance region for a given application. Combined effects of wall shear stress and flow acceleration may be incorporated into single apparent or average Fanning friction coefficient cf, app . Apparent or friction factor cf, app includes the effect of the developing region. The dimensionless pressure drop equation may be expressed in terms of the apparent Fanning friction factor, as discussed in [3] ( ) ΔP z (8.13a) = 4c f,app DH ρ V2 ∕2 or ΔP = (f ReDH )z∗ (8.13b) ρ V2 ∕2 where Re = ρVDH /μ is the Reynolds number based on the hydraulic diameter, and z∗ = z∕DH ReDH is the dimensionless axial distance. The pressure drops from the entrance at z = 0 to any position z can be calculated using this equation. An alternative equation, as discussed in [6], is given in the following form: ) ] ( [ ΔP z + K (8.13c) + (4c ) = K c f e DH ρ V2 ∕2 where Kc is the entrance loss coefficient, Ke is the expansion loss coefficient, and cf is the friction factor for fully developed flow. Parallel plates We will study laminar, incompressible, steady flow between two infinite parallel plates. Figure 8.3 shows both the developing and fully developed region of flow between two infinite parallel plates. We will derive velocity distribution and develop pertinent parameters. Both the plates are stationary. The basic governing equations of momentum and continuity will be used. Hydrodynamic entrance region Consider viscous incompressible constant property flow between parallel plates. Here, we have chosen the x-axis along the channel. Fluid enters the duct at reference Section 8.1 with uniform velocity V. A velocity boundary layer develops along the duct, as illustrated in Figure 8.7. In the entrance region of the parallel plate channel, there is a developing velocity boundary layer, and this region is called hydrodynamic entrance region. The distance between the inlet and the point where flow becomes fully developed is called hydrodynamic entrance length and is denoted by LH . This region has the following characteristic futures:

(a) The vertical velocity component v does not vanish (v ≠ 0), and for this reason, streamlines are not parallel to channel walls. (b) Core velocity Vc increases with axial distance x. (c) Pressure decreases with axial distance (dp/dx) < 0. (d) Velocity boundary layer thickness is less than channel half thickness (δ < b). Boundary layer y (1)

Core

(2)

u=V

0 δ(x)

x

2b

0 LH Figure 8.3

Fully developed region

Development of the velocity profile for laminar flow between parallel plate channels.

317

318

8 Laminar Momentum and Heat Transfer in Channels

Hydrodynamic entrance length Chen [5] studied the flow of Newtonian fluid in the entrance region by the integral method without the use of boundary layer assumptions. In the analysis, Chen included the axial molecular transport of momentum and the variation of pressure across the flow. For flow between parallel plates, Shah and London [3] report the following correlation attributed to Chen [5]: LH 0.315 = + 0.011 ReDH (8.14) DH 1 + 0.0175 ReDH

where ReDH = ρ V DH ∕μ is the Reynolds number. Fully developed region Eventually, the boundary layer fills the parallel plate channel, the boundary layer thickness δ approaches the same magnitude as half thickness b (δ ≈ b), and flow becomes fully developed at reference Section 8.2. This region has the following characteristics:

(a) For two-dimensional incompressible flow, the velocity profile u does not change with the axial distance, i.e. 𝜕u/𝜕x = 0. (b) Flow is essentially streamlined flow, and the streamlines are parallel to duct surface, v = 0.

8.2.2

Fully Developed Laminar Flow in Circular Tube

The tube is long enough so that any entrance or exit effects can be disregarded in the fully developed viscous flow region. Therefore, the velocity profile will be the same at various sections of the duct, the streamlines will be parallel straight lines, and fluid particles will move with speed on any of the streamlines, i.e. inertia force will be zero. The equilibrium is, thus, realized between pressure force and tangential shear force. The following assumptions were made for fully developed flow in a tube: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Steady laminar flow Incompressible flow Constant fluid properties Axisymmetric flow (𝜕/𝜕θ = 0) No θ velocity (no swirl), vθ = 0 Constant pressure gradient in the z-direction No body forces No end effects, fully developed flow with respect to z, 𝜕vz /𝜕z = 0 Streamlines are parallel, vr = 0 Newtonian fluid Continuum

We are interested in fully developed velocity distribution in a tube with constant fluid properties. Based on Figure 8.4, somewhat simplified forms of governing equations of fluid motions are given as: Continuity: 𝜕 vz 1 𝜕 (rv ) + =0 (8.15) r 𝜕r r 𝜕z r-momentum ) { ( ( ) 𝜕2v } 𝜕v 𝜕v 𝜕p 𝜕 1 𝜕 +μ (rvr ) + 2r (8.16a) ρ vr r + vz r = − 𝜕r 𝜕z 𝜕r 𝜕r r𝜕r 𝜕z r

τw τ(r) 0

D = 2R

z

vz (r)

τ(0) = 0

Vc Figure 8.4

Fully developed velocity profile and shear stress distribution in a circular tube.

8.2 Momentum Transfer

z-momentum: ) { ( ) } ( 𝜕v 𝜕 2 vz 𝜕v 𝜕v 𝜕p 1 𝜕 +μ r z + . ρ vr z + vz z = − 𝜕r 𝜕z 𝜕z r𝜕r 𝜕r 𝜕z2

(8.16b)

Fully developed velocity field with respect to z -direction assumption implies that 𝜕vz =0 𝜕z and from continuity, we have 1 𝜕 (r vr ) = 0 ⇒ r vr = const r𝜕r

(8.17)

At r = R vr = 0 (solid wall or pipe wall is impermeable). Thus, from the continuity equation, we have vr = 0. This implies that the streamlines are parallel along the tube axis. The continuity equation is satisfied. Based on the assumptions, the momentum equation reduces to the following form: 𝜕p =0 𝜕r

(8.18)

𝜕p 1 d =μ 𝜕z r dr

( r

dvz dr

) = g(r).

(8.19)

Integrating 𝜕p/𝜕z = g(r) with respect to z, we get p = g(r)z + C0 where C0 is an integration constant. We now consider the momentum equation in the r-direction. Upon integration, the momentum equation in the r-direction gives since vr = 0 𝜕p = 0 ⇒ p = f(z) 𝜕r where f(z) is an integration constant. We now have two solutions for the pressure p and equate them to get g(r)z + C0 = f(z). But this is not possible, and this is only possible if we write g(r) = C where C is a constant. Finally, we get ( ) dv 𝜕p 1 d =μ r z = C. 𝜕z r dr dr

(8.20)

We conclude that the pressure is only a function of z. Consequently, we write 𝜕p dp = . 𝜕z dz

(8.21)

On the other hand, boundary conditions are given as vz = 0 dvz =0 dr

r = R (No slip at the wall) r = 0 (Symmetry at pipe centerline)

(8.22a) (8.22b)

Notice that the pressure gradient is always a negative number in the flow direction. We can now return the momentum equation in the z-direction, where the left-hand side is a function of r and the right-hand side is a function of z, and both the sides are equal to each other ( ) dv r dp d r z = . (8.23) dr dr μ dz

319

320

8 Laminar Momentum and Heat Transfer in Channels

Solving this ordinary differential equation (ODE) along with the given boundary conditions, we obtain the fully developed velocity profile, and this flow is called Hagen–Poiseuille flow: ) ( )( dp r2 R2 vz = (8.24) − 1− 2 . 4μ dz R It is more meaningful to express Eq. (8.24) in terms of mean velocity V. Mean velocity V for incompressible flow is defined as follows: 1 ρ vz dAc ρAc ∫

V=

Ac

1 V= v dA Ac ∫ z c

(8.25)

Ac

where Ac is the cross-sectional area of the duct. For axisymmetric flow in a circular tube, dAc = 2πrdr and Ac = πR2 . Thus, R

2 vz rdr. R2 ∫0

V=

(8.26)

In order to find the average velocity V, we substitute the velocity distribution, Eq. (8.24), into Eq. (8.26) and, after integration, we obtain the following result: ( ) dp R2 V= − . (8.27) 8μ dz Then, the velocity profile becomes ) ( r2 vz = 2 V 1 − 2 . R

(8.28)

On the other hand, to determine the pressure drop, it is convenient to work with the Moody or Darcy friction factor in ducts, which is a dimensionless parameter defined as f=

(−Δp∕L)D 2

ρV ∕2 64 f= ReD

=

(8μV∕R2 )(2R) ρV2 ∕2

=

64μ 32μ = ρVR ρV(2R) (8.29)

where ReD = ρVD/μ is the Reynolds number and f is the Moody or Darcy friction factor. Notice that Eq. (8.29) does not include the entrance effects. The maximum velocity Vc occurs at the centerline of the pipe, at r = 0, and from Eq. (8.24), we have ( ) dP R2 Vc = − . (8.30) 4μ dz Velocity vz , in terms of maximum velocity Vc , takes the following form: ( ) r2 vz = Vc 1 − 2 . R

(8.31)

We see that the relation between average velocity V and maximum or centerline velocity Vc is Vc = 2V. The shear stress at the surface can be evaluated from the gradient of the velocity profile at the surface: ) )] ( [ ( 𝜕vz 4Vμ 2R τw = μ . (8.32) =− = μ 2V − 2 𝜕 r r=R R R We may use an alternative procedure to evaluate wall shear stress τw . Consider a stationary control volume, as shown in Figure 8.5. Consider the momentum equation ⃗s + F

∭ C∀

⃗fρd∀ =

∬ CS

⃗ V ⃗ • dA) ⃗ + 𝜕 ⃗ V(ρ (ρV)d∀. 𝜕t ∭ ∀

(8.33)

8.2 Momentum Transfer

r τ r

D = 2R

z

O

τ δz

p

Figure 8.5

p+

dp δz dz

Control volume to study fully developed flow in a circular tube.

Let us apply the momentum equation, in the z-direction, noting that, because of the steady fully developed nature of the flow, there is no net change in momentum flux and there is no momentum storage. Body forces are neglected. Under these conditions, the momentum equation reduces to ∑ Fx = 0. Thus,

) ( dp δz π r2 − τ(2π r δz) 0 = p(π r2 ) − p + dz ( ) dp r τ= − 2 dz

and τw =

R 2

( −

dp dz

(8.34)

) (8.35)

where τw is the wall shear stress. Equations (8.34) and (8.35) are equally applicable to a fully developed turbulent flow, as long as it is understood that τ refers to the total shear stress that is the linear combination of the viscous stress and the turbulent shear stress. Also, we may write r τ = . τw R

(8.36)

The shear stress distribution for fully developed flow is shown in Figure 8.4. Note that in a fully developed pipe flow, whether laminar or turbulent, the shear stress varies linearly from a maximum at the surface to zero at the pipe or tube centerline. We can express the surface shear stress in terms of a nondimensional skin friction coefficient (Fanning friction factor) cf for fully developed flow. We will base the definition of the skin friction coefficient cf on the mean velocity V. Considering the absolute value of the shear stress, to preserve the fact that surface shear is always opposite to the flow, for the skin friction coefficient cf , we get cf =

τw ρ V2 ∕2

=

4Vμ∕R ρ V2 ∕2

=

8μ 16 = . ρVR ρV(2R)∕μ

(8.37)

This result indicates that the friction factor may be interpreted as a dimensionless wall shear stress. The nondimensional group of variables in the denominator is the Reynolds number, ReD = ρVD/μ. Here, D = 2R is the pipe diameter, V is the mean velocity, ρ is the fluid density, and μ is the fluid dynamic velocity. Thus, the skin or Fanning friction coefficient cf is given as cf =

16 . ReD

(8.38)

321

322

8 Laminar Momentum and Heat Transfer in Channels

Equation (8.38) does not include the entrance effects. The Darcy or Moody friction factor f and the skin friction coefficient cf are related as 64 . (8.39) f = 4cf = ReD Wall shear stress τw can also be expressed in terms of the Moody friction factor f as follows: 2

f R Δp D f ρV = = ρV2 . (8.40) 2 L 4D 2 8 Shah and London [3] proposed an excellent curve fit expression for apparent or average Fanning friction factor cf, app for flow in a tube 3.44 1.25 + 16 − √ ∗ 4z z∗ 3.44 (8.41) cf,app ReD = √ + 0.00021 z∗ 1+ z∗2 * where z = L/(DReD ) and ReD = ρ V D/μ is the Reynolds number. Eq. (8.41) is recommended for the entire range of (L/D)/Re for a circular tube. Equation (8.41) includes the entrance effects. The apparent or average Moody friction factor that includes the developing region for a circular tube may be written, using Eq. (8.41), as 0.31 3.44 ⎡ + 16 − √ ⎤ ∗ ⎢ z z∗ ⎥⎥ 4 ⎢ 3.44 (8.42) fapp = √ + 0.00021 ⎥ ReD ⎢ z∗ 1 + ⎢ ⎥ z∗2 ⎣ ⎦ τw =

where z* = L/(DReD ) is the appropriate dimensionless distance for hydrodynamically developing flow in a tube. Studies indicate that the friction factor in the entrance region of a duct is higher than that in fully developed flow. This is due to very high velocity gradients in the tube entrance and acceleration of the fluid core outside the boundary layer in accordance with the conservation of mass. The ratio of the apparent factor f to the fully developed friction factor f is independent of Reynolds number, and it is given by 0.31 3.44 ⎡ + 16 − √ ⎤ ∗ ⎢ z fapp z∗ ⎥⎥ 1 ⎢ 3.44 . (8.43) = √ + 0.00021 ⎥ 16 ⎢ z∗ f 1 + ⎢ ⎥ z∗2 ⎣ ⎦ The ratio of the average (apparent) friction factor to fully developed friction factor is plotted in Figure 8.6 as a function of dimensionless length z+ . The apparent friction factor fapp approaches to the fully developed friction factor f as the dimensionless distance z+ approaches a value of 0.2. Example 8.1 Ethylene glycol at 37 ∘ C flows inside a 5-cm-diameter tube with at a rate of 0.12 kg/s. Determine the average friction factor for a tube length of 1 m. Solution The physical properties of ethylene glycol at 37 ∘ C are given below: ρ = 1103.7 kg∕m3 μ = 1.07 × 10−2 N.s∕m2 . The next average velocity V is computed as ṁ 4 × 0.12 4ṁ V= = = 0.0553 m∕s. = ρA ρπD2 1103.7 × π × (0.05)2 The Reynolds number ReD is ρVD 1103.7 × 0.0553 × 0.05 ReD = = = 285.5. μ 1.07 × 10−2 Since ReD < 2300, flow is laminar.

8.2 Momentum Transfer

10

8

6 f f 4

2

0 10−3

10−2

10−1

100

101

102

z* Figure 8.6

The ratio of the apparent Moody friction factor to fully developed Moody friction factor for flow in a pipe.

We now compute hydrodynamic entry length LH to check whether the entrance region effects are important LH = 0.05ReD D = 0.05 × 285.5 × 0.05 = 0.713 m. It appears that entrance effects are important, and Eq. (8.42) will be used. The average (Apparent) friction factor fapp is L 1 = 0.07 = DReD 0.05 × 285.5 3.44 0.31 ⎡ + 16 − √ ⎤ ∗ ⎢ z z∗ ⎥⎥ 4 ⎢ 3.44 = = 0.2811. √ + 0.00021 ⎥ ReD ⎢ z∗ 1+ ⎢ ⎥ ∗2 z ⎣ ⎦

z∗ =

fapp

The fully developed friction factor f is computed from Eq. (8.39) for comparison purposes f=

64 64 = 0.224. = ReD 285.5

8.2.3 Fully Developed Flow Between Two Infinite Parallel Plates The two-dimensional, steady laminar flow of a viscous incompressible fluid between two parallel walls is considered. The coordinate system is shown in Figure 8.7. The width of plates parallel to the z-direction is very large compared with the distance 2b between the plates. The gravity term is neglected. Such a flow can be produced by a pressure gradient. The following assumptions are made: (1) The flow is steady, 𝜕/𝜕 t = 0. (2) Constant fluid properties.

323

324

8 Laminar Momentum and Heat Transfer in Channels

τw

y u(y) 0

Figure 8.7

(3) (4) (5) (6) (7) (8)

τ 2b

x

Fully developed velocity profile and shear stress distribution for flow between two parallel plates.

The flow is laminar. No edge effects in the z-direction, 𝜕/𝜕z = 0. No end effects in the x-direction, flow is fully developed along the x-direction, 𝜕/𝜕x = 0. Parallel streamlines, v = 0. Newtonian fluid. Continuum.

The continuity and Navier–Stokes equations in two-dimensional flow, in the absence of body forces, reduce to Continuity 𝜕u 𝜕v + =0 𝜕x 𝜕y x-momentum u

𝜕u 1 𝜕p 𝜕u +v =− +ν 𝜕x 𝜕y ρ 𝜕x

y-momentum u

𝜕v 𝜕v 1 𝜕p +v =− +ν 𝜕x 𝜕y p 𝜕y

(8.44) (

(

𝜕2u 𝜕2u + 2 𝜕x2 𝜕y

𝜕2 v 𝜕2 v + 𝜕x2 𝜕y2

) (8.45a)

) .

(8.45b)

We assume steady fully developed flow. This means that that the velocity profile does not change along the flow direction. That is, 𝜕u/𝜕x = 0. Since we have two infinite parallel walls, we assume no fluid motion in the z-direction, w = 0. The continuity equation reduces to 𝜕v = 0 ⇒ v = C. 𝜕y

(8.45c)

Since wall is impermeable, at y = b v = 0, and for this reason, C = 0. Therefore,v = 0, and the continuity equation is identically satisfied. On the other hand, from the y-component of the momentum equation, we write 𝜕p dp 𝜕p = 0 ⇒ p = p(x) ⇒ = . 𝜕y 𝜕x dx

(8.46)

This means that pressure depends on x only, and we conclude that pressure p must be constant across any cross section. On the other hand, the x-component of the momentum equation reduces to 1 dp d2 u . = 2 μ dx dy

(8.47)

The fully developed flow requires the following boundary conditions: y=bu=0

(8.48a)

y = −b u = 0.

(8.48b)

Solving this ODE along with boundary conditions, we obtain the velocity profile for fully developed flow between parallel plates: ( ) 1 dp u=− (b2 − y2 ). (8.49) 2μ dx

8.2 Momentum Transfer

If the flow is in the positive x-direction, then the pressure gradient is negative. The resulting velocity profile is thus parabolic, as illustrated in Figure 8.7. The maximum velocity Vc occurs at the channel center, and substituting y = 0 into Eq. (8.49), we get ( ) b2 dp . (8.50) Vc = − 2μ dx Then, the velocity distribution (8.49) becomes [ ( y )2 ] . u = Vc 1 − b Taking a unit depth in the z-direction, the total volume flow rate Q between the walls is given by ( ) b dp 2b3 − . Q= udy = ∫−b 3μ dz The mean flow velocity V in the channel can be expressed as ( ) dp Q Q b2 2 = − = Vc . V= = Ac 2b × 1 3μ dx 3 Velocity distribution in terms of average velocity V is [ ( y )2 ] 3 u= V 1− . 2 b Pressure drop in terms of average velocity V is ( ) dp 3μ V − = 2 . dx b Using the velocity profile given in Eq. (8.54), the wall shear stress can be given as ( ) 3μV du =− τw = μ . dy y=b b Taking the absolute value of wall shear stress, the (Fanning) skin friction coefficient cf is ) ( ) ( τw μ 24 6μ = cf = = ρbV ρV (4b) ρV2 ∕2 24 = ReDH

(8.51)

(8.52)

(8.53)

(8.54)

(8.55)

(8.56)

(8.57)

where DH = 4b is the hydraulic diameter. Recall that the hydraulic diameter is defined as DH =

4 × Ac 4 × (W × 2b) 2 × (1 × 2b) = lim = lim ( ) = 4b. W→0 2(W + 2b) W→∞ 2b P 1+ W

The Moody (Darcy) friction factor f is obtained as follows: f=

(−dp∕dz)DH 2

ρV ∕2 96 = ReDH

=

(3μV∕b2 )(4b) 2

ρV ∕2

=

96μ 24μ = ρVb ρV (4b) (8.58)

where DH = 4b is the hydraulic diameter. Shah and London [3] give an equation for the apparent Fanning friction factor for flow between parallel plates

capp ReDH

0.674 3.44 ⎤ ⎡ −√ 24 + ⎢ 4x∗ x∗ ⎥⎥ 3.44 ⎢ = √ + 0.000029 ⎥ ⎢ x∗ 1+ ⎥ ⎢ x∗2 ⎦ ⎣

(8.59)

325

326

8 Laminar Momentum and Heat Transfer in Channels

10

5

cf, app cf

1

0.5

10−4

10−3

10−2

10−1

100

101

102

x* Figure 8.8

Ratio of the apparent Fanning friction factor to the fully developed Fanning friction factor for flow between parallel plates.

where x* is a dimensionless length for the hydrodynamically developing region of a parallel plate channel, and it is defined as L∕DH x∗ = ReDH where DH = 4b is the hydraulic diameter. Shah and London recommend Eq. (8.59) for the entire x+ range. The ratio of the apparent Fanning friction factor to the fully developed Fanning friction factor for flow between parallel plates is given by capp cf

0.674 3.44 ⎤ ⎡ −√ 24 + ⎢ 4x∗ x∗ ⎥⎥ 1 ⎢ 3.44 . = √ + 0.000029 ⎥ 24 ⎢ x∗ 1 + ⎢ ⎥ x∗2 ⎣ ⎦

(8.60)

Figure 8.8 illustrates the ratio of the apparent Fanning friction factor to the fully developed Fanning friction factor as a function of dimensionless distance x+ . An examination of Figure 8.8 reveals that the apparent Fanning friction factor approaches to the fully developed Fanning friction factor when x+ ≈ 0.2.

8.3

Thermal Considerations in Ducts

We are interested in heat transfer from the walls of a tube to a fluid flowing in the tube. Attention will be paid only to situations in which the flow is laminar. Laminar tube flows may occur in some engineering applications such as oil coolers. Laminar flow is sometimes necessary for passages having small diameters because of high pumping losses of turbulent flow. As shown in Figure 8.9, the fluid is assumed to enter the tube at a uniform temperature Ti and a uniform velocity V. The tube surface may be at uniform temperature Tw or uniform wall heat flux q′′w . We will study three different cases of heat transfer in tube flow: (a) Flow is HFD and thermally fully developed (TFD) (fully developed heat transfer). (b) The velocity profile is fully developed, and the temperature profile is developing (thermal entrance region problem). (c) Velocity and temperature profiles are developing simultaneously starting from the inlet of the tube (combined [hydrodynamic and thermal] entrance region problem).

8.3 Thermal Considerations in Ducts

Edge of thermal boundary layer

r (1)

(2)

Core

Ti

z

Δ(z)

Developing temperature profile

Tw

Fully developed region D = 2R

Fully developed temperature profile

LT Developing region Figure 8.9

Thermal entrance region.

Thermal entrance region The region in which the thermal boundary layer grows is called thermal entrance region. The

length of this region is called thermal entrance length and is denoted by LT . At z = LT , the thermal boundary layer thickness Δ(z) becomes equal to D/2. The thermal entrance length LT is measured from the tube inlet. In the entrance region, the core temperature Tc is uniform and is equal to the inlet temperature Ti , i.e. Tc = Ti . The thermal boundary layer thickness Δ(z) is less than the tube radius R. TFD region If either a uniform temperature Tw or uniform heat flux (UHF) q′′ w is applied to a surface, then, eventually, TFD

condition is reached. A dimensionless temperature ϕ is defined such that 𝜕ϕ/𝜕z = 0, and this point will be discussed later in detail. The shape of the fully developed temperature profile depends on the surface boundary condition imposed. Thermal entrance length The thermal entrance length LH in laminar flow is discussed in [1] and [2]. Consider viscous flow in a pipe having diameter D. Both the velocity boundary layer thickness δ and the thermal boundary layer thickness Δ increase with the axial distance in internal flow, and eventually, both reach to the center of the duct. Therefore, the mean fluid velocity V can be used as velocity scale for all the Prandtl numbers. Again, we use the thermal boundary layer thickness for external flow. Scaling is discussed in Chapter 3:

Δ −1∕2 ∼ ReL Pr −1∕2 . L Applying this result at L = LT where Δ ∼ D, we obtain −1∕2

D ∼ LT ReL

T

Pr −1∕2 .

(8.61)

(8.62)

The Reynolds number for internal flow is based on the pipe internal diameter D. Therefore, we express the Reynolds −1∕2 number ReL in terms of the pipe diameter D T

V LT L V D LT = = ReD T ν ν D D where ReD = ρ V D/μ is the Reynolds number. Substituting Eq. (8.63) into Eq. (8.62) yields ) ( L −1∕2 −1∕2 Pr D ∼ LT ReD T D ReLT =

(8.63)

(8.64a)

and rearranging yields LT ∼ ReD Pr . D Recall that we found LH ∼ ReD . D

(8.64b)

(8.64c)

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8 Laminar Momentum and Heat Transfer in Channels

We can now combine Eqs. (8.64a) and (8.64b) and conclude that LT ∼ Pr . LH

(8.65)

Various authors, such as Kays et al. [1], Shah and London [3], and Shah [4], have given the hydrodynamic and the thermal entrance lengths for laminar flow inside ducts. The thermal entrance length LT for laminar flow in a tube may be defined as the point where z+ = 2(z∕D)∕ReD Pr ≈ 0.1, as discussed in [1]. Thus, LT ≈ 0.05 ReD Pr D

(8.66)

where Pr = μcp /k is the Prandtl number, cp is the constant pressure specific heat, and k is the thermal conductivity of the fluid. If we compare hydrodynamic and thermal entry lengths, we reach to the following conclusions: (1) If Pr > 1, the hydrodynamic boundary layer develops faster than the thermal boundary layer (LH < LT ). For large-Prandtl-number fluids such as oils, LH ≪ LT , and it is reasonable to assume HFD flow in the entrance region. Most practical problems of forced convection involving high viscous fluids, such as oils, may be modeled as HFD and thermally developing flow. (2) If Pr < 1, the thermal boundary layer develops faster than the hydrodynamic boundary layer (LH > LT ). The thermal entry length is short for liquid metals. Steady flow thermal energy equation Consider incompressible constant property fluid flow in a duct. Steady flow energy

equation is written as ̇ p (Tmo − Tmi ). q = mc

(8.67)

The right-hand side of this equation gives the net rate of thermal energy transport for incompressible liquid or enthalpy (thermal energy plus flow work) for ideal gas. Thermal energy is carried by the fluid. In this equation, it is assumed that Tmi is the mean inlet temperature and Tmo is the mean outlet temperature. Next, the mean fluid temperature will be defined. The mean fluid temperature Mean (or bulk temperature) is used as a reference temperature in internal flow heat transfer. Consider incompressible constant property fluid flow in a tube. The mixed-mean fluid temperature, Tm , is defined such ̇ p Tm , equals the integrated enthalpy flow over the duct cross section that mean (bulk) enthalpy or mean thermal energy, mc

Tm =

1 ρ cp vz T dAc ̇ p∫ mc

(8.68a)

Ac

or assuming ρ and cp are constant ∫ vz TdAc Tm =

Ac

∫ vz dAc

.

(8.68b)

Ac

Since Tm is averaged over the duct cross section, it is a function of only axial position z. For a circular pipe, with dAc = 2πrdr, the above equation becomes R

Tm =

2 vz T r dr. VR2 ∫0

(8.69)

Newton’s Law of cooling For internal flow heat transfer, Newton’s law of cooling is expressed as

q′′w = h(Tw − Tm ) where h is the local convection heat transfer coefficient and q′′w is the local heat flux.

(8.70)

8.3 Thermal Considerations in Ducts

r Tw R

Flow 0

Figure 8.10

z

ϕ (r)

Fully developed thermal flow in a pipe.

Fully developed conditions Consider incompressible laminar steady motion of fluid flowing in a pipe. See Figure 8.10. We

are talking about forced convection heat transfer in a fluid through the tube produced by some external means such as a fan or pump. A dimensionless temperature ϕ is defined as ϕ=

Tw (z) − T(z, r) 𝜕ϕ ⇒ =0 Tw (z) − Tm (z) 𝜕z

(8.71a)

where Tm (z) is the mean fluid temperature, Tw (z) is the wall temperature, and T(r, z) is the local fluid temperature distribution. Fully developed conditions are discussed in [7]. Eq. (8.71a) tells that a dimensionless temperature ϕ is independent of z and depends only on radial position r, and it is given as ϕ = ϕ(r), and this is called fully developed temperature profile. The temperature profile T(z, r) continues to change with z, but the relative shape of the profile does not change. Flow is said to be TFD. The condition specified by Eq. (8.71a) is eventually reached in a tube subjected to uniform surface heat flux q′′w (q′′w = constant) or a uniform surface temperature Tw (Tw = constant). The mean fluid temperature Tm (z) varies with the axial distance along the tube since energy is added or removed from the tube wall. We may also define a fully developed temperature profile in terms of centerline temperature Tc as given below: ϕ=

Tw (z) − T(z, r) . Tw (z) − Tc (z)

(8.71b)

These two definitions are identical. The statement that the dimensionless profile is invariant with respect to axial distance z can be expressed as [ ] 𝜕ϕ 𝜕 Tw (z) − T(r, z) = =0 (8.72a) 𝜕z 𝜕z Tw (z) − Tm (z) or } } { { Tw − T 𝜕Tw 𝜕T 𝜕Tw 𝜕Tm 1 − − − = 0. (8.72b) Tw − Tm 𝜕z 𝜕z 𝜕z 𝜕z (Tw − Tm )2 and using definition of ϕ, we get Upon solving Eq. (8.72b) for 𝜕T 𝜕z [ ] [ ] T (z) − T(z, r) dTw T (z) − T(z, r) dTm 𝜕T dTw = − w + w 𝜕z dz Tw (z) − Tm (z) dz Tw (z) − Tm (z) dz or using Eq. (8.71a), we obtain [ ] dTw dTm 𝜕T dTw = − ϕ(r) − 𝜕z dz dz dz or ( ) ( ) dTw dTm 𝜕T = [1 − ϕ(r)] + ϕ(r) . 𝜕z dz dz

(8.72c)

(8.72d)

(8.72e)

We are interested in the nature of the heat transfer coefficient in the TFD region. The heat flux at the solid–liquid interface is given by 𝜕T | (8.73) q′′w = k || . 𝜕r |r=R The positive sign arises since r is measured from the pipe center toward the pipe wall. The heat flux q′′w is taken as positive in the inward direction, i.e. from the wall to fluid direction. Using Eq. (8.71a), we have T(z, r) = Tw (z) − ϕ(r)[Tw (z) − Tm (z)].

(8.74)

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8 Laminar Momentum and Heat Transfer in Channels

Differentiating Eq. (8.74) with respect to r yields dϕ | 𝜕T || = −[Tw − Tm ] || . | 𝜕 r |r=R dr |r=R

(8.75)

Thus, the heat flux becomes dϕ | q′′w = −k [Tw − Tm ] || . (8.76) dr |r=R We now define the local heat transfer coefficient at any axial position z in terms of the mean fluid temperature Tm h(z) =

q′′w . Tw − Tm

(8.77)

The local heat transfer coefficient may be written as dϕ | ( ) −k [Tw − Tm ] dr | q′′w dϕ |r=R = = −k = constant. h(z) = Tw − Tm Tw − Tm dr r=R

(8.78)

Since the temperature profile is fully developed, we find that (dϕ/dr)r = R = const. Thus, for a TFD flow of fluid having constant properties, the heat transfer coefficient h is constant and independent of z in the TFD region. This concept is shown in Figure 8.11. Since the thermal boundary layer thickness is zero at the tube inlet, the heat transfer coefficient is very large at the tube inlet where z = 0. The heat transfer coefficient decays rapidly along the tube. This is true since the boundary layer develops along the tube. Before we consider specific problems, we will study the two fundamental boundary conditions. These are: (a) Constant wall heat flux (b) Constant wall temperature (a) Uniform surface heat flux Consider Newton’s cooling law q′′w = h(Tw − Tm ).

(8.79)

Both heat flux q′′w and heat transfer coefficient h are constant in the fully developed region. Thus, we can write q′′w = const. h Hence, we may state that Tw − Tm =

(8.80)

dTw dTm ′′ = qw = const. dz dz

(8.81)

Then, we can combine Eqs. (8.72d) and (8.81); thus, we express

𝜕T 𝜕z

as

dTm ′′ 𝜕T dTw = = qw = constant. 𝜕z dz dz

(8.82) Figure 8.11

h

O

Developing region

Fully developed region

z

Variation of heat transfer coefficient with axial position.

8.3 Thermal Considerations in Ducts

Here, the axial temperature gradient is independent of radial location. The constant wall heat flux boundary condition applies to a wall subjected to radiant heat flux or electrical resistance heating. When the wall heat flux is constant, the mean fluid temperature of a constant property fluid varies linearly in the flow direction. (b) Uniform surface temperature For the case of constant wall temperature Tw = constant, dTw /dz = 0 and we can write [ ] Tw (z) − T(z, r) dT 𝜕T = = ϕ(r) m Tw = const. 𝜕z Tw (z) − Tm (z) dz

𝜕T 𝜕z

from Eq. (8.72d) as follows: (8.83)

The constant wall temperature boundary condition applies to a wall subjected phase change such as condensation or boiling on a surface. From the above discussion, we can conclude that the forms of the velocity and temperature profiles are not changing with the axial distance along the duct for fully developed heat transfer. Consider fully developed flow in a tube. The velocity and temperature profiles may also be expressed as ( ) vz r (8.84a) =f Vc R ( ) Tw − T r (8.84b) =g Tw − Tc R where R is the pipe radius, Vc is the centerline velocity, Tc is the centerline temperature, and Tw is wall temperature. The energy balance in tubes We will consider energy balance in tubes for both UHF and uniform wall temperature (UWT) boundary conditions. Energy balance may be applied to flow in a duct to determine the variation of mean fluid temperature Tm with axial position along the duct. Tubes with uniform surface heat flux Consider a fluid flowing in a of tube. The coordinate system is depicted in Figure 8.12.

Tube can have any arbitrary cross section. Uniform heat flux q′′w is applied at the tube surface and, heat is added to fluid. The mean fluid temperature at z = 0 is T = Tmi , and the outlet mean fluid temperature at z = L is T = Tmo . Under these conditions, we can determine (a) The heat transferred to the tube at any position z as q = q′′w P z, where P is the tube perimeter. (b) Variation of mean fluid temperature Tm (z) along the tube. (c) Variation of tube surface temperature Tw (z). Consider the control volume between z = 0 and any position z. The application of first law of thermodynamics yields q′′w P z = ṁ cp [Tm (z) − Tmi ]

(8.85)

where ṁ is the mass flow rate and cp is the constant pressure specific heat. Energy transported by the fluid is equal to energy supplied by the heat flux applied to surface. Solving this equation, we obtain the variation of mean fluid temperature with axial position: q″w

Tmi

Tmo

0

q″w

z L

Figure 8.12

Control volume in a duct.

D = 2R

Control volume

331

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8 Laminar Momentum and Heat Transfer in Channels

Figure 8.13

Entrance region

Axial variation of mean fluid temperature.

Fully developed region Tw (z)

Tw – Tm

Tm0

Tm (z) q″w = const

Tmi

( Tm (z) = Tmi +

P q′′w ṁ cp

) z.

(8.86)

This relation is derived from the first law of thermodynamics and is valid for (a) For the entrance region (b) For fully developed region Equation (8.86) is also valid for both laminar and turbulent flow. Eq. (8.86) shows that the mean fluid temperature varies linearly with z along the tube. See Figure 8.13. Temperature difference Tw − Tm varies also with axial distance z. The temperature difference Tw − Tm is small in the entrance region since the heat transfer coefficient is large. Notice that the temperature difference Tw − Tm is also independent of z in the fully developed region. Tube surface temperature is determined using Newton’s cooling law as follows: q′′w = h(z)[Tw (z) − Tm (z)].

(8.87a)

Solving Eq. (8.87a) for the wall temperature Tw (z), we obtain q′′w . h(z) We may combine Eqs. (8.86) and (8.87b) to get [ ] 1 zP ′′ Tw (z) = Tmi + qw + . ṁ cp h(z) Tw (z) = Tm (z) +

(8.87b)

(8.87c)

We now need to determine the heat transfer coefficient h(z). The determination of the heat transfer coefficient depends on (a) whether the flow is laminar or turbulent (b) entrance region or fully developed region (c) boundary conditions We can determine heat transfer coefficient experimentally or theoretically depending on the problem in hand. Example 8.2 Water at 12 ∘ C flows with a velocity of 0.06 m/s in a tube having an internal diameter of 2 cm. After a calming section, flow becomes fully developed at a point, and after this point, 20 000 W/m2 heat flux is applied to the tube wall. How much farther down the tube, will the water reach to 62 ∘ C? Solution The physical properties of water will be evaluated at the average temperature Tm T + Tmo 12 + 60 = = 36 ∘ C = 310 K Tm = mi 2 2 Tmi = 12 ∘ C = 285 K (Inlet mean water temperature) Tmo = 62 ∘ C = 335 K (Outlet mean water temperature)

8.3 Thermal Considerations in Ducts

The properties of water are ρ = 993 kg∕m3 cp = 4178 J∕kg.K μ = 695 × 10−6 N.s∕m2 Pr = 4.62 k = 0.628 W∕m.K. ρVD 993 × 0.06 × 0.02 The Reynolds number is ReD = = = 1714. μ 695 × 10−6 Flow is laminar since ReD < 2300. The mean fluid temperature Tm (z) is Tm (z) = Tmi +

P q′′w πD q′′w z = Tmi + z ṁ cp ρV π4 D2 cp

4 q′′w z ρVD cp 4 × 20 000 z. 335 = 285 + 993 × 0.06 × 0.02 × 4179 Solving this equation yields = Tmi +

z = 3.11 m. Tubes with uniform surface temperature We now again consider fluid flowing in a tube. A uniform surface temperature is

applied on the tube surface. This condition can be achieved by condensing steam on the external tube surface. We can now determine: (a) Mean fluid temperature Tm (z) variation along the channel (b) Total heat transferred up to position z (c) Tube surface heat flux variation q′′w (z) Consider the control volume between z and z + dz. The application of the first law of thermodynamics yields d ̇ p Tm )dz − dq′′c ̇ p Tm + (mc ṁ cp Tm = mc dz dT ̇ p m dq′′c = mc dz where q′′c is the convective heat flux. On the other hand, the application of Newton’s cooling law yields dq′′c = h(z)P[Tw − Tm (z)] where P is the channel perimeter. Combining Eqs. (8.88) and (8.89) yields dT ̇ p m = h(z)P[Tw − Tm (z)]. mc dz The initial condition of this ODE is Tm (0) = Tmi . The solution of this differential equation gives the mean fluid temperature at any position z [ ] z Tm (z) − Tw P = exp − h(z) dz . ̇ p ∫0 mc Tmi − Tw We can now introduce a mean heat transfer coefficient h z 1 h= h(z) dz z ∫0 and solve the equation for the mean fluid temperature Tm (z) [ ( ) ] Ph Tm (z) = Tw + (Tmi − Tw ) exp − z ̇ p mc

(8.88)

(8.89)

(8.90)

(8.91)

(8.92)

(8.93)

(8.94)

where h is the average value of h from inlet to position z. Again, we need to determine the heat transfer coefficient. The determination of heat transfer coefficient depends on

333

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8 Laminar Momentum and Heat Transfer in Channels

Tw

Tmi

dq″c″

d m·cp Tm + (m·cp Tm) dz dz

m·cp Tm 0

z

dz

Tmo

Control volume

L Figure 8.14

Control volume in a channel.

(a) whether the flow is laminar or turbulent (b) entrance region or fully developed region (c) boundary conditions We can determine the heat transfer coefficient experimentally or theoretically depending on the problem in hand. Next, we need to determine an equation for the total heat transfer. Consider Figure 8.14. The application of the conservation of energy to a control volume between the inlet and the outlet gives ̇ p [Tmo − Tmi ]. q = mc

(8.95)

Now, we wish to determine a different expression for total heat transfer. The application of the conservation of energy to a control volume between the inlet and the outlet yields ̇ p [(Tw − Tmi ) − (Tw − Tmo )] ̇ p [Tmo − Tmi ] = mc q = mc ̇ p (ΔTi − ΔTo ) = mc

(8.96)

where ΔTi = (Tw − Tmi ) and ΔT0 = (Tw − Tmo ). The application of Eq. (8.94) to a control volume between the inlet and the outlet of a tube having a length L gives [ ] PLh Tmo = Tw + (Tmi − Tw ) exp − . (8.97) ̇ p mc We can now rearrange Eq. (8.97) as given below ) ( ) ( ΔTo Tw − Tm0 PL = ln =− h ln ̇ p mc Tw − Tmi ΔTi

(8.98a)

where h is the average heat transfer coefficient, and it is given as L

h=

1 h dz. L ∫0

(8.98b)

We now solve Eq. (8.98a) for ṁ cp and substitute into Eq. (8.96), and after algebra, we obtain an expression for the total heat transfer q=

As h(ΔTo − ΔTi ) ) ( ΔT0 ln ΔTi

(8.99)

where As = PL is the surface area of the pipe. We may now introduce the log mean temperature difference, ΔTLMTD . The log mean temperature difference is defined as

ΔTLMTD

⎤ ⎡ ⎢ ΔT − ΔT ⎥ o = ⎢ (i ) −⎥ ⎥ ⎢ ΔTi ⎥ ⎢ ln ΔTo ⎦ ⎣

8.4 Heat Transfer in the Entrance Region of Ducts

Tw ΔTo = Tw – Tmo

ΔTi = Tw – Tmi Tmo Tmi

Figure 8.15

The log mean temperature difference (LMTD).

and see Figure 8.15. Total heat transfer will be given as q = hAs ΔTLMTD Tw = const.

(8.100a)

where As is the channel surface area. Eq. (8.100a) is a special form of Newton’s cooling law for the entire tube and ΔTLMTD is a special average of temperature difference over the tube length. In many engineering applications, the external fluid temperature T∞ instead of tube surface temperature is fixed. In this case, we may write ( ) U As T∞ − Tmo = exp − (8.100b) ṁ cp T∞ − Tmi where U is the overall heat transfer coefficient. The overall heat transfer coefficient U may be expressed in terms of inside and outside heat transfer coefficients for the tube U=

1 1 + . hi ho

In this expression, tube wall conduction resistance is neglected since tube wall normal has high thermal conductivity. Overall heat transfer is q = UAs ΔTLMTD Tw = const.

(8.101)

8.4 Heat Transfer in the Entrance Region of Ducts Duct shape and thermal boundary conditions will affect the heat transfer. We will study convection heat transfer in a circular pipe and between two infinite parallel plates. The two most common boundary conditions are constant temperature and constant heat flux. The laminar flow heat transfer is not affected by the surface roughness. These boundary conditions will be considered in the study of convection heat transfer. In many engineering devices, the fluid enters the constant area duct with a uniform velocity and temperature. This constant area duct is heated over the entire length. For this reason, the thermal boundary layer starts at the entrance of the duct, and a simultaneous development of hydrodynamic and thermal boundary layers is observed. This type of forced convection heat transfer problem is called as the combined hydrodynamic and thermal entry length problem. In short, sometimes, it is called combined entry length problem. The presence of a developing velocity profile greatly enhances the heat transfer rate. This problem is not amenable to exact analytical analysis due to the complex nature of the governing differential equations. Heat transfer in the combined entrance region is important due to its practical importance in engineering such as in the design of compact heat exchangers. The knowledge of the velocity profile development is essential in the entrance region. The Prandtl number becomes an important parameter for the combined hydrodynamic and thermal entrance problem. The velocity profile development is independent of the Prandtl number. On the other hand, temperature profile development depends on the Prandtl number. A dimensional analysis of the entrance region shows that the Nusselt number can be expressed as ) ( z NuD = f ReD , Pr , . (8.102) D

335

336

8 Laminar Momentum and Heat Transfer in Channels

The above equation gives a general functional relationship that would be expected in correlation of experimental data or theoretical study. This relation is valid for both the entrance region and the fully developed region. It is also valid for laminar flow as well as turbulent flow. Qualitative behavior of heat transfer in the entrance region Consider viscous flow in a pipe having diameter D. We will study heat transfer in the entrance region of a pipe in a qualitative manner. The local heat transfer coefficient for internal flow heat transfer is defined as q′′w (8.103) h= (Tw − Tm ) where q′′w is the local surface heat flux and Tw and Tm are the local wall and mean fluid temperatures, respectively. The local Nusselt number is based on the tube diameter, and it is given as hD (8.104) k where k is the thermal conductivity of the fluid. A thermal boundary layer develops in the entrance region, and the thermal boundary layer may be modeled as a conduction resistance to heat transfer. We may then estimate the heat flux as NuD =

k (T − Tm ). Δ w Combining Eqs. (8.103) and (8.105), we get q′′w ≈

h=

1 k k (Tw − Tm ) ≈ (Tw − Tm ) Δ Δ

(8.105)

(8.106)

and D . (8.107) Δ Equation (8.107) tells us that the heat transfer coefficient and the Nusselt number will be large in the entrance region and will decrease until it reaches a constant value. The average Nusselt number will approach the local Nusselt number as the flow becomes TFD. When the flow becomes TFD, the thermal boundary layer thickness Δ is fixed by geometric length scale D. Therefore, the Nusselt number NuD and the heat transfer coefficient h become constant. The heat transfer coefficient and the Nusselt number defined by Eqs. (8.103) and (8.104) are local values. We may define an average heat transfer coefficient h NuD ≈

L

h=

1 h dz. L ∫0

(8.108)

We may also define an average Nusselt number NuD for internal flow hD . k Substituting Eq. (8.108) into Eq. (8.109) yields L L( ) D1 D1 k NuD dz NuD = h dz = k L ∫0 k L ∫0 D NuD =

(8.109)

(8.110a)

and we express Eq. (8.110a) in terms of the local Nusselt number as L

NuD =

1 NuD dz. L ∫0

(8.110b)

The qualitative behavior of the average heat transfer coefficient and the Nusselt number is illustrated in Figure 8.16. The average Nusselt number will approach the local Nusselt number in the TFD region. The invariance of the Nusselt number in the downstream of the flow implies TFD flow. Now, it is obvious that the geometric length scale D has significant physical importance in internal flows, and for this reason, it is common to use hydraulic diameter DH for the length scale for noncircular ducts. A reference temperature is needed to evaluate the physical properties of fluid. An average of mean bulk temperature Tm is used to (T + Tmo ) (8.110c) Tm = mi 2

8.4 Heat Transfer in the Entrance Region of Ducts

Figure 8.16 coefficient.

Qualitative behavior of the Nusselt number or the heat transfer

NuD or h

NuD or h 0 Thermal enreance region

Thermally fully developed region

z

where Tmi and Tmo are the inlet and outlet mean fluid temperatures, respectively. Recall that for a constant property fluid, the bulk temperature Tm is defined as Tm =

1 v T dAc Ac V ∫ z

(8.110d)

A

where Ac is the cross-sectional area of the tube and V is the average fluid velocity. The average velocity was defined as V=

1 v dAc . Ac ∫ z

(8.110e)

A

Bulk temperature is a measure of the fluid temperature averaged over the cross section of a duct or channel. This definition does not explain the difference between the cases Tw > Tm or Tw < Tm where Tw is the pipe wall temperature. We know that viscosity is temperature dependent and cold or hot walls will influence the viscosity. For the case of liquids, especially viscous oils, the viscosity ratio term is needed to account for the effect of viscosity. For situations where the viscosity does not vary too much across the fluid, the viscosity ratio may be neglected. In this case, the fluid properties may be evaluated at an average film temperature Tf defined as [ ( )] Tmi + Tmo Tw + 2 Tf = (8.110f) 2 This definition is meaningful for the evaluation of physical properties of gases. First, we will study heat transfer in the entrance region of a duct for inviscid flow due to its simplicity. The velocity profile is approximated as vz ≈ V. Constant area duct is heated over its entire length starting from the inlet. For this reason, the thermal boundary layer starts at the entrance of the duct and begins to develop along the duct. We will study parallel flows in both the circular tube and the infinite parallel plates prescribed by the limit δ/Δ → 0. We can develop some analytical solutions under special conditions. These solutions will shed light on the physical aspects of viscous flow heat transfer in the ducts. We will have an overview of the typical behavior of the Nusselt number NuD in the entrance region. Later, we will consider empirical heat transfer in the entrance region of ducts for viscous flow, which has a great deal of practical importance.

8.4.1 Circular Pipe: Slug Flow Heat Transfer in the Entrance Region Suppose that the unheated entrance length of tube is short, and the Prandtl number is very low (Pr ≪ 1). For low-Prandtl-number fluids, δ/Δ → 0 indicates that the momentum boundary layer may be ignored relative to the thermal boundary layer. In other words, the temperature profile develops faster than the velocity profiles. It is reasonable to assume that the axial velocity is constant and is approximated as vz ≈ V where V is the average fluid velocity. This assumption is valid in the entrance region and is not valid to large distances down the tube since the velocity profile will eventually develop. 8.4.1.1 Heat Transfer for Low-Prandtl-Number Fluid Flow (Slug Flow) in the Entrance Region of Circular Tube Subjected to Constant Wall Temperature

Consider a two-dimensional, incompressible, steady laminar flow of a low-Prandtl-number fluid flowing in the entrance region of a pipe. See Figure 8.17. The fluid properties are constant. Viscous dissipation is negligible, and there is no internal energy generation. The pipe radius is R. Fluid enters the pipe with an initial temperature Ti and uniform velocity V. For

337

338

8 Laminar Momentum and Heat Transfer in Channels

r

Tw T(r, z)

T = Ti vz = V

D = 2R

0

z

Tw Ti

(a) Figure 8.17

0

z (b)

(a) Slug flow in circular tube. (b) Constant wall temperature.

low-Prandtl-number fluids, δ/Δ → 0, and for this reason, the velocity profile is approximated as vz ≈ V. Notice that the thermal diffusivity is larger than the momentum diffusivity. The temperature profiles develop more rapidly than the velocity profile. The coordinate system is located at the inlet of duct where heating begins. Heat transfers into slug flow from isothermal pipe wall and pipe wall is at constant temperature Tw . In other words, constant area duct is heated over its entire length. For this reason, the thermal boundary layer starts at the entrance of the duct and begins to develop along the duct. Axial conduction and viscous dissipation are neglected. There is no energy generation. We wish to determine the temperature distribution and the Nusselt number. The energy equation to be solved in the entrance region is given by [ 2 ] 𝜕T 𝜕 T 1 𝜕T V =α . (8.111) + 𝜕z r 𝜕r 𝜕r2 The boundary conditions are T(0, r) = Ti

(8.112a)

T(z, R) = Tw

(8.112b)

𝜕T(z, 0) = 0. 𝜕r

(8.112c)

Define the following dimensionless variables: θ=

T − Tw Ti − Tw

(8.113)

η=

r R

(8.114)

z+ =

z z0

where the parameter z0 is an unknown parameter and will be determined now: T − Tw 𝜕θ 𝜕T 𝜕T 𝜕θ 𝜕z+ = (Ti − Tw ) = (Ti − Tw ) + = i 𝜕z 𝜕z 𝜕z 𝜕z z0 𝜕z+ 𝜕θ 𝜕θ 𝜕η 𝜕T = (Ti − Tw ) = (Ti − Tw ) 𝜕r 𝜕r 𝜕η 𝜕r (T − Tw ) 𝜕θ = i R 𝜕η (T − T 1 𝜕T 1 i w ) 𝜕θ = r 𝜕r r R 𝜕η r = ηR (T − T ) 1 𝜕θ 1 (Ti − Tw ) 𝜕θ 1 𝜕T = = i 2 w 2 r 𝜕r η 𝜕η η 𝜕η R R [ ] [ ] ( ) 2 (T (T (Ti − Tw ) 𝜕 2 θ − T ) 𝜕 𝜕 T 𝜕 i w 𝜕θ i − Tw ) 𝜕θ 𝜕η = = = 𝜕r R 𝜕η 𝜕η R 𝜕η 𝜕r 𝜕r2 𝜕η2 R2

(8.115)

8.4 Heat Transfer in the Entrance Region of Ducts

Substituting these terms into the partial differential equation (PDE), we obtain ( ( ) ) ] [ Ti − Tw 𝜕θ Ti − Tw 𝜕 2 θ (Ti − Tw ) 1 𝜕θ V + . = α z0 𝜕z+ η 𝜕η 𝜕η2 R2 R2 The unknown parameter z0 is determined next α VR2 V . = 2 ⇒ z0 = z0 α R We now express the axial dimensionless distance as 2(z∕R) 4(z∕D) 4 zα = = z+ = 2 = RePr Pe Gz RV ( ) 2 where Gz = Dz RePr = DαzV = z4+ is the Graetz number, as discussed in [9], and Pe is the Peclet number, and it is defined as Pe = ReD Pr. It is obvious now that the dimensionless temperature distribution is θ = θ(z+ , η). Therefore, the equation to be solved in dimensionless form becomes along with the boundary conditions ( ) 𝜕θ 1 𝜕 𝜕θ η (8.116) = 𝜕z+ η 𝜕η 𝜕η θ(0, η) = 1

(8.117a)

𝜕θ + (z , 0) = 0 𝜕η

(8.117b)

θ(z+ , 1) = 0.

(8.117c)

To solve this PDE, we will employ the separation of variables. Assume a product solution of the form θ = Z(z+ )R(η).

(8.118)

Substituting Eq. (8.118) into the PDE, Eq. (8.116), we get two ODEs: 2 + dZ + λ2 Z = 0 ⇒ Z(z+ ) = e−λ z + dz d2 ℜ 1 dℜ + λ2 ℜ = 0. + η dη dη2 This differential equation, Eq. (8.120a), can be transformed into the following form: ( ) dℜ d η + λ2 ηℜ = 0. dη dη

(8.119) (8.120a)

(8.120b)

The boundary conditions for Eq. (8.120b) are ℜ′ (0) = 0;

(8.121a)

ℜ(1) = 0

(8.121b)

This form of differential equation is called the Sturm–Liouville system, and the weighting function is w = η. See Section 8.10 for a discussion on regular Sturm–Liouville systems. Using Maple 2020, we obtain the following solution to Eq. (8.120a): restart; de ≔ diff(R(𝛈), 𝛈, 𝛈) + de ≔

d𝟐 R(𝛈) + d𝛈𝟐

( ) 𝟏 • diff(R(𝛈), 𝛈) + 𝛌𝟐 • R(𝛈) = 𝟎; 𝛈

d R(𝛈) d𝛈

𝛈

+ 𝛌𝟐 R(𝛈) = 𝟎

sol ≔ dsolve(de, R(𝛈)); sol ≔ R(𝛈) = C1 BesselJ(𝟎, 𝛌𝛈) + C2 BesselY(𝟎, 𝛌𝛈) >

339

340

8 Laminar Momentum and Heat Transfer in Channels

We express the Maple 2020 solution in the following form: ℜ(η) = AJ0 (λη) + BY0 (λη).

(8.122)

Taking the derivative of Eq. (8.122) with respect to η yields ℜ′ (η) = −AλJ1 (λη) − BλY1 (λη).

(8.123)

Using the boundary condition ℜ′ (0) = 0, we obtain ℜ′ (0) = 0 as η → 0 J1 → 0, Y1 → ∞.

(8.124)

Thus, we find that B = 0. Eq. (8.122) becomes ℜ = AJ0 (λη).

(8.125)

Using the second boundary condition ℜ(1) = 0, we find the eigenvalues AJ0 (λ) = 0.

(8.126)

Here, A ≠ 0, and then, we have J0 (λn ) = 0.

(8.127)

The roots of Eq. (8.127) will give characteristic values. There will be an infinite number of values for λn , that is, λ1 , λ2 , . ……λn . For any value of λ, say λn , we may write ℜn = An J0 (λn η). We will present a few roots of Eq. (8.127) using Maple 2020 > evalf(BesselJZeros(𝟎, 𝟏..𝟓)); 𝟐.𝟒𝟎𝟒𝟖𝟐𝟓𝟓𝟓𝟖, 𝟓.𝟓𝟐𝟎𝟎𝟕𝟖𝟏𝟏𝟎, 𝟖.𝟔𝟓𝟑𝟕𝟐𝟕𝟗𝟏𝟑, 𝟏𝟏.𝟕𝟗𝟏𝟓𝟑𝟒𝟒𝟒, 𝟏𝟒.𝟗𝟑𝟎𝟗𝟏𝟕𝟕𝟏 > We have the eigenvalues and eigenfunctions, there are infinite number of solutions, and some of the eigenvalues and eigenfunctions are presented as follows: λ1 = 2.40483 ℜ1 = J0 (λ1 η) λ2 = 5.52008 ℜ2 = J0 (λ2 η) λ3 = 8.653727913 R3 = J0 (λ2 η) From this set, say λn , the following solution is constructed: ) ( θn = An J0 (λn η) exp −λ2n z+

(8.128)

n = 1,2, 3, . … and this solution satisfies the PDE and the boundary conditions. Since any linear combination of these solutions is also a solution, we construct the solution as follows: θ(z+ , η) =

∞ ∑

2 +

An J0 (λn η)e−λn z .

(8.129)

n=1

Using the boundary condition θ(0, η) = 1, we obtain the unknown constant An 1=

∞ ∑ An J0 (λn η). n=1

(8.130)

8.4 Heat Transfer in the Entrance Region of Ducts

This is a Fourier Bessel series. Multiplying this equation by ηJ0 (λn η) and integrating from 0 to 1, we get 1

∫0

1

ηJ0 (λn η)dη = An

∫0

ηJ20 (λn η)dη.

(8.131)

Therefore, the integration of Eq. (8.131) by Maple 2020 yields > > A[n] ≔

) ( 1 ∫0 η • BesselJ(𝟎, 𝛌𝛈)dη

; 1 ∫0 η • (BesselJ(𝟎, 𝛌𝛈))2 dη BesselJ(1, λ) An ≔ ( ) BesselJ(0, λ)2 BesselJ(1, λ)2 + λ 2 2 But BesselJ(0, λ) = 0 for all the values of λ, and we have 2 1 An = . λn J1 (λn ) Thus, the temperature distribution in the pipe is given by θ(z+ , η) = 2

∞ ∑ ( ) 1 Jo (λn η) exp −λ2n z+ . λ J (λ ) n=1 n 1 n

(8.132)

(8.133)

We need to find the mean fluid temperature Tm . Recall that we have vz = V T − Tw Ti − Tw r η= . R The mean fluid temperature is expressed as θ=

∫ vz TdA Tm =

Tm =

A

R

=

∫0 VT(2π r)dr

1

=

∫0 T η dη

1 R ∫ vz dA ∫0 η dη ∫0 V(2π r dr) A { ]} [ ∞ ) ( ∑ 1 Jo (λn η) 1 ∫0 Tw + (Ti − Tw ) 2 ηdη exp −λ2n z+ n=1 λn J1 (λn ) 1

∫0 ηdη ∑1 ) ( Tm = Tw + 4(Ti − Tw ) exp −λ2n z+ . 2 n=1 λn ∞

(8.134)

The dimensionless mean fluid temperature is θm = or

∞ ∑ ) ( Tm − Tw 1 =4 exp −λ2n z+ 2 Ti − Tw n=1 λn

] [ ∞ ∑ 1 2 4(z∕D) exp −λ θm = 4 n 2 Pe n=1 λn

(8.135a)

(8.135b)

where Pe = ReD Pr is the Peclet number. The local heat transfer coefficient in the entrance region of the circular pipe for slug flow subject to constant wall temperature is ( ) 𝜕T ( ) ( ) k k(Ti − Tw ) 𝜕θ −k 𝜕θ 𝜕r r=R h= = = (Tm − Tw ) 𝜕η η=1 Tw − Tm R(Tw − Tm ) 𝜕η η=1 R (Ti − Tw ) ( ) k 1 𝜕θ =− . (8.136) R θm 𝜕η η=1

341

342

8 Laminar Momentum and Heat Transfer in Channels

dJ0 (λn η) = −λn J1 (λn η), the wall temperature gradient (𝜕θ/𝜕η)η = 1 becomes dη [ ( ∞ )] ( ) ∞ 2 + ∑ 1 Jo (λn η) 2 + ∑ 𝜕 1 e−λn z 𝜕θ −λn z e (−λn J1 (λn )) = =2 2 𝜕η η=1 𝜕η λ J (λ ) λ J (λ ) n=1 n 1 n n=1 n 1 n

Since

η=1

= −2

∞ ∑

) ( exp −λ2n z+ .

(8.137)

n=1

Finally, the local Nusselt number is obtained as ∞ ) ( ∑ exp −λ2n z+ h(2R) n=1 NuD = = ) ( ∞ exp −λ2 z+ k ∑ n λ2n

n=1

or

(8.138a)

] 4λ2n (z∕D) exp − Pe n=1 hD = NuD = [ ]. k ∞ 4λ2n (z∕D) ∑ 1 exp − 2 Pe n=1 λn ∞ ∑

[

(8.138b)

The Nusselt number NuD is plotted as a function of dimensionless axial distance z+ , as depicted in Figure 8.18. Both series are convergent for z+ → ∞, and the local Nusselt number is lim NuD = λ21 = (2.40483)2 = 5.7831.

(8.139)

ζ→∞

This result may be seen from an examination of Figure 8.18. Let us now obtain average Nusselt number NuD . Consider an energy balance on control volume (CV), as depicted in Figure 8.19. 20

15

NuD 10

5

0 0.05

0.10

0.15 z+

0.20

0.25

0.30

Figure 8.18 Nusselt number as a function of dimensionless axial distance for laminar slug flow in the entrance region of the isothermal pipe.

8.4 Heat Transfer in the Entrance Region of Ducts

hPdz(Tw – Tm)

m·cp

d m·cp Tm + (m· cp Tm) dz dz

z Figure 8.19

dz

Control volume.

The energy balance on the control volume is d (ṁ cp Tm ) + hP(Tw − Tm ) = 0 dz dT −ρVπ R2 cp m + h(2π R)(Tw − Tm ) = 0 dz −

(8.140)

Tm (0) = Ti

(8.141)

and Eq. (8.140) is expressed in terms of dimensionless variables as Tm = Tw + (Ti − Tw )θm α(Ti − Tw ) dθm dTm dθ dξ = (Ti − Tw ) m = . dz dξ dz dξ VR2 After some manipulation, we obtain −

dθm = NuD dz+ θm

(8.142)

at z+ = 0, θm = 1

(8.143) [

θm =

+

]

z Tm − Tw = exp − NuD dz+ . ∫0 Ti − Tw

(8.144)

NuD is the local Nusselt number based on the pipe diameter. Define now the average Nusselt number as z+

NuD =

1 hD = + NuD dz+ . k z ∫0

(8.145) z+

We take natural log of Eq. (8.144), using ln(θm ) = −∫0 NuD dz+ ; we can obtain the average Nusselt number NuD as NuD = −

ln (θm ) . z+

(8.146)

The local Nusselt number NuD , the mean Nusselt number NuD , and the mean fluid temperature θm are calculated as functions of the Graetz number and presented in Table 8.1. Let us now introduce the logarithmic mean fluid temperature difference. Consider the energy balance dTm + hP(Tw − Tm ) = 0. (8.147) dz Let us introduce the definition ΔT = Tw − Tm . With definition, the temperature at the tube inlet is ΔTi = Tw − Tmi , and the temperature at the tube outlet is ΔT0 = Tw − Tmo . Eq. (8.147) can now be put in the following form: −ṁ cp

−

dTm dΔT hPΔT . = = ̇ p mc dz dz

Separating variable and integrating from the tube inlet to the tube outlet, we have ΔTo

∫ΔTi

L

P dΔT =− h dz ̇ p ∫0 mc ΔT

(8.148a)

343

344

8 Laminar Momentum and Heat Transfer in Channels

Table 8.1 Solutions for slug flow in circular duct under constant wall temperature. z+

𝟏 Gz

NuD

𝛉m

NuD

0.004

0.001

19.5308

0.8613

37.3223

0.01

0.0025

13.0687

0.78452

24.2675

0.02

0.005

9.8838

0.701436

17.7312

0.04

0.01

7.7441

0.59040

13.1737

0.1

0.025

6.1789

0.394175

9.3095

0.2

0.05

5.8167

0.2178524

7.6196

0.4

0.1

5.7834

0.06843129

6.7048

1.0

0.25

5.7831

0.0021295

6.15184

2.0

0.5

5.7831

6.55663 × 10−6

5.96751

4.0

1.0

5.7831

6.215410 × 10−11

5.87535

or

( ln

ΔT0 ΔTi

) =−

PL ̇ p mc

(

) L PLh 1 h dz = − . ̇ p mc L ∫0

Using the definition of the average heat transfer coefficient h, Eq. (8.148b) becomes [ ] ) ( ΔT0 PLh = exp − ̇ p mc ΔTi

(8.148b)

(8.148c)

where the average heat transfer coefficient h is L

h=

1 h dz. L ∫0

(8.149)

Total energy transported by the fluid is q = ṁ cp (Tmo − Tmi ).

(8.150)

By adding and subtracting Tw on the right-hand side of Eq. (8.150), we obtain q = ṁ cp [(Tw − Tmi ) − (Tw − Tmo )] = ṁ cp (ΔTi − ΔTo ).

(8.151)

Solving Eq. (8.148c) for ṁ cp , we get ṁ cp =

PLh ) ( ΔT0 ln ΔTi

and substituting into Eq. (8.151), we obtain (ΔT − ΔTo ) q = PLh (i ) ΔT0 ln ΔTi

(8.152)

(8.153)

or average heat transferred qw is qw = hAs ΔTLMTD

(8.154)

where ΔTLMTD is the logarithmic mean fluid temperature, and As = πDL is the pipe surface area; L is the pipe length. The logarithmic mean fluid temperature ΔTLMTD is ΔTLMTD =

(Tw − Tmi ) − (Tw − Tmo ) . ] [ (Tw − Ti ) ln (Tw − Tm )

(8.155)

8.4 Heat Transfer in the Entrance Region of Ducts

r q″w T(r, z)

T = Ti vz = V 0

D = 2R

z

Ti (a)

Figure 8.20

q″w

0

(b)

z

Slug flow in entrance region of circular tube heated by uniform heat flux.

8.4.1.2 Heat Transfer to Low-Prandtl-Number Fluid Flow (Slug Flow) in the Entrance Region of the Circular Tube Subjected to Constant Heat Flux

Consider laminar flow of a low-Prandtl-number fluid flowing in the entrance region of a pipe. The pipe radius is R. The inlet fluid temperature is Ti . Fluid enters the pipe with uniform velocity V. For low-Prandtl-number fluids, δ/Δ → 0, and for this reason, the velocity profile is approximated as vz ≈ V. See Figure 8.20. Notice that the thermal diffusivity is larger than the momentum diffusivity. The temperature profile develops quicker than the velocity profile. The coordinate system is located at the pipe inlet. Heat transfer begins into slug flow from pipe wall and pipe wall is at UHF q′′w . In other words, constant area duct is heated over its entire length. For this reason, the thermal boundary layer starts at the entrance of the duct and begins to develop along the duct. Axial conduction and viscous dissipation are neglected. There is no energy generation. We wish to determine the temperature distribution and the Nusselt number. The governing equation and its boundary conditions are [ 2 ] 𝜕T 𝜕 T 1 𝜕T =α (8.156) + V 𝜕z r 𝜕r 𝜕r2 T(0, r) = Ti T(z, 0) is finite or

(8.157a) 𝜕T(z, 0) =0 𝜕r

𝜕T(z, R) q′′w = . 𝜕r k

(8.157b) (8.157c)

Let us define the following dimensionless variables: θ=

T − Ti T0

(8.158)

η=

r ; R

(8.159)

z+ =

4(z∕D) 4(z∕D) 4 αz = = = 2 RePr Pe Gz VR

(8.160)

where T0 is the unknown parameter. The unknown parameter T0 will be determined next T = T0 θ + Ti 𝜕T 𝜕θ 𝜕θ 𝜕η T0 𝜕θ = T0 = T0 = . 𝜕r 𝜕r 𝜕η 𝜕r R 𝜕η

(8.161)

Using the wall boundary condition at R T0 𝜕θ q′′ T k 𝜕θ = w ⇒ 0 ′′ = 1. R 𝜕η k Rqw 𝜕η

(8.162)

Next, we choose T0 k Rq′′w . = 1 ⇒ T0 = ′′ k Rqw

(8.163)

345

346

8 Laminar Momentum and Heat Transfer in Channels

Dimensionless temperature θ becomes T(z+ , r) − Ti θ(z+ , η) = . q′′w R∕k Using these dimensionless variables, the PDE and its boundary conditions are given below ( ) 𝜕θ 1 𝜕 𝜕θ η = 𝜕z+ η 𝜕η 𝜕η

(8.164a)

or 𝜕 2 θ 1 𝜕θ 𝜕θ = + 𝜕z+ 𝜕η2 η 𝜕η

(8.164b)

z+ = 0 θ = 0

(8.165a)

𝜕θ =1 (8.165b) 𝜕η 𝜕θ = 0 or θ is finite. (8.165c) z+ > 0 η = 0 𝜕η The PDE is not homogeneous, and the solution can be obtained by applying a special procedure, as discussed in [9] z+ > 0 η = 1

θ(z+ , η) = ψ(z+ , η) + ⏟⏟⏟ dicaying initial transient

ϕ(η) ⏟⏟⏟ radial temperature variation to let wall heat flux into fluid

+

φ(z+ ) ⏟⏟⏟

(8.166)

axial temperature rise due to accumulated heat flux

Now, the formulation of the problem yields a PDE and two ODEs. The PDE is 𝜕 2 ψ 1 𝜕ψ 𝜕ψ = + + 𝜕z η 𝜕η 𝜕η2 ψ(0, η) = −ϕ(η) − φ(0) 𝜕ψ =0 𝜕η 𝜕ψ z+ > 0 η = 1 = 0. 𝜕η The ODEs are d2 ϕ 1 dϕ dφ = + + dη2 η dη dz z+ > 0 η = 0

η = 0 ϕ = finite dϕ = 1. dη Since ϕ(η) and φ(z+ ) can vary independently, equation holds when it is equal to a constant C4 ( ) dϕ d2 ϕ 1 dϕ dφ 1 d = η = C4 . = + η dη dη dη2 η dη dz+ η=1

(8.167) (8.168a) (8.168b) (8.168c)

(8.169) (8.170a) (8.170b)

(8.171)

The general solution of equations can be obtained as φ(z+ ) = C4 z+ + C5 1 C η2 + C6 ln(η) + C7 . 4 4 Using the boundary conditions, θ(z+ , 0)= finite requires that C6 = 0

(8.172)

ϕ(η) =

(8.173)

φ(z+ ) = C4 z+ + C5

(8.174)

ϕ=

1 C η2 + C7 . 4 4

(8.175)

8.4 Heat Transfer in the Entrance Region of Ducts

Using the boundary condition (dϕ/dη)η = 1 = 1, we find that C4 = 2. Hence, equations become φ(z+ ) = 2z+ + C5

(8.176)

1 2 η + C7 . (8.177) 2 Since the solution of ψ(z+ , η) depends on ϕ(η) and φ(z+ ), the constants C5 and C7 may be set equal to zero. Thus, ϕ(η) =

φ(z+ ) = 2z+

(8.178)

1 2 η. 2 On the other hand, assuming a product solution of the form ϕ(η) =

(8.179)

ψ = ℜ(η)Z(z+ )

(8.180)

and substituting this product solution into PDE, Eq. (8.167), and dividing by ℜ(η)Z(z+ ) and separating the variables, we obtain ( ) dℜ 1 dZ 1 d η = = −λ2 (8.181a) η dη dη Z dz+ or dℜ d2 ℜ +η + λ2 η 2 ℜ = 0 dη dη2 dZ + λ2 Z = 0. dz+ The solutions for Eqs. (8.181b) and (8.182) are obtained by Maple 2020 η2

ℜ = C1 J0 (λη) + C2 Y0 (λη) Z = C3 e−λ

2 + z

.

(8.181b) (8.182)

(8.183) (8.184)

Using boundary conditions, θ(z+ , 0) = finite gives that ℜ(0) = finite. we see that C2 = 0 since Y0 → − ∞ for η → 0. Then, the solution is ℜ = C1 J0 (λη).

(8.185)

The boundary condition at η = 1 ℜ′ = 0 requirement gives an infinite number of roots of d dℜ = C1 [J0 (λη)]η=1 = 0 dη dη or J1 (λn ) = 0. Using Maple 2020, we obtain a few roots of Eq. (8.186). These roots are the characteristic values. > restart; > interface(displayprecision = 𝟒); 𝟒 > evalf(BesselJZeros(𝟏, 𝟎..𝟓)); 𝟎.𝟎𝟎𝟎𝟎, 𝟑.𝟖𝟑𝟏𝟕, 𝟕.𝟎𝟏𝟓𝟔, 𝟏𝟎.𝟏𝟕𝟑𝟓, 𝟏𝟑.𝟑𝟐𝟑𝟕, 𝟏𝟔.𝟒𝟕𝟎𝟔 This means we the following roots of Eq. (8.186): λ0 = 0 λ1 = 3.8317

(8.186)

347

348

8 Laminar Momentum and Heat Transfer in Channels

λ2 = 7.0156 λ3 = 10.1735 λ4 = 13.3237 λ5 = 16.4706. Thus, we may now construct the product solution ψ=

∞ ∑

) ( an exp −λ2n z+ J0 (λn η)

(8.187)

n=0

where an = C1 C3 . We now apply the initial condition ψ(0, η) = − ϕ(η) − φ(0) ∞ ∑ 1 an J0 (λn η). − η2 = a0 + 2 n=1

We will find the integrals with Maple 2020 as follows: ) 1 ( η2 1 dη = − η − a0 = 2 ∫0 2 4 ( ) 1 1 ∫0 η − η2 J0 (λn η)dη 2 an = . 1 ∫0 ηJ20 (λn η)dη

(8.188)

(8.189)

(8.190)

Let us evaluate the integrals on the numerator and the denominator >

𝟏

∫𝟎

𝛈 • (BesselJ(𝟎, 𝛌n • 𝛈))𝟐 d𝛈;

J𝟎 (𝛌n )𝟐 J𝟏 (𝛌n )𝟐 + 𝟐 𝟐 ( 𝟐) 𝟏 −𝛈 BesselJ(𝟎, 𝛌n • 𝛈)d𝛈; > 𝛈• ∫𝟎 𝟐 −

J𝟏 (𝛌n )𝛌𝟐n + 𝟐𝛌n J𝟎 (𝛌n ) − 𝟒J𝟏 (𝛌n ) 𝟐 𝛌𝟑n

Knowing the fact that J1 (λn ) = 0, the above integral becomes ( ) 1 ∫0 η − 12 η2 J0 (λn η)dη 2 =− 2 an = n ≠ 0. 1 2 λ J ∫0 ηJ0 (λn η)dη n 0 (λn )

(8.191)

The final result for dimensionless temperature is ∞ ∑ ( ) J (λ η) 1 1 θ = 2z+ + η2 − − 2 exp −λ2n z+ 0 n 2 . 2 4 J0 (λn )λn n=1

The mixing cub fluid temperature is ]} [ { ∞ ∑ q′′w′′ R ( 2 +) J0 (λn η) 1 2 1 1 + ∫0 V Ti + (2πηdη) 2z + η − − 2 exp −λn z 2 k 2 4 n=1 λn J0 (λn ) Tm = . 1 ∫0 V(2πη)dη

(8.192)

(8.193)

We carry out this integration and algebraic manipulation with Maple 2020; then, we obtain the mean fluid temperature ( ′′ ) qw R z+ + T i (8.194) Tm = 2 k

8.4 Heat Transfer in the Entrance Region of Ducts

or in dimensionless form T − Ti θm = ( m ) = 2z+ . q̇′′ R∕k

(8.195)

w

The wall temperature is obtained by setting η = 1 ) ( ∞ ∑ exp −λ2n z+ 1 . θw = θ(z+ , 1) = 2z+ + − 2 4 λ2n n=1

(8.196)

The difference between θm and θw is θw − θm =

∞ ∑ ( ) 1 1 −2 exp −λ2n z+ . 2 4 n=1 λn

(8.197)

Next, we define local heat transfer coefficient as discussed in [9] h(Tw − Tm ) = q′′w or h(Tw − Tm ) = k

𝜕T || 𝜕r ||r=R

(8.198a)

or in terms of dimensionless variables 𝜕θ || hR (θ − θm ) = k w 𝜕η ||η=1 h(2R) (θw − θm ) 𝜕θ || = k 2 𝜕η ||η=1 We now evaluate (𝜕θ/𝜕η)η = 1 , and we find that ( )| 𝜕θ | = 1. | 𝜕η ||η=1

(8.198b) (8.198c)

(8.198d)

The local Nusselt number NuD is NuD =

2 hD = . k (θw − θm )

(8.199)

Using Eqs. (8.197) and (8.199), we get an equation for the local Nusselt number NuD 8

NuD =

∑ ∞

1−8

n=1

1 λ2n

( ) exp −λ2n z+

.

(8.200)

For large values of z+ , we have lim NuD = 8.

(8.201)

z+ →0

The average Nusselt number NuD is usually defined based on the average of difference between wall and mixing cup temperatures, as discussed in [9] z+

1 Tw − Tm = + (Tw − Tm )dξ z ∫0

(8.202)

where ξ is a dummy variable for integration. Then, the average heat transfer coefficient h is q′′w

h=

Tw − Tm

.

(8.203)

The average Nusselt number NuD becomes ′′

NuD =

2 hD D qw = = . k k T −T θw − θm w m

(8.204)

349

350

8 Laminar Momentum and Heat Transfer in Channels

Table 8.2

The values for NuD , NuD , and θm .

z+

NuD

NuD

𝛉w − 𝛉m

0.004

30.5550

44.8367

0.06545

0.01

20.3789

29.4790

0.098140

0.02

15.3304

21.8027

0.130459

0.04

11.8841

16.4650

0.168291

0.1

9.1606

11.9570

0.218326

0.2

8.2382

9.98112

0.242770

0.4

8.0122

8.92762

0.249616

1.0

8.0000

8.34782

0.249999

2.0

8.0000

8.17021

0.25

4.0

8.0000

8.08421

0.25

8.0

8.0000

8.04188

0.25

20.0

8.0000

8.04188

0.25

On the other hand, θw − θm is z+

θw − θm =

1 (θw − θm )dξ z+ ∫0

We substitute Eq. (8.197) into Eq. (8.205) and obtain [ ] ∞ −λ2 ξ z+ ∑ 1 e n 1 −2 dξ θw − θm = + 2 z ∫0 4 n=1 λn [ ) ]z+ ( ∞ ∑ 1 − exp −λ2n z+ 1 z+ −2 θw − θm = + z 4 λ2n λ2n n=1 0 [ ( )] ∞ 2 + ∑ 1 − exp −λ z n 1 = −2 . ( 4) + 4 λ z n=1 n

(8.205)

(8.206)

Then, the average Nusselt number becomes NuD =

∑ ∞

1−8

n=1

{[

8 )] } . ( 1 − exp −λ2n z+

(8.207)

λ4n z+

The manner in which the local Nusselt number NuD , average Nusselt number NuD , and the difference between wall temperature and mean temperature, θw − θm asymptotically approach their limiting values far down the pipe is shown in Table 8.2. 8.4.1.3 Empirical and Theoretical Correlations for Viscous Flow Heat Transfer in the Entrance Region of the Circular Tube

Consider viscous flow heat transfer in the entrance region of a pipe. In the entrance region of a pipe, the flow is not parallel, the radial velocity vr is not zero, and the axial velocity vz is a function of spatial coordinates z and r. Both heat and momentum boundary layers form in the entrance region of the pipe and this simultaneous development of the hydrodynamic and thermal boundary layers is called combined entrance region problem. Combined entrance region problem has been studied extensively due to its practical importance in engineering applications. Early works in combined entrance region were reviewed by Shah and London [3]. We need to use special correlations to take into account the influence of flow disturbances at the pipe inlet. We will present some common correlations valid for combined entrance region.

8.4 Heat Transfer in the Entrance Region of Ducts

Constant wall temperature Sieder and Tate [10] proposed an empirical correlation for developing velocity and temperature

profiles (combined entrance) in short tubes as given below for constant wall temperature. This equation may be used for both liquids and gases in the laminar flow region [( ) ]1∕3 ( μ )0.14 D hD = 1.86 ReD Pr NuD = . (8.208) k L μw Flow is laminar (ReD < 2300) NuD =

hD = mean Nusselt number k

Tw = const 0.48 < Pr < 16700 ( ) μ < 9.75 0.0044 < μw [( ) ]1∕3 ( μ )0.14 D m ReD Pr ≥2 L μw where L is the pipe length from pipe entrance. Here, μw is evaluated at wall surface temperature Tw , and all other fluid properties are evaluated at Tm = (Tmi + Tmo )∕2 where Tmi and Tmo are the mean fluid temperature at the pipe inlet and outlet, respectively. If the outlet temperature is not known, an iterative procedure is required. Baehr and Stephan [11] recommend the following approximate correlation of the average Nusselt number for constant wall temperature in the combined entry region of the circular pipe ( ) 0.0499 3.657 tanh Gz−1 [ ]+ D −1 −1∕3 −2∕3 GzD tanh 2.264GzD + 1.7GzD (8.209) NuD = [ ( )1∕6 ] tanh 2.432 Pr Gz−1 D Tw = const 0.1 ≤ Pr ≤ ∞ ( ) D ReD Pr ≤ ∞ L ρVD ReD = μ 0 ≤ GzD =

NuD =

hD . k

Fluid properties are evaluated at average mean fluid temperature Tm = (Tmi + Tmo )∕2. It is reported that the deviation from numerically calculated mean Nusselt numbers amounts to less than 5% for 1 ≤ Pr ≤ ∞ and increases for small Prandtl numbers 0.1 ≤ Pr ≤ 1 to around 10%. Bennett [13] reports that Stephan recommends the following correlation for the average Nusselt number for circular pipe. The pipe surface temperature Tw is constant. ( )1.33 0.0677 DL ReD Pr NuD = 3.66 + (8.210) )0.83 ( D 0.17 1 + 0.1 Pr ReD Pr L Pr ≥ 1

351

352

8 Laminar Momentum and Heat Transfer in Channels

Tw = const hD . k This correlation falls within −15% to 8.9% of results presented in a table reported by Bennett [13]. Nellis and Klein [14] suggest the following correlation based on the numerical data provided by Hornbeck [15]: ]( )1.12 [ 0.020 D ReD Pr 0.049 + hD Pr L = 3.66 + NuD = (8.211) ( )0.7 k D Re Pr 1 + 0.065 L D Tw = const NuD =

Pr ≥ 0.7. This correlation falls within −15% to 0.8% of results reported by Bennett [13]. Churchill and Ozeo [16] proposed the following correlation for the local Nusselt number. This correlation is developed under simultaneously developing flow conditions in a circular tube. Tube surface is kept under constant wall temperature NuD + 1.7 { )( )]8∕9 }3∕8 [( D π RePr 5.357 1 + 388 L 3∕8

4∕3 ( ) ⎛ ⎡ ⎞ ⎤ π D RePr ⎜ ⎢ ⎟ ⎥ 284 L ⎟ ⎥ = ⎢1 + ⎜ { } { } )2∕3 1∕2 )( )]8∕9 3∕4 ⎟ ⎥ ( [( ⎜ ⎢ D Pr π ⎜ 1+ ⎢ ⎟ ⎥ RePr 1+ ⎣ ⎝ ⎠ ⎦ 0.0468 388 L

(8.212)

TW = const hD . k This correlation falls within −0.7% to 8% of the results presented by Bennet [13]. Kays [18] proposed the following empirical equations due to Hausen for constant wall temperature ] [( ) 0.0668 DL ReD Pr hD NuD = = 3.66 + [( ) ]2∕3 k 1 + 0.04 DL ReD Pr NuD =

(8.213)

Tw = const Pr ≥ 5. This equation gives the average Nusselt number over the length of the tube including the entry region. In this relation, the fluid properties are evaluated at the mean fluid temperature. The pipe diameter and pipe length are D and L, respectively. Kays [18] also proposed the following correlation, based on numerical solution, for thermally and hydrodynamically developing laminar flow of air in a circular pipe with UWT ] [ ReD Pr 0.104 (L∕D) hD = 3.66 + NuD = ] [ k ReD Pr 0.8 1 + 0.016 (L∕D) Pr = 0.7.

(8.214)

The simultaneous development of velocity and temperature profiles result in higher Nusselt numbers compared to the fully developed velocity profile. Fluid properties are evaluated at the average mixing cup or bulk temperature of Tm = (Tmi + Tmo )/2, where Tmi and Tmo are inlet and outlet temperatures of the pipe.

8.4 Heat Transfer in the Entrance Region of Ducts

Bennett [13] reports the following correlation for the average Nusselt number for flow in an isothermal pipe: )0.295 ( + 146.9 − 0.701 5.079Gz1.13 hD L NuD = = { [ ]0.5758 } ) ) ( ( k −1 1∕6 −1 0.2894 tanh 2.434 Pr GzL 1 + 0.4022 Pr GzL

(8.215)

Gz−1 L = L∕(D ReD Pr ) ( GzL =

D L

) ReD Pr

Tw = const. Bennett [13] reports that this equation exhibits error of −3% to 1.2% with respect to exact numerical solution. Gnielinski [19] reported the following correlation for the local Nusselt number: { }1∕3 ( )1∕6 √ [ ( ) ]3 2 1 D D 3 3 Re Pr + 1.077 Re Pr − 0.7 + 0.7 + 3.66 (8.216a) NuD = 2 1 + 22 Pr z D z D D 0 < ReD Pr < ∞ z Tw = constant where z is the distance where heat transfer begins. Gnielinski [19] reported the following correlation for the average Nusselt number: [ ]1∕3 )1∕6 √( ) ( ( )3 2 D 1∕3 3 3 ReD Pr + 1.615GzL − 0.7 + 0.7 + 3.66 NuD = 1 + 22 Pr L D GzL = ReD Pr L

(8.216b)

Tw = constant where Eq. (8.216b) represents over a heated length L of isothermal pipe. Bennett [13] reports that this equation exhibits error of 1.8–8.5% with respect to the exact numerical solution presented in his paper. Constant wall heat flux Churchill and Ozeo [17] also proposed the following correlation for the local Nusselt number NuD .

This correlation is developed under simultaneously developing flow conditions in a circular tube for all (z/D)/Pe and Prandtl numbers. Tube surface is kept under constant wall heat flux NuD + 1 = { )( )]10∕9 }3∕10 [( D π RePr 5.364 1 + 220 z 3∕10

5∕3 ) ⎞ ⎤ ⎡ ⎛ (π) ( D RePr ⎟ ⎥ ⎢ ⎜ 115.2 z ⎟ ⎥ = ⎢1 + ⎜ { )2∕3 }1∕2 { )( )]10∕9 }3∕5 ⎟ ⎥ ( [( ⎢ ⎜ D Pr π ⎟ ⎥ ⎢ ⎜ 1+ RePr 1+ ⎣ ⎠ ⎦ ⎝ 0.0207 220 z

(8.217)

q′′w = const hD NuD = . k Bennett [13] reports that this correlation falls within −5.8% to 3.1% of exact numerical results. Kays [18] also reports the following correlation due to Hausen, for thermally and hydrodynamically developing laminar flow of air in a circular pipe with uniform wall heat flux: ] [( ) D ReD Pr 0.023 hD z NuD = = 4.36 + (8.218) [( ) ]. D k ReD Pr 1 + 0.0012 z

353

354

8 Laminar Momentum and Heat Transfer in Channels

The velocity profile is assumed to be parabolic. Fluid properties are evaluated at average mixing cup or bulk temperature of Tm = (Tmi + Tmo )/2, where Tmi and Tmo are inlet and outlet temperatures of the pipe. Gnielinski [19] recommends the following correlation for local and average Nusselt numbers for pipes of constant wall heat flux: Local Nusselt number. { }1∕3 √( ) [ ]3 √ 3 hD D 1∕3 = 0.462 Pr ReD + 1.302Gzz − 1 + 13 + (4.354)3 (8.219a) NuD = k z D Gzz = ReD Pr z Pr > 0.7 Average Nusselt number. }1∕3 { √( ) [ ]3 √ 3 hD D 1∕3 = 0.924 Pr ReD + 1.953GzL − 0.6 + 0.63 + (4.364)3 NuD = k L D GzL = ReD Pr L

(8.219b)

q′′w = const. Bennett [13] reports that this correlation is within −2.4% to 6.7% of exact results presented in his paper. Bennett [13] reports the following correlation for the average Nusselt number for pipe flow with constant heat flux: ( )0.2908 6.664Gz1.146 + 240.1 − 0.559 hD L = (8.220) NuD = [ ( )0.5670 ] ( ( ) ) k −1 1∕6 −1 0.2940 tanh 2.524 Pr GzL 1 + 1.415 Pr GzL Gz−1 L = L∕(D Re Pr ) where D is the tube diameter and L is the tube length over which convection is averaged. Bennett [13] reports that this equation exhibits errors of −2.4% to 5.2%. Example 8.3 Water flows through a 20-mm tube. The tube length is 100 cm. Mass flow rate of water is 0.0025 kg/s.The mean inlet temperature of the oil is 10 ∘ C and it is required to heat to water 54 ∘ C. The surface of the tube is maintained at a constant temperature of 80 ∘ C. Determine the heat transfer. Solution The average bulk temperature Tm = (Tmi + Tmo )/2. Tm = (10 + 54)∕2 = 32 ∘ C = 305 K. We will estimate the physical properties of water at 305 K. ρ = 995 kg∕m3 cp = 4178 J∕kg.K k = 0.620 W∕m.K Pr = 5.20 μ = 769 × 10−6 N.s∕m2 μw ≈ 351.8 × 10−6 N.s∕m2 First, we calculate the Reynolds number ReD =

4 × 0.025 4ṁ = ≈ 2069. πμD π × 769 × 10−6 × 0.02

We should ascertain whether the flow is fully developed or not. For this reason, we calculate the hydrodynamic and thermal entrance lengths LH = 0.05ReD D = 0.05 × 2069 × 0.02 = 2.069 LT = 0.05ReD D Pr = 0.05 × 2069 × 0.02 × 5.2 = 10.76 m.

8.4 Heat Transfer in the Entrance Region of Ducts

Thus, LH > L LT > L. Both the velocity and thermal boundary layers are developing, and this is a combined entrance length problem. We will use Eq. (8.208). Sieder and Tate empirically developed this equation for short tubes for constant wall temperature [( ) ]1∕3 ( μ )0.14 D hD = 1.86 ReD Pr NuD = k L μw )0.14 [( ) ]1∕3 ( hD 0.02 769 × 10−6 = 1.86 (2069)(5.2) NuD = = 2.38 k 1 351.8 × 10−6 ) ( k 0.620 (2.38) = 74 W∕m2 K. h = NuD = D 0.02 We now calculate the ΔTLMTD (T − T ) − (Tw − Tmi ) (80 − 54) − (80 − 10) ΔTLMTD = w ( mo = ) ) = 44.42 ( (Tw − Tmo ) (80 − 54) ln ln (Tw − Tmi ) (80 − 10) Heat transported by water is q = hAs ΔTLMTD = h(πDL)ΔTLMTD = (74) × π × 0.02 × 1 × 44.42 = 413.6 W.

8.4.2 Parallel Plates: Slug Flow Heat Transfer in the Entrance Region Again, suppose that the unheated entrance length of the parallel plate channel is short and the Prandtl number is very low (Pr ≪ 1). Under these conditions, the temperature profile develops faster than the velocity profiles. It is reasonable to assume that the axial velocity is constant and is approximated as u≈V where V is the average fluid velocity. This assumption is valid in the entrance region and is not valid to large distances down the tube since the velocity profile will eventually develop. 8.4.2.1 Heat Transfer to a Low-Prandtl-Number Fluid (Slug Flow) in the Entrance Region of Parallel Plates: Both Plates Are Subjected to Constant Wall Temperatures

Low-Prandtl-number fluid flows steadily in the entrance region of two parallel plates. The flow is two-dimensional and incompressible. The fluid properties are constant. The separation distance between the parallel plates is 2b. See Figure 8.21. Both the plates are kept at uniform temperature Tw . Fluid enters the channel with initial temperature Ti and with uniform velocity V. For low-Prandtl-number fluids, δ/Δ → 0, and for this reason, the velocity profile is approximated as u ≈ V. Notice that the thermal diffusivity is larger than the momentum diffusivity. The temperature profile develops quicker than the velocity profile. The coordinate system is located at the channel inlet. Constant area duct is heated over its entire length. For this reason, the thermal boundary layer starts at the entrance of the duct and begins to develop along the duct. Neglecting axial conduction and viscous dissipation, we wish to determine the temperature distribution and the Nusselt number. Under the stated assumptions, the energy equation reduces to 𝜕2T 𝜕T =k 2. 𝜕x 𝜕y The boundary conditions are ρ cp V

(8.221)

T(0, y) = Ti

(8.222a)

𝜕 T(x, 0) =0 𝜕y

(8.222b)

355

356

8 Laminar Momentum and Heat Transfer in Channels

y

Tw

u=V 0

2b

x

T = Ti

Δ(x) (a)

Figure 8.21

Tw

Tw Ti (b)

x

Slug flow between parallel plates. (a) Thermal boundary layer development. (b) Temperature distribution.

T(x, b) = Tw .

(8.222c)

We wish to put the governing equation and its boundary conditions in dimensionless form. For this purpose, we will define the following dimensionless variables: θ=

T(x, y) − Tw Ti − Tw

(8.223)

η = y∕b x+ =

(8.224)

4(x∕b) 4(x∕b) = ReDH Pr Pe

(8.225)

where ReDH = ρVDH ∕μ is the Reynolds number, Pr = μcp /k is the Prandtl number, DH = 4b is the hydraulic diameter, and Pe = ReDH Pr is the Peclet number. The axial conduction for Pe ≥ 100 may be neglected, as discussed in [1]. Now, we can put the energy equation in the following dimensionless form: 𝜕2 θ 𝜕θ = 𝜕x+ 𝜕η2

(8.226)

and the boundary conditions take the following form: x+ = 0

θ=1

x+ > 0

η=0

x+ > 0

η = 1 θ = 0.

(8.227a) 𝜕θ = 0 (Symmetry) 𝜕η

(8.227b) (8.227c)

The solution of this PDE, Eq. (8.225), is obtained by separation of variables. Assuming a product solution in the following form: θ = X(x+ )Y(η) and substituting into PDE, Eq. (8.225), and after dividing both sides by X(x+ )Y(η), we get 1 d2 Y 1 dX = = −λ2 . + X dx Y dη2

(8.228)

From Eq. (8.228), we get d2 Y + λ2 Y = 0. dη2

(8.229)

The boundary conditions of Eq. (8.229) are Y′ (0) = 0

(8.230a)

Y(1) = 0.

(8.230b)

The eigenvalues and nontrivial eigenfunctions are Yn = Bn cos(λn η)

(8.231)

8.4 Heat Transfer in the Entrance Region of Ducts

π n = 0,1, 2,3. … 2 The differential equation in the x+ -direction is cos(λn ) = 0 λn = (2n + 1)

1 dX = −λ2 X dx+

(8.232)

and the solution is ) ( Xn = An exp −λ2n x+ .

(8.233)

Thus, the product solution is θ(ξ, η) =

∞ ∑

) ( an exp −λ2n x+ cos(λn η)

(8.234)

n=0

where an = An Bn . Finally, we use the nonhomogeneous boundary condition to evaluate the unknown constant an 1=

∞ ∑

an cos(λn η).

(8.235)

n=0

Equation (8.235) is a Fourier cosine series expansion of 1 over the interval (0, 1). The coefficient an is evaluated in the usual manner, and the result is 1

an =

∫0 cos(λn η)dη 1 ∫0 cos2 (λn η)dη

= (−1)n

2 . λn

(8.236)

Introducing this equation into the solution will give temperature distribution as ∞ ∑ ) ( (−1)n θ(ξ, η) = 2 exp −λ2n x+ cos(λn η). λn n=0

(8.237)

The mean fluid temperature Tm is evaluated as follows: b

Tm =

∫0 uTdy b ∫0

udy

b

=

1 Tdy. b ∫0

(8.238a)

Substituting temperature T = Tw + (Ti − Tw )θ into Eq. (8.238a) to evaluate the mean fluid temperature b

Tm =

1 [Tw + (Ti − Tw )θ]dy b ∫0

(8.238b)

or in terms of dimensionless variables, the dimensionless form of mean temperature is θm =

1 Tm − Tw = θ dη. ∫0 Ti − Tw

Substituting dimensionless temperature distribution θ into Eq. (8.238c), we get ) ( ∞ ∑ exp −λ2n x+ Tm − Tw θm = =2 . Ti − Tw λ2n n=0

(8.238c)

(8.239)

Using the temperature distribution and mean fluid temperature, we can now determine the Nusselt number based on the hydraulic diameter DH = 4b as ( ) [( )] k(Ti − Tw ) 𝜕T 𝜕θ −k − 𝜕y y=b b 𝜕η η=1 q′′w = = (8.240) h= (Tw − Tm ) (Tw − Tm ) (Tw − Tm ) ( ) ∞ ∑ ( ) (−1)n 𝜕θ =2 exp −λ2n x+ [−λn sin(λn η)]η=1 𝜕η η=1 λn n=0 = −2

∞ ∑ n=0

) ( exp −λ2n x+

(8.241)

357

358

8 Laminar Momentum and Heat Transfer in Channels

30

25

20 NuDH 15

10

5 0

0.1

0.2 4(x/b) x+ = RePr

0.3

0.4

Figure 8.22 Nusselt number as a function of dimensionless axial position for plug flow between parallel plates of uniform wall temperature.

The local Nusselt number NuDH is ) ( exp −λ2n x+ hDH n=0 = = ) . ( ∞ exp −λ2 x+ k ∑ n 4

NuDH

∞ ∑

n=0

(8.242)

λ2n

The variation of the Nusselt number with dimensionless axial position is plotted, and the result is depicted in Figure 8.22. The inspection of Figure 8.22 reveals that for large values of ξ, the Nusselt number approaches its asymptotic value of Nu = π2 . 8.4.2.2 Heat Transfer for Low-Prandtl-Number Fluid Flow (Slug Flow) in the Entrance Region of Parallel Plates: Both Plates Are Subjected to UHF

A low-Prandtl-number fluid flows steadily in the entrance region of two parallel plates. The flow is two-dimensional and incompressible with constant properties. The separation distance between the parallel plates is 2b. See Figure 8.23. Both the plates are kept at UHF q′′w . Fluid enters the channel with initial temperature Ti and with uniform velocity V. The velocity profile is approximated as u = V. Notice that the thermal diffusivity is larger than the momentum diffusivity. The temperature profiles develop more rapidly than the velocity profile. The viscous dissipation is negligible, and there is no energy generation. The coordinate system is located at the channel inlet. Constant area duct is heated over its entire length. For this reason, the thermal boundary layer starts at the entrance of the duct and begins to develop along the duct. Neglecting axial conduction, we wish to determine the temperature distribution and the Nusselt number. The energy equation to be solved in the entrance region with uniform velocity distribution between the two parallel plates is given by V

𝜕T 𝜕2T =α 2. 𝜕x 𝜕y

(8.243)

The initial and the boundary conditions are T(0, y) = Ti

(8.244a)

8.4 Heat Transfer in the Entrance Region of Ducts

q″w

y u=V 0

2b

x

T = Ti

Ti

Δ(x)

q″w x

0

q″w (a)

(b)

Figure 8.23 Slug flow between parallel plates of constant heat flux. (a) Thermal boundary layer development. (b) Heat flux distribution.

k

𝜕T(x, b) = q′′w 𝜕y

(8.244b)

𝜕T(x, 0) = 0 (Symmetry) 𝜕y

(8.244c)

Notice that heat flow into fluid is in the negative y-direction. Let us introduce the dimensionless parameters as follows: (T − T ) θ = ( ′′ ) i b qw ∕k 4(x∕b) + x = Pe

(8.245) (8.246)

η = y∕b.

(8.247)

The PDE is expressed in terms of dimensionless variables 𝜕2θ 𝜕θ = 2 + 𝜕x 𝜕η

(8.248)

and the boundary conditions and initial condition takes the following form: x+ = 0 θ = 0

(8.249a)

x+ > 0 η = 0

𝜕θ =0 𝜕η

(8.249b)

x+ > 0 η = 1

𝜕θ = 1. 𝜕η

(8.249c)

The PDE, Eq. (8.248), has a nonhomogeneous boundary condition, and the separation of variables cannot be applied directly. The solution can be obtained by applying a special procedure discussed in [9]: θ(z+ , η) =

ψ(x+ , η) ⏟⏞⏟⏞⏟

+

decaying entrance region transient

ϕ(η) ⏟⏟⏟ transverse temperature variation to let wall heat flux into fluid

+

φ(x+ ) ⏟⏟⏟ axial temperature rise due to accumulated heat flux

We now substitute Eq. (8.250) into the PDE and its boundary conditions. This process of substitution yields dφ 𝜕 2 ψ d2 ϕ 𝜕ψ + = + 2 𝜕x+ dx+ 𝜕η2 dη θ(0, η) = ψ(0, η) + ϕ(η) + φ(0) = 0 𝜕θ(x+ , 0) 𝜕ψ(x+ , 0) dϕ(0) = + =0 𝜕η 𝜕η dη 𝜕θ(x+ , 1) 𝜕ψ(x+ , 1) dϕ(1) = + = 1. 𝜕η 𝜕η dη

(8.250)

359

360

8 Laminar Momentum and Heat Transfer in Channels

We now separate the problem into the following simpler problems 𝜕2ψ 𝜕ψ = 𝜕x+ 𝜕η2

(8.251) (8.252a)

ψ(0, η) = −ϕ(η) − φ(0) η=0 η=1

𝜕ψ =0 𝜕η 𝜕ψ = 0. 𝜕η

(8.252b) (8.252c)

The next set of differential equations is d2 ϕ dφ = + = C. dη2 dx

(8.253)

Functions produced by differential equations must match each other. Since ϕ(η) and φ(x+ ) vary independently, Eq. (8.252a) should be equal to a constant C. Then, we may write two separate ODEs. The first ODE is d2 ϕ = C. dη2

(8.254)

The boundary conditions of Eq. (8.254) are given by dϕ(0) =0 dη dϕ(1) = 1. dη

(8.255a) (8.255b)

Notice that the nonhomogeneous boundary condition on the tube wall is satisfied by ϕ(η). The second ODE is dφ = C. dx+ We now integrate Eqs. (8.254) and (8.256)

(8.256)

φ = Cx+ + C1 ϕ=

Cη2 + C2 η + C3 . 2 ′

The application of the first boundary condition ϕ (0) = 0 yields C2 = 0. Next, we apply the second boundary condition ϕ (1) = 1. The application of this boundary condition yields ′

C = 1. Hence, the solutions are written as φ = x+ + C 1 η2 + C3 2 where C1 and C3 are the remaining unknown constants. Since the solution ψ(x+ , η) depends on φ(x+ ) and ϕ(η), we may arbitrarily set these constants equal to zero. Thus, we write ϕ=

φ = x+ η2 . 2 We now solve Eq. (8.251) by assuming a product solution in the form

(8.257)

ϕ=

(8.258)

ψ = X(x+ )Y(η).

(8.259)

8.4 Heat Transfer in the Entrance Region of Ducts

The substitution of Eq. (8.259) into Eq. (8.251) yields two ODEs. The first differential equation is d2 Y + λ2 Y = 0. dη2

(8.260)

The boundary conditions are Y′ (0) = 0

(8.261a)

Y′ (1) = 0.

(8.261b)

The second ODE is dX + λ2 X = 0. dx+ The solution of Eq. (8.260) is

(8.262)

Yn = An cos(λn η) + Bn sin(λn η). We take the first derivative of Yn as Y′n = −An λn sin(λn η) + Bn λn cos(λn η). Application of Eq. (8.261a) yields Bn = 0. Application of Eq. (8.261b) gives An λn sin(λn ) = 0. Finally, we get the following solution: Yn (η) = An ϑn (η) ϑn (η) = cos(λn η) characteristic functions λn = nπ n = 0,1, 2,3 …… characteristic values. The general solution of Eq. (8.262) is ) ( Xn (x+ ) = Cn exp −λ2n x+ .

(8.263)

Thus, the product solution becomes ψ(x+ , η) = a0 +

∞ ∑

) ( an exp −λ2n x+ cos(λn η)

n=1

where a0 = A0 C0 and an = An Cn . We now apply the initial condition of Eq. (8.251), and this will give us −ϕ(η) − φ(0) = a0 +

∞ ∑

an cos(λn η)

n=1

or ∞ ∑ η2 = a0 + an cos(λn η) 2 n=1 ( 2) 1 ∫0 − η2 dη 1 a0 = =− 1 6 ∫ dη (0 2 ) 1 ∫0 − η2 cos(nπη)dη 2 an = = −(−1)n 2 2 . 1 2 n π ∫0 cos (nπη)dη

−

Equation (8.264) becomes ∞ ∑ ) ( (−1)n 1 ψ(x+ , η) = − − 2 exp −λ2n x+ cos(λn η). 2 6 n=1 λn

(8.264)

361

362

8 Laminar Momentum and Heat Transfer in Channels

The final solution is expressed as θ(x+ , η) = x+ +

∞ ∑ ) ( η2 1 (−1)n − −2 exp −λ2n x+ cos(λn η). 2 2 6 n=1 λn

(8.265)

The mean fluid temperature Tm is b

∫0 uTdy

b

1 Tdy ∫0 b udy ] b[ 1 bq′′ bq′′ 1 Ti + w θ dy = Ti + w θ dη Tm = b ∫0 k k ∫0 ] [ ∞ ∑ ( 2 +) bq′′w 1 + η2 1 (−1)n = Ti + − −2 exp −λn x cos(λn η) dη x + 2 k ∫0 2 6 n=1 λn } { ∞ ∑ b q′′w (−1)n −λ2 x+ 1 1 1 + e n cos(λn η)dη = Ti + x + − −2 2 ∫0 k 6 6 n=1 λn } { ∞ ∑ b q′′w (−1)n −λ2 x+ sin(λn ) 1 1 + e n = Ti + . x + − −2 2 k 6 6 λn n=1 λn

Tm =

b ∫0

=

sin(λn ) = sin(n π) is zero for any value of n and the mean temperature becomes Tm = Ti +

b q′′w + x k

or in dimensionless form θm =

k(Tm − Ti ) = x+ . b q′′w

(8.266a)

The wall temperature Tw is ( ′′ ) qw b θ|η=1 Tw = Ti + k } ( ′′ ) { ∞ ∑ ( 2 +) bqw η2 1 (−1)n + − −2 Tw = Ti + exp −λn x cos(λn η) x + 2 k 2 6 n=1 λn

η=1

(−1)n .

where cos(λn ) = cos(nπ) = Therefore, the wall temperature becomes } ( ′′ ) { ∞ ∑ ( 2 +) bqw (−1)n 1 + n x + −2 exp −λn x (−1) Tw = Ti + 2 k 3 n=1 λn } ( ′′ ) { ∞ ∑ ( 2 +) bqw 1 1 + Tw = Ti + exp −λn x x + −2 2 k 3 n=1 λn Recall that the local heat transfer coefficient is defined as h=

q′′w . Tw − Tm

We can evaluate the temperature difference Tw − Tm as )} ( ( ′′ ) { ∞ ∑ exp −λ2n x+ bqw 1 −2 Tw − Tm = . k 3 λ2n n=1 The local heat transfer coefficient h is h=

q′′w q′′w = ( { )} ( ) ∞ Tw − Tm ∑ exp −λ2n x+ bq′′w 1 −2 k 3 λ2n n=1

(8.266b)

8.4 Heat Transfer in the Entrance Region of Ducts

h=

q′′w k = { Tw − Tm b 1

1 )}. ( ∞ ∑ exp −λ2n x+ −2 3 λ2n n=1

The local Nusselt number based on the hydraulic diameter DH = 4b is NuDH =

(4b)h 4 = { ( 2 +) } ∞ k ∑ exp −λn x 1 −2 3 λ2n n=1

(8.267)

and the axial distance x+ for parallel plates is subjected to UHF. It is noted that the Nusselt number is presented in a manner appropriate for internal flow. The local Nusselt number is plotted as a function of dimensionless distance x+ and depicted in Figure 8.24. The inspection of Figure 8.24 reveals that the asymptotic value of the Nusselt number is NuDH = 12. 8.4.2.3 Heat Transfer for Low-Prandtl-Number Fluid Flow (Slug Flow) in the Entrance Region of Parallel Plates: Upper Plate Is Insulated While the Lower Plate Is Subjected to Constant Wall Temperature

A low-Prandtl-number fluid flow steadily in the entrance region of two parallel plates. The flow is laminar, two-dimensional, and incompressible with constant properties. Viscous dissipation is negligible, and there is no energy generation. The separation distance between the parallel plates is b. The coordinate system is shown in Figure 8.25. The upper plate is insulated, while the lower plate is kept at constant temperature Tw . Fluid enters the channel with initial temperature Ti and with uniform velocity V. For low-Prandtl-number fluids, δ/Δ → 0, and for this reason, the velocity profile is approximated as u ≈ V. Heat is transferred from the lower plate into the fluid. In other words, constant area duct is heated over its entire length. For this reason, the thermal boundary layer starts at the entrance of the duct and begins to develop along the duct. Notice that the coordinate system is located at entrance on the lower plate. Axial conduction and viscous dissipation are neglected. We wish to determine the temperature distribution and the Nusselt number. Let us examine the physics of the problem. Near the origin of the thermal boundary layer, the thermal boundary layer thickness Δ(x) is very thin compared with the channel separation distance b. The insulated surface is unfelt by the heat 30

25

20 NuDH 15

10

5 0

Figure 8.24

0.1

0.2 X+

0.3

Nusselt number as a function of dimensionless.

0.4

363

364

8 Laminar Momentum and Heat Transfer in Channels

Y

Insulated wall

T = Ti u=V

Δ(x)

Thermal boundary layer

0

b

Tw

Figure 8.25

Thermal boundary layer development for slug flow between parallel plates.

transfer process until the thermal boundary layer reaches to the insulated plate. Prior to the boundary layer reaching to the insulated upper plate, the heat transfer problem can be solved using the boundary layer concept. As may be seen from Figure 8.25, the origin of the coordinate system is located on the lower plate and y is measured from the lower plate. A thermal boundary layer develops along the plate starting from the inlet of the parallel plate channel. The problem is clearly a boundary layer problem. This is true until the thermal boundary layer fills the channel. At this point, flow becomes fully developed. The energy equation to be solved in the entrance region with uniform velocity distribution between the two parallel walls separated by a distance b is given by V

𝜕T 𝜕2T =α 2 𝜕x 𝜕y

(8.268)

where V is the average velocity in the channel formed by two parallel plates and α = k/ρcp is the thermal diffusivity. The boundary conditions are T(0, y) = Ti

(8.269a)

T(x, 0) = Tw .

(8.269b)

We will now assume that the temperature of the fluid is not affected by the heating or cooling of the plate. This means that we can safely assume the temperature field to extend the space over the lower plate. Then, we can assume a boundary condition as follows: T(x, ∞) = Ti .

(8.269c)

With this new boundary condition, we can now obtain a similarity solution. The similarity solution will be obtained by dimensional analysis, and the method of obtaining the similarity variable is explained in [21]. The formulation of the problem in terms of a new temperature θ = T − Tw is V

𝜕θ 𝜕2 θ =α 2 𝜕x 𝜕y

(8.270)

θ(0, y) = θi where θi = Ti − Tw

(8.271a)

θ(x, 0) = 0

(8.271b)

θ(x, ∞) = θi .

(8.271c)

We wish to obtain a similarity solution of the problem, and for this purpose, we will nondimensionalize x, y, and θ in terms arbitrarily selected reference lengths x0 , y0 and the characteristic temperature θi , respectively, as follows: x x0 y y∗ = y0 x∗ =

θ∗ =

θ . θi

(8.272) (8.273) (8.274)

8.4 Heat Transfer in the Entrance Region of Ducts

Next, we will nondimensionalize the PDE, Eq. (8.270), as follows: θ 𝜕θ∗ 𝜕θ 𝜕θ∗ 𝜕θ∗ 𝜕x∗ = θi = θi ∗ = i ∗ 𝜕x 𝜕x 𝜕x 𝜕x x0 𝜕x ∗ ∗ 𝜕y∗ θ 𝜕θ∗ 𝜕θ 𝜕θ 𝜕θ = θi = θi ∗ = i ∗. 𝜕y 𝜕y 𝜕y 𝜕y y0 𝜕y ( ) ( ) ( ) θi 𝜕θ∗ θi 𝜕 2 θ∗ θi 𝜕θ∗ 𝜕y∗ 𝜕 𝜕 𝜕2 θ = = ∗ = . 𝜕y y0 𝜕y∗ 𝜕y y0 𝜕y∗ 𝜕y 𝜕y2 y20 𝜕y∗2

(8.275a) (8.275b) (8.275c)

After some algebra, the PDE, Eq. (8.270), becomes V y20 𝜕θ∗ 𝜕 2 θ∗ = ∗2 . ∗ α x0 𝜕x 𝜕y Equation (8.276) implies that ( ) 2 x y V y0 θ =f , , . θi x0 y0 α x0

(8.276)

(8.277)

The use of θi incorporates the effect boundaries into Eq. (8.277). We have the freedom to transform the variables in Eq. (8.277) in any way we wish without changing the number of these variables. First of all, x0 and y0 are rejected by the physics of the problem as being characteristic lengths. We introduce a new variable in place of V y20 ∕α x0 . This new variable is obtained dividing V y20 ∕α x0 by x/x0 : V y20 ∕α x0 x∕x0

=

Vy20 Vy20 x0 = . × αx0 x αx

The result of transformation is ) ( 2 x y V y0 θ . = f1 , , θi x 0 y0 α x

(8.278)

Equation (8.278) has significance only if the left- and right-hand sides are dimensionally homogeneous. Since there is no characteristic length in the x-direction, temperature and, therefore, the right-hand side of Eq. (8.278) are independent of x0 . This reduces Eq. (8.278) to the following form: ) ( y V y20 θ . (8.279) = f1 , θi y0 α x We again transform the independent variables of Eq. (8.279) such that only one term remains depending on y0 . This is done by introducing a new variable in place of V y20 ∕α x. This new variable is obtained by multiplying V y20 ∕α x by y2 ∕y20 : V y20 αx

×

y2 y20

=

V y2 . αx

The result of transformation is ) ( y V y2 θ . = f2 , θi y0 α x

(8.280)

There is no characteristic length in the y-direction; therefore, temperature and, consequently, the right-hand side of Eq. (8.280) must be dimensionally homogeneous. This means that Eq. (8.280) is independent of y0 . Eq. (8.280) reduces to the following form: ( 2) Vy θ . (8.281) = f2 θi αx Equation (8.281) can be recast as θ = f(η) θi

(8.282)

365

366

8 Laminar Momentum and Heat Transfer in Channels

√ √ where η = y V∕α x = y∕ α x∕V is the similarity variable. Notice that we combined two independent variables into one variable. Now, we will express the PDE, Eq. (8.270), in terms of η ⎤ ⎡ ⎥ yV 𝜕θ 𝜕f df 𝜕η df ⎢ = θi = θi = θi − ( )1∕2 ⎥ 𝜕x 𝜕x dη 𝜕x dη ⎢⎢ V 2 ⎥ ⎦ ⎣ 2 αx α x ( )1∕2 ⎤ ⎡ y αVx ⎥ V df ⎢ − = θi ⎥ ⎢ ( ) ( ) 1∕2 1∕2 dη ⎢ V V 2 ⎥ ⎦ ⎣ 2 αx α x αx ⎤ ⎡ ( )1∕2 ⎥ θ η df df ⎢ V V = θi =− i − ( )y ⎥ ⎢ V dη ⎢ 2 αx2 αx 2x dη ⎥ αx ⎦ ⎣ ⎤ ⎡ 𝜕θ 𝜕f df 𝜕η df ⎢ 1 ⎥ = θi = θi = θi √ 𝜕y 𝜕y dη 𝜕y dη ⎢⎢ α x ⎥⎥ V ⎦ ⎣ ⎞ ⎞ ⎛ ⎛ θi d2 f d ⎜ df 1 ⎟ 𝜕η 𝜕 ⎜ df 1 ⎟ 𝜕2θ = θ θ = = . √ √ ⎟ ⎟ ⎜ ⎜ i i α x 𝜕y ⎜ dη α x ⎟ dη ⎜ dη α x ⎟ 𝜕y 𝜕y2 dη2 V V ⎠ V ⎠ ⎝ ⎝

(8.283)

(8.284)

(8.285)

We substitute Eqs. (8.283) and (8.285) into Eq. (8.270), and after algebra, the PDE is now converted to an ODE as given below d2 f η df + = 0. dη2 2 dη

(8.286)

The boundary conditions are f(0) = 0

(8.287a)

f(∞) = 1.

(8.287b)

This is a boundary value problem, and the solution may be obtained by bvp4c in MATLAB 2021a or Maple 2020. The solution of Eq. (8.286) is obtained by Maple 2020 > restart; > interface(displayprecision = 4); > de ≔ diff(f(𝜼), 𝜼, 𝜼) + de ≔

d𝟐 𝟏 f(𝛈) + 𝛈 𝟐 𝟐 d𝛈

(

(𝜼) 𝟐

•

diff(f(𝜼), 𝜼) = 0;

) d f(𝛈) = 𝟎 d𝛈

> sol ≔ dsolve({de, f(0) = 𝟎, f(∞) = 1}, f(𝜼)); sol ≔ f(𝛈) = erf

(

𝟏 𝛈 𝟐

)

The solution can be expressed as ) ( T − Tw y θ . = = erf √ θi Ti − Tw 2 α x∕V

(8.288)

8.4 Heat Transfer in the Entrance Region of Ducts

Let us select Tw − Ti as the appropriate temperature difference. We can determine the Nusselt number based on this temperature difference: ) ) ( ( 𝜕T 𝜕T [ (2b) −k (2b) − √ ] 𝜕y y=0 𝜕y y=0 DH q′′w hDH 2b V = = = √ NuDH = k k (Tw − Ti ) k (Tw − Ti ) (Tw − Ti ) α x π √ √ √ DH V DH 1 1 1 4 b2 V 2b V 2b = √ = √ = √ 1 α x x α x α π π π √ DH √ 1 Pe (8.289) = √ x π where DH = 2b is the hydraulic diameter and Pe = V DH /α is the Peclet number. 8.4.2.4 Heat Transfer for Low-Prandtl-Number Fluid Flow (Slug Flow) in the Entrance Region of Parallel Plates: Upper Plate Is Insulated While the Lower Plate Is Subjected to Constant Heat Flux

A low-Prandtl-number fluid flows steadily in the entrance region of two parallel plates. Flow is laminar, two-dimensional, and incompressible with constant properties. Axial conduction and viscous dissipation are negligible, and there is no internal energy generation. The separation distance between the parallel plates is b. The coordinate system is shown in Figure 8.26, where origin is located at the corner of lower plate. The y-coordinate is measured from the lower plate and fluid enters with uniform velocity u = V as well as with uniform temperature Ti . The lower plate delivers a UHF q′′w to the fluid and the upper plate is insulated. Constant area duct is heated over its entire length. For this reason, the thermal boundary layer starts at the entrance of the duct and begins to develop along the duct. For low-Prandtl-number fluids, δ/Δ → 0, and for this reason, the velocity profile is approximated as u ≈ V. Let us examine the physics of the problem just as we have done before. The opposing insulated surface is unfelt by the heat transfer process until the thermal boundary layer reaches to insulated surface. Prior to reaching to insulated surface, the heat transfer problem can be solved using the boundary layer concept. A thermal boundary layer will develop along the plate starting from the inlet of the parallel plate channel. The problem is clearly a boundary layer problem. This is true until the thermal boundary layer fills the channel. At this point, flow becomes fully developed. The energy equation to be solved with uniform velocity distribution between the two parallel plates separated by a distance b is given by V

𝜕2 T 𝜕T =α 2 𝜕x 𝜕y

(8.290)

where V is the velocity in the channel formed by two parallel plates. The boundary conditions are (8.291a)

T(0, y) = Ti −k

𝜕T(x, 0) = q′′w . 𝜕y

(8.291b)

We will now assume that core temperature outside the thermal boundary layer is not affected by the heating or cooling of the plate. This means that we can safely assume the temperature field to extend the space over the lower plate. Then, we can assume a boundary condition as follows: at y = ∞ T = Ti .

(8.291c)

y

Insulated wall

T = Ti b

u=V

Δ(x)

0 Figure 8.26

x

q″w

Thermal boundary layer development for slug flow between parallel plates.

367

368

8 Laminar Momentum and Heat Transfer in Channels

With this new boundary condition, we can now obtain a similarity solution. The similarity solution will be obtained by dimensional analysis. This method of obtaining the similarity variable is explained in [21]. The formulation of the problem in terms of a new temperature θ = T − Ti is V

𝜕2 θ 𝜕θ =α 2 𝜕x 𝜕y

θ(0, y) = 0 −k

𝜕θ(x, 0) = q′′w 𝜕y

θ(x, ∞) = 0.

(8.292) (8.293a) (8.293b) (8.293c)

Let us make x and y dimensionless in terms of reference lengths x0 and y0 . We make also temperature θ dimensionless by reference temperature θR . This means we have the following dimensionless variables: x x0 y y∗ = y0 θ . θ∗ = θR x∗ =

(8.294) (8.295) (8.296)

Then, the differential equation and the heat flux boundary condition can be expressed as VθR 𝜕θ∗ 𝜕θ∗ 𝜕θ∗ 𝜕x∗ 𝜕θ = VθR = VθR ∗ = 𝜕x 𝜕x 𝜕x 𝜕x x0 𝜕x∗ ∗ ∗ 𝜕y∗ αθ 𝜕θ∗ 𝜕θ 𝜕θ 𝜕θ α = α0R = αθR ∗ = R ∗ 𝜕y 𝜕y 𝜕y 𝜕y y0 𝜕y ( ) ( ) ) ( αθR 𝜕 2 θ∗ αθR 𝜕θ∗ 𝜕y∗ 𝜕2θ 𝜕 𝜕 αθR 𝜕θ∗ α 2 = = ∗ = 𝜕y y0 𝜕y 𝜕y y0 𝜕y 𝜕y 𝜕y y20 𝜕y∗2 kθ 𝜕θ∗ 𝜕θ∗ 𝜕θ∗ 𝜕y∗ 𝜕θ = kθR = θR ∗ = R ∗. k 𝜕y 𝜕y 𝜕y 𝜕y y0 𝜕y V

(8.297a) (8.297b) (8.297c) (8.297d)

We substitute Eqs. (8.297a)–(8.297c) into Eq. (8.292) and obtain αθ 𝜕 2 θ∗ VθR 𝜕θ∗ = 2R ∗2 . ∗ x0 𝜕x y0 𝜕y

(8.298)

From the second boundary condition, Eq. (8.293b), we obtain −

kθR 𝜕θ∗ = q′′w y0 𝜕y∗

(8.299)

−

y0 q′′w 𝜕θ∗ = . 𝜕y∗ kθR

(8.300)

or

Equations (8.298) and (8.300) imply that ) ( 2 ′′ θ x y Vy0 qw y0 . =f , , , θR x0 y0 αx0 kθR

(8.301)

Physics of the problem indicates that θR , x0 , and y0 are not characteristic properties, and for this reason, we will eliminate these arbitrary quantities. Let us eliminate θR . We will transform Eq. (8.301) such that only one term remains depending on θR . To eliminate θR , we multiply the left-hand side of Eq. (8.301) by kθR ∕q′′w y0 kθR θ θ × = ′′ θR q′′w y0 qw y0 ∕k

8.4 Heat Transfer in the Entrance Region of Ducts

and the result of transformation is ( ) 2 ′′ θ x y Vy0 qw y0 , , , = f1 . x0 y0 αx0 kθR q′′w y0 ∕k This mathematical expression has physical importance in θ when it is independent of θR . Thus, we have ( ) 2 θ x y Vy0 . = f1 , , x0 y0 αx0 q′′w y0 ∕k Let us eliminate x0 and y0 as we have done before. Finally, we have ( 2) Vy θ . = f2 αx q′′w y∕k

(8.302)

(8.303)

(8.304)

Equation (8.304) can be expressed as θ = yf(η) (8.305a) q′′w ∕k √ where η = y∕ αx∕V is the similarity variable. We now rearrange Eq. (8.305a) using similarity variable η. New form of Eq. (8.305a) is √ θ αx ηf(η). (8.305b) = V q′′w ∕k Let us introduce F = ηf. Then, Eq. (8.305b) takes the following form: ( ′′ ) √ qw0 αx F(η). θ= k V Now, let us express the PDE, Eq. (8.292), in terms of new variables √ (√ ) ] ′′ [ 𝜕θ qw 𝜕 αx α x dF 𝜕η = F+ 𝜕x k 𝜕x V V dη 𝜕x ] ′′ [√ q y dF 𝜕θ αx F = w − 𝜕x k V 2x 2x dη [( ′′ ) √ ] [√ ] ( ) qw q′′ q′′ dF 𝜕 αx α x dF 𝜕η 𝜕θ = F(η) = w = w 𝜕y 𝜕y k V k V dη 𝜕y k dη [ ′′ ( )] [ ′′ ( )] )−1∕2 ( 2 ) ′′ ( 2 q q q 𝜕η 𝜕 θ 𝜕 dF dF d dF w αx w0 w0 = . = = 𝜕y k dη dη k dη 𝜕y k V 𝜕y2 dη2

(8.306)

(8.307a)

(8.307b)

Substituting Eqs. (8.307a) and (8.307b) into Eq. (8.292) and after some algebra, we can transform a PDE to an ODE d2 F η dF 1 − F=0 + dη2 2 dη 2

(8.308)

F′ (0) = −1

(8.309a)

F(∞) = 0.

(8.309b)

The solution of Eq. (8.308) may be obtained by bvp4c in MATLAB 2021a or Maple 2020. The solution of this boundary value problem is obtained by Maple 2020. > restart; > ( ) (𝛈) 𝟏 • diff(F(𝛈), 𝛈) − F(𝛈) = 𝟎; > de ≔ diff(F(𝛈), 𝛈, 𝛈) + 𝟐 𝟐 ) ( d F(𝛈) 𝛈 d𝛈 𝟐 F(𝛈) d de ≔ − =𝟎 F(𝛈) + 𝟐 𝟐 d𝛈𝟐

369

370

8 Laminar Momentum and Heat Transfer in Channels

> sol ≔ dsolve({de, F′ (𝟎) = −𝟏, F(∞) = 𝟎}, F(𝛈)); √ sol ≔ F(𝛈) =

𝛑 erf

( ) 𝛈 𝟐

𝛈𝟐 𝛈−𝛈 𝛑+𝟐e 𝟒 √ 𝛑 √

−

The final solution can be expressed as ( θ(x, η) =

) q′′w k

√(

⎡√ π erf ) αx ⎢ ⎢ V ⎢ ⎣

( ) η 2

( 2) √ ⎤ η η − η π + 2 exp − 4 ⎥ √ ⎥. π ⎥ ⎦

Let us determine the Nusselt number. First, we express the temperature as follows: ) ( ) ( √ √ ( ′′ ) √( ) ⎡ π erf η η − η π + 2 exp − η2 ⎤ qw 2 4 ⎥ αx ⎢ T − Ti = √ ⎥. k V ⎢⎢ π ⎥ ⎣ ⎦ Next, we evaluate the wall temperature Tw by setting y = 0 on the lower plate ] ( ′′ ) √( ) [ qw 2 αx Ty=0 = Tw = Ti + √ . k V π

(8.310a)

(8.310b)

(8.311)

From this relation, we derive the local Nusselt number using the definition of heat transfer coefficient h h=

q′′w k = (√ ) √ ( ) Tw − Ti 2 αx π

(8.312)

V

where the heat transfer coefficient is based on the temperature difference (Tw − Ti ). We express α x/V as (

αx V

) =

=

2 (x∕b) (2b2 ) α (x∕b) (2b ) α x (2b) ) ( ) = ( ) = ( 1 V 2b 1 V 2b V DH υ υ υ υ υ υ α

(x∕b) (2b2 ) (x∕b) (2b2 ) = . ReDH Pr Pe

Now, the heat transfer coefficient becomes √ √ √ π π π k k k h= = = √ √ √ 2 2 2 2 2 2(x∕b) (x∕b) (2b ) 2(x∕b) (4b ) (2b) Pe Pe Pe Finally, the Nusselt number becomes √ h(2b) π 1 NuDH = = √ k 2 2(x∕b) Pe DH = 2b

(8.313)

(8.314)

(8.315)

8.4.2.5 Empirical and Theoretical Correlations for Viscous Flow Heat Transfer in the Entrance Region of Parallel Plates

Consider viscous flow heat transfer in the combined entrance region of a parallel plate channel. This problem has been studies extensively due to their practical importance in engineering applications. We need to use special correlations to take into account the influence flow disturbances at the channel inlet. Correlations are also available for hydrodynamically and thermally simultaneously developing flows (combined entrance region). We will present some common correlations valid for combined entrance region.

8.4 Heat Transfer in the Entrance Region of Ducts

Constant wall temperature Tw Stephan, as reported by Bennett [13], proposed a correlation for the mean Nusselt number for the combined entry problem between isothermal parallel plates )1.14 ( DH ReDH Pr 0.024 L NuDH = 7.55 + (8.316) ( )0.64 DH ReDH Pr 1 + 0.0358 Pr 0.17 L hDH VDH ReDH = 0.1 ≤ Pr ≤ 1000 NuDH = k ν

Tw = const. Equation (8.316) is a curve fit to some numerical calculations. It is reported by Bennett [13] that this correlation falls within −11% to 2.4% of exact numerical results presented by Bennett. Shah and Bahatti [22] derived the following local Nusselt number correlation based on the average correlation of Stephan } [( ) [( ) ]1.14 { ]0.64 D D ReDH Pr ReDH Pr 0.0179Pr 0.17 0.024 − 0.14 x x h DH NuDH = = 7.55 + (8.317) { }2 [( ) ]0.64 k D 0.17 1 + 0.0358 Pr ReDH Pr x 0.1 < Pr < 1000 Tw = const ρV DH . ReDH = μ It is reported by Bennett [13] that this correlation falls within −13% to 7.4% of exact results. Bennett [13] gives the following expression for the average Nusselt number: ( )0.2734 9.469Gz1.219 + 1561 + 0.076 h DH L = NuDH = [ ( )0.1221 ] ( ( ) ) k 1∕6 −1 1.356 tanh 2.786 Pr Gz−1 1 + 404.6 Pr Gz L L

(8.318)

Tw = const Gz−1 L =

L 1 . DH ReDH Pr

It is reported that this equation exhibits errors of −2.0% to 2.6% compared to the exact solution presented by Bennett [13]. Gnielinski [19] recommends the following correlation for average Nusselt number for parallel plates:

NuDH

1∕3 √( √( ⎧ ⎫ ) ) ( ) 1∕6 DH DH ⎪ ⎪ 2 ReDH Pr + ReDH Pr ⎬ . = ⎨7.541 + 1.841 3 L 1 + 22 Pr L ⎪ ⎪ ⎩ ⎭

(8.319)

Constant wall heat flux q′′ w Bennett [13] gives the following correlation for the average Nusselt number for the parallel plates

with UHF: NuDH

h DH = = k

( )0.2698 12.74Gz1.235 + 2243 + 0.215 L [ ( )0.09116 ] )1∕6 ) ( ( −1 1.828 tanh 2.889 Pr Gz−1 1 + 1392 Pr Gz L L

q′′w = const Gz−1 L =

L 1 . DH ReDH Pr

It is reported by Bennett [13] that this correlation exhibits errors of −2.9% to 3.2%.

(8.320)

371

372

8 Laminar Momentum and Heat Transfer in Channels

8.5

Fully Developed Heat Transfer

Under certain heating conditions, we may imagine that the possibility of dimensionless temperature profile does not vary with tube length, and analytical solutions of energy equation may be obtained. Consider heat transfer between two infinite parallel plates. At a distance far from where the heating begins, the heat transfer coefficient is constant for incompressible constant property fluid flow, and the flow is said to be TFD. For fully developed heat transfer, the heat transfer coefficient is independent of axial position, say x. Then, we may write h=

8.5.1

q′′w k 𝜕T || =− ≠ f(x). Tw − Tm Tw − Tm 𝜕y ||y=0

Circular Tube

In this section, steady fully developed heat transfer will be considered in a circular pipe. There is no internal energy generation, and viscous dissipation is neglected. We assume that fluid properties are constant, flow is incompressible, and swirl velocity is neglected. 8.5.1.1 HFD and TFD Laminar Forced Convection Heat Transfer for Slug Flow in a Circular Pipe Subjected to Constant Wall Heat Flux

Consider axisymmetric, steady, laminar, and incompressible Newtonian fluid with constant properties. The fluid is flowing in a tube having a diameter of D = 2R. Flow is TFD. See Figure 8.27. Notice that for Pr ≪ 1, the thermal diffusivity α is larger than the momentum diffusivity ν. The temperature profile develops much faster than the velocity profile. The velocity profile is approximated as vz ≈ V, and this is called slug flow. Axial conduction and viscous dissipation are neglected. There is no internal energy generation. A UHF q′′w is applied to tube surface. We wish to determine the temperature distribution, and the Nusselt number when the thermal field is fully developed. We will list here certain assumptions to simplify the energy equation. (1) (2) (3) (4) (5) (6) (7) (8) (9) (10)

Steady laminar flow. Low velocities (negligible viscous dissipation). Incompressible flow (ρ = constant). Constant fluid properties (μ, cp , k) Axisymmetric flow 𝜕/𝜕θ = 0. No θ-velocity (vθ = 0). Negligible axial conduction, Pe = RePr > 100. TFD, 𝜕ϕ/𝜕z = 0. Fully developed velocity field with respect to the z-direction, 𝜕/𝜕 z = 0. No internal energy generation.

Under these assumptions, the energy equation becomes ( ) 𝜕T 𝜕T α 𝜕 = r V 𝜕z r 𝜕r 𝜕r

q″w

r vz = V T = Ti

Figure 8.27

2R 0

Slug flow in a pipe.

(8.321)

z

T(r)

8.5 Fully Developed Heat Transfer

where α = k/ρcp is the thermal diffusivity of the fluid. Boundary conditions are At r = 0 T = finite or At r = R k

𝜕T = 0. 𝜕r

(8.322a)

𝜕T = q′′w . 𝜕r

(8.322b)

For TFD flow, T (z) − T(z, r) ϕ= w Tw (z) − Tm (z) where Tm (z) is the mixing cup temperature, and under UHF boundary condition, we have shown that temperature gradient 𝜕T/𝜕z can be written as dTm 𝜕T dTw = = . 𝜕z dz dz Writing the energy balance for a differential control volume, as shown in Figure 8.28, we get the following relation: 2q′′w dTm = = constant. dz R ρ cp V

(8.323)

Thus, for this case, we find that 2q′′w dTm 𝜕T dTw = = = = const 𝜕z dz dz R ρ cp V

(8.324)

where α = k/ρcp is the thermal diffusivity. We see that 𝜕T/𝜕z is a constant, and the energy equation now takes the following form along with boundary conditions: ( ) ( ) dT V dTm 1 d r = . (8.325) r dr dr α dz Boundary conditions are dT = 0 (symmetry) dr dT = q′′w . At r = R k dr Let us integrate Eq. (8.325) twice to obtain the temperature distribution. First integration yields ( ) dT V dTm r2 r = + C1 . dr α dz 2 At r = 0

Dividing Eq. (8.327) by r yields ( ) dT V dTm r C1 = + . dr α dz 2 r

q″w (2π Rdz)

ρcp V π R2 Tm

ρcp V π R2 Tm +

z Figure 8.28

Control volume.

(8.326b)

(8.327)

(8.328)

The first boundary condition, Eq. (8.326a), gives C1 = 0. The energy equation, Eq. (8.328), becomes ( ) dT V dTm r = . dr α dz 2

2R

(8.326a)

dz z + dz

d (ρc V π R2 Tm) dz dz p

(8.329)

373

374

8 Laminar Momentum and Heat Transfer in Channels

Integrating Eq. (8.329) gives ( ) V dTw r2 T= + C2 . α dz 4

(8.330)

The second boundary condition, Eq. (8.326b), cannot be used to obtain the unknown constant C2 since Eq. (8.330) satisfies this boundary condition. Let us define a wall temperature Tw as at r = R

T = Tw

(8.331)

and using this new boundary condition, Eq. (8.331), the temperature distribution becomes ] ( ) ( dT ) [ r2 V m 2 T = Tw − R 1− 2 . 4α dz R

(8.332)

We may express the temperature distribution in terms of heat flux q′′w by substituting the value of dTm /dz into Eq. (8.332) ] ( ′′ ) [ qw R r2 1− 2 . (8.333) T = Tw − 2k R We can obtain the pipe center temperature Tc by setting r = 0 in Eq. (8.333) ( ′′ ) qw R . Tc = Tw − 2k Therefore, the dimensionless temperature distribution can be written as ( )2 Tw − T r =1− . Tw − Tc R

(8.334)

(8.335)

Next, we will calculate the bulk temperature Tm . Recall that the mean fluid temperature Tm can be written as R

Tm ≡

R

1 1 2 v TdAc = V T(2πrdr) = 2 T r dr. Ac V ∫ z πR2 V ∫0 R ∫0

(8.336a)

Ac

The mixed-mean fluid temperature depends on axial position, i.e. Tm = Tm (z). This is also referred to as the bulk fluid temperature or mixing cup temperature; it is the temperature that characterizes the average convected thermal energy at any axial position in the tube. Substituting temperature distribution, Eq. (8.333), into Eq. (8.336a) yields mean fluid temperature Tm )] ( ′′ ) ( R[ qw R r2 2 1− 2 Tw − r dr. (8.336b) Tm = 2 2k R ∫0 R Carrying out the integration with Maple 2020, we obtain the mean fluid temperature ′′

R qw . (8.337) 4 k We will now introduce the heat transfer coefficient h based on the difference between the wall temperature Tw and the mean fluid temperature Tm Tm = Tw −

h=

q′′w q′′ 8k 8k = . = ′′ w = Tw − Tm D qw R∕4 k 2R

Thus, the Nusselt number is NuD =

hD =8 k

(8.338)

where NuD is the mean Nusselt number based on the difference between wall temperature and mean fluid temperature, and tube diameter D = 2R. We can express the temperature distribution in terms of inlet temperature. We eliminate Tw between Eqs. (8.333) and (8.337). This will give us [ ( )2 ] q′′ R 1 r T = Tm + w − + . (8.339) 2k 2 R

8.5 Fully Developed Heat Transfer

Next, we integrate and rearrange Eq. (8.323) ) ( ′′ ) ( 4qw z Tm = Tmi + k ReD Pr where Tmi is the temperature at z = 0. We now combine this last equation with Eq. (8.330) to get ) [ ( ′′ ) ( ( )] q′′w D 4qw 1 1 r 2 z + − + T = Tmi + k ReD Pr k 8 4 R [ ] ( ) 2 T − Tmi 4(z∕D) 1 1 r + − + . = ReD Pr 8 4 R q′′w D∕k

(8.340)

8.5.1.2 HFD and TFD Laminar Forced Convection Heat Transfer for Viscous Flow in a Circular Tube Subjected to Constant Wall Heat Flux

Consider axisymmetric, steady, laminar, constant property, and incompressible fully developed viscous flow in a circular pipe having a diameter D = 2R far from the entrance of the pipe. Both the velocity and temperature profiles are fully developed. See Figure 8.29. Axial conduction is neglected. There is no internal energy generation and viscous dissipation is neglected. Tube surface is subjected to constant surface heat flux q′′w . We wish to determine the temperature distribution and the Nusselt number. We now list the following assumptions. (1) (2) (3) (4) (5) (6) (7) (8) (9)

Steady laminar flow. Low velocities (negligible viscous dissipation). Incompressible flow (ρ = constant). Constant fluid properties (μ, k, c). Axisymmetric flow (𝜕/𝜕θ = 0). No θ-velocity (vθ = 0). Negligible axial conduction, Pe = ReD Pr > 100. Fully developed velocity field with respect to the z-direction 𝜕/𝜕z = 0, vr = 0. TFD, 𝜕ϕ/𝜕z = 0.

Under the stated assumptions, the energy equation becomes ( ) 𝜕T 𝜕T k 𝜕 = r . ρcp vz 𝜕z r 𝜕r 𝜕r The boundary conditions are dT = 0 (Symmetry or T is finite at pipe center.) dr 𝜕T = q′′w . r = R; k 𝜕r Heat is flowing from the pipe wall into fluid. Velocity field for HFD viscous flow is ) ( r2 vz = 2V 1 − 2 . R r = 0;

(8.341)

(8.342a) (8.342b)

For TFD conditions under UHF boundary condition, we have shown that the temperature gradient 𝜕T/𝜕z can be written as dTm 𝜕T dTw = = . (8.343) 𝜕z dz dz r

q″w

T(r)

2R z

Figure 8.29

Flow in a circular pipe.

vz(r)

375

376

8 Laminar Momentum and Heat Transfer in Channels

Writing the energy balance for a differential control volume in pipe as we have done before 2q′′w dTm = = constant. dz R ρ cp V

(8.344)

Thus, for the temperature gradient 𝜕T/𝜕z, we have 2q′′w dTm 𝜕T dTw = = = = const. (8.345) 𝜕z dz dz kR V This means that since the wall heat flux is constant, 𝜕T/𝜕z is constant, and the mean fluid temperature varies linearly along the pipe axis. The energy equation, Eq. (8.341), is converted to the ODE with the help of Eq. (8.345) ) ]( ( ) V [ dTm dT r2 1 d c r = 1− 2 (8.346) r dr dr α dz R where α = k/ρcp is the thermal diffusivity. Rearranging Eq. (8.346) gives ) ]( ( ) V [ dTm dT r3 d r = c r− 2 . dr dr α dz R Integrating Eq. (8.347) once and after simplification, we get ) ]( ( ) V [ C dTm dT r3 c r = − + 1. dr α 2 4R2 dz r

(8.347)

(8.348)

The natural logarithm function becomes infinite at r = 0. Using the first boundary condition, Eq. (8.342a), we find that C1 = 0. The energy equation, Eq. (8.348), now becomes ]( ) ( ) V [ dTm r3 dT r = c − . (8.349) dr α 2 4R2 dz Integrating Eq. (8.349) gives ] [ ( )] [ 2 Vc dTm r4 r + C2 . T(r, z) = − α dz 4 16 R2

(8.350)

We cannot use the second boundary condition, Eq. (8.342b), to find C2 since temperature distribution, Eq. (8.350), satisfies this second boundary condition. We now define a wall temperature Tw , where Tw varies with z r = R T = Tw .

(8.351)

Using the wall boundary condition, Eq. (8.351), temperature distribution becomes ] )] [ 2 [ ( Vc dTm R4 R + C2 − Tw = α dz 4 16 R2 )] [ 2 ] [ ( Vc dTm 3R + C2 . Tw = α dz 16 Thus, C2 becomes )] [ 2 ] [ ( Vc dTm 3R . C2 = Tw − α dz 16 Using C2 , Eq. (8.350) becomes [ ( )] [ 2 )] [ 2 ] ] [ ( Vc dTm Vc dTm r4 r 3R T(r, z) = − . + Tw − 2 α dz 4 α dz 16 16 R Simplifying Eq. (8.352), we get the temperature distribution )] [ [ ( ( )2 ( )] Vc dTm r 1 r 4 3 T(r, z) = Tw (z) − R2 − . + 4α dz 4 R 4 R

(8.352)

(8.353)

We may express this temperature distribution in terms of heat flux q′′w . Noting that Vc = 2V and substituting value of dTm /dz into Eq. (8.353), we get ( ′′ ) [ ( )2 ( )] qw R r 1 r 4 3 T(r, z) = Tw (z) − − . (8.354) + k 4 R 4 R

8.5 Fully Developed Heat Transfer

We may define a centerline temperature Tc as at r = 0 T = Tc . Substituting r = 0 into temperature distribution, Eq. (8.354), we get ( ) ( q′′ R ) 3 w . Tc = Tw − 4 k At this point, we may write the dimensionless temperature in terms of Tc and Tw [ ( ) ( )] Tw − T 4 r 2 1 r 4 . = 1− + Tw − Tc 3 R 3 R

(8.355)

(8.356)

We now wish to determine bulk temperature Tm . Recall that bulk fluid temperature Tm can be written as R

Tm ≡

1 2 v TdAc = 2 vz T r dr. Ac V ∫ z R V ∫0

(8.357)

Ac

Substituting temperature distribution into Eq. (8.357), we get ( )] { ( ′′ ) [ R[ ( )2 ( ) ]} qw R r r2 2 1 r 4 3 − 2V 1 − 2 Tw − rdr. + Tm = 2 k 4 R 4 R R V ∫9 R Thus, after integration, we obtain ( ′′ ) 11 qw R . Tm (z) = Tw (z) − 24 k If we wish, we may define the dimensionless temperature in terms of Tm and Tw [ ( )2 ( )] Tw − T 24 3 1 r 4 r − . = + Tw − Tm 11 4 R 4 R

(8.358)

(8.359)

(8.360)

The heat transfer coefficient h based on (Tw − Tm ) can be written as h=

q′′w q′′w 24 k = . ( ′′ ) = Tw − Tm 11 R 11 qw R 24 k

(8.361)

The Nusselt number is hD 48 = = 4.364 q′′w = const HFD, TFD. (8.362) k 11 We may also express the temperature distribution in terms of inlet temperature. First, we eliminate Tw between Eqs. (8.354) and (8.360), and after simplification, we get ( ′′ ) [ ( )2 ( )] qw R 7 r 1 r 4 . (8.363a) − + − T(r, z) = Tm (z) + k 24 R 4 R NuD =

Next, we integrate Eq. (8.344). This integration process gives ( ) 4q′′w Tm = Tmi + z. ρ VD cp Now, we combine these last two equations, Eqs. (8.363a) and (8.363b), to get ( ) ( ′′ ) [ ( )2 ( )] 4q′′w qw D 7 r 1 r 4 − + . − z+ T(r, z) = Tmi + ρ VD cp 2k 24 R 4 R We may rearrange Eq. (8.363c) to get the dimensionless temperature distribution [ ( ) ( )] T(r, z) − Tmi 4(z∕D) 7 1 r 2 1 r 4 + − + = − ReD Pr 48 2 R 8 R q′′w D∕k where Pe = ReD Pr is the Peclet number.

(8.363b)

(8.363c)

(8.363d)

377

378

8 Laminar Momentum and Heat Transfer in Channels

Example 8.4 Air at atmospheric pressure flows through a tube at a rate of 2.5 × 10−5 kg/s. The internal diameter of the tube is 5 mm. The tube length is 0.75 m. A constant heat flux is applied over the entire tube surface. It is required to heat the air from 17 ∘ C to 75 ∘ C. Determine: (a) The heat transfer coefficient at the tube exit (b) The constant heat flux required to heat the air Solution The average temperature Tm is Tmi + Tmo 17 + 75 = = 46 ∘ C = 319 K. 2 2 The physical properties of air are evaluated at the average temperature Tm ≈ 319 K: Tm =

ρ = 1.091kg∕m3

cp = 1007 J∕kg.K

μ = 193.5 × 10−7 N.s∕m2

Pr = 0.704 k = 0.0277 W∕m.K First, we evaluate the Reynolds number ReD ReD =

4 × 2.5 × 10−5 4ṁ = = 329. πμD π × 193.5 × 10−7 × 0.005

Since ReD < 2300, the flow is laminar. In order to check whether the flow is fully developed or not, let us calculate the hydrodynamic and thermal entrance lengths: LH = 0.05 ReD D = 0.05 × 329 × 0.005 = 0.082 m LT = 0.05 ReD D Pr = 0.05 × 329 × 0.005 × 0.704 = 0.057 m. Both LH and LT are much less than the tube length, and we may assume that flow is both HFD and TFD. The velocity and temperature profiles are fully developed. (a) For fully developed flow in tube with UHF boundary condition, the Nusselt number is NuD =

hD = 4.364 k

) ( k 0.0277 hD ⇒ h = NuD = 4.364 = 24.17 W∕m2 K. k D 0.005 (b) The required heat flux is calculated next. We write an energy balance on the tube NuD =

πDLq′′w = ṁ cp (Tmo − Tmi ) q′′w =

ṁ cp (Tmo − Tmi ) πDL

=

2.5 × 10−5 × 1007(75 − 17) = 124 W∕m2 . π × 0.005 × 0.75

8.5.1.3 HFD and TFD Laminar Forced Convection Heat Transfer for Viscous Flow in a Circular Tube Subjected to Constant Wall Temperature

Consider axisymmetric, steady, laminar, and incompressible flow with constant properties in a tube having a diameter of D = 2R. Fluid is Newtonian. The velocity distribution and temperature distribution are fully developed in the tube. A UWT is applied to the tube surface. See Figure 8.30. Axial conduction and viscous dissipation are negligible along with internal energy generation. We wish to determine the temperature distribution and the Nusselt number. We make the following assumptions: (1) (2) (3) (4)

Steady laminar flow Low velocities (negligible viscous dissipation) Incompressible flow (ρ = constant) Constant properties (μ, k, c)

8.5 Fully Developed Heat Transfer

r

Tw

2R

T (r) z

Figure 8.30

(5) (6) (7) (8) (9) (10)

vz (r)

Flow in a pipe.

Axisymmetric flow 𝜕/𝜕θ = 0 No θ-velocity (vθ = 0) Negligible axial conduction, Pe = RePr > 100 Fully developed velocity field with respect to the z-direction 𝜕/𝜕 z = 0, vr = 0 TFD flow, 𝜕ϕ/𝜕z = 0 There is no internal energy generation

Under the assumptions stated above, the governing differential equation for T(r, z) is ( ) 𝜕T 𝜕T k 𝜕 = r . ρcp vz 𝜕z r 𝜕r 𝜕r Velocity field from the solution of momentum equation is given as ) ( r2 vz = 2V 1 − 2 . R

(8.364)

The boundary conditions are r=R

T = Tw

(8.365a)

𝜕T(0, z) = 0 or T = finite. (8.365b) 𝜕r Flow is TFD. Since the wall temperature is constant, then dTw /dz = 0, and the axial temperature gradient 𝜕 T/𝜕 z is in the following form: [ ]( ) ( ) Tw (z) − T(r, z) dTm dTm 𝜕T = = θ(r). (8.366) 𝜕z Tw (z) − Tm (z) dz dz r=0

Using this relation, we can eliminate the axial temperature gradient 𝜕T/𝜕z from the energy equation. Then, the energy equation, Eq. (8.364), becomes ( )( ) ( ) 𝜕T 2V dTm r2 1 𝜕 r = 1− 2 θ (8.367) r 𝜕r 𝜕r α dz R where α = k/ρcp is the thermal diffusivity. A successive approximation method will be used in the solution of the problem. First iteration: Let us take as a first approximation the temperature distribution obtained for constant wall heat flux case for fully developed heat transfer in a tube under HFD condition: [ ( ) ] ( ) T − Tw 96 1 r 4 1 r 2 3 θ= . (8.368) = − + Tm − Tw 11 16 R 4 R 16 We will substitute this temperature distribution in the energy equation, Eq. (8.367), and rearrange Eq. (8.367) to get )[ ( ) ( )( ] ( ) )( ( ) 1 𝜕 𝜕T V dTm r2 3 1 r 4 1 r 2 192 r = 1− 2 . (8.369) − + r 𝜕r 𝜕r α dz 11 16 R 4 R 16 R First, multiplying Eq. (8.369) by r yields ] )[ ( ) ( ) ( )( ) ( dT ) ( ( ) 𝜕T V 192 r2 3 𝜕 1 r 4 1 r 2 m r = (r) 1 − 2 . − + 𝜕r 𝜕r α 11 dz 16 R 4 R 16 R

(8.370)

379

380

8 Laminar Momentum and Heat Transfer in Channels

Define a new constant denoted by A ( ] )[ ( ) ( ) ( ) 𝜕T r2 3 𝜕 1 r 4 1 r 2 r = Ar 1− 2 − + 𝜕r 𝜕r 16 R 4 R 16 R ( )( ) dT 192 where A = Vα dzm . Integrating Eq. (8.371) with respect to r with Maple 2020, we obtain 11 ) ( 1 r8 5 r6 7 r4 3 2 𝜕T =A − + − + r + C1 . r 𝜕r 128 R6 96 R4 64 R2 32

= 0, we find that C1 = 0. Thus, we have Using the boundary conditions r = 0, 𝜕T 𝜕r ( ) 1 r8 5 r6 7 r4 3 2 𝜕T =A − r . + − + r 𝜕r 128 R6 96 R4 64 R2 32 Dividing Eq. (8.373) by r gives ( ) 1 r7 5 r5 7 r3 3 𝜕T =A − + − + r 𝜕r 128 R6 96 R4 64 R2 32 and integrating Eq. (8.374) with respect to r yields ( ) 1 r8 5 r6 7 r4 3 2 T=A − + − + r + C2 . 1024 R6 576 R4 256 R2 64

(8.371)

(8.372)

(8.373)

(8.374)

(8.375)

Using the boundary conditions r = R and T = Tw , we obtain the constant C2 C2 = Tw −

251 AR2 . 9216

(8.376)

Substituting this equation, Eq. (8.376), into temperature distribution, we obtain the final form of temperature the distribution ) ( 1 r8 5 r6 7 r4 3 r2 251 T = AR2 − + Tw . + − + − (8.377) 1024 R8 576 R6 256 R4 64 R2 9216 We will now calculate the mean fluid temperature Tm R

Tm =

2 Vz Trdr R2 V ∫0

(8.378)

and substituting Eq. (8.377) into Eq. (8.378) and integration with Maple 2020 yields Tm = −

59 AR2 + Tw . 3840

Using Eqs. (8.377) and (8.379), we will now generate new dimensionless temperature θ1 ) )( ( 1 r8 5 r6 7 r4 3 r2 251 3840 − θ1 = − + − + − 59 1024 R8 576 R6 256 R4 64 R2 9216 or factoring out by 64, we obtain [ ] T − Tw 1 r8 60 5 r6 7 r4 r2 251 − =− + − + 3 − θ1 = Tm − Tw 59 16 R8 9 R6 4 R4 R2 144

(8.379)

(8.380)

(8.381)

Direction of heat flow is in the negative r-direction, which is heat is flowing into the fluid and heat transfer coefficient h ( ) 𝜕T ) ( k ) ( 𝜕θ1 k220 k 220 110 𝜕r r=R = = . (8.382) = −k = −k − h= Tw − Tm 𝜕r r=R 59R 59(2R) D 59 The Nusselt number is NuD =

hD = 3.728813 k

(8.383)

8.5 Fully Developed Heat Transfer

Second iteration: Dimensionless temperature θ1 is given as [ ( 8) ( ) ( ) ] ( 2) T − Tw 1 5 r6 7 r4 251 60 r r − . + − +3 − θ1 = =− Tm − Tw 59 16 R8 9 R6 4 R4 144 R2 and θ1 is substituted into the energy equation, Eq. (8.366), and the result of substitution is ( )( )( ] )[ ( ) 𝜕T 2V R dTm r2 5 r6 7 r4 r2 251 1 𝜕 60 1 r8 r = 1− 2 . − + −3 2 + r 𝜕r 𝜕r α dz 59 16 R8 9 R6 4 R4 144 R R First, multiplying Eq. (8.385) by r, we obtain ( )( ] )[ ( ) ) ( 𝜕T V dTm r2 5 r6 7 r4 r2 251 𝜕 120 1 r8 r = (r) 1 − 2 . − + −3 2 + 𝜕r 𝜕r α dz 59 16 R8 9 R6 4 R4 144 R R To make the life easy with integration, we define a new constant B ( ] )[ ( ) 𝜕T r2 5 r6 7 r4 r2 251 𝜕 1 r8 r = Br 1− 2 − + − 3 + 𝜕r 𝜕r 16 R8 9 R6 4 R4 R R2 144 ( )( ) dT 120 . Now, integrate Eq. (8.387) with respect to r to obtain where B = Vα dzm 59

[ ] ( ) 𝜕T 1 r12 89 r10 83 r8 19 r6 683 r4 251 2 r =B − r + C1 + − + − + 𝜕r 192 R10 1440 R8 288 R6 24 R4 576 R2 288

(8.384)

(8.385)

(8.386)

(8.387)

(8.388)

From the boundary conditions at r = 0 𝜕 T/𝜕 r = 0, we find that C1 = 0. This last equation, Eq. (8.388), now becomes [ ] ) ( 1 r12 𝜕T 89 r10 83 r8 19 r6 683 r4 251 2 =B − r (8.389) r + − + − + 𝜕r 192 R10 1440 R8 288 R6 24 R4 576 R2 288 Dividing Eq. (8.389) by r yields [ ] ( ) 1 r11 89 r9 83 r7 19 r5 683 r3 251 𝜕T =B − r + − + − + 𝜕r 192 R10 1440 R8 288 R6 24 R4 576 R2 288 Now, we integrate Eq. (8.390) with respect to r and obtain ] [ 89 r10 83 r8 19 r6 683 r4 251 2 1 r12 r + C2 + − + − + T=B − 2304 R10 14400 R8 2304 R6 144 R4 2304 R2 576

(8.390)

(8.391)

At r = R T = Tw and therefore, unknown constant C2 becomes C2 = Tw −

4627 BR2 . 19200

Temperature distribution now takes the following form: [ ] 1 r12 89 r10 83 r8 19 r6 683 r4 251 r2 2 + Tw . + − + − + T = BR − 2304 R12 14400 R10 2304 R8 144 R6 2304 R4 576 R2

(8.392)

(8.393)

We will now calculate the mean fluid temperature Tm R

Tm =

2 vz Trdr R2 V ∫0

(8.394)

and next we substitute temperature distribution, Eq. (8.393), into Eq. (8.394) and integrating with respect to r with Maple 2020 yields Tm = −

129737 2 BR + Tw . 967680

Dimensionless temperature θ2 ] )[ ( 1 r12 89 r10 83 r8 19 r6 683 r4 251 r2 4627 967680 − . θ2 = − + − + − + − 129737 2304 R12 14400 R10 2304 R8 144 R6 2304 R4 576 R2 19200

(8.395)

(8.396)

381

382

8 Laminar Momentum and Heat Transfer in Channels

Direction of heat flow is in the negative r-direction, that is, heat is flowing into the fluid and heat transfer coefficient h ( ) ( ) 𝜕T ) ( k 𝜕T k(T − T ) m w 𝜕θ2 𝜕r r=R 𝜕r r=R h= = = −k Tw − Tm Tw − Tm 𝜕r r=R ( )( ) ( ) ( ) 237888 2 237888 k = −k − = (2) (8.397) 129737 2R 129737 D The Nusselt number becomes NuD =

hD = 3.6672 k

(8.398)

Example 8.5 Water flows in a tube having an internal diameter of 0.05 m. Tube surface temperature is 120 ∘ C. The mass flow rate of the water is 0.014 kg/s. It is required to heat the water from an inlet mean temperature of 20 ∘ C to an outlet mean temperature of 100 ∘ C. Estimate the necessary tube length. Solution Estimated water properties at Tm = (Tmi + Tmo )/2 = (20 + 100)/2 = 60 ∘ C = 333 K ρ = 983 kg∕m3 , μ = 487.5 × 10−6 m2 . k = 0.650 W∕m K, cp = 4.184 kJ∕kg K, Pr = 2.98 The Reynolds number is (4) × (0.014kg∕s) 4ṁ = ≈ 731 < 2300 ∴ flow is laminar πDμ π × 0.05m × 487.5 × 10−6 N.s∕m2 Assume that flow is HFD and TFD. The hydrodynamic entrance length LH is ReD =

LH = 0.05ReD = 0.05 × 731 × 0.05 = 1.82 m The thermal entrance length LT is LT = 0.05ReD Pr = 0.05 × 731 × 0.05 × 2.98 = 5.44 m The heat transfer coefficient h is 0.65 W∕m.K k × 3.667 = 47.67 W∕m2 K h = NuD = D 0.05 Energy required to heat the water is q = ṁ cp (Tmo − Tmi ) = (0.014 kg∕s)(4184 J∕kg.K)(373 − 293)K = 4686 J∕s The log mean temperature ΔTLMTD is (120 − 20) − (120 − 100) = 49.70 ∘ C ΔTLMTD = ) ( 120 − 20 ln 120 − 100 q = h π D L ΔTLMTD 4686 W = (47.67 W∕m2 K)π(0.05 m) L(49.70 K) L ≈ 12.58 m Since LH and LT are much less than L, our TFD and HFD assumption is valid.

8.5.2

Infinite Parallel Plates

In this section, steady fully developed heat transfer in a very wide duct is considered. Since the duct is very wide, changes in the flow properties in the z-direction are neglected. There is no viscous dissipation as well as energy generation in the flow. Body forces are not important. We then model the problem as flow between two infinite parallel plates.

8.5 Fully Developed Heat Transfer

y

q″w u(y)

Flow 0

2b

x q″w

Figure 8.31

Laminar flow between parallel plates.

8.5.2.1 HFD and TFD Laminar Forced Convection Heat Transfer for Viscous Flow Between a Parallel Plate Channel. Both Plates Are Subjected to Constant Wall Heat Flux Boundary Condition

Consider HFD and TFD, two-dimensional, steady flow of a viscous fluid between two infinite parallel plates spaced a distance 2b apart, and each plate is subjected to a constant heat flux q′w . The coordinate system is shown in Figure 8.31. Fluid properties are constant, and flow is incompressible. There is no energy generation in the flow, and viscous dissipation is neglected. Axial conduction is also negligible. We wish to determine the fully developed temperature distribution and the heat transfer coefficient. Notice that flow is essentially two dimensional. The velocity profile for HFD flow is ( y )2 u =1− Vc b The velocity profile in fully developed plane duct flow is parabolic, and streamlines are parallel to plane walls and v = 0. There is no motion in the z-direction and w = 0. The energy equation for incompressible, constant property flow is uncoupled from the momentum equation, and velocity distribution is known. We can neglect the axial conduction for Pe > 100. The governing equation for the temperature distribution in the flow is ρcp u

𝜕2T 𝜕T =k 2 𝜕x 𝜕y

(8.399)

The flow is symmetrical about the centerline of the channel, so the boundary conditions are 𝜕T =0 𝜕y 𝜕T y = b; k = q′′w 𝜕y

y = 0;

(8.400a) (8.400b)

Here, b is the half-width of the duct. Since we are considering HFD and TFD flow, the form of the temperature profile does not change with distance along the duct, i.e. Tw − T = ϕ(η) Tw − Tc

(8.401)

where η = y/b. This time we are using a little different form of the dimensionless temperature profile ϕ, and the function ϕ is independent of the distance x along the duct. Here, Tc is the temperature on the channel centerline, and Tw is the wall temperature to be determined later. Now, since ϕ does not depend on x, it follows that for TFD flow, [ [ ] ] ][ Tw − T 𝜕Tw dTw dTc 𝜕ϕ 𝜕 Tw − T = − − =0 (8.402a) = 𝜕x 𝜕x Tw − Tc 𝜕x Tw − Tc dx dx or

[ ] ][ Tw − T dTw dTc 𝜕T 𝜕Tw = − − 𝜕x 𝜕x Tw − Tc dx dx

(8.402b)

The heat transfer rate at the wall is obtained using Fourier’s law q′′w = +k

𝜕T || 𝜕y ||y=b

(8.403)

Heat flows in the negative y-direction and y is measured from the centerline of the duct toward the wall. For this reason, we have a positive sign in front of Fourier’s Law. Next, combining this equation with dimensionless temperature profile ϕ,

383

384

8 Laminar Momentum and Heat Transfer in Channels

we obtain q′′w = −k(Tw − Tc )

−k(Tw − Tc ) dϕ | dϕ || | = dy ||y=b b dη ||η=1

(8.404)

Because (dϕ/dη)η = 1 and q′′w are constant, this equation shows that Tw − Tc = constant. Thus, we conclude that dTw dTc = (8.405) dx dx The substitution of Eq. (8.405) into Eq. (8.402b) then shows that for the constant wall heat flux, the boundary condition 𝜕 T/𝜕 x is dTc 𝜕T dTw = = 𝜕x dx dx We now substitute Eq. (8.406) into the energy equation, Eq. (8.399), to get ) ( dTw 𝜕2T =α 2 u dx 𝜕y

(8.406)

(8.407)

The fully developed velocity profile is [ ( y )2 ] u = uc 1 − b where uc is the centerline temperature. If the velocity profile is substituted into Eq. (8.407), the following differential equation is obtained after rearrangement: ) [ ( 2 )] ( dTw y 𝜕2T (8.408) α 2 = uc 1 − dx 𝜕y b2 The integration of Eq. (8.408) once yields [ ( )] 1 y3 𝜕T uc dTw y− + C1 = 𝜕y α dx 3 b2

(8.409)

Now, using the boundary conditions y = 0 and 𝜕T/𝜕y = 0 gives C1 = 0. The energy equation, Eq. (8.409), becomes [ ( )] 𝜕T uc dTw 1 y3 = y− (8.410) 𝜕y α dx 3 b2 Integrating Eq. (8.410), we get [ 2 ( )] u dT y 1 y4 − + C2 T= c w α dx 2 12 b2

(8.411)

We cannot use the second boundary condition to evaluate the unknown constant C2 ; therefore, we define a wall temperature Tw y = b, T = Tw Using this new boundary condition, we get C2 u dT ( 5 2 ) b C2 = Tw − c w α dx 12 Substituting C2 into Eq. (8.411) and after rearrangement, the temperature distribution becomes )( [( )] [ ( ) ( )] dTw u c b2 1 y 2 1 y 4 5 − T = Tw − + α dx 12 2 b 12 b

(8.412)

(8.413)

(8.414)

We have obtained the temperature distribution in fully developed laminar flow between two infinite parallel plates for UHF boundary condition. We wish to express the temperature distribution in terms of the specified wall heat flux, q′′w . This is done as follows: ( ) ( u ) dT 𝜕T | 2b c w (8.415) =k q′′w = +k || 𝜕y |y=b 3 α dx

8.5 Fully Developed Heat Transfer

or

(

)

dTw dx

=

′′ 3 α qw 2 k uc b

(8.416)

Substituting Eq. (8.416) into Eq. (8.414) then allows us to express the temperature distribution as ( ′′ ) [ ( ) ( )] qw b 5 3 y 2 1 y 4 T = Tw − − + k 8 4 b 8 b

(8.417)

From this equation, we obtain the centerline temperature Tc . The centerline temperature Tc is the temperature at y = 0, and it is given as ( ′′ ) 5 qw b Tw − Tc = (8.418) 8 k Now, in duct flows, it is common to utilize the mean fluid temperature,Tm , in defining the heat transfer coefficient. The mean or bulk temperature is Tm given by ∫ ρuTdA A

Tm =

∫ ρudA A

b

=

∫−b uTdy b

∫−b udy

(8.419)

where A is the cross-sectional area of the duct. Substituting temperature distribution, Eq. (8.417), and the velocity profile u into Eq. (8.419) gives { [ ( ′′ ) [ ( ) ( ) ]} ( y )2 ]} { qw b 5 3 y 2 1 y 4 b ∫−b uc 1 − − Tw − dy + b k 8 4 b 8 b Tm = (8.420) b ∫−b {uc [1 − (y∕b)2 ]}dy ( ′′ ) 17 qw b (8.421) Tm = Tw − 35 k Equation (8.412) can be rearranged to obtain the Nusselt number NuDH =

q′′w (DH ) h(4b) 140 = = = 8.2352 k k(Tw − Tm ) 17

(8.422)

where NuDH is the Nusselt number based on the difference between the wall temperature and the mean temperature, i.e. on (Tw − Tm ) and on the duct hydraulic diameter DH . In TFD and HFD regions, the local Nusselt number is equal to the average Nusselt number. The duct hydraulic diameter is DH = 4b. Nusselt number relations for HFD and TFD viscous laminar flow in noncircular ducts are given in Table 8.3. The Reynolds and Nusselt numbers for flow in these noncircular tubes are based on the hydraulic diameter DH = 4A/P. Example 8.6 Air at atmospheric pressure flows with a velocity of 0.3 m/s in a tube having an internal diameter of 2 cm. The tube length is 2 m. The mean inlet and outlet temperatures of air are 17 ∘ C and 137 ∘ C. Consider the following two cases: (a) Heating is done by condensing steam on the tube surface. This means tube surface is kept at constant temperature. (b) Heating is done by heating tube surface by electric resistance heating. This means a uniform surface heat flux is applied to tube surface. We wish to determine the heat transfer coefficient for each case. Solution First, we evaluate the air properties at mean air temperature Tm Tm =

Tmi + Tmo 17 + 137 = = 77 ∘ C = 350 K 2 2

ν = 20.92 × 10−6 m2 ∕s k = 0.030 W∕m.K Pr = 0.7 The Reynolds number ReD is ReD =

0.3 × 0.02 VD = = 286.8 ν 20.92 × 10−6

385

386

8 Laminar Momentum and Heat Transfer in Channels

Table 8.3 Nusselt number for HFD and TFD laminar flow heat transfer in ducts with other cross-sectional shapes. Nusselt number

Circle D

′′

a/b or 𝛉′′

Tw = Const

qw = Const

Friction factor f Re

–

3.66

4.36

64

–

3.862

3.34

15.054

a/b 1

2.98

3.61

56.92

2

3.39

4.12

62.20

3

3.96

4.79

68.36

4

4.44

5.33

72.92

6

5.14

6.05

78.80

8

5.60

6.49

82.32

∞

7.54

8.24

96

a/b 1

3.66

4.36

64

2

3.74

4.56

67.28

4

3.79

4.88

72.96

Hexagonal b a

b a

Isosceles triangle θ

8

3.72

5.09

76.60

16

3.65

5.18

78.16

θ 10∘ 30∘

1.61

2.45

50.80

2.26

2.91

52.28

60∘ 90∘

2.47

3.11

53.32

2.34

2.98

52.60

120∘

2.00

2.68

50.96

Source: Rathore and Kapuno [54].

Flow is laminar since ReD < 2300. We now evaluate hydrodynamic entrance length and thermal entrance length: LH = 0.05Re D = 0.05 × 286.8 × 0.02 = 0.286 m = 28.6 cm LT = 0.05Re D Pr = 0.05 × 286.8 × 0.02 × 0.7 = 0.2 m = 20 cm Both LH and LT are much less then tube length. We can assume that flow is both HFD and TFD. (a) Constant surface temperature NuD = 3.66 ⇒ h = 3.66 ( h = 3.66

0.03 0.02

) = 5.49

( ) k D W m2 K

8.6 Heat Transfer in the Thermal Entrance Region

(b) UHF

( ) k D ) ( W 0.03 = 6.54 2 h = 4.364 0.02 mK NuD = 4.364 ⇒ h = 4.364

8.6 Heat Transfer in the Thermal Entrance Region It is of interest to know the variation of the convective heat transfer coefficient with distance from the entrance. In this way, the best use may be made of the materials of design. For example, if most of the heat transfer occurs within a certain range of duct length, we do not gain much by extending the duct length beyond this length. Assume that for flow in a tube, a hydrodynamic starting length is provided, and the velocity profile is fully developed. Assume that at the point where the velocity profile is fully developed, heat transfer begins. The temperature of the fluid is uniform over the flow cross section at the point where heat transfer begins, and temperature field begins to develop. In other words, thermal conditions begin to develop in the presence of the fully developed velocity profile. We say that heat transfer takes place under HFD and thermally developing flow, and this type of problem is called thermal entry length problem. Under these conditions, the heat transfer coefficient varies along the tube axis. Thermal entry length solutions are reasonable approximation for large-Prandtl-number fluids such as oils since the velocity profile develops much faster than the temperature profile. In other words, the hydrodynamic entrance length is very small compared to the thermal entrance length. Since Graetz was the first person to investigate the problem, this type of problem is called Graetz problem. Now, let us now consider the HFD and thermally developing flow in a circular tube. The velocity is assumed to be fully developed and parallel to the tube axis, i.e. 𝜕 vz /𝜕 z = 0. Fluid properties are assumed to be constant. Assume that a solution for the velocity profile is available. The energy equation is { 2 } 𝜕T 𝜕 T 1 𝜕T =k + ρ cp v z 𝜕z r 𝜕r 𝜕r2 We will utilize dimensional reasoning to gain some insight into the form of the solution. Eq. (8.341) is nondimensionalized by letting v (8.423) u= z V T (8.424) θ= Ti r η= (8.425) D z ξ= (8.426) L where L is the pipe length and Ti is the inlet temperature. Substituting these dimensionless variables into the energy equation gives ] [ { 2 } ρ cp D2 V 𝜕θ 𝜕 θ 1 𝜕θ u = (8.427) + kL 𝜕ξ 𝜕η2 η 𝜕η The coefficient in bracket is dimensionless, and it is called Graetz number, Gz. It may be written in the following form: ρ cp D2 V

μcp ρ D V D D = Pr ReD kL k μ L L The dimensionless temperature and dimensionless temperature gradient at the pipe wall depend on the Graetz number. It is expected that the solution may be expressed in terms of Nusselt number NuD = hD/k and Graetz number GzD correlations. In other words, we may express the solution in the following form: ( ) D (8.428) = f(GzD ) NuD = f ReD Pr L Equation (8.428) gives a general functional relationship that one might expect in the correlation of experimental data. GzD =

=

387

388

8 Laminar Momentum and Heat Transfer in Channels

8.6.1

Circular Tube

In this section, we will study HFD thermally developing flow in a circular tube for both the UWT and UHF boundary conditions. 8.6.1.1 Graetz Problem: HFD and Thermally Developing Flow in a Circular Tube under Constant Wall Temperature Boundary Condition

In many engineering problems, the flow in the duct hydrodynamically fully developed and thermally developing as depicted in Figure 8.32a. For example, oil flow generally has a fully developed velocity field due to its viscosity and a developing temperature field because of its low thermal conductivity. Consider axisymmetric, steady, and incompressible fully developed laminar flow of a Newtonian viscous fluid in a tube with constant properties. Fluid enters the tube at uniform velocity V, and eventually, a fully developed velocity profile is established before heating begins. We assume that z = 0 is the location where the velocity profile is fully developed. The fully developed velocity profile for the circular tube is given as [ ( )2 ] r vz (r) = 2V 1 − R The fluid inlet temperature is Ti , and the temperature of the fluid is uniform over the flow cross section until the heat transfer begins. Heating begins at the point where the fully developed velocity profile is established. The upstream half of the tube is at fluid inlet temperature Ti , while the downstream half is kept at temperature Tw . The wall temperature profile is a step function, as described in Figure 8.32b. The viscous dissipation and energy generation along with axial conduction are neglected, and the thermally developing flow problem is found to be adequate for many practical engineering systems. These solutions are particularly good approximations for high-Prandtl-number fluids. For example, we may see this case if the fluid is highly viscous, such as oil. The velocity profile develops much faster than the temperature profile if the Prandtl number is very large. We wish to determine the temperature distribution and the Nusselt number. Note that this problem is traditionally called the Graetz problem. The temperature profile is changing with the axial distance z along the tube. The radial velocity component is zero since velocity field is fully developed. The energy equation can be written as { 2 } 𝜕T 𝜕 T 1 𝜕T 𝜕 2 T ρ cp v z =k + (8.429) + 𝜕z r 𝜕r 𝜕r2 𝜕z2

r

Tw

Ti T = Ti vz = V

Developing temperature profile T (r, z)

vz (r) O

D = 2r0

z

Ti

(a)

Tw

Tw Ti 0 Figure 8.32

(b)

z

(a) Parabolic velocity profile and developing temperature profile in circular tube, (b) Step change in surface temperature.

8.6 Heat Transfer in the Thermal Entrance Region

The boundary conditions are At z = 0

T = Ti

(8.430a)

For z > 0

T = Tw at r = R

(8.430b)

𝜕T =0 (8.430c) 𝜕r The first boundary condition states that the entrance temperature is uniform at Ti . The second boundary condition sets the fluid interface temperature equal to the pipe wall temperature. Finally, the last boundary condition specifies that heat flux at the tube center is equal to zero or the temperature profile is symmetrical with respect to r. We will put the energy equation and the associated boundary conditions in dimensionless form and identify the important dimensionless parameters. Let us introduce the dimensionless variables T − T(r, z) (8.431) θ(r, z) ≡ w Tw − Ti r η= (8.432) R v u= z (8.433) V 2(z∕D) z+ = (8.434) ReD Pr At r = 0

T = finite or

where Tw is the constant surface temperature and Ti is the fluid inlet temperature. Notice that we have used a different notation for the axial position. The fully developed velocity profile can be expressed in terms of dimensionless variables as u = 2[1 − η2 ]

(8.435)

where u = vz /V is the dimensionless velocity. The substitution of dimensionless variables into the energy equation yields 1 𝜕 2 θ 1 𝜕θ 𝜕2θ u 𝜕θ + = + 2 𝜕z+ 𝜕η2 η 𝜕η (ReD Pr )2 𝜕z+2

(8.436a)

where Pe = ReD Pr is the Peclet number. Eq. (8.436a) includes axial conduction in the axial direction. At large values of the Peclet number, the axial conduction term is negligibly small, and it can be neglected for Pe > 100. Neglecting axial conduction, the energy equation reduces to [1 − η2 ]

𝜕θ 𝜕 2 θ 1 𝜕θ = 2 + + 𝜕z η 𝜕η 𝜕η

(8.436b)

θ(0, η) = 1

(8.437a)

𝜕θ(z+ , 0) = 0 or θ(z+ , 0) is finite 𝜕η

(8.437b)

θ(z+ , 1) = 0

(8.437c)

where θ(z+ , η) is dimensionless temperature. The solution of the above PDE can be obtained by the separation of variables. The boundary conditions in the coordinate η direction are homogeneous. We assume a product solution of the form θ = ℜ(η)Z(z+ ) The substitution of this equation, Eq. (8.438), into Eq. (8.436b) results in two ODEs [ ] 1 1 dZ 1 dℜ d2 ℜ = −λ2 + = Z dz+ ℜ(1 − η2 ) η dη dη2 The first ODE is dZ + λ2 Z = 0 dz+ The general solution of this differential equation, Eq. (8.440), is ( ) Zn = Bn exp −λ2n z+

(8.438)

(8.439)

(8.440)

(8.441)

389

390

8 Laminar Momentum and Heat Transfer in Channels

where Bn is an arbitrary constant. The second ODE and its boundary conditions are η

d2 ℜ dℜ + η (1 − η2 )λ2n ℜ = 0 + dη dη2

(8.442)

ℜ′ (0) = 0

(8.443a)

ℜ(1) = 0

(8.443b)

This differential equation, Eq. (8.442), and the boundary conditions, Eqs. (8.443a) and (8.443b), constitute a Sturm–Liouville system, as discussed in Section 8.10. Detailed information is also given by Arpaci and Larsen [21]. The eigenvalues λn are a positive, real numbers, and the eigenfunctions φn (η) form a orthogonal set with respect to the weighting function η(1 − η2 ), as discussed in [21] and [24–26]. Perhaps, first time, Graetz discussed this problem, as reported in [27]; Graetz used a series method to solve this differential equation, and he was able to obtain first two eigenfunctions in 1885. Brown [25] used the series expansion method and obtained on a computer the first 10 eigenvalues and corresponding eigenfunctions. The eigenfunctions defined by the differential equation are sometimes called cylindrical Graetz functions. Graetz functions are not widely available in a convenient form. First, we wish to study the nature of solution using Maple 2020, and for this purpose, we will obtain a series solution > restart; > Order ≔ 𝟏𝟎; Order ≔ 𝟏𝟎 > de ≔ 𝛈 • R′′ (𝛈) + R′ (𝛈) + 𝛈 • 𝛌𝟐 • (𝟏 − 𝛈𝟐 ) • R(𝛈) = 𝟎; de ≔ 𝛈 D(𝟐) (R)(𝛈) + D(R)(𝛈) + 𝛈 𝛌𝟐 (−𝛈𝟐 + 𝟏)R(𝛈) = 𝟎 > sol ≔ dsolve({de}, R(𝛈), series) ∶ > F ≔ rhs(convert(sol, polynom)) ∶ > G ≔ collect(F, { C1, C2}); ) ( 𝛌𝟐 𝛈𝟐 (𝛌𝟐 + 𝟒)𝛌𝟐 𝛈𝟒 𝛌𝟒 (𝛌𝟐 + 𝟐𝟎)𝛈𝟔 (𝛌𝟒 + 𝟓𝟔𝛌𝟐 + 𝟏𝟒𝟒)𝛌𝟒 𝛈𝟖 + − + G ≔ C1 𝟏 − 𝟒 𝟔𝟒 𝟐𝟑𝟎𝟒 𝟏𝟒𝟕𝟒𝟓𝟔 ) ( ( 𝟐 𝟐 𝟐 𝟐 𝟒 𝟒 𝟐 𝟒 𝟔 𝛌 𝛈 (𝛌 + 𝟒)𝛌 𝛈 𝛌 (𝛌 + 𝟐𝟎)𝛈 (𝛌 + 𝟓𝟔 𝛌𝟐 + 𝟏𝟒𝟒)𝛌𝟒 𝛈𝟖 + C2 ln(𝛈) 𝟏 − + − + 𝟒 𝟔𝟒 𝟐𝟑𝟎𝟒 𝟏𝟒𝟕𝟒𝟓𝟔 ( 𝟐 ) ( 𝟐 𝟐 𝟐 𝟐) 𝟒 𝟐 𝟒 𝛌 𝛈 𝟑(𝛌 + 𝟒)𝛌 𝟏𝟏 𝛌 (𝛌 + 𝟐𝟎) 𝛌 𝛌 + + − 𝛈𝟒 + − + 𝛈𝟔 𝟒 𝟏𝟔 𝟏𝟐𝟖 𝟏𝟗𝟐 𝟏𝟑𝟖𝟐𝟒 ) ) ( (𝟐𝟒 𝛌𝟐 + 𝟏𝟗𝟐)𝛌𝟐 𝟐𝟓(𝛌𝟒 + 𝟓𝟔 𝛌𝟐 + 𝟏𝟒𝟒)𝛌𝟒 − 𝛈𝟖 + 𝟏𝟒𝟕𝟒𝟓𝟔 𝟏𝟕𝟔𝟗𝟒𝟕𝟐 Based on this solution, we may implicitly express the general solution as (o) ℜ(η) = AG(e) 0 (λη) + BG0 (λη)

(8.444)

(o) (e) where G(e) 0 and G0 represent the Graetz functions of first kind and second kind, order zero. Let us assume that G0 and (o) G0 denote the even and odd functions, respectively. These functions are expressed by infinite series, as discussed in [29]. The first boundary condition, Eq. (8.443a), implies that we choose B = 0; in other words, we reject the odd function since characteristic function must be symmetric. The characteristic functions (eigenfunctions) φn are

φn = G(e) 0 (λn η)

(8.445)

The second boundary condition, Eq. (8.443b), gives the characteristic values G(e) 0 (λn ) = 0, n = 0,1, 2,3. …

(8.446)

8.6 Heat Transfer in the Thermal Entrance Region

The characteristic values (eigenvalues) are the roots of Eq. (8.446). We may now construct the product solution of the problem in the following form: θ(z+ , η) =

∞ ∑

) ( Cn exp −λ2n z+ φn (λn η)

(8.447)

n=0

We now use the nonhomogeneous boundary condition θ(0, η) = 1 to evaluate Cn as given 1=

∞ ∑

Cn φn (λn η)

(8.448)

n=0

where Cn is given by 1

Cn =

∫0 η(1 − η2 )φn (λn η) dη

(8.449)

1

∫0 η(1 − η2 )[φn (λn η)]2 dη

It is possible to express Cn in a convenient form by mathematical manipulations, as discussed in [28]. First, let us evaluate the numerator of Eq. (8.449). We know that φn (λn η) is a solution of Eq. (8.442); therefore, we may replace ℜ by φn (λn η) in this equation and write it in the following form: ( ) dφ d η n + λ2n η (1 − η2 )φn = 0 (8.450) dη dη Next, we integrate Eq. (8.450) over the interval (0, 1), and we get ( ) 1 1 dφn d 2 2 λn η dη η (1 − η )φn dη = − ∫0 dη ∫0 dη dφ | dφ |η=1 = − n || = − η n || dη |η=0 dη |η=1 Finally, we get 1

1 η (1 − η )φn dη = − 2 ∫0 λn 2

(

dφn dη

)| | | | |η=1

(8.451)

The rearrangement of the denominator of Eq. (8.449) will be carried out as follows. First, we multiply Eq. (8.450) by dφn /dλn and integrate over the interval (0, 1) with respect to η ( )( ) ( ) 1 1 dφ dφn dφn d η n dη + λ2n dη = 0 (8.452) η (1 − η2 )φn ∫0 dη ∫0 dη dλn dλn The first integral in Eq. (8.542) is evaluated using the method of integrating by parts ) ( )( ) ( )( ) η=1 ) ( 1 1( dφ || dφ dφ dφn dφn d d dφn dη = η n | − η n dη η n ∫0 dη dη dλn dλn dη ||η=0 ∫0 dη dη dλn ) ) )( ( 1 ( dφ || dφn 2 dφn 1 d η n | − η dη = dλn dη ||η=1 2 dλn ∫0 dη

(8.453)

Note that the order of the differentiation with respect to η and λn may be interchanged. We rearrange the second integral in Eq. (8.452) as follows: ( ) 1 1 dφn λ2 d λ2n dη = n η (1 − η2 )φn η (1 − η2 )φ2n dη (8.454) ∫0 dλn 2 dλn ∫0 Using Eqs. (8.453) and (8.454), we now have for the final result ( ) ) )( 1 ( 1 dφ || λ2 d dφn dφn 2 1 d η n | − η dη + n η(1 − η2 )φ2n dη = 0 dλn dη ||η=1 2 dλn ∫0 dη 2 dλn ∫0 Next, we multiply Eq. (8.450) by φn (λn η) and integrate over the interval (0, 1) with respect to η to get ( ) 1 1 dφ d η n φn dη + λ2n η (1 − η2 )φ2n dη = 0 ∫0 ∫0 dη dη

(8.455)

(8.456)

391

392

8 Laminar Momentum and Heat Transfer in Channels

The first integral in Eq. (8.456) is evaluated again using integral by parts, and Eq. (8.456) becomes ) |η=1 )( ) ( 1( 1 dφn dφ dφ | dη + λ2n η n η (1 − η2 )φ2n dη = 0 η n φn | − | ∫0 ∫0 dη dη dη |η=0 or

( φn

dφ η n dη

)| ) 1 ( 1 dφn 2 | η dη + λ2n η (1 − η2 )φ2n dη = 0 | − | ∫0 ∫0 dη |η=1

(8.457)

The differentiation of Eq. (8.457) with respect to λn yields ( ) )| ) )( ( 1( 1 dφn dφn || dφn || dφn 2 d d η η η − dη + 2λn η (1 − η2 )φ2n dη | + φn | ∫0 dλn dη ||η=1 dλn dη | dλn ∫0 dη |η=1 1

+ λ2n

d η (1 − η2 )φ2n dη = 0 dλn ∫0

Noting that φn |η = 1 = 0 and dividing Eq. (8.458) by 2 yields ( ) ) )( 1 ( 1 dφ || dφn 2 1 d 1 dφn η n | − η dη + λn η (1 − η2 )φ2n dη ∫0 2 dλn dη ||η=1 2 dλn ∫0 dη 1 λ2 d η (1 − η2 )φ2n dη = 0 + n 2 dλn ∫0 Subtracting Eq. (8.549) from Eq. (8.455) yields ) ( ) ( 1 dφn 1 dφn η (1 − η2 )φ2n dη = − ∫0 2λn dλn η=1 dη η=1

(8.458)

(8.459)

(8.460)

Substituting Eqs. (8.460) and (8.451) into Eq. (8.449), we obtain (

Cn = − λn

2 ) 𝜕φn 𝜕λn η=1

(8.461)

The eigenvalues λn and constants Cn may be evaluated using the series solution. The solution of Eq. (8.442) will be obtained in terms of Kummer functions using Maple 2020. Maple 2020 gives two linearly independent solutions of the differential equation in terms of Kummer functions. Let us represent the dependent variable by R. We may obtain the solution directly using Maple 2020. > restart; > Order ≔ 𝟏𝟎; Order ≔ 𝟏𝟎 > de ≔ 𝛈 • R′′ (𝛈) + R′ (𝛈) + 𝛌𝟐 • 𝛈 • (𝟏 − 𝛈𝟐 ) • R(𝛈) = 𝟎; de ≔ 𝛈D(𝟐) (R)(𝛈) + D(R)(𝛈) + 𝛌𝟐 𝛈(−𝛈𝟐 + 𝟏)R(𝛈) = 𝟎 > #Let us obtain the solution using first boundary condition > sol ≔ rhs(dsolve{de, R′ (𝟎) = 𝟎}, R (𝛈))); ( ) 𝛌 𝛈𝟐 𝟏 𝛌 sol ≔ C𝟏 e− 𝟐 KummerM − , 𝟏, 𝛌𝛈𝟐 𝟐 𝟒 The solutions of the differential equation are the eigenfunctions to the Graetz problem. If we examine the solution, we see that we have confluent hypergeometric functions. Detailed information about confluent hypergeometric functions is given by Abramowitz and Steigun [29]. Only one of the independent solutions is finite at η = 0, i.e. R(0) = 0 must be bounded. The characteristic function is ) ( ) ( λ η2 1 λn φn = Mn = exp − n KummerM − , 1, λη2 (8.462) 2 2 4

8.6 Heat Transfer in the Thermal Entrance Region

is symmetric and bounded η = 0. The boundary condition ℜ(1) = 0 leads to the following transcendental equation for the eigenvalues: ( ) ( ) λ 1 λn exp − n KummerM − , 1, λn = 0 (8.463) 2 2 4 This equation has infinitely many discrete solutions for λn , n = 1, 2, 3, …∞. Corresponding to each eigenvalue λn , there is there is an eigenfunction φn = Mn given by ) ( ) ( λ η2 1 λn KummerM − , 1, λn η2 (8.464) φn = Mn = exp − n 2 2 4 As we discussed before, the eigenvalues λn and eigenfunctions Mn form a complete orthogonal set with respect to the weighting function η(1 − η2 ). The first three eigenvalues λn and eigenfunctions Mn (λn η) are obtained using Maple 2020 as follows. > restart; > #Characteristic function is given as ( ( ) ) 𝟏 𝟏 𝛌 > M ≔ exp − • 𝛌 • 𝛈𝟐 • KummerM − , 𝟏, 𝛌, 𝛈𝟐 ; 𝟐 ( )𝟐 𝟒 1 2 1 1 − λ, 1, λ η2 M ≔ e− 2 λ η KummerM 2 4 > #We will use the second boundary condition > eq ≔ subs(𝛈 = 𝟏, M); ( ) 1 1 1 eq ≔ e− 2 λ KummerM − λ, 1, λ 2 4 > with(plots) ∶ > Roots of this equation will give us the eigenvalues and we now plot this equation to see location of a few eigenvalues.

plot(eq, 𝛌 = 𝟎.𝟎..𝟏𝟓, gridlines = true, labels = [′′ λ′′ , ′′ eq′′ ], labelfont = [′′ Helvetica′′ , 𝟏𝟓]); Figure 8.33 shows the location of eigenvalues. > #First eigenvalue > 𝛌 [𝟎] ≔ fsolve(eq, 𝛌 = 𝟐..𝟑); λ0 ≔ 2.704364420 > 𝛌 [𝟏] ≔ fsolve(eq, 𝛌 = 𝟔..𝟕); λ1 ≔ 6.679031449 > 𝛌 [𝟐] ≔ fsolve(eq, 𝛌 = 𝟏𝟎..𝟏𝟏); λ2 ≔ 10.67337954 > #First Graetz function

) 𝟏 𝛌[𝟎] − , 𝟏, 𝛌[𝟎] • 𝛈𝟐 ; 𝟐 𝟒 2 M0 ≔ e−1.352182210 η KummerM(−0.1760911050, 1, 2.704364420 η2 ) > M [𝟎] ≔ e−

𝛌 [𝟎] • 𝛈𝟐 𝟐

(

KummerM

> Second Graetz function Second Graetz function > M [𝟏] ≔ e−

𝛌 [𝟏] • 𝛈𝟐 𝟐

KummerM

(

) 𝟏 𝛌 [𝟏] − , 𝟏, 𝛌 [𝟏] • 𝛈𝟐 ; 𝟐 𝟒

393

394

8 Laminar Momentum and Heat Transfer in Channels

1

eq

0.5

0

5

10

15

λ

–0.5 Figure 8.33

Location of eigenvalues. 2

M1 ≔ e−3.339515724 η KummerM(−1.169757862, 1, 6.679031449 η2 ) > Third Graetz function Third Graetz function

) 𝟏 𝛌 [𝟐] 𝟐 • − , 𝟏, 𝛌 [𝟐] 𝛈 ; KummerM > M [𝟐] ≔ e 𝟐 𝟒 2 M2 ≔ e−5.336689770 η KummerM(−2.168344885, 1, 10.67337954 η2 ) • 𝛈𝟐

− 𝛌 [𝟐]𝟐

(

> #We now plot the Graetz function ( > plot [M [𝟎], M [𝟏], M [𝟐]], 𝛈 = 𝟎..𝟏, gridlines = true, labels = [′′ η′′ , ′′ Characteristic functions′′ ] , ) labeldirections = [′′ horizontal′′ , ′′ vertical′′ ], linestyle = [solid, longdash, dash] ; > First three of the characteristic functions are depicted in Figure 8.34. Let us evaluate first three constants s.Cn s > #Let us evaluate the constants C𝟎 , C𝟏 , C𝟐 ( 𝟏 ) > NUM ≔ evalf 𝛈 • (𝟏 − 𝛈𝟐 ).M [𝟎]d𝛈 ; ∫𝟎 NUM ≔ 𝟎.𝟏𝟑𝟖𝟔𝟖𝟕𝟏𝟒𝟎𝟖 ( 𝟏 ) > DENOM ≔ evalf 𝛈 • (𝟏 − 𝛈𝟐 ).M [𝟎] • M [𝟎]d𝛈 ; ∫𝟎 DENOM ≔ 𝟎.𝟎𝟗𝟑𝟗𝟑𝟑𝟕𝟔𝟕𝟗𝟐 NUM ; > C[𝟎] ≔ DENOM C𝟎 ≔ 𝟏.𝟒𝟕𝟔𝟒𝟑𝟓𝟒𝟎𝟔 ) ( 𝟏 𝛈 • (𝟏 − 𝛈𝟐 ).M [𝟏]d𝛈 ; > NUM ≔ evalf ∫𝟎 NUM ≔ −𝟎.𝟎𝟑𝟎𝟐𝟒𝟓𝟔𝟑𝟖𝟐𝟒 ) ( 𝟏 𝛈 • (𝟏 − 𝛈𝟐 ).M [𝟏] • M [𝟏]d𝛈 ; > DENOM ≔ evalf ∫𝟎

8.6 Heat Transfer in the Thermal Entrance Region

1 First eigen function 0.8

Characteristic functions

0.6 Second eigen function 0.4

0.2

0

Third eigen function 0.4

0.2

0.6

0.8

1

η –0.2

–0.4 Figure 8.34

Graetz functions as function of dimensionless distance η.

DENOM ≔ 𝟎.𝟎𝟑𝟕𝟓𝟏𝟗𝟖𝟑𝟖𝟑𝟒 NUM ; > C[𝟐] ≔ DENOM C𝟏 ≔ −𝟎.𝟖𝟎𝟔𝟏𝟐𝟑𝟖𝟗𝟓𝟓 ) ( 𝟏 𝛈 • (𝟏 − 𝛈𝟐 ).M[𝟐]d𝛈 ; > NUM ≔ evalf ∫𝟎 NUM ≔ 𝟎.𝟎𝟏𝟑𝟖𝟎𝟏𝟖𝟑𝟓𝟔𝟐 ) ( 𝟏 𝟐 • • 𝛈 (𝟏 − 𝛈 ).M[𝟐] M[𝟐]d𝛈 ; > DENOM ≔ evalf ∫𝟎 DENOM ≔ 𝟎.𝟎𝟐𝟑𝟒𝟒𝟐𝟏𝟐𝟒𝟑𝟑 NUM ; > C[𝟏] ≔ DENOM C𝟐 ≔ 𝟎.𝟓𝟖𝟖𝟕𝟔𝟐𝟏𝟓𝟒𝟐 Since the temperature distribution is known, we can now determine local and mean surface heat flux, mean temperature, and local and mean Nusselt numbers. Local heat flux We now have the solution, and we can calculate the local heat flux at any z+ along the axial distance

(

q′′w (z+ ) = k

𝜕T 𝜕r

) =− r=R

k(Tw − Ti ) R

(

𝜕θ 𝜕η

)

(8.465) η=1

where we assume that the heat flows from the wall into fluid. Dimensionless surface temperature gradient 𝜕θ(z+ , 1)/𝜕η is obtained from Eq. (8.447) as ) ( ∞ ∑ ( ) dφn 𝜕θ + Cn exp −λ2n z+ (z , 1) = (8.466) 𝜕η dη η=1 n=0 where φn = Mn (λn η) is the KummerM function. Let us define a new constant Gn as follows: ( ) C dφn Gn = − n 2 dη η=1

(8.467)

395

396

8 Laminar Momentum and Heat Transfer in Channels

Using these definitions, the dimensionless temperature gradient becomes ∞ ∑ ) ( 𝜕θ + (z , 1) = −2 Gn exp −λ2n z+ 𝜕η n=0

(8.468)

and local heat flux takes the following form: ( ) 2k(Tw − Ti ) ∑ Gn exp −λ2n z+ R n=0 ∞

q′′w (z+ ) =

(8.469)

Local mean fluid temperature The local mean fluid temperature θm (z+ ) is obtained by substituting temperature and velocity

distributions into R

Tm =

∫0 vz T (2πr dr)

(8.470)

R

∫0 vz (2πr dr)

and using the dimensionless variables defined earlier, the mean fluid temperature becomes 1

Tm =

∫0 [2V(1 − η2 )] [TW − θ(Tw − Ti ) ](η dη) 1

∫0 [2V(1 − η2 )] (η dη)

(8.471a)

or dimensionless mean fluid temperature θm θm =

1 (Tw − Tm ) =4 η(1 − η2 ) θ dη ∫0 (Tw − Ti )

(8.471b)

Now, substituting dimensionless temperature θ, Eq. (8.447), into Eq. (8.471b), we obtain 1

θm = 4

∫0

∞ ∑ ) ( η(1 − η2 ) Cn exp −λ2n z+ Mn (λn η) dη

(8.471c)

n=0

or θm = 4

∞ ∑ ) ( Cn exp −λ2n z+ n=0

1

∫0

η(1 − η2 )Mn (λn η) dη

(8.471d)

and it is possible, by mathematical trick, to evaluate the integral. See [28] for a discussion. Note that φn (λn η) = Mn (λn η) is a solution of the differential equation, and here, we will use φn (λn η) to develop our equation ( ) dℜ d η + λ2n η[1 − η2 ]ℜ = 0 (8.472) dη dη Let us replace ℜ by φn (λn η) in this differential equation, rearrange it, and integrate the result ( ) 1 1 dφ 1 dφn (λn η) || d η n =− 2 η[1 − η2 ]φn (λn η) dη = − ∫0 ∫0 dη dη dη ||η=1 λn

(8.473)

Using Eq. (8.473), the dimensionless mean temperature becomes θm = −4

∞ ∑ ( ) Cn dφ′n || | exp −λ2n z+ 2 dη | n=0 λn |η=1

Introducing the constant Gn = −

θm = 8

∞ ∑ Gn 2 n=0 λn

) ( exp −λ2n z+

(8.474)

Cn dφn || in Eq. (8.474), we get the final form of the mean fluid temperature 2 dη ||η=1 (8.475)

8.6 Heat Transfer in the Thermal Entrance Region

Local Nusselt number The local heat transfer coefficient, by definition, is

h=

q′′w (z) Tw − Tm

Using the definition of dimensionless temperature θ Tw − Tm = θm (Tw − Ti )

(8.476)

we may express the local heat transfer coefficient in terms of dimensionless variable h=

q′′w (z+ ) 1 (Tw − Ti ) θm

(8.477)

Using Eq. (8.465) for local heat flux q′′w , the local heat transfer coefficient, Eq. (8.477), becomes ( ) k(T −T ) 𝜕θ ( ) − wR i 𝜕η η=1 1 k 𝜕θ 1 =− h= (Tw − Ti ) θm R 𝜕η η=1 θm or local Nusselt number is h (2R) h D −2 NuD = = = k k θm

(

𝜕θ 𝜕η

(8.478)

) (8.479) η=1

Substituting the equation for θm , Eq. (8.475), and the equation for (𝜕θ/𝜕η)η = 1 , Eq. (8.468), the local Nusselt number takes the following form: ∞ ∑

NuD =

hD = k

n=0

2

) ( Gn exp −λ2n z+

) ∞ ( ∑ Gn

n=0

λ2n

(8.480)

) ( exp −λ2n z+

where D = 2R is the tube diameter. This equation gives the variation of the Nusselt number in the thermal entrance region of circular pipe with the fully developed velocity profile under UWT boundary condition. The constants and characteristic values for these infinite series are presented in Table 8.4. We now consider Eq. (8.480) to estimate the thermal entrance length. We see that as z+ increases, the series becomes more convergent and the Nusselt number approaches a fully developed value of 3.6672. For z+ > 0.1, only the first term is Table 8.4

The constants needed for the solution of the problem with uniform wall temperature.

N

𝛌n

0

2.704364420

1.476435406

1.01430045

0.7487745550

1

6.679031449

−0.806123895

1.349241623

0.5438279565

2

10.67337954

0.5887621542

−1.57231934

0.4628610608

3

14.67107846

−0.475850425

1.746004333

0.4154184528

4

18.66987186

0.4050218101

−1.89085712

0.3829191869

5

22.66914336

−0.355756506

2.016466653

0.3586855662

6

26.66866200

0.3191690536

–2.12816475

0.3396221650

7

30.66832334

−0.290735829

2.229255418

0.3240622110

8

34.66807382

0.2678911823

−2.32194333

0.3110140728

9

38.66788335

−0.249062532

2.407781139

0.2998440342

10

42.66773381

0.2332277937

−2.48790826

0.2901246773

Cn

Mn (1)

Shah and Bhatti [22] gives the following equations for large values of n. n > 10 Gn =

2.0256 2(λn )1∕3

λn = 4n +

8 . 3

397

398

8 Laminar Momentum and Heat Transfer in Channels

important, and the contribution of the remaining terms is insignificant. Considering the first term of Eq. (8.480), we obtain ( ) G0 exp −λ20 λ20 (2.704364420)2 NuD = ( ) = ≈ 3.6567 = ( 2) G0 2 2 2 exp −λ 0

λ20

This approximation is within 0.05% of the fully developed value, and the thermal entry length may be estimated to be equal to 2(z∕D) ReD Pr ( ) 0.1 (z∕D) ≈ ReD Pr = 0.05ReD Pr 2 The local and mean Nusselt numbers and mean fluid temperature will be presented using the eigenvalues. z+ = 0.1 =

Mean heat transfer coefficient By definition, z

h≡

1 h(z)dz z ∫0

(8.481)

where from 0 to any value of z. To evaluate the average heat transfer coefficient h, consider the energy balance for the differential control volume shown in Figure 8.35. ( ′′ ) qw (2π R)dz = ṁ cp dTm (8.482) and introducing the definition of local heat transfer coefficient h, q′′w = h(Tw − Tm )

(8.483)

We combine Eqs. (8.482) and (8.483); thus, we get (2π R)h(z)[Tw − Tm ]dz = ṁ cp dTm

(8.484)

and separating variables and integrating Eq. (8.484), we obtain z Tm dTm 2π R h(z)dz = ∫Tmi Tw − Tm ṁ cp ∫0 [ ] z Tm dTm 2π R z 1 h(z)dz = ∫Tmi Tw − Tm ṁ cp z ∫0

(8.485)

and the average heat transfer coefficient h is introduced z

h=

1 h(z)dz z ∫0

(8.486)

Thus, Eq. (8.485) is expressed as ( ) T − Tm 2πR h z = −ln w = −ln θm ṁ cp Tw − Tmi q″w . m cp Tm

d . . m cp Tm + (m cp Tm) dz dz

2R

z=0 z Figure 8.35

dz

Control volume for energy balance.

(8.487)

8.6 Heat Transfer in the Thermal Entrance Region

and finally, an expression for average heat transfer coefficient h is obtained ) ( ̇ m cp 1 ln θm h=− 2πR z

(8.488)

We can define the mean Nusselt number as ṁ cp 1 h (2R) =− ln θm k πk z

NuD =

(8.489)

Using the following dimensionless variables, we wish to obtain an expression for ṁ cp ∕π k given in Eq. (8.489): z = z+ (Re Pr )R ReD =

ρ V 2R = (ρV)(πR2 ) μ

(8.490) (

2R πR2

)( ) 2ṁ 1 = μ πRμ

(8.491a)

or πRμ ReD 2

ṁ =

(8.491b)

and after a little algebra ṁ cp

=

πk

R Re Pr 2 D

Thus, the Nusselt number, Eq. (8.489), now becomes ] [ 1 1 1 NuD = − + ln θm (z+ ) = + ln 2z 2z θm (z+ )

(8.492)

(8.493)

Substituting the expression for θm , Eq. (8.474), the mean Nusselt number NuD takes the following form: ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ 1 1 NuD = ln ( ))⎥ ( 2 + 2 z+ ⎢⎢ ∑ ∞ Gn exp −λn z ⎥ ⎢8 ⎥ 2 λn ⎣ n=0 ⎦

(8.494)

Mean temperature, local Nusselt number, and mean Nusselt number for thermal entry length in circular tube are presented in Table 8.5. Example 8.7 Water flows at a rate of 5 cm3 /s through a tube of 20-mm inside diameter. Water enters the tube at a uniform mean temperature of 20 ∘ C. The tube has a length of 2 m. The tube wall is maintained at a constant temperature of 50 ∘ C. Determine the following quantities: (a) Calculate the local heat transfer coefficient at a distance of 1 m from the entrance of tube. (b) Calculate the average heat transfer coefficient between x = 0 and x = 1 m. (c) Determine the amount of heat transfer from the tube wall to water over the first 1 m length of the tube. Solution Average velocity V is V=

4 × 5 × 10−6 4∀̇ = = 0.0159 m∕s 2 π × (0.020)2 πD

We do not know the exit temperature. Estimate the fluid properties at Tm =

Tw + Tmi 50 + 20 = = 35 ∘ C = 308 K 2 2

399

400

8 Laminar Momentum and Heat Transfer in Channels

Table 8.5 Mean temperature and Nusselt numbers for thermal entry length in circular tube: Parabolic velocity profile and uniform wall temperature. z+ = 2(z/D)/ReD Pr

𝛉m

NuD

NuD

0

1

∞

∞

0.00001

0.99814

61.877

93.334

0.00002

0.99705

48.914

73.869

0.00010

0.99147

28.254

42.813

0.00018

0.98756

23.099

35.047

0.0008

0.96668

13.842

21.049

0.0010

0.96175

12.824

19.501

0.0012

0.95698

12.050

18.321

0.0016

0.94826

10.926

16.603

0.0040

0.90736

8.0362

12.152

0.008

0.85723

6.4296

9.6280

0.010

0.83622

6.0015

8.9432

0.018

0.76623

5.0584

7.3963

0.030

0.68436

4.4406

6.3211

0.04

0.62802

4.1724

5.8146

0.06

0.53487

3.8942

5.2145

0.08

0.45901

3.7689

4.8668

0.1

0.39529

3.7100

4.6406

0.2

0.18971

3.6580

4.1556

0.3

0.09129

3.6568

3.9895

0.4

0.04393

3.6568

3.9063

∞

0

3.66

3.66

Source: Shah and London [3]/with permission of Elsevier.

The properties of water are estimated at 308 K are ρ = 994 kg∕m3 cp = 4178 J∕kg.K μ = 698 × 10−6 N.s∕m2 k = 0.620 W∕m.K Pr = 4.85 The Reynolds number is ρVD 994 × 0.0159 × 0.02 = = 453 μ 698 × 10−6 The hydrodynamic entrance length is ReD =

LH = 0.05ReD D = 0.05 × 453 × 0.02 = 0.453 m The thermal entrance length is LT = 0.05ReD Pr D = 0.05 × 453 × 4.85 × 0.02 = 2.19 m We see that LH ≪ L and LT > L We may assume that flow is hydrodynamically is fully developed and thermally developing. Hence, the distance is dimensionless.

8.6 Heat Transfer in the Thermal Entrance Region

We will determine the amount of heat transferred to water in the first 1 m of the tube, and we take z = 1 m: z+ =

2(1∕0.02) 2(z∕D) = ≈ 0.0455 ReD Pr 453 × 4.85

From Table 8.5, we have: 0.04

4.1724

5.8146

0.06

3.8942

5.2145

Interpolation gives NuD ≈ 4.09 and NuD ≈ 5.64. Thus, we have the following. (a) Local heat transfer coefficient ) ( ( ) 0.624 k NuD = × 4.09 ≈ 127.0 W∕m2 K h= D 0.02 (b) Average heat transfer coefficient ) ( ) ( k 0.624 × 5.64 ≈ 174.8 W∕m2 K h= NuD = D 0.02 (c) Heat transferred We will determine the amount of heat transferred to water in the first 1 m of the tube. First let us determine the temperature at z = 1 m [ ] Ph z Tm (x) = Tw + (Tmi − Tw ) exp − ṁ cp P = πD •

•

m = ρ∀ = 994.0 × 5 × 10−6 = 0.00497 kg∕s ) ( π × 0.02 × 174.8 × 1 = 32.3 ∘ C Tm0 = 50 + (20 − 50) exp − 0.00497 × 4178 Hence, the rate of heat transferred to water . ̇ p (Tmo − Tmi ) = 0.00497 × 4178(32.3 − 20) ≈ 257.4 W q = mc 8.6.1.2 The Leveque Solution: UWT Boundary Condition

The Leveque solution discusses the initial behavior of the Graetz problem. The infinite series solution of the Graetz problem, we have just considered, converges very rapidly for large values of z+ . However, this same infinite series solution has a slow convergence rate for small values of z+ . We need a special solution for the small values of z+ . Leveque, as reported by Bennett [30], noticed that for flows of large Prandtl number, Pr ≫ 1, most of the heat transfer takes place across a very thin thermal region close to the surface. For ξ < 10−4 , the thermal boundary layer is very thin compared to the hydraulic diameter, and we can assume that velocity distribution in the thermal boundary layer is linear. In order to obtain the solution of energy equation, we can use the wall tangent of the fully developed velocity profile in place of the Poiseuille parabola. This solution of the energy equation is especially useful for Pr ≫ 1 fluids. Notice that the velocity boundary layer thickness is much larger than the thermal boundary layer thickness for Pr ≫ 1 fluids. Consequently, Leveque’s method is valid in the thermal entrance region of pipe very close to the origin of the heat exchange section as long as the hydrodynamic boundary layer extends the thermal boundary layer. The method cannot be used for low-Prandtl-number fluids such as liquid metals (Pr ≪ 1). The energy equation and its boundary conditions for the thermal entrance problem are repeated here for convenience { 2 } 𝜕T 𝜕 T 1 𝜕T =k (8.495) + ρcp vz 𝜕z r 𝜕r 𝜕r2 At z = 0 T = Ti

(8.496a)

401

402

8 Laminar Momentum and Heat Transfer in Channels

For z > 0 T = Tw at r = R

(8.496b)

𝜕T = 0 at r = 0 (8.496c) 𝜕r For small value of z+ = (z/R)/Pe, heat propagates only in a thin layer of fluid near the wall. The thickness of the thermal boundary layer is much less than the pipe radius, i.e. Δ ≪ R. The Leveque solution is based on this assumption. We may now make certain simplifications. ( ) (1) Curvature effects of the pipe can be neglected in the radial conduction term. This means that 1r 𝜕r𝜕 r 𝜕𝜕rT ≈ ( ) 2 1 𝜕 𝜕T R = 𝜕𝜕 rT2 . R 𝜕r 𝜕r (2) We are only interested in the velocity distribution in the thermal boundary layer, and since the thermal boundary layer is very thin, we may safely assume that velocity distribution is linear within the thermal boundary layer. (3) Constant fluid properties. (4) Axisymmetric temperature field 𝜕T/𝜕θ = 0. Based on these facts, we may assume reversed coordinate based on the tube wall as given below y+r=R⇒r=R−y

(8.497)

where y is measured from the pipe wall. See Figure 8.36 for notation. Next, we can write the velocity profile as [ ] (R − y) 2 u = 2V 1 − R2 We can expand this equation, Eq. (8.498) to obtain the linearized form of the velocity profile ] [ y (R2 − 2 Ry + y2 ) ≃ 4V = βy u = 2V 1 − 2 R R

(8.498)

(8.499)

where β = 4 V/R is the velocity gradient near the pipe wall. This approach results in a very good approximate solution down to Prandtl numbers of 0.7. We are interested in the near wall region where (r/R) ≪ 1. Due to the thinness of this region, we can neglect the curvature of the pipe. The neglect of surface curvature and linearization of the velocity profile is called Leveque approximation. If we use this approximation, the energy equation takes the following form: βy

𝜕T 𝜕2 T =α 2 𝜕z 𝜕y

(8.500)

where α = k/ρcp is the thermal diffusivity. The inlet and boundary conditions are at z = 0, y > 0 T = Ti

(8.501a)

at z > 0 y = 0 T = Tw

(8.501b)

We need another boundary condition for the problem. We will now assume that centerline temperature is not affected by the heating or cooling of the pipe wall. We can now state that far from the wall, the fluid temperature approaches the inlet temperature of the fluid. Then, we can write the boundary condition as at y = ∞ T = Ti

(8.501c) Pipe center line

r R y

Pipe wall

Figure 8.36

Schematic drawing for the coordinate system.

8.6 Heat Transfer in the Thermal Entrance Region

With this new boundary condition, we can now obtain a similarity solution. The similarity solution will be obtained by dimensional analysis. We will employ the method of similarity solution, as explained in [21]. The formulation of the problem in terms of a new temperature θ = T − Tw gives βy

𝜕2 θ 𝜕θ =α 2 𝜕z 𝜕y

θ(0, y) = θi

where θi = Ti − Tw

(8.502) (8.503a)

θ(z, 0) = 0

(8.503b)

θ(z, ∞) = θi

(8.503c)

We will now nondimensionalize z, y, and θ in terms of arbitrarily selected reference lengths z0 , y0 and the characteristic temperature θi as follows z (8.504a) z∗ = z0 y (8.504b) y∗ = y0 θ f= (8.504c) θi Then, the PDE, Eq. (8.502), becomes β y30 y 𝜕f 𝜕2 f = α z0 y0 𝜕z∗ 𝜕y∗2 Equation (8.505) implies that ) ( 3 θ z y β y0 =f , , θi z0 y0 α z0

(8.505)

(8.506)

We introduce a new variable in place of β y30 ∕α z0 . This new variable is obtained by dividing β y30 ∕α z0 by z/z0 . The result is ) ( 3 z y β y0 θ (8.507) = f1 , , θi z0 y0 α z First of all, z0 and y0 are rejected by the physics of the problem as being characteristic lengths. Since there is no characteristic length in the z-direction, temperature and, therefore, the right-hand side are independent of z0 . This reduces the equation to the following form: ( ) y β y30 θ = f1 , (8.508) θi y0 α z Next, again introduce a new variable in place of β y30 ∕α z. This new variable is obtained multiplying β y30 ∕α z by y2 ∕y20 . The result is ) ( y β y3 θ (8.509) = f2 , θi y0 α z There is no characteristic length in the y-direction; therefore, temperature and, consequently, the right-hand side of the equation are independent of y0 . The equation now becomes ( 3) βy θ (8.510) = f2 θi αz This equation can be cast as θ = f(η) θi

(8.511)

403

404

8 Laminar Momentum and Heat Transfer in Channels

where η = y(β/α z)1/3 is the similarity variable. We now have the following variables: ( )1∕3 T − Tw β 4V θ = = f(η), η = y , β= θi Ti − Tw αz R

(8.512)

where θi = Ti − Tw . We can now convert the PDE to an ODE as follows: ⎤ [ ]⎡ θ0 df ⎥ yβ 𝜕θ 𝜕f df 𝜕η df ⎢ 1 = θ0 = θ0 = θ0 − ⎥ = − 3 z dη 𝜕z 𝜕z dη 𝜕z dη ⎢⎢ 3 ( β )2∕3 (α z) ⎥⎦ ⎣ αz ( )1∕3 β 𝜕θ 𝜕f df 𝜕η df = θ0 = θ0 = θ0 𝜕y 𝜕y dη 𝜕y αz dη [ ( ) ] [ ( ) ] ( )2∕3 2 1∕3 1∕3 β β β d 𝜕 df df 𝜕η df 𝜕2θ θ = θ = θ0 = 𝜕y 0 α z dη dη 0 α z dη 𝜕y αz 𝜕y2 dη2 Using these derivatives, the PDE, Eq. (8.502), reduces to the following ODE: d2 f η2 df =0 + 3 dη dη2

(8.513)

f(0) = 0

(8.514a)

f(∞) = 1

(8.514b)

The solution of Eq. (8.513) is obtained as follows: Let p =

dp . dη

Then, Eq. (8.513) can be expressed as dp η2 =− p dη 3 dp η2 = − dη p 3 The solution is p = C1 exp(−η3 ∕9) or df = C1 exp(−η3 ∕9) dη Integration yields η

f = C1

∫0

exp(−x3 ∕9)dx + C2

Using the first boundary condition, Eq. (8.514a), yields C2 = 0. The application of the second boundary condition, ∞ Eq. (8.514b), gives C2 = ∫0 exp(−x3 ∕9)dx ( 3) η ∫0 exp − x9 dx θ (8.515) = ( 3) ∞ θi ∫0 exp − x9 dx where x is a dummy variable. Equation (8.515) is the solution obtained by Leveque [30]. The integral in the numerator is evaluated by Maple 2020 ( 3) ∞ x dx; exp − > ∫0 9

8.6 Heat Transfer in the Thermal Entrance Region

√

2 91∕3 π 3 ( ) 9 Γ 23 > evalf(%); 1.857472237 The value of integral is ( 3) ∞ x dx = 1.857472 exp − ∫0 9 Let us determine the Nusselt number based on the tube diameter ) ( ( ) ( )1∕3 (Ti − Tw ) β df 𝜕η 𝜕T = (θi ) = ( 3) 𝜕y y=0 dη 𝜕y η=0 ∫ ∞ exp − u du αz 0 9 Using Eq. (8.515), the local Nusselt number becomes ( ) ( ) 𝜕T 𝜕T −(2R)k −(2R)k 𝜕y y=0 𝜕y y=0 q′′w (2R) hD NuD = = = = k k(Tw − Tm ) k(Tw − Tm ) k(Tw − Ti ) We replaced the mean fluid temperature Tm with the inlet temperature Ti . This is true, since the nonisothermal region is very thin, almost all of the fluid is still at the inlet fluid temperature Ti ( )1∕3 ( )1∕3 β β (2R) −(2R)(k)(Ti − Tw ) αz αz = NuD = ( 3) ( 3) x x ∞ ∞ dx ∫0 exp − dx k(Tw − Ti ) ∫0 exp − 9 9 )1∕3 ) ( ( 3 3 μ cp 1∕3 ( D )1∕3 2 R ρ cp 4V 1∕3 ρ V D 8 . αz k R μ k z = = ( 3) ( 3) x x ∞ ∞ ∫0 exp − ∫0 exp − dx dx 9 9 and finally, we get ( )1∕3 D NuD = 1.077 (Pe)1∕3 z

(8.516)

where Pe = RePr is the Peclet number. The average heat transfer coefficient h is ] L L[ ( )1∕3 D 1 1 k (1.077) h= h dz = (Pe)1∕3 dz L ∫0 L ∫0 D z L

h=

k 1 dz (1.077) × D1∕3 × (Pe)1∕3 × D L ∫0 z1∕3

h=

k 3 (1.077) × D1∕3 × (Pe)1∕3 × L2∕3 D 2

The average Nusselt number is ] [( )1∕3 D NuD = 1.615 (Pe)1∕3 L Gz−1 D ≤ 0.01 −6 This result is reported to be very accurate for Gz−1 L = L∕(D Re Pr ) ≤ 5 × 10 .

(8.517a)

405

406

8 Laminar Momentum and Heat Transfer in Channels

r Ti

q″w

T = Ti

T(r, z)

vz = V

0

z

D = 2R

q″w q″w = 0 z

0 (a) Figure 8.37

(b)

(a) Poiseuille flow in circular tube being heated by a uniform heat flux, (b) Step change in surface heat flux.

8.6.1.3 Graetz Problem: HFD and Thermally Developing Flow for Viscous Flow in Circular Tube Under Uniform Wall Heat Flux Boundary Condition

We will repeat the Graetz thermal entrance problem here for UHF boundary condition. Consider again axisymmetric, fully developed low-velocity laminar flow of a viscous fluid in a tube. The temperature of the fluid is uniform over the flow cross section until the heat transfer begins. Fluid enters the tube at uniform velocity, and eventually, a fully developed velocity profile is established before heating begins. We assume that z = 0 is the location where the velocity profile is fully developed. See Figure 8.37a. On the other hand, Figure 8.37b shows a step change in heat flux at the point where heating begins. The Fully developed velocity profile for the circular tube is given as [ ( )2 ] r vz (r) = 2V 1 − (8.517b) R The inlet fluid temperature is Ti for z < 0. The downstream (z > 0) half is at uniform wall heat flux q′′w . The axial heat conduction and viscous dissipation effects are neglected. There is no internal energy generation. Physical properties of fluid are constant. We wish to determine the temperature distribution and the Nusselt number. The energy equation and its boundary conditions are given as ] ( ) [ 2 r2 𝜕T 𝜕 T 1 𝜕T (8.518) 2V 1 − 2 + =α 𝜕z r 𝜕r 𝜕r2 R T(0, r) = Ti

(8.519a)

𝜕T(z, 0) = 0 (Symmetry) (8.519b) 𝜕r 𝜕T(z, R) q′′w = (8.519c) 𝜕r k We will use the dimensionless variables defined earlier to nondimensionalize the energy equation and its boundary conditions. The dimensionless variables are r (8.520a) η= R 2(z∕D) 1 z z+ = = (8.520b) ReD Pr R ReD Pr T(z, r) − Ti (8.520c) θ(z+ , η) = q′′w R∕k Using these dimensionless variables, we obtain the dimensionless form of the energy equation and its boundary conditions 𝜕θ 𝜕 2 θ 1 𝜕θ (1 − η2 ) + = 2 + (8.521a) 𝜕z η 𝜕η 𝜕η z+ = 0 θ = 0 𝜕θ =1 𝜕η 𝜕θ = 0 or θ is finite z+ > 0 η = 0 𝜕η

z+ > 0 η = 1

(8.521b) (8.521c) (8.521d)

8.6 Heat Transfer in the Thermal Entrance Region

Partial differential equation, Eq. (8.521a), has a nonhomogeneous boundary condition, Eq. (8.521c) and separation of variables cannot be applied directly. The solution can be obtained by applying a special procedure, as described in [9] θ(z+ , η) = ψ(z+ , η) + ⏟⏟⏟ decaying initial transient

ϕ(η) ⏟⏟⏟ radial temperature variation to let wall heat flux into fluid

φ(z+ ) ⏟⏟⏟

+

(8.522)

axial temperature rise due to accumulated heat flux

Notice that the sum of ϕ(η) and φ(z+ ) is the TFD temperature and ψ(z+ , η) is the thermal entrance region solution of the problem. We are actually using the superposition principle to obtain the solution. The substitution of the assumed solution, Eq. (8.522), into partial differential equation, Eq. (8.521a), yields the following differential equations: (1 − η2 )

𝜕ψ 𝜕 2 ψ 1 𝜕ψ = + 𝜕z+ η 𝜕η 𝜕η2

ψ(0, η) = −ϕ(η) − φ(0) 𝜕ψ =0 𝜕η 𝜕ψ =0 z+ > 0 η = 1 𝜕η

z+ > 0 η = 0

(8.523) (8.524a) (8.524b) (8.524c)

where ψ represents the developing temperature distribution. Thus, we see that ψ satisfies the PDE and the inlet boundary condition. We can state that ψ → 0 as z+ → ∞. When the heat flux is uniform, it is known that for large values of z+ , that is z+ → ∞, there is a fully developed temperature distribution. First, we will determine this fully developed temperature distribution. Since ψ satisfies the PDE, the sum of ϕ(η) and φ(z+ ) also satisfies the PDE. The next set of differential equations is dφ d2 ϕ dϕ 1 1 = + = C0 + (1 − η2 ) dη2 η(1 − η2 ) dη dz

(8.525)

Functions produced by the differential equations must match each other. For this reason, both the differential equations are set equal to the same constant. This constant is denoted by C0 and the differential equations for ϕ(η) and φ(z+ ) are given below along with their boundary conditions d2 ϕ dϕ 1 1 = C0 + (1 − η2 ) dη2 η(1 − η2 ) dη dϕ =0 η = 0 ϕ = finite or dη dϕ =1 η=1 dη

(8.526) (8.527a) (8.527b)

and dφ = C0 (8.528) dz+ The nonhomogeneous boundary condition on the tube wall is satisfied by ϕ(η). Integrating the differential equation for ϕ, we get η4 C0 η2 C0 + + C1 ln(η) + C2 (8.529) 16 4 The application of the boundary condition at η = 0 ϕ = finite or dϕ/dη = 0 yields C1 = 0. The application of the second boundary condition η = 1 dϕ/dη = 1 indicates that C0 = 4. Now, the temperature distribution ϕ(η) becomes ϕ=−

η4 + η 2 + C2 4 On the other hand, the integration of the differential equation for φ(z+ ) yields ϕ(η) = −

(8.530)

φ(z+ ) = C0 z+ + C3 = 4z+ + C3

(8.531)

407

408

8 Laminar Momentum and Heat Transfer in Channels

Combining Eqs. (8.530) and (8.531), fully developed temperature distribution θfd becomes η4 + η 2 + C4 (8.532) 4 where C4 = C3 + C2 is a new constant to be determined. It is not necessary to determine the constant C4 in order to calculate the Nusselt number. But we will determine this constant to complete the temperature distribution. In order to evaluate this constant, we will use the bulk temperature Tm . Heat flux at the tube wall will cause axial variation in the mean or bulk temperature. Consider the bulk temperature Tm θfd → 4z+ −

R

Tm =

∫0 vz Tr dr

(8.533)

R

∫0 vz r dr

Using the dimensionless variables, the bulk temperature becomes ( R R[ 1[ ( )2 )] ( )] VR2 r 2V 1 − η2 (Rη) Rdη = 2V 1 − r dr = vz r dr = ∫0 ∫0 ∫0 R 2 ( ] R 1[ ( )2 )] [( q′′ D ) r w 2V 1 − v Tr dr = θ + Ti (Rη) Rdη ∫0 ∫0 z R k ) ] [( 1[ 1[ ( )] ( )] [ ] q′′w D θ (Rη) Rdη + = 2V 1 − η2 2V 1 − η2 Ti (Rη) Rdη ∫0 ∫0 k ( ′′ ) 1( ) ) qw D ( 2 ) 1 ( = [2V] 1 − η2 θ η dη + (2V) R2 Ti 1 − η2 η dη R ∫0 ∫0 k ( ′′ ) ) qw D ( 2 ) 1 ( VR2 Ti 1 − η2 θ η dη + R = [2V] ∫ k 2 ( ′′ ) ( ) 0( ) 2 qw D 1 VR T 2 R ∫0 1 − η2 θ η dη + 2 i [2V] k Tm = 2 VR 2

Finally, after simplification, we obtain the dimensionless mean temperature θm θm =

1 Tm − Ti = 4 (1 − η2 ) θ η dη ∫0 q′′w R∕k

We substitute Eq. (8.532) into Eq. (8.534). Evaluating the bulk temperature, we find that ) 1( η4 7 + 2 + η + C4 η(1 − η2 )dη = + 4z+ + C4 4z − θm = 4 ∫0 4 24

(8.534)

(8.535)

as z+ → ∞. This is the bulk temperature for the fully developed region. We need another expression for the bulk temperature. This is obtained by writing an energy balance for a control volume in pipe. This energy balance is q̇ ′′ 2πR dTm = w ̇ p mc dz

(8.536)

Tm (0) = Ti

(8.537)

Again, using dimensionless variables, the differential equation, Eq. (8.536), and its boundary condition, Eq. (8.537), become dθm =4 (8.538) dξ θm (0) = 0

(8.539)

This differential equation, Eq. (8.538), has the following solution: θm =

Tm − Ti = 4z+ q′′w R∕k

(8.540)

7 . Notice that far downstream from the Comparing two equations for the bulk temperature θm , we obtain that C4 = − 24 beginning of the heated section, constant heat flux will cause a rise in the fluid temperature, which is linear in z+ . The

8.6 Heat Transfer in the Thermal Entrance Region

shape of the profiles as a function of η will not change with increase in z+ . Now, for z+ → ∞, fully developed temperature distribution is η4 7 θfd = φ(ξ) + ϕ(η) = 4z+ + η2 − − (8.541a) 4 24 or ] [ Tfd − Ti η4 (z∕R) 7 2 + η − (8.541b) − = 4 Re Pr 4 24 q′′w R∕k In addition to satisfying the conservation of energy, Eq. (8.541b) includes the following conditions: ( ) ) ( 𝜕Tfd 𝜕Tfd 𝜕Tfd q′′ = const =0 = w = const 𝜕z 𝜕r r=0 𝜕r r=R k Now, we can construct the dimensionless temperature distribution for any value of z+ as follows: η4 7 − (8.542) 4 24 At this point, we can compute the Nusselt number NuD for large z+ . The Nusselt number is determined from its definition θ(z+ , η) = ψ(z+ , η) + 4z+ + η2 −

NuD =

q′′w h(2R) 2R = k Tw − Tm k

(8.543)

Using dimensionless variables, we can write the following equations for wall temperature and mean fluid temperature: ( ′′ ) qw R θw + Ti (8.544) Tw = k ( ′′ ) qw R θm + Ti (8.545) Tm = k Now, the Nusselt number becomes 2 2 48 NuD = = ( = 4.3636 (8.546) = ) 1 7 θw (z+ , 1) − θb (z+ ) 11 + + 4z + 1 − − − 4z 4

24

It remains to determine the ψ(z+ , η) solution. This is obtained as follows: 𝜕ψ 𝜕 2 ψ 1 𝜕ψ = + 𝜕z+ η 𝜕η 𝜕η2 [ ] 2 η 7 2 ψ(0, η) = − η − − 4 24 𝜕ψ z+ > 0 η = 0 =0 𝜕η 𝜕ψ z+ > 0 η = 1 =0 𝜕η We can now apply the separation of variables and apply a product solution of the form (1 − η2 )

ψ = Z(z+ )ℜ(η)

(8.547) (8.548a) (8.548b) (8.548c)

(8.549)

Noting that only the radial direction will give us a characteristic value problem, with the proper selection of separation constant, the product solution will result in two differential equations as follows: dZ + λ2 Z = 0 (8.550) dz+ The differential equation for the characteristic value problem is d2 ℜ 1 dℜ + λ2 (1 − η2 )ℜ = 0 + η dη dη2 dℜ =0 η=0 dη dℜ =0 η=1 dη

(8.551) (8.552a) (8.552b)

409

410

8 Laminar Momentum and Heat Transfer in Channels

The solution of Eq. (8.550) is Z = Ce−λ

2

z+

(8.553)

The solution of Eq. (8.551) will be presented next. We know that we may implicitly express the general solution as (o) R(η) = A G(e) 0 (λ η) + B G0 (λ η)

(8.554)

G(e) 0

(e) and G(e) 0 represent Graetz functions first kind and second kind, order zero, respectively. Function G0 is an even (o) and G0 is an odd function. Taking the first boundary condition into consideration and noting that function G(o) 0

where function is unbounded at η = 0, we choose B = 0. Let us represent the characteristic functions (eigenfunctions) by ϕn = G(e) 0

(8.555)

The nonhomogeneous condition is left to the end of the problem. The solution of this characteristic value problem, Eq. (8.551), is determined by Maple 2020, and we obtain the characteristic functions. > restart; > > #HFD thermally developing Graetz prublem under uniform heat flux boundary condition > #Velocity profile is parabolic > #We represent the dependent variable by R >

( ) 𝟏 • ′ R (𝛈) + 𝛌𝟐 • (𝟏 − 𝛈𝟐 ) • R(𝛈) = 𝟎; > de ≔ R (𝛈) + 𝛈 D(R)(𝛈) de ≔ D(𝟐) (R)(𝛈) + + 𝛌𝟐 (−𝛈𝟐 + 𝟏)R(𝛈) = 𝟎 𝛈 > #𝛌 is characteristic value to be determined ′′

> #We now use the first boundary condition > > sol ≔ rhs(dsolve({de, R′ (𝟎)}, R(𝛈))); ) ( 𝛌 𝛈𝟐 𝟏 𝛌 − , 𝟏, 𝛌 𝛈𝟐 sol ≔ C1 e− 𝟐 KummerM 𝟐 𝟒 > > #Characteristic functions are ( ) 𝛌 𝛈𝟐 𝟏 𝛌 − , 𝟏, 𝛌 𝛈𝟐 ; > M ≔ e− 𝟐 KummerM ) ( 𝟐 𝟒 𝛌 𝛈𝟐 𝟏 𝛌 M ≔ e− 𝟐 KummerM − , 𝟏, 𝛌 𝛈𝟐 𝟐 𝟒 The eigenfunction is now defined as ) ( λ η2 1 λn − n2 2 φn = G(e) (8.556) = M(λ η) = e KummerM η − , 1, λ n n 0 2 4 ) ( λ where M(λn η) = KummerM 12 − 4n , 1, λn η2 is the hypergeometric Kummer function. We are now in a position to determine the characteristic values of the Graetz function. This will be done by Maple 2020. We will use the boundary conditions η = 1 dℜ/dη = 0 to obtain the characteristic values. > restart; > #Characteristic values of Eq.(𝟖.𝟓𝟓𝟏) > Graetz prublem under uniform heat flux boundary condition

8.6 Heat Transfer in the Thermal Entrance Region

> #Velocity profile is parabolic > #We represent the dependent variable by R > > #𝛌 is characteristic value to be determined > > #Characteristic values and characteristic function are ) ( 𝛌 𝛈𝟐 𝟏 𝛌 > M ≔ e− 𝟐 KummerM − , 𝟏, 𝛌 𝛈𝟐 ; ) ( 𝟐 𝟒 𝟐 𝟏 𝛌 − 𝛌 𝟐𝛈 − , 𝟏, 𝛌 𝛈𝟐 KummerM M≔e 𝟐 𝟒 > #We now wish to evaluate characteristic values.We now use second boundary condition 𝛈 = 𝟏 R′ = 𝟎 > > eq ≔ subs(𝛈 = 𝟏, diff(M, 𝛈)); ) ) ( ( (( 𝛌 𝛌 𝟏 𝛌 𝟏 𝛌 𝟑𝛌 𝟏 − , 𝟏, 𝛌 + 𝟐 e− 𝟐 − KummerM − , eq ≔ −𝛌 e− 𝟐 KummerM 𝟐 𝟒 𝟒 𝟐 𝟐 𝟒 ) ( )) ( 𝟏 𝛌 𝟏 𝛌 + KummerM − − , 𝟏, 𝛌 𝟏, 𝛌) + 𝟐 𝟒 𝟐 𝟒 Roots of this nonlinear equation will give us the eigenvalues. The graphical method is frequently used to find the real roots of a function. It does not give information about the complex roots. The method is fairly simple and involves sketching the function in the desired range. We will use the graphical method with the command fsolve in MAPLE 2020. Plot of this nonlinear equation is shown in Figure 8.38. The intersection point of the curve with x-axis shows the location of the root. We get now an estimate of the root from the figure to use with the fsolve command.

1

0.5

0

2

4

λ

6

–0.5

–1 Figure 8.38

Location of characteristic values.

8

10

411

412

8 Laminar Momentum and Heat Transfer in Channels

Plot of this nonlinear equation is shown in Figure 8.38. Figure 8.38 shows the location of a few eigenvalues. > with(plots) ∶ > > plot(eq, 𝛌 = 𝟎..𝟏𝟎); > #First eigenvalue > 𝛌[𝟏] ≔ fsolve(eq, 𝛌 = 𝟒..𝟔); 𝛌𝟏 ≔ 𝟓.𝟎𝟔𝟕𝟓𝟎𝟓𝟓𝟎𝟏 > #Second eigenvalude > 𝛌[𝟐] ≔ fsolve(eq, 𝛌 = 𝟖..𝟏𝟎); 𝛌𝟐 ≔ 𝟗.𝟏𝟓𝟕𝟔𝟎𝟔𝟒𝟐𝟔 > #Third eigenvalue > 𝛌[𝟑] ≔ fsolve(eq, 𝛌 = 𝟖..𝟏𝟎); 𝛌𝟑 ≔ 𝟗.𝟏𝟓𝟕𝟔𝟎𝟒𝟐𝟔 In general, we may denote the characteristic functions as ) ( λ η2 1 λn − n2 2 − , 1, λn η φn = M(λn η) = e KummerM 2 4

(8.557)

We are now in a position to write the product solution ψ(ζ, η) =

∞ ∑ ) ( an φn (λn η) exp −λ2n z+

(8.558)

n=1

Finally, we will use the inlet condition of the problem in order to determine an ] [ ∞ ∑ η2 7 − = − η2 − an φn (λn η) 4 24 n=1

(8.559)

The characteristic functions are orthogonal with respect to the weighting function η(1 − η2 ). Therefore, unknown constant an is [ ] 2 1 7 ∫0 η2 − η4 − 24 η(1 − η2 )φn (λn η)dη (8.560) an = 1 ∫0 η(1 − η2 )[φn (λn η)]2 dη We can now write the complete temperature distribution as ∑ ( ) η4 7 − + an φn (λn η) exp −λ2n ξ 4 24 n=1 ∞

θ(ξ, η) = 4z+ + η2 −

(8.561)

An interest is the longitudinal variation of pipe wall temperature. Pipe wall temperature distribution can be obtained by setting η = 1 in this equation, and the result is ) ( 11 ∑ + a φ | exp −λ2n z+ 24 n=1 n n |η=1 ∞

θw (z+ , 1) = 4z+ + or Tw − Ti =4 q′′w R∕k

(

z∕R ReD Pr

)

) ( ∞ λ2n z 11 ∑ | + + a φ exp − 24 n=1 n n |η=1 Red Pr R

(8.562)

(8.563)

Using Maple 2020, the first 10 eigenvalues and eigenfunctions are obtained. Then, the an was evaluated from Eq. (8.560). The eigenvalues and other important quantities are given in Table 8.6.

8.6 Heat Transfer in the Thermal Entrance Region

Table 8.6

Eigenvalues and eigenfunctions.

n

𝛌n

𝛗n (1)

an

1

5.067505501

−0.4925165736

0.4034832180

2

9.157606426

0.3955084753

−0.1751100009

3

13.19722474

−0.3458736768

0.1055917224

4

17.22022936

0.3140464817

−0.0732824041

5

21.23551728

−0.2912514573

0.0550365066

6

25.24653118

0.2738069527

−0.0434843844

7

29.25490555

−0.2598530271

0.03559511179

8

33.26152373

0.2483319661

−0.0299084743

9

37.26690821

−0.2385903994

0.02564012046

10

41.27138935

0.2301992523

−0.0223337051

Shah and London [3] report approximate formulas for higher eigenvalues and constants 4 3

λn = 4 n +

(8.564a) −5∕3

an φn (1) = −2.401006045 λn

(8.564b)

The local Nusselt number may be determined from the definition NuD =

q′′w h (2R) h D D = = k k Tw − Tm k

(8.565)

Using dimensionless temperature θ, this equation may be expressed as NuD =

2 θw − θm

(8.566)

where θw and θm are the wall and mean fluid temperatures, respectively. The mean fluid temperature is [∞ ] 1 1 [ ] ∑ ( 2 +) 7 2 + 2 2 η(1 − η ) 4z + η − η(1 − η ) an exp −λn z φn (λn ) dη θm = 4 dη + 4 ∫0 ∫0 24 n=1 = 4z+

(8.567)

Substituting wall temperature and mean fluid temperature equations into Eq. (8.562), we get the local Nusselt number 2

NuD = 11 24

∑ ∞

+

n=1

) ( an φn ||η=1 exp −λ2n z+

(8.568)

The average Nusselt number NuD is in general based on the average difference between the wall and mixing cup temperatures, as discussed in [9] z+

1 Tw − Tm = + (Tw − Tm )dξ z ∫0 where ξ is the dummy variable for integration. The average heat transfer coefficient h is defined as q′′w

h=

(Tw − Tm ) where q′′w is the constant heat flux and the average Nusselt number NuD is NuD =

q′′w (2R) h(2R) 2 = = k k(Tw − Tm ) θw − θm

413

414

8 Laminar Momentum and Heat Transfer in Channels

Next, we evaluate θw − θm [ ] ∞ z+ ( 2 ) 1 11 ∑ | + θw − θm = + a φ exp −λn ξ dξ z ∫0 24 n=1 n n |η=1 { [ ) ]} ( ∞ ∑ 1 − exp −λ2n ξ 1 11 | θw − θm = + an φn |η=1 ξ+ z 24 λ4n n=1 ξ=z+ { [ ) ]} ( ∞ 2 1 − exp −λn ξ 11 ∑ + θw − θm = an φn ||η=1 24 n=1 λ4n z+ Finally, the average Nusselt number becomes NuD =

48∕11 {[ ( )] } 1 − exp −λ2n z+

24 ∑ 1+ a φ | 11 n=1 n n |η=1 ∞

(8.569)

λ4n z+

The variation of the local Nusselt number NuD and the mean Nusselt number NuD along the tube has been presented in Table 8.7. Sellars et al. [31] give the following correlation for the local Nusselt number NuD for short tubes. The tubes are HFD and thermally developing laminar flow with UHF heating ) ) ( ( ReD Pr 1∕3 ReD Pr −1 ; ≤ 0.0005 NuD ≈ 1.301 z∕D z∕D q′′w = const Table 8.7

(8.570)

Nusselt numbers for uniform heat flux and parabolic velocity profile.

z+ = 2(z/D)/ReD Pr

NuD

NuD

0.000002

129.203

194.2

0.000004

102.360

153.9

0.00001

75.190

113.2

0.000014

67.124

101.1

0.00002

59.51

89.68

0.00004

47.077

71.01

0.00008

37.224

56.20

0.0001

34.511

52.12

0.0002

27.275

41.24

0.0004

21.555

32.61

0.001

15.813

23.90

0.002

12.538

18.91

0.003

10.967

16.50

0.006

8.7724

13.10

0.01

7.4937

11.08

0.016

6.5359

9.537

0.02

6.1481

8.896

0.04

5.1984

7.241

0.08

4.6213

6.043

0.1

4.5139

5.747

0.2

4.3748

5.082

0.3

4.3645

4.845

0.4

4.3637

4.724

Source: Shah and London [3]/with permission of Elsevier.

8.6 Heat Transfer in the Thermal Entrance Region

8.6.1.4 Empirical and Theoretical Correlations for Viscous Flow in the Thermal Entrance Region of the Pipe

Classical solutions to the Graetz problem are obtained by the separation of variables and separation of variables yields series solutions. The thermal entrance problem was investigated by several researchers, such as Shah and London [3], Brown [25], Sellars et al. [31], and Shah [32], and their work is well known and is recorded in the literature. The complexity of the solution of the Graetz problem has motivated the researchers to develop simple correlations. Constant wall temperature Bennett [33] reports that Hausen proposed an empirical single relation for laminar flow in an

isothermal pipe for the thermal entrance problem. The average Nusselt number in thermally developing flow can be determined from 0.0668∕Gz−1 hD L = 3.657 + [ NuD = (8.571a) ( −1 )2∕3 ] k 1 + 0.04∕ GzL TW = const D = 2R Gz−1 L =

L D ReD Pr

Properties are evaluated at Tm = (Tmi + Tmo )/2 Tmi = mean inlet temperature Tmo = mean outlet temperature Bennet [13] reports that Eq. (8.571a) deviates from the exact solution by 0.4–13%. We may modify Eq. (8.571a) to take care of variations in viscosity 0.0668 ( )0.14 Gz−1 μ L NuD = 3.657 + [ (8.571b) ] μw 0.04 1+ ( )2∕3 Gz−1 L where μw is the viscosity evaluated at the wall temperature. Shah [32] presented the following correlations as alternative to Graetz solutions for local and mean Nusselt numbers: Local Nusselt number ⎧ 1.077 −1 ⎪( )1∕3 − 0.7 for Gzz ≤ 0.01 ⎪ Gz−1 z NuD = ⎨ ] [ 6.874 for Gz−1 exp −57.2Gz−1 3.657 +( ⎪ ) z z > 0.01 0.488 −1 3 ⎪ 10 Gzz ⎩ Tw = const D = 2R Gz−1 z =

(8.572)

z D ReD Pr

Mean Nusselt number ⎧ 1.615 ⎪ ( −1 )1∕3 − 0.7 ⎪ GzL ⎪ 1.615 NuD = ⎨ ( −1 )1∕3 − 0.2 ⎪ GzL ⎪3.657 + (0.0499) ⎪ Gz−1 L ⎩

for Gz−1 L ≤ 0.005 for 0.005 < Gz−1 L < 0.03 for Gz−1 L ≥ 0.03

Tw = const D = 2R Gz−1 L =

L DReD Pr

This correlation deviates from the exact solution by −2.2% to 2% as reported by Bennett [33].

(8.573)

415

416

8 Laminar Momentum and Heat Transfer in Channels

Churchill and Ozeo [16] proposed a single correlation for local Nusselt number in the thermal entrance region of an isothermal pipe: ]3∕8 [ ) ( 388 −1 −8∕9 Gzz − 1.7 (8.574) NuD = 5.357 1 + π Tw = const z Gz−1 z = D ReD Pr The magnitude of the offset 1.7 is decided empirically. The authors report agreement between this correlation and Graetz solution to within 6%. Bennett [32] reports that this correlation is in error by −4.7% to 8.7% for 10−6 ≤ Gz−1 z ≤ 1. The calculation of the Nusselt number for small values of the dimensionless length requires several terms. Stephan [34] presented correlation approximating the Graetz solution for convection in an isothermal pipe: ( −1 ) 0.0499 3.657 NuD = [ ( −1 )1∕3 ( −1 )2∕3 ] + (Gz−1 ) tanh GzL tanh 2.264 Gz + 1.7 Gz L L

L

Tw = const L 1 Gz−1 L = D ReD Pr

(8.575)

Stephan asserts that the exact values in the range 0 ≤ Gz−1 L ≤ ∞ will be reproduced with the largest error of 1%. It has −6 ≤ 5 × 10 , this equation reduces to been asserted by Stephan that for values of Gz−1 L ( −1 )−1∕3 NuD = 1.615 GzL Tw = const Equation (8.576) is valid short tubes. Gnielinski [19] proposed a correlation for local Nusselt numbers for isothermal pipe flow. His correlation is }1∕3 {{ }3 1∕3 3 3 1.077Gzz − 0.7 + 0.7 + (3.66) NuD =

(8.576)

(8.577)

( ) D ReD Pr Gzz = z Tw = const This correlation is within 0.5–5.7% of the exact solution, as reported by Bennett [33], and a correlation for the average Nusselt number is also proposed by Gnielinski as }1∕3 {[ ]3 1∕3 NuD = 1.615 GzL − 0.7 + 0.73 + (3.66)3 (8.578) GzL =

( ) D ReD Pr L

Tw = const This correlation is within −0.7% to 1.4% of the exact solution, as reported by Bennett [33]. Shome and Jensen [36] made use of their numerical solution to the Graetz problem and proposed the following correlation for the average Nusselt number for the isothermal pipe: −3 ⎧−0.5632 + ( 1.571 10−6 < Gz−1 )0.3351 L < 10 ⎪ Gz−1 L ⎪ 1.129 −2 ⎪0.9828 + ( 10−3 < Gz−1 ) L < 10 NuD = ⎨ −1 0.3686 ] [GzL ⎪ ) ( 0.1272 ⎪ −1 −2 Gz−1 L > 10 ⎪3.6568 + ( −1 )0.7373 exp −3.1563 GzL GzL ⎩

Bennet indicates that this isothermal pipe correlation agrees within −0.3% to 0.5% of the exact solution.

(8.579)

8.6 Heat Transfer in the Thermal Entrance Region

Bennet [33] proposed the following correlation for heat transfer in a circular pipe with constant wall temperature. The Average Nusselt number is [ ]0.295 NuD = 5.079 Gz1.13 + 146.9 − 0.701 L Tw = const

(8.580)

where Gz−1 L = L∕(D ReD Pr ). Bennet indicates that this isothermal pipe correlation agrees within −0.9% to 1.4% of exact solutions. Uniform heat flux Churchill and Ozeo [17] proposed the following correlation for the local Nusselt number for laminar flow heat transfer in a pipe. The pipe has a constant heat flux wall boundary condition: ]3∕10 [ ( ) NuD + 1 220 −1 −10∕9 = 1+ Gzz 5.364 π

q′′w = const Gz−1 z

(8.581)

z 1 D ReD Pr

where = . Bennet [33] reports that this relation differs from the exact solution by −3.4% to 4%. Shah and London [3] recommend the following equations for constant wall heat flux q′′w . These equations are for local Nusselt numbers in thermally developing laminar flow in a pipe ) ( −5 ⎧1.302 Gz−1 −1∕3 − 1 for Gz−1 z ≤ 5 × 10 ⎪ ( z−1 )−1∕3 ( −1 ) −5 (8.582) NuD = ⎨1.302 Gzz − 0.5 for 5 × 10 ≤ Gzz ≤ 0.0015 ⎪4.364 + 8.68(103 Gz−1 )−0.506 exp [−41 (Gz−1 )] for (Gz−1 ) > 0.0015 ⎩ z z z q′′w = const Gz−1 z =

z D ReD Pr

Bennett [13] reports that this correlation agrees within −1% to 0.9% of the exact solution. Gnielinski [19] proposed the following correlations for the local and average Nusselt numbers for pipes of constant wall heat flux q′′w : Local Nusselt number { ( )3 }1∕3 [ ]3 48 1∕3 3 NuD = 1.302 (Gzz ) − 1 + 1 + (8.583) 11 q′′w = const Gz−1 z =

z D ReD Pr

Average Nusselt number {[ ]3 ( )3 }1∕3 48 1∕3 NuD = 1.953 GzL − 0.6 + 0.63 + 11 L −1 GzL = D ReD Pr

(8.584)

Recall that Nu∞ = 48/11 is the fully developed value of the Nusselt number for pipes with walls of constant heat flux q′′w . Bennett reports that local Nusselt number correlation is within −4% to 2.8% and average Nusselt number is within 3.2–14% of the exact solution. Shah [32] recommends the following relations for the mean Nusselt number for the thermal entrance region problem. Laminar flow in a tube is subjected to UHF ) ( ⎧ ( 1.953 ≤ 0.03 for Gz−1 ) L 1∕3 ⎪ Gz−1 L NuD = ⎨ (8.585) ) ( 0.0722 > 0.03 ⎪4.364 + ( −1 ) for Gz−1 L ⎩ GzL Bennett [33] reports that this correlation is in error by −8.3% to 13% of the exact solution.

417

418

8 Laminar Momentum and Heat Transfer in Channels

Bennett [33] proposes the following correlation for the average Nusselt number for convective heat transfer in a pipe with walls of constant heat flux q′′w : ( )0.2908 NuD = 6.664Gz1.146 + 240.1 − 0.559 (8.586) L D Re Pr L D where GzL is the Graetz number, and it is asserted that this correlation agrees within –1.2% to 1% of exact solutions. GzL =

Example 8.8 Engine oil flows through a tube at a rate of 5 × 10−3 kg/s. The internal diameter of tube is 0.5 cm. The tube length is 2 m. A constant heat flux is applied over the entire tube surface. It is required to heat the engine oil from 10 ∘ C to 64 ∘ C. Determine: (a) the heat transfer coefficient at the tube exit (b) the constant heat flux required to heat the engine oil (c) the wall temperature at the exit Solution The average temperature Tm is Tmi + Tmo 10 + 64 = = 37 ∘ C = 310 K 2 2 The physical properties of engine are evaluated at the average temperature Tm = 310 K. Tm =

ρ = 877.9kg∕m3 cp = 1951 J∕kg.K μ = 25.3 × 10−2 N.s∕m2 Pr = 3400 k = 0.145W∕m.K First, we evaluate the Reynolds number ReD 4 × 5 × 10−3 4ṁ = = 5.0 πμD π × 25.3 × 10−2 × 0.005 Since ReD < 2300, the flow is laminar. In order to check whether the flow is fully developed or not, let us calculate the hydrodynamic and thermal entrance lengths ReD =

LH = 0.05 ReD D = 0.05 × 5 × 0.005 = 0.00125 m LT = 0.05 ReD D Pr = 0.05 × 5 × 0.005 × 3400 = 4.27 m The tube length L = 2 m. LH ≪ L and LT ≫ L may assume that flow is HFD and thermally developing. In a unique way of saying, we have assumed that the velocity is fully developed, and the temperature profile is developing. (a) Gnielinski correlation, Eq. (8.584), for the average Nusselt gives {[ ]3 ( )3 }1∕3 48 1∕3 3 NuD = 1.953 GzL − 0.6 + 0.6 + 11 where Gz−1 L = L∕(D ReD Pr ) ) ( D 0.005 ReD Pr = × 5 × 3400 = 42.77 L 2 { ( )3 }1∕3 48 NuD = [1.953 (42.77)1∕3 − 0.6]3 + 0.63 + 11 GzL =

NuD = 6.87 The heat transfer coefficient h is NuD =

) ( k 0.145 hD ⇒ h = NuD = (6.87) = 199.41 W∕m2 K k D 0.005

8.6 Heat Transfer in the Thermal Entrance Region

(b) The required heat flux is calculated next. We write an energy balance on the tube πDLq′′w = ṁ cp (Tmo − Tmi ) ṁ cp (Tmo − Tmi )

5 × 10−3 × 1951 × (64 − 10) = 16 767 W∕m2 πDL π × 0.005 × 2 (c) The tube wall temperature at the pipe wall is calculated next. Let Twe represent the pipe wall temperature at the exit q′′w =

Twe = Tmi +

q′′w

=

= 283 +

h

16 767 = 367 K = 94 ∘ C 199.41

8.6.2 Two Infinite Parallel Plates This problem was investigated by Leveque [30]. In the present analysis, laminar viscous fully developed laminar flow between parallel flat plates with a prescribed wall heat flux as well as constant surface temperature will be considered. These are the most common boundary conditions encountered in practice. Heat transfer occurs as a result of interaction between the velocity and temperature profiles of the fluid. For this reason, the rate of heat transfer depends on the shapes of the two profiles. Recently, the need for compact heat exchangers in industry has accelerated the work related to forced convection heat transfer in noncircular ducts. In this respect, heat transfer between two infinite parallel plates is important. 8.6.2.1 Graetz Problem: HFD and Thermally Developing Flow Between Parallel Plates Subjected to Constant Wall Temperature

Incompressible, constant property fluid enters steadily and uniformly at a velocity V between two parallel plates. The fluid is at an initial temperature Ti , and there is no heat transfer at the surface of the duct until a fully developed velocity profile is established. The temperature of the fluid is uniform over the channel flow cross section at the point where heat transfer begins. Fluid and channel walls are at the same temperature for x < 0. At some point x > 0, the wall temperature of the plate is changed to Tw . See Figure 8.39. At the point where the heating begins, the velocity profile is fully developed and is equal to [ ( y )2 ] 3 u= V 1− 2 b and the temperature profile begins to develop. We are considering laminar, incompressible, low-velocity flow with constant physical properties. Axial conduction and viscous dissipation are neglected, and there is no internal energy generation. We wish to determine temperature distribution and Nusselt number in the thermal entrance region. We will make the following assumptions. (1) (2) (3) (4) (5) (6)

The y-components of the velocity, v, is everywhere zero. ρ, cp , k is uniformly constant throughout the fluid. Ti is constant. T = Ti if x < 0, T = Tw if x > 0. The temperature of the fluid in contact with the plates is the same as that of the plates for x < 0. Assume that Pe > 100 and this means axial conduction is zero in the direction of fluid motion. y

Ti

u(y)

Ti u=V

Tw

O

2b

Ti Ti

Figure 8.39

Tw

x

Graetz problem for parallel plates.

Tw

0

x

419

420

8 Laminar Momentum and Heat Transfer in Channels

(7) Viscous dissipation is neglected. (8) There is no energy generation. The energy equation to be solved in the thermal entrance region with parabolic velocity distribution between the two parallel walls separated by a distance 2b is given by [ ( y )2 ] 𝜕2 T 𝜕T 3 V 1− =α 2 (8.587) 2 b 𝜕x 𝜕y where V is the mean flow velocity. The boundary conditions are x = 0; T = Ti

(8.588a)

𝜕T =0 𝜕y

(8.588b)

y = b; T = Tw

(8.588c)

y = 0;

Introducing the following dimensionless quantities: T − Tw Ti − Tw y η= b 4(x∕b) x+ = ReDH Pr θ=

(8.589a) (8.589b) (8.589c)

where ReDH = VDH ∕ν is the Reynolds number and DH = 4b is the hydraulic diameter, obtain the dimensionless form of the energy equation: 𝜕θ 𝜕2 θ 3 (1 − η2 ) + = 2 2 𝜕x 𝜕η

(8.590)

The boundary conditions in dimensionless form are given by θ(0, η) = 1

(8.591a)

𝜕θ(x+ , 0) = 0 or θ(x+ , 0) = finite x+ > 0 𝜕η

(8.591b)

θ(x+ , 1) = 0

(8.591c)

This problem can be solved by the method of separation of variables. Sellars et al. [30] presented a detailed solution of the problem. Prins et al. [36] presented a series solution and were able to obtain first three eigenvalues and eigenfunctions assuming a product solution θ = X(x+ ) Y(η)

(8.592)

and the application of Eqs. (8.590)–(8.592) yields d2 Y 1 3 1 dX = = −λ2 + 2 X dx Y(1 − η2 ) dη2

(8.593)

Selecting the proper sign for separation constant λ, we obtain two ODEs. The first differential equation is 2 dX + λ2 X = 0 dx+ 3 and the second differential equation and its boundary conditions are d2 Y + λ2 (1 − η2 )Y = 0 dη2 dY(0) =0 dη Y(1) = 0

(8.594)

(8.595) (8.596a) (8.596b)

8.6 Heat Transfer in the Thermal Entrance Region

The solution of the first differential equation, Eq. (8.594), is ) ( 2 X = A exp − λ2 x+ 3

(8.597)

The second ODE, Eq. (8.595), is of the Sturm–Liouville type eigenvalue problem. The weighting function is (1 − η2 ). Let us study the nature of the solution with Maple 2020. For this purpose, we generate a series solution. > restart; > > #Graetz problem for laminar flow between parallel plates UWT - Series >

#We will take only first 10 terms of the series solution to study the nature of the solution

> > Order ≔ 𝟏𝟎; Order ≔ 𝟏𝟎 > #Differential equation > de ≔ Y′′ (𝛈) + (𝟏 − 𝛈𝟐 ) • 𝛌𝟐 • Y(𝛈) = 𝟎; de ≔ D(𝟐) (Y)(𝛈) + (−𝛈𝟐 + 𝟏)𝛌𝟐 Y(𝛈) = 𝟎 > #We now obtain series solution > > sol ≔ dsolve({de, Y′ (𝟎) = a[𝟎], Y(𝟎) = a[𝟏]}, Y(𝛈), series) ∶ > F ≔ convert(sol, polynom) ∶ > F ≔ rhs(collect(F, {a[𝟎], a[𝟏]})); ( ( ) ( 𝟏 𝟒 𝟏 𝟏 𝟔 𝟏 𝛌 + 𝛌𝟐 𝛈𝟓 + − 𝛌 F ≔ 𝛈 − 𝛌𝟐 𝛈𝟑 + 𝟔 𝟏𝟐𝟎 𝟐𝟎 𝟓𝟎𝟒𝟎 ( ) 𝟏𝟑 𝟒 𝟕 𝟏 𝟏𝟕 𝟔 𝛌 𝛈 + 𝛌𝟖 + 𝛌 − 𝟐𝟓𝟐𝟎 𝟑𝟔𝟐𝟖𝟖𝟎 𝟗𝟎𝟕𝟐𝟎 ) ) ( ( 𝟏 𝟒 𝟗 𝟏 𝟏 𝟒 𝛌 𝛈 a𝟎 + 𝟏 − 𝛌𝟐 𝛈𝟐 + 𝛌 + 𝟏𝟒𝟒𝟎 𝟐 𝟐𝟒 ( ( ) ) 𝟏 𝟏 𝟔 𝟕 𝟒 𝟔 𝟏 + 𝛌𝟐 𝛈𝟒 + − 𝛌 − 𝛌 𝛈 + 𝛌𝟖 𝟏𝟐 𝟕𝟐𝟎 𝟑𝟔𝟎 𝟒𝟎𝟑𝟐𝟎 ) ) 𝟏𝟏 𝟔 𝟏 𝟒 𝟖 𝛌 + 𝛌 𝛈 a𝟏 + 𝟏𝟎𝟎𝟖𝟎 𝟔𝟕𝟐 We now see that the solution is composed of an even function and an odd function. We may implicitly express the general solution as (e) Y(η) = A G(e) 0 (λη) + B G0 (λη)

(8.598)

(0) where G(e) 0 and G0 denote the even and odd functions, respectively. The first boundary condition (symmetry boundary condition) implies that B = 0, and thus, we reject the odd function. The characteristic functions are

φn (η) = G(e) 0 (λn η)

(8.599)

The second boundary condition gives the characteristic values G(e) 0 (λn ) = 0

(8.600)

421

422

8 Laminar Momentum and Heat Transfer in Channels

The eigenvalues are the roots of this equation, Eq. (8.600). Let us evaluate the first few eigenvalues using Maple 2020. > restart; > #Eigenvalues of Eq.(8.595) > #Graetz problem for laminar flow between parallel plates UWT-Series >

#We will take only first 10 terms of the series solution to study the nature of the solution

> > Order ≔ 𝟒𝟎; Order ≔ 𝟒𝟎 > #Differential equation > de ≔ Y′′ (𝛈) + (𝟏 − 𝛈𝟐 ) • 𝛌𝟐 • Y(𝛈) = 𝟎; de ≔ D(𝟐) (Y)(𝛈) + (−𝛈𝟐 + 𝟏)𝛌𝟐 Y(𝛈) = 𝟎 > #We now obtain series solution > > sol ≔ dsolve({de, Y′ (𝟎) = a[𝟎], Y(𝟎) = a[𝟏]}, Y(𝛈), series) ∶ > F ≔ convert(sol, polynom) ∶ > F ≔ rhs(collect(F, {a[𝟎], a[𝟏]})) ∶ #We see that solution consists of odd and even functions.We reject the odd function.To reject and function we set a [𝟎] ≔ 𝟎; > #We now plot the first two eignen functions. >

> #We choose > a[𝟏] ≔ 𝟏 ∶ a[𝟎] ≔ 𝟎 ∶ > #Ge is the even function of the solution > #let us represent Ge by G; >G≔F∶ > eq ≔ subs(𝛈 = 𝟏, G) ∶ > plot(eq, 𝛌 = 𝟎.𝟏𝟎); Figure 8.40 Liouville type eigenvalue of the first few eigenvalues. > #Estimate of first eigen value > 𝛌[𝟎] ≔ fsolve(eq, 𝛌 = 𝟏..𝟐); 𝛌𝟎 ≔ 𝟏.𝟔𝟖𝟏𝟓𝟗𝟓𝟑𝟐𝟐 > #Estimate of second eigen value > 𝛌[𝟏] ≔ fsolve(eq, 𝛌 = 𝟒..𝟔); 𝛌𝟏 ≔ 𝟓.𝟔𝟔𝟗𝟖𝟓𝟕𝟑𝟗𝟕 > #Estimate of third eigen value > 𝛌[𝟐] ≔ fsolve(eq, 𝛌 = 𝟗..𝟏𝟎); 𝛌𝟐 ≔ 𝟗.𝟔𝟐𝟖𝟓𝟖𝟓𝟔𝟖𝟒

8.6 Heat Transfer in the Thermal Entrance Region

1.5

1

0.5

0 2

4

6

8

10

λ –0.5

–1

Figure 8.40

Location of eigenvalues.

A general solution is usually composed in the following form: θ(ξ, η) =

( ) 2 Cn φn (λn η) exp − λ2n x+ 3 n=0

∞ ∑

(8.601)

where λn and φn (λn η) are the eigenvalues and eigenfunctions. Next, we will use the nonhomogeneous inlet boundary condition of the problem in order to evaluate Cn 1=

∞ ∑

Cn φn (λn η)

(8.602)

n=0

We will multiply both sides of Eq. (8.602) by (1 − η2 )φn (λn η), and integrating from η = 0 to η = 1 yields 1

Cn =

∫0 (1 − η2 )φn (λn η)dη 1

∫0 (1 − η2 )[φn (λn η)]2 dη

(8.603)

This completes the solution, but some simplifications are possible. The eigenfunction φn (λn η) is a solution of governing ODE, Eq. (8.595); replacing Y by φn (λn η) in this equation, we obtain d2 φ n + λ2 (1 − η2 )φn = 0 dη2

(8.604)

We now integrate Eq. (8.604), and the result over the interval (0, 1) gives a simpler equation for the numerator of Eq. (8.603) in the following form: 1

∫0

1 d2 φn (λn η) 2 dη + λ (1 − η2 )φn (λn η)dη = 0 n ∫0 dη2

(8.605)

423

424

8 Laminar Momentum and Heat Transfer in Channels

Integrating the first term of Eq. (8.605), we get 1 dφn (λn η) || dφn (λn η) || − + λ2n (1 − η2 )φn (λn η)dη = 0 | | ∫0 dη |η=1 dη |η=0

Thus, from this last equation, we get 1

(1 − η2 )φn (λn η)dη = −

∫0

1 dφn (λn η) || | dη ||η=1 λ2n

(8.606)

Next, we need to rearrange the denominator of Eq. (8.603). For this purpose, we multiply the Eq. (8.604) expressed in terms of φn (λn η) by dφn (λn η)/dλn and integrate the result over (0, 1): ( ) ( ) 1 2 1 dφn d φn dφn 2 2 dη + λ dη = 0 (8.607) (1 − η )φ n n∫ ∫0 dη2 dλn dλn 0 The first integral on the left-hand side of Eq. (8.606) is integrated by parts. The second integral on the left-hand side of Eq. (8.606) is rearranged. Then, Eq. (8.606) becomes ( ) ) ( ) 1( 1 dφn dφn || dφn d dφn λ2 d dη + n (1 − η2 )[φn ]2 dη (8.608) | − dλn dη ||η=1 ∫0 dη dη dλn 2 dλn ∫0 Next, we multiply Eq. (8.604) expressed in terms of φn by φn and integrate over (0,1): ( ) 1 1 d2 φn 2 2 2 dη + λ φ n n ∫ (1 − η )[φn ] dη = 0 ∫0 dη2 0 Integrate the first part of integral in Eq. (8.606) by parts to get ) 1( 1 dφn 2 dφn || φn − dη + λ2n (1 − η2 )[φn ]2 dη = 0 | ∫0 dη |η=1 ∫0 dη

(8.609)

(8.610)

Next, we differentiate this equation, Eq. (8.610), with respect to λn ) ) ) ( ( 1( dφn dφn || dφn || dφn dφn d d dη + φn | −2 ∫0 dλn dη ||η=1 dλn dη ||η=1 dη dλn dη 1

+2λn

∫0

1

(1 − η2 )[φn ]2 dη + λ2n

d (1 − η2 )[φn ]2 dη = 0 dλn ∫0

(8.611)

Dividing Eq. (8.611) by 2, we get )] ) ) ( ( [ 1( dφn || dφn dφn 1 d d 1 dφn dφn || dη + φ (1) | − | ∫0 2 dλn dη ||η=1 2 n dλn dη dη dλn dη |η=1 1

+ λn

(1 − η2 )[φn ]2 dη +

∫0

1 λ2n d (1 − η2 )[φn ]2 dη = 0 2 dλn ∫0

(8.612)

The second term in Eq. (8.612) is zero since φn (1) = 0. Thus, we have ( ) ) 1( dφn dφn 1 dφn dφn || d dη − 2 dλn dη ||η=1 ∫0 dη dλn dη 1

+ λn

∫0

(1 − η2 )[φn ]2 dη +

1 λ2n d (1 − η2 )[φn ]2 dη = 0 2 dλn ∫0

(8.613)

We next subtract Eq. (8.613) from Eq. (8.608), and we get 1

∫0

(1 − η2 )[φn ]2 dη = −

1 dφn dφn || 2λn dλn dη ||η=1

(8.614)

Inserting Eqs. (8.606) and (8.614) into Eq. (8.603), we get (

Cn = − λn

2 dφn dλn

)| | | |η=1

(8.615)

8.6 Heat Transfer in the Thermal Entrance Region

Let us evaluate the first three constants C0 , C1 , and C2 . > restart; > #Evaluation of constants C > #Graetz problem for laminar flow between parallel plates UWT-Series >

#We will take only first 𝟏𝟎 terms of the series solution to study the nature of the solution

> > Order ≔ 𝟓𝟎; Order ≔ 𝟓𝟎 > #Differential equation > de ≔ Y′′ (𝛈) + (𝟏 − 𝛈𝟐 ) • 𝛌𝟐 • Y(𝛈) = 𝟎; de ≔ D(𝟐) (Y)(𝛈) + (−𝛈𝟐 + 𝟏)𝛌𝟐 Y(𝛈) = 𝟎 > #We now obtain series solution > > sol ≔ dsolve({de, Y′ (𝟎) = a[𝟎], Y(𝟎) = a[𝟏]}, Y(𝛈), series) ∶ > F ≔ convert(sol, polynom) ∶ > F ≔ rhs(collect(F, {a[𝟎], a[𝟏]})) ∶ #We see that solution consists of odd and even functions.We reject the odd function.To reject and function we set a [𝟎] ≔ 𝟎; > #We now plot the first two eignen functions. >

> #We choose > a[𝟏] ≔ 𝟏 ∶ a[𝟎] ≔ 𝟎 ∶ > #Ge is the even function of the solution > #let us represent Ge by G; >G≔F∶ > eq ≔ subs(𝛈 = 𝟏, G) ∶ > #Estimate of first eigen value > 𝛌[𝟎] ≔ fsolve(eq, 𝛌 = 𝟏..𝟐); 𝛌𝟎 ≔ 𝟏.𝟔𝟖𝟏𝟓𝟗𝟓𝟑𝟐𝟐 > #Estimate of second eigen value > 𝛌[𝟏] ≔ fsolve(eq, 𝛌 = 𝟒..𝟔); 𝛌𝟏 ≔ 𝟓.𝟔𝟔𝟗𝟖𝟓𝟕𝟑𝟒𝟔 > #Estimate of third eigen value > 𝛌[𝟐] ≔ fsolve(eq, 𝛌 = 𝟗..𝟏𝟎); 𝛌𝟐 ≔ 𝟗.𝟔𝟔𝟖𝟐𝟕𝟑𝟑𝟓𝟐

425

426

8 Laminar Momentum and Heat Transfer in Channels

#We now evaluate the denominator of Eq.( ) in text.We evaluate the function G for 𝛈 = 𝟏 and take the derivative with respect to 𝛌 > #Then evaluate this derivative for 𝛌 = 𝛌𝟎 .Then multiply by 𝛌𝟎 >

> > DENOM ≔ 𝛌[𝟎] • subs (𝛌 = 𝛌[𝟎], diff(eq, 𝛌)); DENOM ≔ −𝟏.𝟔𝟔𝟓𝟓𝟏𝟒𝟏𝟔𝟎 𝟐 ; DENOM C𝟎 ≔ 𝟏.𝟐𝟎𝟎𝟖𝟑𝟎𝟑𝟕𝟗

> C[𝟎] ≔ −

> #Evaluation of C𝟏 > DENOM ≔ 𝛌[𝟏] • subs(𝛌 = 𝛌[𝟏], diff(eq, 𝛌)); DENOM ≔ 𝟔.𝟔𝟖𝟓𝟑𝟕𝟎𝟒𝟓𝟗 𝟐 ; DENOM C𝟏 ≔ −𝟎.𝟐𝟗𝟗𝟏𝟔𝟎𝟔𝟖𝟒𝟐

> C[𝟏] ≔ −

> #Evaluatyion of C𝟐 > DENOM ≔ 𝛌[𝟐] • subs(𝛌 = 𝛌[𝟐], diff(eq, 𝛌)); DENOM ≔ −𝟏𝟐.𝟒𝟑𝟒𝟓𝟐𝟗𝟑𝟕 𝟐 ; DENOM C𝟐 ≔ 𝟎.𝟏𝟔𝟎𝟖𝟒𝟐𝟒𝟑𝟔𝟓

> C[𝟐] ≔ −

> Eigenvalues and other constants are given in Table 8.8. Table 8.8 The first 10 eigenvalues and the constants of the Cartesian Graetz problem with constant wall temperature boundary condition. n

𝛌n

Cn

Gn

0

1.681595322

1.200830379

0.8580866740

1

5.669857346

–0.299160683

0.5694628760

2

9.668242462

0.1608265322

0.4760626180

3

13.66766144

–0.1074364037

0.4240085108

4

17.66737357

0.07963433031

0.3908850294

5

21.66720532

–0.0637751556

0.3619336332

6

25.66709649

0.5175931584 × 10−1

0.3451140388

7

29.66702104

–0.4324393733 × 10−1

0.3252947984

8

33.66696607

0.3778583868 × 10−1

0.3158138077

9

37.66692446

−1

–0.3266889710 × 10

0.2998111116

Shah and London [3] give equations for large n values. We can use the following relations for large values of n. λn ≈ 4n +

5 1.01279 2.27114 Gn = . C = (−1)n 7∕6 1∕3 3 n λn λn

8.6 Heat Transfer in the Thermal Entrance Region

Local heat flux

(

q′′w (x+ ) = k

𝜕T 𝜕y

)

Temperature gradient

= y=b

( ) 𝜕θ 𝜕η

( ) k 𝜕θ (Ti − Tw ) b 𝜕η η=1

η=1

(8.616)

is obtained from Eq. (8.601) as

) ( ∑ dφn (λn η) | 𝜕θ(ξ, η) || | exp − 2 λ2 x+ = C n n | 𝜕η ||η=1 n=0 dη 3 |η=1 ∞

(8.617)

We now define a new constant Gn Gn = −

Cn dφn (λn η) || | 2 dη |η=1

(8.618)

Substituting Eq. (8.618) into Eq. (8.617), the temperature gradient becomes ∞ ( ) ∑ 𝜕θ(ξ, η) || 2 = −2 Gn exp − λ2n x+ | 𝜕η |η=1 3 n=0 We now substitute expression for temperature gradient into Eq. (8.616), and local heat flux now becomes ∞ ( ) ∑ 2k 2 q′′w (x+ ) = (Tw − Ti ) Gn exp − λ2n x+ b 3 n=0

(8.619)

Mean temperature Tm

Mean temperature Tm is defined as [ ( )] b 3 y2 b ∫0 uTdy ∫0 2 V 1 − b2 [Tw + (Ti − Tw )θ]dy = Tm = b [ ( )] 2 b ∫0 udy ∫0 32 V 1 − by2 dy After integration, we get dimensionless mean temperature θm 1

θm =

3 (1 − η2 )θdη 2 ∫0

We now substitute Eq. (8.601) for temperature θ into expression for mean temperature θm , and this will give us ∞ 1 ( ) ∑ 3 2 θm = (1 − η2 ) Cn φn (λn η) exp − λ2n x+ dη 2 ∫0 3 n=0 ∞ 1 ) ( ∑ 2 3 C exp − λ2n x+ (1 − η2 )φn (λn η)dη (8.620) = ∫0 2 n=0 n 3 But we have already evaluated the integral in Eq. (8.620). See Eq. (8.606). We just presented Eq. (8.606) here for convenience 1 1 dφn || (1 − η2 )φn (λn η)dη = − 2 | ∫0 λn dη |η=1 We now substitute Eq. (8.606) into Eq. (8.620), and the mean temperature θm takes the following form: [ ] ∞ ) ( 2 3 ∑ Cn dφn || exp − λ2n x+ θm = − | 2 2 n=0 λn dη |η=1 3 or in terms of Gn , the mean fluid temperature θm is ( ) ∞ ( ) ∑ Tm − Tw Gn 2 θm = =3 exp − λ2n x+ 2 Ti − Tw 3 λn n=0 The local heat transfer coefficient h is defined as q′′w q′′w h= = Tw − Tm (Tw − Ti )θm

(8.621a)

(8.621b)

(8.622a)

427

428

8 Laminar Momentum and Heat Transfer in Channels

We substitute expressions for q′′w and θm into Eq. (8.622a), and the local heat transfer coefficient h becomes ( ( ) ) ∞ ∞ ∑ ∑ 2k 2 2 + 2 2 + (T − T ) G exp − λ x G exp − λ x w i n n n n b 3 3 n=0 2k n=0 h= = ) ( ) ( ) ( ) ∞ ( ∞ 3b ∑ Gn ∑ Gn 2 2 + 2 2 + (Tw − Ti )3 λ x λ x exp − exp − 2 2 n n λ 3 λ 3 n=0

n=0

n

n

Let us define DH = 4b. The local Nusselt number NuDH based on the hydraulic diameter DH is ( ) ∞ ∑ Gn exp − 23 λ2n x+ h DH 8 n=0 NuDH = = ( ) ∞ k 3 ∑ Gn 2 2 + exp − λ x λ2 3 n n=0

(8.622b)

(8.623)

n

and the values of Gn are given in Table 8.8. The mean Nusselt number based on the mean logarithmic temperature difference is given as NuDH =

ReDH Pr (4b)h DH h = =− ln(θm ) k k (z∕b)

Substituting Eq. (8.621b) into Eq. (8.624a), we get an expression for the average Nusselt number [ ∞ ] ( ) ∑ Gn 4 2 2 + NuDH = − + ln 3 exp − λn x 2 x 3 n=0 λn

(8.624a)

(8.624b)

8.6.2.2 Graetz Problem: HFD and Thermally Developing Flow Between Parallel Plates Subjected to Constant Wall Heat Flux

Assume that viscous fluid steadily and uniformly enters the channel at a velocity V and temperature Ti . The fluid and the channel walls are at the same temperature for x < 0, and for this reason, there is no heat transfer at the surface of the duct for the upstream half of the channel. The temperature of the fluid is uniform over the duct cross section at the point where the heat transfer begins. At some point x > 0, a UHF is applied to both plates. We assume that at the point where heat transfer begins, the velocity profile is fully developed and is equal to [ ( y )2 ] 3 u= V 1− 2 b and temperature profile begins to develop. See Figure 8.41a. On the other hand, Figure 8.41b shows step change in heat flux where heating begins. We will consider steady, two-dimensional laminar, incompressible, low-velocity flow. Axial conduction is negligible, and there is no viscous dissipation as well as internal energy generation. We wish to determine the temperature distribution and the Nusselt number in the thermal entrance region of the channel formed by two infinite parallel plates. We make the following assumptions: (1) (2) (3) (4) (5)

The y-components of the velocity, v, is everywhere zero. Constant fluid properties. (3) q′′w = 0 if x < 0, q′′w = q′′0 if x > 0. The temperature of the fluid in contact with the plates is the same as that of the plates for x < 0. Assume that Pe > 100, and this means axial conduction is zero in the direction of fluid motion.

The energy equation to be solved in the thermal entrance region with parabolic velocity distribution between the two parallel walls separated by a distance 2b, as shown in Figure 8.41, is given by [ ( y )2 ] 𝜕2 T 𝜕T 3 V 1− =α 2 (8.625) 2 b 𝜕x 𝜕y where V is the mean velocity. The boundary conditions are 𝜕T(x, 0) = 0 or T(x, 0) = finite 𝜕y 𝜕T(x, b) = q′′0 k 𝜕y T(0, y) = Ti

(8.626a) (8.626b) (8.626c)

8.6 Heat Transfer in the Thermal Entrance Region

y

q″w = 0

q″0 q″w (x)

u(y) Ti vz = V

2b

q″0

x

O

(a) Figure 8.41

x

q″w = 0

q″0

q″w = 0

(b)

(a) Graetz problem between parallel plates, (b) Step change in heat flux at the point where heating begins.

Using the following dimensionless variables: θ=

T(x, y) − Ti q′′0 b∕k

(8.627)

η = y∕b x+ =

(8.628)

4(x∕b) ReDH Pr

(8.629)

The PDE becomes 3 𝜕θ 𝜕2θ [1 − η2 ] + = 2 2 𝜕x 𝜕η

(8.630)

and the boundary conditions takes the following form: x+ = 0 θ = 0

(8.631a)

𝜕θ =1 𝜕η 𝜕θ = 0 or θ is finite. x+ > 0 η = 0 𝜕η

x+ > 0 η = 1

(8.631b) (8.631c)

The PDE has a nonhomogeneous boundary condition, and the separation of variables cannot be applied directly. The solution can be obtained by applying a special procedure described in [9] and [24]. We assume a solution in the following form: θ(x+ , η) = ψ(x+ , η) + ⏟⏞⏟⏞⏟ decaying initial transient

ϕ(η) ⏟⏟⏟ temperature variation in y direction to let wall heat flux into fluid

+

φ(x+ ) ⏟⏟⏟

(8.632)

axial temperature rise due to accumulated heat flux

The substitution of the assumed solution into the PDE, Eq. (8.630), yields the following PDE, Eq. (8.633), and two other ODEs, Eq. (8.635) and Eq. (8.636): 𝜕ψ 𝜕2 ψ 3 (1 − η2 ) + = 2 𝜕x 𝜕η2 ψ(0, η) = −φ(η) − φ(0) 𝜕ψ =0 𝜕η 𝜕ψ =0 η=1 𝜕η

η=0

(8.633) (8.634a) (8.634b) (8.634c)

where ψ(ξ, η) represents the developing temperature distribution. The other set of differential equations is dφ = C0 dx+

(8.635)

429

430

8 Laminar Momentum and Heat Transfer in Channels

d2 φ 3 = C0 (1 − η2 ) 2 dη2 dφ η=0 =0 dη dφ =1 η=1 dη The integration of the ODE, Eq. (8.636), once yields ) ( dϕ 3 η3 + C1 = C0 η − dη 2 3

(8.636) (8.637a) (8.637b)

(8.638)

The application of the first boundary condition, Eq. (8.637a), yields C1 = 0 and the differential equation, Eq. (8.638), becomes ( ) η3 dϕ 3 = C0 η − (8.639) dη 2 3 Next, we apply the second boundary condition, Eq. (8.637b), and we find that C0 = 1. After one more integration, the final form of the differential equation, Eq. (8.639), becomes ) ( 3 2 η4 η − + C2 (8.640) ϕ= 4 8 The integration of the ODE, Eq. (8.635), gives φ = x+ + C3

(8.641)

Combining Eqs. (8.640) and (8.641) gives the fully developed temperature distribution η4 3 θfd = x+ + η2 − + C4 (8.642) 4 8 where C4 = C2 + C3 . In order to evaluate the unknown constant C4 , we will use the mean fluid temperature Tm . The mean fluid temperature Tm is b

Tm =

∫0 u T dy b

∫0 u dy

The mean temperature Tm is expressed in terms of dimensionless variables θm =

1 Tm − Ti 3 = (1 − η2 ) θ dη ′′ q b∕k 2 ∫0

(8.643)

The evaluation of bulk temperature θm yields ) ( 1 Tm − Ti 3 2 η4 3 2 + + C4 dη (1 − η ) x + η − = θm = ′′ 2 ∫0 4 8 q0 b∕k 39 = (8.644) + x+ + C 4 280 We need another expression for the bulk temperature θm . This is obtained by writing an energy balance for control volume in parallel plates. The energy balance is q′′0 dTm = dx ρ cp V b

(8.645)

Tm (0) = Ti We convert Eq. (8.644) using dimensionless variables into dimensionless form, and we obtain dθm =1 dξ

(8.646)

θm (0) = 0

(8.647)

8.6 Heat Transfer in the Thermal Entrance Region

The solution of this differential equation, Eq. (8.646), is θm = x+

(8.648)

Comparing two equations, Eq. (8.644) and Eq. (8.648), for the bulk temperatures, we obtain that C4 = − 39/280. Fully developed temperature is θfd =

Tfd − Ti η4 39 3 − = x+ + η2 − ′′ 4 8 280 q0 b∕k

(8.649)

We may obtain the asymptotic Nusselt number as follows: NuDH =

q′′w h DH 4b = k (Tw − Tm ) k

(8.650)

Using the dimensionless variables, we write NuDH =

4 = θw (x+ , 1) − θm (x+ )

17 35

4 = 8.235 + x+ − x+

(8.651)

Now, we can generate the dimensionless temperature distribution as follows: η4 39 3 θ(x+ , η) = ψ(x+ , η) + x+ + η2 − − 4 8 280 It remains to determine the temperature distribution ψ(ξ, η): 𝜕ψ 𝜕2 ψ 3 (1 − η2 ) + = 2 𝜕x 𝜕η2 η4 3 39 ψ(0, η) = − η2 + + 4 8 280 𝜕ψ η=0 =0 𝜕η 𝜕ψ =0 η=1 𝜕η

(8.652)

(8.653) (8.654a) (8.654b) (8.654c)

We can now apply the separation of variables to solve Eq. (8.653). We assume a product solution of the form ψ = X(x+ ) Y(η)

(8.655)

Notice that the y-direction will give us the characteristic value problem. With the proper selection of separation constant, the product solution will result in two ODEs dX 2 + λ2 X = 0 dx+ 3 The solution of this differential equation, Eq. (8.656), is ) ( 2 X = C exp − λ2 x+ 3 The differential equation for the characteristic value problem is d2 Y + λ2 (1 − η2 )Y = 0 dη2

(8.656)

(8.657)

(8.658)

The boundary conditions are dY =0 dη dY η=1 =0 dη η=0

(8.659a) (8.659b)

The solution of this differential equation, Eq. (8.658), will give the characteristic values and characteristic functions. We wish to study the nature of the solution. For this purpose, let us generate a series solution with 10 terms using Maple 2020.

431

432

8 Laminar Momentum and Heat Transfer in Channels

#Each solution is Gaetz function and we reject the odd functoin because of boundary condition > restart; >

> #Series solution > Order ≔ 𝟏𝟎; Order ≔ 𝟏𝟎 > > de ≔ Y′′ (𝛈) + 𝛌𝟐 • (𝟏 − 𝛈𝟐 ) • Y(𝛈) = 𝟎; de ≔ D(𝟐) (Y)(𝛈) + 𝛌𝟐 (−𝛈𝟐 + 𝟏)Y(𝛈) = 𝟎 > > > sol ≔ rhs(dsolve({de, Y′ (𝟎) = a[𝟎], Y(𝟎) = a[𝟏]}, Y(𝛈), series)); ( ) 𝟏 𝟏 𝟏 𝟒 𝟏 sol ≔ a𝟏 + a𝟎 𝛈 − 𝛌𝟐 a𝟏 𝛈𝟐 − 𝛌𝟐 a𝟎 𝛈𝟑 + 𝛌 a𝟏 + 𝛌𝟐 a𝟏 𝛈𝟒 𝟐 𝟔 𝟐𝟒 𝟏𝟐 ( ( ( ) ) 𝟏 𝟒 𝟏 𝟏 𝟔 𝟕 𝟒 𝟏 𝟔 𝛌 a𝟎 + 𝛌𝟐 a𝟎 𝛈𝟓 + − 𝛌 a𝟏 − 𝛌 a 𝟏 𝛈𝟔 + − 𝛌 a𝟎 + 𝟏𝟐𝟎 𝟐𝟎 𝟕𝟐𝟎 𝟑𝟔𝟎 𝟓𝟎𝟒𝟎 ( ) ) 𝟏𝟑 𝟒 𝟏 𝟏𝟏 𝟔 𝟏 𝟒 𝛌 a𝟎 𝛈𝟕 + 𝛌𝟖 a𝟏 + 𝛌 a𝟏 + 𝛌 a 𝟏 𝛈𝟖 − 𝟐𝟓𝟐𝟎 𝟒𝟎𝟑𝟐𝟎 𝟏𝟎𝟎𝟖𝟎 𝟔𝟕𝟐 ( ) 𝟏 𝟏𝟕 𝟔 𝟏 𝟒 𝛌𝟖 a𝟎 + 𝛌 a𝟎 + 𝛌 a𝟎 𝛈𝟗 + O(𝛈𝟏𝟎 ) + 𝟑𝟔𝟐𝟖𝟖𝟎 𝟗𝟎𝟕𝟐𝟎 𝟏𝟒𝟒𝟎 > F ≔ convert(sol, polynom); ( ) 𝛌𝟐 a𝟏 𝛈𝟐 𝛌𝟐 a𝟎 𝛈𝟑 ( 𝟏 𝟒 𝟏 𝟏 𝟒 F ≔ a𝟏 + a𝟎 𝛈 − − + 𝛌 a𝟏 + 𝛌𝟐 a𝟏 𝛈𝟒 + 𝛌 a𝟎 𝟐 𝟔 𝟐𝟒 𝟏𝟐 𝟏𝟐𝟎 ( ( ) ) 𝟏 𝟏 𝟔 𝟕 𝟒 𝟏 𝟔 𝛌 a𝟏 − 𝛌 a𝟏 𝛈𝟔 + − 𝛌 a𝟎 + 𝛌𝟐 a𝟎 𝛈𝟓 + − 𝟐𝟎 𝟕𝟐𝟎 𝟑𝟔𝟎 𝟓𝟎𝟒𝟎 ( ) ) 𝟏𝟑 𝟒 𝟏 𝟏𝟏 𝟔 𝟏 𝟒 𝛌 a𝟎 𝛈𝟕 + 𝛌𝟖 a𝟏 + 𝛌 a𝟏 + 𝛌 a 𝟏 𝛈𝟖 − 𝟐𝟓𝟐𝟎 𝟒𝟎𝟑𝟐𝟎 𝟏𝟎𝟎𝟖𝟎 𝟔𝟕𝟐 ( ) 𝟏 𝟏𝟕 𝟔 𝟏 𝟒 𝟖 𝟗 𝛌 a𝟎 + 𝛌 a𝟎 + 𝛌 a𝟎 𝛈 + 𝟑𝟔𝟐𝟖𝟖𝟎 𝟗𝟎𝟕𝟐𝟎 𝟏𝟒𝟒𝟎 > G ≔ collect(F, {a[𝟎], a[𝟏]}); (

( ) ) 𝛌𝟐 𝛈𝟑 ( 𝟏 𝟒 𝟏 𝟏 𝟔 𝟏𝟑 𝟒 𝟕 + 𝛌 + 𝛌𝟐 𝛈𝟓 + − 𝛌 − 𝛌 𝛈 𝟔 𝟏𝟐𝟎 𝟐𝟎 𝟓𝟎𝟒𝟎 𝟐𝟓𝟐𝟎 ( 𝟐 𝟐 ( ( ) ) 𝛌 𝛈 𝟏 𝟒 𝟏 𝟏𝟕 𝟔 𝟏 𝟒 𝟗 𝛌𝟖 + 𝛌 + 𝛌 𝛈 a𝟎 + 𝟏 − + 𝛌 + 𝟑𝟔𝟐𝟖𝟖𝟎 𝟗𝟎𝟕𝟐𝟎 𝟏𝟒𝟒𝟎 𝟐 𝟐𝟒 ( ( ) ) 𝟏 𝟏 𝟔 𝟕 𝟒 𝟔 𝟏 𝟏𝟏 𝟔 + 𝛌𝟐 𝛈𝟒 + − 𝛌 − 𝛌 𝛈 + 𝛌𝟖 + 𝛌 𝟏𝟐 𝟕𝟐𝟎 𝟑𝟔𝟎 𝟒𝟎𝟑𝟐𝟎 𝟏𝟎𝟎𝟖𝟎 ) ) 𝟏 𝟒 𝟖 𝛌 𝛈 a𝟏 + 𝟔𝟕𝟐 The solution of the differential equation, Eq. (8.658), is composed of an odd and an even function G≔

𝛈−

(o) Y(η) = AG(e) 0 (λη) + BG0 (λη) (0) where G(e) 0 (λη) and G0 (λη) denote the even and odd functions, order zero, respectively. We reject the odd function because ′ of boundary condition Y (0) = 0. Then, the solution is ( 4 ) ) ) ) ( ( ( 8 λ2 η2 λ2 7 λ4 11 λ6 λ4 λ6 λ λ 4 6 + + η − η + + η8 (8.660a) = 1 − + − + G(e) 0 2 24 12 720 360 40320 10080 672

8.6 Heat Transfer in the Thermal Entrance Region

or φn = G(e) o (λn η)

(8.660b)

To get an accurate result, we need to take enough number of terms. The Graetz function will give us the eigenvalues. Next, the calculation of a sample eigenvalue will be presented. > restart; #Thermal entrance problem between parallel plates - Uniform heat flux boundary condition > #Series solution. Here 𝟓𝟎 terms of series are used. >

> #Solution of Eq.(8.658) Order ≔ 𝟓𝟎 > Order ≔ 𝟓𝟎; > #There is an eigenvalue 𝛌n for each eigen function > de ≔ Y′′ (𝛈) + 𝛌𝟐 • (𝟏 − 𝛈𝟐 ) • Y(𝛈) = 𝟎; de ≔ D(𝟐) (Y)(𝛈) + 𝛌𝟐 (−𝛈𝟐 + 𝟏)Y(𝛈) = 𝟎 > sol ≔ rhs(dsolve({de, Y′ (𝟎) = a[𝟎], Y(𝟎) = a[𝟏]}, Y(𝛈), series)) ∶ > F ≔ convert(sol, polynom) ∶ > FF ≔ collect(F, {a[𝟎], a[𝟏]}) ∶ a𝟏 ≔ 𝟏 > #To choose even function we set a[0] = 0 and a[1] = 1 > a[𝟎] ≔ 𝟎; a𝟎 ≔ 𝟎 > a[𝟏] ≔ 𝟏; a𝟏 ≔ 𝟏 > #Graetz function G > G ≔ eval(FF) ∶ > #Next we use the second BC to evaluate eigenvalues > eq ≔ subs(𝛈 = 𝟏, diff(G, 𝛈)) ∶ > with(plots) ∶ > plot(eq, 𝛌 = 𝟎..𝟏𝟎); Figure 8.42 shows the location of a few eigenvalues. > #We now estimate the first two eigenvalues > #First eigenvalue > fsolve(eq, 𝛌 = 𝟒..𝟓); 𝟒.𝟐𝟖𝟕𝟐𝟐𝟒𝟗𝟒𝟔 > #Second eigenvalue > fsolve(eq, 𝛌 = 𝟖..𝟏𝟎); 𝟖.𝟑𝟎𝟑𝟕𝟐𝟕𝟕𝟏𝟏 >

433

434

8 Laminar Momentum and Heat Transfer in Channels

4

2

0 2

4

λ

6

8

10

–2

–4

–6

Figure 8.42

Location of eigenvalues.

First two eigenvalues are λ1 = 4.287224 and λ2 = 8.303727, respectively. We are now in a position to write the solution for Eq. (8.653)

ψ(x+ , η) =

( ) 2 Cn φn (λn η) exp − λ2n x+ 3 n=1

∞ ∑

(8.661)

Finally, we use the inlet boundary condition of the problem to determine an ∑ η4 39 3 + = − η2 + C φ (λ η) 4 8 280 n=1 n n n ∞

(8.662)

The characteristic functions are orthogonal with respect to the weighting function (1 − η2 ). Therefore, the unknown constant an is given by ( ) 4 1 39 ∫0 (1 − η2 ) − 34 η2 + η8 + 280 φn (λn η)dη (8.663) Cn = 1 ∫0 (1 − η2 )[φn (λn η)]2 dη The complete dimensionless temperature distribution θ(x+ , η) can be written as ( ) ∑ T − Ti 39 3 2 η4 2 2 + + η − + λ + − C φ (λ η) exp − x = x 4 8 280 n=1 n n n 3 n q′′0 L∕k ∞

θ(x+ , η) =

(8.664)

and the mean fluid temperature is given by θm (x+ , η) =

Tm − Ti = x+ q′′0 b∕k

(8.665)

The first ten characteristic values and the constants φn (λn , η)η=1 and Cn are given in Table 8.9. The local wall temperature θw is obtained by setting η = 1 ( ) Tw − Ti 17 ∑ + | exp − 2 λ2 x+ + + C φ (λ , η) = x n n n n | η=1 35 n=1 3 q′′0 b∕k ∞

θw (1, x+ ) =

(8.666)

8.6 Heat Transfer in the Thermal Entrance Region

Table 8.9

Characteristic values and the constants.

n

𝛌n

Cn

−Cn 𝛗n (𝛌n , 𝛈)|𝛈 = 1

1

4.287768222

0.1750186040

0.2222234857

2

8.304780656

–0.05166744979

0.07244918373

3

12.31217301

0.02508186303

0.03741231716

4

16.31659875

–0.01488123817

0.02321649969

5

20.31968413

0.01000198546

0.01616476341

6

24.32203822

–0.00712488929

0.01185460389

7

28.32394487

0.005448693899

0.009292576427

8

32.32555565

–0.004210180569

0.007336420422

9

36.32695950

0.003463229588

0.006150420006

10

40.32821232

–0.002792614789

0.005045148442

Shah and London [3] give for large n values. −5∕3 λn = 4n + 13 Cn Gn (1) = −2.401006λn .

The local Nusselt number is defined as NuDH =

q′′0 (4b) h(4b) 4 = = k Tw − Tm k θw − θm

(8.667a)

Using the wall and mean fluid temperatures, the local Nusselt number is NuDH =

hDH = k

140∕17 ∞ ∑ ( ) 35 1+ Cn φn (λn , η)||η=1 exp −2λ2n x+ ∕3 17 n=1

(8.667b)

where DH = 4b is the hydraulic diameter. As we discussed before, the average Nusselt number is usually based on the average difference between the wall and mixing cup temperatures z+

Tw − Tm =

1 (Tw − Tm )dξ z+ ∫0

where ξ is a dummy variable in the integration process. The average heat transfer coefficient h is q′′0

h=

(Tw − Tm ) and the average Nusselt number is NuDH =

4 θw − θm

The average difference between the wall and mixing cup temperatures is ] [ ∞ x+ ( ) 1 2 2 17 ∑ | + θw − θm = + C φ (λ , η) exp − λn ξ dξ x ∫0 35 n=1 n n n |η=1 3 ) ( ⎧ ⎡ 2 2 + ⎤⎫ ∞ λ x 1 − exp − ∑ ⎢ ⎥⎪ 3 n 1 ⎪ 17 Cn φn (λn , η)||η=1 ⎢ = + ⎨ x+ + ⎥⎬ 2 2 x ⎪ 35 λ n=1 ⎢ ⎥⎪ 3 n ⎣ ⎦⎭ ⎩ ( ) ⎧ ⎡ 2 2 + ⎤⎫ ∞ ⎪ 17 ∑ ⎢ 1 − exp − 3 λn x ⎥⎪ | =⎨ + Cn φn (λn , η)|η=1 ⎢ ⎥⎬ 2 2 + λ x ⎪ 35 n=1 ⎢ ⎥⎪ 3 n ⎣ ⎦⎭ ⎩

(8.668a)

435

436

8 Laminar Momentum and Heat Transfer in Channels

Table 8.10

Local and average Nusselt numbers.

x+ = 4(x∕DH )∕ReDH Pr

NuDH

NuDH

0.000004

148.773

223.2

0.000016

93.673

140.6

0.000020

86.954

130.5

0.000036

71.477

107.3

0.00004

69.011

103.55

0.00016

43.521

65.24

0.00020

40.419

60.57

0.00036

33.290

49.82

0.0004

32.153

48.11

0.0016

20.512

30.46

0.0020

19.113

28.33

0.0036

15.928

23.43

0.004

15.427

22.65

0.016

10.516

14.80

0.020

9.9878

13.89

0.036

8.9374

11.88

0.04

8.8031

11.58

0.16

8.2368

9.145

0.20

8.2355

8.963

0.36

8.2353

8.640

0.4

8.2353

8.599

0.6

8.2353

8.478

0.8

8.2353

8.417

Source: Shah and London [3]/with permission of Elsevier..

and finally the average Nusselt number is NuDH =

hDH = k

140∕17

⎧ ( )⎫ 2 ⎪ 1 − exp −2λn x+ ∕3 ⎪ 35 ∑ 1+ Cn φn (λn , η)||η=1 ⎨ ⎬ 2 2 + 17 n=1 ⎪ ⎪ λn x ⎩ ⎭ 3 Local and mean Nusselt numbers are tabulated in Table 8.10.

(8.668b)

∞

8.6.2.3 Empirical and Theoretical Correlations for Viscous Flow in Thermal Entrance Region of Parallel Plates

Constant wall temperature Shah [32] recommended the following equations as alternative to Graetz solution for both the local and mean Nusselt numbers for thermally developing flows between parallel plates with constant wall temperature: { )−1∕3 ( + 0.4 Gz−1 1.233 Gz−1 x x ≤ 0.001 (8.669a) NuDH = ( 3 −1 )−0.488 ( ) −1 7.541 + 6.874 10 Gzx exp −245 Gzx Gz−1 x > 0.001 Tw = const h DH V DH ReDH = k ν x 1 = DH ReDH Pr

NuDH = Gz−1 x

8.6 Heat Transfer in the Thermal Entrance Region

Bennett [33] reports that this correlation agrees within −0.7% to 1.5% of the exact solution. A correlation reported by Shah [32] for the average Nusselt number for thermally developing flows between parallel plates with constant wall temperature is given as ( −1 )−1∕3 ⎧ Gz−1 ⎪1.849(GzL ) L ≤ 0.0005 −1∕3 ⎪1.849 Gz−1 + 0.6 0.0005 < Gz−1 L L > 0.006 NuDH = ⎨ 0.0245 −1 ⎪7.541 + ( Gz > 0.006 ) L ⎪ Gz−1 L ⎩ L 1 Gz−1 L = DH ReDH Pr

(8.669b)

Tw = const h DH V DH NuDH = ReDH = k ν Bennett [33] proposed the following correlation for the average Nusselt number NuDH for flows between parallel plates of constant temperature ]0.2734 hDH [ = 9.469Gz1.219 + 1561 + 0.076 L k D GzL = ReDH Pr L and Bennet reports that this correlation agrees within −0.8% to 1.4% of the exact solution. NuDH =

(8.670)

Uniform heat flux Shah [32] recommended the following equations as alternative to the Graetz solution for both the local and mean Nusselt numbers: Local Nusselt number ( ) ⎧1.409 Gz−1 −1∕3 Gz−1 z ≤ 0.0002 ⎪ ( z−1 )−1∕3 NuDH = ⎨1.490 Gzz (8.671a) − 0.4 0.0002 < Gz−1 z ≤ 0.001 ⎪8.235 + 8.68(103 × Gz−1 )−0.506 exp (−164 × Gz−1 ) Gz−1 > 0.001 ⎩ z z z x 1 −1 Gzz = DH ReDH Pr

q′′w = const Average Nusselt number

NuDH

( −1 )−1∕3 ⎧ Gz−1 ⎪2.236(GzL ) L ≤ 0.001 −1∕3 ⎪2.236 Gz−1 + 0.9 0.001 < Gz−1 L L > 0.01 =⎨ 0.0364 −1 ⎪8.235 + GzL ≥ 0.01 ⎪ Gz−1 L ⎩

(8.671b)

q′′w = const Gz−1 L =

L 1 DH ReDH Pr

Bennett [33] recommends the following correlation for thermally developing flow between parallel plates under constant heat flux boundary condition: [ ]0.2698 NuDH = 12.74Gz1.235 + 2243 + 0.215 (8.672) L ( ) D ReD Pr L where GzL is the Graetz number. GzL =

437

438

8 Laminar Momentum and Heat Transfer in Channels

8.7 Circular Pipe with Variable Surface Temperature Distribution in the Axial Direction We will now study arbitrary surface temperature variation on the tube surface with the axial distance. This problem was discussed in [1] and [9]. Since the energy equation is linear in temperature, the superposition principle can be used to construct solutions to problems with variable wall temperatures from simple step function solutions, which we have obtained in previous sections. For example, the sum of any two solutions will be a solution provided that it matches the boundary conditions. First, we will consider the thermal entrance problem, as illustrated in Figure 8.43b. Constant property, incompressible fluid enters the duct at uniform temperature Ti . Flow is steady and laminar, and axial conduction is neglected. Viscous dissipation and internal energy generation are neglected. The fluid and duct surface temperatures are at the same temperature until z+ = 0 is reached. The velocity profile is fully developed at the point z+ = 0 where heat transfer begins. At the point z+ = 0, the surface temperature steps to a different temperature Tw0 different than inlet temperature Ti and remains constant thereafter. In other words, a step change in surface temperature occurs at z+ = 0, as illustrated in Figure 8.43b. The energy equation and its boundary conditions are repeated here for convenience here (1 − η2 )

𝜕θ 𝜕 2 θ 1 𝜕θ = + 𝜕z+ 𝜕η2 η 𝜕η

(8.673)

θ(0, η) = 1

(8.674a)

θ(z+ , 1) = 0

(8.674b)

𝜕θ + (z , 0) = 0 𝜕η

(8.674c)

where we have used the following notation: Tw0 − T = θ(z+ , η) Tw0 − Ti

(8.675a)

η=

r R

(8.675b)

u=

vz V

(8.675c)

z+ =

2(z∕D) ReD Pr

(8.675d)

dTw

Ti

ΔTw2

Tw(z+)

Tw0 ≠ Ti

Ti

Tw

vz(r) ΔTw1 Tw0

Tw0 Ti

Ti 0

z+ = ξ1 z+ = ξ2 (a)

Figure 8.43

z+

0

ΔTw0 = Tw0 – Ti z+ = 0

z+

dξ (b)

(a) Pipe surface temperature variation. (b) Step change in pipe surface temperature.

8.7 Circular Pipe with Variable Surface Temperature Distribution in the Axial Direction

The solution of Eq. (8.673) is θ(z+ , η) =

∞ ∑

) ( Cn φn exp −λ2n z+

(8.676)

n=0

where φn is the characteristic function, and recall that Eq. (8.676) is the Graetz solution for the thermal entry length problem. We are interested in effect of arbitrary wall temperature distribution, Tw (x), on the fluid. We now assume that surface temperature varies in arbitrary manner starting at z+ = 0, as illustrated in Figure 8.43a. The variable surface temperature can be represented by a number of infinitesimal steps or finite steps or both. Fluid temperature in the duct at any position z+ and η can be obtained by summing up the contribution of each step, either infinitesimal or finite. Here, z+ is treated as a constant, and a dummy length variable ξ is used to designate the location of each step. Therefore, ξ varies from 0 to z+ . We will employ Eq. (7.366a) and it is repeated here with new notation. ) ( N z+ ∑ dTw T − Ti = dξ + f(z+ − ξ, η) f(z+ − ξi , η) ΔTw,i (8.677) ∫0 dξ i=0 How do we find f(z+ − ξ, η)? In order to determine f(z+ − ξ, η), we construct the following auxiliary problem: (1 − η2 )

𝜕f 𝜕 2 f 1 𝜕f = 2 + + 𝜕z η 𝜕η 𝜕η

(8.678)

f(0, η) = 0

(8.679a)

f(z+ , 1) = 1

(8.679b)

𝜕f + (z , 0) = 0 𝜕η

(8.679c)

where f(z+ , η) is the solution of auxiliary problem. How do we get f(z+ , η)? In order to obtain the unit step solution f(z+ , η), we will rearrange the dimensionless temperature distribution θ(z+ , η) as follows: Tw0 − T = θ(z+ , η) Tw0 − Ti

(8.680a)

T − Ti = [1 − θ(z+ , η)] Tw0 − Ti

(8.680b)

T − Ti = (Tw0 − Ti )f(z+ , η)

(8.680c)

or

or

where f(z+ , η) is the desired unit step solution and f(z+ , η) is given by f(z+ , η) = [1 − θ(z+ , η)]

(8.681)

where θ(z+ , η) is given by θ(z+ , η) =

∞ ∑

) ( Cn φn exp −λ2n z+

n=0

On the other hand, the solution at z+ = ξ is given by f(z+ − ξ, η) = [1 − θ(z+ − ξ, η)] where θ(z+ − ξ, η) is given by θ(z+ − ξ, η) =

∞ ∑ n=0

[ ] Cn φn exp −λ2n (z+ − ξ)

(8.682)

439

440

8 Laminar Momentum and Heat Transfer in Channels

Example 8.9 Consider viscous fluid flow in a tube. Temperature distribution on a tube surface is shown in Figure 8.E9. We wish to obtain the temperature distribution in the fluid.

Tw1

(Tw1 – Tw0)

Tw0 (Tw0 – Ti) 0 z + = ξ0 = 0

Ti Figure 8.E9

z+ = ξ1

z+

Stepwise temperature distribution on tube surface in axial direction.

Solution Temperature distribution for z+ > ξ1 is T − Ti = (Tw0 − Ti )f(z+ , η) + (Tw1 − Tw0 )f(z+ − ξ1 , η) or

[ T − Ti = (Tw0 − Ti ) 1 −

∞ ∑

[ ] ] ∞ ∑ ( 2 +) [ 2 + ] Cn φn exp −λn z Cn φn exp −λn (z − ξ1 ) + (Tw1 − Tw0 ) 1 −

n=0

n=0

Let us return to Eq. (8.677) and notice that Eq. (8.677) may be written as ) ( z+ dTw dξ f(z+ − ξ, η) T − Ti = (Tw0 − Ti )f(z+ , η) + ∫0 dξ +

N ∑

f(z+ − ξi , η) ΔTw,i

(8.683a)

i=1

( ) ( ) where ΔTw,i = T ξ+i − T ξ−i and 𝛥Tw, i is the temperature difference across the discontinuity at ξi and N is the number of finite wall temperature step changes. Recall that the combination of Riemann integral and the summation is represented by the Stieltjes integral, and information about the Stieltjes integral is given by Klein and Tribus [37]. We may express Eq. (8.683a) in terms of the Stieltjes integral as follows: ) | ( z+ dTw | dξ | f(z+ − ξ, η) (8.683b) T − Ti = | ∫0 dξ |Stieltjes We are interested in heat flux distribution on the tube surface. In the next step, we evaluate the heat flux at the tube surface at z+ as ( ) ( ) 𝜕T k 𝜕T q′′w (z+ ) = k = (8.684) 𝜕r r=R R 𝜕η η=1 We differentiate the temperature distribution, Eq. (8.683a), with respect to η and evaluate at η = 1. This will give us ) ( z+ dTw 𝜕f(z+ − ξ, η) || 𝜕f(z+ , η) || 𝜕T || dξ = (Tw0 − Ti ) + | 𝜕η ||η=1 𝜕η ||η=1 ∫0 𝜕η |η=1 dξ N ∑ 𝜕f(z+ − ξi , η) || (8.685a) + | ΔTw,i | 𝜕η i=1 |η=1 We now express Eq. (8.685a) in terms θ(z+ , η) and θ(z+ − ξ, η) using Eqs. (8.681) and (8.682). First, we write the following relations using Eqs. (8.681) and (8.682):

8.7 Circular Pipe with Variable Surface Temperature Distribution in the Axial Direction

𝜕θ(z+ , η) || 𝜕f(z+ , η) || = − 𝜕η ||η=1 𝜕η ||η=1 + 𝜕θ(z+ − ξ, η) || 𝜕f(z − ξ, η) || =− | | 𝜕η 𝜕η |η=1 |η=1 Using these relations, we find that Eq. (8.685a) becomes 𝜕θ(z+ , η) || 𝜕T || = −(Tw0 − Ti ) − | 𝜕η |η=1 𝜕η ||η=1 ∫0 −

z+

𝜕θ(z+ − ξ, η) || | 𝜕η |η=1

(

dTw dξ

) dξ

N ∑ 𝜕θ(z+ − ξi , η) || | ΔTw,i | 𝜕η i=1 |η=1

and substituting Eq. (8.685b) into Eq. (8.684), we obtain an expression for heat flux distribution k(Tw0 − Ti ) 𝜕θ(z+ , 1) q′′w (z+ ) = − R 𝜕η [ + ] ) ( N z ∑ dTw k 𝜕 𝜕 + + θ(z − ξ, 1) dξ + θ(z − ξi , 1)ΔTw,i − R ∫0 𝜕η dξ 𝜕η i=1

(8.685b)

(8.686a)

Equation (8.686a) may be written in the following form: [ + ] ) ( N z ∑ dT k 𝜕 𝜕 w θ(z+ − ξ, 1) dξ + θ(z+ − ξi , 1)ΔTw,i (8.686b) q′′w (z+ ) = − R ∫0 𝜕η dξ 𝜕η i=0 ( ) ( ) ( ) ( ) where ΔTw,i = T ξ+i − T ξ−i ,T ξ−0 = Ti , T ξ+0 = Tw0 , ΔTw,0 = Tw0 − Ti , ΔTw,i is the temperature difference across the discontinuity at ξi , and N is the number of finite wall temperature steps. For the circular tube, the derivative 𝜕 θ/𝜕η|η = 1 can be obtained from Eq. (8.676). For this purpose, Eq. (8.676) is differentiated with respect to η and evaluated at the wall for η = 1. The result is ∞ ∞ ∑ dφn (λn η) || 2 + 𝜕θ(z+ , 1) ∑ −λ2n z+ = Cn e = −2 Gn e−λn z (8.687a) | 𝜕η dη | η=1 n=0 n=0 ( ) dφ where the values of Gn = − 12 Cn dηn were tabulated in Table 8.4. Then, we can write the following equation at z+ = ξ: η=1

∑ [ ] 𝜕 θ(z+ − ξ, 1) = −2 Gn exp −λ2n (z+ − ξ) 𝜕η n=0 ∞

Heat flux distribution now becomes ∞ [ ] 2k ∑ Gn exp −λ2n (z+ ) (Tw0 − Ti ) q′′w (z+ ) = R n=0 { +[∞ ]( ) z ∑ [ 2 + ] dTw 2k + dξ Gn exp −λn (z − ξ) ∫0 R dξ n=0 (∞ } ) N ∑ ∑ [ 2 + ] + Gn exp −λn (z − ξi ) ΔTw i i=1

(8.687b)

(8.688)

n=0

The local mixing cup temperature Tm is evaluated as follows: z

q′′ (z)dz = (πR2 )(ρ cp V)[Tm (z) − Ti ] (2 π R) ∫0 w

(8.689a)

Notice that all the heat entering into the fluid must be convected by fluid moving downstream. The dimensionless distance z+ is z 1 dz 1 z+ = ⇒ dz+ = R RePr R RePr Using the relations for z+ , the rearrangement of Eq. (8.689a) yields Tm (z+ ) − Ti =

4R k ∫0

z+

q′′w (ξ)dξ

(8.689b)

441

442

8 Laminar Momentum and Heat Transfer in Channels

We can determine the local Nusselt number NuD q′′w = h(TW − Tm ) NuD =

q′′w D q′′w D hD = = k k(Tw − Tm ) k[(Tw − Ti ) − (Tm − Ti )]

(8.690)

Example 8.10 Consider fully developed laminar flow in a circular tube, in which the surface temperature varies according to relation Tw (z+ ) − Ti = Az+ where A is a constant. Derive an expression for heat flux q′′w (z+ ) and local Nusselt number. Solution q′′w (z+ )

] [ + ) ( N z ∑ dTw k 𝜕 𝜕 + + θ(z − ξ, 1) dξ + θ(z − ξi , 1)ΔTw,i =− R ∫0 𝜕η dξ 𝜕η i=0

The second term is zero since there is no jump.

𝜕 θ(z+ 𝜕η

− ξ, 1) is given as

∞ ∑ [ ] 𝜕 θ(z+ − ξ, 1) = −2 Gn exp −λ2n (z+ − ξ) 𝜕η n=0

We now substitute this expression into the equation for q′′w (z+ ) ) ( ∞ z+ ∑ [ ] dTw 2k dξ Gn exp −λ2n (z+ − ξ) q′′w (z+ ) = R ∫0 n=0 dξ dTw =A dξ Thus, we get q′′w (z+ ) =

∞ z+ ∑ [ ] 2kA Gn exp −λ2n (z+ − ξ) dξ ∫ R 0 n=0

Integration yields

( ) ∞ 2 + 2 k A∑ 1 − e−λn z = G R n=0 n λ2n [∞ ] ∞ ∑ Gn ∑ G 2 + 2 k A n − e−λn z q′′w (z+ ) = 2 2 R n=0 λn n=0 λn q′′w (z+ )

But we can show by numerical calculation that ] ∞ 1 ∑ Gn −λ2n z+ − e 8 n=0 λ2n ∞ k A 2 k A ∑ Gn −λ2n z+ + q′′w (z+ ) = e 4R R n=0 λ2n [

q′′w (z+ )

∞ G ∑ n 2 n=0 λn

≈

1 8

2kA = R

We now integrate this expression from 0 to z+ to get total heat transfer up to this point and write an energy balance to get mean fluid temperature 4πR3 ρVcp k

z+

∫0

q′′w dx = πR2 ρVcp (Tm − Ti )

where x is a dummy variable. Integration yields 11 11 ∑ ∑ ) ( Gn Gn + 8A exp −λ2n Tm − Ti = Az+ − 8A 4 4 n=0 λn n=0 λn

8.8 Circular Pipe with Variable Surface Heat Flux Distribution in the Axial Direction

The local heat transfer coefficient is defined as [ ] ∞ 2 k A 1 ∑ Gn −λ2n z+ − e R 8 n=0 λ2n q′′w = h= [ ] 11 Tw − Tm 11 G ∑ ( 2) Gn ∑ n + + Az − Az − 8A + 8A exp −λn 4 4 n=0 λn n=0 λn [ [ ] ] ∞ ∞ 1 ∑ Gn −λ2n z+ 2 k A 1 ∑ Gn −λ2n z+ − − e e R 8 n=0 λ2n 8 n=0 λ2n 2 kA h= ] = ] [ [ 11 11 R 11 G 11 G ∑ ∑ ( 2) ( 2) Gn Gn ∑ ∑ n n 8A − exp −λn 8A − exp −λn 4 4 4 4 n=0 λn n=0 λn n=0 λn n=0 λn ] [ ∞ 1 ∑ Gn −λ2n z+ − e 8 n=0 λ2n h(2R) = NuD = [ ] 11 k 11 G ∑ ( 2) Gn ∑ n 2 − exp −λn 4 4 n=0 λn n=0 λn

8.8 Circular Pipe with Variable Surface Heat Flux Distribution in the Axial Direction Finding the temperature distribution is an important problem when the wall surface heat flux varies arbitrarily in the axial direction. Consider thermal entry length problem with variable surface heat flux distribution in axial direction for a laminar duct flow. We assume that constant property Newtonian viscous fluid flows in a pipe. Flow is steady and axisymmetric. There is no internal energy generation in the flow, and axial conduction and viscous dissipation are negligible. Since the energy equation is linear, the superposition principle can be used to construct solutions for variable surface heat flux. Therefore, constant heat flux thermal entry length problem solutions are used to construct solutions for any arbitrary axial variation in surface heat flux. Siegel et al. [35], Hanna and Myers [38], and Hanna [39] studied the problem. + First approach: We assume an axial heat flux variation q′′ w (z ), as shown in Figure 8.44a. First, we consider a process where

the heat flux is zero up to a position z+ = 0 and then takes on heat flux q′′0 for the remainder of the tube length, as shown in Figure 8.44b. We will employ Eq. (7.390) developed in Chapter 7. For an arbitrary variation of axial heat flux on a tube, we are interested in the wall temperature distribution Tw (z+ ), and Eq. (7.390) can be written with new notation as ( ′′ ) N z+ ∑ dqw + + Tw (z ) − Ti = dξ + F(z − ξ) Δq′′w,i F(z+ − ξ) (8.691) ∫0 dξ i=0 The next step is to find the function F(z+ − ξ). To find F(z+ − ξ), we will use the constant heat flux solution, which we have obtained in Section 8.6.1.3. In this case, we use Eq. (8.563), and we have )]} ( { [ ( ) ∞ λ2n z z∕R 11 ∑ R ′′ | Tw − Ti = qw0 4 + + (8.692) a φ exp − k ReD Pr 24 n=1 n n |η=1 Red Pr R or

{ Tw − Ti =

q′′w0

]} [ ∞ ( 2 +) 11 ∑ R + | 4z + + a φ exp −λn z k 24 n=1 n n |η=1

(8.693)

where z+ is given by z∕R 2(z∕D) = z+ = ReD Pr ReD Pr We may use Eq. (8.692) or Eq. (8.693) to obtain the function F(z+ − ξ). First, the function F(z+ ) is obtained from Eq. (8.693), and it is given as ] [ ∞ ( 2 +) R 11 ∑ + + | (8.694) 4z + + a φ exp −λn z F(z ) = k 24 n=1 n n |η=1

443

444

8 Laminar Momentum and Heat Transfer in Channels

q″w(z+) q″w2

q″w(z+)

q″w2 q″w1

q″w0

q″w0 0 ξ 1

ξ2

ξ3

z+ (a)

Figure 8.44

z+

0 (b)

(a) Arbitrary variation of surface heat flux. (b) Heat flux step change at the wall.

Then, we write for F(z+ − ξ) { } ∞ [ 2 + ] R 11 ∑ + + 4(z − ξ) + + a φ | exp −λn (z − ξ) F(z − ξ) = k 24 n=1 n n |η=1

(8.695)

Using Eq. (8.695), for arbitrary wall heat flux variation, the wall temperature distribution can be expressed in the following form: { }( ) | ∞ z+ | [ 2 + ] dq′′w R 11 ∑ + | + dξ|| 4(z − ξ) + Tw − Ti = an φn |η=1 exp −λn (z − ξ) (8.696) ∫ k 0 24 n=1 dξ | |Stieltjes We have developed a relation for thermal entrance problem under arbitrary axial heat flux distribution. This equation gives us the total effect of heat flux variation. Integration may be performed for a specified heat flux q′′w (z+ ) since we know the eigenvalues and eigenfunctions. Second approach: We now wish to proceed in a little different way to find F(z+ − ξ). We know that the solution for HFD and

TFD flow in a pipe under UHF q′′w = const is 48 = 4.364 11 Kays et al. [1] define the dimensionless temperature θ as Nu∞ =

(8.697)

Ti − T q′′w D∕k r η= R 2(z∕D) z+ = ReD Pr θ=

and Kays et al. [1] give the local Nusselt number NuD in a compact form [ ( 2 + ) ]−1 ∞ q′′w D hD 1 ∑ exp −γm z 1 NuD = − = = k k(Tw − Tm ) Nu∞ 2 m=1 Am γ4m

(8.698)

where constants and eigenvalues are given in Table 8.11. Recall that for the constant heat flux problem, the mixing cup temperature Tm varies with z+ . The surface temperature Tm and the mean fluid temperature are related as follows: Tw − Tm =

q′′w q′′ D = w h NuD k

Since we have an expression for NuD , we can express the temperature difference Tw − Tm as follows: ( 2 +) ] ( ′′ ) [ ∞ qw D 1 ∑ exp −γn z 1 − Tw − Tm = k Nu∞ 2 n=1 Am γ4m

(8.699)

(8.700)

8.8 Circular Pipe with Variable Surface Heat Flux Distribution in the Axial Direction

Table 8.11 Eigenvalues and constants for Eqs. (8.699) and (8.705). m

𝛄2m

Am

1

25.68

7.630 × 10−3

2

83.86

2.053 × 10−3

3

174.2

0.903 × 10−3

4

296.5

0.491 × 10−3

5

450.9

0.307 × 10−3

For larger m, γm = 4 m + Source: Kays et al. [1].

4 3

−7∕3

Am = 0.4165γm .

The mean temperature is θm = 4z+

(8.701a)

or Tm − Ti =

q′′w R + (4z ) = k

(

q′′w D k

) (2z+ )

(8.701b)

We now express the temperature difference Tw − Ti as follows: Tw − Ti = (Tw − Tm ) + (Tm − Ti )

(8.702)

We now substitute Eqs. (8.700b) and (8.701b) into Eq. (8.702). Then, temperature difference Tw − Ti can be expressed as ] [ 2 +] ( ′′ ) [ ∞ qw D 1 ∑ exp −λn z 1 + − + 2z (8.703a) Tw − Ti = k Nu∞ 2 n=1 Am γ4m or Tw − Ti = q′′w F(z+ )

(8.703b)

where F(z+ ) is the solution for unit heat flux, and it is given as [ ] [ 2 +] ∞ 1 D 1 ∑ exp −γn z + + − + 2z F(z ) = k Nu∞ 2 n=1 Am γ4m We can now express the wall temperature distribution for an arbitrary variation in heat flux q′′w (z+ ) as ( ′′ ) N z+ ∑ dqw + + Tw (z ) − Ti = dξ + F(z − ξ) Δq′′w,i F(z+ − ξ, 1) ∫0 dξ i=0

(8.704)

where F(z+ − ξ) is given as ] [ [ 2 + ] ∞ D 1 ∑ exp −γn (z − ξ) 1 + + − + 2(z − ξ) F(z − ξ) = k Nu∞ 2 n=1 Am γ4m + Third approach: We will now present another method described in [1]. Suppose that q′′ w (z ) is the known axial heat flux on

the pipe surface. The surface temperature can be calculated from the following relation: z+

R g(z+ − ξ) q′′w (ξ) dξ Tw (z ) − Ti = k ∫0 +

where g(z+ − ξ) is given as +

g(z − ξ) = 4 +

[ ] N ∑ exp −γ2m (z+ − ξ)

m=1

Am γ2m

(8.705)

445

446

8 Laminar Momentum and Heat Transfer in Channels

Here, the Kernel function g(z+ − ξ) is the wall temperature at x due to unit flux at ξ. The eigenvalues and constants for Eq. (8.705) are given in Table 8.11. The mixed-mean fluid temperature is determined from Tm (z+ ) − Ti =

4R k ∫0

z+

q′′w (ξ)dξ

(8.706)

Example 8.11 Surface heat flux on a tube varies q′′w = Cz+ . Obtain the surface temperature distribution. Solution We will use Eq. (8.704). The second term is zero since there is no jump in surface temperature ( ′′ ) ξ dqw + + dξ Tw (z ) − Ti = F(z − ξ ) ∫0 dξ dq′′w =C dξ ] [ [ 2 + ] ∞ z+ D 1 ∑ exp −λn (z − ξ) 1 + + − + 2(z − ξ) (C)dξ Tw (z ) − Ti = k ∫0 Nu∞ 2 n=1 Am γ4m [ ] ] ∞ z+ [ z+ ∑ z+ exp −λ2n (z+ − ξ) CD1 CD 2CD 1 + dξ − dξ + (z+ − ξ)dξ Tw (z ) − Ti = 4 k ∫0 Nu∞ k 2 ∫0 n=1 k ∫0 Am γm Integration yields

( ) ⎧ ⎫ ⎡ λ2n z+ ⎤ 1 − exp − ⎪ ⎪ ⎥ 4 ∞ ⎢ ( )⎪ + Am γm ⎥ 2 (z+ )2 ⎪ z D 1∑ ⎢ + − Tw (z ) − Ti = C ⎢ ⎥+ 2 ⎬ k ⎨ λ2n ⎪ Nu∞ 2 n=1 ⎢ ⎪ ⎥ ⎪ ⎪ ⎢ ⎥ ⎣ ⎦ ⎩ ⎭

8.9

Short Tubes

Recall that the thermal boundary conditions for laminar flow in tubes are very important. The developing thermal and hydrodynamic boundary layers will cause a significant increase in the heat transfer coefficient in entrance of short tubes. Rohsenow and Choi [40] present average Nusselt numbers for the case of constant wall temperature boundary condition based on the work of Goldberg [41]. The data are represented by an equation of the form as reported in [42]: ( )m a ReD Pr DL NuD = Nu∞ + (8.707) ( )n 1 + b ReD Pr DL D 1 < ReD Pr < 1000 L Nu∞ = 3.7 The constants a, b, m, and n are given in Table 8.12. Table 8.12

Constants for Eq. (8.707).

Pr

a

b

m

n

0.7

0.0791

0.0331

1.15

0.82

10

0.0534

0.0335

1.15

0.82

∞

0.0461

0.0316

1.15

0.84

8.9 Short Tubes

Table 8.13

Constants for Eq. (8.708).

Pr

a

b

m

n

0.7

0.00398

0.0114

1.66

1.12

10

0.00236

0.00857

1.66

1.13

∞

0.00172

0.00281

1.66

1.29

Heaton et al. [43] provide data for local Nusselt numbers under UHF boundary condition. Their data are represented by an equation of the form as reported in [42]: ( )m a ReD Pr DL (8.708) NuD = Nu∞ + ( )n 1 + b ReD Pr DL Nu∞ = 4.4 The constants a, b, m, and n are given in Table 8.13. For short tubes with L/D > 1 and with a sharp-edged entry, McAdams [44] recommends that the Nusselt number be calculated from [ ( )0.7 ] Nushort D = 1+ Nulong L Example 8.12 Water flows in a smooth tube at a mass flow rate of 0.015 kg/s. Tube is subjected to constant wall temperature. The mean water temperature is 280 K. The tube has a length of 2 m, and its internal diameter is 10 mm. Calculate the average heat transfer coefficient. Solution Water properties at 280 K are ρ = 1000kg∕m3 μ = 1422 × 10−6 N.s∕m2 k = 0.582W∕m K Pr = 10.26 The Reynolds number is 4 × 0.015 4ṁ ReD = = = 1343 πDμ π × (10∕1000) × 1422 × 10−6 Flow is laminar, and we use Eq. (8.708) ( ) D 0.01 ReD Pr = 1343 × 10.26 × ≈ 68.8 L 2.0 For Pr = 10.29, from Table 8.13, we have a = 0.0534 b = 0.0335 m = 1.15 n = 0.82 ( )m a ReD Pr DL NuD = Nu∞ + ( )n = 7.0 1 + b ReD Pr DL ] [ k 0.582 × (7.0) = 409.8W∕m2 K h = NuD = D (10∕1000)

447

448

8 Laminar Momentum and Heat Transfer in Channels

8.10 Effect of Property Variation For engineering calculations, constant property solutions obtained by analytic or experimental methods is corrected to account for property variation with temperature. Kays et al. [1], Thomas [6], Ghiaasiaan [47] discuss the property variation with temperature. In general, For liquids:

( )n μw St Nu = = Nucp Stcp μm ( )m μw cf = cf cp μm

(8.710a) (8.710b)

where Tw is the surface temperature. Tm is the mean fluid temperature. cp refers to constant property solutions For gases:

) ( Tw n Nu St = = Nucp Stcp Tm ) ( Tw m cf = cf cp Tm

Liquids

(8.711a) (8.711b)

(Tw > T∞ )

(Tw < T∞ )

Heating

Cooling

m 0.58

n

m

n

−0.14

0.5

−0.14

Cooling or heating Gases

m

n

1

0.0

All other properties are evaluated at mean temperature for internal flows. Kays et al. [1] report that the exponents m and n are functions of geometry and type of flow and give detailed information about the exponents m and n. Example 8.13 Engine oil at 40 ∘ C is flowing in a 1-cm tube. The tube length is 1 m, and the mass flow rate is 0.25 kg/s. Tube surface is maintained at 100 ∘ C by condensing steam on the external surface of the tube. The exit temperature is 60 ∘ C. Determine the average heat transfer coefficient. Solution Mean oil temperature Tm is Tm =

Tmi + Tmo 44 ∘ C + 70 ∘ C = = 57 ∘ C = 330 K 2 2

ρ = 865.8 kg∕m3 ν = 96.6 × 10−6 m2 ∕s Pr = 1205

μ = 8.36 × 10−2 N.s∕m2 k = 0.41 W∕m.K

8.11 Regular Sturm-Liouville Systems

Dynamic viscosity at 100 ∘ C μw = 0.01725 N.s∕m2 The Reynolds number ReD ReD =

4 × 0.25 4ṁ = ≈ 381 πμD π × 8.36 × 10−2 × 0.01

Flow is laminar. LH = 0.05 ReD D = 0.05 × 381 × 0.01 = 0.1903 m LH = 0.05 ReD D Pr = 0.05 × 381 × 0.01 × 1205 = 229.4 Flow HFD and thermally developing flow. ( ) ( ) 1 L 0.01 D = = 0.000217 We now calculate Gz−1 L = ReD Pr 381 × 1205 We use Eq. (8.517) to compute Nusselt number [( [( )1∕3 ] )1∕3 ] 0.01 D × ReD Pr × 381 × 1205 = 1.615 = 26.83 NuD = 1.615 L 1 We now correct Nusselt number for property variation ( )−0.14 ( )−0.14 μ μw NuD = ⇒ NuD = Nucp w Nucp μm μm ( )0.14 μ NuD = Nucp m μw ) ( 8.36 0.14 NuD = 26.85 = 33.47 1.725 Heat transfer coefficient h k 0.141 × 33.47 = 471.84 W∕m2 K h = × NuD = D 0.01

8.11 Regular Sturm-Liouville Systems Consider the linear homogenous second-order differential equation d2 y 2

dx

+ f1 (x)

] dy [ + f2 (x) + λ2 f3 (x) = 0 dx

This differential equation is converted into the general Sturm-Liouville problem [ ] [ ] dy d p (x) + q (x) + λ2 w (x) y (x) = 0 dx dx where p(x), q(x) and w(x) are defined as [ ] p(x) = exp f1 (x)dx ∫

(8.712)

(8.713)

449

450

8 Laminar Momentum and Heat Transfer in Channels

q(x) = f2 (x)p(x) w(x) = f3 (x)p(x) The eigenvalues and eigenfunction of the Sturm Liouville system have the following properties: (a) The eigenvalues are all real and positive. (b) The eigenvalues can be arranged to form an increasing sequence, i.e., 0 ≤ λ1 < λ2 < λ3 . …………. Additionally, λn → ∞ as n → ∞. (c) Corresponding to each eigenvalue λn , there is an eigenfunction denoted by ϕn (x) and ϕn (x) is unique to within an arbitrary multiplicative constant. (d) Eigenfunctions belonging to different eigenvalues are orthogonal relative to the weighting function w (x). This means that we can write {

b

∫a

w(x)ϕn (x)ϕm (x) dx =

0 N(λ)

m≠n m=n

(8.714)

The λ symbol in N (λ) is related to the nature of orthogonal function. Eq. (c) implies a set of functions ϕn (x) that is orthogonal over the interval a ≤ x ≤ b with the weighting function w (x). (e) Any piecewise smooth function f (x) can be represented by Fourier series of eigenfunctions: f (x) =

∞ ∑

(8.715)

an ϕn (x)

n=0

The constant an is given by b

an =

∫a w (x) ϕn (x) f (x) dx

(8.716)

b

∫a w (x) ϕ2n (x) dx

See Arpaci [28] as well as Ozisik and Hahn [46] for more information.

Problems 8.1

A certain constant property (ρ, cp , μ) fluid is flowing in a tube. The tube mean inlet temperature is Tmi . See Figure 8.P1. Assume that flow is laminar and fully developed. The pipe surface is subjected to sinusoidal heat flux given as ( ) πx q′′w = q′′0 sin L The tube length is L, the tube diameter is D, and the tube wall thickness is neglected. The fluid average velocity is V. Heat transfer coefficient between inside surface of pipe and fluid is h. Develop expression for: (a) the total rate of heat transfer from the pipe surface to fluid (b) the mean fluid temperature variation along the tube q″w

Tmi 0 x L

Figure 8.P1

Geometry and problem description for Problem 8.1.

D

Problems

(c) the fluid mean outlet temperature (d) the pipe surface temperature variation along the tube Consider a 5-cm-diameter and 120-cm-long circular tube. The tube surface is heated by an electric resistance heating element. A certain fluid is heated from 10 ∘ C to 66 ∘ C. The mass flow rate of fluid 15 × 10−4 kg/s. At a mean temperature of 38 ∘ C, the fluid properties are given as

8.2

Pr = 0.5 ρ = 900 kg∕m3 cp = 2.1 kJ∕kg.K k = 0.140 W∕m.K μ = 670 × 10−6 N.s∕m2 Assume Hagen–Poiseuille flow. We wish to plot both the tube surface temperature and the mean fluid temperature as a function of the tube length. 8.3

Consider a 2.5-cm-diameter and sufficiently long circular tube. It is desired to heat water from 10 ∘ C to 64 ∘ C. The mass flow rate of water is 2.5 × 10−3 kg/s. Flow is HFD when the heating begins, and the heated section is 1 m long. The tube surface is heated by an electric resistance heating element. We wish to calculate both the mean fluid temperature and the tube surface temperature along the heated section of the tube.

8.4

Water flows at a bulk temperature of 300 K in tube of 2-cm internal diameter. The mass flow rate of water is 2.5 × 10−2 kg/s. A UHF of 1300 W/m2 is applied to the pipe surface with an electrical heating element. We wish to determine: (a) the wall-to-tube centerline temperature difference, Tw − Tc (b) the difference between the bulk temperature Tm and the centerline temperature Tc (c) the rate of change of axial temperature, 𝜕T/𝜕x

8.5

A low-Prandtl-number fluid flows steadily in the entrance region of two parallel plates. Flow is laminar, two-dimensional, and incompressible with constant properties. Axial conduction and viscous dissipation are negligible, and there is no internal energy generation. The separation distance between the parallel plates is b. The coordinate system is shown in Figure 8.P5, where origin is located at the corner of the lower plate. The y-coordinate is measured from the lower plate and fluid enters with uniform velocity u = V as well as with uniform temperature Ti . The lower plate delivers a UHF q′′w to the fluid and the upper plate is insulated. Constant area duct is heated over its entire length. Study heat transfer and develop an expression for the Nusselt number using integral method. The penetration distance is defined as the distance required for the thermal boundary layer reaches the upper plate. Estimate the penetration distance. Consider only the first domain of the problem. y

Insulated wall

T = Ti b

u=V

Δ(x) 0

Figure 8.P5

x

q″w

Thermal boundary layer development in the entrance region of a parallel plate channel.

451

452

8 Laminar Momentum and Heat Transfer in Channels

A low-Prandtl-number fluid flows steadily in the entrance region of two parallel plates. The flow is laminar, two-dimensional, and incompressible with constant properties. Viscous dissipation is negligible, and there is no energy generation. The separation distance between the parallel plates is b. The coordinate system is shown in Figure 8.P6. The upper plate is insulated, while the lower plate is kept at constant temperature Tw . Fluid enters the channel with initial temperature Ti and with uniform velocity V. Study heat transfer and develop an expression for the Nusselt number using integral method. The penetration distance is defined as the distance required for the thermal boundary layer reaches the upper plate. Estimate the penetration distance. Consider only the first domain of the problem.

8.6

y

Insulated wall

T = Ti b

u=V

Δ(x) Thermal boundary layer 0

Tw

Figure 8.P6

8.7

Thermal boundary layer development in the entrance region of a parallel plate channel.

Consider axisymmetric, steady, laminar, and incompressible flow with constant properties in a tube having a diameter of D = 2R. Fluid is Newtonian. The velocity distribution and temperature distribution are fully developed in the tube. A UWT is applied to tube surface. See Figure 8.P7. Axial conduction and viscous dissipation are negligible along with internal energy generation. Consider now the heating of water flowing with a mass flow rate of ṁ = 0.01 kg/s inside the tube of 2.5 cm in diameter. Heating is done by condensing steam on the outer surface of the tube. Calculate the heat transfer coefficient by assuming water properties can be evaluated at 350 K. r

Tw

T(r)

2R z

Figure 8.P7

8.8

vz(r)

Geometry and problem description for Problem 8.7.

Assume that viscous fluid steadily and uniformly enters the channel at a velocity V and temperature Ti . The fluid and the channel walls are at the same temperature for x < 0, and for this reason, there is no heat transfer at the surface of the duct for the upstream half of the channel. The temperature of the fluid is uniform over the duct cross section at the point where the heat transfer begins. At some point x > 0, a UHF is applied to both plates. See Figure 8.P8. We assume that at the point where heat transfer begins, the velocity profile is fully developed and is equal to [ ( y )2 ] 3 u= V 1− 2 b and the temperature profile begins to develop. We will consider steady, two -dimensional laminar, incompressible, and low-velocity flow. Axial conduction is negligible, and there is no viscous dissipation as well as internal energy generation. Assume that water flows with a mass flow rate of 0.01 kg/s between the parallel plates. An electrical heat tape is placed on each surface of heated section of the parallel plate channel. The electrical heat tape provides a uniform heat flux of 750 W/m2 at each surface. The temperature at the inlet is 17 ∘ C. The heated section of the plates is 1 m and the spacing between parallel plates is 4 mm. The properties of water are ρ = 1000 kg/s, cp = 4200 J/kg.K, k = 0.598 W/m.K, μ = 1080 × 10−6 Pa.s, Pr = 7.56. (a) Estimate the average heat transfer coefficient. (b) Estimate the outlet temperature of the heated section.

Problems

y q″0

q″w = 0 u(y)

Ti 2b

vz = V

q″w (x) x

O

q″w = 0

q″0

x

q″w = 0

(a) Figure 8.P8

q″0

(b)

Geometry and problem description for Problem 8.8.

8.9

Consider viscous fluid flow in a pipe. Flow is both HFD and TFD flow. Flow is laminar and steady. UHF is applied to pipe surface. (a) Express the temperature distribution in terms of centerline temperature Tc . (b) Express the temperature distribution in terms of mean inlet temperature Ti .

8.10

Consider a laminar incompressible Newtonian fluid flowing in a tube having a diameter of D = 2R. The velocity profile is approximated as vz = V, and this is called slug flow. See Figure 8.P10. Axial conduction and viscous dissipation are neglected. A UHF q′′w is applied to tube surface. Define a dimensionless temperature θ θ=

T0 − T T0 − Tc

where T0 is the tube surface temperature and Tc is the tube center temperature. (a) Show that the temperature distribution is ( ) [ ( )2 ] V dT0 r 2 T = T0 − R 1− 4α dz R (b) Obtain the following expression: ) ( 2αq′′0 1 dT0 = dz r=R kV R (c) Show that the centerline temperature is Tc = T0 −

q′′0 R2

2k (d) Show that the mean fluid temperature is given by q′′0 R

Tm = T0 −

4k (e) Show that the Nusselt number is NuD = 8 We wish to determine the temperature distribution, and the Nusselt number in slug flow in the tube when the thermal field is fully developed. r

vz = V T = Ti

Figure 8.P10

2R

O

q″0

z

Geometry and problem description for Problem 8.10.

T(r)

453

454

8 Laminar Momentum and Heat Transfer in Channels

8.11

We know that for small z+ , heat transfer into fluid from the pipe wall affects only a very thin region near the wall. Problem description is given in Figure 8.P11. We can now make the following assumptions: (a) Curvature effects can be neglected, and y is measured from the pipe wall. (b) Fluid outside the thermal boundary layer will not feel the effect of wall, and for this reason, fluid will be treated as if it were extending from y = 0 to y = ∞. (c) The velocity profile will be y 4V vz ≈ 4V = βy β = R R We are stating that wall is flat, the velocity profile is linear, and fluid is infinite in extent. Under these conditions, the energy equation becomes βy

𝜕2 T 𝜕T =α 2 𝜕z 𝜕y

The boundary conditions are T(0, y) = Ti −k

𝜕T(z, 0) = q′′w 𝜕r

T(z, ∞) = Ti The formulation of the problem in terms of θ = T − Ti gives βy

𝜕2 θ 𝜕θ =α 2 𝜕z 𝜕y

The boundary conditions are θ(0, y) = 0 −k

𝜕θ(z, 0) = q′′w 𝜕r

θ(z, ∞) = 0 We will now nondimensionalize z, y, and θ in terms of arbitrarily selected reference lengths z0 , y0 and the characteristic temperature θR as follows: z∗ =

y z θ , y∗ = , f = z0 y0 θR

Develop a similarity solution and study heat transfer.

r y=R–r

Pipe center

Figure 8.P11

Geometry and problem description Leveque problem.

R

y q″w

8.12

Air at 1-atm pressure and 80 ∘ C enters a tube having an internal diameter of 5 mm. The average velocity of the air is 1 m/s. The tube length is 1 m, and a constant heat flux is applied to entire tube surface. An exit mean temperature of 130 ∘ C is required. Determine the required constant surface heat flux.

8.13

Engine oil is flowing in 60-mm-circular tube. The tube length is 20 m. The surface temperature of tube is maintained at 140 ∘ C. The inlet temperature of the engine oil is 17 ∘ C, and the mass flow rate of engine oil is 0.6 kg/s. Determine the exit temperature of the tube. Calculate the heat transfer.

Problems

8.14

Consider steady, laminar constant property, viscous, incompressible fully developed viscous flow in a circular pipe having a diameter D = 2R. Flow is asymptotically fully developed. Tube surface is subjected to constant temperature T0 . (Figure 8.P14). The fully developed velocity profile is given by [ ( )2 ] r vz = Vc 1 − R where Vc is the centerline velocity. Including the effect of viscous dissipation, we wish to determine the temperature distribution for very large distance from the tube entrance. r

2R

Figure 8.P14

8.15

T0

0

z

T(r) vz (r)

Geometry and problem description for Problem 8.14.

Consider steady, laminar constant property, viscous, and incompressible fully developed viscous flow in a circular pipe having a diameter D = 2R far from the entrance of the pipe. Both the velocity and temperature profiles are fully developed. See Figure 8.P15. The tube surface is subjected to constant surface heat flux. The fully developed velocity profile is given by [ ( )2 ] r vz = Vc 1 − R where Vc is the centerline velocity. Including the effect of viscous dissipation, we wish to determine the temperature distribution and the Nusselt number. r

q″0

T(r)

2R

Figure 8.P15

8.16

z

vz(r)

Geometry and problem description for Problem 8.15.

Consider steady, laminar constant property, viscous, and incompressible fully developed viscous flow in a circular pipe having a diameter D = 2R far from the entrance of the pipe. The tube surface is subjected to constant surface heat flux. See Figure 8.P16. The fully developed velocity profile is given by [ vz = Vc

( )2 ] r 1− R

where Vc is the centerline velocity. Consider the fluid far downstream of the heated section. It is intuitively expected that the constant heat flux through the wall will result in a rise in fluid temperature that is linear with respect to x. It is also expected that the shape of the radial temperature profile will not change with the increase in the axial distance x. Assume that the temperature profile is given by TFD (z, r) = B z + T(r) We wish to determine the Nusselt number.

455

456

8 Laminar Momentum and Heat Transfer in Channels

r

T(r)

2R

Figure 8.P16

8.17

q″0

z

vz(r)

Geometry and problem description for Problem 8.16.

Consider HFD and TFD steady flow of a viscous fluid between two infinite parallel plates spaced a distance 2L apart and upper plate is insulated, while the lower plate is subjected to a constant heat flux q′′0 . The coordinate system is shown in Figure 8.P17. Fluid properties are constant, and flow is steady and incompressible. There is no energy generation in the flow, and viscous dissipation is neglected. Axial conduction is negligible. The velocity profile for HFD flow is given by ( y )2 u =1− Vc L where Vc is the centerline velocity. We wish to determine the fully developed temperature distribution. Determine also the Nusselt number based on the hydraulic diameter. y

insulation u(y)

2L

x q″0

Figure 8.P17

8.18

Geometry and problem description for Problem 8.17.

Consider HFD steady flow of a viscous fluid between two infinite parallel plates spaced a distance 2H apart; the upper plate is insulated, while the lower plate is subjected to a constant temperature T0 . Flow is also asymptotic TFD. The coordinate system is shown in Figure 8.P18. Fluid properties are constant, and flow is steady and incompressible. There is no energy generation in the flow, and viscous dissipation is not negligible. Axial conduction is negligible. The velocity profile for HFD flow is given by u(y) =

[ ( y )2 ] 3 V 1− 2 H

We wish to determine the asymptotic TFD temperature distribution (the temperature distribution for very large x). y

insulation u(y)

2H

x T0

Figure 8.P18

Geometry and problem description for Problem 8.18.

Problems

8.19

Consider HFD and TFD steady flow of a viscous fluid between two infinite parallel plates spaced a distance 2L apart; the upper plate is made of glazing material and is adiabatic, while the lower plate is subjected to a constant heat flux q′′0 . The coordinate system is shown in Figure 8.P19a. Fluid properties are constant, and flow is steady and incompressible. There is no energy generation in the flow, and viscous dissipation is neglected. Axial conduction is negligible. The velocity profile for HFD flow is given by ( y )2 u =1− Vc L (a) We wish to determine the fully developed temperature distribution. Determine also the Nusselt number based on the hydraulic diameter. y

adiabatic u(y)

2H

0

x q″0 (a)

Figure 8.P19a

Geometry and problem description for viscous flow between parallel plates.

(b) Consider a flat plate solar collector, as shown in Figure 8.P19b. Water is the working fluid in the collector, and it has a mass flow rate of ṁ = 0.06 kg∕s. The mean inlet water temperature is Tmi = 27 ∘ C All the solar energy is absorbed by the absorber plate. The solar collector is perfectly insulated, and the spacing between the horizontal parallel plates is 2L = 5 cm. Determine the outlet water temperature and heat transfer coefficient. Glass

q″0 = 1200 W/m2 Black absorber plate

2m Water channel

1m Insulation (b)

Figure 8.P19b

Solar collector.

8.20

Engine oil at a mean temperature 57 ∘ C enters through a 1-m-long equilateral triangular duct. One side of the duct is 0.6 cm, and the duct surface is kept at 100 ∘ C by condensing steam on the duct surface. The mass flow rate of engine oil is 10−3 kg/s. Assuming that the flow is both HFD and TFD, estimate the outlet temperature of the duct.

8.21

Water at temperature of 14 ∘ C enters the 2-m-long heated section of a tube after passing through an isothermal calming section. The average water velocity is 3 cm/s. The internal diameter of the tube is 2.5 cm. The heated part of the tube wall is maintained at 100 ∘ C by condensing steam on the external surface of the tube. Problem description is given in Figure 8.P21. We wish to determine the exit temperature of the water.

457

458

8 Laminar Momentum and Heat Transfer in Channels

Fully developed velocity profile r Tmi = 17 °C

100 °C

0

D = 2.5 cm

z 2m

Figure 8.P21

8.22

Geometry and problem description for Problem 8.21.

Water at temperature of 14 ∘ C enters the 2-m-long heated section of a tube after passing through an isothermal calming section. The average water velocity is 3 cm/s. The internal diameter of the tube is 2.5 cm. The velocity profile is fully developed at the point where the heating begins. An electric resistance heater is used to apply a UHF on the external surface of the tube so that the outlet temperature of water becomes 43 ∘ C. See Figure 8.P22. Calculate the required UHF to accomplish this task. We also wish to determine tube surface temperature at the exit. Fully developed velocity profile q″0

r Tmi = 14 °C

D = 2.5 cm 0

z

Tmo = 43 °C

2m Figure 8.P22

Geometry and problem description for Problem 8.22.

8.23

Air is flowing with a mean velocity of 0.5 m/s through a circular tube. The internal diameter of the tube is 3 cm. Air is heated from 290 to 310 K. The tube wall is maintained at a temperature of 320 K. The tube length cannot exceed 2 m. We wish to determine the tube length required to accomplish this task.

8.24

It is desired to heat water from 20 ∘ C to 55 ∘ C in a tube. The mass flow rate of water is 0.015 kg/s. The internal diameter of the tube is 3 cm. Flow is both HFD and TFD. Heating can be done by condensing steam on tube, thus keeping the tube surface at Tw = 100 ∘ C or by an electrical heating element. The heating element provides a uniform surface heat flux of 5500 W/m2 . Determine the length of the tube necessary to accomplish each task.

8.25

Air at 1-atm pressure is flowing with a mass flow rate of 5 × 10−5 kg/s through a circular tube. The internal diameter of the tube is 4 mm. Air enters the tube at a temperature of 700 K. The tube wall is maintained at a temperature of 950 K. The tube length is 50 mm. We wish to determine the tube outlet temperature.

8.26

Water enters a tube at uniform temperature of 27 ∘ C. The internal diameter of the tube is 3 cm. The mass flow rate of water is 0.035 kg/s. It is desired to heat water in a tube. Heating can be done by an electrical heating element. The heating element provides a uniform surface heat flux of q′′0 (W∕m2 ), and heat flux is unknown. The tube surface at the outlet can be no larger than 40 ∘ C. The heated section of the tube is 100 cm. Determine the heat flux necessary to accomplish this task as well as the mean outlet fluid temperature.

8.27

Consider HFD and TFD steady flow of a viscous fluid between two infinite parallel plates spaced a distance 2b apart, and each plate is subjected to a constant surface temperature Tw . The fully developed velocity profile is given by [ ( y )2 ] u = uc 1 − b

Problems

where uc is the velocity on the centerline. See Figure 8.P27. Dimensionless temperature is defined in terms of centerline temperature Tc , i.e. Tw − T y = θ(η) η = Tw − Tc b We wish to determine the fully developed temperature distribution and Nusselt number. y

Tw u(y)

2b

0

x Tw

Figure 8.P27

8.28

Geometry and problem description for Problem 8.27.

Consider a channel formed by two parallel 1-m-wide plates. The separation distance between the plates is 2 b = 4 m m. Air at 1-atm pressure with a mean velocity of 3.5 m/s enters the channel. After a calming section, the velocity profile becomes fully developed. Heating begins at the location where the velocity profile is fully developed. Air is heated by raising the surface temperature of the plates to 60 ∘ C, as shown in Figure 8.P28. Determine (a) mean bulk temperature of air, (b) the local Nusselt number, (c) the mean Nusselt number at the exit of this heated section, and (d) heat transfer to the air. 17 °C

Tw = 60 °C

y

u(y) Tmi = 17 °C V = 3.5 m/s

0

2b = 4 mm

x

17 °C Tw = 60 °C 5 cm

Figure 8.P28

Geometry and problem description for Problem 8.28.

8.29

Air flows in a tube with a mean velocity of 1.2m/s. The mean inlet and exit temperatures are 17 ∘ C and 87 ∘ C. The internal diameter of the tube is 3 cm and a UHF of 5 00 W/m2 is applied on the tube surface. Estimate the tube surface temperature at the tube outlet.

8.30

A Newtonian constant property viscous fluid flows in a tube. The flow is laminar and HFD and z+ = 0 is the point where heat transfer starts. See Figure 8.P30. The heat flux into fluid at the wall is specified as (

) πz L where L is the tube length, and the heat flux varies axially along the tube. Obtain an expression for the wall surface temperature variation along the tube. q′′w = q′′0 sin

η

2R

Figure 8.P30

q″0

vz(η)

z+

Geometry and problem description for Problem 8.30.

459

460

8 Laminar Momentum and Heat Transfer in Channels

Consider steady flow of a liquid metal between two infinite parallel plates. The spacing between the plates is H. The upper plate is insulated while the lower plate is subjected to a constant surface temperature Tw . Velocity distribution may be approximated by u ≈ V. See Figure 8.P31. Using the integral method, we wish to determine the Nusselt number in the entrance region based on Tw − Ti .

8.31

insulation

Ti

y

H

Δ (x)

u=V

Tw

x

0 Figure 8.P31

Thermal boundary layer development in the entrance region of parallel plate channel.

An incompressible viscous fluid with constant properties (ρ, μ, cp , k) is flowing in a tube. See Figure 8.P32. Flow is laminar and HFD. For z < 0, the fluid temperature is uniform at the inlet temperature Ti . For z > 0, the tube is externally insulated. Viscous dissipation is not negligible. Axial conduction is omitted since it is usually small in comparison with convection in the axial direction. At a far downstream distance from the beginning of the heated section, the temperature profile in the fluid may be represented in the form

8.32

T(r, z) = T(r) + A z where A is a constant. This tells us that temperature profile does not change with z. (a) Obtain the temperature profile T(r) in terms of the tube centerline temperature Tc . 𝜕T 8 V ν = . (b) Show that 𝜕z cp R2 V is the average velocity, R is the tube radius, cp is the constant pressure specific heat, and ν is the kinematic viscosity. In this problem, we are seeking an asymptotic solution for large z. We expect that the shape of temperature profiles as a function of r will not undergo any more change with z.

r

insulation 2R

Ti z

0

Figure 8.P32

8.33

Geometry and problem description for Problem 8.32.

Consider steady, laminar constant property, viscous, and incompressible viscous flow in a circular pipe having a diameter D = 2R, as shown in Figure 8.P33. Fluid enters the tube as HFD. The tube surface is subjected to variable surface heat flux, and the heat flux into the fluid at the wall is given by q′′w = q0 sin(π z∕L). We wish to determine the surface temperature distribution.

r

q″w = q0 sin

⎛ πz ⎛ ⎝ L ⎝

vz (r) z

D = 2R

Figure 8.P33

Flow in a circular pipe.

Problems

8.34

Consider steady, laminar constant property, viscous, and incompressible HFD viscous flow in a circular pipe having a diameter D = 2R far from the entrance of the pipe. The temperature profile on the pipe surface is given in Figure 8.P34. Develop an expression for the wall heat flux and bulk fluid temperature. Tw (z+)

η

0

2R

z+

Tw (z+) Tw2 Tw1 Tw0 Ti z+ = ξ1

z+ = 0

Figure 8.P34

8.35

z + = ξ2

Geometry and problem description for Problem 8.34.

Consider steady, laminar constant property, viscous, and incompressible HFD viscous flow in a circular pipe having a diameter D = 2R far from the entrance of the pipe. Heat added at constant rate along the tube, as shown in Figure 8.P35. Develop an expression for the wall temperature distribution. q″w ⎛⎝ z+ ⎛⎝

r

2R

0

z

q″w ⎛⎝z+ ⎛⎝ q′′w1 q″w0 Ti z+ = 0 Figure 8.P35

8.36

z + = ξ1

Geometry and problem description for Problem 8.35.

Consider steady, laminar, constant property, viscous, and incompressible HFD flow in a circular pipe having a diameter D = 2R far from the entrance of the pipe. Heat added along the tube to fluid obeys the relation given by q′′w = Cz+ Develop an expression for the wall temperature distribution. Also determine the mixed-mean temperature.

461

462

8 Laminar Momentum and Heat Transfer in Channels

8.37

Consider steady, laminar, constant property, viscous, and incompressible HFD flow in a circular pipe having a diameter D = 2R. See Figure 8.P37. Surface temperature along the tube is given by the relation Tw = Ti + exp(z+ ) Develop an expression for the Nusselt number. Tw

r

0

2R

Figure 8.P37

8.38

z

Geometry and problem description for Problem 8.37.

Consider HFD thermally developing laminar flow between two infinite parallel plates, as shown in Figure 8.P38. The surface temperature of the plate is given by Tw − Ti = A + Bx where Ti is the fluid inlet temperature, and A and B are constants. Develop an expression for the local heat flux distribution. Ti

2b

y

Tw u(y)

Ti 0 Ti

Figure 8.P38

8.39

x Tw

Geometry and problem description for Problem 8.38.

Consider the Leveque solution in a circular pipe. We have obtained the following governing equation and its boundary conditions 𝜕T 𝜕2 T =α 2 𝜕z 𝜕y T(0, y) = Ti βy

θ(z, 0) = Tw θ(z, ∞) = Ti Assume a dimensionless temperature θ defined as θ=

T − Tw Ti − Tw

The PDE and its boundary conditions become 𝜕θ 𝜕2 θ =α 2 𝜕z 𝜕y θ(0, y) = 1

βy

References

θ(z, 0) = 0 θ(z, ∞) = 1 To find a similarity variable, assume θ = θ(η), where η is the similarity variable, and it is defined as y η= δ(z) Find the unknown parameter δ(z) and show that PDE and its boundary conditions reduce to dθ d2 θ + 3η2 =0 dη dη2 The boundary conditions are θ(0) = 0 θ(∞) = 1.

References 1 Kays, W.M., Crawford, M.E., and Weigand, B. (2005). Convective Heat and Mass Transfer. McGraw-Hill. 2 White, F.M. and Majdalani, J. (2022). Viscous Fluid Flow, 4e. McGraw Hill. 3 Shah, R.K. and London, A.L. (1978). Laminar Flow Forced Convection in Ducts: A Source Book for Compact Heat Exchanger Analytical data. New York: Academic Press. 4 Shah, R.K. (1978). A correlation for laminar hydrodynamic entry length solutions for circular and noncircular ducts. J. Fluids Eng. 100: 177–179. 5 Chen, R.Y. (1973). Flow in the entrance region at low Reynolds numbers. J. Fluids. Eng. 95: 153–158. 6 Thomas, L.C. (1999). Heat Transfer, 2e. Capstone Publishing Company. 7 Incropera, F.P., Dewitt, D.P., Bergman, T.L., and Lavine, A.S. (2013). Foundations of Heat Transfer, 6e. Wiley. 8 Nellis, G. and Klein, S. (2021). Introduction to Engineering Heat transfer, Cambridge University. 9 Burmeister, L.C. (1993). Convective Heat Transfer, 2e. Wiley. 10 Sieder, E.N. and Tate, G.E. (1943). Heat transfer and pressure drop of liquids in tubes. Ind. Eng. Chem. 28: 1429. 11 Baehr, H.D. and Stephan, K. (2011). Heat and Mass Transfer, Revised Third Edition, Springer-Verlag. 12 Bennett, T.D. (2013). Transport by Advection and Diffusion. Wiley. 13 Bennett, T.D. (2019). Correlations for convection in hydrodynamically developing laminar duct flow. ASME J. Heat Transfer 141: 111701–111711. 14 Nellis, G. and Klein, S. (2009). Heat Transfer. Cambridge University. 15 Hornbeck, R.W. (1965). An all-numerical method for heat transfer in the inlet of a tube. American Society of Mechanical Engineering, Paper 65-WA/HT-36. 16 Churchill, S.W. and Ozeo, H. (1973). Correlations for laminar forced convection in flow over an isothermal flat plate and in developing and fully developed flow in an isothermal tube. J. Heat Transfer 95 (3): 416–419. 17 Churchill, S.W. and Ozeo, H. (1973). Correlations for laminar forced convection with uniform heating in flow over a plate and in developing and fully developed flow in tube. ASME J. Heat Transfer 95 (1): 78–84. 18 Kays, W.M. (1955). Numerical solutions for laminar flow heat transfer in circular tubes. Trans. ASME 77: 1265–1274. 19 Gnielinski, V. (2010). Heat Transfer in Pipe Flow, VDI – Heat Atlas, Chapter G1 and Chapter G2. Berlin: Springer. 20 Prins, J.A., Mulder, J., and Schenk, J. (1956). Heat transfer in laminar flow between parallel plates. Appl. Sci. Res. A2: 431–438. 21 Arpaci, V.S. and Larsen, P.S. (1984). Convection Heat Transfer. Prentice-Hall. 22 Shah, R.K. and Bhatti, M.S. (1987). Laminar convective heat transfer in ducts. In: Handbook of Single-Phase Convective Heat Transfer (ed. S. Kakac, R.K. Shah and W. Aung) Chapter 3. Wiley-Interscience. 23 Shah, R.K. and London, A.L. (1974). Thermal boundary conditions and some solutions for laminar duct flow forced convection. ASME J. Heat Transfer 96 (2): 159–165. 24 Deen, W.M. (2013). Analysis of Transport Phenomena. Oxford University Press. 25 Brown, G.M. (1960). Heat or mass transfer in a fluid in laminar flow in a circular or flat conduit. AIChE J. 6: 179–183. 26 Davis, E.M. (1973). Exact solutions for a class of heat and mass transfer problems. Canadian J. Chem. Eng. 51: 562–572. 27 Boelter, L.M.K., Cherry, V.H., Johnson, H.A., and Martinelli, R.C. (1965). Heat Transfer Notes. McGraw Hill Book Company. 28 Arpaci, V.S. (1966). Conduction Heat Transfer. Addison-Wesley.

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8 Laminar Momentum and Heat Transfer in Channels

29 Abramowitz, M. and Steigun, I. (1972). Handbook of Mathematical Functions with Formulas, Graphs, and, Textbook Mathematical Tables, 9th Printing. New York: Dover Publications. 30 Bennett, T.D. (2021). Generalized Leveque solution for ducts of arbitrary cross section, ASME. J. Heat Transfer 143 (4): 041801 31 Sellars, J.A., Tribus, M., and Klein, J.S. (1956). Heat transfer to laminar flow in a round tube or flat conduit – the Graetz problem extended. Trans. ASME 78: 441–448. 32 Shah, R.K. (1975). Thermal entry length solutions for circular tube and parallel plates. Proc. Nat. Heat Mass Transfer Conf. 3rd, Volume I, Paper No. HMT- 11-11-75, Indian, Institute of Technology, Bombay. 33 Bennett, T.D. (2019). Correlations for the Graetz problem in convection-Part 1: for round pipes and parallel plates. Int. J. Heat Mass Transfer 136: 832–841. 34 Stephan, K. (1994). Thermodynamics, in: Dubbel, Handbook of Mechanical Engineering, Volume 1, Edited by W. Beitz and K.-H. Kuttner English Edition edited by B.J. Davies Translation by M.J. Shields C30–C31. Berlin: Springer-Verlag. 35 Siegel, R., Sparrow, E.M., and Hallman, T.M. (1958). Steady laminar heat transfer in a circular tube with prescribed wall heat flux. Appl. Sci. Res. Sect. A 7: 386–392. 36 Shome, B. and Jensen, M.K. (1993). Correlations for simultaneously developing laminar flow and heat transfer in a circular tube. Int. J. Heat Mass Transf. 36 (6): 2710–2713. 37 Tribus, M. and Klein, J. (1952). Forced Convection from non-isothermal surfaces, Engineering Research Institute, Project M992-B. Ann Arbor: University of Michigan. 38 Hanna, O.T. and Myers, J.F. (1962). Heat transfer in boundary layer flows past a flat plate with a step wall-heat flux. Chem. Eng. Sci. 17: 1053–1055. 39 Hanna, O.T. (1962). Step-wall heat flux superposition for heat transfer in the boundary layer flows. Chem. Eng. Sci. 17: 1041–1051. 40 Rohsenow, W.M. and Choi, H. (1961). Heat Mass and Momentum Transfer. Englewood Cliffs, NJ: Prentice-Hall. 41 Goldberg, P. (1958). M.S. Thesis, Mechanical Engineering Department. M.I.T. 42 Duffie, J.A. and Beckman, W.A. (2013). Solar Engineering of Thermal Processes, 4e. Wiley. 43 Heaton, H.S., Reynolds, W.C., and Kays, W.M. (1961). Heat transfer in annular passages. simultaneous development of velocity and temperature fields in laminar flow. Int. J. Heat Mass Transfer 7: 763–781. 44 McAdams (1954). Heat Transmission, 3e. New York: McGraw Hill. 45 Jacob, M. (1949). Heat Transfer. Wiley. 46 Hahn, D.W. and Ozisik, M.N. (2012). Heat Conduction, 3e. Wiley. 47 Ghiaasian, S.M. (2013). Convective Heat and Mass Transfer, 2e. CRC Press. 48 White, R.E. and Subramanian, V.R. (2010). Computational Methods in Chemical Engineering with Maple. Springer. 49 Rice, R.G. and Do, D.D. (2013). Applied Mathematics and Modelling for Chemical Engineers, 2e. Wiley. 50 Aydin, O. and Avci, M. (2006). Viscous dissipation effects on heat transfer in a Poiseuille flow. Appl. Energy 83: 495–512. 51 Ou, J.W. and Cheng, K.C. (1973). Viscous dissipation effects on thermal entrance region heat transfer in pipes with uniform wall heat flux. Appl. Sci. Res. 28: 289–301. 52 Hausen, H. (1943). Darstellung des Wärmeuberganges in Rohren durch verallgemeinerte Potenzbeziehungen (Translation: Representation of the heat transfer in tubes by generalized power relationships.). Z VDI Beah Verfahrenstech 4: 91–98. 53 Deisler, R.G. and Eian, C.S. (1952). Analytical and experimental investigation of heat transfer with variable fluid properties. NACA TN 2629, 1952. 54 Rathore, M.M. and Kapuno, R.R.A. (2011). Engineering Heat Transfer, 2ee. Jones and Bartlett. 55 Magrab, E.B., Azarm, S., Balachandran, B., Duncan, J.H., Herold, K.E. and Walsh, G.C. (2011). An Engineer’s Guide to MATLAB, Pearson Education, Inc. 56 Meade, D.B., Michael May, S.J., Cheung, C.K. and Keough, G.E. (2009). Getting Started with MAPLE, John Wiley and Sons, Inc. 57 Gilbert, R.P., Hsiao, G.C. and Ronkese, R.J. (2021). Differential Equations-A Maple Supplement 2e, CRC Press.

465

9 Foundations of Turbulent Flow 9.1 Introduction Turbulent flow is very important in engineering applications. It involves in a large majority of fluid flow and heat transfer problems encountered engineering. Velocity, temperature, and pressure along with other properties change in time at every point in flow field. The governing equations of mass, momentum, and energy developed for laminar flows are also valid for turbulent flow; however, all three components of the velocity vector will be nonzero functions of the spatial coordinates and time. Velocity, temperature, and pressure along with other properties are instantaneous values in the equations. This makes the problem very complicated in turbulent flows. Turbulent flows are characterized by highly irregular fluctuations in velocity, temperature, pressure, and other properties. At each point, these properties fluctuate around a mean value. For this reason, mean and fluctuating components of velocity, temperature, pressure, and other properties are defined to simplify the problem. Turbulence is a very difficult topic to tackle; although it has been under study for more than a century, there is no single universal solution to problems encountered in turbulent flows. On the other hand, direct numerical simulation (DNS) is very expensive, and at present, it is possible to use only in simple flow configurations. Key tools to study the turbulence are the time-averaged equations of motion, continuity, and energy and experimental methods. We need to model the fluctuating components in turbulent flows. There are numerous models; each model has its own area of applicability in engineering.

9.2 The Reynolds Experiment Reynolds performed a series of experiments in 1883, and these experiments were the first to illustrate the difference between laminar and turbulent flows. See Figure 9.1. Water was allowed to flow in a glass pipe. A thin stream of dye was injected at a point in the tank, and the motion of the dye was observed as it moves along the pipe. (a) At low velocities, the dye moved in a straight line along the tube indicating laminar flow. In general, laminar flow occurs when adjacent layers of fluid move relative to each other, forming smooth streamlines, not necessarily straight, without macroscopic mixing. It is the mode that occurs when viscous shear, which is caused by molecular momentum exchange between fluid layers, is the predominant influence in establishing the flow field. (b) As the velocity increased, the dye line became thinner and began a sinuous motion. This is probably the onset of turbulence, and it is called transitional flow. (c) A further increase in velocity caused the dye to break up into smaller segments, and these are the turbulent eddies. The turbulent flow is characterized by fluid particles having irregular, near random, fluctuating motions and erratic paths. Macroscopic mixing occurs in both the lateral and main flow directions. The value of the Reynolds number for a given flow is an indication of whether the flow may be laminar or turbulent, and for flow in a pipe, it is defined as ReD =

ρVD μ

(9.1)

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

466

9 Foundations of Turbulent Flow

Dye injection

Air

V

D

Water Laminar flow

Figure 9.1

Turbulence

Reynolds experiment.

where V is the mean fluid velocity, D is the tube internal diameter, ρ is the fluid density, and μ is the fluid dynamic viscosity. Based on the experimentally determined value of the Reynolds number ReD , flow in a tube is classified as follows: ReD < 2300 laminar flow 2300 < ReD < 10 000 transitional flow ReD > 10 000 fully turbulent flow. On the other hand, the critical Reynolds number Rex,c for turbulent flow over a flat plate is Rex,c =

9.3

U∞ xc = 3.5 × 105 to 106 . ν

Nature of Turbulence

Turbulence is all around us, and it is a common occurrence in nature. Experiments indicate that turbulence arises from instabilities within the laminar flows. Turbulence needs energy from its environment; otherwise, it dies out. A common energy source for turbulent velocity fluctuations is shear in the mean flow. Turbulent flows are often shear flows. In a turbulent flow, fluid motions vary irregularly. For this reason, quantities such as velocity, pressure, and temperature show a random variation with time and space coordinates. Turbulence can be detrimental or beneficial. Examples of detrimental cases: (a) Any situation when accurate predictions are needed such as weather predictions (b) Drag on rapidly moving objects (c) Cases where noise is an issue (Fan noise). Examples of beneficial cases: (a) Dilution of pollution (Air pollution) (b) Chemical reactions. We can ask the question: What is turbulence? The answer is that there is no clear and precise definition of turbulent flow. We can list some of the characteristics of turbulent flow. Tennekes and Lumley [1], Ting [2], Wilcox [3], and Pope [4] discuss in depth the characteristics of turbulent flow. We will list some common characteristics of the turbulent flow. Irregularity: All turbulent flows are irregular and chaotic and unsteady but governed by the Navier–Stokes equation. The flow may be steady in the mean and have a coherent structure. Diffusivity: In turbulent flow, the diffusivity increases. The turbulence increases the momentum exchange and delays the separation over bluff bodies, such as cylinders, airfoils, and cars. The increased diffusivity also increases resistance (wall resistance). The flow disturbance spreads over large distances. All the turbulent flows exhibit faster heat, momentum, and mass transfer and greater mixing. The diffusivity of turbulence is an important feature for engineering applications.

9.4 Time Averaging and Fluctuations

Figure 9.2a

Distribution of eddies.

Channel wall

Channel wall (a)

Large Reynolds number: Turbulent flows occur at high Reynolds number. The turbulence can be traced back to instabilities in the laminar flow region. The growth of these instabilities is amplified. These growing instabilities eventually lead to turbulent flow. Three-dimensional vorticity fluctuations: The flow is rotational, three dimensional, and unsteady. Turbulence has high levels of fluctuating vorticity. Even a time-averaged fully developed flow has fluctuating velocity components. Eddies are defined as swirling patches of fluid. Eddies are formed and destroyed continually. Different types of eddies occur in channel flow, and this occurrence of different types of eddies is illustrated in Figure 9.2a. Large eddies are generated at some distance from the walls, and large- and medium-size eddies contain most of the kinetic energy of the fluid. Viscous dissipation is negligible in the motion of large eddies, and large eddies transfer their energy to smaller eddies, and in smaller eddies, kinetic energy is consumed by viscous dissipation. Figure 9.2a represents an oversimplification of the physical problem. Actually, there is a spectrum of eddy sizes, and viscous dissipation takes place in the lower range of this spectrum, as discussed in [1, 2] and [4]. For most turbulent flows, an eddy is created as a disturbance near the wall. A rotating fluid element appears, and a small vortex filament rolls along the wall and in a line perpendicular to flow. There is a tendency that filament lifts up from the wall. The lifted region of the vortex filament eventually stretches and takes the shape of a hairpin vortex. This vortex stretching increases the kinetic energy of the vortex, and it is believed to be the mechanism for the main flow to transfer energy to the turbulence. See [4] for a detailed discussion on the origin of turbulence. Dissipation: All turbulent flows are dissipative. There is a large energy loss in turbulent flow. The kinetic energy of turbulence is converted to heat through friction (viscous dissipation). Turbulence needs a continuous supply of energy, which comes from mean flow. Otherwise, it dies out. Continuum: Turbulence is a continuum phenomenon and is a feature of the flow. It is not a feature of the fluid. Dynamics of the turbulence is the same for all the fluids if the Reynolds number is the same. Similar classes of flow have similar turbulence characteristics.

9.4 Time Averaging and Fluctuations To provide a sound engineering background for the analysis of turbulent flows, we will examine some experimental results. A simple imaginary picture of turbulent flow over a flat plate is illustrated in Figure 9.2b. The instantaneous velocity profile is time averaged, and it has a mean profile like dashed line shown in Figure 9.2b. In this way, turbulent velocity has a mean velocity u and a fluctuating component u′ , as shown in Figure 9.2c. Mean turbulent velocity relative to laminar flow near a wall is sketched in Figure 9.2d. The turbulent velocity profile has a steeper velocity profile. This means that the velocity gradient is large. This increases the skin friction coefficient in turbulent flow. Turbulent fluctuations increase momentum transfer between the wall and the flow. Turbulent fluctuations also increase heat transfer between surface and fluid. We will follow Reynold’s decomposition approach to study turbulent flow. Since we are interested in deriving time-averaged forms of transport equations, we will introduce some definitions about time averages. We define the time average of some function S at time interval Δt

S(x, y, z) =

1 𝚫t ∫0

𝚫t

S(x, y, z, t) dt.

(9.2)

467

468

9 Foundations of Turbulent Flow

Figure 9.2b

Instantaneous velocity profile in channel flow.

y

u instantaneous velocity profile

0

u (b) Figure 9.2c

Variation of u with time t.

u(t) uʹ

u

0

t

(c)

Figure 9.2d Comparison of laminar and time-averaged turbulent velocity profiles in a boundary layer.

y Laminar velocity u Mean turbulent velocity u 0

Wall surface (d)

Using instantaneous velocity u as an example, the time-averaged velocity u is defined as u(x, y, z) =

1 𝚫t ∫0

𝚫t

u(x, y, z, t) dt

(9.3)

where 𝚫t is a time increment large enough to eliminate all time dependency from mean velocity u. In turbulent flow, it convenient to express instantaneous velocity u as the sum of its time average part u and fluctuating part u′ , as illustrated in Figure 9.2c u(x, y, z, t) = u(x, y, z) + u′ (x, y, z, t).

(9.4a)

9.4 Time Averaging and Fluctuations

We may write a similar equation for the y-component and z-component of velocity v(x, y, z, t) = v(x, y, z) + v′ (x, y, z, t)

(9.4b)

w(x, y, z, t) = w(x, y, z) + w′ (x, y, z, t)

(9.4c)

where v and w are the mean velocity and v′ and w′ are the fluctuating components of velocity. The fluctuating component of each dependent variable integrates to zero overtime interval Δt. To see this fact, we substitute Eq. (9.4a) into Eq. (9.3): Δt

u(x, y, z) = =

1 [u(x, y, z) + u′ (x, y, z, t)]dt Δt ∫0 1 Δt ∫0

Δt

u(x, y, z) dt +

1 Δt ∫0

Δt

u′ (x, y, z, t)dt.

(9.5)

The first term of Eq. (9.5) on the right-hand side is independent of time, and therefore, Eq. (9.5) becomes u(x, y, z) = u(x, y, z) +

1 Δt ∫0

Δt

u′ (x, y, z, t) dt

(9.6)

and we see from Eq. (9.6) that 1 Δt ∫0

Δt

u′ (x, y, z, t) dt = 0.

(9.7)

All the fluctuating quantities oscillate about their mean value and integrate to zero over a sufficiently long period of time. Example 9.1 Determine the time average of u(x, y)u′ (x, y, t). Solution 1 Δt ∫0

Δt

[

1 u(x, y)u (x, y, t) dt = u(x, y) Δt ∫0 ′

Δt

] u (x, y, t) dt = 0. ′

We see that the time average of a fluctuating component is zero (i.e. u′ = 0, v′ = 0); however, the quantities u′ u′ , u′ v′ u′ w′ , u′ T′ , etc. which are the time averages of the products of any two fluctuating components, will not necessarily be equal to zero. These quantities appear in the governing equations of turbulent flow. They are used as a measure of the magnitude of the turbulent fluctuations at any point in the turbulent flow. These quantities have been measured in turbulent boundary layer on a smooth flat plate by Klebanoff [5]. The quantities such as ρ u′ v′ , ρ w′ u′ have dimensions of stress, and these are called the Reynolds stresses. The Reynolds stresses represent the momentum transfer in turbulent flow due to the turbulent fluctuations. The average of the square of the fluctuation is not zero and the quantity √ u′ 2 V is used as a measure of the magnitude of the turbulent fluctuations, and this quantity is known as the intensity of turbulence. Here, V is the free-stream velocity over the flat plate. The intensity I of turbulence at a given point is defined as √ (u′ 2 + v′ 2 + w′ 2 )∕3 (9.8a) I= V where V is the free-stream velocity at the same point. The intensity of turbulence ranges from approximately 0.01–0.10 for most turbulent flows. The percentage turbulence is defined by √ (u′ 2 + v′ 2 + w′ 2 )∕3 (9.8b) percentage turbulence = 100 × V and is thus about 1–10%. The measured quantities, as discussed in [5], are shown in Figure 9.3. The measurements of Klebanoff [5] indicate that turbulent eddies can penetrate very close to the wall surface. Studies of Corino and Brodkey [6] for flow in a pipe show that the region near the wall is periodically disturbed by fluid elements. Corino and Brodkey [6]

469

470

9 Foundations of Turbulent Flow

0.1 0.12

0.08

1

0.10

1

0.08

0.06

2

0.06 2

0.04

0.04

3

0.02 3 0

0.02

0

0.2

0.4

(uʹ)2/V

0.6

y δ1

0.8

(vʹ )2/V

0.05

1.0

0.015

0.025

1.2

1.4

y δ1

(wʹ )2/V

Figure 9.3 Relative turbulence intensities in the flow along a smooth flat plate. The insert shows values very near the plate. Source: Klebanoff [5]/U.S. Department of Commerce.

indicate that the ejections and the resulting fluctuations are the most important feature of the flow field in the region near the wall. They believe that this is a factor in the generation and maintenance of turbulence. The critical Reynolds number, Rexc , is a function of the turbulence intensity, and Burmeister reports a correlation of Sucker and Brauer √ Rexc = 3.78 × 105 exp(−6 I).

9.5

(9.8c)

Isotropic Homogeneous Turbulence

Isotropic turbulence is discussed in depth by McComb [7]. In isotropic turbulence, the intensity components are, in all the directions, are equal. This may be expressed as u′ 2 = v′ 2 = w′ 2 . Regardless of the orientation of the coordinate axis in space, the values of the quantities are the same. The velocity fluctuations are perfectly random, and there is no correlation between the components of fluctuations in different directions. This means that u′ v′ = v′ w′ = u′ w′ . Homogeneity in turbulence implies that the intensity components are not a function of position in space. We conclude that in isotropic turbulence, statistical properties around a point are same in all directions. Velocity fluctuations are invariant to axis rotation and reflection as well.

9.6

Reynolds Averaging

Most of the flows of practical interest are highly irregular flows known as turbulent flows. Turbulent flows are unsteady as well as three dimensional. All the flow properties at any given spatial location fluctuate at every instant. Much effort has been spent on the onset of turbulence. We will assume that turbulent flow exists, and the question arises how to describe the turbulent flow. At this point, we indicate that mass, momentum, and energy equations that govern the laminar flows also govern the turbulent flows. We did not make any assumption about the flow being laminar or turbulent in the derivation of these fundamental equations. Our need for a theory for turbulence originates from the fact that: (a) we are unable to study analytically the full Navier–Stokes and energy equations. Numerical solutions to these equations using the DNS method even with very powerful computers are extremely difficult except in a few simple cases (b) we wish to work with mean values of physical quantities measured experimentally.

9.6 Reynolds Averaging

The transport phenomena properties such as u, v, w, T, p always vary with time in a turbulent flow. If the time-averaged quantity is independent of time, then we speak of steady mean flow, as depicted in Figure 9.2b. We will now present the method that has been developed for the analysis and prediction of the effects of turbulence, as discussed in [1] as well as [8]. Reynolds proposed that, at any instant of time, an instantaneous flow property, such as temperature, can be written as the sum of a mean value and a fluctuation about this mean value. The concept of partitioning of instantaneous value is illustrated in Figure 9.2c. Averaging rules are very important in turbulence theory, and for this reason, Reynolds developed some rules for calculating averages. We simply assume that these rules apply to problems in turbulence. We will denote the mean value of a quantity by overbar and the fluctuating component by a prime. Let us present some of these rules: (1) F′ = 0

(9.9a)

(2) F = F

(9.9b)

(3) FG = FG

(9.9c)

(4) F′ G = 0

(9.9d)

(5) F + G = F + G

(9.9e)

(6)

𝜕F 𝜕F = 𝜕x 𝜕x

(9.9f)

(7) FG = FG + F′ G′

(9.9g)

F dx =

(9.9h)

(8)

∫

∫

F dx

(9) FG = FG

(9.9i)

(10)

𝜕F =0 𝜕t

(9.9j)

(11)

𝜕F =0 𝜕t

(9.9k)

(12) F2 = (F)2 + (F′ )2

(9.9l)

(13) αF = αF, α is a constant.

(9.9m)

Example 9.2 It is often necessary to take average of an averaged quantity F. We will denote the average of an averaged quantity with a double bar superscript, e.g. F Δt

F=

1 F dt. Δt ∫0

Since F is a constant with respect to time, we write ( Δt ) 1 dt = F =F Δt ∫0 F =F Therefore, we can state that quantities that are averaged can be considered as constants in subsequent averaging. Example 9.3 If a fluctuating quantity is time averaged, this average is zero. Consider the temporal mean value of F Δt

F=

Δt

Δt

Δt

1 1 1 1 F dt = (F + F′ ) dt = F dt + F′ dt. Δt ∫0 Δt ∫0 Δt ∫0 Δt ∫0

471

472

9 Foundations of Turbulent Flow

But we know that F is constant, and for this reason, we take F out of integral sign ( ) Δt Δt Δt 1 1 1 dt + F′ dt = F + F′ dt =F Δt ∫0 Δt ∫0 Δt ∫0 Δt

F=F+

1 F′ dt. Δt ∫0

Thus, Δt

F′ =

1 F′ dt = 0. Δt ∫0

Therefore, we find that the time average of the fluctuating component is zero. Example 9.4 The quantities u′ 2 , v′ 2 , and w′ 2 are often computed for turbulent flows. The equation for u′ 2 , sometimes called the mean square of u′ , is determined as u′ 2 =

1 Δt ∫0

Δt

(u′ )2 dt =

1 Δt ∫0

Δt

(u − u)2 dt.

For an equally spaced data, this integral can be estimated as 1∑ ′ 2 1∑ (u ) = (u − u)2 N i=1 N i=1 N

u′ 2 =

N

where N is the number of available data points. Example 9.5 A certain fluid is moving in a pipe. The flow is turbulent. A hot wire anemometer is used to measure the local velocities at a certain point in the flow. Assume that you obtain the following readings at equal time intervals for a very short time. Calculate (a) the mean velocity, (b) velocity vector V, (c) u′ 2 , v′ 2 , w′ 2 , (d) the intensities of turbulence, and (e) u′ v′ for the flow. Instantaneous velocity data t(s)

u(m/s)

v(m/s)

w(m/s)

0

3.9936

0.4472

0.1976

0.0100

3.6400

0.2184

0.1664

0.0200

3.9520

0.1872

0.1768

0.0300

3.7440

0.3120

0.1352

0.0400

4.3680

0.3744

0.0936

0.0500

4.1600

0.2912

0.1040

0.0600

3.1200

0.3640

0.1664

0.0700

3.3280

0.2808

0.1560

0.0800

3.5360

0.2184

0.1352

0.0900

3.1200

0.2288

0.1872

0.1000

3.6400

0.2392

0.1768

0.1100

4.4720

0.3744

0.1872

0.1200

3.9520

0.3640

0.1768

Solution (a) The time-averaged mean velocity u is calculated using u=

1 𝚫t ∫0

𝚫t

u(t) dt.

9.6 Reynolds Averaging

Using trapz (t, u) command in MATLAB 2021a, we calculate the integral as 0.12

u(t) dt = 0.4505.

∫0

The time-averaged velocity u is u=

0.4505 = 3.7544 m∕s. 0.12

For equally spaced data, the time-averaged velocity u can also be estimated as N ∑

u=

ui

i=1

=

N

49.0256 ≈ 3.7712 m∕s. 13

Similarly, N ∑

v=

i=1

vi

N ∑

w=

≈ 0.30 m∕s

N

wi

i=1

N

≈ 0.1584 m∕s.

(b) Using Trapz command in MATLAB 2021a, the magnitude of the velocity vector is given as √ √ V = (u)2 + (v)2 + (w)2 = (3.7712)2 + (0.2912)2 + (0.1560)2 = 3.7856 m∕s. (c) Using Trapz command in MATLAB 2021a, we can evaluate the integral u′ 2 =

1 Δt ∫0

Δt

1 Δt ∫0

(u′ )2 dt =

Δt

(u − u)2 dt

u′ 2 = 0.1839 m2 ∕s2 . In a similar way v′ 2 =

1 𝚫t ∫0

𝚫t

(v′ )2 dt =

1 𝚫t ∫0

𝚫t

(v − v)2 dt

v′ 2 = 0.0052 m2 ∕s2 w′ 2 =

1 𝚫t ∫0

𝚫t

(w′ )2 dt =

1 𝚫t ∫0

𝚫t

(w − w)2 dt

w′ 2 = 9.6443 × 10−4 m2 ∕s2 . (d) The intensity of turbulence in the x-direction is √ √ 100 u′ 2 100 0.1839 = ≈ 11.32 percent. Ix = V 3.7856 The intensity in the y-direction is √ √ 100 v′ 2 100 0.0052 Iy = = = 1.90 percent. V 3.7856 The intensity in the z-direction is √ √ 100 w′ 2 100 9.6443 × 10−4 Iz = = = 0.82 percent. V 3.7856

473

474

9 Foundations of Turbulent Flow

(e) The cross-turbulence term is obtained using MATLAB 2021a. First, let us obtain u′ and v′ using u′ = u − u and v′ = v − v. The result is given in the following table. Fluctuating velocity components t(s)

0

u′ (m/s)

v′ (m/s)

0.2224

0.1560

0.0100

–0.1312

–0.0728

0.0200

0.1808

–0.1040

0.0300

–0.0272

0.0208

0.0400

0.5968

0.0832

0.0500

0.3888

0

0.0600

–0.6512

0.0728

0.0700

–0.4432

–0.0104

0.0800

–0.2352

–0.0728

0.0900

–0.6512

–0.0624

0.1000

–0.1312

–0.0520

0.1100

0.7008

0.0832

0.1200

0.1808

0.0728

u ′ v′ =

9.7

1 𝚫t ∫0

𝚫t

u′ v′ dt = 0.012 m2 ∕s2 .

Governing Equations of Incompressible Steady Mean Turbulent Flow

In turbulent flows, although the mean flow is steady, the flow variables such as velocity, temperature, and pressure fluctuate randomly with time due to the presence of turbulent eddies on the mean flow. The steady turbulent flow implies that after the averaging process, values of the mean quantities u, v, w, p, T at all the points in the flow are constant. These mean values are time-averaged quantities. In our discussion of turbulent heat transfer, we will assume that temperature fluctuations, T′ , follow the same pattern as the velocity fluctuations. Equations of motion and energy for turbulent flows are discussed in detail in [1] and [8]. We will limit the analysis for convenience to two-dimensional, steady, and incompressible turbulent flow. Continuity equation We will first consider the continuity equation. The two-dimensional form of the continuity equation is 𝜕u 𝜕v + = 0. (9.10) 𝜕x 𝜕y The following variables are substituted into the equation of the conservation of mass, Eq. (9.10), u = u + u′

(9.11a)

v = v + v′

(9.11b)

and we obtain 𝜕(u + u′ ) 𝜕(v + v′ ) + = 0. 𝜕x 𝜕y

(9.12)

We will now take time average of each term in Eq. (9.12) over a time interval Δt. For example, let us consider the first term of Eq. (9.12), and by definition, we have ] Δt [ 𝜕(u + u′ ) 𝜕(u + u′ ) 1 = dt 𝜕x Δt ∫0 𝜕x

9.7 Governing Equations of Incompressible Steady Mean Turbulent Flow

Integration with respect to time is independent of differentiation with respect to space coordinates. Therefore, the above equation can be written as follows: [ ] Δt 𝜕(u + u′ ) 𝜕 1 = (u + u′ ) dt 𝜕x 𝜕x Δt ∫0 [ ] Δt Δt 1 𝜕 1 u dt + u′ dt = 𝜕x Δt ∫0 Δt ∫0 [ [ ] ] Δt Δt 𝜕 1 𝜕 1 u dt + u′ dt = 𝜕x Δt ∫0 𝜕x Δt ∫0 [ ] [ ] 𝜕 𝜕 = u + u′ 𝜕x 𝜕x since u = u and the spatial gradient of a fluctuating velocity is zero. We also note that u′ = 0, and finally, we now have 𝜕(u + u′ ) 𝜕u = . 𝜕x 𝜕x We can write a similar equation for the y-component of the continuity equation as 𝜕(v + v′ ) 𝜕v = . 𝜕y 𝜕x

(9.13)

(9.14)

Therefore, the time average of the continuity equation for steady mean incompressible turbulent flow can be written as 𝜕u 𝜕v + = 0. 𝜕x 𝜕y

(9.15)

We may also write 𝜕u′ 𝜕v′ + = 0. 𝜕x 𝜕y

(9.16)

On the other hand, Eq. (9.16) indicates that mass is conserved at any instant of time. The contributions of velocity fluctuations are averaged out during the time averaging. Notice that both the time-averaged velocity field and the fluctuating velocity components satisfy the same form of conservation of mass, and notice that there are no new terms in Eq. (9.15). Momentum equation In developing momentum equations for turbulent flow, we will follow the procedure outlined in [8–10]. We are considering incompressible constant property flow. Recall that the two-dimensional form of the x- and y-components of the momentum equation is ( ) ) ( 2 𝜕p 𝜕u 𝜕u 𝜕u 𝜕 u 𝜕2 u ρ (9.17) +u +v =− +μ + 2 𝜕t 𝜕x 𝜕y 𝜕x 𝜕x2 𝜕y ) ( 2 ) ( 𝜕p 𝜕v 𝜕v 𝜕 v 𝜕2 v 𝜕v +u +v =− +μ . (9.18) + ρ 𝜕t 𝜕x 𝜕y 𝜕x 𝜕x2 𝜕y2 We will assume that mean flow is steady, that is, u, v, and w do not change with time. However, the overall velocity contains a fluctuating component, and for this reason, u, v, and w still vary with time. To develop the time-averaged momentum equation, we will write the momentum equation in a slightly different form. Consider the x-momentum equation. The terms u𝜕 u/𝜕 x and v𝜕 u/𝜕 x will be replaced by the following relations: 𝜕u 𝜕u2 𝜕u = −u 𝜕x 𝜕x 𝜕x 𝜕u 𝜕u v 𝜕v v = −u . 𝜕x 𝜕x 𝜕x Substituting Eqs. (9.19) and (9.20) into Eq. (9.17), we get ) ) ( ( 2 𝜕p 𝜕u 𝜕 u v 𝜕v 𝜕u 𝜕u2 𝜕 u 𝜕2u =− + ρ + −u + −u +μ 𝜕t 𝜕x 𝜕x 𝜕y 𝜕y 𝜕x 𝜕x2 𝜕y2 u

or

[ ρ

𝜕u 𝜕u2 + −u 𝜕t 𝜕x

(

𝜕u 𝜕v + 𝜕x 𝜕y

) +

] ( 2 ) 𝜕p 𝜕uv 𝜕 u 𝜕2 u =− +μ . + 𝜕y 𝜕x 𝜕x2 𝜕y2

(9.19) (9.20)

(9.21)

(9.22)

475

476

9 Foundations of Turbulent Flow

The third term on the left-hand side of Eq. (9.22) is zero by continuity. Then, Eq. (9.22) reduces to ] ) [ ( 2 𝜕p 𝜕u 𝜕u2 𝜕 u v 𝜕 u 𝜕2u =− . + ρ + + +μ 𝜕t 𝜕x 𝜕y 𝜕x 𝜕x2 𝜕y2

(9.23)

We can now take the time average of each term in Eq. (9.23). Averaging each term over time and applying averaging rules to Eqs. (9.9f), (9.9j), and (9.9k), we get [ ] ) ( 2 𝜕p 𝜕u2 𝜕 u v 𝜕 u 𝜕2 u (9.24) + 2 . ρ + =− +μ 𝜕x 𝜕y 𝜕x 𝜕x2 𝜕y Now, applying the time-averaging rules to Eqs. (9.9g) and (9.9l), we get [ 2 ) ] ( 2 𝜕p 𝜕 𝜕uv 𝜕 𝜕u 𝜕 u 𝜕2 u ρ − (u′ )2 − (u′ v′ ). + =− +μ + 𝜕x 𝜕y 𝜕x 𝜕x 𝜕y 𝜕x2 𝜕y2 The terms in bracket on the left-hand side of Eq. (9.25) can be written as [ )] ( 2 ) ( 𝜕p 𝜕 𝜕u 𝜕u 𝜕 𝜕 u 𝜕2 u 𝜕u 𝜕v ρ u +v +u + =− +μ + 2 − (u′ )2 − (u′ v′ ). 𝜕x 𝜕y 𝜕x 𝜕y 𝜕x 𝜕x 𝜕y 𝜕x2 𝜕y Making use of the continuity equation, the left-hand side of Eq. (9.26) can be simplified to the following form: [ ] ( 2 ) 𝜕p 𝜕 𝜕u 𝜕u 𝜕 𝜕 u 𝜕2 u ρ u +v =− +μ − ρ (u′ )2 − ρ (u′ v′ ). + 𝜕x 𝜕y 𝜕x 𝜕x 𝜕y 𝜕x2 𝜕y2 We may also write the y-component of the momentum equation as ) ( 2 ( ) 𝜕p 𝜕 𝜕v 𝜕v 𝜕 𝜕 v 𝜕2 v =− +μ ρ u +v + 2 − ρ (v′ u′ ) − ρ (v′ )2 . 2 𝜕x 𝜕y 𝜕y 𝜕x 𝜕y 𝜕x 𝜕y Now, we may put the equation of motion in the following form: x-momentum: ( ) [ [ ] ] 𝜕p 𝜕 𝜕u 𝜕u 𝜕 𝜕u 𝜕u ρ u +v =− + μ − ρu′ u′ + μ − ρv′ u′ . 𝜕x 𝜕y 𝜕x 𝜕x 𝜕x 𝜕y 𝜕y y-momentum: ( ) [ [ ] ] 𝜕p 𝜕 𝜕v 𝜕v 𝜕 𝜕v 𝜕v ρ u +v =− + μ − ρ u′ v′ + μ − ρv′ v′ . 𝜕x 𝜕y 𝜕y 𝜕x 𝜕x 𝜕y 𝜕y

(9.25)

(9.26)

(9.27)

(9.28)

(9.29)

(9.30)

Equations (9.29) and (9.30) have the same form as the steady laminar flow momentum equations except for the fact that these equations have additional terms on the right-hand side. These extra terms are caused by the presence of fluctuating velocity components. These fluctuating velocity components cause net transport of momentum. As far as the turbulent mean flow is concerned, these are additional forces. These additional forces per unit area are called Reynolds stresses, eddy stresses, apparent stresses, virtual stresses, or turbulent stresses due to turbulent fluctuations in the flow. The nine eddy stresses form the components of a second-order tensor. Turbulent stress tensor may be shown as ⎡ρu′ u′ .. … … ρu′ w′ ⎤ ⎢ ⎥ . t ⎥. τ =⎢ ⎢. ⎥ ⎢ρw′ u′ … … ρw′ w′ ⎥ ⎣ ⎦ The turbulent stress tensor components may be measured experimentally. The fluctuations of velocity lead to the generation of terms like (ρu′ 2 ) and (ρu′ v′ ). The first of these terms is a normal stress, and the second one is called shear stress. These turbulent stresses represent large amount of momentum transfer because of velocity fluctuations in the flow. Property fluctuations produce apparent momentum flux or stresses. The turbulent stresses are unknown a priori and must be modeled experimentally. In duct flows as well as boundary layer flows, the turbulent stress −ρu′ v′ in the y-direction normal to flow is dominant compared to other turbulent stress terms. We have more unknowns than the number of equations. To close the system and obtain the same number of equations as unknowns, turbulent shear stresses need to be related to other flow properties or come up with ways to model turbulence. We need empirical formulas for the turbulent stresses so that time-averaged equations can be solved.

9.8 Turbulent Momentum Boundary Layer Equation

We assumed that flow is steady on average. The fluctuating terms represent the transport of momentum by turbulent eddy motion. We are essentially assuming that w = 0, u = u(x, y), and v = v(x, y). However, even in this case, u′ = u′ (x, y, z, t), v′ = v′ (x, y, z, t), and w′ ≠ 0. We assume that u′ = u′ (x, y, t), v′ = v′ (x, y, t), and w′ = 0. This point is discussed in depth by Kakac et al. [11]. t Finally, notice that for turbulent flow, the relationship between the components of stress tensor τ and the time-smoothed velocity gradients is complex and is a property of the turbulent flow. On the other hand, for laminar flow, the relationship l between the components of laminar stress tensor τ and the velocity gradients is well defined. The constant of proportionality is viscosity, and it is a property of the fluid.

9.8 Turbulent Momentum Boundary Layer Equation Consider turbulent flow near a solid plane wall. See Figure 9.4. We will assume that turbulent flow is two-dimensional, steady, and incompressible with constant properties. The fluid has free-stream velocity U∞ and free-stream temperature T∞ . We will make the following assumptions and approximations: (1) The boundary layer is thin; this means that δ ≪ L or δ ≪ 1. L u ∼ U∞ . x∼L y ∼ δ. 𝜕2 u 𝜕2u ≪ 2. 2 𝜕x 𝜕y (2) The pressure gradient in the y-direction is negligible 𝜕p ≈ 0. 𝜕y (3) The pressure gradient in the x-direction can be expressed as 𝜕p dp = . 𝜕x dx (4) On the other hand, we can imagine that u′ and v′ are the velocity fluctuations created by an eddy as it rides along the mean flow. Then, we can assume that u′ and v′ are comparable orders of magnitude. In other words, fluctuating components have no preferred direction. This is the same as assuming that turbulence is isotropic. Thus, u′ ∼ v′ . y

U∞

U∞ u

δ(x) x

L Figure 9.4

Turbulent flow over flat plate.

477

478

9 Foundations of Turbulent Flow

This means that u′ 2 ∼ u′ v′ . We can now compare the relative magnitudes of 𝜕u′ 2 ∕𝜕x and 𝜕u′ v′ ∕𝜕y using scale analysis as we have done in laminar boundary layer flow u′ 2 𝜕u′ 2 ∼ 𝜕x L u ′ v′ u′ 2 𝜕u′ v′ ∼ ∼ . 𝜕y δ δ Since the boundary layer is thin, that is δ ≪ L, we conclude that 𝜕u′ v′ 𝜕u′ 2 ≪ . 𝜕x 𝜕y In other words, in the turbulent boundary layer, we can neglect 𝜕u′ 2 ∕𝜕x relative to 𝜕u′ v′ ∕𝜕y. Using these simplifications, we see that the x-momentum equation for the turbulent boundary layer becomes ) ( ) ( dp 𝜕 𝜕u 𝜕u 𝜕u ′ ′ +v =− + μ − ρu v . (9.31) ρ u 𝜕x 𝜕y dx 𝜕y 𝜕y The ρu′ v′ term is called turbulent shear stress, and it is denoted by τt = −ρu′ v′ . To determine the velocity distribution from this mean momentum equation, the turbulent shear stress τt needs to be modeled. Notice that the turbulent shear stresses τt are created by the momentum transfer due to velocity fluctuations.

9.9

Turbulent Energy Equation

Consider the energy equation under steady incompressible turbulent flow. We assume that there is no energy generation term. The viscous dissipation term is neglected. Under these assumptions, the energy equation becomes ] [ 2 [ ] 𝜕T 𝜕T 𝜕 T 𝜕2 T 𝜕T +u +v =k . (9.32) + ρcp 𝜕t 𝜕x 𝜕y 𝜕x2 𝜕y2 This time, we will follow a different procedure. The following variables are substituted into the conservation of the energy equation, Eq. (9.32), given above u = u + u′ v = v + v′ T = T + T′ and Eq. (9.32) takes the following form: [ 2 ] ] [ 𝜕 𝜕 k 𝜕2 𝜕 𝜕 ′ ′ (T + T′ ) + (u + u′ ) (T + T′ ) + (v + v′ ) (T + T′ ) = (T + T ) + (T + T ) . 𝜕t 𝜕x 𝜕y ρcp 𝜕x2 𝜕y2 The application of the averaging rules to each term in Eq. (9.33) yields ] [ ] [ k 𝜕2 T 𝜕2 T 𝜕T 𝜕T 𝜕T′ 𝜕T′ ′ ′ +v = +v . u + 2 − u 𝜕x 𝜕y ρcp 𝜕x2 𝜕x 𝜕y 𝜕y

(9.33)

(9.34)

Consider now the last two terms in bracket on the right-hand side of Eq. (9.34). We will develop an expression for each of the terms as follows: 𝜕 ′ ′ 𝜕T′ 𝜕u′ (u T ) = u′ + T′ 𝜕x 𝜕x 𝜕x

(9.35a)

9.10 Turbulent Boundary Layer Energy Equation

𝜕 ′ ′ 𝜕T′ 𝜕v′ (v T ) = v′ + T′ . 𝜕y 𝜕y 𝜕y Add Eqs. (9.35a) and (9.35b); rearrange them and take the time average of each term [ ] ] [ 𝜕 𝜕T′ 𝜕T′ 𝜕u′ 𝜕v′ 𝜕 ′ ′ ′ ′ +v (u T ) + (v′ T′ ) − T′ + T′ . u = 𝜕x 𝜕y 𝜕x 𝜕y 𝜕x 𝜕y The energy equation now becomes ]} [ ] { [ ′ ′ k 𝜕2 T 𝜕2 T 𝜕T 𝜕T 𝜕 ′ ′ 𝜕 ′ ′ ′ 𝜕u ′ 𝜕v +v = (u T ) + (v T ) − T +T . u + 2 − 𝜕x 𝜕y ρcp 𝜕x2 𝜕x 𝜕y 𝜕x 𝜕y 𝜕y Next, we multiply the continuity equation, Eq. (9.13), by T = T + T′ [ ] 𝜕 𝜕 (u + u′ ) + (v + v′ ) = 0. (T + T′ ) 𝜕x 𝜕y Rearranging the terms in square brackets and taking the time average of each term yields [ ] [ ] 𝜕u 𝜕v 𝜕u′ 𝜕v′ +T + T′ + T′ = 0. T 𝜕x 𝜕y 𝜕x 𝜕y Consider the first bracket in Eq. (9.39). We now take T outside of the square bracket [ ( )] [ ] ′ ′ 𝜕u 𝜕v ′ 𝜕u ′ 𝜕v + + T +T = 0. T 𝜕x 𝜕y 𝜕x 𝜕y Using the continuity equation, Eq. (9.15) in Eq. (9.40) yields [ ] ′ ′ ′ 𝜕u ′ 𝜕v +T = 0. T 𝜕x 𝜕y The substitution of Eq. (9.41) into Eq. (9.37) gives [ ] [ ] [ 2 ] 𝜕T 𝜕T 𝜕 ′ ′ 𝜕 ′ ′ 𝜕 T 𝜕2 T ρcp u +v =k (u T ) + (v T ) . + 2 − ρcp 𝜕x 𝜕y 𝜕x 𝜕y 𝜕x2 𝜕y

(9.35b)

(9.36)

(9.37)

(9.38)

(9.39)

(9.40)

(9.41)

(9.42)

The first term on the right-hand side of Eq. (9.42) represents the transfer of energy by conduction, and the remaining terms in the bracket on the right-hand side of Eq. (9.42) represents turbulent or eddy transfer of energy. Property fluctuations produce turbulent energy flux. These are called the turbulent heat flux gradients. We need empirical formulas for the turbulent heat flux terms so that time-averaged equations can be solved. Finally, this equation can be expressed in the following form: ] ]} ) {[ [( ) [( ) ( 𝜕T 𝜕T 𝜕 𝜕 𝜕T 𝜕T ρcp u . (9.43) +v = k − (ρcp u′ T′ ) + k − (ρcp v′ T′ ) 𝜕x 𝜕y 𝜕x 𝜕x 𝜕y 𝜕y We have neglected the heat generation term and assumed constant thermal properties. The energy flux terms have laminar and turbulent components.

9.10 Turbulent Boundary Layer Energy Equation Consider turbulent flow over a smooth flat plate. See Figure 9.5. The incompressible fluid is flowing over the plate. The approaching fluid has a free-stream temperature of T∞ . The plate has a wall temperature of Tw . We will assume the following scaling arguments for the thermal boundary layer: (1)

x ∼ L. y∼Δ ΔT ∼ (Tw − T∞ ).

479

480

9 Foundations of Turbulent Flow

y

T∞

U∞ Δ(x)

T

T∞

L Figure 9.5

Tw

x

Temperature profile in turbulent flow.

(2) Axial conduction is negligible compared to conduction in the y-direction 𝜕2 T 𝜕2T ≪ 2 2 𝜕x 𝜕y (3) We need to compare relative magnitudes of 𝜕u′ T′ ∕𝜕x and 𝜕v′ T′ ∕𝜕y. We assume that fluctuating velocity components u′ and v′ do not have preferred directions. That is turbulence is isotropic u ′ ∼ v′ . We can then argue that u′ T′ and v′ T′ have the same order of magnitude: u′ T′ ∼ v′ T′ . Then, we can conclude that 𝜕v′ T′ 𝜕 u′ T′ ≪ . 𝜕x 𝜕y This means that we can neglect 𝜕u′ T′ ∕𝜕x relative to 𝜕v′ T′ ∕𝜕y. After some rearrangement, we obtain the energy equation for steady two-dimensional turbulent flow ( ) 𝜕2 T 𝜕T 𝜕 𝜕T +v = k 2 + (−ρcp v′ T′ ) (9.44a) ρcp u 𝜕x 𝜕y 𝜕y 𝜕y or ) ( ( ) 𝜕 𝜕T 𝜕T 𝜕T ′ ′ +v = k − ρcp v T (9.44b) ρcp u 𝜕x 𝜕y 𝜕y 𝜕y or ) ( ( ) 𝜕 𝜕T 𝜕T +v = −q′′y − ρcp v′ T′ (9.44c) ρcp u 𝜕x 𝜕y 𝜕y where the first term q′′y = −k𝜕T∕𝜕y on the right-hand side represents the heat transfer by molecular conduction within the fluid. The second term ρ cp v′ T′ in the equation on the right-hand side represents the turbulent or eddy transfer of heat within the fluid.

9.11 Closure Problem of Turbulence We will now summarize the steady turbulent boundary layer equations as follows: Continuity: 𝜕u 𝜕v + =0 𝜕x 𝜕y x-component of the momentum equation and its boundary conditions: ) ( ) ( dp 𝜕 𝜕u 𝜕u 𝜕u +v =− + μ − ρu′ v′ ρ u 𝜕x 𝜕y dx 𝜕y 𝜕y

(9.45)

(9.46)

9.12 Eddy Diffusivity of Momentum

u(x, 0) = 0 v(x, 0) = 0 u(x, ∞) = U∞ u(0, y) = U∞ . The energy equation and its boundary conditions are ) [( ( )] 𝜕 𝜕T 𝜕T 𝜕T ρcp u +v = k − ρcp v′ T′ 𝜕x 𝜕y 𝜕y 𝜕y T(x, 0) = Tw or − k

(9.47)

𝜕T(x, 0) = q′′w 𝜕y

T(x, ∞) = T∞ T(0, y) = T∞ To solve these equations, we must have the same number of equations as the unknowns. Here, the pressure term can be expressed as dU dp dp∞ ≈ = −ρ U∞ ∞ . (9.48) dx dx dx We can determine the pressure using the velocity field outside the boundary layer. When we examine the energy and momentum equations, we have three equations and five unknowns: u, v, T, u′ v′ and v′ T′ We see that we have more unknowns than the number of equations. This shortage of governing equations creates what is called the closure problem of turbulence. To close the system, we must obtain the same number of equations as unknowns. Thus, turbulent shear stresses, u′ v′ , and turbulent heat fluxes, v′ T′ , need to be related to smoothed flow variables or come up with ways to model turbulence. We need empirical formulas for the turbulent stresses and turbulent heat fluxes so that time-averaged equations can be solved. The best-known approach involves the eddy diffusivity models. We will express the turbulent shearing stress and turbulent heat transfer in terms of mean velocity and temperature gradients. This will be done in terms of eddy diffusivity of momentum and heat. The eddy diffusivity approach is widely used in engineering. Even if we model the new unknowns, we still do not have exact solutions for these equations. To solve these equations, we use numerical methods or some kind of approximate method. A workable and valid theory of turbulent boundary layers is essential for the prediction of: (a) (b) (c) (d)

drag flow rates flow separation heat transfer

9.12 Eddy Diffusivity of Momentum An examination of the turbulent momentum boundary layer equation shows that the product u′ v′ survives the simplification process. To obtain a solution, turbulent shear stress τt = −ρu′ v′ has to be related to the mean flow field. Turbulent shear stress may be evaluated on the basis of experimental measurements for u′ and v′ . Turbulent shearing stress is expressed in terms of the mean velocity gradient by Boussinesq, as discussed in [8, 12] and [13]. Boussinesq, in 1877, proposed the following expression: −ρu′ v′ = ρεM

𝜕u . 𝜕y

(9.49)

481

482

9 Foundations of Turbulent Flow

The turbulent shear stress (eddy shear stress) τt then is expressed as τt = ρεM

𝜕u 𝜕y

(9.50)

where εM is the eddy diffusivity of momentum and the term τt is the turbulent shear stress. This is a close approximation for many practical engineering problems. The eddy diffusivity of momentum εM is often referred to as the eddy viscosity and εM has the dimension (m2 /s). We still must evaluate εM . The eddy diffusivity of momentum εM , particularly near a solid wall, depends on the velocity and geometry of the flow, and it is not a fluid property, that is, it is a flow property and depends on the turbulent fluid motion, hence varies throughout the flow. Finding ways to evaluate εM is the main goal of turbulence research. We are now in a position to write the momentum boundary layer equation in the following form: ( ) [ ] dp 𝜕u 𝜕u 𝜕u 𝜕 ρ u +v =− + (μ + ρεM ) . (9.51) 𝜕x 𝜕y dx 𝜕y 𝜕y It is sometimes convenient to introduce a turbulent or eddy viscosity μt or apparent turbulent viscosity as follows: (9.52)

μt = ρεM

where μt is a flow property that varies across the flow field. Eq. (9.50) is very important since it provides a systematic framework to study turbulent flow. Then, the momentum boundary layer equation can be written in terms of the turbulent viscosity μt as ) [ ] ( dp 𝜕 𝜕u 𝜕u 𝜕u +v =− + (μ + μt ) . (9.53) ρ u 𝜕x 𝜕y dx 𝜕y 𝜕y Notice that the terms in bracket on the right-hand side of Eq. (9.53) represent total shear stress τ = ρ(ν + εM )

𝜕u 𝜕u = (μ + μt ) . 𝜕y 𝜕y

The momentum equation can be expressed in terms of total shear stress as follows: ( ) dp 𝜕 τ 𝜕u 𝜕u ρ u +v =− + . 𝜕x 𝜕y dx 𝜕y

(9.54)

(9.55)

9.13 Eddy Diffusivity of Heat We can model the Reynolds heat flux, eddy heat flux or turbulent heat flux as −ρcp v′ T′ = ρcp εH

𝜕T 𝜕y

(9.56a)

where εH is called the eddy diffusivity of heat or eddy thermal conductivity. Sometimes we talk about total heat flux −q′′ = k

𝜕T 𝜕T − ρcp v′ T′ = ρcp (α + εH ) . 𝜕y 𝜕y

(9.56b)

Negative sign in Eq. (9.56b) indicates the direction to heat flow. We can write the boundary layer energy equation as ( ) [ ] 𝜕 𝜕T 𝜕T 𝜕T +v = (k + ρcp εH ) . (9.57) ρcp u 𝜕x 𝜕y 𝜕y 𝜕y The product of ρcp εH has the unit of fluid thermal conductivity, and it is called eddy or turbulent conductivity and represented by kt = ρcp εH . The eddy conductivity kt is a flow parameter. The energy equation now takes the following form: ) [ ] ( 𝜕 𝜕T 𝜕T 𝜕T +v = (k + kt ) . ρcp u 𝜕x 𝜕y 𝜕y 𝜕y

(9.58)

(9.59)

9.14 Transport Equations in the Cylindrical Coordinate System

It is common to simplify Eq. (9.57) by dividing with ρcp . The energy equation, Eq. (9.57), can now be written in the following form: [ ] 𝜕 𝜕T 𝜕T 𝜕T +v = (α + εH ) (9.60) u 𝜕x 𝜕y 𝜕y 𝜕y where α = k/ρcp is the molecular thermal diffusivity and εH is called the eddy diffusivity of heat. Note that εM and εH are flow properties, not fluid properties. With the present understanding of the thermal energy transport, we relate the eddy diffusivity of heat εH to the eddy diffusivity of momentum εM . We assume that mechanism of momentum and energy transport in turbulent flow is similar. Then, we can define the turbulent Prandtl number Prt as Prt =

εm εH

(9.61)

and we normally assume that the turbulent Prandtl number Prt is constant. The measurements of temperature distribution in turbulent boundary layers of Pr ≥ 1 indicate that Prt ≈ 0.85 − 0.9, as discussed in [14] and [15]. The energy equation can be written in terms of the turbulent Prandtl number Prt as [( ] ) εM 𝜕T 𝜕 𝜕T 𝜕T +v = α+ (9.62) u 𝜕x 𝜕y 𝜕y Prt 𝜕y or since Pr = μcp /k = ν/α the energy equation can also be written as [( ] ) ε 1 𝜕 𝜕T 𝜕T 1 𝜕T +v =ν + M . u 𝜕x 𝜕y 𝜕y Pr ν Prt 𝜕y

(9.63)

We assume that there exists similarity between momentum and energy transfer. Thus, we obtain information about heat transfer using the available information on momentum transfer.

9.14 Transport Equations in the Cylindrical Coordinate System Since there are many similarities between turbulent flow over a flat plate and turbulent flow in a pipe, we will use the following notation for turbulent flow in a pipe. We will represent axial velocity by u and radial velocity by v. We will now consider the equations of motion and energy equation turbulent flow in cylindrical coordinate system. The coordinate system is as given in Figure 9.6, and x = 0 is chosen arbitrarily and x is extended to both −∞ and +∞. Detailed information about equations of motion is given by Kays et al. [14] and Oosthuizen and Naylor [16]. We are interested in two-dimensional continuity, momentum, and energy equations. In turbulent pipe flow, it is convenient to write turbulence quantities in terms of eddy diffusivity heat and momentum. Continuity, momentum, and energy equations are as follows: Continuity equation: 𝜕u 1 𝜕 + (rv) = 0 𝜕x r𝜕r Momentum equation: x-component ) [ ] ( 1 dp 1 𝜕 𝜕u 𝜕u 𝜕u +v =− + r(ν + εM ) u 𝜕x 𝜕r ρ dx r 𝜕 r 𝜕r Energy equation: ) [ ] ( 1 𝜕 𝜕T 𝜕T 𝜕T +v = r(α + εH ) . u 𝜕x 𝜕r r𝜕r 𝜕r The energy equation can also be written as ) [ ( ] ) ( ε 1 ν 𝜕 𝜕T 𝜕T 1 𝜕T +v = r + M . u 𝜕x 𝜕r r 𝜕r Pr ν Prt 𝜕r

(9.64)

(9.65)

(9.66)

(9.67)

483

484

9 Foundations of Turbulent Flow

r

θ

r

v y

u 0

x

0

Figure 9.6

D = 2R

Coordinate system for flow in a circular pipe.

It is observed that as discussed by Bejan [9], the turbulent flow becomes HFD and TFD after a short distance from the tube entrance and hydrodynamic and thermal entrance lengths may be estimated as L LH ≈ T ≈ 10 D D This estimate is particularly applicable to fluids with Prandtl numbers of order unity.

9.15 Experimental Work on the Turbulent Mean Flow The eddy diffusivity information allows us to calculate the turbulent velocity profiles. Computed profiles can be compared with measurements. We may measure wall shear stress τw for a steady flow in a long pipe. See Figure 9.7. Consider steady fully turbulent flow in a long pipe. As a first approximation, a fully developed turbulent flow profile may be observed after a length of 10 tube diameters [9]. Most of the information about turbulent pipe flow comes from experimental data. A force balance yields a relationship between pressure drop and wall shear stress ) ( dp R τw = − (9.68a) dx 2 or we can also write an equation for shear stress at any position r ( ) dp r τ= − . dx 2 From Eqs. (9.68a) and (9.68b), we obtain τw R = τ r or since y + r = R y τ =1− τw R

(9.68b)

(9.68c)

(9.68d)

and we see that apparent or total shear stress τ varies linearly. Total shear stress τ at any radial location r comprises both viscous stress and turbulent stress. Consider Figure 9.7 again. Integrating Eq. (9.68a) between reference section 1 and reference section 2 in the flow, we get τw = p1

D (p1 − p2 ) D Δp = . 4L 4L

(9.69a) p2

r

τw

y Flow 0

τapp

x

D = 2R

L

Figure 9.7

Apparent shear stress variation in a fully developed turbulent flow in a pipe.

9.15 Experimental Work on the Turbulent Mean Flow

It is common to introduce a dimensionless pipe resistance coefficient f to give the pressure drop in a pipe 2 dp Δp p1 − p2 f ρV = = = dx L L D 2

or f=

(p − p ) D 1 2 . L ρ V2 ∕2

(9.69b)

Using the definition of the Moody (Darcy) friction coefficient f, Eq. (9.69a) becomes f ρ V2 . 8 As reported in [17], Blasius experimentally formulated the following expression for f: τw =

f=

0.3164 1∕4

ReD

4 × 103 ≤ ReD ≤ 4 × 105

(9.69c)

(9.70)

and if we can find the Moody friction factor f, then we can compute wall shear stress τw . The nature of turbulent flow is very complex, and for this reason, the analysis of turbulent heat, mass, and momentum transfer processes has largely been of an empirical nature. We will review some experimental information related to turbulent flow in pipes as well as turbulent flow over flat plates.

9.15.1 Turbulent Flow in Pipe: Velocity Profiles To provide a sound engineering background for the analysis of turbulent flows, we will examine some experimental work. It is important to consider the general features to be expected in turbulence near a fixed channel wall. The tangential velocity drops to zero on the wall, and due to action of turbulence, we see a very high velocity gradient on the channel wall. Since it is required that velocity vanishes at the wall, the turbulent activity is suppressed near the surface, and the extraction of energy from the mean flow is limited. The turbulence structure varies across the pipe. The most rapid change is observed very close the wall. An examination of Figure 9.3 reveals that the intensity reaches a peak and then falls off rapidly. Experimentally obtained velocity data in turbulent pipe flow are usually correlated in one of three ways as given below: (y) u (1) = f1 (9.71a) Vc R (2) u+ = f2 (y+ ) (3) Vc − u = f3

(9.71b) (y) R

(9.71c)

where u+ = u∕uτ is the dimensionless velocity and y+ = yuτ√ /ν is the dimensionless distance measured from the wall and notice that y+ has the form of Reynolds number. Here, uτ = τw ∕ν is called the wall friction velocity. The pipe radius is R, Vc is the pipe centerline velocity, and the distance y is measured from the pipe wall. Researchers tried to find a universal velocity profile for turbulent flow. For this reason, a very large part of the experimental work on turbulent flow has been done in circular pipes, mainly because of the simplicity of the geometry and the flow symmetry. It is also easy to control the flow conditions. Velocity profiles in fully turbulent flow in smooth pipes have been measured for different flow rates, pipe diameters, and fluids by many researchers. If one tries to find a universal velocity profile for turbulent flow in a pipe, one approach is to plot u∕Vc as a function of y/R in the form of Eq. (9.71a). As reported in [19], Nikuradse [18] carried out extensive experimental study on turbulent flow in pipes, and a few profiles from his data are shown in Figure 9.8. The symbol y represents distance from the pipe wall. A careful inspection of Figure 9.8 reveals that as the Reynolds number increases, the profile gets flatter. The extremities of the velocity profile curves were brought into coincidence, but the middle portions of the curves are separated. This plot in Figure 9.8 tells us that such a dimensionless plot does not have a universal character. Prandtl [20, 21] did extensive work on turbulent velocity profiles. Prandtl proposed the following equation for the velocity profile for turbulent flow in a pipe: (y) Vc − u (9.72) = −2.5 ln uτ R

485

486

9 Foundations of Turbulent Flow

1.0

0.9

u Vc

Re = 4000

0.8

Re = 23300 Re = 725000 0.7 Nikuradse's data 0.6

0.1

Figure 9.8 1932.

0.2

0.3

0.4 0.5 y/R

0.6

0.7

0.8

Dimensionless velocity profile for different Reynolds numbers. Source: Data from Nikuradse, J., VDI-Forschungsheft 356,

where Vc is the velocity at the pipe center. Eq. (9.72) is in the form of Eq. (9.71c) and is plotted in Figure 9.9 and is compared with the experimental data of Nikuradse. Prandtl’s velocity distribution equation represents experimental data very well. The velocity defect law of the Prandtl applies to turbulent core outside the viscous sublayer and the buffer zone. von Karman [22–25] did extensive research on turbulence and presented a different expression for the point velocity defect law as given below [ ( ) √ ] √ Vc − u y y 1 ln 1 − 1 − + 1− . (9.73) =− uτ κ R R Conditions here are that du =∞ At y = 0 dy At y = R u = Vc

(9.74) (9.75)

The constant κ = 0.4 is called von Karman constant, and this equation is also plotted in Figure 9.9. Wang [26] developed an equation for a velocity distribution. This equation agrees with experimental data better than equation of Prandtl or von Karman equation. However, it is very complex and less useful than either Prandtl’s or von Karman’s equation. The defect velocity plot looks pretty good, but it does not give much information near the wall. The shape of the velocity curve is very important near the wall since it is the slope of the velocity profile at the wall that determines the wall shear stress. In other words, much of the important process in a turbulent flow occurs very near the wall. For this reason, researchers have devoted much work to the behavior of velocity and eddy diffusivity in the vicinity of a pipe wall. Universal velocity profile Consider turbulent flow in a horizontal pipe. If we consider flow very close to the wall, then the ratio r/R ≈ 1, and it is not important anymore. The equations for the velocity profile in turbulent flow are based on both theory and experimental data. Thus, they are semiempirical in nature. Laufer [27–29] performed extensive work on turbulent flow. Experimental data confirm the existence of a universal velocity profiles, and Figure 9.10 shows such a plot, which contains experimental data of Laufer. Researchers plotted the available experimental data related to velocity profiles in terms of u+ = u∕uτ and y+ = y uτ /ν in the form of Eq. (9.71b). They discovered that, close to the wall, velocity profiles measured at different Reynolds numbers collapse onto a single curve. This is

9.15 Experimental Work on the Turbulent Mean Flow

9 8 Nikuradse data for Re = 4000 7 6 (Vc – u)/u

τ

Prandtl equation

5 von Karman Equation

4 3 2 1 0 0

0.4

0.2

0.6

0.8

1

r R Figure 9.9 Comparison of different velocity distribution equations a function of y/R. Source: Data from Nikuradse, J. VDI-Forschungsheft 356, 1932.

30 Tubulent core

Laminar Buffer sublayer layer

20

box:Re = 500 000 circle:Re = 50 000

u+

u+ = 2.5 ln(y+) + 5.5 10 u+ = 5 ln(y+) – 3.05 u+ = y+

Laufer Data

0 101

102 y+

103

104

Figure 9.10 The universal velocity profile plotted in inner variable coordinates in a fully developed turbulent flow in a smooth pipe. Source: Data from Laufer, NACA Technical Report 2123, 1950.

called the universal velocity profile. An examination of Figure 9.10 shows that Eqs. (9.76)–(9.78) are plotted with experimental data. Data are well represented by these equations. Three different equations are used for modeling instead of smooth single curve. Based on the available experimental data, Karman [24] proposed the following equations for the velocity distribution for the whole cross section of a circular tube:

487

488

9 Foundations of Turbulent Flow

(a) Viscous sublayer: 0 ≤ y+ ≤ 5: This zone is a very thin layer in the immediate vicinity of the wall. Essentially, there are no turbulent fluctuations, turbulent stress u′ v′ is negligible, and only viscous stress is important. The viscous stress predominates over turbulent shear stress u + = y+ .

(9.76)

(b) Buffer layer: 5 ≤ y+ ≤ 30: In this zone, the turbulent stresses and the viscous stresses are of the same order of (comparable) magnitude and are equally important u+ = 5 ln(y+ ) − 3.05.

(9.77)

This equation provides a smooth transition between the laminar sublayer and the turbulent layer. It is the best curve fit to data. At the edge of the laminar sublayer (y+ = 5), Eqs. (9.76) and (9.77) intersect and are continuous in slope, since du+ /dy+ is the same for both equations at (y+ = 5). (c) Fully turbulent layer: The turbulent layer is the region from y+ > 30. In this zone, turbulence develops sufficiently, and viscous stresses are neglected. The turbulent diffusion of momentum dominates the molecular diffusion. The mean velocity varies logarithmically in this region, and it is often called logarithmic layer. It appears that experimental velocity data are in good agreement with the equations proposed by von Karman. The universal velocity distribution is the result of both the theoretical and experimental efforts to correlate the available measured data. Data are represented by the equation u+ = 2.5 ln(y+ ) + 5.5

(9.78)

and this equation will indicate a finite value for the velocity slope du∕dy at the tube center, whereas a zero slope is expected at the tube center. Thus, we may say that universal velocity profiles are valid near the wall and lead to an unrealistic situation at the centerline of the tube. At the intersection between the buffer zone and the turbulent core (y+ = 30), Eqs. (9.77) and (9.78) intersect, but with a discontinuity in slope at (y+ = 30). Eqs. (9.76)–(9.78) are compared with experimental data in Figure 9.10. However, Knudsen and Katz [35] indicate some inconsistencies in the universal velocity distribution. Most important inconsistency is the fact that at the tube axis, Eq. (9.78) does not give a zero-velocity gradient. Even though the universal velocity profile is developed for flow in tubes, this velocity distribution is found to represent the flow between parallel plates and in boundary layer on smooth flat plates, as discussed in [8]. However, the universal velocity distribution equations have been used extensively to study the relation between momentum and heat transfer since most of both pressure drop and heat transfer are related to the happenings near the wall. Sometimes several regions can be distinguished in a turbulent channel flow, as discussed in [4] and [30]. The viscous sublayer and the buffer layer occupy only a very small fraction of radius, but they influence the entire flow. A significant change in velocity variation occurs here. The structure of wall layer is nearly the same for many smooth pipe and channel flows, developing boundary layers on smooth walls. Shear stress does not vary across the wall layer, and we will make use of this fact to get solutions turbulent flow problems. The total shear stress τ across the inner region can be assumed to be constant and equal to the wall shear stress τw , i.e. τ ≈ τw = const. Power law In engineering calculations, an expression is needed for the velocity profile that is valid over most of the pipe cross section. This will be very useful in the determination of the mean fluid temperature in the turbulent heat transfer in a pipe. As reported in [17], Blasius developed a model using the dimensional analysis for the Fanning friction factor cf based on the experimental pipe data −1∕4

cf ≈ 0.0791 ReD

(9.79)

4 × 103 < ReD < 4 × 105 where cf = 2 τw /ρ V2 and V is the average velocity in the pipe and Reynolds number ReD = ρ V D/μ. Eq. (9.79) was one of the best applications of dimensional analysis. Based on this empirical correlation of Blasius, Prandtl [20] and von Karman [23] were able to represent the turbulent velocity distribution in a smooth pipe by a power law. The Moody friction factor f and the Fanning friction factor cf are related as f = 4 cf . From Eq. (9.69c) for τw and Eq. (9.70) for f, we have τw = 0.03955 ρ V7∕4 ν1∕4 D−1∕4 .

(9.80)

9.15 Experimental Work on the Turbulent Mean Flow

Table 9.1

Empirical exponents for the power law equation.

ReD

4 × 103

2.3 × 104

1.1 × 105

1.1 × 106

2 × 106

3.2 × 106

n

6

6.6

7

8.8

10

10

Source: Daily and Harleman [57].

Using τw = ρ u2τ and Eq. (9.80), after some algebra, we get ( ) ) ( ( ) ( )7∕4 uτ D 1∕4 u R 1∕7 1 V V = 6.99 τ = ⇒ . uτ 0.03955 ν uτ ν

(9.81)

As discussed in [17], based on the experimental work of Nikuradse, experimental studies indicate that V ≈ 0.8 Vc , and Eq. (9.81) may be expressed as ( ( ) ) u R 1∕7 Vc (9.82) = 8.74 τ uτ ν where Vc is the maximum velocity in the pipe. We may assume that Eq. (9.82) is valid for any distance y measured from the pipe wall; then, we have ( ) ( u y )1∕7 u ≈ 8.74 τ (9.83a) uτ ν or (u+ ) = 8.74(y+ )1∕7 . If we take the ratio of Eqs. (9.83a)–Eq. (9.82), we obtain ( ) ( ) y 1∕7 u = Vc R or

(

u Vc

)

) ( r 1∕7 = 1− . R

We may express Eq. (9.83d) as ( ) ( ) r 1∕n u = 1− Vc R

(9.83b)

(9.83c)

(9.83d)

(9.83e)

where y = R − r and y is measured from the wall. We should notice that the power law profile has a nonzero gradient at the pipe centerline, i.e. 𝜕u∕𝜕y||r=0 ≠ 0. This is the power law for velocity distribution of turbulent flow in a pipe, and it has been verified by experiments up to ReD = 105 . This power law fits the logarithmic form up to at least y+ = 1500. On the other hand, n is the exponent, which varies with the Reynolds number. A few values of n for specific values of the Reynolds number are listed in Table 9.1. The value of 7 is usually employed for the exponent n, and hence, the name seventh-root law is used. On the other hand, the power law breaks down in the laminar sublayer. The power law velocity profile is not valid near the wall since it does not predict the correct velocity gradient at the wall, and rather, it predicts (du∕dy) → ∞. The power law profile is also not valid near the centerline because it does not give du∕dr = 0 at r = 0. Note that the power law profile cannot be used to obtain the slope at the pipe wall since du/dy|y = 0 = ∞ for all values of n. Potter et al. [31] report that the value of n is related to the Moody friction factor f by the empirical relation 1 n≈ √ . f

(9.83f)

Example 9.6 Water at 17 ∘ C flows in a pipe with an average velocity of 3 m/s. The pipe diameter is 50 mm. Estimate the wall shear stress and friction velocity.

489

490

9 Foundations of Turbulent Flow

Solution The physical properties of ware are: ρ = 999 kg∕m3

μ = 1080 × 10−6 N.s∕m2 ν = 1.08 × 10−6 m2 ∕s.

The Reynolds number ReD is ReD =

3 m∕s × 0.05 m VD = = 138 888. ν 1.08 × 10−6 m2 ∕s

From Table 9.1, we estimate n as n = 7. From Eq. (9.79) cf =

0.0791 1∕4 ReD

=

0.0791 = 0.00409. (138 880)1∕4

The wall shear stress τw ) ( cf 2 ( 0.00409 ) kg ( m )2 3 999 3 τw = ρV = = 18.43 N∕m2 . 2 2 s m The friction velocity uτ is √ √ τw 18.43 N∕m2 = = 0.1357 m∕s. uτ = ρ 999 kg∕m3 Example 9.7 Engine oil flows at 350 K through a smooth pipe having an internal diameter of 60 mm. Pressure drop over a 4-m segment is 200 kPA. Flow is turbulent. Determine: (a) the centerline velocity of the pipe (b) thickness of the viscous sublayer. Solution Assume that engine oil is incompressible. The physical properties of engine oil at 350 K are ρ = 853 kg∕m3

μ = 3.56 × 10−2 N.s∕m2 ν = 41.7 × 10−6 m2 ∕s.

(a) The wall shear stress τw is calculated using Eq. (9.69a) τw =

ΔP D 200 × 103 N∕m2 0.06 m = = 750 N∕m2 L 4 4m 4

The friction velocity uτ is √ √ τw 750 N∕m2 = 0.9376 m∕s. = uτ = ρ 853 kg∕m3 The centerline velocity is the maximum velocity when y = R = 30 mm. y is measured from the pipe wall toward the pipe center y = 30 mm = 0.030 m Next, we calculate the dimensionless distance y+ as y+ =

y uτ 0.03 m × 0.9376 m∕s = = 674.5. ν 41.7 × 10−6 m2 ∕s

Since y+ > 30, flow is in the turbulent region, and we may use the log law u+ = 2.5 ln(y+ ) + 5.5

9.15 Experimental Work on the Turbulent Mean Flow

Vc = 2.5 ln(y+ ) + 5.5 ⇒ Vc = uτ [2.5 ln(y+ ) + 5.5] uτ Vc = 0.9376 m∕s × [2.5 ln(674.5) + 5.5] = 20.42 m∕s. (b) The viscous sublayer extends up to y+ = 5 yuτ 5 ν 5 × 41.7 × 10−6 m2 ∕s = 5 ⇒ δv = = 0.000222 m ≈ 0.22 mm = ν uτ 0.9376 m∕s

Example 9.8 Find a relation among average velocity V, centerline velocity Vc , and exponent n. Solution Let us integrate Eq. (9.83e) and determine the flow for any value of n. We have R ) ( r 1∕n n2 Vc 1 − (2 π r) dr = (2 π R2 ) Vc ∀̇ = ∫0 R (n + 1)(2 n + 1) Also, since the volume flow rate Q is ∀̇ = V(π R2 ). Then, the average velocity V is related to the pipe centerline velocity Vc as follows: ] [ 2n2 . V = Vc (n + 1)(2 n + 1) Example 9.9 Air at 20 ∘ C and 1-atm pressure is flowing in a smooth tube. The tube diameter is 0.1 m. If the average velocity is 30 m/s, determine the centerline velocity Vc . Solution For air at 20 ∘ C, ν = 15.46 × 10−6 m2 /s and ReD =

(30 m∕s)(0.1 m) VD = = 196 502. ν 15.46 × 10−6 m2 ∕s

From Table 9.1, we choose n = 7. From Example 9.9, we have ] [ 2n2 . V = Vc (n + 1)(2 n + 1) Then, the centerline velocity Vc is ) ( 8 × 15 (30 m∕s) = 36.73 m∕s. Vc = 2 × 72

9.15.2 Turbulent Flow over a Flat Plate: Velocity Profiles The difference between the velocity profile in pipe flow and external flow takes place far from the wall. It is expected that free-stream conditions will influence the boundary layer. On the other hand, experiments indicate that two flows are very similar near the wall since wall effects predominate. Schultz-Grunow [49] has obtained large amount of data for the flat plate velocity profile. In addition to experiments carried out for turbulent flow in pipes, experiments were also carried out for turbulent flow on flat plates. The velocity profiles in neighborhood of the flat plates have been measured. Experimentally obtained velocity data in turbulent boundary layer flow over a flat plate are usually correlated in one of three ways given below: (y) u (9.84) = f1 U∞ δ

491

492

9 Foundations of Turbulent Flow

u+ = f2 (y+ )

(9.85)

U+∞ − u+ = f3 (y∕δ) (9.86) √ where u+ = u∕ τw ∕ρ is the dimensionless velocity, τw is the wall shear stress, and ρ is the fluid density. On the other hand, √ y+ = y uτ ∕ν = (y τw ∕ρ)∕ν is the dimensionless distance measured from the wall. The distance y is measured from the plate wall, ν is the kinematic viscosity of the fluid, δ is the boundary layer thickness, and U+∞ is the dimensionless free-stream velocity. Experimental data plotted in the form of Eq. (9.84) are most satisfactory in the outer portion of the boundary layer. It has been observed experimentally that the correlating velocity profile in the form of Eq. (9.85) works very nicely near the wall and represents the universal velocity profile. Free-stream conditions have the least influence near the wall. Data plotted in the form of Eq. (9.86) are often called as the universal velocity defect law and correlate quite well in the outer region of the boundary layer. This type of plot represents the attempts for flow over a flat plate to tie the wall effects to the free-stream effects. Clauser [32, 33] plotted the velocity profile u∕U∞ as a function of y/δ in the form of Eq. (9.84) and obtained Figure 9.11. The inspection of Figure 9.11 reveals that turbulent profiles do not coincide and form a family of curves for different Reynolds numbers. Clauser also plotted the data in the form of Eq. (9.86). Clauser chose (u − U∞ )∕uτ for the ordinate and plotted the data as a function of y/δ and obtained the curve for the velocity data, as shown in Figure 9.12. Now, the turbulent flow velocity profiles have been reduced to a single universal curve. This plot is commonly called the defect law plot and the defect law is represented in the form (u − U∞ )∕uτ = f(y∕δ), and this correlation does not extend down to very small values of y/δ in the near wall region. Velocity changes very rapidly in this region, down to zero at the wall. The defect law plot does not show enough detail near the wall. Clauser indicates that this universal profile is not valid near the wall. Ludwig Prandtl and Theodore von Karman have given the theoretical basis for the universal velocity profile, as discussed in [8]. Researchers plotted u+ = u∕uτ versus y+ = y uτ /ν in the form of Eq. (9.85). Figure 9.13 shows a plot of u+ versus y+ near smooth walls using data from [32, 33] and [49]. These wall coordinates collapsed the velocity boundary layer data into a single curve, as depicted in Figure 9.13. Velocity profiles near a flat plate have been found to be very similar to the velocity profiles found in turbulent pipe flow. This is true at least very near the wall. Prandtl and von Karman, based on the available experimental information, deduced that the velocity profile consists of different regions. We may imagine, for practical purposes, the fully turbulent flow field may be divided into the following zones, as discussed in [4, 13] and [34].

0.9 0.8 Increasing cf

0.7 0.6 u U∞

0.5

Klebanoff and Diehl smooth wall

0.4 0.3

Hama 28 mesh screen

0.2 Hama 1 mesh screen 0.1 0

Figure 9.11

0.1

0.2

0.3

0.4 0.5 y/δ

0.6

0.7

0.8

0.9

Turbulent boundary layer velocity profile on a smooth and rough flat plate. Source: Data from Clauser [32, 33].

9.15 Experimental Work on the Turbulent Mean Flow

0

–5

( u – U∞) uτ

open circle: Klebanoff and

–10

solid circle: Schulz and Grunow –15

–20 0

0.2

0.4

0.6

0.8

1

y/δ Figure 9.12

Defect law plot of turbulent velocity profiles in the outer region of the boundary layer. Source: Data from Clauser [32, 33].

INNER REGION Viscous sublayer

Buffer layer

Wake Turbulent layer (log layer)

30

u+ = 2.5 In (y+) –5.5 u+ 20 u+ = 5 In (y+) –3.05

10

Ludwieg and Tillmann Klebanoff and Diehl Shultz -Grunow

u + = y+

y+ = 5

y+ = 30

0 101

Figure 9.13

102 y+

103

104

Universal velocity distribution for turbulent velocity profiles. Source: Data from Clauser [32, 33].

Inner region (y/δ ≤ y1 /δ) u = f(τw , ρ, ν, y) where τw is shear stress on the smooth wall and the equation is recast in dimensionless form as u+ = f(y+ )

493

494

9 Foundations of Turbulent Flow

y+ = u∕uτ uτ =

√

τw ∕ρ

1 ln(y+ ) + B k The y1 /δ ratio generally lies between 0.15 and 0.20. In other words, the inner layer represents typically 15–20% of the thickness of the boundary layer. The velocity profile is strongly influenced by viscosity and wall shear stress. The effect of outer layer on the velocity profile is negligibly small. In the inner region, inertia effects are negligible (the Couette flow approximation), and total shear stress is constant, as discussed in [4]. In the inner region, there are usually three different regions suggested by von Karman: u+ =

(1) Viscous Sublayer (inner wall region): 0 ≤ y+ ≤ 5 u + = y+ .

(9.87)

In this region, viscous effects are influential and turbulent fluctuations die out as we approach the wall. Flow is essentially laminar. This is the Prandtl’s hypothesis, and it is based on experimental data. (2) Buffer layer: 5 ≤ y+ ≤ 30 u+ = 5 ln(y+ ) − 3.05

(9.88)

In this region, viscous and turbulent effects are both equally important. Based on the pipe data, Karman [25] took y+ = 30 as the outer edge of the buffer layer. It appears reasonable to take y+ = 30 as the outer edge of the buffer layer for turbulent flow over a flat plate. (3) Fully turbulent (overlap) layer y+ ≥ 30 This region is also called logarithmic or overlap layer. In the fully turbulent region, turbulent eddies are dominant and viscous effects are negligible u+ = 2.5 ln(y+ ) + 5.5.

(9.89)

Equation (9.89) is often called the law of the wall. The examination of Figure 9.13 indicates that Eqs. (9.87)–(9.89) are in good agreement with available experimental data. Some researchers proposed a single formula to describe the whole inner region. A few of these equations will be presented here. Deissler [53] suggested that the laminar sublayer and the buffer layer may be modeled by a single equation y+ +

u =

∫0

dy+ 1 + n2 u+ y+ [1 − exp(−n2 u+ y+ )]

(9.90)

0 ≤ y+ ≤ 26. This equation contains u+ on both the left- and right-hand sides; for this reason, it is an implicit equation and must be solved iteratively. Reichardt [36] proposed a single formula for the entire inner region. His formula is ( + )] [ ( +) ( +) y y y u+ = 2.5ln(1 + 0.4y+ ) + 7.8 1 − exp − − exp − . (9.91) 11 11 3 This equation compares nicely with experimental data in the near wall region. Spalding [37] proposed a single composite formula that covered the entire inner region ] [ (κ u+ )2 (κ u+ )3 y+ = u+ + exp(−κ B) exp(κ u+ ) − 1 − (κ u+ ) − − . (9.92) 2 6

9.15 Experimental Work on the Turbulent Mean Flow

It is reported that this equation is in good agreement with experimental data all the way from wall to the point for y+ > 300. Spalding formula is implicit function of u+ and κ = 0.40, B = 5.5. Rannie [38] proposed the following correlation: ( + ) y + 0 ≤ y+ ≤ 27.5 (9.93) u = 14.53 tanh 14.53 u+ = 2.5 ln y+ + 5.5 y+ ≥ 27.7. Yu and Yoon [39] proposed a velocity formula that describes the entire inner region ( ) 1 ln y+ + B [1 − exp(−Dy+ )] (9.94) u+ = κ where D ≈ 0.14. This equation is based on the data provided by Laufer. Wake region (outer layer) Couette flow approximation is not valid outside the inner layer, and the shear stress is not constant anymore. For the region (y/δ) > 0.2, the mean velocity profile deviates from the law. The velocity distribution does not collapse into a single curve when expressed in terms of the inner variables. This deviation is seen in Figure 9.13. The behavior of turbulent flow in the wake region is correlated in terms of both inner and outer coordinates. In the wake region, Karman postulated that the velocity defect (U∞ − u) for the outer region is a function of (τw , ρ, y, δ), as discussed in [34] (U∞ − u) = g(τw , ρ, y, δ) and it may be expressed as (U∞ − u) = g(η) η = y∕δ uτ where δ is the boundary layer thickness and U∞ is the free-stream velocity. This equation is usually called the velocity defect law. Experimental studies of Coles indicated that the mean velocity profile in the outer region may be represented by two functions: (a) log law (b) law of the wake. The inner region has a length scale y and a velocity scale uτ . The wake region’s length scale is boundary layer thickness δ, and its velocity scale is free-stream velocity U∞ . Coles [40] introduced the concept of law of the wake to describe the velocity profiles in the wake region. The law of the wake can be written as (y) 1 Π (9.95) u+ = ln(y+ ) + C + W κ κ δ ( y) (y) π = 2 sin2 (9.96) W δ 2δ or (y) ( y) W = 1 − cos π (9.97) δ δ where κ = 0.41 and C ≈ 5.5. The parameter Π is the wake parameter. The wake function W(y/δ) is assumed to be the same for all boundary layers. The parameter Π is about 0.5 for a turbulent flow with no pressure gradient. Power law Based on experimental pipe data of Blasius, Prandtl suggested that the actual mean flow profile may be approximated by a power law similar to that used to describe turbulent pipe flow. Then, the velocity profile in a turbulent boundary layer may be approximated by ( y )1∕7 u = (9.98) U∞ δ where δ is the boundary layer thickness and U∞ is the free-stream velocity. Experimental studies indicate that the simplified velocity profile applies to 90% of the boundary layer thickness. Eq. (9.98) is usually valid up to Re = ρ U∞ x/μ = 107 . Experimental studies indicate that the power law velocity profile, Eq. (9.98), represents the velocity distribution in turbulent boundary layers on a smooth flat plate at zero incidence fairly well. Eq. (9.98) is known as the Prandtl’s one-seventh

495

496

9 Foundations of Turbulent Flow

power law. It is difficult to obtain the analytical solution for the velocity profile for the entire flow field. The best method is to identify the key variables in turbulent flow and obtain a functional relationship using dimensional analysis. Then, the experimental data are used to determine the numerical values of the constants. The division of the turbulent boundary layer into different regions is really not realistic, but the concept is very useful and has been used widely. We are accepting that the laminar sublayer region is free of turbulent eddies. This is not exactly true since eddies gradually become zero as the wall is approached. The presence of the wall quells wall-normal velocity fluctuations. This point is discussed in [41]. The universal velocity profile idea is very useful since both heat transfer and momentum transfer take place near the wall.

9.16 Transition to Turbulent Flow We have covered both the internal and external turbulent flows. Engineers need accurate methods for predicting transition, but this is a difficult task. Transition is very sensitive to disturbances, and for this reason, there is no sound theory. The current research indicates that we must rely on empirical data rather than in analysis. White and Majdalani [13] discuss the stability of laminar flows in depth and give some engineering methods to locate transition point. Mitchel [42] presented a method for predicting transition. He correlated transition for low-speed flows with the Reynolds number based on momentum thickness δ2 . The Reynolds number based on δ2 is defined as Reδ2 =

ρ U∞ δ2 . μ

(9.99)

Then, the equation for transition is given by Reδ2 ,tr ≈ 2.9Re0.40 x,tr .

(9.100)

The momentum thickness δ2 (x) can be calculated by Thwaites’ method, and δ2 (x) can be expressed as Reδ2 , and this is inserted into Eq. (9.100). When the expression is satisfied, transition at that Rex is obtained. Example 9.10 Estimate the transition Reynolds Rex,tr on a flat plate. Solution For a laminar flow on flat plate, the momentum thickness δ2 is given by √ νx δ2 = 0.664 U∞ δ2 0.664 = √ x Rex .We convert this to Reδ2 ( ) )( δ2 U∞ 0.664 ν = √ ν x U∞ Rex √ Reδ2 = 0.664 Rex .We set this equation equal to Eq. (9.100) √ 0.664 Rex,tr = 2.9Re0.40 x,tr

(a)

Let Rex,tr = u

√ > eq ≔ 𝟎.𝟔𝟔𝟒• u − 𝟐.𝟗•u𝟎.𝟒 ; √ eq ≔ 𝟎.𝟔𝟔𝟒• u − 𝟐.𝟗u𝟎.𝟒

We need to solve this nonlinear equation. The graphical method is frequently used to find the real roots of a function. It does not give information about the complex roots. The method is fairly simple and involves sketching the function in

9.16 Transition to Turbulent Flow

250

200

150

100

50

0 2. × 106

4. × 106

u

6. × 106

8. × 106

1. × 107

–50

Figure 9.E16

Plot of Eq. (a).

the desired range. We will use the graphical method with the command fsolve in MAPLE 2020. The fsolve command uses Newton-Raphson method to find the roots of a general nonlinear equation. Plot of this nonlinear equation is shown in Figure 9.E16. The intersection point of the curve with x-axis shows the location of the root. We get now an estimate of the root from the figure to use with the fsolve command. > plot(eq, u = 𝟏𝟎𝟎𝟎.𝟎..𝟏𝟎𝟕 ); > fsolve(eq, u = 𝟐•𝟏𝟎𝟔 ..𝟑•𝟏𝟎𝟔 ); 𝟐.𝟓𝟐𝟓𝟐𝟏𝟗𝟕𝟗𝟔 × 𝟏𝟎𝟔 Rex,tr = 2.525 × 106 . Cebeci and Smith [43] suggested the following correlation: ) ( 22400 Re0.46 Reδ2 ,tr ≈ 1.174 1 + x,tr . Rex,tr

(9.101)

In this method, the boundary layer development on the surface is calculated for laminar flow starting from the leading edge of the flow. This method is based on a combination of Michel’s and Smith’s e9 correlation. In this method, laminar flow starts from the leading edge of the plate. Example 9.11 Estimate the transition Reynolds Rex,tr on a flat plate. Solution For a laminar flow on flat plate, the momentum thickness δ2 is given by √ νx δ2 = 0.664 . U∞

497

498

9 Foundations of Turbulent Flow

We convert this to Reδ2 ( ) )( δ2 U∞ 0.664 ν = √ ν xU∞ Rex √ Reδ2 = 0.664 Rex . We set this equation equal to Eq. (9.101) ) ( √ 22400 Re0.46 0.664 Rex,tr = 1.174 1 + x,tr . Rex,tr Let u = Rex,tr > restart; > ) ( √ 𝟐𝟐𝟒𝟎𝟎 • 𝟎.𝟒𝟔 > eq ≔ 𝟎.𝟔𝟔𝟒• u = 𝟏.𝟏𝟕𝟒• 𝟏 + u ; u ) ( √ 𝟐𝟐𝟒𝟎𝟎 u𝟎.𝟒𝟔 eq ≔ 𝟎.𝟔𝟔𝟒 u = 𝟏.𝟏𝟕𝟒 𝟏 + u > fsolve(eq, u) 𝟐.𝟎𝟐𝟔𝟖𝟗𝟒𝟔𝟖𝟔 × 𝟏𝟎𝟔 > Thus, we see that Rex,tr = 2.02 × 106 .

Example 9.12 Wazzan et al. [44, 45] proposed the following correlation for predicting the transition point. First, we calculate the shape factor H by the Thwaites method. Transition occurs when log10 (Rex,tr ) = −40.4557 + 64.8066H − 26.7538H2 + 3.3819H3 for 2.1 < H < 2.8. For a flat H = 2.59, this method gives Rex,tr ≈ 4.8226 × 106 .

Problems 9.1

A certain fluid is moving in a horizontal pipe. The flow is turbulent. A hot wire anemometer is used to measure the local velocity at a certain point in the flow. Assume that you obtained the following readings at equal time intervals for a very short time: t(s) u(m/s)

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

46.0

44.62

44.52

43.24

40.81

39.22

39.75

40.28

41.34

44.20

We wish to determine the intensity of turbulence at this particular point.

Problems

9.2

A certain fluid is flowing in a horizontal pipe having an internal diameter of 6 cm. Flow is turbulent. The centerline velocity (maximum velocity occurs at the pipe center) is 10 m/s. Using the power law velocity profile ) ( r 1∕7 u = Vc 1 − R estimate the average velocity of the fluid. Here, Vc is the maximum velocity and R is the radius of pipe.

9.3

Water at a mean temperature of 17 ∘ C is flowing in a horizontal pipe with a volume flow rate of 5 × 10−2 m3 /s. The pressure drop in the pipe is 2.80 kPa/m. The pipe diameter is 5 cm. Assume steady fully developed turbulent flow. (a) Calculate the viscous sublayer thickness −δv . (b) Calculate the turbulent shear stress τw . (c) Calculate V/Vc . (d) Calculate the centerline velocity Vc .

9.4

Engine oil flows turbulently at a mean temperature of 320 K through the 60-mm diameter smooth horizontal pipe. Pressure is measured at two different locations, as shown in Figure 9.P4. Engine oil is assumed to be incompressible. (a) Calculate the thickness of viscous sublayer δv . (b) Calculate the maximum shear stress in the oil. (c) Calculate the maximum velocity. p2 = 150 kPa

p1 = 200 kPa (1)

(2)

L=3m Figure 9.P4

Geometry and problem description for Problem 9.4.

9.5

Water at 17 ∘ C flows in a 30 mm smooth horizontal pipe with a velocity of 3 m/s. Using the power law, approximate: (a) the Moody friction factor f (b) the wall shear stress (c) the centerline velocity (d) the friction velocity (e) the pressure drop over per unit length.

9.6

Water at 17 ∘ C flows in a horizontal pipe with an average velocity of 6 m/s. The internal diameter of pipe is 10 cm. The pressure gradient in the flow is 3 kPa/m. Using the power law velocity profile, determine: (a) the centerline velocity (b) the total shear stress at r = 0.030 m (c) the laminar shear stress at r = 0.030 m (d) the turbulent shear stress at r = 0.030 m.

9.7

Consider fully developed turbulent flow in a smooth pipe. The velocity profile may be represented by the power law ( y )1∕n u = Vc R where Vc is the centerline velocity and y is measured from the pipe wall. Eddy viscosity εM is defined by τ = ρεM

du dy

499

500

9 Foundations of Turbulent Flow

(a) Derive an expression for average velocity V in the pipe flow. (b) Derive an expression for εM /uτ R making use of the relation given below: τw f = V2 ρ 8 where V is the average velocity and f is the Moody friction factor. Reduce this equation using 1 n= √ . f Assume n = 7. Develop an expression for εM /uτ R as a function of (y/R). Develop an expression for eddy diffusivity of heat, εH . 9.8

Consider fully developed turbulent flow in a smooth pipe. The velocity profile may be represented by the log law ( yu ) 1 u τ +C = ln uτ κ ν √ where κ = 0.41 is von Karman constant, uτ = τw ∕ρ, and C is a constant to be determined experimentally. The distance y is measured from the pipe wall. (a) Derive an expression for average velocity V in the pipe flow. (b) Eddy viscosity εM is defined by: τ = ρεM

du . dy

Derive an expression for εM /uτ R making use of the relation given below: τw f = V2 ρ 8 where V is the average velocity and f is the Moody friction factor. Reduce this equation using 1 n= √ . f Assume n = 7. Develop an expression for εM /uτ R as a function of (y/R). Develop an expression for eddy diffusivity of εH . 9.9

Consider stagnation point flow. Velocity distribution is given by U∞ = Cx. Using Michel method, estimate the Rex, tr where transition takes place.

9.10

Using Figure 9.3, estimate the values of u′ 2 ∕k, v′ 2 ∕k, w′ 2 ∕k for dimensionless distance y/δ = 0.2 from the surface. Here, k is the turbulent kinetic energy, and it is defined as k=

9.11

1 ′2 (u + v′ 2 + w′ 2 ). 2

The following turbulent -flow velocity data is based on work of John Laufer [47]. Measurements, principally with a hot-wire anemometer, were made in fully developed turbulent flow in a pipe at ReD = 500 000. The internal diameter of the pipe was 24.70 cm. The centerline velocity was 30.50 m/s. Data were collected from Figure 3 given in the report NACA TN 2954,1953, using WebPlotDigitizer. The pipe radius r is measured from pipe center towards to the pipe wall; r = 0 corresponds to the pipe center and R is the pipe radius.

Problems

y/R

u∕Vc

0.0116

0.6059

0.0240

0.6659

0.0623

0.7356

0.0945

0.7794

0.1566

0.8211

0.2147

0.8467

0.2807

0.8803

0.3907

0.9157

0.4927

0.9391

0.5926

0.9685

0.6965

0.9798

0.8044

0.9952

0.9062

0.9966

1.0080

1.0000

Assume that air is at 1 atm pressure and 21 ∘ C. Determine (a) the average velocity by numerical integration (b) the wall shear stress from the log law approximation. 9.12

The following turbulent-flow velocity data is based on work of Reynolds et.al. [56]. Velocity profiles for turbulent incompressible flow of air aver a flat plate with a costant surface temperature have been measured at Reynolds number of Rex = 2.32 × 106 . The plate used in the experiments had an active length of 153.67 cm. The free stream velocity was measured with a pitot-static tube. Velocity surveys in the boundary layer were made with a specially designed and constructed pitot-tube. The turbulence intensity over most of the plate during typical runs was of the order of 2 to 3 percent. An analysis of experimental uncertainity indicated that the proble error in the local Reynolds number is ±1 percent. The velocity survey data are tabulated below. Distance From plate y in inches

Rex = 0.731 × 106 U∞ = 41.2 ft/s 𝛒∞ = 0.0770 lbm /ft3

Distance From plate y in inches

u∕U∞

𝛒∞ = 0.0770 lbm /ft3 Rex = 0.731 × 106 U∞ = 41.2 ft/s

u∕U∞

0

0

0.300

0.848

0.013

0.405

0.350

0.890

0.015

0.443

0.400

0.896

0.020

0.506

0.450

0.925

0.025

0.550

0.500

0.950

0.030

0.578

0.550

0.960

0.040

0.598

0.600

0.962

0.050

0.632

0.650

0.970

0.060

0.642

0.700

0.972

0.070

0.659

0.750

0.975

0.080

0.675

0.800

0.988

0.100

0.702

0.850

0.994

0.125

0.719

0.900

0.998

0.150

0.750

0.95

1.000

0.175

0.760

1.00

1.00

0.200

0.778

0.250

0.812

501

502

9 Foundations of Turbulent Flow

Assume that air is at 1 atm pressure and the free stream temperature of air is T∞ = 65.3 ∘ F. The plate surface temperature is Tw = 89.5 ∘ F. (a) Assume that the turbulent velocity profile can be represented by an equation of the form ( y )1∕m u = U∞ δ (b) (c) (d) (e) (f) 9.13

Obtain a relation between the boundary-layer thickness δ and the displacement thickness δ1 . Plot u as function of y Estimate the boundary layer thickness δ, Calculate displacement thickness δ1 and momentum thickness δ2 . Estimate the exponent m Estimate the average wall shear stress.

Pai [59] has derived a theoretical equation for Reynolds shearing stress given below. [ ] ( ( y )30 y) u′ v′ 1− 1− = 0.9835 1 − R R u2τ The data given below are obtained by Laufer [29] for flow of air in a pipe for ReD = 50 000. (R − r) R

u′ v′ u2𝛕

0.0786

0.9753

0.1821

0.9191

0.2839

0.8225

0.3821

0.7573

0.4821

0.7146

0.5893

0.6360

0.6875

0.5685

0.7946

0.5236

0.8982

0.4449

Plot both the theoretical equation and experimental data. Examine your plot and discuss your findings. 9.14

Chanson [60] reports that in a water tunnel, turbulent velocity distributions were conducted in a developing boundary layer along a flat plate in the absence of pressure gradient. Density ρ = 998.2 kg/m3 and dynamic viscosity μ = 1 × 10−3 Pa.s. y(mm) x(m) =

u (m∕s) 0.25

u (m∕s) 0.50

u (m∕s) 0.75

0

0

0

0

0.5000

5.5700

3.9400

4.1400

1.0000

6.2300

5.3000

4.8200

2.0000

6.6000

6.3800

5.8900

3.5000

7.3000

6.6100

6.1200

6.0000

7.7200

7.0200

6.3000

8.0000

7.7700

7.2000

6.8300

10.2000

7.7000

7.4200

7.0700

14.0000

7.6400

7.6100

7.1300

18.0000

7.6900

7.8700

7.3800

Problems

y(mm) x(m) =

u (m∕s) 0.25

u (m∕s) 0.50

u (m∕s) 0.75

21.0000

7.6800

7.6600

7.8700

35.0000

7.8400

7.6700

7.6900

41.0000

7.6900

7.7200

7.7100

58.0000

7.7600

7.6900

7.6600

60.5000

7.6700

7.7200

7.7200

66.0000

7.6800

7.7000

7.6800

70.0000

7.7700

7.7300

7.7000

90.0000

7.8000

7.6500

7.6900

100.0000

7.6900

7.6800

7.6800

Assume that water is at 17 ∘ C. (a) Plot the velocity distribution at each station. (b) Calculate the boundary layer thickness δ, displacement thickness δ1 and momentum thickness δ2 at x = 0.25 m, 0.5 m and 0.75 m. (c) Estimate the average wall shear stress. 9.15

Chanson [58] reports that velocity measurements in a developing, turbulent boundary layer along a smooth flat plate yield the data set given in the table below. (a) On a graph paper plot ln(u) versus ln(y) at x = 0.4 m (b) Estimate the shear velocity and the boundary shear stress at x = 0.4 m (c) Using integral momentum equation, calculate the boundary shear stress and shear force between x = 0.4 m and 0.6 m. Compare your results with shear velocity estimate and the Blasius formula for smooth turbulent flows. Discuss your findings. The fluid is air at 25 ∘ C and standard pressure.

x mm

y mm

u m/s

u′ m/s

400

2

8.111

1.33

4

8.859

1.362

6

9.68

1.502

8

10.283

1.476

10

10.714

1.297

15

10.756

1.366

20

10.77

1.426

30

10.8

1.286

40

10.799

1.301

50

10.9

1.03

70

10.789

0.965

90

10.98

1.037

110

10.81

0.923

130

10.78

0.919

503

504

9 Foundations of Turbulent Flow

u′ m/s

x mm

y mm

u m/s

600

2

8.591

1.159

4

8.508

1.342

6

9.392

1.249

8

10.241

1.184

10

10.327

1.448

15

10.814

1.368

20

11.12

1.341

30

11.23

1.301

40

11.157

1.102

50

11.162

1.168

70

11.219

0.943

90

11.239

0.969

110

11.186

1.056

130

11.359

0.927

References 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Tennekes, H. and Lumley, J.L. (1972). A First Course in Turbulence. MIT Press. Ting, S.K.D. (2016). Basics of Engineering Turbulence. Elsevier. Wilcox, D.C. (2006). Turbulence Modelling for CFD, 3e. DCW Industries. Pope, S.B. (2000). Turbulent Flows. Cambridge University Press. Klebanoff, P.S. (1954). Characteristics of Turbulence in a Boundary Layer with Zero Pressure Gradient. NACA Tech. Note 3178. Corino, E.R. and Brodkey, R.S. (1969). A visual investigation of the wall region in turbulent flow. J. Fluid Mech. 37 (1): 1–30. McComb, W.D. (2014). Homogeneous, isotropic turbulence: Phenomenology, renormalization and statistical closures. Oxford University Press. Schlichting, H. and Gersten, K. (2017). Boundary Layer Theory, 9e. Springer Verlag. Bejan, A. (2013). Convection Heat Transfer, 4e. Wiley. Jiji, L.M. (2009). Heat Convection, 2e. Springer Verlag. Kakac, S., Yener, Y., and Pramuanjaroenkij, A. (2014). Convective Heat Transfer. CRC Press. Burmeister, L.C. (1993). Convective Heat Transfer. Wiley. White, F.M. and Majdalani, J. (2021). Viscous Flow, 4e. McGraw Hill. Kays, W., Crawford, M., and Weigand, B. (2005). Convective Heat and Mass Transfer, 4e. McGraw Hill. Mills, A.F. (1999). Heat Transfer, 2e. Prentice Hall. Oosthuizen, P.H. and Naylor, D. (1999). Introduction to Convective Heat Transfer Analysis. Mc Graw Hill. Pai, S. (1957). Viscous Flow Theory II – Turbulent Flow. D. Von Nostrand Company. Nikuradse, J. (1950). Laws of Rough Pipes. NACA TM 1292. Parker, J.D., Boggs, J.H., and Blick, E.F. (1970). Introduction to Fluid Mechanics and Heat Transfer. Addison Wesley. Prandtl, L. (1933). Recent Results of Turbulence Research. NACA TM 720. Prandtl, L. (1949). Report on Investigation of Developed Turbulence. NACA TM 1231. von Karman, T. (1931). Mechanical Similitude and Turbulence. NACA TM 611. von Karman, T. (1946). On Laminar and Turbulent Friction. NACA TM 1092. von Karman, T. (1934). Turbulence and skin Friction. J. Aeronaut. Sci. 1 (1): 1–20. von Karman, T. (1939). The analogy between fluid friction and heat transfer. Trans. ASME 61: 705–710.

References

26 Wang, C. (1946). On the velocity distribution of turbulent flow in pipes and channels of constant cross section. J. Appl. Mech. 68: A85. 27 Laufer, J. (1950). Investigation of Turbulent Flow in a Two-Dimensional Channel. NACA TN 2123. 28 Laufer, J. (1951). Investigation of Turbulent Flow in a Two-Dimensional Channel. NACA TR 1053. 29 Laufer, J. (1953). The Structure of Turbulence in Fully Developed Pipe Flow. NACA TN 2954. 30 Reynolds, A.J. (1974). Turbulent Flows in Engineering. Wiley. 31 Potter, M.C., Wiggert, D.C., and Ramadan, B.H. (2017). Mechanics of Fluids, 5e. CENGAGE. 32 Clauser, F.H. (1954). Turbulent boundary layers in adverse pressure gradients. J. Aeronaut. Sci. 21 (2): 91–108. 33 Clauser, F.H. (1956). The Turbulent Boundary Layers, in Advances in Applied Mechanics (ed. I.V. Vol), 1–51. 34 Kundu, P.K., Cohen, I.M., Dowling, D.R., and Tryggvason, G. (2016). Fluid Mechanics, 6e, Elsevier, Inc. 35 Knudsen, J.G. and Katz, D.L. (1958). Fluid Dynamics and Heat Transfer. McGraw Hill. 36 Reichardt, H. (1951). J. Appl. Math. Mech. (ZAMM) 31 (7): 208–219. (Wiley Online Library). 37 Spalding, D.B. (1961). A single formula for the law of the wall. J. Appl. Mech. 28 (3): 455–458. 38 Rannie, W.D. (1956). Heat transfer in turbulent shear flow. J. Aeronaut. Sci. 23 (5): 485–489. 39 Kwonkyu, Yu and Yoon, B. (2005). A velocity profile formula valid in the inner region of turbulent flow. Proceedings of 31st Biennial IAHR Congress, Seul, Korea (ed. B.H. Jun, S.I. Lee, I.W. Seo, and G.W. Choi), Korea Water Resources Association: 378–385. 40 Coles, D.E. (1956). The Law of the wake in the turbulent boundary layer. J. Fluid Mech. 1: 191–226. 41 Kline, S.J., Reynolds, W.C., Schraub, F.A., and Bunstadler, P.W. (1967). The Structure of turbulent boundary layers. J. Fluid Mech. 30: 741–773. 42 Mitchel, R. (1952). Study of Transition on Wing Sections; Establishment of a Criterion for the Determination Point of Transition and Calculation of the Wake of an Incompressible Profile. ONERA, Report 1/1578A. 43 Cebeci, T. (2004). Analysis of Turbulent Flows. Elsevier. 44 Wazzan, A.R., Gazley, C., and Smith, A.M.O. (1979). Tollmien Schlichting waves and transition: heated and adiabatic wedge flows with application to bodies of revolution. Progress Aerospace Sci. 18 (2): 351–392. 45 Wazzan, A.R.C.G. and Smith, A.M.O. (1981). H-Rx method for predicting transition. AIAA J. 19 (6): 810–812. 46 Smith, A.O.M. (1956). Transition, pressure gradient and stability theory. Proc. IX Int. Congress Appl. Mech. Brussels 4: 234–244. 47 Laufer, J. (1953). The structure of turbulence in fully developed pipe flow, NACA TN.2954. 48 Reynolds, W.C., Kays, W.M. and Kline, S.J. (1958). Heat transfer in the turbulent incompressible boundary layer I-Constant wall temperature, NASA MEMO 12-1-58W. 49 Schultz-Grunow, F. (1941). New Frictional Resistance Law for Smooth Plates. NACA TM-986, Wasingtron. 50 Yuan, S.W. (1967). Foundations of Fluid Mechanics. Prentice-Hall. 51 Hinze, J.O. (1959). Turbulence. McGraw-Hill. 52 Weigand, B. (2015). Analytical Methods for Heat Transfer and Fluid Flow Problems, 2e. Springer-Verlag. 53 Deissler, R.G. (1950). Analytical and Experimental Investigation of Adiabatic Turbulent Flow in Smooth Tubes. NACA TN 2138. 54 Thomas, L.C. (1999). Heat Transfer, 2e. Capstone Publishing Corporation. 55 Pfeil, H. and Sticksel, W.J. (1982). Influence the pressure gradient on the law of the wall. AIAA J. 20 (3): 434–436. 56 Pfeil, H. and Sticksel, W.J. (1981). About the influence of the pressure gradient on the law of the wall. AIAA 19th Aerospace Sciences Meeting. St Louis, MO, U.S.A. (12 January 1981–15 January 1981). 57 Daily, J.W. and Harleman, D.R.F. (1966). Fluid Dynamics. Addison-Wesley Publishing Company, Inc. 58 Reynolds, W.C., Kays, W.M. and Kline, S.J. (1958). Heat transfer in the turbulent incompressible boundary layer I-Constant wall temperature. NASA MEMO 12-1-58W, NASA. 59 Pai, S.I. (1953). On turbulent flow in a circular pipe, J. Of Franklin Institute 256 (4): 337–352. 60 Chanson, H. (2014). Applied Hydrodynamics, An Introduction, CRC Press.

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer 10.1 Introduction In this chapter, we will discuss the characteristics of momentum and heat transfer in the turbulent boundary layer. The concept of viscous sublayer and the law of the wall for the near wall region are studied. The Prandtl mixing-length theory is introduced in order to get a solution for the turbulent boundary layer equations. Approximate solutions to momentum and energy equations are introduced. A valid reasonable and workable theory of turbulent boundary layers is essential for the prediction of (a) drag force (b) flow separation (c) heat transfer.

10.2 Turbulent Momentum Boundary Layer We will now consider a flat plate in turbulent flow, as depicted in Figure 10.1. We assume steady-state, two-dimensional, and incompressible turbulent flow with constant properties. Flow over the flat plate has constant velocity U∞ . Governing equations are given as follows: Continuity: 𝜕u 𝜕v + =0 𝜕x 𝜕y

(10.1)

Momentum ( ) [ ] dp 𝜕 𝜕u 𝜕u 𝜕u ρ u +v =− + (μ + μt ) . 𝜕x 𝜕y dx 𝜕y 𝜕y

(10.2)

This equation is subjected to the following boundary conditions: u(x, 0) = 0

(10.3a)

v(x, 0) = 0

(10.3b)

u(x, ∞) = U∞

or

u(x, δ) = U∞

(10.3c)

u(0, y) = U∞

(10.3d)

where μt = ρεM is the eddy viscosity or turbulent viscosity. Formulation is completed by specification of eddy viscosity μt , which is valid within the entire inner region. Engineers are interested in the mean flow field, and the key problem is the specification of eddy viscosity μt . The eddy viscosity is a flow property and depends on flow characteristics. How do we model turbulent viscosity μt ?

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

y

U∞

δ(x) u x

O Figure 10.1

Velocity profile on a flat plate.

10.3 Turbulence Models We learned that Eqs. (10.1) and (10.2) need to be supplemented by a model for the Reynolds stress before any solution can be obtained. The Reynolds stress must be specified by theoretical and/or empirical inputs. The Reynolds stress τt = −ρu′ v′ can be evaluated using experimental measurements for u′ and v′ , but the usual approach is the Boussinesq or mean-field method. In this method, the turbulence parameter is related to the local mean axial velocity u by τt = μt

𝜕u 𝜕y

(10.4a)

where turbulent viscosity (or eddy viscosity) μt must be specified. We need a model for turbulent viscosity μt . On the other hand, εM = μt /ρ is the eddy diffusivity or eddy kinematic viscosity. The eddy viscosity μt has the same dimension as μ. The turbulence models are discussed by several researchers, such as Wilcox [1], Schlichting and Gersten [2], and Reynolds [3]. The turbulence models can be classified as follows: (1) (2) (3) (4) (5) (6)

Zero-equation model One-equation model Two-equation model Second -moment closure model Large eddy simulation Direct numerical simulation

10.3.1

Zero-Equation Models

These are also called algebraic turbulence models. To model turbulent flow, recall that one needs to express the turbulent transport due to fluctuations in terms of time-averaged quantities. The transport quantities for turbulent flow are expressed as the sum of molecular and eddy effects. The contributions from eddy-level activities are proportional to the gradients of the time-averaged quantities. We will only talk about algebraic (zero-equation) model. In this model, the approach suggested by Boussinesq, Boussinesq eddy viscosity approximation, is used, and the constitutive relation of the isotropic Newton–Stokes law is imitated. For computational purposes, the eddy viscosity is computed in terms of mixing length. Eddy viscosity (and the mixing length) depends on the flow. Researchers are interested in finding a workable expression for μt . Such s relationship relies on the empirical or semiempirical approach. 10.3.1.1 Boussinesq Model

In this model, μt = ρεM and εM is constant, as discussed in [4]. Notice that εM is a flow property. But constant εM does not allow u′ v′ approach zero at the wall. Turbulent fluctuations are expected to damp out very close to wall. 10.3.1.2 Prandtl’s Mixing-Length Model

Prandtl’s mixing-length model is an important model, and Wilcox [1], Schlichting and Gersten [2], and Reynolds [3] give detailed information about this model. Prandtl’s mixing-length theory states that eddies travel an average distance (mixing

10.3 Turbulence Models

length 𝓁) before they collide and lose their identity. The turbulent eddies defined as a bunch of fluid particles with similar flow characteristics. In 1925, Prandtl came up with the following expression for turbulent eddy viscosity μt : μt = ρ𝓁 2

𝜕u 𝜕y

(10.4b)

where μt is a flow property and 𝓁 is called the mixing length. We now need a model for mixing length 𝓁. The model depends on the type of flow. For a wall-bounded flow in the inner region, Prandtl stated that mixing length 𝓁 is given by 𝓁=κy

(10.4c)

where κ is determined experimentally from data on turbulent flow in tubes by Prandtl and Nikuradse, and it is approximately equal to 0.4. Notice that the distance y is measured from the wall. We will talk more about the Prandtl’s mixing length 𝓁 in Chapter 11 since the mixing-length model is based on pipe flow data provided by Nikuradse. Equation (10.4c) is not valid in the viscous sublayer and in the outer layer. Equation (10.4a) now becomes ( )2 2 𝜕u (10.4d) τt = ρ𝓁 𝜕y or ( )2 𝜕u τ t = ρ κ 2 y2 . (10.4e) 𝜕y The turbulent viscosity μt is related to the eddy diffusivity of momentum (or kinematic eddy viscosity νT ) εM by μt = ρεM .

(10.4f)

Here, the normally absolute value is used on derivative to make sure that the eddy diffusivity εM (or νT ) remains positive. Using Eqs. (10.4b), (10.4c), and (10.4e), we may write an expression for the eddy diffusivity of momentum, εM | du | (10.4g) ρεM = ρ𝓁 2 || || . | dy | We should note that mixing length 𝓁 may be determined from the measurements of turbulent fluctuations and velocity gradient using the following relation: ( )2 2 du ′ ′ −u v = 𝓁 . (10.4h) dy 10.3.1.3 Van Driest Model

The mixing-length model needs modification to damp the eddy viscosity to zero as the flow approaches a solid wall, and also to account for the outer intermittent region. Van Driest [5] introduced a damping factor into the mixing length, Eq. (10.4c). Actually, close to the wall, the mixing length is smaller than indicated by Eq. (10.4c) since viscosity dampens the velocity fluctuations. Based on this fact, Van Driest proposed the following relation: [ ( + )] y (10.4i) 𝓁 + = (κ y+ ) 1 − exp − + A where A+ is the empirically determined parameter and y+ = yuτ /ν. The term in bracket is a damping factor and is derived from oscillating laminar flow near wall. The constant A+ depends on flow conditions such as pressure gradient, wall roughness, and blowing or suction, as discussed in [6] and [7]. For a smooth, impermeable flat plate with zero-pressure gradient, we may take A+ = 26. It is experimentally observed that Eq. (10.4i) gives satisfactory results for parallel flow (no mass transfer) and with moderate-pressure gradients. Introducing Eq. (10.4i) into Eq. (10.4g) and using wall coordinates u+ = u∕uτ and y+ = yuτ /ν, we get [ ( + )]2 ( + ) εM y du + 2 = (κ y ) 1 − exp − + . (10.4j) ν A dy+ We should notice that εM is an empirical function. These equations may be used to develop analytical and numerical solutions to obtain mean velocity distribution and mean wall shear stress. We are interested in developing simplified

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analytical solutions for dimensionless velocity u+ and the local friction coefficient cf for a fully developed turbulent boundary layer flow over a flat plate with constant free-stream velocity U∞ .

10.4 Turbulent Flow over a Flat Plate with Constant Free-Stream Velocity: Couette Flow Approximation The Reynolds averaged momentum equation for steady flow will be simplified in the region very near the wall. Recall that very near the wall (within about first 20% of the boundary layer thickness), the inertial terms on the left-hand side of Eq. (10.2) may be neglected relative to molecular and turbulent momentum transport.

10.4.1

Inner Region

To study the boundary layer problem near the wall, usually the following assumptions are made: 1) The flow over the flat plate has a constant free-stream velocity. This means dp∕dx = 0. 2) The flat plate is a solid wall. The principal direction of the flow is parallel to flat plate, that is, u = u(y), and for this reason, v = 0. The conservation of mass then implies that 𝜕u∕𝜕 x ≈ 0. Then, in a region very close to the wall, ρu(𝜕u∕𝜕 x) and ρv(𝜕v∕𝜕y) terms can be neglected. We conclude from this discussion that in the proximity to the wall (within approximately first 20% of boundary layer thickness), inertia terms are negligible relative to the molecular and turbulent transport of momentum, and the momentum equation becomes [ ] dτ dτw d du (μ + μt ) = = ≈0 (10.5) dy dy dy dy where τ is the total shear stress, and it is constant and equal to wall shear stress τw . The reduction of Eqs. (10.2)–(10.5) is called as the Couette flow approximation. The relationship given above indicates that the total shear stress is approximately constant with respect to y within the inner region. This total shear stress does not vary with y in the inner region and is given as (μ + μt )

du ≈ τ ≈ τw = constant. dy

(10.6a)

The total shear stress is the sum of the molecular transport of momentum and the turbulent transport of momentum. Here, notice that viscous shear stress in the viscous laminar sublayer is constant, and the total shear stress is constant in an appropriate portion of the turbulent boundary layer. This idea is supported by experimental evidence, as discussed in [8]. The term τw is the wall shear stress exerted by the fluid on the wall. The substitution of Eq. (10.4d) into Eq. (10.6a) yields ( )2 ( ) du 2 du + ρ𝓁 . (10.6b) τw = μ dy dy Let us introduce the following dimensionless wall coordinates. First recall that the definition of the skin friction coefficient cf is given as (c ) 2 τw τw f 2 = U . (10.7) cf = ⇒ ∞ 2 ρ 2 ρ U∞ Very close to the wall, the most important scaling parameters are kinematic viscosity ν and wall shear stress τw . We may now define characteristic velocity uτ and viscous length scales δv : √ √ τw cf uτ = = U∞ (10.8) ρ 2 ν δv = . (10.9) uτ Next, the following dimensionless wall coordinates are introduced: √ u∕U∞ u u 2 + u = = = √ uτ U c cf ∕2 ∞ f

(10.10)

10.5 The Universal Velocity Profile

and y+ =

√ √ y τw ∕ρ yU∞ cf ∕2 y uτ y = = = δv ν ν ν

(10.11)

where u+ is the dimensionless velocity and y+ is the dimensionless distance; it is a sort of Reynolds number, and it represents the relative importance of viscous and turbulent transport at different distances from the solid wall. At very large y+ , the effect of kinematic viscosity ν on momentum transport is negligible, and the height scale on the boundary layer depth δ is y (10.12) η= δ and the quantity uτ δ = δ+ (10.13) ν is a special kind of Reynolds number. We now return to the momentum equation. Substituting the dimensionless wall coordinates, the momentum equation, Eq. (10.6a), becomes ( ε ) du+ =1 (10.14) 1+ M ν dy+ Reτ =

u+ (0) = 0.

(10.15)

The ratio εM /ν must vary with the distance measured from the wall. Given a model of eddy diffusivity of momentum, εM , it is possible to obtain velocity distribution near the wall. Ordinary derivatives are used instead of partial derivatives since u+ is only a function of y+ in the constant-shear-stress region.

10.5 The Universal Velocity Profile There are several velocity profiles proposed by researchers, and we will see some of these models. A useful and concise discussion of the universal velocity profiles can be found in [2, 6, 9, 10].

10.5.1 Three-Layer (von Karman) Model for the Velocity Profile Based on the experimental data of Nikuradse for velocity distribution for turbulent flow in pipes, von Karman [11] divided the turbulent boundary layer into three distinct regions, and this velocity profile is called three-layer model. The three-layer model is also discussed in [12]. However, the division of the turbulent boundary layer into three distinct regions is not really realistic, but the concept is very useful and has been used widely in engineering community. Here, we will basically follow the idea proposed by von Karman. We will assume that the following regions exist in a complete velocity profile, as discussed in [6]. (a) Inner region • Viscous sublayer • Buffer layer • Turbulent layer (b) Outer layer (a) Inner region The inner region (here the Couette approximation is valid, and sometimes, it is called near wall region) can be approximately broken down into three regions: The viscous sublayer: In this region, the molecular momentum diffusion dominates the turbulence, and we have ν ≫ εm . Therefore, Eq. (10.14) reduces to du+ = 1. (10.16) dy+

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

Integrating with boundary condition u+ = 0 at y+ = 0, the velocity profile becomes u + = y+ ,

0 < y+ < 5.

(10.17)

The velocity distribution fits the experimental data very nicely from y+ = 0 to y+ = 5. In a different way of saying, the second term of Eq. (10.6b) is neglected within this region. The buffer layer: In this region, both molecular and turbulent momentum diffusion are equally important. The linear or log-law distribution is not adequate. This region lies approximately within 5 < y+ < 30, and it is reported that the experimental data from various investigators are approximated by the following empirical velocity distribution: u+ = 5 ln y+ − 3.05,

5 < y+ < 30.

(10.18)

The velocity profiles are in good agreement with the measured values. This relation is applicable to turbulent boundary layers as well as turbulent pipe flows. The turbulent layer In this region, the turbulent momentum diffusion is much more important than the molecular diffusion, and we have εm ≫ ν. The Couette flow approximation is still valid in the turbulent region. Under this condition, Eq. (10.14) reduces to εM du+ = 1. ν dy+

(10.19)

Equation (10.19) is valid outside the viscous sublayer and the buffer region. To work with Eq. (10.19), we need an expression for εM /ν. This expression is ( ) 𝜕u (10.20) εM = 𝓁 2 𝜕y where 𝓁 is called the mixing length, and it is a measure of the size of the turbulent eddies. There is no universal expression for mixing length, and it varies from flow to flow. This is called Prandtl’s mixing-length model. This is a robust turbulence model, and this mixing-length model relates the eddy viscosity or eddy diffusivity εM to velocity gradient. Prandtl’s mixing-length model provides an empirical expression for the mixing length 𝓁, and it is a model for the turbulent transport term in the Reynolds averaged momentum equation. The mixing length 𝓁 is the mixing length, which represents the distance traveled by a particle of fluid before its momentum is changed by the new environment. The model assumes that eddies are constrained by the presence wall and mixing length decreases to zero at y = 0 𝓁=κy

(10.21)

where κ is called Karman constant. Equation (10.21) indicates that the mixing length is proportional to the distance from the wall. Substituting Eq. (10.21) into Eq. (10.20) and using wall coordinates, Eq. (10.20) becomes ( +) εm du 2 + 2 = κ (y ) . (10.22) ν dy+ Substituting Eq. (10.22) into Eq. (10.19) yields ( + )2 2 + 2 du = 1. κ (y ) dy+

(10.23)

Solving Eq. (10.23) for du+ /dy+ and integrating, we obtain 1 ln y+ + B (10.24) κ where κ and B are near-universal constants for turbulent flow over smooth, impermeable flat plate. This equation cannot hold down to y = 0 since very near the wall, laminar shear stress cannot be negligible with respect to turbulent shear stress. This equation is called law of the wall, and the constant κ is called von Karman’s constant and is obtained experimentally. u+ =

10.5 The Universal Velocity Profile

As discussed in [12], based on experiments of Nikuradse, researchers found that curve fitting for y+ ≥ 30 gives u+ = 2.5 ln y+ + 5.5

(10.25a)

where B = 5.5 and κ = 0.40. Recently, Coles and Hirst, as reported [6], adjusted these constants to B = 5.0 and κ = 0.41, and with these constants, Eq. (10.24) becomes u+ = 2.44 ln y+ + 5.

(10.25b)

> 30. It is important to mention that the three-region velocity profile has been validated experimentally and may also be applied for turbulent flow in a tube. We should notice that equations developed here are valid for turbulent boundary layer flows with constant free-stream velocity U∞ , and there is no suction or blowing over the plate. Equations (10.25a) or (10.25b) applies to y+

(b) Wake layer: Outside the inner region, there is a region called wake region, and shear stress is not constant and Couette flow assumption is not valid anymore. Velocity distribution does not collapse on a single curve when expressed in terms wall variables. Reynolds stresses are much larger than the viscous stresses and influence the velocity profile. The law of the wall is increasingly inadequate to match experimental data. The velocity distribution in the wake region is correlated in terms of both an inner and outer coordinate. The outer coordinate is the ratio of the position y to the boundary layer thickness δ. The velocity in this region, which for a flat plate with zero-pressure gradient, is represented by a dimensionless relation similar to the following: (y) U∞ − u =f . (10.26) uτ δ At the edge of the boundary layer, y = δ and (u − U∞ )∕uτ = 0, so we have u = U∞ . However, the functional relationship needs to be modified for flows with free-stream pressure gradients. The empirical equation of the velocity profile in the wake region is proposed by Coles [13], and it is given as 1 2Π ln y+ + B + f(η) κ κ where f(η) is some S-shaped function with f(0) = 0, f(1) = 1 and popular forms are u+ =

(10.27)

f(η) = 3η2 − 2η3

(10.28)

or f(η) = sin2

(

) π η 2

(10.29)

y and κ = 0.41, B ≈ 5.0, and Π = 0.45 for a turbulent flow with no pressure gradient over a flat plate. δ ∏ ∏ The deviations from the log law are quantified by the Coles wake parameter , and in general, is function of pressure gradient. By adding the wake law to the log law, an accurate approximation of both overlap and outer layers is obtained, and a combination of log-law region and outer layer is sometimes called defect law region. Equation (10.27), a composite profile, is a good approximation right across the turbulent boundary layer, and a valuable expression for wall friction coefficients can be derived. where η =

Example 10.1 Air at 1-atm pressure and 27 ∘ C flows over a flat plate at 20 m/s. Assume that the measured shear stress at 1.6 m is given as 0.70 Pa. At this position, estimate (a) the friction velocity uτ and (b) velocity u at y = 5 mm from the wall. Solution The physical properties of air at 300 K are ν = 15.89 × 10−6 m2 ∕s

ρ = 1.1614 kg∕m3

μ = 184.6 × 10−7 N.s∕m2 .

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

First, we calculate the Reynolds number Rex U∞ x 20 × 1.6 = 2.01 × 106 . = ν 15.89 × 10−6 Flow is turbulent. Rex =

(a) The friction velocity uτ is √ √ τw 0.7 = = 0.776 m∕s. uτ = ρ 1.1614 (b) Velocity at y = 0.5 cm is u y 0.776 × 0.005 = 244.17. y+ = τ = ν 15.89 × 10−6 We expect that this position is in the turbulent region, and we use Eq. (10.25a) u+ = 2.5 ln y+ + 5.5 u+ = 2.5 ln(244.17) + 5.5 = 19.24 u = 19.24 × 0.776 ≈ 14.93 m∕s.

10.5.2

Other Velocity Models

There are several proposed models for the velocity distribution within the inner region. A few of these models are given in Table 10.1.

10.6 Approximate Solution by the Integral Method for the Turbulent Momentum Boundary Layer over a Flat Plate The need to calculate the skin friction coefficient for turbulent boundary layer on a flat plate is encountered in engineering, e.g. ships and aircrafts. We will now present the application of the integral equation for turbulent flow over a flat surface. Assume that we have a constant density steady turbulent flow over a flat solid surface with no suction or blowing. Turbulent flow starts from the leading edge of the plate. The free-stream velocity of the flow is constant. The momentum integral equation for this case is given as τw ρU2∞

=

dδ2 . dx

(10.30)

Prandtl’s power law is a good approximation for turbulent boundary layer over a flat plate with zero-pressure gradient except the wake region ( y )1∕n u = . U∞ δ In Chapter 9, in terms of inner variables, the power law was shown to be u+ = 8.74(y+ )1∕7 . Kays et al. [7] give the following equation for the power law: u+ = 8.75(y+ )1∕7 .

(10.31)

Equation (10.31) was specifically developed for flow over a flat plate with constant free-stream velocity. Extremely near the wall, in the laminar viscous sublayer, the simple power law velocity distribution is not valid. For this reason, we cannot relate the wall shear stress τw to gradient of this last equation at y = 0. The wall shear stress τw must be related to the

10.6 Approximate Solution by the Integral Method for the Turbulent Momentum Boundary Layer over a Flat Plate

momentum integral equation using an experimental correlation. This is done by borrowing a wall shear correlation from pipe flow data. Consider now Eq. (9.80), which is repeated here for convenience τw = 0.03955 ρ V7∕4 ν1∕4 D−1∕4 . We have shown in Chapter 9 that the average velocity is related to the centerline velocity by ] [ 2n2 . V = Vc (n + 1)(2 n + 1) We choose n = 7, and this will give us V ≈ 0.81 Vc . We wish to estimate the wall shear stress on a flat plate. To do this, we now replace D/2 by δ, V by 0.8 Vc , and Vc by U∞ . Then, we have ( )1∕4 ( )1∕4 ν 7∕4 ν 2 = 0.0225 ρU∞ . (10.32) τw = 0.0225 ρU∞ δ δ U∞ This is the equation given by Von Karman [17]. Next, displacement δ1 and momentum thickness δ2 are evaluated. The results of evaluations are ] δ[ δ[ ( y )1∕7 ] δ u δ1 = 1− dy = 1− dy = (10.33) ∫0 ∫0 U∞ δ 8 δ

δ2 =

∫0

(

u U∞

)[ 1−

] δ ( )1∕7 [ ( y )1∕7 ] y 7 u δ. dy = 1− dy = ∫0 δ U∞ δ 72

Table 10.1 Selected universal velocity distribution models for the inner region of a turbulent boundary layer. Model

Expression

Prandtl–Taylor

⎧ + ⎪y u =⎨ ⎪2.44 ln(y+ ) + 5.5 ⎩

Von Karman

Spalding

Reference

+

for 0 < y+ < 11.5

Rubesin et al. [14]

for y+ > 11.5

⎧ + for 0 < y+ < 5 ⎪y ⎪ u+ = ⎨5 ln(y+ ) − 3.05 for 5 < y+ > 30 ⎪ ⎪2.5 ln(y+ ) + 5.5 for y+ > 30 ⎩ [ 1 y+ = u+ + A exp(κ u+ ) − 1 − κ u+ − (κ u+ )2 2 ] 1 1 − (κ u+ )3 − (κ u+ )4 6 24 A = 0.1108

Von Karman [11]

Spalding [15]

κ = 0.4 all y+ in the inner region Van Driest

du+ = dy+ 1+ y+

Deisler

u+ =

2 [ ( + )]2 y 1 + 4(κ y+ )2 1 − exp − 26

√

∫0

dy+ 1 + n2 u+ y+ [1 − exp(−n2 u+ y+ )]

Van Driest [5]

0 ≤ y+ ≤ 26 Deisler [16]

n = 0.124 u+ = 2.78 ln y+ + 3.8

y+ ≥ 26

(10.34)

515

516

10 Turbulent External Boundary Layers: Momentum and Heat Transfer

The shape factor H is H=

δ1 = 1.29. δ2

(10.35)

We can express shear stress τw in terms of the momentum thickness δ2 τw =

0.02246 ρU2∞ 0.01254 ρU2∞ = . [(72 δ2 ∕7)(U∞ ∕ν)]1∕4 (U∞ δ2 ∕ν)1∕4

(10.36)

Remember that wall shear stress τw is based on experimental data. The next step is to evaluate momentum thickness δ2 as a function of x along the flat plate using the momentum integral equation. We substitute wall shear stress τw into the integral momentum equation ( ) U∞ δ2 −1∕4 dδ2 0.01254 (10.37a) = ν dx at x = 0, δ2 = 0. Equation (10.37a) can be rearranged to the following form: ( ) U∞ −1∕4 1∕4 dδ2 δ2 = 0.01254 dx ν

(10.37b)

at x = 0, δ2 = 0. The solution of this differential equation with Maple 2016 is δ2 0.03598 = x Re0.2 x

(10.38)

where Rex = U∞ x/ν is the local Reynolds number. Knowing the momentum thickness δ2 , we can now evaluate the turbulent boundary layer thickness δ(x) as follows: ) ( )( 0.37 x 72 0.03598 x = (10.39a) δ= 7 Re0.2 Re0.2 x x or δ(x) 0.37 = , x Re0.2 x

5 × 105 ≤ Rex ≤ 107 .

(10.39b)

The displacement thickness δ1 is δ1 0.046 = . x Re0.2 x

(10.40)

We express the local friction coefficient cf in terms of momentum thickness δ2 using Eq. (10.36) τ cf −1∕4 = w2 = 0.01254 Reδ 2 2 ρU∞

(10.41)

where Reδ2 = δ2 x∕ν is the Reynolds number based on the momentum thickness δ2 . The local friction coefficient cf is expressed in terms of the local Reynolds number Rex using expression for momentum thickness δ2 cf 0.02879 = , 2 Re0.2 x

5 × 105 < Rex < 5 × 107 .

(10.42)

The average friction coefficient obtained by the power law is given as cf =

τw ρU2∞ ∕2

=

0.072625 , Re0.2 L

5 × 105 < Rex < 5 × 107 .

(10.43a)

Compared with available experimental data, it was recommended that the coefficient of Eq. (10.43b) was changed to 0.074, as discussed in [18]. Hence, we have a formula for the skin friction coefficient over a smooth flat plate: cf =

0.074 , Re0.2 L

5 × 105 < Rex < 5 × 107 .

(10.43b)

10.6 Approximate Solution by the Integral Method for the Turbulent Momentum Boundary Layer over a Flat Plate

0.01 0.008 0.006

0.004

Schultz-Grunow′s experimental

cf

0.002

Prandtl’s 1/7 th power law solution 0.001 106

Figure 10.2

107 Rex

108

109

Local friction coefficient for turbulent flow over a flat plate.

Shultz-Grunow [19] measured velocity profiles and local friction factors at Reynolds numbers up to 109 . Then, they used the momentum integral equation to find the friction factor. The experimental correlation of Shultz-Grunow [19] for the local friction coefficient is cf = 0.185(log10 Rex )−2.584 , 5 × 105 ≤ Rex ≤ 109 (10.44) 2 where this result is valid from Rex = 5 × 105 up to remarkably high Reynolds numbers. This equation is for turbulent flow from the leading edge of the plate and is in good agreement with experimental data, as seen in Figure 10.2. Experimental results deviate above Rex ≈ 107 . Experimental correlation for the average friction coefficient cf proposed by Shultz-Grunow is 0.427 cf = , 106 < ReL < 109 (10.45) [log10 ReL − 0.407]2.64 Schlichting and Gersten [2] examined the available experimental data and recommended the following correlation for the local friction coefficient cf for turbulent flow over a flat plate , cf = 0.0592 Re−0.2 x

5 × 105 ≤ Rex ≤ 108

(10.46)

Schlichting and Gersten [2] developed an expression for the turbulent friction coefficient using the integral equation and the logarithmic velocity distribution law and suggested the following semiempirical equation for average friction coefficient cf : cf =

0.455 , [log10 (ReL )]2.58

106 < ReL < 109 .

(10.47)

White and Majdalani [6] recommend the following equation for the local skin friction coefficient cf : cf =

0.455 [ln(0.06Rex )]2

(10.48a)

and for average friction coefficient cf cf =

0.523 . [ln(0.06ReL )]2

Equation (10.48b) is shown to be in good agreement with experimental data.

(10.48b)

517

518

10 Turbulent External Boundary Layers: Momentum and Heat Transfer

Consider the log law (u y) 2.3 u τ log10 +C = uτ κ ν

(10.49)

where uτ is given as √ √ uτ = τw ∕ρ = U∞ cf ∕2 in which cf is the local friction coefficient. Using uτ and assuming that u = U∞ at y = δ, Eq. (10.49) becomes ( ) √ U∞ δ √ 2.3 log10 cf ∕2 = cf ∕2 + C. κ ν

(10.50a)

Von Karman [20] converted Eq. (10.50a) and evaluated the constants from measurements of local shear stress and proposed the following equation for local skin friction coefficient: √ 1∕ cf = 4.15log10 (Rex cf ) + 1.7. (10.50b) This equation is useful over flat plate where the laminar boundary layer is negligible. Example 10.2 Consider incompressible constant property turbulent flow over a flat plate with no suction or blowing. Develop a relationship between the skin friction coefficient cf and the Reynolds number Rex = U∞ x/ν using the (Coles law) wall wake law for external turbulent flow over a flat plate. Solution First notice that the contribution to momentum thickness δ2 from the buffer layer and the viscous sublayer is negligible. This is true since constant shear in the inner region implies negligible change in fluid momentum. For this reason, we take ( y) ) ( 1 π 2Π sin2 . u+ = ln y+ + B + κ κ 2δ This equation may also be expressed using the wall coordinates ( ) √ ( ) ( y) yU∞ cf ∕2 u∕U∞ 2Π 1 π +B+ sin2 = ln √ κ ν κ 2δ c ∕2 f

at y = δ u = U∞ and therefore, ) ( √ 1 1 2Π = ln(Reδ cf ∕2) + B + √ κ c ∕2 κ f

where Reδ = δU∞ /ν is the Reynolds number based on boundary layer thickness δ and Π = 0.45, C = 5, and κ = 0.41. With this information, the equation becomes √ √ 2 = 2.44 ln(Reδ cf ∕2) + 7.1951. cf This gives us cf implicitly, and we will now solve this implicit equation by Maple 2020 for different values of Reynolds number Reδ , and friction coefficient cf as a function of Reδ is presented in a table. Reδ

103

104

105

106

107

108

cf

0.0068238

0.0040897

0.0026984

0.0019032

0.0014094

0.0010833

Next, these data will be fitted to a power law approximation over a range of values from Reδ = 103 to 108 using Maple 2020. A suitable power law fit is given below cf =

0.02353 Re0.1829 δ

This approximate equation is useful if Reδ is known. This Reδ can be obtained by the integral method. By combining δ2 0.03598 = x Re0.2 x

10.7 Laminar and Turbulent Boundary Layer

and 7 δ 72 the following relation for the turbulent boundary layer thickness was obtained: δ2 =

0.37 δ = x Re0.2 x Next, this relation is combined with 0.02353 Re0.1829 δ

cf =

to obtain a relation for the skin friction coefficient cf in terms of the local Reynolds number 0.02822 Re0.14585 x

cf =

10.7 Laminar and Turbulent Boundary Layer We now consider a boundary layer flow along a flat plate such that the flow is laminar over the region 0 ≤ x ≤ xc and turbulent over the region xc ≤ x ≤ L. We are interested in the average friction coefficient for the entire plate. The average friction coefficient on a plate must include the contribution from the laminar boundary layer as well as the turbulent boundary layer. For this reason, the actual friction coefficient of the plate will be less than that given by equations presented before. The effect of the laminar boundary layer portion of the plate may be taken into account approximately in the following manner. We assume that there is a transition point instead of the transition region. Before the transition point, the flow is laminar, and after the transition point, the flow is turbulent. Next, we assume that transition point occurs at a definite value of Rex = Rexc . We now assume that transition occurs at xc . The local friction coefficient is defined as τw (10.51a) cf = ρ U2∞ ∕2 The average friction coefficient is defined as 2 τw

cf =

(10.51b)

ρ U2∞

where τw is the average shear stress on the plate having length L, and it is defined as L

τw =

1 τ dx L ∫0 w

(10.52)

Substituting the definition of the local friction coefficient cf , Eq. (10.51a), into Eq. (10.52), we obtain τw =

ρ U2∞ L c (x) dx 2 L ∫0 f

(10.53)

Next, we substitute this equation, Eq. (10.53), into Eq. (10.51b) for the average friction coefficient cf , and we get cf =

2 L L 2 ρ U∞ 1 cf (x) dx = c (x) dx 2 2L ∫ L ∫0 f ρ U∞ 0

(10.54)

Thus, we see that the average friction factor cf is the average of the local friction factor cf (x) over the plate surface. The local friction coefficient is correlated in terms of Reynolds number Rex , and it is better to transform the integration from x to Rex . For this purpose, we do the following: x=

μ Rex ρ U∞

Then, we write μ dx = dRex ρ U∞

(10.55)

(10.56)

519

520

10 Turbulent External Boundary Layers: Momentum and Heat Transfer

We can now write the average friction coefficient as cf = =

1 L ∫0

ReL

cf

1 ReL ∫0

ReL μ μ dRex = cf dRex ρ U∞ ρ U∞ L ∫0

ReL

cf dRex

(10.57)

where ReL = ρ U∞ L/μ is the Reynolds number evaluate using the entire length of the plate. Equation (10.57) is independent of specific correlations. The average friction coefficient must include both the laminar and turbulent regions of the plate. We can now write for the average friction coefficient cf [ Rexc ] ReL 1 (10.58a) cf = [cf ]Laminar dRex + [c ] dRex ∫Rexc f Turbulent ReL ∫0 Recall that the local friction coefficient cf is −1∕2

cf = 0.664 Rex

Substituting the local friction coefficient cf and Eq. (10.42) into Eq. (10.58a), we have [ ] Rexc ReL 1 0.664 −0.2 cf = dRex + 0.05758 Rex dRex √ ∫Rexc ReL ∫0 Re

(10.58b)

x

or after integration, we obtain an expression for average friction coefficient cf [ ( )] 1 1∕2 0.8 cf = − Re 1.328 Rexc + 0.074 Re0.8 xc L ReL which may be rearranged as [ ( ) ) ] ( Rexc 4∕5 1.328 Rexc −1∕5 + 0.074ReL 1− cf = √ ReL Rexc ReL

(10.58c)

(10.58d)

where Eq. (10.58d) is valid for ReL < 107 and ReL = U∞ L/ν and Rexc is the critical Reynolds number for transition to turbulent flow. If we choose Rexc = 5 × 105 , after simplification, we obtain 0.074 1743 cf = − 1∕5 ReL Re L

ReL < 108

(10.58e)

Clearly, the mean friction coefficient cf over the region where the flow is partly laminar and partly turbulent depends on the value of the critical Reynolds number Rexc , and Eq. (10.58e) can be written compactly in the following form: cf =

0.074 A − Re Re0.2 L L

(10.58f)

where this equation is valid for the laminar and turbulent flow over a flat plate. The value of A is rounded and depends on the choice of the value of Rexc and is given in Table 10.2. There are several empirical formulas for cf that account for both laminar and turbulent effects. Nellis and Klein [10] reports an empirical correlation for the local friction coefficient cf cf =

0.027 1∕7

Rex

.

(10.59)

If we use Eq. (10.59) in Eq. (10.58a) along with a local friction coefficient cf , then [ ] Rexc ReL 1 0.664 0.027 cf = dRex + dRex √ ∫Rexc Re1∕7 ReL ∫0 Re x

Table 10.2

x

Values of A for different values of Rexc .

Rexc

2 × 105

3 × 105

5 × 105

1 × 106

A

700

1050

1740

3340

(10.60a)

10.8 Other Eddy Diffusivity Momentum Models

Table 10.3

Values of A for different values of Rexc .

Rexc

3 × 105

5 × 105

106

3 × 106

A

1050

1700

3300

8700

or cf =

( [ )] 1 1∕2 6∕7 6∕7 1.328 Rexc + 0.0315 ReL − Rexc ReL

Rex,c ≤ ReL ≤ 1 × 1010 .

(10.60b)

Mills [8] integrated Eq. (10.48a) numerically, and the result is curve fitted. Then, he used the result in Eq. (10.58a), and the following equation for the average friction coefficient is given: ( ) ) ( Rexc Rexc 0.523 0.523 −1∕2 + cf = 1.328 Rexc − ReL ReL [ln(0.06 Rexc )]2 [ln(0.06 ReL )]2 ReL < 109

(10.61)

and Eq. (10.61) is accurate for ReL < 109 . An expression for the average friction coefficient cf that fits the experimental data over a wider range is reported by Pai [18]. cf =

0.455 A − [log10 (ReL )]2.58 ReL

ReL < 109

(10.62)

where constant A depends on the critical Reynolds number for transition Rexc . Computed values of A for different Rexc values are given in Table 10.3. This expression accounts for both the laminar and turbulent effects. Total viscous drag force on a flat plate of length L and width W is D = (τw )(W × L) = cf (1∕2)ρU2∞ (W × L) where τw is the average wall shear stress.

10.8 Other Eddy Diffusivity Momentum Models Velocity distribution we discussed is a segmental representation, and it produces discontinuities in (du∕dy) at the intersections, and this results in awkward expressions for heat transfer. Some researchers proposed models with continuous derivatives to represent the velocity distribution, which is valid over the whole range of dimensionless distance y+ . Researchers developed different models εM /ν. We will study heat transfer in Section 10.8. In heat transfer calculations, we need eddy diffusivity models. Eddy diffusivity models are used in determining the temperature distribution. Any of the velocity distributions may be used to provide a model for the eddy diffusivity εM . Eddy diffusivity models obtained from velocity distributions are expected to be more accurate than Prandtl’s mixing-length model. There are several models for the eddy diffusivity of momentum, and some of these models are given in Table 10.4. In the Couette flow region, we can solve Eq. (10.6a) for εM /ν: τ εM = w −1 ν μ du dy or in terms of wall coordinates εM 1 = − 1. ν du+ ∕dy+

(10.63a)

(10.63b)

Equations (10.63a) or (10.63b) relates information about the average velocity data to the eddy diffusivity of momentum.

521

522

10 Turbulent External Boundary Layers: Momentum and Heat Transfer

Table 10.4

Selected eddy diffusivity models.

Model

Expression

Prandtl–Taylor

εM = ν

Reference

{ 0

for 0 < y+ < 11.5

κ y+ − 1

for y+ > 11.5

⎧ for 0 < y+ < 5 ⎪0 εM ⎪ y+ =⎨ −1 for 5 < y+ > 30 ν ⎪5 for y+ > 30 ⎪κy+ − 1 ⎩ εM = ν [ ] (κu+ )2 (κu+ )3 0.0526 exp(κu+ ) − 1 − κu+ − − 2 6

Von Karman

Spalding

where κ = 0.40 [ ( { )]}2 | + | εM y+ | du | = κy+ 1 − exp − | +| | dy | ν 24.7 | |

Van Driest

where κ = 0.40 εM = n2 u+ y+ [1 − exp(−n2 u+ y+ )] ν (du∕dy)3 εM = κ 2 2 y+ > 26 (d u∕dy2 )2 n = 0.124

Deisler

y+ < 26

Rubesin et al. [14]

Von Karman [11]

Spalding [15]

Van Driest [5]

Deisler [16]

Example 10.3 Obtain eddy diffusivity model for the velocity profiles given below { y+ 0 < y+ < 11.5 + (a) u = y+ > 11.5 2.44 ln(y+ ) + 5.5 ⎧y+ ⎪ (b) u+ = ⎨5 ln(y+ ) − 3.05 ⎪ + ⎩2.5 ln(y ) + 5.5

for 0 < y+ < 5 for 5 < y+ > 30 for y+ > 30

Solution 1 1 −1= − 1 = 0.4y+ − 1 2.44∕y+ du+ ∕dy+ { 0 y+ > 11.5

(a) εM ∕ν = εM = ν

0.40y+ − 1

(b) εM /ν = 0, εM ∕ν = ⎧0 ε M ⎪ y+ =⎨ −1 ν ⎪ 5+ ⎩κy − 1

y+ > 11.5

5 1 − 1 = + − 1, εM /ν = 0.4y+ − 1 y+ ∕5 y for 0 < y+ < 5 for 5 < y+ > 30 . for y+ > 30

10.9 Turbulent Heat Transfer Consider two-dimensional, constant property, incompressible, and steady turbulent flow over a flat plate without suction or blowing. Free-stream velocity U∞ and free-stream temperature T∞ are constant. We assume that a uniform heat flux q′′w is applied to the flat plate. We also assume that turbulence starts from the leading edge of the plate. We will use

10.9 Turbulent Heat Transfer

the time-averaged energy equation to study heat transfer in the turbulent boundary layer. We would like to determine the mean temperature distribution T and the Nusselt number for fully turbulent flow along a flat plate. The classical differential formulation of mean energy for the turbulent boundary layer is ( ) [ ] 𝜕 𝜕T 𝜕T 𝜕T ρ cp u +v = (k + kt ) . (10.64) 𝜕x 𝜕y 𝜕y 𝜕y The boundary conditions are x=0

T=0

y=0

−k

y→∞

𝜕T = q′′w 𝜕y T → T∞ .

(10.65a) (10.65b) (10.65c)

The first term on the right-hand side describes heat flux by conduction, and the second term on the right-hand side describes the heat flux created by the turbulent fluctuations. Formulation is closed by the specification of turbulent thermal conductivity kt . We may write the energy equation in the following form: ) [( ] ε 𝜕T 𝜕T 𝜕T 𝜕 u +v = α+ M . 𝜕x 𝜕y 𝜕y Prt 𝜕y The boundary conditions are the same as before. At this point, recall that the turbulent Prandtl number Prt is μt cp ρεM cp ε = = M Prt = kt ρεH cp εH

(10.66)

where εM is the eddy diffusivity for momentum and εH is the eddy diffusivity for heat. We should keep in mind that thermal eddy diffusivity εH is an empirical function. It is important to specify the variation of turbulent Prandtl number across the boundary layer. However, it should be noted that the eddy diffusivity εM or eddy diffusivity of heat εH is not a fluid property, and it is a function of wall distance, Reynolds number, Prandtl number, and flow conditions. For most fluids, Kays et al. [7] assumed that Prt ≈ 0.85. On the other hand, Mills [8] recommends that the turbulent Prandtl number is Prt ≈ 0.9 for fluids Pr > 0.7 except liquid metals. With εM /Prt known, we can compute the turbulent temperature profile from the energy equation. The success of this method depends on the eddy diffusivity models as well as the existence of a similarity between energy transfer and momentum transfer. Using the similarity concept, we may obtain the heat transfer from the available information on momentum transport. A special case occurs for flow over a flat plate, as discussed in [22, 23]. In the inner region of the turbulent boundary layer, (y/δ) ≤ 0.2, the mean fluid velocities are very small, and convective terms of Eq. (10.65a) can be neglected. In other words, because of the proximity to the wall, the effects of convection on heat transfer can be neglected. Thus, we conclude that sufficiently close to the wall, in the inner and overlap regions, total or apparent heat flux q′′ is constant. Keeping in mind the Couette flow assumption, we neglected the left-hand side of the energy equation, Eq. (10.65b), and we get [( ] ) εM dT d α+ ≈0 (10.67) dy Prt dy and Eq. (10.67) implies that apparent heat flux is approximately constant and independent of position y. We may safely state that sufficiently close to wall, the total heat flux does not depend on y. Integrating once leads to ( ) ε dT α+ M = C1 . (10.68) Prt dy To determine C1 , evaluate this equation, Eq. (10.68), at y = 0: ( )( ) ) ( ) ( −q′′w q′′ εM k dT =− w = C1 = α + Prt y=0 dy y=0 ρcp k ρcp

(10.69)

where q′′w is the wall heat flux. We can now say that apparent or total heat flux q′′ is equal to wall heat flux q′′w , i.e. q′′ ≈ q′′w . Thus, for the inner region, the energy equation becomes ) ( q′′ ε dT =− w. (10.70) α+ M Prt dy ρcp At this point, we will introduce the nondimensional temperature T+ √ √ τ c (Tw − T)ρcp ρw (Tw − T)ρcp U∞ 2f u (T − T) ρc p τ w + T = = = . q′′w q′′w q′′w

(10.71)

523

524

10 Turbulent External Boundary Layers: Momentum and Heat Transfer

We will use this dimensionless temperature T+ and the dimensionless distance y+ to get the nondimensionalized form of the energy equation. The substitution of these dimensionless variables into the energy equation, Eq. (10.70), will give us the following equations: 1 dT+ = ( ) dy+ 1 εM 1 + Pr Prt ν at y+ = 0

(10.72)

T+ = 0

(10.73)

or +

T+ =

y (Tw − T) dy+ = ). ( q′′w ∕ρcp uτ ∫0 1 εM 1 + Pr Prt ν

(10.74)

Assuming that Pr and Prt are constant, Eq. (10.72) or Eq. (10.74) may be integrated using appropriate correlations for the eddy diffusivity of momentum. This is the temperature law of the wall, and it is the thermal analog of the velocity law of the wall. We need an expression for εM /ν and Prt to evaluate numerically Eq. (10.72) or Eq. (10.74). For example, we may choose Prt = 0.9 and take εM /ν from the Van Driest formula and carry out integration for various values of Pr. For fluids having moderate values of the Prandtl number, as a first approximation, we may set Prt ≈ 1. The results are shown in Figure 10.3 for Prt ≈ 0.9 and for various Prandtl numbers. A family of curves are shown in Figure 10.3 for different Prandtl numbers. Just like the universal velocity profile, a universal temperature profile in turbulent flow provides much physical insight. Using Eq. (10.74), we will present temperature law of the wall. Equation (10.74) is a general expression in wall coordinates. An examination of Figure 10.9 indicates that the profiles in Figure 10.9 have a linear sublayer very near the wall. Then, the figures bend over through a buffer layer. Finally, figures go to a logarithmic line. We now will look each region separately. Viscous sublayer: 0 ≤ y+ ≤ 5 Very close to the wall, in the viscous sublayer, (εM /ν Prt ) is negligible, and molecular heat transfer dominates. We can write 1 1 εM 1 + ≈ Pr Prt ν Pr 100 90 80 Pr = 15

70 60

Pr = 10

T + 50 40

Pr = 5

30 20

Pr = 1

10 Pr = 0.72

1 Figure 10.3

5

10

y+

50

100

500

The temperature law of the wall, computed from Eq. (10.76a) with the Van Driest model.

10.9 Turbulent Heat Transfer

and the thermal sublayer is obtained from Eq. (10.74), and it has the following form: T+ = Pr y+ + C. Since at y+ = 0, T+ = 0, we find that C = 0 T+ = Pr y+ ,

y+ < 5.

(10.75a)

Buffer region: 5 ≤ y+ ≤ 30 In the buffer zone, the velocity profile is u+ = 5 ln y+ − 3.05. Knowing the velocity profile, we obtain an expression for εM /ν εM 1 = ( +) −1 ν du dy+ y+ 1 − 1. = ( ) −1= 5 5 y+ Then, the temperature profile is obtained from Eq. (10.74) y+

= 5Pr +

∫5

dy+ ) ( 1 1 1 + − Prt 5 Pr Prt y+

Integration yields )]} ( { [ Pr y+ −1 . T+ = 5 Pr + Prt ln 1 + Prt 5

(10.75b)

Kader [24] proposed the following expression for the buffer region: T+ ≈

Prt ln (y+ ) + A(Pr) κ

Prt ≈ 0.85

κ = 0.4

(10.75c)

where A(Pr) is a strong function Prandtl number, and it is given as A(Pr) ≈ (3.85Pr1∕3 − 1.3)2 + 2.12 ln(Pr) 6 × 10−3 ≤ Pr ≤ 40 × 103 .

(10.75d)

This expression, Eq. (10.75d), is valid even for liquid metals with low Prandtl numbers ranging from 0.001 to 0.03. Kader reports that this expression is in good agreement with temperature profiles measured in air, water, ethylene glycol, and oil over 0.7 ≤ Pr ≤ 170. Fully turbulent region: y+ > 30 In the fully turbulent region, molecular fluid properties do not influence the dimensionless temperature distribution. In this region, εH ≫ α. Then, we can write ε 1 εM 1 + ≈ M . Pr Prt ν νPrt Notice that the turbulent Prandtl number Prt is empirical quantity. This approximation is not valid for liquid metals. For fully turbulent flow, we have u+ =

1 ln y+ + B κ

κ = 0.4,

B = 5.5.

525

526

10 Turbulent External Boundary Layers: Momentum and Heat Transfer

From this equation, we can write 1 du+ = +. κy dy+ Recall that we can now obtain εM /ν εM 1 = ( + ) − 1 = κy+ − 1 ≈ κy+ . ν du dy+ From Eq. (10.74) { [ ( )]} y+ Prt dy+ Pr + + . T = 5 Pr + Prt ln 1 + 5 ∫30 κy+ Prt Integration yields

) ( Pr Pr 5 Pr + t ln(y+ ) − t ln(30) T+ = 5Pr + 5 Prt ln 1 + Prt κ κ

and this gives us ( +) { [ ( )]} Pr y Pr . + t ln T+ = 5 Pr + Prt ln 1 + 5 Prt 5κ 30

(10.75e)

We may use Prt ≈ 0.9 and κ = 0.4 in Eq. (10.75e). Stanton number St is defined as St =

q′′w 1 1 = ( = + + ) ρcp U∞ (Tw − T∞ ) U∞ (Tw − T∞ ) T∞ U∞ uτ q′′w ∕ρcp uτ

U+∞ =

U∞ U U = √ ∞ = √ ∞ . uτ τw ∕ρ cf ∕2

From these last two equations, we get √ cf ∕2 1 1 + St = + + ⇒ T∞ = . = St T∞ U∞ StU+∞

(10.75f)

(10.75g)

(10.75h)

White and Majdalani [6] based on the numerical integration of Eq. (10.74) with a proper expression for μt /μ proposed the following universal temperature profile: Prt ln (y+ ) + A(Pr). (10.76) κ Equation (10.76) was developed for the region outside the viscous sublayer. The parameter A(Pr) is a strong function Prandtl number. Experimental data or numerical simulations may be used to determine A(Pr). White and Majdalani [6] give a curve fit for A(Pr) T+ ≈

A(Pr) ≈ 13Pr2∕3 − 7,

Pr ≥ 0.7.

(10.77)

On the other hand, for Pr ≥ 0.5, Lienhard IV and Lienhard V [23] give A(Pr) ≈ 12.7Pr2∕3 − 7.2.

(10.78)

The theory behind the universal temperature profile is in good agreement with available experimental data, as reported in [6, 22]. For this reason, we may use this law-of-the-wall correlation with confidence to predict turbulent heat transfer coefficients. Other researchers such as Sucec [25] proposed models for T+ (y+ ). Sucec [25] uses the power law expression for T+ (y+ ) T+ (y+ ) ≈ 6(y+ )0.17 + 13.2Pr − 9.37.

(10.79)

We are now in position to develop an expression for the heat transfer coefficient. A general formula can be obtained from the temperature law of the wall, Eq. (10.75e), as discussed in [6, 23], as well as in [26]. To obtain this equation, we choose

10.9 Turbulent Heat Transfer

Prt = 1 for convenience since at the outer edge of the log layer, the temperature and velocity are close to the free-stream values. We may estimate velocity and temperature log laws at the edge of the boundary layer. In other words, we set At y = δ,

u ≈ U∞

(10.80a)

At y = Δ,

T ≈ T∞ .

(10.80b)

Thus, we obtain the following relations: ( ) uτ δ U∞ 1 +B = ln uτ κ ν ( ) uτ Δ (Tw − T∞ ) 1 + A. ( ′′ ) = ln κ ν qw ∕ρcp uτ We subtract Eq. (10.81) from Eq. (10.82) ( ) ( ) uτ Δ uτ δ (Tw − T∞ ) U∞ 1 1 − ln + A − B. = ln ( ′′ )− uτ κ ν κ ν qw ∕ρcp uτ

(10.81) (10.82)

(10.83)

We now rearrange Eq. (10.83), and this process of rearrangement will give us q′′w 1 = . U∞ 1 ( Δ ) (Tw − T∞ )ρcp uτ + ln +A−B uτ κ δ

(10.84)

Next, we replace the friction velocity uτ in favor of the skin friction coefficient cf by using the following definition: √ √ uτ τw cf 1 = . (10.85) = U∞ U∞ ρ 2 Hence, substitution of Eq. (10.85) into Eq. (10.84) yields q′′w

√

(Tw − T∞ )ρcp U∞

cf 2

= √

1 ( ) ]. 2 Δ 1 ln +A−B + cf κ δ [

(10.86)

The rearrangement of Eq. (10.86) gives q′′w = (Tw − T∞ )ρcp U∞

cf ∕2 ( ) ]√ . Δ 1 cf ln +A−B 1+ 2 κ δ [

(10.87)

The term ln(Δ/δ)/κ is very small compared to (A − B) for all the Prandtl numbers and is neglected. Introducing A(Pr) from Eq. (10.78) and substituting B = 5.5, we obtain an expression for local Stanton number for turbulent flow past a flat plate St =

cf ∕2 h = √ ρcp U∞ 1 + 12.7[Pr2∕3 − 1] cf ∕2

(10.88)

Pr ≥ 0.5 Tw = const

or q′′w = const.

The temperature gradient in turbulent flow is confined to the viscous sublayer, and the turbulent flow Nusselt number is insensitive to thermal boundary conditions. Thus, we may use Eq. (10.88) for uniform heat flux or uniform wall temperature boundary conditions. We can use Eq. (10.88) with Eq. (10.48a) or with any other equation for cf to calculate the local heat transfer coefficient in a turbulent boundary layer. Equation (10.88) is valid for smooth walls with zero- or mild-pressure gradients. Equation (10.88) is not valid for liquid metals since the temperature law of the wall is not valid for liquid metals. Equation (10.88) is accurate for any turbulent Reynolds number and for any Prandtl number about Pr ≥ 0.5. The integration of Eq. (10.88) over the length of the plate shows that the average heat transfer is about 15% higher than the trailing edge local Nusselt number. Based on this, we may estimate the average Stanton number St St = 1.15 Stx=L .

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

104 Eqn. (10.88) with eqn. (10.48a) Seban & Doughty Reynolds et al., Run 1 Reynolds et al., Runs 2–4 Reynolds et al., Runs 5–7 Reynolds et al., Run 8 Junkhan & Serovy, high u′r /u∞ Kestin et al., low u′r /u∞ Kestin et al., high u′r /u∞ Blair, u′r /u∞ ≃ 0.25% Blair, u′r /u∞ ≃ 1.0% Blair, u′r /u∞ ≃ 2.0%

Nux

103

Pr = 0.71 (air) Tw or qw constant

5% + 1 5% –1 102

105

106

107

Rex Figure 10.4 Comparison of Eq. (10.88) with experimental data. Both constant Tw and constant q′′w are shown. u′r is the root-mean square turbulent fluctuation in the free stream. Source: Lienhard [27]/ASME/CC BY 4.0.

Lienhard IV and Lienhard V [23] and Lienhard V [27] plotted experimental data for air from five independent studies. See Figure 10.4. Lienhard IV and Lienhard V report that Eq. (10.88) predicts those data with a standard deviation of ±5.5%. Equation (10.88) also displays excellent agreement with data for water and high-Prandtl-number oils. Example 10.4 Air at a film temperature of 290 K and 1 atm flows over a smooth flat plate at 20 m/s. The plate length in the flow direction is 2 m and the width of the plate is 1 m. The surface temperature of the plate is 300 K. The measured shear stress is 0.65 Pa, and the air temperature is 295 K at y = 1 cm. Estimate, at this position, wall heat transfer for y = 1 cm. Solution Air properties at Tf = 290 K are ρ = 1.208 kg∕m3

ν = 15 × 10−6 m2 ∕s

Pr = 0.709

cp = 1.007 kJ∕kg.K

U∞ L (20 m∕s)(1.5 m) = ≈ 2.66 × 106 > 500 000. Thus, flow is turbulent ν 15 × 10−6 m2 ∕s The friction velocity uτ is √ √ τw 0.65 N∕m2 uτ = = = 0.73 m∕s. ρ 1.22kg∕m3

ReL =

The dimensionless distance y+ is y+ =

y uτ 0.01m × 0.73 m∕s = ≈ 486. ν 15 × 10−6 m2 ∕s

We see that y+ = 486 is in a fully turbulent region. We now use Eq. (10.75e). Let us choose Prt = 0.9 ( +) { [ ( )]} Pr y Pr + t ln T+ = 5 Pr + Prt ln 1 + 5 Prt 5κ 30

10.10 Analogy Between Momentum and Heat Transfer

( { [ ( )]} ) 486 0.709 0.9 ln T+ = 5 0.709 + 0.9 × ln 1 + 5 + ≈ 12 0.9 5 × 0.4 30 q′′w =

(Tw − T)ρcp uτ +

=

T Suppose we use Eq. (10.76)

(300 − 295)(1.208 )(1007 )(0.73) ≈ 372.0 W∕m2 . 12

Prt ln (y+ ) + A(Pr) κ White [22] gives an accurate curve fit to the computed values of A(Pr) 0.7 ≤ Pr ≤ 105 T+ ≈

A(Pr) ≈ 12.7 Pr2∕3 = 7.7 A(Pr) ≈ 12.7 Pr2∕3 − 7.7 = 12.7(0.709)2∕3 − 7.7 ≈ 2.4 ) 0.9 ln (489) + 2.4 ≈ 16.31 0.40 The estimated heat flux q′′w is T+ =

q′′w =

(

(Tw − T)ρcp uτ +

T

=

(300 − 295)(1.208 kg∕m3 )(1007 J∕kg K)(0.73) ≈ 272.11 W∕m2 . 16.31

10.10 Analogy Between Momentum and Heat Transfer There are no exact solutions to momentum and energy equations; for this reason, we will rely on semiempirical solution techniques based on momentum heat transfer analogy. Extensive literature is available on the friction factor, which may be used to solve heat transfer problems. The problem here is that the true nature of the problem is lost due to the oversimplification. Churchill [28] and Webb and Crosse [29] present a critical evaluation of analogy-based correlations for heat transfer with constant properties.

10.10.1 Reynold’s Analogy Reynolds [30] analogy provides a relation between momentum and heat transfer for turbulent boundary layer flow. There is considerable body of experimental knowledge that the mechanism of heat transfer in turbulent flow near wall is similar to the mechanism of momentum transfer near wall. For this reason, several researchers, such as Schlichting and Gersten [2], tried to relate the wall shear stress to wall heat transfer rate. Consider two-dimensional, steady fully turbulent boundary layer flow over an isothermal plane wall with constant free-stream velocity. For constant free-stream velocity, the pressure gradient dp∕dx is zero. We assume an incompressible and constant property fluid. Free-stream temperature and velocity are T∞ and U∞ , respectively. Under these conditions, temperature and velocity profiles will be similar if the boundary conditions are similar. See Figure 10.5. We are interested in developing a relation between the Nusselt number and the Reynolds number. The starting point of the Reynolds analogy is the definition of total fluxes τ = ρ(ν + εM )

du dy

q′′ = −ρcp (α + εH )

(10.89) dT . dy

(10.90)

We will take the ratio of momentum and heat flux, at any location: τ ρ(ν + εM ) du∕dy τ = w =− q′′ ρcp (α + εH ) dT∕dy q′′w τw ρ(ν + εM ) du =− . ρ cp (α + εH ) dT q′′w

(10.91)

Experimental studies indicate that if the plate is isothermal, the heat flux, like the shear stress, is constant in the inner region. Therefore, it is reasonable to assume that the ratio of fluxes is constant across the layer. Reynolds assumed that the

529

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

energy and momentum take place in the turbulent core and neglect the transport process in the sublayers. We now take the integral across the layer from the wall (u = 0, T = Tw ) all the way to free-stream (u = U∞ , T = T∞ ) ( ) T∞ τw ∕ρ (ν + εM ) U∞ dT = − du (10.92) ′′ (α + εH ) ∫0 qw ∕ρ cp ∫Tw or

(

τw ∕ρ ′′ qw ∕ρ cp

) =

U∞ (ν + εM ) . (α + εH ) (Tw − T∞ )

(10.93)

We will rearrange Eq. (10.93). using the following definition of the heat transfer coefficient h and the skin friction coefficient cf : h= cf =

q′′w Tw − T∞ 2τw ρU2∞

(10.94)

.

(10.95)

After algebraic manipulation the above equation now becomes ] [ c α + εH St = f 2 ν + εm

(10.96)

where St is the Stanton number, and it is given as St =

h . ρ cp U∞

(10.97)

In turbulent flow, we will assume that εM ≫ ν and εH ≫ α. Then, Eq. (10.96) becomes c 1 St = f 2 Prt

(10.98)

where the turbulent Prandtl number Prt is assumed to be Prt ≈ 1

(10.99)

Under these conditions, the Stanton number St becomes c (10.100) St = f . 2 Equation (10.100) is the Reynolds analogy between wall friction and heat transfer. Reynolds argued that there is a relationship between heat flux and wall shear stress. Reynold’s analogy is valid for both Pr = 1 and Prt = 1. Schlichting and y

y U∞

T∞ ν ≪ εM T(y)

u(y)

Turbulent core

τw Figure 10.5

q′′ w

Coordinate system for turbulent flow over an external surface.

α ≪ εH εM = εH

10.10 Analogy Between Momentum and Heat Transfer

Gersten [2] suggest that for moderate values of the Reynolds number for turbulent flow over a flat plate, the skin friction coefficient is 0.0592 cf = Re0.2 x 5 × 105 < Rex < 108 where Rex = x U∞ /ν is the local Reynolds number. Recall that we may express Stanton number St as St =

Nux . Rex Pr

(10.101)

It was assumed that Pr ≈ Prt ≈ 1. Then, we can write the final form of the Reynolds analogy in the following form: h x cf = Rex . Nux = (10.102) k 2 If we substitute the skin friction coefficient cf into the above equation, the result is Nux = 0.0296Re0.8 x 5 × 105 < Rex < 108 .

(10.103)

This relation is a good approximation for fluids Pr ≈ 1, and the equation fails for fluids with Prandtl numbers significantly different from unity. This equation provides information about heat transfer from friction measurements only. Experimental information on fluid mechanics of a problem can be related through the Reynolds analogy to heat transfer of the same problem.

10.10.2 Chilton–Colburn Analogy Colburn [31] has suggested the following relation based on experimental data: St Pr2∕3 = cf ∕2 0.6 ≤ Pr ≤ 60.

(10.104a)

If we substitute the expression for the skin friction coefficient developed by Schlichting, we get St Pr2∕3 = 0.0296Re−0.2 x 0.6 ≤ Pr ≤ 60 5 × 105 ≤ Rex ≤ 108 .

(10.104b)

This is actually a modified version of Reynolds analogy. Colburn applied this expression to a wide range of data for flow and geometries of different types and found it to be quite accurate for conditions where drag forces are viscous and no form drag exist. The Prandtl number range is extended to include gases, water, and several other liquids. This analogy is an empirical modification of Reynolds analogy. Unfortunately, power laws like Eq. (10.104b) cannot cover a wide range of Reynolds numbers. Lienhard V [27] and Gnielinski [32] discuss this point with reference to available experimental data for pipe flow and boundary layer flow. Equations having the form of Eq. (10.88) cover full range of Prandtl numbers. Colburn considered a wide spectrum of data limited only to gases. We can express the Stanton number in terms of the Nusselt number since St = Nu/RePr. In this case, we get 4∕5

Nux = (h x∕k) = 0.0296Pr1∕3 Rex 0 ≤ Pr ≤ 60 5 × 105 ≤ Rex ≤ 106 .

(10.104c)

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

In Eq. (10.104c), it assumed that turbulence starts from the leading edge of the plate. The average heat transfer coefficient h for a turbulent flow starting from the leading edge of the plate is L

1 h dx. L ∫0 Using Eqs. (10.104c) and (10.104d), we obtain the average Nusselt number for the entire plate h=

(10.104d)

4∕5

NuL = 0.037Pr1∕3 ReL 0 ≤ Pr ≤ 60 5 × 105 ≤ ReL ≤ 106 .

(10.105)

Notice that Eq. (10.104c) is valid for turbulent flow starting from the leading edge of the plate. Mixed boundary layer: We assume that there is no transition region in turbulent flow over the flat plate. A typical turbulent flow over a flat plate will be laminar up to assuming a transition Reynolds number of Rexc and turbulent afterward. In this case, we are interested in the average turbulent heat transfer coefficient h or the average Nusselt number NuL . The average rate of heat transfer for the entire surface is given by { xc } L 1 qw = q′′wL dx + q′′wT dx ∫xc L ∫0 where q′′wL is the local rate of heat transfer at any point in the laminar portion of the flow and q′′wT is the local rate of heat transfer at any point in turbulent portion of the boundary layer. The local Nusselt number for laminar flow over a flat plate is √ Nux = 0.664 Rex Pr1∕3 . Using expression for the local Nusselt number and Eq. (10.104c) for turbulent flow, we obtain an expression for mixed boundary layer over the flat plate { ) )0.8 } ( ( xc L q′′W L U∞ 1∕2 Pr1∕3 0.0296 U∞ = 0.332 dx + dx . ∫0 ∫xc x0.2 k(Tw − T∞ ) L xν ν After some algebra, we get an expression for average NuL { } 1∕3 NuL = 0.037Re0.8 L − A Pr 0.6 ≤ Pr ≤ 60 5 × 105 ≤ ReL ≪ 107 1∕2

A = 0.664Rexc − Re0.8 xc . If Rexc = 500 000, then Eq. (10.106) becomes [ ] 1∕3 NuL = 0.037 Re0.8 . L − 871 Pr

(10.106)

(10.107)

10.10.3 Prandtl–Taylor Analogy The Prandtl and Taylor analogy is discussed in [12]. Prandtl and Taylor divided the flow region into two parts: (a) The viscous sublayer. In this sublayer, the flow is laminar and there is no turbulent fluctuations. (b) Turbulent region. In this region, εM and εH are large and (ν + εM ) = (α + εH ) is valid. Analogy is integrated over each region separately, and the integration results are coupled at the interface of two regions. We now have du τ = ρ(ν + εM ) dy q′′ = −ρcp (α + εH )

dT dy

10.10 Analogy Between Momentum and Heat Transfer

Laminar sublayer: In this region ν ≫ εM and α ≫ εH . We also assume that τ ≈ τw and q′′ ≃ q′′w . The velocity and temperature distribution in the laminar sublayer is assumed to be linear since this layer is very thin. Based on the assumption, we may write q′′w q′′ ≈ s τw τs

(10.108)

where s refers to the interface of the laminar sublayer and the turbulent layer. We have τw = ρν

du dy

(10.109)

q′′w = −ρcp α

dT . dy

(10.110)

We take the ratio of Eqs. (10.109) and (10.110) τw ν du − . ′′ c qw p α dT

(10.111)

We rearrange Eq. (10.111) and integrate from wall to sublayer edge where y = s. Notice that we have q′′ ≈ q′′w and u = 0 q′′ ≈ q′′s and u = us .

at y = 0, at y = s,

Then, we write from Eq. (10.111) Ts

τw cp α

∫Tw

us

dT = −q′′w ν du ∫0

or

( (Tw − Ts ) =

q′′w τw

)

Prus . cp

(10.112a)

(10.112b)

Turbulent region: In this region, εM ≫ ν and εH ≫ α. For the turbulent region, we may write τw = ρεM

du dy

q′′w = −ρcp εH

(10.113) dT . dy

(10.114)

We take the ratio of Eqs. (10.113) and (10.114) τw ρε du∕dy ε du =− M =− M . c q′′w p εH dT ρcp εH dT∕dy

(10.115)

We now assume that εM ≈ εH . Thus, Eq. (10.115) becomes τw 1 du =− . cp dT q′′w We integrate Eq. (10.116) from to edge of viscous sublayer to outer edge turbulent layer U∞ τ w T∞ 1 dT = − du cp ∫us q′′w ∫Ts

or

( (Ts − T∞ )

cp τw q′′w

(10.116)

(10.117)

) = (U∞ − us )

(10.118)

where s refers to the interface of the laminar sublayer and the turbulent layer. We eliminate Ts between Eqs. (10.112b) and (10.118). Temperature distribution then becomes [ ] q′′ U u Tw − T∞ = w ∞ 1 + s (Pr − 1) . (10.119) τw cp U∞

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

We now rearrange Eq. (10.119) as τw cp q′′w 1 = [ ]. (Tw − T∞ ) U∞ u 1 + s (Pr − 1) U∞

(10.120)

The heat transfer coefficient h and the skin friction coefficient cf are defined as h=

q′′w Tw − T∞

(10.121)

and cf =

τw ρU2∞ ∕2

.

(10.122)

Using Eqs. (10.121) and (10.122), we now develop the following equation for the Nusselt number Nu: cf ∕2

]. us 1+ (Pr − 1) U∞

h = (ρcp U∞ ) [

(10.123)

Multiplying both sides of Eq. (10.123) by x/k, we obtain Nu =

h x cp = (ρU∞ x) [ k k

cf ∕2 ]. us 1+ (Pr − 1) U∞

Multiplying and dividing the right-hand side of Eq. (10.124) by dynamic viscosity μ, we obtain ( )( ) μcp ρU∞ x cf ∕2 hx = Nu = [ ]. k k μ uδ 1+ (Pr − 1) U∞

(10.124)

(10.125)

Finally introducing the Prandtl number Pr = μcp /k and Rex = ρU∞ x/μ, we obtain an expression for Nusselt number Nux (c ∕2)(PrRex ) Nux = [ f ]. us 1+ (Pr − 1) U∞

(10.126)

The next step is to determine the value of velocity ratio us ∕U∞ . We have an expression for the wall shear stress τw due to Blasius )1∕4 ( ν 2 τw = 0.0225ρ U∞ (10.127) U∞ δ and shear stress τw within the sublayer is τw = μ

du . dy

(10.128)

The integration of Eq. (10.128) from wall to edge of the laminar sublayer yields τw = μ

us . s

We now combine Eqs. (10.127) and (10.129) to get ( ) us U∞ δ 1∕4 μ s 1 . = δ 0.0225 U∞ ρU∞ δ ν

(10.129)

(10.130)

At the interface of the turbulent layer and the laminar sublayer, we may use the one-seventh power law of Prandtl ( ) ( u )7 s s = . (10.131) δ U∞

10.10 Analogy Between Momentum and Heat Transfer

From Eqs. (10.130) and (10.131), we get ( ( ) )7 us us U∞ δ 1∕4 μ 1 = U∞ 0.0225 U∞ ρU∞ δ ν or

(

us U∞

)6 =

1 0.0225

(

ν U∞ δ

(10.132a)

)3∕4 .

(10.132b)

For turbulent flow over a flat plate, we have 0.375 δ = . x (U∞ x∕ν)1∕5 Substituting Eq. (10.133) into Eq. (10.132b) and evaluating, we obtain ( ) us 2.1275 = . U∞ Re0.1 x

(10.133)

(10.134)

We now substitute Eq. (10.134) into Eq. (10.126) to get (cf ∕2)(PrRe) Nu = [ ( ) ] 2.1275 1+ (Pr − 1) Re0.1 x or in terms of the Stanton number (cf ∕2) St = [ ( ) ]. 2.1275 1+ (Pr − 1) Re0.1 x

(10.135a)

(10.135b)

For Pr = 1, we get the Reynolds analogy. We need an expression for the skin friction coefficient cf . For example, we may use the Schultz-Grunow [19] equation cf = 0.37[ln(Rex )]−2.584 Rex ≥ 5 × 105 .

10.10.4 Von Karman Analogy Von Karman [11] divided the boundary layer into three different regions and used three- layer velocity distribution to determine an expression for the local Nusselt number for turbulent flow over a flat plate. In other words, Von Karman considered all the sublayers (viscous, buffer, and turbulent sublayers). Heat fluxes q′′ ≈ q′′w and τ ≈ τw are assumed to be constant throughout the boundary layers. This assumption is supported by experimental studies. The derivation of analogy is similar to derivation of the temperature law of the wall. Laminar sublayer: Molecular effects dominate the turbulence effects. In this region, the velocity profile is given by u+ = y+ ,

0 < y+ < 5.

(10.136)

Then, we have du+ =1 dy+ and eddy diffusivity of momentum becomes εM = 0. ν Thus, the temperature distribution becomes dT+ = Pr. dy+

(10.137)

(10.138)

(10.139)

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

Integration yields T+ = Pry+ + C since at y+ = 0, T+ = 0 , we have C = 0. Thus, 0 < y+ < 5

(10.140a)

ΔT+sub = T+ (5) − T+ (0) = 5Pr − 0 = 5Pr.

(10.140b)

T+ = Pry+ or

Buffer layer The velocity distribution in the buffer layer is approximated by the empirical relation 5 < y+ < 30.

u+ = 5 ln y+ − 3.05,

(10.141)

The eddy diffusivity of momentum becomes εM y+ = − 1. ν 5 Then, the temperature distribution takes the following form: dT+ 1 = ) ( + + y dy 1 1 + −1 Pr Prt 5

(10.142)

(10.143)

T+ (5) = 5 Pr.

(10.144)

We solve this ordinary differential equation with Maple 2016. Solution may be put in the following form after some algebra: )] ( [ Pr y+ + −1 (10.145a) T = 5Pr + 5 Prt ln 1 + Prt 5 and we evaluate Eq. (10.145a) at the upper edge of the buffer layer. [ ( )] Pr 30 + −1 T = 5Pr + 5 Prt ln 1 + Prt 5 ] [ 5Pr . = 5Pr + 5 Prt ln 1 + Prt Thus, ΔT+buffer becomes ΔT+buffer

] 5Pr + Prt . = T (30) − T (5) = 5Prt ln Prt +

[

+

(10.145b)

Kader [24] developed the following improved expression for the buffer region: Prt ln y+ + A(Pr) κ where A(Pr) is T+ =

(10.146)

A(Pr) = [3.85Pr1∕3 − 1.3]2 + 2.12 ln(Pr). Fully turbulent core: Consider fully turbulent region, y+ > 30. Here, we will follow a different approach. In this region, the molecular effects can be neglected. Thus, for this region, we have 1 εM 1 εM 1 + ≈ . (10.147) Pr Prt ν Prt ν This approximation is not valid for liquid metals since the 1/Pr term needs to be kept in the derivations. The velocity profile for y+ > 30 is 1 u+ = ln y+ + B. (10.148) κ

10.10 Analogy Between Momentum and Heat Transfer

From (10.148), we get du+ 1 = . κ y+ dy+

(10.149)

The eddy diffusivity of momentum becomes εM 1 1 = −1= − 1 = κ y+ − 1 ≈ κ y+ . ν 1∕κ y+ du+ dy+

(10.150)

Then, the energy equation can be expressed in the following form: Prt 1 dT+ == = + 1 κ y+ dy κ y+ Prt

(10.151)

and its initial condition is

) ( 5 Pr . T+ (30) = 5Pr + 5 Prt ln 1 + Prt

Then, the solution of the ordinary differential equation, Eq. (10.151), gives the temperature distribution ) ( Pr Pr 5 Pr + T = 5Pr + 5 Prt ln 1 + + t ln(y+ ) − t ln(30). Prt κ κ

(10.152)

(10.153)

This is the temperature law of the wall. We know that the local Stanton number St may be expressed as St =

q′′w . ρ cp U∞ (Tw − T∞ )

(10.154)

Using the definition of dimensionless temperature T+ , we may write an expression for free-stream temperature √ (Tw − T∞ )ρcp U∞ cf ∕2 + (10.155) T∞ = q′′w and in an analogous way, we write an expression for free-stream velocity √ U∞ U∞ 2 + = √ . = U∞ = uτ cf τ ∕ρ

(10.156)

w

Then, using Eqs. (10.155) and (10.156), the Stanton number St may be expressed as √ cf ∕2 1 . St = + + = T∞ U∞ T+∞

(10.157)

We can now obtain a relation between St and cf . The next step is to evaluate the dimensionless free-stream temperature T+∞ , and it is evaluated from the temperature law of the wall, Eq. (10.153), as ] [ ( ) Pr Pr 5 Pr + (10.158) + t ln y+∞ − t ln(30). T∞ = 5Pr + 5 Prt ln 1 + Prt κ κ ( ) As we have done before, to evaluate ln y+∞ , we will use the law of the wall for velocity, Eq. (10.24), and the result is ) ln y+∞ ( + = U∞ − B . κ

(10.159)

If we substitute Eq. (10.159) into Eq. (10.158), dimensionless free-stream temperature distribution T+∞ becomes ] [ ( ) Pr 5 Pr (10.160) + Prt U+∞ − B − t ln(30). T+∞ = 5Pr + 5 Prt ln 1 + Prt κ √ U Since U+∞ = u∞ = 2∕cf , Eq. (10.160) is expressed as τ ] [ √ Pr 5 Pr (10.161) + Prt ( 2∕cf − B) − t ln(30). T+∞ = 5Pr + 5 Prt ln 1 + Prt κ

537

538

10 Turbulent External Boundary Layers: Momentum and Heat Transfer

We should note that if the flow has significant adverse or favorable pressure gradients, then these temperature profiles are not valid. Using the equation for T+∞ , the Stanton number can be obtained as √ cf ∕2 . (10.162) St = ] [ √ Pr 5 Pr 5Pr + 5 Prt ln 1 + Pr + Prt ( 2∕cf − B) − κ t ln(30) t

The denominator can be viewed as the sum of three resistances due to the viscous sublayer, the buffer layer, and the turbulent layer. Choosing κ = 0.4, B = 5 and Prt ≈ 1, we obtain an equation for Stanton number St St =

1+

√

cf ∕2 cf ∕2[5Pr + 5 ln(5Pr + 1) − 14]

0.5 < Pr < 30.

(10.163)

Equation (10.163) is in good agreement with experimental data for Prandtl numbers in the range of 0.5 < Pr < 30. Equation (10.163) reduces to the Reynolds analogy for Prandtl number of Pr = 1. Example 10.5 Air at 12 ∘ C and 1 atm flows over a flat plate with a free-stream velocity of 40 m/s. The plate surface temperature is maintained at 42 ∘ C. At a distance of 100 cm measured from the leading edge of the plate, determine: (a) (b) (c) (d)

skin friction coefficient cf the wall shear stress τw the friction velocity uτ the Stanton number St

Assume Prt ≈ 1. Solution The film temperature Tf T∞ + Tw 12 + 42 = = 27 ∘ C = 300 K. 2 2 The physical properties of air at this temperature are: Tf =

ρ = 1.1614 W∕m.K

cp = 1007 J∕kg.K

ν = 15.89 × 10−6 m2 ∕s

k = 0.0263 W∕m.K

μ = 184.6 × 10−7 N.s∕m2 Pr = 0.707.

The Reynolds number Rex is Rex =

U∞ x 40 × 1 = 2.51 × 106 . = ν 15.89 × 10−6

Flow is turbulent. The local skin friction coefficient cf is cf 0.02879 = 2 Re0.2 x (0.02879)(2) cf = ≈ 0.003020. (2.51 × 106 )0.20 Eq. (10.42) We can now compute wall stress τw ) (c ) ( 0.003020 (1.1614 × 402 ) ≈ 2.80 N∕m2 . τw = f ρU2∞ = 2 2 The friction velocity uτ is √ √ τw 2.80 uτ = = ≈ 1.55 m∕s. ρ 1.1614

10.11 Some Other Correlations for Turbulent Flow over a Flat Plate

The Local Stanton number St is evaluated using Eq. (10.163) St =

1+

St = 1+

√ √

cf ∕2 cf ∕2[5Pr + 5 ln(5Pr + 1) − 14] 0.003020∕2

= 0.001702.

0.003020∕2 × [5 × 0.707 + 5 ln(5 × 0.707 + 1) − 14]

The local heat transfer coefficient h is h = (ρcp U∞ )St = 1.1614 × 1007 × 40 × 0.001702 ≈ 79.64 W∕m2 K.

10.11 Some Other Correlations for Turbulent Flow over a Flat Plate Suppose that we assume a critical Reynolds number of Rexc = 500 000. For larger Reynolds numbers, a transition from laminar to turbulent flow takes place. The location of the transition point depends on the intensity of free-stream turbulence, as discussed in [33, 34]. The effect of free-stream turbulence on turbulent boundary layer heat transfer and the mean velocity profile is discussed in [35–38]. Reynolds et al. [39] performed extensive experiments for air flow over an isothermal plate. They suggested the following empirical correlation: StPr0.4 = 0.0296 Re−0.2

for gases

10 ≤ Rex ≤ 10 5

7

0.5 ≤ Pr ≤ 1 Tw = const or q′′w = const.

(10.164)

On the basis of pipe flow data provided by Blasius, Reynolds et al. [39] report an empirical correlation for the skin friction coefficient cf cf ∕2 = 0.0296Re−0.2 . x

(10.165)

The comparison of Eqs. (10.164) and (10.165) shows that, in this range, c (10.166) StPr0.4 = f . 2 Equation (10.166) represents a modification of the Colburn analogy. Temperature-dependent fluid property effects are considered, and a new correlation was proposed by Reynolds et al. [39] ( ) Tw −0.4 StPr0.4 = 0.0296 Re−0.2 for gases x T∞ 0.5 ≤ Pr ≤ 1 5 × 105 ≤ Rex ≤ 5 × 106 .

(10.167)

Reynolds et al. [39] report that the probable error in the local Stanton number and the local Reynolds number is ±3% and ±1%, respectively. Consider Colburn analogy: c StPr2∕3 = f . 2 For a smooth flat plate, the local Nusselt number Nux can be computed using the Colburn analogy together with a suitable correlation for the friction coefficient cf . If we substitute Eq. (10.46) into the Colburn equation, we get 4∕5

Nux = 0.0296 Rex Pr1∕3 105 ≤ Rex ≤ 6 × 108 0.6 < Pr < 60.

(10.168)

539

540

10 Turbulent External Boundary Layers: Momentum and Heat Transfer

Suppose that we use Eq. (10.59) with the Colburn analogy. The local Nusselt number becomes 6∕7

Nux = 0.0135Rex Pr1∕3 105 ≤ Rex ≤ 1 × 108 0.6 < Pr < 60.

(10.169)

We will now follow a different approach to determine the average Nusselt number that covers both the laminar and turbulent regions over a flat plate. First, let us recall that the average Nusselt number NuL is defined as NuL =

hL k

where h is the average heat transfer coefficient and defined as L

h=

1 h(x) dx. L ∫0

Substituting this equation into the equation for average NuL , we obtain NuL =

L L k Nux Nux L1 dx = dx. ∫0 k L ∫0 x x

(10.170)

Nusselt number correlations are expressed in terms of the Reynolds number Rex , and for this reason, it is convenient to change the integration coordinates from x to Rex . Thus, we use Rex =

μ Rex ρ U∞ x →x= μ ρ U∞

and from this equation, we have dx =

μ dRex . ρ U∞

(10.171)

Substituting these into the average Nusselt number, we obtain L

NuL =

ReL Nux dx = ∫0 x

∫0

Nux μ dRex . μ Rex ρ U∞ ρ U∞

(10.172)

Notice that x = 0, Rex = 0 Rex = ReL .

x = L,

Finally, we have ReL

NuL =

∫0

Nux dRex . Rex

(10.173)

This equation can be integrated using any of the correlations we discussed. Recall that Churchill and Ozeo [42] give an empirical correlation for laminar flow over an isothermal flat plate for all the Prandtl numbers claimed to be accurate to ±1%: √ 0.3387 Pr1∕3 Rex Nux = [ . (10.174) ( ) ]1∕4 0.0468 2∕3 1+ Pr We will use this relation in the laminar portion of the following equation: ReL

NuL =

∫0

Rexc ReL Nux Nux Nux dRex = dRex + dRex . ∫ ∫ Rex Rex 0 Rexc Rex

(10.175)

10.11 Some Other Correlations for Turbulent Flow over a Flat Plate

Substituting equations for laminar and turbulent flows, we obtain 0.3387 Pr1∕3 NuL = [ ( ) ]1∕4 ∫0 0.0468 2∕3 1+ Pr

Rexc

−1∕2

Rex

dRex + 0.0296 Pr1∕3

ReL

∫Rexc

Re−0.2 dRex x

or after carrying out the integration, we have √ 0.6774 Pr1∕3 Rexc [ ] 0.8 NuL = [ + 0.037 Pr1∕3 Re0.8 L − Rexc . )2∕3 ]1∕4 ( 0.0468 1+ Pr

(10.176)

(10.177a)

If we use Eqs. (10.174) and (10.169) in Eq. (10.175), we get 0.3387 Pr1∕3 NuL = [ ) ]1∕4 ∫0 ( 0.0468 2∕3 1+ Pr

Rexc

−1∕2

Rex

dRex + 0.0135 Pr1∕3

ReL

∫Rexc

−1∕7

Rex

dRex .

After integration, we have an expression for the average Nusselt number NuL √ [ ] 0.6774 Pr1∕3 Rexc 6∕7 6∕7 1∕3 Re NuL = [ + 0.0158 Pr − Re xc L ( ) ]1∕4 0.0468 2∕3 1+ Pr Rexc ≤ ReL ≤ 1 × 108 0.6 ≤ Pr ≤ 60.

(10.177b)

Example 10.6 Air at 1 atm and 17 ∘ C is flowing over a flat plate. The plate is 1 m wide and 2 m long in the flow direction. The bottom of the plate is insulated. The free-stream velocity of air is 50 m/s. The plate surface temperature is uniform and equal to 57 ∘ C. Calculate the heat loss from the plate. Solution We now use Eq. (10.177a). First, we evaluate the film temperature Tf Tf =

Tw + T∞ 17 + 57 = = 37 ∘ C = 310 K. 2 2

Air properties at this temperature are: ν = 16.89 × 10−6 m2 ∕s

k = 27 × 10−3 W∕m.K

Pr = 0705.

The Reynolds number ReL is ReL =

U∞ L 50 × 2 = = 5.92 × 106 . ν 16.89 × 10−6

The flow is turbulent. We will use Eq. (10.177a) to find the average heat transfer coefficient. Assume that critical Reynolds number is 500000. √ 0.6774 (Pr)1∕3 Rexc [ ] 0.8 NuL = [ + 0.037 Pr1∕3 Re0.8 L − Rexc ( )2∕3 ]1∕4 0.0468 1+ Pr √ 0.6774 (0.705)1∕3 500 000 NuL = [ + 0.037 (0.705)1∕3 [(5.92 × 106 )0.8 − (500 000)0.8 ] = 7834. )2∕3 ]1∕4 ( 0.0468 1+ 0.705

541

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

The average heat transfer coefficient is ) ( k 0.027 (7834) = 105.75 W∕m2 K. h = NuL = L 2 The convective heat loss is q = hA(Tw − T∞ ) = 105.95 × (2 × 1)(330 − 290) = 8460 W.

10.12 Turbulent Flow Along a Semi-infinite Plate with Unheated Starting Length: Constant Temperature Solution Turbulent flow along a semi-infinite plate with unheated starting length is a good approximation of actual engineering problems encountered in practice. The solution of this problem will provide a basic building block for constructing solutions for variable surface temperature (nonisothermal surfaces) problems. Since the energy equation for the constant property turbulent boundary layer is linear, the solution to the problem depicted in Figure 10.6 may be used to build up solutions for variable surface temperature Tw (x). Consider incompressible, two-dimensional, and steady turbulent boundary layer flow over an impermeable flat plate. Free-stream velocity U∞ is constant. A portion of the plate surface starting from the leading edge is unheated. The remaining part of the plate is kept at constant temperature Tw . We are interested in heat transfer from the plate to fluid. The algebraic details are very long, and a detailed development appears in [4, 7, 39, 43, 44]. Ameel [45] presented a single general closed-form analytical solution. The solution of this problem is also discussed in [46]. The solution will be presented briefly. We will assume a power law profiles for both velocity and temperature. Very near the wall, in the laminar viscous sublayer, the simple power law velocity distribution is not valid. Temperature distribution follows one-seventh power law except at very close to the wall. The power law velocity and temperature profiles give infinite velocity and temperature gradients at the plate surface. Therefore, the wall shear stress τw as well as wall heat flux q′′w in the momentum and energy integral equations cannot be determined from the one-seventh power law profiles at y = 0. We also assume that Δ ≤ δ, as shown in Figure 10.6. Assumed velocity and temperature profiles are ( y )1∕7 u = (10.178) U∞ δ and ( y )1∕7 Tw − T = . Tw − T∞ Δ

(10.179)

We also assume that local shear stress τ and heat flux q′′ are given 𝜕u τ = (ν + εM ) ρ 𝜕y

(10.180)

y δ

T∞ U∞

ξ Figure 10.6

Δ

T∞

x Tw

Boundary layer on a plate with an unheated starting length.

10.12 Turbulent Flow Along a Semi-infinite Plate with Unheated Starting Length: Constant Temperature Solution

and q′′ 𝜕T = −(α + εH ) . ρcp 𝜕y

(10.181)

The molecular Prandtl number Pr = 1, and therefore, 𝛼 = 𝜈. It is also assumed that the turbulent Prandtl number Prt ≈ 1, so that εH = εM . From this, we conclude that the total diffusivities α + εH and ν + εM are the same for both heat and momentum transfer in the sublayer as well as the fully developed turbulent region. First, the energy integral equation is introduced for constant property, incompressible flow ( ) Δ d 𝜕T = u(T − T∞ )dy (10.182) −α 𝜕y y=0 dx ∫0 where α = k/ρ cp . We substitute Eqs. (10.178) and (10.179) into the right-hand side of the integral equation, Eq. (10.182). The evaluation of the integral yields ( ) q′′w d 7 Δ8∕7 . (10.183) = U∞ (Tw − T∞ ) ρ cp dx 72 δ1∕7 We cannot use the one-seventh power law temperature profile to evaluate the left-hand side of Eq. (10.183). The problem here is that the gradient of the one-seventh power law for temperature goes to infinity at the wall (y = 0). We faced the same problem in evaluating the wall shear stress in the momentum integral equation. The problem was solved by using empirical information due to Blasius. To overcome the difficulty in Eq. (10.183), we follow a different approach. Consider the apparent heat flux q′′ q′′ 𝜕T = −(α + εH ) . ρcp 𝜕y

(10.184)

We assumed that α = ν (Pr = 1) and εH = εM (Prt = 1), and for this reason, we can write Eq. (10.184) as q′′ 𝜕T = −(ν + εM ) . ρcp 𝜕y

(10.185)

Now, we combine these two equations, Eqs. (10.180) and (10.185), to eliminate (ν + εM ). Thus, we get q′′ τ 𝜕T∕𝜕y . =− ρcp ρ 𝜕u∕𝜕y

(10.186)

Substituting the one-seventh power law for temperature, Eq. (10.179), and one-seventh power law velocity, Eq. (10.178), into Eq. (10.186), we obtain ( ) q′′ τ δ 1∕7 Tw − T∞ =− . (10.187) ρcp ρ Δ U∞ Notice that variable y does not appear anymore. We now need an expression for the shear stress τ in the boundary layer at any point. Equation (10.187) provides a finite heat flux at the wall. The integral form of the momentum equation is ν

d 𝜕 u || = U∞ 𝜕 y ||y=0 dx ∫0

δ(x)

[u(1 − u)]dy.

(10.188)

Equation (10.188) gives an expression for the wall shear stress τw , and for this reason, we cannot use Eq. (10.188). We need an expression for the apparent shear stress τ within the boundary layer at any position y. A more general integral form of the turbulent boundary layer equation can be obtained by integration in the vertical direction from any position y to δ. First, we will consider the conservation of mass. The conservation of mass is applied to the control volume in Figure 10.7 ∑ ∑ ṁ i = ṁ o (10.189) in

out

ṁ x + dṁ y + dṁ e = ṁ x +

dṁ x dx dx

(10.190)

543

544

10 Turbulent External Boundary Layers: Momentum and Heat Transfer

Figure 10.7

Control volume for the conservation of mass.

or d (ṁ )dx − dṁ y dx x ( δ ) d dṁ e = ρu dy dx − ρvdx. dx ∫y

dṁ e =

(10.191)

Next, we will apply the momentum theorem in the x-direction. Consider the Figure 10.8 given below. ∑ Fx = Mout − Min ∑

d (M )dx − [U∞ dṁ e + u(ρvdx) + Mx ]. dx x The momentum flux at position x is Fx = Mx +

(10.192) (10.193)

δ

Mx =

∫y

ρ(u)2 dy.

(10.194)

We now substitute the expression for momentum flux Mx and conservation of mass flow rate dṁ e into the above equation and obtain ( δ [ ( δ ) ) ] ∑ d d ρ(u)2 dy dx − u(ρvdx) − U∞ ρu dy dx − ρvdx (10.195a) Fx = dx ∫y dx ∫y or

( δ ) ) d ρ(u)2 dy dx − U∞ ρu dy dx + (U∞ − u)ρvdx. ∫y dx ∫y ∑ Next, we evaluate the forces Fx acting on the left-hand side of this equation. Consider Figure 10.9 ) ( ∑ dp dx ( dδ ) Fx = p(δ − y) + p + dx dx 2 dx [ ( ) ][ ( ) ] dp dδ − p+ dx (δ − y) + dx − τdx. dx dx ∑

Fx =

d dx

(

δ

(10.195b)

(10.196) Figure 10.8

Control volume for the application of the momentum theorem.

10.12 Turbulent Flow Along a Semi-infinite Plate with Unheated Starting Length: Constant Temperature Solution

Figure 10.9

Control volume for the force balance.

Equation (10.196) is expanded and then simplified by eliminating second-order differential terms to obtain ( ) ( ) ∑ dp dp dx − τ dx + y dx. Fx = −δ dx dx We can now write momentum equation in the following form: ( ) ( ) ( δ ( δ ) ) dp dp d d 2 −δ −τ+y = ρ(u) dy − U∞ ρu dy + (U∞ − u)ρv. dx dx dx ∫y dx ∫y

(10.197)

(10.198)

Since we are considering incompressible turbulent flow over an impermeable flat plate with constant free-stream velocity U∞ dU 1 dp = −U∞ ∞ = 0. ρ dx dx

(10.199)

Then, the momentum equation, Eq. (10.198), reduces to the following form: ( δ ( δ ) ) d d 2 ρ(u) dy − U∞ ρu dy + (U∞ − u)ρv. −τ = dx ∫y dx ∫y

(10.200)

We now need to evaluate velocity v. The continuity equation for incompressible, two-dimensional flow is 𝜕u 𝜕v + = 0. 𝜕x 𝜕y

(10.201)

Integrating in the y-direction ) ) y( y( 𝜕v 𝜕u dy = − dy ∫0 ∫0 𝜕y 𝜕x or y

v(y) − v(0) = − ∫0

(

𝜕u 𝜕x

(10.202a)

) dy.

(10.202b)

To evaluate the integral on the right-hand side of Eq. (10.202b), Leibniz’s rule for differentiation under the integral will be used. The Leibniz rule as discussed in [31] is given as b b 𝜕 f(x, y) d db da f(x, y)dy = dy + f(b, x) − f(a, x) . ∫a dx ∫a 𝜕x dx dx

(10.203)

We now set f = u , a = 0, and b = y to get y

v=−

d u dy. dx ∫0

Using the one-seventh power law for velocity in Eq. (10.204) yields [ y ( ) ] U ( y )8∕7 ( y )1∕7 d d 7yU∞ y 8∕7 dδ . = ∞ v=− U∞ dy = − dx ∫0 δ dx 8 δ 8 δ dx

(10.204)

(10.205)

545

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

Substituting Eq. (10.205) for v into Eq. (10.200) eliminates the velocity v, and the result is ( δ ( δ ) ) U ( y )8∕7 dδ d d 2 τ=− . ρ(u) dy + U∞ ρu dy − ρ(U∞ − u) ∞ dx ∫y dx ∫y 8 δ dx

(10.206)

Substituting the one-seventh power law for velocity into Eq. (10.206) yields an expression for the apparent shear stress τ: [ ( y )9∕7 ] 7 2 τ dδ = U∞ 1 − . (10.207) ρ 72 δ dx Since the viscous sublayer is very thin compared to the boundary layer, small error in the momentum integral equation is negligible for shear stress calculation. At y = 0, wall shear stress τw can now be evaluated τw 7 2 dδ = U . (10.208) ρ 72 ∞ dx We can also express the shear stress as follows by combining the last two equations, Eqs. (10.207) and (10.208). The result is ( y )9∕7 τ =1− . (10.209) τw δ The substitution of the above equation, Eq. (10.209), and the one-seventh power law velocity profile into the relation for apparent shear stress τ = ρ(ν + εm ) gives

𝜕u 𝜕y

( ν + εm = 7 δ

τw ρ U∞

)( ) [ ( y )9∕7 ] y 6∕7 1− . δ δ

(10.210)

Notice also that this equation, Eq. (10.210), is not valid in the viscous sublayer since it indicates that total diffusivity goes to zero for y = 0. However, we may neglect this error since the viscous sublayer is very thin. We are now in a position to evaluate the heat flux q′′w ( y )9∕7 ] ( )1∕7 ( T − T ) τ [ q′′ δ w ∞ w 1− . (10.211) = ρcp Δ U∞ ρ δ Again, notice that this equation is not valid in the viscous sublayer. The friction factor cf can be written as τ 1 U∞ cf = w 2 ρ and substituting Eq. (10.212) into Eq. (10.211), we get [ ( y )9∕7 ] ( )1∕7 ( T − T ) ( c) q′′ δ w ∞ U2∞ f 1 − . = ρcp Δ U∞ 2 δ The next step is to evaluate the heat flux q′′w at the wall where y = 0 ( )1∕7 ( c ) q′′w δ f . = U∞ (Tw − T∞ ) ρcp Δ 2

(10.212)

(10.213)

(10.214)

Finally, we can now substitute expression for heat flux, Eq. (10.214), into the integral energy equation, Eq. (10.183), and after algebra, we obtain ( ) ( )1∕7 ( ) cf δ d 7 Δ8∕7 . (10.215) = dx 72 δ1∕7 Δ 2 We can now use an equation for cf . For this purpose, the Prandtl–von Karman equation for the friction factor cf will be used. The friction factor cf is given as cf 0.02968 = . (10.216) 1∕5 2 Re x

10.12 Turbulent Flow Along a Semi-infinite Plate with Unheated Starting Length: Constant Temperature Solution

The substitution of Eq. (10.216) into the energy integral equation, Eq. (10.215), yields ( ) ( )1∕7 ( x U )−1∕5 δ d 7 Δ8∕7 ∞ = 0.02968 . dx 72 δ1∕7 Δ ν Define now r = Δ/δ and writing the equation, Eq. (10.217) in terms of r gives ( )−1∕5 7 d 8∕7 −1∕7 x U∞ (δ r ) = 0.02968(r) . 72 dx ν Using the product rule for differentiation, this equation, Eq. (10.218), can be written as ) ( ( ) ( ) x U∞ −1∕5 72 8 1∕7 dr 8∕7 dδ −1∕7 r +r = 0.02968 (r ) δ 7 dx dx 7 ν and multiply both sides by r1/7 and rearrange this equation, Eq. (10.219), as ( ) ( x U )−1∕5 ( ) 72 8 2∕7 dr ∞ 9∕7 dδ r +r = 0.02968 δ . 7 dx dx 7 ν

(10.217)

(10.218)

(10.219)

(10.220)

We can use the Prandtl–von Karman expression for boundary layer thickness δ, and this is given as δ 0.3816 = . 1∕5 x Rex Substitution of Eq. (10.221) into Eq. (10.220) gives ) ( ( ) ( x U )−1∕5 ( ) 4 8 0.3816 x ∞ 2∕7 dr 9∕7 x r + r (0.3816) 1∕5 7 dx 5 ν Rex ( ) ( x U )−1∕5 72 ∞ = 0.02968 . 7 ν After simplification, this equation, Eq. (10.222), reduces to the following form: ( ) 0.02968 72 −1 4 8 2∕7 dr r + r9∕7 x−1 = x . 7 dx 5 0.3816 7 Note that by product rule, 7 dr9∕7 dr = . dx 9 dx Equation (10.223) becomes after algebra ( )( ) 0.0296 72 9 9 −1 d r9∕7 + r9∕7 x−1 = x dx 10 0.3818 7 8 r2∕7

Δ = 0 at x = ξ. The solution of this differential equation can be obtained using Maple 2016. The solution is given as [ ( )9∕10 ]7∕9 ξ Δ = 1− . δ x Equation (10.227) is used in expression for heat flux q′′w , and the result is [ ( )9∕10 ]−1∕9 (c ) q′′w Nux ξ h f = = 1− = St = . ρcp U∞ (Tw − T∞ ) ρcp U∞ Rex Pr 2 x

(10.221)

(10.222)

(10.223)

(10.224)

(10.225) (10.226)

(10.227)

(10.228)

Equation (10.228) gives the local heat flux for a flat plate with unheated stating length of ξ. The bracketed term represents the effect of unheated starting length. The main objection to the integral methods is the assumption that the velocity and temperature profiles are similar. This assumption is approximately true for an isothermal

547

548

10 Turbulent External Boundary Layers: Momentum and Heat Transfer

plate, but it is not true for nonisothermal plates. In order to take care of the effects of Prandtl number differing from unity, Eq. (10.228) is written in a slightly different form [ ( )9∕10 ]−1∕9 (c ) q′′w Nux ξ h f 0.4 = StPr = = . (10.229) = 1− 0.6 ρcp U∞ (Tw − T∞ ) ρcp U∞ 2 x Rex Pr This modification is supported by experimental work of Scesa [47]. Kays et al. [7] recommend the following relation for cf : . cf ∕2 = 0.0287 Re−0.2 x Combining Eqs. (10.230) and (10.229) yields [ ( )9∕10 ]−1∕9 ( ) ξ 0.8 0.6 Nux = 0.0287 Rex Pr . 1− x

(10.230)

(10.231)

Let us denote StT as the Stanton number for an isothermal plate without an unheated length (ξ = 0); Eq. (10.228) becomes c (10.232) St = StT = f . 2 Note that we may choose a suitable expression for StT . With this notation, Eq. (10.228) can be written as [ ( )9∕10 ]−1∕9 ξ St = 1− . (10.233) StT x In Eq. (10.233), we may use any suitable expression for StT . Note that StT is the Stanton number for the completely heated isothermal plate. Using the definition of the heat transfer coefficient h and the Stanton number St, we can write an expression for the heat transfer coefficient h for a flat plate with unheated starting length from Eq. (10.233) [ ( )9∕10 ]−1∕9 ξ x > ξ. (10.234) h(ξ, x) = ρ cp U∞ StT 1 − x Klein and Tribus [48] examined experimental data of Maisel and Sherwood [49]. Maisel and Sherwood carried out mass transfer experiments from a plate. Based on these experimental data, Klein and Tribus proposed the following empirical relation: [ ( )8∕10 ]−11∕10 ξ St = 1− . (10.235) StT x The mean Stanton number at the end of a plate of length L is defined as the average Stanton number over heated portion of the plate: L

St =

1 St dx. L − ξ ∫ξ

(10.236)

Using an expression for local Stanton number St, Reynolds et al. [44] obtained an expression for mean Stanton number St ( )−0.2 L∕ξ ξ ξ St = σ−0.2 [1 − σ−9∕10 ]−1∕9 dσ. ∫ξ StT L−ξ L Integral is evaluated by setting σ9/10 = z. The final result is ]8∕9 ( )2∕10 [( )9∕10 5 L L −1 4 ξ ξ St = . L StT −1 ξ

(10.237)

As (L/ξ) → ∞, 5 St = . StT 4

(10.238)

10.12 Turbulent Flow Along a Semi-infinite Plate with Unheated Starting Length: Constant Temperature Solution

Ameel [45] has obtained an expression for the average Nusselt number for an isothermal plate with an unheated starting length )[ ( ( )9∕10 ]8∕9 ξ L NuL = NuL,ξ=0 (10.239) 1− L−ξ L where NuL,ξ=0 is the average Nusselt number for the plate of length L and heating begins at the leading edge of the plate (there is no unheated length). For example, if we assume that flow is turbulent over the entire surface, then NuL,ξ=0 may be given by 4∕5

NuL,ξ=0 = 0.037ReL Pr1∕3 .

(10.240)

Example 10.7 Air at 1 atm and 17 ∘ C flows over a flat plate. The free-stream velocity of the air is 50 m/s. The plate length in the flow direction is 1 m and first 25 cm of the plate is unheated. The heated portion of the plate surface is at 37 ∘ C. Assume that turbulence starts from the leading edge of the plate. Plot the local heat flux along the plate. Solution First, we evaluate the film temperature Tf T∞ + Tw 17 + 37 = = 27 ∘ C = 300 K. 2 2 The physical properties of air are Tf =

ρ = 1.1614 kg∕m3

cp = 1007 J∕kg K

μ = 184.6 × 10−7 N.s∕m2

ν = 15.89 × 10−6 m2 ∕s

k = 0.0263 W∕m K

Pr = 0.707.

The Reynolds number based on the entire plate length ReL is U∞ L 50 × 1 = = 3.14 × 106 . ν 15.89 × 10−6 Flow is turbulent over the plate. Consider now Eq. (10.232) and Eq. (10.233) [ ( )9∕10 ]−1∕9 (c ) ξ f St = 1− . 2 x ReL =

Since St = Nux /Rex Pr, we can write this equation in the following form: [ ( )9∕10 ]−1∕9 (c ) ξ f Nux = Rex Pr 1− . 2 x Next, we substitute Eq. (10.46) for the skin friction coefficient cf and the definition of Nux into this equation. This will give us [ ( )9∕10 ]−1∕9 ( ) ξ hx −0.2 . = Rex Pr 0.0296Rex 1− k x The heat transfer coefficient h is defined as q′′w . h= (Tw − T∞ ) Using the definition of the heat transfer coefficient, we get [ ( )9∕10 ]−1∕9 q′′w x ξ 0.8 = 0.0296Rex Pr 1 − k(Tw − T∞ ) x or

[ ( ) ( )9∕10 ]−1∕9 q′′w x U∞ x 0.8 ξ = 0.0296 Pr 1 − . k(Tw − T∞ ) ν x

549

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

3200

3000

2800

2600 q′′w(x) 2400

2200

2000

1800 0.3

Figure 10.E7

0.4

0.5

0.6 x (m)

0.7

0.8

0.9

1.0

Variation of local heat flux along the plate.

Substituting numerical values into this equation ( )0.8 [ ) ]−1∕9 ( q′′w x 0.25 9∕10 50x = 0.0296 (0.707) 1 − . (0.0263)(310 − 290) x 15.89 × 10−6 Solving for q′′w , we obtain q′′w =

( x1∕5

1737.69 . ( )9∕10 )1∕9 1 1 − 0.2871 x

We now plot heat flux as a function position x. See Figure 10.E7.

10.13 Flat Plate with Arbitrarily Specified Surface Temperature Consider arbitrary surface temperature variation over a flat plate. Assume that the flow over the impermeable flat plate is steady and fully turbulent. Free-stream temperature T∞ and velocity U∞ are constant. It is also assumed that fluid properties are constant. The viscous dissipation is neglected. The eddy diffusivity of heat εH is assumed to be independent of temperature. A comprehensive study of turbulent flat plate flow with variable surface temperature is given in a series of reports by Reynolds et al. [50]. Spalding [51–53], Rubesin [54], Biot [55], Tribus and Klein [56], and Smith and Shah [57] presented papers concerning heat transfer from surfaces of nonuniform temperature. As we have discussed in Chapter 7, Rubesin [54] has shown that the heat transfer rate for an arbitrary surface temperature variation could be determined by superposing several step temperature distributions. The summation of these steps will

10.13 Flat Plate with Arbitrarily Specified Surface Temperature

give us the actual surface temperature distribution. Heat transfer at any point will be the sum of the heat transfer rates attributed to all steps upstream of the point considered. The method of superposition will be used to study the problem of flat plate with arbitrarily specified surface temperature. We will employ Eq. (7.366) to determine the heat transfer from nonisothermal surface. Recall from Eq. (7.366) that, if the wall temperature distribution varies continuously except for a finite number of step discontinuities, we obtain an integral in the Stieltjes sense ) | ( x | dTw (ξ) dξ|| T − T∞ = f(x − ξ, y) . (10.241) ∫0 dξ | |Stieltjes Notice that f(x − ξ, y) is the solution of the energy equation for the case of a unit step-wall temperature distribution at x = ξ and the integral is taken in the Stieltjes sense. Wall heat flux q′′w (x) is obtained as ) | ( x | 𝜕 f(x − ξ, 0) dTw (ξ) ′′ dξ|| qw = −k . (10.242) ∫0 𝜕y dξ | |Stieltjes The heat transfer coefficient h is defined as ) ( 𝜕T h(Tw − T∞ ) = −k . 𝜕y y=0 Recall that we have Eq. (7.361). From Eq. (7.361), we write T − T∞ = (Tw − T∞ )f(x − ξ, y). Then, we express the heat transfer coefficient as [ ] 𝜕 f(x − ξ, y) . h(Tw − T∞ ) = −k(Tw − T∞ ) 𝜕y y=0 Thus, h is given as ] [ 𝜕 f(x − ξ, y) . h = −k 𝜕y y=0 Introducing Eq. (10.243) into Eq. (10.242), we obtain the local heat transfer rate q′′w (x) ( ) | x | dTw (ξ) ′′ h(ξ, x) . qw = dξ|| ∫0 dξ | |Stieltjes

(10.243)

(10.244)

Equation (10.244) is interpreted in the Stieltjes sense since the specified temperature may have a finite discontinuity. Recall that dTw (ξ)∕dξ is undefined at some point, and h(x, ξ) is the heat transfer coefficient for a step-wall temperature distribution. Finally, we may write Eq. (10.244) as ) ( N x ∑ dTw (ξ) ′′ dξ + h(ξ, x) h(ξn , x) ΔTw,n . (10.245) qw = ∫0 dξ n=0 The first integral on the right is Riemann integral. The heat transfer coefficient h(ξ, x) is given by [ ( )9∕10 ]−1∕9 ξ h(ξ, x) = ρ cp U∞ StT 1 − x>ξ x where StT is the local Stanton number for heat transfer from a plate at constant temperature and StT may be chosen as StT = 0.0296Re−0.2 Pr−2∕3 . x Here, we should note that the terms ρ cp U∞ StT can be taken out of integral since integration is performed with respect to ( ) ( ) variable ξ. Here, N is the number of discontinuities and ΔTw,n = Tw ξ+n − Tw ξ−n denotes the temperature jump across the ( +) ( ) nth discontinuity. Notice that Tw ξn is the wall temperature immediately downstream of temperature jump and Tw ξ−n is the temperature immediately upstream of the jump. Each discontinuity gives an additional contribution to the local heat flow. This integration is carried out in the Stieltjes sense. This is done because the specified surface temperature may have

551

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

finite discontinuity, so that dTw is undefined at some point. The Stieltjes integral can be expressed as the sum of an ordinary or Riemann integral and a term that accounts for the effect of the finite discontinuity, as explained in [56]. Maisel and Sherwood [49] carried out mass transfer experiments and obtained an empirical equation for the heat transfer coefficient. This equation is best fit to their data: [ ( )0.8 ]−0.11 ( ) ξ k 0.8 Rex 1 − . h(ξ, x) = 0.035 x x Example 10.8 Consider the following temperature distribution on a flat plate as given in Figure 10.E8. Obtain the heat transfer for x > ξ. Figure 10.E8

Tw(x)

Stepwise temperature distribution.

Tw1 Tw0 T∞ 0

ξ1

x

Solution We have two steps and N = 2 n = 0; h(0, x) = ρ cp U∞ StT (x) ΔTw0 = Tw (0+ ) − Tw (0− ) = Tw0 − T∞ n=1

[

(

h(ξ1 , x) = ρ cp U∞ StT (x) 1 −

ξ1 x

)9∕10 ]−1∕9

( ) ( ) ΔTw1 = Tw ξ+1 − Tw ξ−1 = Tw1 − Tw0 . Thus, for heat transfer x > ξ1 q′′w

[ ( )9∕10 ]−1∕9 ⎤ ⎡ ξ1 ⎥. = ρ cp U∞ StT (x) ⎢(Tw0 − T∞ ) + (Tw1 − Tw0 ) 1 − ⎥ ⎢ x ⎦ ⎣

StT is evaluated with Tw = Tw0 . Example 10.9 Consider turbulent flow over a flat plate. Turbulence starts from the leading edge of the plate. This is possible if the boundary layer is tripped at the leading edge of the plate. Assume that the surface temperature variation over a flat plate is given as in Figure 10.E9. Tw − T∞ = Bx. Find an expression for the heat flux at any point along the plate. Figure 10.E9

Tw – T∞ B

0

x

Linear temperature distribution.

10.14 Constant Free-Stream Velocity Flow Along a Flat Plate with Uniform Heat Flux

Solution (

x

q′′w =

∫0

h(ξ, x)

dTw (ξ) dξ

) dξ +

N ∑

h(ξn , x) ΔTw,n .

n=1

The second term on the right-hand side is zero since there is no jump in temperature distribution. The heat transfer coefficient is [ ( )9∕10 ]−1∕9 ξ h = ρ cp U∞ StT (x) 1 − x where StT is the Stanton number for isothermal plate. We also have dTw = B. dξ We substitute these into the equation to get [ ( )9∕10 ]−1∕9 ⎤ x⎡ ′′ ⎢ρ c U St 1 − ξ ⎥ (B)dξ. qw = ∫0 ⎢ p ∞ T ⎥ x ⎣ ⎦ Note that the terms ρ cp U∞ StT and B may be taken outside the integral since the integration is performed with respect to variable ξ [ ( )9∕10 ]−1∕9 x ξ ′′ dξ. qw = (ρcp U∞ )(StT )B 1− ∫0 x Integral is evaluated by Maple 2020 ( >

x ∫𝟎 (

𝟏𝟎 × 𝚪

𝛏 x ) ( )

𝟏− 𝟏 𝟗

𝟏 ( ) 𝟗 )− 𝟗

𝚪

𝟏𝟎

d𝛏;

𝟖 𝟗

𝟖𝟏

> evalf(%); 1.134000299 x q′′w = (ρcp U∞ )(StT )(1.134 )(B x).

10.14 Constant Free-Stream Velocity Flow Along a Flat Plate with Uniform Heat Flux Based on the experimental work of Reynolds, Kays and Kline [70], the following empirical correlation is proposed by Kays et al. [7]: . StPr0.4 = 0.030Re−0.2 x

(10.246a)

Nellis and Klein [10] report a correlation for turbulent flow over a flat plate subjected to uniform heat flux: hx 1∕3 = 0.0308Re0.8 Nux = x Pr k 0.6 ≤ Pr ≤ 60 5 × 105 ≤ Pr ≤ 107

(10.246b)

where Nux is the local Nusselt number and turbulent flow starts from the leading edge of the plate. Notice that surface temperature Tw (x) varies along the plate but turbulent boundary layer is much less sensitive to variation of surface temperature. The variation of Tw (x) is determined by Newton’s law of cooling using the local heat transfer coefficient h(x) is determined from Eq. (10.246a). The fluid properties are determined at the film temperature Tf = (Tw + T∞ )∕2 and Tw is the average surface temperature.

553

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

The total rate of heat transfer from the surface is q = A q′′w .

(10.247)

Therefore, the average heat transfer coefficient is defined in terms of the average wall temperature Tw L

Tw =

1 T (x) dx. L ∫0 w

(10.248a)

We may also determine the average temperature difference, Tw − T∞ , between Tw and T∞ from Tw − T∞ =

L q′′ L x 1 (Tw − T∞ )dx = w dx L ∫0 L ∫0 k Nux

(10.248b)

where Nux is the appropriate correlation. The average heat transfer coefficient is defined as q′′w

h=

Tw − T∞ and an average Nusselt number NuL is hL . k If the heat flux is known, the heat transfer coefficient may be used to obtain the local surface temperature Tw (x) NuL =

Tw (x) = T∞ +

q′′w . h

(10.249)

The average temperature difference Tw − T∞ can be written in terms of the average Nusselt number NuL Tw − T∞ =

q′′w L

(10.250a)

k NuL

where NuL is given by 4∕5

NuL ≈ 0.038ReL Pr1∕3 .

(10.250b)

In laminar boundary layer flow, the heat transfer coefficient with uniform heat flux is about 36% higher than with uniform wall temperature. With turbulent boundary layers, this difference is less than 4%, as reported in [50]. This suggests that any expression for NuL obtained for uniform surface temperature may be used to estimate the average Nusselt number for the uniform heat flux case. We may then determine the average temperature difference given by Eq. (10.250a) using correlations for either the isothermal surface or uniform heat flux surface. Note that the task of determining whether a surface is isothermal or isoflux is not easy, and surface temperature measurements may be required.

10.15 Turbulent Flow Along a Semi-Infinite Plate with Arbitrary Heat Flux Distribution Consider turbulent flow over a flat plate. The free-stream velocity U∞ and temperature T∞ are constant. Fluid properties are assumed to be constant. A comprehensive study of turbulent flat plate flow with variable surface heat flux is given by Reynolds et al. [50]. Smith and Shah [57], Hanna [60], Hanna and Myer [61], Shuh and Pop [62], Antonia et al. [63], Smith and Shah [57, 64], and Taylor et al. [65] published papers related to the turbulent flow along a semi-infinite plate with arbitrary heat flux. Chen [66] presented an alternative procedure for evaluating the surface heat flux of a turbulent boundary layer with an unheated starting length. Following Hanna [60] and Hanna and Myers [61], solution will be presented for arbitrary heat flux distribution over a flat plate in a steady and incompressible turbulent boundary layer. Arbitrary heat flux distribution is treated by superposition assuming that the solution is obtainable for a step change in heat flux applied downstream of leading edge. In principle, the step-function solution may be derived by employing semiempirical methods of turbulent flow. In this problem, we are usually interested in wall temperature distribution. We will employ Eqs. (7.391) and (7.391) is presented here for convenience ( ′′ ) x dqw dξ|Stieltjes . F(x − ξ) (10.251) Tw (x) − T∞ = ∫0 dξ

10.15 Turbulent Flow Along a Semi-Infinite Plate with Arbitrary Heat Flux Distribution

Figure 10.10

(a) Single step in heat flux at x = 0. (b) Single step in heat flux at x = ξ.

Equation (10.251) can be written as ( ′′ ) N x ∑ dqw dξ + Tw (x) − T∞ = F(x − ξ) F(x − ξn )Δq′′w,n (10.252) ∫0 dξ n=0 ( ) ( ) where Δq′′w,n = q′′w ξ+n − q′′w ξ−n and step in heat flux at the leading edge is included in the summation. We need an expression for F(x − ξ). First, we define the heat transfer coefficient h as q′′w = h(Tw − T∞ ).

(10.253)

The heat flux distribution on a flat plate is shown in Figure 10.10b. Notice that arbitrary heat flux distribution is treated by superposition assuming that the solution is available for a step change in heat flux at x = ξ. Temperature distribution in terms of unit step solution was defined as ] [ f(x − ξ, y) T(x − ξ, y) − T∞ = q′′w0 . (10.254) k Temperature distribution is evaluated at wall by setting y = 0. This will give us [ ] f(x − ξ, y)| y=0 T(x − ξ, y)|y=0 − T∞ = q′′w0 k or

[ Tw − T∞ = q′′w0

f(x − ξ, 0) k

(10.255)

] (10.256)

where Tw = T(x − ξ, 0). We may express Eq. (10.256) in terms of F(x − ξ) as follows: Tw − T∞ = q′′w0 F(x − ξ) where F(x − ξ) = f(x − ξ, 0)/k. Using Eqs. (10.253) and (10.257), we may write 1 F(x − ξ) = . h(ξ, x) Thus, Eq. (10.251) for wall temperature distribution takes the following form: ] ( ′′ ) x[ dqw 1 dξ|Stieltjes Tw (x) − T∞ = ∫0 h(ξ, x) dξ

(10.257)

(10.258)

(10.259)

where h(x, ξ) is the heat transfer coefficient for a single step in wall heat flux. We may express this integral in terms of an ordinary Riemann integral and a summation [ ( +) ( )] ] ( ′′ ) N x[ ∑ q′′w ξn − q′′w ξ−n dqw 1 dξ + . (10.260) Tw (x) − T∞ = ∫0 h(ξ, x) dξ h(ξn , x) n=0 Notice that integral takes care of continuous variation of wall heat flux and summation takes care of stepwise discontinuities.

555

556

10 Turbulent External Boundary Layers: Momentum and Heat Transfer

We need an expression for the heat transfer coefficient h(ξ, x). Let us see some expressions for h(ξ, x). The heat transfer coefficient given by Hanna [60] and Hanna and Myers [61] is ]−1∕9 [ ( ) ξ k 0.8 0.49 Rex Pr 1− . (10.261) h(ξ, x) = 0.03 x x Smith and Shah [57] give an expression for the local heat transfer coefficient. This expression was developed using the one-seventh power law for temperature and velocity distribution )−1∕9 ( ( ) ξ k 0.6 h(ξ, x) = 0.03035 Re0.8 1 − Pr . (10.262) x x x Equations (10.261) and (10.262) are relatively simple since they do not contain beta function. Reynolds et al. [70], based on their experimental work, give [ ( )9∕10 ]−1∕9 ξ St = StT 1 − x

(10.263)

where St is the local Stanton number and StT denotes the Stanton number, which would be obtained if the plate were isothermal over its entire length. The Stanton number StT is given by ( ) Tw 0.4 = 0.0296Re−0.2 . (10.264) StT Pr0.4 x T∞ The factor (Tw /T∞ )0.4 accounts for variable gas properties, and this factor will be omitted in favor of a reference temperature to evaluate the fluid properties. Using the usual definition of Stanton number St, we have Nux St = . (10.265) Rex Pr Equations (10.264) and (10.265) are combined to get the heat transfer coefficient h [ ( )9∕10 ]−1∕9 ( )[ ] ξ k 0.8 0.6 0.0296Rex Pr . 1− h= x x

(10.266)

The surface temperature distribution for a step change in heat flux applied at ξ is [ ( ′′ ) ( )9∕10 ]1∕9 qw x∕k ξ Tw − T∞ = 1− 0.8 0.6 x 0.0296 Rex Pr x > ξ.

(10.267)

Reynolds et al. give the following expression for the arbitrary heat flux distribution: Tw − T∞ =

x qw (ξ) dξ 0.0975 . [ ( )9∕10 ]8∕9 ∫ ρcp U∞ StT 0 ξ 1− x

(10.268)

Spalding and Lin [71] give an expression for the surface temperature distribution corresponding to a step change in heat flux at x = ξ { [ ( )]1∕9 } ξ x 1 ′′ 1− Tw − T∞ = qw 0.8 0.6 k 0.0296 Rex Pr x x > ξ.

(10.269)

From Eq. (10.269), we obtain F(x − ξ) =

[ ( )]1∕9 ξ 1 x 1 − . 0.6 k 0.0296 Re0.8 x Pr x

For arbitrary heat flux distribution, we write the following expression: ( )]1∕9 ( ′′ ) x[ dqw ξ x 1 dξ|Stieltjes . 1 − Tw (x) − T∞ = 0.8 0.6 k 0.0296 Rex Pr ∫0 x dξ

(10.270)

(10.271)

10.15 Turbulent Flow Along a Semi-Infinite Plate with Arbitrary Heat Flux Distribution

We now do integration by parts to express this equation in a more convenient form. Let u = [1 − ξ/x]1/9 and dv = dq′′w [ ]8∕9 ξ 1 1− du = − , v = q′′w . 9x x Then, we get the expression given by Sparrow and Lin Tw − T∞ =

x

(1∕9k) 0.6 ∫ 0.0296Re0.8 0 x Pr

(

q′′ (ξ) dξ )8∕9 . ξ 1− x

(10.272)

As usual, this equation applies to constant property flow, and viscous dissipation is neglected. Example 10.10 Consider now double-step heat flux distribution shown in Figure 10.E10. Obtain the surface temperature distribution ) ( Tw (x) − T∞ = q′′w0 F(x) + q′′w1 − q′′w0 F(x − ξ1 ) ) ( ′′ ]1∕9 [ qw1 − q′′w0 q′′w0 ξ 1− + . Tw (x) − T∞ = ( ) ( ) x 0.03 k Re0.8 Pr0.49 0.03 k Re0.8 Pr0.49 x

x

x

x

q″w q″w1 q″w0

0 Figure 10.E10

ξ1

X

Stepwise heat flux distribution.

Consider heat flux distribution in Figure 10.11. Using the integral method, Smith and Shah [57] presented a simple equation for surface temperature distribution. In many engineering problems given the heat flux, the temperature distribution of wall is required. Smith and Shah considered a step heat flux q′′0 over a turbulent boundary layer at x = ξ. Using the integral method and assuming a one-seventh power law for velocity and temperature profiles, they developed the following relations for temperature distribution and local Nusselt number: [ ( )]1∕9 32.95 q′′0 0.2 0.4 ξ Tw (x) − T∞ = Rex Pr 1 − . (10.273) ρ cp U∞ x The local Nusselt number is given by [ ]−1∕9 ξ 0.8 0.6 Nux = 0.03035Rex Pr 1 − x

(10.274)

where q′′0 is the magnitude of the step heat flux applied on the surface at x = ξ. This equation is simple, and it may be used to determine surface temperature distribution for arbitrary heat flux distribution. Using Eq. (10.273), Smith and Shah give the following relations given an arbitrary heat flux distribution. Figure 10.11

Step heat flux distribution on a surface.

q″w q″0 q″w = 0 0

ξ

x

557

558

10 Turbulent External Boundary Layers: Momentum and Heat Transfer

Surface temperature distribution. Tw − T∞ =

32.95 Re0.2 Pr0.4 ∫0 ρ c p U∞ x

x[

1−

( )]1∕9 ( ′′ )| dqw | ξ dξ | x dξ ||Steltjes

(10.275)

Local Nusselt number Nux =

0.6 ′′ 0.03035Re0.2 x Pr qw (x) . [ ] ( )1∕9 ( ′′ )| dqw | x ∫0 1 − xξ dξ dξ | |Steltjes

(10.276)

Notice that in these equations, the integrals must be taken in the Stieltjes sense. Example 10.11 Consider a heat flux distribution as given in Figure 10.E11. Obtain the temperature distribution. Figure 10.E11

q″w(x)

Linear heat flux distribution.

q″w (x) = mx

Solution q′′w (x) = m x dq′′w =m dξ

( )]1∕9 ( ′′ ) dqw ξ dξ x dξ ( )]1∕9 x[ ξ 32.95 0.4 1 − Tw − T∞ = Re0.2 Pr (m)dξ ∫0 ρ c p U∞ x x 29.655 0.4 (m x)Re0.2 Tw − T∞ = x Pr . ρ c p U∞ Tw − T∞ =

32.95 Re0.2 Pr0.4 ∫0 ρ c p U∞ x

x[

1−

Kays et al. [7] give the following relation to calculate the variable wall temperature Tw (x): x

Tw (x) − T∞ =

∫0

g(ξ, x) q′′w (ξ) dξ

(10.277a)

where g(ξ, x) is [ ( )9∕10 ]−8∕9 (9∕10)Pr−0.6 Re−0.8 ξ x g(ξ, x) = ( ) ( ) 1− 8 1 x Γ (0.0287 k) Γ 9 9 where Γ is the Gamma function. Substituting g(ξ, x) into Eq. (10.277a) and evaluating the Gamma function yields [ ( )9∕10 ]−8∕9 x ) ( ξ 3.4139 −0.6 −0.8 Pr Rex 1− q′′w (ξ) dξ. (10.277b) Tw (x) − T∞ = ∫0 k x

10.16 Turbulent Transition and Overall Heat Transfer We know how to calculate the heat transfer coefficient for the laminar region as well the fully turbulent region of the boundary layer. There is a lengthy transition region between these two regions. The location of the transition point depends

10.16 Turbulent Transition and Overall Heat Transfer

c=

1.7

5

Nusselt number, Nux

103

Eqn. (10.282), ReI = 68000, c = 1.75 Eqn. (10.88) with eqn. (10.48a) NuIam = 0.4587 Re1/2 Pr1/3 Eqn. (10.279) Ž&Š, Tab. 22, Run 7 Ž&Š, Tab. 22, Run 8 t len bu r tu

102

r

ina

lam

Pr = 6.6 (water) qw = constant

104

105 Reynolds number, Rex

Figure 10.12 Nusselt number across the transition region for water flowing over constant heat flux plate (data of Zukauskas and Slanciaauskas). Source: Lienhard [27]/ASME/CC BY 4.0.

on several factors such as free stream turbulence and surface roughness. There is no basic theory for transition but there are experimental studies. Heat transfer in the transition region is studied by several researchers. Lienhard IV and Lienhard V [23] and Lienhard V [27] have presented some especially useful engineering correlations. Data for water and air in turbulent transition are shown in Figures 10.12 and 10.13. Lienhard also plotted equations for laminar and fully developed turbulent flow on these figures. An examination of Figures 10.12 and 10.13 reveals that the local Nusselt number Nux rises steeply in the transition region, and it varies nearly as a straight line. Lienhard noticed that the Nusselt number, Nutrans , in the transition region can be represented as Nutrans ≃ aRecx .

(10.278)

The value of c is obtained by plotting Eq. (10.278) on the log–log plot. The value of c is the slope and the slope is c = 1.75 in Figure 10.12. Lienhard V indicates that Eq. (10.278) intersects the laminar curve at Rel , and in Figure 10.12, the intersection is at Rel ≈ 68 000. Lienhard V modified Eq. (10.278) by examining the data in Figure 10.12 as ) ( Rex c (10.279) Nutrans = Nulam (Rel , Pr) Rel where Rel is the lower Reynolds number of transition region. The values of c and Rel may be estimated for a particular data set by plotting. The values of c and Rel are given in the legends of Figures 10.12 and 10.13, which fit to neach data set. The values of Rel depend on flow configuration, and it is difficult to predict. Lienhard V examined many experimental data sets and proposed the following correlation for c within ±8%: c = 0.9922 log10 (Rel ) − 3.013 30 000 ≤ Rel ≤ 500 000. Lienhard V [27] reports an expression for Rel ) ( 100 u′r −5∕4 Rel = (3.6 × 105 ) U∞

(10.280)

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

104

Nusselt number, Nux

560

103

Eqn. (10.282), ReI = 60000, c = 1.7 Eqn. (10.282), ReI = 140000, c = 2 Eqn. (10.282), ReI = 235000, c = 2.5 Eqn. (10.88) with eqn. (10.48a) 1/3 NuIam = 0.332 Re1/2 x Pr Reynolds et al., natural transition Reynolds et al., Run 1 Reynolds et al., Run 8 Kestin et al., low u′r /u∞ Kestin et al., high u′r /u∞

nt

le bu tur

102 Pr = 0.71 (air) Tw = constant

nar

i lam 104

105 Reynolds number, Rex

106

Figure 10.13 Nusselt numbers measured across the transition region for air flowing over constant temperature plates. Source: Lienhard [27]/ASME/CC BY 4.0.

for a laminar boundary starting at the leading edge under zero-pressure gradient. This equation requires that the experimental conditions must be defined clearly. The effect of free-stream turbulence and unheated length are shown in Figure 10.14. Lienhard reports following working scenarios. Disturbance Rel 100 u′r ≥ 3%; 4 × 104 ≤ Rel ≤ 105 U∞ 100 u′r ≤ 0.5%; Rel ≈ 106 U∞ 100 u′r ≤ 0.1%; Rel ≈ 2.8 × 106 U∞ The heat transfer for laminar flow with uniform heat flux boundary condition with an unheated length may be predicted as follows: √ 0.4587 Rex Pr1∕3 . (10.281) Nux = [ ( )3∕4 ]1∕3 ξ 1− x Experimental data agree quite well with Eqs. (10.88) and (10.281). Lienhard V [27] proposed a combined correlation for the laminar, transitional, and turbulent flow [ ( )−1∕2 ]1∕5 −10 + Nu . (10.282) Nux (Rex , Pr) = Nu5x,lam + Nu−10 trans x,turb Equation (10.282) is continuous over three regions as can be seen in Figures 10.12 and 10.13.

10.16 Turbulent Transition and Overall Heat Transfer

Nusselt number, Nux

104 Eqn. (10.88) with eqn. (10.48a) 0.0296 Re0.8 Pr0.6 Eqn. (10.281) for x0 = 4.3 cm 0.4587 Re1/2 Pr1/3 u′r / u∞ = 0.25% u′r / u∞ = 1.0% u′r / u∞ = 2.0%

nt

le bu tur

103

inar

lam

102

Pr = 0.71 (air) qw = constant

Increased h caused by unheated starting length

105

106 Reynolds number, Rex

107

Figure 10.14 Comparison of Eq. (10.88) to constant wall heat flux data of Blair [49] for three nominal levels of free-stream turbulence in air. u′r is the root-mean-square velocity fluctuation. Source: Lienhard [27]/ASME/CC BY 4.0.

Laminar region

[ 1∕2 0.332Rex Pr1∕3 Nux (Rex , Pr) = 1∕2 0.453Rex Pr1∕3

UWT UHF

With an unheated starting length of x = ξ [ ( )3∕4 ]−1∕3 ξ Nulam (Rex , Pr) × 1 − x Transitional region

(

Nutrans (Rex , Pr) = NuLAM (Rel , Pr) ×

Rex Rel

)c

Rel is the Reynolds number at the onset transition region c = 0.9922log10 (Rel ) − 3.013 Rel < 500 000 Turbulent region Nuturb (Rex , Pr) =

Rex Pr(cf ∕2)

. √ 1 + 12.7(Pr2∕3 − 1) cf ∕2

The Fanning friction factor may be evaluated by White’s formula. The accuracy is around 1–2% cf (Rex ) =

0.455 . [ln(0.06Rex )]2

White recommends this equation for zero-pressure gradient turbulent flow over a plate. For gases, Reynolds et al. [39] recommend the following equation: 0.6 Nuturb (Rex , Pr) = 0.0296Re0.8 x Pr .

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

Lienhard V reports that these equations are in good agreement with air, water, and oil data in the range of 0.7 ≤ Pr ≤ 257 and 4000 ≤ Rex ≤ 4.3 × 106 with a free-stream turbulence level of up to 5%. Lienhard [27] reports an empirical correlation for proposed by Zukauskas and Slanciauskas for all the Prandtl numbers 0.43 . Nuturb = 0.032Re0.8 x Pr

This equation correlates their data roughly to ±15%. Average Nusselt number for the entire plate Lienhard IV and Lienhard V [23] obtained the average heat transfer coefficient for the plate including the laminar, transitional, and turbulent regions. They approximated the integral in three distinct regions [ L [ L [ xl ] ] ] xu L q 1 1 1 1 h= w = q′′w dx = h dx = hlam dx + htrans dx + hturb dx (10.283) ∫xl ∫xu ΔT ΔT L ∫0 L ∫0 L ∫0 Tw = const const ΔT = Tw − T∞ ( ) ( ) where xl = Uν Rel is the start of transition and xu = Uν Reu is the end of transition. ∞ ∞ For gases, using Eq. (10.164a) for gases, Lienhard IV and Lienhard V [23] obtained an expression for the average Nusselt number NuL : ( ) hL 1∕2 0.8 = 0.037Pr0.6 Re0.8 + 0.664Rel Pr1∕3 L − Reu k ( ) 1 1∕2 0.6 (10.284) 0.0296Re0.8 + − 0.332Rel Pr1∕3 u Pr c where ReL is the Reynolds number based on plate length, Rel is the Reynolds for the lower end of transition, and Reu is the Reynolds number for the upper end of transition. Notice that the last term in Eq. (10.284) is the contribution of the transition region. Lienhard IV and Lienhard V [23] report the following correlation developed by Zhukauskas and Ambrazyavichyus [40] for nonmetallic liquids NuL =

0.43 Nuturb = 0.032 Re0.8 . x Pr

It is reported that this correlation fits the experimental data of Zhukauskas and Ambrazyavichyus about ±15%. The equation over predicts the air by 15–25%. For liquids, we may use the following correlations to get the average Nusselt number NuL . Again, we are considering uniform wall temperature boundary condition 1∕2

NuLAM = 0.332Rex Pr1∕3 0.43 Nuturb = 0.032 Re0.8 x Pr

( Nutrans = Nulam

Rex Rel

)c .

We substitute these equations into Eq. (10.283) to get NuL ( ) hL 1∕2 0.8 = 0.037Pr0.6 Re0.8 + 0.664Rel Pr1∕3 L − Reu k ( ) 1 1∕2 0.43 0.032Re0.8 + − 0.332Rel Pr1∕3 . u Pr c More information about heat transfer in the transition region is available in [23, 27]. NuL =

(10.285)

Example 10.12 Air at 270 K and 1-atm pressure flows over a 2-m-long flat plate. The width of the plate is 1 m. The free-stream velocity of the air is 30 m/s. The flat plate is maintained at a uniform temperature of 330 K. Assume that transition occurs at Rex, cr = 500 000.

10.16 Turbulent Transition and Overall Heat Transfer

(a) Calculate the heat loss from the plate using Eq. (10.107). (b) Calculate the heat loss from the plate using Eq. (10.284). Solution The properties of air are evaluated at the film temperature Tf Tw + T∞ 270 + 330 = = 300 K 2 2 kg N.s μ = 184.6 × 10−7 2 ρ = 1.1614 3 m m (a) We first calculate the Reynolds number Tf =

ReL =

k = 0.0263

W m.K

Pr = 0.707.

ρU∞ L 1.1614 × 30 × 2 = = 3.77 × 106 . μ 184.6 × 10−7

The average Nusselt number NuL is determined using Eq. (10.107) [ ] 1∕3 NuL = 0.037Re0.8 L − 871 Pr NuL = [0.037 × (3.77 × 106 )0.8 − 871] × (0.707)1∕3 = 5243. The average heat transfer coefficient over the plate is ) ( W k 0.0263 × (5243) ≈ 68.94 2 . h = NuL = L 2 mK The total heat transfer q is q = hA(Tw − T∞ ) = 68.94 × (2 × 1) × (330 − 270) ≈ 8273.66 W. (b) We now use Eq. (10.284)

NuL =

( ) hL 1∕2 0.8 + 0.664Rel Pr1∕3 = 0.037Pr0.6 Re0.8 L − Reu k ( ) 1 1∕2 0.6 0.0296Re0.8 + − 0.332Rel Pr1∕3 . u Pr c

We assume that the Reynolds number at the beginning of the turbulent transition regime is Rel = 500 000. We now need the Reynolds number Reu at the end of the turbulent transition regime. To find Reu , we proceed as follows. Equation (10.279). Intersects the curve for gasses, Eq. (10.164a), at Reu . Now, Eq. (10.164) is StPr0.4 = 0.0296Re−0.2 x or

(

Nux Rex Pr

) Pr0.4 = 0.0296Re−0.2 x

0.6 Nux = 0.0296Re0.8 x Pr .

Thus, we equate these equations at Reu . This will give us ( ) Reu c 1∕2 0.6 0.332Rel Pr1∕3 = 0.0296Re0.8 u Pr . Rel Exponent c is estimated from Eq. (10.280) c = 0.9922log10 (Rel ) − 3.013 c = 0.9922log10 (500 000) − 3.013 = 2.64 )2.64 ( √ Reu 0.6 0.332 500 00(0.707)1∕3 = 0.0296Re0.8 u (0.707) . 500000

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

We solve this equation for Reu Re2.64−0.8 = u

0.0296 × (0.707)0.6 (500 000)2.64 √ 0.332 500 00 × (0.707)1∕3

Reu = 1.085 × 106 . Next, we calculate the average Nusselt number. Note that ReL = 3.77 × 106 hL = 0.037(0.707)0.6 [(3.77 × 106 )0.8 − (1.085 × 106 )0.8 ] k √ √ 1 (0.0296(1.085 × 106 )0.8 (0.707)0.6 − 0.332 5 × 105 (0.707)1∕3 ) + 0.664 5 × 105 (0.707)1∕3 + 2.64 NuL = 3457.9 + 418.27 + 533.78 = 4409.95. NuL =

The average heat transfer coefficient is ) ( W k 0.0263 × (4409.95) ≈ 58 2 . h = NuL = L 2 mK The total heat loss q is q = hA(Tw − T∞ ) = 58 × 2 × 1(330 − 270) ≈ 6959 W.

10.17 Property Variation Based on their numerical solutions of turbulent boundary layer equations with variable properties, Kays et al. [7] proposed the following method for correcting the constant property solutions for the effects of property variations. For Liquids St Nu = = Nucp Stcp ( )m μw cf = cfcp μ∞

(

μw μ∞

)n (10.286a) (10.286b)

For Gases

) ( Tw n Nu St = = Nucp Stcp T∞ ) ( Tw m cf = cfcp T∞

(10.287a) (10.287b)

Heating

m

n

n

0.26

−0.23

−0.12

0.26

−0.58

−0.55

−0.47

−0.39

Liquids Gases

Cooling

m

All properties are evaluated at free stream temperature for external flows.

Problems 10.1

Von Karman proposed the following expression for the mixing: | du∕dy | | | 𝓁 =κ| 2 |. | d u∕dy2 | | | Using this relation, develop an expression for the velocity profile near the wall of a flat plate.

Problems

10.2

Evaluate u+ at y+ = 40 using the following velocity profiles: ⎧y+ ⎪ u+ = ⎨5 ln(y+ ) − 3.05 ⎪ ⎩2.5 ln(y+ ) + 5.5

Von Karman

for 0 < y+ < 5 for 5 < y+ > 30 for y+ > 30

[ 1 y+ = u+ + A exp(B u+ ) − 1 − B u+ − ( B u+ )2 2 ] 1 1 − ( B u+ )3 − ( B u+ )4 6 24 A = 0.1108

Spalding

B = 0.4 all y+ in inner region [ ( +) ( ) ( + )] + y − yy+ exp y3 u+ = 2.5 ln(1 + κy+ ) + 7.8 1 − exp + 1 y1

Reichardt

y+1 = 11

10.3

κ = 0.4

Air is flowing over a flat plate with a free-stream velocity of 35 m/s. The free-stream temperature of air is 300 K. At a position of 100 cm measured from the leading edge, the velocity boundary layer thickness is 20 mm. Assuming that the mean velocity in the turbulent boundary layer is represented by the Coles’ law ( yu ) ( ) ( y) 2Π 1 π u τ +B+ sin2 = ln uτ κ ν κ 2δ where κ = 0.41, B = 5, and Π = 0.45, calculate: (a) the friction velocity (b) the wall shear stress (c) the thickness of the viscous sublayer.

10.4

Consider fully developed turbulent flow over a flat plate. The velocity profile can be represented by ( y )1∕7 u = U∞ δ where U∞ is the free-stream velocity and δ is the boundary layer thickness. Using integral momentum equation, develop a relation for δ(x)/x, which is valid for the range of Reynolds numbers, 5 × 105 ≤ Rex ≤ 107 .

10.5

Consider fully developed turbulent flow over a flat plate. Assuming Pr = Prt = 1, according to Prandtl, the velocity and temperature profiles may be represented as ( y )1∕7 u = U∞ δ ( y )1∕7 T − Tw = T∞ − Tw Δ where U∞ is the free-stream velocity, δ is the boundary layer thickness, Tw is the wall temperature, T∞ is the free-stream temperature, and Δ is the thermal boundary layer thickness. The boundary layer thickness is given by 10235 δ 0.3805 = − . 1∕5 x Rex Re x

Using the integral momentum equation, develop a relation for Stanton number Stx , which is valid for 5 × 105 ≤ Rex ≤ 107 . 10.6

Air at 1-atm pressure and 350 K flows over a flat plate with a free-stream velocity of 21 m/s. Assume that the measured shear stress at x = 1.6 m measured from the leading edge of the plate is given as 0.70 Pa. At this position, estimate (a) the friction velocity uτ and (b) velocity u at y = 1 cm from the wall.

565

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

10.7

Air at 350 K and 1-atm pressure flows over a flat plate. The free-stream velocity of the air is 100 m/s. At a distance of 2 m from the leading edge of the plate, estimate the boundary layer thickness δ using ( +) πy u+ = 2.434 ln(y+ ) + 5.0 + 2.19sin2 2 δ+ yu u where u+ = and y+ = τ uτ ν

10.8

A flat plate is heated by an electrical heating element. Heat applied to the plate is 5 kW. The bottom side of the plate is insulated. The plate is 2 m long in the flow direction and 1 m wide. Atmospheric air is flowing over the plate at 300 K with a free-stream velocity of 15 m/s. Estimate the plate surface temperature.

10.9

Air at 1 atm and 17 ∘ C flows over a flat plate. The free-stream velocity of the air is 50 m/s. The plate length in the flow direction is 1 m. The plate surface is maintained at 37 ∘ C. Assume a transition Reynolds number of 5 × 105 . Plot the local heat flux along the plate.

10.10

Air at 1 atm and 17 ∘ C flows over a flat plate. The free-stream velocity of the air is 100 m/s. The plate length in the flow direction is 2 m and first 50 cm of the plate is unheated. The heated portion of the plate surface is at 37 ∘ C. Assume that turbulence starts from the leading edge of the plate. Plot the local heat flux along the plate.

10.11

Consider heat transfer to a turbulent boundary layer flow over a flat plate. Working fluid is air. There is no pressure gradient. The surface temperature variation over a flat plate is given in Figure 10.P11. Find an expression for the heat flux at any point along the plate. Tw(x) Tw2 Tw1 Tw0

T∞ 0 Figure 10.P11

10.12

ξ1

ξ2

x

Stepwise temperature distribution along the plate.

Consider heat transfer to a turbulent boundary layer flow over a flat plate. Working fluid is air. There is no pressure gradient. Assume that the surface temperature variation over a flat plate is given as Tw − T∞ = A +

N ∑

Bn x n .

n=1

Find an expression for the heat flux at any point along the plate. 10.13

Consider heat transfer to a turbulent boundary layer flow over a flat plate. Working fluid is air. There is no pressure gradient. Assume that the surface temperature variation over a flat plate is given as in Figure 10.P13. Tw − T∞ = (Tw0 − T∞ ) + mx. Find an expression for the heat flux at any point along the plate.

10.14

Consider heat transfer to a turbulent boundary layer flow over a flat plate. Working fluid is air. There is no pressure gradient. Assume that the surface temperature variation over a flat plate is given as in Figure 10.P14.

Problems

Figure 10.P13

Step-ramp temperature distribution.

Tw(x)

m

Tw0 T∞ 0

Tw − T∞ = m(x − a),

x

x > a.

Find an expression for the heat flux at any point along the plate. Figure 10.P14

Delayed ramp wall temperature.

Tw(x)

m T∞ 0

10.15

x

a

Consider uniform step heat flux distribution on a flat plate as shown in Figure 10.P15. We wish to determine the temperature distribution along the flat plate.

Figure 10.P15

Step heat input.

q″w q″0 q″w = 0 ξ

0

10.16

x

Consider a flat plate subjected to a delayed ramp heat input as shown in Figure 10.P16. Flow is turbulent over the plate. Heat flux is represented by q′′w = m(x − a). We wish to determine the temperature distribution along the flat plate.

Figure 10.P16

Delayed ramp heat input.

q″w m q″w = 0 0

10.17

a

x

Air at 300 K moves over a flat plate with a velocity of 50 m/s. The flat plate consists of five strip heaters. Each heater is 50 mm wide and has a heating length of (into page) of 120 mm. It is desired to always keep surface of the strip heaters at a temperature of 400 K by adjusting the power supply for each strip heater. Assume the critical Reynolds number of 500 000. Estimate the energy required by each heater.

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10 Turbulent External Boundary Layers: Momentum and Heat Transfer

10.18

Air at 300 K moves over a flat square smooth plate with a velocity of 50 m/s. One side of the plate is 1 m and its surface temperature is 400 K everywhere. At x = 0.5 m from the leading edge of the plate, using the velocity and temperature laws of the wall calculate the following quantities. (a) Wall shear stress (b) The thickness of viscous and buffer layers (c) The wall heat flux (d) Air temperature at y = 25 μm, 60 μm, and 2 mm. The distance y is measured from the wall. Assume that flow is turbulent starting from the leading edge of the plate and take Prt ≈ 1 for simplicity.

10.19

Consider air flow over a flat plate. The Prandtl–Taylor model of the velocity law of the wall for turbulent flow over a flat plate is u + = y+ ,

0 ≤ y+ ≤ 11.5

u+ = 2.5 ln y+ + 5.5, y+ > 11.5. (a) Obtain an expression for eddy diffusivity of momentum. (b) Obtain an expression for the law of the temperature. (c) Obtain an expression for the Stanton number. 10.20

Reynolds et al. [30] performed experiments on a smooth flat plate in the Wind tunnel. Free-stream velocity is lb U∞ = 113 ft/s and free-stream temperature is T∞ = 74.7 ∘ F. The density of the air is ρ = 0.0746 ftm3 . The plate surface temperature Tw ≈ 98 ∘ F. They give the following measurements. Develop a power law relation in the following form: Nux = mRenx and compare this relation with Eq. (10.88). Rex × 10−6

St × 103

Rex × 10−6

St × 103

Rex × 10−6

St × 103

0.235

2.56

1.247

2.05

2.25

1.78

0.386

2.48

1.398

1.84

2.40

1.74

0.528

2.23

1.539

1.95

2.54

1.72

0.676

2.13

1.682

1.85

2.68

1.71

0.816

2.12

1.828

1.83

2.80

1.72

0.960

2.03

1.978

1.81

2.95

1.66

1.108

2.01

2.11

1.77

3.10

1.53

10.21

Incompressible atmospheric air at 17 ∘ C flows with a velocity of 45 m/s over a flat plate having a length of 1 m. The plate surface is maintained at 137 ∘ C. Turbulent transition begins at 400000. Determine the average heat transfer coefficient.

10.22

Incompressible atmospheric air at 300 K flows with a velocity of 15 m/s over a flat plate having a length of 1 m. A uniform heat flux of 2000 W/m2 is applied to the plate surface. Flow is fully turbulent starting from the leading edge of the plate. (a) Develop an expression for the average surface temperature difference for the plate and evaluate the average surface temperature of the plate. (b) Determine the plate surface temperature at the trailing edge of the plate.

10.23

Develop Eq. (10.284).

References

10.24

Ethylene glycol at a film temperature of 310 K flows over a flat plate with a free-stream velocity of 10 m/s, and at a distance x = 50 cm from the leading edge of the plate, the local shear stress is measured to be τw = 70 N/m2 . Estimate the local heat transfer coefficient at the same location.

10.25

Water is flowing over a square plate with a free-stream velocity of 3 m/s. The free-stream temperature of water is 300 K. The side of the square plate is 1 m, and the plate is kept at temperature of 400 K. Assuming that the back side of the plate is insulated, we wish to estimate the heat loss from the plate.

10.26

Water at 300 K flows over a flat plate at 6 m/s. At 0.5 m from the leading edge of the plate, estimate the following quantities. (a) Boundary layer thickness, 𝛿 (b) Displacement thickness, 𝛿1 (c) Momentum thickness, 𝛿2 (d) Skin friction coefficient, cf (e) Friction velocity, u𝜏 (f) The value of velocity at y+ = 4, 15 and 150.

References 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Wilcox, D.C. (2006). Turbulence Modelling for CFD. DCW Industries. Schlichting, H. and Gersten, K. (2017). Boundary Layer Theory, 9e. Springer. Reynolds, A.J. (1974). Turbulent Flows in Engineering. Wiley. Burmeister, L.C. (1993). Convective Heat Transfer, 2e. Wiley. Van Driest, E.R. (1956). On turbulent flow near a wall. J. Aerosp. Sci. 23: 1007–1011. White, F.M. and Majdalani, J. (2021). Viscous Flow, 4e. McGraw-Hill. Kays, W., Crawford, M., and Weigand, B. (2005). Convective Heat and Mass Transfer, 4e. McGraw-Hill. Mills, A.F. (1999). Heat Transfer, 2e. Prentice Hall. Tennekes, H. and Lumley, J.L. (1972). A First Course in Turbulence. MIT Press. Nellis, K. and Klein, S. (2021). Introduction to Engineering Heat Transfer. Cambridge University Press. Von Karman, T. (1934). The Analogy Between Fluid Friction and Heat Transfer, vol. 1, 1–20. Knudsen, J.G. and Katz, D.L. (1958). Fluid Dynamics and Heat Transfer. McGraw-Hill. Coles, D.E. (1956). The law of the wake in the turbulent boundary layer. J. Fluid Mech. 1: 191–226. Rubesin, M.W., Inouye, N., and Parikh, P.G. (1998). Forced Convection External Flows, in the Handbook of Heat Transfer (ed. W.M. Rohsenow, J.P. Harnett and Y.I. Cho). Boston: McGraw-Hill. Spalding, D.B. (1961). A single formula for the law of the wall. J. Appl. Mech. 28 (3): 455–457. Deisler, R.G. (1950). Analytical and Experimental Investigation of Adiabatic Turbulent Flow in Smooth Tubes. NACA TN 2138. Von Karman, T. (1946). On Laminar and Turbulent Friction. NACA 1092. Pai, S.I. (1957). Viscous Flow Theory II-Turbulent Flow. Van Nostrand Company. Schultz-Grunow, F. (1941). New frictional resistance law for smooth plates. NACA TM-986. Von Karman, T. Turbulence and skin friction. J. Aeronaut. Sciences 1 (1): 1934. Daily, J.W. and Harleman, D.R.F. (1966). Fluid Dynamics. Addison-Wesley Publishing Company. White, F.M. (1991). Heat and Mass Transfer. Addison Wesley. Lienhard, J.H. IV, and John H. Lienhard, V. (2020). Heat Transfer, 5e. Phlogiston Press. Kader, B.A. (1981). Temperature, and concentration profiles in fully turbulent boundary layers. Int. J. Heat Mass Transfer 24: 1541–1554. Sucec, J. (1999). Prediction of heat transfer in turbulent, transpired boundary layers. J. Heat Transfer 121 (1): 186–190. Petukov, B.S. and Kirillov, V.V. (1958). Heat exchange for turbulent flow of liquids in tubes. Soviet Teploenergetika 5: 63–68.

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27 John H. Lienhard, V. (2020). Heat transfer in flat-plate boundary layers: a correlation for laminar, transitional, and turbulent flow. J. Heat Transfer 142 (6): 061805. 28 Churchill, S.W. (1997). Critique of the classical algebraic analogies between heat, mass, and momentum transfer. Ind. Eng. Chem. Res. 36: 3866–3878. 29 Webb, R.L. and Crosse, L. (1971). A critical evaluation of analytical solutions and Reynolds analogy equations for turbulent heat and mass transfer in smooth tubes. Wärme-Und Stoffübertragung 4: 197–204. 30 Reynolds, O. (1874). On the extent and action of the heating surface for steam boilers. Proc. Manchester Literary Philos. Soc. 14: 7–12. 31 Colburn, A.P. (1933). A method of correlating forced convection heat-transfer data and a comparison with fluid friction. Trans. AIChE 29: 174. 32 Gnielinski, V. (1976). New equations for heat and mass transfer in turbulent pipe and channel flow. Int. Chem. Eng. 16 (2): 359–368. 33 Bejan, A. (2013). Convective Heat Transfer, 4e. Wiley. 34 Schubauer, G.B. and Skramstad, H.K. (1949). Laminar Boundary Layer Oscillations and Transition on a Flat Plate. NACA Rept. 909. 35 Blair, M.F. (1983). Influence of free stream turbulence on turbulent boundary layer heat transfer and mean profile development, part I-experimental data. J. Heat Transfer 105: 41–47. 36 Blair, M.F. (1983). Influence of free stream turbulence on turbulent boundary layer heat transfer and mean profile development, part I-analysis of results. J. Heat Transfer 105: 33–40. 37 Junkhan, G.H. and Serovy, G.K. (1967). Effects of free stream turbulence and pressure gradient on flat plate boundary layer velocity profiles and on heat transfer. ASME J. Heat Transfer 89 (2): 169–175. 38 Kestin, J., Maeder, P.F., and Wang, H.E. (1961). The influence of turbulence on the transfer of heat from plates with and without a pressure gradient. Int. J. Heat Mass Transfer 3: 113–154. 39 Reynolds, W.M., Kays, W.M., and Kline, S.C. (1958). Heat Transfer in the Turbulent Incompressible Boundary Layer I: Constant Wall Temperature Distribution. NASA MEMO 12-1-58W. 40 Zhukauskas, A.A. and Ambrazyavichyus, A.B. (1961). Heat transfer of a plate in a liquid flow. Int. J. Heat Mass Transfer 3: 305. 41 Whitaker, S. (1977). Fundamental Principles of Heat Transfer. Pergamon Press. 42 Churchill, S.W. and Ozeo, H. (1973). Correlations for laminar forced convection with uniform heating in flow over a plate and in developing and fully developed flow in a tube. ASME J. Heat Transfer 95 (1): 78–84. 43 Jiji, L.M. (2009). Heat Convection, 2e. Springer Verlag. 44 Reynolds, W.M., Kays, W.M., and Kline, S.C. (1958). Heat Transfer in the Turbulent Incompressible Boundary Layer II: Step Wall Temperature Distribution. NASA MEMO 12-2-58W. 45 Ameel, T.A. (1997). Average effects of forced convection over a flat plate with an unheated starting length. Int. Commun. Heat Mass Transfer 24 (8): 1113–1120. 46 Ghiaasiaan, S.M. (2018). Convective Heat and Mass Transfer, 2e. CRC Press. 47 Scesa, S. (1949). Experimental investigation of convective heat transfer to air from a flat plate with a stepwise discontinuous surface temperature. M.S. Thesis. University of California, Berkeley, CA, USA. 48 Klein, J. and Tribus, M. (August 1952), Forced convection from non-isothermal surfaces, Eng. Res. Inst. University of Michigan. 49 Maisel, D.S. and Sherwood, T.K. (1950). Evaporation of liquids to turbulent gas streams. Chem. Engr. Prog. 46: 131. 50 Reynolds, W.M., Kays, W.M., and Kline, S.C. (1958). Heat Transfer in the Turbulent Incompressible Boundary Layer III: Arbitrary Wall Temperature and Heat Flux. NASA MEMO 12-3-58W. 51 Spalding, D.B. (1958). Heat transfer from surfaces non-uniform temperature. J. Fluid Mech. 4 (1): 22–32. 52 Spalding, D.B. (1961). Heat transfer to a turbulent stream from a surface with a stepwise discontinuity in wall temperature. International Developments in Heat transfer (Proceedings of the 1961-1962 Heat Transfer Conference, Boulder, CO), Part II, 439–446. 53 Spalding, D.B. (1964). Contribution to theory of heat transfer across turbulent boundary layer. Int. J. Heat Mass Transfer 7: 743–161. 54 Rubesin, M.W. (1952). The effect of an arbitrary surface temperature variation along a flat plate on convective heat transfer in an incompressible turbulent boundary layer. NACA TN 2345.

References

55 Biot, M.A. (1962). Fundamentals of boundary-layer heat transfer with streamwise temperature variations. J. Aerosp. Sci. 29 (5): 558–567. 56 Tribus, M. and Klein, J. (1955). Forced convection through a laminar boundary layer over an arbitrary surface with an arbitrary temperature variation. J. Aeronaut. Sci. 22: 62–64. 57 Smith, A.G. and Shah, V.L. (1962). Study of heat and mass transfer from non-isothermal surfaces. The College of Aeronautics, Cranfield, Contract Af 61(052)-267. 58 Cengel, Y.A. and Ghajar, A.J. (2015). Heat and Mass Transfer, 5e. McGraw-Hill. 59 Thomas, L.C. and Al-Sharif, M.M. (1981). An integral analysis for heat transfer in turbulent incompressible boundary flow. J. Heat Transfer 103: 772–777. 60 Hanna, O.T. (1962). Step-wall heat flux superposition for heat transfer in boundary-layer flows. Chem. Eng. Sci. 17: 1041–1051. 61 Hanna, O.T. and Myers, J.F. (1962). Heat transfer in boundary layer flows past a flat plate with a step wall-heat flux. Chem. Eng. Sci. 17: 1053–1055. 62 Shu, J.J. and Pop, I. (1998). On thermal boundary layers on a flat plate subjected to a variable heat flux. Int. J. Heat Fluid Flow 19: 79–84. 63 Antonia, R.A., Danh, H.Q., and Prabhu, A. (1977). Response of turbulent boundary layer to a step change in surface heat flux. J. Fluid Mech. 80 (Part 1): 153–177. 64 Smith, A.G. and Shah, V.L. (2012). Heat transfer in incompressible boundary layer on a flat plate with arbitrary heat flux. J. Aerosp. Sci. 28 (9), https://doi.org/10.2514/8.9163. 65 Taylor, R.P., Love, P.H., Coleman, H.W., and Hosni, M.H. (1990). Step heat flux effects on turbulent boundary-layer heat transfer. J. Thermophys. Heat Transfer 4 (1), 121–123. 66 Chen, K. (1996). An alternative procedure for evaluating the surface heat flux of a turbulent boundary layer with an unheated starting length. Int. Commun. Heat Mass Transfer 23 (2): 163–168. 67 Reichardt, H. (1951). Vollstandige Darstellung der turbulenten Geschwindigkeitsverteulung in glatten Leitungen. ZAMM J. Appl. Math. Mech. 31 (7): 208–219. 68 Levich, V.G. (1962). Physicochemical Hydrodynamics, translated by Scripta technical Inc. Prentice Hall. 69 Ambrok, G.S. (1957). Approximate solution of equations for the thermal boundary layer with variations in the boundary layer structure. Soviet Phys. 2 (9): 1979–1986. 70 Reynolds, W.C., Kays, W.M., and Kline, S.J. (1960). A summary of experiments on turbulent heat transfer from a non-isothermal flat plate. J. Heat Transfer 82: 341–348. 71 Sparrow, E.M. and Lin, S.H. (1965). Boundary layers with prescribed heat flux-application to simultaneous convection and radiation. Int. J. Heat Mass Transfer 8: 437–448. 72 Lienhard, J.H. IV, and John H. Lienhard, V. (2020). A Heat Transfer Textbook, 5e. Cambridge, Massachusetts: Phlogiston Press. 73 Chanson, H. (2014). Applied Hydrodynamics, An Introduction, CRC Press.

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11 Turbulent Internal Flow: Momentum and Heat Transfer 11.1 Introduction A common problem encountered by engineers is the heat transfer to turbulent fluid flow in a tube. This may occur in heat exchangers, boilers, condensers, evaporators, etc. For this reason, it is useful to know how to estimate the heat transfer coefficient. We will study fully developed turbulent momentum and heat transfer for internal flows in this chapter. Many engineering devices have elements made up of uniform cross section. Flows in uniform channels are the simplest turbulent motions. After a development period, the mean flow and turbulence remain the same at successive sections; then, the flow is said to be fully developed. Experimental studies for friction factor and heat transfer coefficient indicate that momentum transfer and heat transfer are fairly insensitive to the geometry for fully turbulent flow in internal flow systems having uniform cross section. For this reason, the analysis of forced convection heat transfer for fully developed turbulent flow will be presented in the context of parallel plate channel and circular pipe systems. The experimental measurements of turbulent flows in smooth pipes indicate that the near wall velocity profiles of flat plates can be used in internal flows. In other words, the turbulence structure of turbulent flow in the inner region (region very close to tube wall) of the tube wall is essentially identical to the turbulent flow along a flat plate. This means that velocity profile u+ (y+ ), mixing length 𝓁 + , and eddy viscosity variation εM can be used in both the turbulent flow in a tube and the turbulent flow over a flat plate. For these reasons, we will borrow ideas from the external turbulent flow over a flat plate whenever needed.

11.2 Momentum Transfer We are interested in fully developed turbulent flow in a pipe or parallel plates having uniform cross section. These are of great importance in engineering applications such as heat exchangers. Flows in uniform channels are important in another way: in some ways, they are the simplest turbulent flows. Conservation laws governing the fluid motion are greatly simplified. Following a period of development, the flow is said to be fully developed. Engineers often ask questions about the flowing fluids: What is the pressure drop? What is the pumping power? What is the flow rate? These questions need to be answered. The turbulent flow in a parallel plate channel is not much different than turbulent flow over a flat plate, and for this reason, whenever needed, the material of external turbulent flow will be utilized. The close relationship between theory and experimental results is also presented whenever possible.

11.2.1 Momentum Transfer in Infinite Two Parallel Plates Before we proceed, we will present some experimental data for the turbulent flow in a two- dimensional channel. Laufer [1] presented a detailed exploration of mean velocity for a two-dimensional turbulent channel flow. The measurements were reported at three different Reynolds numbers, 12 300, 30 800, and 61 600, based on the half-width of the channel and the maximum mean velocity. Data provided by Laufer are plotted in Figure 11.1. The axes are normalized by plotting dimensionless velocity u∕Vc versus the dimensionless position in the channel y/H, where u is the time smoothed mean velocity. Included are the two velocity profiles for two different Reynolds numbers. Vc is the velocity at the channel centerline. The velocity profile for turbulent flow is quite different from the laminar flow. The profiles in turbulent flow are much steeper in the wall region and are flatter in the core region.

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

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11 Turbulent Internal Flow: Momentum and Heat Transfer

Figure 11.1

1

Mean velocity distribution. Source: Laufer [1].

0.8 Re = 61300 0.6 u Vc

Re = 12300 0.4 Laufer’s data 0.2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 y/H

y

Fully developed region

LH

V u(y) 0 Figure 11.2

H

x

Turbulent flows between infinite parallel plates.

Turbulent flow in a channel has an inner region adjacent to the wall. The turbulence structure near the channel wall is very similar and is essentially identical to that for turbulent flow along a flat plate. The same inner region velocity profile u+ , mixing length 𝓁, and eddy viscosity variation εM can be used. Deviations from the logarithmic velocity profile becomes discernible for (y/H) > 0.15; however, the deviations are observed to be small. For most engineering problems, the use of the logarithmic velocity profile in the turbulent core is adequate, even though a nonzero velocity gradient is observed at the center of the channel. Consider pressure-driven incompressible turbulent flow between smooth two infinite parallel plates separated by a distance H, as illustrated in Figure 11.2. The coordinate system is shown in Figure 11.2, and origin is located on the lower plate. The x-axis is aligned with the flow direction and the y-axis is chosen in the cross-stream direction perpendicular to the plates. The flow first enters at uniform velocity, and after some distance, it becomes fully developed. Both the plates are fixed. We assume that the mean flow is steady. The plates are assumed to be infinitely long and wide, and they are at rest with respect to the coordinate system. The extent of the channel in the z-direction is very large compared to width H so that the flow is independent of z. The mean velocity w = 0. The plates are located at y = 0 and y = H. 11.2.1.1 The Entrance Region

First, we will look at the entrance region. The entrance length in turbulent flows is very short, and the hydrodynamic entrance length LH for turbulent flows is given by different correlations. As a rule of thumb, the hydrodynamic entrance length is approximately LH ≈ 10 DH

(11.1)

where DH is the hydraulic diameter and is defined as DH = 4A/P; A is the cross-sectional area and P is the wetted perimeter.

11.2 Momentum Transfer

Anselmet et al. [2] recommends the following correlation: 1∕4

LH ≈ 1.6 DH ReD , H

ReDH ≤ 107 .

(11.2)

Equation (11.2) is mainly based on computational fluid dynamics (CFD) results. It is experimentally observed that the turbulent flow becomes hydrodynamically fully developed (HFD) after a relatively short distance from the tube entrance. 11.2.1.2 The HFD Region

We will consider HFD, steady, two-dimensional turbulent flow of incompressible fluid between two infinite parallel smooth plates, as shown in Figure 11.2. The fluid properties are constant. At present, there exists no theoretical treatment of turbulent flow, and for this reason, we rely on some experimental information to solve turbulent flow problems. The experimental studies of the friction factor in fully turbulent flows indicate that momentum transfer is insensitive to the geometry in internal flows with a uniform cross-sectional area. We are interested in the Moody friction factor f, the Fanning friction cf , and velocity distribution u(y). This information will be used to develop solutions for the mean temperature distribution and the Nusselt number for fully developed turbulent flow between infinite parallel plates. This problem helps greatly to our understanding of turbulent flow. The two-dimensional form of conservation of mass and momentum equations is Continuity 𝜕u 𝜕v + =0 𝜕x 𝜕y x-momentum ( ) [ ] [ ] 𝜕p 𝜕 𝜕 𝜕u 𝜕u 𝜕u 𝜕u ρ u +v =− + (μ + μt ) + (μ + μt ) 𝜕x 𝜕y 𝜕x 𝜕x 𝜕x 𝜕y 𝜕y y-momentum ( ) [ ] [ ] 𝜕p 𝜕 𝜕 𝜕v 𝜕v 𝜕v 𝜕v ρ u +v =− + (μ + μt ) + (μ + μt ) . 𝜕x 𝜕y 𝜕y 𝜕x 𝜕x 𝜕y 𝜕y

(11.3)

(11.4)

(11.5)

We will make the following assumptions: (1) (2) (3) (4) (5) (6) (7)

Incompressible constant property fluid flow. 𝜕 Fully developed flow, 𝜕x = 0. Parallel flow, v = 0. Two- dimensional flow, no motion in the z-direction, w = 0. Neglect body forces. The fluid flow pressure gradient is a constant. Neglect the pressure variation across the channel.

Mean flow is steady. For HFD flow, the axial velocity component does not change with axial position. This means that 𝜕u = 0. 𝜕x We now introduce Eq. (11.6) into the continuity equation, Eq. (11.3), to obtain 𝜕v = 0. 𝜕y

(11.6)

(11.7)

The integration of this equation, Eq. (11.7), yields v = C.

(11.8)

Since the plate is a solid wall, the no-slip boundary condition v = 0 yields that C = 0. Thus, The y-component of velocity v becomes v = 0.

(11.9)

This means that v = 0 everywhere in the flow, and thus, u = u(y). For HFD steady turbulent flow between two infinite parallel plates, the momentum equation reduces to [ ] dp d du + (μ + μt ) = 0. (11.10) − dx dy dy

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11 Turbulent Internal Flow: Momentum and Heat Transfer

The boundary conditions are: u = 0 at y = 0 (No slip boundary condition at the wall)

(11.11a)

du = 0 at y = H∕2 (Symmetry at the centerline at y = H∕2.) dy

(11.11b)

In Eq. (11.10), ordinary differentials are used since (dp∕dx) is constant and the only variation is with y. Notice that u′ v′ = 0. Unfortunately, there are two unknowns in the one equation, Eq. (11.10), i.e. u and μt . Differential formulation will be completed by specifying an expression for eddy viscosity (turbulence parameter) μt . The turbulence parameter μt is specified by the use of theoretical and/or empirical inputs. The eddy viscosity μt are evaluated on the basis of experimental measure′ ′ ments for u and v . Another term, eddy diffusivity or eddy kinematic viscosity εM = μt /ρ, is used in turbulence literature. The eddy viscosity μt is a characteristic of turbulence and varies across the flow field. In the above equation, the partial derivatives for axial velocity and turbulent stress have been replaced by ordinary derivatives since these are functions of y only. Note that the term in the square bracket is the total shear stress. The total shear stress is made up of the laminar and turbulent stresses. We neglect the pressure variation across the channel, and this is a good assumption for the present problem. Again, notice that the time-averaged pressure is only a function of x. The pressure gradient is assumed to be constant, i.e. it is independent of x: 𝜕p dp = = constant. 𝜕x dx The pressure gradient dp/dx drives the flow against the shear stresses at the channel walls. The total shear stress can be expressed as the sum of laminar and turbulent shear stress τ = (μ + μt )

du . dy

(11.12)

Then, we can express Eq. (11.10) as dτ dp Δp = = . dy dx L

(11.13)

We now integrate this equation, and this will give us (y) τ = Δp + C. L At y = H/2, the total shear stress τ = 0, and this gives us C = − (ΔP H/2 L). A force balance on control volume (CV) in the channel yields Δp 2 = − τw L H and finally, we get.

( τ = τl + τt = τw 1 −

(11.-13)

y H∕2

) (11.14)

where τl is the viscous shear stress and τt is the turbulent shear stress. Experimental studies of turbulent shear stress τt = −ρu′ v′

(11.15)

have been made by several investigators, and illustrative results are shown in Figure 11.3. y centerline Viscous stress

Turbulent stress 0 Figure 11.3

Variation of total shear stress across a turbulent channel flow.

11.2 Momentum Transfer

Thus, we conclude that total shear stress distribution may be determined by the measurement of the pressure drop. Shear stress distribution is linear for turbulent flow between parallel plates, as shown in Figure 11.3. The turbulent shear stress contribution at the wall is negligible, and the total shear at the centerline (y = H/2) is zero for reasons of symmetry. Far from the wall, the turbulent stress is important, while very close to the wall, the viscous stress dominates, and at the wall, the stress is completely viscous. As discussed in [4], the viscous contribution drops from 100% at the wall (y+ = 0) to 50% at y+ ≈ 12 and is less than 10% at y+ = 50. The momentum equation can be expressed as [ ] dp du d (μ + μt ) = . (11.16) dy dy dx The integration of the x-component of the momentum equation yields (μ + μt )

du dp = y + C1 . dy dx

Using now the boundary condition, we get at y = H/2, τ = 0 or du∕dy = 0, and therefore, C1 = −(dp∕dx)(H∕2), and the momentum equation becomes ( ) H du dp (μ + μt ) = y− . (11.17) dy dx 2 The shear stress at the wall is determined by the pressure gradient. Let us now eliminate the pressure gradient in favor of wall shear stress τw . The wall shear stress τw is the stress exerted on the wall by fluid. This can be done by evaluating Eq. (11.17) at y = 0 or writing a force balance on the CV in the parallel plate channel (μ + μt )

du || du | = μ || = τw dy ||y=0 dy |y=0

where turbulent shear stress is nearly zero at the wall and shear stress is purely viscous. We now have ( ) ( ) τ dp dp H τw = − ⇒ =− w . dx 2 dx H∕2

(11.18)

(11.19)

The wall shear stress τw is obtained from pressure loss data and is a known quantity. The momentum equation takes the following form: ( )( ) τ H du = − w y− (11.20) (μ + μt ) dy H∕2 2 Eddy viscosity helps to develop a qualitative feel for the nature of the turbulent flow. We have a postulated relationship between shear stress and the rate of deformation. Experimental studies indicate that μ is constant for a large number of fluids. On the other hand, experiments show that turbulent viscosity μt is not constant and depends on position and on the intensity of turbulence. In other words, μt is the flow property. We need an empirical expression for turbulent viscosity μt . Since μ = ρν and μt = ρεM , the momentum equation becomes ( ) y du τw (ν + εM ) = 1− . (11.21) dy ρ H∕2 We can express the wall shear τw as √ τw τw 2 → uτ = uτ = ρ ρ

(11.22)

where uτ is the wall friction velocity. Using the friction velocity, we define the following dimensionless parameters: u uτ yu y+ = τ ν Hu τ H+ = . ν u+ =

(11.23) (11.24) (11.25)

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11 Turbulent Internal Flow: Momentum and Heat Transfer

Using these dimensionless parameters, the momentum equation becomes ) ( ( ε ) du+ y+ 1+ M = 1 − ν dy+ H+ ∕2 u+ (0) = 0

(11.26) (11.27)

or y+

[1 − y+ ∕(H+ ∕2)] + (11.28) ( ε ) dy ∫0 1+ M ν where y/H = y+ /H+ . Here, H+ = Huτ /ν is the friction Reynolds number based on the full channel height. We have now obtained the dimensionless form of the momentum equation for channel flow. The formulation is closed by specifying εM /ν, and we may integrate Eq. (11.26) or Eq. (11.28) numerically. How do we get εM /ν? We will rely on experimental data. We will simplify our computational task by making practical approximations. We will focus our attention within the inner region (y/(H/2) = y+ /(H+ /2) < 0.2), where the y/(H/2) term is small, and it may be neglected. Then, Eq. (11.25) becomes ( ε ) du+ =1 (11.29) 1+ M ν dy+ u+ =

u+ (0) = 0.

(11.30)

In the region very near the wall, viscous effects are dominant, and this region is called viscous sublayer. In spite of the fact that this region contains turbulent fluctuations, the turbulent stresses are small due to the presence of the wall. In this viscous sublayer region, at high Reynolds number, the viscous sublayer is very thin, and the total shear stress τ is essentially constant and is equal to wall shear stress τw . In other words, in a sufficiently thin flow region close to the wall, τ ≈ τw .

(11.31)

In the viscous sublayer region, εM ≪ ν and (εM /ν) ≪ 1, and (εM /ν) can be neglected. Then, Eq. (11.29) becomes du+ = 1. dy+

(11.32)

Thus, integrating Eq. (11.32) gives u + = y+ + C 1

(11.33)

where C1 = 0 since u+ (0) = 0. Thus, the velocity profile is u+ = y+ , 0 ≤ y+ ≤ 5.

(11.34)

Notice that Eq. (11.34) indicates that the mean velocity distribution near wall is linear in the viscous sublayer and is in good agreement with the available experimental data. As we move away from the wall, the viscous effects gradually begin to diminish, and in turbulent (intermediate) region, εM ≫ ν and (εM /ν) ≫ 1, and Eq. (11.29) becomes εM du+ = 1. ν dy+

(11.35)

We need an empirical expression for εM /ν. At this point, we rely on experimental data. 11.2.1.3 Prandtl’s Mixing-Length Model

Prandtl’s mixing-length theory was one of the first attempts to develop a semitheoretical analysis of turbulent flow. Prandtl postulated that ( ) du (11.36a) εM = 𝓁 2 dy where 𝓁 is the mixing length. The Prandtl’s mixing-length model for eddy diffusivity of momentum in terms of wall coordinates is εM du+ = (𝓁 + )2 + (11.36b) ν dy

11.2 Momentum Transfer

where 𝓁 + = κ y+ and the value of κ comes from the experiment. Using Eq. (11.36b), Eq. (11.35) becomes ( + )2 du = 1. κ 2 (y+ )2 dy+ Equation (11.37) can be written as ( +) 1 du = κ y+ dy+

(11.37)

(11.38)

u+ (0) = 0. Integrating Eq. (11.38) yields u+ =

1 ln y+ + C1 κ

(11.39)

where C1 is the integration constant. Based on experimental data, κ = 0.4 and C1 ≈ 5.5. Then, we have u+ = 2.5 ln y+ + 5.5.

(11.40)

This logarithmic law is in good agreement with experimental data. This equation is also in reasonable agreement with experimental data in the outer region and is called the law of the wall in the turbulence literature. Note that this equation does not represent the data in the viscous sublayer. At the center, the symmetry condition has to be satisfied by Eq. (11.40), i.e. du+ = 0 at y+ = H+ ∕2 dy+

(11.41)

but this is not the case. This situation is expected since we made the approximation that y+ /(H+ /2) − 1 ≈ 1. 11.2.1.4 Buffer Region

Prandtl’s mixing-length model does not work in the buffer layer. In this region, εM ≈ ν; the velocity profile is based on experimental studies in smooth channels, and it is given as u+ = 5 ln y+ − 3.05,

5 < y+ < 30.

(11.42)

Equations (11.34), (11.41), and (11.42) are appropriate for fully developed velocity profiles, i.e. reasonably far from the channel entrance effects. Equation (11.40) applies almost all the way to centerline of the channel. Note that Prandtl assumed that shear stress τ to be equal to wall shear stress τw throughout. But τ/τw is equal to 1 − y/(H/2), and however, Prandtl’s assumption is successful. These equations fairly adequately represent the velocity distribution over most of the flow cross section. Equation (11.40) does not satisfy the boundary condition at the channel centerline. The velocity profile is flat at the channel centerline. The velocity gradient du∕dy is not zero at the centerline. Another problem with these equations is that the derivative is discontinuous at the juncture point of y+ = 30, while the experimental data show a smooth continuous curve without a discontinuity at the juncture point of y+ = 30. The curve is continuous in its first derivative at y+ = 5. Equation (11.40) is quite good for velocity distribution near the centerline of channel flow. However, these equations are used extensively in engineering. We have applied the Prandtl mixing-length theory to the turbulent flow problem in a parallel plate channel. Time-averaged flow was steady, and gravity was neglected. The form chosen for the mixing length is purely intuitive but does satisfy the requirement that the turbulent shear stress become zero at the channel wall. Velocity distribution in turbulent flow in a smooth channel is important and is of great interest because it provides the necessary information for the detailed calculation of convective heat transfer. Velocity distribution is a segmental representation, and it produces discontinuities in (du∕dy) at the intersections, and this results in awkward expressions for heat transfer. Some researchers proposed models with continuous derivatives to represent the velocity distribution from the channel wall to the centerline, which is valid over the whole range of dimensionless distance y+ . Researchers developed different models εM /ν. In addition to eddy diffusivity models given in Table 10.2 for εM /ν, some more selected eddy diffusivity models for εM /ν are given in Table 11.1.

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11 Turbulent Internal Flow: Momentum and Heat Transfer

Table 11.1

Selected eddy diffusivity models and velocity profiles.

[

( + )] ⎧ y + + y+ ≤ 50 ⎪κ y − y1 tanh + y 1 ⎪ [ εM ⎪ ) ( + )2 ] ( = ⎨ κy+ 1 r+ r ν 1 + y+ > 50 + ⎪ 3 + + 2 R R ⎪ ⎪y+ = 11 ⎩ 1

Correlation of Reichardt is reported in [5]

for all y+

The velocity profile is given as [

( +) ( +) ] y y + exp(−0.33y ) u+ = 2.5 ln (1 + 0.4y+ ) + 7.8 1 − exp − + − y1 y+1 + y1 = 11 ( + ) εM y = sin2 h y+ ≤ 27.5 ν 14.5

Rannie [6]

The velocity profile ) as ( is+given y 0 ≤ y+ ≤ 27.5 u+ = 14.53 tanh 14.53 u+ = 2.5 ln y+ + 5.5 y+ > 27.5

Example 11.1 Water at a mean temperature of 330 K flows turbulently between two infinite smooth parallel plates, as shown in Figure 11.2. The spacing between the parallel plates is 1 cm. The mean velocity between the plates is 1 m/s. Flow is fully turbulent. Consider now the eddy viscosity developed by Reichardt ( + )] [ εM y . = κ y+ − 11 tanh ν 11 This function goes to zero for y+ → 0. For large values of y+ , it approaches to εM /ν = κy+ . Determine the dimensionless velocity u+ corresponding to y+ = 200 for a given H+ = 1063. Solution The physical properties of water at 330 K ρ = 984 kg∕m3

cp = 4184 J∕kgK μ = 489 × 10−6 N.s∕m2

k = 0.650W∕mK Pr = 3.15 ν = 4.968 × 10−7 m2 ∕s. The hydraulic diameter for the parallel plate channel is DH = 2H = 2 × 0.01 = 0.02 m ρVDH 984 × 1 × 0.02 = = 40 245. μ 489 × 10−6 Flow is turbulent. We will integrate Eq. (11.26) using Maple 2020. ReDH =

> restart; > #Numerical solution of Eq.(11.26) > > #𝜿 is Karman constant > 𝛋 ≔ 𝟎.𝟒; 𝛋 ≔ 𝟎.𝟒 > #For simplicity > #H represents H+ > #y represents y+

11.2 Momentum Transfer

> #u represents u+ > > > H ≔ 𝟏𝟎𝟔𝟑; H ≔ 𝟏𝟎𝟔𝟑 > > de ≔ diff(u(y), y) = de ≔

𝟏−

y H

( )) ; ( 𝟏 + 𝛋 • y − 𝟓 • tanh y𝟓

y 𝟏 − 𝟏𝟎𝟔𝟑 d u(y) = ( ) dy 𝟏 + 𝟎.𝟒 y − 𝟐.𝟎 tanh y𝟓

> soln ≔ dsolve({de, u(𝟎) = 𝟎}, u(y), type = numeric); soln ≔ proc(x rkf45) … end proc > soln (𝟐𝟎𝟎); [y = 𝟐𝟎𝟎., u(y) = 𝟏𝟒.𝟓𝟏𝟎𝟏𝟒𝟗𝟏𝟕𝟕𝟏𝟔𝟎𝟒] The solution gives us that for y+ = 200 u+ = 14.5101. Example 11.2 If we know the mixing length 𝓁, we can find εM . Develop an expression for εM /ν within the inner region for HFD turbulent flow between parallel plates using the van Driest model [7]. Solution For a boundary layer flow over a plate, shear stress is constant and is equal to wall shear stress τw . The momentum equation becomes du τw = ρ(εM + ν) . dy We can rearrange this equation using u2τ = τw ∕ρ ( ) | | du 2 | du | du 2 = ν+𝓁 | | . uτ = (ν + εM ) dy | dy | dy We can now put this equation in dimensionless form [ ⌊ + ⌋] ( + ) du du 1 + (𝓁 + )2 =1 dy+ dy+ or

(

)2 ( + ) du+ du − 1 = 0. + dy+ dy+ We can now put this equation in the following form: ( + )2 ( + ) dy dy − (𝓁 + )2 = 0. − du+ du+ The solution of this quadratic equation is ( +) √ dy 1 1 + = 1 + 4(𝓁 + )2 . 2 2 du+ We know that du = τw ρ(εM + ν) dy (𝓁 + )2

581

582

11 Turbulent Internal Flow: Momentum and Heat Transfer

and we can write this equation as (εM + ν) dy+ = . ν du+ We can now eliminate dy+ /du+ to get √ εM 1 1 =− + 1 + 4(𝓁 + )2 . ν 2 2 We now may use the Van Driest model [7] given as [ ( + )] y 𝓁 + = κ y+ 1 − exp − + A and this will give us ( ε ) −1 + {1 + 4(κy+ )2 [1 − exp(−y+ ∕A+ )]2 }1∕2 M = , ν 2

A+ = 26 κ = 0.4.

11.2.1.5 The Mean Velocity

The bulk or mean velocity V for incompressible constant property turbulent flow is H

V=

1 1 ρ u(y)dAc = u(y)dy. ρAc ∫ H ∫0

(11.43)

Ac

This equation is put into dimensionless form using u+ = u∕uτ and y+ = yuτ /ν V= (

V+ =

uτ Huτ ν

H+

)

1 H+ ∫0

∫0

u+ dy+ =

uτ

H+ ∫0

H+

u+ dy+

H+

u+ dy+

(11.44a)

or H+ ∕2

2 u+ dy+ H+ ∫0 where H+ = H uν /ν, and we evaluate DH as 2H 4A 4 (H × W) = = ≈ 2H as (H∕W) → ∞ DH = P 2(H + W) (1 + H∕W) where DH is the hydraulic diameter. V+ =

(11.44b)

(11.45)

11.2.1.6 Skin Friction Coefficient or Fanning Friction Factor cf

We can express the skin friction coefficient cf in terms of friction velocity uτ and mean velocity V as given below cf = or

√

2 τw ρV2

=

2 ρ u2τ ρV2

=

2u2τ V2

√ τw cf =V ρ 2

or V+ =

V = uτ

√

(11.46a)

2 . cf

(11.46b)

Using the Fanning friction factor cf , Eqs. (11.44a) and (11.44b) become √ H+ 1 2 = + u+ dy+ cf H ∫0 or

√

2 2 = + cf H ∫0

(11.47a)

H+ ∕2

u+ dy+ .

(11.47b)

11.2 Momentum Transfer

In particular, we should note the fact that the condition for the velocity gradient is zero at the channel centerline because symmetry is not satisfied by the law of the wall. We can develop a simple explicit equation for the Fanning friction factor cf using the simple one-seventh power law developed by Prandtl throughout the entire flow field. The one-seventh power law is u+ = 8.75(y+ )1∕7 .

(11.48)

Equation (11.48) will be discussed later. This equation does not represent the data in the sublayer and the buffer region, but the volume flow rates through these regions are small compared to the volume flow rate in the turbulent region, and this introduces small error. Considering half of the channel and substituting the one-seventh power law into the mean velocity equation, we obtain √ H+ ∕2 ( ) ( + )1∕7 2 7 2 H = + 8.75 (y+ )1∕7 dy+ = (8.75) . (11.49) cf 8 2 H ∫0 Here, H+ is given by

√ √ √ √ ReDH cf uτ H H cf 1 (2H)V cf 1 DH V cf = V = = = H = ν ν 2 2 ν 2 2 ν 2 2 2 +

where ReDH = VDH ∕ν and DH = 2H. Hence, the equation for cf becomes ( ) √ ( ) ReD √ c 1∕7 2 7 H f = (8.75) . cf 8 4 2

(11.50)

Using Maple 2016, we can solve this equation for cf , and the following explicit relation for cf is given: cf =

0.0802

(11.51)

1∕4

ReD

H

where Re = DH V/ν and DH = 2H is the hydraulic diameter. This equation is very close to the empirical equation of Blasius reported in [8] cf =

0.0791 1∕4

ReD

,

5 × 103 ≤ ReDH ≤ 105

(11.52)

H

where ReDH = DH V∕ν and Blasius obtained this equation by curve fitting the experimental data for HFD turbulent flow in a pipe. Based on available data, it appears that the mixing-length model provides quite satisfactory results for fully developed flow in a parallel plate channel and pipe. Suppose that we wish to find the Fanning friction factor cf for larger values of the Reynolds number. For this purpose, we can use the logarithmic correlation for u+ . For flow between parallel plates, cf is given as √ H+ ∕2 2 2 = + u+ dy+ . (11.53) cf H ∫0 The use of the logarithmic law of the wall for u+ in Eq. (11.53) gives √ H+ ∕2 ( ) 2 1 2 = + C + ln y+ dy+ cf κ H ∫0 ( +) ] ]H+ ∕2 [ [ 1 2 1 H 1 ln +C− . = + Cy+ + (y+ ln y+ − y+ ) ≃ κ κ 2 κ 0 H Re √ Substituting H+ = 2DH cf ∕2 into Eq. (11.54) yields ( ] [ √ ) √ ReDH cf 1 1 2 ln +C− . = cf κ 4 2 κ Setting κ = 0.41 and C = 5, we obtain ( √ ) √ cf 2 − 0.82. = 2.44 ln ReDH cf 2

(11.54)

(11.55)

(11.56)

583

584

11 Turbulent Internal Flow: Momentum and Heat Transfer

Dean [9] proposed the following empirical correlation for fully developed turbulent flow between parallel plates. The local skin friction coefficient cf is cf =

0.073 Re1∕4

(11.57)

6 × 103 ≤ Re ≤ 6 × 105 where Re = (V H)/ν and H is the full channel height. Zanoun and Durst [10] proposed the following experimental correlation for local skin friction coefficient cf for fully turbulent flow 0.058 (11.58) cf = Re0.243 2 × 104 ≤ Re ≤ 1.2 × 105 where Re = (V H/2)/ν and H is the full channel height. Example 11.3 Water at a mean temperature of 330 K flows turbulently between two infinite smooth parallel plates, as shown in Figure 11.2. The spacing between the parallel plates is 2 cm. The mean velocity between the plates is 1 m/s. Flow is fully turbulent. (a) Estimate the wall shear stress τw . (b) Estimate the friction velocity uτ . (c) Estimate H+ . Solution The physical properties of water at 330 K ρ = 984 kg∕m3

cp = 4184 J∕kgK μ = 489 × 10−6 N.s∕m2

k = 0.650W∕mK Pr = 3.15 ν = 4.968 × 10−7 m2 ∕s. The hydraulic diameter for the parallel plate channel is DH = 2H = 2 × 0.02 = 0.04 m ReDH =

ρVDH 984 × 1 × 0.04 = = 80 490. μ 489 × 10−6

Flow is turbulent. (a) Wall shear stress from Eq. (11.46a) is τw =

1 c ρV2 . 2 f

We need the skin friction coefficient cf . We will use the experimental equation of Blasius, Eq. (11.52), cf =

0.0791 1∕4 ReD H

ReDH =

VDH ν

0.079 = 0.00468 804901∕4 1 N τw = (0.00468)(984)(1)2 = 2.307 2 . 2 m (b) The friction velocity uτ is cf =

√ uτ =

τw = ρ

√

m 2.307 = 0.0484 . 984 s

11.2 Momentum Transfer

r y

v

τ

V O

u

x

R

u(r)

LH

Figure 11.4

Turbulent flow in a circular pipe

(c) The estimation of H+ is ( ) Huτ (2H)V 1 uτ (2H)V 1 uτ = = ReDH ReDH = ν ν 2V 2 V ν (u ) ( ) 1 1 0.0484 τ ReDH = (80490) = 1948. H+ = 2 V 2 1 H+ =

11.2.2 Momentum Transfer in Circular Pipe Flow Consider the turbulent flow of incompressible fluid in a smooth circular straight pipe of radius R. Flow enters the tube at uniform velocity V. The cylindrical coordinate system is shown in Figure 11.4. We assume that the mean flow is steady and axially symmetric. Let x be the direction of flow along the axis of the flow and let r denote the radial direction measured from the x-axis of the pipe. For internal flow in tubes, the flow is generally classified as (a) transitional turbulent flow for 2000 ≤ Re < 104 (b) fully turbulent flow for ReD ≥ 104 where ReD = ρVD/μ is the Reynolds number based on the pipe diameter. The three-layer turbulent flow model presented before also describes the flow in a pipe. Reynolds numbers are very high, and for this reason, the flow inside the pipe has the characteristics of boundary layer flow, and there are significant gradients near the wall. Flow develops very quickly and becomes fully developed. 11.2.2.1 Entrance Region

First, we will look at the entrance region. The entrance length in turbulent flows is very short, and the hydrodynamic entrance length LH for turbulent flows is given by different correlations. As a rule of thumb, the hydrodynamic entrance length may be approximately found as LH ≈ 10 DH

(11.59)

where DH is the hydraulic diameter. Anselmet et al. [2] recommends the following correlation: 1∕4

LH ≈ 1.6 DH ReD , H

ReDH ≤ 107 .

(11.60)

Equation (11.2) is mainly based on CFD results. It is experimentally observed that the turbulent flow becomes HFD after a relatively short distance from the tube entrance. The entrance length [2] for turbulent flow in a channel may also be estimated by LH 1∕6 = 4.4ReD . H DH

(11.61)

Significant discrepancies have been observed in literature concerning the entrance length and these two relations give close results in range of Reynolds numbers, 105 ≤ ReD ≤ 106 .

585

586

11 Turbulent Internal Flow: Momentum and Heat Transfer

11.2.2.2 The HFD Region

In most engineering problems involving turbulence, the hydrodynamic entrance length is not very important, and we will consider fully developed turbulent flow. The growing boundary layers in the entrance region meet at the centerline of the tube, as shown in Figure 11.4. Very often, the turbulent flow in a circular tube is encountered in engineering problems. Velocity profiles have been determined experimentally by several researchers. A few of the researchers will be listed. The work of Nikuradse [11], Deissler [12], Rothfus and Monrad [13], and Laufer [14, 15] provides experimental data. In this chapter, Laufer’s data are plotted, as shown in Figure 11.5, where u∕Vc (the ratio of the point velocity at a position y from the wall to maximum velocity) as a function of y/R. The dimensionless velocity profiles at two different values of the Reynolds number are shown in Figure 11.5. The maximum or centerline velocity Vc occurs at the axis of the tube. Figure 11.5 indicates that the velocity profile for a turbulent flow is different from the velocity profile for laminar flow in a tube. It can be seen from the figure that the profiles get flatter in the core region with increasing Reynolds number. We are interested in the velocity profile near the wall. Turbulent flow in a pipe has an inner region adjacent to the pipe wall. The turbulence structure is very similar and is essentially identical to the turbulent flow along a flat plate. The same inner region velocity profile u+ , mixing length 𝓁, and eddy viscosity variation εM can be used. Deviations from the logarithmic velocity profile becomes discernible for y/R ≥ 0.15 − 0.2; however, the deviations are observed to be small. For most engineering problems, the use of the logarithmic velocity profile in the turbulent core is adequate, even though a nonzero velocity gradient is observed at the center of the pipe. Consider now the fully developed turbulent flow of incompressible fluid in a straight smooth circular pipe having a radius R, as shown in Figure 11.4. We assume that the mean flow is steady and axially symmetric. We consider a two-dimensional form of the conservation of mass and momentum equations Continuity 𝜕u 1 𝜕 (r v) + =0 r 𝜕r 𝜕x x-component ( ) [ ] 1 𝜕p 1 𝜕 𝜕u 𝜕u 𝜕u +v =− + r(ν + εM ) u 𝜕x 𝜕r ρ𝜕x r𝜕r 𝜕r

(11.62)

(11.63)

where v is the radial velocity and u is the axial velocity. Since flow is assumed to be fully developed with respect to axial direction, 𝜕 u/𝜕 x = 0. Thus, continuity becomes 𝜕 (rv) = 0. 𝜕r

(11.64)

Figure 11.5

1.0

0.9

0.8

Re = 50 000

u 0.7 Vc

Re = 500000

0.6 Laufer’s data for pipe flow 0.5

0.4

0.1

0.2

0.3

0.4

0.5 y/R

0.6

0.7

0.8

0.9

Mean velocity as function of r/R. Source: Laufer [14].

11.2 Momentum Transfer

The integration of this equation yields v=

C r

(11.65)

and no-slip boundary condition yields that C = 0. Thus, the radial component of velocity becomes v = 0.

(11.66)

We conclude that the mean flow has velocity components along the x-axis only and no motion in the r-direction. Under these assumptions, the axial velocity component becomes u = u(r)

(11.67)

and u is a function of r alone. The axial pressure gradient, becomes an ordinary differential equation: [ ] 1 dp du 1 d r(ν + εM ) = r dr dr ρ dx

𝜕p 𝜕x

=

dp , dx

is constant for HFD flow. The momentum equation

(11.68)

and its boundary conditions are at r = R,

u=0

(11.69a)

at r = 0,

du = 0. dr

(11.69b)

Consider the following control volume depicted in Figure 11.6. The control volume has a diameter D = 2R and a length Δx. We wish to express the pressure gradient in terms of shear stress on the wall by taking a macroscopic force balance on the control volume p(x) π R2 = p(x + Δx) π R2 + 2π RΔx τw τw = −

R p(x + Δx) − p(x) . 2 Δx

If Δx → 0, then this equation can be written as τw = −

R dp 2 dx

(11.70)

and shearing stress τw at the pipe wall can also be written in the form ) ( Δp ( R ) τw = L 2

(11.71)

where Δp = p1 − p2 is the pressure difference. Since the value of the right-hand side can be determined directly by measuring the pressure gradient along the pipe, the wall shearing stress is experimentally determined, and now, it is a known quantity. We can also write by a force balance over a control volume at any r position similar to one we discussed τ=−

R dp . 2 dx

r

(11.72)

τw p(x) p (x + Δx) x Δx

Figure 11.6

τw

Control volume for force balance in a pipe.

2R

587

588

11 Turbulent Internal Flow: Momentum and Heat Transfer

Combining these two relations, Eq. (11.70) and Eq. (11.72), we obtain ( ) r . (11.73) τ = τw R Equation (11.73) tells us that the shear stress is a linear function of radius r. We will now change the coordinate system and let y = R − r. The distance y is measured from the wall. Since y + r = R, Eq. (11.73) can also be expressed as ( y) . (11.74) τ = τw 1 − R Combining Eqs. (11.68) and (11.70), we get [ ] 2 r τw du d (ν + εM )r =− . (11.75) dr dr ρR We can now introduce the friction velocity uτ √ τw uτ = ρ and using the friction velocity uτ in Eq. (11.75), we get [ ] ( ) du d r r(ν + εM ) = −2 u2τ . dr dr R We will integrate Eq. (11.76) with respect to r and simplify the result of integration. This will give us [ ] ( ) C du r (ν + εM ) = −u2τ + dr R r

(11.76)

(11.77)

We use the boundary condition, Eq. (11.69b). Thus, we choose C = 0. Now, the momentum equation becomes after rearrangement u2τ r du =− . dr (ν + εM ) R

(11.78)

The boundary condition for this equation is Eq. (11.69a). Experimental studies indicate that the characteristics of the flow near the wall of a pipe are not influenced greatly by the curvature of the radius of the pipe wall. The universal velocity profile in a pipe is very similar to that of flow over a flat plate. Therefore, it is reasonable start to modeling pipe flow like the flow over a flat plate. It is more convenient to define y as the distance measured from the wall so that y = R − r. The momentum equation takes the following form: ( u2τ y) du = 1− . (11.79) dy (ν + εM ) R The boundary condition now is given by y = 0, u = 0.

(11.80)

Let us introduce the following dimensionless quantities using the friction velocity uτ : u+ =

u uτ

√ √ y τw ∕ρ (r − R) τw ∕ρ y uτ = = y = ν ν ν +

or

√ ( y ) Re √ c y uτ y R uτ ( y ) + ( y ) ReD f D f = = R = = y = ν R ν R R 2 8 R 2 2 +

(11.81) (11.82a)

(11.82b)

where ReD = ρ V D/μ and cf = 2τw /ρV2 is the skin fiction coefficient. The friction velocity uτ is constant for a given set of flow conditions. Using the wall variables u+ , y+ , and R+ , we arrive at the following dimensionless velocity profile: ) ( y+ 1 − R+ du+ (11.83) = ( + ε ) dy 1+ M ν

11.2 Momentum Transfer

y+ = 0, u+ = 0

(11.84)

where R+ = uτ R/ν. This differential equation is solved subject to the no-slip boundary condition at the pipe wall. We may write Eq. (11.84) as y+

(1 − y+ ∕R+ ) dy+ (11.85) ( ε ) . ∫0 1+ M ν If we have a model for εM /ν, we can find the velocity profile in the tube. We need an expression for εM /ν to solve this ordinary differential equation, Eq. (11.83). In engineering, we are interested in Moody and Fanning friction factor f. Let us first introduce some definitions. +

u =

11.2.2.3 Average Velocity V

Average velocity V is defined as R

R

1 2 u (2 π r)dr = 2 u r dr 2∫ πR 0 R ∫0

V=

(11.86)

or in terms of dimensionless variables V+ =

2 (R+ )2 ∫0

R+

u+ r+ dr+ =

2 (R+ )2 ∫0

R+

u+ (R+ − y+ )dy+

(11.87)

where V+ is the dimensionless mean velocity, and it is defined as V+ =

V . uτ

(11.88)

11.2.2.4 Skin Friction Factor cf

The skin friction coefficient cf is defined as cf =

2τw ρV2

.

(11.89)

We will now find a relation between V+ and cf . From the definition cf , we have τw 1 = cf V2 . ρ 2 Since τw = ρu2τ , we can write ( )2 2 1 1 V ⇒ = (V+ )2 . u2τ = cf (V)2 ⇒ 1 = cf 2 2 uτ cf From this equation, the relation between V+ and cf is √ 2 V+ = . cf

(11.90)

(11.91)

(11.92)

11.2.2.5 Moody Friction Factor f

Next, let us find the relation between f and V+ . The Darcy friction factor f is defined as f=

−(dp∕dx)D ρV2 ∕2

.

The relation between (−dp∕dx) and τw was found to be ( ) 2τ dp − = w. dx R Thus, by combining Eqs. (11.93) and (11.94), we can write ) 2τ ( 1 2 ρV = w D f 2 R

(11.93)

(11.94)

589

590

11 Turbulent Internal Flow: Momentum and Heat Transfer

or ( ) (f) u2τ

(

V uτ

)2 =

4τw (2R) = 8 R

(

τw ρ

) = 8u2τ .

Thus, from Eq. (11.95), we write ( ) √ 8 V . = V+ = uτ f

(11.95)

(11.96)

The relation among f, V+ , and cf is u2 cf f 1 = = τ2 = + 2 . 2 8 V (V ) The Reynolds number is ( )( ) ( )( ) ρRuτ Ruτ ρV(2R) V V =2 =2 = 2V+ R+ ReD = μ uτ μ uτ ν

(11.97)

(11.98)

where ReD = ρVD/μ is the Reynolds number. The friction factor f, the dimensionless distance R+ , and the Reynolds number ReD can be expressed as √ √ Ruτ ReD 1 (2R) V f f + = = . (11.99) R = ν 2 ν 8 2 8 For a given εM /ν, we can choose a value for R+ and numerically evaluate Eq. (11.83) to get u+ . Using this velocity profile in Eq. (11.87), we can obtain V+ . Knowing V+ now, we can obtain the friction factor f or cf and the Reynolds number ReD . We can repeat the calculation until we obtain a graph of f versus ReD . However, εM /ν is based on experimentally determined velocity profiles for tube flow. For this reason, we can use a selected velocity profile to use in Eq. (11.87) to get average velocity. 11.2.2.6 Prandtl Mixing-Length Model

Eddy diffusivity εM is discussed in Chapter 9 and can be written in terms of wall coordinates ( +) du + 2 εM = (𝓁 ) dy+

(11.100)

where 𝓁 + is the mixing length and Prandtl proposed that 𝓁 + is related to the distance from the wall 𝓁 + = κ y+ where y+ represents the normal dimensionless distance from a solid surface. Launder and Spalding [73] and Weigand [74] report that Prandtl’s mixing-length model can be used successfully to model unidirectional flows such as flow in a straight pipe and parallel plate channel. Weigand reports that the comparison between predicted and measured velocity profiles of fully developed flow in a planar parallel plate channel and pipe is quite satisfactory. It is reported that the comparison of velocities predicted by this model with experimentally determined velocities yields a value of κ = 0.4. The mean velocity profiles in pipes are similar to those external boundary layers, differing slightly in their details. However, we should be careful using the mixing-length model in other flow situations since the mixing-length model has some shortcomings: (1) Specific expression for mixing length may not be available for the specific turbulence problem of interest. (2) The mixing-length model implies zero turbulent viscosity at a position where the velocity gradient is zero. This means no momentum and energy transport due to eddies at the pipe centerline, and this is not true. (3) The mixing-length model does not give satisfactory results for complex turbulent flows. Experimental studies show that for practical purposes, as von Karman recommended [16], the inner region of a fully turbulent pipe flow field can be broken into the following regions: Laminar sublayer 0 < y+ < 5 Buffer layer 5 < y+ < 30 y Turbulent layer y+ > 30, δ ≤ 0.2.

11.2 Momentum Transfer

In the region near the wall, in the inner region, where y+ /R+ ≪ 1, the total shear stress is essentially constant and equal to wall shear stress τw , that is, τ ≈ τw . Then, the momentum equation becomes 1 du+ ≈ ( ε ). dy+ 1+ M ν

(11.101)

11.2.2.7 Laminar Sublayer

In the laminar sublayer region, εM ≪ ν, turbulent shear stress is negligible and as y+ → 0, du+ ≈ 1. dy+

εM ν

≪ 1. Thus, (11.102)

We have the following relation after integration: u+ = y+ + C since at y+ = 0 u+ = 0; thus, C = 0. This is just the no-slip boundary condition. u+ = y+ ,

0 < y+ < 5.

(11.103)

The thickness of the sublayer δv can be estimated by the following relation: δv =

5D 5ν 5ν = = √ √ uτ V f∕8 ReD f∕8

(11.104)

where f is the Moody (Darcy) friction factor to be discussed later. 11.2.2.8 Buffer Region

Prandtl’s mixing-length model does not work in the buffer layer. In this region, εM ≈ ν; the velocity profile is based on experimental studies in smooth tubes, and it is given as 5 < y+ < 30.

u+ = 5 ln y+ − 3.05,

(11.105)

11.2.2.9 Turbulent Region

In this region, εM ≫ ν; at large values of y+ ,

εM ν

≫ 1, and viscous stress is negligible. The momentum equation becomes

+

1 du (11.106) ≈ (ε ). M dy+ ν We can now use the mixing-length model of Prandtl in Eq. (11.106). Then, the momentum equation can be put in the following form: du+ ≈ dy+

1 ( (κy+ )2

du+ dy+

)

or du+ 1 . ≈ (κy+ ) dy+

(11.107)

Equation (11.107) is integrated to give 1 u+ = ln y+ + C. (11.108) κ The constant C is determined experimentally. Based on his pipe flow measurements in 1930, Nikuradse proposed that κ ≈ 0.4 and C ≈ 5.5. Later, in 1968, Coles and Hirst, as reported in [17], adjusted these values to κ = 0.41 and C ≈ 5. The dimensionless velocity profile becomes u+ = 2.43 ln y+ + 5,

y+ > 30.

(11.109)

This is the Nikuradse equation for smooth tubes. The inconsistency with this equation is the fact that at the center of the tube, it does not give a zero-velocity gradient. Despite this inconsistency, this relation holds with good accuracy across almost the entire pipe radius.

591

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11 Turbulent Internal Flow: Momentum and Heat Transfer

11.2.2.10 Moody Friction Factor

Now, let us return to the calculation of the Moody friction factor f. The universal velocity profile (Figure 9.10) shows that turbulent pipe flow has very little wake, and this wake can be neglected. The viscous sublayer is very thin, and contribution can be neglected. For these reasons, we may safely assume that the logarithmic velocity profile is accurate all the way across the pipe. We will use the following velocity profile to determine the dimensionless average velocity V+ : u+ = 2.5 ln y+ + 5.5.

(11.110)

The logarithmic velocity profile, Eq. (11.110), is substituted into Eq. (11.87) V+ =

2 (R+ )2 ∫0

R+

(2.5 ln y+ + 5.5)(R+ − y+ )dy+ .

(11.111)

Integration is carried out using Maple 2016, and the result of integration is V+ = 2.5 ln (R+ ) + 1.75.

(11.112)

The required relations for Eq. (11.112) are given below for convenience 1 f = 8 (V+ )2 ReD R = 2

√

+

f . 8

Substituting these into Eq. (11.112), we obtain a relation between the Moody friction factor f and the Reynolds number ReD : ( √ ) √ ReD 8 f + 1.75 (11.113) = 2.5 ln 2 8 f or converting to base 10 logarithms √ 1 √ = 2.035log10 (ReD f) − 0.9132. f

(11.114)

This relation does not fit the experimental data nicely, and for this reason, in 1935, Prandtl [18] adjusted these constants to give a better fit to experimental data, and he obtained his famous formula ( √) 1 (11.115) √ = 2.0log10 ReD f − 0.8 f 3.1 × 103 ≤ ReD ≤ 3.2 × 106 . Notice that Eq. (11.115) is an implicit relation for the Moody friction factor f. Blasius, as reported in [8], proposed the following correlation for fully developed turbulent flow in smooth pipe: f=

0.3164 Re0.25 D

(11.116)

3 × 103 ≤ ReD ≤ 105 . An explicit equation for the friction factor in fully developed turbulent flow is the Filonenko equation reported in [19] f=

1 [1.82log10 ReD − 1.64]2

(11.117)

3 × 103 < ReD < 5 × 107 . Petukhov [20] presented the following relation for the friction factor for fully developed turbulent flow in a smooth pipe: f=

1 [0.79 ln ReD − 1.64]2

3 × 104 < ReD < 5 × 106 .

(11.118)

11.2 Momentum Transfer

Xiande and Zhou [21] evaluate existing single-phase friction factor correlations and propose the following correlation for turbulent flow in smooth pipes: [ ( )]−2 152.66 150.39 f = 0.25 log10 − (11.119) ReD Re0.98865 D 3 × 103 ≤ ReD ≤ 108 . Since the turbulent entry length is short, most correlations neglect the additional pressure drop due to hydrodynamically developing region. The average friction factor, including the effect of hydrodynamically developing region, is called the apparent friction factor, and we may denote it by fapp . The apparent friction factor fapp may be estimated as follows: [ ( )0.7 ] D (11.120) fapp = f 1 + L where f is the Moody friction factor, as discussed before. A relation for the ratio between the average and maximum velocity V/Vc may be written as 1 V = √ . Vc 1 + 1.33 f See Problem 11.7.

(11.121)

Example 11.4 Water at a mean temperature of 17 ∘ C flows in a smooth tube of 150 mm. Far from the inlet, the average velocity is measured as 1.6 m/s. What is the maximum velocity in the tube? Solution Water properties at 17 ∘ C = 290 K are ρ ≈ 1000 kg∕m3 , μ = 1080 × 10−6 N.s∕m2 . First, we calculate the Reynolds number ReD ReD =

ρVD 1000 kg∕m3 × 1.6 m∕s × 0.15 m = = 2.22 × 105 . μ 1080 × 10−6 N.s∕m2

Flow is turbulent. To find the maximum velocity, we use Eq. (11.121) √ 1 V = √ ⇒ Vc = V(1 + 1.33 f). Vc 1 + 1.33 f Calculate the friction factor with Eq. (11.117) 1 f= [1.82log10 ReD − 1.64]2 3 × 103 < ReD < 5 × 107 1 1 = = 0.0152 [1.82log10 ReD − 1.64]2 [1.82log10 (2.22 × 105 ) − 1.64]2 √ √ Vc = V(1 + 1.33 f) = 1.6 m∕s × (1 + 1.33 0.0152) = 1.86m∕s. f=

11.2.2.11 Fanning Friction Factor

Suppose we are interested in the Fanning friction factor cf in an HFD flow in a smooth tube. We will again use the logarithmic law. First, we combine Eqs. (11.87) and (11.92) to get √ ) R+ ( y+ 2 2 1 − + u+ dy+ . = + (11.122a) cf R ∫0 R The substitution of the expression for u+ , Eq. (11.108), into Eq. (11.122a) yields √ ) R+ ( ) y+ ( 1 2 2 1− + C + ln y+ dy+ = + cf κ R ∫0 R

593

594

11 Turbulent Internal Flow: Momentum and Heat Transfer

) R+ ( y+ y+ 1 2 + + ln y C + dy+ − C − ln y κ R+ ∫0 R+ κ R+ )] [ ( +2 (R ) ln R+ (R+ )2 2 1 + C + 1 + + + − = + C R + (R ln R − R ) − R − κ 2 2 4 R κR+ 3 1 . = ln R+ + C − κ 2κ =

(11.122b)

Equation (11.122b) was first derived by Prandtl in 1935. Based on the experimental measurements of Nikuradse, Prandtl proposed the following correlation: ( √ ) √ cf 2 + 0.292. (11.123) = 2.46 ln ReD cf 2 This equation is in good agreement with experimental data over the entire turbulent flow range.

√ Re c Let us consider Eq. (11.122b). This time, we will use the constants {0.41, 5} in Eq. (11.122b). Substituting R+ = 2D 2f , κ = 0.41, and C = 5.0 into Eq. (11.122b), we obtain ( √ ) √ ReD cf 2 + 1.341. (11.124) = 2.44 ln cf 2 2 Finally, after some rearrangement, Eq. (11.124) becomes ( √ ) √ cf 2 − 0.351. = 2.44 ln ReD cf 2

(11.125)

Recall that in deriving this equation, we neglected the viscous and buffer sublayers. For values of ReD ≥ 104 , Eq. (11.123) may be approximated by the following equation: −1∕5

cf = 0.046ReD

(11.126)

3 × 104 ≤ ReD ≥ 106 . Prandtl’s student H. Blasius curve fitted to smooth wall data collected and proposed the following correlation for skin friction coefficient as reported in [17]: cf ≈

2τw 2

ρV

−1∕4

= 0.0791 ReD

.

(11.127)

Equation (11.127) is based on the experiments of Blasius. Another explicit correlation attributed to Petukhov is cf =

1 . [1.58 ln ReD − 3.28]2

Example 11.5 Air at 27 ∘ C is flowing with an average velocity of 15 m/s in a smooth pipe of 20-cm inside diameter. (a) (b) (c) (d) (e) (f) (g)

Calculate the skin friction coefficient. Calculate the friction velocity. Calculate the maximum velocity in the pipe. Calculate the velocity at a point 5 cm from the pipe wall. Calculate the thickness of the viscous sublayer. Calculate y+ for the value of u+ = 10. Calculate the wall shear stress.

Solution The physical properties of air at 300 K are ρ = 1.1614 kg∕m3 μ = 184.6 × 10−7 N.s∕m2 ν = 15.89 × 10−6 m2 ∕s.

(11.128)

11.2 Momentum Transfer

The Reynolds number ReD is ReD =

ρVD 1.1614 kg∕m3 × 15 m∕s × 0.20m = 1.88 × 105 . = μ 184.6 × 10−7 N.s∕m2

Flow is turbulent. (a) We may calculate the skin friction factor using Eq. (11.123) (

√ ) cf ReD + 0.292 2 ( √ ) √ cf 2 5 + 0.292. = 2.46 ln 1.88 × 10 cf 2

√

2 = 2.46 ln cf

Solving this equation with Maple 2020 gives ( √ √ ) cf 2 5 = 2.46 • ln 1.88 • 10 • >> eq ≔ + 0.292; cf 2 √ √ √ √ 1 = 2.46 • ln (94000.00 2 cf ) + 0.292 eq ≔ 2 cf > fsolve(eq, cf ); 0.003948251249 cf = 0.003948. (b) Friction velocity is √ √ V+ = 2∕cf ⇒ (V∕uτ ) = 2∕cf or

√ √ uτ = V cf ∕2 = 15 0.003948∕2 = 0.666 m∕s.

(c) Maximum velocity is The Moody friction factor f is f = 4cf ( √) √ Vc = V 1 + 1.33 f = 1.6 m∕s × (1 + 1.33 4 × 0.003948) = 17.5m∕s. (d) The velocity at a point 5 cm from the pipe wall. The law of the wall is u+ = 2.5 ln (y+ ) + 5.5 or u = 2.5 ln uτ

( yu ) τ

+ 5.5 ) ] [ ( 0.05 × 0.666 + 5.5 u = 0.666 × 2.5 ln 15.89 × 10−6 u = 16.40 m∕s. ν

(e) The thickness of viscous sublayer can be calculated using Eq. (11.104) δv =

5D √

ReD f∕8

595

596

11 Turbulent Internal Flow: Momentum and Heat Transfer

or δv =

5 × 0.20 5D = √ √ ReD cf ∕2 1.88 × 105 0.00394∕2

δv = 0.000119 m = 0.119 mm. (f)

y+

for value of u+ = 10 is calculated using the velocity profile for the buffer layer u+ = −3.05 + 5 ln (y+ ) 10 = −3.05 + 5 ln (y+ ) ⇒

(10 + 3.05) = ln (y+ ) 5

y+ = 13.59. (g) The wall shear stress is √ τw ⇒ τw = ρu2τ = 1.1614 kg∕m3 × (0.666)2 = 0.515 N∕m2 . uτ = ρ 11.2.2.12 The Power Law Velocity Distribution

The universal velocity profiles that have been discussed are valid near the wall and give a nonzero velocity gradient at the tube axis. This is an unrealistic situation. However, the universal velocity profiles are quite useful since both pressure drop and heat transfer are closely related to physical happenings near the wall. In many engineering computations, it is desirable to have an expression for the velocity profile that is valid over most of the cross section. This is very useful for the determination of bulk temperatures. Based on dimensional analysis and on the work of Blasius, Prandtl proposed u = Vc (y∕R)1∕7

(11.129a)

or if we use V = (49/60)Vc , Eq. (9.69c) and Eq. (9.70), we may obtain ) ( u R 1∕7 Vc = 8.74 τ uτ ν

(11.129b)

If we assume that the maximum values, Vc and R, can be replaced by the local values, u and y, we again obtain the power law for velocity profile, ( u y )1∕7 u = 8.74 τ uτ ν Nikuradse tested Eq. (11.129a) with his experimental data for the velocity distribution. He found that Eq. (11.129a) provides a good representation of turbulent core, and it is valid only for smooth pipes. It is valid only for ReD < 105 . Then, he generalized Eq. (11.129a) as u = Vc (y∕R)1∕n .

(11.129c)

The velocity distribution in a smooth pipe can be approximated by Eq. (11.129c), and Vc is the maximum velocity at the centerline of the pipe and the exponent n varies with Reynolds number. The average velocity is obtained by integration ) ( R( ) Vc r 1∕n (2π r)dr V= 1 − R π R2 ∫0 V 2n2 = 2 . (11.130) Vc 2n + 3n + 1 For the Reynolds number ReD in the range of 104 –105 , an average value of 7 can be used for n. Notice that for n = 7, we obtain the following relation between average velocity V and maximum velocity Vc : 49 V. 60 c The parameter n in Eq. (11.130) depends on the Reynolds number, and it is given in tabular form. V=

ReD

4 × 103

2.3 × 104

1.1 × 105

1.1 × 106

2.0 × 106

3.24 × 106

n

6

6.6

7

8.8

10

10

V/Vc

0.791

0.806

0.817

0.853

0.865

0.865

Source: Daily and Harleman [83].

11.3 Fully Developed Turbulent Heat Transfer

Figure 11.7

Typical turbulent flow velocity profiles.

1

0.8 n=6 0.6 u uτ

n=8 n = 10

0.4

0.2

0 0

0.2

0.4

0.6

0.8

1

r R

Experimental studies indicate that the universal velocity profile can be represented by the one-seventh power law, as reported in [23] u+ = 8.75(y+ )1∕7 .

(11.131)

The power law is not valid extremely near the wall since the velocity gradient is infinite there. In other words, it fails to give a finite gradient at the wall, and it is not also valid near the centerline because it does not give du∕dr = 0 at the pipe center, r = 0. Typical turbulent velocity profiles based on this power law approximation is shown in Figure 11.7. We can see from Figure 11.7 that the turbulent velocity profiles are rather flat, and this flatness increases with the Reynolds number. The velocity defect law is discussed in Chapter 9, and the velocity defect law developed by Prandtl applies to the core of the turbulent flow in a pipe, and it is given as (y) Vc − u . = −2.5 ln uτ R

(11.132)

Equation (11.132) is not valid for the laminar sublayer and the buffer layer.

11.3 Fully Developed Turbulent Heat Transfer The experimental studies of the heat transfer coefficient in fully turbulent flows indicate that heat transfer is insensitive to geometry in internal flows having a uniform cross-sectional area. The hydrodynamic entrance length in turbulent flow is short and depends on the entrance conditions, turbulence intensity, and surface conditions. It is experimentally observed that the turbulent flow becomes thermally fully developed (TFD) after a relatively short distance from the channel entrance. In most engineering applications, the entry-length solution is not very important. Flow may be modeled as both TFD and HFD. One important feature of the turbulent flow in a constant area channel is the fact that the growing boundary layer in the entrance region must meet at the channel center and no further growth is possible. This is true for momentum as well as thermal boundary layers. Another important point is that turbulent flow for the heat transfer mechanism in the thermal entrance and fully developed regions is quite insensitive to the constant temperature and constant heat flux wall boundary conditions for Pr ≥ 0.7, as discussed in [24]. Additionally, notice that the temperature gradient in turbulent flow is confined to the viscous sublayer, and the turbulent flow Nusselt number is insensitive to shape of the duct or thermal boundary conditions. On the other hand, the Nusselt number is strongly affected by the surface roughness. In other words, in contrast to laminar flow, the effect of surface boundary conditions (uniform wall temperature [UWT] or uniform heat flux [UHF]) is unimportant for the turbulent flow of all the fluids except liquid metals. There is no need to study each boundary condition separately. The same correlation can be used for UWT and UHF boundary conditions.

597

598

11 Turbulent Internal Flow: Momentum and Heat Transfer

Figure 11.8 heat flux.

q0ʺ

y

Turbulent flow between parallel plates subjected to uniform

H x q0ʺ

0

11.3.1

TFD and HFD Turbulent Flow Between Parallel Plates Subjected to UHF

We are now interested in the mean temperature distribution T and the Nusselt number NuDH for HFD and TFD steady turbulent flow between two infinite parallel plates. We will consider steady-state heat transfer in an HFD and TFD turbulent flow of incompressible fluid between parallel plates. The plates are subjected to UHF, as depicted in Figure 11.8. Viscous dissipation is neglected. We will assume that axial energy transport by conduction is negligible compared to energy transport in the vertical direction. (1) (2) (3) (4)

Flow is both TFD and HFD. Steady incompressible constant property fluid flow. There is no internal energy generation in the fluid. Viscous dissipation is neglected.

The differential equation of the mean energy transport for turbulent flow between two infinite parallel plates is given as ) [ ] ( 𝜕 𝜕T 𝜕T 𝜕T ρcp u +v = (k + kt ) . (11.133) 𝜕x 𝜕y 𝜕y 𝜕y For HFD turbulent flow, 𝜕u = 0. 𝜕x Therefore, from the continuity equation 𝜕v = 0. 𝜕y

(11.134)

(11.135)

The integration of this equation yields v = C.

(11.136)

Since plates are impermeable, v = 0, and therefore, C = 0. Hence, v = 0.

(11.137)

With this information, the differential energy equation becomes [ ] 𝜕 𝜕T 𝜕T = (k + kt ) . ρcp u 𝜕x 𝜕y 𝜕y

(11.138)

The boundary conditions are y = 0,

−k

y = H∕2,

𝜕T = q′′0 . 𝜕y

(11.139a)

𝜕T = 0. 𝜕y

(11.139b)

Next, we will evaluate the 𝜕T∕𝜕x term for TFD turbulent flow between parallel plates under the UHF boundary condition. TFD conditions can be defined for turbulent flow in terms of the time-averaged temperature T as follows: ( ) T − Tw 𝜕 =0 (11.140) 𝜕x T − T m

w

11.3 Fully Developed Turbulent Heat Transfer

or 𝜕T dTw = − 𝜕x dx

[

T − Tw

]

Tm − Tw

dTm + dx

[

T − Tw

]

Tm − Tw

dTw . dx

(11.141)

The mean temperature Tm is a convenient reference temperature for internal flows, and Newton’s law of cooling can be expressed as q′′0 = h(Tw − Tm )

(11.142)

and for a TFD turbulent flow h is constant and then for uniform constant heat flux q′′0 , we can write dTw dTm = . dx dx Therefore, Eq. (11.141) becomes

(11.143)

dTm ′′ 𝜕T dTw = = q0 = const. 𝜕x dx dx Then, the energy equation becomes ) ( [ ] dTm 𝜕 𝜕T = (k + kt ) . ρcp u dx 𝜕y 𝜕y

(11.144)

(11.145)

Next, consider the control volume shown in Figure 11.9. The dTm ∕dx term can be evaluated by writing an energy balance ̇ p Tm ) + ̇ p Tm ) + q′′0 Pdx = (mc (mc

d ̇ p Tm )dx (mc dx

or q′′ P dTm = 0 . ̇ p mc dx

(11.146)

Therefore, q′′ P (2Z + 2H)q′′0 2 Z(1 + H∕Z)q′′0 dTm = 0 = ≈ lim ̇ p mc dx ρcp V(H × Z) Z→∞ ρcp V(H × Z) ≈

(2)q′′0 ρcp V(H)

≈

4q′′0 ρcp V(2H)

=

4q′′0

(11.147)

ρcp VDH

4q′′0 𝜕T dTm = = 𝜕x dx ρcp VDH

(11.148)

where Z is the depth into paper and DH = 2H for turbulent flow between two infinite parallel plates. Hence, the energy equation takes the following form: [ ] ′′ dT d u 4q0 (k + kt ) = . (11.149) dy dy V DH q0ʺ (mcp Tm) + d [(mcp Tm)]dx dx

(mcp Tm)

q0ʺ dx Figure 11.9

Control volume for energy balance.

H

599

600

11 Turbulent Internal Flow: Momentum and Heat Transfer

The integration of the energy equation from y = 0 to any position y yields ( ) ′′ y dT 1 4q0 dT − (k + kt ) = u dy. (k + kt ) dy dy y=0 V DH ∫0 The following relation is obtained by the application of the boundary condition, Eq. (11.139b): dT || dT || −(k + kt ) | = −k | = q′′0 . dy ||y=0 dy ||y=0

(11.150)

(11.151)

Using this relation in Eq. (11.150), we get ′′

(k + kt )

y

dT 1 4q0 + q′′0 = u dy dy V DH ∫0

(11.152)

or we can express Eq. (11.152) in the following form: [ ] y 1 4 dT −(k + kt ) u dy . = q′′0 1 − dy V DH ∫0 We will now put this energy equation in dimensionless form by using the following dimensionless variables: yu y+ = τ ν T+ =

(Tw − T)(ρcp uτ )

(11.153)

(11.154) (11.155)

q′′0

where Tw is the unknown wall temperature and the turbulent thermal conductivity kt is ε kt = ρcp εH = ρcp M . Prt The result of this nondimensionalization process is [ ] y+ ε 1 dT+ 4 1 + M = 1 − u+ dy+ Pr ν Prt dy+ ReDH ∫0 T+ (0) = 0.

(11.156)

(11.157) (11.158)

The integration of this equation yields y+ + 4 +⎤ ⎡ y+ 1 − Re ∫0 u dy T − T ⎥ ⎢ D w H + = T = ′′ ) ⎥ dy+ . ⎢ (1 ∫ ε q0 ∕ρcp uτ 0 ⎥ ⎢ + νM Pr1 Pr ⎦ ⎣ t

(11.159)

If we specify εM , Prt , and u+ , we can integrate this equation numerically. We will simplify this integral by breaking up into two parts: a) Inner region, y/(H/2) ≤ 0.2. b) Outer region, y/(H/2) > 0.20. y+

In the inner region, y/(H/2) ≤ 0.2, the ∫0 u+ dy+ term is small, and it can be neglected. The temperature distribution becomes dT+ 1 (11.160) = ( ) ε 1 dy+ + M 1 Pr

ν Prt

+

y (0) = 0. Temperature distribution can be obtained by numerical integration. Suppose that we choose the following inputs for Eq. (11.160): Pr = 0.7 Prt = 0.9.

11.3 Fully Developed Turbulent Heat Transfer

The van Driest expression [7] for εM /ν is [ ( + )]2 + εM y du = (κ y+ )2 1 − exp − + . ν A dy+

(11.161)

The relationship between εM /ν and du+ /dy+ may be obtained from the form of the momentum equation, which is valid within the inner region. The applicable momentum equation is 1 du+ . = (1 + εM ∕ν) dy+

(11.162)

We can now combine Eqs. (11.161) and (11.162) to get [ ( + )]2 y 1 + 2 εM ∕ν = (κ y ) 1 − exp − + (1 + εM ∕ν) A or ( ε )2 M

ν

+

(ε ) M

ν

[ ( + )]2 y = 0. − (κ y+ )2 1 − exp − + A

(11.163)

The solution of Eq. (11.163) is √{ −1 + εM = ν

[ ( + )]2 } y 1 + 4(κ y+ )2 1 − exp − A+ (11.164)

2

A+ = 26 κ = 0.4. Substituting Eq. (11.164) into Eq. (11.160), we can now determine temperature distribution numerically. The numerical solution is obtained by Maple 2016. Dimensionless mean temperature is depicted in Figure 11.10 for TFD turbulent flow between infinite parallel plates. An examination of Figure 11.10 reveals that the model agrees well with experimental data of Johnk and Hanratty [25]. In the viscous sublayer, εM ≪ ν, and this approximation gives T+ = Pr y+ , for y+ < 5.

(11.165)

This equation is consistent with available experimental data. Figure 11.10 Dimensionless mean temperature distribution for the parallel plate channel.

100 Pr = 0.2 50 Prt = 0.9 Van Driest Model 10 T

+

5 Data Johnk and Hanratty (25) Re = 24 900 1 0.5 5

10

50 y+

100

601

602

11 Turbulent Internal Flow: Momentum and Heat Transfer

In the fully turbulent layer of the inner region, εM /ν is approximately given as εM ≈ κ y+ ν

(11.166)

and assuming Prt ≈ 1 for simplicity, we obtain dimensionless temperature y+

T+ =

∫0

(

dy+ 1 Pr

+ κ y+

).

(11.167)

Integration by Maple 2020 yields ( ) 1 1 T+ = ln + κ y+ + B κ Pr

(11.168)

where the constant B is a function of the Prandtl number. Since εH > > α in this region for fluids except liquid metals, the above equation can be approximated by T+ =

1 ln y+ + B. κ

(11.169)

White and Majdalani [17] proposed a correlation for B in the following form: B ≈ 13Pr2∕3 − 7

(11.170)

Pr ≥ 0.7. Kader [26] reports a more complicated curve fit B = (3.85Pr1∕3 − 1.3)2 + 2.12 ln (Pr )

(11.171)

6 × 10−3 < Pr < 40 × 103 . It is observed experimentally that the inner law for dimensionless temperature T+ may also be used to approximate the temperature distribution in the outer region. Equation (11.169) may be approximated by the one-seventh power law for temperature T+ = B1 (y+ )1∕7 .

(11.172)

We may estimate the coefficient B1 by equating the law of wall for temperature and power law at some value of y+ within the intermediate region. For example, B1 (y+ )1∕7 =

1 ln y+ + B. κ

(11.173)

Choose y+ = 50, κ = 0.4. B1 (50)1∕7 =

1 ln 50 + B κ

(11.174)

and we find that B1 = (9.78 + B)/1.7486 and B is calculated from White’s equation for a given Prandtl number. 11.3.1.1 Mean Stream Temperature

For incompressible, constant property, turbulent flow between parallel plates, bulk temperature Tm is defined as ∫ u T dAc Tm =

Ac

∫ u dAc

H

=

1 2 u Tdy = HV ∫0 HV ∫0

H∕2

u T dy.

(11.175)

Ac

Equation (11.175) can be expressed in terms of T+ , u+ , and y+ as +

Tm =

(Tw − Tm )(ρcp uτ ) q′′0

( =

2uτ H+ V

)

H+ ∕2

∫0

u+ T+ dy+

(11.176)

11.3 Fully Developed Turbulent Heat Transfer

or solving for q′′0 ∕(Tw − Tm ), one gets q′′0

=

(Tw − Tm )

ρcp V H+ ∕2 + + 2 ∫ u T dy+ H+ 0

.

(11.177)

The heat transfer coefficient h is q′′0

h=

(11.178)

(Tw − Tm ) and the Nusselt number is defined as NuDH =

q′′0 hDH 2H = = k (Tw − Tm ) k

ReDH Pr H+ ∕2 + + 2 ∫ u T dy+ H+ 0

(11.179)

where Nu = DH h/k is the Nusselt number, Pr = μcp /k is the Prandtl number, ReDH = (ρVDH ∕k) is the Reynolds number, and DH = 2H is the hydraulic diameter. We will now approximate u+ and T+ as follows: u+ =

V u ≈ uτ uτ

(11.180)

T+ =

1 ln y+ + B. κ

(11.181)

The Nusselt number now becomes NuDH =

Since

uτ V

hDH = k

√ =

NuDH

cf , 2

H V 2 ∫ uτ H+ 0

+

ReDH Pr . ) ( ∕2 1 + + B dy+ ln y κ

the Nusselt number becomes

hDH = = k

√ ReDH Pr cf ∕2 ) ( H+ ∕2 1 2 + + B dy+ ln y + ∫0 H κ

where cf is the Fanning friction factor. Carrying out the integration by Maple 2020, we obtain √ ReDH Pr cf ∕2 hDH NuDH = = [ ]H+ ∕2 k 2 1 + + − y+ ) + By+ (y ln y + H κ 0 √ Pr ∕2 c Re hDH D f = [ (H + ) . NuDH = ] 1 H k ln −1 +B κ

NuDH

(11.183)

(11.184)

2

The dimensionless distance H+ is √ ReDH cf H+ = 2 2 where ReDH =

(11.182)

(11.185)

DH V ν

and DH = 2H. The Nusselt number now becomes √ ReDH Pr cf ∕2 hDH = = [ ( Re √ ) ] k cf DH − 1 B + κ1 ln 4 2

where we take κ and B as κ = 0.4 B ≈ 12.7Pr2∕3 − 7.7.

(11.186)

603

604

11 Turbulent Internal Flow: Momentum and Heat Transfer

We can eliminate the logarithmic term by utilizing the relationship between the Fanning friction factor cf and the Reynolds number ReDH . This relationship is ( √ ) √ ReDH cf 1 1 2 +C− . (11.187) = ln cf κ 4 2 κ We can now eliminate the logarithmic term, and we obtain NuDH =

ReDH Pr (cf ∕2) hDH = . √ k 1 + cf ∕2(B − C)

(11.188)

Substituting B = 12.7Pr2/3 − 7.7 and C = 5, after some algebra, we obtain NuDH =

ReDH Pr (cf ∕2) . √ 1 + 12.7(Pr2∕3 − 1) cf ∕2

(11.189a)

If we use Eq. (11.170) in Eq. (11.188), we get NuDH =

ReDH Pr (cf ∕2) . √ 1 + (13Pr2∕3 − 12) cf ∕2

(11.189b)

Suppose that we wish to express the Nusselt number in terms of the Moody friction factor f. We know that f = 4cf , and we may write the Nusselt number as NuDH =

ReDH Pr (f∕8)

√ . 1 + 12.7(Pr2∕3 − 1) f∕8

(11.190)

Example 11.6 Water at a mean temperature of 330 K flows turbulently between two infinite smooth parallel plates, as shown in Figure 11.2. The spacing between the parallel plates is H = 1 cm. Mean velocity between the plates is 1 m/s. The plate surface temperature is 360 K. At a location where the mean water temperature is 330 K, calculate the local wall heat flux and estimate the water temperature at 0.5 mm away from the wall. Solution The physical properties of water at 330 K are ρ = 984 kg∕m3

cp = 4184 J∕kgK μ = 489 × 10−6 N.s∕m2

k = 0.650W∕mK Pr = 3.15 ν = 4.968 × 10−7 m2 ∕s. We will use ReDH Pr (cf ∕2) √ 1 + 12.7(Pr2∕3 − 1) cf ∕2 hDH . = k

NuDH = NuDH

The hydraulic diameter for the parallel plate channel is DH = 2H = 2 × 0.01 = 0.02 m ReDH =

ρVDH 984 × 1 × 0.02 = = 40245. μ 489 × 10−6

Flow is turbulent. We need the skin friction coefficient cf . We will use the experimental equation of Blasius, cf = cf =

0.0791 1∕4 ReD H

Re =

VDH ν

0.0791 = 0.00557. 402451∕4

11.3 Fully Developed Turbulent Heat Transfer

We can now calculate the Nusselt number (40245) × (3.15)(0.00557∕2) = 199.73 NuDH = √ 2∕3 1 + 12.7 0.00557∕2 ( ) × (3.15 − 1) hDH k NuDH = ⇒h= NuDH . k DH The heat transfer coefficient is ) ( ) ( W 0.650 k 199.73 ≈ 6491.22 2 . NuDH = h= DH 0.02 mK Heat flux at the wall is q′′w = h(Tw − Tm ) = 6491.22 × (360 − 330) ≈ 194 736 W∕m2 . yu

At y = 5 mm, we calculate y+ = ν τ . First, let us calculate uτ . For this purpose, we can use Eq. (11.46b) √ √ cf V 2 = ⇒ uτ = V uτ cf 2 √ 0.00557 = 0.0528 m∕s uτ = 1 × 2 yu (5∕1000) × 0.0528 y+ = τ = ≈ 531.4. ν 4.968 × 10−7 This point is in fully turbulent region. We will estimate temperature with the following relation: T+ =

1 ln y+ + B κ

B ≈ 12.7Pr2∕3 − 7.7 B ≈ 12.7(3.15)2∕3 − 7.7 ≈ 20 1 ln (531.4) + 20 ≈ 34.89 0.41 Finally, we write T+ =

) ( Tw − T ⇒ T = Tw − q′′w ∕ρcp uτ T+ ′′ qw ∕ρcp uτ ( ) 194 736 T = 360 − (34.89) ≈ 329 K. 984 × 4184 × 0.0528

T+ =

11.3.2 TFD and HFD Turbulent Flow in a Pipe Subjected to UHF Consider steady, TFD and HFD, incompressible, constant property, turbulent flow in a circular tube. See Figure 11.11. Uniform wall heat flux is applied to the pipe surface. Mean velocity of the fluid is in the x-direction and the x-axis is coincident with the centerline of the pipe. Flow is symmetrical about the x-axis 𝜕/𝜕 θ = 0. There is no swirl in the flow, and viscous dissipation is neglected. Axial conduction in the fluid is neglected with respect to bulk transport of energy in the x-direction. It is experimentally observed that the turbulent flow becomes TFD after a relatively short distance from the tube entrance. In most engineering applications, the entry-length solution is not very important. Flow is both TFD and HFD. The hydrodynamically fully development criterion is particularly applicable to fluids having high Prandtl numbers. One important feature of the turbulent flow in a constant area pipe is the fact that the growing boundary layer in the entrance region must meet at the tube center, and no further growth is possible. An important feature of the turbulent flow in a constant area pipe flow different from the turbulent boundary layer flow is that the wake can be neglected. This fact may be seen from an examination of Figure 9.12. Kays et al. [23] report that wake may be neglected due to negative pressure gradient. The energy equation can be expressed as ) [ ] ( 1 𝜕 𝜕T 𝜕T 𝜕T +v = r(α + εH ) . (11.191) u 𝜕x 𝜕r r𝜕r 𝜕r

605

606

11 Turbulent Internal Flow: Momentum and Heat Transfer

q0ʺ

r v

y

u D=2R

x

0

q0ʺ Figure 11.11

Turbulent flow heat transfer a in a circular tube.

Flow is parallel to the axial direction, and for a fully developed flow from continuity, we found that v = 0, and we have already obtained a solution for the momentum equation. The energy equation, Eq. (11.92), becomes [ ] 𝜕T 1 𝜕 𝜕T = r(α + εH ) . (11.192) u 𝜕x r 𝜕r 𝜕r The boundary conditions for the energy equation are r = 0 T = finite

𝜕T =0 𝜕r

(11.193)

𝜕T = q′′w . (11.194) 𝜕r The positive sign appears since r is measured from the pipe center toward the wall, whereas the heat flux q′′w is taken as positive in the inward direction. We are assuming that the velocity profile as well as temperature profile is fully developed. The energy equation can be solved provided that we have a turbulence model for Eq. (11.193). It is more convenient to use the following transformation to rewrite Eq. (11.193) in terms of the distance y measured from the wall. The distance y is related to radius R by r = R;

k

r=R−y dr = −dy. With this transformation, the energy equation now becomes [ ] 1 𝜕 𝜕T 𝜕T = (R − y)(α + εH ) . u 𝜕x R − y 𝜕y 𝜕y

(11.195)

The boundary conditions are y = R,

𝜕T =0 𝜕y

(11.196a)

y = 0,

T = Tw

(11.196b)

where Tw is the unknown wall temperature. The first boundary condition follows from the fact that the temperature profile by symmetrical about the pipe centerline. TFD conditions can be defined for turbulent flow in terms of the time-averaged temperature T as follows: ) ( T − Tw 𝜕 =0 (11.197) 𝜕x T − T m

w

or 𝜕T dTw = − 𝜕x dx

[

T − Tw Tm − Tw

]

dTm + dx

[

T − Tw Tm − Tw

]

dTw . dx

(11.198)

The mean temperature Tm is a convenient reference temperature for incompressible constant property internal flows and Newton’s law of cooling can be expressed as q′′w = h(Tw − Tm )

(11.199)

11.3 Fully Developed Turbulent Heat Transfer

where Tm is the mean fluid temperature defined as ∫ u T dAc Tm =

Ac

∫ u dAc

R

=

Ac

∫0 u T(2π r)dr R ∫0 u(2π r)dr

R

=

1 u T 2π r dr VπR2 ∫0

R

=

2 u T r dr V R2 ∫0

(11.200)

and for a TFD turbulent flow, both h and q′′w are constant; then, from Eq. (11.199), we can obtain dTw dTm = . dx dx

(11.201)

Therefore, we can now write dTm ′′ 𝜕T dTw = = q0 = const. 𝜕x dx dx

(11.202)

The energy equation now becomes ) ( [ ] dTm 𝜕T 𝜕 (R − y)(α + εH ) =u (R − y). 𝜕y 𝜕y dx Next, we may evaluate

dTm dx

(11.203)

by the application of an overall energy balance to the control volume shown in Figure 11.12

̇ p Tm ) + ̇ p Tm ) + q′′0 (Pdx) = (mc (mc

d ̇ p Tm )]dx [(mc dx

P = 2πR or q′′ (2πR) P q′′0 dTm = 0 2 . = ̇ p mc dx (ρVπ R )cp

(11.204)

Therefore, 2q′′0 𝜕T dTm = = . 𝜕x dx (ρcp V R)

(11.205)

Substituting Eq. (11.205) into Eq. (11.203), the energy equation becomes ) ( [ ] 2q′′0 𝜕 𝜕T (R − y)(α + εH ) =u (R − y). 𝜕y 𝜕y ρcp V R

(11.206)

q0ʺ

(mcp Tm) + d [(mcp Tm)]dx dx

(mcp Tm)

q0ʺ dx Figure 11.12

Control volume for energy balance.

2R

607

608

11 Turbulent Internal Flow: Momentum and Heat Transfer

The boundary conditions of Eq. (11.206) are y = R,

𝜕T =0 𝜕y

y = 0,

T = Tw .

We now integrate the energy equation, Eq. (11.206), from the pipe centerline where y = R to any distance y from the wall ( ) y 2q′′0 𝜕T = u(R − y) dy + C1 (11.207) (R − y)(α + εH ) 𝜕y (ρcp V R) ∫R when y = R, 𝜕T∕𝜕y = 0, and thus, C1 = 0. Then, the differential equation becomes ( ) y[ 2q′′0 ] 𝜕T = (R − y)(α + εH ) u (R − y) dy. 𝜕y (ρcp V R) ∫R We now break the integral in Eq. (11.208) into two parts as follows: ( ){ } y[ R[ 2q′′0 ] ] 𝜕T = u(R − y) dy + u(R − y) dy (R − y)(α + εH ) ∫0 ∫y 𝜕y ρcp V R or 𝜕T = (R − y)(α + εH ) 𝜕y or 𝜕T = (R − y)(α + εH ) 𝜕y

(

2q′′0

){ ∫0

ρcp V R (

dTm dx

y

){

y

∫0

[

] u(R − y) dy −

[ ] u(R − y) dy −

R

∫0

R

∫0

[

(11.209a)

}

] u(R − y) dy

(11.209b)

}

[ ] u(R − y) dy

.

To simplify the algebra, define the following quantity: { y } R[ ] ] [ M= u(R − y) dy − u(R − y) dy . ∫0 ∫0 Using Eqs. (11.205) and (11.210), the energy equation, Eq. (11.209c), is expressed in the following form: ( ) d Tm M 𝜕T = . 𝜕y (R − y)(α + εH ) dx Equation (11.211) is integrated from the wall ( ) y d Tm M T= dy + C2 ∫0 (R − y)(α + εH ) dx

(11.208)

(11.209c)

(11.210)

(11.211)

(11.212a)

or T=

2q′′0

y

M dy + C2 . ρcp V R ∫0 (R − y)(α + εH )

(11.212b)

The second boundary condition, y = 0 T = Tw , is used to evaluate the unknown constant C2 . Then, C2 = Tw . Thus, the energy equation becomes ) ( y 2q′′0 M dy. (11.213) T − Tw = ρcp V R ∫0 (R − y)(α + εH ) However, we will estimate the velocity distribution u as u ≈ V since the velocity profile is quite flat in the core region of the turbulent pipe flow except for a thin layer next to the wall. This approximate velocity profile will be used in the solution of the energy equation. We now evaluate M

11.3 Fully Developed Turbulent Heat Transfer

{ M=

y

[

∫0 {

] u(R − y) dy −

R

∫0

y

R

[(R − y)]dy −

=V

[

} ] u(R − y) dy

} [(R − y)]dy

∫0 ∫0 V 2 = − (R − y) . 2 We now substitute this approximation into the energy equation, Eq. (11.213), to get ( ) y q′′0 (1 − y∕R) dy. T − Tw = − ρcp ∫0 (α + εH )

(11.214)

(11.215)

This approximation of velocity leads to simplification and yet gives good results. Equation (11.215) is put in dimensionless form using wall coordinates: yu (11.216) y+ = τ , ν ε Prt = M , (11.217) εH T+ =

T − Tw . q′′0 ∕(ρcp uτ )

(11.218)

Therefore, in terms of wall coordinates, T − TW = −

q′′0

ρcp uτ ∫0

y+

(1 − y+ ∕R+ ) + ( ) dy εM 1 1 + Pr ν Pr

(11.219a)

t

or y+

T+ = − ∫0

(1 − y+ ∕R+ ) + ( ) dy εM 1 1 + Pr ν Pr

(11.219b)

t

where

R+

= R uτ /ν. We need an expression for εM /ν. We know that εM /ν can be written as

εM 1 − (y+ ∕R+ ) − 1. = ν (du+ ∕dy+ )

(11.220)

We now rely on experimentally determined velocity profile, and we assume that the velocity profile can be split into three zones. We will use the von Karman velocity profile. We will obtain temperature distribution as follows: 11.3.2.1 Laminar Viscous Sublayer: 0 < y+ < 5

In this region, the turbulent shear stress and turbulent heat transfer rate are assumed to be negligible. Since the viscous sublayer is very thin, y/R ≪ 1. y+ ≪ R+ → 1 − (y+ ∕R+ ) ≈ 1

(11.221)

u+ = y+ .

(11.222)

Taking the derivative of Eq. (11.222) yields du+ = 1. dy+ Then, we obtain εM ≈ 0. ν Since (y+ /R+ ) ≪ 1 we have 1 − (y+ /R+ ) ≈ 1, and we obtain expression for temperature ) ( q′′0 Pr y+ . T − Tw = − ρcp uτ

(11.223)

(11.224)

(11.225)

609

610

11 Turbulent Internal Flow: Momentum and Heat Transfer

The temperature at the edge of the viscous sublayer (y+ = 5) is denoted by T5 and Eq. (11.225) gives ( ) q′′0 T5 − Tw = −5 Pr . ρcp uτ

(11.226)

11.3.2.2 Buffer Layer: 5 < y+ < 30

Available velocity measurements indicate that the velocity profile is given as u+ = 5 ln (y+ ) − 3.05.

(11.227)

We take the derivative of Eq. (11.227) and note that

y+ /R+

≈ 0 since buffer layer is also very thin

+

5 du = +. + y dy

(11.228)

Then, the eddy diffusivity of momentum becomes εM y+ = − 1. ν 5 In this region, we assume that Prt ≈ 1. Substituting Eq. (11.229) into the energy equation, we now get T − T5 = −

y+

q′′0

ρcp uτ ∫5

Integration yields ( T − T5 = −5

q′′0

[

)

ρcp uτ

1 Pr

+

dy+ ( + y 5

(11.229)

)] . −1

)] [ ( + y −1 . ln 1 + Pr 5

(11.230)

The temperature at the top edge of the buffer layer (y+ = 30) is denoted by T30 . Equation (11.230) gives ( ) )] [ ( q′′0 30 −1 T30 − T5 = −5 ln 1 + Pr ρcp uτ 5 ( ) q′′0 = −5 ln [1 + 5 Pr ]. ρcp uτ

(11.231)

11.3.2.3 Turbulent Region: y+ > 30

In the turbulent region, the velocity profile is u+ = 2.5 ln (y+ ) + 5.5.

(11.232)

We differentiate Eq. (11.232) to get du+ 2.5 = + y dy+ εM 1 − (y+ ∕R+ ) = −1 2.5 ν y+

εM + ν = ν

[1 − (y+ ∕R+ )] 2.5 y+

.

In this region, εM ≫ ν, and for this reason, ( +) εM y ≈ [1 − (y+ ∕R+ )]. ν 2.5 The integral now becomes ( ) y+ q′′0 T − T30 = − ( ρcp uτ ∫30

1 Pr

)

(11.233)

(1 − y+ ∕R+ )dy+ [( + ) ( y + Pr1 1− 2.5 t

y+ R+

)] .

(11.234)

11.3 Fully Developed Turbulent Heat Transfer

In this region, we assume that molecular shearing stress and molecular heat transfer rate are negligible compared to turbulent shearing stress and turbulent heat transfer rate. This means that εM ≫ ν εH ≫ α. Thus, we can neglect the molecular Prandtl number Pr, and the integral takes the following form: ( ) ( ) y+ y+ q′′0 q′′0 (1 − y+ ∕R+ )dy+ dy+ = −2.5Pr T − T30 ≈ − [( + ) ( )] t y y+ ρcp uτ ∫30 1 ρcp uτ ∫30 y+ 1− (

T − T30 ≈ −2.5Prt

Prt

q′′0 ρcp uτ

)

ln

2.5

(

) +

R+

y 30

(11.235)

which is valid from y+ = 30 to the center of the tube where y = R (y+ = R+ ). The temperature at y = R is called the centerline temperature and denoted by Tc , and this temperature is obtained as follows: ( ) ( +) q′′0 R . (11.236) ln Tc − T30 ≈ −2.5Prt ρcp uτ 30 √ Ru Re c But we know that R+ = ν τ = 2D 2f , and the centerline temperature takes the following form: ( ) [ √ ] q′′w ReD cf . (11.237) ln Tc − T30 = −2.5Prt ρcp uτ 60 2 The overall temperature changes from the wall to the tube center is ( ) q′′0 Tw − Tc = [(Tw − T5 ) + (T5 − T30 ) + (T30 − Tc )] ρcp uτ or

( Tw − Tc =

q′′0 ρcp uτ

[ √ ]] ] [ ReD cf 5 Pr + 2.5Prt ln 5 Pr +5Prt ln 1 + . Prt 60 2

(11.238a)

)[

We may also write a relationship for the temperature at any point in the tube ] [ ( +) Tw − T y 5 Pr + 2.5Prt ln . ( ′′ ) = 5 Pr +5Prt ln 1 + Pr 30 q0 ∕ρcp uτ t √ Substituting uτ = V cf ∕2 into Eq. (11.238b), we obtain )√ [ ( [ √ ]] ] [ q′′0 ReD cf 5 Pr 2 . 5 Pr +5Prt ln 1 + + 2.5Prt ln Tw − Tc = ρcp V cf Prt 60 2 We will now determine the heat transfer coefficient. The heat transfer coefficient h is defined as q′′0 h= Tw − Tm

(11.238b)

(11.238c)

(11.239)

(11.240)

and in order to determine the heat transfer coefficient, we need to evaluate the temperature difference (Tw − Tm ). However, to determine (Tw − Tm ), we now need a relationship between (Tw − Tm ) and (Tw − TC ). This could be done by integrating the temperature distribution across each layer. For practical purposes, we will assume the following velocity and temperature profiles: ) ( y )1∕7 ( R − r 1∕7 u ≈ ≈ (11.241) Vc R R ) ( y )1∕7 ( T − Tw R − r 1∕7 ≈ ≈ (11.242) Tc − Tw R R

611

612

11 Turbulent Internal Flow: Momentum and Heat Transfer

where Tc is the centerline temperature and Vc is the centerline velocity in the pipe flow. The mean fluid velocity V R ( ) ∫ ρ u(2πr)dr 1 2 49 r 1∕7 V ρu dAc = 0 = V (r)dr = 1 − c ρAc ∫ R 60 c ρπR2 R2 ∫0 R

V=

(11.243)

Ac

and the mean fluid temperature Tm for incompressible constant property turbulent flow ( ( )1∕7 [ )1∕7 ] R ∫0 Vc 1 − Rr Tw − (Tc − Tw ) 1 − Rr r dr R R ∫0 ρcp u T dAc ∫0 u T(2π r)dr = Tm = = ( )1∕7 R R r ∫0 ρcp u dAc ∫0 u(2π r)dr ∫0 0 Vc 1 − Rr r dr { } R R )1∕7 )2∕7 ( ( 60 r r = Tw V 1− r dr − (Tc − Tw ) V 1− r dr ∫0 c ∫0 c 49 Vc R R { } R R ) ) ( ( r 1∕7 r 2∕7 60 Tw Vc 1 − r dr − (Tc − Tw ) Vc 1 − r dr = ∫0 ∫0 49 Vc R R or after carrying out the integrations, we obtain 5 Tw − Tm = (Tw − Tc ). 6 Then, the temperature difference Tw − Tm becomes {( )√ [ [ √ ]]} ] [ ( ) q′′0 ReD cf 5 Pr 2 5 5 Pr + 5Prt ln 1 + . + 2.5Prt ln Tw − Tm = 6 ρcp V cf Prt 60 2 Let us now assume that Prt ≈ 1. Then, we have {( [ )√ [ √ ]]} ( ) q′′0 ReD cf 5 2 Tw − Tm = 5 Pr + 5 ln [1 + 5 Pr ] + 2.5 ln 6 ρcp V cf 60 2 or in terms of the Moody friction factor {( )√ [ [ √ ]]} ( ) q′′0 ReD 5 8 f . 5 Pr + 5 ln [1 + 5 Pr ] + 2.5 ln Tw − Tm = 6 ρcp V 60 8 f The heat transfer coefficient h is q′′0 q′′0 = ( ) {( ′′ ) √ [ h= [ √ ]]} q0 Tw − Tm ReD cf 5 2 5 Pr +5 ln [1 + 5 Pr ] + 2.5 ln 6 ρcp V cf 60 2 √ ρcp V cf ∕2 = ( ) {[ [ √ ]]} . ReD cf 5 5 Pr +5 ln [1 + 5 Pr ] + 2.5 ln 6 60 2 The Nusselt number NuD is hD NuD = = k

(11.244)

(11.245)

(11.246a)

(11.246b)

(11.247a)

(11.248)

√ ρcp V D cf ∕2

[ √ ]]} ( ) {[ ReD cf 5 5 Pr +5 ln [1 + 5 Pr ] + 2.5 ln 6 60 2 √ (ρVD∕μ)(μcp ∕k) cf ∕2 = ( ) {[ [ √ ]]} ReD cf 5 5 Pr +5 ln [1 + 5 Pr ] + 2.5 ln 6 60 2 √ ReD Pr cf ∕2 = ( )[ [ √ ]] ReD cf 5 5 Pr +5 ln [1 + 5 Pr ] + 2.5 ln 6 60 2 k

or in terms of the Moody friction factor √ ReD Pr f∕8 NuD = ( ) [ [ √ ]] ReD 5 f 5 Pr +5 ln [1 + 5 Pr ] + 2.5 ln 6 60 8 0.5 ≤ Pr ≤ 30.

(11.249a)

(11.249b)

11.3 Fully Developed Turbulent Heat Transfer

Equation (11.249a) or Eq. (11.249b) is in good agreement experimental data for flow of fluids with Prandtl numbers between 0.5 and 30. Example 11.7 Air is heated from 17 to 57 ∘ C in tube having an internal diameter of 60 mm. UHF is applied to tube surface. The mass flow rate of air is 0.04 kg/s. Determine the heat transfer coefficient. Solution The average bulk fluid temperature Tm is T + Tmo 17 + 57 Tm = mi = = 37∘ C = 310 K 2 2 ρ = 1.1281 kg∕m3 cp = 1008 J∕Kg.K μ = 189.3 × 10−7 N.s∕m2 ν = 16.89 × 10−6 m2 ∕s

k = 0.027W∕m.K Pr = 0.705.

The Reynolds number ReD is 4 × 0.04 4ṁ = ReD = = 44 840. πμD π × 189.3 × 10−7 × 0.06 Flow is fully turbulent. We calculate the skin friction coefficient cf by Eq. (11.126) −1∕5

cf = 0.046ReD

= 0.046(44 840)−1∕5 = 0.0054.

The Nusselt number will be calculated from Eq. (11.249a) √ ReD Pr cf ∕2 NuD = ( ) {[ [ √ ]]} Re c 5 5 Pr +5 ln [1 + 5 Pr ] + 2.5 ln 60D 2f 6 √ (44 840) × 0.705 0.0054∕2 NuD = ( ) {[ [ ]]} √ 5 44 840 0.0054 5 × 0.71 + 5 ln [1 + 5 × 0.71] + 2.5 ln 6 60 2 NuD = 97 ) ( k 0.027 (97) ≈ 43.86 W∕m2 K. h = NuD = D 0.06 Example 11.8 (a) Consider fully developed turbulent flow heat transfer in a circular pipe subject to constant heat flux. We wish to study heat transfer without using the one-seventh power law for velocity and temperature profiles. (b) Water flows at 5 m/s in a tube having an internal diameter of 4 cm. At a particular location in the flow, the pipe wall temperature is 360 K, and the mean water temperature is 290 K. (1) Determine the Moody friction factor f. (2) Compute the Nusselt number and compare with an experimental correlation. Solution The energy equation is y+

T+ =

∫0

yuτ , ν ε Prt = M εH

(1 − y+ ∕R+ ) + ( ) dy εM 1 1 + Pr ν Pr

y+ =

T+ =

Tw − T q′′w ∕(ρcp uτ )

R+ = R uτ ∕ν.

t

613

614

11 Turbulent Internal Flow: Momentum and Heat Transfer

For 0 < y+ ≤ 5 In the viscous sublayer, as we have done before u + = y+ du+ =1 dy+ y+ ∕R+ ≈ 0 εM 1 − (y+ ∕R+ ) −1 = ν (du+ ∕dy+ ) εM ∕ν ≈ 0. We assume that Prt ≈ 1 y+

T+ =

∫0

+

y (1 − y+ ∕R+ ) + 1 ( ) dy = ( ) dy+ ∫0 εM 1 1 1 + ν Pr Pr Pr t

T+ = Pr y+ . Temperature at the upper edge of the viscous sublayer is T+ ||y+ =5 = 5 Pr . In this region, turbulent shearing stress and turbulent heat transfer rate are assumed to be negligible For 5 < y+ ≤ 30. In the buffer layer, based on the available experimental data, velocity distribution is u+ = 5 ln y+ − 3.05 5 du+ = + + y dy y+ ∕R+ ≈ 0 εM 1 − (y+ ∕R+ ) − 1. = ν (du+ ∕dy+ ) Thus, we get ( + ) εM y = −1 . ν 5 In this region, we also assume that Prt ≈ 1 y+

y (1 − y+ ∕R+ ) + T = T |y+ =5 + ( ) dy = T+ ||y+ =5 + ∫5 ∫5 ε 1 + νM Pr1 Pr t ) ] [ ( + y −1 +1 . T+ = 5 Pr +5 ln Pr 5 +

+|

+

1 Pr

+

1 ( + y 5

) dy+ −1

Temperature at the upper edge of the buffer layer is ) ] [ ( 30 T+ ||y+ =30 = 5 Pr +5 ln Pr −1 +1 5 T+ ||y+ =30 = 5 Pr +5 ln [5 Pr +1]. For y+ > 30, we are in turbulent core. In turbulent core, the velocity profile is u+ = 2.5 ln y+ + 5.5

11.3 Fully Developed Turbulent Heat Transfer

2.5 du+ = + y dy+ εM 1 − (y+ ∕R+ ) = −1 2.5 ν y+

εM + ν = ν

[1 − (y+ ∕R+ )] 2.5 y+

.

In this region, εM ≫ ν, and for this reason, ( + )[ ] εM y y+ ≈ 1− + ν 2.5 R T+ = T+ ||y+ =30 + ∫

y+ 30

1 Pr

(1 − y+ ∕R+ ) ( + )[ y + Pr1 2.5 1 − t

y+ R+

] dy+ .

In this region, we assume that molecular shearing stress and molecular heat transfer rate are negligible compared to turbulent shearing stress and turbulent heat transfer rate. This means that εM ≫ ν εH ≫ α. Thus, we neglect the (1/Pr) term. Temperature distribution becomes y+

+

dy T+ = T+ ||y+ =30 + 2.5Prt ∫30 y+ [ ( + )] y . T+ = T+ ||y+ =30 + 2.5Prt ln 30 Assume that this time, Prt ≈ 0.9, and temperature distribution becomes for R+ ≫ 30 [ ( + )] y . T+ = 5 Pr +5 ln [5 Pr +1] + 2.25 ln 30 We can now evaluate the centerline temperature T+c at y+ = R+ [ ( + )] R T+c = 5 Pr +5 ln [5 Pr +1] + 2.25 ln 30 √ Re f or using R+ = 2D 8 [ ( √ )] ReD f + Tc = 5 Pr +5 ln [5 Pr +1] + 2.25 ln . 60 8 Next, we write the Stanton number St as St = T+ =

q′′w ρ cp V(Tw − Tm ) Tw − T q′′w = ⇒ ′′ qw ∕(ρcp uτ ) (Tw − Tm )

(

ρcp uτ T+m

Then, the Stanton number becomes 1 1 1 1 ρcp uτ 1 = = + +. St = ρ cp V T+m (V∕uτ ) T+m V Tm We have an expression for V+ √ 8 + V = . f

) .

615

616

11 Turbulent Internal Flow: Momentum and Heat Transfer

The Stanton number St becomes √ f 1 . St = 8 T+m We can express the Stanton number in the following form: √ f 1 1 St = ( ) 8 T+c T+m T+c

T+m

T − Tm = w . + Tw − Tc Tc

We now substitute T+c into the expression for the Stanton number √ f∕8 St = ( ) [ [ ( √ )]] + Tm ReD f 5 Pr +5 ln [5 Pr +1] + 2.25 ln T+ 60 8 c

ReD > 104 0.5 < Pr < 30.

) ( ) ( The next step is to evaluate T+m ∕T+c . For high Prandtl numbers, T+m ∕T+c ≈ 1. The mean temperature Tm is R

Tm =

1 u T 2π(R − y)dy Vπ R2 ∫0

or R+

2 u+ T+ (R+ − y+ )dy+ . V+ × (R+ )2 ∫0 ) ( Let us calculate the T+m ∕T+c ratio for Pr = 10 and Re = 105 . First, we calculate the friction factor f T+m =

f=

1 = 0.01799. [0.79 ln ReD − 1.64]2

Next, we determine V+ √ 8 = 21.086. V+ = f Then, we determine R+ R+ =

ReD 2V+

= 2371.18.

We now find T+c

[

(

√ )] ReD f = 5 Pr +5 ln [5 Pr +1] + 2.25 ln 60 8 )] [ ( √ 105 0.01799 + = 79.4915. Tc = 5 × 10 + 5 ln [5 × 10 + 1] + 2.25 ln 60 8

T+c

Then, we evaluate T+m . We split the integral in three parts [ ] 5 30 R+ 2 + + + + + + + + + + + + + + + + u T (R − y )dy + u T (R − y )dy + u T (R − y )dy Tm = + ∫5 ∫30 V × (R+ )2 ∫0 T+m = 76.5687. ( ) The ratio T+m ∕T+c is (

) 76.5687 = 0.963. T+m ∕T+c = 79.4915

11.3 Fully Developed Turbulent Heat Transfer

) ( This result is 2% different from that given by Mills. The ratio T+m ∕T+c is evaluated and tabulated by Mills [28] for different ( + +) Reynolds number and Prandtl numbers. Mills informs that the ratio Tm ∕Tc is insensitive to the choice of εM /ν. Calculated ) ( ratio of T+m ∕T+c for fully developed turbulent flow in a pipe is presented below in tabular form. Water properties will be evaluated at Tf =

TW + Tm 360 + 290 = = 325 K 2 2

ρ = 987kg∕m3

cp = 4182kJ∕kg.KW∕m.K k = 0.645 Pr = 3.42

μ = 528 × 10−6 N.s∕m2 . The Moody friction factor f The Reynolds number ReD ReD =

ρVD 987 × 5 × 0.04 = = 373 863. μ 528 × 10−6

Flow is turbulent. The Moody friction factor f is obtained using Eq. (11.118) f=

1 = 0.0138. [0.79 ln (ReD ) − 1.64]2

Compute the Nusselt NuD number and compare with an experimental correlation. Gnielinski correlation is an experimental correlation. First, we evaluate the experimental Nusselt number NuD =

(f∕8)(ReD − 1000) Pr = 1321.3. √ 1 + 12.7 (f∕8)(Pr2∕3 − 1)

The Stanton number St is St = (

T+m T+c

√

f∕8 [ ( √ )]] . )[ ReD f 5 Pr +5 ln [5 Pr +1] + 2.25 ln 60 8

Interpolating from Table E11.8, we get ( +) Tm ≈ 0.924 T+c St = 0.001021 NuD = ReD Pr St = 1306. Difference is about 1%, and the result is fairly close to the result obtained by Gnielinski correlation. Table E11.8

Pr

The ratio (Tw − Tm )∕(Tw − Tc ) for fully developed turbulent flow in a pipe. ReD

ReD

ReD

ReD

104

105

106

107

0.7

0.808

0.856

0.883

0.905

1

0.833

0.873

0.896

0.914

5

0.920

0.941

0.949

0.957

10

0.944

0.963

0.968

0.974

100

0.978

0.995

0.999

1.004

1000

0.984

1.001

1.000

1.000

617

618

11 Turbulent Internal Flow: Momentum and Heat Transfer

11.4 HFD Thermally Developing Turbulent Heat Transfer We are interested in flow in a long pipe having an unheated section. The unheated section is long enough for flow to be HFD. Consider now HFD thermally developing incompressible, steady turbulent flow in a pipe. The velocity profile is fully developed before the heating begins. Heating could be under UWT or UHF. When heating begins, the temperature profile begins to develop. Fluid enters the pipe at a uniform and constant temperature Ti , and fluid temperature is uniform at the point where the heating begins. Fluid properties are constant. This is called the thermal entrance problem. See Figure 11.13. Several researchers studied problem. Notter et al. [29], Notter and Sleicher [30], Notter and Sleicher [31], and Sparrow et al. [32] examined the problem. Kays et al. [23] and Ghiaasiaan [5] also presented these solutions in condensed form.

11.4.1

Circular Duct with UWT

Fully developed turbulent pipe flow is under UWT boundary condition for x ≥ 0. Since the velocity profile is fully developed, the radial velocity component v is zero. Axial heat conduction is neglected compared to radial heat conduction. For fully developed flow, the mean velocity u is a function of r. Fluid properties are constant, mean temperature at any radius does not vary with time and viscous dissipation is negligible. The UWT case is important since the “step-function” solution can be used to solve the case of variable wall-temperature distribution by superposition. The energy equation can be put into the following form: [ ] 1 𝜕 𝜕T 𝜕T = r(α + εH ) (11.250) u 𝜕x r 𝜕r 𝜕r T = Ti at x = 0

(11.251a)

𝜕T = 0 at r = 0 𝜕r

(11.251b)

T = Tw at r = R.

(11.251c)

Now, this equation will be put in dimensionless form to obtain a general solution, and for this purpose, the following dimensionless variables are introduced: r η= (11.252) R u (11.253) U= V (x∕R) (11.254) x+ = ReD Pr where ReD = VD/ν is the Reynolds number, V is the average fluid velocity, ν is the fluid kinematic viscosity, ε is the eddy diffusivity of momentum, D is the diameter of the pipe, and U is the dimensionless turbulent velocity profile. The dimensionless wall temperature is defined as θ=

T − Tw Ti − Tw

(11.255)

where Ti is the inlet fluid temperature at the point where the heating begins and Tw is the wall temperature. The mathematical formulation of the problem in dimensionless form is given by [ ( ) ] 𝜕θ Pr εM 𝜕θ 2 𝜕 U + = η 1+ (11.256) 𝜕x η𝜕η Prt ν 𝜕η

Unheated section

Heated section

r

ʺ Tw or qw

u(r) O

x Velocity profile

Figure 11.13

Thermally developing flow.

D=2R

Temperature profile

11.4 HFD Thermally Developing Turbulent Heat Transfer

subject to the boundary conditions θ = 1 at x+ = 0

(11.257a)

𝜕θ = 0 at η = 0 𝜕η

(11.257b)

θ = 0 at η = 1.

(11.257c)

We will briefly present solution of Notter and Sleicher. The application of the separation of variables θ = ℜ(η) X(x+ ) yields the appropriate eigenvalue problem needed for the solution of the problem given by {[ ] } λ2 Pr εM dℜn d 1+ η + n η U(η)ℜn = 0. dη Prt ν dη 2

(11.258)

The boundary conditions for this equation are ℜ′n (0) = 0

(11.259a)

ℜn (1) = 0

(11.259b)

where the prime denotes differentiation with respect to η. Here, ℜn (η) and λn are the eigenfunctions and eigenvalues of the Sturm–Liouville differential equation, respectively. The solution of this Sturm–Liouville problem requires an expression for dimensionless velocity U, the ratio of eddy diffusivity of momentum to kinematic viscosity εM /ν, and the turbulent Prandtl number Prt . Then, Notter and Sleicher introduced the temperature distribution in the form θ=

∞ ∑

) ( Cn ℜn (η) exp −λ2n x+

(11.260)

n=0

where ℜn (η) are eigenfunctions. Cn ’s are constants and are evaluated from (

Cn = λn

2 ) 𝜕ℜn (λn η) || | | 𝜕λn |η=1

(11.261a)

or 1

Cn =

∫0 U(η)η ℜn (λn η)dη

(11.261b)

1

∫0 U(η)η ℜ2n (λn η)dη

The heat flux at the pipe wall is given by ) ( 4 k (Ti − Tw ) ∑ Gn exp −λ2n x+ D n=0 ∞

q′′ (x+ ) = −

(11.262)

where Ti is the inlet temperature and Tw is the wall temperature. The constants Gn are evaluated from dℜn || 1 . Gn = − Cn 2 dη ||η=1

(11.263)

The mean temperature θm is ( ) ∞ ∑ ) ( Gn θm = 8 exp −λ2n x+ . 2 λn n=0 The local Nusselt number is defined as ∞ ) ( ∑ Gn exp −λ2n x+ ′′ + (x )D q n=0 w . NuD (x+ ) = = ∞ ) ( ) k(Tw − Tm ) ∑( Gn ∕λ2n exp −λ2n x+ 2

(11.264)

(11.265)

n=0

The parameters λ0 , λ1 , λ2 and G0 , G1 , G2 are functions of Reynolds number and Prandtl numbers given in Table 11.2.

619

620

11 Turbulent Internal Flow: Momentum and Heat Transfer

Table 11.2

Selected eigenvalues and constants for the turbulent Graetz problem. λ20

ReD

λ21

λ22

G0

G1

G2

C0

−C1

C2

Pr = 0.1 10 × 103

18.66

113.6

296

1.928

1.235

0.965

1.468

0.774

0.540

20 × 103

27.12

171.6

450.7

2.89

1.701

1.304

1.444

0.728

0.499

3

50 × 10

48.05

327.5

876.1

5.34

2.65

1.959

1.398

0.644

0.431

100 × 103

77.13

564.7

1 534

8.79

3.77

2.71

1.361

0.577

0.378

200 × 103

127.4

1 007

2 777

14.79

5.46

3.84

1.325

0.515

0.332

500 × 103

253.6

2 226

6 239

29.9

9.16

6.27

1.284

0.444

0.280

1000 × 103

437.3

4 150

11 750

52.3

14.05

9.45

1.257

0.400

0.249

7.596

1.829

1.217

1.239

0.369

0.227

Pr = 0.72 10 × 103 20 × 10

3

64.38

646.8

1 870

109

1 119

3 240

13.06

2.95

1.784

1.231

0.352

0.208

50 × 103

219.0

2 350

6 808

26.6

5.63

3.32

1.220

0.333

0.193

100 × 103

375.9

4 183

12 130

45.8

9.25

5.48

1.210

0.319

0.185

200 × 103

651.2

7 539

21 940

79.6

15.03

9.10

1.200

0.302

0.177

3

1 357

16 630

48 540

166

28.9

17.50

1.190

0.282

0.165

500 × 10

Pr = 3 10 × 103

119.1

2 736

–

14.5

1.37

–

1.097

0.151

20 × 103

209.4

4 680

–

25.8

2.20

–

1.097

0.146

3

50 × 10

445.6

10 250

–

55.3

4.45

–

1.093

0.139

100 × 103

794.9

18 240

–

98.9

7.71

–

1.093

0.137

3

1 429

33 980

–

178

13.7

–

1.090

0.133

500 × 103

3 147

77 120

–

391

28

–

1.087

0.127

1000 × 103

5 733

144 200

–

713

51.1

–

1.085

0.125

200 × 10

Pr = 8 10 × 103

176.6

–

–

21.6

–

–

1.056

20 × 103

313.5

–

–

38.7

–

–

1.056

3

50 × 10

685.6

–

–

85.4

–

–

1.054

100 × 103

1 232

–

–

154

–

–

1.054

200 × 103

2 271

–

–

284

–

–

1.054

500 × 103

5 020

–

–

625

–

–

1.052

9 369

–

–

1 170

–

–

1.052

λ20

λ21

λ22

G0

G1

G2

C0

1000 × 10

3

ReD

Pr = 20 10 × 103

247.9

30.3

1.033

20 × 103

448.2

55.4

1.033

3

990.6

124

1.032

100 × 103

1 799

225

1.032

200 × 103

3 346

418

1.032

500 × 103

7 509

936

1.031

14 090

1 760

1.031

50 × 10

1000 × 10

3

−C1

C2

11.4 HFD Thermally Developing Turbulent Heat Transfer

Table 11.2

(Continued) Pr = 50

10 × 103

348

42.6

1.019

3

631.1

78.1

1.019

50 × 103

1 393

174

1.018

100 × 103

2 570

321

1.018

200 × 103

4 778

598

1.018

10 800

1 350

1.018

20 420

2 550

1.018

54.5

1.012

20 × 10

500 × 10

3

1000 × 103

Pr = 100 10 × 103

444.6

3

811.1

100

1.012

50 × 103

1 788

223

1.012

100 × 103

3 317

415

1.012

200 × 103

6 129

766

1.012

500 × 103

14 040

1 750

1.012

26 220

3 270

1.012

20 × 10

1000 × 10

3

Pr = 1000 10 × 103

988.4

121

1.003

20 × 103

1 794

222

1.003

50 × 103

4 012

501

1.003

100 × 103

7 383

923

1.003

3

13 820

1 730

1.003

500 × 103

31 380

3 910

1.003

ReD

λ20

200 × 10

λ21

λ22

G0

G1

G2

C0

−C1

C2

Pr = 10 000 10 × 103

2.132

261

1.001

20 × 103

3.907

484

1.001

3

50 × 10

8.695

1 090

1.001

100 × 103

16 140

2 020

1.001

200 × 103

29 840

3 732

1.001

Source: Data from Notter and Sleicher [31].

The formulas for the calculation of the asymptotic values for the eigenvalues and constants are given by Sleicher and Notter ) ( 2 ∕G (11.266) λn = n + 3 ⎫ ⎧ ⎪ 0.897(−1)n H1∕6 ⎪ 1 Cn = [ ( ) ]⎬ √ ⎨ 2∕3 G 2g0 λn ⎪ 1 + 1 2 c ln (Gλn π) − 2 + 12 ⎪ (Gλn ) 2π 3 6π ⎭ ⎩ ( )1∕3 { [ ]} 1 0.201 H 7 c . ( ln (G λ 1− Gn = π) − 1) + n G λn (G λn )2 2π 36π2

(11.267a)

(11.267b)

621

622

11 Turbulent Internal Flow: Momentum and Heat Transfer

Table 11.3

Values of G, g0 , and c.

ReD

G

g0

c

ReD

G

Pr = 0.1

g0

c

Pr = 0.72

10 × 103

0.154

2.51

7

10 × 103

0.0609

19.1

15

20 × 103

0.125

3.98

9

20 × 103

0.0456

33.9

15

3

3

50 × 10

0.0891

8.16

11

50 × 10

0.0311

72.7

11

100 × 103

0.0671

14.8

13

100 × 103

0.0232

131

7.8

3

0.0172

238

3.9

3

0.0497

27.6

14

500 × 103

0.0331

63.7

15

200 × 10

200 × 10

Source: Notter et al. [29].

The parameters G, g0 , and c are functions of Reynolds number and Prandtl numbers, and typical values are given in Table 11.3. The value of H is obtained from (11.268)

H = (ReD ∕32)f

and f is the Moody friction factor. The fully developed Nusselt number for the case of UWT is given by the first term of the solution λ20 (11.269) 2 Far downstream from the thermal entrance region, all terms except the first of Eq. (11.265) become small and (the asymptotic Nusselt number) Eq. (11.269) is obtained. Burmeister [3] gives the mean Nusselt number based on the log-mean temperature difference ΔTLMTD ( ) λ20 λ20 ReD Pr + ln NuD = (11.270) 2 (z∕D) 8 G0 Nu∞ =

or NuD Nu∞

Re Pr 2 =1+ D ln (z∕D) λ2n

(

λ20 8 G0

) .

(11.271)

Example 11.9 Consider HFD thermally developing atmospheric air flow inside a 5-cm tube with a velocity of 16 m/s. Here, it is assumed that tube has an unheated initial section, and the unheated section is long enough for the velocity profile to become fully developed before heating begins. See Figure 11.E9. The atmospheric air at a mean temperature of 27 ∘ C flows in the heated section of the tube. Assume that Pr = 0.72. (a) Plot the local mean fluid temperature θm as a function dimensionless axial distance x+ . (b) Plot the local Nusselt number Nu(x+ ) as a function dimensionless axial distance x+ . (c) Calculate the local Nusselt number at x = 20 cm. Solution Air properties at 300 K are ρ = 1.1614

kg m3

m2 s The Reynolds number is ν = 15.89 × 10−6

ReD =

J N.s μ = 184.6 × 10−7 2 kg K m W m2 k = 0.0263 Pr = 0.707 α = 22.5 × 10−6 . m.K s

cp = 1007

ρVD 1.1614 × 16 × 0.05 = = 50 495. μ 184.6 × 10−7

11.4 HFD Thermally Developing Turbulent Heat Transfer

Flow is turbulent. We assume that the Prandtl number is Pr ≈ 0.72 and ReD ≈ 50 000 so that we can use Tables 11.2 and 11.3 L 200 cm = = 40 < 60. D 5 cm Next, we write the following equations to do calculations: (x∕R) ReD Pr ( ) ∞ ∑ ) ( Gn θm = 8 exp −λ2n x+ . 2 λn n=0 x+ =

The local Nusselt number is defined as ∞ ∑

NuD (x+ ) =

q′′w (x+ )D k(Tw − Tm )

=

n=0

) ( Gn exp −λ2n x+

∞ ( ) ( ) ∑ Gn ∕λ2n exp −λ2n x+ 2

.

n=0

For Re = 50 000 and Pr = 0.72, from Table 11.3, we get G = 0.0311 g0 = 72.7 c = 11. For Re = 50 000 and Pr = 0.72, we take the first three eigenvalues λ0 , λ1 , λ2 and first three constants G0 , G1 , G2 from Table 11.2. Next, we now compute the eigenvalues for n > 3 using ) ( n + 2∕3 . λn = G Next, we calculate Gn for n > 3, and we need Eq. (11.267b) ( )1∕3 { [ ]} 0.201 H 7 c 1 . 1− ( ln (G λn π) − 1) + Gn = G λn (G λn )2 2π 36π2 To calculate Gn , we need friction factor f and H f=

1 = 0.02088. [1.82 log10 (ReD ) − 1.64]2

Then, we calculate H ( ) f H = ReD = 32.395. 32 Finally, we construct the following table:

n

𝛌n

Gn

n

𝛌n

Gn

0

14.798

26.6

6

214.362

3.18

1

48.476

5.63

7

246.517

3.08

2

82.510

3.32

8

278.680

3.00

3

117.899

3.42

9

310.825

2.91

4

150.053

3.36

10

342.979

2.84

5

182.207

3.27

We plot the mean temperature and the local Nusselt number, as shown in Figures 11.E9a and 11.E9b. > #Calculation of Nusselt number at x = 0.2 m > # x plus is represented by X

623

624

11 Turbulent Internal Flow: Momentum and Heat Transfer

Unheated section

Heated section D = 5 cm 0

x 2m

Figure 11.E9

Geometry and problem description for Example 11.9.

1.0 0.9 0.8 0.7

θm(x+)

0.6 0.5 0.4 0.3 0.2

0

0.002

0.004

0.006

0.008

0.010

x 2 D RePr (a)

Figure 11.E9a

Variation of dimensionless mean temperature along the tube.

> x ≔ 0.2; x ≔ 0.2 > d ≔ 0.05; d ≔ 0.05 >X≔

2•

( ) x d

R • Pr

X ≔ 0.0002222222222 ) ( 10 ∑ (G[i]) • exp(−(λ[i])2 • X) i=0 ; > Nu ≔ ) 10 ( ∑ G[i] 2 • • • 2 exp(−(λ[i]) X) (λ[i])2 i=0

N ≔ 126.1929498

11.4 HFD Thermally Developing Turbulent Heat Transfer

126 124 122 120

NuD

118 116 114 112 110 0

0.002

0.004

0.006

0.008

0.010

x 2 D RePr (b)

Figure 11.E9b

Variation of local Nusselt number along the tube.

11.4.2 Circular Duct with Uniform Wall Heat Flux We will now consider the thermal entrance problem for a turbulent flow in a tube under UHF. The fluid enters tube having radius R and the inlet temperature Ti . The velocity profile is fully developed at the point where heating begins. The tube is heated for x ≥ 0 by a UHF q′′0 . We again start with the energy equation for the problem under consideration as given below [ ] 1 𝜕 𝜕T 𝜕T = r(α + εH ) (11.272) u 𝜕x r𝜕r 𝜕r and the boundary conditions are at x = 0 T = Ti at r = 0

𝜕T = 0 or T = finite 𝜕r

(11.273a) (11.273b)

𝜕T (11.273c) = q′′0 . 𝜕r Here, again axial heat conduction is neglected compared to radial heat conduction. The mean velocity u is a function of r and flow is HFD 𝜕u∕𝜕x = 0. Now, this equation can be put in dimensionless form to obtain a general solution, and for this purpose, the following dimensionless variables are introduced: r η= (11.274) R u U= (11.275) V 2(x∕D) (11.276) x+ = ReD Pr at r = R k

is the Reynolds number. Here, V is average fluid velocity, ν is the fluid kinematic viscosity, and ε is the where ReD = VD ν eddy diffusivity of momentum. The dimensionless wall temperature is defined as ( ) θ = (T − Ti )∕ q′′0 R∕k (11.277)

625

626

11 Turbulent Internal Flow: Momentum and Heat Transfer

where Ti is the inlet fluid temperature at the point where the heating begins and Tw is the unknown wall temperature. With these dimensionless variables, the energy equation and its boundary conditions become [ ( ) ] 𝜕θ Pr εM 𝜕θ 2 𝜕 U + = η 1+ (11.278) 𝜕x η𝜕η Prt ν 𝜕η θ = 0 at x+ = 0

(11.279a)

𝜕θ = 0 at η = 0 𝜕η

(11.279b)

𝜕θ = 1 at η = 1. 𝜕η

(11.279c)

Sleicher and Notter [30] and Sparrow et al. [32] solved this problem. Since the wall heat flux is uniform, they split the temperature profile θ(x+ , η) into a sum of a fully developed temperature profile θ1 (x+ , η) and an entry region profile θ2 (x+ , η). The following temperature profile is assumed: θ = θ1 + θ2 .

(11.280)

At large distances, the downstream of the thermal entrance region, θ2 → 0, approaches zero. Substituting Eq. (11.280) into the energy equation results in two problems as discussed below [ ( ) ] 𝜕θ Pr εM 𝜕θ1 2 𝜕 U +1 = η 1+ (11.281) 𝜕x η𝜕η Prt ν 𝜕η 𝜕 θ1 = 1 at η = 1 𝜕η

(11.282a)

𝜕θ1 = 0 at η = 0 𝜕η

(11.282b)

and U

[ ( ) ] 𝜕θ2 Pr εM 𝜕θ2 2 𝜕 η 1 + = 𝜕x+ η𝜕η Prt ν 𝜕η

(11.283)

𝜕θ2 = 0 at η = 0 𝜕η

(11.284a)

𝜕 θ2 = 0 at η = 1. 𝜕η

(11.284b)

In addition to these boundary conditions, both θ1 and θ2 are subject to an unspecified boundary condition at z+ = 0. 11.4.2.1 Solution for Fully Developed Temperature Distribution 𝛉1

̃ a constant for As specified by Sparrow et al. [32], a property of the fully developed profile for UHF case is that 𝜕θ1 ∕𝜕ξ = A, all values of η. Then, the fully developed solution θ1 can be cast into the following form: ̃ + + H(η). θ1 (z+ , η) = Ax

(11.285)

̃ = 4. The first term of Eq. (11.285) on the right-hand side represents the axial variation of mean fluid temperature and A The substitution of this equation into Eq. (11.281) gives the following ordinary differential equation for H(η): [ ( ) ] ̃ Pr εM dH A d η 1+ − Uη = 0 (11.286) dη Prt ν dη 2 subject to boundary conditions H′ (1) = 1

(11.287a)

H′ (0) = 0.

(11.287b)

11.4 HFD Thermally Developing Turbulent Heat Transfer

Sleicher and Tribus have shown that the fully developed Nusselt number for the uniform wall heat flux boundary condition is 1 (11.288) Nu∞ = ∞ ( ) ∑ Gn ∕λ4n 16 n=0

where Gn and λn are functions of Reynolds and Prandtl numbers, respectively. Gn and λn are the same as those for the UWT solution. 11.4.2.2 Solution for the Entry Region Temperature Distribution 𝛉2

The thermal entrance effect of the problem is presented next. The entry region temperature θ2 is [ ( ) ] 𝜕θ2 Pr εM 𝜕θ2 2 𝜕 η 1+ U + = 𝜕x η𝜕η Prt ν 𝜕ξ + 𝜕 θ2 (x , 1) =0 𝜕η 𝜕 θ2 (x+ , 0) = 0. 𝜕η The third boundary condition is

(11.289) (11.290a) (11.290b)

(11.290c)

θ2 (0, η) = −H(η).

The entry region profile θ2 is obtained by solving the above equation. The separation of variables can be used, and the problem reduces to [ ( ) ] ̃2 λ Pr εM dRn d η 1+ + n U η Rn = 0 (11.291) dη Prt ν dη 2 R′n (0) = 0

(11.292a)

R′n (1) = 0

(11.292b)

and normalizing condition Rn (0) = 1. The temperature distribution θ2 is ∞ ( 2 ) ∑ ̃ n ℜn (η) exp −̃ C θ2 = λn x+ .

(11.293)

n=1

̃ n are evaluated from The constants C 1

̃n = C

∫0 −H(η) ηU(η)ℜn dη 1

∫0 ηUℜ2n dη

̃ n is expressed as or C 2 Cn = . 𝜕 R′n (1) λn 𝜕̃ λ

(11.294a)

(11.294b)

n

Local Nusselt numbers for the thermal entrance region are given by Sleicher and Notter as 2 NuD (x+ ) = ( ) ∑ ( 2 ). ∞ 2 + G exp −̃ λn x+ n Nu ∞

(11.295)

n=1

The constant Gn is given by ̃ n Rn (1). ̃n = C G

(11.296)

Some numerical values of eigenvalues and the constants Gn are presented in Table 11.4. The asymptotic values of λn and Gn can be computed from the relations given below } { 0.189 G2∕3 ̃λ = 1 n + 1 − (11.297) n G 3 H1∕3 (n + 1∕3)2∕3

627

628

11 Turbulent Internal Flow: Momentum and Heat Transfer

Table 11.4 Selected eigenvalues and constants for the thermal entrance region problem under uniform heat flux boundary condition. 2 ̃ 𝛌1

ReD

2 ̃ 𝛌2

2 ̃ 𝛌3

̃ −G 1

̃ −G 2

̃ −G 3

0.0286

0.0165

Pr = 0.1 10 × 103

69.52

224.9

3

463

0.737

219.6

695.9

1 421

0.0250

0.0109

0.00667

100 × 103

396.9

1 247

2 531

0.0143

0.00663

0.00427

500 × 103

1 718

5 341

10 750

0.00344

0.00176

0.00122

0.00738

0.00653

50 × 10

Pr = 0.72 10 × 103

519.5

1 624

3

50 × 10

3 202

0.0123

1 952

6 154

12 480

0.00296

0.00147

0.00106

100 × 103

3 510

11 030

22 340

0.00164

0.00081

0.00056

500 × 103

14 310

44 690

89 830

0.000405

0.00020

–

Pr = 3 10 × 103

2 483

0.00338

50 × 103

9 507

0.000646

3

16 980

0.000349

500 × 103

72 190

0.000083

100 × 10

Pr = 8 10 × 103 50 × 10

6 564

3

25 300

100 × 103 500 × 10

0.00144

3

45 100 192 000

Source: Notter et al. [29].

̃n = G

0.762 1∕3

GH

5∕3 ̃ λn

[ 1+ [ 1−

] 0.343 2∕3 H1∕3 ̃ λ n

].

(11.298)

0.343 2∕3 H1∕3 ̃ λ n

The parameters G and H are the same as those for the UWT case. Calculations indicate that Nusselt numbers for UWT and UHF boundary conditions of turbulent pipe flow are nearly same as discussed in [24] NuUHF ≈ NuUWT for Pr ≥ 0.72.

(11.299)

Al-Arabi [33] presented correlation for the mean Nusselt number NuD for thermally developing flow with UWT or UHF boundary conditions: NuD C =1+ Nu∞ (x∕D)

(11.300)

3500 < Re < 105 where Nu∞ stands for the fully developed Nusselt number for UWT or UHF boundary condition and constant C is given as [ ] (x∕D)0.1 3000 0.68 + . (11.301) C= Pr1∕6 Re0.81

11.5 Analogies for Internal Flow

Al-Arabi reports that this equation is valid for air, water, and oil for the values of (x/D) > 3, for the values of Re from 5 × 103 to 105 , and for the values of Pr from 0.7 to 75.

11.5 Analogies for Internal Flow The analogies developed by different investigators have been extensively applied to pipe flows because of importance in engineering. Analogies presented for fully developed turbulent pipe flows may also be applied to fully developed turbulent flow between parallel plates. The fully developed state is attained in a relatively short distance from the entrance. Formulations are presented in a simplified form. The starting point for the analogies is the definition of total heat flux and momentum flux as given below q′′ = −ρcp (α + εH ) τ = ρ(ν + εM )

dT dy

(11.302)

du . dy

(11.303)

Experimental studies show that the heat flux and shear stress are approximately constant in the inner region. For a fully developed turbulent flow in a pipe, pressure gradient dp∕dx along the pipe is nearly constant. Actually, the pressure gradient in a pipe flow destroys the analogy. Experiments indicate that we can safely neglect errors in analogy due to pressure gradient since such departures appear to remain within the limits of experimental errors involved with heat transfer measurements. We can also safely assume that the velocity and temperature profiles are uniform in the turbulent core since profiles are nearly flat if we neglect the viscous sublayer and the buffer layer. The velocity and temperature in a pipe are approximately equal to their respective bulk values and can be estimated as u ≈ V and T ≈ Tm .

11.5.1 Reynolds Analogy The Reynolds analogy [34] provides a relation between momentum and heat transfer. This analogy is the first and most widely known analogy and is postulated by Reynolds in 1874. It is based on the postulate that the mechanism of momentum and energy transport is similar. The Reynolds analogy neglects both the viscous sublayer and the buffer layer, and it is based on the turbulent core of the flow. In other words, the Reynolds analogy assumes that the entire flow field consisted of a single zone of a highly turbulent region. Consider now incompressible constant property HFD and TFD turbulent flow in a pipe. Notice that y + r = R. If we take the coordinate system shown in Figure 11.14 into consideration, the following relations apply: q′′ = −ρcp (α + εH ) τ = ρ(ν + εM )

dT dy

(11.304)

du . dy

(11.305)

The eddy diffusivity of momentum εM and the eddy diffusivity of heat εH are highly complex functions of turbulent flow field. However, both are representations of the turbulent mixing fluctuations. It is expected that they will have similar values, and this is confirmed by experiments. We now assume that in the inner region, q′′ ≈ q′′w and τ ≈ τw . It is also assumed that the distributions of total shear stress and heat transfer rate are similar. Next, we divide Eq. (11.305) by Eq. (11.304). This will give us u≈V r

Flow

y Figure 11.14

R

Pipe center

T ≈ Tm Turbulent core

Coordinate system for turbulent flow in a pipe.

629

630

11 Turbulent Internal Flow: Momentum and Heat Transfer

(ν + εM ) du τw ∕ρ (ν + εM ) du dy = =− . dT (α + εH ) dT q′′w ∕ρcp −(α + εH ) dy

(11.306)

For turbulent core, ν ≪ εM and α ≪ εH . Recall that the viscous sublayer and the buffer layer are neglected. We obtain from Eq. (11.306) dT = −

′′ 1 qw Pr du cp τw t

(11.307)

where Prt = εM /εH is the turbulent Prandtl number. Here, T is the mean temperature at position y and u is the mean velocity at position y. Integrating Eq. (11.307) outward from wall, at y = 0, where u = 0 and T = Tw up to the pipe centerline where y = R, u ≈ V, and T ≈ Tm , we obtain Tm

dT = −

∫ Tw

V ′′ 1 qw Prt du cp τw ∫0

(11.308)

and after performing the integration of Eq. (11.308), we get q′′w cp (Tw − Tm )

Prt =

τw . V

(11.309)

In pipe flow, it is common to work with the Moody friction factor f. We need a relation between wall shear stress τw and friction factor f. This relation is obtained by a force balance on a control volume for fully developed turbulent flow in a pipe as we have done before. The force balance is πDLτw = (p1 − p2 )(πD2 ∕4)

(11.310)

and consequently, the final result can be expressed as f 2 ρV . 8 Using Eq. (11.311), we obtain the following equation: τw =

q′′w

1

(Tw − Tm ) (ρ cp V)

=

1 f . Prt 8

(11.311)

(11.312)

Now, recall the definition heat transfer coefficient for flow in a pipe h=

q′′w Tw − Tm

and using the definition of the heat transfer coefficient h, we obtain an expression for the Stanton number St: q′′w h 1 = St. = ρcp V (T − T ) ρcp V w m

(11.313)

Thus, we may express Eq. (11.313) as St =

1 f . Prt 8

(11.314a)

If we choose Prt ≈ 1, i.e. εM = εH , then we get f . (11.314b) 8 This is the Reynolds analogy. In the Reynolds analogy, it is also expected that α = ν, and this means Pr = 1. Most gases have Prandtl numbers close to unity. The Stanton number is defined in terms of the Nusselt number NuD , the Reynolds number ReD , and the Prandtl number Pr as St = NuD /ReD Pr. We may express the Reynolds analogy in terms of the Nusselt number NuD = hD/k and the Reynolds number ReD = VD/ν as St =

NuD = (f∕8)ReD Pr .

(11.315)

11.5 Analogies for Internal Flow

We made certain assumptions in deriving this relation. Recall that we neglected the viscous sublayer and the buffer layer with respect to the turbulent core as we have done before. In spite of simplifications, the Reynolds analogy gives results that agree with experimental data. Most important restriction concerning Eq. (11.314a) is the Prandtl number limitation. Note that the Reynolds analogy is only valid for fluids having Pr ≈ 1 such as gases. Using the Reynolds analogy, we obtain information about heat transfer from friction measurements. Analogy fails for high or low Prandtl number fluids.

11.5.2 Colburn Analogy The Chilton–Colburn analogy [35] has proved to be useful in engineering since it is based on empirical correlations and is an empirical modification of the Reynolds analogy to include the effect of the Prandtl number. Of course, it is better not to extrapolate the empirical correlations. Colburn suggested that most experimental data can be approximated by introducing a term to allow for the effect of Prandtl number variation. Colburn has proposed a correlation for turbulent flow in a pipe in the form f St Pr2∕3 = (11.316) 8 104 < ReD < 3 × 105 0.6 < Pr < 60 where f is the Moody friction factor and the average Stanton number is q′′w

St =

=

ρcp V(Tw − Tm )

h . ρcp V

(11.317)

This analogy has been found to be quite accurate in the range 0.5 < Pr < 50 provided that the drag forces are wholly viscous in nature.

11.5.3 Prandtl–Taylor Analogy The Prandtl–Taylor analogy is discussed in [36], and this analogy is based on the concept of two layers of turbulent flow: (a) a viscous sublayer (b) a turbulent zone. Consider fully developed turbulent flow in a pipe. See Figure 11.15. The pipe wall temperature Tw is constant. We now assume that turbulent flow comprises a viscous sublayer of thickness δ and the turbulent core. Dimensionless temperature and velocity profiles are similar. Integration over each layer will be performed separately, and the results of integration are coupled at the interface. Now, we will consider each sublayer separately. The apparent (total) shear stress and heat flux in turbulent flow are given by q′′ = −ρcp (α + εH ) τ = ρ(ν + εM )

dT dy

du . dy Pipe center

r y

V uδ

R

Tm Turbulent core Tδ

δ

Viscous sublayer τw

Figure 11.15

ʺ qw

Coordinate system for turbulent flow in a pipe; Prandtl–Taylor analogy.

631

632

11 Turbulent Internal Flow: Momentum and Heat Transfer

11.5.3.1 Laminar Sublayer

We assume that ν ≫ εM , α ≫ εH , τ ≈ τw , and q′′ ≈ q′′w . Within the viscous sublayer, the turbulent diffusion of heat and momentum is neglected. Under these assumptions, the apparent stress and heat flux are τw = μ

du dy

(11.318)

q′′w dT = −α . ρcp dy

(11.319)

We now integrate these equations. The integration of Eq. (11.318) yields uδ

δ

τw

dy = μ du ∫0 ∫0

μuδ δ and the integration of Eq. (11.319) gives τw =

Tδ q′′w δ dy = −α dT ∫Tw ρcp ∫0 q′′w α = (Tw − Tδ ). ρcp δ

Dividing Eq. (11.321) by Eq. (11.320) yields cp q′′w (Tw − Tδ ) = τw uδ Pr ) ( where Pr = ν/α and recall that (q′′ ∕τ) = q′′w ∕τw = constant; we obtain from Eq. (11.322) ( ) uδ Pr q′′w (Tw − Tδ ) = . cp τw

(11.320)

(11.321)

(11.322)

(11.323)

11.5.3.2 Turbulent Core

The eddy coefficient of momentum εM and the eddy coefficient of heat εH are highly complex functions of flow field, and both are representations of the turbulent mixing fluctuations. In the fully turbulent zone, we assume that εM ≫ ν and εH ≫ α. This time, the molecular diffusion of heat and momentum is neglected. With these assumptions, we have τw du = εM ρ dy q′′w dT = −εH . ρcp dy

(11.324) (11.325)

Many measurements indicate that we may assume εM ≈ εH ; thus, Prt = εM /εH ≈ 1 and combine last two equations to get dT = −

q′′w du τw cp

(11.326)

Note that this relation is being applied only in the fully turbulent zone. At y = δ, u = uδ , and T = Tδ , and at the center of the pipe u = V, T = Tm . Also, Prandtl assumed that q′′ ∕τ = q′′w ∕τw = constant. Now, carrying out the integration Tm

V q′′w du. ∫ Tδ τw cp ∫uδ q′′ (Tδ − Tm ) = w (V − uδ ). τw cp

dT = −

Eliminating Tδ from Eqs. (11.327) and (11.323) yields )[ ( ] uδ q′′w V 1 + (Pr −1) . Tw − Tm = τw cp V

(11.327)

(11.328)

11.5 Analogies for Internal Flow

Using the definition of the heat transfer coefficient h and the Moody friction f q′′w = h(Tw − Tm ) τw =

(11.329)

( ) f ρV2 . 8

(11.330)

We obtain the following relation: St =

NuD h f 1 . = = ReD Pr ρcp V 8 1 + (uδ ∕V)(Pr −1)

(11.331)

It is worth to note that we are assuming only Prt = εM /εH ≈ 1 and Prt is a property of the flow field. On the other hand, the Prandtl number Pr = ν/α is a fluid property and may be different from unity. Next, we need to evaluate the unknown velocity uδ . This done by using the universal velocity profile. For turbulent flow in a pipe, the ratio of the velocity at the outer edge of the laminar sublayer to mean velocity V is obtained as follows: For laminar sublayer: u+ = y+

(11.332)

√ where u+ = u∕ τw ∕ρ. We now set y+ = 5 at the edge of the sublayer, then 5= √

uδ

=

τw ∕ρ

uδ √( ) . f V 8

Therefore, uδ =5 V

√

f . 8

(11.333)

Using this result, the Prandtl–Taylor analogy takes the following form: St =

f 8

√ 1+5

1 f (Pr 8

.

(11.334)

−1)

Thus, after replacing St by NuD /ReD Pr, Eq. (11.334) becomes NuD =

(f∕8)ReD Pr . √ 1 + 5 (f∕8)(Pr −1)

(11.335)

Equation (11.335) is obtained by taking the laminar sublayer thickness into consideration. Notice that this equation implies that the buffer layer is included in the turbulent core. In other words, the buffer zone in which εM and εH are of the same order of magnitude is omitted in this analysis.

11.5.4 von Karman Analogy Consider fully turbulent flow in a pipe. See Figure 11.16. The pipe wall temperature Tw is constant. The analogy is based on the concept of three layers in the turbulent flow: (a) a viscous sublayer (b) a buffer layers (c) a turbulent zone. von Karman [16] improved the Prandtl–Taylor analogy by including the buffer layer through the use of the universal velocity distribution. Fluid properties are constant. Throughout the boundary layer, it is assumed that q′′ ≈ q′′w and τ ≈ τw . The coordinate system is shown in Figure 11.6.

633

634

11 Turbulent Internal Flow: Momentum and Heat Transfer

Pipe center

r

V ub

y

Tb

Tm

Buffer layer

b L

Laminar sublayer

τw Figure 11.16

Turbulent core

uL

ʺ qw

TL

Coordinate system for turbulent flow in a pipe; von Karman analogy.

11.5.4.1 Laminar Sublayer: 0 ≤ y+ ≪ 5

We assume that ν ≫ εm and α ≫ εh . In the laminar sublayer, the eddy coefficients εM and εH vanish. There is a linear temperature variation in the laminar sublayer. The shear stress and heat flux are constant and are equal to their wall values τw and q′′w q′′ q′′ dT = w = −α ρcp ρcp dy

(11.336)

and wall shear stress τ du τ = w = −ν . ρ ρ dy

(11.337)

We take the ratio of these two relations as we have done before; then, we obtain q′′w ∕ρcp τw ∕ρ

α dT . ν du

=−

(11.338)

The integration of this relation from the wall up to the edge of the laminar sublayer gives q′′w ∕ρcp

uL

τw ∕ρ ∫0

T

du = −

Tw − TL =

L α dT ∫ ν Tw

′′ Pr qw u . cp τw L

(11.339)

In the laminar sublayer, the velocity profile is given by u + = y+ and at the edge of the sublayer, u = uL and y+ = 5, and we can now evaluate uL at the sublayer as we have done before. The result is √

uL

= 5.

(11.340)

τw ∕ρ

Substituting Eq. (11.340) into Eq. (11.339), we obtain the following relation: ( ′′ ) ( ) √ qw τw Pr Tw − TL = 5 τw cp ρ or after some algebra, we obtain ( )√ q′′w ρ (5 Pr ). Tw − TL = ρcp τw

(11.341)

11.5 Analogies for Internal Flow

11.5.4.2 Buffer Layer: 5 ≤ y+ ≪ 30

In the buffer zone, the eddy diffusivity of heat εH and the eddy diffusivity of momentum εM are of the same order as their molecular counterparts. Further, the shear stress and heat flux are taken to be constant within the buffer zone and are again equal to their wall values τw and q′′w . We assume that Prt ≃ 1, εM ≈ εH , q′′ = q′′w , and τ = τw . Shear stress and heat flux are given as τw = ρ(ν + εM )

du dy

q′′w = −ρcp (α + εM )

(11.342) dT . dy

(11.343)

The velocity profile in the buffer layer is u+ = 5 ln y+ − 3.05

(11.344)

and using the velocity profile in buffer layer, we obtain 5 du+ = +. y dy+ √ √ Using y+ = y τw ∕ρ∕ν and u+ = u∕ τw ∕ρ, after some algebra, we obtain ( ) τw 5 du = . dy ρν y+ We now combine Eq. (11.345) with the stress equation, Eq. (11.342), and we get ) ( + y −1 . εM = ν 5

(11.345)

(11.346)

√ Recalling that Prt ≈ 1, we now combine Eq. (11.346) with the equation for heat flux, Eq. (11.343), using y+ = y τw ∕ρ∕ν to get √ q′′w ρ 1 dT = − (11.347) ( + )] . [ + ρc p τw 1 y dy + −1 Pr 5 We now integrate this relation from y+ = 5 to y+ = 30 as given below √ Tb 30 q′′ dy+ ρ dT = − w ( + )] . [ ∫TL ρc p τw ∫5 y 1 + −1 Pr 5 Using Maple 2020, we carry out the integration √ q′′ ρ TL − Tb = w 5 ln (5 Pr + 1). ρcp τw

(11.348)

11.5.4.3 Turbulent Core: y+ ≥ 30

In the fully turbulent zone, the molecular coefficients ν and α negligible. Further, it is also assumed that εM ≈ εH . We assume that εM ≫ ν and εH ≫ α for the fully turbulent region. It is also assumed that q′′w ≈ q′′ and τw ≈ τ. Then, we have for the shear stress and heat flux τw du = εM ρ dy q′′w dT = −εH . ρcp dy It is assumed that εM ≃ εH , and thus, we have Prt = 1. We now combine these last two equations to get dT = −

q′′w du. τw cp

(11.349)

635

636

11 Turbulent Internal Flow: Momentum and Heat Transfer

Integrating Eq. (11.349) yields Tm

V q′′w du ∫ Tb τw cp ∫ub q′′ Tb − Tm = w (V − ub ) τw cp

dT = −

or

( ) + + q′′w V − ub Tb − Tm = (11.350) √ τw cp τw ∕ρ √ √ + where V+ = V∕uτ = V∕ τw ∕ρ and ub = ub ∕uτ = ub ∕ τw ∕ρ. Using velocity distribution in the buffer layer, we have u+L = 5 ln 5 − 3.05

(11.351)

u+b = 5 ln 30 − 3.05.

(11.352)

Subtracting the first equation, Eq. (11.351), from the second one, Eq. (11.352), we get u+b − u+L = 5 ln

30 = 5 ln 6. 5

We know that u+L = 5; thus, 30 + 5 = 5 ln 6 + 5 = 5(1 + ln 6). 5 On the other hand, we have √ f + V = . 8 u+b = 5 ln

Finally, substituting expressions for u+b and V+ into Eq. (11.350), we obtain ] √ [√ q′′w ρ f − 5(1 + ln 6) . Tb − Tm = τw cp τw 8 We now combine the temperatures in the three regions, and we get ] √ [√ ( ) q′′w ρ 5 Pr +1 f + 5(Pr −1) + 5 ln . Tw − Tm = ρcp τw 8 6

(11.353)

(11.354)

(11.355)

Using the definition of the heat transfer coefficient h for turbulent flow in a pipe, we have q′′w = h(Tw − Tm ) and using the Moody friction factor f, we have ( ) f τw = ρ V2 . 8 We obtain the Stanton number St as St =

11.5.5

NuD h = = ReD Pr ρcp V

f∕8 √ [ ( )] . 1 + 5 8f (Pr −1) + ln 5 Pr6 +1

(11.356)

The Analogy of Kadar and Yaglom

Kadar and Yaglom [37] devised the following expression for a round tube: √ Pr ReD cf ∕2 hD = . NuD = √ k 2.12 ln {ReD cf } + 12.5Pr2∕3 + 2.12 ln (Pr ) − 10.1 They found that Eq. (11.357) is in good agreement with experimental data for Pr ≥ 0.7.

(11.357)

11.5 Analogies for Internal Flow

11.5.6 The Analogy of Yu et al. This analogy is reported in [5, 38] NuD =

1 hD = ( ) ; [ ( )2∕3 ] k Prt Prt 1 1 + 1− Pr Nu1 Pr Nu∞

R+ > 500.

The TFD Nusselt number Nu∞ is √ ( )1∕3 cf Pr Nu∞ = 0.07343 . ReD Prt 2 The turbulent Prandtl number is evaluated from 0.015 Prt = 0.85 + Pr Pr ≥ 0.7. For the UWT boundary condition, Nu1 is given by (c ) ReD f 2 . Nu1 = 145 1 + + 2.5 (V ) For the UHF boundary condition, Nu1 is given by (c ) ReD f 2 . Nu1 = 195 1 + + 2.7 (V ) Finally, the following relation is used to determine V+ : ( )2 ln (R+ ) 227 50 + V+ = 3.2 − + + + 0.436 R R

(11.358)

(11.359)

(11.360)

(11.361)

(11.362)

(11.363)

150 < R+ < 50 000 ( +) (c ) 2 R cf = + 2 ReD f = 2 2 (V ) V+ )√ ( cf ReD R+ = . 2 2 ReD = 2V+ R+

Example 11.10 Water flows at a mass flow rate of 1.5kg/s through a smooth tube having an internal diameter of 25 mm. The pipe wall surface temperature is kept at 100 ∘ C by condensing steam on the external surface. It is desired to heat water from 14 ∘ C to 40 ∘ C. Calculate the tube length to accomplish this task. Solution The properties of water at the average bulk temperature of Tm is T + Tmo 14 ∘ C + 40 ∘ C Tm = mi = = 27 ∘ C = 300K 2 2 are found to be ρ = 997 kg∕m3

cp = 4179 J∕kgK μ = 855 × 10−6 N.s∕m2

k = 0.613W∕mK Pr = 5.83 ρVD 4ṁ 4 × 1.5 = = = 89 350. μ μπD 855 × 10−6 π × 0.025 Flow is fully turbulent. We wish to use the analogy of Yu et al. ReD =

637

638

11 Turbulent Internal Flow: Momentum and Heat Transfer

First, we calculate the mean velocity V as 4 × 1.5 m 4ṁ = = 3.06 . V= s 997 × π × (0.025)2 ρπD2 The skin friction coefficient cf , Eq. (11.126), is 0.046 0.046 cf = = = 0.0047. 1∕5 1∕5 (89350) Re D

The wall shear stress τw is ( ) c 0.0047 N τw = f ρV2 = × 997 × 3.062 = 22.03 2 . 2 2 m The wall friction velocity uτ is √ √ τw m 22.03 = = 0.148 . uτ = ρ 997 s The dimensionless pipe radius R+ is √ √ ReD cf 89 350 0.0047 = = 2166.87. R+ = 2 2 2 2 The dimensionless velocity V+ is V 3.06 = V+ = = 20.61 uτ 0.148 or we may use Eq. (11.363)

( )2 ln (2166.87) 227 50 + + = 20.71. 2166.87 2166.87 0.436 Next, we compute the turbulent Prandtl number Prt as 0.015 0.015 = 0.85 + = 0.8525 Prt = 0.85 + Pr 5.83 √ ( )1∕3 cf Pr ReD Nu∞ = 0.07343 Prt 2 √ )1∕3 ( 5.83 0.0047 ≈ 604. = 0.07343 (89350) 0.8525 2 Since the pipe wall is at constant temperature, we have ( ( ) ) c ReD 2f (89350) 0.0047 2 = ≈ 195.40. Nu1 = 145 145 1 + (V+ )2.5 1 + (20.61)2.5 V+ = 3.2 −

For the Nusselt number NuD , we use Eq. (11.358) 1 hD = ( ) [ ( )2∕3 ] k Prt Prt 1 1 + 1− Pr Nu1 Pr Nu∞ 1 ≈ 514 NuD = ( [ ) ( ) ] 1 0.8525 2∕3 1 0.8525 + 1− 5.83 (195.4) 5.83 604 k 0.613 2 (514) ≈ 12 603 W∕m K. h = NuD = D 0.025 Pipe length L is calculated as follows: [ ( ) ] Tw − Tmo πDL = exp − h ̇ p mc Tw − Tmi [ ( ) ] π × 0.025 × L 100 − 40 = exp − × 12 603 100 − 14 1.5 × 4179 L ≈ 2.27 m. NuD =

11.5 Analogies for Internal Flow

11.5.7 Martinelli Analogy The Martinelli analogy is discussed in [5, 36, 40]. Martinelli studied turbulent flow in tubes and presented relationships, which apply to all the fluids for all the Reynolds numbers. The conditions of flow and heat transfer are as follows: (1) Incompressible fluid flow. Fully developed velocity and temperature profiles in turbulent flow are assumed. (2) Constant fluid properties. This means that the temperature gradients do not influence the velocity gradients. (3) UHF applied to the tube wall. Shear stress and heat flux are τ du = (ν + εM ) ρ dy q′′w dT = −(α + εM ) . ρcp dy Shear stress τ at any point y is related to the wall shear stress τw by ( y) r τ = τw = τw 1 − R R ′′ since r + y = R. The heat flux q at any point y is related to the wall heat flux q′′w q′′ ( q′′ y) . = w 1− ρ cp ρ cp R Martinelli gives the following relations based on Eqs. (11.364a) and (11.364b): ( y) du = (ν + εM ) τw 1 − R dy q′′w ( y) dT = −(α + εM ) . 1− ρ cp R dy

(11.364a) (11.364b)

(11.365a)

(11.365b)

(11.366a) (11.366b)

The Nusselt number may be calculated from ( ) D hD 𝜕T NuD = =− . k 𝜕 y y=0 T −T w

m

Martinelli obtained following temperature distribution by integrating Eq. (11.366b). Martinelli assumed that εH is proportional to εM and εM is determined from Eq. (11.366a) using universal velocity distribution. The main relationships derived by Martinelli are given as: Laminar sublayer (0 ≤ y+ ≤ 5) ( )( ) y Pr Prt y1 Tw − T = (11.367) ( √ ) [ ( )] Tw − Tc ReD cf Pr Pr + 0.5 F ln + ln 1 + 5 Prt Prt 60 2 where y1 is the value of y at y+ = 5. Buffer layer (5 ≤ y+ ≤ 30) [ ( )] ( ) y Pr Pr + ln 1 + −1 Prt Prt y1 Tw − T = ( √ ). [ ( )] Tw − Tc ReD cf Pr Pr + 0.5 F ln + ln 1 + 5 Prt Prt 60 2 Turbulent region (y+ > 30) ( √ ) [ ] ( ) ReD cf ( y ) Pr Pr + ln 1 + 5 + 0.5 F ln Prt Prt 60 2 R Tw − T . = ( √ ) ) ( Tw − Tc ReD cf Pr Pr + 0.5 F ln + ln 1 + 5 Prt Prt 60 2

(11.368)

(11.369)

639

640

11 Turbulent Internal Flow: Momentum and Heat Transfer

The pipe centerline temperature is ( ) √ Tw − Tc Tw − Tc ReD Pr cf ∕2 + . = Tc = ′′ NuD qw ∕(ρ cp uτ ) T −T w

(11.370)

m

The mean fluid temperature is ) ( )( Tw − T u R ∫0 r dr Vc Tw − Tc Tw − Tm . = ( ) u R Tw − Tc ∫0 r dr Vc

(11.371)

The Stanton number is ) ( cf Tw − Tc 2 T −T w m St = { [ √ ]} . [ ( )] ReD cf F Pr Pr + ln 5 + ln 1 + 5 Prt Prt 2 60 2 1 Prt

√

(11.372)

If molecular transport is neglected in the turbulent core, F = 1. When Pr ≪ 1, as in liquid metals, the contribution of molecular diffusivity to conduction heat transport in the turbulent core cannot be neglected. Mc Adams [41] reported corrected values for F and (Tw − Tm )/(Tw − Tc ) in tabular form, as shown in Tables 11.5 and 11.6.

Table 11.5

Values of F in the Martinelli analogy for Prt = 1.

PeD ↓

ReD = 104

ReD = 105

ReD = 106

102

0.18

0.098

0.052

103

0.65

0.45

0.29

4

0.92

0.83

0.65

105

0.99

0.985

0.98

106

1.0

1.0

1.0

10

Source: McAdams [41].

Table 11.6 Pr↓

Values of (Tw − Tm )/(Tw − Tc ) ratio in the Martinelli analogy. ReD = 104

ReD = 105

ReD = 106

ReD = 107

0

0.564

0.558

0.553

0.550

10−4

0.568

0.560

0.565

0.617

10−3

0.570

0.572

0.627

0.728

10−2

0.589

0.639

0.738

0.813

−1

0.692

0.761

0.823

0.864

0.865

0.877

0.897

0.912

10 1 10

0.958

0.962

0.963

0.966

100

0.992

0.993

0.993

0.994

1000

1.00

1.0

1.0

1.0

Source: McAdams [41].

11.6 Combined Entrance Region

The parameter F may be computed from the expression reported in [5] ( ) ⎡ ⎤ ⎧ 1 + √1 + 20ζ 60 − 1 − √1 + 20ζ ⎫ ⎢ ⎥ ⎪ ⎪ 5ζ 1 R+ ln ⎨ ln ⎢ ( )⎥ + √ ( ) √ ⎬ √ 60 ⎢ 30 ⎥ 30 1 + 20ζ ⎪ ⎪ 1 − 1 + 20ζ − 1 + 1 + 20ζ 1 − 5ζ + ⎢ ⎥ ⎭ ⎩ R+ R+ R+0 ⎦ ⎣ F= √ ) ( cf ReD 2 ln 60 2 where R+ is defined as √ ReD cf R+ = 2 2

(11.373)

(11.374a)

and ζ is defined by Prt √

ζ=

ReD Pr

cf 2

.

(11.374b)

It is reported that the Martinelli analogy is better than all other analogies for Pr ≪ 1, and all the fluid properties are bulk temperature properties.

11.6 Combined Entrance Region In the combined entrance region, the velocity and temperature profiles develop simultaneously. The entry length depends on the shape of the inlet. Deisler [42] experimentally investigated the turbulent combined entrance region problem in a circular pipe for both the UWT and UHF boundary conditions. Working fluid was air, and uniform velocity and uniform temperature were assumed at the inlet. It was observed that the difference between local Nusselt numbers for UWT and UHF boundary conditions is negligible. It was also found that the difference between thermal entry lengths for UWT and UHF boundary conditions is negligible. Boelter et al. [43] and Mills [44] experimentally investigated turbulent heat transfer in the combined entrance region of a pipe. Experimental studies indicated that the pipe entrance configuration affects the combined heat transfer. The local Nusselt number was found to be different for each entrance configuration. Bhatti and Shah [27] studied the available experimental work and suggested the following formula for the calculation of the mean Nusselt number in the entrance region of circular pipes NuD

=1+

NuD,fd

C (L∕D)n

(11.375)

L >3 D Pr = 0.7 where Nu∞ denotes the fully developed Nusselt number for UWT or UHF boundary condition. Working fluid was air and Pr = 0.7. The constants C and n have been determined from the measurements given by Mills [44] for UHF boundary conditions. Table 11.7 lists the constants C and n for different entrance configurations. This equation may be used for both UWT and UHF boundary conditions. Equation (11.375) may be used for short tubes as well. Typical errors of less than 15% occur when assuming NuD ≈ Nu∞ for (L/D) > 60. Here, Nu∞ is the fully developed Nusselt number. Note that entry lengths are short for turbulent flow in channels (10 ≤ L/D ≪ 60). Fluid properties are evaluated at mean fluid temperature Tm = (Tmi + Tmo )/2.

641

642

11 Turbulent Internal Flow: Momentum and Heat Transfer

Table 11.7

C and n constants for Eq. (11.375). Configurations for air Pr = 0.72. Uniform heat flux boundary condition.

Entrance configuration

C

n

Long Calming section. Adiabatic surface up to heating point

0.9756

0.760

Sharp square entrance 180∘ Circular bend. Adiabatic surface up to heating point.

2.4254

0.676

0.9759

0.700

90∘ Circular bend. Adiabatic surface up to heating point 90∘ Sharp elbow

1.0517

0.629

2.0152

0.614

Source: Lienhard IV and Lienhard V [82].

11.7 Empirical and Theoretical Correlations for Turbulent Flow in Channels Fully turbulent pipe flow occurs for ReD ≥ 104 , and flow is, in general, transitional in the range of 2 × 103 ≤ ReD ≤ 1 × 104 . Internal turbulent forced convection is very complex, and we rely on empirical correlations to determine the Nusselt numbers. Many correlations are available in the literature. Analytical solutions are always compared with proven reliable empirical correlations under similar flow conditions. In turbulent flow, entry lengths that are required to establish the HFD and TFD regimes are much shorter than compared to entry lengths seen in laminar flows. This is due to the disturbances encountered at the tube entry. Disturbances contribute to the creation of turbulence, and turbulence enhances the mixing of the fluid. Therefore, short entry lengths are seen in turbulent flows. For example, HFD conditions occur for about x/D > 10. We can safely assume that HFD and TFD exist over the entire length of the duct. Therefore, we may accept that the average Nusselt number for the entire duct is approximately equal to the Nusselt number for a fully developed region. If we wish to study the heat exchange at pipe entry, the relations need to be modified to take into account the entry effects. Convective heat transfer correlations in the literature for flow in tubular channels generally depend on: (a) (b) (c) (d)

Reynolds number, ReD = ρ V D/μ Prandtl number, Pr = μcp /k dimensionless distance, z/D geometry and boundary conditions.

Empirical information and dimensional analysis techniques are used to study turbulent heat transfer since mathematical methods are too complicated. In general, recall that if we wish to correlate experimental heat transfer data, we can correlate the data as ( )p D NuD = C1 RenD Prm (11.376) L or ( )p L St = C2 RenD Prm . (11.377) D Notice that the correlations of the form St = f(ReD , Pr) are commonly used since the Stanton number can be obtained directly from experimental measurements. Consider a pipe of length L and diameter D. An energy balance on the fluid flowing through a pipe between inlet and exit gives the following: Heat added at the wall = change of fluid enthalpy ( h π D L(Tw − Tm ) = ρcp V

πD2 4

) (Tmo − Tmi )

or St =

D (Tmo − Tmi ) . 4L (T − T ) w

m

(11.378)

11.7 Empirical and Theoretical Correlations for Turbulent Flow in Channels

We may take Tm as Tm = (Tmo + Tmi )∕2 Notice that no fluid properties are needed to be known. Consider now the determination of the Nusselt number NuD . For the tubes, the relation required to determine the mean heat transfer coefficient for a tube of length L is ) ( π D2 (ρ cp V)(Tmo − Tmi ) = h(π D L)ΔTLMTD (11.379) 4 where ΔTLMTD is given by (T − Tmo ) − (Tw − Tmi ) . ΔTLMTD = w ) ( Tw − Tmo ln Tw − Tmi Here, Tmo and Tmi are the mean outlet and inlet temperatures. From this relation, we may obtain the following equations: ( ) D2 ρ cp V (Tmo − Tmi ) NuD = . (11.380) 4L k ΔTLMTD Notice the difference between Eqs. (11.378) and (11.380). No fluid properties are needed to determine the Stanton number. On the other hand, the expression for the Nusselt number requires the knowledge of specific heat cp , the thermal conductivity k, density ρ, and average fluid velocity V. A reference temperature is needed to evaluate the physical properties of fluid. An average of bulk temperature Tm is used to evaluate the fluid properties, and it is defined as (T + Tmo ) Tm = mi 2 where Tmi and Tmo are the inlet and outlet mean fluid temperatures, respectively. Recall that for a constant property fluid, bulk temperature Tm is defined as Tm =

1 u T dA A V∫ A

where A is the cross-sectional area and V is the average bulk velocity. The bulk velocity was defined as V=

1 u dA. A∫ A

Bulk temperature is a measure of the fluid temperature averaged over the cross section of a duct or channel. This definition does not explain the difference between the cases Tw > Tm or Tw < Tm , where Tw is the pipe wall temperature. We know that viscosity is temperature dependent, and cold or hot walls will influence the viscosity. In other words, viscosity will be light or heavy near the wall depending on the wall temperature. This will affect the heat transfer. For situations where the properties, especially the viscosity, do not change too much across the fluid, viscosity ratio may be neglected. Fluid properties may be evaluated at a specific mean film temperature Tf . For this purpose, a film temperature Tf is defined as: ] [ (T + Tmo ) Tw + mi 2 . Tf = 2 This definition represents a more meaningful temperature for the evaluation of physical properties of gases. For the case of liquids, particularly viscous oils, the viscosity ratio term is needed to take care of the viscosity variation. Finally, we can write for the average heat flux q′′w = h ΔTLMTD .

(11.381)

We will now talk about the hydrodynamic entrance length. There is no definite expression for the hydrodynamic and thermal entrance lengths, and we can approximate the entry lengths as discussed in [23, 42] L 10 ≤ H ≤ 60. D It may be assumed that the hydrodynamic and thermal entry lengths could be estimated to be LH ≈ LT > 10. Notice that equations developed for round pipes may also be used for noncircular ducts by using the hydraulic diameter DH = 4A/P, where A is the cross-sectional area and P is the wetted perimeter of the noncircular duct. In general, the length

643

644

11 Turbulent Internal Flow: Momentum and Heat Transfer

of the thermal entrance region for turbulent flow is relatively short, and because of this, it is common to use the fully developed Nusselt number starting from the beginning of pipe to all the way to end of the pipe. Example 11.11 To characterize flows in noncircular pipes, we use the hydraulic diameter. The hydraulic diameter is defined as DH = 4A∕P where A is the cross-sectional area and P is the wetted perimeter. Let us calculate the hydraulic diameter for a few geometries. See Figure 11.E11. (a) Cylindrical pipe of diameter D 2 4A 4(π D ∕4) = = D. P πD (b) The rectangular duct having sides a and b

DH =

4(a × b) 2(a × b) 4A = = . P 2(a + b) (a + b)

DH =

(c) Concentric annulus having diameters Di and D0 , where Di is the diameter of the inner tube and D0 is the diameter of the outer tube. The thickness of each tube is neglected ( 2 ) D0 − D2i π 4 4A = = D0 − Di . DH = P π(D0 + Di ) (d) Semicircular DH =

πD . π+2

(e) Circular tube with square inserted inside DH =

D(π − 2) √ . π+ 8

(f) Triangular a DH = √ . 3 Figure 11.E11 (c) Concentric annulus. (d) Semicircular duct. (e) Circular tube with square inserted inside. (f) Triangular duct.

Di

D

D0 (c)

(d)

D B

a

(a) a

(e)

(f)

11.7 Empirical and Theoretical Correlations for Turbulent Flow in Channels

11.7.1.1 Colburn Correlation

Colburn correlation [35] is modified from the Reynolds analogy, and it is given by StPr2∕3 = cf ∕2

(11.382)

where cf is the Darcy–Weisbach and f is the Moody friction factor. Recall that the Moody friction factor f was defined as D Δp f= . L ρV2 ∕2 Here, Δp is the pressure drop over length L. We may also write a relation between pressure drop Δp and wall shear stress τw with a force balance: Δp D . L 4 We can now write τw . cf = ρV2 ∕2 τw =

Then, we may write 4cf = f. With this information, Eq. (11.382) can be expressed in the following form for fully developed turbulent flow in a smooth tube: StPr2∕3 = f∕8

(11.383)

where f is the Moody friction factor, and it may be calculated for smooth pipe by the relation f = 0.184 Re−0.2 D ReD < 105 . This is called Colburn correlation. We now substitute the friction factor f, Eq. (11.384), into Eq. (11.383), and this gives StPr2∕3 = 0.023 Re−0.2 D .

(11.384)

The Stanton number is St = NuD /ReD Pr, and Eq. (11.384) is rearranged to yield an equation for turbulent flow in smooth tube. Experimental data have confirmed validity for the following range conditions: 1∕3 NuD = 0.023 Re0.8 D Pr

(11.385)

ReD > 104 0.7 ≤ Pr ≤ 160 L∕D > 60. HFD turbulent flow TFD turbulent flow Smooth tubes where NuD = hD∕k, ReD = V D/ν, and Pr = μ cp /k. Equation (11.384) is valid for fully developed turbulent flow in smooth tubes. Notice that Eq. (11.384b) is of the form NuD = f(ReD , Pr ). Fluid properties are evaluated at an average of bulk temperature Tm defined by Tm = (Tmi + Tmo )∕2 where Tmi is the mean inlet temperature and Tmo is the mean outlet fluid temperature. Error associated with this equation might be as much as 25%. This relation should be used only in the case of small to moderate temperature differences.

645

646

11 Turbulent Internal Flow: Momentum and Heat Transfer

11.7.1.2

Dittus and Boelter Correlation

Dittus and Boelter [45] and Winterton [46] discuss Dittus and Boelter correlation. Based on numerous experimental data, for HFD and TFD turbulent flow in a smooth tube with moderate temperature difference, relation given below may be used to estimate average Nusselt number with the Mc Adams [41] modification: n NuD = 0.023 Re0.8 D Pr

(11.386)

0.7 ≤ Pr ≤ 160 n = 0.4 if Tw > Tm (heating) n = 0.3 if Tw < Tm (cooling) L∕D > 60 from pipe entrance. ReD > 104 q′′w = const. or Tw = const. Smooth tubes HFD turbulent flow TFD turbulent flow. The fluid properties are evaluated at the average of bulk temperature Tm = (Tmi + Tm0 )/2, where Tmi and Tmo are the mean fluid temperatures at the inlet and the outlet of the pipe. Equation (11.386) is confirmed experimentally to within ±25% for UWT as well as UHF conditions. Winterton [47] gives information about the origin of the Dittus and Boelter equation. This correlation may yield errors as large as 25% for some applications. This relation should be used only in the case of small to moderate temperature differences. 11.7.1.3

Sieder–Tate Correlation

An extension of Eq. (11.386) is given by Sieder and Tate [47], and Sieder and Tate correlation is useful for situations involving large property variations across the smooth tube in fully developed turbulent flow. The large property variations across the smooth tube are taken care of by the viscosity ratio especially for high Prandtl numbers 4∕5

NuD = 0.027 ReD Pr1∕3 (μm ∕μw )0.14

(11.387)

ReD ≥ 104 0.7 ≤ Pr ≤ 16 700 L∕D > 60 Smooth pipes q′′w = const. or Tw = const. HFD turbulent flow TFD turbulent flow. The viscosity μw is evaluated at the wall temperature, and all the other properties at the mean temperature Tm = (Tmi + Tmo )∕2. The temperatures Tmi and Tmo are the mean fluid temperatures at the inlet and the outlet of the pipe. This correlation may yield errors as large as 25% for some applications.

11.7 Empirical and Theoretical Correlations for Turbulent Flow in Channels

11.7.1.4 Hausen Correlations

Hausen empirical correlation [48] is for transition and turbulent entrance and fully developed flow forced turbulent convection. This correlation is based on the data collected by Sieder and Tate [47] and was published in 1959. This correlation is the first extension of usual power law correlations to transition flow region [ ( )2∕3 ] ( μ )0.14 [ ] 0.42 D m 1 + NuD = 0.037 Re0.75 − 180 Pr (11.388) D L μw 0 < D∕L < 1 0.6 < Pr < 103 2300 ≤ ReD ≤ 106 with [1 + (D/L)2/3 ] term; Eq. (11.388) takes into account the dependency of heat transfer coefficient on the tube length. This is important for short tubes. This means that Eq. (11.388) takes care of developing hydraulic and thermal boundary layers from the beginning of the tube. It is reported that Eq. (11.388) deviates from experimental data on heat transfer in the transition region, as reported in [71]. Hausen [49] based on new measurements proposed the following correlations in 1974: [ ( )2∕3 ] ( μ )0.14 [ 0.8 ] 0.42 D m NuD = 0.0215 ReD − 230 Pr 1+ (11.389a) L μw 0.5 < Pr < 1.5 [ ( )2∕3 ] ( μ )0.14 [ 0.87 ] 0.42 D m NuD = 0.0107 ReD − 280 Pr 1+ L μw

(11.389b)

1.5 < Pr < 500 [ ( )2∕3 ] ( μ )0.14 [ 0.8 ] D m 0.3 NuD = 0.0235 ReD − 230 (1.8Pr − 0.8) 1 + L μw

(11.389c)

10−1 < Pr < 103 . These correlations are developed for transition, turbulent entrance, and fully developed flow of forced turbulent convection. 11.7.1.5 Petukhov Correlation

Petukhov [20] carried out extensive parametric studies with an eddy diffusivity model of Reichardt, and he assumed Prt ≈ 1. Petukhov proposed the following semiempirical correlation: NuD =

(f∕8)(ReD Pr ) √ K1 + K2 f∕8(Pr2∕3 − 1)

104 ≤ ReD ≤ 5 × 106 Tw = constant or q′′w = constant 0.5 ≤ Pr ≤ 2 × 103 K1 = 1 + 3.4f K2 = 11.7 + 1.8 Pr−1∕3 f=

1 . [1.82log10 ReD − 1.64]2

(11.390)

647

648

11 Turbulent Internal Flow: Momentum and Heat Transfer

A different form of Petukhov correlation is ( )n (f∕8)(ReD Pr ) μ NuD = √ 1.07 + 12,7 f∕8(Pr2∕3 − 1) μw

(11.391)

104 ≤ ReD ≤ 5 × 106 0.5 ≤ Pr ≤ 2000 ( ) μ ≤ 40 0.08 ≤ μw n = 0 Constant heat flux n = 0.11, Tw > Tm n = 0.25, Tw < Tm . This correlation provides fully developed heat transfer coefficients. Petukhov correlation for forced convection heat transfer in smooth pipes gives very good estimates of heat transfer coefficients. On the other hand, it gives reasonable estimates of heat transfer coefficients for rough pipes. In Eq. (11.390) and Eq. (11.391), all properties are evaluated using the average of bulk temperature Tm = (Tmi + Tmo )∕2 except for μw which is evaluated at the pipe wall temperature Tw . Example 11.12 Consider the flow of specific purpose light oil in smooth tube. The tube inside diameter is 5 cm. The tube wall temperature is constant and equal to 300 ∘ C. Flow is HFD at the point where the heating begins. See Figure 11.E12. The mean inlet oil temperature is Tmi = 30 ∘ C and mean oil temperature at z = L is equal to Tmo = 45 ∘ C. The oil flows through the tube with an average velocity of 6 m/s. Estimate length L required to heat the oil from 30 to 45 ∘ C. Is the flow TFD? Fluid properties at mean fluid temperature Tm are given as T + Tmo 30∘ C + 45∘ C = ≈ 310 K Tm = mi 2 2 kg J N.s ρ = 897 3 cp = 1926 μ = 2.28 × 10−2 2 kg K m m W Pr = 340. k = 0.131 m.K Figure 11.E12

Schematic diagram.

L Tm0 = 45°C

Tmi = 30°C HFD Tw = 300°C

Solution The Reynolds number is ReD =

ρVD 897 × 6 × 0.05 = = 11 802. μ 2.28 × 10−2

Since the Reynolds number is sufficiently large, flow is assumed to be turbulent. Let us assume fully turbulent flow. Let us use the Petukhov equation to calculate the Moody friction factor f f=

1 1 = = 0.03. [0.79 ln ReD − 1.64]2 [0.79 ln (11 802) − 1.64]2

11.7 Empirical and Theoretical Correlations for Turbulent Flow in Channels

We choose Petukhov correlation to calculate the heat transfer coefficient h (f∕8)(ReD Pr ) NuD = √ K1 + K2 f∕8(Pr2∕3 − 1) 104 ≤ ReD ≤ 5 × 106 0.5 ≤ Pr ≤ 2 × 103 K1 = 1 + 3.4f K2 = 11.7 + 1.8 Pr−1∕3 . Let us evaluate each term K1 = 1 + 3.4f = 1.1022 K2 = 11.7 + 1.8 Pr−1∕3 = 11.957 (0.03∕8)(11 802 × 340) ≈ 418 √ 1.1022 + 11.957 0.03∕8(3402∕3 − 1) ( ) W k 0.131 × 418 = 1095 2 . h = Nu = D 0.05 mK Next, we write energy balance on fluid flowing through a circular pipe ( 2) πD (Tmo − Tmi ) hπDL(Tw − Tm ) = ρVcp 4 ( ) π (0.05)2 (45 − 30). 1095 × π × (0.05)L(300 − 37.5) = 897 × 6 × 1926 4 NuD =

This gives us L ≈ 6.7 m. Now, we check the

L D

ratio.

L 6.7 = = 135. D 0.05 This shows that since ReD > 10 000 and L/D ≫ 60, flow is both TFD and HFD. 11.7.1.6 Gnielinski Correlation

The empirical correlation of Gnielinski [50] is valid for smooth tubes over a large Reynolds number fully developed turbulent flow including transition region, and it is given as NuD =

(f∕8)(ReD − 1000) Pr √ 1 + 12.7 f∕8(Pr2∕3 − 1)

(11.392a)

0.5 ≤ Pr ≤ 2000 2300 ≤ ReD ≤ 5 × 106 q′′w = const Tw = const. The factor (ReD − 1000) was included to reproduce the measured data in the transition region. Gnielinski calibrated his correlation with experimental values primarily in the nearly fully turbulent flow regime (ReD > 8000), as reported in [71]. Nevertheless, the correlation of Gnielinski was one of the correlations that allowed the calculation of Nusselt numbers in the transition and turbulent regions. Gnielinski recognized that his correlation deviates from experimental results especially in the transition region, and he proposed a new method as we shall see in Section 11.8. To calculate f for smooth tubes, the Filonenko relation [33] may be used f = [1.82log10 (ReD ) − 1.64]−2 .

649

650

11 Turbulent Internal Flow: Momentum and Heat Transfer

Fluid properties are evaluated at mean fluid temperature Tm Tm = (Tmi + Tm0 )∕2 where Tmi and Tmo are the mean fluid temperatures at the inlet and the outlet of the pipe. Gnielinski correlation is based on the correlation of Petukhov. This correlation may be used if the frictional effect on the wall is significant. This equation reduces errors to less than 10%. A detailed discussion of heat transfer in the transition region is provided in Section 11.8. Equation (11.390) is a modification of the Petukhov equation. 11.7.1.7

Gnielinski Correlation with Modification

For the developing region, Gnielinski modified his correlation to account for both the entrance effects and effects of fluid property variations in short tubes. This empirical correlation may be used for smooth and rough tubes. Gnielinski used several data points from different researchers in developing this correlation [ ( )2∕3 ] (ReD − 1000) Pr (f∕8) D NuD = 1+ F (11.392b) √ L 2∕3 H 1 + 12.7 f∕8(Pr − 1) ( ) ⎧( Pr )0.11 Pr < 10 for liquids 0.1 < Pr ⎪ Prw F = ⎨ ( )0.45 ( w) Tm Tm ⎪ T 0.5 < T < 1.5 for gases w ⎩ w Temperatures Tw and Tm are in Kelvin. Tw = constant or q′′w = constant f = [1.8 log10 (ReD ) − 1.5]−2 LH = heated length of tube 0.5 < Pr < 2000 2300 ≤ ReD ≤ 5 × 106 D < 1. L Gnielinski reports that Eq. (11.392b) is in excellent agreement with experimental data and for developing forced convection. Correlation is valid for transition, turbulent entrance, and fully developed flow of forced turbulent convection. Gnielinski used large amount of data points from various researchers to establish the constants. Properties are evaluated at Tm = (Tmi + Tmo )/2, where Tmi and Tmo are the mean fluid temperatures at the inlet and the outlet of the pipe. The Prandtl numbers at bulk fluid temperature Tm and tube wall temperature Tw are used to take care of variable property effects. Equation (11.392b) is valid for transition flow (2300 ≤ ReD < 104 ) and fully developed turbulent flow (ReD > 104 ) regimes. Viscous dissipation corresponds to the energy loss due to friction experienced by the fluid on the inner surface of the pipes. Friction depends on the irregularity on the pipe surface. For the friction factor for rough pipes, the Colebrook correlation may be used ( ) e∕D 2.51 1 + √ = −2log10 √ 3.7 f Re f 0
25 D where NuD = hD∕k is the average Nusselt number. All the fluid properties are evaluated at the bulk temperature except for the variables with subscript w, which is evaluated at the pipe wall temperature. It is reported that Eq. (11.393) correlates well with experimental data. 11.7.1.9 Nusselt Correlation

When the tube is short, the entrance effects are important. For shorter pipes, the entrance effects related to the flow development need to be taken into account. Ozisik [52] reports that Nusselt studied the available experimental data for (L/D) from 10 to 100 and proposed the following local number NuD : ( )0.055 1∕3 D NuD = 0.036 Re0.8 (11.394) D Pr L L < 400 10 < D where L is the length measured from beginning of heat transfer section and NuD = hD∕k. The fluid properties are evaluated at the average of bulk temperature Tm = (Tmi + Tmo )/2. Here, Tmi and Tmo are the fluid temperature at the inlet and the outlet, respectively. Example 11.13 Water flows at a rate of 0.6 kg/s through a 5-m-long tube having an internal diameter of 2 cm. A UHF of 7 × 104 W/m2 is applied to the tube surface. It is assumed that velocity and temperature fields are fully developed. Water properties are evaluated at local mean temperature of 17 ∘ C. Calculate: (a) (b) (c) (d) (e) (f) (g)

the pressure drop over the entire tube length heat transfer coefficient based on the Colburn analogy heat transfer coefficient based on the Dittus–Bolter correlation heat transfer coefficient based on the Gnielinski correlation heat transfer coefficient based on the Kadar and Yoglom analogy the difference between wall and local mean temperature differences increase in mean temperature in axial direction

Solution The physical properties of water are ρ = 999 kg∕m3

cp = 4184 J∕Kg.K μ = 1080 × 10−6 N.s∕m2

k = 0.598W∕m.K Pr = 7.56. The average velocity V is V=

4ṁ 4 × 0.6 = = 1.9117m∕s 999 × π × (0.02)2 ρπD2

The Reynolds number Re is ρVD 999 × 1.9117 × 0.02 = ≈ 35 367. μ 1080 × 10−6 Flow is turbulent. The friction factor f is 1 1 = = 0.0227. f= [0.79 ln (ReD ) − 1.64]2 [0.79 ln (35 367) − 1.64]2 ReD =

651

652

11 Turbulent Internal Flow: Momentum and Heat Transfer

(a) The pressure drop Δp is

( ) [ 999 × (1.9117)2 ] 2 5 L ρV = (0.0226) × ≈ 10360N∕m2 . Δp = f D 2 0.02 2

(b) The Colburn analogy is St =

(0.0227∕2) (f∕8) = = 0.000737 2∕3 Pr (7.56)2∕3

h = (ρcp V) St = (999 × 4184 × 1.9117) (0.000737) ≈ 5889W∕m2 K. (c) The Dittus–Boelter correlation is NuD = 0.023Pr0.4 Re0.8 = 0.023(7.56)0.4 (35 367)0.8 = 224.91 d ( ) ( ) 0.598 k = (224.91) = 6725W∕m2 K. D 0.02 (d) The Gnielinski correlation is h = (224.91)

(f∕8)(ReD − 1000) Pr √ 1 + 12.7 f∕8(Pr2∕3 − 1) (0.0227∕8)(35 367 − 1000) × 7.56 NuD = = 251.8 √ 1 + 12.7 0.0227∕8(7.562∕3 − 1) ) ( 0.598 (251.8) ≈ 7522.84W∕m2 K. h= 0.02 (e) The analogy of Kadar and Yoglom is √ Pr ReD cf ∕2 hD = NuD = √ k 2.12 ln {ReD cf } + 12.5Pr2∕3 + 2.12 ln (Pr ) − 10.1 0.046 ≈ 0.00568 cf = 1∕5 ReD √ (7.56)(35 367) 0.00568∕2 hD ≈ 241.2 = NuD = √ k 12.5(7.56)2∕3 + 2.12 ln {7.56 × 35 367 0.00568} − 10.1 ) ( 0.598 (241.2) = 7213.2W∕m2 K. h= 0.02 (f) The temperature difference between wall and local mean temperatures is NuD =

Tw − Tm =

q′′w

=

7 × 104 ≈ 9.72 K (Gnielinski). 7522.84

h (g) We write an energy balance on a CV in tube

q′′ πD 7 × 104 × π × 0.02 dTm = = w = 1.75 K∕m. ̇ p mc dx 0.6 × 4184

11.8 Heat Transfer in Transitional Flow Heat exchangers have many applications in engineering, and many of these heat exchangers use smooth tubes with fluid such as water. After ReD > 2300, the development of turbulence depends on several factors such as the shape of tube entrance. Heat exchangers may operate in the transitional flow regime. It is important to determine the pressure drop and heat transfer. Normally, transitional flow exists for Reynolds numbers between 2 300 and 10 000. However, Cengel and Ghajar [54] report that in many cases, the flow becomes fully turbulent when ReD ≥ 4000. In the design of heat exchangers, it is advised to remain outside the transitional flow regime. There is little design information about heat transfer and pressure drop in the transitional flow regime. Several researchers such as Ghajar and Tam [55], Ghajar and Tam [56], Ghajar et al. [57], Tam and Ghajar [58], Tam and Ghajar [59], Tam and Ghajar [60], Tam et al. [61], Tam et al. [62], Meyer and

11.8 Heat Transfer in Transitional Flow

Oliver [63], Meyer and Everts [53], Churchill [64], Abraham et al. [65], Everts and Meyer [66], Everts and Meyer [67], Everts and Meyer [66], and Gnielinski [68] worked on the turbulent transition region. Ghajar and co-workers performed extensive experimental studies on transitional flow in tubes. Ghajar and co-workers used a constant heat flux boundary condition. The purpose of this section is to present limited amount of experimental heat transfer and pressure drop information in the transitional flow regime. Heat transfer in transitional flow is discussed in [5, 54].

11.8.1 Friction Factor in the Transitional Flow Tam and Ghajar [58] measured pressure drops in circular tubes for fully developed flows in the transition regime for three types of inlet configurations. See Figure 11.17. They observed experimentally that a single correlation does not represent the data since inlet configuration affects the beginning and end of the transition region. For this reason, a correlation for each inlet was developed. The proposed correlation for the transition region friction factor is of the following form: [ ) ]C ( )m ( μm ReD B (11.395) ftrans = 4cf,trans = 4 1 + A μw m

m = m1 − m2 GrD 3 Prm4

(11.396)

where the coefficients A, B, and C depend on the inlet configuration, where GrD = gβD3 (Tw

)/ν2

− Tm is the Grashof number. All the properties are evaluated at the average of bulk temperature Tm = (Tmi + Tmo )∕2 except μw is evaluated at the wall temperature. Here Tmi and Tmo are the mean inlet and outlet temperatures. The constants in Eqs. (11.395) and (11.396) are given in Table 11.8. The range of application of Eq. (11.395) is as follows: Re-entrant: ( ) μm 5 ≤ 2.13 2700 ≤ ReD ≤ 5500 16 ≤ Pr ≤ 35 7410 ≤ GrD ≤ 1.58 3 × 10 1.13 ≤ μw Start of cooling

Fully developed

Test section Square-edged

Re-entrant

Bell-mouth

Figure 11.17 Table 11.8

Schematic of inlet configurations. Constants for Eqs. (11.395) and (11.396).

Inlet geometry

A

B

C

m1

m2

m3

m4

Re-entrant

5840

−0.0145

−6.23

−1.1

0.460

−0.133

4.10

Square-edged

4230

−0.1600

−6.67

−1.13

0.396

−0.160

5.10

Bell-mouth

5340

−0.0990

−6.32

−2.58

0.420

−0.410

2.46

653

654

11 Turbulent Internal Flow: Momentum and Heat Transfer

Square-edged

(

3500 ≤ ReD ≤ 6900 12 ≤ Pr ≤ 29 6800 ≤ GrD ≤ 1.045 × 105 1.11 ≤ Bell-mouth

(

5900 ≤ ReD ≤ 9600 8 ≤ Pr ≤ 15 11900 ≤ GrD ≤ 3.53 × 105 1.05 ≤

μm μw

μm μw

) ≤ 1.89

) ≤ 1.47

Equation (11.395) is applicable to a fully developed transition region and should be used with appropriate set of constants for each inlet configuration. The following correlation for the friction factor for the transitional developing region (entrance-affected region) is proposed by Tam et al. [60] )]} ( )m ) [( {( ][ ( ) )1∕a ( μm 64 c 0.75 a 1 + 0.0049ReD +b 1+ . (11.397) fapp,trans = ReD x∕D μw The constants for Eq. (11.397) are given in Table 11.9 and m = 0 for isothermal flow.

11.8.2

Heat Transfer in the Transition Region

Flow has both laminar and turbulent characteristics in transition region. Additionally, type of the inlet configuration influences the beginning and end of transition region. The design of a heat exchanger operating in the transition region between laminar and turbulent flow is not recommended. It is better to do this only if it is strictly necessary. 11.8.2.1 Tam and Ghajar Approach

Following correlation for local Nusselt numbers in transition region is reported by Tam and Ghajar [62]. ] }c { [ (a − ReD ) + NucD,turb NuD,trans = NulD,am + exp b

(11.398)

where NuD, lam is the laminar flow Nusselt number and NuD, turb is the turbulent flow Nusselt number ) ]1∕3 ( )0.14 [( μm ReD Pr D 0.75 + 0.025(GrD Pr ) NuD,lam = 1.24 x μw ( )−0.0054 ( μ )0.14 m 0.385 x NuD,turb = 0.023Re0.8 . D Pr D μw Ghajar and Tam used a large amount of data to develop these equations. The constants for Eq. (11.398) are given in Table 11.10. The ranges of validity of Eq. (11.398) are as follows: Re-entrant: ( ) μm 5 1700 ≤ ReD ≤ 9100 5 ≤ Pr ≤ 51 4000 ≤ GrD ≤ 2.1 × 10 1.2 ≤ ≤ 2.2 μw x ≤ 192. 3≤ D Table 11.9

Constants for Eq. (11.397).

Inlet geometry

Re-entrant

a = 0.52 b = −3.47 c = 4.8 m = −1.8 + 0.46 Gr−0.13 Pr0.41 D 1883 ≤(ReD)≤ 3262 19.1 ≤ Pr ≤ 46.5 4560 ≤ GrD ≤ 24339 1.12 ≤

Square-edged

μm μw

≤ 1.54 3
3100, Eq. (11.392b) may be used for both the boundary conditions. For heat exchange device that operate in the low-Reynolds-number end of the laminar-to-turbulent transition regime, Abraham et al. [85] recommends the Gnielinski correlation, Eq. (11.392a), with relationships for friction factors listed in the Table 11.11. These friction factors are developed for fully developed flow in the low-Reynolds number end of the transition regime. Example 11.14 Water at a rate of 0.02 kg/s flows through a smooth 10 mm diameter tube. The tube length is 1000 mm. The inlet water temperature is 10 ∘ C and the tube wall is kept at 100 ∘ C by condensing steam on the tube surface. (a) Estimate the tube exit temperature. (a) Estimate the heat transfer rate to the water. Solution We assume an outlet temperature of 44 ∘ C. Average of bulk temperature Tm is 10 ∘ C + 44 ∘ C Tm = = 27 ∘ C = 300 K 2 We estimate water properties at 305 K ρ = 997 kg∕m3

cp = 4179 J∕kg•K

k = 0.613 W∕m•K

Pr = 5.83

The Reynolds number ReD ReD =

4ṁ ≈ 2978 πμD

μ = 855 × 10−6 Pa•s

Prw = 1.76

11.8 Heat Transfer in Transitional Flow

The Reynolds number is in the transition region. Eq. (11.400) will be used to calculate the Nusselt number NuD . First, we calculate laminar Nusselt number NuD,lam,2300 }1∕3 { [ ]3 3 3 NuD,lam,2300 = Nu1 + 0.7 + Nu2 − 0.7 + Nu3 Nu1 = 3.66 ) ( D 1∕3 Nu2 = 1.615 2300 Pr = 8.26 L √ )1∕6 ( D 2 = 2.55 Nu3 = 2300 Pr 1 + 22 Pr L NuD,lam,2300 = {(3.66)3 + 0.73 + [8.26 − 0.7]3 + 2.55}1∕3 = 7.93 Calculation of turbulent Nusselt number NuD,4000 . We calculate f and NuD,4000 at ReD = 4000 1 = 0.04026 [1.8 log10 (4 000) − 1.5]2 [ ( )2∕3 ] (f∕8) Pr D 1+ = 30.52 NuD,4000 = √ L 1 + 12.7 f∕8(Pr2∕3 − 1)

f=

Calculation of γ ReD − 2300 3311 − 2300 = = 0.399 γ= 4 000 − 2300 4 000 − 2300 Calculation of Nusselt number NuD = (1 − γ)NuD,lam,2300 + γNuD,turb,4000 NuD = (1 − 0.399)(7.93) + (0.399)(30.52) ≈ 16.94 Next, we correct NuD for property variation. ) ( 5.20 0.11 NuD = 16.94 × ≈ 19.33 1.76 Heat transfer coefficient ) ( 0.620 k (19.33) ≈ 1185.17 W∕m2 K h = NuD = D 0.025 Energy balance on the tube ̇ p (Tmo − Tmi ) = h(πDL)ΔTLMTD mc ΔTLMTD =

(Tw − Tmo ) − (Tw − Tmi ) (T − Tmi ) = − [mo [ ] ] (T − T ) (T − T ) ln (Tw − Tmo) ln (Tw − Tmo) w

mi

w

mi

We get now Eq. (8.148c), and we write. ( ) PhL Tmo = Tw − (Tw − Tmi ) exp − ̇ p mc ( ) π × 0.02 × 1185.17 × 1 Tmo = 100 − (100 − 10) exp − 0.02 × 4179 Tmo ≈ 42.00 C Estimated outlet temperature is close enough to the assumed outlet temperature. Therefore, there is no need to repeat the calculation. Heat transfer rate to the water is ̇ p (Tmo − Tmi ) = 0.02 × 4179(42 − 10) ≈ 2704 W q = mc

659

660

11 Turbulent Internal Flow: Momentum and Heat Transfer

11.9 Effect of Property Variation For engineering calculations, constant property solutions obtained by analytic or experimental methods is corrected to account for property variation with temperature. Ghiaasiaan [5], Kays et.al. [23] and Ozisik [52] discuss the property variation with temperature. For liquids: St Nu = = Nucp Stcp ( )m μw cf = cfcp μm

(

μw μm

)n (11.409a) (11.409b)

where Tw is the surface temperature. Tm is the mean fluid temperature. cp refers to constant property solutions For gases: The viscosity, thermal conductivity and density are functions of absolute temperature. ) ( Tw n St Nu = = Nucp Stcp Tm )m ( Tw cf = cfcp Tm Liquid

(Tw > T∞ )

(Tw < T∞ )

Heating

Cooling

m 0.25 Gases

n −0.11

Heating m −0.1

m 0.25

(11.410a) (11.410b)

n −0.25

Cooling n

−0.5

m −0.1

n 0.0

All other properties are evaluated at mean temperature for internal flows.

Problems 11.1

Prandtl’s velocity defect law is given by (y) Vc − u . = 5.75log10 uτ R The defect law applies in the turbulent core far from the wall. Derive the expression.

11.2

Derive an expression for the difference between average velocity V and velocity at any point for a turbulent flow in a pipe.

11.3

Water with a density of 1000 kg/m3 is flowing turbulently in a pipe. Pipe has in internal diameter of 80 mm. The velocities at the pipe center and 30 mm from the pipe center are measured as 1.6 and 1.2 m/s. Estimate the wall shear stress at this point.

11.4

von Karman proposed the following relation for the mixing length 𝓁: | du∕dy | | | 𝓁 =κ| 2 |. | d u∕dy2 | | |

Problems

Figure 11.P4

Geometry and problem description for Problem 11.4.

y

0

x

2H

Develop an expression for a fully developed turbulent flow between two infinite parallel plates. See Figure 11.P4. The pressure gradient in the flow is dp∕dx, the spacing between the plates is 2H, and y is measured from the center of the parallel plate channel. 11.5

Water flows in tube with an average velocity of 0.5 m/s. The tube inside diameter is 15 mm. Mean inlet and outlet temperatures of water are 12 and 32 ∘ C, respectively. The tube surface is maintained at 40 ∘ C. Determine the tube length to heat the water.

11.6

Air at 17 ∘ C is flowing in a tube with a mass flow rate of 0.04 kg/s. The internal diameter of the tube is 2 cm. The tube length is 2 m, and a UHF of 5 kW/m2 is applied at the tube wall. (a) Determine the tube outlet temperature. (b) Determine the variation of bulk temperature. (c) Determine the variation of wall temperature.

11.7 (a) Show that a relation for the ratio between the average velocity V and maximum velocity (centerline velocity) Vc is given by 1 V = √ Vc 1 + 1.33 f where f is the Moody friction factor. (b) Water at a mean temperature of 27 ∘ C flows through a 3-cm tube with an average velocity of 15 m/s. Assume that flow is hydrodynamically fully developed, estimate (a) Moody friction factor; (b) wall shear stress; (c) pressure drop per meter; (d) centerline velocity. 11.8

Consider steady turbulent Couette flow between two infinite parallel plates. See Figure 11.P8. The plates are moving in opposite directions. The pressure drop is zero. There is no body force. The mixing length 𝓁 is given by 𝓁=

κ (H2 − y2 ). 2H

(a) Obtain an expression for the turbulent shear stress τt . (b) Obtain an expression for the velocity distribution u.

U

y

x

0

U Figure 11.P8

Turbulent Couette flow.

2H

661

662

11 Turbulent Internal Flow: Momentum and Heat Transfer

11.9

Sleicher proposed that near the wall, eddy kinematic viscosity obeys the following relationship: εM = ν(cy+ )2 . Considering the turbulent and viscous contribution to shear stress develop the following relationship for the velocity profile: 1 u+ = tan−1 (cy+ ) c If c = 0.088, then the velocity profile gives a good fit of turbulent velocities in laminar and buffer layers.

11.10

Consider fully developed turbulent flow in a pipe. The Reynolds number is 5 × 105 . The velocity profile may be represented by u+ = 8.75(y+ )1∕7 . Show that the Moody friction factor f is given as f=

0.3046 1∕4

ReD where ReD = 11.11

VD ν

is the Reynolds number.

The Taylor–Prandtl velocity profile is given by u+ = y+ , 0 ≤ y+ ≤ 11.5 u+ = 2.5 ln y+ , y+ > 11.5. Assuming constant fluid properties, develop an expression for the Nusselt number for fully developed turbulent flow in a tube. Assume a UHF on tube.

11.12

The following velocity profile holds across a pipe having radius R except very near the wall. The contribution to the volume flow rate from this thin region is negligible 1 u = ln (E y+ ) uτ κ where y = R − r is measured from the wall, κ = 0.41, and E = 7.76. Prandtl proposed the following relation: ) ( ReD √ 1 c . √ ≃ 1.73 ln 1.63 f cf Obtain this relation.

11.13

Water at a mean temperature of 510 K is flowing turbulently in a tube of 3 cm. The viscous sublayer thickness is measured to be 30 μm. Determine: (a) the Reynolds number for the flow (b) εM /ν at position of y = 3 mm from the pipe wall (c) the Nusselt number.

11.14

Water is flowing at 22 ∘ C in a smooth 5-cm-diameter tube. The average velocity of the water is 2 m/s. Blasius found empirically that for ReD ≤ 105 , the Moody friction factor is given by f=

0.3164 1∕4

ReD

.

Experimental studies indicate that turbulent flow in a pipe may be approximated by using the relation ) ( r 1∕7 u = Vc 1 − R where Vc is the tube centerline velocity and R is the pipe velocity. Using this approximate velocity profile:

Problems

(a) show that the wall shear stress in a turbulent flow in a pipe may be estimated by τw =

0.039 ρ V2 1∕4

ReD

(b) determine total shear stress at r/R = 0.9 (c) determine fraction of this shear stress due to turbulent stress. 11.15

Consider steady fully developed turbulent flow between two parallel plates. The separation distance is 2H. See Figure 11.P15.

y

2H

0 Figure 11.P15

x Turbulent flow between parallel plates.

Experimental studies indicate that the velocity distribution in the core of the turbulent flow is approximately represented by ( )1∕7 u u = Vc H where Vc is the mean centerline velocity at y = H. Determine the variation εM and εH as a function of 11.16

y . H

Water at a mean temperature of 80 ∘ C flows inside a smooth tube with a velocity of 0.5 m/s. Flow is fully turbulent. The internal diameter of the tube is 3 cm. Calculate the following quantities: (a) Moody friction factor f (b) The quantity V+ (c) The quantity R+ (d) The Reichardt expression for μt /μ is given by [ ( + )] μt y = κ y+ − 5 tanh . μ 5 Using Reichardt expression, calculate εM /ν for y+ = 15.

11.17

Water is flowing in a parallel plate channel at an average velocity of 3 m/s. At a certain location, the average temperature of water is 17 ∘ C. The spacing between the channel is H = 4 cm. At 1 mm from the wall, determine: (a) y+ (b) u+ (c) H+ .

11.18

Water at a mean temperature of 80 ∘ C flows inside a smooth tube with a velocity of 1 m/s. Flow is fully turbulent. The internal diameter of the tube is 3 cm. Calculate the following quantities: (a) thickness of the viscous sublayer (b) thickness of the buffer layer.

663

664

11 Turbulent Internal Flow: Momentum and Heat Transfer

11.19

Water at a mean temperature of 17 ∘ C enters into a smooth tube with a velocity of 0.6 m/s. A UHF of 4 × 104 W/m2 is applied to tube to heat the water. Tube is 1.5 m long and its internal diameter is 2 cm. Calculate the surface temperature at the tube exit: (a) using the Gnielinski correlation (b) using the von Karman analogy correlation.

11.20

Water at 17 ∘ C with a velocity of 2.5 m/s enters a 2-m-long smooth tube. The inside diameter of the tube is 3 cm and its surface temperature is maintained at 80 ∘ C. Calculate the following quantities: (a) mean water temperature Tm (b) time-averaged velocity u at a distance 0.5 mm measured from the wall (c) time-averaged temperature T at a distance 0.5 mm measured from the wall (d) turbulent thermal eddy diffusivity εH at a distance 0.5 mm measured from the wall and also calculate the eddy thermal conductivity (e) mixing length 𝓁 at a distance 0.5 mm measured from the wall

11.21

Consider HFD thermally developing atmospheric air flow inside a 5-cm tube with a Reynolds number of 50 000. Here, it is assumed that the tube has an unheated initial section. The unheated section is long enough for the velocity profile to become fully developed before heating begins. See Figure 11.P21. The heated section of the tube is maintained at 50 ∘ C. The air at an initial temperature of 10 ∘ C flows into the heated section of the tube. Assume that the Prandtl number is Pr ≈ 0.72. (a) Plot the local mean fluid temperature θm as a function dimensionless axial distance x+ . (b) Plot the local Nusselt number Nu(x+ ) as a function dimensionless axial distance x+ . (c) Calculate the local Nusselt number at x = 20 cm.

Unheated section

Tw = 50 °C

Tmi = 10 °C

D = 5 cm 0

x 2m

Figure 11.P21

11.22

Geometry and problem description for Problem 11.21.

Consider HFD thermally developing atmospheric air flow inside a 5-cm tube with a velocity of 31.6 m/s. Here, we assume that tube has an unheated initial section. The unheated section is long enough for the velocity profile to become fully developed before heating begins. See Figure 11.P22. The heated section of the tube is maintained at 50 ∘ C. The air at an initial temperature of 10 ∘ C flows into the heated section of the tube. Use Notter-Sleicher solution.

Unheated section

Tw = 50 °C D = 5 cm

Tmi = 10 °C 0

Figure 11.P22

x

Geometry and problem description for Problem 11.22.

Problems

(a) Plot the local mean fluid temperature θm as a function of dimensionless axial distance x+ . (b) Plot the local Nusselt number Nu(x+ ) as a function of dimensionless axial distance x+ . (c) Calculate the local Nusselt number at x = 50 cm. 11.23

Consider HFD thermally developing atmospheric air flow inside a 2-cm tube with a Reynolds number of 10 000. Here, it is assumed that tube has an unheated initial section. The unheated section is long enough for the velocity profile to become fully developed before heating begins. A UHF of 2 × 103 W/m2 is applied to the heated section of the tube. See Figure 11.P23. Air temperature at x = 0 is assumed to be 290 K. Air flows in the heated section with a mean of 27 ∘ C. Assume that Prandtl number is Pr ≈ 0.72. (a) Plot the local Nusselt number Nu(x+ ) as a function of dimensionless axial distance x+ . (b) Calculate the local Nusselt number at x = 20 cm. (c) Compute the mean fluid temperature at x = 50 cm (d) Calculate the pipe surface temperature at x = 50 cm. (e) Calculate the local Nusselt number at x = 90 cm and compare with the Gnielinski correlation. (f) Calculate the pipe surface temperature at x = 90 cm.

Unheated section

ʺ = 2 × 103 W/m2 qw D = 2 cm 0

x 1m

Figure 11.P23

Geometry and problem description for Problem 11.23.

Hint: Use Notter-Sleicher solution. 11.24

Air with an average velocity of 30 m/s is flowing inside a tube having an internal of 50 mm. The tube length is 1 m. The air inlet temperature 14 ∘ C and pressure loss in the tube is 75 mm of water column. The external tube surface is kept 100 ∘ C by steam condensation. We wish to estimate the heat transfer using the Colburn analogy. The density of water is ρw = 1000 kg/m3 .

11.25

Water at 17 ∘ C enters a smooth tube. Tube surface temperature is kept at 100 ∘ C. The tube diameter is 5 cm, and its length is 5 m. Determine the mass flow rate so that the exit temperature will be 50 ∘ C. Use the von Karman analogy.

11.26

Water enters a rough pipe at a mass flow rate of 0.13 kg/s. The mean inlet and outlet temperatures are 14 and 70 ∘ C. The inner diameter of the pipe is 20 mm and pipe has a mean roughness height of 0.080 mm. If the pipe surface temperature is 100 ∘ C, calculate the pipe length required to accomplish the heating process. Calculate also the length for a smooth pipe. The Colebrook equation is ( ) ε∕D 2.51 1 + √ . √ = −2log10 3.7 f Re f

11.27

Water enters a smooth tube at a mean temperature 320 K and flows turbulently in the tube with a mass flow rate of 0.050 kg/s. The tube has a length of 2 m, and it has an internal diameter of 20 mm. A heat flux is applied to the tube surface, and the wall heat flux varies with axial distance x according to the relation ( ) W πx q′′w = q′′0 sin 2 m2

665

666

11 Turbulent Internal Flow: Momentum and Heat Transfer

where q′′0 = 60 000 and x is measured from the inlet, and it has the unit m. Plot the mean bulk temperature Tm (x) and the pipe surface temperature Tw (x) as functions of the axial distance. 11.28

Consider a fully developed turbulent pipe flow with UHF q′′w . Suppose that a two-layer universal velocity profile is given by u+ = y+ 0 ≤ y+ < 13.6 u+ = 2.44 ln y+ + 5.0 y+ > 13.6. (a) Obtain an expression for the two-layer temperature profile. (b) Obtain an expression for Stanton number.

11.29

Mikheev, M.A. [84] proposed the following correlation for fully turbulent flow of liquids in straight smooth tubes to calculate the average heat transfer coefficients. It is reported that the experimental data may be correlated with ( )0.25 Pr 0.43 0.43 NuD = 0.021 Re0.8 Pr Pr D Prw hD k ρVD ReD = μ L > 50 D This correlation may used to calculate the mean heat transfer coefficient for short tubes. In this case, correlation is presented in the following form ( )0.25 Pr 0.43 0.43 NuD = 0.021 Re0.8 Pr Pr F D Prw L < 50 D where F is the correction factor, and it is given as ( ) D F≈1+2 L Here L is tube length measured from the tube inlet and D is the tube internal diameter. Properties are evaluated at Tm = (Tmi + Tmo )∕2. The Prandtl number Prw is evaluated at pipe surface temperature. Compare the values of NuD predicted by this correlation with other experimental correlation you may find in literature. NuD =

11.30

Compare Eq. (11.393) with Eq. (11.390).

11.31

Water at 30 ∘ C enters a tube having internal diameter of 2.5 cm. The tube wall temperature is maintained at 100 ∘ C by condensing steam on the pipe. The mass flow rate of water is 0.6 kg/s. It is desired to heat the water to 70 ∘ C. We wish to calculate the required tube length to accomplish this task.

11.32

Water flows in a tube of 25 mm diameter with a mass flow rate of 0.2 kg/s. The tube surface temperature is maintained at 100 ∘ C by condensing steam on the external pipe surface. Water enters the tube at 27 ∘ C and leaves at 57 ∘ C. Flow is fully developed, and we wish to determine the length of the tube to accomplish the heating process.

11.33

Suppose you wish to correct Colburn equation, Eq. (11.385), for the entrance effects and property variations. StPr2∕3 = 0.023 Re−0.2 D .

Problems

For this purpose, you multiply Eq. (11.387b) to correct for the entrance effects when L/D < 10 by the following two expressions: [ [ ( )] ( )0.7 ] D D 1 + 1.1 or 1 + . L L ( )p μ Multiply also by μ to take care of property variations, in particular, viscosity vary significantly with temw perature, where you estimate p. Compare with some other correlation you may choose. How do they compare? Comment on your findings. 11.34

The following correlation is recommended by Hausen [65] to account for higher values of the average heat transfer coefficient in short tubes: [ ( )2∕3 ] ( μ )0.14 [ ] D 2∕3 m NuD = 0.116 ReD − 1 Pr1∕3 1 + . L μw It is reported that it is also applicable in the transition region 2300 ≤ ReD ≤ 6000. Prepare a plot of NuD ∕Pr1∕3 versus ReD for L/D of 20, 50, 70, and 110. Let ReD vary from 2 500 to 10 000. Compare the value of NuD predicted by this equation with some other equations you choose.

11.35

Water at 300 K enters a 2.5-cm tube with an average velocity of 1.2m/s. The tube wall is maintained at 373 K by condensing steam on the tube external surface. We wish to determine tube length required to heat the water to 330 K.

11.36

Engine oil at a mean temperature of 400 K flows in a 5-cm tube with an average velocity of 1.2 m/s. The tube wall is maintained at 430 K. The tube length is 2 m. We wish to estimate the heat transfer coefficient.

11.37

Consider fully developed turbulent flow in a smooth pipe. The pipe is subjected to UHF. Kays et al. [23] give a two-layer velocity profile u+ = y+ ;

0 ≤ y+ ≤ 10.8

u+ = 2.44 ln y+ + 5;

y+ ≥ 10.8.

We wish to derive the corresponding two-layer temperature profile. 11.38

Hausen [75] proposed the following correlation for Nusselt number for transition region: ( )0.14 [ ( )2∕3 ] [ ] μ D 2∕3 NuD = 0.116 ReD − 125 Pr1∕3 1+ μw L 2100 < ReD < 10 000. Compare this correlation with other available correlations.

11.39

Seban and Shimazaki [76] proposed the following correlation for liquid metals flowing turbulently in tubes. Flow is fully developed. NuD = 5 + 0.025(ReD Pr )0.8 Tw = constant PeD > 100 L > 30. D Compare this correlation with other available correlations.

667

668

11 Turbulent Internal Flow: Momentum and Heat Transfer

11.40

Calculation of friction factors for smooth tubes in turbulent flow is important in heat transfer calculations. Recently, Li et al. [77] have provided an explicit correlation for turbulent flow in smooth tubes: ] [ C3 C2 C1 + + f=4 [ ln (ReD )] [ ln (ReD )]2 [ ln (ReD )]3 C1 = −0.0015702 C2 = 0.3942031 C3 = 2.5341533 4000 < Red < 1 × 107 . Compare this correlation with available experimental correlations.

11.41

Sukomel, A.S. et al. [88] proposed the following correlation for turbulent flow of gases in straight smooth tubes to calculate the local heat transfer coefficients. It is reported that the experimental data may be correlated with 0.43 Pr 0.43 F NuD = 0.022 Re0.8 D Pr

ρVD hD L ReD = > 50 k μ D In this correlation, it is assumed that the turbulence starts from the tube inlet. Here F is a correction factor to account for the change of heat transfer coefficient in the thermal entrance region. F is given by ( ) x ⎧1 for > 15 ⎪ D F=⎨ ( )−0.12 ( ) x ⎪1.38 x < 15 ⎩ D D NuD =

Properties are evaluated at Tm = (Tmi + Tmo )∕2. Compare the values of NuD predicted by this correlation with other experimental correlation you find in literature. 11.42

Water at a mass flow rate of 0.44 kg/s is flowing turbulently in a 5 cm smooth tube. Flow is HFD before heating begins. Tube has a short heating section of 1-m length and section is heated uniformly. Water enters the tube at a mean temperature of 300 K. Determine the average heat transfer coefficient.

11.43

Consider air at a mean temperature of 17 ∘ C and one atmospheric pressure. Air flows with an average velocity of 35 m/s into 12.5 mm inside diameter tube having a length of 12 cm. Tube surface temperature is 77 ∘ C. Determine the exit temperature.

11.44

Consider water flow in a 10 cm tube. Tube has a calming section and for this reason flow becomes hydrodynamically fully developed (HFD) before heating begins. Heating section is 1 m, and it is heated uniformly. Tube surface temperature is kept at 100 ∘ C. The water flows at an average velocity of V = 2m/s. Water enters the tube at a mean bulk temperature of Tmi = 27 ∘ C. Estimate the average heat transfer coefficient. Calming section

V = 2 m/s

Heating section

Tmi = 27 °C

L = 1m Figure 11.P44

11.45

Geometry and problem description for Problem 11.44.

Consider water flow in a 2-cm tube. Heating section is 1 m, and it is heated uniformly by applying a UHF of q′′0 = 110kW∕m2 . The water enters the tube at a velocity of V = 1.1m/s and at a mean temperature of Tmi = 17 ∘ C.

Problems

q0ʺ = 110 kW/m2 Tmi = 17 °C D = 2 cm V = 1.1 m/s Figure 11.P45

L=1m Geometry and problem description for Problem 11.45.

See Figure 11.P45. Estimate the outlet temperature Tmo and tube surface temperature Tw at the outlet. Plot the variation of mean fluid temperature along the tube. 11.46

The mixing length 𝓁 is measured by Nikuradse [11] in 1932 for turbulent flows in pipes. Nikuradse found that the mixing length 𝓁 does not depend on the Reynolds number for ReD > 105 . Using his own experimental data, Nikuradse developed an expression for the mixing-length distribution ( )2 ( )4 r r 𝓁 = 0.14 − 0.08 − 0.06 . R R R This equation has the limits of 𝓁 = κ y as (y/R) → 0. Water at 315 K flows in a 15-cm tube with an average velocity of 1.42m/s. Consider now the position where the flow is fully developed. (a) In Chapter 11, an expression for eddy diffusivity of momentum is developed as follows: √ εM 1 + 4(𝓁 + )2 1 =− + . ν 2 2 Calculate the eddy diffusivity of momentum using the mixing length developed by Nikuradse at a distance of 5 mm from the pipe wall. (b) Calculate the eddy diffusivity of momentum using the Reichardt model at a distance of 5 mm from the pipe wall.

11.47

Water at an average velocity of 0.1 m/s flows in a 3 cm smooth tube. Flow is TFD. Assume that at a certain location, the mean water temperature is 310 K and the tube surface temperature is 370 K. Estimate the heat transfer coefficient at this point.

11.48

Levenspiel [70] reports the following correlation for transition region for flow in pipes: [ [ ] ( )2∕3 ] ( μ )0.14 D 2∕3 m NuD = 0.116 ReD − 125 (Pr1∕3 ) 1 + L μw 2100 < ReD < 104 . Compare this correlation with other correlations you choose.

11.49

Water at a mean temperature of 27 ∘ C flows through a 2-cm tube with an average velocity of 1 m/s. Assume that flow is hydrodynamically fully developed, estimate (a) Skin friction factor cf . (b) Moody friction factor f. (c) V+ (d) R+ (e) At y = 0.25 mm from the wall, calculate the eddy diffusivity of momentum εM , using van Driest expression. (f) At y = 0.25 mm from the wall, calculate the eddy diffusivity of momentum εM , using Reichardt expression. (g) At y = 0.25 mm from the wall, calculate the eddy diffusivity of momentum εM , using Spalding expression.

11.50

Water at a mean temperature of 27 ∘ C flows in a 2.5-cm tube with an average velocity of 4 m/s. Assume that flow is both hydrodynamically and thermally fully developed, estimate Nusselt number using Martinelli analogy, Eq. (11.372).

669

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11 Turbulent Internal Flow: Momentum and Heat Transfer

11.51

The solution for UWT can be used to solve the problem of arbitrary wall temperature distribution. We can use the superposition method since the energy equation is linear. If the wall temperature, Tw (x), can be approximated by a series steps, the temperature distribution within the fluid at any point can be written as ) | ( x+ dTw | dξ| T − Ti = [1 − θ(x+ − ξ, η)] | ∫ξ=0 dξ |Stieltjes The heat flux may be written in the following form ( ) ) ]( x+ [ | dTw k 𝜕T k 𝜕θ + q′′w (x) = (x − ξ, η)|| dξ =− R 𝜕η η=1 R ∫0 𝜕η dξ |η=1 Assume that the wall temperature distribution is given by Tw (x+ ) − Ti = Bx+ Obtain an expression for the fluid temperature distribution for turbulent flow in a pipe.

11.52

Sleicher [84] performed experiments for turbulent air flow in a pipe having an internal diameter of 1.5′′ . The mean air temperature may be assumed to be 80 ∘ F. Average air velocity is 15.3 ft/s. The maximum velocity is measured to be 19.6 ft/s. The experimental data may be given as y (inch)

0.746

0.636

0.536

0.436

0.336

0.236

0.136

0.086

0.066

0.046

0.036

0.031

0.026

u (ft/s)

19.6

19.45

19.0

18.25

17.20

16.15

14.75

13.45

12.50

11.15

9.8

9.0

7.8

(a) Prepare a graph of u∕Vc versus r/R. (b) Compare this data with Spalding and von Karman velocity distributions.

References 1 Laufer, J. (1950). Investigation of Turbulent Flow in a Two-Dimensional Channel. NACA TN 2123. 2 Anselmet, F., Ternat, F., Amielh, M. et al. (2009). Axial development of the mean flow in the entrance region of turbulent pipe and duct flows. C. R. Mec. 337: 573–584. 3 Burmeister, L.C. (1993). Convective Heat Transfer, 2e. Wiley. 4 Pope, S.B. (2000). Turbulent Flows. Cambridge University Press. 5 Ghiaasiaan, S.M. (2018). Convective Heat and Mass Transfer, 2e. CRC Press. 6 Rannie, W.D. (1956). Heat transfer in turbulent shear flow. J. Aeronaut. Sci. 23 (5): 485–489. 7 van Driest, E.R. (1956). On turbulent flow near a wall. J. Aerosp. Sci. 23: 1007–10011. 8 Pai, S.I. (1957). Viscous Flow Theory II-Turbulent Flow. D. Van Nostrand, Inc. 9 Dean, R.B. (1978). Reynolds number dependence of skin friction and other bulk flow variables in two-dimensional rectangular duct flow. J. Fluids Eng. 100 (2): 215–223. 10 Zanoun, E.S. and Durst, F. (2003). Evaluating the law of the wall in two-dimensional fully developed turbulent channel flows. Phys. Fluids 15 (10), 3079–3089. 11 Nikuradse, J. (1950). Laws of Flow in Rough Pipes. NACA TN 1292. 12 Deissler, R.G. (1950). Analytical and Experimental Investigation of Adiabatic Turbulent Flow in Smooth Tubes. NACA TN 2138. 13 Rothfus, R.R. and Monrad, C.C. (1955). Correlation of turbulent velocities for tubes and parallel plates. Ind. Eng. Chem. 47 (6): 1147–1149. 14 Laufer, J. (1953). The Structure of Turbulence in Fully Developed Pipe Flow. NACA TN 2954. 15 Laufer, J. (1950). Investigation of Turbulent Flow in a Two-Dimensional Channel. NACA TN 2132. 16 von Karman, T. (1939). The analogy between fluid friction and heat transfer. Trans. ASME 61: 705. 17 White, F.M. and Majdalani, J. (2022). Viscous Fluid Flow, 4e. McGraw Hill.

References

18 Prandtl, L. (1935). The mechanics of viscous fluids. In: Aerodynamic Theory, vol. 3 (ed. W.F. Durand) Berlin: Springer, 155: 162. 19 Taler, D. (2016). Determining velocity and friction factor for turbulent flow in smooth tubes. Int. J. Therm. Sci. 105: 109–122. 20 Petukhov, B.S. (1970). Heat transfer and friction factor in turbulent pipe flow with variable physical properties. In: Advances in Heat Transfer, vol. 6 (ed. J.P. Harnett and T.F. Irvine). New York: Academic Press, 503–564. 21 Xiande, F. and Zhou, Z. (2011). New correlations of single-phase friction factor for turbulent pipe flow and evaluation of existing single-phase friction factor correlations. Nucl. Eng. Des. 241: 897–902. 22 Churchill, S.W. (2001). Turbulent flow, and convection: the prediction of turbulent flow and convection in a round tube. Adv. Heat Transfer 34: 255–361. 23 Kays, W., Craward, M., and Weigend, B. (2005). Convective Heat Transfer, 4e. McGraw-Hill. 24 Siegel, R. and Sparrow, E.M. (1960). Comparison of turbulent heat transfer results for uniform wall heat flux and uniform wall temperature. ASME J. Heat Transfer 82: 152. 25 Johnk, R.E. and Hanratty, T.J. (1962). Temperature profiles for turbulent flow of air in a pipe I – the fully developed heat transfer region. Chem. Eng. Sci. 17: 802. 26 Kader, B.A. (1981). Temperature and concentration profiles in fully turbulent boundary layers. Int. J. Heat Mass Transfer 24 (9): 1541–1544. 27 Bhatti, M.S. and Shah, R.K. (1987). Turbulent and transition flow convective heat transfer in ducts. In: Handbook of Single-Phase Convective Heat Transfer (ed. S. Kakac, R.K. Shah and W. Aung). New York: Wiley Inter Science, 4.1:4.160. 28 Mills, A.F. (1999). Heat Transfer, 2e. Prentice Hall. 29 Notter, R.H., Sleicher, C.A., and Crippen, M.D. (1970). A solution to turbulent Graetz problem by matched asymptotic expansions I – the case of uniform wall temperature. Chem. Eng. Sci. 25: 845–857. 30 Notter, R.H. and Sleicher, C.A. (1970). A solution to turbulent Graetz problem by matched asymptotic expansions II: the case of uniform wall heat flux. Chem. Eng. Sci. 26: 539–565. 31 Notter, R.H. and Sleicher, C.A. (1972). A solution to turbulent Graetz problem III. Fully developed and entry region heat transfer rates. Chem. Eng. Sci. 27: 2073–2093. 32 Sparrow, E.M., Hallman, T.M., and Siegel, R. (1957). Turbulent heat transfer in the thermal entrance region of a pipe with uniform heat flux. Appl. Sci. Res. Sec. A 7: 37–52. 33 Al-Arabi, M. (1982). Turbulent heat transferrin the entrance region of a tube. Heat Transfer Eng. 3 (4): 76–83. 34 Reynolds, A.J. (1974). Turbulent Flows in Engineering. Wiley. 35 Colburn, A.P. (1964). A method of correlating forced convection heat transfer data and comparison with fluid friction. Int. J. Heat Mass Transfer 7 (12): 1359–1384. 36 Knudsen, J.G. and Kats, D.L. (1958). Fluid Dynamics and Heat Transfer. McGraw-Hill. 37 Kadar, B.A. and Yoglom, A.M. (1972). Heat and mass transfer laws for fully turbulent wall flows. Int. J. Heat Mass Transfer 15 (12): 2229–2351. 38 Yu, B., Ozeo, H., and Churchill, S.W. (2001). The characteristics of fully developed turbulent convection in a round tube. Chem. Eng. Sci. 56: 1781–1800. 39 Martinelli, R.C. (1947). Heat transfer to molten metals. Trans. ASME 69: 947–959. 40 Churchill, S.W. (1997). Critique of the classical algebraic analogies between heat, mass, and momentum transfer. Ind. Eng. Res. 36: 3866–3878. 41 McAdams, W.H. (1954). Heat Transmission, 3e. McGraw Hill. 42 Deisler, R.G. (1953). Analysis of Turbulent Heat Transfer and Flow in the Entrance Region of Smooth Passages. NACA TM 3016. 43 Boelter, L., Young, D. and Iversen, H. (1948). An Investigation of Aircraft Heaters XXVII: Distribution of Heat-Transfer Rate in the Entrance Section of a Circular Tube. NACA TN 1451. 44 Mills, A.F. (1962). Experimental investigation of turbulent heat transfer in the entrance region of a circular conduit. J. Mech. Eng. Sci. 4 (1): 63–77. 45 Dittus, F.W. and Boelter, L.M.K. (1930). Heat Transfer in Automobile Radiators of Tubular Tube, 443–461. Berkeley, Publ. Eng.: University of California. 46 Winterton, R.H.S. (1998). Where did the Dittus and Boelter equation come from? Int. J. Heat Mass Transfer 41 (4–5): 809–810. 47 Sieder, E.N. and Tate, G.E. (1943). Heat transfer and pressure drop of liquids in tubes. Ind. Eng. Chem. 28: 1429–1435.

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48 Hausen, H. (1959). New equations for heat transfer in free or forced flow. Allg. Wärmetech. 9 (4/5): 75–79. 49 Hausen, H. (1974). Extended equation for heat transfer in tubes at turbulent flow. Wärme Stoffubertragung 7: 222–225. 50 Gnielinski, V. (1976). New equations for heat and mass transfer in turbulent pipe and channel flow. Int. Chem. Eng. 16 (2): 369–368. 51 Sleicher, C.A. and Rouse, M.W. (1975). A convenient correlation for heat transfer to constant and variable property fluids in turbulent pipe flows. Int. J. Heat Mass Transfer 18: 667–683. 52 Ozisik, M.N. (1985). Heat Transfer, A Basic Approach. McGraw Hill. 53 Meyer, J.P. and Everts, M. (2019). A review of the recent developments in laminar, transitional, quasi-turbulent and turbulent forced and mixed convective flow through horizontal tubes. Adv. Heat Transfer 51: 131–205. 54 Cengel, Y. and Gajhar, A.J. (2020). Heat and Mass Transfer, Fundamentals and Applications, 6e. McGraw Hill. 55 Ghajar, A.J. and Tam, L.M. (1994). Heat transfer measurements and correlations in the transition region for a circular tube with three different inlet configurations. Exp. Therm. Fluid Sci. 8 (1): 79–90. 56 Ghajar, A.J. and Tam, L.M. (1995). Flow regime map for horizontal pipe with uniform heat flux and three inlet configurations. Exp. Therm. Fluid Sci. 10: 287–297. 57 Ghajar, A.J., Tam, L.M., and Tam, S.C. (2004). Improved heat transfer correlation in transition region for a circular tube with three inlet configurations using artificial neural networks. Heat Transfer Eng. 25 (2): 30–40. 58 Tam, L.M. and Ghajar, A.J. (1997). Effect of inlet geometry and heating on the fully developed friction factor in the transition region of a horizontal tube. Exp. Therm. Fluid Sci. 15 (1): 52–64. 59 Tam, L.M. and Ghajar, A.J. (1998). The unusual behavior of local heat transfer coefficient in a circular tube with a bell mouth inlet. Exp. Therm. Fluid Sci. 16: 187–194. 60 Tam, H.K., Tam, L.M., and Ghajar, A.J. (2013). Effect of inlet geometries and heating on the entrance and fully developed friction factors in the laminar and transition regions of a horizontal tube. Exp. Therm. Fluid Sci. 44: 680–696. 61 Tam, L.M., Ghajar, A.J., and Tam, H.K. (2008). Contribution analysis of dimensionless variables for laminar and turbulent flow convection heat transfer in a horizontal tube using artificial neural network. Heat Transfer Eng. 29 (9): 793–804. 62 Tam, L.M. and Ghajar, A.J. (2006). Transitional heat transfer in plain horizontal tubes. Heat Transfer Eng. 27 (5): 23–38. 63 Meyer, J.P. and Oliver, H.A. (2014). Heat transfer and pressure drop characteristics of smooth horizontal tubes in the transitional flow regime. Heat Transfer Eng. 35 (14–15): 1246–1253. 64 Churchill, S.W. (1977). Comprehensive correlating equations for heat, mass, and momentum transfer in fully developed flow in smooth tubes. Ind. Eng. Chem. Fundam. 16 (1): 109–116. 65 Abraham, J.P., Sparrow, E.M., and Tong, J.C.K. (2009). Heat transfer in all pipe flow regimes: laminar, transitional/intermittent, and turbulent. Int. J. Heat Mass. Transfer 52: 557–563. 66 Everts, M. and Meyer, J.P. (2018). Heat transfer of developing and fully developed flow in smooth horizontal tubes in transitional flow regime. Int. J. Heat Mass Transfer 117: 1331–1351. 67 Everts, M. and Meyer, J.P. (2018). Relationship between pressure drop and heat transfer of developing and fully developed flow in smooth horizontal circular tubes in the laminar, transitional, quasi-turbulent and turbulent flow regimes. Int. J. Heat Mass Transfer 117: 1231–1250. 68 Gnielinski, V. (2010). Heat transfer in pipe flow. In: VDI Heat Atlas, 2e., (ed. VDI e.V.) Chapter G1 and G2, Springer-Verlag Berlin Heidelberg, 693: 701. 69 Gnielinski, V. (2013). On heat transfer in tubes. Int. J. Heat Mass Transfer 63: 134–140. 70 Levenspiel, O. (2014). Engineering flow and Heat Exchange, 3e. Springer. 71 Bertsche, D., Knipper, P., and Wetzel, T. (2016). Experimental investigation on heat transfer in laminar, transitional and turbulent circular pipe flow. Int. J. Heat Mass Transfer 95: 1008–1018. 72 Bertsche, D., Knipper, P., and Wetzel, T. (2019). Experimental investigation on heat transfer in laminar, transitional and turbulent circular pipe flow with respect to flow regime boundaries. Int. J. Heat Mass Transfer 145: 118746. 73 Launder, B.E. and Spalding, D.B. (1972). Lectures in Mathematical Models of Turbulence. New York: Academic Press. 74 Weigand, B. (2015). Analytical Methods for Heat Transfer and Fluid Flow Problems, 2e. Springer. 75 Hausen, H. (1943). Darstellung des Wärmeuberganges in Rohren durch verallgemeinerte Potenzbeziehungen (Translation: Representation of the heat transfer in tubes by generalized power relationships.). Z VDI Beih. Verfahrenstech. 4: 91–98. 76 Seban, R.A. and Shimazaki, T.T. (1951). Heat transfer to fluid flowing turbulently in a smooth pipe with walls at a constant temperature. Trans ASME 73: 803–809.

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77 Li, P., Seem, J.E., and Li, Y. (2011). A new explicit equation for accurate friction factor calculation of smooth pipes. Int. J. Refrig. 34 (6): 1535–1541. 78 Molki, M. and Sparrow, E.M. (1986). An empirical correlation for the average heat transfer coefficient in circular tubes. J. Heat Transfer, ASME 108: 482–484. 79 Merker, G.P. (1987). Konvektive Waerme-Ubertragung. Springer. 80 Reichardt, H. (1951). Vollstandige Darstellung der turbulenten Geschwindigkeitsverteulung in glatten Leitungen. ZAMM J. Appl. Math. Mech. 31 (7): 208–219. 81 Sleicher, C.A. (1958). Experimental velocity and temperature profiles for air in turbulent pipe flow. Trans. ASME 80 (3): 693–702. 82 Lienhard, J.H. IV and Lienhard, J.H. V (2020). Heat Transfer, 5e. Phlogiston Press. 83 Daily, J.W. and Harleman, D.R.F. (1966). Fluid Dynamics. Addison-Wesley Company, Inc. 84 Sleicher, C.A. (1955). Heat Transfer in a Pipe with Turbulent Flow and Arbitrary Wall-temperature distribution, University of Michigan, PhD Dissertation. 85 Abraham, J.P., Sparrow, E.M., and Minkowycz, W.J. (2011). Internal-flow Nusselt numbers for the low-Reynolds-number end of the laminar-to-turbulent transition regime, 54: 584–588. 86 Rotta, J.C. (1972). Turbulent Flows-An introduction to the theory and its application, Vieweg. Teubner Verlag Wiesbaden. 87 Jorge, S. and Cartaxo, M. (2022). Tubular Heat Exchangers for Chemical Engineers, Walter de Gruyter GmbH, Berlin, Germany. 88 Sukomel, A.S., Isachenko, V.P., and Osipova, V.A. (1977). Heat Transfer, English translation, MIR Publishers, Moscow.

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12 Free Convection Heat Transfer 12.1 Introduction We have discussed forced convection heat transfer due to greater scope of its application in engineering. However, free convection is also an equally important means to transfer energy. For this reason, we will discuss its characteristics and present its analysis. Heat transfer resulting from a body force field, such as gravitational field, is called natural or free convection and will be studied in this chapter. The motion of the fluid in free convection is created by the buoyancy forces imposed on the fluid due to density variations. For example, density variations may be created between the fluid and the heated solid wall. There are two main characteristics of free convection flows: (1) There is a strong coupling between heat transfer and flow. (2) Buoyancy forces are in general so weak that the resulting velocities are small. Typical examples of free convection heat transfer are: heat transfer from a radiator to heat the room through which hot water circulates, heat transfer from pipes, electric transformers, electric motors, electronic equipment, etc. The accurate prediction of the Nusselt number is very important in engineering. Based on the analysis, we see that the Nusselt number Nu is a function of the Prandtl number Pr and the Rayleigh number Ra: Nu = f(Ra, Pr ).

(12.1a)

The power law is often used to correlate the experimental data in terms of the Rayleigh number and the Prandtl number: Nu ≈ CRam Pr n .

(12.1b)

12.2 Fundamental Equations and Dimensionless Parameters of Free Convection Figure 12.1 illustrates the nature of free convection near the heated vertical plate. Density variations in a fluid create body forces. Such body forces cause a significant effect on the fluid motion, and therefore, energy transport within the fluid is affected. Consider a heated vertical plate in a quiescent fluid. The coordinate system is located with the origin at the lower end of the heated vertical plate. The x-axis is parallel, and the y-axis is normal to the wall. It is assumed that the temperature of the fluid adjacent to the hot plate is higher than the temperature of the undisturbed cold fluid. Notice that the density of the hot fluid is less than the density of the cold undisturbed fluid. This low-density fluid (hot fluid) rises from the hot surface and the low-density fluid is replaced by the high-density fluid (cold fluid); thus, a free convection loop is created. There is a significant fluid motion near the heated plate. This motion leads to the formation of the boundary layer on the vertical hot plate. The momentum boundary layer thickness is represented by δ, and the thermal boundary layer thickness is represented by Δ. The fluid beyond the thermal boundary layer is assumed to be stationary. The fluid velocity in free convection is zero at the solid surface, increases to a maximum value, and then decreases asymptotically toward zero at the outer edge of the boundary layer. The flow field can be divided into two regions: (a) the boundary layer region (b) the undisturbed fluid region

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

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Δ T∞ δ g 𝜌∞

T∞

Tw Tw > T∞ x y Figure 12.1

The formation of velocity boundary layer in free convection.

The thermal boundary layer lies within the hydrodynamic boundary layer (Δ < δ) regardless of the Prandtl number. The fact that Δ < δ is a consequence of the motion producing buoyancy forces acting within the entire thermal boundary layer. In other words, δ and Δ are physically related in free convection such that Δ < δ. Motion is generally produced across the thermal boundary layer. Fluid motion cannot take place outside the thermal boundary layer thickness Δ since we cannot have the region of fluid away from the wall that is heated, and therefore buoyant, but not moving. In summary, we can summarize the important features of free convection as follows: (a) Flow pattern is determined by density changes in the fluid. (b) The velocity and temperature profiles are interdependent, and therefore, the momentum and energy equations are coupled and must be solved simultaneously. In free convection around a vertical isothermal plate, the local Nusselt number depends on the Grashof number Grx and the Prandtl number Pr hx Nux = = f(Grx , Pr ). (12.1c) k The local Grashof number Grx is defined as gβ(Tw − T∞ )x3 ν2 where Tw and T∞ are the surface and quiescent fluid temperatures, respectively. The parameter β is the coefficient of thermal expansion. On the other hand, ν and g are the fluid kinematic viscosity and the gravitational acceleration, respectively. Another important dimensionless number in free convection along a vertical plate is the Rayleigh number Rax , and it is defined as Grx =

Rax = PrGrx .

(12.1d)

The transition from laminar flow to turbulent flow depends on the relative magnitude of buoyancy and viscous forces. For a free convection on an isothermal vertical plate, the transition from laminar to turbulent flow takes place at a critical Raxc gβ(Tw − T∞ )(xc )3 ≈ 109 . να Consider the steady two-dimensional momentum, continuity, and energy equations for incompressible constant property flow over a vertical hot plate. Gravity acts only in the x-direction. The viscous dissipation term is neglected due to low velocities, and there is no energy generation in the free convection flow. Making the usual boundary layer approximations, we obtain the following equations: Continuity Raxc = PrGrxc =

𝜕u 𝜕v + =0 𝜕x 𝜕y

(12.2)

12.2 Fundamental Equations and Dimensionless Parameters of Free Convection

x-momentum: ) ( 𝜕p 𝜕u 𝜕2 u 𝜕u +v = −ρg − +μ 2. ρ u 𝜕x 𝜕y 𝜕x 𝜕x

(12.3)

The body force per unit volume due to gravity force in the negative x-direction is given as −ρg and p is the pressure. Now, if the temperature were to approach a uniform value, say T∞ , convection would stop, and Eq. (12.3) reduces to 𝜕p (12.4) −ρ∞ g − ∞ = 0 𝜕x where p∞ is the hydrostatic pressure corresponding to density ρ∞ and ρ∞ is the fluid density outside the boundary layer. Eq. (12.4) tells us that the pressure variation in the x-direction is caused only by the static elevation head. Notice that ρ∞ is constant since the fluid is considered to be incompressible. We now combine Eqs. (12.4) and (12.3), and we obtain ( ) 𝜕u 𝜕u 𝜕 𝜕2u ρ u +v = −g(ρ − ρ∞ ) − (p − p∞ ) + μ 2 . (12.5) 𝜕x 𝜕y 𝜕x 𝜕y Free convection is characterized by slow motion and small rates of deformation. For free convection flows, p − p∞ is very small and can be neglected. Assuming that the coefficient of thermal expansion β is constant and change in density is small, we may write ρ∞ − ρ ≈ βρ(T − T∞ ). Substituting Eq. (12.6) into Eq. (12.5) and dividing by density p, the x-momentum equation becomes ) ( 2 ) ( 𝜕u 𝜕u 𝜕 u +v = gβ(T − T∞ ) + ν . u 𝜕x 𝜕y 𝜕y2 The boundary conditions are

(12.6)

(12.7)

at y = 0 u = 0 No slip boundary condition.

(12.8a)

at y = 0 v = 0 (Impermeable wall)

(12.8b)

at y = ∞ u = 0 (Fluid is stationary)

(12.8c)

at x = 0 u = 0

(12.8d)

Energy

) 𝜕T 𝜕2 u 𝜕T +v =k 2. ρcp u 𝜕x 𝜕y 𝜕y The boundary conditions are (

(12.9)

at y = 0 T = Tw

(12.10a)

at y = ∞ T = T∞

(12.10b)

at x = 0 T = T∞ .

(12.10c)

Let us now nondimensionalize the continuity and momentum equations along with its boundary conditions. Since there is no explicit velocity scale in free convection, first we need to determine a velocity scale. Let us introduce the following dimensionless quantities: x (12.11) X= L y (12.12) Y= L u U= (12.13) uc v V= (12.14) uc T − T∞ θ= (12.15) Tw − T∞

677

678

12 Free Convection Heat Transfer

where uc is the characteristic velocity in the free convection flow to be determined. Then, the nondimensional momentum equation becomes ( ) ) 2 ( 𝜕U 𝜕U 1 𝜕 U ν U +V = (g β)(Tw − T∞ )L θ + (12.16a) 2 𝜕X 𝜕Y u L uc 𝜕 Y2 c where uc L/ν may be called the Reynolds number for free convection. To determine the velocity scale: (a) We may choose ν/uc L = 1 → uc = ν/L. Suppose that we consider free convection on a 1-m hot vertical plate in water. Assume that Tw = 34 ∘ C and T∞ = 20 ∘ C. We wish to estimate the free convection velocity. We evaluate the water properties at film temperature Tw + T∞ 34 + 20 = = 27 ∘ C. 2 2 At this temperature, we have Tf =

μ = 855 × 10−6 m2 ∕s β = 276.1 × 10−6 1∕K ρ = 997 kg∕m3 ν=

μ = 8.576 × 10−7 m2 ∕s ρ

L = 1 m. We now estimate the characteristic free convection velocity uc =

8.576 × 10−7 (m2 ∕s) ν m = = 8.576 × 10−7 . L 1m s

But this velocity is too small compared to free convection velocities encountered in engineering practice, and we reject this. ( ) √ (b) We may now choose uL2 (g β)(Tw − T∞ ) ≈ 1 → uc = (g β L)(Tw − T∞ ). c We now estimate the characteristic velocity √ √ uc = (g β L)(Tw − T∞ ) = 9.81 × 276.1 × 10−6 × 1 × 14 ≈ 0.194 m∕s. √ We choose uc = (g β L)(Tw − T∞ ) as the characteristic velocity. Let us rearrange this characteristic velocity as follows: √ uc L (g β L3 )(Tw − T∞ ) = ν ν2 where ReL = uc L/ν is the Reynolds number for free convection. Recall that the Reynolds number may be expressed as the ratio of inertia to viscous forces ReL ≈

Inertia force Viscous force

The nondimensional group inside the square root is the Grashof number GrL and is given as (g β )(Tw − T∞ ) L3 ν2 buoyancy force GrL = Viscous force

GrL =

We may form the following ratio: GrL Re2L

∝

Buoyancy force Inertia force

12.3 Scaling in Natural Convection

and with this ratio, convection heat transfer problems can be classified into three categories: (1) GrL ≫ Re2L (Free convection) (2) GrL ≪ Re2L (Forced convection) (3) GrL ≈ Re2L (Mixed convection) We will only consider free convection in this chapter. Momentum and energy equations can now be expressed as 𝜕U 1 𝜕2 U 𝜕U +V =θ+ √ 𝜕X 𝜕Y GrL 𝜕 Y2 𝜕θ 𝜕θ 1 𝜕2θ U +V = √ . 𝜕X 𝜕Y GrL Pr 𝜕 Y2

U

(12.16b) (12.16c)

We can also define the Rayleigh number based on plate height L as given below (g β L3 )(Tw − T∞ ) . ν2 α Consider free convection over a vertical plate. If the disturbance in free convection flow is amplified, the transition from laminar to turbulent flow takes place. Experiments show that free convection flows over isothermal vertical plates become turbulent for RaL between 108 and 109 . For an environment of very low turbulence, the rule for determination of the flow regime is based on the Rayleigh number RaL , as discussed in [1]. The critical Rayleigh number is RaL = GrL Pr =

RaL,c < 109 Laminar flow regime RaL,c > 109 Turbulent flow regime.

12.3 Scaling in Natural Convection The first important scale is the thermal boundary layer thickness Δ. The thermal boundary layer thickness Δ is an important scale to estimate the heat transfer in free convection. The heat transfer coefficient h scales as ) ( ) ( 𝜕T ΔT k k ′′ 𝜕y w q Δ ∼ k. ∼ (12.17) h= w ∼ ΔT ΔT ΔT Δ The second important scale is the velocity boundary layer thickness δ. These scales are related such that Δ ≤ δ. We assume that δ ≈ Δ. Let us consider the governing equations of free convection and scale each one. Continuity v u ∼ . (12.18) x Δ Momentum u u u u , v ∼ ν 2 , gβΔT. (12.19) x Δ Δ Energy equation u

ΔT ΔT ΔT ,v ∼α 2 x Δ Δ

where ΔT ∼ (Tw − T∞ ). From the continuity equation, we have ( ) Δ . v∼u x We now combine Eq. (12.21) with energy and momentum equations to get the following: Momentum ( ) Δ u u u2 , u ∼ ν 2 , gβΔT x x Δ Δ or u u2 u2 , ∼ ν 2 , gβΔT. x x Δ

(12.20)

(12.21)

(12.22)

679

680

12 Free Convection Heat Transfer

The terms on the left-hand side of Eq. (12.22) have similar orders of magnitude. Energy equation ( ) Δ ΔT ΔT ΔT , u ∼α 2 u x x Δ Δ or

( ) α u u , ∼ 2. x x Δ The terms on the left-hand side of Eq. (12.23) have similar orders of magnitude. Eq. (12.23) gives us α u ∼ 2. x Δ The momentum equation, Eq. (12.22), indicates a competition among three forces:

(12.23)

(12.24)

2

(a) Inertia forces: ux (b) Friction or viscous forces: ν Δu2 (c) Buoyancy forces: gβΔT. We may consider the following cases. (a) Buoyancy force is balanced by viscous force. Inertia force is neglected u ν 2 ∼ gβΔT. Δ This occurs when Pr > 1. Using Eqs. (12.24) and (12.25), we get ν αx ∼ gβΔT. Δ2 Δ2 Multiply and divide the right-hand side of this last expression by x3

(12.25)

1 ν αx ∼ (gβΔTx3 ) 3 . x Δ2 Δ2 We now rearrange this last expression keeping in mind that Pr = ν/α, and after some algebra, we obtain Δ −1∕4 ∼ Rax . x Using Eqs. (12.17) and (12.26), we get an expression for the heat transfer coefficient h=

k k ∼ . Δ x Ra−1∕4 x

and the Nusselt number is hx 1∕4 ∼ Rax . Nux = k (b) Buoyancy force is balanced by inertia force u2 ∼ gβΔT. x This takes place when Pr ≪ 1. In a similar manner, we may derive the following relations:

(12.26)

(12.27a)

(12.27b)

(12.28)

Δ ∼ (Rax Pr )−1∕4 x

(12.29a)

Nux ∼ (Rax Pr )1∕4 .

(12.29b)

or

Remember that we cannot neglect the buoyancy force since without the buoyancy force, there would be no flow. More information about scaling is given by Bejan [2] and Ghiaasiaan [3].

12.4 Similarity Solution for Laminar Boundary Layer over a Semi-Infinite Vertical Flat Plate

12.4 Similarity Solution for Laminar Boundary Layer over a Semi-Infinite Vertical Flat Plate 12.4.1 Constant Wall Temperature Consider free convection along a vertical plate in a quiescent fluid at temperature T∞ . The plate temperature is Tw (Tw > T∞ ), and it is constant. The fluid properties are constant except the fluid density near the hot plate. Continuity, momentum, and energy equations will be solved using the similarity method. Using the similarity variables, the partial differential equations will be transformed to ordinary differential equations. The similarity solution is discussed in several manuscripts, such as [4, 5]. Velocity and temperature profiles in different x positions in the boundary layer are geometrically similar, and they differ by a stretching factor in the x-direction, as discussed in [6]. This is the main idea behind the similarity solution. First time, as reported in [7], Pohlhausen, in collaboration with Schmidt and Beckmann, solved this set of boundary-layer differential equations for a vertical flat plate. In an examination of experimental data provided by Schimidt and Beckmann, Pohlhausen noted that the velocity profiles along the plate were similar. He also observed that the temperature profiles were similar. Using the similarity of the velocity profiles and the similarity of temperature profiles at various locations on the vertical flat plate, Pohlhausen was able to obtain parameters to convert the partial differential equations into a pair of ordinary differential equations. The solution involves transformation variables by introducing a similarity variable of the form cy η = 1∕4 (12.30) x where C is given by ) ( g β θw 1∕4 c= 4ν2 where θw = Tw − T∞ . Recall that g is the acceleration of gravity, β is the thermal expansion coefficient, and ν is the kinematic viscosity. The variable η can also be expressed in terms of the Grashof number Grx with the substitution of constant c into Eq. (12.30) ) ( )1∕4 ( ) )1∕4 ( ( g β θw 1∕4 g β θw 1∕4 g β θw x3 y Grx y = y = y = (12.31) η = 1∕4 x 4 4ν2 4ν2 x 4ν2 x4 x where Grx = g β (Tw − T∞ )x3 /ν2 is the local Grashof number. A dimensionless temperature θ is θ=

T − T∞ . Tw − T∞

(12.32a)

It is assumed that the dimensionless temperature θ depends on η θ(x, y) = θ(η).

(12.32b)

Velocity components are represented in terms of the stream function ψ(x, y) u=

𝜕ψ 𝜕y

v=−

𝜕ψ 𝜕x

and the continuity equation is satisfied. The stream function for the problem is defined as [ ( ) ] Grx 1∕4 F(η) ψ(x, y) = 4v 4

(12.33) (12.34)

(12.35)

681

682

12 Free Convection Heat Transfer

where F(η) is the dimensionless velocity function. Introducing Eq. (12.35) into Eqs. (12.33) and (12.34), we get ( )√ √ dF √ dF dF 2ν u = 4νc2 x = 2 x β g (Tw − T∞ ) = Grx dη dη x dη ] [ ]1∕4 [ ] [ g β ν2 (Tw − T∞ ) dF dF v = νc x−1∕4 η − 3F = − 3F η dη 4x dη ] 1∕4 [ (Gr ) dF ν x η − 3F = 1∕4 x dη 4

(12.36)

(12.36)

We can now transform the partial differential equations of momentum and energy to ordinary differential equations. After some algebra, we obtain the following ordinary differential equations: ( )2 d2 F d3 F dF +θ=0 (12.38) Momentum: 3 + 3F 2 − 2 dη dη dη Energy:

d2 θ dθ + 3 Pr F =0 dη dη2

(12.39)

where in these transformed equations, F and θ are now functions of η. Notice that the partial differential equations are transformed into ordinary differential equations. The transformed boundary conditions for the momentum equation are η=0 η→∞

F=0

dF =0 dη

dF = 0. dη

(12.40a) (12.40b)

The transformed boundary conditions for the energy equation are η=0 η=∞

θ=1 θ = 0.

(12.41a) (12.41b)

For a given Prandtl number, we must develop solutions for these coupled differential equations. Pohlhausen carried out a numerical solution for Pr = 0.733, and later, Schuh solved these equations for the Prandtl number of 10, 100, and 1000, as reported in [7]. The approximate solutions of Pohlhausen were improved by Ostrach [4], and Ostrach solved these coupled nonlinear ordinary differential equations numerically for Prandtl numbers from 0.01 to 1000. These equations can be solved by bvp4c in MATLAB 2021 and Maple 2020. We will use Maple 2020 to obtain the solution of these coupled nonlinear ordinary differential equations numerically. Results illustrated in Figures 12.2 and 12.3 show the variation of dimensionless temperature and velocity profiles against dimensionless distance η for various values of the Prandtl number Pr. Some initial values of θ′ (0) and F′′ (0) are listed in Table 12.1 for various values of Prandtl numbers. Also included are 1∕4 Nux ∕Grx . Local heat transfer coefficients can be obtained from these solutions. The local heat flux q′′w (x) at any x-position is ) ( ) ( dθ 𝜕T ′′ −1∕4 qw (x) = −k = kcx (Tw − T∞ ) − (12.42) 𝜕y y=0 dη η=0 ( )1∕4 gβθ where c = 4ν2 w and the local heat transfer coefficient h(x) is ( ) q′′w (x) dθ = −k c x−1∕4 (12.43) h(x) = Tw − T∞ dη η=0 and the local Nusselt number Nux is ) ) ( ) ( ( Grx 1∕4 dθ dθ hx Nux = = c x3∕4 − − = k dη η=0 4 dη η=0 or

) ( Nux dθ 1 = √ − dη η=0 (Grx )1∕4 2

(12.44a)

(12.44b)

12.4 Similarity Solution for Laminar Boundary Layer over a Semi-Infinite Vertical Flat Plate

1 Pr = 0.7 0.8 Pr = 2 Pr = 10 0.6 Pr = 50

θ 0.4

Pr = 100 0.2

0 0

1

2

3

4

η Figure 12.2

Temperature boundary layers of free convection near a constant temperature vertical plate.

Pr = 0.7

0.25

Pr = 2 0.20 Pr = 10 0.15

Pr = 50

F′ (η) 0.10

Pr = 100

0.05

0 0

Figure 12.3

1

2

η

3

4

Velocity boundary layers of free convection near a constant temperature vertical plate.

where Grx is the local Grashof number, and it is given as g β(Tw − T∞ )x3 . ν2 The local heat flux q′′w is integrated over the length L of the plate to obtain the total average heat flux qw per unit width ) ( L 1 4 dθ qw = q′′w (x) dx = k c L−1∕4 (Tw − T∞ ) − (12.45) L ∫0 3 dη η=0 Grx =

683

684

12 Free Convection Heat Transfer

Table 12.1

Computed parameters for free convection on an isothermal vertical plate.

Pr

0.01

F′′ (0)

𝛉′ (0)

Nux ∕Gr1∕4 x

Source

0.9862

–0.0812

0.0571

Ostrach

0.1

0.8591

–0.2301

0.1627

White

0.72

0.6760

–0.5046

0.3568

Ostrach

1

0.6421

–0.5671

0.4010

Ostrach

2

0.5713

–0.7165

0.5066

Ostrach

3

0.5309

–0.8155

0.5767

White

5

0.4804

–0.9510

0.6724

Maple 2016

6

0.4649

–1.0075

0.7124

White

10

0.4192

–1.1693

0.8268

Ostrach

30

0.3312

–1.5891

1.1237

White

50

0.2936

–1.8155

1.2838

Maple 2016

100

0.2517

–2.1913

1.5495

Ostrach

500

0.1703

–3.3066

2.3381

Maple 2016

1000

0.1449

–3.9650

2.8037

Ostrach

and the average heat flux qw is used to define the average heat transfer coefficient h ) ( qw dθ 4 h= = k c L−1∕4 − . Tw − T∞ 3 dη η=0 Finally, the average heat transfer coefficient h is used to obtain the average Nusselt number ( ) ( )1∕4 ( ) dθ dθ 4 4 GrL h L L qw − = = c L3∕4 − NuL = = k k (Tw − T) 3 dη η=0 3 4 dη η=0

(12.46)

(12.47a)

or 4 (12.47b) Nu 3 x=L where GrL = gβ(Tw − T∞ )L3 /ν2 is the Grashof number based on the plate length. Results presented in Table 12.1 are represented to within about 1% by correlations developed by Ede [8] as well as Churchill and Usagi [9]. LeFevre, as reported by Ede [8] using asymptotic results at Pr = 0 and Pr = ∞ as a guide, proposed an expression that fits to the numerical results very closely and proposed the following interpolation equation: [ ] √ 0.75 Pr (12.48) (Grx )1∕4 . Nux = √ (2.43478 + 4.884 Pr + 4.95283 Pr )1∕4 NuL =

The average Nusselt number is [ ] √ Pr hL = NuL = (GrL )1∕4 √ k (2.43478 + 4.884 Pr + 4.95283 Pr )1∕4

(12.49)

Tw = const 104 ≤ RaL ≤ 109 0 ≤ Pr ≤ ∞. It is reported that these equations fit to Ostrach’s results within ±0.5% over the entire Prandtl number range. For air, (Pr = 0.72), and local and average Nusselt numbers are Nux = 0.3568 (Grx )1∕4

(12.50a)

12.4 Similarity Solution for Laminar Boundary Layer over a Semi-Infinite Vertical Flat Plate

NuL = 0.4757 (GrL )1∕4 .

(12.50b)

The asymptotic values of the average Nusselt numbers for isothermal vertical flat plates are given by Le Fevre, as reported in [8] ( ) 1∕4 NuL = 0.800 GrL Pr 1∕2 Pr → 0 (12.51a) ( 1∕4 NuL = 0.670 Grl Pr

1∕4

)

Pr → ∞.

(12.51b)

Kays et al. [6] give the asymptotic values of the local Nusselt numbers for isothermal vertical flat plates Nux = 0.600 (Grx Pr

1∕2 1∕4

)

Nux = 0.503 (Grx Pr )1∕4

Pr → 0

(12.51c)

Pr → ∞.

(12.51d)

Example 12.1 A 40-cm-high and 20-cm-wide flat plate is placed in a vertical position in air at 1-atm pressure and 17 ∘ C. The back side of the plate is insulated, and the thickness of the plate is negligible. The front surface of the plate is at 237 ∘ C. (a) Plot the velocity and temperature profiles at the middle of the vertical plate. (b) Estimate the total heat loss from the plate. Solution First, we will solve Eqs. (12.38) and (12.39) using BVP4c solver given in MATLAB 2021a. The code to solve these equations is given below. % Similarity solution of free convection on an isothermal vertical plate % % Eqs. (12.38) and (12.39) are solved with BVP4c % solinit=bvpinit(linspace(0,8,10) ,[0,0,0,0,0]); sol=bvp4c(@NATCONV_BVP,@BC, solinit); function = NATCONV_BVP(x,y) % Pr=0.69; f=zeros(5,1); f(1)=y(2); f(2)=y(3); f(3)=-3*y(1)*y(3)+2*y(2)*y(2)-y(4); f(4)=y(5); f(5)=-(3*Pr)*y(1)*y(5); end function [res] = BC(ya,yb) % res=zeros(5,1); res(1)=ya(1); res(2)=ya(2); res(3)=ya(4)-1; res(4)=yb(2); res(5)=yb(4); end Numerical solution is given in a table.

685

686

12 Free Convection Heat Transfer

𝛈

F

F′

0

0

0

0.2500

0.0187

0.1402

0.5000

0.0656

0.7500

0.1284

1.0000 1.2500

F′′

𝛉

𝛉′

0.6803

1.0000

−0.4969

0.4470

0.8759

−0.4952

0.2267

0.2515

0.7531

−0.4850

0.2695

0.0981

0.6345

−0.4614

0.1976

0.2793

−0.0126

0.5235

−0.4242

0.2662

0.2665

−0.0838

0.4233

−0.3761

1.5000

0.3297

0.2401

−0.1223

0.3360

−0.3223

1.7500

0.3857

0.2073

−0.1364

0.2622

−0.2677

2.0000

0.4333

0.1733

−0.1341

0.2018

−0.2165

2.2500

0.4725

0.1411

−0.1222

0.1535

−0.1712

2.5000

0.5041

0.1126

−0.1057

0.1156

−0.1330

2.7500

0.5292

0.0883

−0.0882

0.0864

−0.1017

3.0000

0.5487

0.0684

−0.0716

0.0642

−0.0770

3.2500

0.5637

0.0524

−0.0569

0.0475

−0.0577

3.5000

0.5751

0.0397

−0.0445

0.0350

−0.0430

3.7500

0.5838

0.0299

−0.0344

0.0257

−0.0318

4.0000

0.5903

0.0224

−0.0263

0.0188

−0.0235

4.2500

0.5951

0.0166

−0.0199

0.0138

−0.0173

4.5000

0.5987

0.0123

−0.0150

0.0100

−0.0127

4.7500

0.6014

0.0091

−0.0112

0.0073

−0.0093

5.0

0.6033

0.0068

0.0066

0.0084

0.0053

(a) Using this numerical solution, we can compute the velocity and temperature profiles. Similarity solution indicates that the temperature and velocity distributions are given by: ( )√ 2v dF u= Grx x dη θ=

T − T∞ Tw − T∞

The dimensionless distance η and Grashof number Grx are given by: ( )1∕4 y Grx η= x 4 Grx =

gβ(Tw − T∞ )x3 ν2

The air properties are evaluated at the film temperature. The film temperature is Tf = At 400 K ν = 26.41 × 10−6 m2 /s k = 33.8 × 10−3 W/m.K Pr = 0.69 β=

Tw +T∞ 2

1 1 = 2.5 × 10−3 K−1 = Tf 400

We compute the dimensionless distance η at the middle of the plate where x = 0.20 m. ( )1∕4 ( )1∕4 y Grx y 1 gβ(Tw − T∞ )x3 × η= = x 4 x 4 ν2 )1∕4 ( y 1 9.81 × 2.5 × 10−3 (510 − 290) × (0.2)3 × η= = 313.5820731y 0.2 4 (26.41 × 10−6 )2

=

237+17 2

= 127 ∘ C = 400 K.

12.4 Similarity Solution for Laminar Boundary Layer over a Semi-Infinite Vertical Flat Plate

This equation is rearranged to express y in terms of η. η (m) y= 313.5820731 An equation for the velocity u is determined as √ ) ( ) gβ(T − T )x3 ( ( )√ 2ν dF 2ν dF w ∞ = Grx u= x dη x dη ν2 √ ) ) ( ( 9.81 × 2.5 × 10−3 × (510 − 290) × (0.2)3 dF 2 × 26.41 × 10−6 u= 0.2 dη (26.41 × 10−6 )2 ( ) dF u = 2.077594764 (m∕s) dη 0.6 0.5

u (m/s)

0.4 0.3 0.2 0.1 0

0

0.002

0.004

0.006

0.008 y (m)

0.01

0.012

0.014 0.016

0.01

0.012

0.014 0.016

(a) 550 500

T (K)

450 400 350 300 250

0

0.002

0.004

0.006

0.008 y (m) (b)

Figure 12.E1

(a) Variation of u velocity with y. (b) Variation of tempetrature T with y.

687

688

12 Free Convection Heat Transfer

Temperature distribution is determined as follows θ=

T − 290 ⇒ T = 290 + (510 − 290)θ 510 − 290

T = 290 + (510 − 290)θ Since the dimensionless distance η is known, the velocity u and the temperature T can be computed from these equations as function of y. The variations of u and T are given in Figure 12.E1a,b. (b) We now compute the heat transfer from the plate. GrL =

g β(Tw − T∞ ) L3 9.81 × 2.5 × 10−3 (510 − 290)(0.4)3 = = 4.95 × 108 . ν2 (26.41 × 10−6 )2

Flow is laminar since GrL < 109 . Eq. (12.50b) will be used to calculate the average Nusselt number: 1∕4

NuL = 0.4757GrL = 0.4757(4.95 × 108 )1∕4 ≈ 70.95 ) ( k 33.8 × 10−3 (70.95) ≈ 5.99 W∕m2 K h = NuL = L 0.4 q = hA(Tw − T∞ ) = (5.99)(0.4 × 0.20)(510 − 290) = 105.52 W.

12.4.2

Uniform Heat Flux

Consider free convection along a vertical plate in a quiescent fluid at temperature T∞ . The plate is subjected to constant uniform heat flux (UHF) q′′w . The UHF boundary condition is more realistic than the uniform wall temperature (UWT) boundary condition. Constant heat flux q′′w may be generated by electrical heating of a surface. The fluid properties are constant except the fluid density near the plate. The density has been considered as a variable in the formulation of the buoyancy term in the momentum equation. Sparrow and Gregg [10] solved this problem by similarity transformation for various Prandtl numbers. The similarity solution starts by selecting a stream function ψ 𝜕ψ 𝜕y 𝜕ψ v=− 𝜕x and the continuity equation is satisfied. The energy equation and its boundary conditions are given as follows: u=

u

𝜕θ 𝜕2θ 𝜕θ +v = α 2. 𝜕x 𝜕y 𝜕y

(12.52) (12.53)

(12.54)

The boundary conditions are x=0

θ=0

(12.55a) 𝜕θ = q′′w 𝜕y

at

y=0

−k

at

y=∞

θ=0

(12.55b) (12.55c)

where θ = T − T∞ is the temperature. The following variables are introduced to transform partial differential equations to ordinary differential equations: C1 y x1∕5 where C1 is ) ( g β q′′w 1∕5 C1 = 5 k ν2 η=

where η is the similarity variable. The stream function ψ is ψ = C2 x4∕5 F(η)

(12.56)

12.4 Similarity Solution for Laminar Boundary Layer over a Semi-Infinite Vertical Flat Plate

where F(η) is the dimensionless velocity function and C2 is given as )1∕5 ( 4 5 g β q′′w ν3 C2 = . k The dimensionless temperature θ is θ=

C1 (T − T∞ ) ( ). x1∕5 q′′w ∕k

(12.57)

The x- and y-components of velocity are dF dη [ ] dF −1∕5 v = ν C1 x η − 4F . dη

u = 5ν C21 x3∕5

(12.58) (12.59)

Using these equations and similarity variables, momentum and energy equations are transformed into ordinary differential equations. It is found that all the factors containing x cancel out, leaving only functions that depend upon η. The final result is the following: Momentum equation: ( )2 d3 F d2 F dF + 4F − 3 +θ=0 (12.60) dη dη3 dη2 F(0) = 0

(12.61a)

F′ (0) = 0

(12.61b)

F′ (∞) = 0.

(12.61c)

Energy equation: [ ] dθ dF d2 θ + Pr 4F − θ =0 dη dη dη2

(12.62)

θ′ (0) = −1

(12.63a)

θ(∞) = 0.

(12.63b)

The primes indicate differentiation with respect to η. The dependent variables F(η) and θ(η) are coupled. The simultaneous solution of these equations is required. Sparrow and Gregg [10] obtained solutions for Prandtl numbers of 0.1, 1.0, 10, and 100. These equations can be solved numerically with Maple 2020 or bvp4c in MATLAB 2021a. Sample results of numerical solutions are presented in Table 12.2 along with solutions of Sparrow and Gregg [10]. We should recognize that for uniform surface heat flux, the plate surface temperature varies along the plate. We need to determine the surface temperature variation Tw (x) and the local Nusselt number Nux . The surface temperature Tw (x) is obtained by the evaluation of Eq. (12.57) at the wall ( ) q′′ x1∕5 q′′ x1∕5 θ(0 ) θ(0) = w Tw (x) − T∞ = w k C1 k [( g β q′′ )]1∕5 w 5k ν2 q′′ x1∕5 q′′ x1∕5 θ(0) θ(0) (5x4 )1∕5 = w = w [( )] [( )]1∕5 1∕5 k k g β q′′ x4 1 g β q′′ x4 w

k ν2 = 51∕5

w

5x4

q′′w x 1 θ(0) k (Gr∗ )1∕5 x

k ν2 (12.64a)

689

690

12 Free Convection Heat Transfer

Table 12.2

θ (0) and F′′ (0) values for different Prandtl numbers. Sparrow and Gregg [8]

Pr

𝛉(0)

F′′ (0)

0.1

–2.7507

1.6434

–1.3574

0.72196

Maple 2020 𝛉′ (0)

F′′ (0)

1.48785

0.81168

1.3586

0.7201

2

1.1315

0.55910

5

0.90346

0.39739

0.72 1

10

–0.76746

0.30639

0.76955

0.30589

100

–0.46566

0.12620

0.46798

0.12562

where Gr∗x is the modified Grashof number suggested by Sparrow and Gregg, and it is given as g β q′′w x4 . k ν2 In terms of the modified Grashof number Gr∗x , Eq. (12.64a) becomes Gr∗x =

(Tw (x) − T∞ ) ( ′′ ) = 51∕5 θ(0). qw x∕k

(12.64b)

The modified Grashof number is needed since temperature difference [Tw (x) − T∞ ] is unknown a priori and varies with x along the vertical plate. The local heat transfer coefficient h(x) is found in the usual way ( ∗ )1∕5 q′′w q′′w k Grx ∕5 (12.65) = = h= Tw − T∞ x θ(0) q′′ x θ(0) 51∕5 w ( k Gr∗ )1∕5 x

and the local Nusselt number Nux is given by ( ∗ )1∕5 ( ∗ )1∕5 Grx ∕5 1 Grx hx = = 1∕5 . (12.66) Nux = k θ(0) θ(0) 5 The negative sign in results of Sparrow and Gregg is due to the definition of dimensionless temperature. They defined the dimensionless temperature function as θ=

C1 (T∞ − T) ) ( x1∕5 q′′w ∕k

(12.67)

Fuji and Fuji [12] developed an expression that the dimensionless parameter θ(0) can be estimated from the correlation given below ]1∕5 [ √ 4 + 9 Pr + 10 Pr (12.68) θ(0) = 5Pr 2 0.001 < Pr < 1000. The local Nusselt number can now be expressed as ]1∕5 [ ( ∗ )1∕5 Pr Grx Pr Nux = √ 4 + 9 Pr + 10 Pr

(12.69a)

where Gr∗x = g β q′′w x4 ∕k ν2 is the modified Grashof number. Burmeister [13] reported that flow is laminar if the following condition is satisfied: ) ( 0.916 Pr 1∕4 × 1011 . Gr∗x ≤ Pr +0.8

12.4 Similarity Solution for Laminar Boundary Layer over a Semi-Infinite Vertical Flat Plate

) ( Note that the modified Grashof number Gr∗x = g β q′′w x4 ∕kν2 is related to the conventional Grashof number Grx = g β (Tw − T∞ ) x3 /ν2 by Gr∗x = Grx Nux since q′′w = h(Tw − T∞ ). Following the footsteps of La Fevre, White [14] reports a curve fit formula to numerical results accurate to ±0.5 5%: ( ∗ )1∕5 Pr 2∕5 Nux = ( . (12.69b) )1∕5 Grx √ 3.91 + 9.32 Pr + 9.95 Pr Once we have Nux , we can find the local temperature difference Tw (x) − T∞ =

q′′w x 1 . k Nux

(12.69c)

We may use Eq. (12.69a) or Eq. (12.69b) in Eq. (12.69c). Suppose that we use Eq. (12.69a) to find [Tw (x) − T∞ ] ]1∕5 √ ( ′′ ) [ qw x 4 + 9 Pr + 10 Pr 1 Tw (x) − T∞ = ( )1∕5 k Pr Pr Gr∗x [ √ ( ) ( ′′ )4 ]1∕5 qw 4 + 9 Pr + 10 Pr ν2 = x . gβ k Pr 2 Vertical plate q′′w = const 104 < Gr∗x Pr < 109 0 ≤ Pr ≤ ∞. In the literature, average Nusselt numbers are defined based on the average of wall temperature and ambient temperature difference, which is L

Tw − T∞ =

1 5 (Tw − T∞ )dx = (Tw − T∞ )x=L . L ∫0 6

(12.70a)

Next, we evaluate (Tw − T∞ )x = L from Eq. (12.64a) q′′w L 1 θ(0). k (Gr∗ )1∕5

(Tw − T∞ )x=L = 51∕5

L

Now, the average temperature (Tw − T∞ ) becomes ( ) ′′ 5 1∕5 qw L 1 5 (Tw − T∞ ) = θ(0) 6 k (Gr∗ )1∕5 L ( 6∕5 ) ′′ qw L θ(0) 5 . = 6 k (Gr∗ )1∕5

(12.70b)

L

We will now write the average heat transfer coefficient based on the temperature difference (Tw − T∞ ) from Eq. (12.70b) q′′w

h=

Tw − T∞

= (

6∕5

5

6

)

q′′w L θ(0) k (Gr∗ )1∕5

q′′w

L

q′′w

is constant. We can also write the average Nusselt number based on Tw − T∞ from Eq. (12.70b) ] [ q′′w L 6 1 ( ∗ )1∕5 GrL NuL = . = 6∕5 5 θ(0) k(T − T )

Recall that

(12.71)

w

∞

(12.72)

691

692

12 Free Convection Heat Transfer

The modified Grashof number may be expressed in terms of GrL . Here, GrL is the Grashof number defined as gβ(Tw − T∞ )L3 . ν2 The modified Grashof number now becomes GrL =

g β q′′w L4 gβ(Tw − T∞ )L3 hL = GrL NuL . = k k ν2 ν2 Substituting the new form of the modified Grashof number into Eq. (12.72), the Nusselt number now becomes ] [ 6 1 (GrL NuL )1∕5 NuL = 6∕5 5 θ(0) or after the rearrangement of this last equation, we get ]5∕4 [ 6 1 NuL = 6∕5 (GrL )1∕4 . (12.73) 5 θ(0) Gr∗L =

Example 12.2 A vertical plate L = 20 cm high and W = 10 cm wide is thermally insulated on the back and exposed to heat flux of q′′w = 2500 W∕m2 on the front side. Energy is dissipated by free convection. Surrounding quiescent air is at the atmospheric pressure of 1 atm and the temperature of T∞ = 17 ∘ C. Estimate: (a) the maximum temperature on the plate (b) variation of the surface temperature of the plate. (c) average heat transfer coefficient. Solution The surface temperature is not known and the film temperature Tf cannot be determined to evaluate the air properties. We convert the air temperature to Kelvin, and as a first approximation, we evaluate the air properties at this temperature T = 17 ∘ C = 290K. ∞

First iteration: The properties of air at this temperature are taken from the air table ν = 15.0 × 10−6 m2 ∕s k = 25.5 × 10−3 W∕m.K Pr = 0.709. We evaluate the coefficient of thermal expansion at the film temperature T∞ 1 1 = 0.00344 K−1 β= = T∞ 290 Next, we evaluate the modified Grashof number Gr∗x as ( ) 1 9.81 × (2500)x4 ′′ 4 gβqw x 290 Gr∗x = = = 1.473 × 1013 x4 . kν2 (0.0255)(15.0 × 10−6 )2 Next, we compute the local Nusselt number Nux . We will use Eq. (12.69b) ( ∗ )1∕5 Pr 2∕5 Nux = Grx √ (3.91 + 9.32 Pr + 9.95 Pr )1∕5 = 208.45 x4∕5 . Then, we write the surface temperature Tw (x) as q′′w x 1 k Nux 2500 x 1 Tw (x) = 17 + 0.00255 208.45 x4∕5

Tw (x) = T∞ +

Tw (x) ≈ 17 + 470.32 x1∕5

12.4 Similarity Solution for Laminar Boundary Layer over a Semi-Infinite Vertical Flat Plate

An array values of temperature for the first trial is given in a table. x (cm) T ∘C w

0

2

4

6

8

10

12

14

16

18

20

17

232.0

264.0

284.9

300.7

313.7

324.7

334.4

342.9

350.7

357.8

The maximum temperature occurs at the trailing edge, x = 0.20 m: Tw ||max ≈ 17 + 470.32 (0.2)1∕5 ≈ 357.8 ∘ C The average wall temperature is L

0.20

1 1 T (x)dx == L ∫0 w 0.20 ∫0

Tw =

[17 + 470.32 x1∕5 ] dx ≈ 301.0 ∘ C.

Second iteration: We now evaluate the film temperature Tf as Tw + T∞ 301.0 + 17 = = 159.2 ∘ C ≈ 432 K. 2 2 We now evaluate air properties at Tf ≈ 432 K as Tf =

ν = 30.23 × 10−6 m2 ∕s k = 36.0 × 10−3 W∕m.K Pr = 0.687 1 1 = = 0.002314−1 K. Tf 432

β=

The Grashof number is Gr∗x

=

gβq′′w x4 kν2

9.81 × =

(

1 432

)

(2500)x4

(0.0360)(30.23 × 10−6 )2

≈ 1.725 × 1012 x4 .

The Local Nusselt number is Nux = 134.53x4∕5 . Wall temperature distribution is Tw (x) = 17 +

2500 x 1 0.686 134.53 x4∕5

Tw (x) = 17 + 516.20 x1∕5 Values of calculated local temperatures for the second iteration is given in a table. x (cm) T ∘C w

0

2

4

6

8

10

12

14

16

18

20

17

253.0

288.1

311.0

328.4

342.7

354.8

365.3

374.8

383.3

391.1

The maximum temperature occurs at the trailing edge, x = 0.20 m: Tw ||max = 17 + 516.2 (0.2)1∕5 = 391.1 ∘ C. The average surface temperature is L

Tw =

1 1 T (x)dx == L ∫0 w 0.20 ∫0

0.20

[17 + 516.20 x1∕5 ] dx ≈ 328.7 ∘ C.

Third iteration: We now evaluate film temperature Tf as Tf =

Tw + T∞ 328.7 + 17 = = 172.85 ∘ C ≈ 446 K. 2 2

693

694

12 Free Convection Heat Transfer

We evaluate the air properties at this new temperature and repeat the calculation ν = 31.91 × 10−6 m2 ∕s k = 37 × 10−3 W∕m.K Pr = 0.686 β=

1 = 0.002242. Tf

The Grashof number is Gr∗x = 1.459 × 1012 x4 . The Nusselt number is Nux = 130.07x4∕5 . Temperature distribution is Tw (x) = 17 + 519.47 x1∕5 Values of calculated local temperatures for the third iteration is given in a table. x (cm) T ∘C w

0

2

4

6

8

10

12

14

16

18

20

17

254.5

289.8

312.9

330.4

344.7

356.9

367.5

377.0

385.6

393.5

The maximum temperature occurs at the trailing edge, x = 0.20 m: Tw ||max = 17 + 519.47 (0.2)1∕5 = 393.5 ∘ C. The average surface temperature is L

Tw =

1 1 T (x)dx == L ∫0 w 0.20 ∫0

0.20

[17 + 519.55 x1∕5 ] dx = 330.7 ∘ C.

No more iteration is needed. Example 12.3 In many practical engineering problems, surface temperature is nonuniform. Similarity solutions exist for certain surface temperature distributions Tw (x). Sparrow and Gregg [15] have shown that for Tw (x) − T∞ = Axn similarity solutions exist. This power law temperature distribution could be very useful since surfaces are not always isothermal in free convection. In this example, the effect of nonuniform surface temperature upon free convection for a vertical plate is studied. Solution

) ( Grx 1∕4 ψ = 4νF(η) 4 ( )1∕4 y Grx η= x 4 gβ(Tw − T∞ ) Grx = ν2 T − T∞ θ= . Tw − T∞

It is assumed that θ = F(η). The following differential equations for momentum and energy equations: F′′′ + (n + 3)F F′′ − (2n + 2)(F′ )2 + θ = 0 θ′′ + Pr [(n + 3)F θ′ − 4 nF′ θ] = 0.

12.5 Integral Method (von Karman–Pohlhausen Method): An Approximate Analysis of Laminar Free Convection on a Vertical Plate −1∕4

Values of Nux ∕Grx

Table 12.3

for different Prandtl numbers (UHF BC n=1/5).

Pr

0.72

1

2

4

6

8

10

−1∕4 Nux ∕Grx

0.40695

0.45558

0.57252

0.70996

0.8

0.87

0.9279

The boundary conditions are F(0) = F′ (0) = 0 θ(0) = 1 F′ (∞) = 0 θ(∞) = 0. Heat flux at the wall is ) ( ( ) ( ) 𝜕η 𝜕T dθ = −k(Tw − T∞ ) q′′w = −k 𝜕 y y=0 dη η=0 𝜕y ( ) k(Tw − T∞ ) Grx 1∕4 = [−θ′ (0)] x 4 ( )1∕4 ′ 3∕4 gβ θ (0) 5∕4 x = −k 2 √ (Tw − T∞ ) x ν 2 = C1 (Tw − T∞ )5∕4 x−1∕4 . Here, C1 is

(

C1 = −k

gβ ν2

)1∕4

θ′ (0) √ . 2

The Nusselt number is q′′w x −θ′ (0) Nux = = √ (Grx )1∕4 . k(Tw − T∞ ) 2 The solution for n = 0 corresponds to the constant wall temperature boundary condition and n = 1/5 corresponds to the constant heat flux boundary condition q′′w = C1 (Tw − T∞ )5∕4 x−1∕4 = C1 (x1∕5 )5∕4 x−1∕4 = C1 . Physically acceptable solutions may be obtained for −3/5 < n < 1. We will now solve these ordinary differential equations numerically using Maple 2020 or bvp4c in MATLAB 2021a and construct the following table for constant heat flux case (Table 12.3).

12.5 Integral Method (von Karman–Pohlhausen Method): An Approximate Analysis of Laminar Free Convection on a Vertical Plate Here, the idea is to solve the integral momentum and energy equations by assuming velocity and temperature profiles. Consider the integral momentum equation ( ( ) ) ) Y Y Y( U∞ d dU∞ 1 d 2 R R dy. (12.74a) −ρU∞ ρu dy − 𝜌udy + vw ρw U∞ + −τw = ∫0 ∫0 ∫0 R dx R dx dx Following the procedure outlined in [6], we wish to add a body force term to this integral momentum equation. First, we assume that U∞ = 0, vw = 0, and R = constant, and along with an added body force term, the integral equation reduces to the following form: ( Y ) Y dp d + ρu2 dy + Y ρ g dy. (12.74b) −τw = dx ∫0 dx ∫0

695

696

12 Free Convection Heat Transfer

The pressure gradient term is dp = −ρ∞ g. dx

(12.75)

Substituting Eq. (12.75) into Eq. (12.74b) gives ( Y ) Y d 2 ρu dy − Yρ∞ g + ρ g dy −τw = ∫0 dx ∫0 ( Y ) Y Y d ρu2 dy − ρ∞ gdy + ρ g dy = ∫0 ∫0 dx ∫0 ( Y ) Y d ρu2 dy − g (ρ∞ − ρ)dy. = ∫0 dx ∫0

(12.75)

Using Boussinesq approximation, the density difference ρ∞ − ρ is approximated by ρ∞ − ρ ≈ βρ(T − T∞ ).

(12.77)

Substituting Eq. (12.77) into Eq. (12.76) yields ( Y ) Y d ρu2 dy − gβρ (T − T∞ ) dy −τw = ∫0 dx ∫0

(12.78a)

or ρ

d dx

( ∫0

)

δ

u2 dy

( = −μ

𝜕u 𝜕y

)

δ

+ρgβ y=0

∫0

(T − T∞ ) dy.

(12.78b)

This equation is valid for both the laminar and turbulent flows. The velocity gradient at the edge of the boundary layer vanishes, i.e. 𝜕u(x, δ)/𝜕y = 0. This means there is no shear force. Consider the integral energy equation ] [ Y ) ( 1 d q′′w = (12.79a) R ρu h∗ − h∗∞ dy . R dx ∫0 We write down the applicable energy integral equation with the assumptions that h∗ − h∗∞ = cp (T − T∞ ), R = constant and with the Boussinesq approximation [ Y ] d u(T − T∞ ) dy (12.79b) q′′w = ρcp dx ∫0 or ρcp

[ Y ] ( ) d 𝜕T u(T − T∞ ) dy = −k dx ∫0 𝜕y y=0

(12.79c)

where it is assumed that the velocity and thermal boundary layers have the same local thickness. Recall that following assumptions were made in deriving the integral formulation of the conservation of energy. (a) (b) (c) (d)

No changes in kinetic and potential energy Negligible axial conduction No viscous dissipation Constant fluid properties.

In integral analysis, we assumed that δ ≈ Δ. This assumption is justified since the results obtained with this assumption agree quite well with the experimental work of Touloukian et.al. [16]. Sparrow and Gregg [15] performed an analysis for δ ≠ Δ and present results that are also very close to these results. Notice that this approach also requires less work. For a detailed analysis of integral analysis, one may consult the work of other authors such as Kakac et al. [17] and Ghiaasian [3].

12.5 Integral Method (von Karman–Pohlhausen Method): An Approximate Analysis of Laminar Free Convection on a Vertical Plate

12.5.1 Constant Wall Temperature In some technical applications, the surface temperature is known. It is then of interest to determine the heat transfer. Let us now assume laminar boundary layer flow and a third-order polynomial u(x, y) = a0 (x) + a1 (x) y + a2 (x) y2 + a3 (x) y3 .

(12.80)

The boundary conditions are u(x, 0) = 0

(12.81a)

u(x, δ) = 0

(12.81b)

𝜕u(x, δ) =0 𝜕y

(12.81c)

βg 𝜕 2 u(x, δ) = − (Tw − T∞ ). (12.81d) ν 𝜕y2 The last condition is obtained from the x-momentum equation by setting y = 0. Using these boundary conditions, the following coefficients are obtained: a0 = 0 β g (Tw − T∞ ) δ 4ν β g (Tw − T∞ ) a2 = − 2ν β g (Tw − T∞ ) 1 a3 = − 4ν δ Substituting a0 , a1 , a2 and a3 into Eq. (12.80), we now obtain the velocity profile as [ ( y ) ( y )2 ] β g (Tw − T∞ ) δy 1−2 + u= 4ν δ δ a1 =

or Eq. (12.82a) can be written as (y)[ y ]2 1− u = uc δ δ where δ(x) and uc (x) are unknown and uc is given as ] [ β g (Tw − T∞ ) 2 δ . uc = 4ν

(12.82a)

(12.82b)

(12.83)

The next step is to find the temperature profile, and for this purpose, a second-order polynomial is assumed T(x, y) = b0 (x) + b1 (x) y + b2 (x) y2 .

(12.84)

The boundary conditions are T(x, 0) = Tw

(12.85a)

T(x, Δ) = T∞

(12.85b)

𝜕T(x, Δ) = 0. 𝜕y Using the boundary conditions, the unknown coefficients b0 , b1 , b2 , and b3 are evaluated b0 = Tw (Tw − T∞ ) Δ (T − T ) b2 = w 2 ∞ . Δ

b1 = −2

(12.85c)

697

698

12 Free Convection Heat Transfer

Substituting b0 , b1 , b2 , and b3 into Eq. (12.84), the temperature profile can be written as [ y ]2 T(x, y) = T∞ + (Tw − T∞ ) 1 − . Δ Since we assumed that δ ≈ Δ, Eq. (12.86a) is written as [ y ]2 T(x, y) = T∞ + (Tw − T∞ ) 1 − . δ It is assumed that heat transfer coefficient h(x) is defined as ( ) −k 𝜕T ′′ 𝜕y y=0 qw 2k h(x) = = = Tw − T∞ Tw − T∞ Δ(x)

(12.86a)

(12.86b)

(12.87)

or since it is assumed that δ(x) = Δ(x) h=

2k . δ(x)

(12.88)

We only need to determine the velocity boundary layer thickness δ(x). To determine δ(x), we substitute velocity and temperature profiles into the integral form of the momentum equation, Eq. (12.78b) { 2 δ [ } δ [[ uc u y ]]2 y ]4 d 2 −ν c + gβ(Tw − T∞ ) dy = y dy . (12.89) 1− 1 − ∫0 δ δ dx δ2 ∫0 δ Evaluating the integrals and rearranging them yield u 1 d [ 2 ] β g (Tw − T∞ ) δ uc δ = −ν c. 105 dx 3 δ

(12.90)

Next, we substitute velocity and temperature profiles into integral form of the energy equation, Eq. (12.79b). This will give us { } uc δ [ y ]4 1 d y 1− dy . (12.91) 2α(Tw − T∞ ) = (Tw − T∞ ) δ dx δ ∫0 δ Evaluating the integral and rearranging the result yields α 1 d (u δ) = . (12.92) 60 dx c δ We have two first-order differential equations and two dependent variables uc (x) and δ(x). We now assume a solution of the form given below uc (x) = A xm

(12.93)

δ(x) = B xn

(12.94)

where A, B, m, and n are constants. These constants are determined by substituting Eqs. (12.93) and (12.94) into Eqs. (12.78b) and (12.79b). This process of substitution yields 2m + n 2 2m+n−1 1 A A Bx = βg (Tw − T∞ )B xn − ν xm−n 105 3 B α m+n ABxm+n−1 = x−n . 60 B Equations (12.95) and (12.96) are satisfied if the exponents of x in each equation are identical. Thus,

(12.95) (12.96)

2m + n − 1 = n = m − n

(12.97)

m + n − 1 = −n.

(12.98)

Solving these equations yields m=

1 1 , n= . 2 4

(12.99)

12.5 Integral Method (von Karman–Pohlhausen Method): An Approximate Analysis of Laminar Free Convection on a Vertical Plate

Using these values of m and n, after simplification, we obtain the following equations in terms of A and B: 1 A 1 2 A B = βg(Tw − T∞ )B − ν 84 3 B 1 α AB = . 80 B Solving Eqs. (12.100) and (12.101) for A and B, we get ( )−1∕2 [ g β (T − T ) ]1∕2 20 w ∞ A = 5.17ν + Pr 21 ν2 )−1∕4 [ g β(T − T ) ]−1∕4 ( 20 w ∞ B = 3.93 Pr −1∕2 + Pr . 21 ν2

(12.100) (12.101)

(12.102) (12.103)

Substituting Eqs. (12.102) and (12.103) into Eqs. (12.93) and (12.94), introducing the definition of Grashof number Grx , and after some algebra, eventually, we get √ ) ( 20 −1∕2 Grx uc = 5.17ν Pr + (12.104) 21 x ) ( 20 1∕4 −1∕4 δ = 3.93Pr −1∕2 Pr + Grx (12.105a) x 21 or ( ) 20 1 1∕4 −1∕4 δ = 3.93 1 + Rax . (12.105b) x 21 Pr The velocity boundary layer thickness can also be expressed in the following form: ]1∕4 [ δ Pr +0.952 . (12.106) = 3.93 x Pr 2 Grx The local heat transfer coefficient h(x) is 1∕4

h(x) =

2k k 0.508Pr 1∕2 Grx = δ(x) x (0.952 + Pr )1∕4

(12.107)

and the local Nusselt number Nux is 1∕4

Nux =

0.508 Pr 1∕2 Grx [Pr +0.952]1∕4

(12.108)

or in terms of the Rayleigh number Rax Nux = or

0.508 Pr 1∕4 1∕4 Rax [0.952 + Pr ]1∕4

( Nux = 0.508

Pr 0.952 + Pr

)1∕4

(12.109a)

1∕4

Rax

where Rax = g β (Tw − T∞ )x3 /ν α is the Rayleigh number. The average heat transfer coefficient h is [ ] L g β (Tw − T∞ ) 1∕4 4 1 1 h= h(x) dx = 0.508Pr 1∕2 (0.952 + Pr )−1∕4 k L ∫0 3 L1∕4 ν2 and the average Nusselt number in terms of the Rayleigh number RaL ]1∕4 [ Pr 1∕4 NuL = 0.68 RaL 0.952 + Pr

(12.109b)

(12.110)

(12.111)

where RaL = g β (Tw − T∞ )L3 /ν α is the Rayleigh number based on the plate length L. Example 12.4 The surface temperature of a square plate is 400 K. The one side of the plate is 0.5 m, and the plate is in vertical position in quiescent air at 1 atm and 300 K.

699

700

12 Free Convection Heat Transfer

(a) Determine the velocity boundary layer thickness at the trailing edge. (b) Determine the total heat loss from the plate. Solution The film temperature is Tf = (Tw + T∞ )/2 = (400 + 300)/2 = 350 K. Air properties at this temperature are ν = 20.92 × 10−6 m2 ∕s k = 30 × 10−3 W∕m.K Pr = 0.700. 1 The coefficient of thermal expansion β = T1 = 350 K−1 = 0.00285 K−1 . f The Grashof number at the trailing edge of the plate

g β (Tw − T∞ ) x3 9.81 × (0.00285) (400 − 300) (0.5)3 = = 8 × 108 . ν2 (20.92 × 10−6 )2 (a) We will use Eq. (12.106) to estimate the velocity boundary layer thickness at the trailing edge of the plate ]1∕4 ]1∕4 [ [ 0.7 + 0.952 Pr +0.952 x = 3.93 (0.5) = 0.01582 m δ = 3.93 Pr 2 Grx (0.7)2 × (8 × 108 ) Grx =

(b) To calculate the total heat transfer from the plate, we will use Eq. (12.111) RaL = GrL Pr = 8 × 108 × 0.7 = 5.6 × 108 . Equation (12.111) will be used to calculate the average Nusselt number ]1∕4 ]1∕4 [ [ Pr 0.7 1∕4 NuL = 0.68 RaL = 0.68 (2.8 × 108 )1∕4 ≈ 84.41. 0.952 + Pr 0.952 + 0.7 The average heat transfer coefficient h is ) ( W k 30 × 10−3 (84.41) = 5.06 2 . h = NuL = L 0.5 mK The total heat transfer q is q = 2 × h × A(Tw − T∞ ) = 2 × (5.06)(0.5 × 0.5)(400 − 300) = 253.26 W.

12.5.2

Uniform Heat Flux

In some technical applications, heat flux is prescribed over the surface. It is then of interest to determine the temperature distribution on the surface, as discussed in [11]. Let us now assume laminar boundary layer flow over a vertical plate and a third-order polynomial for the velocity profile u(x, y) = a0 (x) + a1 (x) y + a2 (x) y2 + a3 (x) y3 .

(12.112)

The boundary conditions are u(x, 0) = 0

(12.113a)

u(x, δ) = 0

(12.113b)

𝜕u(x, δ) =0 𝜕y gβ 𝜕2u = − (Tw − T∞ ). ν 𝜕y2 The application of the boundary conditions for the assumed velocity profile gives the velocity distribution (y)( y )2 1− u = uc δ δ where uc =

g β(Tw −T∞ ) 2 δ. 4ν2

(12.113c) (12.113d)

(12.114)

12.5 Integral Method (von Karman–Pohlhausen Method): An Approximate Analysis of Laminar Free Convection on a Vertical Plate

Next, we assume a second-order polynomial for the temperature profile T(x, y) = a0 (x) + a1 (x) y + a2 (x) y2 .

(12.115)

The boundary conditions are (12.116a)

T(x, Δ) = T∞ 𝜕T(x, Δ) =0 𝜕y q′′ 𝜕T(x, 0) = − w. 𝜕y k

(12.116b) (12.116c)

The application of the boundary conditions to Eq. (12.115) yields ( ′′ ) qw a0 = T∞ + Δ 2k q′′w a1 = − k ′′ q 1 a2 = w . 2kΔ The substitution of a0 , a1 , and a2 into Eq. (12.115) gives the temperature distribution ( ′′ ) ( qw δ y )2 1− . T − T∞ = 2k Δ

(12.117)

We assumed that δ ≈ Δ. Assuming UHF, these profiles, Eq. (12.114) and Eq. (12.117), are introduced into integral momentum and energy equations, and the results are given as Δ2 Ω 1 d (Ω2 Δ) = − 105 dX 6 Δ 1 d 2 2 [ΩΔ ] = 30 dX Pr where X, Δ, and Ω are )−1∕4 ) ) ( ( ( g βq′′w ν2 g βq′′w 1∕4 g βq′′w 1∕4 X=x , Δ = δ and Ω = u c k k ν2 k ν2

(12.118) (12.119)

Solutions to these equations, Eq. (12.118) and Eq. (12.119), are given as Ω = (6000)1∕5 Pr Δ = (360)1∕5 Pr

−1∕5

−2∕5

(0.8 + Pr )−2∕5 X3∕5

(12.120)

(0.8 + Pr )1∕5 X1∕5

(12.121)

The following expression for the boundary layer thickness is obtained [ ]1∕5 δ 1∕5 0.8 + Pr = (360) . x Pr 2 Gr∗x The wall temperature distribution is obtained from Eq. (12.117) by setting y = 0 ( ′′ ) ( ( ′′ ) qw δ qw δ y )2 || 1− Tw − T∞ = | = | 2k Δ |y=0 2k or

( Tw (x) − T∞ = 1.622

x q′′w k

)[

0.8 + Pr Pr 2 Gr∗x

(12.122)

(12.123)

]1∕5 .

The local Nusselt number is given as ( 2 ∗ )1∕5 x q′′w Pr Grx = 0.62 Nux = k(Tw − T∞ ) 0.8 + Pr where Gr∗x = g βq′′w x4 ∕k ν2 is the modified Grashof number.

(12.124)

(12.125)

701

702

12 Free Convection Heat Transfer

12.6 Turbulent Free Convection Heat Transfer on a Vertical Plate Turbulent free convection flows are quite often encountered in engineering applications. Considerable amount of experimental and analytical work has been done to understand the mechanism of turbulent free convection heat transfer. Experimentally, it is observed that when RaL > 109 , flow over a vertical flat plate is considered to be turbulent. The integral momentum and energy equations can also be used to calculate the turbulent free convection heat transfer over a vertical flat plate. The integral analysis of a turbulent boundary layer over a vertical plate was carried out by Eckert and Jackson [18]. The vertical plate surface temperature Tw is constant. Eckert and Jackson assumed that the velocity boundary layer thickness δ and temperature boundary layer thickness Δ is equal (Δ = δ). They report that experimental data for the velocity and temperature distribution for air agree with following approximate profiles: Velocity profile: ( y )1∕7 ( y )4 u = . (12.126) 1− U∞ δ δ Temperature profile: Θ=

[ ( y )1∕7 ] T − T∞ . = 1− Tw − T∞ δ

(12.127)

Notice that these velocity and temperature profiles are valid within the turbulent region outside the laminar sublayer. The velocity profile cannot be used within the sublayer, and for this reason, these profiles cannot be used to calculate shear at the wall using the expression τw = μ(𝜕u∕𝜕y)y=0 . Likewise, the temperature profile cannot be used to calculate heat flux using q′′w = −k(𝜕T∕𝜕y)y=0 . Flow near the wall in a turbulent free convective boundary layer is similar to that in the turbulent forced convective boundary layer. Therefore, empirical expression for the wall shear stress is used. This expression is obtained from measurements on turbulent flow in circular tubes, as discussed in [19], and it is given as ) ( ν 1∕4 τw = 0.0225 ρV2 . Vδ It is experimentally observed that this equation is also valid for flat plate, and in free convection, velocity V is replaced by characteristic velocity uc , and shear stress becomes ( )1∕4 ν 2 τw = 0.0225 ρuc . uc δ With this information, the integral momentum equation is written as ( δ )1∕4 ( ) δ d ν ρu2 dy = gβρ (T − T∞ ) dy − 0.0225 ρu2c . ∫0 dx ∫0 uc δ

(12.128)

The substitution of the velocity profile, Eq. (12.126), and the temperature profile, Eq. (12.127), into the integral momentum equation yields )1∕4 ( d ( 2 ) ν 2 0.0523 . (12.129) u δ = 0.125 g β (Tw − T∞ )δ − 0.0225 uc dx c uc δ On the other hand, the wall heat flux is obtained as follows. First, consider the Reynolds analogy. Since temperature and velocity profiles are similar for Pr = 1, the Reynolds analogy can be written as cp (Tw − T∞ ) qw = . τw uc The introduction of wall shear stress into the above equation yields ( )1∕4 q′′w ν St = = 0.0225 ρ cp uc (Tw − T∞ ) uc δ

(12.130)

(12.131)

where St is the Stanton number. Experimental investigations show that for fluids with Prandtl numbers varying from approximately 0.5 to 50, the equation given above can be modified as ( )1∕4 q′′w ν = 0.0225 Pr −2∕3 . (12.132) St = ρ cp uc (Tw − T∞ ) uc δ

12.6 Turbulent Free Convection Heat Transfer on a Vertical Plate

Then, using this equation along with temperature and velocity profiles, the integral energy equation becomes )1∕4 ( d ν 0.0366 (uc δ) = 0.0225 uc Pr −2∕3 . (12.133) dx uc δ We have two differential equations from which the two unknowns δ and uc can be determined as functions of x. Again, following Eckert and Jackson [18], we assume that u and δ can be expressed as exponential functions of x. Trying, uc = Axm

(12.134)

δ = Bxn .

(12.135)

Equations (12.134) and (12.135) are introduced into Eqs. (12.130) and (12.133), and this process transforms the equations into the following form: (2 m + n)A2 x2m+n−1 = 2.389(Tw − T∞ )g βxn − 0.435

A7∕4 ν1∕4 7m−n x 4 B5∕4

3m−n ν1∕4 x 4 . Pr 2∕3 A1∕4 B5∕4 Since these equations are valid for any value of x, exponents of x must be identical, and thus, 7m n − (2m + n) − 1 = n = 4 4 3m n − . (m + n) − 1 = 4 4 These equations are satisfied if 7 1 m = and n = 2 10 With these values of m and n, A, and B can be found as gβ(Tw − T∞ ) A2 = 1.406 1 + 0.494Pr 2∕3 ]1∕10 [ 1 + 0.494 Pr 2∕3 . B = 0.678ν1∕5 Pr −8∕15 gβ(Tw − T∞ )

(m + n)xm+n−1 = 0.622

(12.136) (12.137)

(12.138) (12.139)

Next, we use the definition of the Grashof number Grx = gβ(Tw − T∞ )x3 /ν2 , and we obtain the velocity uc ν√ uc = 1.185 Grx [1 + 0.494Pr 2∕3 ]−1∕2 (12.140) x and the boundary layer thickness δ is δ −1∕10 = 0.565 Pr −8∕15 [1 + 0.494Pr 2∕3 ]−1∕10 Grx . (12.141) x The local heat transfer coefficient at any position x is obtained from the Stanton number St by introducing the boundary layer thickness δ and the characteristic velocity uc into Eq. (12.132) { } [gβ(Tw − T∞ ) Pr ∕ν2 ]2∕5 q′′w 1∕15 x1∕5 . h(x) = = 0.0298kPr (12.142a) Tw − T∞ [1 + 0.494Pr 2∕3 ]2∕5 The local Nusselt number Nux is 2∕5

0.0295Grx Pr 7∕15 hx = . (12.142b) k [1 + 0.494Pr 2∕3 ]2∕5 With the assumption that the turbulence starts from the leading edge of the plate, the average heat transfer coefficient is Nux =

x

1 h(x) dx = 0.833 h|x=L x ∫0 and the average Nusselt number is h=

2∕5

NuL =

0.02458 RaL Pr [1 + 0.494Pr

RaL = GrL Pr ≥ 109 .

7∕15

2∕3 ]2∕5

(12.143a)

703

704

12 Free Convection Heat Transfer

A simplified version of Eq. (12.143a) for the average Nusselt number is NuL = 0.0210(RaL )2∕5

(12.143b)

RaL ≥ 109 Pr = 0.7. It is reported in [13] that the agreement of Eq. (12.143a) with experimental data is satisfactory.

12.7 Empirical Correlations for Free Convection The analysis of free convection is very complicated, and it is better to rely on experimental data to develop reliable heat transfer relations. The Grashof number and the Prandtl number are the key dimensionless numbers in free convection, and correlations are functions of the Grashof number GrL and the Prandtl number Pr. We present some of the recommended empirical correlations for both laminar and turbulent free convection under UWT and UHF boundary conditions. Experimental studies indicate that the average Nusselt number in free convection can be represented in the following form for various cases: NuL = C(GrL Pr )m = CRam L

(12.144a) g β(T −T )L3

g β(T −T )L3

w ∞ w ∞ Pr = where RaL is the Rayleigh number, and it is defined as RaL = GrL Pr = ν2 να where L is the characteristic length, and the values of the constants C and m in Eq. (12.144a) depend on the surface geometry, and whether the flow is laminar or turbulent. Fluid properties are usually evaluated at film temperature Tf

Tw + T∞ 2 where Tw and T∞ are the surface and ambient temperatures, respectively. Consider free convection from a vertical plate. The average convective heat transfer coefficient h is defined as Tf =

L

h=

1 h(x) dx L ∫0

where h(x) is the local convective heat transfer coefficient along the vertical hot plate. Assume that the local heat transfer coefficient h(x) depends on x in the following form: h = C x−n . We can now find the average heat transfer coefficient h h=

L | 1 C −n 1 C x−n dx = L = h(x)|| . L ∫0 1−n 1−n |x=L

Thus, we see that the average Nusselt number becomes NuL =

| 1 Nu | 1 − n x ||x=L

(12.144b)

where Nu|x = L is the local Nusselt number evaluated at x = L. The characteristic dimension L used in the Grashof number GrL and the Nusselt number NuL depends on the geometry of the problem. For example, for a vertical plate, it is the height L of the plate, and for a horizontal cylinder, it is the cylinder diameter D.

12.7.1

Vertical Plate

Uniform wall temperature The analysis of free convection heat transfer is often very complicated, and for this reason, experimental data are used to develop reliable equations for design. Geometry and coordinate system for an isothermal vertical plate is shown in Figure 12.4.

12.7 Empirical Correlations for Free Convection

Figure 12.4

Isothermal vertical plate.

Tw L

x 0

McAdams [20] has analyzed data for free convection. The data are based on experiments performed by various researchers. He correlated experimental data for UWT and proposed the following empirical correlations: 1∕4

NuL = 0.59 RaL

104 ≤ RaL ≤ 109

(12.145a)

1∕3 NuL = 0.1 RaL 109 ≤ RaL ≤ 1013

(12.145b)

where NuL = hL∕k, GrL = gβ(Tw − T∞ )L3 /ν2 , and RaL = GrL Pr. The fluid properties are evaluated at film temperature T = (Tw + T∞ )/2. Churchill and Chu [21] recommend a correlation having good accuracy within the laminar region for the average Nusselt number 1∕4

0.67RaL NuL = 0.68 + [ ) ]4∕9 ( 0.492 9∕16 1+ Pr

(12.146)

10−1 ≤ RaL ≤ 109 0 ≤ Pr ≤ ∞ Tw = const where NuL = hL∕k, GrL = gβ(Tw − T∞ )L3 /ν2 , and RaL = GrL Pr. It is reported that this equation gives very accurate results. Churchill and Chu [21] proposed a different and very useful empirical correlation for the average Nusselt number for natural convection over a vertical plate maintained at constant temperature Tw and exposed to ambient at T∞ that applies to laminar, transition, and turbulent flow and valid for all the Prandtl numbers. This mathematical model by Churchill and Chu correlates the extensive laminar and turbulent data with a single correlation 2

⎧ ⎫ ⎪ ⎪ 1∕6 0.387RaL ⎪ ⎪ (NuL ) = ⎨0.825 + [ )9∕16 ]8∕27 ⎬ ( 0.492 ⎪ ⎪ 1+ ⎪ ⎪ Pr ⎩ ⎭

(12.147)

10−1 ≤ GrL ≤ 1012 0 ≤ Pr ≤ ∞ Tw = const where NuL = hL∕k, GrL = gβ(Tw − T∞ )L3 /ν2 , and RaL = GrL Pr. Notice that each regime is fundamentally different from each other, and we may consider Eq. (12.147) as an approximate representation of both the regimes instead of being accurate for each regime. Physical properties are evaluated at the film temperature Tf = (Tw + T∞ )/2.

705

706

12 Free Convection Heat Transfer

Nellis and Klein [22] report a correlation of Churchill and Usagi. The average Nusselt number obtained asymptotically weighting the Nusselt number for laminar and turbulent flows using the empirical formula [ 6 ]1∕6 6 NuL = NuL,lam + NuL,turb (12.148) 0.1 ≤ RaL ≤ 1 × 1012 . The laminar Nusselt number is: 2 NuL,lam = [ ] 2 ln 1 + 1∕4 Clam RaL 0.671 Clam = [ ) ]4∕9 ( 0.492 9∕16 1+ Pr The turbulent Nusselt number is: 1∕3

Cturb RaL NuL,turb = ) ( Pr 1 + (1.4 × 109 ) RaL 0.13Pr 0.22 . Cturb = (1 + 0.61Pr 0.81 )0.42 Uniform wall heat flux Free convection on a vertical plate subject to UHF at the wall surface has been investigated by several investigators such as Sparrow and Gregg [10], Fuji and Fuji [12], Vliet and Liu [23], and Vliet [24]. Geometry and coordinate system for a vertical plate under UHF is shown in Figure 12.5. For the constant heat flux case, the rate of heat transfer is known, and we do not now the surface temperature. In this case, we are interested in determining the surface temperature distribution Tw (x), which varies along the plate. The local Nusselt number Nux is defined as Nux =

q′′w x . Tw (x) − T∞ k

Knowing the local Nusselt number, we may determine temperature distribution Tw (x) − T∞ =

q′′w x . Nu(x) k

(12.149)

In correlations for the average Nusselt number under the UHF boundary condition, the average Nusselt number may be based on the difference between average plate surface temperature Tw and free-stream temperature T∞ NuL =

q′′w

L . [TW − T∞ ] k

(12.150)

Figure 12.5

″ qw L

x 0

Vertical plate under uniform heat flux.

12.7 Empirical Correlations for Free Convection

For air and water Prandtl number range, the transition from laminar to turbulence occurs near Ra∗x ≈ 1013 . Fuji and Fuji [12] developed correlation for the local Nusselt number under the UHF boundary condition for a vertical plate: [ ]1∕5 ( ∗ )1∕5 xh Pr Nux = Rax = (12.151) √ k 4 + 9 Pr + 10 Pr q′′w = const 104 ≤ Ra∗x ≤ 109 0 ≤ Pr ≤ ∞ where Ra∗x is the modified Rayleigh number and modified local Rayleigh number Ra∗x is g β q′′w x4 . kαν On the other hand, Gr∗x is the modified local Grashof number, and it is defined as ][ ] [ q′′w g β q′′w x4 gβ(Tw − T∞ ) x = Gr∗x = Grx Nux = (Tw − T∞ ) k ν2 k ν2 Ra∗x = Gr∗x Pr =

where q′′w is the UHF. If we combine Eqs. (12.149) and (12.151), we obtain an expression for the wall temperature distribution in dimensionless form [ ]−1∕5 Tw (x) − T∞ k (x∕L)1∕5 Pr = . (12.152) √ L (Ra∗ )1∕5 4 + 9 Pr + 10 Pr q′′w L

Vliet and Liu [23] performed experimental studies on turbulent natural convection with water and Vliet [24] performed experiments with air. Based on these experimental data, the following practical relations were proposed for the local Nusselt number under UHF, as discussed in [25]: )1∕5 ( laminar flow (12.153) Nux = 0.6 Pr Gr∗x 105 ≤ Pr Gr∗x ≤ 1011 q′′w = const )0.22 ( turbulent flow Nux = 0.568 Pr Gr∗x

(12.154)

2 × 1013 ≤ Pr Gr∗x ≤ 1016 q′′w = const

( ) where Nux = h x/k and Gr∗x = g β q′′w x4 ∕(k ν2 ) is the modified Grashof number. Holman [26] indicates that the transition begins between Gr∗x Pr = 3 × 1012 − 4 × 1013 and ends between 2 × 1013 − 1 × 1014 . According to Holman, fully developed turbulent flow is present by Gr∗x Pr = 3 × 1014 and recommends the following correlation: )1∕4 ( turbulent flow (12.155) Nux = 0.17 Pr Gr∗x 2 × 1013 ≤ Pr Gr∗x ≤ 1016 q′′w = const

) ( where Nux = h x/k and Gr∗x = g β q′′w x4 ∕(k ν2 ) is the modified Grashof number. Fluid properties are evaluated at the film temperature Tf = (Tw + T∞ )/2. We now wish to determine the average Nusselt number. Based on Eqs. (12.153) and (12.154), the local heat transfer coefficient h(x) depends on x in the following form:

707

708

12 Free Convection Heat Transfer

Laminar region: )0.2 1 4 0.2 1( h(x) ∼ Gr∗x ∼ (x ) ∼ x−0.2 x x n = 0.2. Turbulent region: )0.22 1 4 0.22 1( h(x) ∼ Gr∗x ∼ (x ) ∼ x−0.12 x x n = 0.12. Then, Eq. (12.144b) is used to determine the mean Nusselt number. The mean Nusselt number NuL for Eq. (12.153) is )1∕5 ( 1 [Nux ]x=L = 1.25[Nux ]x=L = 0.75 Ra∗L NuL = (12.156) 1 − 0.2 105 ≤ Pr Gr∗L ≤ 1011 q′′w = const. The mean Nusselt number NuL for Eq. (12.154) is NuL =

)0.22 ( 1 [Nux ]x=L = 1.136[Nux ]x=L = 0.645 Ra∗L 1 − 0.12

(12.157)

2 × 1013 ≤ Pr Gr∗L ≤ 1016 q′′w = const. Example 12.5 A vertical plate having height L = 2 m and width W = 1 m is insulated on one side and exposed to a heat flux of q′′w = 3000 W∕m2 on the other side. The flat plate is dissipating heat by free convection in quiescent air at 1-atm pressure. The air temperature is 300 K. Determine: (a) average surface temperature (b) maximum temperature on the plate. Solution We do not know the surface temperature, and for this reason, we cannot estimate the film temperature Tf to evaluate the air properties. First iteration: We will estimate air properties at T∞ = 300 K as ν = 15.89 × 10−6 m2 ∕s k = 0.0263 W∕m.K Pr = 0.707. The coefficient of thermal expansion β is β=

1 1 1 ≈ 3.33 × 10−3 K−1 Tf ≈ ≈ Tf T∞ 300

The local modified Grashof number is q′′ g β x4 3000 × 9.81 (3.33 × 10−3 ) x4 = 1.47 × 1013 x4 Gr∗x = w 2 = kν (0.0263) (15.89 × 10−6 )2 We expect that flow is turbulent and Eq. (12.154) will be used )0.22 ( = 0.568(0.707 × 1.47 × 1013 x4 )0.22 ≈ 415.0 (x4 )0.22 Nux = 0.568 Pr Gr∗x The position x is in meters. Then, the local wall temperature Tw (x) may be written as Tw (x) = T∞ +

q′′w x 3000x 274.86x = 300 + = 300 + 4 0.22 k Nux 0.0263 × 415.0 (x4 )0.22 (x )

12.7 Empirical Correlations for Free Convection

Maximum temperature occurs at the trailing edge of the plate is TW ||max = 598.6 K The average wall temperature TW is L

TW =

1 T (x) dx = 566.6 K ≈ 566.0 K L ∫0 w

Second iteration: We can now evaluate film temperature Tf as Tw + T∞ 566 + 300 = ≈ 433 K 2 2 We evaluate properties at this new temperature Tf =

ν = 30.35 × 10−6 m2 ∕2 k = 0.03611 W∕m.K Pr = 0.6870 β=

1 1 1 = 0.002309 = Tf 433 K

The Grashof number is q′′w g β x4 = 2.04 × 1012 x4 k ν2 We expect that flow is turbulent, and Eq. (12.154) will be used. The local Nusselt number is ( )0.22 Nux = 0.568 Pr Gr∗x = 267 (x4 )0.22 . Gr∗x =

The local wall temperature is Tw (x) = T∞ +

q′′w x 311.15 x = 300 + 4 0.22 . k Nux (x )

Maximum temperature occurs at the trailing edge of the plate TW ||max = 638.2 K. The average wall temperature TW is L

TW =

1 T (x) dx ≈ 602.0 K. L ∫0 w

Third iteration: We can now evaluate film temperature Tf as Tw + T∞ 602 + 300 = ≈ 451 K. 2 2 We evaluate properties at this new temperature Tf =

ν = 32.51 × 10−6 m2 ∕s k = 0.0373 W∕m.K Pr = 0.685 β=

1 1 1 = 0.002217 . = Tf 451 K

709

710

12 Free Convection Heat Transfer

The Grashof number is Gr∗x =

q′′w g β x4 = 1.65 × 1012 x4 . k ν2

We expect that flow is turbulent, and Eq. (12.154) will be used. The local Nusselt number is ( )0.22 = 254.71 (x4 )0.22 . Nux = 0.568 Pr Gr∗x The local wall temperature is Tw (x) = T∞ +

q′′w x 315.76 x = 300 + 4 0.22 . k Nux (x )

The maximum temperature occurs at the trailing edge of the plate TW ||max = 643.1 K. The average wall temperature TW is L

TW =

1 T (x) dx = 606.4 K. L ∫0 w

This is close enough, and if needed, one more iteration may be carried out. However, there is no need for more iteration. Variation of local surface temperature along the plate is shown in Figure 12.E5.

600

550 Tw(x)

500

450

0 Figure 12.E5

0.5

1 x(m)

1.5

2

Variation of local surface temperature.

Example 12.6 A flat plate measuring 20 cm by 20 cm is oriented vertically in 17 ∘ C quiescent air. The flat plate is dissipating 12 W heat from its surface. The back side of the plate is insulated. Assuming a UHF at the plate surface, we wish to find the average heat transfer coefficient.

12.7 Empirical Correlations for Free Convection

Solution We wish to show a different solution method now. Churchill and Chu showed that Eq. (12.146) could be applied to the UHF case if the average Nusselt number is based on the wall heat flux and the difference between the plate vertical midpoint temperature Tw (L/2) and ambient fluid temperature T∞ . To start the solution, we make a guess. Assume Tw = 70 ∘ C = 343 K at x = L/2. The film temperature Tf is Tf =

343 + 290 = 316.5 K. 2

Air properties at this temperature are ν = 17.54 × 10−6 m2 ∕s k = 0.0275 W∕m.K Pr = 0.704 α = 24.94 × 10−6 m2 ∕s. This time, the Rayleigh number will be based on the difference between the temperature of the vertical midpoint of plate, Tw (L/2), and the ambient air temperature T∞ : ( ) L ΔT = Tw − T∞ . 2 The Rayleigh number is defined as gβ[Tw (L∕2) − T∞ ]L3 να ) ( 1 [70 − 17](0.23 ) 9.81 × 316.5 RaL = ≈ 3.0 × 107 . 17.54 × 10−6 × 24.94 × 10−6 RaL =

Flow is laminar since GrL < 109 . We will use Eq. (12.146) 1∕4

0.67RaL NuL = 0.68 + [ ( ) ]4∕9 0.492 9∕16 1+ Pr 0.67(3.0 × 107 ) NuL = 0.68 + [ ≈ 38.59. ) ]4∕9 ( 0.492 9∕16 1+ 0.704 The average heat transfer coefficient h is ) ( ) ( W k 0.0275 (38.59) = 5.31 2 . h= NuL = L 0.2 mK We now calculate the heat flux as [ ( ) ] L − T∞ = 5.31(70 − 17) = 281.4 W∕m2 . q′′w = h Tw 2 The actual heat flux dissipated from the surface is q′′w =

12 = 300 W. 0.2 × 0.2

The difference between the actual heat flux and the estimated heat flux is small, and there is no need for further iteration.

711

712

12 Free Convection Heat Transfer

12.7.2

Horizontal Plate

The numerical values of Nusselt numbers for heat transfer from horizontal plates depend on whether the plate is hot or cold relative to the ambient fluid. Consider a horizontal hot surface at a uniform temperature Tw , exposed to ambient at temperature T∞ . According to Goldstein et al. [27], good agreement with experimental data can be obtained by using the following characteristic dimension: L=

A P

where A is the surface area (one side) and P is the perimeter of the plate. This is especially useful for irregular surfaces. They showed that the use of this characteristic length makes it possible to correlate data for square plate, circular plate, and rectangular plate by a single equation. For this reason, this characteristic length is commonly used whenever possible in the correlations. The average Nusselt number NuL used in free convection on a horizontal plate is based on whether the hot surface or cold surface is facing up or down, as shown in Figure 12.6. Uniform wall temperature Upper surface of hot plate or lower surface of cold plate Figure 12.7 shows buoyancy driven flows over a horizontal isothermal plate. McAdams [20] recommends the following correlations: 1∕4

NuL = 0.54 RaL

(12.158)

1 × 105 ≤ RaL ≤ 2 × 107 1∕3

NuL = 0.14 RaL

(12.159)

2 × 107 ≤ RaL ≤ 3 × 1010 where NuL = hL∕k is the mean Nusselt number, and RaL = GrL Pr = g β (Tw − T∞ )L3 /ν α is the Rayleigh number. Fluid properties are evaluated at film temperature Tf = (Tw + T∞ )/2. Nellis and Klein [22] report a correlation of Churchill and Usagi. The average Nusselt number is the average of the Nusselt number for laminar and turbulent flows using the empirical formula: [ 10 ]1∕10 10 NuL = NuL,lam + NuL,turb (12.160) 0.1 ≤ RaL ≤ 1 × 1010 Cold surface

Hot surface g T∞

Hot surface

Cold surface Figure 12.6

Fluid

Free convection on the horizontal surface.

Tw > T∞

g

Cold surface down

Hot surface up (a) Figure 12.7

Tw < T∞ (b)

(a, b) Buoyancy driven flows over horizontal plates.

12.7 Empirical Correlations for Free Convection

The laminar Nusselt number is 1.4 NuL,lam = [ ] 1.4 ln 1 + 1∕4 0.835Clam RaL 0.671 Clam = [ ( ) ]4∕9 0.492 9∕16 1+ Pr The turbulent Nusselt number is 1∕3

NuL,turb = Cturb RaL ( Cturb = 0.14

1 + 0.0107 Pr 1 + 0.01 Pr

)

Example 12.7 A horizontal 1 m by 1 m plate with an upper surface temperature of 500 K is in a quiescent air at 300 K and 1-atm pressure. The bottom surface of the plate is insulated. Calculate the convective heat transfer from the plate. Solution We begin by finding the film temperature Tf Tw + T∞ 500 + 300 = = 400 K 2 2 The air properties at 400 K Tf =

ν = 26.41 × 10−6 m2 ∕s k = 33.8 × 10−3 W∕m.K Pr = 0.690 β=

1 1 −1 K = 2.5 × 10−3 K−1 = Tf 400

The characteristic length L is A 1×1 = = 0.25 m P 4 The Grashof number GrL is L=

GrL =

g β(Tw − T∞ )L3 9.81 × (2.5 × 10−3 )(500 − 300)(0.25)3 = = 1.09 × 108 2 ν (26.41 × 10−6 )2

The Rayleigh number RaL is RaL = PrGrL = 0.690 × 1.09 × 108 = 7.58 × 107 The average Nusselt number NuL is 1∕3

NuL = 0.14RaL = 0.14 × (7.58 × 107 )1∕3 = 59.24 ) ( k 33.8 × 10−3 The average heat transfer coefficient is h = NuL = (59.24) = 8.01 W∕m2 K L 0.25 The heat transfer from the plate is given by Newton’s law of cooling q = hA(Tw − T∞ ) = (8.01)(1 × 1)(500 − 300) = 1601 W Lower surface of hot plate or upper surface of cold plate Figure 12.8 shows buoyancy driven flows over a horizontal isothermal plate. McAdams [20] recommends: 1∕4

NuL = 0.27 RaL

3 × 105 ≤ RaL ≤ 3 × 1010 where NuL = hL∕k is the mean Nusselt number, and RaL = GrL Pr = g β (Tw − T∞ )L3 /ν α is the Rayleigh number.

(12.161)

713

714

12 Free Convection Heat Transfer

Hot surface down

Fluid g

Fluid Tw > T∞ (a)

Figure 12.8

Cold surface up Tw < T∞ (b)

(a–b) Buoyancy driven flows over horizontal plates.

Nellis and Klein [22] report a correlation for average Nusselt number {

NuL = ln

2.5 } [ ( )0.9 ]2∕9 2.5 1.9 1+ 1+ Pr 0.527Ra0.2 L

(12.162)

1 × 103 ≤ RaL ≤ 1 × 1010 Uniform wall heat flux Fuji and Imura [28] studied the free convection on a horizontal, vertical, and inclined plates subjected to UHF. Working fluid was air. The plate is heated electrically, and the plate surface is exposed to ambient air. The following correlations are proposed: For the heated surface facing upward NuL = 0.13(RaL )1∕3

(12.163)

RaL < 2 × 108 NuL = 0.16(RaL )1∕3

(12.164)

5 × 108 < RaL < 1 × 1011 For heated surface facing downward NuL = 0.58(GrL Pr )1∕5

(12.165)

106 < GrL Pr < 1011 In these equations, all the physical properties except β are evaluated at the mean temperature Tm defined by Tm = Tw − 0.25(Tw − T∞ ) where Tw is the average surface temperature of the plate. The coefficient of volumetric thermal expansion β is evaluated at (Tw + T∞ )∕2. In these expressions, the Grashof number is defined as GrL =

gβ(Tw − T∞ ) ν2

and the mean Nusselt number is NuL =

q′′w hL L = k (Tw − T∞ ) k

12.7 Empirical Correlations for Free Convection

Cold surface up

θ

Hot surface facing down

L

g

Cold surface L down

Hot surface facing up

θ is negative

θ

L

Fluid at T∞ θ is positive (a) Figure 12.9

(b)

Concept of positive and inclination negative angles from vertical to define the orientation of hot surface.

12.7.3 Inclined Plates For many free convective flows in engineering, the heated surface is inclined relative to the direction of the gravity vector. The orientation of the inclined plate (facing upward or downward) affects the Nusselt number. Following Ozisik [25], sign of the angle θ that the surface makes with vertical is defined as follows: 1) The angle θ is negative if the hot surface is facing up. See Figure 12.9b. 2) The angle θ is positive if the hot surface facing down. See Figure 12.9a. Rich [29] indicated that the heat transfer coefficient in natural convection from an inclined plate can be calculated by vertical plate formulas if the gravitational term g in the Grashof number is adjusted to take into account the effect of inclination. The orientation of the inclined plate, whether the surface is facing upward or downward, is also a parameter that influences the Nusselt number. The transition from the laminar to turbulent flow depends on the angle of inclination. Churchill and Chu [21] presented correlations. Several investigators, such as Vliet [24], Fuji and Imura [28], and Vliet and Ross [30], carried out experiments for inclined plates in water for different angles of inclination. Uniform wall temperature Heated surface facing down or cooled surface facing up Assuming that plate is at uniform temperature, the Nusselt number can be calculated from the correlation given for vertical plates by replacing the gravitational term g by g cos(θ) for 0 < θ < 60∘ , where θ is the angle of inclination of the plate with respect to vertical. Thus, we can use the vertical plate equation, Eq. (12.147), as given below { }2 0.387Ra1∕6 hL = 0.825 + NuL = k [1 + (0.492∕ Pr )9∕16 ]8∕27 10−1 ≤ RaL ≤ 1012 Here, the Rayleigh number is given as RaL =

g cos(θ) β (Tw − T∞ ) L3 να

Uniform wall heat flux We will follow Fuji and Imura to define the sign convention for the orientation of the plate. The sign convention for θ is given as follows: (a) The inclination of the plate is measured with respect to vertical position. (b) The angle θ is considered negative for a hot surface facing up. (c) The angle θ is considered positive for a hot surface facing down.

715

716

12 Free Convection Heat Transfer

Heated surface facing downward Fuji and Imura [28] carried out experiments for natural convection from inclined plates in water under UHF conditions, and the following correlations were proposed: NuL =

hL = 0.56[GrL Pr Cos(θ)]0.25 k

(12.166)

+θ < 88∘ 105 < GrL Pr cos(θ) < 1011 where GrL = (gβ)(Tw − T∞ )L3 /(ν2 ) is the Grashof number. All the physical properties except coefficient of thermal expansion β are evaluated at the mean temperature Tm defined as Tm = Tw − 0.25(Tw − T∞ ) where Tw is the average wall temperature and T∞ is the temperature of quiescent fluid. The coefficient of thermal expansion β is evaluated at reference temperature Tr Tr = T∞ + 0.25(Tw − T∞ ) Hot surface facing upward The empirical correlations are complicated when the heated surface of the inclined plate faces upward. Ozisik [25] recommends that the following correlations can be used for free convection on the inclined plate with the heated surface facing upward: NuL = 0.56(GrL Pr Cos(θ) )1∕4

(12.167)

Pr GrL < Pr Grc where Grc is the critical Grashof number at which the transition from to turbulent flow takes place. Eq. (12.167) is applicable in the laminar flow regime. Consider now turbulent regime. If GrL > Grc , then the following correlation is used: NuL =

] [ hL = 0.145 (GrL Pr )1∕3 − (GrC Pr )1∕3 + 0.56[Grc Pr Cos(θ)]1∕4 k

(12.168)

GrL Pr < 1011 GrL > Grc −15∘ < θ < −75∘ where Grc is the critical Grashof number and GrL = (gβ)(Tw − T∞ )L3 /(ν2 ) is the Grashof number. At this critical number Grc , the transition from laminar to turbulent flow takes place. The transition Grashof number Grc depends on the tilt angle, and it is tabulated in Table 12.4. Fluid properties are evaluated at the mean temperature Tm = Tw − 0.25(Tw − T∞ ) and the coefficient of thermal expansion β is evaluated at reference temperature Tr Tr = T∞ + 0.25(Tw − T∞ ) The characteristic length is the geometrical length of the inclined plate. Table 12.4

Critical angles.

Angle θ in Degrees Grc

–15 5 × 10

Source: Fuji and Imura [28].

–30 9

1 × 10

–60 9

1 × 10

–75 8

1 × 106

12.7 Empirical Correlations for Free Convection

Example 12.8 A plate 1 m by 1 m is insulated on one side and subjected to heat flux of 700 W/m2 on the other surface, as shown in Figure 12.E8. The plate is in quiescent air at T∞ = 27 ∘ C and 1-atm pressure. As can be seen from Figure 12.E8, the hot surface is facing up. Estimate the average surface temperature of the plate.

Figure 12.E8

Geometry and problem description for Example 12.8.

Insulated surface 1m

q″w = 700 W/m2 L

θ θ = –60°

Solution First iteration We do not know the surface temperature, and to start the calculation, we make a guess for the heat transfer coefficient h: h = 7 W∕m2 K Using Newton’s law of cooling, we estimate the surface temperature Tw Tw = T∞ +

q′′w

= 300 +

700 = 400 K 7

h The mean temperature needed to evaluate air properties are Tm = Tw − 0.25(Tw − T∞ ) = 400 − 0.25(400 − 300) ≈ 375 K Air properties are taken at 375 K ν = 23.66 × 10−6 m2 ∕s k ≈ 0.0319 W∕m.K Pr = 0.695 β is evaluated at Tr Tr = T∞ + 0.25(Tw − T∞ ) = 300 + 0.25(400 − 300) = 325 K. β=

1 −1 1 K = 0.00307 K−1 = Tr 325

The Grashof number GrL is gβ(Tw − T∞ )L3 9.81 × (0.00307)(400 − 300)(1)3 = = 5.38 × 109 2 ν (26.41 × 10−6 )2 We obtain Grc = 108 from Table 12.3 for θ = − 60∘ . We will use Eq. (12.168) since GrL > Grc . [ ] NuL = 0.145 (GrL Pr )1∕3 − (GrC Pr )1∕3 + 0.56(GrC Pr Cos(θ) )1∕4 ) )1∕4 ( ( π NuL = 0.145[(5.38 × 109 × 0.695)1∕3 − (108 × 0.695)1∕3 ] + 0.56 108 × 0.695 × Cos −60 × 180 NuL = 208.57 GrL =

717

718

12 Free Convection Heat Transfer

The average heat transfer coefficient h is ) ( k 0.0319 (208.57) = 6.6 W∕m2 K h = NuL = L 1 Second iteration We estimate the wall temperature Tw by taking h = 6.14 W∕m2 K W h = 6.6 2 mK q′′ 700 = 406 K Tw = T∞ + w = 300 + 6.6 h The mean temperature to evaluate physical properties is Tm = Tw − 0.25(Tw − T∞ ) = 406 − 0.25(406 − 300) ≈ 380K We evaluate air properties at this temperature ν = 24.21 × 10−6 m2 ∕s k = 0.0322 W∕m.K k = 0.694 The reference temperature Tr to evaluate the coefficient of thermal expansion β is Tr = T∞ + 0.25(Tw − T∞ ) = 300 + 0.25(406 − 300) = 326.5 K. β=

1 1 K−1 = 0.00306 K−1 = Tr 326.5

GrL =

gβ(Tw − T∞ )L3 = 5.43 × 109 ν2

) )1∕4 ( ( π NuL = 0.145[(5.43 × 109 × 0.696 )1∕3 − (108 × 0.696 )1∕3 ] + 0.56 108 × 0.696 × Cos −60 × 180 NuL = 209.29 k Nu = 6.7 W∕m2 K L L Third iteration The average surface temperature Tw is h=

Tw = T∞ +

q′′w

= 300 +

h

700 = 404.4K 6.7

Tm = Tw − 0.25(Tw − T∞ ) = 404.4 − 0.25(404.4 − 300) ≈ 378 K. Air properties are evaluated at this temperature ν ≈ 24 × 10−6 m2 ∕s k = 0.0321 W∕m.K k = 0.695 Tr = T∞ + 0.25(Tw − T∞ ) ≈ 326 K β = 0.00306 GrL = 5.45 × 109 NuL = 209.49 h = 6.72 W∕m2 K

12.7 Empirical Correlations for Free Convection

We now estimate plate surface temperature q′′w

= 404.16K h The mean surface temperature in the second and third iterations is close enough. There is no need for the iteration. Tw = T∞ +

12.7.4 Vertical Cylinders Natural convection heat transfer from the surface of a cylinder of circular cross section has important applications in engineering. For instance, this problem is encountered in heating and cooling equipment. The cylinder surface can be at UHF or UWT. Constant wall temperature The continuity, momentum, and energy can be written for free convection from a vertical cylinder. These equations can be reduced to boundary layer formulation using the order of magnitude analysis. Consider an isothermal vertical cylinder, as shown in Figure 12.10. If the laminar boundary layer thickness δ is much less than the cylinder diameter D(δ ≪ D), then the vertical isothermal cylinder can be modeled as a vertical flat plate, as discussed in [31]. The general criterion is 35 D ≥ Pr = 0.72 1∕4 L GrL

(12.169)

where D is the cylinder diameter and L is the length of cylinder. This condition is called as the thick cylinder limit, and GrL = (gβ)(Tw − T∞ )L3 /(ν2 ) is the Grashof number based on the height of the cylinder. This limit is derived by Sparrow and Gregg. If this criterion is satisfied, we can use the vertical flat plate equations. Vertical flat plate solutions can be used to estimate the average Nusselt number for vertical cylinders within 5% accuracy as discussed by Sparrow and Gregg. The literature survey indicates that the available correlations concerning free convection from vertical slender cylinders show a widespread, and it should be noticed that the correct length to use in the Grashof number is the vertical dimension of the cylinder. For smaller (D/L) ratios, the Nusselt number is augmented by the effect of cylinder curvature, and the following correlation may be used, as discussed in [22]: NuL =

ξ hL = NuL,nc k ln(1 + ξ)

(12.170)

where NuL,nc is the average Nusselt number obtained using vertical plate correlation and ξ is given by 1.8

ξ=

NuL,nc

L D

Arpaci et al. [32] recommend the following correlation for isothermal vertical cylinder: NuL = C πnN πN =

(12.171)

RaL C

1 + Pr0 gβ(Tw − T∞ )L3 RaL = να where C0 , C, and n are chosen from the table given below. Fluid properties are evaluated at the film temperature Tf = (Tw + T∞ )/2. Here Tw and T∞ are the cylinder surface and free stream temperatures, respectively. Notice that L is the height of the cylinder in the direction of the gravity. Figure 12.10

Vertical cylinder in free convection.

D

L

𝛿

Tw

719

720

12 Free Convection Heat Transfer

C0

C

n

104 ≤ πN ≤ 109

0.492

0.67

1/4

109 < πN ≤ 1012

0.492

0.15

1/3

Gebhart [33] recommended the following correlation for isothermal vertical cylinder: Laminar: Nu =

hL = 0.021(GrL Pr )0.4 k

(12.172)

GrL ≤ 109 35 D ≥ Pr = 0.72 1∕4 L GrL Turbulent: Nu =

hL = 0.55(GrL Pr )0.25 k

(12.173)

GrL ≥ 109 35 D ≥ Pr = 0.72 1∕4 L GrL where GrL = (gβ)(Tw − T∞ )L3 /(ν2 ) is the Grashof number based on the height of the cylinder. The fluid properties are evaluated at the film temperature Tf = (Tw + T∞ )/2. Yang [34] proposed the following correlation for free convection from an isothermal vertical cylinder for laminar region: 1∕6 ⎧ ⎡ ⎤ ⎫ ⎪ √ √ ⎢ ⎥ ⎪ ( ) RaL ⎪ L ⎢ ⎥ ⎪ + 0.387 [ NuL = ⎨0.6 ( )9∕16 ]16∕9 ⎥ ⎬ ⎢ D ⎪ ⎢ 1 + 0.492 ⎥ ⎪ ⎪ ⎣ ⎦ ⎪ Pr ⎭ ⎩ hL NuL = k

(12.174)

RaL < 109 Tw = const where GrL = (gβ)(Tw − T∞ )L3 /(ν2 ) is the Grashof number and RaL = GrL Pr is the Rayleigh number based on the cylinder height. Example 12.9 A 20-cm-diameter, 1-m-long vertical cylinder is in quiescent air at 1-atm pressure and 27 ∘ C. If the cylinder has a surface temperature of 70 ∘ C, estimate the heat loss by free convection from the cylinder. Solution Tw = 70 ∘ C = 343 K T∞ = 27 ∘ C = 300 K We evaluate properties of air at Tf as Tf =

Tw + T∞ 343 + 300 = = 321.5 K 2 2

12.7 Empirical Correlations for Free Convection

ν = 18 × 10−6 m2 /s k = 0.0278 W/m. K Pr = 0.703 Pr = 0.703 1 1 K−1 = 0.00311 K−1 β= = Tf 321.5 First, we check 35 D ≥ Pr = 0.72 1∕4 L GrL The evaluation of the Grashof number GrL yields GrL =

gβ(Tw − T∞ )L3 9.81 × (0.00311) × (343 − 300)(1)3 = ≈ 4 × 109 2 ν (20.92 × 10−6 )2

Then, we have D = 0.2 L 35 = 0.138 1∕4 GrL 35 D > 1∕4 L Gr

(12.174)

L

Thus, the curvature of the cylinder may be neglected safely, and the cylinder can be treated as a vertical wall. The characteristic length is L = 1 m. Flow is turbulent since GrL > 109 , and we can use the Churchill and Chu equation, Eq. (12.147), 2

2

⎫ ⎫ ⎧ ⎧ ⎪ ⎪ ⎪ ⎪ 1∕6 0.387RaL ⎪ ⎪ ⎪ 0.387(0.703 × 4 × 109 )1∕6 ⎪ 0.825 + (NuL ) = ⎨0.825 + [ = = 169.45 ⎨ [ ( ( ) ]8∕27 ⎬ ) ]8∕27 ⎬ 0.492 9∕16 0.492 9∕16 ⎪ ⎪ ⎪ ⎪ 1+ 1+ ⎪ ⎪ ⎪ ⎪ Pr 0.703 ⎭ ⎭ ⎩ ⎩ ) ( W k 0.0278 (169.45) = 4.72 2 h = NuL = L 1 m K Heat loss from the cylinder is q = h(πDL)(Tw − Tw ) = 4.72 × (π × 0.20 × 1)(343 − 300) = 127.69 W Uniform heat flux For the vertical cylinder subjected to UHF, Janna [35] reports the following relation based on experimental work: ( ) hD D 0.25 D = 0.6 RaD ≥ 104 NuD = RaD (12.175) k L L ( ) hD D 0.16 D = 1.37 RaD ≤ 104 NuD = 0.05 ≤ RaD (12.176) k L L ( ) hD D 0.05 D = 0.93 RaD ≤ 0.05 (12.177) NuD = RaD k L L where the Rayleigh number is based on diameter D RaD =

g β (Tw − T∞ ) D3 να

β is evaluated at Tf = (Tw + T∞ )∕2 liquids β is evaluated at β =

1 gases T∞

where Tw is the average surface temperature, and all other fluid properties are evaluated at the film temperature Tf = (Tw + T∞ )∕2.

721

722

12 Free Convection Heat Transfer

Yang [34] studied free convection heat transfer from a vertical cylinder subjected to UHF and proposed the following correlation: ) ( D 1∕4 0.67 RaD hD L = 0.36 + [ NuD = (12.178) )9∕16 ]4∕9 ( k 0.492 1+ Pr 0 < Pr < ∞ RaL < 109 where GrD = (gβ)(Tw − T∞ )D3 /(ν2 ) is the Grashof number based on the cylinder diameter and RaD = GrD Pr is the Rayleigh number based on the cylinder height.

12.7.5

Horizontal Cylinder

Free convection heat transfer over the surface of a horizontal cylinder is similar to the free convection heat transfer from the vertical plate. The difference is that the surface of the cylinder is curved. The Nusselt number and the Grashof number are based on the diameter of the cylinder. See Figure 12.11. The average Nusselt number over the whole cylinder surface is great interest in engineering applications. Uniform wall temperature McAdams [20] gives the following correlations for horizontal cylinders, which are long and end effects are negligible. Mean Nusselt numbers are NuD = 0.53(GrD Pr )1∕4

104 < GrD Pr < 109

(12.179)

NuD = 0.13(GrD Pr )1∕3

109 < GrD Pr < 1012

(12.180)

Tw = const Morgan [36] suggests an empirical relation for an isothermal horizontal cylinder NuD =

hD = C RanD k

(12.181)

Tw = const where constants C and n for Eq. (12.181) are given in Table 12.5. Churchill and Chu [37] recommend a correlation for isothermal horizontal cylinders. The average Nusselt number is given as 1∕4

⎛ ⎞ ⎜ ⎟ (GrD Pr ) hD ⎟ NuD = = 0.36 + 0.518⎜ [ ] 16∕9 )9∕16 ( ⎜ ⎟ k 0.559 ⎜ 1+ ⎟ ⎝ ⎠ Pr

(12.182)

″ Tw or qw

D L

Figure 12.11

Horizontal cylinder in free convection.

T∞

θ

12.7 Empirical Correlations for Free Convection

Table 12.5

Constants C and n for Eq. (12.181).

RaD

10−10 < RaD < 10−2 −2

< RaD < 10

2

C

n

0.675

0.058

1.02

0.148

102 < RaD < 104

0.850

0.188

104 < RaD < 107

0.480

0.250

107 < RaD < 1012

0.125

0.333

10

Source: Morgan [32].

10−6 ≤ GrD Pr ≤ 109 0 ≤ Pr ≤ ∞ Tw = const The Nusselt number and the Rayleigh number are based on the cylinder diameter. Fluid properties are evaluated at the film temperature Tf = (Tw + T∞ )/2. Eq. (12.179) may be used for both the constant surface temperature and constant heat flux boundary conditions. For turbulent flow, Churchill and Chu [38] recommend the following correlation:

(NuD )1∕2

⎛ ⎞ ⎜ ⎟ (Gr Pr ) D ⎟ = 0.60 + 0.387⎜ [ ] ( )9∕16 16∕9 ⎟ ⎜ 0.559 ⎜ 1+ ⎟ ⎝ ⎠ Pr

1∕6

(12.183)

10−5 ≤ GrD Pr ≤ 1012 0 ≤ Pr ≤ ∞ Tw = const Evaluate β at the film temperature Tf = (Tw + T∞ )/2 for liquids. 1 for gases. T∞

Evaluate β =

Uniform heat flux Churchill and Chu [38] proposed the following correlation for the UHF boundary condition: ( )1∕4 ( ∗) RaD hD = 0.579 NuD = k [1 + (0.442∕ Pr )9∕16 ]16∕9

(12.184)

10−6 ≤ GrD Pr ≤ 109 0 ≤ Pr ≤ ∞ q′′w = const where Ra∗D = gβq′′w D4 ∕kνα is the modified Rayleigh number.

12.7.6 Inclined Cylinder Research has been carried out for the experimental determination of free convection heat transfer from inclined circular cylinders. Geometry and coordinate system for free convection around an inclined cylinder is illustrated in Figure 12.12.

723

724

12 Free Convection Heat Transfer

D 180 L

0

θ Figure 12.12

φ

Coordinate system for inclined cylinder in free convection.

Uniform wall temperature Stewart [39] carried out several experiments on the free convective heat loss from inclined circular cylinders in air. The cylinder is inclined with respect to horizontal. Based on his experimental work, Stewart proposed the following relation for the average Nusselt number for small aspect ratio: NuD = 0.53[RaD cos(θ)]0.25 + 0.555[RaD cos(θ)]0.25 [1 − cos (θ)0.25 ]

(12.185)

4 × 104 ≤ RaD ≤ 4 × 108 Pr = 0.7 All the fluid properties are evaluated at the film temperature Tf = (Tw + T∞ )/2. It is reported that data fit the correlation within 10%.

12.7.7

Free Convection from Vertical Cylinders of Small Diameter

The boundary layer thickness of small-diameter cylinders may become large when compared with the radius of curvature of small-diameter cylinders such as wires. Kyte et al. [40] performed experiments to investigate the free convection. They correlated their experimental data with air and proposed the following correlation: NuD =

10−11

2

⎡ ⎤ ⎢ ⎥ 4.47 ln ⎢1 + ( )0.26 ⎥ D ⎢ ⎥ GrD Pr ⎣ ⎦ L ) ( D ≤ 10−4.5 ≤ GrD Pr L

(12.186)

Tw = const where GrD is the Grashof number based on the wire diameter D. Nagendra et al. [41] studied the free convection heat transfer from vertical wires experimentally. The experiments were in water. They proposed an empirical correlation given below ) ( D 0.05 NuD = 0.87 RaD (12.187) L Tw = const ( ) D 10−4 ≤ GrD Pr ≤ 5 × 10−2 L where RaD is the Rayleigh number based on the wire diameter D.

12.8 Free Convection Within Parallel Plate Channels

Example 12.10 A 0.025-mm-diameter wire is heated electrically to 57 ∘ C and exposed to ambient air at 1 atm and 27 ∘ C. The wire is 15 cm long, and it is in vertical position. Estimate the heat loss from the wire. Solution Air properties are evaluated at the film temperature Tf ) ( ( ) Tw + T∞ 330 + 300 = = 315 K Tf = 2 2 ν = 17.4 × 10−6 m2 ∕s k = 27.4 × 10−3 W∕m.K Pr = 0.704 β=

1 = 0.003174 K−1 Tf

The Grashof number GrD is g β(Tw − T∞ )D3 9.81 × (31.74 × 10−4 )(330 − 300)(2.5 × 10−5 )3 = = 4.82 × 10−5 ν2 (17.4 × 10−6 m2 ∕s )2 D 0.000025 = 4.82 × 10−5 × 0.703 × ≈ 5.66 × 10−9 GrD Pr L 15 We use Eq. (12.186) 2 2 = = 0.3107 NuD = ⎡ ⎡ ⎤ ⎤ ⎢ ⎢ ⎥ ⎥ 4.47 4.47 ln ⎢1 + ( ⎥ )0.26 ⎥ ln ⎢1 + ( ) D ⎢ ⎢ ⎥ −5 0.26 ⎥ Gr Pr 2.5 × 10 D ⎣ ⎦ ⎢ ⎥ 4.82 × 10−5 × 0.703 × L ⎣ ⎦ 0.15 ) ( W k 27.4 × 10−3 (0.3107) = 340.55 2 h = NuD = D mK 2.5 × 10−5 GrD =

Heat loss from the wire is q = (πDL)h(Tw − T∞ ) = π × 2.5 × 10−5 × 340.55 × (330 − 300) = 0.12 W

12.8 Free Convection Within Parallel Plate Channels 12.8.1 Vertical Parallel Plate Channel Fin arrays are used to enhance heat transfer by free convection. Fin arrays can be modeled as vertical channels formed by vertical parallel plates. An array of channels formed by parallel plates, which are oriented parallel to gravity, is a typical geometry encountered in engineering applications, as depicted in Figure 12.13. The heated fluid rises and exits from the top of the channel. There is a developing region followed by the fully developed region. Flow will be fully developed if the ratio of the spacing distance S to the channel length L is small, the boundary layers eventually merge, and the flow will be fully developed. If the ratio (S/L) is large (i.e. plates are far apart), they behave as isolated vertical plates in infinite, quiescent fluid. In this case, vertical plate correlations can be used. The following types of boundary conditions may be seen in literature. (a) (b) (c) (d) (e)

Both the plates are subjected to UHF. One plate is subjected to uniform wall heat flux and other plate is insulated. Both the plates are subjected to UWT. One plate is subjected to UWT and other plate is insulated. Plates are at different temperatures.

725

726

12 Free Convection Heat Transfer

Figure 12.13

Channel configuration for isothermal walls.

S

L

u(y)

TH

g

TC

x

y

T∞

Uniform wall temperature Consider free convection between two parallel plates and plates from a channel with height L and with a separation distance S between plates. This configuration is open to ambient at opposite ends. Plates are at constant temperatures TH = TC = Tw . The channel is in a large volume of fluid, where the bulk of fluid is quiescent. Flow enters the channel with a temperature T∞ . It is assumed that the plates are heated (Tw > T∞ ). Free convection motion takes place due to temperature variation, and therefore, momentum and energy equations need to be solved together. There is a developing region on the walls. Free convection boundary layers form on both the channel walls, and if the ratio of the channel length to channel spacing (L/S) is large, the boundary layers on the channel walls will merge, and the flow will be fully developed. We have now buoyant stream. The momentum equation in the x-direction ( ) ) ( 2 𝜕p 𝜕u 𝜕u 𝜕 u 𝜕2u ρ u (12.188) +v =− − ρg + μ + 𝜕x 𝜕y 𝜕x 𝜕x2 𝜕y2 Since we assume that the channel is very long, we may assume that v ≪ u. This means that we have fully developed flow, and we have 𝜕u =0 (12.189) 𝜕x We know that for a fully developed flow, the pressure is a function of x. Both the ends of the channel are open to ambient, and the ambient air density is ρ∞ . Thus, we can write 𝜕p dp = = −gρ∞ (12.190) 𝜕x dx With the help of Eqs. (12.189) and (12.190) and Boussinesq approximation, the momentum equation reduces to the following form: d2 u gβ(T − T∞ ) = (12.191) ν dy2 Notice that 𝜕𝜕xu2 ≪ 𝜕𝜕yu2 since the scale of x is much larger than that of y. In order to carry out the integration, we need temperature T as a function of x. Since the channel is very tall and narrow, the fluid temperature T(x, y) is nearly the same as the wall temperature Tw . We now assume that 2

2

[Tw − T(x, y)] ≪ [Tw − T∞ ]

(12.192)

This implies that flow is thermally fully developed. We also assume that Pr ≥ 1. The momentum equation becomes d2 u gβ(Tw − T∞ ) (12.193) = ν dy2 The boundary conditions are S At y = u=0 2 S At y = − u = 0. 2

(12.194) (12.195)

12.8 Free Convection Within Parallel Plate Channels

The solution of the momentum equation is [ ( )2 ] gβS2 (Tw − T∞ ) x 1− u= 8ν S∕2

(12.196)

The mass flow rate per unit width (perpendicular to the plane of Figure 12.13) is S∕2

ṁ =

∫−S∕2

ρ u dy =

ρgβS3 (Tw − T∞ ) 12 ν

(12.197)

The total energy transport in the channel is q = ṁ cp (Tw − T∞ ) =

ρgβcp (Tw − T∞ )2 S3

(12.198)

12 ν

Since the channel has two sides q′′w =

ρgβcp (Tw − T∞ )2 S3 q = 2L 24 ν L

(12.199a)

or ρgβcp (Tw − T∞ ) S3 q′′w = (Tw − T∞ ) 24 ν L

(12.199b)

The Nusselt number based on the channel height L is ( )( ) q′′w hL L = NuL = k Tw − T∞ k ][ ] [ [ ] [ μ cp ] [ gβ(T − T ) S3 ] ρν cp gβ(Tw − T∞ ) S3 1 w ∞ (L) = = 2 k 24 k 24 ν L ν2 [Pr ][GrS ] RaS = = 24 24

(12.200a)

or ( ) Ra S hS S = k L 24 The Rayleigh number RaS is

(12.200b)

NuS =

g β(Tw − T∞ ) S3 να Bejan [42] gives similar results to Eq. (12.200b) for different channel geometries. See Table 12.6. For each case, NuL ∕NuDH is constant, and DH is the hydraulic diameter of a given geometry. Elenbaas [43] published the benchmark paper about the free convection between vertical parallel plates. Elenbaas correlated the Nusselt number against the Rayleigh number RaS and the ratio of the channel height to channel spacing (L/S) for air. Both the plates are isothermal, and both the plates are at the same temperature Tw . Elenbaas developed a semiempirical correlation for the average Nusselt number RaS = GrS Pr =

Table 12.6

Average Nusselt numbers for flow in a vertical channel.

Cross-sectional shape

Parallel plates Circular duct Square Equilateral triangle

NuL ∕NuDH

1 192 1 128 1 113.6 1 106.4

727

728

12 Free Convection Heat Transfer

( NuS =

RaS 24

)( )[ ( )]3∕4 S 35 1 1 − exp − L RaS (S∕L)

(12.201)

S Ra ≤ 105 L S where the average heat transfer coefficient, the Nusselt number, and the Rayleigh number are defined as ( ) q′′w q = h= A(Tw − T∞ ) Tw − T∞ ) ( hS NuS = k 0.1 ≤

and g β (Tw − T∞ )S3 αν where A is the surface area of one plate and q′′w is the rate of heat flux from one side of the plates. The heat transfer coefficient h is based on the difference between the plate surface temperature and the temperature of the fluid that is entering into channel. It is possible to determine the total heat transfer by knowing the average Nusselt number for one plate. In the fully developed limit (S/L) → 0, this equation reduces to ) ( RaS ( S ) (12.202) NuS,fd = 24 L RaS =

A common configuration is that one plate is isothermal Tw and the other plate is insulated (q′′w = 0). In this case, the fully developed Nusselt number is ) ( Ras ( S ) Nus,fd = 12 L

(12.203)

Bar-Cohen and Rohsenow [44] presented correlations for a vertical parallel plate channel under different boundary conditions. For small values of (S/L), the boundary layers developing on opposite walls eventually merge to form a fully developed condition. On the other hand, for large values of (S/L), the boundary layers on each wall develop independently, yielding a condition that corresponds to an isolated vertical plate in an infinite quiescent fluid. Then, vertical plate correlations can be used. Bar-Cohen and Rohsenow presented the following semiempirical correlation for isothermal plates: }−1∕2 { C1 C2 NuS = + (12.204) [RaS (S∕L)]2 [RaS (S∕L)]1∕2 where constants C1 and C2 are given in Table 12.7. Bar-Cohen and Rohsenow [44] determined the optimum plate spacing Sopt to maximize the heat transfer from plates for different boundary conditions, and their results are given in Table 12.8. The physical properties of the fluid between the plates are evaluated at T = (Tw2 + Tw1 )∕2. Wei [45] proposed a correlation for turbulent flow based on data obtained from direct numerical simulation (DNS) studies at Pr ≈ 0.7 NuS = Table 12.7

Pr Gr0.38 hS = k 1.1 × ln(Pr Gr0.38 ) + 1.19 × Gr0.056 + 1.5

Constants for use in Eq. (12.206).

Surface condition

C1

C2

Sopt

Uniform heat flux q′′w1 = q′′w2 = q′′w

48

2.51

( )−1∕5 2.12 Ra∗S ∕S4 L

Uniform heat flux/Adiabatic q′′w1 , q′′2 = 0

24

2.51

)−1∕5 ( 1.69 Ra∗S ∕S4 L

(12.205)

12.8 Free Convection Within Parallel Plate Channels

Table 12.8

Constants for use in Eq. (12.204).

Surface condition

C1

C2

Sopt

Isothermal plates Tw1 = Tw2 = Tw

576

2.87

2.71(RaS /S3 L)−1/4

Isothermal/Adiabatic Tw , q′′w = 0

144

2.87

2.15(RaS /S3 L)−1/4

103 106 104 102

102 100

105 1010 1015 1020 Versteegh Kis Ng 0.14Gr1/3 New

101

100 103

104

105

106 Grs

107

108

109

Figure 12.14 Prediction of Nusselt number in turbulent differentially heated vertical channel using Eq. (12.205). The inset depicts the prediction over an extended range of Grashof numbers. Source: Wei [45]/with permission of Elsevier.

104 ≤ Gr ≤ ∞ Figure 12.14 shows the Nusselt number as a function of the Grashof number with prediction using Eq. (12.205). The prediction is similar to a power law with an exponent of m = 1/3, as shown in the inset of Figure 12.14, and up to Gr = 1020 . Wei states that Eq. (12.205) agrees with available DNS data, and more DNS data over a wider range of Grashof number are required to evaluate the validity of Eq. (12.205). In Eq. (12.205), the Grashof number is defined as Gr =

gβΘmp δ ν2

=

1 gβΔT(2δ)3 16 ν2

δ = S∕2 and Θmp = 0.5ΔT ΔT = TH − TC Example 12.11 A vertical channel is formed by two parallel plates. The spacing between the plates is 10 cm, and the channel height is 100 cm. Each plate is at 227 ∘ C and is externally insulated. See Figure 12.E11. We wish to determine the heat flux if the channel draws in atmospheric nitrogen at 27 ∘ C. Solution Tf =

Tw + T∞ 227 + 27 = = 127 ∘ C = 400 K 2 2

Properties of nitrogen are ν = 26.16 × 10−6 m2 ∕s k = 0.0327 W∕m.K

729

730

12 Free Convection Heat Transfer

Figure 12.E11

Geometry and problem description for Example 12.11.

S = 10 cm

Tw = 227 °C L = 1m

g x y

T∞ = 27 °C

Pr = 0.704 α = 37.1 × 10−6 m2 ∕s β=

1 = 0.0025 Tf

The Rayleigh number Ras is Ras = GrS Pr =

gβ(Tw − T∞ )S3 9.81 × 0.0025 × (500 − 300)(0.1)3 = = 5 × 106 να (26.16 × 10−6 ) × (37.1 × 10−6 )

We will use Eq. (12.204) }−1∕2 { C1 C2 NuS = + [RaS (S∕L)]2 [RaS (S∕L)]1∕2 For isothermal plates, C1 = 576 C2 = 2.87. We now evaluate the average Nusselt number }−1∕2 { 2.82 576 NuS = + = 15.83 [5 × 106 (0.1∕1)]2 [5 × 106 (0.1∕1)]1∕2 Heat flux q′′w is ) ( ( ) 0.0327 k (Tw − T∞ ) = 15.83 (500 − 300) = 1035.6 W∕m2 q′′w = Nus S 0.1 Uniform heat flux This configuration is also open to ambient at opposite ends. See Figure 12.15. The plates may constitute a fin array used to enhance free convection heat transfer from a base area. The buoyancy acts to induce fluid motion in the streamwise direction. Beginning at x = 0, the boundary layer begins to develop on each plate’s surface. The boundary layers develop independently for large spacing between parallel plates. For small spacing, boundary layers meet at the channel center and yield a fully developed condition. Bar-Cohen and Rohsenow [44] investigated the free convection between vertical parallel plates subjected to UHF. Bar-Cohen and Rohsenow presented the following semiempirical correlation for the Nusselt number: { }−1∕2 C1 C2 (12.206) NuS,x=L = [ ∗ ]+[ ]2∕5 Ras (S∕L) Ra∗s (S∕L) The Nusselt number and modified Rayleigh number are defined as )( ) ( q′′w S NuS,x=L = Tw,x=L − T∞ k Ra∗S =

g β q′′w S4 kνα

12.8 Free Convection Within Parallel Plate Channels

Figure 12.15

Channel configuration under uniform heat flux.

L

u(y)

″ qw

g

″ qw

S x

y

T∞ Table 12.9

Comparison of correlations for horizontal parallel plates.

𝚫T = (Th − T∞ )∘ C)

10

20

30

40

50

60

Experimental Nusselt number (Turgut and Onur)

Nus = 0.1688[(S∕Lc )Ras ]0.3362

1.73

2.07

2.25

2.36

2.43

2.47

Numerical Nusselt number (Turgut and Onur)

Nu = 0.1528(sRa/L)0.3362

1.57

1.88

2.04

2.14

2.20

2.24

1.6

1.8

1.9

1.9

2.1

2.2

1.49

1.79

1.94

2.04

2.10

2.13

Numerical Nusselt number (Masoomi et al.) Numerical Nusselt number (Fu et al.)

Nu = 0.1456(sRa/L)0.3362

where the reference surface temperature Tw, x = L is the temperature at the end of the plate and T∞ is the fluid temperature at the outside of the channel. Notice that the local Nusselt number NuS, x = L is defined in terms of the surface temperature at the top of the channel. Fluid properties are evaluated at T = (Tw,x=L + T∞ )∕2 and β = 1∕T. The maximum surface temperature for symmetric UHF surfaces occurs at the top of the channel. Thus, the device at the top of the channel is the most critical from the reliability point of view. The constants C1 and C2 for Eq. (12.206) are given in Table 12.7. Bar-Cohen and Rohsenow also determined the optimum plate spacing Sopt to maximize the heat transfer from the plates for different boundary conditions, and their results are given in Table 12.7. For symmetric, iso-flux (UHF) plates, the Nusselt number for the fully developed limit is [ ]1∕2 NuS,fd = 0.144 Ra∗S (S∕L) (12.207) and for asymmetric iso-flux conditions with one surface insulated (q′′w2 = 0), the Nusselt number for the fully developed limit is [ ]1∕2 NuS,fd = 0.204 Ra∗S (S∕L) (12.208)

12.8.2 Horizontal Parallel Plate Channel Turgut and Onur [46] studied the free convection heat transfer between horizontal parallel plates experimentally and numerically. See Figure 12.16. The lower plate is isothermally heated, while the upper plate is insulated. The plate separation distance is S. The characteristic length is defined as Lc = A/P. Here, A is the surface area and P is the perimeter of the plate. The experiments were carried out in air, and the air is assumed to have constant thermophysical properties. The experimental correlation for the average Nusselt number is given as Nus = 0.1688[(S∕Lc )Ras ]0.3362

(12.209)

731

732

12 Free Convection Heat Transfer

Tw2

T∞ S

g

Tw1 > Tw2

Tw1 Figure 12.16

Horizontal parallel plates.

103 ≤ RaS ≤ 2.3 × 105 where RaS is the Rayleigh number, and it is defined as g β (T1 − T∞ )S3 αν Here, α = k/ρ cp is the thermal diffusivity. Air properties are evaluated at Tf = (T1 + T∞ )/2. The correlations of Fu et al. [47] and Masoomi et al. [48] and the correlation of Turgut and Onur are presented in Table 12.9. In Table 12.9, a comparison of results is also shown. RaS =

12.8.3

Inclined Parallel Plate Channel

Consider an inclined channel, as shown in Figure 12.17. This configuration is open to ambient at opposite ends. Again, buoyancy acts to induce fluid motion in stream wise direction. Buoyancy force has normal as well as parallel component to streamwise direction. Azevedo and Sparrow [49] performed experiments for inclined channels in water. A channel formed by two inclined plates is considered for 0 ≤ θ ≤ 45∘ with respect to vertical and RaS (S/L) > 200. In experiments, symmetric isothermal plates and isothermal insulated plates were considered. Data for all the experimental conditions were correlated with within ±10% by the following relation: NuS = 0.645[RaS (S∕L)]1∕4

(12.210)

RaS (S∕L) > 200 RaS =

gβ(Tw − T∞ )S3 να Figure 12.17

θ Tw1

g

L S

Tw2

x O

T∞

y

Inclined channel configuration.

12.8 Free Convection Within Parallel Plate Channels

8.0

7.0

Average nusselt number

6.0

5.0

4.0

3.0

30 degree Onur & Aktas

2.0

30 degree Talukdar et al. 30 degree Azevedo & Sparrow 45 degree Onur & Aktas

1.0

45 degree Talukdar et al. 45 degree Azevedo & Sparrow 0.0 100

Figure 12.18

1000 Modified rayleigh number

10 000

Nusselt number versus modified Rayleigh number.

where the Rayleigh number is based on the spacing S, and fluid properties were evaluated at T = (Tw + T∞ )∕2. Onur and Aktas [50] performed an experimental study on natural convection in asymmetrically and isothermally heated open-ended inclined channel with air. Experiments were carried out with various inclination angles, temperature differences, and inner plate spacing for downward facing the heated plate. The Nusselt number correlations were suggested, acceptable within ±15% deviation with the results of Azevedo and Sparrow [49] and Talukdar et al. [51]. The angle of inclination θ is measured with respect to vertical. A comparison of Eq. (12.211) with the results of Azevedo and Sparrow and Talukdar et al. is shown in Figure 12.18. The experimental data were correlated by following expression: Nus = a[(S∕L) Ras ]b

(12.211)

where a and b are given Table 12.10. Here, Ras = g β (Tw − T∞ )S3 /ν α is the Rayleigh number and Nus = h S∕k is the Nusselt number. The thermophysical properties in the Nusselt and Rayleigh numbers were all evaluated at the film temperature (Tw +T∞ )/2. Here, Tw and T∞ are the surface temperature of the heated plate and the ambient air temperature, respectively.

733

734

12 Free Convection Heat Transfer

Table 12.10

Constants a and b for Eq. (12.211).

A

B

𝛉

(S/L) Ras

0.577

0.243

0∘

10−2 − 104

0.542

0.250

10−2 − 104

0.542

0.236

30∘ 45∘

10−2 − 104

Example 12.12 A flat plat solar collector is modeled as two inclined parallel plate channel, as shown in Figure 12.E12. The upper plate is transparent and assumed to be adiabatic with respect to convective heat transfer. The channel is inclined 40∘ with respect to vertical. Water is flowing in the channel by buoyancy forces only and the lower plate is maintained at 67 ∘ C by solar radiation passing through the cover plate. Additional data for the problem are given on Figure 12.E12. Estimate the heat transfer from the absorber plate to water.

Figure 12.E12 Geometry and problem description for Example 12.12.

θ

Transparent cover plate

g L=2m

L

Tw2 S

S = 10 mm Tw = 67 °C = 340 K T∞ = 37 °C = 310 K

Black absorber plate T∞

Solution The average water temperature is T=

Tw + T∞ 340 + 310 = = 325 K 2 2

Water properties are evaluated at this temperature μ = 528 × 10−6 N.s∕m2 k = 0.645 W∕m K ρ = 987 kg∕m3 Pr = 3.42 cp = 4.182 kJ∕kg K

12.9 Rectangular Enclosures

β = 471.2 × 10−6 1∕K μ 528 × 10−6 = = 5.34 × 10−7 m2 ∕s ρ 987 The thermal diffusivity α is 0.640 k = 1.56 × 10−7 m2 ∕s = α= ρcp 987 × 4182 ν=

Next, we calculate the Rayleigh number RaS as gβ(Tw − T∞ )S3 9.81 × 471.2 × 10−6 (340 − 310)(0.01)3 = = 1.65 × 106 να (5.34 × 10−7 )(1.56 × 10−7 ) ( ) S = 8250 RaS L Since RaS (S/L) > 200, we can use Eq. (12.210) RaS =

NuS = 0.645[RaS (S∕L)]1∕4 = 0.645[8250]1∕4 = 6.15 h=

) ( ) ( k 0.645 (7.32) = 397 W∕m2 K NuS = S 0.1

q′′w = h(Tw − T∞ ) = 397(340 − 310) = 11 908 W∕m2

12.9 Rectangular Enclosures Free convection heat transfer in enclosures is very complex and has several applications in engineering such as cooling of electronic equipment, thermal design of furnaces, energy storage systems, and solar collectors. Several researchers, including Ostrach [52] and Catton [53], have studied rectangular enclosures. We will present simple correlations for free convection for enclosures along with complicated empirical correlations. Simple correlations are very useful to obtain a quick estimate of the heat transfer coefficient and correlations may be given in the following form, as discussed in [25]: ( )m hS H NuS = (12.212) = C(RaS )n k S where RaS is given by gβ(T1 − T2 )S3 Pr ν2 where S is the separation distance between hot and cold surfaces. Two opposing walls are kept at temperatures Tw1 and Tw2 . Here, Tw1 is the temperature of the hot wall and Tw2 is the temperature of cold wall. Remaining walls are insulated. Heat flux across the cavity is RaS = GrS Pr =

q′′ = h(Tw1 − Tw2 ) can depend on the aspect ratio (H/L) and the tilt angle θ.

12.9.1 Horizontal Rectangular Enclosure (𝛉 = 0) First, consider horizontal cavity heated from above (Tw2 > Tw1 ). See Figure 12.19. The distance between the hot and cold plates is the characteristic length. In this case, there are no convection currents within the cavity, and heat transfer takes place downward by conduction. Next, consider the case where the hot plate at the bottom (Tw1 > Tw2 ). A critical Raleigh number is defined as RaL,c = 1708 where RaL is defined as RaL =

g β (Tw1 − Tw2 )L3 να

(12.213)

735

736

12 Free Convection Heat Transfer

H

Insulated

Tw2

g

Insulated

L Tw1

Figure 12.19

Horizontal enclosure.

is the Rayleigh number. If RaL, c < 1708, fluid is stagnant, and then, there is no free convection currents within the cavity since buoyancy forces cannot overcome the viscous forces, and heat transfer takes place from the bottom wall to the top wall by conduction only and thus NuL = 1

(12.214)

If RaL, c > 1708, then there is a significant free convection currents within the cavity. Heat transfer can be evaluated using q = h(WL)(Tw1 − Tw2 ) where W is the depth into the paper. Dropkin and Sommerscales [54] presented results of an experimental study of free convective heat transfer in liquids confined by two parallel plates. The liquids used were water, silicone oils of 2 and 1000 centistokes, and mercury. The free convection heat transfer coefficient for a horizontal cavity can be estimated by the following experimental relations given by Dropkin and Sommerscales: NuL =

hL 1∕3 = 0.069RaL Pr 0.074 k

(12.215)

1.5 × 105 ≤ RaL ≤ 7.5 × 108 0.02 < Pr < 11 560 and the hot plate is below the cold plate. All the fluid properties are evaluated at average temperature T = (Tw1 + Tw2 )∕2. The end effects can be neglected if the ratio (L/H) is sufficiently small. Hollands et al. [55], based on experiments with air, recommends the following empirical correlation for horizontal enclosures: ]∗ ) ]∗ [( [ RaL 1∕3 1708 NuL = 1 + 1.44 1 − + −1 (12.216) RaL 5830 1708 ≤ RaL ≤ 108 where []* indicates that if the quantity in the bracket is negative, it should be set equal to zero. Working fluid is air. For water, Hollands et al. give the following relation: ] ]∗ [( [ ) RaL 1∕3 1708 NuL = 1 + 1.44 1 − + −1 RaL 5830 [ + 2.0

[

1∕3 RaL

]

)] ( 1∕3 1−ln RaL ∕140

140

where []* indicates that if the quantity in the bracket is negative, it should be set equal to zero. Example 12.13 Consider a rectangular enclosure containing atmospheric air, as shown in Figure 12.E13.

(12.217)

12.9 Rectangular Enclosures

H = 20 cm

Insulated

Tw2 = 27 °C Insulated

g L = 3 cm

Tw1 = 127 °C Figure 12.E13

Geometry and problem description for Example 12.13.

The bottom surface of the cavity is at 127 ∘ C while the top surface is at 27 ∘ C. Additional data for the problem are given on Figure 12.E13. We wish to determine the heat transfer. Solution T + Tw2 27 + 127 Air properties are evaluated at Tm = w1 = = 77 ∘ C = 350 K 2 2 Air properties at this temperature are: ν = 20.92 × 10−6 m2 ∕s k = 0.03 W∕m.K Pr = 0.70 α = 29.90 × 10−6 m2 ∕s β=

1 Tm

=

1 = 0.00285 350

The Rayleigh number RaL is defined as ( ) 1 9.81 × (127 − 27)(0.03)3 3 gβ(Tw1 − Tw2 )L 350 RaL = = ≈ 120 985 να (20.92 × 10−6 )(29.90 × 10−6 ) We will use Eq. (12.215) to calculate the average Nusselt number hL 1∕3 = 0.069RaL Pr 0.074 = 0.069 × (120985)1∕3 (0.7)0.074 = 3.32 W∕m2 K k ) ( ) ( k 0.03 (3.32) = 3.32 W∕m2 K h= NuL = L 0.03 Heat flux is NuL =

q′′ = h(Tw1 − Tw2 ) = (3.32)(127 − 27) = 332W∕m2

12.9.2 Vertical Rectangular Enclosure Figure 12.20 shows fluid contained between two vertical parallel plates of height H and separated by a distance L. The plates are kept at uniform temperatures Tw1 and Tw2 . The ratio (H/L) is called aspect ratio. While the right vertical surface is hot, the left surface is cold, and the horizontal top and bottom surfaces are insulated. The separation distance is L, and hot and cold walls are at uniform temperatures Tw1 and Tw2 , respectively. The enclosure makes an angle θ = 90∘ with respect to horizontal. Heat transfer across the vertical gap was investigated experimentally. The working fluid was air. Heat transfer is evaluated using q = h(WL)(Tw1 − Tw2 ) where W is the depth into the paper. Fluid properties are evaluated at the average temperature Tm = (Tw1 + Tw2 )∕2 and β = 1∕Tm for gases. The Rayleigh number is defined as g β(Tw1 − Tw2 )L3 g β(Tw1 − Tw2 ) Pr = 2 να ν The aspect ratio A is defined as RaL =

A=

H L

737

738

12 Free Convection Heat Transfer

Figure 12.20

L

Fluid circulation in a vertical cavity.

Insulated

Cold Hot Tw2

Tw1

Tw1 > Tw2

H θ = 90° Insulated

ElSherbiny et al. [56] studied heat transfer by free convection across the gap experimentally for air, covering the range of aspect ratios from A = 5 to 110 and the range of Rayleigh numbers RaL = 102 to 2 × 107 . ElSherbiny et al. recommend the following empirical correlations for the average Nusselt number for the free convection across vertical enclosure: Nu90 = Max[Nu1 , Nu2 , Nu3 ]

(12.218)

102 ≤ RaL ≤ 2 × 107 5 ≤ (H∕L) ≤ 110 where Nu1 , Nu2 , Nu3 are given below 1∕3

Nu1 = 0.0605 RaL

(12.219a) 1∕3

3 ⎧ ⎡ ⎤⎫ ⎪ ⎢ 0.104 Ra0.293 ⎥ ⎪ ⎪ ⎪ L Nu2 = ⎨1 + ⎢ )1.36 ⎥⎥ ⎬ ( ⎢ ⎪ ⎢ 1 + 6310 ⎥⎪ ⎪ ⎣ ⎦⎪ Ra L ⎩ ⎭ [ ]0.272 RaL Nu3 = 0.242 (H∕L)

(12.219b)

(12.219c)

where the Rayleigh number is based on spacing L between hot and cold walls. The Nusselt number can be defined as Nu90∘ =

q′′w L k(Tw1 − Tw2 )

All the fluid properties are evaluated at average temperature T = (Tw1 + Tw2 )∕2. Eq. (12.218) implies that one should select the maximum of three numbers Nu1 , Nu2 , and Nu3 . Example 12.14 A vertical enclosure consists of two vertical parallel glasses. Each plate is 1 m × 1 m and plates are separated by 50 mm air space at atmospheric pressure. See Figure 12.E14. The temperatures of the hot and cold surfaces are Tw1 = 400 K and Tw2 = 300 K, respectively. We wish to find the free convection heat loss from this vertical enclosure. Solution 400 + 300 The physical properties of air at Tm = = 350 K are 2 ν = 20.92 × 10−6 m2 ∕s

12.9 Rectangular Enclosures

Figure 12.E14

Geometry and problem description for Example 12.14.

Insulated L = 5 cm

Tw2 = 300 K Tw1 = 400 K H = 1m θ = 90° Insulated

k = 0.03 W∕m.K Pr = 0.7 α = 29.9 × 10−6 m2 ∕s β=

1 Tm

=

1 = 0.002857 350

The Rayleigh number RaL is gβ(Tw1 − Tw2 )L3 9.81 × (1∕350)(400 − 300)(0.05)3 = 5.6 × 105 = να (20.92 × 10−6 ) × (29.9 × 10−6 )

RaL = GrL Pr = The aspect ratio is A=

1 H = = 20 L 0.05 1∕3

Nu1 = 0.0605Ral

= 0.0605(5.6 × 105 ) = 4.98 1∕3

3 ⎧ ⎡ ⎤⎫ ⎪ ⎢ 0.104Ra0.293 ⎥ ⎪ ⎪ ⎪ L Nu2 = ⎨1 + ⎢ )1.36 ⎥⎥ ⎬ ( ⎢ ⎪ ⎢ 1 + 6310 ⎥⎪ ⎪ ⎣ ⎦⎪ RaL ⎩ ⎭ )0.272 ( RaL Nu3 = 0.242 = 3.92 A

= 5.02

We choose the largest Nusselt number NuL = Nu2 = 5.02 The average heat transfer coefficient h is ) ( ) ( W k 0.03 (5.02) = 3 2 h= NuL = L 0.05 mK Heat flux across the vertical enclosure is q′′w = h(Tw1 − Tw2 ) = 3(400 − 300) = 300 The total heat transfer across the gap is q = Aq′′w = 300 W

W m2

739

740

12 Free Convection Heat Transfer

12.9.3

Inclined Rectangular Enclosure

Air spaces in an inclined enclosure are encountered in flat-plate solar collectors (between glass cover and the absorber plate). Figure 12.21 shows fluid contained within an inclined enclosure. The separation distance is L, and hot and cold walls are at uniform temperatures Tw1 and Tw2 , respectively. The cold plate is above the hot plate. Heat transfer is evaluated using q = h(WL)(Tw1 − Tw2 ) where W is the depth into the paper. Fluid properties are evaluated at the average temperature T = (Tw1 + Tw2 )∕2 and β = 1∕Tm for gases. The Rayleigh number is defined as g β(Tw1 − Tw2 )L3 g β(Tw1 − Tw2 ) Pr = 2 να ν The enclosure makes an angle θ with respect to horizontal. The tilt angle for solar collectors may vary from 0∘ (horizontal) to about 70∘ at high-latitude applications. Heat transfer across an inclined enclosure depends on the aspect ratio (H/L), the tilt angle θ, the Prandtl number Pr, and the Rayleigh number RaL . The functional relationship may be expressed in the following form: ( ) H NuL = f Pr , RaL , , θ (12.220) L In an enclosure, fluid motion consists of a combination of roll structure of horizontal cavities and cellular structure of vertical cavities. The transition between the two types of fluid motion occurs at a critical tilt angle θ* with a change in the value of the Nusselt number NuL . Incropera et al. [1] give the tilt angles less than the critical θ* , and these values are presented in Table 12.11. When the tilt angle is reduced below a critical tilt angle θ* , then the regularly spaced convective cells shown in Figure 12.22 tend to form, provided that the Rayleigh number exceeds a critical value, RaL, c RaL =

1708 cos θ∗ The critical tilt angle is a function of the aspect ratio. For high aspect ratio, (H/L) > 12, the critical tilt angle is approximately θ* ≈ 1.22 rad ≈ 70∘ . If the Rayleigh number RaL is less than the critical Rayleigh number, RaL, c ,(RaL < RaL, c ), then the buoyancy force is sufficient to overcame the viscous force and fluid remains stagnant. Then, the problem becomes a conduction problem, and heat transfer is calculated according to RaL,c =

q′′w =

k (T − Tw2 ) L w1

L Insulation

Tw2

H

Tw1 Tw1 > Tw2 θ Insulation Figure 12.21 Table 12.11

Inclined rectangular enclosure. Critical angle for inclined enclosure.

(H/L)

1

3

6

12

>12

θ*

25∘

53∘

60∘

67∘

70∘

12.9 Rectangular Enclosures

Figure 12.22

(a) Regularly spaced convective cell. (b) Cellular flow.

Cold

Hot

Cold

Hot (a)

(b)

Above the critical Rayleigh numbers, for the large aspect ratio (H/L) > 12 and the tilt angle θ less than critical tilt angle θ* , the following empirical correlations may be used. El- Sherbiny et al. [56] correlated the experimental data with air making θ = 60∘ with horizontal and proposed the following correlations: Inclined layer (θ = 60∘ ) (NuL )θ=60∘ = Max(Nu1 , Nu2 )

(12.221)

where Nu1 , Nu2 are given ⎡ Nu1 = ⎢1 + ⎢ ⎣ G= [

(

(

0.0936 Ra0.314 L

)7

1+G 0.5 RaL 3160

1∕7

⎤ ⎥ ⎥ ⎦

(12.222a)

)20.6 ]0.1

1+ ( ) 0.175 Nu2 = 0.104 + Ra0.283 L (H∕L)

(12.222b)

0 < RaL < 107 The Nusselt number is defined as q′′ L Nu60 = k(Tw1 − Tw2 ) To determine the Nusselt number with inclination in the range 60∘ ≤ θ ≤ 90∘ , the following correlation is used (90∘ − θ)Nu60∘ + (θ − 60∘ )Nu90∘ 30∘ where θ is in degrees. Experimental data agree with Eq. (12.223c) to within 6.5%. Inclined layer (0∘ < θ < 60∘ ) Based on experimental studies with air, Hollands et al. [55] proposed the following correlation: ]∗ ]∗ [ ] [( ) [ RaL cos θ 1∕3 1708(sin 1.8 θ)1.6 1708 + 1− NuL = 1 + 1.44 1 − −1 RaL cos θ RaL cos θ 5830 ( ) H ≥ 12 L Nuθ =

(12.222c)

(12.223)

0 ≤ RaL ≤ 105 The Nusselt number is defined as q′′w L Nuθ = k(Tw1 − Tw2 ) where q′′W is the heat flux (W/m2 ) and []* indicates that if the quantity in the bracket is negative, it should be set equal to zero. This implies that if the Rayleigh number RaL is less than the critical RaL, c , there is no fluid motion within the cavity, as discussed in [1]. It is reported that the experimental data with air agree with Eq. (12.223) to within 5% and Eq. (12.223) may be used up to 75∘ .

741

742

12 Free Convection Heat Transfer

Table 12.12

Constants C and m for Eq. (12.224).

𝛉 (degree) and range of RaS

C

m

0 and 1.5 × 105 < RaL < 7.5 × 108 30∘ and 1.5 × 105 < Ra < 2.5 × 108

0.069

0.074

0.065

0.074

45∘ and 1.5 × 105 < RaL < 2.5 × 108 60∘ and 1.5 × 105 < Ra < 2.5 × 108

0.059

0.074

0.057

0.074

90∘ and 5 × 104 < RaL < 2.5 × 108

0.049

0.074

L

L

Dropkin and Sommerscales [54] performed experiments to investigate convective heat transfer in liquids confined by two parallel plates and inclination at various angles with respect to horizontals. The liquids used were water, silicone oils, and mercury. They proposed the following correlation: NuS = C(RaS )1∕3 Pr m

(12.224)

0.02 < Pr < 11560 The constant C is a function of angle of inclination and is given in Table 12.12. It is reported that the maximum uncertainty for the Nusselt number was ±10%. Buchberg et al. [57] recommend the following three-region semiempirical correlation to estimate the free convective heat transfer coefficient: )+ ( 1708 NuL = 1 + 1.446 1 − (12.225a) RaL cos θ 1708 ≤ RaL cos θ ≤ 5900 The + bracket goes to zero when it is negative. NuL = 0.229(RaL cos θ)0.252

(12.225b)

5900 ≤ RaL cos θ ≤ 9.23 × 104 NuL = 0.157(RaL cos θ)0.285

(12.225c)

9.23 × 104 ≤ RaL cos θ ≤ 106 Example 12.15 A solar collector consists of two parallel glasses. Each plate is 1 m × 1 m and the absorber plate and the glass cover are separated by 15-mm air space at atmospheric pressure. Solar collector is inclined with respect horizontal with an angle of 45∘ . See Figure 12.E15. We wish to find the free convection heat loss from this solar collector. Solution Tw1 + Tw2 60 ∘ C + 30∘ = = 45 ∘ C = 318 K 2 2 Air properties are evaluated at this temperature Tm =

ν = 17.7 × 10−6 m2 ∕s k = 0.0276 W∕m.K Pr = 0.704

12.10 Horizontal Concentric Cylinders

Tw2 = 30 °C

L = 15 mm

Glass cover

Insulation H = 1m Tw1 = 60 °C θ = 45°

Absorber plate

Insulation Figure 12.E15

Geometry and problem description for Example 12.15.

α = 25.16 × 10−6 m2 ∕s β=

1

=

Tm

1 = 0.00314 318

First, we calculate the Rayleigh number RaL as RaL =

gβ(Tw1 − Tw2 )L3 9.81 × (0.00314) × (333 − 303)(0.015)3 Pr = × 0.704 = 7012 2 ν (17.7 × 10−3 )2

The aspect ratio A is H 1000 mm = = 66.66 A= L 15 mm A > 12 and we find from Table 12.11 that the critical tilt angle is nearly 70∘ . Thus, we find that the tilt angle is less than the critical tilt angle. We may use Eq. (12.223) ]∗ [ ] [ 1708(sin 1.8 θ)1.6 1708 1− NuL = 1 + 1.44 1 − RaL cos θ RaL cos θ [( ] ∗ )1∕3 RaL cos θ + −1 5830 ] [ ]∗ [ 1708(sin 1.8 45∘ )1.6 1708 1− NuL = 1 + 1.44 1 − 7012 × cos 45∘ 7012 × cos 45∘ [( ] ∗ ) 1∕3 7012 × cos 45∘ + −1 5830 The third term is negative, and we drop it NuL = 1.62 W∕m2 K ) ( W k 0.0276 (1.62) = 2.98 2 NuL = L 0.05 mK The heat flux across the gap is h=

q′′w = h(Tw1 − Tw2 ) = 2.98(60 − 30) = 89.40

W m2

12.10 Horizontal Concentric Cylinders Raithby and Hollands [58] have studied the free convection heat transfer in the annular space between long, horizontal concentric cylinders. We assume that the inner cylinder is heated (Tw i > Tw 0 ). Assume that both the cylinders have a length L. Heat transfer is 2 π L keff (Ti − T0 ) q= ln(r0 ∕ri )

743

744

12 Free Convection Heat Transfer

where ri is the radius of the inner cylinder and r0 is the radius of the outer cylinder. The effective conductivity keff , the critical length Lc , and the critical Rayleigh number Rac are defined ( )1∕4 keff Pr 1∕4 = 0.386 Rac (12.226) k Pr +0.861 0.7 ≤ Pr ≤ 6000 Properties are evaluated the mean temperature Tm = (Ti + To )∕2. Here, Ti is the temperature of the inner cylinder and T0 is the temperature of the outer cylinder Rac =

g β (Ti − To )L3c αν

2[ln(r0 ∕ri )]4∕3 Lc = ( )5∕3 −3∕5 −3∕5 + r0 ri where this equation can be used for the range 0.7 ≤ Pr ≤ 6000 and Rac < 107 . Fluid properties are evaluated at Tm = (Ti + T0 ). k Take keff = k if eff < 1 k

12.11 Concentric Spheres Raithby and Hollands [58] have also studied the free convection heat transfer in the annular space between concentric spheres. We assume that inner cylinder is heated (Tw i > Tw 0 ). Heat transfer is q=

4 π keff (Ti − T0 ) (1∕ri ) − (1∕r0 )

where ri is the radius of the inner sphere and r0 is the radius of the outer sphere. The effective conductivity keff and Rayleigh Ras and Ls are given below ( )1∕4 keff Pr 1∕4 Ras (12.227) = 0.74 k Pr + 0.861 g β (Ti − To )L3s αν ]4∕3 [ 1 1 − ri r0 Ls = ( )5∕3 −7∕5 −7∕5 21∕3 ri + r0

Ras =

where this equation can be used for the range 0.7 ≤ Pr ≤ 4000 and Rac < 104 . Fluid properties are evaluated at Tm = (Ti + T0 )/2. Here, Ti is the temperature of the inner sphere and T0 is the temperature of the outer sphere. k Take keff = k if eff < 1. k

12.12 Spheres Free convection around a sphere is similar to natural convection around the cylinder. Natural convection from a sphere to a fluid can be formulated starting with the equations of motion and energy. We rely on experimental correlations to estimate the free convection from spheres. The following correlation was proposed by Churchill [59] for spheres in fluids of Pr ≥ 0.7 and for RaD ≤ 1011 . 1∕4

0.589RaD NuD = 2 + [ ) ]4∕9 ( 0.469 9∕16 1+ Pr

(12.228)

Problems

RaD ≤ 1011 Pr ≥ 0.7 Tw = const where the characteristic length, D, is the sphere diameter. All the fluid properties are evaluated at the film temperature Tf .

Problems 12.1

Two vertical plates are suspended in water, as shown in Figure 12.P1. The plates are initially at temperature 53∘ C. What is the minimum distance between plates to prevent interference between their natural convection boundary layers?

Tw = 53 °C

Tw = 53 °C

Quiescent water T∞ = 21 °C

H = 20 cm

x

L 0

Figure 12.P1

Free convection boundary layers between vertical parallel plates.

12.2

Consider a 40-cm-high and 1-m-wide vertical plate. The plate has a surface temperature of 138 ∘ C and is suspended in quiescent air at 1 atm and 17 ∘ C. Using the integral solution, estimate the following quantities: (a) Maximum velocity at 20 cm from the leading edge of the plate (b) Velocity boundary layer thickness at 20 cm from the leading edge of the plate (c) Local heat transfer coefficient at 20 cm from the leading edge of the plate.

12.3

A 0.5 m by 0.5 m square plate is suspended vertically in quiescent air. The back side of the plate is insulated. The plate is maintained at a temperature of 410 K, and quiescent air is at 290 K and 1 atm. We wish to calculate the average heat transfer coefficient and heat loss from the plate. Use the correlation of Mc Adams and Churchill and Chu.

12.4

A vertical wall is covered by a thin square panel. One side of the panel is 3 m. Assume that wall acts as insulation on the back of the panel, and the panel is exposed to a solar radiation flux of 1450 W/m2 . The panel has an absorptivity of αs = 0.90 for solar radiation. The quiescent air is at atmospheric pressure, and it has a temperature of 300 K. We wish to estimate the panel surface temperature.

12.5

Consider a circular plate having a diameter of 0.8 m. The bottom surface of the plate is insulated and the top surface of the plate is maintained at 400 K. The circular plate is in quiescent atmospheric air, and air temperature is 300 K. Calculate the average heat transfer coefficient and heat loss assuming that the plate is horizontal and hot surface faces up.

12.6

Consider a circular plate having a diameter of 2 m. One surface of the plate is insulated, and other surface of the plate is subjected to a solar radiation flux of 1000 W/m2 . The circular plate is in quiescent atmospheric air, and air temperature is 300 K. The circular plate makes an angle −60∘ with the vertical, as illustrated in Figure 12.P6.

745

746

12 Free Convection Heat Transfer

Calculate the equilibrium temperature of the plate assuming hot surface faces up.

Figure 12.P6

Hot surface

Geometry and problem description for Problem 12.6.

θ = –60°

Disk

12.7

A large tank contains engine oil at 17∘ C. A horizontal circular electrical heater is inserted into the tank to heat the engine oil. See Figure 12.P7. The heater has a diameter of 3 cm and a length of 1 m. The surface temperature of the electrical heater is 137∘ C. We wish to estimate the heat transferred to the engine oil. Figure 12.P7

Heater in a tank.

Engine oil D = 3 cm Heater L = 1m

12.8

A vertical enclosure is formed using two vertical parallel square plates, each 1 m by 1 m, and enclosure is filled by atmospheric air. The separation distance between the plates is 3 cm. One plate is kept at 254 ∘ C and the other plate is kept at 100 ∘ C. We wish to calculate the heat transfer by free convection.

12.9

A vertical cylinder has 5-cm diameter and 30-cm height. The side surface temperature of the cylinder is 45∘ C and ambient fluid is water at 1 atm pressure and 5∘ C. Determine the heat lost by free convection from the cylinder.

12.10

A circular plate of 0.50-m diameter is placed horizontally in the water tank. The upper side is insulated, and the lower side is at 95∘ C. The hot side of the plate is facing downward. The water temperature is 17∘ C. Estimate the rate of heat transfer.

Tw2 = 40°

L

L = 3 cm Insulation

H=2m Tw1 = 100 °C θ = 30° Insulation Figure 12.P11

Geometry and problem description for Problem 12.11.

Problems

12.11

A 2-m-long and 1-m-wide rectangular enclosure is inclined at an angle of 30∘ with respect to horizontal, as shown in Figure 12.P11. Atmospheric air is contained in the enclosure. Determine the heat transfer from the hot plate to cold plate.

12.12

Fins have been added to a vertical wall for cooling purposes. Twenty-one vertical fins have been added to the wall. The wall is 20 cm by 20 cm. The fins are 1 mm thick, 20 cm high, and 2 cm wide. They are equally spaced. The fins are at 400 K and air is at 300 K. We wish to calculate the heat loss from the fins.

12.13

Consider a double-pane window consisting of two sheets of glass. The separation distance between the glass sheets is 2 cm. Window is 0.75 m high and 1 m wide. Glass surface temperatures across the air gap are measured to be 17∘ C and −2∘ C. See Figure 12.P13. We wish to estimate the heat loss by free convection from window.

Figure 12.P13

Geometry and problem description for Problem 12.13.

H = 75 cm Tw1 = 290 K

Tw2 = 271 K

L = 2 cm

12.14

Consider the piping system shown in Figure 12.P14. The external diameter of the pipe is 30 cm. Dimensions of the piping system are shown in Figure 12.P14. The temperature of the outside surface of the piping system is 35 ∘ C and that of atmospheric air is 20 ∘ C. We wish to estimate the heat loss from the piping system. Neglect elbow.

Figure 12.P14

Geometry and problem description for Problem 12.14.

2m D = 30 cm

2m

12.15

A spherical thermocouple bead has a diameter of 3 mm. Estimate the maximum power dissipation if the surface temperature of thermocouple bead is not to exceed 130 ∘ C when the air temperature is 22 ∘ C. Consider only heat loss by free convection.

12.16

Consider a vertical channel having a circular cross section with a radius of R. Surface temperature is Tw and the height of the channel is H. Flow is laminar and both thermally fully developed and hydrodynamically fully developed. See Figure 12.P16. Assume that Pr ≥ 1. Determine (a) velocity distribution and (b) Nusselt number.

747

748

12 Free Convection Heat Transfer

Figure 12.P16

Geometry and problem description for Problem 12.16.

R

H Tw

x r

p∞ T ∞

12.17

Churchill and Usagi [9] tested and presented an equation that fits to experimental data over a wide range of Rayleigh numbers, Prandtl numbers, and geometries. This expression can be put in the following form: √

NuL =

√

1∕6

0.387RaL Nu0 + [ ( )9∕16 ]8∕27 0.5 1+ Pr

Geometry

Characteristic length L

Nu0

Vertical cylinder

Height of cylinder L

0.68

Compare this correlation with available experimental data and with other correlations you find in the literature. 12.18

Arpaci et al. [32] have proposed Eq. (12.171) for a vertical cylinder and plate subjected to constant surface temperature Tw in free convection Compare this correlation with other experimental correlations you find in the literature.

12.19

Arpaci et al. [32] have proposed the following correlation for a horizontal cylinder subjected to constant surface temperature Tw in free convection: Nu =

hL = C πnN k

RaL πN = [ ( )] C0 1+ Pr 0 < Pr < ∞

Problems

Horizontal cylinder Diameter of cylinder is the characteristic length

C0

C

n

104 ≤ πN ≤ 109

0.559

0.518

1/4

10 ≤ πN ≤ 10

0.559

0.15

1/3

9

12

Compare this correlation with other experimental correlations you find in the literature. 12.20

Arpaci et al. [32] have proposed the following correlation for a sphere subjected to constant surface temperature Tw in free convection: hL = C πnN k RaL πN = [ ( )] C0 1+ Pr Nu =

0 < Pr < ∞ Sphere Diameter of sphere is the characteristic length

C0

C

n

104 ≤ πN ≤ 109

0.469

0.589

1/4

10 ≤ πN ≤ 10

0.469

0.15

1/3

9

12

Compare this correlation with other experimental correlations you find in the literature. 12.21

A UHF of 2200 W/m2 is applied to a vertical plate by heating electrically. The plate is 20 cm high and 10 cm wide. We wish to determine the temperature distribution over the plate. Both sides of the plate are cooled by free convection. The ambient air temperature is 40 ∘ C.

12.22

In many practical engineering problems, surface temperature is nonuniform. Similarity solutions exist for certain surface temperature distributions Tw (x). Sparrow and Gregg [15] have shown that for Tw (x) − T∞ = Axn similarity solutions exist. In this problem, the effect of nonuniform surface temperature upon free convection for a vertical plate will studied. The following similarity variables are defined: ) ( Grx 1∕4 ψ = 4νF(η) 4 ( )1∕4 y Grx η= x 4 gβ(Tw − T∞ ) Grx = ν2 T − T∞ θ= Tw − T∞ It is assumed that θ = F(η). The following differential equations for momentum and energy equations are obtained: F′′′ + (n + 3)F F′′ − (2n + 2)(F′ )2 + θ = 0 θ′′ + Pr [(n + 3)F θ′ − 4 nF′ θ] = 0

749

750

12 Free Convection Heat Transfer

The boundary conditions are F(0) = F′ (0) = 0 θ(0) = 1 F′ (∞) = 0 θ(∞) = 0 (a) Show that heat flux at the wall is

q′′w = [−θ′ (0)]

k(Tw − T∞ ) x

(

Grx 4

)1∕4

(b) Show that the Nusselt number is

Nux =

q′′w x −θ′ (0) = √ (Grx )1∕4 k(Tw − T∞ ) 2

Using bvp4c in MATLAB 2021a or with any other software, complete the following table. Compare your results with the solution of Sparrow and Gregg [15]. Nux ∕Gr1∕4 x N

Pr 1

4

6

10

4 3

0.7686

2

0.697

1

0.595

0.5

0.519

0.2 0 −0.2 −0.5

12.23

A cylindrical electric heater is immersed horizontally in water tank. The water temperature is 27 ∘ C. The rating of the electrical heater is 1000 W. The electrical heater is 20 mm in diameter, and it has 400-mm length. We wish to estimate the surface temperature of heater under steady conditions.

12.24

Sometimes it is possible to use simplified expressions for quick but approximate estimation of average heat transfer coefficient from isothermal vertical plates in air at atmospheric pressure and moderate temperatures. If we write ) ( )1∕4 ( ( ) [ gβ(T − T )L3 ]1∕4 Tw − T∞ 1∕4 gβ ν k w ∞ h = C1 = C k 1 L α να L ν2 If we now select average property values, average heat transfer coefficient h can be expressed as ) ( Tw − T∞ 1∕4 h = C2 L

Problems

One such correlation for laminar flow is reported by Giedt [76] ) ( T − T∞ 1∕4 BTU h = 0.28 w L hr − ft2 ∘ F Consider free convection from an isothermal vertical square flat plate in a quiescent constant property fluid. The fluid is air. One side of the plate is L = 0.1587 m. The hot plate is at temperature Tw and undisturbed fluid is at constant temperature T∞ = 300 K. Following data were collected as discussed in Onur and Hewitt [77]. 𝚫T = Tw − T∞ (K)

h(W∕m2 K)

3.19

3.09

5.60

3.24

7.94

3.70

10.48

4.06

12.62

4.16

15.45

4.26

17.74

4.46

Plot this data and find a correlation in the form h = m ΔTn where m and n are constants to be determined and compare it with correlation reported by Giedt. 12.25

It is reported that Eq. (12.147) may be used for inclined plates. For this purpose, we may replace g by g cos(φ) in RaL for heated surface up or cooled surface down as discussed in Fuji and Imura [28] and notice that inclination is measured with respect to vertical. Consider free convection from an isothermal square flat plate in a quiescent constant property fluid. The plate is inclined at an angle of 45∘ from vertical. The fluid is air. One side of the plate is L = 0.1587 m.The hot plate is at temperature Tw and undisturbed fluid is at constant temperature T∞ = 300K. Following data are collected by Onur and Hewitt [77]. 𝚫T = Tw − T∞ (K)

h(W∕m2 K)

3.19

3.09

2.96

3.05

5.41

3.72

7.87

3.91

10.66

3.95

12.93

4.14

14.75

4.25

17.48

4.36

Test Eq. (12.147) with this data. 12.26

Consider free convection from a vertical flat plate in a quiescent constant property fluid. The fluid is air. Aluminum flat plate, 4 in. by 16 in, was used as the test plate. The plate was heated through the use of an electro-film resistance heating element applied on one side of the plate. Temperatures were measured using chromel-alumel thermocouples. Following data are collected by Rich [29]. Here, h is local heat transfer coefficient and Nu is the local Nusselt number.

751

752

12 Free Convection Heat Transfer

x (in)

T∞ (∘ F)

Tw (∘ F)

Grx

h(BTU/hr ft2∘ F)

Nu

3

80.2

283.0

4.4 × 106

1.36

16.0

6

79.4

305.5

37.8 × 106

1.14

27.8

6

10

76.6

315.0

178 × 10

0.98

39.8

13

74.5

327.8

424 × 106

0.87

44.5

a) b) c) d)

Test Eq. (12.44b) with this data. Test Eq. (12.48) with this data. Test Eq. (12.147) with this data. Eckert (78) gives the following equation for local Nusselt number. √ 0.508Pr 1∕2 4 RaL Nu = [0.952 + Pr ]1∕4 Test this equation with the given data.

References 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

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755

Index a acceleration convective change 8, 10 local change 8, 10 algebraic turbulence models 508 see also zero-equation turbulence models analogies Chilton–Colburn 631 Kadar and Yoglom 636 Martinelli 639–641 Prandtl–Taylor 531–535, 631–633 Reynolds 120, 529–531, 629–631 von Karman 535–539, 633–636 Yu et al. 637–638 apparent friction factor 322, 323 apparent heat flux 523 apparent shear stresses 474, 482 arbitrary wall heat flux distribution in axial direction over flat plate 270–275, 289–293 over tubes 443–446 arbitrary wall temperature distribution in axial direction over flat plate 266–272, 284–289 over tubes 443–446 Archimedes principle 114 aspect ratio, laminar free convection 724, 735 average friction coefficient 204, 516, 517, 519–521 average friction factor 12, 316, 322, 519, 593 average Nusselt number 15, 129, 132, 727, 733, 219, 336, 354, 532, 540, 736 average velocity 589, 129, 159, 320, 322, 337 see also mean velocity axial heat conduction in internal flow 625, 618 axisymmetric body, boundary layer over 212, 297

b Bessel functions differentiation of 348 first and second kind 339 roots of 340, 347 beta function 556

Blasius solution 191, 204, 214 see also similarity solution 187 blunt bodies, flow past 200 boundary layer approximations, laminar flow momentum boundary layer on flat plate boundary layer concept 70–71 boundary layer thickness 71, 77 continuity equation 33, 38, 41, 72 scaling concept 71–76 thermal boundary layer on flat plate boundary layer concept 76 heat transfer coefficient 14, 15, 21, 79, 119, 122, 128, 220, 227, 235, 371 incompressible laminar flow 76 scaling concept 77–81 thermal boundary layer thickness 17, 77 viscous dissipation 76, 81 turbulent boundary layer closure problem eddy diffusivity of heat 482–483 eddy diffusivity of momentum 481–482 eddy shear stress 482 turbulent Prandtl number 483 continuity 474–475 energy equation energy boundary layer equation 478–480 fluctuations 467–470 intensity 469, 467 momentum equation momentum boundary layer equation 475–478 Reynolds averaging 465–472 transition to turbulent flow 496–498 velocity profiles power law velocity profile 488–489, 495–496 universal velocity profiles buffer layer 486, 488, 494 viscous sublayer 488, 494 wake region 495 body forces 35, 42, 43 boundary conditions 55–727

®

Introduction to Convective Heat Transfer: A Software-Based Approach Using Maple and MATLAB , First Edition. Nevzat Onur. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/introtoconvectiveheattransfer

756

Index

Boussinesq approximation 694–724 buoyancy balanced by friction 678 balanced by inertia 678 Buckingham 𝜋 theorem 102–103 buffer layer 488, 494 see also turbulent flow bulk temperature 21, 337, 377, 385, 408, 602, 643

c Cartesian coordinate system 1, 45–46 characteristic value problem 409–410, 431 Chilton–Colburn analogy see analogies closed system analysis 36 closure problem 480–481 coefficient of thermal expansion 57–58, 114, 123, 674, 675, 692 Colburn correlation 531, 645 combined entrance region circular tubes 337–355 parallel plate channels 355–371 complimentary error function 227 conduction heat transfer Fourier’s law 10–11 conduction thickness see integral method conservation of energy equation in cylindrical coordinates 59 in rectangular coordinates 59 in spherical coordinates 59 conservation of mass 33–34, 38–42, 85–87 conservation of momentum equation in cylindrical coordinates 46 in rectangular coordinates 45 in spherical coordinates 46–47 continuity (conservation of mass) equation in cylindrical coordinates 41 in rectangular coordinates 38 in spherical coordinates 41 continuum 2–3, 467 control surface 23, 26, 37–39 control volume 34–38, 44–45 control volume analysis 34–38 convection 93 see also external flow, internal flow, natural or free convection convective heat transfer coefficient 14, 93, 100, 156, 702 convective terms 58, 523 correlations 128–136 see also external flow correlations, internal flow correlations, natural flow correlations Couette flow 151–156 Couette flow approximation 494, 495, 510–511 critical Reynolds number 69, 313, 466, 470, 520, 539 cylindrical coordinates 40–41, 46, 59, 483–484

d Darcy friction factor 316, 320, 325, 485, 589, 591 derivative material 5, 8, 10 substantial 4–10 total 8 diameter, hydraulic 156, 313, 325, 326, 357, 363, 644, 603, 643 differential equations, solving with Maple 153, 154, 160, 164, 165, 169, 189, 215, 233, 241, 244, 339, 366, 369, 390, 392, 410, 421, 425, 432, 433 diffusivity eddy 481–482, 521–522 heat 482–483 thermal 18, 56, 58, 78, 108, 110, 115, 125, 364, 372, 483, 730, 735 dimensional analysis homogeneity 102 Ipsen method 106–115 nondimensionalization, differential equations boundary layer, forced convection 116- 122 boundary layer, free convection 122–125 tube flow, forced convection 120–122 Rayleigh method 103–104 repeating variables method 105 dimensionless groups interpretation 97 dimensionless numbers 125–127 dimensionless parameters Brinkman number 127 drag coefficient 192 Eckert number 127 friction coefficient 126 friction factor 315, 316, 320, 322 Graetz number 127 Grashof number 127 Knudsen number 2–3 Nusselt number 126 Peclet number 126 Prandtl number 126 Rayleigh number 127, 697 Reynolds number 29, 69, 75, 125, 104, 108, 109, 111, 117, 119, 122, 156, 186, 313, 320, 420, 466, 470, 496, 516 Stanton number 126 dimensions of common variables, table 98–100 displacement thickness, momentum 88–89 dissipation in boundary layers 81 in Couette flow 151–156 function 54, 153, 169 Dittus–Boelter equation 652, 646 duct flow see internal flow Duhamel’s superposition integral, xvii dynamic viscosity 45

Index

e Eckert number 127 Eddy diffusivity of heat 482–483 see also turbulent flow Eddy diffusivity of momentum 481–482 see also turbulent flow Eddy temperature fluctuation 467–470 see also turbulent flow Eddy viscosity 482, 577, 480 eigenfunctions 340, 413 eigenvalues 393–394, 411–413, 422–423, 433–434, 620, 628 empirical correlations for forced convection between parallel plates 370–371, 436–437 for forced convection in tubes 350–355, 415–418 for forced convection on flat horizontal surfaces 217–219, 233–234, 529–564 for free convection on horizontal surfaces 710–712 see also free convection free convection in horizontal enclosures 733–735 see also free convection, 733–741 empirical equations see external flow; internal flow energy conservation of energy 36–38, 47–53 internal 54 kinetic 467 potential 37 energy balance 331 energy equation in differential form for boundary layers 42 in cylindrical coordinates 59 for infinite parallel plates 41–42 in spherical coordinates 59 for tube flow 121 energy generation 51 enthalpy thickness 93 entrance region circular duct prediction of friction coefficient, in laminar flow 322–323 parallel plate channel prediction of friction coefficient, in laminar flow 325–326 entrance region, thermal 387 entry length thermal 327–328, 400 velocity 315, 318, 400 equation of motion Cartesian coordinates 45–46 cylindrical coordinates 46 spherical coordinates 46–47 equilibrium, thermodynamic 2–3 error function 227

Eulerian viewpoint 5–7 external laminar boundary layer, forced flow momentum boundary layers Blasius solution 189–191 integral methods, application axisymmetric body 212 flat plate 201–202, 203–204 Pohlhausen parameter 205 Thwaites’s method Holstein and Bohlen parameter 207–208 shape parameter 207 shear correlation 207 shear thickness 207 2D flow with pressure gradient 204–207 similarity solution flat plate average skin friction coefficient 192 boundary layer thickness 190–191 displacement thickness 192 drag 192 local skin friction coefficient 191 momentum thickness 192–193 wedge flow (Falkner Skan Flow) 195–199 thermal boundary layers correlations, empirical 220 effect of property variation 252–253 integral methods, application flat plate arbitrary wall heat flux distribution, 272–277, 289–293 stepwise variation in surface heat flux 282–284 arbitrary wall temperature distribution, 266–272, 284–289 stepwise wall temperature distribution 278–282 unheated starting length, UHF boundary condition, 262–265 unheated starting length, UWT boundary condition 256–262 one parameter integral method 293–298 similarity solution flat plate average heat transfer coefficient 220, 219 average Nusselt number 220, 219 empirical correlations 220, 221 high Prandtl number effect 228–230 local heat transfer coefficient 217 local Nusselt number 220, 217 low Prandtl number effect 225–228 uniform heat flux 230–237 uniform wall temperature 212–225 variable wall temperature 237–238 wedge flow 249–252

757

758

Index

external laminar boundary layer, forced flow (contd.) slug flow arbitrary wall heat flux distribution 272–278 arbitrary wall temperature distribution 266–272 superposition method 245–248 external turbulent boundary layer, forced flow momentum boundary layer Couette flow approximation inner region 510–511 buffer region 512 turbulent region 512–513 viscous sublayer 511–512 wake region 513 other velocity models 514 universal velocity profile 511–514 integral method, application average friction coefficient 516–517 local friction coefficient 516–517 other eddy diffusivity models 521–522 zero equation models, algebraic equation model Prandtl’s mixing length model 508–509 Van Driest Model 509–510 thermal boundary layer analogy between momentum and heat transfer 529–539 see also analogies correlations, empirical 539 integral method, application arbitrary wall heat flux distribution 554, 555 average heat transfer coefficient 554 local heat transfer coefficient 556 unheated starting length, UHF boundary condition 553–55 unheated starting length, UWT boundary condition 542–549 turbulent transition, overall heat transfer 558–564 uniform heat flux the law of the wall 498, 512, 524 local Nusselt number 553

f Falkner–Skan transformation 195–201, 249–252 Fanning friction factor 316, 582–585 film coefficient see heat transfer coefficient film temperature 14, 220, 222, 337 first law of thermodynamics body forces 52 enthalpy 55 flow work 49 internal energy 54 kinetic energy 54 local specific energy 47 potential energy 52

rate of work, power 49–50 specific energy 47 surface forces 43 flat plate see external flow, forced convection and free convection flow over axisymmetric bodies approximate solution 212 fluctuations temperature 474 velocity 467, 470, 474 fluid motion Eulerian approach 5–7 Lagrangian approach 4–5 forced convection definition 11 forces buoyancy 675, 680 gravity 103 inertia 103 pressure 103 viscous 103 Fourier’s law of heat conduction 10 free convection, heat transfer boundary layer, laminar flow boundary layer equations 677 integral method, application vertical plate UHF boundary condition average Nusselt number 701 local Nusselt number 701 UWT boundary condition average Nusselt number 699 local Nusselt number 699 scaling 679–680 similarity solutions vertical plate 681 UHF boundary condition average Nusselt number 691 local Nusselt number 689–690 UWT boundary condition average Nusselt number 684–685 local Nusselt number 682 boundary layer, turbulent flow integral method, application vertical plate 702–704 correlations, empirical concentric cylinders 743–744 enclosures horizontal enclosure 735–737 inclined enclosure 740–743 vertical enclosure 737–739 plate channels

Index

horizontal parallel plate channel 731–732 inclined parallel plate channel 732–735 vertical parallel plate channel 725–731 plates horizontal plate 712–715 inclined plate 715–719 vertical plate 704–711 plates cylinders horizontal cylinder 722–723 inclined cylinders 723–724 vertical cylinder 719–722 spheres 744–745 vertical thin wires 724 definition 675 free stream velocity 256–265 friction coefficient 126, 191, 192 friction factor in duct flow 315 fully developed pipe flow 317 hydrodynamic entrance region 318 Moody friction factor 316–317 friction velocity 485 fully developed flow pipe or channel flow 323–326 fully developed temperature 626–627

g gasses 58 Gnielinski correlation 649–650 Graetz number definition 127 functions 127 Graetz problem definition 388 dimensionless variables 389 correlations, empirical 415 eigenfunctions 390 eigenvalues 390 laminar flow arbitrary heat flux distribution 443 arbitrary temperature distribution 438 eigenvalues and constants 392 infinite series solutions 390 uniform heat flux 406 uniform wall temperature 388 local Nusselt number 397–398 mean Nusselt number 398, 399 Peclet number 389 slug flow uniform heat flux 272–277, 345–350, 367–370 uniform wall temperature 266–272, 337–344, 355–358, 363–367

Sturm-Liouville system 339, 390, 619 turbulent flow eigenvalues and constants 620 infinite series solutions 619 uniform heat flux 625 uniform wall temperature 618 Grashof number dimensionless parameters 127 modified 690–692, 707–708

h heat flux 230–237, 243, 262–265, 272–277, 282–284, 289–293, 345–350, 353–354, 358–363, 367–370, 371, 372–378, 383–385, 406–415, 417, 428–436, 437, 443–446, 625–629, 686–693, 698–700 heat generation 479 heat transfer coefficient, average 14–15 coefficient, local 13–14 conduction 10–11 external free convection, flat plate 725–735, 679–710 external laminar flow over a wedge 249–252 external laminar flow, parallel to flat plate 212–248 external turbulent flow, parallel to flat plate 542–564, 522–528 fully developed laminar flow in tubes 372–382 fully developed laminar flow in parallel plates 382–385 fully developed turbulent flow in tubes 605–617 fully developed turbulent flow in between parallel plates 598–605 internal free convection 733–742 internal flow 19–22 radiation 22 turbulent transition region 558–564, 654–660 horizontal flat plate, free convection on 712–715 hydraulic diameter of duct 315 hydrodynamic entrance length 314–315, 318 hydrodynamic entrance region 314, 317

i inclined flat surface see free convection inclined rectangular enclosure, free convection see free convection incompressible laminar external boundary layer flows with pressure gradient 204–207 axisymmetric body 212 Falkner–Skan equation 195–201 laminar velocity boundary layer boundary condition 184 boundary layer thickness 184–185, 190 continuity equation 184 momentum equation 184

759

760

Index

incompressible laminar external boundary layer (contd.) momentum thickness 192–193 Reynolds number 184 similarity variable 183, 187 stream function 196 velocity profile 184, 185, 191 local shear stress 209, 191 skin friction coefficient 211, 191 stream function 196 wedge flow 249–252, 195–199 semi-infinite flat plate 183–195, 212–225 thermal boundary layer dimensionless temperature 213 film temperature 220 flat plate 220 kinematic viscosity 75 local heat flux 216 local heat transfer coefficient 217 Newton’s law of cooling 328 Nusselt number, average 219 Nusselt number, local 217 Prandtl number 217 similarity solution 79, 80, 183 thermal boundary layer thickness 82 viscous dissipation 212 inertial system 1 inlet temperature 327, 376, 377, 387, 388, 389, 402, 438, 625, 645 inner region see turbulent flow integral formulation integral form of energy equation conduction thickness 93 control volume 91–92 convection conductance 93 enthalpy thickness 93 heat transfer coefficient 93 integral form of momentum equation displacement thickness 88–89 laminar flow flat plate 94 over body of revolution 91 momentum thickness 89–90 turbulent flow, flat plate 90 integral method, heat transfer free convection heat transfer laminar flow, flat plate 693–699 turbulent flow, flat plate 700–702 laminar forced convection heat transfer flat plate arbitrary varying wall heat flux 272–276, 289–291 arbitrary varying wall temperature 266–272, 284–289

uniform heat flux 230–237 , 253–254 uniform wall temperature 212–230 pipe flow arbitrary varying wall heat flux 443–446 arbitrary varying wall temperature 438–443 turbulent forced convection heat transfer flat plate, uniform heat flux 553–554 Stanton number 94 integration with MATLAB 17 internal laminar flow heat transfer, forced convection average velocity 159, 320, 325 axial conduction 150, 338, 345 bulk temperature 21, 328, 337, 408, 430 correlations, empirical, combined entrance region 350–355, 370–371, 641–644 correlations, empirical, thermal entrance region 415–419, 436–437 Couette flow 151–156 energy balance 331, 343, 373, 376, 377, 398, 408, 430, 442 entrance region 149, 326–327 fully developed conditions 329–331 fully developed heat transfer circular tube, 372–382 parallel plate 382–387 Graetz problem, thermal entrance region, viscous flow circular tube arbitrary surface heat flux distribution 443–446 arbitrary surface temperature distribution 438–443 Leveque solution 401–406 UHF boundary condition 406–414 UWT boundary condition 388–401 parallel plate, thermal entrance, viscous flow UHF boundary condition 428–436 UWT boundary condition 419–428 heat transfer coefficient 330–336, 341, 349, 357, 362, 370, 374, 377, 382, 397, 405, 413 log mean temperature difference 334–335 slug flow circular tube, entrance region 337–350 parallel plate, entrance region 355–370 thermal entry length 328, 335, 387, 398 thermally fully developed flow 334 372- 387 momentum transfer circular tubes apparent Fanning friction factor, 322 apparent Moody friction factor 322, 323 entrance length 315 entrance region 314 Fanning friction factor 316, 321 fully developed flow 318–323 Moody friction factor 316–317, 320

Index

velocity profile 318, 320 parallel plates apparent Fanning friction factor 325, 326 entrance length 318 entrance region 317 Fanning friction factor 325 fully developed flow 323–326 Moody friction factor 325 velocity profile 324 internal turbulent flow correlations, empirical Colburn correlation 645 Dittus–Boelter correlation 646 Gnielinski equation with modification 649–650 Hausen correlation 647 Nusselt correlation 651 Petukhov correlation 647–649 Sieder–Tate correlation 646 Sleicher and Rouse correlation 650–651 variable property effects 660 Graetz problem analogies 629–641 combined entrance region 641–642 circular pipe UHF boundary condition dimensionless axial distance 625 dimensionless energy equation 626 eigenvalues 628 Nusselt number 627 UWT boundary condition dimensionless axial distance 618 dimensionless energy equation 388, 618 eigenvalues 390, 620 Nusselt number 619 heat transfer, forced convection circular tubes, HFD and TFD mean temperature 606, 607 Nusselt number 612 power law, temperature 611 power law, velocity 611 temperature law of the wall 602 UHF boundary condition 605 parallel plates, HFD and TFD mean temperature 602 Nusselt number 603–604 temperature law of the wall 593, 602 UHF boundary condition 598 heat transfer in turbulent transition friction factor 653–654 Nusselt number Abraham et al. correlation 657 Churchill correlation 655 Gnielinski equation 656

Tam and Ghajar correlation 654 momentum transfer circular tubes entrance region 585 HFD region buffer layer 591 Fanning friction factor 593–594 the law of the wall 595 mean or average velocity 589 Moody friction factor 589–590 power law velocity distribution 596–597 Prandtl’s mixing length model 590–591 turbulent layer 591 velocity profile 586 viscous sublayer 592, 591 turbulent hydrodynamic entry length 585 parallel plates entrance region 574 HFD region buffer layer 579–582 Fanning friction factor 582–584 the law of the wall 579 mean velocity 582 Prandtl’s mixing length model 578–579 selected eddy diffusivity models 580 selected velocity profiles 580 turbulent layer 578 velocity profile 574 viscous sublayer 578 turbulent hydrodynamic entry length 575, 576

k kinematic viscosity 12, 17, 56, 58, 99, 108, 492 Knudsen number 2, 3

l Lagrangian viewpoint 4–5 laminar forced flow in ducts fully developed heat transfer circular pipe UHF 372–378 UWT 378–382 parallel plate channel UHF 383–387 laminar flow and heat transfer entrance region entrance region 335–337 circular pipe 337–350 parallel plate channel 355–370 thermal entrance region, Graetz problem entrance region 387 parallel plate channel

761

762

Index

laminar forced flow (contd.) dimensionless axial distance 420, 429 dimensionless temperature 420, 429 Nusselt number 428, 435, 436 UHF 428–436 UWT 419–428 circular pipe dimensionless axial distance 389 dimensionless temperature 389 Nusselt number 388, 397, 399 UHF 406– 414 UWT 388–405 wall heat flux variation 443 wall temperature variation 438 laminar external flow, correlation for 220–221, 252–253 thermal boundary layer superposition principle heat flux evaluation 246 local heat flux 246 temperature distribution 246 wall temperature variation 247 thermal boundary layer, flat plate constant surface heat flux 230 constant surface temperature 212, 225 local heat flux 216 local Nusselt number 217 temperature distribution 217 thermal boundary layer thickness 216 unheated starting length 256–261, 262–264 velocity boundary layer, flat plate free stream velocity 184 local friction coefficient 191, 199 local wall shear stress 191, 197 von Karman–Pohlhausen integral method boundary-layer thickness 77, 204 displacement thickness 88–89 momentum thickness 89–90 velocity profiles 79 Walz or Thwaites approximation 207 laminar slug flow boundary layer local Nusselt number 274 boundary conditions 273, 274, 275, 278, 279, 280 temperature distribution 268, 269 in parallel plate channel UHF 358–363, 367–370 UWT 363–367 in pipe UHF 345–350 UWT 337–345 over flat plate step change UHF

wall heat flux variation 272–277 wall temperature variation 266–272 step change UWT 262–265 laminar sublayer 488, 533, 535–536, 591 Laplace transform 267–268, 273–274 law-of-the wall 494, 512, 524, 526, 579 Leveque solutions 401–406 linear momentum equation differential form 38–64 integral form 33–38 liquid metals 17–18 liquid metals, properties 17–18 liquids, properties 562 log law 490 see also turbulent flow log mean temperature difference 334, 335 local friction coefficient 90, 199, 209, 516, 517, 519–520, 526 local Nusselt number 119, 219, 220, 227, 230, 234, 235, 250, 262, 278, 354, 397–398, 215, 217, 262

m Maple analytical solutions 61, 153, 154, 158, 160, 164, 169 differentiation 689, 155 integration 277, 289, 293, 341, 275, 287, 289, 341 solving algebraic equations 348 solving differential equations 267, 274 Martinelli analogy see analogies mass conservation principle see conservation of mass material derivative 5, 8 Maxwell relations 55, 57 mean free path 2, 3 mean temperature 20–22 mean velocity 19, 320, 479, 481, 582 mechanical energy equation 53 mixing-cup temperature 21, 166, 179, 374 mixing-length theory 507, 508 momentum diffusivity 338, 345, 355, 358 integral formulation 34–35 shape parameter 207 similarity solutions 189, 198 thickness 192–193 momentum equation differential form 42–47 integral form 34–35

n natural convection 679–680 see also free convection for flow classification 679 Rayleigh number 679 Navier–Stokes equation see momentum equation no slip 11

Index

Nusselt numbers for circular ducts entrance region 342–354, 350–354 fully developed heat transfer 372–382, 605–617 thermal entrance region 327, 388–414, 415–518, 618–628 for parallel plate channel entrance region 355–370, 370–371, 436–437 fully developed heat transfer 383–385, 597–605 thermal entrance region 419–436, 436–437

one-dimensional solutions Couette flow energy equation 153 momentum equation 152 Poiseuille flow 156–171 rotating flows 171–174 Ostrach solutions, free convection 682 outer layer of boundary layer 495 overlap layer of boundary layer 494–496

Rayleigh number 674, 677 stream function 679 Poiseuille flow 156–171, 318–326 potential flow 70, 74 power law friction 488 distribution 596–597 one seventh 542, 543 velocity 596–597 Prandtl mixing-length model 590–591 Prandtl number 78, 111, 113, 122, 126 effect on temperature profile 217, 524 turbulent 483 Prandtl’s mixing length 508–509 Prandtl–Taylor analogy 532–535, 631–633 pressure drop, entrance 317 gradient, friction factor 315–317 property extensive property of system 36 intensive property of system 36 property variation 252–253, 448–449, 564, 660

p

r

parallel flow past horizontal flat plate with adiabatic segment, application of integral method 256–261 with arbitrary surface temperature distribution 266–272, 284–289 with arbitrary wall heat flux distribution 272–277, 289–293 parallel plate channels laminar flow, convection 355–370, 370–371, 383–385, 419–436, 436–437 turbulent flow, convection 702–704, 598–604 Peclet number 126 pipe flow die experiment of Reynolds 465–466 Pi theorem 102–103 Pohlhausen solution to forced convection, laminar flow along a plate dimensionless temperature 213 dimensionless velocity 188 Nusselt number, average 219, 221 Nusselt number, local 217, 219 Reynolds number 191 stream function 196 to free convection along vertical plate dimensionless velocity 680, 681, 689 dimensionless temperature 679, 687 Grashof number 681, 682, 688 local heat flux 680, 700–701 Nusselt number, average 682, 689, 690, 701 Nusselt number, local 680, 688

rate of strain 2, 45 Rayleigh number 127, 674 regime the continuum regime 3 free molecule regime 3 the slip flow regime 3 the transition regime 3 Reynolds averaged conservation equations 474–481 number 125 stress 469, 476, 508 transport theorem 22–27 Reynold’s analogy heat transfer 529–531, 629–631 momentum transfer 529, 629 Reynolds averaged equations apparent heat flux 523, 543 apparent shear stress 484, 543, 546 averaged quantities 471 averaging process 474 averaging rules 471 boundary layer approximations 477–480 continuity equation 474–475 fluctuating quantities 475 momentum equation 475–477 Reynolds stress 476 thermal energy equation 478–480, 483 total shear stress 484, 488

o

763

764

Index

s scaling analysis 71–81, 677–678 semiempirical methods, turbulence 508, 517, 554 semi-infinite horizontal plate, similarity solution 183–195 , 212–224, 225–245 semi-infinite vertical plate, similarity solution 679–693 shape parameter, boundary layers 207 shear stress 12, 75, 191, 476, 478, 481–484 Sieder–Tate equation 646 similarity solutions Blasius solution 189, 191 boundary layer thickness based on 190, 191 dimensionless stream function 196 dimensionless temperature 213 Falkner–Skan 195–199, 249–252 friction coefficient based on 191 in forced convection for horizontal flat surface with UWT or UHF boundary conditions 213–237, 249–251 in free convection for vertical flat surface with UWT or UHF boundary conditions 679–693 momentum equation 189 Nusselt number from 217, 218–219, 682 similarity variable 187 simultaneously developing flow 350, 352, 353, 354, 370–371 see also combined entrance region single phase forced convection in ducts laminar flow 337- 371, 372–387 turbulent flow 597–617, 618–628, 642–652 skin friction average 192 local 191 slug flow over flat plate step change in wall heat flux 282–284 step change in wall temperature 278–282 variable wall heat flux 272–277 variable wall temperature 266–272 in parallel plate channel step change in wall heat flux 358–363, 367–370 step change in wall temperature 355–358, 363–367 in tube flow step change in wall heat flux 345–350 step change in wall temperature 337–344 specific heat constant pressure 56–58 constant specific volume 55–56 spherical coordinate system continuity equation 41 energy equation 47–53 equation of motion 45–47

Stanton number 126 steady turbulent flow averaging rules 471 eddy diffusivity of heat 482–483 eddy diffusivity of momentum 481–482 Reynolds-averaged equations 474–481 turbulent heat flux 479, 481 turbulent Prandtl number 483 turbulent shear stress 476 2D continuity 475, 483 2D energy equation 480, 483 2D momentum equation 478, 483 step change in surface heat flux 282 step change in surface temperature 278 stream function 196, 686 stress apparent 476, 632 normal 43 Reynolds 469, 476, 508 shear 191 Sturm–Liouville boundary value problem 449–450 sublayer, viscous 609–610 substantial derivative 7–10 superposition, principle of 245–248 surface forces 35, 37 pressure forces 42, 103 viscous forces 38, 42, 70 surface heat flux 282–284

t temperature bulk 21, 328, 373, 377, 385 dimensionless 118, 119, 120, 144, 232, 246, 329, 346, 387 law-of-the wall 524, 526 mixing 21 thermal boundary layer, on flat plate dimensionless temperature 118, 213, 232 film temperature 220 local heat flux local heat transfer coefficient 14, 217, 220, 227, 235, 236, 242, 250, 251 Nusselt number, average 219, 220, 221, 228, 230, 236, 240, 250, 251 Nusselt number, local 217, 219, 220, 228, 230, 234, 235, 239, 242, 250 Prandtl number 17, 217, 218, 219–220 similarity solution 212–220, 225–245 thermal boundary layer thickness 212, 216, 217, 220, 255, 256, 258, 263, 264, 290 thermal conductivity 10, 99, 109, 166, 482 thermal diffusivity 17, 56, 78, 80, 108, 171, 237, 338, 355, 483

Index

thermal energy equation in terms of enthalpy 55 in terms of internal energy 54 thermal entry length in laminar pipe flow 328, 335, 398, 443 in laminar parallel plate channel 327–328, 335 thermal expansion coefficient 679 thermal fully developed flow UHF boundary condition 372–377, 383–385 UWT boundary condition 378–382 thermal radiation 22 thermodynamic equilibrium 2–3 Thwaites’s integral 207–212 time average 465, 467 total derivative 8 turbulence apparent heat flux 523, 543 apparent shear stress 484, 543, 546 buffer region 512, 525, 536, 579, 591 closure problem 480–481 Couette flow approximation 494, 495, 510 fluctuations 467, 469, 476, 508 Eddy diffusivity models Deisler model 515, 522 of heat 482, 523 of momentum 481, 509 Prandtl–Taylor model 522 Spalding model 515, 522 temperature fluctuation 508 van Driest model 509, 522 von Karman 522 effective turbulent conductivity 482, 523 fluctuations 487, 492, 508 friction coefficient 485, 510, 516–521, 531, 582, 584, 589, 594, 656 friction law 510 friction velocity 485, 514, 527 intensity 469–470, 473, 539 inner region 511 logarithmic law 488, 579 mean values 470, 474 Nusselt number 523, 527, 532, 535, 535, 539, 540, 541, 549, 553, 557–559, 562, 604, 612, 619, 627, 630, 637, 639, 641, 646, 649, 650, 654, 655, 656, 657, 658 power law velocity distribution 514, 542 Prandtl mixing-length-model 508–509 Reynolds averaged equations 465 see also Reynolds averaged equations Reynolds stress 469, 476, 508 selected eddy diffusivity models 521–522 selected universal velocity distributions 515 temperature law-of-the-wall 524

three-layer velocity profile 511 transition to turbulence 496, 519 turbulent boundary layer thickness 519 turbulent Prandtl number 523 turbulent viscosity 482, 507, 509 universal velocity profile 511–514 velocity-defect law 513 velocity distribution 511 viscous sublayer 511–512 wake region 513 transition flow 465, 486 transition flow heat transfer 558–564, 652–659 transition to turbulence along flat plate 494, 496, 519 in a tube flow 496, 652 turbulent boundary layer 519–521 turbulent core 486, 586, 635–636 turbulent external flow, correlations for 11–19, 516–518, 520–521, 527, 531, 532, 5365, 538–541, 548–549, 559–562 turbulent flow 485–498 turbulent flow heat transfer along a flat plate 522 between infinite parallel plates 598 in a tube 605 turbulent flow of incompressible fluid over body of revolution 86 over flat plate 477, 491, 507 turbulent Graetz problem 618 turbulent pipe flow 484, 485 turbulent stress 476 2D turbulent boundary layer 478, 480

u uniform velocity 203–204 uniform wall heat flux (UHF) incompressible fluid flowing in a parallel plate channel 358–363, 367–370, 371, 383–387, 428–436, 437 incompressible fluid flowing in a tube 345–350, 353–354, 372–377, 406–414, 417–418 similarity solution over flat plate, forced convection 230–237 similarity solution over vertical flat plate, free convection 686–693 thermally developed flow heat transfer 372–378, 383–385 uniform wall temperature (UWT) incompressible fluid flowing in a parallel plate channel 151–156, 157–161, 166–171, 170–171, 355–358, 363–367, 371, 419–428, 436–437 incompressible fluid flowing in a tube 161–166, 337–344, 351–353, 377–382, 388–405, 415–417

765

766

Index

uniform wall temperature (UWT) (contd.) similarity solution over flat plate, forced convection 212–230 similarity solution over vertical flat plate, free convection 679–686 thermally developed flow heat transfer 378–382 universal temperature profile, flat plate 526 universal velocity profile 511–514

v Van Driest damping function 509–510 Van Driest eddy diffusivity model 509, 522 variable property effects 252–253, 448, 564, 660 variable surface temperature distribution in axial direction 438–443 see also arbitrary wall temperature distribution in axial direction variable wall heat flux distribution in axial direction 443–446 see also arbitrary wall heat flux distribution in axial direction velocity around a cylinder 200 fluctuating component 465, 469 instantaneous component 467–468 mean velocity 19–20 mean component 468 velocity boundary layer 11–12, 183–195 velocity boundary layer thickness 11, 77, 79, 81, 88, 107, 258 velocity defect law 486, 597

viscosity dynamic 45 eddy 482 kinematic 45 viscous dissipation function cylindrical coordinates 59 rectangular coordinates 59 spherical coordinates 59 viscous flow 256–265, 278–293, 375–382 viscous oils 18 viscous sublayer 21, 486, 488, 494, 511, 524 von Karman analogy 535–539, 633–641 von Karman constant 486, 500, 513 von Karman integral method 695–701 von Karman–Pohlhausen method 201–212, 253–265, 542–550, 693–699

w wake region 495, 513 wall friction 485, 513, 530, 577 wall heat flux 272–277, 282–284, 289–293, 353–355, 443–446, 554–558 wall shear stress 191 wall temperature 266–272, 278–282, 284–289, 337–344, 438–443, 550–554 water 18 wedge flows 193–201, 247–252

z zero-equation turbulence models 508–510

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