Heat and Mass Transfer [7 ed.] 9789352533848

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A Textbook of

Heat and Mass Transfer (SI Units)

(ii)

A Textbook of

Heat and Mass Transfer (SI Units) A textbook for the students of B.E., B.Tech., as per the 2018 AICTE Curriculum ESE, GATE & Other Competitive Examinations

Er. R.K. RAJPUT M.E. (Hons.), Gold Medalist; Grad.-Mech. Engg. & Elec. Engg.; MIE (India) MSESI; MISTE; CE (India) Recipient of : “Best Teacher (Academic) Award” “Distinguished Author Award” “Jawahar Lal Nehru Memorial Gold Medal” for an Outstanding Research Paper (Institution of Engineers—India)

l

Principal (Formerly) Thapar Polytechnic College; l Punjab College of Information Technology; PATIALA

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© 1999, Er. R.K. Rajput All rights reserved. No part of this publication may be reproduced or copied in any material form (including photocopying or storing it in any medium in form of graphics, electronic or mechanical means and whether or not transient or incidental to some other use of this publication) without written permission of the copyright owner. Any breach of this will entail legal action and prosecution without further notice. Jurisdiction : All disputes with respect to this publication shall be subject to the jurisdiction of the Courts, Tribunals and Forums of New Delhi, India only. First Edition 1999; Subsequent Editions and Reprints of the Main Textbook 2001, 2002, 2003, 2004, 2005, 2006, 2007 (Twice), 2008, 2009 (Twice), 2010 (Twice), 2011, 2012, 2013 (Twice), 2014 (Twice); 2015, 2016, 2017 Revised Edition 2018

ISBN : 978-93-525-3384-8

To my wife

Mrs Ramesh Rajput

(vi)

Preface I feel pleasure in presenting “Seventh Edition” of this book. Besides revising the whole book, in this edition, a set of “Typical Questions” (Conventional & Objectives) selected from latest Competitive Examinations has been added to make this book a complete unit in all respects. Suggestions for improvement of the book are most welcome. Er. R.K. RAJPUT (Author)

Disclaimer : While the authors of this book have made every effort to avoid any mistakes or omission and have used their skill, expertise and knowledge to the best of their capacity to provide accurate and updated information. The author and S. Chand does not give any representation or warranty with respect to the accuracy or completeness of the contents of this publication and are selling this publication on the condition and understanding that they shall not be made liable in any manner whatsoever. S. Chand and the author expressly disclaim all and any liability/responsibility to any person, whether a purchaser or reader of this publication or not, in respect of anything and everything forming part of the contents of this publication. S. Chand shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in this publication. Further, the appearance of the personal name, location, place and incidence, if any; in the illustrations used herein is purely coincidental and work of imagination. Thus the same should in no manner be termed as defamatory to any individual.

(vii)

Preface to the First Edition This treatise on “Heat and Mass Transfer” contains comprehensive treatment of the subject matter in simple, lucid and direct language. It envelopes a large number of solved problems, properly graded, including typical examples from examination point of view. The book comprises 13 chapters. All chapters are saturated with much needed text supported by simple and self explanatory figures. A large number of Worked Examples (including Typical Examples selected from various Indian Universities Examinations question papers), Highlights, Objective Type Question, Theoretical Questions and Unsolved Examples have been added to make the book a comprehensive and a complete unit in all respects. The book will prove to be a boon to the students preparing for engineering undergraduate, AMIE, UPSC and other competitive examinations. The author’s thanks are due to his wife Ramesh Rajput for extending all cooperation during preparation of the manuscript and proof reading. In the end the author wishes to express his gratitude to Shri Ravindra Kumar Gupta, Director, S.Chand & Company Ltd., New Delhi for taking a lot of pains in bringing out the book with very good presentation in a short span of time. Although every care has been taken to make the book free of errors both in text as well as in solved examples, yet the author shall feel obliged if errors present are brought to his notice. Constructive criticism of the book will be warmly received. Er. R.K. RAJPUT (Author)

(viii)

Contents Chapters Pages Nomenclature (xvi)—(xvii) 1. BASIC CONCEPTS 1 — 23 1.1. Heat Transfer—General Aspects, 1 1.1.1. Heat, 1 1.1.2. Importance of heat transfer, 2 1.1.3. Thermodynamics, 2 1.1.3.1. Definition, 2 1.1.3.2. Thermodynamic systems, 3 1.1.3.3. Macroscopic and microscopic points of view, 4 1.1.3.4. Pure substance, 4 1.1.3.5. Thermodynamic equilibrium, 5 1.1.3.6. Properties of systems, 6 1.1.3.7. State, 6 1.1.3.8. Process, 6 1.1.3.9. Cycle, 7 1.1.3.10. Point function, 7 1.1.3.11. Path function, 7 1.1.3.12. Temperature, 7 1.1.3.13. Pressure, 7 1.1.3.14. Energy, 8 1.1.3.15. Work, 8 1.1.3.16. Heat, 9 1.1.3.17. Comparison of work and heat, 9 1.1.4 Differences between thermodynamics and heat transfer, 9 1.1.5. Basic laws governing heat transfer, 10 1.1.6. Modes of heat transfer, 11 1.2. Heat Transfer by Conduction, 11 1.2.1. Fourier’s law of heat conduction, 12 1.2.2. Thermal conductivity of materials, 13 1.2.3. Thermal resistance (Rth), 15 1.3. Heat Transfer by Convection, 17 1.4. Heat Transfer by Radiation, 18 Highlights, 21 Theoretical Questions, 23 Unsolved Examples, 23

PART – I : HEAT TRANSFER BY “CONDUCTION” 2. “CONDUCTION” HEAT TRANSFER AT STEADY STATE– ONE DIMENSION 27 — 249

2.1. 2.2. 2.3. 2.4.

Introduction, 27 General Heat Conduction Equation in Cartesian Coordinates, 27 General Heat Conduction Equation in Cylindrical Coordinates, 31 General Heat Conduction Equation in Spherical Coordinates, 34 (ix)

2.5. Heat Conduction Through Plane and Composite Walls, 37 2.5.1. Heat conduction through a plane wall, 37 2.5.2. Heat conduction through a composite wall, 40 2.5.3. The overall heat transfer coefficient, 43 2.6. Heat Conduction Through Hollow and Composite Cylinders, 82 2.6.1. Heat conduction through a hollow cylinder, 82 2.6.1.1. Logarithmic mean area for the hollow cylinder, 87 2.6.2. Heat conduction through a composite cylinder, 88 2.7. Heat Conduction Through Hollow and Composite Spheres, 119 2.7.1. Heat conduction through a hollow sphere, 119 2.7.1.1. Logarithmic mean area for the hollow sphere, 122 2.7.2. Heat condition through a composite sphere, 123 2.8. Critical Thickness of Insulation, 132 2.8.1. Insulation–General aspects, 132 2.8.2. Critical thickness of insulation, 133 2.9. Heat conduction with Internal Heat Generation, 139 2.9.1. Plane wall with uniform heat generation, 140 2.9.2. Dielectric heating, 155 2.9.3. Cylinder with uniform heat generation, 158 2.9.4. Heat transfer through the piston crown, 178 2.9.5. Heat conduction with heat generation in the nuclear cylindrical fuel rod, 179 2.9.6. Sphere with uniform heat generation, 185 2.10. Heat Transfer from Extended Surfaces (Fins), 188 2.10.1. Introduction, 188 2.10.2. Heat flow through “Rectangular fin”, 190 2.10.2.1. Heat dissipation from an infinitely long fin (l → ∞), 192 2.10.2.2. Heat dissipation from a fin insulated at the tip, 197 2.10.2.3. Heat dissipation from a fin losing heat at the tip, 207 2.10.2.4. Efficiency and effectiveness of fin, 215 2.10.2.5. Design of rectangular fins, 221 2.10.3. Heat flow through “straight triangular fin”, 224 2.10.4. Estimation of error in temperature measurement in a thermometer well, 227 2.10.5. Heat transfer from a bar connected to the two heat sources at different temperatures, 232 Highlights, 240 Theoretical Questions, 244 Unsolved Examples, 245

3. CONDUCTION HEAT TRANSFER AT STEADY STATE– TWO DIMENSIONS AND THREE DIMENSIONS 250 — 270 3.1. Introduction, 250 3.2. Two Dimensional Steady State Conduction, 251 3.2.1. Analytical method, 251 3.2.1.1. Two-dimensional steady state heat conduction in rectangular plates, 251 3.2.1.2. Two-dimensional steady state heat con duction in semi-infinite plates, 253 (x)

3.2.2. Graphical method, 258 3.2.3. Analogical method, 265 3.2.4. Numerical methods, 266 3.3. Three-dimensional Steady State Conduction, 268 Highlights, 269 Theoretical Questions, 269 Unsolved Examples, 269

4. HEAT CONDUCTION–TRANSIENT (UNSTEADY STATE)

271 — 315

4.1. Introduction, 271 4.2. Heat conduction in Solids having Infinite Thermal Conductivity (Negligible Internal Resistance) — Lumped Parameter Analysis, 272 4.3. Time constant and Response of Temperature Measuring Instruments, 284 4.4. Transient Heat Conduction in Solids with Finite Conduction and Convective Resistances (0 < Bi < 100), 288 4.5. Transient Heat Conduction in Semi-infinite Solids (h or Bi → ∞), 297 4.6. Systems with Periodic Variation of Surface Tem-perature, 305 4.7. Transient Conduction with Given Temperature Distribution, 307 Typical Examples, 309 Highlights, 312 Theoretical Questions, 313 Unsolved Examples, 313

PART – II : HEAT TRANSFER BY “CONVECTION” 5. HEAT TRANSFER BY “FORCED CONVECTION”

319 — 442

A. LAMINAR FLOW, 319 5.1. Laminar Flow over a Flat Plate, 319 5.1.1. Introduction to boundary layer, 319 5.1.1.1. Boundary layer definitions and characteristics, 320 5.1.2. Momentum equation for hydrodynamic boundary layer over a flat plate, 326 5.1.3. Blasius (exact) solution for laminar boundary layer flows, 328 5.1.4. Van-Karman integral momentum equation (Approximate hydro-dynamic boundary layer analysis), 332 5.1.5. Thermal boundary layer, 342 5.1.6. Energy equation of thermal boundary layer over a flat plat, 343 5.1.7. Integral energy equation (Approximate solution of energy equation), 350 5.2. Laminar Tube Flow, 367 5.2.1. Development of boundary layer, 367 5.2.2. Velocity distribution, 368 5.2.3. Temperature distribution, 371 B. TURBULENT FLOW, 377 5.3. Introduction, 377 5.3.1. Turbulent boundary layer, 377 5.3.2. Total drag due to laminar and turbulent layers, 380 5.3.3. Reynolds analogy, 387 5.4. Turbulent Tube Flow, 396 5.5. Empirical Correlations 403 (xi)

5.5.1. Laminar flow over flat plates and walls, 403 5.5.2. Laminar flow inside tubes, 405 5.5.3. Turbulent flow over flat plate, 409 5.5.4. Turbulent flow in tubes, 409 5.5.5. Turbulent flow over cylinders, 418 5.5.6. Turbulent flow over spheres, 424 5.5.7. Flow across bluff objects, 425 5.5.8. Flow through packed beds, 425 5.5.9. Flow across a bank of tubes, 427 5.5.10. Liquid metal heat transfer, 430 Highlights, 433 Theoretical Questions, 437 Unsolved Examples, 438

6. HEAT TRANSFER BY “FREE CONVECTION”

443 — 474

6.1. Introduction, 443 6.2. Characteristic Parameters in Free Convection, 444 6.3. Momentum and Energy Equation for Laminar Free Convection Heat Transfer on a Flat Plate, 445 6.4. Integral Equations for Momentum and Energy on a Flat Plate, 446 6.4.1. Velocity and temperature profiles on a vertical flat plate, 446 6.4.2. Solution of integral equations for vertical flat plate, 447 6.4.3. Free convection heat transfer coefficient for a vertical wall, 448 6.5. Transition and Turbulence in Free Convection, 449 6.6. Empirical Correlations for Free Convection, 449 6.6.1. Vertical plates and cylinders, 449 6.6.2. Horizontal plates, 449 6.6.3. Horizontal cylinders, 450 6.6.4. Inclined plates, 450 6.6.5. Spheres, 450 6.6.6. Enclosed spaces, 450 6.6.7. Concentric cylinders space 451 6.6.8. Concentric spheres spaces, 451 6.7. Simplified Free Convection Relations for Air, 451 6.8. Combined Free and Forced Convection, 452 6.8.1. External flows, 452 6.8.2. Internal flows, 452 Typical Examples, 469 Highlights, 473 Unsolved Examples, 474

7. BOILING AND CONDENSATION 7.1. Introduction, 475 7.2. Boiling Heat Transfer, 476 7.2.1. General aspects, 476 7.2.2. Boiling regimes, 476 7.2.3. Bubble shape and size consideration, 478 7.2.4. Bubble growth and collapse, 479 (xii)

475 — 509

7.2.5. Critical diameter of bubble, 480 7.2.6. Factors affecting nucleate boiling, 480 7.2.7. Boiling correlations, 481 7.2.7.1. Nucleate pool boiling, 481 7.2.7.2. Critical heat flux for nucleate pool boiling, 482 7.2.7.3. Film pool boiling, 482 7.3. Condensation Heat Transfer, 486 7.3.1. General aspects, 486 7.3.2. Laminar film condensation on a vertical plate, 487 7.3.3. Turbulent film condensation, 493 7.3.4. Film condensation on horizontal tubes, 493 7.3.5. Film condensation inside horizontal tubes, 494 7.3.6. Influence of the presence of non-condensable gases, 494 Highlights, 506 Theoretical Questions, 508 Unsolved Examples, 508

8. HEAT EXCHANGERS

510 — 603

8.1. Introduction, 510 8.2. Types of Heat Exchangers, 510 8.3. Heat Exchanger Analysis, 515 8.4. Logarithmic Mean Temperature Difference (LMTD), 516 8.4.1. Logarithmic mean temperature difference for parallel-flow, 516 8.4.2. Logarithmic mean temperature difference for counter-flow, 518 8.5. Overall Heat Transfer Coefficient, 520 8.6. Correction Factors for Multi-pass Arrangements, 557 8.7. Heat Exchanger Effectiveness and Number of Transfer Units (NTU), 562 8.8. Pressure Drop and Pumping Power, 566 8.9. Evaporators, 593 8.9.1. Introduction, 593 8.9.2. Classification of evaporators, 593 Highlights, 599 Theoretical Questions, 600 Unsolved Examples, 600

PART – III : HEAT TRANSFER BY “RADIATION” 9. HEAT TRANSFER BY RADIATION

9.1. Thermal Radiations–Introduction, 607 9.2. Surface Emission Properties, 608 9.3. Absorptivity, Reflectivity and Transmissivity, 609 9.4. Concept of a Black body, 611 9.5. The Stefan-Boltzmann Law, 612 9.6. Kirchoff’s Law, 612 9.7. Planck’s Law, 613 9.8. Wien Displacement Law, 614 9.9. Intensity of Radiation and Lambert’s Cosine Law, 615 9.9.1. Intensity of radiation, 615 9.9.2. Lambert’s cosine law, 617 9.10. Radiation Exchange Between Surfaces–Introduction, 620 (xiii)

607 — 694



9.11. Radiation Exchange Between Black Bodies Separated by an a Non-absorbing Medium, 620 9.12. Shape Factor Algebra and Salient Features of the Shape Factor, 623 9.13. Heat Exchange Between Non-black Bodies, 639 9.13.1. Infinite parallel planes, 640 9.13.2. Infinite long concentric cylinders, 641 9.13.3. Small gray bodies, 644 9.13.4. Small body in a large enclosure, 644 9.14. Electrical Network Analogy for Thermal Radiation Systems, 645 9.15. Radiation Heat Exchange for Three Gray Surfaces, 648 9.16. Radiation Heat Exchange for Two Black Surfaces Connected by a Single Refractory surface, 648 9.17. Radiation Heat Exchange for Two Gray Surfaces Connected by Single Refractory Surface, 649 9.18. Radiation Heat Exchange for Four Black Surfaces, 650 9.19. Radiation Heat Exchange for Four Gray Surfaces, 651 9.20. Radiation Shields, 671 9.21. Coefficient of Radiant Heat Transfer and Radiation Combined with Convection, 683 9.22. Error in Temperature Measurement due to Radiation, 685 9.23. Radiation from Gases, Vapours and Flames, 688 Highlights, 690 Theoretical Questions, 692 Unsolved Examples, 693

PART – IV : MASS TRANSFER 10. MASS TRANSFER

697 — 736



10.1. Introduction, 697 10.2. Modes of Mass Transfer, 698 10.3. Concentrations, Velocities and Fluxes, 698 10.3.1. Concentrations, 698 10.3.2. Velocities, 699 10.3.3. Fluxes, 700 10.4. Fick’s Law, 702 10.5. General Mass Diffusion Equation in Stationary Media, 707 10.6. Steady-State Diffusion in Common Geometries, 709 10.6.1. Steady state diffusion through a plain membrane, 709 10.6.2. Steady state diffusion through a cylindrical shell, 711 10.6.3. Steady state diffusion through a spherical shell, 713 10.7. Steady-State Equimolar Counter Diffusion, 715 10.8. Steady State Undirectional Diffusion (Steady state Diffusion through a stagnant Gas Film), 718 10.9. Steady State Diffusion in Liquids, 723 10.9.1. Steady-state Equimolar counter-diffusion in liquids, 723 10.9.2. Steady-state uni-directional diffusion in liquids, 724 10.10. Transient Mass Diffusion in Semi-finite Stationary Medium, 724 10.11. Mass Transfer Co-efficient, 726 10.12. Convective Mass Transfer, 728 10.13. Correlations for Connective Mass Transfer, 729 10.14. Reynolds and Colburn Analogies for Mass Transfer-Combined Heat and Mass Transfer, 730 (xiv)



Highlights, 734 Theoretical Questions, 735 Unsolved Examples, 735

PART – V : MISCELLANEOUS 11. INTRODUCTION TO HYDRODYNAMICS

739 — 750

11.1. Introduction, 739 11.2. Ideal and Real Fluids, 739 11.3. Viscosity, 740 11.4. Continuity Equation in Cartesian Coordinates, 741 11.5. Continuity Equation in Polar Coordinates, 742 11.6. Velocity Potential and Stream Function, 743 11.6.1. Velocity Potential, 743 11.6.2. Stream Function, 744 11.7. Laminar and Turbulent Flows, 746 Highlights, 749 Theoretical Questions, 750

12. DIMENSIONAL ANALYSIS

751 — 771

12.1. Introduction, 751 12.2. Dimensions, 752 12.3. Dimensional Homogeneity, 753 12.4. Methods of Dimensional Analysis, 753 12.4.1. Rayleigh’s Method, 753 12.4.2. Buckingham’s Π-Method/Theorem, 755 12.5. Dimensional Analysis Applied to Forced, 761 12.6. Dimensional Analysis Applied to Natural or Free, 763 12.7. Advantages and Limitations of Dimensional Analysis, 764 12.8. Dimensionless Numbers and their Physical Significance, 765 12.9. Characteristic Length or Equivalent Diameter, 768 12.10. Model Studies and Similitude, 770 12.10.1. Model and Prototype, 770 12.10.2. Similitude, 770 Highlights, 770 Theoretical Questions, 771

13. UNIVERSITIES’ EXAMINATIONS QUESTIONS (LATEST-SELECTED) WITH ANSWERS/SOLUTIONS: 772 — 840 A. Short Answer Questions, 772

B. Conventional Questions with Solutions, 777 C. Multiple-choice Questions with Answers, 813

14. GATE AND UPSC EXAMINATIONS’ QUESTIONS (LATEST-SELECTED) WITH ANSWERS/SOLUTIONS: A. Conventional Questions with Answers/Solutions, 841

841 — 912



913 — 915



B. Multiple-choice Questions with Answers and “Explanations”, 882

INDEX

ON THE WEBSITE (www.schandpublishing.com): Chapter on “Radiation Exchange Between surfaces” (xv)

Nomenclature A

Area

Q

Heat transfer rate per unit time

c cp cv C

Specific heat specific heat at constant pressure Specific heat at constant volume Mass concentration

Q´

Total heat transfer

Qg

Heat generated per unit time

Qg´

Total heat generated

R

Characteristic gas constant

Cfx

Local skin friction coefficient

Rth

Thermal resistance

Cf

Average skin friction coefficient

(Rth)cond.

Conductive thermal resistance

D

Diffusion coefficient

(Rth)conv.

Convective thermal resistance

D, d

Diameter

(Rth)rad.

Radiative thermal resistance

E

Total emissive power

t

temperature

Eb

Emissive power of a black body

T

Absolute temperature

E

Monochromatic emissive power

U

Overall heat transfer coefficient, velocity of fluid

f

Interchange factor

x

Mole fraction

F

Correction factor

G

Irradiation

G

Universal gas constant

h

Average heat transfer coefficient

hmc

Mass transfer coefficient based upon concentration

hmp

Mass transfer coefficient based upon pressure

J

Radiosity

k

Thermal conductivity

L

Length

m

Mass

Greek Notations  k     c 

Thermal diffusivity;



Coefficient of volume expansion; Temperature coefficient of thermal conductivity



Time; Transmittivity

th

Thermal time constant



Temperature difference; Momentum thickness

m

Log-mean temperature difference



Absorptivity

m M

Mass flow rate Molecular weight

*

Hydrodynamic boundary layer thickness Displacement thickness

m*

Mass fraction Mass flux of species

e

Energy thickness

N p

Pressure

Thermal boundary layer thickness

P

Perimeter

th 

q

Heat transfer per unit area per unit time Heat generated per unit volume per unit time



Velocity potential



Stefan-Boltzmann constant

.

qg

(xvi)

Stream function

(= 5.67 × 10–8 W/m2K4)

   

Reflectivity; Mass density Emissivity Wavelength Collision integral

Dimensionless Groups  hL  Bi     k   L3 g t  Gr    v2  

Biot number

Grashof number

 mc p  Gz     Lk 

Graetz number

  Le     D

Lewis number

 hx  Nu     h 

Nusselt number

 hx  Nu     k 

Average Nusselt number

LU   Pe   Re.Pr   Peclet number     c p v    Prandtl number Pr     k  Ux  Re       Sh 

Reynolds number

hm . x D

Sherwood number

 v Sc     D  h St    Uc p  i, 1 0, 2 hf cf w

(xvii)

Schmidt number

  

Stanton number

Subscripts

Inner or inlet conditions Outer or outlet conditions Hot fluid Cold fluid Wall conditions

C H A P T E R



Basic Concepts 1.1. HEAT TRANSFER–GENERAL ASPECTS 1.1.1. HEAT “The energy in transit is termed heat”

1.1. Heat transfer – general aspects – Heat – Importance of heat transfer – Difference between thermodynamics and heat transfer – Modes of heat transfer. 1.2. Heat transfer by conduction: Fourier’s law of heat conduction – Thermal conductivity of materials – Thermal resistance (Rth). 1.3. Heat transfer by convection. 1.4. Heat transfer by radiation. Highlights – Theoretical Questions – Unsolved Examples.

While Aristotle was of the opinion that fire was one of the four primary elements, Plato thought that the heat was sort of motion of particles; accordingly there are two theories of heat. Any theory should be able to explain the facts given below : (i) Whenever there is an exchange of heat, heat is consumed (heat lost by the hot body is always equal to heat gained by the cold body). (ii) The heat flow takes place from higher to lower temperature. (iii) The substances expand on heating. (iv) In order to change the state of a body from solid to liquid or liquid to gas without rise in temperature, certain amount of heat is required. (v) When a body is heated or cooled its weight does not change. According to the modern or dynamical theory of heat: “Heat is a form of energy. The molecules of a substance are in parallel motion. The mean kinetic energy per molecule of the substance is proportional to its absolute temperature”. A molecule may consist of one or two or many atoms depending upon the nature of the gas. The force of attraction between the molecules of a perfect gas is negligible. The atoms in a molecule vibrate with respect to one another, consequently a molecule has ‘vibrational energy’. The whole molecule may rotate

2

Chapter : 1

about one or more axes, so it can have ‘rotational energy’. A molecule has ‘translational energy’ due to its motion. Thus kinetic energy of a molecule is the sum of its translational, rotational and vibrational energies. Summarily heat energy given to a substance is used in increasing its internal energy. Increase in internal energy causes increase in kinetic energy or potential energy or increase in both the energies. Due to increase in kinetic energy of a molecule, its translational, vibrational or rotational energy may increase.

1.1.2. IMPORTANCE OF HEAT TRANSFER Heat transfer may be defined as : “The transmission of energy from one region to another as a result of temperature gradient”. In heat transfer the driving potential is temperature difference whereas in mass transfer the driving potential is concentration difference. In mass transfer we concentrate upon mass motion which result in changes in composition, and are caused by the variations in concentrations of the various constituent species. This transfer, in literature, is also known as “diffusion”. The study of heat transfer is carried out for the follows purposes : 1. To estimate the rate of flow of energy as heat through the boundary of a system under study (both under steady and transient conditions). 2. To determine the temperature field under steady and transient conditions. In almost every branch of engineering, heat transfer (and mass transfer) problems are encountered which cannot be solved by thermodynamic reasoning alone but require an analysis based on heat transfer principles. The areas covered under the discipline of heat transfer are : l Design of thermal and nuclear power plants including heat engines, steam generators,

condensers and other heat exchange equipments, catalytic converters, heat shields for space vehicles, furnaces, electronic equipments etc.

l Internal combustion engines. l Refrigeration and air conditioning units. l Design of cooling systems for electric motors, generators and transformers. l Heating and cooling of fluids etc. in chemical operations. l Construction of dams and structures; minimisation of building-heat losses using improved

insulation techniques.

l Thermal control of space vehicles. l Heat treatment of metals. l Dispersion of atmospheric pollutants.

1.1.3. THERMODYNAMICS 1.1.3.1. Definition Thermodynamics may be defined as follows: “Thermodynamics is an axiomatic science which deals with the relations among heat, work and properties of system which are in equilibrium. It describes state and changes in state of physical systems.” Thermodynamic, basically entails four laws or axioms known as Zeroth, First, Second and Third law of thermodynamics. l First law throws light on concept of internal energy. l Zeroth law deals with thermal equilibrium and establishes a concept of temperature.

Basic Concepts

3

l Second law indicates the limit of converting heat into work and introduces the principle of

increase of entropy. l Third law defines absolute zero of entropy. These laws are based on experimental observations and have no mathematical proof. Like all physical laws, these laws are based on logical reasoning.

1.1.3.2. Thermodynamic systems System, boundary and surroundings : System. A system is a finite quantity of matter or a prescribed region of space (Refer Fig. 1.1).

Fig. 1.1. The system.

Fig. 1.2. The real and imaginary boundary.

Boundary. The actual or hypothetical envelop enclosing the system is the boundary of the system. The boundary may be fixed or it may move, as and when a system containing a gas is compressed or expanded. The boundary may be real or imaginary. It is not difficult to envtisage a real boundary but an example of imaginary boundary would be one drawn around a system consisting of a fresh mixture about to enter the cylinder of an I.C engine together with remanants of the last cylinder charge after the exhaust process (Fig 1.2). Refer to Fig. 1.3. If the boundary of the system is impervious to the flow of matter, it is called a closed system. An example of this system is mass of gas or vapour contained in an engine cylinder, the boundary of which is drawn by the cylinder walls, the cylinder head and piston crown. Here the boundary is continuous and no matter may enter or leave.

Fig. 1.3. Closed system.

Fig. 1.4. Open system.

4

Chapter : 1

Open system : Refer figure 1.4. An open system is one in which matter flows into or out of the system. Most of the engineering systems are open. Isolated system : An isolated system is that system which exchanges neither energy nor matter with any other system or with environment. Adiabatic system : An adiabatic system is one which is thermally insulated from its surroundings. It can, however, exchange work with its surroundings. If it does not, it becomes an isolated system. Phase : A phase is a quantity of matter which is homogeneous throughout in chemical composition and physical structure. Homogeneous system : A system which consists of a single phase is termed as homogeneous system. Examples : Mixture of air and water vapour, water plus nitric acid and octane plus heptane. Heterogeneous system : A system which consists of two or more phases is called a heterogeneous system. Examples : Water plus steam, ice plus water and water plus oil.

1.1.3.3. Macroscopic and microscopic points of view Thermodynamic studies are undertaken by the following two different approaches : 1. Macroscopic approach—(Macro mean big or total) 2. Microscopic approach—(Micro means small)

Fusion reaction Helium nucleus (two protons + two neutrons) Four hydrogen nuclei (protons)

Energy

Sun is the major source of energy for the earth.

These approaches are discussed (in a comparative way) below : Note. Although the macroscopic approach seems to be different from microscopic one, there exists relation between them. Hence when both the methods are applied to a particular system, they give the same result.

1.1.3.4. Pure substance A “pure substance” is one that has a homogeneous and invariable chemical composition even though there is a change of phase. In other words, it is a system which is (a) homogeneous in composition, (b) homogeneous in chemical aggregation. Examples : Liquid, water, mixture of liquid water and steam, mixture of ice and water. The mixture of liquid air and gaseous air is not a pure substance.

Basic Concepts S.No.

Macroscopic approach

1.

In this approach a certain quantity of matter is considered without taking into account the events occuring at molecular level. In other words this approach to thermo-dynamics is concerned with gross or overall behaviour. This is known as “Classical thermodynamics”.

The approach considers that the system is matter of a very large number of discrete particles known as molecules. These molecule have different velocities and energies. The values of these energies are constantly changing with time. This approach to thermodynamics which is concerned directly with the structure of the matter is known as “Statistical thermodynamics”.

2.

The analysis of macroscopic system requires simple mathematical formulae.

The behaviour of the system is found by using statistical methods as the number of molecules is very large. So advanced statistical and mathematical methods are needed to explain the change in the system.

3.

The values of the properties of the system are their average values. For example, consider a sample of a gas in a closed container. The pressure of the gas is the average value of the pressure exerted by millions of individual molecules. Similarly the temperature of this gas is the average value of translational kinetic energies of millions of individual molecules. These properties like pressure and temperature can be measured very easily. The changes in properties can be felt by our senses.

The properties like velocity, momentum, impulse kinetic energy, force of impact etc. which describe the molecule cannot be easily measured by instruments. Our senses cannot feel them.

4.

In order to describe a system only a few properties are needed.

Large number of variables are needed to describe a system. So the approach is complicated.

5

Microscopic approach

1.1.3.5. Thermodynamic equilibrium A system is in thermodynamic equilibrium if the temperature and pressure at all points are same; there should be no velocity gradient; the chemical equilibrium is also necessary. Systems under temperature and pressure equilibrium but not under chemical equilibrium are sometimes said to be in metastable equilibrium conditions. It is only under thermodynamic equilibrium conditions that the properties of a system can be fixed. Thus for attaining a state of thermodynamic equilibrium the following three types of equilibrium states must be achieved : 1. Thermal equilibrium. The temperature of the system does not change with time and has same value at all points of the system. 2. Mechanical equilibrium. There are no unbalanced forces within the system or between the surroundings. The pressure in the system is same at all points and does not change with respect to time.

6

Chapter : 1

3. Chemical equilibrium. No chemical reaction takes place in the system and the chemical composition which is same throughout the system does not vary with time.

1.1.3.6. Properties of systems A “property of a system” is a characteristic of the system which depends upon its state, but not upon how the state is reached. There are two sorts of property : 1. Intensive properties. These properties do not depend on the mass of the system. Example : Temperature and pressure. 2. Extensive properties. These properties depend on the mass of the system. Example : Volume. Extensive properties are often divided by mass associated with them to obtain the intensive poroperties. For example, if the volume of a system of mass m is V, then the specific volume of matter within the V v which is an intensive property. system is m

1.1.3.7. State “State” is the condition of the system at an instant of time as described or measured by its properties. Or each unique condition of a system is called a “state”. It follows from the definition of state that each property has a single value at each state. Stated differently, all properties are state or point functions. Therefore, all properties are identical for identical states. On the basis of the above discussion, we can determine if a given variable is property or not by applying the following tests : l A variable is a property, if and only if, it has a single value at each equilibrium state. l A variable is a property, if and only if, the change in its value between any two prescribed

equilibrium states is single-valued. Therefore, any variable whose change is fixed by the end states is a “property”.

1.1.3.8. Process A process occurs when the system undergoes a change in a state or an energy transfer at a steady state. A process may be non-flow in which a fixed mass within the defined boundary is undergoing a change of state. Example : A substance which is being heated in a closed cylinder undergoes a nonflow process (Fig. 1.3.). Closed systems undergo non-flow processes. A process may be a flow process in which mass is entering and leaving through the boundary of an open system. In a steady flow process (Fig. 1.4.) mass is crossing the boundary from surroundings at entry, and an equal mass is crossing the boundary at the exit so that the total mass of the system remains constant. In an open system it is necessary to take account of the work delivered from the surroundings to the system at entry to cause the mass to enter, and also, of the work delivered from the system at surroundings to cause the mass to leave, as well as any heat or work crossing the boundary of the system. Quasi-static process. Quasi means ‘almost’. A quasi-static process is also called a reversible process. This process is a succession of equilibrium states and infinite slowness is its characteristic feature.

1.1.3.9. Cycle Any process or series of processes whose end states are identical is termed a cycle. The processes through which the system has passed can be shown on a state diagram, but a complete section of the path requires in addition a statement of the heat and work crossing the boundary of the system. Fig.1.5 shows such a cycle in which a system commencing at condition ‘1’ changes in pressure and volume through a path 1-2-3 and returns to its initial condition ‘1’.

Basic Concepts

7

1.1.3.10. Point function When two properties locate a point on the graph (coordinate axes) then those properties are called as point function. Examples. Pressure, temperature, volume etc. 2

³1

dV

V2 – V1 (an exact differential )

1.1.3.11. Path function

Fig. 1.5. Cycle of operations. There are certain quantities which cannot be located on a graph by a point but are given by the area or so, on that graph. In that case, the area on the graph, pertaining to the particular process, is a function of the path of the process. Such quantities are called path functions. Examples : Heat, work etc. Heat and work are inexact differentials. Their change cannot be written as differences between their end states. 2

Thus

³ 1 GQ z Q 2

Similarly

³ 1 GW

2

– Q1 and is shown as 1Q2 or Q1–2

z W2 – W1 , and is shown as 1W2 or W1–2

Note. The operator G is used to denote inexact differentials and operator d is used to denote exact differentials.

1.1.3.12. Temperature The temperature is a thermal state of a body which distinguishes a hot body from a cold body. The temperature of a body is proportional to the stored molecular energy i.e., the average molecular kinetic energy of the molecules in a system. (A particular molecule does not have a temperature, it has energy; the gas as a system has temperature). Instruments for measuring ordinary temperatures are known as thermometers and those for measuring high temperatures are known as pyrometers. It has been found that a gas will not occupy any volume at a certain temperature. This temperature is known as absolute zero temperature. The temperatures measured with absolute zero as basis are called absolute temperatures. Absolute temperature is stated in degree centigrade. The point of absolute temperature is found to occur at 273ºC (app.) below the freezing point of water. Then, Absolute temperature = Thermometer reading in ºC + 273. Absolute temperature in degree centigrade is known as degree kelvin, denoted by K (SI units).

1.1.3.13. Pressure The pressure of a system is the force exerted by the system on unit area of boundaries. Units of pressure are : SI Units. N/m2 (sometimes called pascal, Pa) or bar 1 bar = 105 N/m2 = 105 Pa Standard atmospheric pressure = 1.01325 bar = 0.76 m Hg. MKS System. kgf/cm2, mm of mercury, metre or mm of water column. Standard atmospheric pressure = 1.033 kgf/cm2 (= 760 mm of Hg)

8

Chapter : 1

Technical pressure = 1 kgf/cm2. The pressure gauges, vacuum gauges or manometers are used for measuring fluid pressure. These devices indicate pressure relative to atmospheric pressure and this pressure is known as gauge pressure. To get absolute pressure, atmospheric pressure is added to the gauge pressure. In other words, Absolute pressure = Gauge pressure + atmospheric pressure. Vaccum is defined as the absence of pressure. A perfect vacuum is obtained when absolute pressure is zero, at this instant molecular momentum is zero. Energy, Work and Heat

1.1.3.14. Energy “Energy” is a general term embracing energy in transition and stored energy. The stored energy of a substance may be in the forms of mechanical energy and internal energy (other forms of stored energy may be chemical energy and electrical energy). Part of the stored energy may take the form of either potential energy (which is the gravitational energy due to height above a chosen datum line) or kinetic energy due to velocity. The balance part of the energy is known as internal energy. In a nonflow process usually there is no change of potential or kinetic energy and hence change of mechanical energy will not enter the calculations. In a flow process, however, there may be changes in both potential and kinetic energy and these must be taken into account while considering the changes of stored energy. Heat and work are the forms of energy in transition. These are the only forms in which energy can cross the boundaries of a system. Neither heat nor work can exist as stored energy.

1.1.3.15. Work Work is said to be done when a force moves through a distance. If a part of the boundary of a system undergoes a displacement under the action of a pressure, the work done W is the product of the force (pressure × area), and the distance it moves in the direction of the force. Fig. 1.6 (a) illustrates this with the conventional piston and cylinder arrangement, the heavy line defining the boundary of the system. Fig. 1.6 (b) illustrates another way in which work might be applied to a system. A force is exerted by the paddle as it changes the momentum of the fluid, and since this force moves during rotation of the paddle room work is done. Work is a transient quantity which only appears at the boundary while a change of state is taking place within a system. Work is ‘something’ which appears at the boundary when a system changes its state due to the movement of a part of the boundary under the action of a force.

Fig. 1.6.

Sign convention : l If the work is done by the system on the surroundings, e.g., when a fluid expands pushing a

piston outwards, the work is said to be positive.

i.e.,

Work output of the system = + W

Basic Concepts

9

l If the work is done on the system by the surroundings, e.g., when a force is applied to a

rotating handle, or to a piston to compress a fluid, the work is said to be negative.

i.e.,

Work input to system = – W

1.1.3.16. Heat Heat (denoted by the symbol Q), may be, defined in an analogous way to work as follows : “Heat is ‘something’ which appears at the boundary when a system changes its state due to a difference in temperature between the system and its surroundings”. Heat, like work, is a transient quantity which only appears at the boundary while a change is taking place within the system. It is apparent that neither GW or GQ are exact differentials and, therefore, any integration of elemental quantities of work or heat which appear during a change from state 1 to state 2 must be written as 2

³ 1 GW 2

³ 1 GQ

W1– 2 or 1W2 (or W ), and Q1– 2 or 1Q2 (or Q )

Sign convention : If the heat flows into a system from the surroundings, the quantity is said to be positive and conversely, if heat flows from the system to the surroundings it is said to be negative. In other words : Heat received by the system = + Q Heat rejected or given up by the system = – Q.

1.1.3.17. Comparison of work and heat Similarities : (i) Both are path functions and inexact differentials. (ii) Both are boundary phenomenon i.e., both are recognized at the boundaries of the system as they cross them. (iii) Both are associated with a process, not a state. Unlike properties, work or heat has no meaning at a state. (iv) Systems possess energy, but not work or heat. Dissimilarities : (i) In heat transfer temperature difference is required. (ii) In a stable system there cannot be work transfer, however, there is no restriction for the transfer of heat. (iii) The sole effect external to the system could be reduced to rise of a weight but in the case of a heat transfer other effects are also observed.

1.1.4. DIFFERENCES BETWEEN THERMODYNAMICS AND HEAT TRANSFER The fundamental differences between thermodynamics and heat transfer are given below : To understand the difference between thermodynamics and heat transfer, let us consider the cooling of a hot steel bar which is placed in a water bath. Thermodynamics may be used to predict the final equilibrium temperature of the steel bar-water combination; however, it will not help us to find out how long it takes to reach this equilibrium condition or what the temperature of the bar will be

10

Chapter : 1

after a certain length of time before the equilibrium condition is attained. Heat transfer on the other hand, may be used to predict the temperatures of both the bar and the water as a function of time. Heat transfer theory combines thermodynamics and rate equations together (to quantify the rate at which heat transfer occurs in terms of the degree of non-equilibrium). Thermodynamics

Heat transfer

1.

It deals with the equilibrium states of matter, and precludes the existence of a temperature gradient.

It is inherently a non-equlibrium process (since a temperature gradient must exist for exchange of heat to take place).

2.

When a system changes from one equilibrium state to another, thermodynamics helps to determine the quantity of work and heat interactions. It describes how much heat is to be exchanged during a process but does not hint how the same could be achieved.

It helps to predict the distribution of temperature and to determine the rate at which energy is transferred across a surface of interest due to temperature gradients at the s u r f a c e , a n d d i ff e r e n c e o f temperature between different surfaces.

1.1.5. BASIC LAWS GOVERNING HEAT TRANSFER The following are the basic laws which govern heat transfer : 1. First law of thermodynamics : In the early part of nineteenth century the scientists developed the concept of energy and hypothesis that it can neither be created nor destroyed; this came to be known as the ‘law of the conservation of energy’. The first law of thermodynamics is merely one statement of this general law/principle with particular reference to heat energy and mechanical work i.e., work, and is stated as follows : “When a system undergoes a thermodynamic cycle then the net heat supplied to the system from the surroundings is equal to the net work done by the system on its surroundings.

Ü dQ = Ü dW where

Ü represents the sum for the complete cycle”.

The first law of thermodynamics applies to reversible as well as irresversible transformations. For non-cyclic process, a more general formulation of first law of thermodynamics is required. A new concept which involves a term called internal energy fulfills this need. As per this law, the relationships for closed and open systems are as follows : Closed system : The net flow of heat across the system boundary + heat generated inside the system = change in the internal energy of system. Open system : The net energy transported through a control volume + energy generated within the control volume = change in the internal energy in the control volume. 2. Second law of thermodynamics: It states that “heat will flow naturally from one reservoir to another at a lower temperature, but not in opposite direction without assistance”. This law establishes the direction of energy transport as heat and postulates that the flow of energy as heat through a system boundary will always be in the direction of lower temperature (or along negative temperature gradient). 3. Law of conservation of mass. This law is used to determine the parameters of flow. 4. Newton’s laws of motion. These laws are used to determine fluid flow parameters.

Basic Concepts

11

5. The rate equations. These equations are made applicable depending upon the mode of heat transfer being considered.

1.1.6. MODES OF HEAT TRANSFER Heat transfer which is defined as the transmission of energy from one region to another as a result of temperature gradient takes place by the following three modes : (i) Conduction; (ii) Convection; (iii) Radiation. Heat transmission, in majority of real situations, occurs as a result of combinations of these modes of heat transfer. Example : The water in a boiler shell receives its heat from the fire-bed by conducted, convected and radiated heat from the fire to the shell, conducted heat through the shell and conducted and convected heat from the inner shell wall, to the water. Heat always flows in the direction of lower temperature. The above three modes are similar in that a temperature differential must exist and the heat exchange is in the direction of decreasing temperature; each method, however, has different controlling laws. Conduction : “Conduction” is the transfer of heat from one part of a substance to another part of the same substance, or from one substance to another in physical contact with it, without appreciable displacement of molecules forming the substance. In solids, the heat is conducted by the following two mechanisms : (i) By lattice vibration (the faster moving molecules or atoms in the hottest part of a body transfer heat by impacts some of their energy to adjacent molecules). (ii) By transport of free electrons (Free electrons provide an energy flux in the direction of decreasing temperature — For metals, especially good electrical conductors, the electrtonic mechanism is responsible for the major portion of the heat flux except at low temperature). In case of gases, the mechanism of heat conduction is simple. The kinetic energy of a molecule is a function of temperature. These molecules are in a continuous random motion exchanging energy and momentum. When a molecule from the high temperature region collides with a molecule from the low temperature region, it loses energy by collisions. In liquids, the mechanism of heat is nearer to that of gases. However, the molecules are more closely spaced and intermolecular forces come into play.

Convection : “Convection” is the transfer of heat within a fluid by mixing of one portion of the fluid with another. l Convection is possible only in a fluid medium and is directly linked with the transport of medium itself. l Convection constitutes the macroform of the heat transfer since macroscopic particles of a fluid moving in space cause the heat exchange. l The effectiveness of heat transfer by convection depends largely upon the mixing motion of the fluid. This mode of heat transfer is met with in situations where energy is transferred as heat to a flowing fluid at any surface over which flow occurs. This mode is basically conduction in a very thin fluid layer at the surface and then mixing caused by the flow. The heat flow depends on the properties of fluid and is independent of the properties of the material of the surface. However, the shape of the surface will influence the flow and hence the heat transfer.

12

Chapter : 1

Free or natural convection. Free or natural convection occurs when the fluid circulates by virtue of the natural differences in densities of hot and cold fluids; the denser portions of the fluid move downward because of the greater force of gravity, as compared with the force on the less dense. Forced convection. When the work is done to blow or pump the fluid, it is said to be forced convection.

Radiation : “Radiation” is the transfer of heat through space or matter by means other than conduction or convection. Radiation heat is thought of as electromagnetic waves or quanta (as convenient) an emanation of the same nature as light and radio waves. All bodies radiate heat; so a transfer of heat by radiation occurs because hot body emits more heat than it receives and a cold body receives more heat than it emits. Radiant energy (being electromagnetic radiation) requires no medium for propagation and will pass through vaccum. Note: The rapidly oscillating molecules of the hot body produce electromagnetic waves in hypothetical medium called ether. These waves are identical with light waves, radio waves and X-rays, differ from them only in wavelength and travel with an approximate velocity of 3 × 108 m/s. These waves carry energy with them and transfer it to the relatively slow- moving molecules of the cold body on which they happen to fall. The molecular energy of the later increases and results in a rise of its temperature. Heat travelling by radiation is known as radiant heat.

The properties of radiant heat in general, are similar to those of light. Some of the properties are : (i) It does not require the presence of a material medium for its transmission. (ii) Radiant heat can be reflected from the surfaces and obeys the ordinary laws of reflection. (iii) It travels with velocity of light. (iv) Like light, it shows intereference, diffraction and polarisation etc. (v) It follows the law of inverse square. The wavelength of heat radiations is longer than that of light waves, hence they are invisible to the eye.

1.2. HEAT TRANSFER BY CONDUCTION 1.2.1. FOURIER’S LAWS OF HEAT CONDUCTION Fourier’s law of heat conduction is an emperical law based on observation and states as follows : “The rate of flow of heat through a simple homogeneous solid is directly proportional to the area of the section at right angles to the direction of heat flow, and to change of temperature with respect to the length of the path of the heat flow”. Mathematically, it can be represented by the equation : dt Qv A. dx where, Q = Heat flow through a body per unit time (in watts), W, A = Surface area of heat flow ( perpendicular to the direction of flow), m2, dt = Temperature difference of the faces of block (homogeneous solid) of thickness ‘dx’ through which heat flows, °C or K, and dx = Thickness of body in the direction of flow, m. Thus, where,

dt ...(1.1) dx k = Constant of proportionality and is known as thermal conductivity of the body. Q

–k.A

Basic Concepts

13

The – ve sign of k [eqn. (1.1)] is to take care of the decreasing temperature alongwith the direction dt is always negative of increasing thickness or the direction of heat flow. The temperature gradient dx along positive x direction and, therefore, the value of Q becomes + ve.

Assumptions : The following are the assumptions on which Fourier’s law is based : 1. Conduction of heat takes place under steady state conditions. 2. The heat flow is unidirectional. 3. The temperatures gradient is constant and the temperature profile is linear. 4. There is no internal heat generation. 5. The bounding surfaces are isothermal in character. 6. The material is homogeneous and isotropic (i.e., the value of thermal conductivity is constant in all directions).

Some essential features of Fourier’s law : Following are some essential features of Fourier’s law : 1. It is applicable to all matter (may be solid, liquid or gas). 2. It is based on experimental evidence and cannot be derived from first principle. 3. It is a vector expression indicating that heat flow rate is in the direction of decreasing temperature and is normal to an isotherm. 4. It helps to define thermal conductivity ‘k’ (transport property) of the medium through which heat is conducted.

1.2.2. THERMAL CONDUCTIVITY OF MATERIALS From eqn. (1.1), we have k

Q dx . A dt

The value of k = 1 when Q = 1, A = 1 and

dt dx

1

Q dx 1 m (unit of k : W × . × = W/mK. or W/m°C ) 1 dt m 2 K (or °C) Thus, the thermal conductivity of a material is defined as follows : “The amount of energy conducted through a body of unit area, and unit thickness in unit time when the difference in temperature between the faces causing heat flow is unit temperature difference”. It follows from eqn. (1.1) that materials with high thermal conductivities are good conductors of heat, whereas materials with low thermal conductivities are good thermal insulator. Conduction of heat occurs most readily in pure metals, less so in alloys, and much less readily in non-metals. The very low thermal conductivities of certain thermal insulators e.g., cork is due to their porosity, the air trapped within the material acting as an insulator. Thermal conductivity (a property of material) depends essentially upon the following factors : (i) Material structure (ii) Moisture content (ii) Density of the material (iv) Pressure and temperature (operating conditions). Thermal conductivities (average values at normal pressure and temperature) of some common materials are as under :

Now k

14

Chapter : 1 Material

Thermal conductivity (k) (W/mK)

Material

Thermal conductivity (k) (W/mK)

1. Silver

410

8. Asbestos sheet

0.17

2. Copper

385

9. Ash

0.12

3. Aluminium

225

10. Cork, felt

0.05 – 0.10

4. Cast iron

55–65

11. Saw dust

0.07

5. Steel

20–45

12. Glass wool

0.03

6. Concrete

1.20

13. Water

0.55 – 0.7

7. Glass (window)

0.75

14. Freon

0.0083

Following points regarding thermal conductivity – its variation for different materials and under different conditions are worth noting : 1. Thermal conductivity of a material is due to flow of free electrons (in case of metals) and lattice vibrational waves (in case of fluids). 2. Thermal conductivity in case of pure metals is the highest (k = 10 to 400 W/m°C). It decreases with increase in impurity. The range of k for other materials is as follows : Alloys : k = 12 to 120 W/m °C Heat insulating and building materials : k = 0.023 to 2.9 W/m°C Liquids : k = 0.2 to 0.5 W/m°C Gases and vapours : k = 0.006 to 0.05 W/m°C 3. Thermal conductivity of a metal varies considerably when it (metal) is heat treated or mechanically processed / formed. 4. Thermal conductivity of most metals decreases with the increase in temperature (aluminium and uranium being the exceptions). – In most of liquids the value of thermal conductivity tends to decrease with temperature (water being an exception) due to decrease in density with increase in temperature. – In case of gases the value of thermal conductivity increases with temperature. Gases with higher molecular weights have smaller thermal conductivities than with lower molecular weights. This is because the mean molecular path of gas molecules decreases with increase in density and k is directly proportional to the mean free path of the molecule. 5. The dependence of thermal conductivity (k) on temperature, for most materials is almost linear; ...(1.2) k = k0 (1 + Et) where, k0 = Thermal conductivity at 0°C, and E = Temperature coefficient of thermal conductivity, 1/°C (It is usually posi- tive for non-metals and insulating materials (magnesite bricks being the exception) and negative for metallic conductors (aluminium and certain non-ferrous alloys are the exceptions). 6. In case of solids and liquids, thermal conductivity (k) is only very weakly dependent on pressure; in case of gases the value of k is independent of pressure (near standard atmospheric). 7. In case of non-metallic solids :

Basic Concepts

15

–

Thermal conductivity of porous materials depends upon the type of gas or liquid present in the voids. – Thermal conductivity of a damp material is considerably higher than that of the dry material and water taken individually. – Thermal conductivity increases with increase in density. 8. The Wiedemann and Franz law (based on experiment results), regarding thermal and electrical conductivities of a material, states as follows : “The ratio of the thermal and electrical conductivities is the same for all metals at the same temperature; and that the ratio is directly proportional to the absolute temperature of the metal.” Mathematically,

k vT V

k C ...(1.3) VT where, k = Thermal conductivity of metal at temperature T(K), V = Electrical conductivity of metal at temperature T (K), and C = Constant (for all metals), referred to as Lorenz number (= 2.45 × 10–8 W:/K2; : stands for ohms). This law conveys that the materials which are good conductors of electrticity are also good conductors of heat.

or,

1.2.3. THERMAL RESITANCE (Rth) When two physical systems are described by similar equations and have similar boundary conditions, these are said to be analogous. The heat transfer processes may be compared by analogy with the flow of electricity in an electrical resistance. As the flow of electric current in the electrical resistance is directly proportional to potential difference (dV); similarly heat flow rate, Q, is directly proportional to temperature difference (dt), the driving force for heat conduction through a medium. As per Ohm’s law (in elctric-circuit theory), we have Current ( I )

Potential difference (dV ) Electrical resistance (R )

...(1.4)

By analogy, the heat flow equation (Fourier’s equation) may be written as Temperature difference (dt ) ...(1.5) § dx · ¨ ¸ © kA ¹ By comparing eqns. (1.4) and (1.5), we find that I is analogous to, Q, dV is analogous to dt and dx § dx · R is analogous to the quantity ¨ ¸ . The quantity is called © kA ¹ kA thermal conduction resistance (Rth)cond. i.e., dx (R th )cond. Fig. 1.7. kA x The reciprocal of the thermal resistance is called thermal conductance. Heat flow rate (Q)

x It may be noted that rules for combining electrical resistances in series and parallel apply equally well to thermal resistances. The concept of thermal resistance is quite helpful while making calculations for flow of heat.

16

Chapter : 1

Example 1.1. Calculate the rate of heat transfer per unit area through a copper plate 45 mm thick, whose one face is maintained at 350°C and the other face at 50°C. Take thermal conductivity of copper as 370 W/m°C. Solution. Temperature difference, dt (= t2 – t1) = (50 – 350) Thickness of copper plate, L = 45 mm = 0.045 m Thermal conductivity of copper, k = 370 W/m°C Rate of heat-transfer per unit area, q : From Fourier’s law dt (t – t1 ) Q – kA – kA 2 dx L or,

...(Eqn. 1.1) dt q –k dx (50 – 350) – 370 u 0.045 6 = 2.466 × 10 W/m2 or Q A

2.466 MW/m2 (Ans.)

Fig. 1.8.

Example 1.2. A plane wall is 150 mm thick and its wall area is 4.5 m2. If its conductivity is 9.35 W/m°C and surface temperatures are steady at 150°C and 45°C, determine : (i) Heat flow across the plane wall; (ii) Temperature gradient in the flow direction. Solution. Thickness of the plane wall, L = 150 mm = 0.15 m Area of the wall, A = 4.5 m2 Temperature difference, dt = t2 – t1 = 45 – 150 = – 105°C Thermal conductivity of wall material, k = 9.35 W/m°C (i) Heat flow across the plane wall, Q : As per Fourier’s law, Q

(t2 – t1 ) L (–105) 29452.5 W – 9.35 u 4.5 u 0.15

– kA

dt dx

– kA

dt : dx From Fourier’s law, we have dt Q 29452.5 – 700°C/m – dx kA 9.35 u 4.5 Example 1.3. The following data relate to an oven : Thickness of side wall of the oven = 82.5 mm Thermal conductivity of wall insulation = 0.044 W/m°C Temperature on inside of the wall = 175°C

(ii) Temperature gradient,

Basic Concepts

17

Energy dissipated by the electrical coil within the oven = 40.5 W Determine the area of wall surface, perpendicular to heat flow, so that temperature on the other side of the wall does not exceed 75°C. Solution. Given : x = 82.5 mm = 0.0825 m; k = 0.044 W/m°C; t1 = 175°C; t2 = 75°C; Q = 40.5W Area of the wall surface, A : Assuming one-dimentional steady state heat conduction, Rate of electrical energy dissipation in the oven. = Rate of heat transfer (conduction) across the wall i.e.

Q

or,

40.5

or,

A

dt (t – t1 ) kA (t1 – t2 ) – kA 2 dx x x 0.044 A (175 – 75) 0.0825 40.5 u 0.0825 0.759 m 2 0.044 (175 – 75) – kA

1.3. HEAT TRANSFER BY CONVECTION The rate equation for the convective heat transfer (regardless of particular nature) between a surface and an adjacent fluid is prescribed by Newton’s law of cooling (Refer Fig. 1.9) ...(1.6) Q = hA (ts – tf) where, Q = Rate of conductive heat transfer, A = Area exposed to heat transfer, ts = Surface temperature, tf = Fluid temperature, and h = Co-efficient of convective heat transfer. The units of h are, h

Q A (t s – t f )

W m 2 °C

or W/m 2 °C

or, W/m2K The coefficient of convective heat transfer ‘h’ (also known as film heat transfer coefficient) may be defined as “the amount of heat transmitted for a unit temperature difference between the fluid and unit area of surface in unit time.” The value of ‘h’ depends on the following factors : (i) Thermodynamic and transport properties (e.g. viscosity, density, specific heat etc.). (ii) Nature of fluid flow. (iii) Geometry of the surface. (iv) Prevailing thermal conditions. Since ‘h’ depends upon several factors, it is difficult to frame a single equation to satisfy all the variations, however, by dimensional analysis an equation for the purpose can be Fig. 1.9. Convective heat-transfer obtained.

18

Chapter : 1

The mechanisms of convection in which phase changes are involved lead to the important fields ts – t f º 1 ª ...Eqn (1.6) » is «Q of boiling and condensation. Refer Fig. 1.9 (b). The quantity (1/ hA) hA ¬ ¼ called convection thermal resistance [(Rth)conv.] to heat flow. Example 1.4. A hot plate 1m × 1.5 m is maintained at 300°C. Air at 20°C blows over the plate. If the convective heat transfer coefficient is 20W/m2°C, calculate the rate of heat transfer. Solution. Area of the plate exposed to heat transfer, A = 1 × 1.5 = 1.5 m2 Plate surface temperature, ts = 300°C Temperature of air (fluid), tf = 20°C Connvective heat-transfer coefficient, h = 20 W/m2 °C Rate of heat transfer, Q : From Newton’s law of cooling, Q = hA (ts – tf) = 20 × 1.5 (300 – 20) = 8400 W or 8.4 kW Example 1.5. A wire 1.5 mm in diameter and 150 mm long is submerged in water at atmospheric pressure. An electric current is passed through the wire and is increased until the water boils at 100°C. Under the condition if convective heat transfer coefficient is 4500 W/m2°C find how much electric power must be supplied to the wire to maintain the wire surface at 120°C ? Solution. Diameter of the wire, d = 1.5 mm = 0.0015 m Length of the wire, L = 150 mm = 0.15 m ? Surface area of the wire (exposed to heat transfer), A = Sd L = S × 0.0015 × 0.15 = 7.068 × 10–4 m2 Wire surface temperature, ts = 120°C Water temperature, tf = 100°C Convective heat transfer coefficient, h = 4500 W/m2 °C Electric power to be supplied : Electric power which must be supplied = Total convection loss (Q) ? Q = hA (ts – tf) = 4500 × 7.068 × 10–4 (120 – 100) = 63.6 W

1.4. HEAT TRANFER BY RADIATION Laws of Radiation : 1. Wien’s law. It states that the wavelength Om corresponding to the maximum energy is inversely proportional to the absolute temperature T of the hot body. 1 Om v i.e., or, OmT = constant ...(1.7) T 2. Kirchhoff’s law. It states that the emissivity of the body at a particular temperature is numerically equal to its absorptivity for radiant energy from body at the same temperature. 3. The Stefan-Boltzmann law. The law states that the emissive power of a black body is directly proportional to fourth power of its absolute temperature. i.e.,

Q v T4

...(1.8)

Basic Concepts

19

Refer Fig. 1.10 (a) Q = F VA (T14 – T24) where,

...(1.9)

F = A factor depending on geometry and surface properties, V = Stefan-Boltzmann constant = 5.67 × 10–8 W/m2K4, A = Area, m2, and

T1, T2 = Temperatures, degrees kelvin (K). This equation can also be rewritten as : T1 – T2 Fig. 1.10. Heat transfer by radiation. Q ...(1.10) 1/[ F V A (T1  T2 ) (T12  T22 )] where denomenator is radiation thermal resistance, (Rth)rad. [Fig. 1.10 (b)] i.e.,

(Rth)rad = 1/ [F VA (T1 + T2) (T12 + T22)]

The values of F are available for simple configurations in the form of charts and tables. F =1

... for simple cases of black surface enclosed by other surface

F = emissivity (H) ... for non-black surface enclosed by other surface. [Emissivity (H) is defined as the ratio of heat radiated by a surface to that of an ideal surface.] Example 1.6. A surface having an area of 1.5 m2 and maintained at 300°C exchanges heat by radiation with another surface at 40°C. The value of factor due to the geometric location and emissivity is 0.52. Determine : (i) Heat lost by radiation, (ii) The value of thermal resistance, and (iii) The value of equivalent convection coefficient. Solution. Given : A = 1.5 m2; T1 = t1 + 273 = 300 + 273 = 573K; T2 = t2 + 273 = 40 + 273 = 313K; F = 0.52. (i) Heat lost by radiation, Q : Q = F V A (T14 – T24 ) ...[Eqn. (1.9)] –8 2 4 (where V = 5.67 × 10 W/m K ) or, Q = 0.52 × 5.67 × 10–8 × 1.5 [(573)4 – (313)4]

or,

ª§ 573 ·4 § 313 ·4 º u u 0.52 5.67 1.5 «¨ ¸ –¨ ¸ » © 100 ¹ ¼ ¬© 100 ¹ (Please note this step) Q = 4343 W

(ii) The value of thermal resistance, (Rth)rad : We know that, ?

Q

(T1 – T2 ) ( Rth ) rad

( Rth ) rad.

(T1 – T2 ) Q

...[Eqn. (1.10)] (573 – 313) 4343

0.0598 °C/W

20

Chapter : 1 (iii) The value of equivalent convection coefficient, hr : Q = hrA (t1 – t2) hr

or,

Q A (t1 – t2 )

4343 1.5 (300 – 40)

11.13 W/m 2 °C

hr = F V (T1 + T2) (T12 + T22 )

ª Alternatively, « « «¬

= 0.52 × 5.67 ×

10–8

= 11.13 W/m2 °C

...From eqn. (1.10) º » (573 + 313) (5732 + 3132) » »¼

Example 1.7. A carbon steel plate (thermal conductivity = 45 W/m°C) 600 mm × 900 mm × 25 mm is maintained at 310°C. Air at 15°C blows over the hot plate. If convection heat transfer coefficient is 22 W/m2 °C and 250 W is lost from the plate surface by radiation, calculate the inside plate temperature. Solution. Area of the plate exposed to heat transfer, A = 600 mm × 900 mm = 0.6 × 0.9 = 0.54 m2 Thickness of the plate, L = 25 mm = 0.025 m Surface temperature of the plate, ts = 310°C Temperature of air (fluid), tf = 15°C Convective heat transfer coefficient, h = 22 W/m2°C Heat lost from the plate surface by radiation, Qrad. = 250W Thermal conductivity, k = 45 W/m °C

Inside plate temperature, ti : In this case the heat conducted through the plate is removed from the plate surface by a combination of convection and radiation. Heat conducted through the plate = Convection heat losses + radiation heat losses. or, Qcond. = Qconv. + Qrad. – kA

or, or, or, or,

– 45 u 0.54 u – 45 u 0.54 u

dt dx

hA (ts – t f )  F VA (Ts 4 – T f 4 )

(ts – ti ) = 22 × 0.54 (310 – 15) + 250 (given) L

(310 – ti ) 0.025

= 22 × 0.54 × 295 + 250

972 (ti – 310) = 3754.6 ti

3754.6  310 972

313.86°C

Basic Concepts

21

Fig. 1.11. Combination of conduction, convection and radiation heat transfer.

Example 1.8. A surface at 250°C exposed to the surroundings at 110°C convects and radiates heat to the surroundings. The convection coefficient and radiation factor are 75W/m2°C and unity respectively. If the heat is conducted to the surface through a solid of conductivity 10W/m°C, what is the temperature gradient at the surface in the solid ? Solution. Temperature of the surface, Temperature of the surroundings, The convection co-efficient, Radiation factor, Boltzmann constant, Conductivity of the solid,

ts tsur h F V k

= = = = = =

250°C 110°C 5W/m2°C 1 5.67 × 10–8 W/m2K4 10W/m°C

dt : dx Heat conducted through the plate = Convection heat losses + radiation heat losses

Temperature gradient,

Qcond. = Qconv. + Qrad. – kA dt hA (t – t )  F VA (T 4 – T 4 ) s sur s sur dx Substituting the values, we have dt – 10 u = 75 (250 – 110) + 1 × 5.67 × 10–8 [(250 + 273)4 – (110 + 273)4] dx ª§ 523 ·4 § 383 ·4 º dt – 10 u 10500  5.67 «¨ ¸ –¨ ¸ » dx © 100 ¹ ¼ ¬© 100 ¹ = 10500 + 3022.1 = 13522.1 dt 13522.1 – 1352.21 °C/m – ? dx 10 i.e.,

HIGHLIGHTS 1. 2. 3.

The energy in transit is termed heat. Heat transfer may be defined as “The transmission of energy from one region to another as a result of temperature gradient.” The study of heat transfer is carried out for the following purposes : (i) To estimate the rate of flow of energy as heat through the boundary of a system under study (both steady and transient conditions). (ii) To determine the temperature field under steady and transient conditions.

22

Chapter : 1 4.

5. 6.

7.

8.

Thermodynamics is an axiomatic science which deals with the relations among heat, work and properties of system which are in equilibrium. It describes state and changes in state of physical systems. It basically entails four laws or axioms known as Zeroth, First, Second and third law of thermodynamics. Heat transfer theory combines thermodynamics and rate equations together (to quantify the rate at which heat transfer occurs in terms of the degree of non-equilibrium). Basic laws which govern the heat transfer are : (i) First laws of thermodynamics (ii) Second law of thermodynamics (iii) Law of conservation of mass (iv) Newton’s laws of motion (v) The rate equations. Heat transfer takes place by the following three modes : (i) Conduction (ii) Convection (iii) Radiation. ‘Conduction’ is the transfer of heat from one part of a substance to another part of the same substance, or from one substance to another in physical contact with it, without appreciable displacement of molecules forming the substance. ‘Convection’ is the transfer of heat within a fluid by mixing of one portion of the fluid with another. Convection is possible only in a fluid medium and is directly linked with the transport of medium itself. ‘Radiation’ is the transfer of heat through space or matter by means other than conduction or convection. Radiant energy (being electromagnetic radiation) requires no medium for propagation and will pass through a vaccum. Fourier’s law of heat conduction states : “The rate of flow of heat through a single homogeneous solid is directly proportional to the area of the section at right angles to the direction of heat flow, and to change of temperature with respect to the length of path of the heat flow”. Mathematically, where,

Q

– kA

dt dx

Q = Heat flow through a body per unit time, W, A = Surface area of heat flow (perpendicular to the direction of flow), m2, dt = Temperature difference of the faces of the block, dx = Thickness of the body, and k = Thermal conductivity of the body.

9.

The –ve sign of k is to take care of the decreasing temperature along with the direction of increasing thickness or the direction of heat flow. The thermal conductivity (k) of a material is defined as : “The amount of energy conducted through a body of unit area, and unit thickness in unit time when the difference in temperature between the faces causing heat flow is unit temperature difference.”

10. The rate equation for the convective heat transfer (regardless of particular nature) between a surface and an adjacent fluid is prescribed Newton’s law of cooling : Q = hA (ts – tf) where, Q = Rate of convective heat transfer, A = Area exposed to heat transfer, ts = Surface temperature, tf = Fluid temperature, and h = Coefficient of convective heat transfer. Units of h : W/m2°C or W/m2K 11. The Stefan-Boltzmann law states : “The emissive power of a black body is directly proportional to fourth power of its absolute temperature.”

Basic Concepts

23

12. Thermal resistance (Rth) : dx kA 1 Convection thermal resistance, (Rth)conv. = hA

Conduction thermal resistance, (Rth)cond. =

Radiation thermal resistance,

(Rth)rad. =

1 1/[ F V A (T1  T2 ) (T12  T2 2 )]

THEORETICAL QUESTIONS 1.

Define the following terms :

(i) Heat (ii) Heat transfer (iii) Thermodynamics. 2. What is the difference between thermodynamics and heat transfer ? 3. Enumerate the basic laws which govern the heat transfer. 4. Name and explain briefly the various modes of heat transfer. 5. What is conduction heat transfer ? How does it differ from convective heat transfer ? 6. What is the significance of heat transfer ? 7. Enumerate some important areas which are covered under the discipline of heat transfer. 8. What is the difference between the ‘natural’ and ‘forced’ convection ? 9. What is ‘Fourier’s law of conduction’? State also the assumptions on which this law is based. 10. State some essential features of Fourier’s law. 11. How is thermal conductivity of a material defined ? What are its units ? 12. What is thermal resistance ? 13. What is ‘Newton’s law of cooling ? 14.

What is Stefan’s Boltzmann law ?

UNSOLVED EXAMPLES 1.

The inner surface of a plane brick wall is at 40°C and the outer surface is at 20°C. Calculate the rate of heat transfer per m2 of surface area of the wall, which is 250 mm thick. The thermal conductivity of the brick is 0.52 W/m°C. (Ans. 41.6 W/m2)

2. A plane wall (thermal conductivity = 10.2 W/m°C) of 100 mm thickness and area 3m2 has steady surface temperature of 170°C and 100°C. Determine : (i) The rate of heat flow across the plane wall; (ii) The temperature gradient in the flow direction.

[Ans. (i) 21.42 kW; (ii) – 700°C/m]

3. Determine the heat transfer by convection over a surface of 0.75 m2 if the surface is at 200°C and the fluid is at 80°C. The value of convective heat transfer is 25 W/m2 °C. [Ans. 2.25 kW] 4.

A surface of area 3m2 and at 200°C exchanges heat with another surface at 30°C by radiation. If the value of factor due to the geometric location and emissivity is 0.69, determine : (i) The rate of heat transfer, (ii) The value of thermal resistance, and (iii) The equivalent convection coefficient. [Ans. (i) 4885.6W; (ii) 0.0348 °C/W; (iii) 9.58 W/m2 °C]

5. A surface at 200°C exposed to the surroundings at 60°C convects and radiates heat to the surroundings. The convection coefficient and radiation factor are 80W/m2 °C and unity respectively. If the heat is conducted to the surface through a solid of conductivity 15W/m °C what is the temperature gradient at the surface ? (Ans. – 889.4 °C/m)

24

Chapter : 1

PART – I

HEAT TRANSFER BY “CONDUCTION”

C H A P T E R

“Conduction” Heat Transfer at Steady State–One Dimension 2.1. INTRODUCTION

2.1. Introduction. 2.2. General heat conduction equation in cartesian coordinates. 2.3. General heat conduction equation in cylindrical coordinates. 2.4. General heat conduction equation in spherical coordinates. 2.5. Conduction through plane and composite walls. 2.6. Heat conduction through hollow and composite cylinders – heat conduction through a hollow cylinder – logarithmic mean area for the hollow cylinder– heat conduction in a composite cylinder. 2.7. Heat conduction through hollow and composite spheres – heat conduction through hollow sphere – heat conduction through a composite sphere. 2.8. Critical thickness of insulation. 2.9. Heat conduction with internal heat generation – plane wall with uniform heat generation – Dielectric heating – cylinder with uniform heat generation – heat transfer through the piston crown. 2.10. Heat transfer from extended surfaces (fins).

In this chapter an attempt will be made to derive general heat conduction equation and examine the applications of Fourier’s law of heat conduction to the calculation of heat flow in some simple onedimensional systems. Under the category of one-dimensional systems several different physical shapes may fall; when the temperature of the body is a function only of radial distance and is independent of azimuth angle or axial distance cylindrical and spherical systems are treated as onedimensional. In case of problems of two-dimensional nature the effect of a second-space coordinate may be so small that it may be neglected and the heatflow problems of multi-dimensional type may be approximated with a one-dimensional analysis; in such cases the differential equations are simplified and as a consequence of this simplification much easier solution in available.

2.2. GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES Consider an infinitesimal rectangular parallelopiped (volume element) of sides dx, dy and dz parallel, respectively, to the three axes (X, Y, Z) in a medium in which temperature is varying with location and time as shown in Fig. 2.1. Let, t = Temperature at the left face ABCD; this temperature may be assumed uniform over the entire surface, since the area of this face can be made arbitrarily small, and dt = Temperature changes and rate of dx change along X-direction.

Chapter : 2

28 Then,

§ wt · ¨ ¸ dx = Change of temperature through distance dx, and © wx ¹

§ wt · t  ¨ ¸ dx = temperature on the right face EFGH (at a distance dx from the left face © wx ¹ ABCD). Further, let, kx, ky, kz = Thermal conductivities (direction characteristics of the material) along X, Y and Z axes.

Fig. 2.1. Elemental volume for three-dimensional heat conduction analysis - Cartesian coordinates.

If the directional characterisitics of a material are equal/same, it is called an “Isotropic material” and if unequal/different “Anisotropic material’. qg = Heat generated per unit volume per unit time. Inside the control volume there may be heat sources due to flow of electric current in electric motors and generators, nuclear fission etc. (Note : qg may be function of position or time, or both). U = Mass density of material, and c = Specific heat of the material. Energy balance/equation for volume element : Net heat accumulated in the element due to conduction of heat from all the coordinate directions considered (A) + heat generated within the element (B) = Energy stored in the element (C). ...(1) Let,

Q = Rate of heat flow in a direction, and Q' = (Q.dW) = Total heat flow (flux) in that direction (in time dW).

A. Net heat accumulated in the element due to conduction of heat from all the directions considered: Quantity of heat flowing into the element from the left face ABCD during the time interval dW in X-direction is given by :

“Conduction” Heat Transfer at Steady State–One Dimension

29

wt · dW ...(i) wx During the same time interval dW the heat flowing out of the right face of control volume (EFGH) will be : w Q c( x  dx ) Qxc  (Q cx ) dx Heat efflux, ...(ii) wx ? Heat accumulation in the element due to heat flow in X-direction,

Heat influx,

Qxc

– k x ( dy.dz )

w ª º (Q cx ) dx » Qxc – «Qxc  [Subtracting (ii) from (i)] wx ¬ ¼ w – (Qxc ) dx wx w ª wt º – – k x (dy.dz ) . d W » dx « wx ¬ wx ¼ w ª wt º kx dx.dy.dz.d W ...(2.1) wx «¬ wx »¼ Similarly the heat accumulated due to heat flow by conduction along Y and Z directions in time dW will be : w ª wt º dQyc ky dx.dy.dz.d W ...(2.2) wy «¬ wy »¼ dQ cx

w ª wt º kz dx.dy.dz.d W ...(2.3) wz «¬ wz »¼ ? Net heat accumulated in the element due to conduction of heat from all the coordinate directions considered w ª wt º w ª wt º w ª wt º k x » dx.dy.dz.d W  k y » dx.dy.dz.d W  kz dx.dy.dz.d W « « wx ¬ wx ¼ wy ¬ wy ¼ wz «¬ wz »¼ ª w § wt · w § wt · w § wt · º « wx ¨© k x wx ¸¹  wy ¨ k y wy ¸  wz ¨© k z wz ¸¹ » dx.dy.dz.d W ...(2.4) © ¹ ¬ ¼ B. Total heat generated within the element (Qg' ) : The total heat generated in the element is given by Qg' = qg (dx.dy.dz) dW ...(2.5) C. Energy stored in the element : The total heat accumulated in the element due to heat flow along coordinate axes (Eqn. 2.4) and the heat generated within the element (Eqn. 2.5) together serve to increase the thermal energy of the element/lattice. This increase in thermal energy is given by wt U (dx.dy.dz ) c. . d W ...(2.6) wW [Q Heat stored in the body = Mass of the body × specific heat of the body material × rise in the temperature of body]. Now, substituting eqns. (2.4), (2.5), (2.6), in the eqn. (1), we have dQzc

ª w § wt · w § wt · w § wt · º « wx ¨ k x wx ¸  wy ¨ k y wy ¸  wz ¨ k z wz ¸ » dx.dy.dz.d W  qg (dx.dy.dz.)d W ¹ © ¹¼ © ¹ ¬ © Dividing both sides by dx.dy.dz.dW, we have w § wt · w § wt · w § wt · wt  ¨ k z ¸  qg U . c. ¨ kx ¸  ¨ k y ¸ wx © wx ¹ wy © wy ¹ wz © wz ¹ wW

U (dx.dy.dz ) c.

wt . dW wW

...(2.7)

30

Chapter : 2 or, using the vector operator ’, we get

wt ...[2.7 (a)] wW This is known as the general heat conduction equation for ‘non-homogeneous material’, ‘self heat generating’ and ‘unsteady three-dimensional heat flow’. This equation establishes in differential form the relationship between the time and space variation of temperature at any point of solid through which heat flow by conduction takes place. General heat conduction equation for constant thermal conductivity : In case of homogeneous (in which properties e.g., specific heat, density, thermal conductivity etc. are same everywhere in the material) and isotropic (in which properties are independent of surface orientation) material, kx = ky = kz = k and diffusion equation Eqn. (2.7) becomes ’ .(k ’t )  qg

w 2t wx

2



w 2t wy

2



w 2t wz

2



U . c.

qg

U.c wt · k wW

k

k U.c

D

where,

1 wt · D wW

...(2.8)

Thermal conductivity Thermal capacity

k is known as thermal diffusivity. U.c – The larger the value of D, the faster will the heat diffuse through the material and its temperature will change with time. This will result either due to a high value of thermal conductivity k or a low value of heat capacity U.c. A low value of heat capacity means the less amount of heat entering the element, would be absorbed and used to raise its temperature and more would be available for onward transmission. Metals and gases have relatively high value of D and their response to temperature changes is quite rapid. The non-metallic solids and liquids respond slowly to temperature changes because of their relatively small value of thermal diffusivity. – Thermal diffusivity is an important characteristic quantity for unsteady conduction situations. Eqn. (2.8) by using Laplacian ’2, may be written as : qg 1 wt ’ 2t  . ...[2.8 (a)] D wW k Eqn. (2.8), governs the temperature distribution under unsteady heat flow through a material which is homogeneous and isotropic. Other simplified forms of heat conduction equation in cartesian coordinates : (i) For the case when no internal source of heat generation is present, Eqn. (2.8) reduces to The quantity,

D

w 2t wx

2



w 2t wy

2



w 2t wz

2

1 wt § wt · · z 0 ¸ heat flow with no D wW [Unsteady state ¨© wW ¹ internal heat generation]

1 wt (Fourier’s equation) ...(2.9) · D wW (ii) Under the situations when temperature does not depend on time, the conduction then takes wt § · 0 ¸ and the eqn. (2.8) reduces to place in the steady state ¨ i.e., wW © ¹ 2 2 2 qg w t w t w t  2  2  0 2 k wx wy wz

or,

’ 2t

“Conduction” Heat Transfer at Steady State–One Dimension

’2t 

qg

0 (Poisson’s equation) k In the absence of internal heat generation, Eqn. (2.10) reduces to or,

w 2t

w 2t



w 2t

0 wx 2 (v) Steady state, two dimensional, without internal heat generation w 2t

w 2t

0 wx 2 wy 2 (vi) Unsteady state, one dimensional, without internal heat generation 

...(2.10)

w 2t

0 wy 2 wz 2 or, ’2t = 0 (Laplace equation) (iii) Steady state and one-dimensional heat transfer: qg w 2t  0 2 k wx (iv) Steady state, one-dimensional, without internal heat generation wx 2



31

w 2t wx 2

1 wt · D wW

...(2.11)

...(2.12)

...(2.13)

...(2.14)

...(2.15)

2.3. GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES While dealing with problems of conduction of heat through systems having cylindrical geometries (e.g., rods and pipes) it is convenient to use cylindrical coordinates. Consider an elemental volume having the coordinates (r, I, z), for three-dimensional heat conduction analysis, as shown in Fig. 2.2.

Fig. 2.2. Elemental volume for three-dimensional heat conduction analysis - Cylindrical coordinates.

The volume of the element = rdI.dr.dz Let, qg = Heat generation (uniform) per unit volume per unit time.

32

Chapter : 2

Further, let us assume that k (thermal conductivity), U (density), c (specific heat) do not alter with position. A. Net heat accumulated in the element due to conduction of heat from all the coordinate directions considered : Heat flow in radial direction (x–I) plane : Heat influx,

– k ( rd I.dz )

Qrc

wt . dW wr

...(i)

w (Qr ) dr ...(ii) wr ? Heat accumulation in the element due to heat flow in radial direction, dQ'r = Q'r – Q'(r + dr) [subtracting (ii) from (i)]

Heat efflux,

Q(cr  dr )

Qrc 

w wr w – wr

–

(Qrc ) dr

wt ª º «¬ – k (rd I.dz ) wr . d W »¼ dr w § wt · k (dr .d I.dz ) ¨r · ¸ dW wr © wr ¹ § w 2t wt · k (dr .d I.dz ) ¨ r 2  ¸ dW wr ¹ © wr ª w 2 t 1 wt º k (dr.rd I.dz ) « 2  ...(2.16) » dW r wr ¼ ¬ wr Heat flow in tangential direction (r–z) plane : wt dW Heat influx, Q'I – k (dr.dz ) ...(iii) r.wI w Q(cI  d I) QIc  (QIc ) rd I Heat efflux, ...(iv) r.wI Heat accumulated in the element due to heat flow in tangential direction, dQ'I = Q'I – Q'(I + dI) [subtracting (iv) from (iii)] w – (QI ) r.d I r.wI w ª wt º – – k (dr.dz ) . d W » r.d I r.wI «¬ r.wI ¼ w § 1 wt · k (dr .d I.dz ) · dW wI ¨© r wI ¸¹

k (dr.rd I.dz )

1

w 2t

. dW r 2 wI2 Heat flow in axial direction (r-I plane) : wt Qzc – k (r.d I.dr ) dW Heat influx, wz w (Qzc ) dz Heat efflux, Q(cz  dz ) Qzc  wz Heat accumulated in the element due to heat flow in axial direction, ·

...(2.17)

...(v) ...(vi)

“Conduction” Heat Transfer at Steady State–One Dimension

33

dQ'z = Q'z – Q'(z + dz) [subtracting (vi) from (v)] w ª wt º – – k (r.d I.dr ) . d W » dz wz «¬ wz ¼ w 2t k (dr.rd I.dz ) 2 . d W ...(2.18) wz Net heat accumulated in the element ª w 2 t 1 wt w 2t º 1 w 2t  2 · 2  2 » dW k .dr .rd I.dz « 2  · ...(2.19) r wr r wI wz ¼ ¬ wr B. Heat generated within the element (Q'g) : The total heat generated within the element is given by ...(2.20) Q'g = qg (dr.rdI.dz).dW C. Energy stored in the element : The increase in thermal energy in the element is equal to wt U (dr.rd I.dz ). c. . d W ...(2.21) wW Now, (A) + (B) = (C) ... Energy balance/equation ª w 2 t 1 wt w 2t º 1 w 2t  2 · 2  2 » d W  q g (dr.rd I.dz ).d W k .dr .rd I.dz « 2  · r wr r wI wz ¼ ¬ wr wt U (dr.rd I.dz ).c. . d W wW Dividing both sides by dr.rdI.dz.dW, we have

?

ª w 2 t 1 wt w 2t º wt 1 w 2t  2 · 2  2 »  q g U.c. k « 2  · wW r wr r wI wz ¼ ¬ wr 2 2 2 ª w t 1 wt w t º qg U c wt 1 w t 1 wt  2 · 2  2»  · · « 2  · or, ...(2.22) r wr r k k wW D wW wI wz ¼ ¬ wr Equation (2.22) is the general heat conduction equation in cylindrical coordinates. In case there are no heat sources present and the heat flow is steady and one-dimensional, then eqn. (2.22) reduces to w 2 t 1 wt  · 0 ...(2.23) wr 2 r wr w 2 t 1 dt  · 0 or, wr 2 r dr 1 d § dt · · or, ¨r · ¸ 0 r dr © dr ¹ 1 Since z 0, therefore, r d § dt · dt constant ...(2.24) ¨r · ¸ or r · dr © dr ¹ dr Equation (2.22) can also be derived by transformation of coordinates, as follows : x = r cos I, y = r sin I and z = z Now, by chain rule : wt wt wx wt wy wt wt ·  · cos I  sin I wr wx wr wy wr wx wy

Chapter : 2

34

cos I

or,

wt wr

wt wI sin I wt · r wI

Also, or,

wt wt  sin I . cos I . ...(i) wx wy (Multiplying both sides by cos I) wt wx wt wy wt wt ·  · (– r sin I)  (r cos I) wx wI wy wI wx wy wt wt  sin I . cos I . – sin 2 I ...(ii) wx wy

cos 2 I ·

(Multiplying both sides by From Eqns. (i) and (ii), we have wt ª wt wt º sin I wt  «cos I . · – sin 2 I – cos 2 I » wI wx ¬ wr wx ¼ r wt wt –  cos I wx wr wt wt sin I wt cos I – · ? wx wr r wI Differentiating both sides with respect to x, we have w § wt · w ª wt sin I wt º cos I · – · » ¨ ¸ « wx © wx ¹ wx ¬ wr r wI ¼ or,

w 2t

wx

2

w § wt · sin I w § wt · · ¨ ¸– ¨ ¸ wr © wx ¹ wI © wx ¹ r w § wt sin I wt · sin I w § wt sin I wt · cos I . ¨ cos I . – · – · cos I · – · wr © wr wI ¸¹ wI ¨© wr wI ¸¹ r r r

cos 2 I .

Similarly,

wy 2

...(iii)

cos I .

[Substituting the value of

w 2t

sin I ) r

sin 2 I.

w 2t wr

2

w 2t wr 2

–



wt from (iii)] wx

cos I . sin I wt sin 2 I wt sin 2 I w 2 t sin I .cos I wt  · ·  · 2  · wI r wr wI r2 r2 dI r2 ...(iv) cos 2 I wt cos I . sin I wt cos 2 I w 2 t cos I .sin I wt · – ·  2 · 2 – · 2 wr wI r wI wI r r r2 ...(v)

By adding (iii) and (iv), we get w 2t



w 2t

w 2t



1 wt 1 w 2t  2 · 2 · r wr r wI

wx 2 wy 2 wr 2 Substituting it in eqn (2.8), we get, ª w 2 t 1 wt w 2t º qg 1 w 2t  2 · 2  2»  « 2  · r wr r k wI wz ¼ ¬ wr which is the same as eqn. (2.22)

1 wt · D wW

2.4. GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES Consider an elemental volume having the coordinates (r, I, T), for three dimensional heat conduction analysis, as shown in Fig. 2.3.

“Conduction” Heat Transfer at Steady State–One Dimension

35

Fig. 2.3. Elemental volume for three-dimensional heat conduction analysis - Spherical coordinates.

The volume of the element = dr.rdT.rsin T dI Let, qg = Heat generation (uniform) per unit volume per unit time. Further let us assume that k (thermal conductivity), U (density), c (specific heat) do not alter with position. A. Net heat accumulated in the element due to conduction of heat from all the coordinate directions considered : Heat flow through r–T plane; I-direction : wt QIc – k ( dr.rd T) dW Heat influx, ...(i) r.sin T. wI w Q(cI  d I ) QIc  (QIc ) r sin T .d I Heat efflux, ...(ii) r.sin T. wI ? Heat accumulated in the element due to heat flow in the I-direction, dQ'I = Q'I – Q' (I + dI) [subtracting (ii) from (i)] w 1 – · (QIc ) r sin T . d I r sin T wI w ª wt 1 1 º – · – k (dr.rd T) · . d W » r sin T . d I « r sin T wI ¬ r sin T wI ¼ k (dr . rd T.r sin T . d I)

Heat flow in r–I plane, T-direction : Heat influx,

QTc

– k (dr. r sin T. d I)

1 2

2

·

w 2t

r sin T wI2 wt · dW r wT

dW

...(2.25)

...(iii)

w (QTc ) rd T ...(iv) r wT ? Heat accumulated in the element due to heat flow in the T-direction, dQ'T = Q'T – Q'(T + dT) [subtracting (iv) from (iii)]

Heat efflux,

Q(cT  d T)

QTc 

–

w (QTc ) r.d T r.wT

Chapter : 2

36

w ª wt º – k (dr . r sin T. d I) . d W » r. d T r.wT «¬ r.wT ¼ wt º k dr.rd I. rd T w ª sin T · » d W r r wT «¬ wT ¼ w ª wt º 1 k (dr. rd T. r sin T.d I) 2 · sin T · » d W ...(2.26) wT ¼ r sin T wT «¬ Heat flow in T-I plane, r-direction : wt · wW Qrc – k ( rd T.r sin T. d I) Heat influx, ...(v) wr w (Qrc ) dr Heat efflux, Q(cr  dr ) Qrc  ...(vi) wr ? Heat accumulation in the element due to heat flow in the r-direction, dQ'r = Q'r – Q'(r + dr) [subtracting (vi) from (v)] w – (Qrc ) dr wr w ª wt º – – k (rd T . r sin T. d I) · d W » dr « wr ¬ wr ¼ w ª 2 wt º k d T. sin T. d I dr r · » dW wr «¬ wr ¼ 1 w ª 2 wt º k (dr . rd T . r sin T . d I) 2 · r · » dW ...(2.27) wr ¼ r wr «¬ Net heat accumulated in the element –

ª w 2t w § wt · 1 w § 2 wt · º 1 1  2 k dr. rd T. r sin T. d I « 2 · · ¨ sin T . ¸  2 · ¨ r . ¸ » d W 2 2 wT ¹ r wr © wr ¹ ¼ r sin T wT © ¬ r sin T wI ...(2.28) B. Heat generated within the element (Q'g) : The total heat generated within the element is given by, Q'g = qg (dr.rdT.r sin T.dI) dW ...(2.29) C. Energy stored in the element : The increase in thermal energy in the element is equal to wt U (dr.rd T . r sin T. d I) c. . dW ...(2.30) wW Now, (A) + (B) = (C) ...Energy balance/equation ª w 2t w § wt · 1 w § wt · º 1 1  2 k dr .rd T . r sin T. d I « 2 · · ¨ sin T . ¸  2 · ¨ r 2 . ¸ » . d W 2 2 wT ¹ r wr © wr ¹ ¼ r sin T wT © ¬ r sin T wI wt  q g (dr.rd T.r sin T. d I) d W U (dr.rd T.r sin T.d I) c . . d W wW Dividing both sides by k.(dr.rdT. r sin T.dI)dW, we get ª w 2t w § wt · 1 1 1 w § 2 wt · º q g  2 · · « 2 ¨ sin T . ¸  2 · ¨ r . ¸ »  2 2 wT ¹ r wr © wr ¹ ¼ k r sin T wT © ¬ r sin T wI

?

U c wt 1 wt · · ...(2.31) k wW D wW Equation (2.31) is the general heat conduction equation in spherical coordinates.

“Conduction” Heat Transfer at Steady State–One Dimension

37

In case there are not heat sources present and the heat flow is steady and one-dimensional, then eqn. (2.31) reduces to 1 d § 2 dt · · ¨r · ¸ 0 ...(2.32) dr ¹ r 2 dr © Equation (2.31) can also be derived by transformation of coordinates as follows : x = r sin T sin I ; y = r sin T cos I ; z = r cos T

2.5. HEAT CONDUCTION THROUGH PLANE AND COMPOSITE WALLS 2.5.1. HEAT CONDUCTION THROUGH A PLANE WALL Case I : Uniform thermal conductivity Refer to Fig. 2.4 (a) Consider a plane wall of homogeneous material through which heat is flowing only in x-direction. Let, L = Thickness of the plane wall, A = Cross-sectional area of the wall, k = Thermal conductivity of the wall material, and t1, t2 = Temperatures maintained at the two faces 1 and 2 of the wall, respectively. The general heat conduction equation in cartesian coordinates is given by qg w 2t w 2t w 2t 1 wt    · 2 2 2 D wW k dx dy dz ...[Eqn. 2.8] If the heat conduction takes place under the Fig. 2.4. Heat conduction through § wt · 0 ¸ , one-dimensional conditions, steady state ¨ a plane wall. © wW ¹ ª w 2t º w 2t q § · 0 » and with no internal heat generation ¨ g 0 ¸ then the above equation is « 2 2 y z w w k © ¹ ¬ ¼ reduced to w 2t d 2t 0, 0 or ...(2.33) dx 2 dx 2 By integrating the above differential twice, we have dt C1 and t = C1x + C2 ...(2.34) dx where C1 and C2 are the arbitrary constants. The values of these constants may be calculated from the known boundary conditions as follows : At x = 0 t = t1 At x = L t = t2 Substituting the values in the eqn. (2.34), we get t1 = O + C2 and t2 = C1L + C2

Chapter : 2

38

After simplification, we have, C2 = t1 and

C1

t2 – t1 L

Thus, the eqn. (2.34) reduces to : § t2 – t1 · ¨ ¸ x  t1 ...(2.35) © L ¹ The eqn. (2.35) indicates that temperature distribution across a wall is linear and is independent of thermal conductivity. Now heat through the plane wall can be found by using Fourier’s equation as follows : dt dt Q – kA (where, = Temperature gradient) ... dx dx [Eqn.(1.1)] t

º t2 – t1 d ª§ t2 – t1 · ¨ ¸ x  t1 » « dx ¬© L ¹ L ¼ (t2 – t1 ) kA (t1 – t2 ) Q – kA ? ...(2.36) L L Eqn (2.36) can be written as : (t1 – t2 ) (t1 – t2 ) Q ...(2.37) ( L / kA) ( Rth )cond. where, (Rth)cond. = Thermal resistance to heat conduction. Fig. 2.4 (b) shows the equivalent thermal circuit for heat flow through the plane wall. Let us now find out the condition when instead of space, weight is the main criterion for selection of the insulation of a plane wall.

But,

dt dx

Thermal resistance (conduction) of the wall, (Rth)cond. =

L kA

...(i)

Weight of the wall, W = UA L ...(ii) Eliminating L from (i) and (ii), we get ...(2.38) W = UA. (Rth)cond. kA = (U.k)A2.(Rth)cond. The eqn., (2.38) stipulates the condition that, for a specified thermal resistance, the lightest insulation will be one which has the smallest product of density (U) and thermal conductivity (k). Case II. Variable thermal conductivity A. Temperature variation in terms of surface temperatures (t1, t2) : Let the thermal conductivity vary with temperature according to the relation k = k0 (1 + Et) ...(2.39) [In most of the cases, the thermal conductivity is found to vary linearly with temperature] where, k0 = Thermal conductivity at zero temperature. When the effect of temperature on thermal conductivity is considered, the Fourier’s equation, Q Q

or, or,

Q · dx A Q L dx A ³0

dt is written as : dx dt – k0 (1  E t ) .A dx

– kA

– k0 (1  E t ) dt – k0

t2

³t

1

(1  E t ) dt

...(2.40)

“Conduction” Heat Transfer at Steady State–One Dimension or, or,

Q.L A Q.L A

E º ª – k0 « t  t 2 » ¬ 2 ¼ t1

E ª º – k0 «(t2 – t1 )  (t22 – t12 ) » ¬ ¼ 2 E ª º k0 «(t1 – t2 )  (t1 – t2 ) (t1  t2 ) » ¬ ¼ 2 E ª º k0 «1  (t1  t2 ) » (t1 – t2 ) ¬ ¼ 2 k0 [1  E tm ] (t1 – t2 )

?

Q

39

t2

k0 (1  E tm ).

...(2.41)

t1  t2 2

where tm

A (t1 – t2 ) L

From eqn. (2.39) t is replaced by tm, then km = k0 (1 + Etm) ...(2.42) ª t – t2 º Q km A « 1 ? ...(2.43) ¬ L »¼ where km is known as mean thermal conductivity of the wall material. Further, if t is the temperature of the surface at a distance x from the left surface (Fig. 2.5), then eqn. (2.41) becomes E Qx ª º – k0 « (t – t1 )  (t 2 – t12 ) » ...(2.44) A ¬ ¼ 2 Form eqns. (2.41) and (2.44), we have E 2 E 2 ª ª 2 º X 2 º «¬(t2 – t1 )  2 (t2 – t1 ) »¼ L «¬(t – t1 )  2 (t – t1 ) »¼ [Equating the values of Q and rearranging] Solving the above equation for t, we get 1/ 2

xº 1 ª (1  E t1 ) 2 – {(1  E t1 ) 2 – (1  E t2 ) 2 } » « E ¬ L¼ B. Temperature variation in terms of heat flux (Q) : Fourier’s equation for heat conduction is given by dt dt Q – kA. – k0 (1  E t ) A. dx dx or, Q.dx = – k0 (1 + Et) A.dt Integrating both sides, we get E · § Q.x – k0 A ¨ t  t 2 ¸  C © 2 ¹ (where, C = Constant of integration) To evaluate C, applying the condition : At x = 0, t = t1, we get E · § C k0 A ¨ t1  t12 ¸ © 2 ¹ Substituting the values of the constant C in (i), we get E · E · § § Q.x – k0 A ¨ t  t 2 ¸  k0 A ¨ t1  t12 ¸ © ¹ © 2 2 ¹ Dividing both sides by k0A and rearranging, we obtain, t

–

1 E

...(2.45)

...(i)

Chapter : 2

40

E 2 ª Q.x t t« – 2 ¬ k0 A By solving the above quadratic equation, we have – 1 1 – 4 u t

?

E 2 ·º § ¨ t1  t1 ¸ » © 2 ¹¼

ª Q.x «k A – ¬ 0 §E· 2u¨ ¸ ©2¹

E 2

0

E 2 ·º § ¨ t1  t1 ¸ » © 2 ¹¼

1/ 2

t

or,

E ·º 1 ª1 2 § Q.x – « 2 – ¨ – t1 – t12 ¸ » E ¬E E © k0 A 2 ¹¼ 1/ 2

–

1 ª1 2 2Q.x º  – t1  t12 – E «¬ E2 E Ek0 A »¼ 1/ 2

2 1 ª§ 1· 2Q.x º –  «¨ t1  ¸ – » E ¬«© E¹ Ek0 A ¼»

1/ 2

2 1 ª§ 1· 2Q.x º t –  «¨ t1  ¸ – » Hence, ...(2.46) E «¬© E¹ Ek0 A »¼ In most of the practical applications where the variation of temperature is small, the average value of k for the given temperature range is commonly used as given in eqn. (2.42).

If the variation of k with temperature is not linear, then k = k0 f (t), and Q A

or, But,

L

³ 0 dx Q

–

t2

³t

1

A L

[k0 f (t ) dt

ª– «¬

t2

³t

1

Q

[k0 f (t ) dt º »¼

...(2.47)

§ t – t2 · km A ¨ 1 ¸ © L ¹

...[Eqn. (2.43)]

Equating these eqns. (2.47) and (2.43), we have km

1 ª t2 [k f (t ) dt º 0 »¼ (t1 – t2 ) ¬« ³ t1 1 (t1 – t2 )

t1

³ t [k0 f (t ) dt ] 2

...(2.48)

Fig. 2.5.

The effect of + E and – E on temperature is depicted in Fig. 2.5.

2.5.2. HEAT CONDUCTION THROUGH A COMPOSITE WALL Refer to Fig. 2.6 (a). Consider the transmission of heat through a composite wall consisting of a number of slabs. Let,

LA, LB, LC = Thicknesses of slabs A, B and C respectively (also called path lengths), kA, kB, kC = Thermal conductivities of the slabs A, B, and C respectively,

“Conduction” Heat Transfer at Steady State–One Dimension

41

t1, t4 (t1 > t4) = Temperatures at the wall surfaces 1 and 4 respectively, and t2, t3 = Temperatures at the interfaces 2 and 3 respectively. Since the quantity of heat transmitted per unit time through each slab/layer is same, we have, Q

k A . A (t1 – t2 ) LA

k B . A (t2 – t3 ) LB

kC . A (t3 – t4 ) LC

(Assuming that there is a perfect contact between the layers and no temperature drop occurs across the interface between the materials).

Fig. 2.6. Steady state conduction through a composite wall.

Rearranging the above expression, we get Q . LA t1 – t2 kA . A

...(i)

t2 – t3

Q . LB kB . A

...(ii)

t3 – t4

Q . Lc kC . A

...(iii)

Adding (i), (ii) and (iii), we have

(t1 – t4 )

LC º LB ª L  Q« A  » ¬ k A . A k B . A kC . A ¼

Chapter : 2

42 or,

or,

Q

Q

A (t1 – t4 ) LC º ª LA LB «k  k  k » B C ¼ ¬ A

(t1 – t4 ) LC º LB ª LA «k . A  k . A  k . A» B C ¬ A ¼

>Rth – A

...(2.49)

(t1 – t4 )  Rth – B  Rth – C @

...[2.49 (a)]

If the composite wall consists of n slabs/layers, then Q

[t1 – t( n  1) ] n

¦ 1

L kA

...(2.50)

In order to solve more complex problems involving both series and parallel thermal resistances, the electrical analogy may be used. A typical problem and its analogous electric circuit are shown in Fig. 2.7. Q

' toverall ¦ Rth

...(2.51)

Fig. 2.7. Series and parallel one-dimensional heat transfer through a composite wall and electrical analog.

Thermal contact resistance. In a composite (multi-layer) wall, the calculations of heat flow are made on the assumptions : (i) The contact between the adjacent layers is perfect, (ii) At the interface there is no fall of temperature, and (iii) At the interface the temperature is continuous, although there is discontinuity in temperature gradient. In real systems, however, due to surface roughness and void spaces (usually filled with air) the contact surfaces touch only at discrete locations. Thus there is not a single plane of contact, which means that the area available for the flow of heat at the interface will be small compared to geometric face area. Due to this reduced area and presence of air voids, a large resistance to heat flow at the interface occurs. This resistance is known as thermal contact resistance and it causes temperature drop between two materials at the interface as shown in Fig. 2.8.

“Conduction” Heat Transfer at Steady State–One Dimension

Fig. 2.8. Temperature drops at the interfaces.

Refer to Fig. 2.8. The contact resistances are given by (t2 – t3 ) ( Rth – AB )cont . and Q/ A

( Rth – BC )cont .

(t4 – t5 ) Q/ A

2.5.3. THE OVERALL HEAT-TRANSFER COEFFICIENT While dealing with the problems of fluid to fluid heat transfer across a metal boundary, it is usual to adopt an overall heat transfer coefficient U which gives the heat transmitted per unit area per unit time per degree temperature difference between the bulk fluids on each side of the metal. Refer to Fig. 2.9 Let, L = Thickness of the metal wall, k = Thermal conductivity of the wall material, t1 = Temperature of the surface–1, t2 = Temperature of the surface–2, thf = Temperature of the Fig. 2.9. The overall heat transfer through a plane wall. hot fluid, tcf = Temperature of the cold fluid, hhf = Heat transfer coefficient from hot fluid to metal surface, and hcf = Heat transfer coefficient from metal surface to cold fluid. (The suffices hf and cf stand for hot fluid and cold fluid respectively.) The equations of heat flow through the fluid and the metal surface are given by

43

44

Chapter : 2 Q = hhf. A (thf – t1) Q

k . A (t1 – t2 ) L

Q = hcf. A (t2 – tcf) By rearranging (i), (ii) and (iii), we get Q thf – t1 hhf . A QL t1 – t2 k .A Q t2 – tcf kcf . A Adding (iv), (v) and (vi) we get 1 º L ª 1   thf – tcf Q « » ¬ hhf . A k . A hcf . A ¼ A (thf – tcf ) or, Q 1 1 L   hhf k hcf If U is the overall coefficient of heat transfer, then A (thf – tcf ) Q U . A (thf – tcf ) 1 1 L   hhf k hcf 1 or, U 1 1 L   hhf k hcf

...(i) ...(ii) ...(iii)

...(iv) ...(v) ...(vi)

...(2.52)

...(2.53)

It may be noticed from the above equation that if the individual coefficients differ greatly in magnitude only a change in the least will have any significant effect on the rate of heat transfer. Example 2.1. Discuss the effects of various parameters on the thermal conductivity of solids. (AMIE Summer, 2001) Solution. The following are the effects of various parameters on the thermal conductivity of solids. 1. Chemical composition. Pure metals have very high thermal conductivity. Impurities or alloying elements reduce the thermal conductivity considerably. [Thermal conductivity of pure copper is 385 W/m° C, and that for pure nickel is 93 W/m° C. But monel metal (an alloy of 30% Ni and 70% Cu) has k of 24 W/m° C. Again for copper containing traces of Arsenic the value of k is reduced to 142 W/m° C.] 2. Mechanical forming. Forging, drawing and bending or heat treatment of metals cause considerable variation in thermal conductivity. For example, the thermal conductivity of hardened steel is lower than that of annealed state. 3. Temperature rise. The value of k for most metals decreases with temperature rise since at elevated temperatures the thermal vibrations of the lattice become higher that retard the motion of free electrons. 4. Non-metallic solids. Non-metallic solids have k much lower than that for metals. For many of the building materials (concrete, stone, brick, glass wool, cork etc.) the thermal conductivity may vary from sample to sample due to variations in structure, composition, density and porosity.

“Conduction” Heat Transfer at Steady State–One Dimension

45

5. Presence of air. The thermal conductivity is reduced due to the presence of air filled pores or cavities. 6. Dampness. Thermal conductivity of a damp material is considerably higher than that of dry material. 7. Density. Thermal conductivity of insulating powder, asbestos etc. increases with density growth. Thermal conductivity of snow is also proportional to its density. Example 2.2. The inner surface of a plane brick wall is at 60°C and the outer surface is at 35°C. Calculate the rate of heat transfer per m2 of surface area of the wall, which is 220 mm thick. The thermal conductivity of the brick is 0.51 W/m°C. (AMIE Winter, 2000) Solution. Temperature of the inner surface of the wall, t1 = 60°C Temperature of the outer surface of the wall, t2 = 35°C The thickness of the wall, L = 220 mm = 0.22 m Thermal conductivity of the brick, k = 0.51 W/m°C Rate of heat transfer per m2, q : Fig. 2.10. Rate of heat transfer per unit area, q

Q A

k (t1 – t2 ) L

or 0.51 u (60 – 35) 57.95 W/m 2 (Ans.) 0.22 Example 2.3. Consider a slab of thickness L = 0.25 m. One surface is kept at 100°C and the other surface at 0°C. Determine the net flux across the slab if the slab is made from pure copper. Thermal conductivity of copper may be taken as 387.6 W/m K. (AMIE Winter, 1998) Solution. Given : L = 0.25 m; t 1 = 100°C; t 2 = 0°C; k = 387.6 W/m K. Fig. Fig.2.11. 2.11. From Fourier’s law, dt Q  kA ...[Eqn. (1.1)] dx Q (t – t1 ) q –k· 2 Net flux, A L (0 – 100) – 387.6 u 0.25 = 1.55 × 105 W/m2 (Ans.) Example 2.4. A reactor’s wall, 320 mm thick, is made up of an inner layer of fire brick (k = 0.84 W/m°C) covered with a layer of insulation (k = 0.16 W/m°C). The reactor operates at a temperature of 1325°C and the ambient temperature is 25°C. q

46

Chapter : 2

(i) Determine the thickness of fire brick and insulation which gives minimum heat loss; (ii) Calculate the heat loss presuming that the insulating material has a maximum temperature of 1200°C. If the calculated heat loss is not acceptable, then state whether addition of another layer of insulation would provide a satisfactory solution. Solution. Refer to Fig. 2.12. Given : t1 = 1325°C; t2 = 1200°C, t3 = 25°C LA + LB = L = 320 mm or 0.32 m ? LB = (0.32 – LA); ...(i) kA = 0.84 W/m°C; kB = 0.16 W/m°C. (i) LA : LB : The heat flux, under steady state conditions, is constant throughout the wall and is same for each layer. Then for unit area of wall, t1 – t3 q LA / k A  LB / k B

Fig. 2.12.

t1 – t2 LA / k A

t2 – t3 LB / k B

Considering first two quantities, we have (1325 – 25) (1325 – 1200) LA / 0.84  LB / 0.16 LA / 0.84 1300 105 or, 1.190 L  6.25 (0.32 – L ) LA A A 1300 105 or, 1.190 LA  2 – 6.25 LA LA 1300 105 or, 2 – 5.06 LA LA or, 1300 LA = 105 (2 – 5.06 LA) or, 1300 LA = 210 – 531.3 LA 210 0.1146 m or 114.6 mm LA or, (1300  531.3) ? Thickness of insulation

LB = 320 – 114.6 = 205.4 mm (Ans.)

(ii) Heat loss per unit area, q : t1 – t2 LA / k A

1325 – 1200 916.23 W/m 2 (Ans.) 0.1146 / 0.84 If another layer of insulating material is added, the heat loss from the wall will reduce; consequently the temperature drop across the fire brick lining will drop and the interface temperature t2 will rise. As the interface temperature is already fixed, therefore, a satisfactory solution will not be available by adding another layer of insulation.

Heat loss per unit area,

q

Example 2.5. A wall of a furnace is made up of inside layer of silica brick 120 mm thick covered with a layer of magnesite brick 240 mm thick. The temperatures at the inside surface of silica brick wall and outside surface of magnesite brick wall are 725°C and 110°C respectively. The contact

“Conduction” Heat Transfer at Steady State–One Dimension

47

thermal resistance between the two walls at the interface is 0.0035°C/W per unit wall area. If thermal conductivities of silica and magnesite bricks are 1.7 W/m°C and 5.8 W/m°C, calculate. (i) The rate of heat loss per unit area of walls, and (ii) The temperature drop at the interface. Solution. Refer Fig. 2.13. Given : LA = 120 mm = 0.12 m; LB = 240 mm = 0.24 m; kA = 1.7 W/m°C; kB = 5.8 W/m°C The contact thermal resistance (Rth)cont. = 0.0035°C/W The temperature at the inside surface of silica brick wall, t1 = 725°C The temperature at the outside surface of the magnesite brick wall, t4 = 110°C (i) The rate of heat loss per unit area of wall, q : 't 't q 6 Rth Rth – A  ( Rth )cont .  Rth – B (t1 – t4 ) LA / k A  0.0035  LB / k B (725 – 110) 0.12 /1.7  0.0035  0.24 / 5.8 615 0.0706  0.0035  0.0414



Fig. 2.13.

5324.67 W/m2

? The rate of heat loss per unit area of wall, q = 5324.67 W/m2 (ii) The temperature drop at the interface, (t2 – t3) : As the same heat flows through each layer of composite wall, therefore, t3 – t4 t1 – t2 q LA / k A LB / k B 5324.67

or, or, t2

725 – 5324.67 u

Similarly,

0.12 1.7

5324.67

(725 – t2 ) 0.12 /1.7 349.14qC (t3 – 110) 0.24 / 5.8

0.24 330.33qC 5.8 Hence, the temperature drop at the interface = t2 – t3 = 349.14 – 330.33 = 18.81°C (Ans.)

or, t3

110  5324.67 u

(Ans.)

48

Chapter : 2

Example 2.6. An exterior wall of a house may be approximated by a 0.1 m layer of common brick (k = 0.7 W/m°C) followed by a 0.04m layer of gypsum plaster (k = 0.48 W/m°C). What thickness of loosely packed rock wool insulation (k = 0.065 W/m°C) should be added to reduce the heat loss or (gain) through the wall by 80 per cent ? (AMIE Summer, 1999) Solution. Refer to Fig. 2.14. Thickness of common brick, LA = 0.1 m Thickness of gypsum plaster, LB = 0.04 m Thickness of rock wool, LC = x (in m) =? Thermal conductivities : Common brick, kA = 0.7 W/m°C; Gypsum plaster, kB = 0.48 W/m°C; Rock wool, kC = 0.065 W/m°C. Case I. Rock wool insulation not used : A (' t ) A (' t ) Q1 LA LB 0.1 0.04 ...(i)   Fig. 2.14. kA kB 0.7 0.48 Case II. Rock wool insulation used : A (' t ) A (' t ) Q2 L x LA L 0.1 0.04 ...(ii)    B  C 0.7 0.48 0.065 kA kB kC But, Q2 = (1 – 0.8) Q1 = 0.2 Q1 ...(given) A (' t ) A (' t ) ? 0.2 u x 0.1 0.04 0.1 0.04    0.7 0.48 0.065 0.7 0.48 x º 0.1 0.04 ª 0.1 0.04    0.2 « or, 0.7 0.48 ¬ 0.7 0.48 0.065 »¼ or, 0.1428 + 0.0833 = 0.2 [0.1428 + 0.0833 + 15.385x] or, 0.2261 = 0.2 (0.2261 + 15.385 x) or, x = 0.0588 m or 58.8 mm Thus, the thickness of rock wool insulation should be 58.8 mm (Ans.) Example 2.7. A furnace wall consists of 200 mm layer of refractory bricks, 6 mm layer of steel plate and a 100 mm layer of insulation bricks. The maximum temperature of the wall is 1150°C on the furnace side and the minimum temperature is 40°C on the outermost side of the wall. An accurate energy balance over the furnace shows that the heat loss from the wall is 400 W/m2. It is known that there is a thin layer of air between the layers of refractory bricks and steel plate. Thermal conductivities for the three layers are 1.52, 45 and 0.138 W/m°C respectively. Find : (i) To how many millimeters of insulation brick is the air layer equivalent? (ii) What is the temperature of the outer surface of the steel plate? (AMIE) Solution. Refer Fig. 2.15. Thickness of refractory bricks,

“Conduction” Heat Transfer at Steady State–One Dimension

49

LA = 200 mm = 0.2 m Thickness of steel plate, LC = 6 mm = 0.006 m Thickness of insulation bricks, LD = 100 mm = 0.1 m Difference of temperature beetween the innermost and outermost sides of the wall, ' t = 1150 – 40 = 1110°C Thermal conductivities : kA = 1.52 W/m°C; kB = kD = 0.138 W/m°C; kC = 45 W/m°C Heat loss from the wall, q = 400 W/m2 (i) The value of x ( = LC) : We know,

Q

Q A

or

A · 't L 6 k q

Fig. 2.15.

't L 6 k

or,

400

or,

400

0.8563  0.0072 x

or,

1110 L LA LB L   C  D kA kB kC kD 1110 0.2 ( x /1000) 0.006 0.1    1.52 0.138 45 0.138 1110 0.1316  0.0072 x  0.00013  0.7246 1110 400

2.775

2.775 – 0.8563 266.5 mm (Ans.) 0.0072 (ii) Temperature of the outer surface of the steel plate tso: (t so – 40) q 400 LD / kD (tso – 40) or, 400 1.38 (tso – 40) (0.1/ 0.138)

or,

x

or,

tso

400  40 1.38

329.8°C (Ans.)

1110 0.8563  0.0072 x

50

Chapter : 2

Example 2.8. A furnace wall is composed of 220 mm of fire brick, 150 mm of common brick, 50 mm of 85% magnesia and 3 mm of steel plate on the outside. If the inside surface temperature is 1500°C and outside surface temperature is 90°C, estimate the temperatures between layers and calculate the heat loss in kJ/h- m2. Assume, k (for fire brick) = 4kJ/mh-°C, k(for common brick) = 2.8 kJ/ m-h-°C, k (for 85% magnesia) = 0.24 kJ/m-h-°C, and k (steel) = 240 kJ/m-h-°C. (AMIE, Winter, 1997) Solution. Given : LA = 220 mm = 0.22 m; LB = 150 mm = 0.15 m; LC = 50 mm = 0.05m; LD = 3 mm = 0.003 m t1 = 1500°C, t5 = 90°C; kA = 4 kJ/mh°C; kB = 2.8 kJ/mh°C Fig. 2.16. kC = 0.24 kJ/mh°C; kD = 240 kJ/mh°C. Heat loss in kJ/hm2 : The equivalent thermal resistances of various layers are : LA 0.22 Rth–A 0.055 m 2 h°C/kJ kA 4 LB 0.15 Rth–B 0.05357 m 2 h°C/kJ kB 2.8 LC 0.05 Rth–C 0.2083 m 2 h°C/kJ kC 0.24 LD 0.003 Rth–D 1.25 u 10 –5 m 2 h°C/kJ kD 240 Total thermal resistance, (Rth)total = 0.055 + 0.05357 + 0.2083 + 1.25 × 10–5 = 0.3169 m2h°C/kJ Heat loss,

q

(t1 – t5 ) ( Rth ) total

(1500 – 90) 0.3169

4449.35 kJ/hm 2

(Ans.)

Temperatures between layers : t4 – t5 q Also, Rth–D or t4 = t5 + q Rth–D = 90 + 4449.35 × 1.25 × 10–5 = 90.056°C Similarly, t3 = t4 + q Rth–C = 90.056 + 444.35 × 0.2083 = 1016.86°C and t2 = t3 + q Rth–B = 1016.86 + 4449.35 × 0.05357 = 1255.2°C [Check t1 = t2 + q Rth–A = 1255.2 + 4449.35 × 0.55 ~ 1500°C]

(Ans.)

“Conduction” Heat Transfer at Steady State–One Dimension Example 2.9. A metal piece of length l has a cross-section of a sector of a circle of radius r and included angle of T. Its two ends are maintained at temperatures t1 and t2 (t1 > t2). Find the expression for heat flow through the metal piece, assuming that the conductivity of metal varies with temperature according to relation, k = k0 (1 – Et). Also assume that

wt wT

0 and

51

Fig. 2.17.

wt wr

0 and outer surfaces of the slab except the end surfaces are

completely insulated. What will be the rate of heat transfer if l = 600 mm, r = 120 mm, T = 60°, t1 = 125°C, t2 = 25°C and k0 = 115 W/m°C and E = 10–4 ? Solution. As per given conditions, kA(t1 – t2 ) Q l The area through which heat is flowing is given by, A

Sr 2 u

T 2S

r 2T 2

where T is in radians. km = k0 (1 – Etm) where tm as its variation is linear. Q

? Rate of heat transfer, Q :

t1  t2 2

km § r 2 ˜ T · ¨ ¸ (t1 – t2 ) l © 2 ¹

Given : l = 600 mm = 0.6 m; r = 120 mm = 0.12 m, T = 60° k0 = 115 W/m°C and E = 10–4.

...(i) S rad., t = 125°C, t = 25°C, 1 2 3

ª § 125  25 · º 115 «1 – 10 –4 ¨ ¸ » 114.14 W/m°C © ¹¼ 2 ¬ Subsituting the proper values in the expression (i), we have S· § 0.122 u ¸ 114.14 ¨ 3 (125 – 25) 143.43 W (Ans.) Q ¨ ¸ ¹ 0.6 © 2 Example 2.10. (i) Derive an expression for the heat loss per m2 of the surface area for a furnace wall (Fig. 2.18), when the thermal conductivity varies with temperature according to the relation : k = (a + bt2) W/m°C, where t is in °C (ii) Find the rate of heat transfer through the wall, if L = 0.2 m, t1 = 300°C, t2 = 30°C and a = 0.3 and b = 5 × 10–6. (Maharashtra University) 2 Solution. (i) The rate of heat transfer through the wall per m is given by

? km

k0 (1 – E tm )

q

km (t1 – t2 ) L

Chapter : 2

52

t2 –1 k . dt , where k = f (t) ...[Eqn. (2.47)] ³ t (t1 – t2 ) 1 t2 –1 (a  bt 2 ) dt ³ (t1 – t2 ) t1

where km

t

–1 (t1 – t2 )

2 ª bt 3 º  at « » ¬ 3 ¼ t1

–1 (t1 – t2 ) –1 (t1 – t2 )

b 3 ª 3 º «¬ a (t2 – t1 )  3 (t2 – t1 ) »¼ b 2 ª 2 º (t2 – t1 ) « a  (t2  t1 t2  t1 ) » ¬ ¼ 3

a

b ª 2 2 ¬t1  t1 t 2  t2 º¼ 3

Fig. 2.18.

b 2 ª 2 º ª t1 – t 2 º «¬ a  3 (t1  t1 t 2  t2 ) »¼ «¬ L »¼ ...Required expression. (Ans.) (ii) Rate of heat transfer per m2, q : Thickness of wall, L = 0.2m; t1 = 300°C; t2 = 30°C; a = 0.3 and b = 5 × 10–6. Substituting these values in the said equation, we have ª º ª (300 – 30) º 5 u 10 –6 q «0.3  (3002  300 u 30  302 ) » « »¼ ¬ ¼¬ 3 0.2

?

q

§ · 5 u 10 –6 u 99900 ¸ u 1350 629.77 W/m 2 ¨ 0.3  © ¹ 3 Hence, rate of heat transfer per m2 through the wall = 629.77 W/m2 (Ans.) Example 2.11. The surfaces of a plane wall of thickness L are maintained at temperatures t1 and t2. The thermal conductivity of wall material varies according to the relation: k = k0t2. (i) Derive an expression to find the steady state conduction through the wall. (ii) Find the temperature at which mean thermal conductivity be evaluated in order to get the same heat flow by its substitution in the simplified Fourier’s equation. Solution. Thickness of wall = L Temperatures of surfaces = t1, t2 Relation of variation of thermal conductivity k = k0t2 (i) Expression for heat conduction through wall : Heat conduction through a plane wall is given by (Fourier's law) dt Q –k A dx dt 2 – k0 t ˜ A dx By rearranging and integrating, we get L

³ 0 Q.dx

– k0 A

t2 2

³t

t dt

1

t

Q

L x0

ªt3 º 2 – k0 A « » ¬ 3 ¼ t1

Fig. 2.19.

“Conduction” Heat Transfer at Steady State–One Dimension

53

– k0 A 3 t2 – t13 3 k0 A 3 t1 – t23 ...Required expression. Q or, 3L (ii) Temperature, tm : If the above heat flow is to be obtained by substituting mean value of thermal conductivity in the simplified Fourier's equation, we have k0 A 3 t1 – t23 km A (t1 – t2 ) L 3L QL

tm 2

or,

k0 tm2 A (t1 – t2 ) L L ª k0 A 3 ª º 3 º «¬ 3L (t1 – t2 ) »¼ u « k A (t – t ) » 1 2 ¼ ¬ 0 3 3 2 2 t1 – t2 (t1 – t2 ) (t1  t1t2  t2 ) 3(t1 – t2 ) 3(t1 – t2 )

2

t1  t1 t2  t2 3

2

t12  t2 2  t1 t2 ...Required temperature. (Ans.) 3 Example 2.12. The variation of thermal conductivity of a wall material is given by k = k0 (1 + Dt + Et2) If the thickness of the wall is L and its two surfaces are maintained at temperatures t1 and t2, find an expression for the steady state one-dimensional heat flow through the wall. Solution. The rate of heat transfer through a wall per unit area is given by tm

?

q

–k˜

dt dx

...Fourier’s equation

dt dx or, q. dx = –k0 (1 + Dt + Et2) . dt Integrating both sides we get – k0 (1  Dt  Et 2 ) ˜

L

q ³ dx 0

quL quL q q

Fig. 2.20.

t2

– k0 ³ (1  Dt  Et 2 ) dt t1

t

2 ª§ t2 t 3 ·º – k 0 «¨ t  D  E . ¸ » 2 3 ¹ ¼ t1 ¬©

D 2 E 3 ª 2 3 º – k0 « (t2 – t1 )  (t2 – t1 )  (t2 – t1 ) » ¬ ¼ 2 3 k0 ª D E º 2 2 – «(t2 – t1 )  (t2 – t1 ) (t2  t1 )  (t2 – t1 ) (t1  t2  t1 t2 ) » L ¬ ¼ 2 3 k0 (t2 – t1 ) ª D E 2 º 2 – «¬1  2 (t1  t2 )  3 (t1  t2  t1 t2 ) »¼ ...Required expression. L (Ans.)

54

Chapter : 2

Example 2.13. It is proposed to carry pressurized water through a pipe imbeded in a 1.2 m thick wall whose surfaces are held at constant temperatures of 200°C and 60°C respectively. It is desired to locate the pipe in wall where the temperature is 120°C, find how far from the hot surface should the pipe be imbedded ? The thermal conductivity of the wall material varies with the temperature according to the relation, k = 0.28 (1 + 0.036t) where t is in degree celsius and k is in W/m°C. Solution. Thickness of wall, L = 1.2 m Temperatures of wall surfaces t1 = 200°C; t2 = 60°C Fig. 2.21. Temperature, t = 120°C Relation for conductivity k = 0.28 (1 + 0.036t) The rate of heat transfer through a plane wall of variable thermal conductivity is given by A (t1 – t2 ) L D ª º A k0 «1  (t1  t2 ) » (t1 – t2 ) ¬ ¼ L 2 Now, when t2 = 60°C, L = 1.2 m and, when t2 = 120°C, L = x (unknown) Substituting the values and equating the two expressions, we have Q

km .

0.036 0.036 ª º A ª º A 0.28 «1  (200  60) » (200 – 60) 0.28 «1  (200  120) » (200 – 120) ¬ 2 ¼ 1.2 ¬ 2 ¼ x 151.42 151.42 185.54 0.816 m or x 185.54 Hence the pipe should be imbedded 0.816 m from the hot wall surface. (Ans.) Example 2.14. Find the steady state heat flux through the composite slab as shown in the Fig. 2.22 and the interface temperature. The thermal conductivities of the two materials vary with temperature as given below : kA = 0.05 (1 + 0.0065t) W/m°C; kB = 0.04 (1 + 0.0076t ) W/m°C, where temperatures are in °C. [M.U.] Solution. t1 = 600°C; t3 = 300°C LA = 50 mm = 0.05m LB = 100 mm = 0.1m kmA

kmB

ª § t  t ·º kOA «1  D A ¨ 1 2 ¸ » © 2 ¹¼ ¬ ª § t  t ·º 0.05 «1  0.0065 ¨ 1 2 ¸ » © 2 ¹¼ ¬ ª § t2  t3 · º kOB «1  D B ¨ ¸» © 2 ¹¼ ¬ ª § t  t ·º 0.04 «1  0.0075 ¨ 2 3 ¸ » © 2 ¹¼ ¬

Fig. 2.22.

“Conduction” Heat Transfer at Steady State–One Dimension Interface temperature, t2 : Rate of heat transfer per m2,

q

Q A

(t1 – t2 ) ( LA / kmA )

(t2 – t3 ) ( LB / k mB )

...(1)

Now substituting the values of kmA and kmB in eqn (1), we get

(600 – t2 ) 0.05 ª § 600  t2 · º 0.05 «1  0.0065 ¨ ¸» © ¹¼ 2 ¬

(t2 – 300) 0.1 ª § t  300 · º 0.04 «1  0.0075 ¨ 2 ¸» © ¹¼ 2 ¬

ª ª § 600  t2 · º § t2  300 · º ¸ » 0.4 (t2 – 300) «1  0.0075 ¨ ¸» or, (600 – t2 ) «1  0.0065 ¨ © ¹ © ¹¼ 2 2 ¬ ¼ ¬ ª 5.9  0.0065 t2 º ª 4.25  0.0075 t2 º (600 – t2 ) « (t2 – 300) « or, » »¼ u 0.4 ¬ ¼ ¬ 2 2 or, (600 – t2) (5.9 + 0.0065 t2) = (t2 – 300) (1.7 + 0.003 t2)

or,

3540 + 3.9t2 – 5.9t2 – 0.0065t22 = 1.7 t2 + 0.003 t22 – 510 – 0.9 t2

or,

0.0095 t22 + 2.8 t2 – 4050 = 0

or,

t22 + 294.7 t2 – 426315 = 0

– 294.7 r 1338.7 – 294.7 r 294.72  4 u 426316 2 2 ª § 600  522 · º kmA 0.05 «1  0.0065 ¨ ¸ » 0.2323 W / mqC ? © ¹¼ 2 ¬ Rate of heat transfer per m2, q : The steady state heat flow through the composite slab, (t1 – t2 ) (600 – 522) q 362.39 W/m 2 (Ans.) ( LA / kmA ) (0.05 / 0.2323)

or,

t2

Example 2.15. The composite wall of a furnace is made up with 120 mm of fire clay [k = 0.25 (1 + 0.0009 t) W/m°C] and 600 mm of red brick (k = 0.8 W/m°C). The inside surface temperature is 1250°C and the outside air temperature is 40°C. Determine : (i) The temperature at the layer interface, and (ii) The heat loss for 1m2 of furnace wall. Solution. Refer Fig. 2.23. LA = 120 mm = 0.12 m; LB = 600 mm = 0.6 m; kA = 0.25 [1 + 0.0009 t]; kB = 0.8 W/m°C; 't = (t 1 – t air) = 1250 – 40 = 1210°C.

Fig. 2.23.

55

522q C

Chapter : 2

56

(i) The temperature at layer interface, t2 : Average/mean thermal conductivity of fire clay, ª § 1250  t2 · º 0.25 «1  0.0009 ¨ ¸» © ¹¼ 2 ¬ = 0.25 [1 + 0.00045 (1250 + t2)] ? Thermal resistance of fire clay, LA 0.12 Rth – A (k A ) m A 0.25 >1  0.00045(1250  t2 )@ u 1 (k A )m

1 2.083  0.000937 (1250  t2 )

Similarly, thermal resistance of red brick, LB 0.6 Rth – B 0.75 k B ˜ A 0.8 u 1 Heat loss for 1m2 of furnace wall, 't 't Q 6Rth Rth – A  Rth – B 1210 1 ...(i)  0.75 2.083  0.000937 (1250  t2 ) Under steady state conditions the same amount of heat flows through each layer. Then considering heat flow through the red brick, we have (t2 – 40) (t2 – 40) Q ...(ii) kB 0.8 From expression (i) and (ii), we obtain

or, or, or, or,

(t2 – 40) 1210 1 0.8  0.75 2.083  0.000937 (1250  t2 ) t2 – 40 1210[2.083  0.000937 (1250  t2 )] 1  0.75[2.083  0.000937 (1250  t2 )] 0.8 1210[3.254  0.000937 t2 ] (t2 – 40) 1  1.562  0.878  0.000703 t2 0.8 3937.34  1.134 t2 (t2 – 40) 3.44  0.000703 t2 0.8 0.8 (3937.34 + 1.134 t2) = (t2 – 40) (3.44 + 0.000703 t2) 2

3149.87 + 0.907 t2 = 3.44 t2 + 0.000703 t2 – 137.6 – 0.0281 t2

or, or, or,

0.000703 t22 + 2.505 t2 – 3287.47 = 0

t2

– 2.505  (2.505) 2  4 u 0.000703 u 3287.47 2 u 0.000703

(ii) Heat loss, Q : Heat loss for 1 m2 of the furnace wall, Q

(t2 – 40) Rth – B

(1020.24 – 40) 0.75

1306.98 W (Ans.)

1020.24°C (Ans.)

“Conduction” Heat Transfer at Steady State–One Dimension Example 2.16. Find the heat flow rate through the composite wall as shown in Fig. 2.24 Assume one dimensional flow. kA = 150 W/m°C, kB = 30 W/m°C, kC = 65 W/m°C, and kD = 50 W/m°C (M.U.) Solution. The thermal circuit for heat flow in the given composite system (shown in Fig. 2.24) has been illustrated Fig. 2.24. in Fig. 2.25. Thickness : LA = 3 cm = 0.03 m; LB = LC = 8 cm = 0.08 m; LD = 5 cm = 0.05 m Areas : AA = 0.1 × 0.1 = 0.01m2 ; AB = 0.1 × 0.03 = 0.003m2 AC = 0.1 × 0.07 = 0.007m2 ; AD = 0.1 × 0.1 = 0.01m2 Heat flow rate, Q : The thermal resistances are given by, Rth – A

LA k A AA

0.03 150 u 0.01

0.02

Rth – B

LB k B AB

0.08 u 30 0.003

0.89

Fig. 2.25. Thermal circuit.

Rth – C

LC kC AC

0.08 65 u 0.007

0.176

57

Chapter : 2

58

Rth – D

LD k D AD

0.05 50 u 0.01

0.1

The equivalent thermal resistance for the parallel thermal resistance Rth – B and Rth – C is given by 1

1

( Rth )eq

Rth – B



1 Rth – C

1 1  0.89 0.176

6.805

1 0.147 6.805 Now, the total thermal resistance is given by

?

( Rth )eq.

(Rth)total = Rth – A + (Rth)eq. + Rth – D = 0.02 + 0.147 + 0.1 = 0.267 ?

Q

('t )overall ( Rth )total

(400 – 60) 0.267

1273.4 W (Ans.)

Example 2.17. The insulation boards for air-conditioning purposes are made of three layers, middle being of packed grass 10 cm thick (k = 0.02 W/m°C) and the sides are made of plywood each of 2 cm thickness (k = 0.12 W/m°C). They are glued with each other. (i) Determine the heat flow per m2 area if one surface is at 35°C and other surface is at 20°C. Neglect the resistance of glue. (ii) Instead of glue, if these three pieces are bolted by four steel bolts of 1 cm diameter at the corner (k = 40 W/m°C) per m2 area of the board then find the heat flow per m2 area of the combined board. (M.U.) Solution. (i) When the layers are glued : Refer to Fig. 2.26. Thickness of each of the plywood layer, LA = LC = 2 cm = 0.02 m Thickness of grass layer, LB = 10 cm = 0.1 m Thermal conductivities : kA = kC = 0.12 W/m°C; kB = 0.02 W/m°C; Temperatures : t1 = 35°C; t4 = 20°C Heat flow per m2 area, q : (t1 – t4 ) q Rth – A  Rth – B  Rth – C (t1 – t4 ) L LA L  B  C k A ˜ A k B ˜ A kC ˜ A (35 – 20) 0.02 0.1 0.02  u u u 0.12 1 0.02 1 0.12 u 1 15 0.167  5.0  0.167

2.81 W/m 2 (Ans.)

Fig. 2.26.

“Conduction” Heat Transfer at Steady State–One Dimension (ii) When the layers are joined by steel bolts : Refer to Fig. 2.27. Number of steel bolts used = 4 Diameter of each bolt, db = 1cm = 0.01 m S u 0.012 7.854 u 10 –5 m 2 ? Area of each bolt, Ab 4 Thermal conductivity of bolt material, kD = 40 W/m°C The equivalent thermal resistance (Rth)eq. of the thermal circuit for the system is given by 1 1 4  ( Rth )eq ( Rth – A  Rth – B  Rth – C ) Rth – D where

Rth – D 1

? or,

( Rth )eq

( LA  LB  LC ) 0.02  0.1  0.02 44.56qC/W k D ˜ Ab 40 u 7.854 u 10 –5 1 4  0.187  0.089 0.276 (0.167  5.0  0.167) 44.56

( Rth )eq. or ( Rth )total

1 0.276

3.623q C/W

Fig. 2.27.

59

60

Chapter : 2

Heat flow per m2 area, q : (t1 – t4 ) (35 – 20) q 4.14 W/m 2 (Ans.) ( Rth )total 3.623 Example 2.18. Two slabs, each 120 mm thick, have thermal conductivities of 14.5 W/m°C and 210 W/m°C. These are placed in contact, but due to roughness, only 30 percent of area is in contact and the gap in the remaining area is 0.025 mm thick and is filled with air. If the temperature of the face of the hot surface is at 220°C and the outside side surface of other slab is at 30°C, determine : (i) Heat flow through the composite system. (ii) The contact resistance and temperature drop in contact. Assume that the conductivity of air is 0.032 W/m°C and that half of the contact (of the contact area) is due to either metal. LA1 0.025mm = 0.000025 m Solution. LA = 120 mm = 0.12 m; LB = 120 mm = 0.12 m;

LB1 = 0.025 mm = 0.000025 m

Fig. 2.28.

“Conduction” Heat Transfer at Steady State–One Dimension

61

LC = 0.025 mm = 0.000025 m; kA = k A1 = 14.5 W/m°C; kB = k B1 = 210 W/m°C; kC = 0.032 W/m°C; t1 = 220°C; t2 = 30°C (i) Heat flow through the system, Q : 0.12 Rth – A 14.5 u 1 0.000025 0.000025 Rth – A1 , Rth – C 14.5 u 0.15 0.032 u 0.7 0.000025 0.12 Rth – B1 ; Rth – B 210 u 0.15 210 u 1 1 1 1 1   Rth – A1 Rth – C Rth – B1 ( Rth )eq 1

14.5 u 0.15 0.032 u 0.7 210 u 0.15   ( Rth )eq 0.000025 0.000025 0.000025 or, (Rth)eq = 7.419 × 10–7 or, (Rth)total = Rth – A + (Rth)eq + Rth – B 0.12 0.12  7.419 u 10 –7  ; 8.84 u 10 –3 14.5 u 1 210 u 1 ('t )overall (220 – 30) Q Hence, = 21493 W or 21.493 kW (Ans.) ( Rth )total 8.84 u 10 –3 (ii) The contact resistance and temperature drop in contact : The contact resistance = 7.419 × 10–7 °C/W (Ans.) The temperature drop in contact = Q × contact resistance = 21493 × 7.419 × 10–7 = 0.0159°C (Ans.) Example 2.19. A mild steel tank of wall thickness 12 mm contains water at 95°C. The thermal conductivity of mild steel is 50 W/m°C, and the heat transfer coefficients for the inside and outside the tank are 2850 and 10 W/m 2 °C, respectively. If the atmospheric temperature is 15°C, calculate : (i) The rate of heat loss per m2 of the tank surface area. (ii) The temperature of the outside surface of the tank. Solution. Refer to Fig. 2.29. Fig. 2.29. Thickness of mild steel tank wall L = 12 mm = 0.012 m Temperature of water, thf = 95°C Temperature of air, tcf = 15°C

Chapter : 2

62

Thermal conductivity of mild steel, k = 50 W/m°C Heat transfer coefficients : Hot fluid (water), hhf = 2850 W/m2°C Cold fluid (air), hcf = 10 W/m2°C. (i) Rate of heat loss per m2 of the tank surface area, q: Rate of heat loss per m2 of tank surface, q = UA (thf – tcf) The overall heat transfer coefficient, U is found from the relation, 1 1 1 L 1 0.012 1     U hhf k hcf 2850 50 10 = 0.0003508 + 0.00024 + 0.1 = 0.1006 1 U 9.94 W/m 2 ° C ? 0.1006 ? q = 9.94 × 1 × (95 – 15) = 795.2 W/m2 (Ans.) (ii) Temperature of the outside surface of the tank, t2 : We know that, or,

q = hcf × 1 × (t2 – tcf) 795.2 = 10 (t2 – 15)

795.2 (Ans.)  15 94.52°C 10 Example 2.20. An electric hot plate is maintained at a temperature of 350°C, and is used to keep a solution boiling at 95°C. The solution is contained in a cast-iron vessel of wall thickness 25 mm, which is enamelled inside to a thickness of 0.8 mm. The heat transfer coefficient for the boiling solution is 5.5 kW/m2K, and the thermal conductivities of the cast iron and enamel are 50 and 1.05 W/mK, respectively. Calculate : (i) The overall heat transfer coefficient. (ii) The rate of heat transfer per unit area. (GATE) Solution. Given : theater = 350°C; tsolution = 95°C; ('x)C.I. = 25 mm = 0.025 m; ('x)enamel = 0.8 mm = 0.8 × 10–3 m; hsolution = 5.5 kW/m2K; kC.I. = 50 W/mK; kenamel = 1.05 W/mK. Refer to Fig. 2.30.

or,

t2

Fig. 2.30.

“Conduction” Heat Transfer at Steady State–One Dimension

63

(i) The overall heat transfer coefficient, U : 1 ('x)C .I . ('x)enamel 1   U kC . I . kenamel hsolution § 0.025 0.8 u 10 –3 · 1 –3  ¸ 1.444 u 10 W 3 ¸ ¨¨ 50  1.05 u 5.5 10 ¹ © ? U = 692.5 W/m2K (Ans.) (ii) The rate of heat transfer per unit area, Q : Q = UA (theater – tsolution) = 692.5 × 1 × (350 – 9.5) = 176587.5 W/m2 ; 176.6 kW/m2 (Ans.) Example 2.21. The maximum operating temperature of a kitchen oven is set at 310°C. Due to seasonal variations, the kitchen temperature may vary from 12°C to 32°C. If the average heat transfer coefficient between the outside oven surface and kitchen air is 12 W/ m2°C, determine the necessary thickness of fibre glass (k = 0.036 W/m°C) insulation to ensure that the outside surface temperature of oven does not exceed 45°C. Assume that the steady state conditions prevail and the thermal resistance of metal wall is negligible. Solution. Refer to Fig. 2.31. Maximum temperature of kitchen oven, ti = 310°C Outside surface temperature of oven, t0 = 45°C Kitchen air temperature, tair = 12°C to 32°C Thermal conductivity of insulating material Fig. 2.31. (fibre glass) k = 0.036W/m°C Heat transfer coefficient, ho = 12W/m2°C. Thickness of insulation (fibre glass), L : The rate of heat transfer per unit area of the wall is given as ti – tair Q q A L / k  1/ h0 Further, as the steady state conditions prevail, heat flow through each section is same. ti – tair t0 – tair ? L k h / 1/ 0 1/ h0 1 or, h (ti – tair ) (t0 – tair ) [ L / k  1/ h0 ] 0 L 1 (t0 – tair )  (t0 – tair ) k h0

64

Chapter : 2 or or

1 [(ti – tair ) – (t0 – tair )] h0 1 (ti – t0 ) h0

L (t0 – tair ) k L (t0 – tair ) k

k ª ti – t0 º h0 «¬ t0 – tair »¼ The thickness of insulation (fibre glass) will be large for tair = 32°C. or

L

0.036 ª 310 – 45 º = 0.06115 m or 61.15 mm (Ans.) 12 «¬ 45 – 32 »¼ Example 2.22. Hot gases at 1020°C flow past the upper surface of a gas turbine blade (the blade to be considered as a flat plate 1.2 mm thick) and the lower surface is cooled by air bled off the compressor. The thermal conductivity of blade material is 12 W/m°C and the heat transfer coefficients (convective) at the upper and lower surfaces are 2750 W/m2°C and 1400 W/m2°C respectively. Assuming steady state conditions have reached and the metallurgical considerations limit the blade temperature to 900°C, estimate the temperature of coolant-air. Solution. Temperature of hot gases (fluid) thf = 1020°C Thickness of blade, L = 1.2 mm = 0.0012 m Thermal conductivity of blade material, k = 12 W/m°C Convective heat transfer coefficients : Upper surface, hhf = 2750 W/m2°C Lower surface, hcf = 1400 W/m2°C Temperature at the upper surface of the blade, t1 = 900°C.

?

L

Fig. 2.32.

Temperature of the coolant air, tcf : The rate of heat transfer per unit area, Q = hhf · A (thf – t1) = 2750 × 1 (1020 – 900) = 330000 W/m2 Since the heat transfer takes place under steady state conditions, therefore, this heat would be conducted across the gas turbine blade. Using Fourier's law of heat conduction, we have kA (t1 – t2 ) Q L where, t2 = Temperature of the lower surface. 12 u 1 u (900 – t2 ) 330000 or, 0.0012

“Conduction” Heat Transfer at Steady State–One Dimension or,

t2

900 –

330000 u 0.0012 12

65

867q C

As the heat conducted across the blade would be transferred to the coolant-air, therefore, Q = hcf · A (t2 – tcf) 330000 = 1400 × 1 (867 – tcf) 330000 631.28°C (Ans.) 1400 Example 2.23. A metal plate of 4mm thickness (k = 95.5 W/m°C) is exposed to vapour at 100°C on one side and cooling water at 25°C on the opposite side. The heat transfer coefficients on vapour side and water side are 14500 W/m2°C and 2250 W/m2°C respectively. Determine : (i) The rate of heat transfer, (ii) The overall heat transfer coefficient, and (iii) Temperature drop at each side of heat transfer.

?

tcf

867 –

Fig. 2.33.

Solution. Thickness of metal plate, Thermal conductivity of plate material, Temperature of vapour (hot fluid), Temperature of water (cold fluid), Heat transfer coefficients : Vapour side, Water side,

L k hhf tcf

= 4 mm = 0.004 m = 95.5 W/m°C = 100°C = 25°C

hhf = 14500 W/m2°C hcf = 2250 W/m2°C

66

Chapter : 2 (i) The rate of heat transfer per m2, q : (thf – tcf ) (' t )overall q ( Rth )total ( Rth )total (thf – tcf )

( Rth )conv.– hf  ( Rth )1 – 2  ( Rth )conv.– cf (100 – 25) 1 1 L   hhf k hcf 75 1 0.004 1   14500 95.5 2250 75 6.896 u 10 –5  4.188 u 10 –5  44.444 u 10 –5 = 1.35 × 105 W/m2 Hence, rate of heat transfer, q = 1.35 × 105 W/m2 (Ans.) (ii) The overall heat transfer coefficient, U: The rate of heat transfer through a composite system is given by Q = U.A. ('t)overall

Q q 1.35 u 105 1800 W/m 2 ° C (Ans.) A .(' t ) ' t (100 – 25) (iii) Temperature drop at each side of heat transfer : We know that q = qhf = q1 – 2 = qcf = 1.35 × 105 W/m2 (' t ) hf qhf Now, (R ) or,

U

th conv .– cf

1 9.31q C 14500 i.e., Temperature drop in vapour film = 9.31°C (Ans.) (' t )1 – 2 0.004 q1 – 2 (' t )1 – 2 1.35 u 105 u Similarly, ( Rth )1 – 2 or 95.5 i.e., Temperature drop in the metal = 5.65°C (Ans.) (' t )cf qcf and, ( Rth )conv – cf

or,

(' t )hf

1.35 u105 u

5.65q C

1 60q C 2250 i.e., Temperature drop in the water film = 60°C (Ans.)

or,

(' t )cf

1.35 u 105 u

Example 2.24. The interior of a refrigerator having inside dimensions of 0.5 m × 0.5 m base area and 1m height, is to be maintained at 6°C. The walls of the refrigerator are constructed of two mild steel sheets 3mm thick (k = 46.5 W/m°C) with 50 mm of glass wool insulation (k = 0.046 W/ m°C) between them. If the average heat transfer coefficients at the outer and inner surfaces are 11.6 W/m 2°C and 14.5 W/m 2°C respectively, calculate : (i) The rate at which heat must be removed from the interior to maintain the specified temperature in the kitchen at 25°C, and (ii) The temperature on the outer surface of the metal sheet.

“Conduction” Heat Transfer at Steady State–One Dimension

67

Solution. Refer to Fig. 2.34. LA = LC = 3mm = 0.003 m LB = 50 mm = 0.05 m kA = kC = 46.5 W/m°C; kB = 0.046 W/m° C h0 = 11.6 W/m2°C; hi = 14.5 W/m2°C t0 = 25°C; ti = 6°C. The total area through which heat is coming into the refrigerator, A = 0.5 × 0.5 × 2 + 0.5 × 1 × 4 = 2.5 m2

Fig. 2.34.

(i) The rate of removal of heat, Q : A (t0 – ti ) Q L LA L 1 1   B  C  h0 kA kB kC hi 2.5(25 – 6) 38.2 W (Ans.) 1 0.003 0.05 0.003 1     11.6 46.5 0.046 46.5 14.5 (ii) The temperature at the outer surface of the metal sheet, t : Q = h0A (25 – t1) or, 38.2 = 11.6 × 2.5 (25 – t1) 38.2 23.68° C (Ans.) t1 25 – or, 11.6 u 2.5 Example 2.25. Calculate the rate of heat flow per m2 through a furnace wall consisting of 200 mm thick inner layer of chrome brick, a centre layer of kaolin brick 100 mm thick and an outer layer of masonry brick 100 mm thick. The unit surface conductance at the inner surface is 74 W/m2°C and

68

Chapter : 2

the outer surface temperature is 70°C. The temperature of the gases inside the furnace is 1670°C. What temperatures prevail at the inner and outer surfaces of the centre layer ? Take : kchrome brick = 1.25 W/m°C; kkaolin brick = 0.074 W/m°C; kmasonry brick = 0.555 W/m°C Assume steady heat flow. (M.U.) Solution.

Thickness of chrome bricks, LA Thickness of kaolin bricks, LB Thickness of masonry bricks, LC Thermal conductivities : kA kB The unit surface conductance, hhf Temperature of hot fluid, thf (= tg) Temperature of the outer surface, t4

= = = = = = = =

200 mm = 0.2 m 100 mm = 0.1 m 100 mm = 0.1m 1.25 W/m°C; 0.074 W/m°C; kC = 0.555 W/m°C; 74 W/m2°C 1670°C 70°C

Fig. 2.35

(i) Rate of heat flow per q

m2,

q:

(thf – t4 ) L L L 1  A  B  C hhf kA kB kC (1670 – 70) 1 0.2 0.1 0.1    74 1.25 0.074 0.555 1600  0.0135 0.16  1.351  0.1802

938.58 W/m2 (Ans.)

“Conduction” Heat Transfer at Steady State–One Dimension (ii) Temperatures; t2, t3 : The heat flow is given by (thf – t1 ) (t1 – t2 ) q 1/ hhf LA / k A ? or

938.58

69

(t2 – t3 ) ( LB / k B )

1670 – t1 1/ 74

t1 1670 – 938.58 u

1 74

1657.3q C

(1657.3 – t2 ) or t2 1657.3 – 938.58 u 0.2 1507.1° C (Ans) 0.2 /1.25 1.25 (1507.1 – t3 ) 0.1 938.58 238.7° C (Ans.) or t3 1507.1 – 938.58 u (0.1/ 0.074) 0.074 Example 2.26. A cold storage room has walls made of 220 mm of brick on the outside, 90 mm of plastic foam, and finally 16 mm of wood on the inside. The outside and inside air temperatures are 25°C and –3°C respectively. If the inside and outside heat transfer coefficients are respectively 30 and 11 W/m2°C, and the thermal conductivities of brick, foam and wood are 0.99, 0.022 and 0.17 W/m°C respectively, determine : (i) The rate of heat removal by refrigeration if the total wall area is 85 m2; (ii) The temperature of the inside surface of the brick. Solution. Refer Fig. 2.36.

Similarly,

938.58

Fig. 2.36.

Thickness of brick wall, LA Thickness of plastic foam, LB Thickness of wood, LC Temperature of hot fluid (air), thf Temperature cold fluid (air), tcf Heat transfer coefficients : Hot fluid (air), hhf Cold fluid (air), hcf Thermal conductivities : Brick, kA

= = = = =

220 mm = 0.22 m 90 mm = 0.09 m 16 mm = 0.016 m 25°C –3°C

= 11 W/m2°C = 30 W/m2°C = 0.99 W/m°C

70

Chapter : 2 Foam, Wood, Total wall area, (i) Rate of heat transfer, Q :

kB = 0.022 W/m°C kC = 0.17 W/m°C A = 85 m2

Q = UA (thf – tcf) The overall heat transfer co-efficient (U) may be found from the following relation : 1 U

1 0.22 0.09 0.06 1 L L L 1 1      A  B  C  11 0.99 0.022 0.17 30 hhf kA kB kC hcf = 0.091 + 0.222 + 4.091 + 0.094 + 0.033 = 4.531

1 0.2207 W/m 2 °C 4.531 ? Q = 0.2207 × 85 [25 – (–3)] = 525.26 W (Ans.) (ii) Temperature of inside surface of the brick, t2 : Q = U.A (thf – t2)

?

U

or,

525.26

1 ª º A (thf – t2 ) « 1 LA »  « » k A »¼ «¬ hhf

1 ª º u 85 (25 – t2 ) «1 0.22 » «  » ¬ 11 0.99 ¼

271.45(25 – t2 )

525.26 23.06°C (Ans.) 271.45 Example 2.27. A furnace wall is made of composite wall of total thickness 550 mm. The inside layer is made of refractory material (K = 2.3 W/mK) and outside layer is made of an insulating material (K = 0.2 W/mK). The mean temperature of the glass inside the furnace is 900°C and interface temperature is 520°C. The heat transfer coefficient between the gases and inner surface can be taken as 230 W/m2 °C and between the outside surface and atmosphere as 46 W/m2°C. Taking air temperature = 30°C, calculate : (i) Required thickness of each layer, (ii) The rate of heat loss per m2 area, and (iii) The temperatures of surface exposed to gases and of surface exposed to atmosphere. (B.U.) Solution. Given : Total thickness of wall, L A + L B = 550 mm = 0.55 m; k A = 2.3 W/mK; kB = 0.2 W/mK; thf = 900°C; t2 = 520°C; hhf = 230 W/m2°C; hcf = 46 W/m2°C; tcf = 30°C.

?

t2

25 –

Fig. 2.37.

“Conduction” Heat Transfer at Steady State–One Dimension Thickness of each layer, LA, LB : The heat flow rate, Q q A

thf – t2 L 1  A hhf kA

Equating (i) and (ii) we get, thf – t2 L 1  A hhf kA

t2 – tcf 0.55 – LA 1  kB hcf

900 – 520 L 1  A 230 2.3 1º ª 0.55 – LA 380 «  0.2 46 »¼ ¬

520 – 30 0.55 – LA 1  0.22 46 L º ª 1 490 «  A ¬ 230 2.3 »¼

71

t2 – tcf 0.55 – LA 1  kB hcf

1045 – 1900 LA + 8.26 = 2.13 + 213 LA 2113 LA = 1051.13 or, and,

(Ans.) LA = 0.497 m or 497 mm LB = 550 – 497 m = 53 mm (Ans.)

(i) The rate of heat loss per m2 area, q : thf – t2 900 – 520 q LA 1 0.497 1   hhf kA 230 2.3 380 (Ans.) 1724.5 W/m 2 0.004348  0.216 Example 2.28. The inside temperature of furnace wall, 200 mm thick, is 1350°C. The mean thermal conductivity of wall material is 1.35 W/m°C. The heat transfer coefficient of the outside surface is a function of temperature difference and is given by h = 7.85 + 0.08 't where 't is the temperature difference between outside wall surface and surroundings. Determine the rate of heat transfer per unit area if the surrounding temperature is 40°C. Solution. Thickness of wall, L = 200 mm = 0.2m Temperature of inner surface of wall, t1 = 1400°C Temperature of air (cold fluid), tcf = 40°C Mean thermal conductivity of wall material, k = 1.35 W/m°C Fig. 2.38.

72

Chapter : 2 Rate of heat transfer per unit area, q :

q

(t1 – t2 ) L/k

(t2 – tcf ) 1/ h

or,

(1350 – t2 ) h (t2 – 40) 0.2 /1.35 6.75 (1350 – t2) = [7.85 + 0.08 (t2 – 40)] (t2 – 40)

or,

9112.5 – 6.75t2 = 7.85 (t2 – 40) + 0.08 (t2 – 40)2

or,

9112.5 – 6.75t2 = 7.85t2 – 314 + 0.08 (t22 – 80t2 + 1600)

or, or,

9112.5 – 6.75 t2 – 7.85 t2  314 0.08 = 117831 – 182.5 t2 t22 + 102.5t2 – 116231 = 0 t2 2 – 80 t2  1600

–102.5 r (102.5) 2  4 u 116231 2 –102.5 r 689.5 293.5q C 2 (1350 – 293.5) q 7131.37 W/m 2 (Ans.) ? 0.2 /1.35 Exampel 2.29. The furnace wall consists of Refractory bricks 120 mm wide refractory brick and 120 mm wide Air gap Insulating fire bricks insulating fire brick separated by an air gap. The Plaster outside wall is covered with a 12 mm thickness of plaster. The inner surface of the wall is at 1090°C t1(=thf) and the room temperature is 20°C. The heat transhcf t2 = 1090°C fer coefficient from the outside wall surface to t3 the air in the room is 18 W/m2°C, and the resistance to heat flow of the air gap is 0.16 K/W. If the thermal conductivities of the refractory brick, insulating fire brick, and plaster are 1.6, 0.3 kB kA t5 t4 and 0.14 W/mK, respectively calculate : t =20°C 2 (i) Rate at which heat is lost per m of the wall surface; LC LA LB (ii) Each interface temperature; and = 120 mm = 120 mm = 12 mm (iii) Temperature of the outside surface of the Fig. 2.39. wall. Solution. Refer to Fig. 2.39. Thickness of refractory brick, LA = 120 mm = 0.12 m Thickness of insulating fire brick, LB = 120 mm = 0.12 m Thickness of plaster, LC = 12 mm = 0.012 m Heat transfer coefficient from the outside wall surface to the air in the room, hcf = 18 W/m2°C Resistance of air gap to heat flow = 0.16 K/W Thermal conductivities : Refractory brick, kA = 1.6 W/m°C or,

t2

cf

“Conduction” Heat Transfer at Steady State–One Dimension

73

Insulating fire brick, kB = 0.3 W/m°C Plaster, kC = 0.14 W/m°C. Temperatures : thf = 1090°C; tcf = 20°C Consider 1m2 of surface area. (i) Rate of heat loss per m2 of surface area, q : (thf – tcf ) q LA L L 1  air gap resistance  B  C  kA kB kC hcf (1090 – 20) 0.12 0.12 0.012 1  0.16    1.6 0.3 0.14 18 1070 1378.5 W or 1.3785 kW 0.075  0.16  0.4  0.0857  0.0555 i.e., Rate of heat loss per m2 of surface area = 1.3785 kW (Ans.) (ii) Temperatures at interfaces, t2, t3, t4: 1090 – t2 1090 – t2 1090 – t2 Q 1378.5 LA / k A 0.12 /1.6 0.075 ? t2 = 1090 – 1378.5 × 0.075 = 986.6°C (Ans.) t2 – t3 986.6 – t3 Q 1378.5 Also, air gap resistance 0.16 ? t3 = 986.6 – 1378.5 × 0.16 = 766.04°C (Ans.) t3 – t4 766.04 – t4 766.04 – t4 Q 1378.5 Again, LB / k B 0.12 / 0.3 0.4 ? t4 = 766.04 – 1378.5 × 0.4 = 214.64°C (Ans.) (iii) Temperature of the outside surface of the wall, t5 : t4 – t5 214.64 – t5 214.64 – t5 Q 1378.5 LC / kC 0.012 / 0.14 0.0857 ? t5 = 214.64 – 1378.5 × 0.0857 = 96.5°C (Ans.) Example 2.30. A furnace wall is made up of three layers of thicknesses 250 mm, 100 mm and 150 mm with thermal conductivities of 1.65, k and 9.2 W/m°C respectively. The inside is exposed to gases at 1250°C with a convection coefficient of 25 W/m2°C and the inside surface is at 1100°C, the outside surface is exposed to air at 25°C with convection coefficient of 12 W/m2°C. Determine : (i) The unknown thermal conductivity ‘k’; (ii) The overall heat transfer coefficient; (iii) All surface temperatures. Solution. LA = 250 mm = 0.25 m; LB = 100 mm = 0.1 m; LC = 150 mm = 0.15 m; kA = 1.65W/m°C; kC = 9.2 W/m°C; thf = 1250°C, t1 = 1100°C; hhf = 25 W/m2°C; hcf = 12 W/m2°C.

74

Chapter : 2 (i) Thermal conductivity, k (= kB) : The rate of heat transfer per unit area of the furnace wall, q = hhf (thf – t1) = 25(1250 – 1100) = 3750 W/m2 (' t )overall q Also, ( Rth ) total (thf – tcf ) q or ( Rth )conv – hf – Rth – A  Rth – B  Rth – C  ( Rth )conv – cf

Fig. 2.40.

or,

or,

or, or, ?

3750

3750

0.1 · § 3750 ¨ 0.289  k B ¸¹ © 0.1 kB kB

(1250 – 25) L LA LB 1 1    C  hhf kA kB kC hcf 1225 1 0.25 0.1 0.15 1     25 1.65 k B 9.2 12 1225 0.1  0.0163  0.0833 0.04  0.1515  kB 1225 1225 – 0.2911 0.0355 3750 0.1 k = 2.817 W/m°C 0.0355

(Ans.)

1225 0.2911 

0.1 kB

“Conduction” Heat Transfer at Steady State–One Dimension

75

(ii) The overall heat transfer coefficient, U : 1 ( Rth ) total 1 0.25 0.1 0.15 1     ( Rth ) total 25 1.65 2.817 9.2 12 = 0.04 + 0.1515 + 0.0355 + 0.0163 + 0.0833 = 0.3266 °C m2/W

The overall heat transfer coefficient, U Now,

1 1 ( Rth ) total 0.3266 (iii) All surface temperatures; t1, t2, t3, t4 : q = qA = qB = qC

?

U

or,

3750

(t1 – t2 ) LA / k A

or,

3750

(1100 – t2 ) 0.25 /1.65

or,

t2

Similarly, or, and or,

3750 t3 3750 t4

(t2 – t3 ) LB / k B

1100 – 3750 u (531.8 – t3 ) 0.1/ 2.817

0.25 1.65

531.8 – 3750 u

0.1 2.817

(t3 – t4 ) LC / kC

531.8° C (Ans.)

398.6° C (Ans.)

(398.6 – t4 ) (0.15 / 9.2) 398.6 – 3750 u

[Check using outside convection, q

3.06 W/m 2 ° C (Ans.)

(337.5 – 25) 1/ hcf

0.5 9.2

337.5° C (Ans.)

(337.5 – 25) 1/12

3750 W/m 2 ]

Example 2.31. A square plate heater (15 cm × 15 cm) is inserted between two slabs. Slab A is 2 cm thick (k = 50 W/m°C) and slab B is 1 cm thick (k = 0.2 W/m°C). The outside heat transfer coefficients on side A and side B are 200 W/m2°C and 50 W/m2°C respectively. The temperature of surrounding air is 25°C. If rating of heater is 1kW, find : (i) Maximum temperature in the system; (ii) Outer surface temperature of two slabs. Draw an equivalent electrcal circuit. (M.U., 2001) Solution. Refer to Fig. 2.41. Thickness of slab A, LA = 2cm = 0.02m Thickness of slab B, LB = 1cm = 0.01m Thermal conductivities, kA = 50 W/m°C; kB = 0.2 W/m°C Overall heat transfer co-efficients : h1 = 200 W/m2°C; h2 = 50 W/m2°C Area of the plate = 0.15 × 0.15 = 0.0225 m2 Rating of heater = 1kW = 1000 W Temperature of surrounding air, ta = 25°C

76

Chapter : 2

Fig. 2.41.

(i) Maximum temperature in the system, tmax : For steady state heat flow, we have Q = Heat flow through slab A(QA) + heat flow through slab B(QB) (tmax – ta ) (tmax – ta ) 1 1 ª   A (tmax – ta ) « LA LB LA LB 1 1 1 1     « k A . A h1 . A k B . A h2 . A h1 kB h2 ¬« k A Substituting the values in the above equation, we get, 1 1 ª º  1000 0.0225(tmax – 25) « 0.02 1 0.01 1 »   « » 200 0.2 50 ¼ ¬ 50 1 1 ª º  0.0225(tmax – 25) « » ¬ (0.0004  0.005) (0.05  0.02) ¼ = 0.0225 (tmax – 25) × 199.47 100 0.0225 u 199.47 (ii) Outer surface temperature of two slabs; t1, t2 :

?

tmax

QA

or, or,

25 

k A . A (tmax – t1 ) LA

50 (247.8 – t1 ) 200 (t1 – 25) 0.02 2500 (247.8 – t1) = 200 (t1 – 25)

247.8° C (Ans.)

hi . A (t1 – ta )

º » » ¼»

“Conduction” Heat Transfer at Steady State–One Dimension or, or,

200 (t1 – 25) 2500 1.08 t1 = 249.8

247.8 – t1

t1

? Similarly, or, or, or,

QB

249.8 1.08

77

0.08 t1 – 2

231.3° C (Ans.)

k B . A (tmax – t2 ) LB

h2 . A (t2 – ta )

0.2 (247.8 – t2 ) 50 (t2 – 25) 0.01 20 (247.8 – t2) = 50 (t2 – 25)

50 (t2 – 25) 20 3.5 t2 = 310.3

(247.8 – t2 )

or,

2.5 t2 – 62.5

310.3 88.6° C (Ans.) 3.5 Equivalent electrical/thermal circuit is shown in Fig. 2.41 (b).

?

t2

Example 2.32. The following data relate to furnace of a steam boiler : Temperature of gases in the furnace ............. 1300°C Temperature of air in the boiler room ............. 30°C Thickness of refractory material ............. 250 mm The heat transfer coefficient from gases to refractory wall ......... 30 W/m2°C The heat transfer coefficient from outside surface to surrounding air ......... 10W/m2°C Thermal conductivity of refractory material : k = 0.28 (1 + 0.000833 t) W/m°C Thermal conductivity of diatomite layer : k = 0.113 (1 + 0.000206 t) W/m°C Estimate the thickness of the diatomite layer of setting so that the loss of heat to the surroundings should not exceed 750 W/m2. Solution. Refer to Fig. 2.42. Given : thf = 1300°C; tcf = 30°C; LA = 250 mm = 0.25 m; hhf = 30 W/m2°C; hcf = 10 W/m2°C. Thermal conductivities : kA = 0.28(1 + 0.000833 t)W/m°C kB = 0.113(1 + 0.000206 t)W/m°C Loss of heat to the surroundings, q = 750 W/m2 Thickness of diatomite layer, LB (=x) : The rate of heat transfer per m2 through the composite furnace wall is given by,

Fig. 2.42.

78

Chapter : 2

q

(thf – t1 )

(t3 – tcf )

(t1 – tcf )

1/ hhf

1/ hcf

LA / kmA

(t2 – t3 ) LB / kmB

where,

kmA

ª § t  t ·º koA «1  D A ¨ 1 2 ¸ » © 2 ¹¼ ¬

and,

kmB

ª § t  t ·º koB «1  D B ¨ 2 3 ¸ » © 2 ¹¼ ¬

?

750

(1300 – t1 ) 1/ 30

or

t1 1300 – 750 u

Again,

750

(t3 – 30) 1/10

or

t3

30  750 u

1 30

1 10

1275q C 105q C

Now, using the following equation,

(t3 – tcf )

kmA (t1 – t2 ) LA

1/ hcf or,

hcf (t3 – tcf )

kmA (t1 – t2 ) LA

Substituting the value of kmA in the above equation, we have

hcf (t3 – tcf )

10 (105 – 30)

or,

ª § t  t ·º ko A «1  D A ¨ 1 2 ¸ » © 2 ¹¼ ¬ (t1 – t2 ) LA ª § 1275  t2 · º 0.28 «1  0.000833 ¨ ¸» © ¹¼ 2 ¬ (1275 – t2 ) 0.25

750

ª § 1275  t2 · º 1.12 «1  0.000833 ¨ ¸ » (1275 – t2 ) © ¹¼ 2 ¬

750 1.12

[1  0.0004165 (1275  t2 )](1275 – t2 )

669.6 = (1275 – t2) + 0.0004165 [(1275)2 – (t2)2] 669.6 = (1275 – t2) + 677 – 0.0004165 t22 0.0004165 t22 + t2 – 1282.5 = 0 ?

t2 kmB

–1 r 1  4 u 0.0004165 u 1282.5 2 u 0.0004165 ª § 925.6  105 · º 0.113 «1  0.000206 ¨ ¸» © ¹¼ 2 ¬

ª § 1275  925.6 · º 0.28 «1  0.000833 ¨ ¸» © ¹¼ 2 ¬ Now using the following equation, we have : kmA

(t1 – t2 ) LA / kmA

(t2 – t3 ) LB / k mB

925.6q C (ignoring –ve sign) 0.125 W / mq C 0.536 W/m°C

“Conduction” Heat Transfer at Steady State–One Dimension kmA (t1 – t2 ) LA

or, or,

0.536 (1275 – 925.6) 0.25

kmB (t2 – t3 ) LB 0.125(925.6 – 105) x

102.57 102.57 ? x 749.11 x Hence, thickness of diatomite layer = 137 mm. (Ans.)

or,

79

749.11

0.137 m or 137 mm

Example 2.33. Fig. 2.43 shows the temperature distribution through a furnace wall consisting of fire brick and high temperature block insulation and steel plate. If the thermal conductivity of the fire brick is 1.13 W/m°C, determine : (i) Rate of heat per unit area of furnace wall; (ii) Thermal conductivities of block insulation and steel; (iii) Combined convective and radiative heat transfer coefficient for the outside surface of the furnace wall; (iv) Heat exchange by radiation between the hot gases and inside surface of furnace wall. The absorptivity and emissivity of the fire brick wall surface is 0.82. (v) Convective heat transfer coefficient for the inside surface of the furnace wall. Solution. Refer to Fig. 2.43.

Fig. 2.43.

Given :

thf = 810°C; t1 = 808°C; t2 = 777°C; t3 = 78.5°C; t4 = 78.4°C; tcf = 26°C LA = 6.5 cm = 0.065 m; LB = 12 cm = 0.12 m; LC = 0.65 cm = 0.0065 m kA = 1.13 W/m°C Hfire brick = 0.82 V = 5.67 × 10–8 W/m2K4

80

Chapter : 2 (i) Rate of heat transfer per unit area of furnace wall, q : For steady state heat flux for resistances in series, q = qhf = qA = qB = qC = qcf t1 – t2 LA / k A

(808 – 777) 538.9 W/m 2 0.065 1.13 (ii) Thermal conductivities of block insulation (kB) and steel (kC) : q

?

or,

qA

qB

(t2 – t3 ) LB / k B

k B (t2 – t3 ) LB

kB

qB LB (t2 – t3 )

538.9 u 0.12 (777 – 78.5)

0.0926 W/m° C (Ans.)

qC LC 538.9 u 0.0065 35 W/m° C (Ans.) (t3 – t4 ) (78.5 – 78.4) (iii) Combined convective and radiative coefficient on the outside surface, h0 :

Similarly,

kC

(hrad )0  (hconv )

q t4 – tcf

h0

538.9 2 78.4 – 26 = 10.28 W/m ° C (Ans.)

(iv) Heat exchange by radiation, qrad. : Assuming the emissivity of hot gases as unity, the net radiation heat gain of wall per unit area is given by

qrad .



HV (Thf )4 – (T1 )4



= 0.82 × 5.67 × 10–8 [(810 + 273)4 – (808 + 273)4] ª§ 810  273 ·4 § 808  273 ·4 º 0.82 u 5.67 «¨ ¸ –¨ ¸ » ¹ © 100 ¹ ¼ ¬© 100

471.2 W/m 2 (Ans.)

(v) Convective heat transfer coefficient for the inside surface of the furnace wall (hconv.) : As convective and radiative heat transfers between gases and wall are in parallel, therefore q = (qrad) + (qconv.)i or

(qconv.)i = q – qrad. = 538.9 – 471.2 = 67.7 W/m2°C

? Convective heat transfer coefficient on the inside surface, (hconv. )i

(qconv. )i thf – t1

67.7 810 – 808

33.85 W/m 2 ° C

(Ans.)

Example 2.34. The following data relate to a large rectangular combustion chamber for a furnace made of 220 mm common brick, lined on the inside with 220 mm thick layer of magnesite brick : Temperature of gases = 1300°C; Temperature of surrounding air = 40°C; Radiation coefficient, inside surface = 17.5 W/m2°C; convection coefficient, inside surface = 16.4 W/m2°C; Radiation coefficient, outside surface = 7.2 W/m2 °C; convection co-efficient, outside surface = 11.5 W/m2°C; Thermal conductivity of common brick = 0.65 W/m°C; Thermal conductivity of magnesite brick = 3.5 W/m°C.

“Conduction” Heat Transfer at Steady State–One Dimension

81

Fig. 2.44.

Determine the following : (i) Rate of heat transfer through the wall per unit area; (ii) Maximum temperature to which common brick is subjected. Solution. Given : LA = 220 mm = 0.22 m; LB = 220 mm = 0.22 m kA = 3.5 W/m°C; kB = 0.65 W/m°C thf = 1300°C; tcf = 40°C; (hconv.)i = 16.4 W/m2°C; (hconv.)0 = 11.5 W/m2°C; (hrad.)i = 17.5 W/m2°C; (hrad.)0 = 7.2 W/m2°C; (i) Rate of heat transfer through the wall per unit area, q : Since the convective and radiative resistances are in parallel, therefore : q = (qconv. + qrad.)i = qA = qB = (qconv. + qrad.)o (thf – tcf ('t )overall Also,

( Rth )total ( Rth )total hi = (hconv. + hrad.)i = 16.4 + 17.5 = 33.9 W/m2°C ho = (hconv. + hrad.)o = 11.5 + 7.2 = 18.7 W/m2°C L L 1 1  A  B  ( Rth )total hi kA kB ho

1 0.22 0.22 1    33.9 3.5 0.65 18.7 = 0.0295 + 0.0628 + 0.3385 + 0.0535 = 0.4843 m2°C/W Substituting the values in (i), we get q

(1300 – 40) 0.4843

2601.7 W/m 2

(Ans.)

...(i)

82

Chapter : 2 (ii) Maximum temperature to which common brick is subjected, t2 : (t1 – t2 ) q hi (thf – t1 ) LA / k A or, 2601.7 = 33.9 (1300 – t1) 2601 t1 1300 – 1223.27qC or, 33.9 (1223.27 – t2 ) and, 2601.7 (0.22 / 3.5) 0.22 t2 1223.27 – 2601.7 u or, 3.5 = 1059.73°C (Ans.)

2.6.

HEAT CONDUCTION THROUGH HOLLOW AND COMPOSITE CYLINDERS

2.6.1. HEAT CONDUCTION THROUGH A HOLLOW CYLINDER Case I. Uniform conductivity : Refer to Fig. 2.45. Consider a hollow cylinder made of material having constant thermal conductivity and insulated at both ends.

Fig. 2.45.

Let, r1, r2 = Inner and outer radii; t1, t2 = Temperatures of inner and outer surfaces, and k = Constant thermal conductivity within the given temperature range.

“Conduction” Heat Transfer at Steady State–One Dimension

83

The general heat conduction equation in cylindrical coordinates is given by, ª w2 1 wt wt w 2t º qg 1 wt 1 ˜  2 ˜ 2  2»  « 2  ˜ ...(Eqn. 2.22) D wW k w r r wI wz ¼ r ¬ wr § wt · 0 ¸ , unidirectional [t z f (I, x)] heat flow in radial direction and with no For steady state ¨ © wW ¹ internal heat generation (qg = 0), the above equation reduces to d 2t

1 dt ˜ 0 r dr dr 1 d ª dt º ˜ r˜ 0 or, r dr «¬ dr »¼ d § dt · 1 Since, z 0, therefore, ¨r ˜ ¸ 0 dr © dr ¹ r dt or, r˜ C (a constant) dr Integrating the above equation, we get t = C ln (r) + C1 (where C1 = Constant of integration) Using the following boundary conditions, we have : At r = r1, t = t1; At r = r2, t = t2 ? t1 = C ln (r1) + C1 t2 = C ln (r2) + C1 From (i) and (ii), we have 2



t –t (t1 – t2 ) and C1 t1  1 2 ln ( r1 ) ln ( r2 / r1 ) ln ( r2 / r1 ) Substituting the values of these constants in eqn. (2.55), we have t –t (t – t ) t t1  1 2 ln ( r1 ) – 1 2 ˜ ln (r ) ln (r2 / r1 ) ln (r2 / r1 ) [Equation 2.57 is the expression for temperature distribution in a hollow cylinder]. C

or,

–

...(2.54) ...(2.55)

...(i) ...(ii) ...(2.56)

...(2.57)

(t – t1) ln (r2 / r1) = (t1 – t2) ln (r1) – (t1 – t2) ln (r) = (t2 – t1) ln (r) – (t2 – t1) ln (r1) = (t2 – t1) ln (r/r1)

t – t1 ln (r / r1 ) (Dimensionless form) t2 – t1 ln (r2 / r1 ) From the above equation, the following points are worth noting :

or,

...(2.58)

(i) The temperature distribution is logarithmic (not linear as in the case of plane wall). (ii) Temperature at any point in the cylinder can be expressed as a function of radius only. Isotherms (or lines of constant temperatures) are then concentric circles lying between the inner and outer boundaries of the hollow cylinder. (iii) The temperature profile [Eqn. (2.57)] is nearly linear for values of (r2/r1) of the order of unity, but decidely non-linear for large values of (r2/r1) Determination of conduction heat transfer rate (Q) : The conduction heat transfer rate is determined by utilizing the temperature distribution [Eqn. (2.57)] in conjunction with Fourier's equation as follows :

84

Chapter : 2 dt dr t –t d ª (t – t ) º – kA «t1  1 2 ln (r1 ) – 1 2 ln ( r ) » dr ¬ ln (r2 / r1 ) ln (r2 / r1 ) ¼ [Substituting the value of t from Eqn. (2.57)] ª – (t1 – t2 ) º – k (2Sr . L ) « » ¬ r ln (r2 / r1 ) ¼ (t – t ) (t1 – t2 ) ª ' t º ln ( r2 / r1 ) 2S k L 1 2 (where, Rth ) « » r r ln ( / ) ln (r2 / r1 ) 2S k L 2 1 ¬ Rth ¼ 2S k L (t1 – t2 ) ln (r2 / r1 ) ...(2.59) 2S k L

Q

Hence,

– kA

Q

Alternative method : Refer to Fig. 2.45 Consider an element at radius ‘r’ and thickness ‘dr’ for a length of the hollow cylinder through which heat is transmitted. Let dt be the temperature drop over the element. Area through which heat is transmitted, A = 2Sr. L. Path length = dr (over which the temperature falls is dt) ?

§ dt · – kA ˜ ¨ ¸ © dr ¹

Q

– k .2Sr . L dr r Integrating both sides, we get Q.

or,

Q

r2

³r

1

dr r r

dt per unit time dr

– k .2S L . dt

t2

– k .2SL ³ dt t1

t

or,

Q >In (r ) @r2

or,

Q.ln(r2/r1) = – k.2SL(t2 – t1) = k.2SL(t1 – t2)

1

Q

?

– k .2 S L >t @t2 1

k .2S L (t1 – t2 ) ln (r2 / r1 )

(t1 – t2 ) ª ln (r2 / r1 ) º « 2S k L » ¬ ¼

...(2.60)

Case II. Variable thermal conductivity : A. Temperature variation in terms of interface temperatures (t1, t2) : The heat flux equation is given by Q

– kA

dt dr

[where k = k0 (1 + Et)]

– k0 (1  E t ) 2S . r . L . dr r Integrating both sides, we have

or,

Q.

– k0 .2 S L (1  Et ) dt

dt dr

...(2.61)

“Conduction” Heat Transfer at Steady State–One Dimension Q

r2

³r

1

dr r

85

t2

– k0 .2SL ³ (1  Et ) dt t1

t

Q >ln (r ) @

r2 r1

2 ª t2 º – k0 .2S L «t  E . » ¬ 2 ¼ t1

E ª º – k0 .2S L «(t2 – t1 )  (t22 – t12 ) » ¬ ¼ 2 E ª º – k0 .2S L « (t2 – t1 )  (t2  t1 ) (t2 – t1 ) » ¬ ¼ 2 E ª º k0 .2S L «1  (t1  t2 ) » [t1 – t2 ] ¬ ¼ 2 E ª º k0 .2S L «1  (t1  t2 ) » [t1 – t2 ] ¬ ¼ 2 ? Q ln (r2 / r1 ) Integrating between r1 and r, we obtain E ª º k0 .2SL «1  (t1  t ) » [t1 – t ] ¬ ¼ 2 Q ln (r / r1 ) Equating eqns. (2.62) and (2.63), we get Q .ln ( r2 / r1 )

E ª º k0 .2S L «1  (t1  t2 ) » [t1 – t2 ] ¬ ¼ 2 ln (r2 / r1 )

or, or,

...(2.62)

...(2.63)

E ª º k0 .2S L «1  (t1  t ) » [t1 – t ] ¬ ¼ 2 ln (r / r1 )

E 2 ª 2 º ln (r / r1 ) « (t1 – t2 )  (t1 – t2 ) » ¬ ¼ 2 E 2 2 (t1 – t )  (t1 – t ) 2

E 2 ª º ln (r2 / r1 ) « (t1 – t )  (t1 – t 2 ) » ¬ ¼ 2 E 2 ln (r / r1 ) ª 2 º (t1 – t2 )  (t1 – t2 ) » ¼ ln ( r2 / r1 ) «¬ 2

E 2 E 2 ln ( r / r1 ) ª E 2 2 º t1 – t (t1 – t2 )  (t1 – t2 ) » « ¼ 2 2 ln (r2 / r1 ) ¬ 2 E 2 ª E 2 ln (r / r1 ) E 2 2 º t  t – «t1  t1 – (t1 – t2 )  (t1 – t2 ) » 2 2 ln (r2 / r1 ) 2 ¬ ¼ (t1 – t ) 

or,

^

or,

or,

or,

t

^

`

º ln (r / r1 ) Eª E E t1  t12 – (t1 – t2 )  (t12 – t22 » 2 «¬ 2 ln (r2 / r1 ) 2 ¼ E 2u 2 1/ 2 ln (r / r1 ) E º 1 1ª 2 ˜ 2E (t1 – t2 )  (t12 – t22 » – r «(1  2 Et1  2Et1 ) – E E¬ ln (r2 / r1 ) 2 ¼ 1/ 2 r r ln ( / ) 1 1ª 1 ^2E t1 – 2E t2  Et12 – Et22 `º» – r «(1  Et1 )2 – E E¬ ln (r2 / r1 ) ¼ –1 r 12  4.

t

`

^

`

1/ 2

–

ln (r / r1 ) 1 1ª ^E t12  2E t1  1 – 1 – 2Et2 – Et2 2 `º» r «(1  E t1 ) 2 – E E¬ ln (r2 / r1 ) ¼ [Note this step please] 1/ 2

–

ln (r / r1 ) 1 1ª r ^(1  E t1 )2 – (1  E t2 )2 `º» (1  E t1 ) 2 – E E «¬ ln (r2 / r1 ) ¼

86

Chapter : 2 1/ 2

ln (r / r1 ) 1 1ª ...(2.64) r « (1  E t1 ) 2 – ^(1  E t1 )2 – (1  E t2 )2 `º» E E¬ ln (r2 / r1 ) ¼ Since, t = t2 where r = r2, therefore, only the +ve sign in the above expression can be used.

i.e.,

t

–

1/ 2

ln (r / r1 ) 1ª ^(1  E t1 )2 – (1  E t2 )2 `º» (1  E t1 )2 – E «¬ r r ln ( 2 / 1 ) ¼ B. Temperature variation in terms of heat. flux (Q) : Fourier’s equation for heat conduction is given as

Hence,

t

or,

–

1 E

...(2.65)

dt dr

Q

– kA .

Q

– k0 (1  Et ).2S r . L

dt dr

Rearranging the above equation, we get dr Q. – k0 .2S L (1  Et ) dt r Integrating both sides, we obtain § Er 2 · Q ln (r) – k0 .2S L ¨ t  ¸C © 2 ¹ (where C = Constant of integration) To evaluate the constant of integration (C), using the following boundary condition, we have : At r = r1, t = t1 § Et 2 · C k0 .2S L ¨ t1  1 ¸  Q In (r1 ) © 2 ¹ § § E t12 · Et2 · ? Q ln (r) – k0 .2S L ¨ t1  ¸  Q ln ( r1 ) ¸  k0 .2 S L ¨ t1  © © 2 ¹ 2 ¹ § § E t12 · Et 2 · ¸ or, Q ln (r) – Q ln (r1) – k0 .2S L ¨ t  ¸  k0 .2S L ¨ t1  © © 2 ¹ 2 ¹ ª§ Et 2 · § E t 2 ·º k0 .2S L «¨ t1  1 ¸ – ¨ t  ¸» 2 ¹ © 2 ¹¼ ¬© E t12 · § Et2 · Q ln ( r / r1 ) §   t t – ¨ ¸ ¨ ¸ or 1 © k0 .2S L 2 ¹ © 2 ¹ ª Q ln (r / r1 ) § E t12 Et2 t t    – « ¨ or, 1 © 2 2 «¬ k0 .2 S L Solving the above quadratic equation for ‘t’, we get

or,

Q ln (r/r1)

–1  1 – 4 u t

–

1  E

–

1  E

1 E2 1 E

2

·º ¸» ¹ »¼

0

E ª u ^Q .ln (r / r1 ) /( k2 .2S L)` – ^t1  (E t12 / 2`º¼ 2 ¬ E 2u 2

–

2 E t ·º 2 ª Q ln (r / r1 ) § – ¨ t1  1 ¸ » « E ¬« k0 .2SL © 2 ¹ ¼»

–

Q ln (r / r1 ) 2 t1  t12 – E E . k0 . S L

“Conduction” Heat Transfer at Steady State–One Dimension –

i.e.,

t

1  E

87

2

Q ln (r / r1 ) 1· § ¨ t1  E ¸ – E k ˜ S L © ¹ 0

1

2 Q ln (r / r1 ) º 2 1 ª§ 1· ˜ –  «¨ t1  ¸ – » E «¬© E¹ E k0 S L »¼

...(2.66)

2.6.1.1. LOGARITHMIC MEAN AREA FOR THE HOLLOW CYLINDER Invariably it is considered convenient to have an expression for the heat flow through a hollow cylinder of the same form as that for a plane wall. Then thickness will be equal to ( r2 – r1) and the area A will be an equivalent area Am as shown in the Fig. 2.46. Now, expressions for heat flow through the hollow cylinder and plane wall will be as follows :

Fig. 2.46.

(t1 – t2 ) ... Heat flow through cylinder. ln (r2 / r1 ) 2S k L (t1 – t2 ) ... Heat flow through plane wall. Q (r2 – r1 ) k Am Am is so chosen that heat flow through cylinder and plane wall will be equal for the same thermal potential. (t1 – t2 ) (t1 – t2 ) r r r2 – r1 ) In ( / ) ( 2 1 ? k Am 2S k L Q

or,

ln (r2 / r1 ) 2Sk L

or,

Am

(r2 – r1 ) k Am 2 S L ( r2 – r1 ) ln ( r2 / r1 )

2 S L r2 – 2 S L r1 ln (2S L r2 / 2S L r1 )

A0 – Ai ln ( A0 – Ai ) where Ai and Ao are inside and outside surface areas of the cylinder.

or,

Am

...(2.67)

88

Chapter : 2

The expression is known as logarithmic mean area of the plane wall and the hollow cylinder. By the use of this expression a cylinder can be transformed into a plane wall and the problem can be solved easily. A0  2, then we can take, If, Ai Ai  A0 which is within 4% of Am (where, Aav. = Average area) Aav. 2 2SL (r2 – r1 ) Am 2S rm L Further, ln (r2 / r1 ) Obviously, logarithmic mean radius of the hollow cylinder is (r2 – r1 ) rm ..(2.68) ln (r2 / r1 )

2.6.2. HEAT CONDUCTION THROUGH A COMPOSITE CYLINDER Consider flow of heat through a composite cylinder as shown in Fig. 2.47. Let, thf = The temperature of the hot fluid flowing inside the cylinder, tcf = The temperature of the cold fluid (atmospheric air), kA = Thermal conductivity of the inside layer A, kB = Thermal conductivity of the outside layer B, t1, t2, t3 = Temperatures at the points 1, 2, and 3 (see Fig. 2.47) L = Length of the composite cylinder, and hhf, hcf = Inside and outside heat transfer coefficients. The rate of heat transfer is given by k A .2SL (t1 – t2 ) Q hhf .2S r1 . L (thf – t1 ) ln (r2 / r1 ) k B .2SL (t2 – t3 ) hcf .2S r3 . L (t3 – tcf ) ln (r3 / r2 )

Fig. 2.47.

“Conduction” Heat Transfer at Steady State–One Dimension Rearranging the above expression, we get Q thf – t1 hhf . r1 .2S L t1 – t2

...(i)

Q k A .2 S L ln (r2 / r1 )

...(ii)

Q k B .2S L ln (r3 / r2 ) Q t3 – tcf hcf . r3 .2S L Adding (i), (ii), (iii) and (iv), we have Q ª 1 1 1 1 º    kA kB 2SL « hhf . r1 hcf . r3 » « » ln ( r2 / r1 ) ln (r3 / r2 ) ¬« ¼» t 2 – t3

?

Q

or,

Q

89

...(iii)

...(iv)

thf – tcf

2S L (thf – tcf ) 1 1 1 º ª 1   « h .r  kA kB hcf . r3 » « hf 1 » ln ( r2 / r1 ) ln ( r3 / r2 ) «¬ »¼ 2S L (thf – tcf )

ln (r2 / r1 ) ln (r3 / r2 ) 1 º ª 1   « h .r  » k k h A B cf . r3 ¼ ¬ hf 1 If there are ‘n’ concentric cylinders, then

...(2.69)

2S L (thf – tcf )

...(2.70) n n ª 1 º 1 1  ¦ ln ^r( n  1) / rn `  « » hcf . r( n  1) »¼ «¬ hhf . r1 n 1 kn If inside and outside heat transfer coefficients are not considered then the above equation can be written as 2 S L ªt1 – t( n  1) ¼º ...(2.71) Q n n ¬ 1 ¦ ln ª¬ r(n  1) / rn º¼ n 1 kn

Q

Example 2.35. A thick walled tube of stainless steel with 20 mm inner diameter and 40 mm outer diameter is convered with a 30 mm layer of asbestos insulation (k = 0.2 W/m°C). If the inside wall temperature of the pipe is maintained at 600°C and the outside insulation at 1000°C, calculate the heat loss per metre of length. (AMIE) Solution. Refer to Fig. 2.48. Given:

r1

r2

20 10 mm 2 = 0.01 m 40 2

20 mm

90

Chapter : 2

= 0.02 m r3 = 20 + 30 = 50 mm = 0.05m t1 = 600° C t3 = 1000° C kB = 0.2 W/m°C Heat transfer per metre of a length, Q/L : 2SL (t1 – t3 ) Q ln ( r2 / r1 ) ln (r3 / r2 )  kA kB Since the thermal conductivity of stainless steel is not given, therefore, neglecting the resistance offered by stainless steel to heat transfer across the tube, we have Q L

2S (t1 – t3 ) ln (r3 / r2 ) kB

2S (600 – 1000) ln (0.05 / 0.02) 0.2

Fig. 2.48.

– 548.57 W/m (Ans.) Negative sign indicates that the heat transfer takes place radially inward. Example 2.36. A steel pipe with 50 mm OD is covered with a 6.4 mm asbestos insulation [k = 0.166 W/mK] followed by a 25 mm layer of fiber-glass insulation [k = 0.0485 W/mK]. The pipe wall temperature is 393 K and the outside insulation temperature is 311 K. Calculate the interface temperature between the asbestos and fiber-glass. (G.U.) Solution. Given : 50 25 mm = 0.025 m; 2 r2 = r1 + 6.4 = 25 + 6.4 = 31.4 mm or 0.0314 m; r3 = r2 + 25 = 31.4 + 25 = 56.4 mm = 0.0564 m; T1 = 393 K; T3 = 311 K kA = 0.166 W/mK; kB = 0.0485 W/mK. Interface temperature between the asbestos and fiber-glass, t2 : , We know that r1

Q

2SL (T1 – T3 ) ln (r2 / r1 ) ln (r3 / r2 )  kA kB Fig. 2.49.

“Conduction” Heat Transfer at Steady State–One Dimension Q L

Also, Q L or, 38.31

38.31

91

2S (T1 – T3 ) ln ( r2 / r1 ) ln ( r3 / r2 )  kA kB 2S (393 – 311) ln (0.0314 / 0.025) ln (0.0564 / 0.0314)  0.166 0.0485 2S (T1 – T2 ) ln (r2 / r1 ) kA

515.22 1.373  12.075

38.31W/m

2S (393 – T2 ) ª ln (0.0314 / 0.025) º «¬ »¼ 0.166 2S (393 – T2 ) 1.373

38.31 u 1.373 384.6 K 2S t2 = 384.6 – 273 = 111.6° C (Ans.)

?

T2

or,

393 –

Example 2.37. A gas filled tube has 2 mm inside diameter and 25 cm length. The gas is heated by an electrical wire of diameter 50 microns (0.05 mm) located along the axis of the tube. Current and voltage drop across the heating element are 0.5 amps and 4 volts, respectively. If the measured wire and inside tube wall temperatures are 175°C and 150°C respectively, find the thermal conductivity of the gas filling the tube. (GATE) Solution. Given : Inside radius of the tube, Length of the tube,

rt = 2 mm L = 25 cm = 0.25 m

Radius of the electric wire, rw = 0.025 mm Inside tube temperature,

tt = 150°C

Wire temperature,

tw = 175°C

Current through the element

= 0.5 A

Voltage across the element

= 4V

Thermal conductivity of the gas, k : Heat transferred through a cylinder, Q

2S k L (tw – tt ) ln (rt / rw ) 2Sk u 0.25(175 – 150) ln (1/ 0.025)

Also,

Q = VI = 4 × 0.5 = 2.0 W

10.645 k = 2.0 k = 0.188 W/m°C.

...(i) ...(ii) Fig. 2.50.

From (i) and (ii), we get or,

10.645 kW

(Ans.)

92

Chapter : 2

Example 2.38. A standard cast iron pipe (inner diameter = 50 mm and outer diameter = 55 mm) is insulated with 85 percent magnesium insulation (k = 0.02 W/m°C). Temperature at the interface between the pipe and insulation is 300°C. The allowable heat loss through the pipe is 600 W/m length of pipe and for the safety, the temperature of the outside surface of insulation must not exceed 100°C. Determine : (i) Minimum thickness of insulation required, and (ii) The temperature of inside surface of the pipe assuming its thermal conductivity 20 W/m°C. [M.U.] Solution. Refer to Fig. 2.51.

Fig. 2.51.

50 25 mm 0.025 m 2 55 r2 27.5 mm 0.0275 m 2 kA = 20 W/m°C, kB = 0.02W/m°C t2 = 300°C, t3 = 100°C Heat loss per metre length of pipe Q/L = 600 W/m (i) Minimum thickness of insulation required, (r3 – r2) : 2S L (t1 – t2 ) 2S L (t2 – t3 ) Q ln (r2 / r1 ) ln (r3 / r2 ) kA kB r1

or,

? or, or, ? (ii)

2S (t1 – t2 ) 2S (t2 – t3 ) ln (r2 / r1 ) ln (r3 / r2 ) kA kB 2S (t1 – 300) 2S (300 – 100) 600 ln (0.0275 / 0.025) ln (r3 / 0.0275) 20 0.02 ln (r3 / 0.0275) 2S (300 – 100) 0.02 600 ª 2S (300 – 100) º ln (r3 / 0.0275) 0.02 « »¼ 0.04188 ¬ 600 r3 / 0.0275 = 1.0427 or r3 = 0.0287 m or 28.7 mm Minimum thickness of insulation required = r3 – r2 = 28.7 – 27.5 = 1.2 mm (Ans.) The temperature of inside surface of the pipe, t1 : 2S (t1 – 300) 2S (t1 – 300) 600 ln (0.0275 / 0.025) 0.00476 20 Q L

“Conduction” Heat Transfer at Steady State–One Dimension

93

600 u 0.00476  300 300.45° C (Ans.) 2S Example 2.39. A pipe (k = 180 W/m°C) having inner and outer diameters 80 mm and 100 mm respectively is located in a space at 25°C. Hot gases at temperature 160°C flow through the pipe. Neglecting surface heat transfer coefficients, calculate : (i) The heat loss through the pipe per unit length, (ii) The temperature at a point halfway between the inner and outer surfaces, and (iii) The surface area normal to the direction of heat flow so that the heat transfer through the pipe can be determined by considering material of pipe as a plane wall of the same thickness. 80 40 mm 0.04 m ri Solution. Inner diameter of the pipe, 2 100 50 mm 0.05 m ro Outer diameter of the pipe, 2 Temperature of hot gases, ti = 160°C Temperature of space in which the pipe is located, to = 25°C. Thermal conductivity of pipe material, k = 180 W/m°C (i) The heat loss through the pipe per unit length, Q : (160 – 25) 135 't 684229 W (Ans.) Q Rth ª ln (ro / ri ) º ln (0.05 / 0.04) « 2S k L » 2S u 180 u 1 ¬ ¼ t1

(ii) The temperature at a point halfway between the inner and outer surfaces, t : Radius at halfway through the pipe wall, ri  r0 40  50 r 45 mm 0.045 m 2 2 Thermal resistance of the pipe upto its mid-plane ln (r / ri ) ln (0.045 / 0.04) 0.0414 u 10– 4 qC/W 2S k L 2S u 180 u 1 As same heat flows through each section (ti – t ) (160 – t ) 684229 1.0414 u 10 – 4 1.0414 u 10 –4 ? t = 160 – 684229 × 1.0414 × 10–4 = 88.74°C (Ans.) Alternatively : (t – 160) (25 – 160)

t – t1 t2 – t1

ln (r / ri ) ln ( ro / ri )

...[Eqn. 2.58]

ln (0.045 / 0.04) 0.5278 ln (0.05 / 0.04) or, t = 160 + (25 – 160) × 0.5278 = 88.74°C (iii) Equivalent log-mean area, Am : Ao – Ai 2S L (r0 – ri ) 2S u 1 u (0.05 – 0.04) Am 0.2816 m 2 ln ( Ao / Ai ) ln (ro / ri ) ln (0.05 / 0.04) k Am (ti – t0 ) 180 u 0.2616 (160 – 25) Q 684288 W Check : (ro – ri ) (0.05 – 0.04) which is approximately same as calculated above.

94

Chapter : 2

Example 2.40. A 240 mm steam main, 210 metres long is covered with 50 mm of high temperature insulation (k = 0.092 W/m°C) and 40 mm of low temperature insulation (k = 0.062 W/m°C). The inner and outer surface temperatures as measured are 390°C and 40°C respectively. Calculate : (i) The total heat loss per hour, (ii) The heat loss per m2 of pipe surface, (iii) The total heat loss per m2 of outer surface, and (iv) The temperature between two layers of insulation. Neglect heat conduction through pipe material. Solution. Refer to Fig. 2.52.

Fig. 2.52.

240 120 mm = 0.12 m 2 r2 = 120 + 50 = 170 mm = 0.17 m r3 = 120 + 50 + 40 = 210 mm = 0.21 m kA = 0.092 W/m°C; kB = 0.062 W/m°C t1 = 390°C; t3 = 40°C Length of steam main, L = 210 m. (i) Total heat loss per hour : 2S L (t1 – t3 ) Q [Eqn. (2.71)] ª ln (r2 / r1 ) ln ( r3 / r2 ) º  « k » kB ¬ ¼ A

Given :

r1

2S u 210 (390 – 40) ª ln (0.17 / 0.12) ln (0.21/ 0.17) º  0.092 0.062 ¬« ¼» 64194.3 u 3600 231099.5 kJ/h 1000 i.e., The total heat loss per hour = 231099.5 kJ/h (Ans.)

461814 (3.786  3.408)

64194.3 W

“Conduction” Heat Transfer at Steady State–One Dimension

95

m2

(ii) Total heat loss per of the pipe surface : Total heat loss per m2 of the surface 231099.5 231099.5 2Sr1 . L 2S u 0.12 u 210 (iii) Total heat loss per m2 of the outer surface : Total heat loss per m2 of the outer surface 231099.5 231099.5 2Sr3 .L 2S u 0.21 u 210 (iv) The temperature between two layers, t2 : 2S L (t1 – t2 ) Q ª ln (r2 / r1 ) º « k » A ¬ ¼

1459.55 kJ/h (Ans.)

834.03 kJ/h (Ans.)

2S u 210 (390 – t2 ) 348.5(390 – t2 ) ª ln (0.17 / 0.12) º «¬ »¼ 0.092 64194.3 205.8° C (Ans.) t2 390 – ? 348.5 Example 2.41. A steam pipe of outer diameter 120 mm is covered with two layers of lagging, inside layer 45 mm thick (k = 0.08W/m°C) and outside layer 30 mm thick (k = 0.12 W/m°C). The pipe conveys steam at a pressure of 20 bar with 50°C superheat. The outside temperature of lagging is 25°C. If the steam pipe is 30m long, determine : (i) Heat lost per hour, and (ii) Interface temperature of lagging. The thermal resistance of steam pipe may be neglected. Solution. Refer to Fig. 2.53. 64194.3

Fig. 2.53.

96

Chapter : 2 120 60 mm 0.06 m 2 r2 = 60 + 45 = 105 mm =0.105 m r3 = 105 + 30 = 135 mm = 0.135 m kA = 0.08 W/m°C; kB = 0.12 W/m°C t3 = 25°C; L (length of pipe) = 30 m Corresponding to 20 bar (from steam tables), tsat (saturation temp.) = 212.4°C ? Temperature of steam, t1 = tsat + 50 = 212.4 + 50 = 262.4°C r1

(i) Heat lost per hour : The rate of heat transfer is given by 2S L (t1 – t3 ) ln (r2 / r1 ) ln (r3 / r2 )  kA kB

2S u 30 (262.4 – 25) ln (0.105 / 0.06) ln (0.135 / 0.105) = 4923 W (or J/s)  0.08 0.12 4923 u 3600 = 17722.8 kJ/h (Ans.) ? Heat lost per hour 1000 (ii) Interface temperature of lagging, t2 : 2S L (t1 – t2 ) Q ln (r2 / r1 ) kA 2S u 30 (262.4 – t2 ) or, 4923 ln (0.105 / 0.06) 0.08 4923 ln (0.105 / 0.06) or, (262.4 – t2 ) u 182.69 2S u 30 0.08 ? t2 = 262.4 – 182.69 = 79.71°C (Ans.) Example 2.42. A steam pipe of outside diameter 80 mm and 25 m long conveys 800 kg of steam per hour at a pressure of 22 bar. The steam enters the pipe with a dryness fraction of 0.99 and is to leave the other end of the pipe with the minimum dryness fraction of 0.97. This is to be accomplished by using a lagging material (k = 0.2 W/m°C), determine its minimum thickness to meet the necessary conditions, if the temperature of the outside surface of lagging is 25°C. Q

Assume that there is no pressure drop across the pipe and the resistance of the pipe material is negligible. Solution. Refer to Fig. 2.54. 80 r1 mm 0.04 m, 2 k = 0.2 W/m°C Length of pipe, L = 25m, t2 = 25°C

Fig. 2.54.

“Conduction” Heat Transfer at Steady State–One Dimension Minimum thickness of insulation, (r2 – r1) : From steam tables, corresponding to 22 bar pressure : tsat (= t1) = 217.2°C, hfg = 1868.1 kJ/kg ? Heat loss per kg of steam passing through the pipe = (0.99 – 0.97) × 1868.1 = 37.36 kJ/kg Total heat loss through the pipe per second 800 = 8.302 kJ/s or 8302 J/s or 8302 W 3600 Heat loss through the pipe (neglecting pipe thermal resistance) is given by 2SL (t1 – t2 ) 2S u 25 u (217.2 – 25) Q ln (r2 / r1 ) ln (r2 / r1 ) k 0.2 6038.14 6038.14 8302 ln ( r2 / r1 ) 0.72731 or, or ln (r2 / r1 ) 8302 r2 2.069 ? or r2 = 40 × 2.069 = 82.76 mm r1 ? Minimum thickness of insulation = r2 – r1 = 82.76 – 40 = 42.76 mm (Ans.) Example 2.43. A steam pipe (inner diameter = 150 mm and outer diameter = 160 mm) having thermal conductivity 58 W/m°C is covered with two layers of insulation, of thickness 30 mm and 50 mm respectively and thermal conductivities 0.18 W/m°C and 0.09 W/m°C respectively. The temperature of inner surface of steam pipe is 320°C and that of the outer surface of the insulation layers is 40°C. (i) Determine the quantity of heat lost per metre length of steam pipe and layer contact temperature, and (ii) If the condition of the steam is dry and saturated, find the quality of the steam coming out of one metre pipe assuming the quantity of steam flowing is 0.32 kg/min. Solution. Refer to Fig. 2.55. 37.36 u

150 75 mm = 0.075 m 2 160 r2 80 mm = 0.08 m 2 r3 = 80 + 30 = 110 mm = 0.11 m r4 = 110 + 50 = 160 mm = 0.16 m r1

Fig. 2.55.

97

98

Chapter : 2 t1 = 320° C, t4 = 40°C kA = 58 W/m°C kB = 0.18 W/m°C, kC = 0.09 W/m°C (i) Quantity of heat lost per meter (Q) and layer contact temperatures (t2, t3) : Quantity of heat lost is given by 2S L (t1 – t4 ) Q ln (r2 / r1 ) ln (r3 / r2 ) ln ( r4 / r3 )   kA kB kC 2S u 1 u (320 – 40) 296.5 W/m (Ans). ln (0.08 / 0.075) ln (0.11/ 0.08) ln (0.16 / 0.11)   58 0.18 0.09 2S u 1(t1 – t2 ) 2S u 1(t2 – t3 ) 2S u 1(t3 – t4 ) Also, Q ln (r2 / r1 ) ln (r3 / r2 ) ln (r4 / r3 ) kA kB kC S (320 – t2 ) 2 ln ( r2 / r1 ) 296.5 ? or u t2 320 – 296.5 ln (r2 / r1 ) S kA 2 kA 296.5 ln (0.08 / 0.75) u 319.95° C (Ans.) 2S 58 2S (319.95 – t3 ) Similarly, 296.5 ln (r3 / r2 ) kB 296.5 ln (r3 / r2 ) 296.5 ln (0.11/ 0.08) u u 236.5° C (Ans.) t3 319.95 – 319.95 – or, kB 2S 2S 0.18 (ii) Quality of steam coming out of one metre pipe, x : Total heat of steam when it is saturated at 320°C = 2703 kJ/kg ...from steam tables Heat carried by steam per minute after losing heat in the pipe 320 –

0.32 (kg / min) u 2703 (kJ / kg) –

296.5 u 60 (kJ / min) 1000

847.17 kJ / min

Now 847.17 = 0.32 (hf + xhfg) Corresponding to 320°C saturation temperature, from steam tables, we have hf = 1463 kJ/kg, hfg = 1240 kJ/kg ? 847.17 = 0.32 (1463 + x × 1240) = 468.16 + 396.8x (847.17 – 468.16) 0.955 (Ans.) 396.8 Example 2.44. Thermal conductivity, k, of a certain material is given by k = a + bT + cT2 where a, b, c are constants and T is the absolute temperature. Derive an expression for heat flow per unit length of a hollow cylinder made of this material. Assume that inner and outer radii of the cylinder are r1 and r2 respectively and the cylinder ends are perfectly insulated. (M.U.) Solution. Refer to Fig. 2.56. Consider a hollow ring at radius r, and thickness dr of the hollow cylinder. The radius heat flow across the ring, per unit length, is given by

or,

x

Q

– kAr

dT dr

– k u 2Sr u

dT dr

“Conduction” Heat Transfer at Steady State–One Dimension

99

Fig. 2.56.

dr – 2S (a  bT  cT 2 ) dT r Integrating from inner to outer radius, we get Q

or,

Q

³

r2 r1

dr r

– 2S

³

T2

T1

2 (a  bt  cT ) dT

§r · Q In ¨ 2 ¸ © r1 ¹

b 2 c 3 ª 2 3 º 2S « a (T1 – T2 )  (T1 – T2 )  (T1 – T2 ) » 2 3 ¬ ¼ b c ª 2 2 º 2S (T1 – T2 ) « a  (T1  T2 )  (T1  T1 T2  T2 ) » 2 3 ¬ ¼ or, Q § r2 · ln ¨ ¸ © r1 ¹ which is the required expression. (Ans.) Example 2.45. At a certain time, the temperature distribution in a long cylindrical fire tube, inner radius 30 cm and outer radius 50 cm, is given by t = 800 + 1000 r – 5000 r2 where (t) is in (°C) and (r) in (m). The thermal conductivity and thermal diffusivity of tube material are 58 W/m-K °C and 0.004 m2/h, respectively. Find : (i) Rate of heat flow at inside and outside surfaces per unit length; (ii) Rate of heat storage per unit length; (iii) Rate of change of temperature at inner and outer surfaces. (AMIE) Solution. Given : r1 = 30 cm = 0.3 m; r2 = 50 cm = 0.5 m, k = 58 W/m° C; D = 0.004 m2/h (i) Rate of heat flow per unit length : Temperature distribution, t = 800 + 1000 r – 5000 r2 ...(Given) dt 1000 – 10000 r dr Rate of heat flow at inside surface (1), per unit length, dt dt Q1 – k A1 – k 2 S r1 ˜ r r1 dr dr r 0.3

or,

= – k × 2Sr1 (1000 – 10000 r1)

100

Chapter : 2

= – 58 × 2S × 0.3 (1000 – 10000 × 0.3) = 2.1865 × 105 W/m, in outward direction (as it is positive) (Ans.) Similarly, rate of heat flow at outer surface (2), per unit length, Q2 = – k × 2Sr2 (1000 – 10000 r2) = –58 × 2S × 0.5 (1000 – 10000 × 0.5) = 7.2885 × 105 W/m, outward direction (as it is positive) (Ans.) (ii) Rate of heat storage per unit length : As the heat leaving Q2 is greater than heat entering Q1, there is decrease in the heat contents. In other words, the rate of heat storage is negative. It is equal to

Fig. 2.57.

Q1 – Q2 = 2.1865 × 105 – 7.2885 × 105 = – 5.102 × 105 W/m (Ans.) (iii) Rate of change of temperature at inner and outer surfaces : We know that,

dt dW dt dr d § dt · ¨r ¸ dr © dr ¹ dt dW

D d § dt · ¨r ¸ r dr © dr ¹

...(i)

1000 – 10000 r d (1000 r – 10000 r 2 ) 1000 – 20000 r dr D (1000 – 20000 r ), by using (i) and (ii) r § 1000 · D¨ – 20000 ¸ © r ¹

...(ii)

Rate of change of temperature : (a) At the inner surface, r = 0.3 m dt dW

§ 1000 · § 1000 · D¨ – 20000 ¸ 0.004 ¨ – 20000 ¸ © 0.3 ¹ © 0.3 ¹ = – 66.67°C/h, (i.e., decrease) (Ans.)

(b) At the outer surface, r = 0.5 m dt § 1000 · 0.004 ¨ – 20000 ¸ = –72°C/h, (i.e., decrease) (Ans.) dW © 0.5 ¹ Example 2.46. A steam pipe of 220 mm outer diameter is carrying steam at 280°C. It is insulated with a material having thermal conductivity k = 0.06 (1 + 0.0018 t) where k is in W/m°C and t is °C. If the insulation thickness is 50 mm and the temperature of the outer surface is 50°C, determine:

(i) The heat flow per metre length of the pipe, and (ii) The temperature at the mid thickness.

“Conduction” Heat Transfer at Steady State–One Dimension

101

Fig. 2.58.

Solution. Refer to Fig. 2.58. 220 110 mm 2 = 0.11 m; r2 = 110 + 50 = 160 mm = 0.16 m k = 0.06 (1 + 0.0018 t) (i) The heat flow per metre length of pipe, Q : 't Q ln (r2 / r1 ) 2S km L r1

...(i)

ª § 280  50 · º 0.06 «1  0.0018 ¨ ¸ » = 0.0778 W/m°C © ¹¼ 2 ¬ (280 – 50) 300.06 W/m (Ans.) Q ? ln (0.16 / 0.11) 2S u 0.778 u 1 (ii) The temperature at the mid thickness, tmt : ª where rmt radius at mid thickness/plane º (280 – tmt ) Q « » ln (rmt / r1 ) « §¨ r1  r2 ·¸ §¨ 0.11  0.16 ·¸ 0.135 m » 2S k m u 1 ¹ 2 ¬« © 2 ¹ © ¼» ª º § 280  tmt · 2S u 0.06 «1  0.0018 ¨ ¸ (280 – tmt ) » 2S km (280 – tmt ) © ¹ 2 ¬ ¼ 300.06 ln (0.135 / 0.11) 0.2048

where

km

102 or, or,

Chapter : 2 300.06 u 0.2048 2S u 0.06 (1 + 0.252 + 0.0009 tmt) (280 – tmt) = 163

[1  0.0009 (280  tmt )](280 – tmt )

or,

(1.252 + 0.0009 tmt) (280 – tmt) = 163

or,

350.56 – 1.252 tmt + 0.252 tmt – 0.0009 tmt2 = 163

163

0.0009 tmt2 + tmt – 187.56 = 0

or,

–1  1  4 u 0.0009 u 187.56 2 u 0.0009

163.5q C

or,

tmt

?

Temperature at the mid thickness, tmt = 163.5°C

(Ans.)

ª (280 – 163.5) ­ § 280  163.5 · ½ u 2S u 0.06 ®1  0.0018 ¨ ¸¾ «Check Q © ¹¿ In (0.135 / 0.11) 2 ¯ ¬ Example 2.47. An insulated steam pipe having outside diameter of 30 mm is to be covered with two layers of insulation, each having thickness of 20 mm. The thermal conductivity of one material is 5 times that of the other.

º 300.06 W » ¼

Assuming that the inner and outer surface temperatures of composite insulation are fixed, how much will heat transfer be increased when better insulation material is next to the pipe than it is outer layer ? (M.U.) Solution. Case I. When better insulation is inside : Refer to Fig. 2.59. 30 r1 15 mm 0.015 m; 2 r2 = 15 + 20 = 35 mm = 0.035 m; r3 = 35 + 20 = 55 mm = 0.055m kB = 5kA

Fig. 2.59.

Heat lost through the pipe is given by 2S L (t1 – t3 ) 2S L (t1 – t3 ) Q1 ln (r2 / r1 ) ln (r3 / r2 ) ln (0.035 / 0.015) ln (0.055 / 0.035)   kA 5 kA kA kB k A. 2S L (t1 – t3 ) k A .2S L (t1 – t3 ) Q1 1.066 2S L k A (t1 – t3 ) or,  0.8473 0.0904 0.9377 Case II. When better insulation is outside : Refer to Fig. 260. 2 S L (t1 – t3 ) 2 S L (t1 – t3 ) Q2 ln (0.035 / 0.015) ln (0.055 / 0.035) ln (r2 / r1 ) ln (r3 / r2 )   kA 5 kA kB kA

...(i)

“Conduction” Heat Transfer at Steady State–One Dimension

103

Fig. 2.60.

k A .2S L (t1 – t3 ) k A .2S L (t1 – t3 ) 1.609 2S L . k A (t1 – t3 ) ...(ii) 0.1694  0.452 0.6214 From expression (i) and (ii), we have, Q2 1.609 u 2S L . k A (t1 – t3 ) 1.509 Q1 1.066 u 2S L . k A (t1 – t3 ) As Q2 > Q1, therefore, putting the better insulation next to the pipe decreases the heat flow. ? Percentage decrease in heat transfer

or,

Q2

Q2 – Q1 Q2 – 1 1.509 – 1 0.509 or 50.9% (Ans.) Q1 Q1 Example 2.48. Hot air at a temperature of 65°C is flowing through a steel pipe of 120 mm diameter. The pipe is covered with two layers of different insulating materials of thickness 60 mm and 40 mm, and their corresponding thermal conductivities are 0.24 and 0.4 W/m°C. The inside and outside heat transfer coefficients are 60 W/m°C and 12 W/m°C respectively. The atmosphere is at 20°C. Find the rate of heat loss from 60 m length of pipe. Solution. Refer to Fig. 2.61.

Given :

r1

r2 r3 kA hhf thf

= = = = =

120 60 mm 0.06 m 2 60 + 60 = 120 mm = 0.12 m 60 + 60 + 40 = 160 mm = 0.16 m 0.24 W/m°C ; kB = 0.4 W/m°C 60 W/m2°C ; hcf = 12 W/m2°C 65°C ; tcf = 20°C

104

Chapter : 2

Length of pipe, L = 60 m

Fig. 2.61.

Rate of heat loss, Q : Rate of heat loss is given by

Q

2S L (thf – tcf )

[Eqn. (2.69)] ln (r2 / r1 ) ln ( r3 / r2 ) 1 º ª 1    « h .r kA kB hcf . r3 »¼ ¬ hf 1 2S u 60 (65 – 20) 1 ln (0.12 / 0.06) ln (0.16 / 0.12) 1 ª º   « 60 u 0.06  0.24 0.4 12 u 0.16 »¼ ¬ 16964.6 3850.5 W 0.2777  2.8881  0.7192  0.5208 i.e., Rate of heat loss = 3850.5 W (Ans.) Example 2.49. Calculate the overall heat transfer coefficient (based on inner diameter) for a steel pipe covered with fiber glass insulation. The following data are given : ID of pipe = 2 cm Thickness of pipe = 0.2 cm Thickness of insulation = 2 cm Heat transfer coefficient (inside) = 10 W/m2 K Heat transfer coefficient (outside) = 5 W/m2 K Conductivity of insulation = 0.05 W/m K Conductivity of steel = 46 W/m K Inside fluid temperature = 200°C Ambient temperature = 30°C Also find net heat loss from the pipe. (PTU)

“Conduction” Heat Transfer at Steady State–One Dimension

105

2 = 1 cm = 0.01 m; r2 = r1 + 0.2 = 1 + 0.2 = 1.2 cm or 0.012 m, 2 r3 = r2 + 2 = 1.2 + 2 = 3.2 cm or 0.032 m, hhf = 10 W/m2 K; hcf = 5 W/m2 K; kA = 46 W/m K; kB = 0.05 W/m K; thf = 200°C; tcf = 30°C. Overall heat transfer co-efficient, Ui : Q = Ai Ui 't = 2Sr1 LUi (thf – tcf) where Ui = Overall heat transfer coefficient based on inner diameter.

Solution. Given : r1

Fig. 2.62.

1 r r 1 1 §r · r §r ·  1 ln ¨ 2 ¸  1 ln ¨ 3 ¸  1 ˜ hhf k A © r1 ¹ k B © r2 ¹ r3 hcf 1 1 0.01 § 0.012 · 0.01 § 0.032 · 0.01 1  u ln ¨ ln ¨ ¸ ¸ 10 46 © 0.01 ¹ 0.05 © 0.012 ¹ 0.032 5 1 –5 0.1  3.96 u 10  0.19616  0.0625 1 2.788 W/m 2 K (Ans.) 0.3587 Heat loss/m length, Q/L : Q 2S r1 u U i u (thf – tcf ) L = 2S × 0.01 × 2.788 (200 – 30) = 29.78 W/m (Ans.) Example 2.50. An aluminium pipe carries steam at 110°C. The pipe (k = 185 W/m°C) has an inner diameter of 100 mm and outer diameter of 120 mm. The pipe is located in a room where the ambient air temperature is 30°C and the convective heat transfer coefficient between the pipe and air is 15 W/m2°C. Determine the heat transfer rate per unit length of pipe. To reduce the heat loss from the pipe, it is covered with a 50 mm thick layer of insulation (k = 0.20 W/m°C). Determine the heat transfer rate per unit length from the insulated pipe. Assume that the convective resistance of the steam is negligible. (PU) Ui

106

Chapter : 2

Solution. Case I. Refer to Fig. 2.63. 100 r1 50 mm Given : 2 = 0.05 m 120 r2 60 mm 2 = 0.06 m Temperature of steam (hot fluid), thf = 110°C Temperature of ambient air (cold fluid), tcf = 30°C Thermal conductivity of pipe material, k = 185 W/m°C Heat transfer coefficient between the pipe and air, hcf = 15 W/m2°C

Fig. 2.63.

Heat transfer rate per unit length of pipe, Q/L : Heat transfer rate is given by,

Q

or,

Q L

2S L (thf – tcf ) 1 º ª ln (r2 / r1 )  « k » h A cf . r2 ¼ ¬ 2S L (thf – tcf ) 1 º ª ln ( r2 / r1 )  « k hcf . r2 »¼ A ¬

[Eqn. (2.69)]

2S (110 – 30) ln (0.06 / 0.05) 1 ª º  « » u 185 15 0.06 ¬ ¼

i.e., Heat transfer rate per unit length of pipe = 451.99 W/m (Ans.) Case II : Refer to Fig. 2.64. r1 = 50 mm = 0.05 m; r2 = 60 mm = 0.06 m r3 = 60 + 50 = 110 mm = 0.11 m; kA = 185 W/m°C kB = 0.20 W/m°C; hcf = 15 W/m2°C Heat transfer rate per unit length from the insulated pipe, Q/L : Heat transfer rate in this case will be given by

Fig. 2.64.

451.99 W/m

“Conduction” Heat Transfer at Steady State–One Dimension

Q

Q L

107

2SL (thf – tcf ) 1 º ª ln (r2 / r1 ) ln (r3 / r2 )   « k » k h A B cf . r3 ¼ ¬ 2S (thf – tcf )

1 º ª ln (r2 / r1 ) ln (r3 / r2 )   « k » k h A B cf . r3 ¼ ¬ Substituting the given data is the above equation, we have, Q L

2S (110 – 30) ln (0.06 / 0.05) ln (0.11/ 0.06) 1 ª º   « 185 0.20 15 u 0.11 »¼ ¬ 502.65 138.18 W/m 0.000985  3.030679  0.606060 i.e., Heat transfer rate per unit length of insulated pipe = 138.18 W/m (Ans.) Example 2.51. A 150 mm steam pipe has inside diameter of 120 mm and outside diameter of 160 mm. It is insulated at the outside with asbestos. The steam temperature is 150°C and the air temperature is 20°C. h (steam side) = 100 W/m2°C, h (air side) = 30 W/m2°C, k (asbestos) = 0.8 W/m°C and k (steel) = 42 W/m°C. How thick should the asbestos be provided in order to limit the heat loses to 2.1 kW/m2 ? (MDU Haryana) Solution. Refer to Fig. 2.65. 120 r1 60 mm 0.06 m Given : 2 160 r2 80 mm 0.08 m 2

Fig. 2.65.

kA thf hhf Heat loss

= = = =

42 W/m°C; 150°C; 100 W/m2°C; 2.1 kW/m2

kB = 0.8 W/m°C tcf = 20°C hcf = 30 W/m2°C

108

Chapter : 2

Thickness of insulation (asbestos), (r3 – r2) : Area for heat transfer = 2Sr L (where L = length of the pipe) ? Heat loss = 2.1 × 2SrL kW = 2.1 × 2S × 0.075 × L = 0.989 L kW = 0.989 L × 103 watts 150 75 mm or 0.075 m ... given] (where r, mean radius 2 Heat transfer rate in such a case is given by 2 S L (thf – tcf ) Q [Eqn. 2.69] ln (r2 / r1 ) ln ( r3 / r2 ) 1 º ª 1   « h .r  » kA kB hcf . r3 ¼ ¬ hf 1

0.989 L u 103

2S L (150 – 20) 1 ln (0.08 / 0.06) ln (r3 / 0.08) 1 º ª   «100 u 0.06  u r3 »¼ 42 0.8 30 ¬

816.81 ln (r3 / 0.08) 1 º ª  «0.16666  0.00685  r3 »¼ 0.8 30 ¬ ln (r3 / 0.08) 1 816.81  or, – (0.16666  0.00685) 0.6524 0.8 30 r3 0.989 u 103 1 or, 1.25 ln (r3 / 0.08)  – 0.6524 0 30 r3 Solving by hit and trial, we get r3 ; 0.105 m or 105 mm ? Thickness of insulation = r3 – r2 = 105 – 80 = 25 mm (Ans.) 0.989 u 103

Example 2.52. A 160 mm diameter pipe carrying saturated steam is covered by a layer of lagging of thickness of 40 mm (k = 0.8 W/ m°C). Later, an extra layer of lagging 10 mm thick (k = 1.2 W/m° C) is added. If the surrounding temperature remains constant and heat transfer coefficient for both the lagging materials is 10 W/m2°C, determine the percentage change in the rate of heat loss due to extra lagging layer. (M.U.) Solution. Case I. Without extra layer of lagging : Refer to Fig. 2.66 (a). 160 r1 80 mm 0.08 m 2 r2 = 80 + 40 = 120 mm = 0.12 m kA = 0.8 W/m°C, h0 = 10 W/m2°C Let, ts = Temperature of steam, and ta = Temperature of air.

(a) Fig. 2.66.

“Conduction” Heat Transfer at Steady State–One Dimension

109

Considering unit length of pipe in both the cases (Case I and Case II) and neglecting internal heat transfer coefficient (being not given) and also neglecting the resistance of the pipe (as thickness and conductivity of the pipe are not given), the heat flow rate is given as 2 S (t s – t a ) Q1 ln (r2 / r1 ) 1  kA h0 . r2 2S (t s – ta ) ln (0.12 / 0.08) 1  0.8 10 u 0.12 2 S (t s – t a ) 1.340

...(i)

Case II. With extra layer of lagging : Refer to Fig. 2.66 (b). r3 = 120 + 10 = 130 mm = 0.13 m kB = 1.2 W/m°C, h0 = 10 W/m2°C

Fig. 2.66.

2 S (t s – t a ) ln (r2 / r1 ) ln (r3 / r2 ) 1   kA kB h0 . r3 2 S (t s – t a ) ln (0.12 / 0.08) ln (0.13 / 0.12) 1   0.8 1.2 10 u 0.13 2 S (t s – t a ) ...(ii) 1.343 The percentage decrease in heat flow due to extra addition of insulation can be calculated using eqns. (i) and (ii) as follows : Q1 – Q2 ª (1/1.34) – (1/1.343) º « » 0.00223 or 0.223  (Ans.) Q1 (1/1.34) ¬ ¼ Q2

Example 2.53. A steam pipe (k = 45 W/m°C) having 70 mm inside diameter and 85 mm outside diameter is lagged with two insulation layers; the layer in contact with the pipe is 35 mm asbestos (k = 0.15 W/m°C) and it is covered with 25 mm thick magnesia insulation (k = 0.075 W/m°C). The heat transfer coefficients for the inside and outside surfaces are 220 W/m2°C and 6.5 W/m2°C respectively. If the temperature of steam is 350°C and the ambient temperature is 30°C, calculate : (i) The steady loss of heat for 50 m length of the pipe; (ii) The overall heat transfer coefficients based on inside and outside surfaces of the lagged steam main. Solution. Refer to Fig. 2.67. r1 r2

70 2 85 2

35 mm or 0.035 m 42.5 mm or 0.0425 m

110

Chapter : 2 r3 = 42.5 + 35 = 77.5 mm or 0.0775 m r4 = 77.5 + 25 = 102.5 mm or 0.1025 m L = 50 m kA = 45 W/m°C kB = 0.15 W/m°C kC = 0.075 W/m°C

Temperature of steam, thf = 350°C Ambient temperature, tcf = 30°C hhf = 220 W/m2°C, hcf = 6.5 W/m2°C. (i) Loss of heat, Q :

Fig. 2.67.

2S L (thf – tcf ) Q In (r2 / r1 ) In (r3 / r2 ) In (r4 / r3 ) 1 1     hhf . r1 kA kB kC hcf . r4 2S u 50 (350 – 30) 1 ln (0.0425 / 0.035) ln (0.0775 / 0.0425) ln (0.1025 / 0.0775) 1     220 u 0.035 45 0.15 0.075 6.5 u 0.1025 100530.96 10731.23 W 0.129870  0.00431  4.00516  3.72779  1.50094 i.e., Loss of heat for 50 m of length = 10731.23 W (Ans.) (ii) The overall heat transfer coefficients, Uo, Ui : The loss of heat can also be expressed as follows : Q = Uo Ao 't = Ui Ai 't Where Uo and Ui are the overall heat transfer co-efficients based on the outside area Ao and inside area Ai respectively. Q 10731.23 U0 ? Ao . ' t 2Sr4 L u ' t 10731.23 1.0414 W/m 2 ° C (Ans.) 2S u 0.1025 u 50 (350 – 30) Q 10731.23 10731.23 Similarly, Ui Ai . ' t 2 S r1 L u ' t 2 S u 0.035 u 50 (350 – 30) = 3.05 W/m2°C (Ans.)

“Conduction” Heat Transfer at Steady State–One Dimension

111

Example 2.54. A steel pipe is carrying steam at a pressure of 30 bar. Its outside diameter is 90 mm and is lagged with a layer of material 45 mm thick (k = 0.05 W/m°C). The ambient temperature is 20°C and the surface of the lagging has a heat transfer coefficient of 8.4 W/m2°C. Neglecting resistance due to pipe material and due to steam film on the inside of steam pipe, find the thickness of the lagging (k = 0.07 W/m°C) which must be added to reduce the steam condensation rate by 50 percent if the surface coefficient remains unchanged. Solution. Case I. Single layer of insulation : Refer to Fig. 2.68.

Fig. 2.68.

Fig. 2.69.

90 45 mm 0.045 m, r2 = 45 + 45 = 90 mm = 0.09 m 2 h0 = 8.4 W/m2°C, kA = 0.05 W/m°C The total thermal resistance is given by 6Rth = Rth – A + (Rth)conv. ln (r2 / r1 ) 1  2S k A . L 2S r2 L . h0 ln (0.09 / 0.045) 1  2S u 0.05 u 1 2S u 0.09 u 1 u 8.4 r1

= 2.206 + 0.210 = 2.416°C/W per metre length Case II. Two layers of insulation : Refer Fig. 2.69. 6Rth = Rth – A + Rth – B + (Rth)conv. ln ( r2 / r1 ) ln ( r3 / r2 ) 1   S S S k L k L r 2 A. 2 B. 2 3 . L . ho ln ( r3 / 0.09) ln (0.09 / 0.045) 1   2S u 0.05 u 1 2S u 0.07 u 1 2S r3 u 8.4 0.019 2.206  2.274 ln (r3 / 0.09)  r3 When condensation is to be reduced by 50 percent, the thermal resistance to heat flow must become two times,

112 i.e.,

Chapter : 2 2.206  2.274 ln (r3 / 0.09) 

0.019 r3

2 u 2.416

0.019 2.274 ln ( r3 / 0.09)  2.626 r3 By hit and trial, we get r3 ; 0.275 m or 275 mm Hence thickness of insulation (B) = (r3 – r2) = 275 – 90 = 185 mm (Ans.) Example 2.55. 55 kg/s of steam is flowing through a convective steam superheater 35/45 mm in diameter made of steel (k = 38.5 W/m°C). The pressure of dry saturated steam at the inlet of the superheater is 120 bar. The temperature of the steam leaving the superheater is 480°C. The heat transfer coefficients from the gas to wall and from wall to steam are 82 W/m2°C and 1120 W/m2°C respectively. If the mean flue gas temperature is 920°C, determine the outer heating surface of the superheater. Take: cps (for steam) = 1.92 kJ/kg°C. Solution. Refer to Fig. 2.70. 35 r1 17.5 mm 0.0175 m 2 45 r2 22.5 mm 0.0225 m 2 k = 38.5 W/m°C ho = 82 W/m2°C hi = 1120 W/m2°C to = 920°C cps (for steam) = 192 kJ/kg°C Mass of steam flowing through

the superheater, m& s

55 kg/s

Condition of steam at inlet of the superheater; p = 120 bar, dryness fraction (x) = 1 Fig. 2.70. The temperature of steam leaving the superheater, tsup = 480°C. Outer heating surface of the superheater : From steam tables, corresponding to 120 bar, Saturation temperature, tsat. = 324.6°C The heat flow through the tube is given by 2S L (t0 – ti ) Q ln (r2 / r1 ) 1 1 ...(i)   h0 r2 k hi r1 tsup  tsat 480  324.6 402.3qC ti where, 2 2 and, Q = m& s × cps × (tsup – tsat) = 55 × 1.92 × (480 – 324.6) = 16410.24 kJ/s Substituting the proper values in eqn. (i), we have 2S L (920 – 402.3) 16410.24 u 103 1 ln (0.0225 / 0.0175) 1   82 u 0.0225 38.5 1120 u 0.0175 3252.8 L 5425.8 L 0.5995

“Conduction” Heat Transfer at Steady State–One Dimension

113

3

16410.24 u 10 3024 m 5425.8 ? Outer surface area of the superheater = 2Sr2L = 2S × 0.0225 × 3024 = 427.5 m2 (Ans.) L

or,

Example 2.56. A pipe having outer diameter 250 mm is insulated by a material of thermal conductivity of 0.48 W/m°C. The insulation of outside diameter 500 mm, due to restriction of space, is placed with an eccentricity of 60 mm. Determing the heat loss for a length of 10 m if inner and outer surfaces are at temperatures of 280°C and 50°C respectively. Solution. Refer to Fig. 2.71. 250 125 mm 0.125 m; 2 500 r2 250 mm 0.25 m 2 Eccentricity, e = 60 mm = 0.06 m Length of pipe, L = 10 m Thermal conductivity of insulation, k = 0.48 W/m°C Fig. 2.71. Heat loss, Q : 't Q Rth The thermal resistance (Rth) in this case is given by (from hand book) r1

Rth

...(i)

ª ^(r  r )2 – e2 `1/ 2  ^(r – r )2 – e2 `1/ 2 º 1 2 1 » ln « 2 1 1/ 2 « 2 2 2 2 1/ 2 » 2S k L ¬ ^(r2  r1 ) – e ` – ^(r2 – r1 ) – e ` ¼ ª{(0.250  0.125)2 – (0.06)2 }1/ 2  {(0.250 – 0.125)2 – (0.06)2 }1/ 2 º 1 ln « » 2S u 0.48 u 10 ¬ {(0.250  0.125)2 (0.06)2 }1/ 2  {(0.250 – 0.125)2 – (0.06)2 }1/ 2 ¼

1 ª 0.3702  0.1096 º ln 0.0202q C/W 2 S u 0.48 u 10 «¬ 0.3702 – 0.1096 »¼ Substituting the proper values in expression (i), we have (280 – 50) 11386 W Q (Ans.) 0.0202

Example 2.57. A current of 950 amperes is flowing through a long copper rod of 25 mm diameter, having an electrical resistance of 22 × 10–6 ohm per metre length. The rod is insulated to a radius of 17 mm with fibrous cotton (k = 0.058 W/m°C) which is further covered by a layer of plastic (k = 0.42 W/m°C). The heat transfer coefficient between the plastic and the surroundings is 20.5 W/m2°C and the temperature of surroundings is 15°C. Determine : (i) The thickness of the plastic layer which gives minimum temperature in a cotton insulation. (ii) The temperature of copper rod and the maximum temperature in the plastic layer, for the condition as at (i).

114

Chapter : 2

Solution. Refer to Fig. 2.72.

Fig. 2.72 25 12.5 mm 0.0125 m 2 r2 = 17 mm = 0.017 m r1

kA = 0.058 W/m°C; kB = 0.42 W/m°C ho = 20.5 W/m2°C; tsurr. = 15°C (i) The thickness of plastic layer which gives minimum temperature in cotton insulation : For the given system the heat transfer rate is given by 't 't Q ln (r2 / r1 ) ln (r3 / r2 ) 1 6 Rth   2S k A . L 2S k B . L 2S r3 L . ho The effect of insulation (plastic) can be studied by differentiating the total thermal resistance (6Rth) w.r.t. r3 and setting the derivative equal to zero. d (6 Rth ) d ª ln (r2 / r1 ) ln (r3 / r2 ) 1 º   0 i.e., « dr3 dr3 ¬ 2S k A L 2S k B L 2S r3 L . ho »¼ 1 1 0 or, 2S k L . r – 2 S L 2 . h0 . r3 3 B kB r3 or, ho Let us find second derivative to determine whether the foregoing result maximises or minimises the total resistance. d 2 (6 Rth ) 1 1  – 2 2 S h0 L r33 dr3 2S k B L . r3 kB r3 At h0 2 3 d 2 (6 Rth ) 1 1 ª h0 º ª h0 º  – . S ho L «¬ k B »¼ which is obviously +ve. 2S k B L «¬ k B »¼ dr32

“Conduction” Heat Transfer at Steady State–One Dimension

115

kB Thus r3 represents the condition for minimum thermal resistance (i.e., maximum heat flow ho rate) and hence minimum temperature in the cotton insulation. k B 0.42 r3 0.02048 m or 20.48 mm ? ho 20.5 Hence, thickness of plastic insulation = 20.48 – 17 = 3.48 mm (Ans.)

(ii) Temperature of copper rod (t1) and maximum temperature in the plastic layer (t2) : Heat generated in the copper rod due to flow of current = I 2 R = 9502 × 22 × 10–6 = 19.855 W/m (Q R = 22 × 10–6 ohm per metre length Total thermal resistance, ln (r2 / r1 ) ln (r3 / r2 ) 1   (6 Rth ) 2S k A . L 2S k B . L 2Sr3 L . h0

... given)

ln (0.017 / 0.0125) ln (0.02048 / 0.017) 1   2S u 0.058 u 1 2S u 0.42 u 1 2S u 0.02048 u 1 u 20.5

= 1.293°C/W t1 – tsurr . t1 – 15 't 6 Rth 6 Rth 1.293 This heat flow, under steady conditions, equals the heat generated in the copper rod due to flow of current.

Heat flow through the composite system, Q

t1 – 15 19.855 or t1 = 19.855 × 1.293 + 15 = 40.67°C (Ans.) 1.293 (t1 – t2 ) (40.67 – t2 ) Similarly, 19.855 or 19.855 ln (r2 / r1 ) ln (0.017 / 0.0125) 2S u 0.058 u 1 2S k A . L ln (0.017 / 0.0125) 23.9° C (Ans.) t2 40.67 – 19.855 u or, 2 S u 0.058 Example 2.58. (i) Find an expression for distribution of temperature and heat flow due to conduction in a circular conical rod with diameter at any section given by D = cx where x is the distance measured from the apex of the cone and c is a certain numerical constant. Assume that lateral surface is well insulated, there is no internal heat generation and heat flow takes place under steady state conditions. (ii) What will be the heat flow rate if the smaller and longer ends are located at x 1 = 50 mm and x2 = 250 mm and have temperatures 400°C and 200°C respectively ? Take : c = 0.22 and k (average thermal conductivity) = 3.6 W/m°C. Solution. Refer to Fig. 2.73. Since the lateral surface is well insulated, the conduction of heat is one-dimensional (x-direction), using Fourier's equation, we have S dt dt Q –k.A – k . D2 u dx dx 4 S dt Q – k u u c2 x 2 u or (Q D = cx ...given) dx 4 Rearranging the above equation, we get 4Q dx – kdt S c2 x2

?

116

Chapter : 2

Fig. 2.73.

Integrating both sides, we have t 4Q – k ³ dt t1 Sc 2

x

dx

1

x2

³x

x

or

– k (t – t1 )

4Q ª x – 2  1 º « » Sc 2 ¬ – 2  1¼ x

1

x

4Q ª 1 º – S c 2 «¬ x »¼ x1

4Q § 1 1· or – k (t – t1 ) – ¸ 2 ¨x Sc © 1 x ¹ 4Q § 1 1· or t t1 – 2 ¨ – ¸ Sc k © x1 x ¹ At x = x2, where t = t2, the above expression becomes 4Q § 1 1 · t2 t1 – – 2 ¨x x2 ¸¹ Sc k © 1 2 S c k (t1 – t2 ) or Heat flow rate, Q 1 · §1 4¨ – ¸ x x © 1 2 ¹

...(i)

...(ii)

...(iii)

By substituting Q in (i), we get the temperature distribution as follows : ª §1 1·º «¨x – x ¸» 1¹ » ...Required expression. (Ans.) t t1  (t1 – t2 ) « © «§ 1 1 ·» «¨ x – x ¸» 2 ¹¼ ¬© 1 (ii) Given : x1 = 50 mm = 0.05 m; x2 = 250 mm = 0.25 m; t1 = 400°C; t2 = 200°C; c = 0.22; k = 3.6 W/m°C. Heat flow rate, Q : Substituting the various values in expression (iii), we have S u 0.222 u 3.6 (400 – 200) 109.48 1.71 W (Ans.) Q 1 · 64 § 1 4¨ – ¸ © 0.05 0.25 ¹

“Conduction” Heat Transfer at Steady State–One Dimension

117

Example 2.59. Heat is conducted through a tapered circular rod of 200 mm length. The ends A and B having diameters 50 mm and 25 mm are maintained at 27°C and 227°C respectively. k (rod material) = 40 W/m°C. Find : (i) Heat conducted through the rod. (ii) The temperature at the mid-point of the end. Assume there is no temperature gradient at a particular cross-section and there is no heat transfer through the peripheral surface. (M.U.) Solution. Given : L = 200 mm = 0.2 m; D1 = 50 mm = 0.05 m; D2 = 25 mm = 0.025 m; t1 = 227°C; t2 = 27°C; k = 40 W/m°C. (i) Heat conducted through the rod, Q :

Fig. 2.74.

The heat flow through the rod is given by k S R1 R2 (t1 – t2 ) Q ...(i) L 40 u S u 0.025 u 0.0125 u (227 – 27) 39.27 W (Ans.) 0.2 (ii) The temperature at the mid-point of the rod, t : x Rx R2  ( R1 – R2 ) Now, L R R 1 § R  R2 · ( Rx ) 1 R2  ( R1 – R2 ) u R2  1 – 2 ¨ 1 ¸ x 2 2 2 © 2 ¹ 2 Now substituting the value of ( Rx )

x

L 2

in the eqn. (i) for R1 and t for t1 and L

L , we get 2

§ R  R1 · kS¨ 2 ¸ R2 (t – t2 ) k S ( R  R ) R (t – t ) © 2 ¹ 1 2 2 2 Q L/2 L L where t is the temperature at x 2 Now substituting the values in the above equation, we get 39.27

or,

t

40 u S (0.0125  0.025) u 0.0125(t – 27) 0.2 39.27 u 0.2  27 160.33° C 40 u (0.0125  0.025)

Note. For steady state heat flow, Q remains constant at all sections.

(Ans.)

118

Chapter : 2

Example 2.60. The following data relate to a gas turbine rotor : Radius of the rotor = 550 mm; Thickness of rotor- varies linearly from 160 mm at the centre to 60 mm at the outer periphery where blades are attached; Temperatures indicated by the thermocouples attached to the rotor at radial distances of 90 mm and 450 mm = 330°C and 645°C respectively; Thermal conductivity of rotor material = 0.345 W/m°C. (i) Starting from the basic principles, determine the radial heat flow rate through the runner; (ii) Determine the percentage change in heat flow rate that would occur if the rotor has a uniform thickness of 60 mm.

Fig. 2.75. Gas turbine rotor.

Solution. Refer to Fig. 2.75. Radius of the rotor, R = 550 mm = 0.55 m Thickness of the rotor – varies linearly : At the centre, xc = 160 mm = 0.16 m At the outer periphery, xop = 60 mm = 0.06 m Temperatures indicated by the thermocouples : At radial distance, r1 = 90 mm or 0.09m ; t1 = 330°C At radial distance, r2 = 450 mm or 0.45 m; t2 = 645°C Thermal conductivity of rotor material, k = 0.345 W/m°C. (i) Radial heat flow rate through the runner, Q : The thickness of rotor changes from xc to xop from the centre to outer periphery; therefore, the thickness of rotor at a distance r from the centre, ª xc – xop º x xc – « » r xc – zr ¬ R ¼ xc – xop z ( a constant) where, R Heat conducted at radius r, dt dt – k .2Sr ( xc – zr ) Q – k A. dr dr Rearranging and integrating within the given limits, we have t2 r2 Q dr – ³ dt ³ t1 r 1 r ( xc – zr ) 2S k The right hand integral is of the form dx 1 § x · ³ x (a – bx) a ln ©¨ a – bx ¹¸ ?

t1 – t2

Q 2Sk

r

ª 1 § r ·º 2 « x ln ¨ x – zr ¸ » ¹ ¼ r1 ¬ c © c

“Conduction” Heat Transfer at Steady State–One Dimension ª § r2 · § r1 · º « ln ¨ ¸ – ln ¨ x – zr ¸ » x zr – 2 ¹ 1 ¹¼ © c ¬ © c ª r2 ­ xc – zr1 ½ º Q ln « ® ¾» 2S kxc ¬ r1 ¯ xc – zr2 ¿ ¼ 2 S kxc (t1 – t2 ) ª r ­ x – zr1 ½º ln « 2 ® c ¾» ¬ r1 ¯ xc – zr2 ¿¼

119

Q 2S kxc

?

Q

xc – xop

0.16 – 0.06 0.55 R Substituting the given data in the expression (i), we get Now,

z

Q

...(i)

0.182

2S u 0.345 u 0.16 (330 – 645) ª 0.45 ­ 0.16 – 0.182 u 0.09 ½ º ln « ® ¾» ¬ 0.09 ¯ 0.16 – 0.182 u 0.45 ¿ ¼ –109.252 0.1436 º ª ln «5 u 0.0781 »¼ ¬

– 49.25 W (Ans.)

Negative sign indicates that the heat flow is radially inward. (ii) Percentage change in heat flow rate if the rotor has uniform thickness of 60 mm : For a rotor of uniform thickness xop dt Q – k .2 Sr . xop . dr Rearranging and integrating within the given limits, we have t2 r2 dr Q – ³ dt ³ t1 r 2S k xop 1 r or

(t1 – t2 ) Q

Q ln (r2 / r1 ) 2S k xop 2S k xop (t1 – t2 )

...(ii) ln (r2 / r1 ) This expression (ii) is identical to the heat flow equation through a hollow cylinder of length xop. Substituting the given data in the above expression, we have 2S u 0.345 u 0.06 (330 – 645) Q – 25.45 W In (0.45 / 0.09) ? Percentage change in heat flow rate ?

49.25 – 25.45 49.25

0.4832

or

48.32%

(Ans.)

2.7. HEAT CONDUCTION THROUGH HOLLOW AND COMPOSITE SPHERES 2.7.1. HEAT CONDUCTION THROUGH HOLLOW SPHERE Case I. Uniform conductivity : Refer Fig. 2.76. Consider a hollow sphere made of material having constant thermal conductivity.

120

Chapter : 2

Fig. 2.76. Steady state conduction through a hollow sphere.

Let,

r1, r2 = Inner and outer radii, t1, t2 = Temperatures of inner and outer surfaces, and k = Constant thermal conductivity of the material with the given temperature range. The general heat conduction equation in spherical coordinates is given as follows : 1 w r 2 wr

w 2t w § wt · q g 1 1 § 2 wt · · 2  2 ¨r ¸ 2 ¨ sin T ¸ 2 wT ¹ k © wr ¹ r sin I wI r sin T wT ©

1 wt · D wW

...[Eqn. 2.31]

§ wt · 0 ¸ , unidirectional heat flow in the radial direction {t z f (T, I)} and with For steady state ¨ wW © ¹ no heat generation (qg = 0), the above equation reduces to 1 d § 2 dt · ¨r ˜ ¸ 0 dr ¹ r 2 dr © d § 2 dt · 1 z0 or, as ¨r ˜ ¸ 0 dr © dr ¹ r2 dt r2 ˜ C (a constant) or, dr Integrating the above equation, we obtain C t –  C1 r (where C1 = a constant of integration) Using the following boundary conditions, we have At r = r1, t = t1; At r = r2, t = t2 C t1 –  C1 ? r1

...(2.72)

...(2.73)

...(i)

“Conduction” Heat Transfer at Steady State–One Dimension C t2 –  C1 r2 From (i) and (ii), we have (t1 – t2 ) r1 r2 C r1 – r2 (t – t ) r r C1 t1  1 2 1 2 and, r1 (r1 – r2 ) Substituting the values of these constants in eqn. (2.73), we get (t – t ) r r (t – t ) r r t – 1 2 1 2  t1  1 2 1 2 r (r1 – r2 ) r1 (r1 – r2 )

121 ...(ii)

(t1 – t2 ) (t1 – t2 )  t1  r (1/ r2 – 1/ r1 ) r1 (1/ r2 – 1/ r1 ) (t1 – t2 ) ª 1 1 º t t1  – or, ...(2.74) (1/ r2 – 1/ r1 ) «¬ r1 r »¼ t – t1 1/ r – 1/ r1 or, t2 – t1 1/ r2 – 1/ r1 t – t1 r2 ª r – r1 º or, [Dimensionless form] ...(2.75) t2 – t1 r «¬ r2 – r1 »¼ From the eqn. (2.75) it is evident that the temperature distribution associated with radial conduction through a sphere is represented by a hyperbola. Determination of conduction heat transfer rate, Q : The conduction heat transfer rate is determined by using the temperature distribution expression [Eqn. (2.75)] in conjuction with Fourier's equation as follows : dt Q – kA dr (t1 – t2 ) § 1 1 · º d ª – k .4S r 2 . «t1  – dr ¬ (1/ r2 – 1/ r1 ) ¨© r1 r ¸¹ »¼ t –t § 1· u – ¨– 2 ¸ – k .4Sr 2 . 1 2 (1/ r2 – 1/ r1 ) © r ¹ 1 2 (t1 – t 2 ) u – k .4S r . § r1 – r2 · r 2 ¨ r .r ¸ © 1 2 ¹ (t1 – t2 ) r1 r2 4S k (t1 – t2 ) r1 r2 (t1 – t2 ) – 4S k (r1 – r2 ) (r2 – r1 ) (r2 – r1 ) / 4S k r1 r2 ( t – t ) ' t ª º 1 2 i.e. Q ...(2.76) ª (r2 – r1 ) º «¬ Rth »¼ « 4S k r r » ¬ 1 2¼

or,

t

–

where the term (r2 – r1) / 4Sk r1 r2 is the thermal resistance (Rth) for heat conduction through a hollow sphere. Alternative method : Refer Fig. 2.76. Consider a small element of thickness dr at any radius r. Area through which the heat is transmitted, A = 4Sr2 dt Q – k .4S r 2 . ? dr

122

Chapter : 2

Rearranging and integrating the above equation, we obtain r2 dr t2 Q ³ 2 – 4S k ³ dt r1 r t1 r

or, or, or,

ª r–2 1 º 2 Q« » ¬ – 2  1 ¼ r1 1· §1 –Q¨ – ¸ © r2 r1 ¹ Q (r2 – r1 ) r1 r2

or,

Q

– 4 S k [t ]tt12 – 4 S k (t2 – t1 ) 4S k (t1 – t2 )

4S k r1 r2 (t1 – t2 ) (r2 – r1 )

t1 – t2 r ( ª 2 – r1 ) º « 4S k r r » 1 2¼ ¬

... (2.77)

Case II. Variable conductivity : A. Temperature variation in terms of surface temperatures, t1, t2 : By adopting the similar procedure as was followed in case of a hollow cylinder, we would obtain the following expressions for the rate of heat transfer (Q) and temperature variation in a hollow sphere in terms of surface temperatures (t1,t2). E 4S k0 r1 r2 ª º Q 1  (t1  t2 ) » (t1 – t2 ) ...(2.78) « ¼ (r2 – r1 ) ¬ 2 and,

t

1/ 2

1ª § r – r1 · § r2 · 2 2 º (1  E t1 )2 – ¨ ¸ ¨© r ¸¹ (1  E t1 ) – (1  E t2 ) » r r E «¬ – © 2 1¹ ¼

^

`

–

1 E

...(2.79) B. Temperature variation in terms of heat flux (Q) : By using the same approach as was adopted in case of hollow cylinder, we would obtain the following expression 1/ 2 2 Q 1 § 1 1 ·º 1 ª§ 1· ˜ t –  «¨ t1 – ¸ – – » ...(2.80) E «¬© E¹ E k0 2S ©¨ r1 r ¹¸ »¼

2.7.1.1. Logarithmic mean area for the hollow sphere Adopting the same concept as used for hollow cylinder, we can write (t1 – t2 ) Qsphere ª (r2 – r1 ) º « 4S k r r » 1 2¼ ¬ (t1 – t2 ) Q plane wall § (r2 – r1 ) · ¨ kA ¸ m ¹ © Am is so chosen that the heat flow through cylinder and plane wall will be equal for the same thermal potential. ? Qsphere = Qplane wall (t1 – t2 ) (t1 – t2 ) ª ( r2 – r1 ) º ª r2 – r1 º « 4S k r r » « k A » 1 2¼ m ¼ ¬ ¬ r2 – r1 r2 – r1 or, k Am 4S k r1 r2

“Conduction” Heat Transfer at Steady State–One Dimension or,

Am = 4Sr1 r2

or,

Am2

(4 S r1 r2 ) 2

or,

Am2

Ai u A0

or,

Am

Ai A0

Further,

Am = 4S

or,

rm

rm2

2

123

2

(4 S r1 ) u (4Sr2 )

...(2.81)

= 4 S r1 r2

r1 r2

(where rm = logarithmic mean radius of hollow sphere).

2.7.2. HEAT CONDUCTION THROUGH A COMPOSITE SPHERE Considering Fig. 2.77 as cross-section of a composite sphere, the heat flow equation can be written as follows 4S k A r1 r2 (t1 – t2 ) 4S k B r2 r3 (t2 – t3 ) Q hhf .4S r12 (thf – t1 ) (r2 – r1 ) (r3 – r2 ) hcf .4S r32 (t3 – tcf ) By rearranging the above equation we have Q thf – t1 hhf .4S r12 Q (r2 – r1 ) t1 – t2 4S k A . r1 r2 Q (r3 – r2 ) t2 – t3 4S k B . r2 r3 Q t3 – tcf hcf .4S r32

Fig. 2.77. Steady state conduction through a composite sphere.

...(i) ...(ii) ...(iii) ...(iv)

124

Chapter : 2

Adding (i), (ii), (iii), and (iv), we get

(r – r ) (r – r ) Qª 1 1 º  2 1  3 2  « » thf – tcf 2 4S « hhf ˜ r1 k A ˜ r1 r2 k B ˜ r2 r3 hcf ˜ r32 ¼» ¬ 4S (thf – tcf ) Q (r2 – r1 ) (r3 – r2 ) 1 º ª 1 « h ˜r2  k ˜r r  k ˜r r  » hcf ˜ r32 ¼ A 1 2 B 2 3 ¬ hf 1

?

If there are n concentric spheres then the above equation can be written as follows 4S (thf – tcf ) Q n n ª 1 º ­° r( n  1) – rn ½° 1  ¦® « » ¾ 2 2 n 1° ¯ kn ˜ rn ˜ r( n  1) °¿ hcf ˜ r( n  1) ¼» ¬« hhf ˜ r1

...(2.82)

...(2.83)

If inside and outside heat transfer coefficients are not considered, then the above equation can be written as follows : 4S (t1 – tn  1 ) ...(2.84) Q n n ª r( n  1) – rn º ¦« » n 1 ¬ kn ˜ rn ˜ r( n  1) ¼ Example 2.61. A spherical shaped vessel of 1.4 m diameter is 90 mm thick. Find the rate of heat leakage, if the temperature difference between the inner and outer surfaces is 220°C. Thermal conductivity of the material of the sphere is 0.083 W/m°C. Solution. Refer to Fig. 2.78. 1.4 r2 0.7 m. 2 90 r1 0.7 – 0.61m 1000 t1 – t2 = 220°C; k = 0.083 W/m°C The rate of heat transfer/leakage is given by

Q

(t1 – t2 ) ...Fig. (2.76) ª (r2 – r1 ) º «4Sk r r » 1 2¼ ¬ 220 (0.7 – 0.61) ª º « 4 S u 0.083 u 0.61 u 0.7 » ¬ ¼ Fig. 2.78.

= 1088.67 W i.e., Rate of heat leakage = 1088.67 W

(Ans.)

Example 2.62. A spherical thin walled metallic container is used to store liquid N2 at – 196° C. The container has a diameter of 0.5 m and is covered with an evacuated reflective insulation composed of silica powder. The insulation is 25 mm thick and its outer layer is exposed to air at 27°C. The convective heat transfer coefficient on outer surface = 20 W/m2° C. Latent heat of evaporation of N2 = 2 × 105 J/kg. Density of N2 = 804 kg/m3. k (silica powder) = 0.0017 W/m°C. Find out the rate of heat transfer and rate of N2 boil-off. (N.U.)

“Conduction” Heat Transfer at Steady State–One Dimension

125

Solution. Given : 0.5 0.25 m; 2 r2 = r1 + 0.025 = 0.25 + 0.025 = 0.275 m;

t1 = – 196°C; t2 = 27°C; r1 h0 = 20 W/m2°C; h fg N2

= 2 × 105 J/kg; UN2 = 804 kg/m3; k = 0.0017 W/m°C. Rate of heat transfer, Q : The heat flow is given by (t1 – ta ) Q (r2 – r1 ) 1  4S k r1 r2 h0 u 4 S r22 (–196 – 27) (0.275 – 0.25) 1  4S u 0.0017 u 0.25 u 0.275 20 u 4S u 0.2752 – 223 –13.1W 17.022  0.0526 The – ve sign indicates that the heat flows in.

mN 2 u h fg

?

Fig. 2.79.

13.1

13.1 u 3600 kg/h 0.2358kg/h (Ans.) 2 u 105 Example 2.63. Determine the rate of heat flow through a spherical boiler wall which is 2 m in diameter and 2 cm thick steel (k = 58 W/m K). The outside surface of boiler wall is covered with asbestos (k = 0.116 W/m K) 5mm thick. The temperature of outer surface and that of fluid inside are 50°C and 300°C respectively. Take inner film resistance as 0.0023 K/W. (N.U.) 2 1m; Solution. Given : r1 2 2 r2 1  1.02 m; 100 kA = 58 W/m K; kB = 0.116 W/m k; 5 r3 r2  1.02  0.005 100 = 0.025 m Q = h1 A1 (ti – t1) as heat flows from fluid to inner surface by convection only. mN 2

or

or, Q

where,

ti – t1 1 hi A1

1 is inner film resistance. hi A1

Fig. 2.80.

126

Chapter : 2 Q

?

(ti – t3 ) (r – r ) (r2 – r1 ) 1   3 2 hi A1 4S k A r1 r2 4S k B r2 r3

(300 – 50) (1.02 – 1.0) (1.025 – 1.02)  0.0023  4S u 58 u 1.0 u 1.02 4S u 0.116 u 1.02 u 1.025 250 44581W 4.581kW (Ans.) 0.0023  2.6902 u 10 –5  0.0032808 Example 2.64. A spherical container having outer diameter 500 mm is insulated by 100 mm thick layer of material with thermal conductivity k = 0.03 (1 + 0.006 t) W/m°C, where t is in °C. If the surface temperature of sphere is – 200°C and temperature of outer surface is 30°C, determine the heat flow in.

Solution. Refer to Fig. 2.81. Given : r1 = 250 mm = 0.25m, r2 = 250 + 100 = 350 mm = 0.35 m k = 0.03 (1 + 0.006 t) Heat flow, Q : 't Q ª (r2 – r1 ) º « 4S k r r » m 1 2¼ ¬ where,

km

ª § – 200  30 · º 0.03 «1  0.006 ¨ ¸» © ¹¼ 2 ¬

0.0147 W/m° C

Fig. 2.81. ?

Q

30 – (– 200) (0.35 – 0.25) ª º « 4S u 0.0147 u 0.25 u 0.35 » ¬ ¼

37.17 W (Ans.)

“Conduction” Heat Transfer at Steady State–One Dimension

127

Example 2.65. The inside and outside surfaces of a hollow sphere of radii r1 and r2 are maintained at t1 and t2 respectively. The thermal conductivity of sphere material varies with temperature as given below : k = k0 (1 + Dt + Et2) Derive an expression for total heat flow rate through the sphere. (P.U.) Solution. Considering steady state conduction through a hollow sphere r = r and of thickness dr , we can write dt dt dt Q – kA – k u 4Sr 2 u dr dr Substituting the given value of k in the above equation, we get Q

– k0 (1  D t  E t 2 ) u 4S r 2 u

dt dr

Q dr u – k0 (1  Dt  E t 2 ) dt 4S r 2 Integrating the above equation in the given range, we get

?

Q 4S ?

Q 4S

³

r2 r1

dr r2 r

ª 1º 2 – ¬« r »¼ r1

– k0

³

t2

t1

2 (1  Dt  Et ) dt t

2 ª§ t2 t 3 ·º – k0 «¨ t  D  E ¸ » 2 3 ¹ ¼ t1 ¬©

Fig. 2.82.

ª º § t22 – t12 · E 3 Q§1 1· 3 k t t  D – ( – ) « ¨ ¸  (t2 – t1 ) » 0 2 1 ¨ ¸ 4S © r2 r1 ¹ 2 ¹ 3 © ¬ ¼ D 2 E 3 Q § r1 – r2 · ª 2 3 º – k0 « (t1 – t2 )  t1 – t2  (t1 – t2 ) » or, ¬ ¼ 4S ¨© r1 r2 ¸¹ 2 3 D E 2 Q § r2 – r1 · ª 2 º – k0 (t1 – t2 ) «1  (t1  t2 )  t1  t1 t2  t2 » or, ¬ ¼ 4S ¨© r1 r2 ¸¹ 2 3 D E 4S r1 r2 ª º u k0 (t1 – t2 ) «1  (t1  t2 )  (t12  t1 t2  t2 2 ) » Q or, r2 – r1 ¬ ¼ 2 3 ...Required expression (Ans.) Example 2.66. The inside and outside surfaces of a hollow sphere, having inner and outer radii r1 and r2 respectively, are maintained at uniform temperatures t1 and t2. Find the rate of heat transfer through the sphere if the conductivity of the material of which the sphere is made varies according to relation : k = k1 + (k2 – k1) [(t – t1) / (t2 – t1)] Solution. Fourier’s equation for unidirectional steady state heat conduction is given as : dt Q – kA ...(i) dr where A = 4 Sr2 (area normal to radial direction). Now, by substituting the values of A and k in eqn. (i), we get dt Q –[k1  (k2 – k1 ){(t – t1 ) /(t2 – t1 )}] u 4Sr 2 u dr

128

Chapter : 2 dr

– 4S >k1  (k2 – k1 ) {(t – t1 ) /(t2 – t1 )}@ dt r2 Integrating both sides, we get Q.

or,

Q³

r2

dr

r1

2

r

r

ª1º 2 –Q« » ¬ r ¼ r1

ª 1 1º –Q« – » ¬ r2 r1 ¼ Q˜

?

(r2 – r1 ) r1 r2

Q

– 4S

t2

³t

1

ª ­ (t – t1 ) ½ º « k1  (k2 – k1 ) ® t t ¾ » dt ¯ ( 2 – 1 ) ¿¼ ¬ t

2 ª ½º k – k1 ­ t 2 ® – t ˜ t1 ¾» – 4S « k1t  2 ¿¼t t2 – t1 ¯ 2 ¬ 1

ª k –k – 4S «k1 (t2 – t1 )  2 1 t2 – t1 ¬«

º § t22 – t12 · §k – k · ¨ ¸ – t1 ¨ 2 1 ¸ (t2 – t1 )» © 2 ¹ »¼ © t2 – t1 ¹

ª º  k – k1 – 4S « k1 (t2 – t1 )  2 (t2  t1 ) – t1 (k2 – k1 ) » ¬ ¼ 2 (k – k1 ) ª º – 4S « k1 (t2 – t1 )  2 (t2  t1 – 2 t1 ) » ¬ ¼ 2 k – k1 ª º – 4S « k1 (t2 – t1 )  2 (t2 – t1 ) » ¬ ¼ 2 k k – ª 1º – 4S (t2 – t1 ) « k1  2 ¬ 2 »¼ § k  k2 · – 4S (t1 – t2 ) ¨ 1 ¸ © 2 ¹ § k  k2 ·§ t1 – t2 · 4S r1 r2 ¨ 1 ¸ ...Required expression. © 2 ¹ ¨© r2 – r1 ¸¹

(Ans.)

Example 2.67. Show that for small objects transferring heat to the surrounding, the minimum Nusselt number is equal to 2. (M.U.) Solution. Consider a small body as sphere. The heat flow by conduction through a sphere is given by t1 – t2 t1 – t2 Q R2 – R1 Rth – cond. 4S k R1 R2 The heat conducted at r = R2 is further convected to the surrounding air and it is given by t 2 – ta t 2 – ta Q 1 Rth – conv. 4S R22 h where Rth-cond. and Rth-conv. are conduction and convection resistances respectively. 1 § 1 1 · Rth – cond. – 4S k ¨© R1 R2 ¸¹ ...(i)

Fig. 2.83.

Now the surrounding fluid will be considered a spherical shell of radius r and infinite outside radius. If we neglect the motion of the fluid, the only mechanism of heat transfer will be conduction through the small sphere and the resistance of this shell to heat flow [by substituting R1 = r and R2 = f] in eqn. (i)], will be

“Conduction” Heat Transfer at Steady State–One Dimension

129

1 §1 1 · 1 ¨ – ¸ 4S k © r f ¹ 4S r k t2 – ta Q ; where t2 is the surface temperature. 1 4S r k

Rth – cond.

?

The heat flow is also given by introducing h as t2 – ta Q 1 4 S r 2h Now equating (i) and (ii), we get t2 – ta t2 – ta 1 1 4Sr k 4 S r2h or, 4 S r2h = 4 Srk or, ?

h hd k

k r

...(ii)

2k d

2

hd is known as Nusselt number (Nu). k ? Nu = 2 (Ans.) Example 2.68. A Cylindrical tank of 1.0 m diameter and 5 m total length has hemispherical ends. It contains liquid oxygen which has boiling point and heat of vaporisation –180°C and 210 kJ/ kg respectively. It is required to insulate the tank so as to reduce the boil-off rate of oxygen in steady state to 14 kg/h. Determine the thermal conductivity of the insulating material if its maximum thickness is limited to 70 mm. Assume room temperature outside the insulation as 25°C. Solution. Refer to Fig. 2.84.

where

Fig. 2.84.

130

Chapter : 2

70 1.0 0.57 m; ti = – 180°C; t0 = 25°C 0.5 m; r2 0.5  1000 2 Boil-off rate of liquid oxygen = 14 kg/h Heat of vaporisation of liquid oxygen, hfg = 210 kJ/kg Thermal conductivity of insulating material, k : For the cylindrical section of the tank, (t0 – ti ) Qcyl . ln (r2 / r1 ) k ˜ 2S L ln (r2 / r1 ) u Qcyl . (t0 – ti ) or, ...(i) 2S k L For the hemispherical ends, (t0 – ti ) Qends (r2 – r1 ) k .4S r1 r2 r2 – r1 u Qends (t0 – ti ) or, ...(ii) 4S k r1 r2 Equating expressions (i) and (ii), we have r1

ln (r2 / r1 ) u Qcyl 2Sk L Qends

or,

r2 – r1 u Qends 4S k r1 r2 ln ( r2 / r1 ) 4S k r1 r2 u u QcyL 2Sk L (r2 – r1 ) 2 r1 r2 ln (r2 / r1 ) u Qcyl . (r2 – r1 ) L

Also or, or,

Qcyl. + Qends = Qboil = 14 × 210 = 2940 2 r1 r2 ln (r2 / r1 ) u Qcyl . (r2 – r1 ) L

2940

2 r r ln (r2 / r1 ) º ª Qcyl . «1  1 2 (r2 – r1 ) L »¼ ¬

2940

Qcyl . 

or,

Also, ?

2940 2 r1 r2 ln (r2 / r1 ) º ª «1  (r – r ) L » 2 1 ¬ ¼ L + 2r1 = 5 or L + 2 × 0.5 = 5 Qcyl .

Qcyl .

2940 u u 2 0.5 0.57 u ln (0.57 / 0.5) º ª «1  » (0.57 – 0.5) u 4 ¬ ¼ 2320.9 kJ/h

From expression (i), we have k

? L = 4.0 m

2320.9 u 1000 W 3600

ln (r2 / r1 ) u Qcyl . 2S L (to – ti )

644.69 W

“Conduction” Heat Transfer at Steady State–One Dimension ln (0.57 / 0.5) u 644.69 2S u 4[25 – (–180)]

131

0.0164 W / mqC (Ans.)

Example 2.69. A cylindrical tank with hemispherical ends is used to store liquid oxygen at –183° C. The diameter of the tank is 1.5 m and the total length is 8 m. The tank is covered with a 10 cm thick layer of insulation. Determine the thermal conductivity of the insulation, so that the boil-off rate does not exceed 10.8 kg/h. The latent heat of vapourisation of liquid oxygen is 214 kJ/kg. Assume that the outer surface temperature of the insulation is 27°C and that the thermal resistance of the wall of the tank is negligible. (U.P.S.C.) Solution. Heat generated during boiling of oxygen, Qboil = 10.8 × 214 = 2311.2 kJ/h Let t0 be the room temperature outside of insulation and ti be the temperature of liquid oxygen inside the tank. For the cylindrical section of the tank, 2Sk Lcy1 (t0 – ti ) Qcy1 ln (r2 / r1 ) or,

(t0 – ti )

ln (r2 / r1 ) u Qcy1 2S k Lcy1

Qends

t0 – ti (r2 – r1 ) / (4S k r1 r2 )

For the two spherical ends,

(t0 – ti )

or,

4S k r1 r2 (t0 – ti ) (r2 – r1 ) r2 – r1 u Qends 4S k r1 r2

...(i)

...(ii)

From eqns. (i) and (ii), we get ln (r2 / r1 ) u Qcy1 2S k Lcy1

2 r1 r2 u ln (r2 / r1 ) u Qcy1 (r2 – r1 ) Lcy1 Qcy1 + Qends = Qboil = 2311.2 Qends

? Now or, Qcy1 

r2 – r1 u Qends 4S k r1 r2

2 r1 r2 u In (r2 / r1 ) u Qcy1 (r2 – r1 ) Lcy1 Qcy1

?

From the given geometry of the tank,

2311.2 2311.2 ª 2 r r u ln (r2 / r1 ) º 1 « 1 2 » ¬ (r2 – r1 ) Lcy1 ¼

1.5 0.75 m; r2 = 0.75 + 0.10 = 0.85 m; Lcyl = 8 – 1.5 = 6.5 m 2 Substituting the values in eqn. (iii), we get 2311.2 2311.2 Qcy1 u u u 2 0.75 0.85 ln (0.85 / 0.75) ª º 1  0.2455 1 « » u (0.85 – 0.75) 6.5 ¬ ¼ r1

...(iii)

132

Chapter : 2 = 1855.6 kJ/h = 515.44 J/s

Now, from eqn. (i), we have k

ln (r / r1 ) u Qcy1 (t0 – ti ) 2 S Lcy1 ln (0.85 / 0.75) u 515.44 [27 – (–183)]2S u 6.5

= 7.522 × 10–3 W/m K Hence, the thermal conductivity of the insulation, k = 7.522 × 10–3 W/m K (Ans.)

2.8. CRITICAL THICKNESS OF INSULATION 2.8.1. INSULATION-GENERAL ASPECTS Definition. A material which retards the flow of heat with reasonable effectiveness is known as ‘Insulation’. Insulation serves the following two purposes : (i) It prevents the heat flow from the system to the surroundings; (ii) It prevents the heat flow from the surroundings to the system. Applications : The fields of application of insulations are : (i) Boilers and steam pipes; (ii) Airconditioning systems; (iii) Food preserving stores and refrigerators; (iv) Insulating bricks (employed in various types of furnaces); (v) Preservation of liquid gases etc. Factors affecting thermal conductivity : Some of the important factors which affect thermal conductivity (k) of the insulators (the value of k should be always low to reduce the rate of heat flow) are as follows : 1. Temperature. For most of the insulating materials, the value of k increases with increase in temperature. 2. Density. There is no mathematical relationship between k and U (density). The common understanding that high density insulating materials will have higher values of k is not always true. 3. Direction of heat flow. For most of the insulating materials (except few like wood) the effect of direction of heat flow on the values of k is negligible. 4. Moisture. It is always considered necessary to prevent ingress of moisture in the insulating materials, during service. it is, however, difficult to find the effect of moisture on the values of k of different insulating materials. 5. Air pressure. It has been found that the value of k decreases with decrease in pressure. 6. Convection in insulators. The value of k increases due to the phenomenon of convection in insulators.

“Conduction” Heat Transfer at Steady State–One Dimension

133

2.8.2. CRITICAL THICKNESS OF INSULATION The addition of insulation always increases the conductive thermal resistance. But when the the total thermal resistance is made of conductive thermal resistance [(Rth)cond.] and convective thermal resistance [(Rth)conv.], the addition of insulation in some cases may reduce the convective thermal resistance due to increase in surface area, as in the case of a cylinder and sphere, and the total thermal resistance may actually decrease resulting in increased heat flow. It may be shown that the thermal resistance actually decreases and then increases in some cases. “The thickness upto which heat flow increases and after which heat flow decreases is termed as Critical thickness. In case of cylinders and spheres it is called ‘Critical radius’. A. Critical thickness of insulation for cylinder : Fig. 2.85. Critical thickness of insulation for cylinder. Consider a solid cylinder of radius r1 insulated with an insulation of thickness (r2 – r1) as shown in Fig. 2.85. Let, L = Length of the cylinder, t1 = Surface temperature of the cylinder, tair = Temperature of air, ho = Heat transfer coefficient at the outer surface of the insulation, and k = Thermal conductivity of insulating material. Then the rate of heat transfer from the surface of the solid cylinder to the surroundings is given by 2SL (t1 – tair ) Q ...(2.85) ln (r2 / r1 ) 1  k h0 ˜ r2 ln (r2 / r1 ) From Eqn. (2.85) it is evident that as r2 increases, the factor increases but the factor k 1 1 º ª ln ( r2 / r1 ) decreases. Thus Q becomes maximum when the denominator «  » becomes h0 ˜ r2 k h o ˜ r2 ¼ ¬ minimum. The required condition is d ª ln (r2 / r1 ) 1 º  0 (r2 being the only variable) « dr2 ¬ k ho ˜ r2 »¼ 1 1 1§ 1 · ˜  – 0 ? k r2 ho ¨© r22 ¸¹ or,

1 1 – k ho ˜ r2

0

or,

r2 ( rc )

k ho

or

ho . r2 = k ...(2.86)

134

Chapter : 2

The above relation represents the condition for minimum resistance and consequently*maximum heat flow rate. The insulation radius at which resistance to heat flow is minimum is called the ‘critical radius (rc). The critical radius rc is dependent on the thermal quantities k and ho and is independent of r1 (i.e. cylinder radius). *It may be noted that if the second derivative of the denominator is evaluated, it will come out to be positive; this would verify that heat flow rate will be maximum, when r2 = rc. In eqn. (2.85) ln (r2 / r1) / k is the conduction (insulation) thermal resistance which increases with increasing r2 and 1/ho.r2 is convective thermal resistance which decreases with increasing r2. At r2 = rc the rate of increase of conductive resistance of insulation is equal to the rate of decrease of convective resistance thus giving a minimum value for the sum of thermal resistances. In the physical sense we may arrive at the following conclusions : (i) For cylindrical bodies with r1 < rc, the heat transfer increases by adding insulation till r2 = rc as shown in Fig. [2.86 (a)]. If insulation thickness is further increased, the rate of heat loss will decrease from this peak value, but until a certain amount of insulation denoted by r2c at b is added, the heat loss rate is still greater for the solid cylinder. This happens when r1 is small and rc is large, viz., the thermal conductivity of the insulation k is high (poor insulating material) and ho is low. A practical application would be the insulation of electric cables which should be a good insulator for current but poor for heat. (ii) For cylindrical bodies with r1 > rc, the heat transfer decreases by adding insulation [Fig.2.86 (b)]. This happens when r1 is large and rc is small, viz., a good insulating material is used with low k and h0 is high. In steam and refrigeration pipes heat insulation is the main objective. For insulation to be properly effective in restricting heat transmission, the outer radius must be greater than or equal to the critical radius.

Fig. 2.86. Dependence of heat loss on insulation thickness.

B. Critical thickness of insulation for sphere : Refer to Fig. 2.87. The equation of heat flow through a sphere with insulation is given as

“Conduction” Heat Transfer at Steady State–One Dimension

135

Fig. 2.87. Critical thickness of insulation for sphere.

(t1 – tair ) 1 ª r2 – r1 º « 4S k r r »  2 4S r2 ˜ ho 1 2¼ ¬ Adopting the same procedure as that of a cylinder, we have d ª r2 – r1 1 º  0 « 2 dr2 ¬ 4S k r1 r2 4S r2 ˜ ho »¼ Q

or, or,

d dr2

1 1 º ª 1 «k r – k r  2 » r2 ho ¼ 2 ¬ 1 1 2 – k r2 2 r23 ho

or,

r23 h0

or,

r2 ( rc )

0 0 2 k r22 2k h0

...(2.87)

Example 2.70. Calculate the critical radius of insulation for asbestos [k = 0.172 W/m K] surrounding a pipe and exposed to room air at 300 K with h = 2.8 W/m K. Calculate the heat loss from a 475 K, 60 mm diameter pipe when covered with the critical radius of insulation and without insulation. (D.U.) Solution. Given : k = 0.172 W/m K; T1 = 475 K; T2 = 300 K; h0 = 2.8 W/m2 K; r1 The critical radius of insulation, rc

k h0

0.172 2.8

60 2

30 mm = 0.03 m.

0.06143m or

61.43 mm

(Ans.)

136

Chapter : 2

Q (with insulation)

2S (T1 – T2 ) ln (rc / r1 ) 1  k h0 rc

2S (475 – 300) 1099.56 110.16 W/m (Ans.) ln (0.06143 / 0.03) 1 4.167  5.814  0.172 2.8 u 0.06143 Q (without insulation) = h0 × 2Sr1 (T1 – T2) = 2.8 × 2S × 0.03 (475 – 300) = 92.36 W/m (Ans.) Example 2.71. A 10 mm cable is to be laid in atmosphere of 20°C with outside heat transfer coefficient 8.5 W/m2°C. The surface temperature of cable is likely to be 65°C due to heat generation within. Will the rubber insulation, k = 0.155 W/m °C, be effective ? If yes how much ? (AMIE) Solution. Refer to Fig. 2.85. 10 5 mm; t1 = 65° C; tair = 20° C; k = 0.155 W/m°C; h0 = 8.5 W/m2°C 2 For a cable (cylinder), k 0.155 rc 0.018235 m 18.235 mm, h0 8.5 rc is greater than the radius of the cable. Hence, the rubber insulation upto a thickness of 13.235 mm (18.235 – 5) will be effective in heat dissipation. (Ans.) The maximum heat dissipation per m length of the cable will be Qmax 2S (t1 – tair ) ln (rc / k1 ) 1 L  k h0 rc 2S (65 – 20) 282.74 19.1 W / m (Ans.) ln (18.235 / 5) 1 8.348  6.452  0.155 8.5 u 0.018235

Given : r1

Example 2.72. A small electric heating application uses wire of 2 mm diameter with 0.8 mm thick insulation (k = 0.12 W/m°C). The heat transfer coefficient (ho) on the insulated surface is 35 W/m2°C. Determine the critical thickness of insulation in this case and the percentage change in the heat transfer rate if the critical thickness is used, assuming the temperature difference between the surface of the wire and surrounding air remains unchanged. Solution. Refer to Fig. 2.88. 2 r1 1mm 0.001m 2 r2 = 1 + 0.8 = 1.8 mm = 0.0018 m k = 0.12 W/m°C, ho = 35 W/m2 °C. Critical thickness of insulation : The critical radius of insulation is given by

Fig. 2.88.

“Conduction” Heat Transfer at Steady State–One Dimension

137

k 0.12 3.43 u 10 –3 m or 3.43 mm. h0 35 Percentage change in heat transfer rate : Case I. The heat flow through an insulated wire is given by 2S L (t1 – tair ) 2S L (t1 – tair ) 2S L (t1 – tair ) ...(i) Q1 ln (r2 / r1 ) ln L (0.0018 / 0.001) 1 1 20.77   k h0 ˜ r2 0.12 35 u 0.0018 Case II. The heat flow through an insulated wire, when critical thickness is used, is given by 2S L (t1 – tair ) 2S L (t1 – tair ) Q2 ln (rc / r1 ) ln (0.00343 / 0.001) 1 1   0.12 35 u 0.00343 k h0 ˜ rc rc

2S L (t1 – tair ) 18.6 ? Percentage increase in heat flow by using critical thickness of insulation 1 1 – 18.6 20.77 u 100 11.6% (Ans.) 1 20.77 Example 2.73. A wire of 6.5 mm diameter at a temperature of 60°C is to be insulated by a material having k = 0.174 W/m°C. Convection heat transfer coefficient (ho) = 8.722 W/m2°C. The ambient temperature is 20°C. For maximum heat loss, what is the minimum thickness of insulation and heat loss per metre length? Also find percentage increase in the heat dissipation too. (M.U.) Solution. Refer to Fig. 2.89. 6.5 r1 3.25 mm 0.00325 m 2 k = 0.174 W/m°C, ho = 8.722 W/m2°C. For maximum heat loss, the minimum insulation thickness corresponds to critical radius of insulation which is given by k 0.174 rc 0.01995 m or 19.95 mm h0 8.722 ? Minimum insulation thickness = rc – r1 = 19.95 – 3.25 = 16.7 mm (Ans.) Heat loss per metre length : Case I. Without insulation : 2S L (t1 – tair ) 2S u 1(60 – 20) Q1 1 1 Fig. 2.89. ho ˜ r1 8.722 u 0.00325 = 7.124 W/m Case II. With insulation when critical thickness is used : 2S L (t1 – tair ) 2S u 1 u (60 – 20) Q2 ln (rc / r1 ) ln (0.01995 / 0.00325) 1 1   0.174 8.722 u 0.01995 k ho ˜ rc = 15.537 W/m Q2 – Q1 u 100 Q1

138

Chapter : 2

? Percentage increase in heat dissipation

Q2 – Q1 u 100 Q1

(15.537 – 7.124) (Ans.) u 100 118.09% 7.124 Example 2.74. A refrigerant suction line having outer diameter 30 mm is required to be thermally insulated. The outside air film coefficient of heat transfer is 12 W/m2°C. The thermal conductivity of insulation is 0.3 W/m°C. (i) Determine whether the insulation will be effective; (ii) Estimate the maximum value of thermal conductivity of insulating material to reduce heat transfer; (iii) Determine the thickness of cork insulation to reduce the heat transfer to 22 percent if the thermal conductivity of cork is 0.038 W/m°C.

Solution. Given : r0 0.038 W/m°C.

30 2

15 mm

0.015 m; ho = 12 W/m2 °C; kinsu. = 0.3W/m°C; kcork =

kinsu. 0.30 0.025 m or 25 mm or 25 mm ho 12 Since ro = 15 mm < rc, therefore, heat transfer will increase by adding this insulation and thus it is not effective. (Ans.) (ii) For insulation to be effective : ro t rc k 0.015 t insu. or, 12 or, kinsu d (0.015 × 12) = 0.18 W/m°C (Ans.) (iii) Consider unit length of pipe. For base pipe Q = ho.A.'t = ho × 2S ro × 1 × 't 't For pipe with cork-insulation, QC1 0.22 Q ln (rCI / ro ) 1  2S kCI 2S rCI ˜ ho (where rCI = radius of cork-insulation layer)

(i) The critical radius, rc

or,

't ln ( rCI / ro ) 1  2S kCI 2S rCI · ho

or,

ln (rCI / ro ) 1  2S kCI 2S rCI ˜ ho

or,

ln (rCI / ro ) 1  12 rCI kCI

or,

ln (rCI / 0.015) 1  0.038 12 rCI

0.22 u h0 u 2S r0 u 1 u ' t

1 0.22 u ho u 2S ro 1 0.22 u 12 u 0.015

25.25

25.25

Solving by trial and error, we have ? Thickness of cork-insulation

rCI = 0.036 m or 36 mm = rCI – ro = 36 – 15 = 21 mm (Ans.)

“Conduction” Heat Transfer at Steady State–One Dimension Example 2.75. A uniform sheathing of plastic insulation (k = 0.18 W/m°C) is applied to an electric cable of 8mm diameter. The convective film coefficient on the surface of bare cable as well as insulated cable was estimated as 12.5 W/m2°C and a surface temperature of 45°C was observed when the cable was directly exposed to ambient air 20°C. Determine : (i) The thickness of insulation to keep the wire as cool as possible; (ii) The surface temperature of insulated cable if the intensity of current flowing through the conductor remains unchanged. Solution. Refer to Fig. 2.90. 8 r1 4 mm 0.004 m; 2 k = 0.18 W/m°C ho = 12.5 W/m2°C

139

Fig. 2.90.

(i) Thickness of insulation : In order to keep the wire as cool as possible, the required condition corresponds to that for critical radius of insulation, k 0.18 i.e., r2 rc 0.0144 m or 14.4 mm ho 12.5 ? Thickness of insulation, = r2 – r1 = 14.4 – 4 = 10.4 mm (Ans.) (ii) Surface temperature of insulated cable, t2 : For a bare wire, the heat flow (per metre length) is given by Q1 = ho A 't = 12.5 × (2S × 0.004 × 1) × (45 – 20) = 7.85 W/m In case of the sheathed (insulated) cable, the heat flow (per metre length) is given as 2S L (t2 – 20) 2S u 1(t2 – 20) Q2 0.495 (t2 – 20) ln (r2 / r1 ) 1 ln (0.044 / 0.004) 1   ho ˜ r2 k 2.5 u 0.0144 0.18 Since the intensity of current flowing through the conductor remains unaltered, therefore, Q1 = Q2 = (I2R) 7.85 = 0.495 (t2 – 20) or

t2

20 

7.85 0.495

35.86qC

(Ans.)

2.9. HEAT CONDUCTION WITH INTERNAL HEAT GENERATION Following are some of the cases where heat generation and heat conduction are encountered : (i) Fuel rods – nuclear reactor; (ii) Electrical conductors; (iii) Chemical and combustion processes; (iv) Drying and setting of concrete.

140

Chapter : 2

It is of paramount importance that the heat generation rate be controlled otherwise the equipment may fail (e.g., some nuclear accidents, electrical fuses blowing out). Thus, in the design of the thermal systems temperature distribution within the medium and the rate of heat dissipation to the surroundings assumes ample importance / significance.

2.9.1. PLANE WALL WITH UNIFORM HEAT GENERATION Refer to Fig. 2.91. Consider a plane wall of thickness L (small in comparison with other dimension) of uniform thermal conductivity k and in which heat sources are uniformly distributed in the whole volume. Let the wall surfaces are maintained at temperatures t1 and t2. Let us assume that heat flow is onedimensional, under steady state conditions, and there is a uniform volumetric heat generation within the wall. Consider an element of thickness at a distance x from the left hand face of the wall. Heat conducted in at distance x, dt Qx – kA dx Heat generated in the element, Qg = A . dx . qg (where qg = heat generated per unit volume per unit time in the element) Heat conducted out at distance d Fig. 2.91. Plane wall uniform heat generation. Both the (Qx ) dx (x + dx), Q( x  dx ) Qx  surfaces maintained at a common temperature. dx As Qg represents an energy increase in the volume element, an energy balance on the element of thick dx is given by Qx + Qg = Q(x + dx) d (Qx ) dx Qx  dx d (Qx ) dx Qg or, dx d ª dt º – k A » dx q g . A . dx or, « dx ¬ dx ¼ – k A. d 2t

qg

d 2t dx 2

. dx

0 ...(2.88) k dx Eqn. (2.88) may also be obtained from eqn. (2.8) by assuming one-dimensional steady state conditions. The first and second integration of Eqn. (2.88) gives respectively

or,

2



“Conduction” Heat Transfer at Steady State–One Dimension

dt dx t

–

qg k qg

x  C1

141 ...(2.89)

˜ x 2  C1 x  C2 ... (2.90) 2k Case I. Both the surfaces have the same temperature : Refer to Fig. 2.92. At x = 0 t = t1 = tw, and At x = L t = t2 = tw (where tw = temperature of the wall surface). Using these boundary conditions in eqn. (2.90), we get qg ˜L C2 = tw and C1 2k Substituting these values of C1 and C2 in eqn. (2.90), we have qg 2 qg ˜ L ˜ x  tw t – x  2k 2k qg ( L – x) x  t w t or, ...(2.91) 2k In order to determine the location of the maximum temperature, differentiating the eqn. (2.91) w.r.t x and equating the derivative to zero, we have dt qg ( L – 2 x) 0 dx 2k qg Since, z 0, therefore, 2k L x L – 2x = 0 or 2 Thus the distribution of temperature given by eqn. (2.91) is the parabolic and symmetrical about L the midplane. The maximum temperature occurs at x and its value equals 2 ª qg º  tw tmax « ( L – x) x » ¬ 2k ¼x L –

2

ª qg § L · Lº or, « ¨ L – ¸ »  tw 2 ¹ 2¼ ¬ 2k © qg 2 ˜ L  tw tmax i.e. ...(2.92) 8k Heat transfer then takes place towards both the surfaces, and for each surface it is given by § dt · Q – kA ¨ ¸ © dx ¹ x 0 or x L ª qg º – kA « ( L – 2 x) » ¬ 2k ¼ x 0 or x L AL ˜ qg Q i.e., ...(2.93) 2 When both the surfaces are considered, AL Q 2u q g A . L . qg ...[2.93 (a)] 2 Also heat conducted to each wall surface is further dissipated to the surrounding atmosphere at temperature ta,

142 Thus,

Chapter : 2 AL ˜ qg 2

ta 

qg

˜L 2h Substituting this value of tw in eqn. (2.91), we obtain qg qg ˜L  ( L – x) x t ta  2h 2k At x = L/2 i.e., at the midplane : qg q g L2 t tmax ta  .L  2h 8k 2º ªL L  tmax ta  q g « or, » ¬ 2 h 8k ¼ or,

tw

hA (t w – ta )

...(2.94)

...(2.95)

...[2.95 (a)]

Fig. 2.92. Heat conduction in an insulated wall. Fig. 2.93. Plane wall with uniform heat generation— Both the surfaces of the wall having different temperatures.

The eqn. (2.95) also works well in case of conduction in an insulated wall Fig. (2.92). The following boundary conditions apply in the full hypothetical wall of thickness 2L : dt 0 At x = L dx At x = 2L t = tw The location x = L refers to the mid-plane of the hypothetical wall (or insulated face of given wall). Eqns. (2.91) and (2.92) for temperature distribution and maximum temperature at the mid- plane (insulated end of the given wall) respectively can be written as qg (2 L – x) x  tw t ...(2.96) 2k qg 2 tmax L  tw ...(2.97) 2k [Substituting L = 2L in eqn. (2.91) and (2.92)]

“Conduction” Heat Transfer at Steady State–One Dimension

143

Case II. Both the surfaces of the wall have different temperatures : Refer to Fig. 2.93 The boundary conditions are : At x = 0 t = tw1 At x = L t = tw2 Substituting these values in eqn. (2.90), we obtain the values of constant C1 and C2 as : qg t –t C2 = tw1; ˜L C1 w2 w1  L 2k Inserting these values in eqn. (2.90), we get qg 2 tw 2 – tw1 qg t – x  x L . x  tw1 2k 2k L qg qg 2 x L˜ x – x  (t w2 – t w1 )  t1 2k 2k L ª qg tw2 – tw1 º t « ( L – x)  or, ...(2.98) » x  tw1 L ¬ 2k ¼ The temperature distribution, in dimensionless form can be obtained by making the following transformations : qg 2 ª x § x ·2 º x t – tw2 L « – ¨ ¸ »  (tw2 – tw1 )  (tw1 – tw2 ) 2k ¬ L © L ¹ ¼ L ª x § x ·2 º x qg t – tw 2 L2 ˜ « – ¨ ¸ » – 1 or, t w1 – t w2 2k (t w1 – t w2 ) ¬ L © L ¹ ¼ L or,

t – tw 2 t w1 – t w2

L2 x ˜ 2k (t w1 – t w2 ) L

qg

˜

Replacing the parameter t – tw 2 t w1 – t w2

xº ª xº ª «¬1 – L »¼  «¬1 – L »¼

L2 (a constant) by a factor Z, we have 2k (t w1 – t w2 ) qg

˜

xª xº ª xº Z . «1 – »  «1 – » L¬ L¼ ¬ L¼

t – tw2 x º ªZ x ª º  1» 1 – »« « ¼ t w1 – t w2 ¬ L¼¬ L In order to get maximum temperature and its location, differentiating Eqn. (2.99) w.r.t x and equating the derivative to zero, we have dt x· § § Zx ·  1¸ (–1) 0 ¨1 – ¸ Z  ¨ © L ¹ d ( x / L) © L¹ Zx Zx – –1 0 or, Z – L L 2 Zx Z –1 or, L x Z – 1 ...(2.100) or, L 2Z Thus the maximum value of temperature x Z –1 occurs at and its value is given by: L 2Z

or,

...(2.99)

Fig. 2.94. Effect of factor Z on the temperature distribution in the plane wall.

144

or,

Chapter : 2 tmax – t w2 t w1 – t w2 tmax – tw2 tw1 – tw2

º Z – 1º ª ª § Z – 1· «1 – 2 Z » ¬« Z u ©¨ 2 Z ¹¸  1¼» ¬ ¼ § Z  1 ·§ Z  1 · ¸ ¨ ¸¨ © 2 Z ¹© 2 ¹

( Z  1) 2 ...(2.101) 4Z Fig. 2.94 shows the effect of factor Z on the temperature distribution in the plane wall.The following points emerge : a As the value of Z increases the slope of the curve changes; obviously the direction of heat flow can be reversed by an adequately large value of qg. a When Z = 0, the temperature distribution is linear (i.e., no internal heat generation). a When the value of Z is negative, qg represents absorption of heat within the wall/body. Case III. Current carrying electrical conductor : When electrical current passes through a conductor, heat is generated (Qg) in it and is given by UL Qg = I2R, where R A where, I = Current flowing in the conductor, R = Electrical resistance, U = Specific resistance or resistivity, L = Length of the conductor, and A = Area of cross-section of the conductor. Also, Qg = qg × A × L UL UL I 2U 1 u qg I 2 u qg u A u L I 2 u ? or, A A AL A2 2 2 J §I · qg ¨ ¸ U J 2U or, ...(2.102) © A¹ ke where, J = Current density; ke = Electrical conductivity (reciprocal of U). Example 2.76. The rate of heat generation in a slab of thickness 160 mm (k = 180 W/m°C) is 1.2 × 106 W/m3. If the temperature of each of the surface of solid is 120°C, determine : (i) The temperature at the mid and quarter planes; (ii) The heat flow rate and temperature gradients at the mid and quarter planes. Solution. Refer to Fig. 2.95. Thickness of slab, L = 160 mm = 0.16 m The rate of heat transfer, qg = 1.2 × 106 W/m3 Fig. 2.95.

“Conduction” Heat Transfer at Steady State–One Dimension

145

Thermal conductivity of slabs, k = 180 W/m°C The temperature of each surface, t1 = t2 = tw = 120°C (where tw = temperature of the wall surface) The temperature at the mid and quarter planes : The temperature distribution is given by:

t At mid plane: x

L 2

tmp

?

qg 2k

( L – x) x  tw

...[Eqn. (2.91)]

1.2 u 106 § L· L ¨ L – ¸ u  120 2 u 180 © 2¹ 2

1.2 u 106 § 0.16 · 0.16  120 ¨ 0.16 – ¸u 2 u 180 © 2 ¹ 2 At quarter planes : x = L/4 and x = 3L/4 or,

tmp

141.33° C (Ans.)

1.2 u 106 § L· L ¨ L – ¸ u  120 L / 4) 2 u 180 © 4¹ 4 6 1.2 u 10 § 0.16 · 0.16 or,  120 136° C (Ans.) tqp( x L / 4) ¨ 0.16 – ¸u u © 2 180 4 ¹ 4 1.2 u 106 § 3L · 3L  120 136° C (Ans.) Similarly tqp( x 3 L / 4) ¨L – ¸u 2 u 180 © 4 ¹ 4 (ii) The heat flow rate and temperature gradients at the mid and quarter planes : For unit area, i.e., 1m2 Heat flow : Q(x=L/2) = qg × A × x = 12 × 106 × 1 × (0.16/2) = 96000 W/m2 (Ans.) Q(x=L/4) = 1.2 × 106 × 1 × (0.16/4) = 48000 W/m2 (Ans.) dt dt Q Q – kA ˜ – Temperature gradients : or dx dx kA

?

tqp( x

§ dt · ¨ ¸ © dx ¹ x §d · ¨ ¸ © dx ¹ x

96000 180 u 1 48000 – 180 u 1

– L/2

L/4

– 533.3° C/m

(Ans.)

– 266.67° C/m

(Ans.)

Example 2.77. A meat slab of 25 mm thickness having k = 1W/m°C is heated with the help of microwave heating for roasting the meat slab. The centre temperature of the slab is maintained at 100°C when surrounding temperature is 30°C. The heat transfer coefficient on the surface of meat slab is 20W/m2°C. Find out the microwave heating capacity in W/m3. [N.U. Summer, 1998] Solution. Thickness of meat slab, L = 25 mm = 0.025 m Thermal conductivity of meat, k = 1W/m°C The centre temperature of the slab, tmax = 100°C Surrounding temperature, ta = 30°C The heat transfer coefficient; h = 20 W/m2°C.

146

Chapter : 2

The microwave heat capacity qg : The maximum temperature occurs at the centre of the slab only and is given by ªL L2 º  tmax ta  q g « ...[Eqn. 2.95 (a)] » ¬ 2 h 8k ¼ Substituting the proper values, we obtain ª 0.025 0.0252 º  100 30  qg « » 30  0.000703 qg 8u1 ¼ ¬ 2 u 20 100 – 30 qg 99573 W/m3 or 99.573 kW/m 3 ? (Ans.) 0.000703 Example 2.78. A plate 2 cm thick and 10 cm wide is used to heat a fluid at 30°C. The heat generation rate inside the plate is 7 × 106 W/m3. Determine the heat transfer coefficient to maintain the temperature of the plate below 180°C. Given k(plate) = 26 W/m°C. Neglect heat losses from the edge of the plate. (D.U.) Solution. Thickness of the plate, b = 2 cm = 0.02 m Temperature of fluid to be heated, ta = 30°C Heat generation rate, qg = 7 × 106 W/m3 Maximum temperature of the plate, tmax = 180°C Thermal conductivity of plate material, k = 26 W/m°C. Heat transfer coefficient, h : Using the following relation, we have ªL L2 º  tmax ta  q g « ...[Eqn. 2.95 (a)] » ¬ 2 h 8k ¼ ª 0.02 0.022 º  180 30  7 u 106 « » 8 u 26 ¼ ¬ 2h or,

0.02 2h

(180 – 30) 7 u 106

–

0.022 8 u 26

0.02 2 u 1.95 u 10 – 5 Example 2.79. The temperatures on the two surfaces of a 25 mm thick steel plate, (k = 48 W/m°C) having a uniform volumetric heat generation of 30 × 106 W/m3, are 180°C and 120°C. Neglecting the end effects, determine the following : (i) The temperature distribution across the plate; (ii) The value and position of the maximum temperature, and (iii) The flow of heat from each surface of the plate. Solution. Refer to Fig. 2.96. L = 25 mm = 0.025 m; tw1 =180°C; tw2 = 120°C qg = 30 × 106 W/m3; k = 48 W/m°C. h

1.95 u 10 – 5

512.8 W/m 2 ° C (Ans.)

Fig. 2.96.

“Conduction” Heat Transfer at Steady State–One Dimension

147

(i) The temperature distribution across the plate : The temperature distribution, when both the surfaces of the wall have different temperatures, is given by ª qg (t – t º t « ( L – x)  w2 w1 » x  tw1 ...[Fig. (2.98)] L ¬ 2k ¼ Substituting the values, we have

t

ª 30 u 106 (120 – 180) º (0.025 – x)  « » x  180 0.025 ¼ ¬ 2 u 48 = [312500 (0.025 – x) – 2400]x + 180 = [7812.5 – 312500x – 2400]x + 180

or,

t = 180 + 5412.5x – 312500 x2 ...Required temperature distribution. (Ans.)

(where temperature t is in °C and the distance x is in metres) The temperature distribution is parabolic. (ii) The value and position of the maximum temperature; tmax, x : In order to determine the position of maximum temperature, differentiating the above expression and equating it to zero, we obtain dt dx

5412.5 – 625000 x

0

5412.5 0.00866m or 8.66mm (Ans.) 625000 The value of maximum temperature, tmax = 180 + 5412.5 × 0.00866 – 312500 × 0.008662 = 203.44°C (Ans.) (iii) The flow of heat from each surface of the plate, q1, q2 : The heat flow at the left face (x = 0) § dt · q1 – k A ¨ ¸ © dx ¹ x 0 = – 48 × 1 × (5412.5 – 625000 x)x = 0 = – 259800 W/m2 (Ans.) The negative signs signifies that the heat flow at the left face is in a direction opposite to that of measurement of the distance. The heat flow at the right face, q2 = 48 × 1 × (5412.5 – 625000 x)x = 0.025 = 48 × 1(5412.5 – 625000 × 0.025) = – 490200 W/m2 (Ans.) Check : The sum of q1 and q2 must be equal to total heat generated per unit length of plate. Now q1 + q2 = 259800 + 490200 = 750000 W/m2, and Qg = 30 × 106 × (0.025 × 1) = 750000 W/m2 i.e. q1 + q2 = Qg

?

x

Example 2.80. A plane wall is 1m thick and it has one surface (x = 0) insulated while the other surface (x = L) is maintained at a constant temperature of 350°C. The thermal conductivity of wall is 25 W/m°C and a uniform heat generation per unit volume of 500 W/m3 exists throughout the wall. Determine the maximum temperature in the wall and the location of the plane where it occurs. (D.U.)

148

Chapter : 2

Solution. Refer Fig. 2.97. L = 1 m; t1 = 350°C; k = 25 W/m°C, Heat generated per unit volume per unit time, qg = 500 W/m3. Maximum temperature and its location, tmax, x: The differential equation controlling heat flow by conduction when the body is generating heat is given by d 2t qg  0 ...(i) k dx 2 Integrating eqn. (i) twice, we have dt qg  ˜ x C1 ...(ii) dx k qg x 2 ˜ t C1 x  C2 ...(iii) k 2 (where C1, C2 = constants of integration). In order to evaluate C1 and C2, using the following boundary conditions, we have dt 0 At x = L, dx ? C1 = 0 At x = L, t = t1 q g L2 ˜ C2 t1  ? [From eqn. (iii)] k 2 Substituting the values of C1 and C2 in eqn. (iii), we have qg x 2 q g L2 ˜ ˜ t t1  k 2 k 2 q g ª L2 x2 º t t1  – or, « » k ¬2 2¼ 2 q g L2 ª §x· º ˜ «1 – ¨ ¸ » ©2¹ ¼ k 2 ¬ For the maximum temperature in the wall, q g L2 ª 2 x º dt ˜ 0 – dx k 2 «¬ L2 »¼

or,

and

t

Fig. 2.97.

[From eqn. (ii)]

t1 

d 2t 2

–

...(iv)

or

x=0

qg

k dx Hence temperature will be maximum at x = 0 Substituting x = 0 in eqn. (iv), we have 500 1 u 360° C 25 2 Example 2.81. A plane wall 90 mm thick (k = 0.18 W/m°C) is insulated on one side while the other side is exposed to environment at 80°C. The rate of heat generation within the wall is 1.3 × 105 W/m3. If the convective heat transfer coefficient between the wall and the environment is 520 W/ m2°C, determine the maximum temperature to which the wall will be subjected. tmax

350 

“Conduction” Heat Transfer at Steady State–One Dimension

149

Solution. Refer Fig. 2.98 L = 90 mm = 0.09 m, k = 0.18 W/m°C h = 520 W/m2°C Temperature of environment, ta = 80°C The rate of heat generation, qg = 1.3 × 105 W/m3 Maximum temperature, tmax : One dimensional, steady state heat conduction equation is given by d 2t qg  0 ..(i) k dx 2 Integrating the above equation twice, we have qg dt – ˜ x  C1 ...(ii) dx k Fig. 2.98.

qg x 2  C1 x  C2 ...(iii) t – ˜ k 2 (where C1, C2 = constants of integration) In order to evaluate C1 and C2, using the following boundary conditions, we have dt 0 (i) At x = 0, ? C1 = 0 dx (ii) At x = L, the heat conduction equals the convective heat flow to the environment. dt h A[t( L ) – ta ] – kA i.e dx x L dt h – [t( L ) – ta ] ...(iv) dx x L k Again from eqn (ii), qg dt – .L ...(v) dx k x

L

From eqns. (iv) and (v), we obtain h – ª¬t( L ) – ta º¼ k

t( L)

–

qg k

ta 

˜L qg h

.L

Substituting into eqn. (iii), we have

t( L)

ta 

qg h qg

˜L

–

qg 2k

qg

˜ L2  C2

˜ L2 h 2k Inserting the values of constants C1 and C2 in eqn. (iii), we get qg qg qg 2 ˜L  ˜L t – ˜ x 2  ta  h 2k 2k

?

C2

ta 

˜L 

...(vi)

150 or,

Chapter : 2

t

ta 

qg

˜L 

qg

( L2 – x 2 )

h 2k The maximum temperature occurs at the insulated wall boundary i.e., x = 0 qg qg 2 ˜L tmax ta  .L  ? h 2k Substituting the proper values, we have 1.3 u 105 1.3 u 105 tmax 80  u 0.09  u (0.09)2 520 2 u 0.18 = 3027.5°C (Ans.)

...(vii)

...(viii)

Example 2.82. A plane wall ‘X’ (k = 75 W/m°C) is 60 mm thick and has volumetric heat generation of 1.5 × 106 W/m3. It is insulated on one side while the other side is in contact with surface of another wall ‘Y’ (k = 150 W/m°C) which is 30 mm thick and has no heat generation. The non-contact surface of wall ‘Y’ is exposed to a cooling fluid at 20°C. If the convective heat transfer coefficient between wall ‘Y’ and fluid is 950 W/m2°C, determine : (i) The temperature at the insulated surface; (ii) The temperature at the cooled surface of the composite wall. Solution. Refer Fig. 2.99. kX = 75 W/m°C; LX = 60 mm = 0.06 m kY = 150 W/m°C; LY = 30 mm = 0.03 m h = 950 W/m2°C; tcf = 20°C. (i) The temperature at the insulated surface, to : The heat generated in wall X per unit area, Qg = qg . A . LX = 1.5 × 106 × 1 × 0.06 = 90000 W/m2 As the left face of the wall X is insulated, the flow of heat through this surface is nil and all the heat generated would pass through wall Y and finally dissipated to cooling fluid. i.e. 90000 = h.A (t2 – tcf) = 950 × 1 × (t2 – 20) 90000  20 114.7q C t2 or, 950

Fig. 2.99.

“Conduction” Heat Transfer at Steady State–One Dimension

151

Cosidering heat flow through wall Y, we have kY ˜ A (t1 – t2 ) Q LY 150 u 1 u (t1 – 114.7) 5000(t1 – 114.7) 0.03 90000  114.7 132.7qC t1 or, 5000 The temperature of the insulated surface of the wall X is given by, qg ˜ LX2  t1 to ...[Refer eqn. (2.97)] 2k X 6 1.5 u 10 u (0.06)2  132.7 168.7° C to or, (Ans.) 2 u 75 Example 2.83. The inside surface (x = 0) of a flat plate is insulated and the outside surface (x = L) is maintained at a uniform temperature T2, and the heat generation term is in the form of g (x) = g0 e–Jx W/m3 where go and J are constants and x is measured from the insulated inside surface. Develop : (i) An expression for the temperature distribution in the plate, and (ii) An expression for the temperature at the insulated surface (i.e., x = 0) of the plate. (PTU) Solution. The applicable differential equation is d 2t qg  0 ...[Eqn. (2.88)] k dx 2 g 0 e – Jx d 2t  0 or, k dx 2 g 0 e – Jx d 2t or, k dx 2 Integrating w.r.t. x, we get g dt e – Jx  C1 – 0 u –J dx k dt 0 (insulated) At x = 0, dx g C1 – 0 ? kJ g g dt e – Jx – 0 u – 0 ? Fig. 2.100. –J dx k kJ Integrating again w.r.t. x, we have g g e – Jx – 0 x  C2 t  0 u k J (– J ) k J At x = L, t = T2 Substituting the values, we get g g T2 – 02 ˜ e – JL – 0 L  C2 J k kJ g g or, C2 T2  02 e – JL  0 L J k kJ

or,

90000

152

Chapter : 2

g0 g g x  T2  02 e – JL  0 L kJ kJ kJ kJ (i) Thus, expression for this temperature distribution in the plate, g g t T2  0 ( L – x)  02 (e – JL – e – Jx ) (Ans.) kJ kJ (ii) The expression for temperature at the insulated surface (x = 0) is : g g t0 T2  0 L  02 (e – JL – 1) (Ans.) kJ kJ Example 2.84. A sheet of glass coated on one side with an exceedingly thin but transparent electrically conducting layer is mounted vertically in a room where air is at 20°C. An electric current is passed through the conducting layer to generate heat at the rate of 2.5 kW/m2 of conducting material. If the mean heat transfer coefficient on both sides of the sheet is 10 kW/m2 °C, find the temperature on both sides of the glass. The glass thickness is 6 mm and neglect the thickness of conducting layer. Take k (glass) = 0.7 W/m K. (A.U.) 2 2 2 Solution. Given : ta = 20°C; q = 2.5 kW/m = 2500 W/m ; h = 10 W/m °C; k = 0.7 W/m K; L = 6 mm = 0.006 m Temperatures, t1, t2 : Consider surface area of 1 m2. h (t1 – ta) + h (t2 – ta) = 2500 ...(i) or, h (t1 + t2 – 2 ta) = 2500 k (t1 – t2 ) h (t1 – ta )  2500 ...(ii) L Substituting the values in eqn. (ii), we get 0.7 (t1 – t2 ) 10 (t1 – 20)  2500 0.006 700 (t1 – t2 ) 2500 or, 10 (t1 – 20)  6 or, (t1 – 20) + 11.67 (t1 – t2) = 250 or, 12.67 t1 = 270 + 11.67 t2 or, t1 = 21.3 + 0.92 t2 Substituting the value of T1 in eqn. (i), we get 10 [(21.3 + 0.92 t2) + t2 – 2 × 40] = 2500 or, 1.92 t2 – 58.7 = 250 Fig. 2.101. 250  58.7 t2 160.78° C (Ans.) or, 1.92 t1 = 21.3 + 0.92 × 160.78 = 169.22°C (Ans.) Example 2.85. A plane wall is a composite of three materials a, b and c. The wall of materials generates heat at a rate of 2 × 106 W/m2. The other details are : La = 5 cm; Lb = 3 cm; Lc = 1.5 cm; ka = 190 W/mK; kb = 150 W/m K and kc = 50 W/m K. The inner surface of material ‘a’ is well insulated and outer surface of material ‘c’ is cooled by water stream with tw = 50°C, and h = 2000 W/m2°C. The wall materials of ‘b’ and ‘c’ have no heat generation. Determine the temperature of insulated surface and temperature of cooled surface. (N.M.U.)

?

L

–

g0

2

˜ e – Jx –

“Conduction” Heat Transfer at Steady State–One Dimension 106

153

W/m2;

Solution. Given : qg = 2 × La = 5 cm = 0.05 m; Lb = 3 cm = 0.03 m; Lc = 1.5 cm = 0.015 m; ka = 190 W/m K; kb = 150 W/m K; kc = 50 W/m K; tw = 50°C; h = 200 W/m2°C. Under steady state condition, the heat generated in slab is passed through a composite slab. Q qg ? = Heat generated per unit surface area of a. A ? Volume of slab ‘a’ per m2 of surface area, V = 1 × 0.05 = 0.05 m3

Fig. 2.102.

? Heat generated V × qg = 0.05 × 2 × 106 = 105 W/m2 (t 2 – t w ) Q Lb L 1  c  kb kc h0 t2 – 50 (t2 – 50) (t2 – 50) 105 –4 –4 –4 0.03 0.015 1 2 u 10  3 u 10  5 u 10 10 – 3   150 50 2000 ? t2 = 105 × 10–3 + 50 = 150°C We can also write t2 – t3 t3 – t4 q Lb Lc kb kc L 0.03 t3 t2 – 105 u b 150 – 105 u 150 – 20 130q C ? kb 150 L 0.015 100° C (Ans.) t4 t3 – 105 u c 130 – 105 u ? kc 50 The temperature distribution in a generating slab is given by d 2t qg  0 [Eqn. (2.88)] k dx 2 The first and second integration of the above equation gives respectively, dt qg  ˜ x C1 ...(i) dx k

154

Chapter : 2

qg x2 ˜ C1 x  C2 k 2 The boundary conditions are : t

1. 2.

...(ii)

dt 0 at x = 0 ? C1 = 0 dx t = 150°C at x = 0.05 m

2 u 106 0.052 u 163.16 190 2 ? The temperature distribution in wall ‘a’ is given by qg x 2 u 163.16 t [From eqn. (ii)] 2 ka The temperature at insulated surface can be calculated by substituting x = 0 in the above eqn. ? t1 = 163.16° C. (Ans.) Example 2.86. A copper bar (conductor) 80 mm × 6 mm in cross-section (k = 370 W/m°C) is lying in an insulation trough so that the heat transfer from one face and both the edges is negligible. It is observed that when a current of 5000A flows through the conductor, the bare face has a constant temperature of 45°C. If the resistivity of copper is 2 × 10–8 :m, determine : (i) The maximum temperature which prevails in the bar and its location; (ii) The temperature at the centre of the bar. Solution. Refer Fig. 2.103. Given : Cross-section of the conductor = 80 mm × 6 mm or 0.08 m × 0.006 m Thermal conductivity of copper, k = 370 W/m°C Resistivity of copper, U = 2 × 10–8 :m Temperature of bare face, t = 45°C Current flowing in the conductor, I = 5000 A (i) The value of maximum temperature and its location : In case of one-dimensional and steady state heat flow, the heat conduction equation may be written as : d 2t qg  0 ...(i) k dx 2 (where qg = rate of heat generation per unit volume per unit time) Integrating eqn. (i), twice, we have Fig. 2.103. qg dt – x  C1 ...(ii) dx k qg x 2  C1 x  C2 t – ˜ ...(iii) k 2 (where C1, C2 = constants of integration). The values of the constants C1 and C2 can be found out by using the following boundary conditions :

?

C2

150 

“Conduction” Heat Transfer at Steady State–One Dimension

155

dt 0 ?C =0 (ii) At x = 0.006 m, t = 45°C 1 dx Amount of heat generated per unit time, Ul Qg I 2 R I 2 u a 2 Qg 1 §I· 2 § Ul · u u q I ¨ ¸ ¨ ¸ ˜U ? g aul © a ¹ al © a ¹ (where a = cross-sectional area of the conductor through which current is flowing.) 2 5000 § · –8 qg ¨ 2.17 u 106 W/m3 or, ¸ u 2 u 10 © 0.08 u 0.006 ¹ ? From eqn. (iii), we have qg x 2 2.17 u 106 0.0062 ˜ u C2 t  45  45.105 k 2 370 2 Substituting the values of C1 and C2 in eqn. (iii), we get the expression for temperature distribution as follows : qg x 2  45.105 t – ˜ ...(iv) k 2 The maximum temperature occurs at the insulated face (x = 0) and its value is tmax = 45.105°C [Substituting x = 0 in eqn. (iv)] (ii) The temperature at the centre of the bar, tmid : 0.006 x 0.003m At the midpoint, 2 2.17 u 106 0.0032 u  45.105 45.078° C tmid – ? (Ans.) 370 2 [Substituting x = 0.003 m in eqn (iv)]

(i) At x = 0,

2.9.2. DIELECTRIC HEATING Dielectric heating is a method of quickly heating insulating materials packed between the plates (of an electric condenser) to which a high frequency, high voltage alternating current is applied. The temperature rise is uniform and the internal heat generated is of equal intensity on the surface and the core. The heat generated : – is directly proportional to the area of condenser plates, voltage and frequency;

Fig. 2.104. Dielectric heating.

156

Chapter : 2

– is inversely proportional to the distance between the plates; – per unit volume of the material is constant. The method of dielectric heating entails the following advantages : (i) More economical, (ii) No pollution (iii) High efficiency and (iv) Greater safety. Dielectric heating refers to the situation that corresponds to the conduction in a plane wall with uniform internal heat source. Consider an insulating material (say wool) placed between the two plates of an electric condenser as shown in the Fig. 2.104. Let, L = Thickness of slab (insulating material), T1 = Temperature of electrode (1) above ambient temperature [= (tw1 – ta)], T2 = Temperature of electrode (2) above the ambient temperature [= (tw2 – ta)], h1, h2 = Heat transfer coefficients on the surfaces of plates of the electric condenser, and qg = Heat generated per unit volume of dielectric medium (wool). Consider an element of thickness dx at a distance x from the left hand plate (see Fig. 2.104); T being the difference between the temperature of wool and the ambient temperature. dT dx Heat generated due to dielectric heating, Qg = qg A. dx

Heat conducted in at a distance x, Qx

Heat conducted out at distance ( x  dx), Q( x  dx ) For steady state heat conduction, we have

or,

Qx  qg . A . dx

or,

q g . A dx

d 2T

qg

– k . A.

Qx 

d (Qx ) dx dx

Qx + Qg = Q(x + dx) d (Qx ) dx Qx  dx d ª dTº –k A dx dx «¬ dx »¼ d 2T – k A 2 dx dx

0 ...(i) k dx Integrating eqn. (i) twice, we have d T qg  . x C1 ...(ii) dx k 2 qg x T ˜ C1 x  C2 ...(iii) k 2 (where C1 and C2 = constants of integration). C1 and C2 are evaluated from known boundary conditions as follows : Heat conducted at x = 0 = heat convected to the surrounding at ambient temperature ta dT h1 A (tw1 – ta ) – kA i.e. dx x 0 kAC1 = h1AT1 [Using Eqn. (ii)] h1 T1 C1 or, k q g x 2 h1 T1 T ˜ ˜ x  C2 ? k 2 2

or,

2



“Conduction” Heat Transfer at Steady State–One Dimension Also, at x = 0, T = T1 ? C2 = T1 Thus temperature distribution may be written as qg x 2 h T T – ˜  1 1 ˜ x  T1 k 2 k [substituting the values of C1 and C2 in eqn (iii)] Again, At x = L, T = T2 q g L2 h1T1 T2 – ˜  ˜ L  T1 ? k 2 k

157

...(iv) ...Electrode (2) ...(2.103)

[Substituting x = L in eqn. (iv)] Under steady state conditions : Total heat generated within insulating material = surface heat loss from both the electrodes ? qgA.L = h1A (tw1 – ta) + h2A (tw2 – ta) ...(2.104) or, qg L = h1T1 + h2T2 The values of the temperature tw1 and tw2 can be found out by solving eqns. (2.103) and (2.104). Example 2.87. A 70 mm thick slab of insulation material (k = 0.42 W/m°C) is placed between and is in contact with two parallel electrodes, and is then subjected to dielectric heating (high frequency) at a uniform rate of 40800 W/m3. On the attainment of steady state conditions, the coefficients of combined radiation and convection for the exposed surfaces are 12.5 W/m2°C and 14.5 W/m 2 °C respectively. If the ambient temperature is 20°C, determine : (i) Surface temperatures; (ii) Location and magnitude of maximum temperature in the system. Assume the flow of heat to be unidirectional and each electrode to be at a uniform temperature equal to that of the slab with which it is in contact. Solution. Refer to Fig. 2.104. L = 70 mm = 0.07 m; k = 0.42 W/m°C; qg = 40800 W/m3 h1 = 12.5 W/m2°C; h2 = 14.5 W/m2°C; ta = 20°C. (i) Surface temperatures tw1, tw2 : The temperature distribution, in case of dielectric heating, is given by q g x 2 h1T1 T – ˜  ˜ x  T1 k 2 k 40800 x 2 12.5 T1 ˜  x  T1 – 0.42 2 0.42 or, T = – 48571.4x2 + 29.76 T1x + T1 ...(i) At x = 0.07m, T = T2 ? T2 = – 48571.4 × (0.07)2 + 29.76 × 0.07 T1 + T1 or, T2 = – 238 + 3.08 T1 ...(ii) Under steady state conditions, Heat generated due to dielectric heating = heat lost due to convection from the electrode surfaces or, qg.A.L = h1.A.T1 + h2.A.T2 or, qg.L = h1.T1 + h2.T2 or, 40800 × 0.07 = 12.5 T1 + 14.5 T2

158

Chapter : 2

or, 196.96 = 0.862 T1 + T2 or, T2 = 196.96 – 0.862T1 ...(iii) From eqns. (ii) and (iii), we have – 238 + 3.08 T1 = 196.96 – 0.862 T1 (238  196.96) T1 110.34q C or, (3.08  0.862) and T2 = – 238 + 3.08 × 110.34 = 101.85°C [From eqn. (ii)] Thus, the electrode temperatures are : T1 = tw1 – ta ? tw1 = T1 + ta = 110.34 + 20 = 130.34°C (Ans.) T2 = tw2 – ta ? tw2 = T2 + ta = 101.85 + 20 = 121.85°C (Ans.) (ii) Location and magnitude of maximum temperature; x, tmax : The location of maximum temperature can be obtained by differentiating eqn. (i) w.r.t. x and equating the derivative to zero. Thus, or, or, x ?

29.76 T1 48571.4 u 2

dT d ª 2 ¬ – 48571.4 x  29.76 T1 ˜ x  T1 º¼ dx dx dT – 48571.4 u 2 x  29.76 T1 0 dx 29.76 u 110.34 0.0338 m or 33.8 mm (from left hand electrode) 48571.4 u 2 tmax = – 48571.4 × 0.03382 + 29.76 × 110.4 × 0.0338 + 110.4 = 165.96°C (Ans.)

2.9.3. CYLINDER WITH UNIFORM HEAT GENERATION Refer to Fig. 2.105. Consider a cylindrical rod in which one-dimensional radial conduction is taking place under steady state conditions. Let, R = Radius of the rod, L = Length of the rod, k = Thermal conductivity (uniform), qg = Uniform volumetric heat generation per unit volume per unit time, h = Heat transfer coefficient, and ta = Ambient temperature. In order to obtain temperature distribution, consider an element of radius r and thickness dr as shown in Fig. 2.105. Heat conducted in at radius r, dt Qr – k .2S r L . dr Heat generated in the element, Qg = qg.2Sr.dr.L Heat conducted out at radius, r + dr, d Q( r  dr ) Qr  (Qr ) dr dr Under steady state conditions, Qr + Qg = Q(r + dr) d (Qr ) dr Qr  dr

“Conduction” Heat Transfer at Steady State–One Dimension

159

d (Qr ) dr dr d ª dt º – k .2 S r L . » dr q g .2S r . dr . L « dr ¬ dr ¼ qg d ª dt º – ˜r r. or ...(2.105) dr «¬ dr »¼ k [Eqn. (2.105) may also be obtained from eqn. 2.22 assuming steady state uni-directional heat conduction in radial direction]. Intergrating the above equation twice, we obtain qg r 2 dt  C1 r. – ˜ dr k 2 qg r C dt – ˜  1 or, ...(2.106) dr k 2 r qg r 2  C1 log e r  C2 t – ˜ k 4 ...(2.107) (where C1 and C2 = constants of integration). The constants C1 and C2 are evaluated from the boundary conditions, as follows : (i) At r = R, t = tw (ii) Heat generated = Heat lost by conduction at the rod surface Fig. 2.105. Heat conduction in a i.e. solid cylinder with heat generation. ª dt º q g u (S R 2 u L ) – k u 2SRL u « » ¬ dr ¼ r R dt 0 Also, at r = 0, dr Since in case of a cylinder, centre line is line of symmetry for temperature distribution and as dt (temperature gradient) must be zero. such dr § dt · The temperature gradient ¨ ¸ at the surface (i.e. at r = R) is given by © dr ¹ qg R C ª dt º – ˜  1 «¬ dr »¼ k 2 R r R

?

Qg

Also from boundary condition (ii), we have qg R ª dt º – ˜ «¬ dr »¼ k 2 r

R

qg R C qg R – ˜  1 – ˜ ? or C1 = 0 k 2 R k 2 Applying the boundary condition (i) [i.e. at r = R, t = tw] to eqn. (2.107), we obtain, tw

–

qg R 2 ˜  C2 k 4

160

Chapter : 2

qg R 2 ˜ k 4 Substituting the values of C1 and C2 in eqn. (2.107), we have the general solution for temperature distribution as qg r 2 qg R 2  tw  ˜ t – ˜ k 4 k 4 qg 2 t tw  [R – r2 ] or, ...(2.108) 4k It is evident from eqn. (2.108) that temperature distribution is parabolic and the maximum temperature occurs at the centre of the rod (r = 0) and its value is given by qg 2 ˜R tmax tw  ...(2.109) 4k By combining eqns. (2.108) and (2.109), we arrive at the following dimensionless form of temperature distribution : qg 2 2 (R – r 2 ) t – tw R2 – r 2 §r· 4k 1 – ¨ ¸ qg 2 ©R¹ tmax – tw R2 ˜R 4k 2 t – tw §r· 1–¨ ¸ i.e. ...(2.110) ©R¹ tmax – t w Also, energy generated within the rod (per unit time) = Energy dissipated (per unit time) by convection at the rod boundary i.e. qg × (SR2 × L) = h × 2SRL (tw – ta) qg .R t w ta  or, ...(2.111) 2h Inserting the value of tw in eqn. (2.108), we obtain the temperature distribution (in terms of ta) as qg qg 2 ˜R  [R – r 2 ] t ta  ...(2.112) 2h 4k The value of tmax, at r = 0, is given by qg qg 2 ˜R  ˜R tmax ta  ...(2.113) 2h 4k Example 2.88. A current of 300 amperes passes through a stainless steel wire of 2.5 mm diameter and k = 20 W/m°C. The resistivity of the wire is 70 × 10–8 :m and the length of the wire is 2m. If the wire is submerged in a fluid maintained at 50°C and convective heat transfer coefficient at the wire surface is 4000 W/m2°C, calculate the steady state temperature at the centre and at the surface of wire. (M.U)

or,

C2

tw 

Solution. Refer to Fig. 2.106. 2.5 R 1.25 mm 0.00125 m; 2 k = 20 W/m°C, resistivity, U = 70 × 10–8 :m L = 2m, ta = 50°C, Current, I = 300 amp.

Fig. 2.106.

“Conduction” Heat Transfer at Steady State–One Dimension

161

Temperature at the surface of wire (tw) and at the centre of wire (tmax) : Rate of heat generation, UL Qg I 2 Re I 2 u A (where Re = electrical resistance) Rate of heat generation per unit volume, 2 Qg UL 1 §I · u U¨ ¸ qg I2u © A¹ AL A AL 2 300 ª º 70 u 10 – 8 « 2» ¬ S u 0.00125 ¼ = 26.14 × 108 W/m3 Temperature at the surface of wire is given by qg ˜R t w ta  ...[Eqn. (2.111)] 2h 26.14 u 108  u 0.00125 458.44° C (Ans.) t 50 or, w 2 u 4000 Temperature at the centre of wire is given by qg 2 ˜R tmax tw  [Eqn. (12.109)] 4k 26.14 u 108 tmax 458.44  u (0.00125)2 509.5° C (Ans.) or, 4 u 20 Example 2.89. A 3 mm diameter stainless steel wire (k = 20 W/m°C, resistivity, U = 10 × 10–8 :m) 100 metres long has a voltage of 100 V impressed on it. The outer surface of the wire is maintained at 100°C. Calculate the centre temperature of the wire. If the heated wire is submerged in a fluid maintained at 50°C, find the heat transfer coefficient on the surface of the wire. (M.U.) 3 1.5 mm 0.0015 m Solution. Radius of stainless steel wire, R 2 Length of the wire, L = 100 m Voltage impressed = 100 V Thermal conductivity, k = 20 W/m°C Resistivity, U = 10 × 10–8 :m The temperature of the outer surface of the wire, tw = 100°C Fluid temperature, ta = 50°C. Centre temperature of the wire, tmax : UL 10 u 10 – 8 u 100 1.415 : A S u 0.00152 V 2 1002 Q VI 7067 W Rate of heat generation, g Re 1.415 ? Rate of heat generation per unit volume Qg 7067 9.998 u 106 W/m3 qg AL S u 0.00152 u 100

Electrical resistance of the wire, Re

162

Chapter : 2

The centre temperature is given by

tmax

qg

˜ R2 4k 9.998 u 106 u 0.00152 100.28q C 100  4 u 20

tw 

...[Fig. (2.109)]

Heat transfer coefficient, h :

tw 100

or,

(100 – 50)

ta 

qg

˜R 2h 9.998 u 106 u 0.0015 50  2h 7498.5 h

...[Eqn. (2.111)]

7498.5 (Ans.) 149.97 W/m 2 ° C 50 Example 2.90. One metre long Nichrome wire of resistivity 1z:m is to dissipate power of 10 kW in the surrounding fluid which is at 80°C. Find the diameter of the wire if the maximum operating temperature of the wire is 1000°C. (M.U.) Take h = 1000 W/m2°C and k (wire) = 60 W/m°C. Solution: Length of wire, L = 1 m Resistivity of wire material, U = 1z:m Power to be dissipated, P = 10 kW Temperature of surrounding fluid, ta = 80°C Maximum operating temperature of the wire, tmax = 1000°C Heat transfer coefficient, h = 1000 W/m2 °C Thermal conductivity of wire, k = 60 W/m2 °C. Diameter of the wire, D : The maximum temperature is given by qg qg uR u R2 tmax ta  ...[Eqn.(2.113)] 2h 4k (where, R = radius of the wire). The rate of heat generation per unit volume P 10 u 1000 10 u 1000 10000 qg Volume of wire S R2 u L S u R2 u 1 S R2 Substituting the relevant values in the above equation, we have 10000 10000 uR u R2 1000 80  2 SR u 2 u 1000 S R 2 u 4 u 60

?

h

1.59  13.26 R or, R = 1.75 × 10–3 m or 1.75 mm ? Diameter of the wire, D = 2 × 1.75 = 3.5 mm (Ans.) Example 2.91. The meat rolls of 25 mm diameter having k = 1 W/m°C are heated up with the help of microwave heating for roasting. The centre temperature of the rolls in maintained at 100°C when the surrounding temperature is 30°C. The heat transfer coefficient on the surface of the meat roll is 20 W/m2 °C. Find the microwave heating capacity required in W/m3. (1000 – 80)

“Conduction” Heat Transfer at Steady State–One Dimension

163

25 12.5 mm or 0.0125 m; k = 1W/m°C; tmax = 100° C; ta = 30°C; 2 h = 20 W/m2°C. Microwave heating capacity, qg : The maximum temperature occurs at the centre and is given by qg qg uR u R2 tmax ta  ...[Eqn. (2.113)] 2h 4k qg qg u 0.0125  u 0.01252 100 30  2 u 20 4u1 (100 – 30) = 0.0003125 qg + 0.00003906 qg (100 – 30) qg 1.991 u 105 W/m 3 or, (0.0003125  0.00003906) Example 2.92. (a) Prove that the maximum temperature at the centre of wire, carrying electrical current, is given by the relation :

Solution. R

J2 ˜ R2 4k . ke where tw = surface temperature; J = current density; k, ke = thermal and electrical conductivities of the wire material respectively; R = radius of the wire. (b) A 3 mm dia. copper wire 10 m long, is carrying electric current and has a surface temperature of 25°C. The thermal and electrical conductivities of copper are 375 W/m°C and 5.1 × 107 :m respectively. Determine the voltage if the temperature rise at the wire axis is limited to 15°C. Solution. (a) The maximum temperature in case of a cylindrical wire conductor occurs at the centre and is given by [Eqn. (2.109)]. qg 2 ˜R tmax tw  ...(i) 4k U L § I 2L · 1 2 2 Total volumetric heat generated I Re I u ¨ ¸ A © A ¹ ke 1 [where, Re = electrical resistance of the conductor, U electrical resistivity ] ke ? Heat generated per unit volume, tmax

tw 

§ I 2L · 1 ¨ ¸ © A ¹ ke AL

qg (where J = I/A) From Eqns. (i) and (ii), we have

tmax 3 1.5 mm 2 ke = 5.1 × 107 :m

(b) Given : R

Voltage drop :

J2

tw 

2

1 §I· ¨ ¸ u © A ¹ ke

J2 ˜ R2 4 k ke

J2 ke

...(ii)

(Proved)

...(iii)

0.0015 m; L = 10 m; tmax – tw = 15°C; k = 375 W/m°C (tmax – tw ) u 4 k ke R2 15 u 4 u 375 u 5.1 u 107 (0.0015)

2

[From eqn. (iii)] 5.1 u 1017

164

Chapter : 2 J = 7.14 × 108 amperes/m2

? Current density,

JL 7.14 u 108 u 10 §UL· ( J . A). ¨ ¸ J UL © A ¹ ke 5.1 u 107 = 140 volts (Ans.) Example 2.93. The heat generating rate per unit volume, qg, at any radius, r, of a solid rod is given by 2 ª §r· º qg q0 «1 – ¨ ¸ » ©R¹ ¼ ¬ where q0 is the heat generation rate at the centre of the rod of radius R. Find out an expression for maximum temperature at the centre of a rod 40 cm diameter, when heat generating rate at the centre is 24 × 106 kJ/m3-h and temp. at the surface of the rod is 20°C. Assume K for the material of the rod as 200 kJ/m-h-°C. (AMIE) Solution. The governing differential equation is qg d ª dt º r ˜ » – ˜r [Eqn. (2.105)] « dr ¬ dr ¼ k

Voltage drop

IRe

2 d ª dt º r ­ §r· ½ ® ˜  r q 1 – ¨ ¸ ¾ 0 dr «¬ dr »¼ k ¯ ©R¹ ¿

0

(substituting for qg)

Integrating, we get r˜

ª r2 r4 º « – »  C1 4 R2 ¼ ¬2 q ªr r 3 º C1 – 0« – » k ¬ 2 4 R2 ¼ r

dt dr

–

dt dr

q0 k

dt 0 ?C =0 1 dr Integrating again, we get

Since at r = 0,

q0 ª r 2 r4 º – « »  C2 k ¬ 4 16 R 2 ¼ At r = R, t = ts where ts is surface temperature. q0 ª R 2 R4 º t – – « »  C2 s ? k ¬ 4 16 R 2 ¼ q 3 2 C2 t s  0 u R or, k 16 From (i) as occurs at r = 0 tmax = C2 t

or,

–

tmax

ts 

Substituting the numerical values,

...(i)

3 q0 2 is the required expression. R 16 k 2

(Ans.)

3 24 u 106 § 40 · u u¨ ¸ 920° C. (Ans.) 16 200 © 2 u 100 ¹ Example 2.94. A long hollow cylinder has inner and outer radii 50 mm and 150 mm resepctively. It generates heat at a rate of 1 kW/m3 (k = 0.5 W/m°C). If the maximum temperature occurs at radius of 100 mm and temperature of outer surface is 50°C, find : (i) Temperature at inner surface, and (ii) Maximum temperature in the cylinder. (P.U.) tmax

20 

“Conduction” Heat Transfer at Steady State–One Dimension

165

Solution. Refer to Fig. 2.107. r1 = 50 mm = 0.05 m; r2 = 150 mm = 0.15 m qg (rate of heat generation) = 1 kW/m3 = 1000 W/m3 k = 0.5 W/m°C; t2 = 50°C. t1, tmax : Consider an element of hollow cylinder at a radius r and thickness dr and length L. Heat conducted at radius r, dt Qr – k u 2S r L u dr Heat generated in the element, Qg = qg × 2SrL × dr Heat conducted at radius (r + dr), d Q( r  dr ) Qr  (Qr ) dr dr For steady state conduction of heat flow, Qr + Qg = Q(r + dr) d (Qr ) dr or, Qr  Qg Qr  dr d (Qr ) dr Qg or, dr or, Fig. 2.107. Heat generation in hollow cylinder. d ª dt º S u – 2 q g u 2SrL u dr rLk dr dr «¬ dr »¼ rq g – k d r dt dr dr d § dt · k ¨ r ¸  r qg 0 or, dr © dr ¹ ª d 2t dt º k « r 2  »  rq g 0 or, dr ¼ ¬ dr d 2t dt q g r  r 2  0 or, dr k dr d § dt · qg r 0 or, ...(i) ¨r˜ ¸  dr © dr ¹ k Integrating eqn. (i) twice, we have dt q g r 2  ˜ r C1 dr k 2 dt qg r C1  ˜ or, ...(ii) dr k 2 r qg r 2 ˜ t C1 ln r  C2 and, ...(iii) k 4 (where C1, C2 = constants of integration) For evaluating the constants, let us use the following boundary conditions as follows : (i) At r = 0.1 m, dt 0 (condition for maximum temperature) ...(Given) dr



166

Chapter : 2

qg r C1 ˜  0 [From Eqn. (ii)] k 2 r 2 2 qg r 1000 0.1 ˜ u C1 10 ? k 2 0.5 2 (ii) At r = r2 = 0.15 m, t = t2 = 50°C q g 0.152 u t2  10 ln (0.15)  C2 [From eqn. (iii)] k 4 2 1000 0.15 u 50  10 ln (0.15)  C2 or, 0.5 4 or, C2 = 80.22 Substituting the values of C1 and C2 in eqn. (iii), we have, at r = r1 = 0.05 m 1000 0.052 u t1  10 ln (0.05)  80.22 0.5 4 ? t1 = 49°C (Ans.) Again, substituting the values in eqn. (iii) at r = 0.1 m, we have 1000 0.12 u tmax  10 ln (0.1)  80.22 0.5 4 ? tmax = 52.2°C (Ans.) Example 2.95. A chemical reaction takes place in a packed bed (k = 0.6 W/m°C) between two coaxial cylinders with radii 15 mm and 45 mm. The inner surface is at 580°C and it is insulated. Assuming the reaction rate of 0.55 MW/m3 in the reactor volume, find the temperature at the outer surface of the reactor. Solution. Refer to Fig. 2.108. r1 = 15 mm = 0.015 m; r2 = 45 mm = 0.045 m, qg (heat generation rate) = 0.55 MW/m3; t1= 580°C. Temperature at the outer surface, t2 : For the given problem, the controlling differential equation is given by d ª dt º qg  r r 0 dr «¬ dr »¼ k ...[Eqn. (2.105)] Integrating the above equation twice, we obtain dt q g r 2  ˜ r C1 dr k 2 dt qg r C1  ˜ or, ...(i) dr k 2 r qg r 2 ˜ t C1 ln (r )  C2 or, k 4 Fig. 2.108. ...(ii) (where C1, C2 = constants of integration) In order to evaluate C1 and C2, using the following boundary conditions, we have dt 0 At r = r1, dr q g r12 ˜ C1 ? [From eqn (i)] k 2 dt dr

–

“Conduction” Heat Transfer at Steady State–One Dimension At

167

r = r1, t = t1 q g r12 q g r12 ˜ ˜ t1  ln ( r )  C2 ? k 4 k 2 [Substituting the values in eqn. (ii)] q g r12 q g r12 q g r12 ª 1º  ˜ ˜ C t – ˜ ln (r1 ) – » or, ln (r ) t1 – 2 1 k 4 k 2 k 2 «¬ 2¼ Substituting the values of the integration constants in eqn. (ii), we have q g r 2 q g r12 q g r12 ª 1º ˜ ˜ ˜ « ln (r1 ) – » t ln (r )  t1 – k 4 k 2 k 2 ¬ 2¼ qg r 2 q g r12 q g r12 ˜ ˜ t t1  · ln (r / r1 )  k 4 k 2 k 4 2 2 qg r1 qg r12 ª §r· º  ˜  ˜ ln ( / ) 1 – t t r r « or, ...(iii) 1 1 ¨r ¸ » k 2 k 4 ¬« © 1 ¹ ¼» Substituting the values in eqn. (iii) as t = t2 when r = r2, we have 2 0.55 u 106 0.0152 0.55 u 106 0.0152 ª § 0.045 · º u u t2 580  ln (0.045/0.015) + «1 – ¨ ¸ » © 0.015 ¹ ¼ 0.6 2 0.6 4 ¬ or, t2 = 580 + 113.29 – 412.5 = 280.79°C (Ans.) Example 2.96. A nuclear fuel element is in the form of hollow cylinder insulated at the inner surface. Its inner and outer radii are 50 mm and 100 mm respectively. Outer surface gives heat to the fluid at 50°C where the unit surface conductance is 100 W/m2 °C. The thermal conductivity of material is 50 W/m°C. Find the rate of heat generation so that maximum temperature in the system will not exceed 200°C. (P.U.) Solution. Refer to Fig. 2.109. Given : r1 = 50 mm = 0.05 m; r2 = 100 mm = 0.1 m; ta = 50°C, k = 50 W/m°C, h = 100 W/m2 °C; t1 (tmax) = 200°C.

Fig. 2.109.

168

Chapter : 2

Rate of heat generation, qg : For the given problem the controlling differential equation is d § dt · qg ˜r 0 ¨r˜ ¸  dt © dr ¹ k Integrating the above equation twice, we have dt q g r 2  ˜ r˜ C1 dr k 2 dt qg r C1  ˜ or, dr k 2 r qg r 2 t C1 ln (r )  C2 . and, k 4 (where C1, C2 = constants of integration).

...[Eqn. 2.105]

...(i) ...(ii)

dt 0, we have, dr q g r12 ˜ C1 k 2 Applying the following boundary condition, we have,

Applying boundary condition; at

(Heat conducted) r or,

or,

r2

r = r1,

...[From Eqn. (i)]

= Heat convected from the outer bondary to the surrounding fluid.

§ dt · – k u (2S r2 u 1) u ¨ ¸ © dr ¹r

r2

qg r º ªC ˜ » –k « 1 – k 2 ¼r ¬r

r2

h u (2Sr2 u 1) u (t2 – ta )

(considering unit length of cylinder) h (t2 – ta )

ª q g r 2 q g r2 º –k « ˜ 1 – » h (t2 – ta ) 2k ¼ ¬ 2k r2 2 qg r2 ª §r · º or, «1 – ¨ 1 ¸ » h (t2 – ta ) 2 «¬ © r2 ¹ »¼ Substituting the value of C1 in eqn. (ii), we have

or,

...(iii)

q g r 2 q g r12 ˜ ˜ ln ( r )  C2 k 4 k 2 Now, applying boundary condition, at r = r1, t = t1, we have t

t1 

q g r12 ˜ k 4

...(iv)

q g r12 ln (r1 )  C2 k 2

qg r12 ª 1 º – ln ( r1 ) » ¼ 2k «¬ 2 Inserting the values of C1 and C2 in eqn. (iv), we have C2

?

t or,

qg r 2 ˜ k 4 t – t1

t1 

q g r12 q g r12 ˜ ln (r )  t1  2k k 2 qg r12 2k

ln (r ) 

qg r12 4k

–

ª1 º «¬ 2 – ln (r1 ) »¼

qg r12 2k

ln (r1 ) –

qg r 2 4k

“Conduction” Heat Transfer at Steady State–One Dimension

qg r12

2º

qg r12

ª §r· ln ( r / r1 ) «1 – » 4 k «¬ ¨© r1 ¸¹ »¼ 2k At r = r2, t = t2; the eqn. (v) becomes 2 2 qg r12 ª § r2 · º qg r1 ln (r2 / r1 ) t2 – t1 «1 – ¨ ¸ »  4 k ¬« 2k © r1 ¹ ¼» qg r12 ª§ r2 · 2 º qg r12 – 1» – ln (r2 / r1 ) t1 – t2 « ? 4 k ¬«¨© r1 ¸¹ 2k ¼» º q g r12 ª§ r2 · 2 t1 – t2 «¨ ¸ – 2 ln (r2 / r1 ) – 1» or, 4 k ¬«© r1 ¹ ¼» Substituting the value of t2 from eqn. (iii) in eqn. (vi), we have 2 2 2 ª º q g r2 ­° § r · ½° º q g r1 ª§ r2 · t1 – «ta  «¨ ¸ – 2 ln ( r2 / r1 ) – 1» ®1 – ¨ 1 ¸ ¾ » 2h ¯° 4k ¬«© r1 ¹ © r2 ¹ ¿° ¼» ¼» ¬« Now, inserting the proper values in (vii), we obtain 2 2 2 ª º q g u 0.1 ­ § 0.05 · ½º q g u 0.05 ª§ 0.1 · ®1 – ¨ 200 – «50  «¨ ¸ ¾» ¸ – 2 ln (0.1/ 0.05) – 1» © 0.1 ¹ ¿»¼ 2 u 100 ¯ 4 u 50 ¬© 0.05 ¹ «¬ ¼ 200 – [50 + 0.000375 qg] = 0.0000125 qg [4 – 1.386 – 1] or, 150 – 0.000375 qg = 0.00002 qg 150 qg ? (0.000375  0.00002) = 3.797 × 105 W/m3 (Ans.) Example 2.97. (a) Establish a relation for the temperature as the function of the radial coordinate r, for a hollow cylinder (thermal conductivity k) having inner and outer radii r1 and r2 respectively; it is perfectly insulated at its outer radius and is held at a temperature t1 by a coolant at the inner radius. Electrical energy is dissipated at the constant rate of qg per unit volume. Assume that the temperature distribution is primarily radial and heat transfer takes place under steady state conditions. (b) A hollow conductor (k = 18 W/m°C) of inner and outer radii 50 mm and 65 mm has an electrical resistance of 0.024 ohm per metre. It is perfectly insulated at its outer radius.The cooling fluid at the inside is at 40°C. Determine the maximum allowable current if the temperature Fig. 2.110. is not to exceed 52°C. or,

Solution.

or

t – t1

(a) The heat conduction equation is given by d 2 t 1 dt q g  ˜  0 k dr 2 r dr qg d ª dt º r˜ » – r « dr ¬ dr ¼ k

169 ...(v)

...(vi)

...(vii)

170

Chapter : 2

Integrating the above equation twice, we have qg r 2 dt  C1 r – ˜ dr k 2 qg r C dt – ˜  1 or, ...(1) dr k 2 r qg r 2  C1 ln (r )  C2 t – ˜ and ...(2) k 4 (where C1, C2 = constants of integration). The values of C1 and C2 are determined from the following boundary conditions as follows : (i) At r = r1, t = t1 (ii) At r = r2, the heat flow is zero (since the conductor is insulated at the outer radius) i.e.

dt dr

ª From Fourier's law « dt 0 « Q – kA ˜ «¬ dr

º » » »¼

q g r2 C1 ...[From eqn. (1)] ˜  k 2 r2 q g r2 2 ˜ C1 or, k 2 Now, applying boundary condition (i) [i.e., at r = r1, t = t1] to eqn. (2), we have q g r 2 q g r2 2 ˜ t1 – ˜ 2  ln (r1 )  C2 k 4 k 2 ª q g r2 2 º ˜ » «Substituting C1 k 2 ¼ ¬ º q g ª r12 2 C2 t1  – r2 ln (r1 ) » or, « ¼ 2k ¬ 2 Inserting the values of C1 and C2 in eqn. (2), we have the general solution for temperature distribution as º qg r 2 q g r2 2 q g ª r12 2 ˜  ˜ t – ln ( r )  t1  – r2 ln ( r1 ) » « ¼ k 4 k 2 2k ¬ 2 2 2 ª º § · qg r –r 2 ˜ «¨ 1 t t1  ¸  r2 ln (r / r1 ) » or, ...(3) 2 k ¬© 2 ¹ ¼ (b) Given : r1 = 5 mm = 0.005 m; r2 = 6.5 mm = 0.0065 m, k = 18 W/m°C t1 = 40°C, tmax = 52°C, Re = 0.024 ohms per metre. Maximum allowable current, I : The maximum temperature occurs at the insulated boundary, i.e., at r = r2. The eqn. (3) then takes the form

?

0

tmax 52

or,

52

–

º q g ª§ r12 – r2 2 · 2 ˜ «¨ ¸  r2 ln (r2 / r1 ) » ¹ 2k ¬© 2 ¼ 2 2 ª º qg § 0.005 – 0.0065 · 2 40  «¨ ¸  0.0065 u ln (0.0065 / 0.005) » ¹ 2 u 18 ¬© 2 ¼ qg ª¬ – 8.625 u 10 –6 u 11.085 u 10 –6 º¼ 40  36

t1 

“Conduction” Heat Transfer at Steady State–One Dimension or,

12

or,

qg

qg

171

u 2.46 u 10 –6

36 12 u 36 2.46 u 10 –6

175.61 u 106 W/m3

S ª¬(0.0065) 2 – (0.005) 2 º¼ 4 = 2379.2 W per metre length of the conductor. Also heat generated = I2Re = I2 × 0.024 watts per metre length ? I2 × 0.024 = 2379.2

Total volumetric heat generation

175.61 u 106 u

1/ 2

§ 2379.2 · 314.85 A ¨ ¸ © 0.024 ¹ Example 2.98. A steel rod (k = 40 W/m°C) 30 mm diameter and 300 mm long separates two large steel plates maintained at 100°C and 75°C . The space between the plates is filled with insulation which also insulates the circumference of the rod. Voltage difference exits between the two plates due to which current flows through the rod and the electrical energy is dissipated at a rate of 12 W. Determine : (i) The maximum temperature in the rod, and its point of occurence; (ii) The heat flux at each end. Solution. Refer to Fig. 2.111. t 1 = 100°C; t 2 = 75°C; k = 40 W/m°C. The rate of dissipation of electrical energy = 12W

or

I

(Ans.)

Fig. 2.111.

12 56588 W/m3 S 2 u 0.03 u 0.30 4 (i) The maximum temperature in the rod (tmax) and its point of occurence (x) : The heat conduction equation for steel rod is given by d 2t qg  0 k dx 2 Integrating eqn. (i) twice, we obtain

? Heat generated per unit volume,

qg

dt qg  ˜ x C1 dx k qg x 2 ˜ t C1 x  C2 k 2

...(i)

...(ii) ...(iii)

172

Chapter : 2

(where C1, C2 = constants of integration). In order to evaluate C1 and C2, using the following boundary conditions, we have (i) At x = 0, t = t1 ? C2 = t1 (ii) At x = L, t = t2 qg L q g L2 t –t ˜ C1 2 1  ˜ t2  C1 L  t1 or ? L k 2 k 2 Substituting the values of C1 and C2 in eqn. (iii), we get q g x 2 ª t2 – t1 q g L º ˜ «  ˜ » x  t1 k 2 k 2¼ ¬ L dt 0 The maximum temperature occurs where dx qg qg ª t – t qg L º dt ˜ » x  C1 – ˜ x « 2 1  – ? dx k k k 2¼ ¬ L qg t2 – t1 qg L ˜x  ˜ or, k L k 2 k (t2 – t1 ) L  x or, qg ˜ L 2 Inserting the value of x in eqn. (iv), we have t

tmax

–

...(iv)

0

...from eqn. (ii)

...(v)

qg ª k (t2 – t1 ) L º 2 t2 – t1 ª k (t2 – t1 ) L º qg L ª k (t2 – t1 ) L º  »   »  »  t1 2k «¬ qg ˜ L 2¼ 2¼ 2k «¬ qg ˜ L 2¼ L «¬ qg ˜ L 2 qg ª k (t2 – t1 ) L º ª k (t – t ) L º ª qg ˜ L t2 – t1 º  » « 2 1  »«  t1 – » « 2k ¬ q g ˜ L 2¼ 2 ¼ ¬ 2k L ¼ ¬ qg ˜ L

–

qg ª k (t2 – t1 ) L º 2 qg ª k (t2 – t1 ) L º 2  »   » 2k «¬ qg ˜ L 2¼ 2¼ k «¬ qg ˜ L 2 qg ª k (t2 – t1 ) L º  » t1  2k «¬ qg ˜ L 2¼ t1 –

or,

tmax

...(vi)

Substituting the proper values in eqn. (vi), we get 56588 ª 40 (75 – 100) 0.3 º  2 u 40 «¬ 56588 u 0.3 2 »¼ The distance x at which tmax occurs is given by k (t2 – t1 ) L  x qg ˜ L 2 tmax

100 

40 (75 – 100) 0.3  56588 u 0.3 2 (ii) The heat flux at each end : The heat flow from the rod at x = 0, dt Q1 – kA dx x 0

2

105.8° C

...[Eqn. (v)]

0.09109 m or 91.09mm

ª q g ˜ x t2 – t1 qg L º   ˜ » – kA « – k L k 2 ¼x ¬

(Ans.)

0

(Ans.)

“Conduction” Heat Transfer at Steady State–One Dimension – 40 u

S ª 75 – 100 56588 0.3 º u 0.032 «  u 4 40 2 »¼ ¬ 0.3

173

– 3.64 W

(–ve sign indicates that the heat flows in a direction opposite to that of x-direction). The heat flow from the rod at x = L (=0.3m), Q2

–k A

dt dx

x

L

ª q g ˜ L t2 – t1 q g L º   ˜ » – kA « – k L k 2¼ ¬ qg L º ªt – t ˜ » – kA « 2 1 – k 2¼ ¬ L S ª 75 – 100 56588 0.3 º u 0.032 « u – 4 40 2 »¼ ¬ 0.3 [Check : The sum of Q1 and Q2 should be equal to Qg, i.e., Q1 + Q2 = 3.64 + 8.36 = 12W i.e., Qg] – 40 u

8.36 W

(Ans.)

Example 2.99. The rate of heat generation per unit volume in a long cylinder of radius R is given by qg = a + br2 where a and b are constants and r is any radius. The cylinder is undergoing heat transfer with a medium at a temperature t a and surface heat transfer coefficient is h. Find the steady state temperature distribution in the solid. (M.U.) Solution. For the given problem, the controlling differential equation is given by d § dt · qg ˜r 0 ¨r˜ ¸  dr © dr ¹ k

Fig. 2.112.

...[Eqn. (2.105)] Substituting the value of qg, we have d § dt · r 2 ¨ r ¸  (a  br ) dr © dr ¹ k

0

Integrating the above equation, we get r

or,

dt 1 § ar 2 br 4 ·  ¨  ¸ dr k © 2 4 ¹ dt 1 § ar br 3 ·  ¨  ¸ dr k © 2 4 ¹

C1

...(i)

C1 r

Integrating again, we get t

1 ª ar 2 br 4 º  « » k¬ 4 16 ¼

C1 ln ( r )  C2

(where C1, C2 = constants of integration).

...(ii)

174

Chapter : 2

The boundary conditions for finding C1 and C2 are : dt 0 (i) At r = 0, ? C1 = 0 dr dt 1 § ar br 3 ·  ¨  ? ¸ 0 dr k © 2 4 ¹ and

t

1 ª a r 2 br 4 º  « » k¬ 4 16 ¼

[from eqn. (i)] ...(iii) ...(iv)

C2

[From eqn. (ii)] (ii) or,

(Heat conducted)r = R = (Heat convected)r = R

ª § dt · º « – kA ˜ ¨ dr ¸ » © ¹¼r ¬

R

dt º ª «¬ – k ˜ dr »¼ r

R

>h ˜ A (t – ta )@r >h (t

– ta )@r

R

Substituting the values from eqns. (iii) and (iv) for we get

ª 1 § aR bR3 · º  – k «– ¨ ¸» 4 ¹¼ ¬ k© 2 ?

R ª bR 2 º «a  » 2h ¬ 2 ¼

or,

C2

R

dt and t in the above equation for r = R, dr

ª º 1 § aR 2 bR 4 ·  h «C2 – ¨ ¸ – ta » k© 4 16 ¹ ¬ ¼ C2 – ta 

R2 ª bR 2 º «a  » – ta 4k ¬ 4 ¼ R ª bR 2 º R 2 ª b R2 º «a  » «a  » 2h ¬ 2 ¼ 4k ¬ 4 ¼

Now, substituting this value in eqn. (iv), we obtain t

? or,

r2 4k

ª br 2 º «a  » ¬ 4 ¼

t – ta t – ta

ta 

R 2h

ª bR 2 º R 2 «a  » ¬ 2 ¼ 4k

ª bR 2 º «a  » ¬ 4 ¼

R ª bR 2 º ª aR 2 bR 4 º ª ar 2 br 4 º   «a  »« »–« » 2h ¬ 2 ¼ ¬ 4k 16k ¼ ¬ 4k 16 k ¼

2 4 4 R ª bR 2 º aR 2 ª § r · º bR ª §r· º «1 – ¨ ¸ »  «1 – ¨ ¸ » «a  » © R ¹ ¼ 16 k ¬ ©R¹ ¼ 2h ¬ 2 ¼ 4k ¬

...(Temperature distribution) (Ans.) –8 Example 2.100. A copper conductor (k = 380 W/m°C, resistivity U = 2 × 10 W m) having inner and outer radii 1.0 cm and 2.25 cm respectively is carrying a current density of 4800 amperes/ cm2. The conductor is internally cooled and a constant temperature of 65°C is maintained at the inner surface and there is no heat transfer through insulation surrounding the conductor. Determine: (i) The maximum temperature of the conductor and the radius at which it occurs, and (ii) The internal heat transfer rate. Solution. Refer to Fig. 2.113. r1 = 1.0 cm; r2 = 2.25 cm; J (current density) = 4800 amp/cm2 or 4800 × 104 amp/m2 k = 380 W/m°C; U = 2 × 10–8 :m; to = 65°C.

“Conduction” Heat Transfer at Steady State–One Dimension

175

Fig. 2.113.

(i) The maximum tem-perature of the conductor and the radius at which it occurs; tmax, r : Total volumetric heat generated I 2R

I2 ˜

UL A

Heat generated per unit volume, UL 2 I2 ˜ UL 1 §I· A u U¨ ¸ qg I2 ˜ © A¹ Au L A AL = UJ2 = 2 × 10–8 × [4800 × 104]2 = 46.08 × 106 W/m3 The differential equation describing the temperature distribution through a cylindrical conductor is given by

d § dt · qg ˜r ¨r ¸ dr © dr ¹ k Integrating eqn. (i) twice, we get r

or, and, or,

dt q g r 2  ˜ dr k 2 dt qg r  ˜ dr k 2 t

qg r 2 ˜ k 4 t

0

C1

...(ii)

C1 r C1 ln (r )  C2 –

qg r 2 ˜  C1 ln ( r )  C2 k 4

...(iii)

(where C1, C2 = constants of integration). The values of C1 and C2 are found from the following boundary conditions : (i) At r = r2 = 2.25 cm (or 0.0225 m), dt dr transfer).

0 (since the face is insulated and there is no heat

176

Chapter : 2

q g r 2 46.08 u 106 0.02252 ˜ u k 2 380 2 (ii) At r = r1 = 1.0 cm (or 0.01 m), t = t1 = 65°C C1

?

30.69. ...[From eqn. (ii)]

46.08 u 106 0.012 u  30.69 ln (0.01)  C2 380 4 46.08 u 106 0.012 C2 65  u – 30.69 ln (0.01) 209.36q C ? 380 4 Substituting the values of C1 and C2 in eqn. (iii) we get the temperature distribution through the conductor as 46.08 u 106 2 t – r  30.69 ln (r )  209.36 4 u 380 or, t = – 30315.8 r2 + 30.69 ln (r) + 209.36 ...(iv) It is evident from eqn. (iv), that the temperature distribution is parabolic. Maximum temperature occurs at the insulated face (at r = r2 = 0.0225 m) and its value equals, tmax = – 30315.8 × 0.02252 + 30.69 ln (0.0225) + 209.36 = – 15.35 – 116.44 + 209.36 = 77.57°C (Ans.) (ii) The internal heat transfer rate, Q : dt Q – kA dr qg r C dt – ˜  1 But, [From eqn. (ii)] dr k 2 r dt 46.08 u 106 0.01 30.69 u  – 2462.68 ? dr r 0.01 380 2 0.01 65

–

? Q = – 380 × (2S × 0.01× 1) × 2462.68 ; 58800 W/m (– ve sign indicates that heat flow is radially inwards.) [Check : Internal heat transfer = qg × [S (0.02252 – 0.012) × 1] = 46.08 × 106 [ S (0.02252 – 0.012) × 1] ; 58800 W/m] Example 2.101. A hollow cylinder having inner and outer radii r1 and r2 respectively is developing heat uniformly, qg per unit volume per unit time. The conductivity of the cylinder material is given by; k = ko (1 + Et). If the outside surface temperature is tw, prove that the temperature distribution in cylinder is given by 2

2 qg r1 1 §1 ·  ¨  tw ¸ – E 2E ko ©E ¹ Solution. Refer to Fig. 2.114. Heat generation rate between r = r 1 and r = r = (qg)r = qg × S (r2 – r12) × 1

t

–

– k0 (1  Et ) 2Sr u 1 u § r 2 – r12 · q g ˜ dr ¨ ¸˜ © r ¹ 2 ko

– (1  Et ) dt

2 § r –¨r – 1 © r

(1  Et ) dt

· q ˜ dr ¸˜ ¹ 2 ko

2 ª§ r · 2 §r · º «¨ ¸  2 ln ( r2 / r ) – ¨ 2 ¸ » © r1 ¹ ¼» ¬«© r1 ¹

heat conducted at r = r

dt (Considering unit length of the cylinder) dr

“Conduction” Heat Transfer at Steady State–One Dimension

177

Fig. 2.114.

Integrating both sides, we have

³ (1  E t ) dt

–

qg

³

2 ko

2 § r · ¨ r – 1 ¸ dr © r ¹

qg ª r 2 º t2 2 – « – r1 ln ( r ) »  C ¼ 2 2k o ¬ 2 (where C = constant of integration). Using boundary condition (for finding the constant C), at r = r, t = tw, we have or,

t  E.

2 º E 2 q g ª r2 2 ˜ tw  – r1 ln (r2 ) » « ¼ 2 2ko ¬ 2 Substituting the value of C in eqn. (i), we have qg ª r 2 qg ª r22 º º E E 2 2    – ln ( ) · « – r12 ln (r2 ) » t  ˜t 2 – r r t t « » w w 1 ¼ 2 2 k0 ¬ 2 2 2k0 ¬« 2 ¼»

C

tw 

º qg ª r 2 q g ª r2 2 º E 2 E 2 2 2 ˜t  t  – r1 ln (r ) » – t w – ˜ t w – – r1 ln (r2 ) » « « ¼ ¼ 2 2 ko ¬ 2 2 2 ko ¬ 2 2 2 ª º ­ ½ qg r2 – r E 2 E ® ˜ t  t – « ˜ tw2  t w  – r12 ln (r2 / r ) ¾» 0 or, 2 2 2 2 k ¿¼» o ¯ ¬«

or,

or,

or,

–1  1  4 u t

t

½º q g ­ r2 2 – r 2 E ªE 2 2 ® – r1 ln (r2 / r ) ¾» « tw  tw  ¿¼» 2 ¬« 2 2k o ¯ 2 E

–

1  E

­°§ 1 ·2 2 ½° 2 qg 2 ®¨ ¸  tw  tw ¾  ˜ E °¿ E 2 ko ¯°© E ¹

–

1  E

2 qg §1 ·  t w¸ – ¨E E ko © ¹

­ r22 – r 2 ½ ® – r12 ln ( r2 / r ) ¾ ¯ 2 ¿

ª r2 r2 2 º 2  r r r ln ( / ) – « » 1 2 ¬2 2 ¼

..(i)

178 or,

Chapter : 2

t

2 q g r12 1 §1 · –  ¨  tw ¸ – E 2E ko ©E ¹

2 ª § 2 «§ r ·  2 ln (r2 / r ) – ¨ r2 ¨ ¸ ¨ © r1 ¬«© r1 ¹

· ¸¸ ¹

2º

» ¼»

....(Proved)

2.9.4. HEAT TRANSFER THROUGH A PISTON CROWN Let,

b = Thickness of crown, qg = Quantity of heat given by the gases to the piston by convection and radiation per unit area per unit time, k = Thermal conductivity of piston material, R = Outer radius of the piston, and to = Outer surface temperature of the piston. Consider an element of thickness dr at radius r as shown in Fig. 2.115. dt . dr Heat given by the gases to the piston, Qg = qg × 2Sr.dr Heat conducted out at radius (r + dr), Q(r+dr) d (Qr )dr Qr  dr For steady state heat conduction, we have Qr + Qg = Q(r + dr)

Heat conducted in at radius r , Qr

Qr 

or, or, or, or,

– k .2S r b .

d (Qr ) dr dr

d (Qr ) dr dr d ª dt º – k u 2S rb ˜ » dr q g u 2Sr . dr dr «¬ dr ¼ d ª dt º – kb ˜ r ˜ » qg u r dr «¬ dr ¼ d ª dt º qg  ˜r 0 r dr «¬ dr »¼ kb Qg

Fig. 2.115. Heat transfer through a piston crown.

...(i)

Integrating eqn. (i) twice, we obtain dt q g r 2  ˜ r. C1 ...(ii) dr kb 2 dt qg r C1  ˜ or, dr kb 2 r qg r 2 ˜ t C1 ln (r )  C2 and, ...(iii) kb 4 (where C1 and C2 = constants of integration). The values of C1 and C2 can be found out from the following boundary conditions, as follows : dt 0 (i) At r = 0, ? C1 = [From eqn. (ii)] dr qg R 2 ˜ ? C2 t0  [From eqn. (iii)] (ii) At r = R, t = to kb 4

“Conduction” Heat Transfer at Steady State–One Dimension

179

Substituting the values of C1 and C2 in eqn (iii), we have qg r 2 qg R 2 ˜ ˜ t to  kb 4 kb 4 qg t to  (R2 – r 2 ) or, ...(2.114) 4k b It is evident from eqn. (2.114) that the temperature distribution is parabolic and the maximum temperature which occurs at the centre of the piston crown (r = 0), is given by qg tmax to  . R2 ... (2.115) 4k b If the total heat given by the gases to the piston crown is denoted by Q, then Q Q q g u SR 2 or q g SR 2 Q 1 u u R2 tmax t0  ? 2 4kb SR Q tmax t0  or, ...(2.116) 4S k b From eqn. (2.114) the thickness of the piston crown can be calculated. i.e.,

b

Q 4S k (tmax – t0 )

...(2.117)

2.9.5. HEAT CONDUCTION WITH HEAT GENERATION IN NUCLEAR CYLINDRICAL FUEL ROD A. Nuclear cylindrical fuel rod without cladding : The internal heat generation for a cylindrical fuel rod is generally given by (not being uniform) 2 ª § r · º q0 «1 – ¨ ¸ » «¬ © R fr ¹ »¼ where, qg = Heat generation rate at radius r, qo = Heat generation rate at the centre of the rod (r = 0), and Rfr = Outer radius of the fuel rod. One-dimensional, steady state heat transfer in radial direction is given by d ª dt º qg  ˜r 0 r˜ dr «¬ dr »¼ k

qg

2 d ª dt º q0 ª § r · º ˜  r 1 – « »r ¨ ¸ dr «¬ dr »¼ k «¬ © R fr ¹ »¼ Integrating eqn. (i), we have

or

r˜

dt qo ª r 2 r4 º  « – » dr k ¬2 4 R fr 2 ¼

Fig. 2.116. Nuclear cylindrical fuel rod.

0

...(i)

C1

...(ii)

180

Chapter : 2

dt qo ª r r3 º  – « » or, dr k «¬ 2 4 R 2fr »¼ Integrating again, we obtain

C1 r

ª r2 r4 º C1 ln (r )  C2 « – 2 » «¬ 4 16 R fr »¼ (where C1, C2 = constants of integration). Using the following boundary conditions, we have t

qo k

...(iii)

dt 0 ? C1 = 0 dr (ii) At r = 0, t = tmax ? C2 = tmax Substituting these values of the constants in eqn. (iii), we get

(i) At r = 0,

t

qo k

ª r2 r4 º – « 2 » ¬« 4 16 R fr ¼»

[From eqn. (ii)] [From eqn. (iii)]

tmax

qo ª r 2 r4 º « – » k «¬ 4 16 R 2fr »¼ If tw is the temperature at the outer surface of the rod i.e., at r = Rfr, then

or,

or,

t – tmax

–

tw – tmax

–

tw – tmax

–

ª R 2fr R 4fr º « » – 16 R 2fr »¼ «¬ 4 3q0 R 2fr qo k fr

–

qo k

...(iv)

ª R 2fr R 2fr º « » – ¬ 4 16 ¼

...(2.118)

16 k fr

(where kfr = thermal coductivity of fuel rod material). Also, the rate of heat transfer at the surface of the rod, dt Q – k fr A dr r R fr

ª q k fr A « – o «¬ k fr

r 3 °½º °­ r ® – 2 ¾» °¯ 2 4 R fr °¿»¼ r

R fr

[Substituting the value of

dt from eqn. (ii)] dr

ª q ­° R fr R3fr ½° º q0 A R fr 0 « Q – k fr A – – ® ¾» or, ...(2.119) 4 «¬ k fr °¯ 2 4 R 2fr °¿ »¼ This heat would be convected from the outside surface of the rod, under steady state conditions. qo A R fr

h A (t w – ta ) 4 [where h = convective heat transfer coefficient and ta = ambient temperature.] qo A R fr qo R fr tw ta  ta  or, 4hA 4h Inserting this value of tw in eqn. (2.118), we obtain

?

...(v)

“Conduction” Heat Transfer at Steady State–One Dimension qo R fr º ª «ta  » – tmax 4h ¼ ¬

or,

tmax – ta

or,

tmax – ta

–

181

3qo R 2fr 16 k fr

3qo R 2fr 16 k fr



qo R fr 4h

qo R fr ª 3 R fr 1º  » « 4 «¬ 4 k fr h »¼

...(2.120)

B. Nuclear cylindrical fuel rod with cladding : When nuclear fuel elements happen to come in contact with the cooling medium they are likely to get damaged owing to oxidation. In order to check this damage the fuel elements are usually covered with pretoctive material known as ‘cladding’. Refer to Fig. 2.117. The heat generated per unit volume in the nuclear fuel rod is given by 2

r º ª qo «1 – » R fr ¼ ¬ where, qg = Heat generation rate at radius r, qo = Heat generation rate at the centre of the rod Fig. 2.117. Nuclear cylindrical fuel (r = 0), and rod with cladding. Rfr = Outer radius of fuel rod. Let, Rcl = Outer radius of cladding, qcl = Heat generation rate in the cladding, qfr = Heat generation rate in the fuel rod, kfr = Thermal conductivity of fuel rod material, and kcl = Thermal conductivity of cladding material. One-dimensional steady state heat conduction in radial direction is given by d ª dt º qg  r r 0 ...(i) dr «¬ dr »¼ k § Q· ¸ is given by Also, heat flow rate per unit area, q ¨ © A¹ dt dt q q –k˜ – or dr dr k dt q Substituting – in eqn. (i) we have dr k d ª rq º qg –  ˜r 0 (assuming k to be constant) dr «¬ k »¼ k d (rq ) q g . r or, ...(ii) dr d (r . q fr ) q g . r or, ...for fuel rod dr qg

182

Chapter : 2

2 ª § r · º q 1 – « o ¨ R ¸ »r or, «¬ © fr ¹ »¼ Since there is no internal heat generation in cladding, therefore, d (r . qcl ) 0 dr Integrating eqn (iii), we obtain ª r2 r4 º r.q fr qo « – »  C1 4 R 2fr ¼» ¬« 2

d (r . q fr ) dr

ªr r 3 º C1  qo « – 2 » r «¬ 2 4 R fr »¼ Now, integrating eqn. (iv), we have r.qcl = C2

or,

q fr

...(iii)

...(iv)

...(v) ...(vi) ...(vii)

C2 ...(viii) r In order to evaluate C1 and C2 (integration constants), using the following boundary conditions, we have At r = 0, qfr = finite ? C1 = 0 ...[From eqn. (v)] At r = Rfr qfr = qcl ? C2 = qfr . Rfr ....[from eqn. (vii)]

or,

qcl

ª R fr R 3fr º » R fr qo « – 4 R 2fr »¼ «¬ 2 [substititing the value of qfr from eqn. (vi)] qo 2 ˜ R fr C2 or, 4 dt fr ªr r3 º ˜ – – q k q ? The rate of heat flow through fuel rod, fr ...(ix) fr o« 2 » dr ¬« 2 4 R fr ¼» dtcl qo 2 ˜ R fr and, the rate of heat flow through the cladding, qcl – kcl ˜ ...(x) dr 4r By integrating eqns. (ix) and (x), we get the temperatures tfr and tcl as follows :

ª r4 r2 º – « »  C3 2 ...(xi) 4 »¼ «¬16 R fr q tcl – o R 2fr ln (r )  C4 and, ...(xii) 4 kcl (where C3, C4 = constants of intergration) (i) Applying the boundary condition : At r = Rcl, tcl = tw (outside surface temperature of the system), qo ˜ R 2fr ln ( Rcl ) C4 tw  we have [From eqn. (xii)] 4 kcl q q tcl – 0 R 2fr ln (r )  tw  o ˜ R 2fr ln ( Rcl ) ? 4 kcl 4 kcl qo 2 ˜ R fr ln ( Rcl / r ) (tcl – t w ) or, ...(xiii) 4 kcl t fr

qo k fr

“Conduction” Heat Transfer at Steady State–One Dimension

183

(ii) Applying boundary condition, at r = Rfr, tcl = tfr to eqn. (xi), we have t fr

qo k fr

ª R 4fr R 2fr º « »  C3 – 2 4 »¼ ¬«16 R fr

tcl

or,

qo k fr

ª R 4fr R 2fr º « »  C3 – 2 4 »¼ «¬16 R fr

–

qo R 2fr ln (R fr )  C4 4 kcl

or,

qo k fr

ª R 4fr R 2fr º « »  C3 – 2 4 »¼ «¬16 R fr

–

qo q ˜ R 2fr ln ( R fr )  t w  o ˜ R 2fr ln ( Rcl ) 4 kcl 4 kcl

(substituting the value of C4) 2 qo ª R fr

R 2fr

or,

º « »  C3 – 4 ¼ k fr ¬ 16

tw 

qo ˜ R 2fr ˜ ln ( Rcl / R fr ) 4 kcl

or,

qo R 2fr ª 3 º  C3 – 4 k fr «¬ 4 »¼

tw 

qo ˜ R 2fr ˜ ln ( Rcl / R fr ) 4 kcl

qo R 2fr ª 3 1 º  ˜ ln ( Rcl / R fr ) » « kcl 4 ¬ 4 k fr ¼ Subsituting the value of C3 in eqn. (xi), we get C3

or,

tw 

qo R 2fr ª 3 ª r4 r2 º 1 º    ˜ ln ( Rcl / R fr ) » ...(xiv) t – « » w « 2 k k 4 4 4 R 16 cl fr ¬ fr ¼ ¬« ¼» The maximum value of temperature (tmax) which occurs at the centre of the rod at (at r = 0), is given by qo R 2fr ª 3 1 º  ˜ ln ( Rcl / R fr ) » tmax tw  ...(2.121) « kcl 4 ¬ 4 k fr ¼ t fr

qo k fr

Example 2.102. In a cylindrical fuel rod of nuclear reactor, the internal heat generation is given by 2 ª § r · º qo «1 – ¨ ¸ » «¬ © R fr ¹ »¼ where Rfr is the radius of the fuel rod. Calculate the temperature drop from the centre to the surface of a 25 mm diameter fuel rod having k = 20 W/m°C when the rate of heat generation from the surface = 0.25 MW/m2. (P.U.) Solution. Radius of the fuel rod, 25 12.5 mm 0.0125 m R fr Fig. 2.118. 2 Thermal conductivity, kfr = 20 W/m°C The rate of heat generation from the surface = 0.25 MW/m2

qg

184

Chapter : 2

Temperature drop from the centre to the surface of the rod : Let, tmax = Maximum temperature at the centre of the rod (r = 0), tw = Temperature at the surface of the fuel rod, qo = Heat generation rate at the centre (r = 0), and qg = Heat generation rate at a radius r. Heat generation rate from the surface = Heat transfer rate from the surface of the rod, Q = 0.25 MW/m2 qo AR fr Q But, ...[Eqn. (2.119)] 4 q0 u 1 u 0.0125 0.25 u 106 or, 4 0.25 u 106 u 4 qo 80 u 106 W/m3 ? 0.0125 3qo R 2fr tmax – tw Also, ...[Eqn. (2.118] 16 k fr

3 u 80 u 106 u 0.01252 (Ans.) 117.2q C 16 u 20 Example 2.103. The rate of heat generation in a cylindrical fuel rod of a nuclear reactor is 2 given by q g [1 – ( r / R fr ) ], where qg and qo are the heat generation rates at any radius r and at the centre respectively and Rfr is the outside radius of the fuel rod. Derive a relation for the temperature drop from the centre line to the outside surface of the rod. Determine this temperature drop for 35 mm outside diameter rod (k = 24 W/m°C) if heat is removed from the outside surface at the rate of 2.5 × 106 W/m2. What will be surface heat transfer coefficient if the maximum wall surface temperature is limited to 180°C and the temperature of the fluid surrounding the rod is 70°C. or,

(tmax – tw )

35 2 Thermal conductivity of fuel rod, kfr = 24 W/m°C

Solution. Outside radius of the fuel rod, R fr

17.5 mm

0.0175 m

Q 2.5 u 106 W/m 2 A Maximum wall surface temperature, tw = 180°C Temperature of fluid surrounding the rod, ta = 70°C Temperature drop, (tmax – tw) : The heat flow at the outside surface of the fuel rod is given by qo A R fr Q ...[Eqn. (2.119)] 4 Q qo R fr or, 4 A Q 4 4 u qo 2.5 u 106 u 5.714 u 108 W/m3 ? A R fr 0.0175 The temperature drop from the centre line (t = tmax) to the outside surface (t = tw) of the fuel rod is given by

Rate of heat removal,

tmax – tw

3 qo R 2fr 16 k fr

...[Eqn. (2.118)]

“Conduction” Heat Transfer at Steady State–One Dimension 8

3 u 5.714 u 10 u 0.0175 16 u 24 Surface heat transfer coefficient, h : Q = hA (tw – ta) tmax – tw

or,

h

Q 1 ˜ A (tw – ta )

185

2

2.5 u 106 u

1367q C

1 (180 – 70)

(Ans.)

22727 W/m 2 q C

(Ans.)

2.9.6. SPHERE WITH UNIFORM HEAT GENERATION Consider one dimensional radial conduction of heat, under steady state conditions, through a sphere having uniform heat generation. Let, R = Outside radius of sphere, k = Thermal conductivity (uniform), qg = Uniform heat generation per unit volume, per unit time within the solid, tw = Temperature of the outside surface (wall) of the sphere, and ta = Ambient temperature. Consider an element at radius r and thickness, dr as shown in Fig. 2.119. Heat conducted in at radius r, dt dt – k u 4Sr 2 ˜ dr dr Fig. 2.119. Sphere with uniform heat generation. Heat generated in the element, 2 Qg = qg × A × dr = qg × 4Sr × dr Heat conducted out at radius (r + dr) d Q( r  dr ) Qr  (Qr ) dr dr Under steady state conditions, we have Qr + Qg = Q(r + dr) Qr

– kA

= Qr  or,

Qg

or,

qg u 4Sr 2 u dr

or,

qg u 4S r 2 u dr

or,

1 d ª 2 dt º qg «r ˜ »  k r 2 dr ¬ dr ¼

d (Qr ). dr dr

d (Qr ) dr dr d ª dt º – 4S k r 2 ˜ » dr « dr ¬ dr ¼ d ª 2 dt º – 4S k « r ˜ » dr dr ¬ dr ¼

0

...(i) (Heat flow equation)

186

Chapter : 2

dt º q g 1 ª 2 d 2t r ˜ 2  2r ˜ »  0 2 « dr ¼ k r ¬ dr d 2 t 2 dt q g  u  0 or, r dr k dr 2 d 2t dt q g r  r 2 2 0 or, (multiplying both sides by r) dr k dt d 2t dt dt q g r   r 2  0 or, dr dr k dr dt § dt · dt qg r 0  or, ¨r ¸  dr © dr ¹ dr k Integrating both sides, we have qg r 2 dt t ˜ r C1 ...(ii) dr k 2 qg r 2 d ˜ C1 (rt )  or, dr k 2 Integrating again, we have qg r 3 ˜ rt  C1r  C2 ...(iii) k 6 (where C1, C2 = constants of integration). [From eqn. (iii)] At the centre of sphere, r = 0 ? C2 = 0 Applying boundary condition, at r = R, t = tw to eqn. (iii), we have qg R3 ˜ Rtw  C1 R (Q C2 = 0) k 6 qg 2 C1 t w  R or, 6k By substituting the values of C1 and C2 in eqn. (iii), we have the temperature distribution as qg r 3 ª qg 2 º ˜ rt  R »r «t w  k 6 ¬ ¼ 6k qg 2 qg 2 ˜r t tw  R or, 6k 6k qg 2 t tw  (R – r2 ) or, ...(2.122) 6k From eqn. (2.122) it is evident that the temperature distribution is parabolic; the maximum temperature occurs at the centre (r = 0) and its value is given by qg 2 tmax tw  R ...(2.123) 6k

or,

From eqns. (2.122) and (2.123), we have

i.e.,

t – tw tmax – tw

R2 – r 2

t – tw tmax – t w

§r· 1–¨ ¸ ©R¹

§r· 1–¨ ¸ ©R¹

R2

2

2

...(2.124)

(temperature distribution in dimensionless form)

“Conduction” Heat Transfer at Steady State–One Dimension

187

Invoking Fourier’s equation (to evaluate heat flow), we have § dt · Q – kA ¨ ¸ © dr ¹r R – k u 4SR 2 u

qg 2 º d ª ( R – r 2 )» «t w  dr ¬ ¼r 6k

R

[substituting the value of t from eqn. (2.122)] ª qg º – k u 4S R 2 « (– 2r ) » ¬ 6k ¼r

k u 4S R 2 u R

qg 3k

˜R

4 3 SR u qg 3 (= volume of sphere × heat generation capacity) ...(iv) Thus heat conducted is equal to heat generated. Under steady state conditions the heat conducted (or generated) should be equal to the heat convected from the outer surface of the sphere. Q

or,

i.e., or,

qg u

4 S R3 3

tw

h u 4S R 2 (tw – ta )

ta 

qg R

...(2.125) 3h Inserting this value of tw in eqn. 2.122, we have qg R qg 2  t ta  (R – r 2 ) ...(2.126) 3h 6k qg qg 2 The maximum temperature, tmax ta  (at r = 0) ...(2.127) ˜R  ˜R 3k 6k Example 2.104. An opproximately spherical shaped orange (k = 0.23 W/m°C), 90 mm in diameter, undergoes riping process and generates 5100 W/m3 of energy. If external surface of the orange is at 8°C, determine : (i) Temperature at the centre of the orange, and (ii) Heat flow from the outer surface of the orange. Solution. Outside radius of the orange, R =

90 = 45 mm = 0.045 m 2

Rate of heat generation, qg = 5100 W/m3 The temperature at the outer surface of the orange, tw = 8°C (i) Temperature at the centre of the orange, tmax : qg 2 tmax tw  R ...[Eqn. (2.123)] 6k 5100 u (0.045) 2 15.48q C (Ans.) tmax 8  or, (6 u 0.23) (ii) Heat flow from the outer surface of the orange, Q : Heat conducted = Heat generated 4 S R3 3

?

Q

qg u

or,

Q

5100 u

4 S u (0.045)3 3

1.946 W (Ans.)

188

Chapter : 2

Example 2.105. Write down Fourier equation for heat conduction in spherical coordinate system. Hence, deduce an expression for steady state heat conduction in radial direction through a solid sphere of radius R with a uniform volumetric heat generation of qg W/m3 at the centre. Assume thermal conductivity of the material of the cylinder to be constant. (NU) Solution. For constant thermal conductivity k, the general heat conduction equation in spherical coordinates is given as ª w 2t w wt wt · º q g 1 1 1 w § ˜  2 ˜ ¨ r 2 ˜ ¸»  sin T˜ « 2 2 ˜ 2  2 wT wr ¹ ¼ k r sin T wT r wr © ¬ r sin T wI

^

`

For steady state heat conduction in radial direction, wt wt wt { 0, { 0, and {0 wI wT wW Hence, the above equation reduces to 1 d § 2 dt · qg 0 ¨r ¸ k r 2 dr © dr ¹ d § 2 dt · qg 2 0 r or ¨r ¸ dr © dr ¹ k Integrating both sides, we get dt q g r 3  ˜ r2 C1 dr k 3 dt 0 At the centre, r = 0, dr ? C1 = 0 The eqn. (i), therefore, reduces to qg r dt – ˜ dr k 3 Heat conduction in radial direction, at any radius r, is dt Q – kAr dr where, Ar = surface area of a sphere = 4Sr2 § qg r · Q – k (4S r 2 ) ¨ – ˜ ¸ , using eqn. (i) ? © k 3¹ 4 3 S r q g , which is the required expression. 3 4 Hence, heat conducted at the surface (Ans.) S R 3 qg 3

1 wt ˜ D wW ...[Eqn. (2.31)]

...(i)

...(ii)

(Here r = R)

2.10. HEAT TRANSFER FROM EXTENDED SURFACES (FINS) 2.10.1. INTRODUCTION Whenever the available surface is found inadequate to transfer the required quantity of heat with the available temperature drop and convective heat transfer coefficient, extended surfaces or fins are used. This practice, Invariably, is found necessary in heat transfer between a surface and gas as the convective heat transfer coefficient is rather low in these situations. The finned surfaces are widely used in :

“Conduction” Heat Transfer at Steady State–One Dimension

189

(i) Economisers for steam power plants; (ii) Convectors for steam and hot-water heating systems; (iii) Radiators of automobiles; (iv) Air-cooled engine cylinder heads; (v) Cooling coils and condenser coils in refrigerators and air conditioners; (vi) Small capacity compressors; (vii) Electric motor bodies; (viii) Transformers and electronic equipments etc. In practice all kinds of shapes and sizes of fins are employed; some common types of fin configurations are shown in Fig. 2.120.

Fig. 2.120. Common types of fin configurations.

For the proper design of fins, the knowledge of temperature distribution along the fin is necessary. In this article the mathematical analysis for finding out the temperature distribution and heat flow from different types of fins is dealt with. The following assumptions are made for the analysis of heat flow through the fin :

190 1. 2. 3. 4. 5. 6. 7.

Chapter : 2 Steady state heat conduction. No heat generation within the fin. Uniform heat transfer coefficient (h) over the entire surface of the fin. Homogeneous and isotropic fin material (i.e. thermal conductivity of material constant). Negligible contact thermal resistance. Heat conduction one-dimensional. Negligible radiation.

2.10.2. HEAT FLOW THROUGH “RECTANGULAR FIN” Consider a rectangular fin protruding from a wall surface as shown in Fig. 2.121.

Fig. 2.121. Rectangular fin of uniform cross-section.

Let,

l = Length of the fin (perpendicular to surface from which heat is to be removed), b = Width of the fin (parallel to the surface from which heat is to be removed), y = Thickness of the fin, P = Perimeter of the fin [=2(b + y)], Acs = Area of cross-section (=by), to = Temperature at the base of the fin, and ta = Temperature of the ambient/surrounding fluid, k = Thermal conductivity (constant), and h = Heat transfer coefficient (convective). In order to determine the governing differential equation for the fins, shown in Fig. 2.121, consider the heat flow to and from an element dx thick at a distance x from the base. Heat conducted into the element at plane x, ª dt º Qx – k Acs « » ...(i) ¬ dx ¼ x Heat conducted out of the element at plane (x + dx) ª dt º Q( x  dx ) – k Acs « » ...(ii) ¬ dx ¼ x  dx Heat convected out of the element between the planes x and (x + dx), Qconv = h (P . dx) (t – ta)

“Conduction” Heat Transfer at Steady State–One Dimension

191

Applying an energy balance on the element, we can write Qx = Q(x+dx) + Qconv. ª dt º – k Acs « » ¬ dx ¼ x

ª dt º – k Acs « »  h ( P . dx) (t – ta ) ¬ dx ¼ x  dx

...(2.128)

Making a Taylor’s expansion of the temperature gradient at (x + dx) in terms of that at x, we get d § dt · d2 § dt · ¨ ¸  ¨ ¸ dx  2 © dx ¹ x dx © dx ¹ x dx Substituting this in eqn. (2.128), we have § dt · ¨ ¸ © dx ¹ x  dx

2 § dt · (dx)  ... ¨ ¸ © dx ¹ 2!

ª d 3t º ( dx) 2 ª d 2t º ª dt º ª dt º  ..  h ( P. dx) (t – ta ) dx – k Acs « 3 » – k Acs « » – k Acs « » – kACS « » ¬ dx ¼ x ¬ dx ¼ x ¬ dx ¼ 2! ¬ dx 2 ¼ x Neglecting higher terms as dx o 0, we have ª d 2t º ª dt º ª dt º – k Acs « » – k Acs « » – k Acs « 2 » dx  h ( P. dx) (t – ta ) ¬ dx ¼ ¬ dx ¼ ¬ dx ¼ ª d 2t º k Acs « 2 » dx – h ( P . dx) (t – ta ) 0 ¬ dx ¼ Dividing both sides by Acs dx, we get,

k

d 2t dx

d 2t

2

hP (t – ta ) Acs

0

hP (t – t a ) k Acs

0 ...(2.129) dx Eqn. (2.129) is further simplified by transforming the dependent variable by defining the temperature excess T as, T(x) = t(x) – t(a) As the ambient temperature ta is constant, we get by differentiation or,

2

dT dx 2

Thus, where

d T dx 2

– m2 T m

dt ; dx 0

–

–

d 2T

d 2t

dx 2

dx 2

...(2.130) hP k Acs

Eqns. (2.129) and (2.130) represent a general form of the energy equation for one-dimensional heat dissipation from an extended surface (fin). The parameter m, for a given fin, is constant provided the convective film coefficient h is constant over the whole surface and the thermal conductivity k is constant within the temperature range considered. Then the general solution of this linear and homogeneous second order differential equation is of the form : T = C1emx + C2 e–mx ...(2.131) or, [t – ta = C1emx + C2 e–mx] where C1 and C2 are the constants, these are to be determined by using proper boundary conditions. One boundary condition is : T = To = to – ta at x = 0

192

Chapter : 2

The other boundary condtition depends on the physical situation. The following cases may be considered : Case I. The fin is infinitely long and the temperature at the end of the fin is esentially that of the ambient/surrounding fluid. Case II. The end of the fin is insulated. Case III. The fin is of finite length and loses heat by convection.

2.10.2.1. Heat dissipation from an infinitely long fin (l o f ) : Refer to Fig. 2.122. In this case the boundary conditions are :

Fig. 2.122. Infinitely long fin (Case I).

(i) At

x = 0,

t = to

(Temperature at the base of fin equals the temperature of the surface to which fin is attached.) (in terms of excess temperature)

t – ta = to – ta or, At x = 0, T = To (ii) At x = f, t = ta (Temperature at the end of an infinitely long fin equals that of the surroundings) At x = f, T = 0 (in terms of excess temperature) Substituting these boundary conditions in eqn. (2.131), we get, C1 + C2 = T0 ...(i) C1em(f) + C2 e–m(f) = 0 ...(ii) or, C1em(f) + 0 = 0 ?C1 = 0 and, C2 = T0 [From eqn. (i)] Inserting these values of C1 and C2 in eqn. (2.131), we get the temperature distribution along the length of the fin,

T

T0 e – mx ; (t – ta )

ª t – ta (to – ta ) e – mx «or ¬ to – ta

º e – mx » ¼

...(2.132)

“Conduction” Heat Transfer at Steady State–One Dimension

193

ª t – ta º The dependence of dimensionless temperature « t – t » along the fin length for different values ¬ o a¼ of parameter m (m1 < m2 < m3) is shown in Fig. 2.123; the plot indicates : (i) As the value of m increases, the dimensionless temperature falls; t – ta (ii) As the length of fin increases to infinity all the curves approach 0 asymptotically. to – t a

Fig. 2.123. Temperature distribution in a fin.

The heat flow rate can be determined in either of the two ways : (a) By considering the heat flow across the root (or base) by conduction; (b) By considering the heat which is transmitted by convection from the surface of the fin to the surrounding fluid. (a) The rate of heat flow across the base of the fin is given by (Fourier’s equation) ª dt º Q fin – k Acs « » ¬ dx ¼ x 0 ª dt º ª¬ – m (t0 – ta ) e – mx º¼ – m (t0 – ta ) [From eqn. (2.132)] «¬ dx »¼ x 0 x 0 ? Qfin = – k Acs × [– m (to – ta)] = k Acs m (to – ta) i.e., Qfin = k Acs m (to – ta) ...(2.133) or,

Q fin

or,

Q fin

k Acs

Ph (t0 – ta ) kAcs

Ph k Acs (t0 – ta )

(Substituting for m) ...[2.133 (a)]

(b) Alternatively : Q fin

f

h ( P . dx ) (t – ta )

f

h P (to – t a ) e – mx dx

³0

³0

...convective rate of heat flow [From Eqn. (2.132)]

f

h P (to – t a ) ³ e – mx dx 0

1 h P (to – ta ) m ( to – ta )

h P (to – ta )

or,

Q fin

Ph kAcs

k Acs (Substituting for m) Ph ...[Same as Eqn. 2.133 (a)]

194

Chapter : 2

[An infinitely long fin is one for which ml of, and this condition may be approached when ml > 5] From the Eqn. (2.132) it is evident that the temperature falls towards the tip of the fin, thus the area near the fin tip is not utilised to the extent as the lateral area near the base. Hence beyond a certain point the increase in the length of the fin does not contribute much in respect of increase in the dissipation of heat. Consequently a tapered fin is considered to be a better design since its lateral area is more near the base/root where temperature difference is high. Example 2.106. Calculate the amount of energy required to solder together two very long pieces of bare copper wire 1.5 mm in diameter with solder that melts at 190°C. The wires are positioned vertically in air at 20°C. Assume that the heat transfer coefficient on the wire surface is 20W/m2°C and thermal conductivity of wire alloy is 330 W/m°C. Solution. Given : d = 1.5 mm = 0.0015 m; to = 190°C; ta = 20°C; h = 20 W/m2°C; k = 330 W/m°C. Energy required to solder : S 2 S u 0.00152 1.767 u 10–6 m 2 Acs d Area of cross-section, 4 4 Perimeter, P = Sd = S × 0.0015 = 4.712 × 10–3 m Heat dissipation from a long fin is given by, Qfin = k Acs m (to – ta) ...[Eqn. (2.133)]

hP 20 4.712 u 10 –3 u 12.71 m –1 k Acs 330 1.767 u 10 –6 ? Qfin = 330 × 1.767 × 10–6 × 12.71 (190 – 20) = 1.26 W ? Total energy required for two wires = 2 × 1.26 = 2.52 W (Ans.) Example 2.107. It is required to heat oil to about 300°C for frying purpose. A laddle is used in the frying. The section of the handle is 5 mm × 18 mm. The surroundings are at 30°C. The conductivity of the material is 205 W/m°C. If the temperature at a distance of 380 mm from the oil should not reach 40°C, determine the convective heat transfer coefficient. Solution. Refer to Fig. 2.124. to = 300°C; b = 18 mm = 0.018 m; y = 5 mm = 0.005 m; l = 380 mm = 0.38 m; k = 205 W/m°C; ta = 30°C. where,

m

Fig. 2.124.

Convective heat transfer coefficient, h : Assumming the fin to be long one, we have t – ta e – mx to – t a to – ta emx or, t – ta 300 – 30 e m u 0.38 or, 40 – 30

...[Eqn. (2.132)]

(Q x = 380 mm = 0.38 m)

“Conduction” Heat Transfer at Steady State–One Dimension e0.38m

or,

m

But, or, or,

hP k Acs h u [0.018  0.005) u 2] 205 u (0.018 u 0.005)

= 27 hP k Acs

or

195

m = 8.673 8.673

75.22 75.22

75.22 u 205 u (0.018 u 0.005) 30.17 W/m 2 q C (Ans.) (0.018  0.005) u 2 Example 2.108. A temperature rise of 60°C in a circular shaft of 60 mm diameter is caused by the amount of heat generated due to friction in the bearing mounted on the crankshaft. The thermal conductivity of the shaft material is 50 W/m°C and the heat transfer coefficient is 6.5 W/m2°C. (i) Develop an expression for the temperature distribution; (ii) Determine amount of heat transferred through the shaft. Assume that the shaft is a rod of infinite length. Solution. Temperature rise, To = 60°C Diameter of the shaft, d = 60 mm = 0.06 m Thermal conductivity of material, k = 50 W/m°C Heat transfer coefficient, h = 6.5 W/m2 °C (i) Expression for temperature distribution : The general expression for temperature distribution for an extended surface is given by ...[Eqn. (2.131)] T = C1emx + C2e–mx Using the following boundary conditions, we have At x = 0, T = To At x = f, T=0 C1 = 0; C2 = To T = T o e–mx ...Required expression. (Ans.) ? (ii) Amount of heat transferred, Q : dT Q – k Acs ˜ dx x 0 ªd º – k Acs . « (To e – mx ) » ¬ dx ¼x 0 or, Q = kAcs mT0

or,

h

Sd h 4h 4 u 6.5 u 2.944 k S d 2 kd 50 u 0.06 4 S § · 2 ? (Ans.) Q 50 u ¨ u 0.06 ¸ u 2.944 u 60 24.97 W ©4 ¹ Example 2.109. A very long 25 mm diameter copper rod (k = 380 W/m°C) extends horizontally from a plane heated wall at 120°C. The temperature of the surrounding air is 25°C and the convective heat transfer coefficient is 9.0W/m2°C. (i) Determine the heat loss; (ii) How long the rod be in order to be considered infinite?

where,

m

hP k Acs

196

Chapter : 2

Solution. d = 25 mm = 0.025 m; k = 380 W/m°C; to = 120°C; ta = 25°C; h = 9.0 W/m2°C. (i) Heat loss: Heat dissipation from an infinitely long fin is given by Qfin = k Acs m (to – ta) ...[Eqn. (2.133)] h u Sd 4h 4 u 9.0 1.946 S kd 380 u 0.025 k u d2 4 S 4h ? Q fin k u d 2 u (t o – t a ) kd 4 S 380 u u 0.0252 u 1.946 (120 – 25) 34.48 W (Ans.) 4 (ii) Length of the rod (to be considered infinite) : Since there is no heat loss from the tip of an infinitely long rod and as such it behaves as if the tip were insulated; therefore, an estimate of the validity of this approximation can be made by comparing the following two expressions for the fin : Qfin = kAcs m (to – ta) ... infinitely long fin; Qfin = kAcs m (to – ta) tanh (ml) ... fin with tip insulated. Equivalent results are obtained if tanh (ml) t 0.99 or ml t 2.646. Hence the rod can be considered infinite if 2.646 2.646 1.36 m lt (Ans.) m 1.946 Example 2.110. Two long rods of the same diameter, one made of brass (k = 85 W/m°C) and other made of copper (k = 375 W/m°C) have one of their ends inserted into the furnace. Both of the rods are exposed to the same environment. At a distance 105 mm away from the furnace end, the temperature of the brass rod is 120°C. At what distance from the furnace end the same temperature would be reached in the copper rod ? (IES) Solution. k1 = 85 W/m°C; k2 = 375 W/m°C The controlling differential equation for the heat flow in the rod is given by d 2T – m2 T 0 dx 2 The general solution is T = C1emx + C2 e–mx ...(1) The boundary conditions are : (i) At x = 0, T = To (ii) At x = f, T=0 Using the boundary condition (ii), we get 0 = C1emx ? C1 = 0 –mx ? T = C2 e ...(2) [Substituting C1 = 0 in eqn. (1)] Now using boundary condition (i), we get To = C2 ? T = To e–mx [Substituting C2 = To in eqn (2)] t – ta T e – mx or, T0 to – ta ? t = ta + (to – ta)e–mx ...(3)

where,

m

hP k Acs

“Conduction” Heat Transfer at Steady State–One Dimension

197

Now using eqn. (3) for brass rod, when t = 120°C at x = 105 mm = 0.105 m, we have

Fig. 2.125.

? 120 = ta + (to – ta) e(– m1 u 0.105) Now using eqn. (3) for copper rod, when t = 120°C at x = l, we have ? 120 ta  (t0 – ta ) e – m2l as ta and to are same for both the rods. Equating (4) and (5), we get or 0.105 m1 = m2l e – m1 u 0.105 e – m2 l m1 u 0.105 l ? m2 hP hP m1 m2 But, and k1 Acs k2 Acs and, ?

k2 Acs m1 k2 hP 375 u 2.1 m2 k1 Acs hP k1 85 l = 2.1 × 0.105 = 0.22 m or 220 mm (Ans.)

2.10.2.2. Heat dissipation from a fin insulated at the tip : Fig. 2.126 illustrates a fin of finite length with insulated end (i.e. no heat loss from the end of the fin). The boundary conditions are : (i) At x = 0, T = To (ii) At x = l, dt 0 dx Applying these boundary conditions to eqn. (2.131), we have

Fig. 2.126. The fin with insulated end (Case II).

...(4) ...(5)

...(6)

198

Chapter : 2 C1 + C2 = To t – ta = C1 emx + C2 e–mx

Further

dt dx ª dt º «¬ dx »¼ x l ml ? C1 e – C2 e–ml = Solving eqns. (i) and (ii), we have C2 = C1eml – (To – C1) e–ml = or, C1eml – To e–ml + C1 e–ml = or, C1 (eml + e–ml) =

...(i) [Eqn. 2.131]

m C1 emx – m C2 e – mx m C1 eml – m C2 e – ml

0 To – C1 0 0 To e–ml

0

[As per boundary condition (ii)] ...[From eqn. (i)]

ª e – ml º To « ml – ml » ¬e  e ¼ ª °­ e – ml °½º C2 To – «To ® ml ? ...[From eqn. (i)] – ml ¾» «¬ ¯° e  e ¿°»¼ ª ª º e – ml º e ml To « ml C2 To «1 – ml or, – ml » – ml » e e ¼ ¬ ¬e  e ¼ Inserting the values of C1 and C2 in eqn. (2.131), we have C1

or,

ª e – ml º mx ª º – mx e ml To « ml  T e e » « o – ml – ml » ml ¬e  e ¼ ¬e  e ¼ ª e m( x – l )  e m ( l – x) º ª e m( l – x)  e[– m( l – x )] º T « » « » or, To ¬ e ml  e – ml e ml  e – ml ¼ ¬ ¼ The above expression, in terms of hyperbolic functions, can be expensed as t – ta T cosh{m(l – x)} ..(2.134) cosh (ml ) To to – ta ...Expression for temperature distribution ( – ) [– m l x ª  e m(l – x)] e e ml  e – ml º , and cosh (ml ) «Q cosh {m (l – x)} » ¬ ¼ 2 2 The rate of heat flow from the fin is given by ª dt º Q fin – k Acs « » ¬ dx ¼ x 0 ª cosh{m (l – x )} º ...[From eqn. (2.134)] t – ta (to – ta ) « Now, » ¬ cosh (ml ) ¼ dt ª sinh{m (l – x )} º d ª º (t o – t a ) « [cosh (mx)] m sinh (mx) » » (– m) «Q cosh ( ) dx ml dx ¬ ¼ ¬ ¼ dt ª º – m (to – ta ) tanh (ml ) «¬ dx »¼ x 0 ? Qfin = kAcs m (to – ta) tanh (ml) ...(2.135) (Substituting for m) T

or,

Q fin

PhkAcs (to – ta ) tanh (ml )

...[2.135 (a)]

“Conduction” Heat Transfer at Steady State–One Dimension

199

Example 2.111. Aluminium fins of rectangular profile are attached on a plane wall with 5 mm spacing. The fins have thickness y = 1 mm, length l = 10 mm, and the thermal conductivity, k = 200 W/m K. The wall is maintained at a temperature 200°C, and the fins dissipate heat by convection into the ambient air at 40°C, with heat transfer coefficient h = 50 W/m2K. Determine the heat loss. (AMIE Winter, 1998) Solution. Given : t = 1 mm = 0.001 m; l = 10 mm = 0.01 m; y = 1 mm = 0.001 m; k = 200 W/mK; to = 200°C; ta = 40°C; h = 50 W/m2 K. Heat loss, Q : m

hP kAcs

h (b  y ) u 2 k (b u y )

h u 2b , assuming b >> y k u by 2h ky

2 u 50 200 u 0.001

22.36

For unit width of the fin, (b = 1 m)

Fig. 2.127.

Q

PhkAcs (t0 – ta ) tanh (ml ) (2 u 1) u 50 u 200 u (1 u 0.001) u (200 – 40) u tanh (22.36 u 0.01)

= tanh (0.2236) = 157.38 W/m (Ans.) Example 2.112. Find out the amount of heat transferred through an iron fin of length 50 mm, width 100 mm and thickness 5 mm. Assume k = 210 kJ/mh°C and h = 42 kJ/m2h°C for the material of the fin and the temperature at the base of the fin as 80°C. Also determine the temperature at tip of the fin, if the atmosphere temperature is 20°C. (GATE) Solution. Given : l = 50 mm = 0.05 m; b = 100 mm = 0.1 m; y = 5 mm = 0.005 m; k = 210 kJ/mh°C; h = 42 kJ/m2h°C; t0 = 80°C; ta = 20°C.

200

Chapter : 2

Amount of heat transferred through the fin, Q : Perimeter, P = 2 (b + y) = 2 (0.1 + 0.005) = 0.21 m Area, Acs = b × y = 0.1 × 0.005 = 0.0005 m2 hP kAcs

m

?

Qfin

42 u 0.21 210 u 0.0005

9.165

PhkAcs (t0 – ta ) tanh (ml )

...[Eqn. (2.135)] 0.21 u 42 u 210 u 0.0005 u (80 – 20) tanh (9.165 u 0.05)

= 0.9263 × 60 × 0.4286 = 24.75 kJ/h (Ans.) Temperature at the tip of the fin, QL : We know that, t – ta t0 – ta

T T0

cosh{m (l – x)} cosh (ml ) ...[Eqn. (2.134)]

Fig. 2.128.

At x = l, we have t – 20 1 80 – 20 cosh ml or,

t

60  20 cosh (9.165 u 0.05)

74.21q C (Ans.)

Example 2.113. A carbon steel (k = 54 W/m°C) rod with a cross-section of an equilateral triangle (each side 5 mm) is 80 mm long. It is attached to a plane wall which is maintained at a temperature of 400°C. The surrounding environment is at 50°C and unit surface conductance is 90 W/m2°C. Compute the heat dissipated by the rod. (M.U., 1997) Solution. Refer to Fig. 2.129. a = 5 mm = 0.005 m; l = 80 mm = 0.08 m; to = 400°C; ta = 50°C; h = 90 W/m2°C; k = 54 W/m°C.

Fig. 2.129.

Heat dissipated by the rod, Q : The heat flow from the rod (considering tip of the fin to be insulated) is given by Q = kAcs m (to – ta) tanh (ml) ...[Eqn. (2.135)] where,

m

hP ;P kAcs

3a, Acs

§ 3 · 1 uau¨ a¸ © 2 ¹ 2

3 2 a 4

“Conduction” Heat Transfer at Steady State–One Dimension

201

3a 4 3 6.93 1386 a 0.005 3 2 ua 4 90 u 1386 48.06 m and, 54 Substituting the values in the above equation, we have ª 3 º u (0.005)2 » u 48.06 u (400 – 50) tanh(48.06 u 0.08) 9.82 W (Ans.) Q 54 u « ¬ 4 ¼ Example 2.114. The aluminium square fins (0.5 mm × 0.5 mm) of 10 mm length are provided on a surface of semiconductor electronic device to carry 1W of energy generated by electronic device. The temperature at the surface of the device should not exceed 80°C when surrounding temperature is 40°C. Taking the following data, find the number of fins required to carry out above duty. Neglect the heat loss from the end of fins. k (aluminium) = 200 W/m°C; h = 15 W/m2°C (M.U.) Solution. Given : Acs (area of cross-section of fin) = 0.5 mm × 0.5 mm = 2.5 × 10–7 m2; l = 10 mm = 0.01 m; Q (rate of heat transfer) = 1 W; t0 = 80°C; ta = 40°C; k = 200 W/m°C; h = 15 W/m2°C. Number of fins required, n : The heat carried by n number of fins is given by ?

P Acs

Qfins = n [k Acs m (to – ta) tanh (ml)] Where,

m

hP kAcs

0.5 · § 15 u ¨ 4 u ¸ © 1000 ¹ 200 u 2.5 u 10 –7

...[Refer Eqn. (2.135)]

24.5

0.5 º ª «¬ where P (perimeter) 4 u 1000 m »¼ and ml = 24.5 × 0.01 = 0.245 Substituting the values in the above equation, we have l = n [ 200 × 2.5 × 10–7 × 24.5 (80 – 40) tanh (0.245)] = n (200 × 2.5 × 10–7 × 24.5 × 40 × 0.2402) = 0.01177 1 n ; 85 ? (Ans.) 0.01177 Example 2.115. Pin fins are provided to increase the heat transfer rate from a hot surface. Which of the following arrangement will give higher heat transfer rate ? (i) 6 – fins of 10 cm length. (ii) 12 – fins of 5 cm length. Take k (fins material) = 200 W/m°C, h = 20 W/m2°C Cross-sectional area of fin = 2 cm2 Perimeter of fin = 4 cm Fin base temperature = 230°C Surrounding air temperature = 30°C (Shivaji University) Solution. Given : Acs = 2 × 10 –4 m 2; P = 0.04 m, t o = 230°C; t a = 30°C; k = 200 W/m°C; h = 20 W/m 2°C. The heat flow through ‘n’ number of fins is given by Q = n [k Acs m (t0 – ta) tanh (ml)]

202

Chapter : 2 hP kAcs

m

20 u 0.04

4.47 200 u 2 u 10–4 (i) Case I. n = 6 and l = 10 cm = 0.1 m ml = 4.47 × 0.1 = 0.447 ? Q1 = 6 [200 × 2 × 10–4 × 4.47 × (230 – 30) tanh (0.447)] = 89.99 W (ii) Case II. n = 12 and l = 5 cm = 0.05 m ml = 4.47 × 0.05 = 0.2235 ? Q2 = 12 [200 × 2 × 10–4 × 4.47 (230 – 30) tanh (0.2235)] = 94.34 W This shows that the rate of heat transfer is higher in second case, therefore, this arrangement (Case II) is better. Example 2.116. One end of a long rod, 35 mm in diameter, is inserted into a furnace with the other end projecting in the outside air. After the steady state is reached, the temperature of the rod is measured at two points 180 mm apart and found to be 180°C and 145°C. The atmospheric air temperature is 25°C. If the heat transfer coefficient is 65 W/m2 °C, calculate the thermal conductivity of the rod. Solution. Diameter of the rod, d = 35 mm = 0.035 m The atmospheric air temperature, ta = 25°C Heat transfer coefficient; h = 65 W/m2°C The starting point x = 0 is considered at the first point where the temperature is measured; x = l is considered at the outer point. Assume that the end of the fin is insulated. For insulated end, we have t – ta T cosh{m (l – x)} ...[Eqn. (2.134] cosh ml To to – ta At x = l, this equation, reduces to Tl 1 or, ...(i) To cosh ml Here,Tl = 145 – 25 = 120°C and To = to – ta = 180 – 25 = 155°C 120 1 ? 155 cosh ml 155 cosh ml 1.292 or or, ml = 0.747 120 0.747 m or, l hP m But, kAcs

where,

?

hP kAcs

0.747 l

or,

hP kAcs

0.7472

or,

or,

h Sd ˜ k S d2 4 k

l2

0.747 2 l

2

4 hl 2 0.558 d

or

h 4 u k d

0.558 l2

4 u 65 u 0.182 0.558 u 0.035

[where, l = 180 mm = 0.18 m (given)]

431.34 W/mq C

(Ans.)

“Conduction” Heat Transfer at Steady State–One Dimension

203

Example 2.117. An electric motor drives a centrifugal pump which circulates a hot liquid metal at 480°C. The motor is coupled to the pump impeller by a horizontal steel shaft (k = 32 W/m°C) 25 mm in diameter. If the ambient air temperature is 20°C, the temperature of the motor is limited to a maximum value of 55°C and the heat transfer coefficient between the steel shaft and the ambient air is 14.8 W/m2°C, what length of shaft should be specified between the motor and the pump? Solution. Given : d = 25 mm = 0.025 m; to = 480°C; ta = 20°C; tl = 55°C k = 32 W/m°C; h = 14.8 W/m2°C. (The shaft conducts heat from the pump (to = 480°C) towards motor (tl = 55°C) and also loses energy from its surface to the surroundings (ta = 20°C) by convection.) The temperature distribution, treating the shaft as a fin insulated at the tip, is given by T To

t – ta to – ta

cosh[m (l – x )] cosh (ml )

[Eqn. 2.134)]

At x = l, t = tl; the above equation reduces to or,

tl – ta to – t a

1 cosh (ml )

or,

55 – 20 480 – 20

1 cosh (ml )

?

ml = 3.267

But,

m

?

hP kAcs

or

cosh (ml )

h u Sd S k u d2 4

4h kd

(480 – 20) (55 – 20) 4 u 14.8 32 u 0.025

13.14

8.6 m –1

8.6 l = 3.267

3.267 or 379.9 mm (Ans.) 0.3799 m 8.6 Example 2.118. The heat exchange, in a certain chemical process, is involved from metal surface to distilled water. A number of thin metal fins, each 70 mm long and 3.5 mm thick, are provided to increase the heat transfer rate and their ends are attached to an insulated wall. In order to prevent ionization of water the metal fins are coated with 0.12 mm thick layer of plaster. The mean water temperature and temperature at the base of the fin are 30°C and 90°C respectively. The thermal conductivities of the fin material and plastic are 205 W/m°C and 0.52 W/m°C respectively. If the heat transfer coefficient between the plastic coating and water is 250 W/m2°C, calculate :

or

l

(i) The temperature at the tip of fin, and (ii) The fin efficiency. Solution. Length of a fin, l = 70 mm = 0.07 m Thickness of fin, y = 3.5 mm = 0.0035 m Thickness of plastic coating, Gp = 0.12 mm = 0.00012 m Mean water temperature, ta = 30°C Temperature at the base of fin, to = 90°C Thermal conductivity of fin material, kf = 205 W/m°C Thermal conductivity of plastic, kp = 0.52 W/m°C Heat transfer coefficient; h = 250 W/m2°C.

204

Chapter : 2

(i) Temperature at the tip of fin, tl : hP , the value of kAcs h is replaced by U as a result of resistance to heat transfer due to layer of plastic coating on the metallic fins; thus 1 1 Gp 1 0.00012 0.00023   250 0.52 U h kp

When making calculations for the value of m given by the expression m

or,

U = 236.4 W/m2°C

For fin of rectangular cross-section, 2U kf y

m

236.4 u 2 205 u 0.0035

25.67 (when 2 y  2b)

For a fin with insulated tip, t – ta to – ta

T To

cosh m(l – x) cosh ml

[Eqn. (2.134)]

At x = l, the above equation reduces to Tl To

?

tl – 30 90 – 30

1 cosh ( ml )

1 cosh (25.67 u 0.07)

0.3227

tl = 30 + ( 90 – 30 ) × 0.3227 = 49.36°C (Ans.)

(ii) Fin efficiency, Kfin : K fin

tanh (ml ) ml

...[Eqn. (2.139)]

tanh (25.67 u 0.07) 0.5267 or 52.67% (Ans.) (25.67 u 0.07) Example 2.119. A heating unit is made in the form of a vertical tube fitted with rectangular section steel fins. The tube height is 1.2 m and its outer diameter is 60 mm. The fins are 50 mm in height and their thickness is 3 mm. The total number of the fins used is 20. The temperature at the base of a fin is 80°C and surrounding air temperature is 18°C. The heat transfer coefficient on the fin surface and tube surface to the surrounding air is 9.3 W/m2°C. k (fin material) = 55.7 W/m°C

Calculate the amount of heat transferred from the tube with and without fin. Solution. Given : Height of the tube, H = 1.2 m Outer diameter of the tube, do = 60 mm = 0.06 m Height of the fin, l = 50 mm = 0.05 m Width of the fin, b (= H) = 1.2 m Thickness of the fin, y = 3 mm = 0.003 m Total number of fins, n = 20 Temperature at the base of the fin, to = 80°C Surrounding air temperature, ta = 18°C Heat transfer coefficient, h = 9.3 W/m2°C Thermal conductivity of fin material, k = 55.7 W/m°C. Heat transferred from the tube with and without fin : Let Q1 = Rate of heat flow from the tube surface without fins. Q2 = Rate of heat flow from the tube surface when fins are fitted,

“Conduction” Heat Transfer at Steady State–One Dimension

205

Qb = Rate of heat convected from the base, and Qf = Rate of heat convected from the fins. Now, Q1 = h (SdoH) (to – ta) =9.3 × (S × 0.06 × 1.2) (80 – 18) = 130.42 W (Ans.) Qb = h Ab (to – ta) where, Ab = (S × 0.06 × 1.2) – 20 (1.2 × 0.003) = 0.1542 m2 ? Qb = 9.3 × 0.1542 (80 – 18) = 88.91 W Qf = n kAcs m (to – ta) tanh (ml) ...[Eqn. (2.135) (Assuming fin end is insulated) where,

m

hP kAcs

h (b  y ) u 2 u k by

9.3 (1.2  0.003) u 2 u 55.7 1.2 u 0.003

10.56

and, ml = 10.56 × 0.05 = 0.53 ? Qf = 20 × 55.7 × (1.2 × 0.003) × 10.56 (80 – 18) tanh (0.53) = 1274.5 W and, Q2 = Qb + Qf = 88.91 + 1274.5 = 1363.41 W (Ans.) Example 2.120. A metal tank containing cooling oil is to have its dissipation rate by convection, increased by 70% by adding the fins to the wall surface. The fins will be 5 mm thick and spaced 100 mm apart between the centres. The surface temperature of the tank is 95°C and surrounding atmosphere temperature is 15°C. The natural convection heat transfer coefficient of the surface is 40 W/m2°C. Determine the length (height) of each fin on the assumption that the convection coefficient remains unchanged and surface temperature of the tank is dropped to 90°C when fins are fitted. Take k (metal tank and fin) = 240 W/m°C. Neglect the heat transfer from the tips of the fin. (M.U.) Solution. Consider tank surface of 1m × 1 m as shown in Fig. 2.130. Let, Q1 = The rate of heat transfer from the tank surface when fins are not fitted, Q2 = The amount of heat which will be dissipated per unit time after fitting the fins, and [= 1.7 Q1 .... given] l = Length of each fin.

Fig. 2.130.

206

Chapter : 2

Now,

Q1 = hA (to – ta) = 40 × 1 × (95 – 15) = 3200 W Q2 = 1.7 Q1 = 1.7 × 3200 = 5440 W Out of Q2, part of the heat is dissipated from the base surface and remaining from the fin surfaces. The number of fins per metre width of the tank = 10 ...(See fig. 2.130) § 5 · u 1 ¸ 0.95 m 2 1 u 1 – 10 ¨ © 1000 ¹ ? Qbase (the rate of heat transfer from the base surface) = 40 × 0.95 (90 – 15) = 28.50W ? Qfins (the rate of heat transfer from the fin surfaces) = Q2 – Qbase = 5440 – 2850 = 2590 W Also, Qfins = nkAcs m (to – ta) tanh (ml) ...[Eqn. (2.135]

? The base surface left after fitting the fins

where,

m

hP P where kAcs Acs

(b  y ) u 2 by

(1  0.005) u 2 1 u 0.005

402

40 u 402 8.185 240 Substituting the values in the above equation, we get

?

m

or,

tanh (8.185 l )

or,

2590 10 u 240 u (1 u 0.005) u 8.185 u (90 – 15) 8.185 l = 0.37

0.351

0.37 0.0452 m or 45.2 mm (Ans.) 8.185 Example 2.121. In the Fig. 2.131 are shown copper tubes soldered to steel plate panel (1mm thick); the tubes are parallel to each other and kept 450 mm apart. The steam at 100°C saturation temperature is condensed to water at 100°C passing through the copper tubes. Determine : (i) The heat loss per metre length of each tube, and (ii) The mass of steam condensed per hour if the panel contains 25 tubes. Take h (air) = 25 W/m2°C; k (steel) = 48 W/m°C; Air temperature = 26°C.

?

l

Fig. 2.131.

“Conduction” Heat Transfer at Steady State–One Dimension

207

Solution. Refer to Fig. 2.131. Thickness of steel plate, y = 1 mm = 0.001 m Length of the fin, l = 225 mm = 0.225 m Temperature at the base of the fin, to = 100°C Temperature of air, ta = 26°C Thermal conductivity of steel, k = 48 W/m°C Heat transfer coefficient for air, h = 25 W/m2°C (i) Heat loss per metre length (b = 1 m) of each tube : If the panel between the two tubes is considered, the coldest spot will be middle of the two tubes § dt · (at Z–Z) due to symmetry of the tubes arrangement. Therefore, the temperature gradient ¨ ¸ at this © dx ¹ point will be zero. The half length of the panel between the two tubes becomes a fin for the tube and each tube will have two fins each of 250 mm length/height. The heat flow though the panel for each tube is given by Q = [k Acs m (to – ta) tanh (ml)] × 2 where,

m

hP kAcs

h (b  y ) u 2 u k b˜ y

25 (1  0.001) u 2 u 48 1 u 0.001

32.29

Q = [48 × (1 × 0.001) × 32.29 (100 – 26) tanh (32.29 × 0.225) ] × 2 = 229.38 W = 229.38 J/s per metre length of each tube. (Ans.) (ii) The mass of steam condensed per hour : Number of tubes = 25 ...(Given) ? The mass of steam condensed per hour (229.38 u 3600) u 25 9.146 kg/h (Ans.) 2257 u 1000 [where latent heat (hfg) = 2257 kJ/kg at 100°C ....from steam tables] ?

2.10.2.3. Heat dissipation from a fin losing heat at the tip Fig. 2.132 illustrates a fin of finite length losing heat at the tip.

Fig. 2.132. A fin of finite length losing heat at tip (case III).

208

Chapter : 2

The boundary conditions are : (i) At x = 0, T = T0 (ii) Heat conducted to the fin at

x=l = Heat convected from the end to the surroundings.

ª dt º – k Acs « » h Asu (t – ta ) ¬ dx ¼ x l where Acs (cross-sectional area for heat conduction) equals Asu (surface area from which the convective heat transport takes place), at the tip of the fin; i.e. Acs = Asu.

i.e.,

dt hT – at x = l dx k Applying these boundary conditions to eqn. (2.131), we get C1 + C2 = T0 Further t – ta = C1emx + C2 e–mx Differentiating this expression w.r.t. x, we have

Thus

dt dx ª dt º «¬ dx »¼ x

l

...(i) [Eqn. 2.131]

m C1 emx – m C2 e – mx m C1 e ml – m C2 e – ml

–

hT k

hT km h – ml ml C1 e – C2 e – [C1 eml  C2 e – ml ] or, km [Q T(x = l) = C1 eml + C2 e–ml] Solving eqns. (i) and (ii), we have C2 = To – C1

or,

C1 e ml – C2 e – ml

C1 e ml – (To – C1 ) e – ml C1 e ml – To e – ml  C1 e ml h ml h – ml º ª C1 «(eml  e – ml )  e – e » km km ¬ ¼ h ª º C1 «(e ml  e – ml )  (eml – e – ml ) » km ¬ ¼

?

and,

C1

C2

–

...(ii)

...[From eqn. (i)]

h [C1 eml  To – C1 ) e – ml ] ...[From eqn. (ii)] km h h h ˜ C1e ml – ˜ C1 ˜ e – ml – · To ˜ e – ml  km km km h To e – ml – T0 ˜ e – ml km h º ª To e – ml «1 – km »¼ ¬ –

ª ml «¬(e

h º – ml ª T0 «1 – e km »¼ ¬ h ml º  e – ml )  (e – e – ml ) » km ¼

ª º h · – ml § T0 ¨ 1 – ¸e « » km © ¹ To – « » « (eml  e – ml )  h (eml – e – ml ) » «¬ »¼ km

“Conduction” Heat Transfer at Steady State–One Dimension ª « To «1 – « (e ml  «¬

209

º h · – ml § ¨1 – ¸e » km ¹ © » h ml – ml – ml » e ) (e – e ) »¼ km

h ml h – ml º ª ml – ml – ml – ml « (e  e )  km (e – e ) – e  km e » To « » h ml « » (e ml  e – ml )  (e – e – ml ) «¬ »¼ km h ml h – ml h – ml º ª ml – ml ˜e » – e – ml  « e  e  km e – km e km To « » h ml « » (eml  e – ml )  (e – e – ml ) km ¬« ¼» h º ml ª To «1  e km »¼ ¬ or, C2 h ml ª ml – ml – ml º «¬ (e  e )  km (e – e ) »¼ Substituting these values of constants C1 and C2 in eqn. (2.131), we get T = C1emx + C2 e–mx T

ª º h · – ml § To ¨ 1 – ¸e « » mx km ¹ © « »e  « (eml  e – ml )  h (eml – e – ml ) » «¬ »¼ km

ª § To ¨ 1  « © « « (eml  e – ml )  «¬

...[Eqn (2.129)] º h · ml ¸e » – mx km ¹ »e h ml (e – e – ml ) » »¼ km

h m (l – x ) [e – e – m (l – x ) ] km or, h ml [(e ml  e – ml )  (e – e – ml )] km h cosh[m (l – x)]  [sinh{m (l – x)}] t – ta T km or, h To to – ta cosh (ml )  [sinh (ml )] km The rate of heat flow from the fin is given by ª dt º Q fin – k Acs « » ¬ dx ¼ x 0 T To

[e m ( l – x )  e – m ( l – x ) ] 

h ª º « cosh{m (l – x)}  km [sinh m (l – x)] » Now, t – ta (to – ta ) « » h « » cosh (ml )  {sinh (ml )} «¬ »¼ km Differentiating the above expression w.r.t. x, we get dt dx

ª ª h ºº « – m sinh{m(l – x)} – m «¬ km {cosh[m (l – x)]}»¼ » (to – ta ) « » h « » cosh (ml )  {sinh (ml )} km ¬« ¼»

...(2.136)

210

Chapter : 2

ª dt º «¬ dx »¼ x

?

or,

0

Q fin

Q fin

ª « sinh (ml )  – (to – ta ) m « « cosh (ml )  «¬

h º {cosh (ml ) » km » h {sinh (ml )} » »¼ km

h ª º « sinh (ml )  km {cosh (ml )} » k Acs m (to – ta ) « » « cosh ( ml )  h {sinh ( ml )} » «¬ »¼ km h ª º « sinh (ml )  km {cosh (ml )} » P h k Acs (to – ta ) « » « cosh (ml )  h {sinh (ml )} » (Substituting for m) «¬ »¼ km h ª º « tanh (ml )  km » P h k Acs (to – ta ) « » ...(2.135) «1  h ˜ tanh (ml ) » km ¬« ¼»

Example 2.122. A motor body is 360 mm in diameter (outside) and 240 mm long. Its surface temperatere should not exceed 55°C when dissipating 340 W. Longitudinal fins of 15 mm thickness and 40 mm height are proposed. The convection coefficient is 40 W/m2°C. Determine the number of fins required. Atmospheric temperature is 30°C. Thermal conductivity = 40 W/m°C. Solution. Refer to Fig. 2.133, l = 40 mm = 0.04 m; b = 240 mm = 0.24 m; y = 15 mm = 0.015 m; k = 40 W/m°C; h = 40 W/m2°C, t0 = 55°C; ta = 30°C; Qtotal = 340 W. The given problem is a case of heat dissipation from a fin losing heat at the tip (short fin situation). The rate of heat transfer in such a case is given by

Q fin

h º ª « tanh (ml )  km » P h k Acs (to – ta ) « » «1  h u tanh (ml ) » «¬ »¼ km

Fig. 2.133

...[Eqn. (2.137)]

“Conduction” Heat Transfer at Steady State–One Dimension h (b  y ) u 2 k (b u y )

hP kAcs

where,

m

and,

ml = 11.9 × 0.04 = 0.476 Q fin

?

? Number of fins required

40 u (0.24  0.015) u 2 40 u (0.24 u 0.015)

211 11.9

[(0.24  0.015) u 2] u 40 u 40 u (0.24 u 0.015) u (55 – 30) 40 ª º « tanh (0.476)  40 u 11.9 » » u« 40 «1  u tanh (0.476) » «¬ »¼ 40 u 11.9

ª 0.443  0.084 º 1.714 u 25 u « » ¬1  0.084 u 0.443 ¼ Qtotal 340 16 fins (Ans.) 21.77 Q fin

21.77 W

Example 2.123. A turbine blade made of stainless steel (k = 29 W/m°C) is 60 mm long, 500 mm2 cross-sectional area and 120 mm perimeter. The temperature of the root of blade is 480°C and it is exposed to products of combustion passing through the turbine at 820°C. If the film coefficient between the blade and the combustion gases is 320 W/m2°C, determine : (i) The temperature at the middle of the blade; (ii) The rate of heat flow from the blade. Solution. Given : l = 60 mm = 0.06 m; Acs = 500 mm2 = 500 × 10–6 m2; P = 120 mm = 0.12 m; to = 480°C; ta = 820°C; k = 29 W/m°C; h = 320 W/m2°C. (i) The temperature at the middle of the blade : The temperature distribution for a fin losing heat at the tip is given by, T To

where,

m

hP kAcs

t – ta to – ta

h [sinh{m(l – x )}] km h cosh (ml )  [sinh (ml )] km

cosh{m(l – x )} 

320 u 0.12

29 u 500 u 10 ml = 51.46 × 0.06 = 3.087 h 320 0.214 km 29 u 51.46

–6

51.46 m –1

l (middle of the blade), the above equation reduces to 2 h § ml · § ml · t(l / 2) – ta cosh ¨© 2 ¸¹  km sinh ¨© 2 ¸¹ h to – ta cosh (ml )  sinh (ml ) km Substituting the values, we have

At x

t(l / 2) – 820 480 – 820

§ 3.087 · § 3.087 · cosh ¨ ¸  0.214 sinh ¨ ¸ © 2 ¹ © 2 ¹ cosh (3.087)  0.214 sinh (3.087)

...[Eqn. (2.136)]

212

Chapter : 2

2.925 0.219 13.318 ? t(l/2) = 820 + 0.219 (480 – 820) = 745.54°C (Ans.) (ii) The rate of heat flow from the blade, Q :

Q

h º ª « tanh (ml )  km » k Acs m (to – ta ) « » «1  h tanh (ml ) » km ¬« ¼»

...[Eqn. (2.137)]

ª tanh (3.087)  0.214 º 29 u 500 u 10 –6 u 51.46 (480 – 820) « » ¬1  0.214 tanh (3.087) ¼ = – 253 W (Ans.) The –ve sign indicates that heat flows from the combustion gases to the turbine blade. Example 2.124. A fin 5 mm thick and 45 mm long has its base on a plane plate which is maintained at 125°C. The ambient temperature is 25°C. The conductivity of fin material is 55W/m°C and the heat transfer coefficient is 145 W/m2°C. Determine : (i) Temperature at the end of the fin, (ii) Temperature at the middle of the fin, and (iii) Heat dissipated by the fin (per metre width).

Solution. Refer to Fig. 2.134. l = 45 mm = 0.045 m; b = 1m; y = 5mm = 0.005m; k = 55 W/m°C; h = 145 W/m2°C; to = 125°C; ta = 25°C. (i) Temperature at the end of the fin, tl : Assuming heat loss by convection from the end of the fin; under this condition, temperature at the end of the fin is given by T To

or,

t – ta to – ta tl – ta to – t a

h ª º « cosh{m (l – x )}  km [sinh{m (l – x )} » « » h « » cosh ( ml )  [sinh (ml )] »¼ x km ¬« 1 h cosh (ml )  [sinh (ml )] km

Fig. 2.134.

...[Refer Eqn. 2.136] l

“Conduction” Heat Transfer at Steady State–One Dimension where,

m

hP kAcs

h u (2b  2 y ) k u (b u y )

h 2 u k y

145 2 u 55 0.005

213

32.47

(Q 2y 1 or Biot number defined as

where Lc

h Lc  1. In other words fin is only effective when ‘h’ is small or k is large. In boiling and k condensation, when ‘h’ is quite large the fins may actually produce a reduction in heat transfer rate.

“Conduction” Heat Transfer at Steady State–One Dimension

219

Example 2.129. A composite fin consists of a cylindrical rod (3 mm diameter and 100 mm length) of one material, uniformly covered with another material forming outer diameter 10 mm and length 100 mm. k (inner material) = 15 W/m°C; k (outer material) = 45 W/m°C. h (surface heat transfer coefficient) = 12 W/m2°C. Fig. 2.136. (i) Determine the effectiveness of the composite fin. Assume no temperature gradient along the radial direction and end is insulated for the composite fin. (ii) Also find out the expression for the efficiency of this fin and its value for the given data. Solution. Refer to Fig. 2.136. The composite fin consists of solid fin (cylindrical rod) in which heat flows along the axis by conduction only and hollow fin exposed to surroundings. (i) Effectiveness of the composite fin, H fin : The rate of heat transfer for the solid fin is given by (Q ) solid fin

(to – ta ) l 2 S § 3 · (to – ta ) 15 u u ¨ ¸ u 4 © 1000 ¹ (100 /1000)

k Acs u

0.00106 (to – ta )

...(i)

(Q)hollow fin = kAcs m (to – ta) tanh (ml) where,

m

S u 0.01 12 u 45 [S / 4 (0.012 – 0.0032 )]

10.827

ml = 10.827 × 0.1 = 1.0827

? ?

hP kAcs

(Q )hollow fin

S (0.012 – 0.0032 ) u 10.827 (to – ta ) tanh (1.0827) 4 = 0.03482 (to – ta) × 0.794 = 0.0276 (to – ta) 45 u

...(ii)

The total rate of heat transfer from the composite fin is given by (Q)composite fin = (Q)solid fin + (Q)hollow fin = [0.00106 (to – ta)] + [0.0276 (to – ta)] = 0.02866 (to – ta)

...(iii)

The rate of heat transfer from the surface when there is no composite fin, (Q)without composite fin = hAccs (to – ta) 12 u

2

S § 10 · u¨ ¸ u (to – ta ) 4 © 1000 ¹

0.000942 (to – ta )

...(iv)

220

Chapter : 2

? Effectiveness of the composite fin, (Q)composite fin H fin (Q) without composite fin 0.02866 (to – ta ) 0.000942 (to – ta )

30.42

(Ans.)

Fig. 2.137.

(ii) Expression for efficiency of this fin and its value for the given data : Let the suffix sf stands for solid fin and suffix hf for hollow fin. Refer to Fig. 2.137. Consider an element of thickness dx at a distance x from the base of the composite fin. Then, Heat conducted in, Qx Heat conducted out,

dt · § dt · § ¨ – k sf Asf ¸  ¨ – k hf ˜ Ahf ¸ dx ¹ © dx ¹ ©

^

`

ª º ª dt d dt dt d § dt · º   dx »  « – khf Ahf – ksf Asf ¨ – khf Ahf ¸ dx « – ksf Asf dx dx dx dx dx © dx ¹ »¼ ¬ ¼ ¬ Heat transfer from the surface, Q( x  dx )

Qconv. = h (P × dx) (t – ta) For steady state :

Qx = Q (x + dx) + Qconv.

2

or, k sf Asf

d t dx 2

 khf Asf ˜

or,

(k sf Asf  khf Ahf )

or,

(k sf Asf  khf Ahf )

or,

where,

dx 2 d 2t dx 2 d 2T dx 2 d 2T dx

2

or,

d 2t

d T dx 2

2

– m2 T m2

Efficiency of the fin is given by

hP (t – ta ) hPT hPT

k sf

hPT Asf  k hf Ahf

0 hP k sf Asf  khf Ahf

(Q T = t – ta)

“Conduction” Heat Transfer at Steady State–One Dimension K fin

Now

or, ?

tanh (ml ) ml

221

...[Eqn. (2.139)]

12 u (S u 0.01) 0.377 S  0.0032 0.000106 §S · 15 u ¨ u 0.0032 ¸  45 u (0.012 – 0.0032 ) ©4 ¹ 4 m = 10.68 tanh (10.68 u 0.1) K fin ; 0.74 or 74% (Ans.) (10.68 u 0.1) m2

114.03

2.10.2.5. Design of Rectangular fins The design of fins is considered to be optimum when the fins (i) offer minimum resistance to flow of fluid, (ii) are easy to manufacture, (iii) require minimum cost of manufacture, and (iv) are light in weight. By the use of a fin, surface area is increased due to which heat flow rate increases. Increase in surface area decreases the surface convection resistance, whereas the conduction resistance increases. The decrease in convection resistance must be greater than the increase in conduction resistance in order to increase the rate of heat transfer from the surface. In practical applications of fins the surface resistance must be the controlling factor (the addition of fins might decrease the heat transfer rate under some situations). Design of the dimensions for a given profile area, to get maximum heat transfer : Let, b = Face width of the fin, y = Thickness of the fin, and l = Length of the fin. Then the perimeter, P is expressed as P = (2b + 2y) ; 2b ...(2.146) (in most practical applications, the width is large as compared to thickness) The cross-sectional area of the fin is given by ...(2.147) Acs = b.y 2h hP h u 2b ; ; m and, ...(2.148) kAcs k by ky Consider the case of heat dissipation from a fin insulated at the tip. The expression for heat flow rate in this case is given by Q fin PhkAcs (to  ta ) tanh (ml ) [Eqn. (2.135)] PhkAcs To .tanh (ml )

or,

Q fin b

ª 2h º 2bhkby To .tanh « . l» ¬ ky ¼ ª 2h ly º u » b 2hky .To .tanh « y¼ ¬ ky ª 2h º To 2hky tanh « Ap . » ky 3 ¼» ¬« (where Ap = profile area = ly)

(Q T = to – ta)

222

Chapter : 2

Q fin

q fin

or,

b

= Heat transfer rate per unit width

ª 2h º 2hky tanh « Ap . » ky3 ¼» ¬« For maximum heat transfer, the required condition is dq fin 0 (since qfin is dependent only on y, Ap being constant). dy dq fin ­° d ª 2h ½°º «To 2hky tanh ® Ap . ¾» 0 i.e., dy dy ¬« ky3 ¿°¼» ¯° q fin

or,

or,

? or, or,

To

dq fin

ª 1 § · § 2h 2h 3/ 2 · º To 2hk « tanh ¨ Ap . y 3/ 2 ¸  y1/ 2 sech 2 ¨ Ap .y ¸» dy k k © ¹ © ¹ »¼ «¬ 2 y º d ª 2h . y 3/ 2 » 0 u « Ap dy ¬ k ¼ 1 ª º ª ºª 1 2h 2h 2 h § 3 · 5 / 2 º tanh « Ap . y 3/ 2 »  y 2 sech 2 « Ap . y 3/ 2 » « Ap ¨ ¸ y » 0 k k k © 2¹ 2 y ¬ ¼ ¬ ¼¬ ¼ ª 2h 3/ 2 º tanh « Ap .y » k ¬ ¼ 2 y ª º 2h tanh « Ap . y 3/ 2 » 3 Ap k ¬ ¼ 1

3 1 Ap 2 y

ª 2h 3/ 2 2h 3/ 2 º .y . sech 2 « Ap .y » k k ¬ ¼

2h . y 3/ 2 . sech 2 k

ª « Ap ¬

º 2h . y 3/ 2 » k ¼

ª º 2h sinh « Ap . y 3/ 2 » 2h 1 k ¬ ¼ 3A . y 3/ 2 . p or, k ª º ª º 2h 2h cosh « Ap . y 3/ 2 » cosh 2 « Ap . y 3/ 2 » k k ¬ ¼ ¬ ¼ ª º ª º 2h 2h 2h sinh « Ap . y 3/ 2 » cosh « Ap . y 3/ 2 » 3 Ap . y 3/ 2 or, k k k ¬ ¼ ¬ ¼ ª º h 2h 2 sinh « 2 Ap . y 3/ 2 » 6 Ap . y 3/ 2 or, k k ¬ ¼ [Multiplying both sides by 2 and recalling that sinh (2x) = 2 sinh (x) cosh (x)] Equation (2.149) can be solved graphically and its solution yields : 2h . y 3/ 2 k 2h l. y . y 3/ 2 k l 2h y k Ap

or, or, or,

l 2hy y k

0

1.419

...(2.149)

...(2.150)

1.419 1.419

1.419 (Multiplying numerator and denomenator by y )

Let,

hy 2k

Bi , where B is referred to as Biot number. i

“Conduction” Heat Transfer at Steady State–One Dimension 2l Bi y

Then,

l y

or,

223

1.419

1.419

0.7095

2 Bi

Bi

...(2.151)

l where, y is known as the optimum ratio. Now, consider the case of heat dissipation from a fin losing heat at the tip. The expression for heat flow rate in this case is given by,

ª  h º « tanh ml  km » Q fin PhkAcs (to  ta ) « ...[Eqn. (2.137)] » «1  h . tanh (ml ) » km ¬« ¼» h º ª « tanh( ml )  km » PhkAcs . To « » (Q To = to – ta) «1  h . tanh (ml ) » km ¬« ¼» The limit of conditions for which addition of fins will increase the rate of heat transfer is given by dQ fin 0 (since h, k, m, Acs, P and To are constants). dl ª h ½º ­ dQ fin ° tanh (ml )  km °» d « « PhkAcs . To ® ¾» i.e., dl dl « °1  h . tanh (ml ) °» «¬ km ¯ ¿»¼ dQ fin or,

dl

h m h º h ª º ª «¬1  km tanh (ml ) »¼ cosh 2 (ml )  «¬ tanh (ml )  km »¼ k cosh 2 ( ml ) h ª º «¬1  km tanh ( ml ) »¼

2

0

h m h ºª h ª ºª º ª º «¬1  km tanh (ml ) »¼ « »  «¬ tanh (ml )  km »¼ « k » 0 2 2 h ml ml cosh ( ) cosh ( ) ¬ ¼ ¬ ¼ h h h2 m tanh (ml )  tanh (ml )  2 0 or, km k k m 2 h m 2 0 or, ...(2.152) k m 2h m; Also, ...[from Eqn. (2.148)] ky 2 ky 2h h  u 0 Then, ky k 2 2h 2h h 2 or, or 2hk2 = ky h2 ky k 2 hy 1 or, (i.e., Bi = 1) ...(2.153) 2k On rearranging this equation, we get

?

224

Chapter : 2

1 ( y / 2) ...(2.154) h k 1 = Surface convection resistance per unit area, and where, h (y/2)/k = Conduction resistance for a plane wall whose thickness is onehalf the fin thickness. x The heat flow rate will be minimum when these resistances are equal. x The heat flow rate will be minimum when Biot number (Bi) is equal to unity and will increase when Bi < 1. Thus the conditions for fins to be effective are : 1. Thermal conductivity (k) should be large. 2. Heat transfer coefficient (h) should be small. 3. Thickness of the fin (y) should be small. It is advantageous to use a large number of fins of smaller thickness.

2.10.3. HEAT FLOW THROUGH “STRAIGHT TRIANGULAR FIN” The tapered fin is of paramount practical importance since it yields the maximum heat flow per unit weight. Fig. 2.138. shows a straight triangular fin. Let, l = Length of the fin, between the base and the origin/tip, b = Width of the fin (per-pendicular to the paper), y = Thickness at the base of the fin (increasing uniformly from 0 at the tip to y at the base), to = Temperature of the base of the fin, Fig. 2.138. ta = Temperature of the ambient/surrounding fluid, k = Thermal conductivity of the material of the fin (constant), and h = Heat transfer coefficient (convective). Assume the fin to be sufficiently thin i.e., (y CLa LT ± b ML± c ML±T± d @ )RUGLPHQVLRQDOKRPRJHQHLW\WKHH[SRQHQWVRIHDFKGLPHQVLRQRQERWKVLGHVRIWKHHTXDWLRQ PXVWEHLGHQWLFDO7KXV )RU 0 cd i )RU / ab±c±d ii )RU 7± ±b±d iii 7KHUHDUHfour unknownsabcd EXWequations are threeLQQXPEHU7KHUHIRUHLWLVQRW SRVVLEOHWR¿QGWKHYDOXHVRIabcDQGd+RZHYHUWKUHHRIWKHPFDQEHH[SUHVVHGLQWHUPVRI IRXUWKYDULDEOHwhich is most important7KHUHWKHUROHRIYLVFRVLW\LVYLWDORQHDQGKHQFHabcDUH H[SUHVVHGLQWHUPVRIdieSRZHUWRYLVFRVLW\  ? c ±d IURPi b ±d IURPiii 3XWWLQJWKHVHYDOXHVLQii ZHJHW  a ±bcd ±d±d d   ±d±dd ±d 6XEVWLWXWLQJWKHVHYDOXHVRIH[SRQHQWVLQHTQ ZHJHW  F C >D±d V±d U±d Pd @  

d ⎡ ⎛ μ ⎞ ⎤ = C [ D 2 V 2 ρ ( D – d . V – d . ρ– d . μ d )] = C ⎢ρ D 2 V 2 ⎜ ⎥ ⎝ ρ VD ⎟⎠ ⎥⎦ ⎢⎣ ⎛ μ ⎞ = ρ D2 V 2 φ ⎜ ⎝ ρ VD ⎟⎠

([DPSOH7KHHI¿FLHQF\Kof a fan depends on the densityUthe dynamic viscosityPof WKHÀXLGthe angular velocityZdiameter D of the rotor and the discharge QExpressKin terms of dimensionless parameters                              $0,( 6ROXWLRQ7KHHI¿FLHQF\KRIDIDQLVDIXQFWLRQRI  i  'HQVLW\U ii  9LVFRVLW\P iii  $QJXODUYHORFLW\Z iv  'LDPHWHUDDQG  v  'LVFKDUJHQ 0DWKHPDWLFDOO\ K f UPZDQ RU K C Ua Pb Zc Dd Qe   ZKHUHCLVDQRQGLPHQVLRQDOFRQVWDQW 8VLQJ0±/±7V\VWHPWKHFRUUHVSRQGLQJHTXDWLRQIRUGLPHQVLRQVLV  MƒLƒTƒ &>ML± aML±T± bT± cL dLT± e@ )RUGLPHQVLRQDOKRPRJHQHLW\WKHH[SRQHQWVRIHDFKGLPHQVLRQRQERWKVLGHVRIWKHHTXDWLRQ PXVWEHLGHQWLFDO7KXV

Dimensional Analysis

755

)RU 0 ab )RU / ±a±bde )RU 7 ±b±c±e 7KHUHDUH¿YHYDULDEOHVDQGZHKDYHRQO\three equations([SHULHQFHKDVVKRZQWKDWUHFRJQL]HG GLPHQVLRQOHVVJURXSVDSSHDULIWKHH[SRQHQWVRIDZDQGUDUHHYDOXDWHGLQWHUPVRIbDQGeH[SRQHQWV RIYLVFRVLW\DQGGLVFKDUJHZKLFKDUHmore important  ? a ±bc ±be   d ab±e ± b b±e ±b±e ±be 6XEVWLWXWLQJWKHVHYDOXHVRIH[SRQHQWVLQHTQ ZHJHW  K  C U±b Pb Z±be D ±be Qe  CU±b Pb Z±b Z±e'±b D±e Qe 

⎡⎛ μ ⎞ b ⎛ Q ⎞ e ⎤ = C ⎢⎜ ⎥ 2⎟ ⎜ 3⎟ ⎢⎣⎝ ρ ω D ⎠ ⎝ ω D ⎠ ⎥⎦



⎡⎛ μ ⎞ ⎛ Q ⎞ ⎤ = φ ⎢⎜ , 2⎟ ⎜ 3⎟⎥  ⎣⎝ ρ ω D ⎠ ⎝ ω D ⎠ ⎦

12.4.2. BUCKINGHAM’S 3-METHOD/THEOREM :KHQ D ODUJH QXPEHU RI SK\VLFDO YDULDEOHV DUH LQYROYHG 5D\OHLJK¶V PHWKRG RI GLPHQVLRQDO DQDO\VLVEHFRPHVincreasingly laboriousDQGcumbersome%XFNLQJKDP¶VPHWKRGLVDQimprovement RYHUWKH5D\OHLJK¶VPHWKRG%XFNLQJKDPGHVLJQDWHGWKHGLPHQVLRQOHVVJURXSE\WKH*UHHNFDSLWDO OHWWHUS3L ,WLVWKHUHIRUHRIWHQFDOOHGBuckinghamSmethod7KHadvantageRIWKLVPHWKRGRYHU 5D\OHLJK¶VPHWKRGLVWKDWLWOHWVXVNQRZLQDGYDQFHRIWKHDQDO\VLVDVWRhow many dimensionless groups are to be expected 7KH%XFNLQJKDP¶VSWKHRUHPVWDWHVDVIROORZV ³If there are n variablesdependent and independent variables in a dimensionally homogeneous equation and if these contain m fundamental dimensionssuchas MLTetc then the variables are arranged intonm dimensionless termsThese dimensionless terms are calledSterms.´ 0DWKHPDWLFDOO\ LI DQ\ YDULDEOH X GHSHQGV RQ LQGHSHQGHQW YDULDEOHV X X X  Xn WKH IXQFWLRQDOHTXDWLRQPD\EHZULWWHQDV  X fXXXXn   (TXDWLRQ FDQDOVREHZULWWHQDV  fXXXXn    ,WLVDGLPHQVLRQDOO\KRPRJHQHRXVHTXDWLRQDQGFRQWDLQVnYDULDEOHV,IWKHUHDUHmIXQGDPHQWDO GLPHQVLRQVWKHQDFFRUGLQJWR%XFNLQJKDP¶VSWKHRUHPLW>(TQ @FDQEHZULWWHQLQWHUPVRI QXPEHURISWHUPVGLPHQVLRQOHVVJURXSV LQZKLFKQXPEHURISWHUPVLVHTXDOWRn±m +HQFH (TQ EHFRPHVDV  fSSSSn±m    (DFK GLPHQVLRQOHVV SWHUP LV IRUPHG E\ FRPELQLQJ m YDULDEOHV RXW RI WKH WRWDO n YDULDEOHV ZLWKone of the remainingn±m variables ieHDFKSWHUPFRQWDLQVm  YDULDEOHV7KHVHm YDULDEOHVZKLFKDSSHDUUHSHDWHGO\LQHDFKRISWHUPVDUHFRQVHTXHQWO\FDOOHGrepeating variablesDQG DUHFKRVHQIURPDPRQJWKHYDULDEOHVVXFKWKDWWKH\WRJHWKHUinvolve all the fundamental dimensions and they themselves do not form a dimensionless parameter/HWLQWKHDERYHFDVHXXDQGXDUH WKHUHSHDWLQJYDULDEOHVLIWKHIXQGDPHQWDOGLPHQVLRQVmMLT  7KHQHDFKWHUPLVZULWWHQDV

756

Chapter : 12

⎫ ⎪ a b c ⎪ π 2 = X 2 2 . X 3 2 . X 42 . X 5  ⎪ :  ⎬ ⎪ : ⎪ a b c π n – m = ( X 2 n – m . X 3 n – m . X 4n – m . X n ⎪⎭ ZKHUHabcabcHWFDUHWKHFRQVWDQWVZKLFKDUHGHWHUPLQHGE\FRQVLGHULQJGLPHQVLRQDO KRPRJHQHLW\7KHVHYDOXHVDUHVXEVWLWXWHGLQ(TQ DQGYDOXHVRISSSSn±mDUHREWDLQHG 7KHVHYDOXHVRIS¶VDUHVXEVWLWXWHGLQ(TQ 7KH¿QDOJHQHUDOHTXDWLRQIRUWKHSKHQRPHQRQ PD\WKHQEHREWDLQHGE\H[SUHVVLQJDQ\RQHRIWKHSWHUPVDVDIXQFWLRQRIWKHRWKHUDV π1 = φ (π 2 , π3 , π 4 , ... π n – m )⎫⎪   ⎬ π 2 = φ (π1 , π3 , π 4 , ... π n – m )⎪⎭ a

b

c

π1 = X 21 . X 31 . X 41 . X1

Selection of repeating variables : 7KHIROORZLQJSRLQWVVKRXOGEHNHSWLQYLHZZKLOHVHOHFWLQJmUHSHDWLQJYDULDEOHV  mUHSHDWLQJYDULDEOHVPXVWFRQWDLQjointlyDOOWKHIXQGDPHQWDOGLPHQVLRQVLQYROYHGLQWKH SKHQRPHQRQ8VXDOO\WKHIXQGDPHQWDOGLPHQVLRQVDUHMLDQGT+RZHYHULIRQO\WZR GLPHQVLRQVDUHLQYROYHGWKHUHZLOOEHUHSHDWLQJYDULDEOHVDQGWKH\PXVWFRQWDLQtogether WKHWZRGLPHQVLRQVLQYROYHG   7KHUHSHDWLQJYDULDEOHVmust notIRUPWKHQRQGLPHQVLRQDOSDUDPHWHUVDPRQJWKHPVHOYHV   $VIDUDVSRVVLEOHWKHGHSHQGHQWYDULDEOHshould notEHVHOHFWHGDVUHSHDWLQJYDULDEOH   1RWZRUHSHDWLQJYDULDEOHVVKRXOGKDYHWKHVDPHGLPHQVLRQV   7KHUHSHDWLQJYDULDEOHVVKRXOGEHFKRVHQLQVXFKDZD\WKDWRQHYDULDEOHFRQWDLQVJHRPHWULF SURSHUW\eglengthldiameterdheightHHWF RWKHUYDULDEOHFRQWDLQVÀRZSURSHUW\ egvelocityVaccelerationHWF DQGWKLUGYDULDEOHFRQWDLQVÀXLGSURSHUW\egmass densityUweight densitywdynamic viscosityPHWF  7KHFKRLFHRIUHSHDWLQJYDULDEOHVLQPRVWRIÀXLGPHFKDQLFVSUREOHPVPD\EH  i  lVU ii  dVU iii  lVP iv  dVP 7KH SURFHGXUH IRU VROYLQJ SUREOHP E\ %XFNLQJKDP¶V SWKHRUHP LV RXWOLQHG LQ WKH ([DPSOHEHORZ 

([DPSOHThe resistance R experienced by a partially submerged body depends upon the velocity Vlength of the body lYLVFRVLW\RIWKHÀXLGPGHQVLW\RIWKHÀXLGUand gravitational acceleration gObtain a dimensionless expression for R 6ROXWLRQ6WHS7KHUHVLVWDQFHRLVDIXQFWLRQRI  i  9HORFLW\V ii  /HQJWKl iii  9LVFRVLW\PDQG iv  'HQVLW\U  v  *UDYLWDWLRQDODFFHOHUDWLRQg 0DWKHPDWLFDOO\ R f VlPUg  i RU fRVlPUg   ii ? 7RWDOQXPEHURIYDULDEOHVn  ⎧m is obtained by writing dimensions of each variables as ⎫ ⎪ ⎪ –2 –1 –1 –1 –3 –2 ⎨ R = MLT , V = LT , μ = ML T , ρ = ML , g = LT . Thus the⎬  ⎪fundamental dimensions in the problem are M , L, T and hence m = 3. ⎪ ⎩ ⎭

Dimensional Analysis

757

1XPEHURIGLPHQVLRQSWHUPV n±m ±  7KXVWKUHHS±WHUPVVD\SSDQGSDUHIRUPHG 7KHHTQii PD\EHZULWWHQDV  fSSS   iii 6WHS Selection of repeating variables2XWRIVL[YDULDEOHVRVlPUgWKUHHYDULDEOHVDVm  DUHWREHVHOHFWHGDVrepeating variablesRLVDGHSHQGHQWYDULDEOHDQGVKRXOGnotEH VHOHFWHGDVDUHSHDWLQJYDULDEOH2XWRIWKHUHPDLQLQJ¿YHYDULDEOHVRQHYDULDEOHVKRXOG KDYHgeometric propertyVHFRQGVKRXOGKDYHÀRZSURSHUW\DQGWKLUGRQHVKRXOGKDYHÀXLG propertyWKHVHUHTXLUHPHQWVDUHPHWE\VHOHFWLQJlVDQGUDVrepeating variables7KH UHSHDWLQJ YDULDEOHV WKHPVHOYHV VKRXOG QRW IRUP D GLPHQVLRQOHVV WHUP DQG PXVW FRQWDLQ jointly all fundamental dimensions equal to m ieKHUH'LPHQVLRQVRIlVDQGUDUH LLT±ML±DQGKHQFHWKHWKUHHIXQGDPHQWDOGLPHQVLRQVH[LVWLQlVDQGUDQGDOVRno dimensionless group is formed by them 6WHS (DFKS±WHUP mYDULDEOHV LVZULWWHQDVJLYHQLQHTQ ie π1 = l a1 . V b1 . ρc1 . R ⎫ ⎪⎪ iv a b c  π 2 = l 2 . V 2 . ρ 2 . μ ⎬  ⎪ π3 = l a3 . V b3 . ρc3 . g ⎪⎭ 6WHS (DFKSWHUPLVVROYHGE\WKHprinciple of dimensional homogeneityDVIROORZV  S±WHUP a b c  π1 = l 1 . V 1 . π 1 . R a 0 0 0 –1 b –3 c –2  M L T = L 1 . ( LT ) 1 . ( ML ) 1 . ( MLT ) 

(TXDWLQJWKHH[SRQHQWVRIMLDQGTUHVSHFWLYHO\ZHJHW )RU0  c )RU/  ab±c )RU7  ±b± ? c ±b ± DQG a ±bc± ±± ± 6XEVWLWXWLQJWKHYDOXHVRIabDQGcLQSZHJHW R ? S l±V±U± R = 2 2  l V ρ S±WHUP π 2 = l a2 . V b2 . ρc2 . μ

 

M 0 L0T 0 = La2 .( LT –1 )b2 . ( ML–3 )c2 . ( ML–1T –1 )

(TXDWLQJWKHH[SRQHQWVRIMLDQGTUHVSHFWLYHO\ZHJHW )RU0  c )RU/  ab±c± )RU7  ±b± ? c ±b ± DQG a ±bc ± ± 6XEVWLWXWLQJWKHYDOXHVRIabDQGcLQSZHJHW μ ? S l ±V ±U± μ = lV ρ

v

758

Chapter : 12

S±WHUP  

π3 = D a3 . V b3 . ρc3 . k M 0 L0T 0 = La3 . ( LT –1 )b3 . ( ML–3 )c3 . ( LT –2 )

(TXDWLQJWKHH[SRQHQWVRIMLDQGTUHVSHFWLYHO\ZHJHW )RU0  c )RU/  ab±c )RU7  ±b± ? c b ± DQG a ±bc± ±  6XEVWLWXWLQJWKHYDOXHVRIabDQGcLQSZHJHW lg π3 = D1 . V −2 . ρ0 . g = 2 ? V 6WHS 6XEVWLWXWHWKHYDOXHVRISSSLQHTQiii 7KHIXQFWLRQDOUHODWLRQVKLSEHFRPHV μ lg ⎞ ⎛ R f1 ⎜ 2 2 , , =0  ⎝ l V ρ lV ρ V 2 ⎟⎠ RU 

lg ⎞ ⎛ μ =φ⎜ , 2⎟  ⎝ lV ρ V ⎠ l V ρ R

2

2

=φ

⎛ ρ Vl V ⎞ , ⎜⎝ μ lg ⎟⎠

7KHDERYHVWHSKDVEHHQPDGHRQWKHSRVWXODWHWKDWreciprocal of piterm and its square root is nondimensional ⎛ ρ Vl V ⎞ R = l 2V 2 ρφ ,  ⎜⎝ μ lg ⎟⎠ ⎛ ρ Vl ⎞ ⎛ V ⎞ DQG)URXGH¶VQXPEHU ⎜  7KHUHVLVWDQFHRLVWKXVDIXQFWLRQRI5H\QROGVQXPEHU ⎜ ⎟ ⎝ μ ⎠ ⎝ lg ⎟⎠ ([DPSOH)ULFWLRQDO/RVVLQ3LSHV The pressure difference'pin a pipe of diameter 'DQGOHQJWKOGXHWRWXUEXOHQWÀRZGHSHQGVRQWKHYHORFLW\9viscosityPdensityUand roughness kUsing Buckingham¶sStheoremobtain an expression for'p 6ROXWLRQ7KHSUHVVXUHGLIIHUHQFH'pLVDIXQFWLRQRIDlVPUk 'p f DlVPUk  i 0DWKHPDWLFDOO\ RU f'pDlVPUk   ii ? 7RWDOQXPEHURIYDULDEOHVn  :ULWLQJGLPHQVLRQVRIHDFKYDULDEOHZHKDYH  'pGLPHQVLRQVRISUHVVXUH  ML±T±D Ll LV LT±  S ML±T±U ML±k L 7KXVQXPEHURIIXQGDPHQWDOGLPHQVLRQVm  ? 1XPEHURISWHUP n±m ±  (TXDWLRQii FDQEHZULWWHQDV  fSSSS   iii (DFKSWHUPFRQWDLQVm YDULDEOHVZKHUHm DQGLVDOVRHTXDOWRUHSHDWLQJYDULDEOHV &KRRVLQJDVDQGUDVUHSHDWLQJYDULDEOHVZHJHWIRXUSWHUPVDV

Dimensional Analysis a1

b1

c1



π1 = D . V . ρ . Δ p



π 2 = D a2 . V b2 . ρc2 . l



π3 = D a3 . V b3 . ρc3 . μ



π 4 = D a4 . V b4 . ρc4 . k

S±WHUP  

π1 = D a1 . V b1 . ρc1 . Δp M 0 L0T 0 = La1 . ( LT –1 )b1 . ( ML–3 )c1 . ( ML–1 T –2 )

(TXDWLQJWKHH[SRQHQWVRIMLDQGTUHVSHFWLYHO\ZHJHW )RU0  c )RU/  ab±c± )RU7  ±b± ? c ±b ±  a ±bc ±  6XEVWLWXWLQJWKHYDOXHVRIabDQGcLQSZHJHW Δp π1 = D 0 . V –2 . ρ–1 . Δ p =  ρV 2 S±WHUP π 2 = D a2 . V b2 . ρc2 . l  

M 0 L0T 0 = La2 . ( LT –1 )b2 . ( ML–3 )c2 . L

(TXDWLQJWKHH[SRQHQWVRIMLDQGTUHVSHFWLYHO\ZHJHW )RU0  c )RU/  ab±c )RU7  ±b ? c b  DQG a ±bc± ± 6XEVWLWXWLQJWKHYDOXHVRIabDQGcLQSZHJHW 

π 2 = D –1 . V 0 . ρ0 . l =

S±WHUP 

π3 = D a3 . V b3 . ρc3 . μ



l D

M 0 L0T 0 = La3 . ( LT –1 )b3 . ( ML–3 )c3 . ( ML–1T –1 )

(TXDWLQJWKHH[SRQHQWVRIMLDQGTUHVSHFWLYHO\ZHJHW )RU0  c )RU/  ab±c± )RU7  ±b± ? c ±b ± DQG a ±bc ± ± 6XEVWLWXWLQJWKHYDOXHVRIabDQGcLQSZHJHW μ π3 = D –1 . V –1 . ρ–1 . μ =  DV ρ

759

760

Chapter : 12

S±WHUP 

π 4 = D a4 . V b4 . ρc4 . k M 0 L0T 0 = La4 . ( LT –1 )b4 . ( ML–3 )c4 . L 





   'LPHQVLRQRIk L

(TXDWLQJWKHH[SRQHQWVRIMLDQGTUHVSHFWLYHO\ZHJHW )RU0  c )RU/  ab±c )RU7  ±b ? c b  DQG a ±bc± ± ± 6XEVWLWXWLQJWKHYDOXHVRIabDQGcLQSZHJHW k π 4 = D –1 . V 0 . ρ0 . k =  D 6XEVWLWXWLQJWKHYDOXHVRISSSDQGSLQHTQiii ZHJHW μ k⎞ ⎛ Δp l f1 ⎜ , , , ⎟ =0  2 ⎝ ρV D DV ρ D ⎠ Δp

μ k⎤ ⎡l =φ⎢ , , ⎥ ρV ⎣ D DV ρ D ⎦ ([SUHVVLRQIRUGLIIHUHQFHRISUHVVXUHKHDGhf  RU

2

l  $VREVHUYHGIURPH[SHULPHQWV'pLVDOLQHDUIXQFWLRQRI WKHUHIRUHWDNLQJWKLVRXWRI D IXQFWLRQZHKDYH Δp l k⎤ ⎡ μ = φ⎢ , ⎥  2 D ⎣ DV ρ D ⎦ ρV l k⎤ Δp ⎡ μ = V2. φ ⎢ , p D ⎣ DV ρ D ⎦⎥

RU 'LYLGLQJERWKVLGHVE\gZHJHW

Δ p V2 l k⎤ ⎡ μ = . φ⎢ , ⎥ g D ⎣ DV ρ D ⎦ ρg



k⎤ ⎡ μ , ⎥ FRQVLVWVRIIROORZLQJWZRWHUPV 1RZ φ ⎢ ⎣ DV ρ D ⎦ μ 1 1 which is or i  DV ρ Reynold number Re ii 

k  D

FDOOHGURXJKQHVVIDFWRU

⎡ 1 k⎤ φ ⎢ , ⎥ LVSXWHTXDOWRf ⎣ Re D ⎦ ZKHUHf &RHI¿FLHQWRIIULFWLRQIXQFWLRQRI5H\QROGQXPEHUDQGURXJKQHVVIDFWRU  Δ p 4 f V 2l ⎡ k ⎞⎤ ⎛ μ = . ∵f=φ⎜ , ⎟⎥ ?  ⎢ ρg 2 gD ⎝ DV ρ D ⎠ ⎦ ⎣ 0XOWLSO\LQJRUGLYLGLQJE\DQ\FRQVWDQWGRHVQRWFKDQJHWKHFKDUDFWHURISWHUPV Δp  f lV  = hf = ? ρg D î g

?

Dimensional Analysis

761

12.5. DIMENSIONAL ANALYSIS APPLIED TO FORCED CONVECTION HEAT TRANSFER /HWXVDVVXPHWKDWWKHKHDWWUDQVIHUFRHI¿FLHQWLQDIXOO\GHYHORSHGIRUFHGFRQYHFWLRQLQDWXEH LVDIXQFWLRQRIWKHIROORZLQJYDULDEOHV  h f UDVPcpk  i RU fhUDVPcpk  ii 7KHSK\VLFDOTXDQWLWLHVZLWKWKHLUGLPHQVLRQVDUHDVXQGHU SNo       

Variables +HDWWUDQVIHUFRHI¿FLHQW )OXLGGHQVLW\ 7XEHGLDPHWHU )OXLGYHORFLW\ )OXLGYLVFRVLW\ 6SHFL¿FKHDW 7KHUPDOFRQGXFWLYLW\

Symbols h U D V P cp k

Dimensions MT±T± ML± L LT± ML±T± LT±T± MLT±T±

 7RWDOQXPEHURIYDULDEOHVn  )XQGDPHQWDOGLPHQVLRQVLQWKHSUREOHPDUHMLTTDQGKHQFHm  1XPEHURIGLPHQVLRQOHVVSWHUPV n±m  ±  7KHHTQii PD\EHZULWWHQDV  fSSS   :HFKRRVHhUDVDVWKHFRUHJURXSUHSHDWLQJYDULDEOHV ZLWKXQNQRZQH[SRQHQWV7KH JURXSVWREHIRUPHGDUHQRZUHSUHVHQWHGDVWKHIROORZLQJSJURXSV 

π1 = h a1 . ρb1 . D c1 . V d1 . μ



π 2 = h a2 . ρb2 . D c2 . V d 2 . c p



π3 = h a3 . ρb3 . D c3 . V d3 . k

S±WHUP 



M 0 L0T 0 = ( MT –3 θ –1 ) a1 . ( ML–3 )b1 . ( L)c1 . ( LT –1 ) d1 . ( ML–1 T –1 )

(TXDWLQJWKHH[SRQHQWVRIMLTDQGTUHVSHFWLYHO\ZHJHW )RU0  ab )RU/  ±bcd± )RU7  ±a±d± )RUT  ±a 6ROYLQJWKHDERYHHTXDWLRQVZHKDYH a b ±c ±d ± ? RU S±WHUP  )RU0 )RU/

S U±D±V±P μ π1 = ρ DV M 0 L0T 0 = ( MT –3 θ –1 ) a2 . ( ML–3 )b2 . ( L)c2 . ( LT –1 ) d 2 . ( L2 T –2 θ –1 )  ab  ±bcd

762

Chapter : 12

)RU7

 ±a±d±

)RUT

 ±a±

6ROYLQJWKHDERYHHTXDWLRQVZHKDYH a ±b c d 

 ?

S h±UV cp

RU

π2 =



π3 = ( MT –3 θ –1 ) a3 . ( ML–3 )b3 . ( L)c3 . ( LT –1 ) d3 . ( MLT –3 θ –1 )

c p ρV

h k 6LQFHGLPHQVLRQVRIhDQG DUHWKHVDPHKHQFH D c p ρVD π2 =  k S±WHUP (TXDWLQJWKHH[SRQHQWVRIMLTDQGTUHVSHFWLYHO\ZHJHW )RU0



 a±b

)RU/  ±bcd )RU7  ±a±d± )RUT  ±a± 6ROYLQJWKHDERYHHTXDWLRQVZHJHW a ±b c ±d  ? S h±D±k k π3 =  RU hD $FFRUGLQJWRS±WKHRUHPS ISS n′

k ⎡ μ ⎤ =C⎢ ? ⎥ hD ⎣ ρ DV ⎦ ZKHUHm’DQGn’DUHFRQVWDQWV ,Im’!n’WKHQ

m′

⎡ c p ρ DV ⎤ ⎢ ⎥ ⎣ k ⎦

n′



k ⎡ μ ⎤ =C⎢ ⎥ hD ⎣ ρ DV ⎦

⎡ c p ρ DV ⎤ ⎢ ⎥ ⎣ k ⎦

⎡ μ ⎤ =C⎢ ⎥ ⎣ ρ DV ⎦

m′ – n′



c p ρ DV ⎤ ⎡ μ . ⎢ ⎥ k ⎣ ρ DV ⎦

⎡ μ ⎤ =C⎢ ⎥ ⎣ ρ DV ⎦

m′ – n′



⎡ μc p ⎤ ⎢ ⎥ ⎣ k ⎦

m

⎡ μc p ⎤ ⎢ ⎥ ⎣ k ⎦

RU

hD ⎡ ρ DV ⎤ =C⎢ ⎥ k ⎣ μ ⎦

RU

Nu C Re mPr n

n′

⎡ μ ⎤ ⎢ ⎥ ⎣ ρ DV ⎦

m′ – n′

n′

n′

n



Dimensional Analysis

763

ZKHUHCmDQGnDUHFRQVWDQWVDQGHYDOXDWHGH[SHULPHQWDOO\ hD ⎤ ⎡ ⎢ where Nu = Nusselt number = k ⎥ ⎢ ⎥ ⎢ Re = Reynolds number = ρ DV ⎥  ⎢ ⎥ μ ⎢ ⎥ μc p ⎢ ⎥ ⎢⎣ Pr = Prandtl number = k ⎥⎦ ,WLVZRUWKQRWLQJWKDWLIVmUcpZHUHFKRVHQDVWKHFRUHJURXSUHSHDWLQJYDULDEOHV WKHQWKH DQDO\VLVZRXOGKDYH\LHOGHGWKHIROORZLQJQRQGLPHQVLRQDOJURXSV μ cp ρ VD h ; Pr = ; St = Re =  μ k ρ Vc p



 ZKHUHSt 6WDQWRQQXPEHU 6RDQRWKHUIRUPRIFRUUHODWLQJKHDWWUDQVIHUGDWDLV St I RePr 



12.6. DIMENSIONAL ANALYSIS APPLIED TO NATURAL OR FREE CONVECTION HEAT TRANSFER 7KHKHDWWUDQVIHUFRHI¿FLHQWLQFDVHRIQDWXUDORUIUHHFRQYHFWLRQOLNHIRUFHGFRQYHFWLRQKHDW WUDQVIHUFRHI¿FLHQWGHSHQGVXSRQWKHYDULDEOHVVUkPcpDQGLRUD6LQFHWKHÀXLGFLUFXODWLRQLQ IUHHFRQYHFWLRQLVRZLQJWRGLIIHUHQFHLQGHQVLW\EHWZHHQWKHYDULRXVÀXLGOD\HUVGXHWRWHPSHUDWXUH JUDGLHQWDQGQRWE\H[WHUQDODJHQF\WKHUHIRUHYHORFLW\VLVQRORQJHUDQLQGHSHQGHQWYDULDEOHEXW GHSHQGVXSRQWKHIROORZLQJIDFWRUV  i  'ti.eWKHGLIIHUHQFHRIWHPSHUDWXUHVEHWZHHQWKHKHDWHGVXUIDFHDQGWKHXQGLVWXUEHGÀXLG  ii  EieFRHI¿FLHQWRIYROXPHH[SDQVLRQRIWKHÀXLG  iii  gieDFFHOHUDWLRQGXHWRJUDYLW\  Eg 'tLVFRQVLGHUHGDVRQHSK\VWLFDOIDFWRU 7KXVKHDWWUDQVIHUFRHI¿FLHQWµh¶PD\EHH[SUHVVHGDVIROORZV  h f ULPcpkEg't  i  fULPkhcpEg't  ii >7KHSDUDPHWHUEg't UHSUHVHQWVWKHEXR\DQWIRUFHDQGKDVWKHGLPHQVLRQVRILT±@ 7RWDOQXPEHURIYDULDEOHVn  )XQGDPHQWDOGLPHQVLRQVLQWKHSUREOHPDUHMLTTDQGKHQFHm  1XPEHURIGLPHQVLRQOHVVSWHUPV n±m  ±  7KHHTXDWLRQii PD\EHZULWWHQDV  fSSS   :HFKRRVHULPDQGkDVWKHFRUHJURXSUHSHDWLQJYDULDEOHV ZLWKXQNQRZQH[SRQHQWV7KH JURXSVWREHIRUPHGDUHQRZUHSUHVHQWHGDVWKHIROORZLQJSJURXSV π1 = ρa1 . Lb1 . μ c1 . k d1 . h

 

π 2 = ρa2 . Lb2 . μ c2 . k d 2 . c p



π3 = ρa3 . Lb3 . μ c3 . k d3 . β g Δt

SWHUP M 0 L0T 0 θ0 = ( ML–3 ) a1 . ( L)b1 . ( ML–1 T –1 )c1 . ( MLT –3 θ –1 ) d1 . ( ML–3 θ –1 ) 

Chapter : 12

764

(TXDWLQJWKHH[SRQHQWVRIMLTDQGTUHVSHFWLYHO\ZHJHW )RU0  acd )RU/  ±ab±cd )RU7  ±c±d± )RUT  ±d± 6ROYLQJWKHDERYHHTXDWLRQVZHJHW  a b c d ± hL π1 = Lk –1 h or π1 = ? k SWHUP

M 0 L0T 0 θ0 = ( ML–3 ) a2 . ( L)b2 . ( ML–1 T –1 )c2 . ( MLT –3 θ –1 ) d 2 . ( L2 T –2 θ –1 )



(TXDWLQJWKHH[SRQHQWVRIMLTTUHVSHFWLYHO\ZHJHW )RU0  acd )RU/  ±ab±cd )RU7  ±c±d± )RUT  ±d± 6ROYLQJWKHDERYHHTXDWLRQVZHJHW  a b c d ± μ cp π2 = ? S Pk±cp RU k S±WHUP

M 0 L0T 0 θ0 = ( ML–3 ) a3 . ( L)b3 . ( ML–1 T –1 )c3 . ( MLT –3 θ –1 ) d3 . ( LT –2 )



(TXDWLQJWKHH[SRQHQWVRIMLTTUHVSHFWLYHO\ZHJHW )RU0  acd )RU/  ±ab±cd )RU7  ±c±d± )RUT  ±d 6ROYLQJWKHDERYHHTXDWLRQVZHJHW  a b c ±d  ? S ULP±Eg't RU

π3 =

(β g Δt )ρ2 L3 μ2

=

(β g Δt ) L3 v2



RU 1X IPr Gr RU 1X CPr nGr mZKHUHGr *UDVKRIIQXPEHU   +HUHCnDQGmDUHFRQVWDQWVDQGPD\EHHYDOXDWHGH[SHULPHQWDOO\





12.7. ADVANTAGES AND LIMITATIONS OF DIMENSIONAL ANALYSIS Advantages :   

 ,WH[SUHVVHVWKHIXQFWLRQDOUHODWLRQVKLSEHWZHHQWKHYDULDEOHVLQGLPHQVLRQOHVVWHUPV  %\WKHSURSHUVHOHFWLRQRIYDULDEOHVWKHGLPHQVLRQOHVVSDUDPHWHUVFDQEHXVHGWRPDNH FHUWDLQORJLFDOGHGXFWLRQVDERXWWKHSUREOHP  'HVLJQFXUYHVE\WKHXVHRIGLPHQVLRQDODQDO\VLVFDQEHGHYHORSHGIURPWKHH[SHULPHQWDO GDWDRUGLUHFWVROXWLRQRIWKHSUREOHP

Dimensional Analysis   



765

 ,WHQDEOHVJHWWLQJXSDWKHRUHWLFDOVROXWLRQLQDVLPSOL¿HGGLPHQVLRQOHVVIRUP  'LPHQVLRQOHVVDQDO\VLVSURYLGHVSDUWLDOVROXWLRQVWRWKHSUREOHPVWKDWDUHWRRFRPSOH[ WREHGHDOWZLWKPDWKHPDWLFDOO\  'LPHQVLRQDODQDO\VLVLVDXVHIXOWRROLQWKHDQDO\VLVDQGFRUUHODWLRQRIH[SHULPHQWDOGDWD LQWKHSODQQLQJRIH[SHULPHQWVDQGLQWKHIRUPXODWLRQRIHPSLULFDOFRUUHODWLRQGHVFULELQJ DSDUWLFXODUSKHQRPHQRQ  7KHUHVXOWVRIRQHVHULHVRIWHVWVFDQEHDSSOLHGZLWKWKHKHOSRIGLPHQVLRQDODQDO\VLVWR DODUJHQXPEHURIRWKHUVLPLODUSUREOHPV

Limitations : 



 

 'LPHQVLRQDODQDO\VLVGRHVQRWJLYHDQ\FOXHUHJDUGLQJWKHVHOHFWLRQRIYDULDEOHV,IWKH YDULDEOHVDUHZURQJO\WDNHQWKHUHVXOWLQJIXQFWLRQDOUHODWLRQVKLSLVHUURQHRXV,WSURYLGHV WKHLQIRUPDWLRQDERXWWKHJURXSLQJRIYDULDEOHV,QRUGHUWRGHFLGHZKHWKHUVHOHFWHG YDULDEOHVDUHSHUWLQHQWRUVXSHUÀXRXVH[SHULPHQWVKDYHWREHSHUIRUPHG  7KHFRPSOHWHLQIRUPDWLRQLVQRWSURYLGHGE\GLPHQVLRQDODQDO\VLVLWRQO\LQGLFDWHVWKDW WKHUHLVVRPHUHODWLRQVKLSEHWZHHQWKHSDUDPHWHUV,WGRHVQRWJLYHWKHYDOXHVRIFRHI¿FLHQWV LQWKHIXQFWLRQDOUHODWLRQVKLS7KHYDOXHVRIFRHI¿FLHQWVDQGKHQFHWKHQDWXUHRIIXQFWLRQV FDQEHREWDLQHGRQO\IURPH[SHULPHQWVRUIURPPDWKHPDWLFDODQDO\VLV  1RLQIRUPDWLRQLVJLYHQDERXWWKHLQWHUQDOPHFKDQLVPRIWKHSK\VLFDOSKHQRPHQRQ  *HQHUDOO\LWLVGHVLUHGWR¿QGWKHHIIHFWRIRQHSK\VLFDOTXDQWLW\XSRQDQXPEHURIRWKHU SK\VLFDOTXDQWLWLHVWKDWDUHVXSSRVHGWRHQWHULQWRDSUREOHP,WLVQRWSRVVLEOHWRJHWVXFK LQIRUPDWLRQZLWKWKHKHOSRIGLPHQVLRQDODQDO\VLV

12.8. DIMENSIONLESS NUMBERS AND THEIR PHYSICAL SIGNIFICANCE 1. Reynolds number (Re):     

 ,WLVGH¿QHGDVWKHratio of the inertia force to the viscous force Inertia force ρ U 2 L2 ρ UL UL Re = = = =  Viscous force μ UL μ ν



y 5H\QROGVQXPEHUVLJQL¿HVWKHUHODWLYHSUHGRPLQDQFHRIWKHLQHUWLDWRWKHYLVFRXVIRUFHV RFFXULQJLQWKHÀRZV\VWHPV y 7KHKLJKHUWKHYDOXHRIReWKHJUHDWHUZLOOEHWKHUHODWLYHFRQWULEXWLRQRILQHUWLDHIIHFW7KH VPDOOHUWKHYDOXHRIReWKHJUHDWHUZLOOEHWKHUHODWLYHPDJQLWXGHRIWKHYLVFRXVVWUHVVHV y 5H\QROGVQXPEHULVWDNHQDVDQLPSRUWDQWFULWHULRQRINLQHPDWLFDQGG\QDPLFVLPLODULWLHV LQIRUFHGFRQYHFWLRQKHDWWUDQVIHU

2. Prandtl number (Pr):     

 ,WLVWKHratio of kinematic viscosityQ to thermal diffusivityD  μ cp ρ ν cp ν ν Pr = = = =  k k (k / ρc p ) α



 Kinematic viscosity indicates the impulse transport through molecular friction whereas thermal diffusivity indicates the heat energy transport by conduction process y 3UDQGWOQXPEHUSURYLGHVDPHDVXUHRIWKHUHODWLYHHIIHFWLYHQHVVRIWKHPRPHQWXPDQGHQHUJ\ WUDQVSRUWE\GLIIXVLRQ y 3UDQGWOQXPEHULVDFRQQHFWLQJOLQNEHWZHHQWKHYHORFLW\¿HOGDQGWHPSHUDWXUH¿HOGDQG LWVYDOXHVWURQJO\LQÀXHQFHVUHODWLYHJURZWKRIYHORFLW\DQGWKHUPDOERXQGDU\OD\HUV

766

Chapter : 12

3. Nusselt number (Nu):  

 1XVVHOWQXPEHUFDQEHGH¿QHGLQVHYHUDOZD\V i  ,WLVWKHUDWLRRIKHDWÀRZUDWHE\FRQYHFWLRQSURFHVVXQGHUDXQLWWHPSHUDWXUHJUDGLHQW WRWKHKHDWÀRZUDWHE\FRQGXFWLRQSURFHVVXQGHUDXQLWWHPSHUDWXUHJUDGLHQWWKURXJK a stationary thickness of L metres7KXV Nu =

 



ii  ,WLVWKHratio of heat transfer rateQ to the rate at which heat would be conducted ZLWKLQWKHÀXLGXQGHUDWHPSHUDWXUHJUDGLHQWRI'TL7KXV Nu =

 

Qconv. h hL = =  Qcond . k /L k

Q Q L hL = . =  ( Δ θ . k )/L Δ θ k k



iii  ,WLVWKHratio of characteristic length L to the thickness'[RIDVWDWLRQDU\ÀXLGOD\HU conducting the heat at the same rate under the same temperature difference as in the case of convection process7KXV Q=k



Δt = h . Δt Δx

k h L L hL  Nu = = = ?  Δx k / h k 7KH1XVVHOWQXPEHULVDFRQYHQLHQWPHDVXUHRIWKHFRQYHFWLYHKHDWWUDQVIHUFRHI¿FLHQW)RUD JLYHQYDOXHRIWKH1XVVHOWQXPEHUWKHFRQYHFWLYHKHDWWUDQVIHUFRHI¿FLHQWLVGLUHFWO\SURSRUWLRQDO WRWKHUPDOFRQGXFWLYLW\RIWKHÀXLGDQGLQYHUVHO\SURSRUWLRQDOWRWKHVLJQL¿FDQWOHQJWKSDUDPHWHU RU

Δx =

4. Stanton number (St): 

 ,WLVWKHUDWLRRIKHDWWUDQVIHUFRHI¿FLHQWWRWKHÀRZRIKHDWSHUXQLWWHPSHUDWXUHULVHGXHWR WKHYHORFLW\RIWKHÀXLG

 St = RU  St =

h  ρ Uc p



hL /k Nu =  μ c Re × Pr ⎡ ⎤ ⎡ ρ UL ⎤ p ⎢ μ ⎥⎢ k ⎥ ⎣ ⎦⎣ ⎦





 7KXV6WDQWRQQXPEHUPD\DOVREHGH¿QHGDVWKHratio of Nusselt number and the product of Reynolds number and Prandtl number



 ,WLVZRUWKQRWLQJWKDW6WDQWRQQXPEHUFDQEHused only in correlating forced convection dataVLQFHWKHH[SUHVVLRQFRQWDLQVWKHYHORFLW\U 

5. Peclet number (Pe): 

 ,WLVWKHUDWLRRIPDVVKHDWÀRZUDWHE\FRQYHFWLRQWRWKHÀRZUDWHE\FRQGXFWLRQXQGHUD unit temperature gradient and through a thickness L

 Qconv UUcp

Qcond . =

k 1 k × = 1 L L

Dimensional Analysis Pe =

Qconv . ρ Uc p ρ c p LU LU . = = =  1 α Qcond . k /L k

767 



 ?



 7KH3HFOHWQXPEHUFDQEHZULWWHQDV



 



 ,WLQGLFDWHVWKDWWKH3HFOHWQXPEHUFDQEHZULWWHQDVDIXQFWLRQRIReDQGPrDOVR

ρ LU μ c p ρ c p LU LU × = = . μ k k α 1

Pe = Re . Pr =

6. Graetz number (G): 

 ,WLVUHODWHGRQO\IRUWKHKHDWÀRZWRWKHÀXLGÀRZLQJWKURXJKFLUFXODUSLSH,WLVGH¿QHGDV WKHUDWLRRIKHDWFDSDFLW\RIÀXLGÀRZLQJWKURXJKWKHSLSHSHUXQLWOHQJWKRIWKHSLSHWRWKH conductivity of the pipe



G=

 

k

=

mc p Lk





⎛π 2 ⎞ cp ⎜⎝ D . V . ρ⎟⎠ π ρ VD μ c p D L 4 = G= . . k k L 4 μ

 RU



⎛ mc p ⎞ ⎜⎝ L ⎟⎠



π D Re . Pr .  4 L  ZKHUHDDQGLDUHWKHGLDPHWHUDQGOHQJWKRIWKHSLSHUHVSHFWLYHO\



 7KLVFDQEHZULWWHQLQWKHRWKHUIRUPDV



 



G=

 RU



G=

mc p Lk

=

ρ AUc p Lk

=



AU αL

π 2 U ⎛ UD ⎞ ⎛ π D ⎞ D × =⎜ ⎟ ⎜ ⎟ 4 α L ⎝ α ⎠ ⎝ 4L ⎠

RU

G =

RU

⎛ π D⎞ G = Pe ⎜  ⎝ 4 ⎟⎠



7KLVVKRZVWKDW*UDHW]QXPEHULVPHUHO\DSURGXFWRI3HFOHWQXPEHUDQGDFRQVWDQW

7. Grashoff number (Gr): 

 *UDVKRIIQXPEHULVUHODWHGZLWKnatural convection heat transfer,WLVGH¿QHGDVWKHratio of the product of inertia force and buoyancy force to the square of viscous force7KXV



 

Gr =



 RU

Gr =

(Inertia force) × (Buoyancy force) (Viscous force) 2 ρ2β g Δt L3 μ2



=

(ρ′ U 2 L2 ) × (ρβ g . Δt L3 ) (μ UL) 2 

*UDVKRIIQXPEHUKDVDUROHLQIUHHFRQYHFWLRQVLPLODUWRWKDWSOD\HGE\5H\QROGVQXPEHULQ IRUFHGFRQYHFWLRQ)UHHFRQYHFWLRQLVXVXDOO\VXSSUHVVHGDWVXI¿FLHQWO\VPDOOGrEHJLQVDWVRPH FULWLFDOYDOXHRIGrGHSHQGLQJXSRQWKHDUUDQJHPHQWDQGWKHQEHFRPHVPRUHDQGPRUHHIIHFWLYHZLWK LQFUHDVLQJGr

768

Chapter : 12

([DPSOHShow by dimensional analysis that for natural convection heat transferNusselt number is a function of Grashoff number and Prandtl number$0,(6XPPHU 6ROXWLRQ5HIHUWR$UWLFOH ([DPSOHIt is required to estimate the heat transfer from a cylinder 50 mm diameter and of surface temperature 140ƒCwhen kept in crossÀRZRIDLUDWDYHORFLW\RIPs and temperature 20ƒCFor this purposescale model experiments are performed using a 15 scalemodel with the same surface and air temperature but different velocitiesThe following results are obtained from experiments on the model Velocity of airms   2.0 5.0 10.0 20.0   +HDWWUDQVIHUFRHI¿FLHQWWm K   39.5 71.2 106.5 165.3 Calculate i  7KHKHDWWUDQVIHUFRHI¿FLHQW  ii  The rate of heat transfer per metre length of actual cylinder*$7( 6ROXWLRQGivenDp PP Dm = 50 ×

1 = 10 mm; Vp PVt ƒ&t ƒ& 5

i  7KHKHDWWUDQVIHUFRHI¿FLHQW )RUIRUFHGFRQYHFWLRQPRGHOWHVWLQJ5H\QROGVQXPEHURIWKHSURWRW\SHDQGPRGHOPXVWEHequal ⎛ UD ⎞ ⎛ UD ⎞ ? ⎜⎝ ⎟⎠ = ⎜⎝ ⎟ ν m ν ⎠p 6LQFHWKHPRGHOWHVWLQJLVFRQGXFWHGDWVDPHVXUIDFHDQGDLUWHPSHUDWXUHQFDQEHFRQVLGHUHG WREHconstant ? UD m UD p ie RU

 

UmDm UpDp Um =

U p Dp Dm

=

4 × 50 = 20 m/s 10

7KHKHDWWUDQVIHUFRHI¿FLHQWDWPVRQWKHPRGHOLV:P.$QV ii  7KHUDWHRIKHDWWUDQVIHUSHUPHWUHOHQJWKRIDFWXDOF\OLQGHU4 Q hAt±t  Sîî î±  :$QV

12.9. CHARACTERISTIC LENGTH OR EQUIVALENT DIAMETER ,Q WKH QRQGLPHQVLRQDO QXPEHU GLVFXVVHG LQ $UW   H[SUHVVLRQV WKHUH KDV DSSHDUHG D FKDUDFWHULVWLFOHQJWKLRUGLDPHWHUD7KHSLSHDQGÀDWSODWHDUHWKHPRVWVLPSOHJHRPHWULHVIRUWKH RFFXUUHQFHRIDÀRZ+RZHYHULQPDQ\LQVWDQFHVVRPHFRPSOLFDWHGJHRPHWULHVDUHDOVRXVHGDQG WKHQDOOWKHFDOFXODWLRQVFRQFHUQLQJÀRZDQGFRQYHFWLYHKHDWWUDQVIHUEHFRPHPXFKPRUHFRPSOLFDWHG DQGGLI¿FXOW,QRUGHUWRDYRLGVXFKGLI¿FXOWLHVWKHFRQFHSWRIDQequivalent circular tube is used 7KLVLVDtubeZKLFKZRXOGSUHVHQWWKHsame resistance againsttheÀRZRUZRXOGVHFXUHWKHsame heat transfer as the duct usually used under equal or comparable conditions7KHGLDPHWHURIDQ HTXLYDOHQW WXEH LV NQRZQ DV equivalent diameter De RU characteristic length Le7KH HTXLYDOHQW GLDPHWHULVXVXDOO\GH¿QHGDV ⎛π ⎞ 4 × ⎜ D2 ⎟ ⎝4 ⎠ 4 Ac = = D LQQHUGLDPHWHURIWKHWXEH  De =  π P D ZKHUHAc FURVVVHFWLRQDODUHDDQGP SHULPHWHU 

Dimensional Analysis

769

7KLVSULPDULO\KROGVZKHQLWLVUHTXLUHGWR¿QGWKHSUHVVXUHGURS+RZHYHULWLVDOVRXVHGLQ problems of heat transfer by convectionEHFDXVHRIWKHH[LVWLQJVLPLODULW\EHWZHHQPRPHQWXP WUDQVIHUDQGKHDWWUDQVIHU 7KHHTXLYDOHQWGLDPHWHURUFKDUDFWHULVWLFOHQJWKRIIHZJHRPHWULHVDUHJLYHQEHORZ  For rectangular duct>)LJi @ 4 Ac 4lb 2lb  De = = =   P 2 (l + b) l + b

)LJ(TXLYDOHQWGLDPHWHUDe IRUYDULRXVJHRPHWULHV

 For rectangular annulus>)LJii @ 4 Ac 4 × (l1b1 – l2 b2 ) 2 (l1b1 – l2 b2 ) = =  P 2 [(l1 + b1 ) + (l2 + b2 )] [(l1 + b1 ) + (l2 + b2 )] :KHQl bDQGl b 2 (l12 – l22 ) De = = (l1 – l2 )   2l1 + 2l2  De =





 For annulus>)LJiii @ 

π ⎞ ⎛ 4 ⎜ lb − d 2 ⎟ ⎝ 4 Ac 4 ⎠ De = =  P [2 (l + b) + πd ]



 For annulus>)LJiv @ 

π 4 × (D2 – d 2 ) 4 Ac 4 De = = = (D – d )  P π (D + d )



770

Chapter : 12

12.10. MODEL STUDIES AND SIMILITUDE 12.10.1. MODEL AND PROTOTYPE ,QRUGHUWRNQRZDERXWWKHSHUIRUPDQFHRIODUJHPDFKLQHVHTXLSPHQWVRUVWUXFWXUHV LQYROYLQJ ÀXLGÀRZEHIRUHDFWXDOO\PDQXIDFWXULQJRUFRQVWUXFWLQJ WKHPWKHLUPRGHOVDUHPDGHDQGWHVWHGWR JHWWKHUHTXLUHGLQIRUPDWLRQ7KHPRGHOLVWKHsmall scale replica of the actual machineequipment or structureThe actual machineequipment or structure is called3URWRW\SH7KHPRGHOVDUHQRW DOZD\VVPDOOHUWKDQWKHSURWRW\SHLQVRPHFDVHVDPRGHOPD\EHHYHQODUJHURURIWKHVDPHVL]HDV SURWRW\SHGHSHQGLQJXSRQWKHQHHGDQGSXUSRVHegWKHZRUNLQJRIDZULVWZDWFKRUFDUEXUHWWRU FDQEHVWXGLHGLQDODUJHVFDOHPRGHO 

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Chapter : 13 772 C H A P T E R

13

Universities’ Examinations Questions with Answers/Solutions: (Latest-Selected)

A. Short Answer Questions B. Conventional Questions with Solutions C. Multiple-choice Questions with Answers. A. SHORT ANSWER QUESTIONS 

4 What is thermal conductivity of a material? On what factors does it depend?

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Universities’ Exams. Questions (Latest)...

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 4 What is dielectric heating?  $QV Dielectric heatingLVDPHWKRGRITXLFNO\KHDWLQJLQVXODWLQJPDWHULDOVSDFNHGEHWZHHQWKH SODWHVRIDQHOHFWULFFRQGHQVHU WRZKLFKDhigh frequency, high voltage alternating current is applied7KH temperature rise is uniform and the internal heat generated is of equal intensity on the surface and the core   7KHKHDWJHQHUDWHG   ²LVdirectly proportionalWRWKHarea of condenser platesvoltageDQGfrequency   ²LVLQYHUVHO\SURSRUWLRQDOWRWKHGLVWDQFHEHWZHHQWKHSODWHV   ²SHUXQLWYROXPHRIWKHPDWHULDOLVFRQVWDQW   7KHPHWKRGRI'LHOHFWULFKHDWLQJHQWDLOVWKHIROORZLQJadvantagesi 0RUHHFRQRPLFDO  ii 1RSROOXWLRQiii +LJKHIIFLHQF\DQGiv *UHDWHUVDIHW\   'LHOHFWULF KHDWLQJ UHIHUV WR WKH VLWXDWLRQ WKDW FRUUHVSRQGV WR WKH FRQGXFWLRQ LQ D SODQH  ZDOOZLWKuniform internal heat source.  4 What is a thermometer well?  $QV )RUHVWLPDWLQJHUURULQWKHYDOXHRIWHPSHUDWXUHPHDVXUHGE\DWKHUPRPHWHUGLSSHGLQD WKHUPRPHWHU ZHOO WKH WKHRU\ RI H[WHQGHG VXUIDFHV LV YHU\ KHOSIXO$ thermometer well LV GH¿QHG DV D VPDOO WXEH ZHOGHG UDGLDOO\ LQWR D SLSHOLQH WKURXJK ZKLFK D ÀXLG ZKRVH WHPSHUDWXUHLVWREHPHDVXUHGLVÀRZLQJ  4 :ULWHGRZQ¿YHH[DPSOHVRIPXOWLGLPHQVLRQDOKHDWFRQGXFWLRQ  $QV ([DPSOHVRIPXOWLGLPHQVLRQDOKHDWFRQGXFWLRQDUH  i  &RROLQJRILQWHUQDOFRPEXVWLRQHQJLQHEORFN  ii  $LUFRQGLWLRQLQJGXFWV  iii  &RPSRVLWHERGLHV  iv  +HDWWUHDWPHQWRIPHWDOOLFSDUWVRIGLIIHUHQWVKDSHV  v  &KLPQH\V  4 Name the four methods, in general, prevalent for the solution of the multi-dimensional heat conduction problems.  $QV 7KHYDULRXVPHWKRGVDUH   Analytical methods:7KHVHPHWKRGVDUHDSSOLHGRQO\WRVLPSOHSUREOHPVDVWKHVHDUH TXLWHFXPEHUVRPHDQGLWLVGLI¿FXOWWRJHWWKHVROXWLRQVDOZD\V   Graphical methods:7KHVHPHWKRGVDUHDOVRXVHGIRUVLPSOHSUREOHPVRQO\7KH\JLYH DURXJKHVWLPDWHRIWKHWHPSHUDWXUH¿HOG   Analogical methods7KHVHPHWKRGVFDQEHHPSOR\HGIRUFRPSOLFDWHGJHRPHWULHVDV WKHVHDUHEDVHGRQFHUWDLQVLPLODULWLHVEHWZHHQWZRW\SHVRI¿HOGHTXDWLRQV   Numerical methods:7KHVHPHWKRGVFDQEHDSSOLHGWRDQ\PXOWLGLPHQVLRQDOSUREOHP  4 What is ‘conduction shape factor Sfc?’ M  i.e QXPEHU RI ÀRZ ODQHVWXEHV GLYLGHG E\ WKH QXPEHU RI WHPSHUDWXUH  $QV 7KH UDWLR N LQFUHPHQWV LVFDOOHGWKHconduction shape factor Sfc  4 What do you mean by conduction of heat in unsteady state?  $QV ,IWKHWHPSHUDWXUHRIDERG\GRHVQRWvarywith timeLWLVVDLGWREHLQDsteady state%XW LI WKHUH LV DQ abrupt change LQ LWV VXUIDFH WHPSHUDWXUH LW ERG\  DWWDLQV DQ HTXLOLEULXP WHPSHUDWXUHRUDVWHDG\VWDWHDIWHUVRPHSHULRG'XULQJWKLVSHULRGWKHWHPSHUDWXUHvaries with timeDQGWKHERG\LVVDLGWREHLQDQunsteady or transient state.7KHWHUPWUDQVLHQWRU XQVWHDG\GHVLJQDWHVDSKHQRPHQRQZKLFKLVWLPHGHSHQGHQW7KHVWHDG\VWDWHLVWKXVWKH limitRIWUDQVLHQWWHPSHUDWXUHGLVWULEXWLRQIRUODUJHYDOXHVRIWLPH Conduction of heat in unsteady stateUHIHUVWRWKHWUDQVLHQWFRQGLWLRQVZKHUHLQWKHKHDWÀRZ and the temperature distribution at any point of the system vary continuously with time

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Chapter : 13

 4 What is ‘characteristic length or equivalent diameter’?  $QV ,Q WKH QRQGLPHQVLRQDO QXPEHU H[SUHVVLRQV WKHUH DSSHDUV D FKDUDFWHULVWLF OHQJWK L RU GLDPHWHUD7KHSLSHDQGÀDWSODWHDUHWKHPRVWVLPSOHJHRPHWULHVIRUWKHRFFXUUHQFHRID ÀRZ+RZHYHULQPDQ\LQVWDQFHVVRPHFRPSOLFDWHGJHRPHWULHVDUHDOVRXVHGDQGWKHQDOOWKH FDOFXODWLRQVFRQFHUQLQJÀRZDQGFRQYHFWLYHKHDWWUDQVIHUEHFRPHPXFKPRUHFRPSOLFDWHG DQGGLI¿FXOW,QRUGHUWRDYRLGVXFKGLI¿FXOWLHVWKHFRQFHSWRIDQequivalent circular tube is used7KLVLVDtubeZKLFKZRXOGSUHVHQWWKHsame resistance againsttheÀRZRUZRXOG VHFXUHWKHsame heat transfer as the duct usually used under equal or comparable conditions 7KHGLDPHWHURIDQHTXLYDOHQWWXEHLVNQRZQDVequivalent diameterDeRUcharacteristic lengthLe7KHHTXLYDOHQWGLDPHWHULVXVXDOO\GH¿QHGDV π 4 × D2 4 Ac 4    DLQQHUGLDPHWHURIWKHWXEH    De  P πD   ZKHUHAc FURVVVHFWLRQDODUHDDQGP SHULPHWHU    7KLVSULPDULO\KROGVZKHQLWLVUHTXLUHGWR¿QGWKHSUHVVXUHGURS+RZHYHULWLVDOVRXVHG in problems of heat transfer by convection EHFDXVH RI WKH H[LVWLQJ VLPLODULW\ EHWZHHQ PRPHQWXPWUDQVIHUDQGKHDWWUDQVIHU  4 What is boundary layer thickness G ?  $QV 7KHYHORFLW\ZLWKLQWKHERXQGDU\OD\HULQFUHDVHVIURP]HURDWWKHERXQGDU\VXUIDFHWRWKH YHORFLW\RIWKHPDLQVWUHDPDV\PSWRWLFDOO\7KHUHIRUHWKHWKLFNQHVVRIWKHERXQGDU\OD\HU LVDUELWUDULO\GH¿QHGDVthat distance from the boundary in which the velocity reaches per centof the velocityof the free streamu U It is denoted by the symbolGThis GH¿QLWLRQhowevergives anDSSUR[LPDWHYDOXHRIWKHERXQGDU\OD\HUWKLFNQHVVDQGKHQFH GLVJHQHUDOO\WHUPHGDVnominal thicknessof the boundary layer.   7KHERXQGDU\OD\HUWKLFNQHVVIRUgreater accuracyLVGH¿QHGLQWHUPVRIFHUWDLQPDWKHPDWLFDO H[SUHVVLRQVZKLFKDUHWKHPHDVXUHRIWKHERXQGDU\OD\HURQWKHÀRZ7KHFRPPRQO\DGRSWHG GH¿QLWLRQVRIWKHERXQGDU\OD\HUWKLFNQHVVDUH   'LVSODFHPHQWWKLFNQHVVG   0RPHQWXPWKLFNQHVVT   (QHUJ\WKLFNQHVVGe   4 What is displacement thickness G ?  $QV 7KHdisplacement thicknessFDQEHGH¿QHGDVIROORZV   ³It is the distance, measured perpendicular to the boundary, by which the main/free stream is displaced on an account of formation of boundary layer´ 2U   ³It is an additional³wall thickness´that would have to be added to compensate for the UHGXFWLRQLQÀRZUDWHRQDQDFFRXQWRIERXQGDU\OD\HUIRUPDWLRQ´   7KHGLVSODFHPHQWWKLFNQHVVLVGHQRWHGE\G  4 What is momentum thickness T ?  $QV Momentum thicknessLVGH¿QHGDVWKHGLVWDQFHWKURXJKZKLFKWKHWRWDOORVVRIPRPHQWXP per second be equal to if it were passing a stationary plate,WLVGHQRWHGE\T   ,WPD\DOVREHGH¿QHGDVWKHdistancemeasured perpendicular to the boundary of the solid body, by which the boundary should be displaced to compensate for reduction in momentum RIWKHÀRZLQJÀXLGRQDQDFFRXQWRIERXQGDU\OD\HUIRUPDWLRQ  4 'H¿QHHQHUJ\WKLFNQHVVGe  $QV Energy thicknessLVGH¿QHGDVWKHGLVWDQFHPHDVXUHGSHUSHQGLFXODUWRWKHERXQGDU\RIWKH solid body by which the boundary should be displaced to compensate for the reduction in .(RIWKHÀRZLQJÀXLGRQDQDFFRXQWRIERXQGDU\OD\HUIRUPDWLRQ,WLVGHQRWHGE\Ge

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Universities’ Exams. Questions (Latest)...

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776 

Chapter : 13

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Universities’ Exams. Questions (Latest)...

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B. CONVENTIONAL QUESTIONS WITH SOLUTIONS 4 A refrigerant suction line of 25 mm OD (outer diameter) is to be insulated using a PDWHULDORIWKHUPDOFRQGXFWLYLW\N :P.7KHVXUIDFHKHDWWUDQVIHUFRHI¿FLHQWK is 10W/m2K. 9HULI\LIWKHLQVXODWLRQLVHIIHFWLYHRUQRW:KDWVKRXOGEHWKHPD[LPXPYDOXHRIWKHUPDOFRQGXFtivity of insulation to reduce the heat transfer? 8378 6ROXWLRQGiven: 25 r1    PPk  :P.; h :P2. 2 7RYHULI\LIWKHLQVXODWLRQLVHIIHFWLYHRUQRW 7KH critical radiusRILQVXODWLRQ k 0.25    P PP rc  h0 10 rcLVgreaterWKDQWKHUDGLXVRIWKHFDEOH +HQFHWKHinsulation upto a thickness of 12.5 PP i.e. PP±PP will be effective in heat dissipation 0D[LPXPYDOXHRIWKHUPDOFRQGXFWLYLW\(kPD[ WRUHGXFHWKHKHDWWUDQVIHU k⎞ ⎛ 7KHYDOXHRIkPD[UHTXLUHGIRUrcWREHHTXDOWRrLVJLYHQE\kPD[ rîh ⎜ from rc = ⎟  h ⎝ a⎠ i.e.kPD[ rh i.e.kPD[ î±î :P. $QV 4A wire of 8 mm diameter at a temperature of 60°C is to be insulated by a material KDYLQJN :Pƒ&+HDWWUDQVIHUFRHI¿FLHQWRQWKHRXWVLGHK0 = 8W/m2°C. Ambient temSHUDWXUH ƒ&)RUPD[LPXPKHDWORVVZKDWLVWKHPLQLPXPWKLFNQHVVRILQVXODWLRQDQGWKH heat loss per metre length? Find the increase in heat dissipation due to insulation. Also, calculate the increase in current carrying capacity due to insulation. '8 8 6ROXWLRQGiven:r  =  PP Pt ƒCk :P°& 2 h :Pƒ&ta ƒ& 7KLFNQHVVRILQVXODWLRQIRUPD[LPXPKHDWORVV k 0.174 = 0.02175 m Critical radiusrc  = h0 8 0D[LPXPKHDWORVVRFFXUVDWWKHFULWLFDOUDGLXV ?7KLFNQHVVRILQVXODWLRQDWWKHFULWLFDOUDGLXV rc±r    ±  PP +HDWORVVSHUPHWUHOHQJWK +HDWORVVSHUPHWUHZLWKRXWLQVXODWLRQ Q hA't hSrL ît±ta    îSîî î±  :P$QV ,QFUHDVHLQKHDWGLVVLSDWLRQGXHWRLQVXODWLRQ ,QFDVHRILQVXODWHGZLUHWKHKHDWÀRZSHUPHWUHLVJLYHQDV 2π l (t1 − ta ) Q    ln (rc / r1 ) 1 + h0 . rc k

Chapter : 13

778  





 





2π × 1 (60 − 25) 1 ln (0.02175 / 0.004) + 8 × 0.02175 0.174 219.91 = 14.207 W/m 5.75 + 9.73

? Increase in heat dissipation due to insulation    Q±Q ± :P$QV ,QFUHDVHLQFXUUHQWFDUU\LQJFDSDFLW\GXHWRLQVXODWLRQ +HDWGLVVLSDWLRQZLWKEDVHZLUHQ IR +HDWGLVVLSDWLRQZLWKLQVXODWHGZLUHQ IR :KHUHIDQGIDUHFXUUHQWVIRUEDVHZLUHDQGZLUHZLWKLQVXODWLRQUHVSHFWLYHO\DQG5LVWKH HOHFWULFDOUHVLVWDQFHRIWKHHOHFWULFZLUH Q2 I2 R I2   22 = 22 ? Q1 I1 R I1 I2   I1

RU

Q2 14.207 = = 1.421 Q1 7.037

?Increases in current carrying capacity due to insulation I − I1 1.421 − 1 × 100 = × 100   $QV     2 I1 1 4A copper conductor (k = 380 W/ m°C), U = 2 × 10–8 (ohm × m), 8 mm diameter and 1m long, connects two large plates. One face is maintained at 30°C and the outer face at 50°C. 6SDFHEHWZHHQWKHSODWHVLV¿OOHGZLWKDQLQVXODWLRQ L :KDWLVWKHPD[LPXPWHPSHUDWXUHDQGLWVORFDWLRQLIWKHPD[LPXPFXUUHQWÀRZLQJLV 150 A ? (ii) Calculate the heat dissipation to L.H.S. and R.H.S. 18 6ROXWLRQ  



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$QVtPD[ ƒ&IURP/+6Q5+6 WQ/+6 ±W 4Meat rolls of 25 mm diameter, k = 1 W/m°C are heated by microwave heating. Centre WHPSHUDWXUHRIWKHUROOLVƒ&6XUURXQGLQJWHPSHUDWXUHLVDWƒ&+HDWWUDQVIHUFRHI¿FLHQWDWWKH 378 surface is 25 :Pƒ&. Find the microwave heating capacity required in :P3. 6ROXWLRQ5HIHUWR([DPSOH$QV î:P 4$OXPLQLXPVTXDUH¿QVPPîPP RIFPOHQJWKDUHSURYLGHGRQWKHVXUIDFH of an electronic semiconductor device to carry 46 mW of energy generated by the electronic device DQGWKHWHPSHUDWXUHDWWKHVXUIDFHRIWKHGHYLFHVKRXOGQRWH[FHHGƒ&7KHWHPSHUDWXUHRIWKH surrounding medium is 40°C. Thermal conductivity of the aluminium = 190 W/m K and heat transfer FRHI¿FLHQWK :P2K. Find the number of pins required to carry out the above duty. Neglect the KHDWORVVIURPWKHHQGRIWKH¿Q                        0'8 6ROXWLRQ5HIHUWR([DPSOH $QV VD\

Universities’ Exams. Questions (Latest)...

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4 A cylinder 5 cm diameter and 50 cm long, is provided with 14 longitudinal straight ¿QVHDFKPPWKLFNDQGPPKHLJKW&DOFXODWHWKHKHDWORVVIURPWKHF\OLQGHUSHUVHFRQGLIWKH surface temperature of the cylinder is 200°C. Take h = 25W/m2K, k = 80:P. and ta = 45°C. 08 6ROXWLRQ5HIHUWR([DPSOH $QVQtotal :  4The temperature of an air stream in a tube is measured by a thermometer placed LQDSURWHFWLYHZHOO¿OOHGZLWKRLO7KHWKHUPRPHWHULVPDGHRIVWHHOWXEHRIPPWKLFNVKHHWRI length 120 mm. The thermal conductivity of steel = 58.8W/mK, and ha = 23.3 W/m2K. If the air temperature record was 84°C, estimate the measurement error, if the temperature at the base of the well was 40°C. 378 6ROXWLRQ5HIHUWR([DPSOH $QV ƒ& 4 A thermocouple (TC) function is in the form of 8mm sphere. Properties of the material cp = 420 J/kg K; U = 500 kg/m3N :P.DQGKHDWWUDQVIHUFRHI¿FLHQWK :P2K. Find, if the junction is initially at a temperature of 28°C and inserted in a stream of hot-air at 300°C: (i) The time constant of the TC. (ii) The TC is taken out from the hot air after 10s and kept in still air at 30°C. Assuming ‘h’ in air as 10W/m2.¿QGWKHWHPSHUDWXUHDWWDLQHGE\WKHMXQFWLRQVDIWHUUHPRYLQJIURPKRW air stream. *78 6ROXWLRQ5HIHUWR([DPSOH $QVƒ& 4$FDUERQVWHHOVKDIWRIPGLDPHWHULVKHDWWUHDWHGLQDJDV¿UHGIXUQDFHZKRVH JDVHV DUHDW.DQGSURYLGHDFRQYHFWLRQFRHI¿FLHQWRI:P2K. If the shaft enters the furnace at 300 K, how long must it remain in the furnace to achieve a centre line temperature of 900 K? Take for the carbon steel : U = 7854 NJP3; k = 48.8W/mK; C = 559 -NJ.. 0'8 0.2  PT KT Kh :P.Ta K 2 U NJPk :P.c -NJ . 6XUIDFHDUHDRIWKHVKDIW SRL 9ROXPHRIWKHVKDIW SRL V πR 2 L R 0.1 = = = = 0.05 m &KDUDFWHULVWLFOHQJWKLc  As 2πRL 2 2 6ROXWLRQ Given :R 

hLc 80 × 0.05 = = 0.082 k 48.8 6LQFHBiLV less thanKHQFHOXPSHGFDSDFLWDQFHPHWKRG1HZWRQLDQKHDWLQJRUFRROLQJ  PD\EHDSSOLHGIRUWKHVROXWLRQRIWKHSUREOHP 7KHWHPSHUDWXUHGLVWULEXWLRQLVJLYHQE\ ⎡ − hAs ⎤ T − Ta τ⎥    exp. ⎢  >(TQ@ Ti − Ta ⎣ ρVc ⎦ %LRWQXPEHU



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⎫ ⎪ −h . τ ⎪ ⎬ ⎛V ⎞ ⎪ × c ⎜⎝ A ⎟⎠ ⎪ s ⎭

Chapter : 13

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6XEVWLWXWLQJWKHYDOXHVZHJHW ⎡ ⎤ 900 − 1200 80 × τ   exp ⎢ −   ⎥ 300 − 1200 ⎣ 7854 × 0.05 × 559 ⎦ ±

RU  e±î W ± ±î RU e W  1 ± = 3.0 RU eî W  0.333 RU î±W  1.098 = 3.016 ×103 s = 0.838 hrs. RU W  –4 3.64 ×10 i.e.,

W KUV$QV

4'U\DLUDWDWPRVSKHULFSUHVVXUHDQGƒ&LVÀRZLQJZLWKDYHORFLW\RIPV DORQJWKHOHQJWKRIDÀDWSODWHVL]HPîP PDLQWDLQHGDWƒ&8VLQJ%ODVLXVH[DFW VROXWLRQ FDOFXODWH WKH KHDW WUDQVIHU UDWH IURP L  WKH ¿UVW KDOI RI WKH SODWH LL  IXOO SODWH DQG LLL QH[WKDOIRIWKHSODWH7KHSURSHUWLHVRIDLUDWPHDQEXONWHPSHUDWXUH  ƒ&DUH U = 1.025 kg/m3; cp = 1017 J/kg K; P = 19.907 × 10–6 kg/ms; k = 0.0279 W/m°C, Pr = 0.71. 8378 6ROXWLRQGiven :tf ƒ&U PVts ƒ&L PB P i +HDWWUDQVIHUUDWHIURP¿UVWKDOIRIWKHSODWH 0.5 = 0.25 m )RU¿UVWKDOIRIWKHSODWH[  2 3 × 0.25 8[ 3 × 0.25 = =   Re[   î ν (μ / ρ) 19.907 × 10 −6 / 1.025 6LQFHRe[îKHQFHWKHÀRZLVlaminar. 7KHORFDO1XVVHOQXPEHULVJLYHQE\   Nu[ Re[ òPr 

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 0.332 (3.862 × 10 ) (0.71) = 58.2 K [ 1X [ × N %XW Nu[  [ or h[ = N [ 58.2 × 0.0279 = 6.495 W/m2 °C  ? h[  0.25 $YHUDJHKHDWWUDQVIHUFRHI¿FLHQWh h[ = × :Pƒ& ?Heat transfer from half of the plate,  

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Q  h As (ts − t∞ )    îî î±  :$QV ii +HDWWUDQVIHUUDWHIURPIXOOSODWH )RUIXOOSODWH[ L P 8[ 3 × 0.5 = = 7.723 × 104  ? ReL  ν (19.907 × 10 −6 / 1.025)

Universities’ Exams. Questions (Latest)...

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hL k k × L

 

Nu   0.664 ( ReL )½ ( Pr )1/3 =

 ?

h   0.664 ( ReL )1/2 ( Pr )1/3

 

4 1/2 1/3   0.664 × (7.723 × 10 ) × (0.71) ×

 

2   9.186 W/m °C

0.0279 0.5

?7KHKHDWWUDQVIHUUDWHIURPIXOOSODWH  

Q  h As (ts − t∞ ) = 9.186 × (0.5 × 0.75) (100 − 20)

   :$QV iii +HDWWUDQVIHUUDWHIURPQH[WKDOIRIWKHSODWH  +HDWWUDQVIHUUDWHIURPQH[WKDOIRIWKHSODWH    Q±Q ± :$QV 4 A refrigerated truck is moving at a speed of 85km/h where ambient temperature is ƒ&7KHERG\RIWKHWUXFNLVRIUHFWDQJXODUVKDSHRIVL]HP/ îP% îP+ $VVXPH the boundary layer is turbulent and the wall surface temperature is at 10°C. Neglect heat transfer IURPWKHYHUWLFDOIURQWDQGEDFNVLGHRIWUXFNDQGÀRZRIDLULVSDUDOOHOWRPORQJVLGH&DOFXODWH WKHKHDWORVVIURPWKHIRXUVXUIDFHV$OVR¿QGRXWWKHSRZHUUHTXLUHGWRRYHUFRPHZLQGUHVLVWDQFH )RUWKHWXUEXOHQWÀRZRYHUÀDWVXUIDFHV Nu = 0.036 Re0.8 Pr0.32. ⎛ 50 + 10 ⎞  U = 1.165 kg/m3, cp = 1.005 kJ/kgK; Average properties of air at 30°C ⎜ ⎝ 2 ⎟⎠ Q = 16 × 10–6 m2/s; Pr = 0.701.

0*8.HUDOD

6ROXWLRQ Given :6L]HRIWKHWUXFNERG\L PB PH PU NPK 

85 × 1000 PV%RXQGDU\OD\HUWXUEXOHQW:DOOVXUIDFHWHPSHUDWXUHts ƒ&tf ƒ& 60 × 60 :HNQRZ Pr 

c p .μ

or k =

c p ρν

k Pr 1005 × 1.165 × 16 × 10 −6 = 0.02672 W/mK RU k  0.701 +HDWORVVIURPWKHIRXUVXUIDFHV  

ReL 

U × L 23.61 × 10 = = 1.476 × 107 ν 16 × 10 −6

6LQFHReL!îWKHUHIRUHWKHÀRZLVturbulent. )RUWXUEXOHQWÀRZZHKDYH 1XVVHOWQXPEHUNu  0.036 Re L RU 

0.8

. Pr 0.33

Nu îî î  î

+HDWWUDQVIHUFRHI¿FLHQW h  

Nu.k 1.74 × 104 × 0.02672 = = 46.49 W/m 2 K L 10

?Heat loss from the four surfaces, Q  h . A (t∞ − ts )

Chapter : 13

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  i.e.,



 h × L ( B + H ) × 2 × (50 − 10) î îî î: Q î:$QV

3RZHUUHTXLUHGWRRYHUFRPHZLQGIULFWLRQ )RU5H\QROGVQXPEHUEHWZHHQDQGWKHIROORZLQJUHODWLRQVKLSVXJJHVWHGE\3UDQGWO DQG6FKOLFKWLQJKROGVJRRG 0.455 $YHUDJHVNLQIULFWLRQFRHI¿FLHQW C f    >(TQ@ (log10 Re L ) 2.58 RU

Cf  

0.455 (log10 1.476 × 107 ) 2.58

= 2.82 × 10 −3

1 ρAU 2  2 1  î±î îî> î@î  2  N FDîU î N:  N:$QV

?  'UDJIRUFHFD  C f × 



  ? 3RZHUUHTXLUHGP i.e., P

4$ÀDWSODWHPZLGHDQGPORQJLVPDLQWDLQHGDWƒ&LQDLUZLWKIUHHVWUHDP WHPSHUDWXUHRIƒ&ÀRZLQJDORQJPVLGHRIWKHSODWH'HWHUPLQHWKHYHORFLW\RIDLUUHTXLUHG to have a rate of energy dissipation as 3.75 kW. Use relations: 978.DUQDWND 13

NuL = 0.664 Re0.5 Pr

IRUODPLQDUÀRZDQG 13

0.8 NuL =>0.036 Re – 836 @ Pr

IRUWXUEXOHQWÀRZ

⎛    ⎞ Take average properties of air at 50°C ⎜⎝ ⎟: 2 ⎠ U = 1.0877 kg/m3; cp = 1.007 kJ/kg K; P = 2.029 × 10–5 kg/ms; Pr = 0.703; k = 0.028 :P. 6ROXWLRQL PB Pts ƒ& t∞  ƒ&Q N: : 9HORFLW\RIDLUU :HNQRZ  RU

Q  h A tV± t∞   h îîî î± h   :P.

h . L 15.625 × 1.5 = = 837.05 k 0.028 1RZ¿UVWRIDOOZH¿QGRXWZKHWKHUWKHÀRZLVODPLQDURUWXUEXOHQW,IRe ZRUNVRXWWREHOHVV WKDQîLWZLOOEHDODPLQDUÀRZLIJUHDWHUWKDQîLWZLOOEHDWXUEXOHQWRQH ?1XVVHOWQXPEHU NuL 

Universities’ Exams. Questions (Latest)...

783

)RUODPLQDUÀRZ 



13 0.5 NuL î ReL . Pr 2

Nu L 837.05 ⎛ ⎞ ⎡ ⎤ =⎢ ReL  ⎜ 13⎥ 0.664 0.703 × ⎝ 0.664 × Pr1 3 ⎟⎠ ⎣ ⎦  6LQFH ReL!î WKHUHIRUHÀRZLVturbulent. RU



2

î

1RZOHWXVFRQ¿UPWKDWWKHÀRZLVWXUEXOHQWE\XVLQJWKHJLYHQIRUPXOD 



0.8 13 NuL  ⎡⎣0.036 ReL – 836⎤⎦ Pr 1

?

⎡ Nu L ⎤ 0.8 ⎢ Pr1/3 + 836 ⎥ ReL = ⎢ ⎥ ⎢ 0.036 ⎥ ⎣⎢ ⎦⎥

$OVR

ReL 

Re μ ρUL or U = L m/s μ ρL

U

7.36 × 105 × 2.029 × 10 –5  PV$QV  1.0877 × 1.5

?





îi.e.!î 

4'HULYHDQH[SUHVVLRQIRUKHDWÀRZWKURXJKDKROORZVSKHUHRILQVLGHUDGLXVU1 and outside radius r2ZKRVHLQWHUQDODQGH[WHUQDOVXUIDFHVDUHPDLQWDLQHGDWWHPSHUDWXUHVW1 and t2 respectively. The thermal conductivity of the sphere material has a quadratic variation with temperature k = k0 DWEt2)                      378 6ROXWLRQ5HIHUWR([DPSOH 4 Estimate heat transfer rate from a 100 W incandescent bulb at 140°C to an ambiHQWDWƒ&$SSUR[LPDWHWKHEXOEDVFPGLDPHWHUVSKHUH&DOFXODWHSHUFHQWDJHSRZHUORVVE\ natural convention. Use the correlation and the air properties as: ⎛    ⎞ Nu = 0.60 (Gr Pr)1/4; Properties of air at 82°C ⎜⎝ ⎟⎠ are : Q = 21.46 × 10–6 m2/s, 2 k = 3.038 × 10–3 W/mK, Pr = 0.699.

0'8+DU\DQD

6ROXWLRQ5HIHUWR([DPSOH 4  Air at 35°C flows across a cylinder of 50 mm diameter at a velocity of 50 m/s. The cylinder surface is maintained at 145°C. Find the heat lost per unit length. Properties at mean temperature of 90°C are: U = 1 kg/m3; P = 20 × 10–6 kg/ms; k = 0.0312 W/m°C; cp = 1.0 kJ/kg °C. Use relation: NuD = 0.027 (ReD)0.805 (Pr)1/3 6ROXWLRQGivenD PP P t∞  ƒ&ts ƒ&V PV +HDWORVWSHUXQLWOHQJWKQ/L 5H\QROGVPXPEHU Re



 3UDQGWOQXPEHU



Pr

PVD 1 × 50 × 0.05 =  î μ 20 × 10 –6 cp . μ k

=

(1.0 × 103 ) × 20 × 10 –6 = 0.641 0.0312

'8

Chapter : 13

784

?1XVVHOWQXPEHUNuD RePr   îî î   1RZKHDWWUDQVIHUFRHI¿FLHQW h  

Nu D .k 295.16 × 0.0312 =  :Pƒ& D 0.05

?+HDWWUDQVIHUUHGQ

 h Asts± t∞



îSîî/ î±



RUKHDWWUDQVIHUUHGSHUXQLWOHQJWK Q  îî î:$QV  L 4:DWHUDWƒ&ÀRZVWKURXJKDWXEHFPGLDPHWHUPORQJWXEHVXUIDFHEHLQJ PDLQWDLQHGDWƒ&7HPSHUDWXUHRIZDWHULQFUHDVHVIURPƒ&WRƒ&)LQGWKHPDVVÀRZUDWH 8VH'LWWXV%RHOWHUHTXDWLRQYL]1XD = 0.023 (ReD)0.8 (Pr)0.4. Take properties of water at mean bulk temperature of 40°C as :  U = 993kg/m3, cp = 4170 J/kg °C; k = 0.64 W/m°C; Q = 0.65 × 10– 6 m2s. 8378 6ROXWLRQGiven:ti ƒ&t ƒ&tb   

60 + 20 = 40°C ; ts ƒ& 2

D FP PL P

0DVVÀRZUDWHPNJV  μ cp

(v × ρ) × c p

(0.65 × 10 –6 × 993) × 4170 = 4.206  k k 0.64 +HDWJDLQHGE\ZDWHU +HDWWUDQVIHUUHGEHWZHHQWKHSLSHVXUIDFHDQGEXONRIZDWHU i.e. mcpt±ti  hAsLMTD Pr 

=

=

RU

0.8 ⎡k ⎤ ⎛ m× D⎞ . Pr 0.4 . As .LMTD ⎥  mcpt±tL   ⎢ × 0.023 ⎜ ⎟ ⎝ Ac × μ ⎠ ⎢⎣ D ⎥⎦

 

⎡ ⎢∵ h = ⎢ ⎢ = ⎢ ⎣



k VD {m / ( Ac × ρ)} × D ⎤ × Nu D ; ReD = = ⎥ D v v ⎥ m×D m×D ⎥ = ⎥ Ac × vρ Ac × μ ⎦

π π 2 × D 2 = × (0.04) = 1.257 × 10 –3 m 2 4 4 DQGVXUIDFHDUHDRIKHDWWUDQVIHUAs SD.L Sîî P /RJULWKPLFPHDQWHPSHUDWXUHGLIIHUHQFH (t – t ) – (ts – t0 )   LMTD  s i ⎡ (t – t ) ⎤ ln ⎢ s i ⎥ ⎣ (ts – t0 ) ⎦ ZKHUHDUHDRIFURVVVHFWLRQAc 

 





(90 – 20) – (90 – 60) = 47.21°C ⎡ (90 – 20) ⎤ ln ⎢ ⎥ ⎣ (90 – 60) ⎦

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785

1

0.8 ⎡k ⎤ 0.2 ⎛ D ⎞ Pr 0.4 × As × LMTD ⎥ ⎢ × 0.023 ⎜ ⎟ ⎝ AC × μ ⎠ ⎢D ⎥ m  ⎢ ⎥ c t t ( – ) p 0 i ⎣ ⎦ 5

0.8 ⎡ 0.64 ⎤ 0.04 ⎧ ⎫ × 0.023 ⎨ (4.206)0.4 × 1.131 × 47.21⎥ ⎢ ⎬ –3 –6 ⎢ 0.04 ⎥ ⎩1.257 × 10 × 0.65 × 10 × 993 ⎭   ⎢ ⎥ 4170 (60 – 20) ⎣ ⎦ 5 ⎡ 0.368 × 5679.2 × 94.85 ⎤   ⎢ ⎥  NJV ⎣ 4170 × (60 – 20) ⎦ 7KXVPDVVÀRZUDWHm  NJV$QV 4$KRWSODWHFPKLJKDQGPZLGHDWƒ&LVH[SRVHGWRDPELHQWDLUDW 20°C. Calculate the following :  L  0D[LPXPYHORFLW\DWFPIURPWKHOHDGLQJHGJHRIWKHSODWH (ii) Boundary layer thickness at 12 cm from the leading edge of the plate;  LLL  /RFDOKHDWWUDQVIHUFRHI¿FLHQWDWFPIURPWKHOHDGLQJHGJHRIWKHSODWH  LY  $YHUDJHKHDWWUDQVIHUFRHI¿FLHQWRYHUWKHVXUIDFHRIWKHSODWH  Y  7RWDOPDVVÀRZWKURXJKWKHERXQGDU\ (vi) Total heat loss from the plate; (vii) Temperature rise of air. 378

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6ROXWLRQ5HIHUWR([DPSOH 4$YHUWLFDOVWHHOSODWHPîPLQVL]HDQGPPWKLFNDWDXQLIRUPWHPSHUDWXUHRIƒ&LVH[SRVHGWRDWPRVSKHULFDLUDWƒ&)LQGWKHDSSUR[LPDWHWLPHUHTXLUHGIRUWKH SODWHWRFRROWRƒ&LIWKHKHDWWUDQVIHUFRHI¿FLHQWLQQDWXUDOFRQYHFWLRQIRUWKHYHUWLFDOSODWHLV given by: h = 1.42 × ('t/L)1/4. For steel: U = 7800 kg/m3 and cp = 473 J/kg °C.

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6ROXWLRQGiven:L PB PT PP Pts ƒ& t∞  ƒ& t ƒ& h î'tL U NJPcp -NJ. 7KHDSSUR[LPDWHWLPHUHTXLUHGIRUWKHSODWHWRFRROWRƒ&W $WDQ\LQVWDQWOHWWKHWHPSHUDWXUHRIWKHSODWHEHt7KHQKHDWEDODQFHHTXDWLRQFDQEHZULWWHQDV Rate of ‘decrease’ RIHQWKDOS\RIWKHSODWH Rate of ‘instantaneous’ KHDWWUDQVIHUIURPWKH SODWHE\FRQYHFWLRQi.e.

dt  hA t± t∞     dτ ALVWDNHQVLQFHWKHSODWHORVHWKHKHDWRQboth the sides RIWKHSODWH 

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Chapter : 13

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 T

dθ  TîAîT dτ 1.785 θ1/4 × 2 Aθ 1.785 × (θ)5/4 × 0.32 dθ =   m . cp 3.744 × 473 dτ î±T 

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3.225 × 10 –4

 îV K

4Saturated steam at a temperature of 65°C condenses on a vertical surface at ƒ&'HWHUPLQHWKHWKLFNQHVVRIWKHFRQGHQVDWH¿OPDWORFDWLRQVPDQGPIURPWKHWRS$OVR FDOFXODWHWKHFRQGHQVDWHÀRZUDWHDQGORFDODQGDYHUDJHKHDWWUDQVIHUFRHI¿FLHQWVDWWKHVHORFDWLRQV Assumse vapour density is much less as compared to that of the condensate. Properties of water at the mean temperature are: Ul = 983.3 kg/m3; kl = 0.654 W/m K; Pl = 4.67 × 10– 4 kg/ms; cpl = 4.85 J/ kg°C; hfg = 2346 3 × 10 J/kg, tsat = 65°C; ts = 55°C; g = 9.81 m/s2; b = 1m (width assumed)              18

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Chapter : 13

788

4A hemispherical furnace of radius 1.0 m has a roof temperature of T1 = 800 K and emissivity H1 7KHÀDWFLUFXODUÀRRURIWKHIXUQDFHKDVDWHPSHUDWXUHRI72 = 600 K and emissivity H2 &DOFXODWHWKHQHWUDGLDQWKHDWH[FKDQJHEHWZHHQWKHURRIDQGÀRRU978 6ROXWLRQGiven:R =PT .H T2 .H  1HWUDGLDQWKHDWH[FKDQJHEHWZHHQWKHURRIDQGÀRRUQ )LJVKRZVWKHUDGLDWLRQQHWZRUNIRUWKHJLYHQSUREOHP +HPLVSKHULFDO IXUQDFH

7 . ε   5 P

 Eb

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Eb1 – Eb 2 R1 + R12 + R2

4πR 2  SR Sî P 2 A SR Sî P A 

$OVR6WHIDQ%ROW]PDQFRQVWDQWV î±:P. View factors: 6LQFHDOOWKHKHDWUDGLDWHGE\VXUIDFHLVLQWHUFHSWHGE\KHPLVSKHULFDOVXUIDFH ?

F 

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A2 .F21 3.142 × 1 = = 0.5 A1 6.283

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1 – ε1 1 – 0.8 =  P± ε1 . A1 0.8 × 6.283

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1 – ε2 1 – 0.5 =  P± ε 2 . A2 0.5 × 3.142

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R 

1 1 =  P± A1. F12 6.283 × 0.5

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 P±

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789

Eb1 – Eb 2 R1 + R12 + R2

σ T14 – σ T24 5.67 × 10 –8 (8004 – 6004 ) = 0.676 0.676 Q î:$QV 

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4

Eb1 – Eb 2 5.67 × 10 –8 (8004 – 6004 ) =  R12 0.318

î:$QV a What are the objectives of conduction analysis? b  ([SODLQ³*XDUGHGKRWSODWHPHWKRG´RIPHDVXULQJWKHUPDOFRQGXFWLYLW\'8

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a  &RQGXFWLRQDQDO\VLVLVFDUULHGRXWWRDFKLHYHWKHIROORZLQJobjectives  7RGHWHUPLQHWKHWHPSHUDWXUHGLVWULEXWLRQYDULDWLRQRIWHPSHUDWXUHZLWKWLPH DQGSRVLWLRQ   7RPDNHFRPSXWDWLRQIRUKHDWWUDQVIHUHWF  b  0HDVXUHPHQWRIWKHUPDOFRQGXFWLYLW\E\µ*XDUGHGKRWSODWHPHWKRG¶ *XDUGHGKRWSODWHPHWKRGLVDVWHDG\VWDWHDEVROXWHPHWKRGVXLWDEOHIRUi PDWHULDOVZKLFKFDQ EHODLGEHWZHHQWZRSDUDOOHOSODWHVii ORRVH¿OOPDWHULDOVZKLFKFDQEH¿OOHGEHWZHHQVXFKSODWHV ,WLVWKHmost dependable and reproducible method for the measurement of thermal conductivity of insulating materials. Out

Coolant in

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i  $RQHGLPHQVLRQDOKHDWÀRZWKURXJKWKHÀDWVSHFLPHQ ii  $QDUUDQJHPHQWIRUPDLQWDLQLQJWKHIDFHVRIWKHVSHFLPHQDWFRQVWDQWWHPSHUDWXUH iii  6RPHPHWHULQJPHWKRGWRPHDVXUHWKHKHDWÀX[WKURXJKDNQRZQDUHD )LJVKRZVWKHVFKHPDWLFDUUDQJHPHQWRIH[SHULPHQWDOVHWXSIRUWKHJXDUGHGKRWSODWH PHWKRG y 7KHPDLQKHDWHULVLQFRUSRUDWHGDWWKHFHQWUHRIWKHXQLWDQGLWLVPDLQWDLQHGDWD¿[HG  temperature E\WKHHOHFWULFDOHQHUJ\ZKLFKFDQEHPHWHUHG   y 7KHJXDUGHGKHDWHUZKLFKVXUURXQGVWKHPDLQKHDWHURQLWVHQGVLVVXSSOLHGDGHTXDWH HOHFWULFDOHQHUJ\WRPDLQWDLQLWVWHPSHUDWXUHVDPHDVWKDWRIWKHPDLQKHDWHU7KLV DUUDQJHPHQWHQVXUHVXQLGLUHFWLRQDOKHDWÀRZDQGHOLPLQDWHVWKHGLVWRUWLRQFDXVHGE\ HGJHORVVHV  ² 7KHPDLQDQGJXDUGHGKHDWHUVDUHPDGHXSRIPLFDVKHHWVLQZKLFK1LFKURPHZLUHLV ZRXQGFORVHO\VSDFHG,QYDULDEO\WKHVHKHDWHUVDUHVXUURXQGHGORQJLWXGLQDOO\E\FRSSHU VXUIDFHSODWHVVRWKDWWHPSHUDWXUHLVXQLIRUPO\GLVWULEXWHG y 7HVWVSHFLPHQVµ¶DQGµ¶DUHSODFHGRQERWKVLGHVRIWKHKHDWHURUFRSSHUVXUIDFH  SODWHVDVWKHFDVHPD\EH$WWKHKRWDQGFROGIDFHVRIWKHVSHFLPHQVWKHUPRFRXSOHV DUHDWWDFKHG 7KHIROORZLQJmeasurements DUHPDGH Q +HDWÀRZIURPWKHPDLQKHDWHUWKURXJKDWHVWVSHFLPHQLWZLOOEHKDOI    RIWKHWRWDOHOHFWULFDOHQHUJ\VXSSOLHGWRWKHPDLQKHDWHU  A $UHDRIKHDWÀRZ $UHDRIPDLQKHDWHUDUHDRIRQHKDOIRIDLUJDS    EHWZHHQLWDQGWKHJXDUGHGKHDWHU  L 7KLFNQHVVRIWKHWHVWVSHFLPHQDQG  th±tc  7HPSHUDWXUHGURSDFURVVWKHVSHFLPHQVXEVFULSWVhDQGcUHIHUWRKRW    DQGFROGIDFHVUHVSHFWLYHO\ 7KHWKHUPDOFRQGXFWLYLW\k LVWKHQIRXQGRXWDVIROORZV (t − hc ) ⎤ Q L dt ⎡ k= = kA h ∵ Q = − kA  ⎢ ⎥⎦ $ (Wh − Kc ) G[ / ⎣ +RZHYHUZKHQWKHVSHFLPHQVDUHRIGLIIHUHQWWKLFNQHVVWKHUHVSHFWLYHWHPSHUDWXUHVDWWKHKRW DQGFROGVXUIDFHVZRXOGEHGLIIHUHQWDQGLQWKDWFDVHKZLOOEHHYDOXDWHGE\WKHIROORZVUHODWLRQ ⎤ L1 L2 Q ⎡ k = +  ⎢ ⎥ A ⎣ (th1 − tc1 ) (th 2 − tc 2 ) ⎦ +HUHVXI¿FHVDQGUHIHUWRWKHXSSHUDQGORZHUVSHFLPHQVDQGQLVWKHWRWDOHQHUJ\VXSSOLHG WRWKHPDLQKHDWHU   4 a  /LVWWKHYDULRXVIDFWRUVZKLFKLQÀXHQFHWKHWKHUPDOFRQGXFWLYLW\RIVXEVWDQFH In what way, the conductivity is affected by the solid, liquid and gaseous phases of the substance? b  7KHÀRZRIKHDWRFFXUVDORQJWKHD[LVRIWKHVROLGZKLFKKDVWKHVKDSHRID truncated cone with circumferential surface insulated. The base is at 360°C and WKHDUHDRIWKHVHFWLRQDWGLVWDQFH[PHDVXUHGIURPWKHEDVHRIWKHFRQHLVJLYHQ E\$ ±[ P2ZKHUH[LVLQPHWUHV ,IWKHSODQHDW[ PLVPDLQWDLQHGDWƒ&DQGWKHWKHUPDOFRQGXFWLYLW\RIWKHVROLG material is 3W/m°C, determine : i  +HDWÀRZ ii  7HPSHUDWXUHDW[ PDQG iii  7HPSHUDWXUHJUDGLHQWDWWKHWZRIDFHVDQGDW[ P       8378

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791

6ROXWLRQa  7KHWKHUPDOFRQGXFWLYLW\RIDPDWHULDOLVLQÀXHQFHGE\WKHIROORZVfactors   +RPRJHQHRXVDQGQRQKRPRJHQHRXVFKDUDFWHURIWKHPDWHULDO   &U\VWDOOLQHDPRUSKRXVDQGSRURXVVWUXFWXUHRIWKHVXEVWDQFH   &KHPLFDOFRPSRVLWLRQRIWKHVXEVWDQFHRUVXEVWDQFHV RIZKLFKLWLVFRPSRVHG   *DVHRXVOLTXLGDQGVROLGSKDVHVLQZKLFKWKHVXEVWDQFHH[LVWV   7HPSHUDWXUHDQGSUHVVXUHWRZKLFKWKHVXEVWDQFHLVVXEMHFWHG 7KHIDFWRUVZLWKWKHJUHDWHVWLQÀXHQFHDUH  y &KHPLFDOFRPSRVLWLRQ  y 3KDVHFKDQJHDQG y 7HPSHUDWXUH  *HQHUDOO\DOLTXLGLVEHWWHUFRQGXFWRUWKDQDJDVDQGDVROLGLVDEHWWHUFRQGXFWRUWKDQDOLTXLG  b  Given t ƒ&t ƒ&A ±[ Pk :Pƒ& 5HIHUWR)LJ i  +HDWÀRZQ  7KHLQVXODWHGFLUFXPIHUHQWLDOVXUIDFHLPSOLHVWKDWWKHVLWXDWLRQFRUUHVSRQGVWRone-dimensional conduction LQWKH[ GLUHFWLRQIRUZKLFKZHKDYH dt dt   − N[1.2(1 − 1.5 [)]        4  − KA G[ G[ t1 = 360°C

Insulated surface t2 = 120°C

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Q

x = 0.2 m x = 0.1 m x=0



Fig. 13.3

6HSDUDWLQJWKHYDULDEOHVDQGXSRQLQWHJUDWLRQZHJHW G[ Q = − 1.2 k dt (1 − 1.5 [)

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⎡ 1 ⎤2 4 ⎢− OQ (1 − 1.5 [) ⎥ = − 1.2 N (W2 − W1 ) ⎣ 1.5 ⎦ [1 Q=−

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=−



1.2 × 1.5 k (t1 − t2 ) OQ {1 − 1.5 ( [2 − [1 )} 1.2 × 1.5 × 3 (360 − 120) = . : $QV ln [1 − 1.5 (0.2 − 0)]

ii  7HPSHUDWXUHDWx P 3633.6 = − ?

1.2 × 1.5 × 3 (360 − t ) 5.4 (360 − t ) = ln {1 − 1.5 (0.1 − 0)} 0.1625

t = 360 −

3633.6 × 0.1625 = ƒ& ( $QV) 5.4

792

Chapter : 13

iii  7HPSHUDWXUHJUDGLHQWDWWKHWZRIDFHVDQGDW[ P 7KHWHPSHUDWXUHJUDGLHQWDWDVHFWLRQPD\EHZULWWHQDV dt Q =−  G[ N$ y $W[ A ±[  P dt 3633.6 =− = ± ƒ&  P ( $QV) ? G[ 3 × 1.2 y $W[ PA ±î  P dt 3633.6 =− =  ƒ&  P ( $QV) ? G[ 3 × 1.02 y $W[ P A ±î  P ?

dt 3633.6 =– = .ƒ&  P $QV G[ 3 × 0.84

  4a How does the heat conduction in insulators and metals takes place? b A 3.6 cm diameter pipe at 105°C is losing heat at the rate of 120 W/m length of pipe to the surrounding air at 15°C. This is to be reduced to a minimum value by providing insulation. The following insulation materials are available : Insulation A : Quantity = 0.00372 m3/m length of pipe; Thermal conductivity = 5 W/m°C Insulation B : Quantity = 0.0048 m3/m length of pipe; Thermal conductivity = 1 W/m°C Determine the position of better insulating layer relative to the pipe and percentage saving in heat loss. 18 6ROXWLRQ a  7KHPHFKDQLVPRIKHDWFRQGXFWLRQLQLQVXODWRUVDQGPHWDOVLVDVIROORZV  y 7KHFRQGXFWLRQRIKHDWLQLQVXODWRUVe.g.ZRRGJODVVDVEHVWRV WDNHVSODFHdue to vibration of atoms about their mean positions:KHQKHDWLVLPSDUWHGWRRQHSDUW RIDQLQVXODWLQJVXEVWDQFHDWRPVEHORQJLQJWRWKDWSDUWDUHSXWLQDYLROHQWVWDWHRI DJLWDWLRQDQGVWDUWYLEUDWLQJZLWKJUHDWHUDPSOLWXGHV&RQVHTXHQWO\WKHVHPRUHDFWLYH SDUWLFOHVFROOLGHZLWKOHVVDFWLYHDWRPVO\LQJQH[WWRWKHPUHVXOWLQJLQOHVVDFWLYH DWRPVJHWWLQJH[FLWHG7KHSURFHVVLVUHSHDWHGOD\HUDIWHUOD\HURIPROHFXOHVDWRPV XQWLOWKHRWKHUSDUWRIWKHLQVXODWRULVUHDFKHG ,QVXODWRUVKDYHDlowYDOXHRIWKHUPDOFRQGXFWLYLW\GXHWRWKHLUSRURVLW\ZKLFKPD\  FRQWDLQDLU  y ,QPHWDOV besides atomic vibrationsDlarge number of free electrons also participate in the heat conduction process. :KHQDGLIIHUHQFHRIWHPSHUDWXUHH[LVWVEHWZHHQWKH GLIIHUHQWSDUWVRIWKHPHWDOWKHVHIUHHHOHFWURQVGULIWLQWKHGLUHFWLRQRIGHFUHDVLQJ WHPSHUDWXUH,WLVGXHWRWKLVdrift of free electrons WKDWPHWDOVEHKDYHEHWWHUFRQGXFWRUV WKDQRWKHUVROLGV7KHREVHUYHGSURSRUWLRQDOLW\EHWZHHQWKHWKHUPDODQGHOHFWULFDO FRQGXFWLYLWLHVRISXUHPHWDOVLVRQDFFRXQWRIWKHVHIUHHHOHFWURQV ±7KHHOHFWURQVGRQRWFRQWULEXWHWRKHDWFRQGXFWLYLW\LQWKHLQVXODWRUVLQFHWKHVHDUHnot free EXW¿[HGLQWKHYDOHQFHEDQG  b  7KHUPDOUHVLVWDQFHGXHWRSLSHPDWHULDO Δt 105 − 15 Rth − p = = = 0.75°C /W  Q 120 )RUDSLSHZLWKWZROD\HUVRILQVXODWLRQ Σ Rth = Rth − p + Rth − 1 + Rth − 2   ⎛r ⎞ ⎛r ⎞ 1 1 = 0.75 + ln ⎜ 2 ⎟ + ln ⎜ 3 ⎟ 2π k1l ⎝ r1 ⎠ 2 π k2l ⎝ r2 ⎠

Universities’ Exams. Questions (Latest)...

793

$UUDQJHPHQW,7KHinsulation material $LVSODFHGLQVLGHLHQH[WWRWKHSLSH r P 2 2 π r − r × 1 = 0.00372  ?r P 1RZ 2 1 $OVR ?

( π (r

2 3

) − r ) × 1 = 0.0048  2 2

?r P 1 1 ⎛ 0.0388 ⎞ ⎛ 0.055 ⎞ Σ Rth = 0.75 + ln ⎜ + ln ⎜ ⎟ ⎟ 2 π × 5 × 1 ⎝ 0.018 ⎠ 2π × 1 × 1 ⎝ 0.0388 ⎠



  ƒ&: Δt 105 − 15 = = 108.45 W ? +HDWORVV Q = Σ Rth 0.8299 $UUDQJHPHQW,,7KHinsulation material B is placed inside, i.e.,QH[WWRWKHSLSH r P π (r22 − 0.0182 ) = 0.0048  ?r P 1RZ $OVR ? 

π (r32 − 0.0432 ) = 0.00372  ?r P 1 1 ⎛ 0.043 ⎞ Σ Rth = 0.75 + ln ⎜ + ln ⎟ 2π × 1 × 1 ⎝ 0.018 ⎠ 2π × 5 × 1

⎛ 0.055 ⎞ ⎜⎝ ⎟ 0.043 ⎠

  ƒ&: 105 − 15 = 100.4 W ? +HDWORVV Q = 0.8964 2EYLRXVO\WKHKHDWORVVLVVPDOOZKHQLQVXODWLRQPDWHULDO%LVSODFHGQH[WWRWKHSLSH 108.45 − 100 × 100 =  ( $QV.) DJHVDYLQJLQKHDWORVV  100   4a List some of the situations where poor conductivity of air helps to restrict the heat transmission by conduction. b  $KHPLVSKHULFDORYHQRIFPLQWHUQDOUDGLXVLVODJJHGZLWKFPWKLFN¿UH bricks covering surrounded by a magnesia layer of thickness 6 cm. An electric heater is placed at the centre and under steady state conditions, the inner surface is maintained at 815°C. The system is a room for which the ambient WHPSHUDWXUHLVƒ&DQGRXWVLGHXQLWFRQYHFWLYHFRHI¿FLHQWLV:P2°C. 7KHUPDO FRQGXFWLYLWLHV RI ¿UH EULFNV DQG PDJQHVLD DUH  :P ƒ& DQG 0.0525 W/°C respectively. Neglecting any thermal resistance due to the oven material, determine : i  7KHKHDWORVVWKURXJKWKHRYHQDQGWKHZDWWDJHUHTXLUHGRIWKHKHDWHU¿ODPHQW to be placed inside to affect the same heat transfer. ii  7KHWHPSHUDWXUHDWDSRLQWKDOIZD\WKURXJKWKH¿UHEULFNFRYHULQJ08

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Chapter : 13

794

b  Given r FPRUPr  FPRUPr  FPRU P   't   ±   ƒ& h  :P ƒ& kfb  :Pƒ& kPDJ   :Pƒ& i  7KHKHDWORVVIURPWKHRYHQDQGWKHZDWWDJHUHTXLUHG 7KHUPDOUHVLVWDQFHIRUDVSKHULFDOERG\LVJLYHQE\ r − ri Rth = 0  4π k r0 ri DQGIRUDKHPLVSKHUHLWHTXDOVKDOIRIWKLVYDOXH 1 (0.84 − 0.72) = 0.02506°C/W ? 5HVLVWDQFHRI¿UHEULFN Rth − fb = × 2 4π × 0.315 × 0.84 × 0.72 

5HVLVWDQFHRIPDJQHVLD Rth − mg =



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=



1 (0.9 − 0.84) × = 0.06015°C/W 2 4 π × 0.0525 × 0.9 × 0.84 1 1 × 2 h0 A0

1 1 × = 0.005614°C/W 2 8.75 × 4π × (0.9) 2

7RWDOWKHUPDOUHVLVWDQFH Σ Rth = 0.02506 + 0.06015 + 0.005614 = 0.0908°C/W Δt 800 = = . : ( $QV.) Σ Rth 0.0908 i.e.5HTXLUHG¿ODPHQWZDWWDJH N:$QV ii  7KHWHPSHUDWXUHDWDSRLQWKDOIZD\WKURXJKWKH¿UHEULFNFRYHULQJt  5DGLXVDWPLGSODQHRIWKH¿UHEULFNOLQLQJ r + r2 0.72 + 0.84 r= 1 = = 0.78 m  2 2 7KHUPDOUHVLVWDQFHRI¿UHEULFNVXSWRLWVPLGSODQH 1 (0.78 − 0.72) = × = 0.0135°C/W  2 4π × 0.315 × 0.78 × 0.72 6LQFHWKHKHDWÀRZLQJWKURXJKHDFKVHFWLRQLVVDPHWKHUHIRUH ?+HDWORVWIURPRYHQ =

815 − t 0.0135 t ±î ƒ&$QV

8810.6 =

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A 7.8 m diameter vertical kiln has a hemispherical dome at the top; the dome is fabricated from a 30 cm thick layer of chrome brick which has a thermal conductivity of 1.16 W/m°C. The kiln dome has inside temperature of 880°C, and 25°C atmospheric air results into 11.4 W/m2 ƒ&KHDWWUDQVIHUFRHI¿FLHQW between the dome and air. Determine : i The outside surface temperature of the dome and the heat loss from the kiln. ii  7KHSHUFHQWDJHUHGXFWLRQLQKHDWORVVWKDWZRXOGUHVXOWE\XVLQJDÀDWGRPH fabricated from the same material and with kiln operating under identical temperature conditions. 378

  4

6ROXWLRQGiven : r1 = t ƒ&ta ƒ&

7.8 = 3.9 m; r2 = 3.9 + 0.3 = 4.2 m; k = 1.16 W/m°C; h :Pƒ& 2

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+HDWORVVIURPWKHGRPH ±  N:$QV ii  3HUFHQWDJHUHGXFWLRQLQKHDWORVWE\XVLQJÀDWGRPH )RUDGRPHZLWKÀDWWRS kA (t1 − t2 ) = hA (t2 − ta )  L 7KHDUHDA IRUFRQGXFWLRQDQGFRQYHFWLRQKHDWÀRZZLOOEHVDPH 1.16 (880 − t2 ) = 11.4 (t2 − 25) ? 0.3 (880 − t2 ) = 2.948 t2 − 73.7 RU ?

t2 =

? +HDWORVVIURPWKHGRPH =

880 + 73.7 = 241.56°C 2.948 + 1 kA (t1 − t2 ) L

1.16 × (π × 3.92 ) × (880 − 241.56) 0.3  :RUN: 258.747 − 117.96 × 100 =  ( $QV.) DJHUHGXFWLRQLQKHDWORVV = 258.747



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  4 The thermal conductivity of an insulating material used over a 200 mm diameter SLSHFDUU\LQJDKRWÀXLGYDULHVDVN W ZKHUHµW¶LVLQµƒ&¶ and ‘k’ is in W/m°C. The temperatures at the pipe surface and at the outside of insulation are 250°C and 60°C respectively. If the thickness of the insulation is 60 mm, determine : i  7KHKHDWÀRZIURPWKHKRWÀXLGDQGWHPSHUDWXUHDWPLGWKLFNQHVVRILQVXODWLRQ ii  7KH VORSHV RI WHPSHUDWXUH SUR¿OH DW LQVLGH VXUIDFH PLG SODQH DQG RXWVLGH surface. '8

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Chapter : 13

200 = 100 mm or 0.1 m; r  Pk W  2 t ƒ&t ƒ& i  7KHKHDWÀRZDQGWKHWHPSHUDWXUHDWWKHPLGWKLFNQHVV $Vthe dependence of thermal conductivity on temperature is linear, the mean thermal conductivity corresponds to thermal conductivity taken at mean wall temperature. ⎡ ⎛ 250 + 60 ⎞ ⎤ km = 0.065 ⎢1 + 0.0015 ⎜  ⎟⎠ ⎥ = 0.08 W/m°C ⎝ 2 ⎣ ⎦ 6ROXWLRQGiven : r1 =

7KHKHDWÀRZWKURXJKDF\OLQGULFDOVXUIDFHZLWKUDGLLrDQGrLVJLYHQE\ (t1 − t2 ) Q=   ⎡ ln (r2 / r1 ) ⎤ ⎢ 2 π kL ⎥ ⎣ ⎦ =



>(TQ @

(250 − 60) =  :  P OHQJWK ( $QV) ⎡ ln (0.16 / 0.1) ⎤ ⎢ 2 π × 0.08 × 1⎥ ⎣ ⎦

/HWµt¶EHWKHWHPSHUDWXUHDWPLGSODQHrm  P ⎡ ⎛ 250 + t ⎞ ⎤ = 0.065 [1 + 0.00075 (250 + t ) ] km = 0.065 ⎢1 + 0.0015 ⎜  ⎝ 2 ⎟⎠ ⎥⎦ ⎣ 

Q=



=

(250 − t ) ⎡ ⎤ ln (0.13 / 0.1) ⎢ 2 π × 0.065 {1 + 0.00075 (250 + t )} × 1⎥ ⎣ ⎦ (250 − t ) [ 2 π × 0.065 {1 + 0.00075 (250 + t )} × 1] ln (0.13 / 0.1)

)RUVWHDG\VWDWHKHDWFRQGXFWLRQKHDWSDVVLQJWKURXJKHDFKVHFWLRQRIWKHSLSHLVVDPH (250 − t ) [2 π × 0.065 {1 + 0.00075 (250 + t ) × 1}] 203.2 = ? ln (0.13 / 0.1) RU   RU RU

(250 − t ) [0.4084 (1 + 0.1875 + 0.00075 t )] 0.2624 53.32 = (250 − t ) (0.485 + 0.0003063 t )

203.2 =

 t±t±t tt±  tt± 

(1333.33) 2 + 4 × 221776 2 − 1333.33 ± 1632.44 = = .°& ( $QV.)  2 ii  7KH6ORSHVRIWKHWHPSHUDWXUHSUR¿OHV 7KHVORSHVRIWKHWHPSHUDWXUHSUR¿OHVDUHFDOFXODWHGIURPWKHIROORZLQJUHODWLRQ dt Q =− (Fourier law)  G[ N$ RU

t=

− 1333.33 ±

Universities’ Exams. Questions (Latest)...

y Inside surface A Srl Sîî ?

y Mid-plane  ?

797

P

k î  :Pƒ& dt 203.2 =− = ± ƒ&  P ( $QV.) G[ 0.0894 × 0.628 A Srml Sîî P k î  :Pƒ& dt 203.2 =− = ± ƒ&  P ( $QV.) G[ 0.0796 × 0.8168

y Outside surface  A = 2π r2l = 2π × 0.16 × 1 = 1.005 m 2 ?

k î  :Pƒ& dt 203.2 =− = ± ƒ&  P ( $QV.) G[ 0.07085 × 1.005

  4 $QKHDWH[FKDQJHURIRXWVLGHUDGLXVPPLVWREHLQVXODWHGZLWKJODVVZRRORI thermal conductivity 0.0825 W/m°C. The temperature at the surface of the shell is 280°C and it can be assumed to remain constant after the layer of insulation has been DSSOLHGWRWKHVKHOO7KHFRQYHFWLYH¿OPFRHI¿FLHQWEHWZHHQWKHRXWVLGHVXUIDFHRIVODJ wool and the surrounding air is estimated to be 8W/m2ƒ&)XUWKHULWLVVSHFL¿HGWKDW WKHWHPSHUDWXUHDWWKHRXWHUVXUIDFHRILQVXODWLRQPXVWQRWH[FHHGƒ&DQGWKHORVV of heat per metre length of the shell should not be greater than 200 W. Would the slag wool serve the intended purpose to restrict the heat loss? If yes, what should be the thickness of insulating material to suit the prescribed conditions. 978 6ROXWLRQ Given : r PP Pk :Pƒ&ti ƒ&to ƒ&ho :P ƒ&Q : 7KHFULWLFDOUDGLXVRILQVXODWLRQLVJLYHQE\ k >(TQ@ rc =   ho 0.0825 = = 0.01031 m or 10.31 mm  8 6LQFHWKHFULWLFDOUDGLXVrc PP LVFRQVLGHUDEO\VPDOOHUWKDQWKHUDGLXVr PPRI WKHVKHOOWKHUHIRUHWKHuse of slag wool as insulating material would serve the intended purpose to restrict the heat loss.$QV /RVVRIKHDWIURPDcylindrical shell LVJLYHQE\ 2 π L (ti − to ) Q=  >(TQ @  ln (ro / ri ) 1 + ho ro k 2 π (ti − to ) Q = RU ln (ro / ri ) 1 L + ho ro k ,QVHUWLQJWKHDSSURSULDWHYDOXHVZHJHW 2π (280 − 30) 200 =  ln (r0 / 0.15) 1 + 8ro 0.0825 1570.8 200 = RU 0.125 + 12.12 ln (ro / 0.15) r0

798

Chapter : 13

0.125 + 12.12 ln (ro / 0.15) = 7.854 r0 %\WULDODQGHUURUZHJHW ro PRUPP ? Thickness of insulation ± PP$QV   4 (Critical thickness of insulation) An electric cable of 6 mm radius is applied an XQLIRUPVKHDWKLQJRISODVWLFLQVXODWLRQN :Pƒ& 7KHFRQYHFWLYH¿OP FRHI¿FLHQWRQWKHVXUIDFHRIEDUHFDEOHDVZHOODVLQVXODWHGFDEOHZDVHVWLPDWHG as 11.7 W/m2°C and a surface temperature of 60°C was noted when the cable was GLUHFWO\H[SRVHGWRDPELHQWDLUDWƒ&'HWHUPLQH i The thickness of insulation, for keeping the wire as cool as possible. ii The surface temperature of insulated cable if the intensity of current carried by the conductor remains unchanged            378 6ROXWLRQ Given: r PP Pk :Pƒ&ho :Pƒ&t ƒ& ta ƒ& i  7KLFNQHVVRILQVXODWLRQ )RUNHHSLQJWKHZLUHas cool as possible WKHUHTXLUHGFRQGLWLRQFRUUHVSRQGVWRWKDWIRUFULWLFDO UDGLXVRILQVXODWLRQ i.e. k 0.187 r2 = rc = = = 0.016 m = 16 mm ho 11.7 RU

? 7KLFNQHVVRILQVXODWLRQ r±r ± PP$QV ii  7KHVXUIDFHWHPSHUDWXUHt  )RUDEDVHXQLQVXODWHG ZLUHWKHKHDWÀRZ Q hAt±ta  îSîî ±  :POHQJWK )RUWKHVKHDWKHGLQVXODWHG FDEOH 2π (t2 − 20) Q2 = W/m length  ln (r2 / r1 ) 1 + h0 r2 k 

=



=

2π (t2 − 20) W/m length ln (0.016 / 0.006) 1 + 11.7 × 0.016 0.187 6.283 (t0 − 20) = 0.593 (t2 − 20) W/m length 5.342 + 5.245

,IWKHLQWHQVLW\RIFXUUHQWFDUULHGE\WKHFRQGXFWRUUHPDLQVXQFKDQJHGWKHQ Q Q IR RU  t± ? t ƒ& $QV 4a Critical thickness of insulation Derive the following relation for heat dissipation from an electric cable which has been provided with critical insulation: 2π k L (ti − to ) Q= 1 + ln (k / ho ri ) The symbols have their usual meanings and subscripts ‘i’ and ‘o’ refers to the conditions at the inner and outer surfaces respectively. b An electric cable of 6 mm outer radius has a sheathing of rubber insulation for ZKLFKWKHUPDOFRQGXFWLYLW\LV:Pƒ&:KHQFXUUHQWÀRZVWKURXJKWKH

Universities’ Exams. Questions (Latest)...   

799

 cable, heat is generated and surface temperature of 75°C is anticipated for the cable. This cable is laid in an environment having a temperature 15°C and the WRWDOFRHI¿FLHQWDVVRFLDWHGZLWKFRQYHFWLRQDQGUDGLDWLRQEHWZHHQWKHFDEOH and environment is 8 W/m2 °C. Determine : i The most economical thickness and corresponding increase in heat dissipation due to insulation. ii The increase in current carrying capacity of the cable by providing critical thickness of insulation.                    18

6ROXWLRQ a 7KHKHDWWUDQVPLWWHGIURPWKHFDEOHWRWKHHQYLURQPHQWLVJLYHQE\ 2 π L (to − ti ) Q=  ln (ro / ri ) 1 + ho ro k :KHQFULWLFDOWKLFNQHVVDWKLFNQHVVZKLFKJLYHVWKHPD[LPXPKHDWGLVVLSDWLRQIURPWKHFDEOH k VXUIDFH LVSURYLGHG rc = ro = ho 7KHQ

   

Q=

=

2 π L (to − ti ) ln (k / ho ri ) 1 + ho × (k / ho ) k 2 π k L (to − ti )  1 + ln (k / ho ri )

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b  Given : ri PP Pk :Pƒ&ti ƒ&ta ƒ&ho :Pƒ& i  7KHPRVWHFRQRPLFDOWKLFNQHVVDQGLQFUHDVHLQKHDWGLVVLSDWLRQ 6XEVWLWXWLQJWKHYDULRXVYDOXHVLQWKHDERYHHTQZHJHW 2π × 0.16 (75 − 15) ⎛ Q⎞ = ⎜⎝ ⎟⎠ L insulated 1 + ln {0.16 / (8 × 0.006)}



:POHQJWK )RUDEDUHFDEOHWKHKHDWGLVVLSDWLRQZRXOGEH ⎛ Q⎞ = 2πri h (ti – to )  ⎜⎝ ⎟⎠ L bare $VVXPLQJWKDWWKHVXUIDFHWHPSHUDWXUHRIWKHFDEOHDQGWKHRXWVLGHFRQQHFWLYHFRHI¿FLHQWLQWKH EDUHFRQGLWLRQDUHVDPHDVWKRVHLQLQVXODWHGFRQGLWLRQZHKDYH ⎛ Q⎞ = 2π × 0.006 × 8 (75 − 15)  ⎜⎝ ⎟⎠ L bare  :POHQJWK DJHincreaseLQKHDWGLVVLSDWLRQ 27.37 − 18.09 = × 100 =  ( $QV)  18.09 7KLVLQFUHDVHLQKHDWGLVVLSDWLRQZRXOGRFFXUZKHQFULWLFDOLQVXODWLRQKDVEHHQDSSOLHGWRWKHFDEOH k 0.16 = = 0.02 m = 20 mm  &ULWLFDOUDGLXV rc = ho 8 ? 7KLFNQHVVRILQVXODWLRQ ± PP$QV ii  3HUFHQWDJHLQFUHDVHLQFXUUHQWFDUU\LQJFDSDFLW\ Q IR ZKHUH I &XUUHQWFDSDFLW\DQG

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 1RZ

2 Q  I 2 R ,QVXODWHGFDEOH

I2 Q2 27.37 = = = 1.23 I1 Q1 18.09 ? DJHincreaseLQFXUUHQWFDUU\LQJFDSDFLW\ 1.23 I1 − I1 = × 100 =  ( $QV.)  I1

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a Critical insulation Derive the following relation which will keep the insulated wire at the same temperature as if it were bare uninsulated : ⎛r ⎞ r⎞ k ⎛ ln ⎜ o ⎟ = 1− i⎟ r ⎠ ⎝ r ⎠ h r ⎜⎝ i

o

i

o

The symbols have their usual meanings and subscripts i and o refer to the conditions at the inside and outside surfaces respectively. b An electric cable of 7.5 mm radius is applied a uniform sheathing 2.5 mm thick of plastic insulation k = 0.15 W/m°C 7KHFRQYHFWLYH¿OPFRHI¿FLHQWEHWZHHQ the surface of plastic and surrounding air is estimated as 10 W/m2 °C. Does the insulation serve to augment heat loss and thus help in cooling of wire? $OVR¿QGRXWWKHRXWHUUDGLXVRILQVXODWLRQZKLFKZLOONHHSWKHLQVXODWLRQDW the same temperature as if it were bare.              '8 6ROXWLRQ a )RUDEDUHXQLQVXODWHG ZLUHWKHKHDWÀRZ 2 π L (ti − to ) Q1 = 2π ri Lho (ti − to ) =  1 ho ri )RUWKHLQVXODWHGZLUH 

Q2 =

2 π L (ti − to ) ln (ro / ri ) 1 + ho ro k

7KHFRQGLWLRQWRNHHSWKHLQVXODWHGZLUHDWWKHVDPHWHPSHUDWXUHDVLILWZHUHEDUHLQKHUHQWO\ VWLSXODWHVWKDWKHDWÀRZIURPWKHZLUHLVsameLQERWKWKHVLWXDWLRQV7KXV Q Q ln (ro / ri ) 1 1 = + RU ho ri ho ro k RU

⎛r ⎞ ⎛ 1 1 ⎞ ln ⎜ o ⎟ = k ⎜ − ho ro ⎟⎠ ⎝ ri ⎠ ⎝ ho ri

RU

r⎞ k ⎛ ⎛r ⎞ ln ⎜ ro ⎟ = l − ri ⎟  ⎜ ⎝ i ⎠ ho ri ⎝ o⎠

b  Given ri PP Pro  PPRUP k :Pƒ&h :Pƒ& 7KHFULWLFDOUDGLXVRILQVXODWLRQ k 0.15 rc = = = 0.015 m = 15 mm  ho 10

3URYHG

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801

6LQFHrcLVgreater than the outer radius roi.e.PP RIWKHLQVXODWHGZLUHWKHsheathinghelps to dissipate more heatDQG WKXVhelps in the cooling of the wire. $QV 6XEVWLWXWLQJDSSURSULDWHYDOXHVLQWKHDERYHUHODWLRQZHJHW r⎞ r⎞ 0.15 ⎛r ⎞ ⎛ ⎛ ln ⎜ o ⎟ = 1 − i ⎟ = 2 ⎜1 − i ⎟  ro ⎠ ro ⎠ ⎝ ri ⎠ 10 × 0.0075 ⎜⎝ ⎝ ro =5 ri

%\WULDODQGHUURU ?

2XWHUUDGLXVro îri î P PP$QV

  4 Fins Determine the energy input required to solder together two very long pieces of bare copper wire 1.7 mm in diameter with a solder that melts at 190°C. The ZLUHVDUHSRVLWLRQHGYHUWLFDOO\LQDLUDWƒ&DQGWKHKHDWWUDQVIHUFRHI¿FLHQWRQ the wire surface is 16 W/m2 °C. For the wire alloy, take the thermal conductivity as 330 W/m°C.                    0'8+DU\DQD 6ROXWLRQ Given d PPRUPto ƒ&ta ƒ&h :Pƒ&k :Pƒ& (QHUJ\LQSXWUHTXLUHGWRVROGHUWZRORQJSLHFHV 7KHSK\VLFDOVLWXDWLRQDSSUR[LPDWHVDVWZRLQ¿QLWH¿QVZLWKDEDUHWHPSHUDWXUHRIƒ&LQDQ HQYLURQPHQWRIƒ&ZLWKWKHJLYHQYDOXHRIVXUIDFHFRHI¿FLHQW Copper wire

Solder at 190°C

∞

∞ [=0 Fig. 13.4

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π 2 π d = × (0.0017) 2 = 2.27 × 10 −6 m 2 4 4 P Sd Sî P



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hP = k Acs

16 × 0.00534 330 × 2.27 × 10 −6

= 10.68

+HDWGLVVLSDWLRQIURPDQLQ¿QLWHO\ORQJ¿Q Q fin = k Acs m (to − ta )    îî±î±  : ?(QHUJ\LQSXWUHTXLUHGIRUWZRZLUHV î :$QV   4

(Fins) The Fig. 13.5 shows a 60 mm diameter rod, 1m long, which is having its ORZHUIDFHJURXQGVPRRWK7KHUHPDLQGHURIWKHURGLVH[SRVHGWRWKHƒ&URRP DLUDQGDVXUIDFHFRHI¿FLHQWKHDWWUDQVIHUHTXDOWR:P2ƒ&H[LVWVEHWZHHQ the rod surface and the room air. The grinder dissipates mechanical energy at WKHUDWHRI:,IWKHUPDOFRQGXFWLYLW\RIURGPDWHULDOLV:Pƒ&¿QGWKH temperature of the rod at the point where grinding is taking place. 8378

6ROXWLRQ Given d    PP   P l  P ta  ƒ& h   :P ƒ& Q¿Q   : k :Pƒ& 7HPSHUDWXUHRIWKHURGDWWKHSRLQWRIJULQGLQJto  ,QWKLVFDVHWKHHQWLUHGLVVLSDWHGHQHUJ\JRHVLQWRWKHURGDWLWVORZHUIDFHDQGLVWKHQWUDQVIHUUHG E\convectionIURPWKHRXWVLGHVXUIDFHDUHDRIWKHURGWRWKHVXUURXQGLQJDLU

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Chapter : 13

7UHDWLQJWKHURGDVD¿QORVLQJKHDWDWWKHWLSWKHKHDWÀRZWKURXJK WKHURGLVJLYHQE\ h ⎤ ⎡ ⎢ tanh (ml ) + km ⎥ (TQ Q fin = Ph k Acs (to − ta ) ⎢ ⎥ h ⎢1 + . tanh (ml ) ⎥ km ⎣ ⎦ +HUH P Sd Sî P π Acs = × (0.06) 2 = 0.00283 m 2  4 hP 7.1 × 0.1885 = = 3.34 m −1 k Acs 42.4 × 0.00283 ml î  h 7.1 = = 0.05  km 42.4 × 3.34 ,QVHUWLQJWKHDSSURSULDWHYDOXHVLQWKHDERYHHTQZHJHW ⎡ tanh (3.34) + 0.05 ⎤ 40 = 0.1885 × 7.1 × 42.4 × 0.00283 (to − 25) ⎢ ⎣1 + 0.05 tanh (3.34) ⎥⎦

ZKHUH

 RU

m=

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⎡ 0.9975 + 0.05 ⎤ = 0.4 (to − 25) ⎢  to± ⎣1 + 0.05 × 0.9975 ⎥⎦ to ƒ&$QV

  4 A cylinder 60 mm in diameter and 1.2 m long is provided with 10 longitudinal VWUDLJKW¿QVRIPDWHULDOKDYLQJWKHUPDOFRQGXFWLYLW\:Pƒ&7KH¿QVDUH mm thick and protrude 15 mm from the cylinder surface. The system is placed LQDWPRVSKHUHDWƒ&DQGWKHKHDWWUDQVIHUFRHI¿FLHQWIURPWKHF\OLQGHUDQG ¿QVWRWKHDPELHQWDLULV:P2 °C. If the surface temperature of the cylinder is 140°C. Determine : i The rate of heat transfer. ii  7KHWHPSHUDWXUHDWWKHHQGRI¿QV   &RQVLGHUWKH¿QWREHRI¿QLWHOHQJWK            978 6ROXWLRQ Given d PP Pb Pn k :Pƒ&y PP  Pl PP Ph :Pƒ&to ƒ&ta ƒ& )RUD¿QRIUHFWDQJXODUFURVVVHFWLRQ Acs bîy î P P by    P m=



hP = k Acs

18 × 2.402 = 18.66 m −1 115 × 0.00108

ml î  h 18 = = 0.00839 km 115 × 18.66

  i  7KHUDWHRIKHDWWUDQVIHU 

Q fin =

Ph k Acs

h ⎤ ⎡ ⎢ tanh (ml ) + km ⎥ (to − ta ) ⎢ ⎥ h ⎢1 + × tanh (ml ) ⎥ km ⎣ ⎦

Universities’ Exams. Questions (Latest)...  

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803

⎡ tanh (0.28) + 0.00839 ⎤ 2.402 × 18 × 115 × 0.00108 (140 − 30) ⎢ ⎣1 + 0.00839 × tanh (0.28) ⎥⎦ ⎡ 0.273 + 0.00839 ⎤ = 2.32 × 110 × ⎢ = 71.65 W per fin ⎣1 + 0.00839 × 0.273 ⎥⎦

+HDWWUDQVIHUIURPXQ¿QQHGEDVH VXUIDFH = h [ π d × 1.2 – n × Acs ] × (to − ta )  = 18 (π × 0.06 × 1.2 − 10 × 0.00108) × (140 − 30) = 426.5 W Q totalIRU¿QV î :$QV ii 7KH7HPSHUDWXUHDWWKHHQGRI¿QV )RUD¿QGLVVLSDWLQJKHDWWRWKHVXUURXQGLQJVIURPLWVWLSHQG h cosh {P (O − [)} + [sinh {P (O − [)}]  >(TQ @ t − ta θ km = =  θ o to − t a h cosh (ml ) + [sinh (ml )] km $W[ lWKHDERYHHTQUHGXFHVWR tl − ta 1 =  to − t a h cosh (ml ) + [sinh (ml )] km tl − 30 1 = = 0.9598 RU 140 − 30 cosh (0.28) + 0.00839 sinh (0.28) ? tl ± î ƒ&$QV   4 The cylindrical head of an engine is 1.2 m long and has an outside diameter of 60 mm. Under typical operating conditions, the outer surface of the head is at a WHPSHUDWXUHRIƒ&DQGLVH[SRVHGWRDPELHQWDLUDWƒ&ZLWKDFRQYHFWLYH FRHI¿FLHQWRIN-P2h°C. The head has been provided with 10 longitudinal VWUDLJKW¿QVZKLFKDUHPPWKLFNDQGSURWUXGHPPIURPWKHF\OLQGULFDO VXUIDFH$VVXPLQJWKDWWKH¿QVKDYHLQVXODWHGWLSVDQGWKDWWKHWKHUPDOFRQGXFWLYLW\ RIWKHF\OLQGHUKHDGDQG¿QPDWHULDOLVN-PKƒ&'HWHUPLQH i  7KHLQFUHDVHLQKHDWGLVVLSDWLRQGXHWRDGGLWLRQRI¿QV ii  7KHWHPSHUDWXUHDWWKHFHQWUHRIWKH¿Q             '8 

6ROXWLRQ Givend PP Pb Pto ƒ&ta ƒ&h N-PKƒ&n  y PP P l PP Pk N-PKƒ& )RUDVWUDLJKWUHFWDQJXODU¿Q P 2 (b + y ) 2b 2 = ≈ = VLQFHyb  Acs b× y b× y y 

m=



=

hP = kAcs

2h ky

2 × 85 = 26.7 m −1 265 × 0.0009

ml î  i  7KHLQFUHDVHLQKHDWGLVVLSDWLRQGXHWRDGGLWLRQRI¿QV )RUD¿QZLWKLQVXODWHGWLS Q¿Q k AFVmto±ta WDQKml 

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6XUIDFHDUHDRIWKHF\OLQGULFDOKHDGQRWRFFXSLHGE\WKH¿QV A = [ π × 0.06 × 1.2 − (10 × 1.2 × 0.0009) ] = 0.2154 m 2



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?DJH increaseLQKHDWGLVVLSDWLRQGXHWRDGGLWLRQRI¿QV 7945.5 − 2211 × 100 = . ( $QV.) 2211 ii  7KH7HPSHUDWXUHDWWKHFHQWUHRIWKH¿Qt  =



7KHH[SUHVVLRQIRUWHPSHUDWXUHGLVWULEXWLRQLVJLYHQE\ t − ta θ cosh {P (O − [)} = = θ o to − t a cosh (ml )

 6XEVWLWXWLQJ [ = 

l = 0.015 m DQGRWKHUDSSURSULDWHYDOXHVZHJHW 2 t − 35 cosh {26.7 (0.03 − 0.015)} 1.08 = = = 0.807 150 − 35 cosh (0.801) 1.338 t = (150 − 35) × 0.807 + 35 = . ƒ& ($QV.)

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6ROXWLRQ 5HIHUWR)LJ &RQVLGHUD¿QKDYLQJFURVVVHFWLRQDODUHDAcs DQGVXUIDFHDUHDAs YDU\LQJDORQJWKHKHDW ÀRZGLUHFWLRQ;GLUHFWLRQ 7KHQDKHDWEDODQFHIRUDQHOHPHQWDOFURVVVHFWLRQRIWKLFNQHVVG[DWD GLVWDQFH[IURPWKHEDVHZDOOLVJLYHQE\ P(x) A(n) Qx+dx

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Q[ Q[G[QFRQY d (4 ) G[ + 4FRQY.  4[ + G[ [ d G[ d G[

(4[ ) G[ + 4FRQY. = 0 dt ⎞ ⎛ ⎜⎝ − N$cs G[ ⎟⎠ G[ + K × $s (W − Wa ) = 0 

+HUHLWKDVEHHQSUHVXPHGWKDWWKHWHPSHUDWXUHRIWKH¿QLVXQLIRUPDQGQRQYDULDQWIRUWKH LQ¿QLWHVLPDOHOHPHQW)XUWKHUDVVXPLQJWKHWKHUPDOFRQGXFWLYLW\RIWKH¿QPDWHULDOWREHFRQVWDQW ZLWKLQWKHFRQVLGHUHGWHPSHUDWXUHUDQJHZHKDYH d Acs dt dt 2 − N $cs . G[ − N . . G[ + K × G$s (W − Wa ) = 0  2 G[ G[ G[ 'LYLGLQJERWKVLGHVE\k AcsG[ZHJHW dA d 2t h 1 dAcs dt i  + − × s (t − ta ) = 0  .  2 $cs G[ G[ N $cs G[ G[ 



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G[ 7KHDERYHHTXDWLRQSURYLGHVDgeneral form of the energy equation for steady, one-dimensional KHDWGLVVLSDWLRQIRUD¿QRIDQ\FURVVVHFWLRQ dA :KHQ WKH ¿Q LV RI FRQVWDQW FURVVVHFWLRQDO DUHD WKH WHUP cs YDQLVKHV DQG WKH HTXDWLRQ G[ UHGXFHVWR d 2θ hP − θ=0  k Acs G[ 2 RU ZKHUH   4

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VDPHDV(TQ

hP kAcs

Both ends of a 6 mm diameter U-shaped copper (k = 285 W/m°C) rod are ULJLGO\¿[HGWRDYHUWLFDOZDOOZKLFKLVDWƒ&WHPSHUDWXUH7KHOHQJWKRI 8VKDSHG URG LV  PP DQG LW LV H[SRVHG WR DLU DW ƒ& 7KH FRPELQHG UDGLDWLYH DQG FRQYHFWLYH KHDW WUDQVIHU FRHI¿FLHQW IRU WKH URG LV  :P2°C. Determine : i The temperature at the centre of U-shaped rod. ii The heat transfer rate.  8378

806

Chapter : 13

6ROXWLRQ 5HIHUWR)LJGiven d PP Pk :Pƒ& tw ƒ&l PP Pta ƒ&h :Pƒ& i  7KHWHPSHUDWXUHDWWKHFHQWUHRI8VKDSHGURGt 7KHWHPSHUDWXUHµt¶FDQEHZRUNHGRXWIURPWKHUHODWLRQ sinh (ml / 2)  θ = 2θ w × sinh (ml ) >)URPHTQIRU [ =

l @ 2

U-shaped rod

tw

d

ZKHUHT  t ±taTw tw±ta 

m =



=

hP = k Acs

h × πd = π k × d2 4

4h kd

4 × 24 = 7.493 m −1 285 × 0.006

(a) ta tw

ml î  ml 4.496 = = 2.248  2 2 sinh (2.248) = 18.8 ? t − 20 = 2 (110 − 20) × sinh (4.496)

l

tw

(b) Fig. 13.7

RU t ƒ&$QV ii  7KHKHDWWUDQVIHUUDWHQ  7KHKHDWWUDQVIHUUDWHIURPWKH¿QFDQEHZRUNHGRXWE\FRQVLGHULQJWZR¿QVRIPOHQJWK HDFKZLWKinsulated tip Q=2

  

=2

h P k Acs (to − ta ) tanh (ml )

⎛π ⎞ 24 × (π × 0.006) × 285 × ⎜ × 0.0062 ⎟ (110 − 20) tanh (4.496) ⎝4 ⎠

îîî :$QV

  4 D  ([SODLQ EULHÀ\ 1HZWRQ5LNKPDQ /DZ :ULWH DOVR WKH IDFWRUV RQ ZKLFK WKH YDOXHRI¿OPFRHI¿FLHQWGHSHQGV  E  ,QDWKHUPDOERXQGDU\OD\HUWKHWHPSHUDWXUHSUR¿OHDWDSDUWLFXODUORFDWLRQ LVJLYHQE\DQH[SUHVVLRQ   W\   $±%\&\2   ZKHUH$%DQG&DUHFRQVWDQWV'HULYHDQH[SUHVVLRQIRUWKHFRUUHVSRQGLQJ KHDWWUDQVIHUFRHI¿FLHQW                  18 6ROXWLRQ a  ,UUHVSHFWLYHRIWKHSDUWLFXODUQDWXUHIUHHRUIRUFHG WKHDSSURSULDWHUDWHHTXDWLRQIRU WKHFRQQHFWLYHKHDWWUDQVIHUEHWZHHQDVXUIDFHDQGDQDGMDFHQWÀXLGLVSUHVHQWHGE\ Newton’s law of cooling, Q = hA (ts − t f ) ZKHUH

Q A ts tf

&RQYHFWLYHKHDWÀRZUDWH 6XUIDFHDUHDH[SRVHGWRKHDWWUDQVIHU 6XUIDFHWHPSHUDWXUHRIVROLG 8QGLVWXUEHGWHPSHUDWXUHRIWKHÀXLGDQG

Universities’ Exams. Questions (Latest)...

807

h &RQVWDQWRISURSRUWLRQDOLW\DOVRUHIHUUHGWRDVFRQYHFWLYHKHDW   WUDQVIHUFRHI¿FLHQWWKHVXUIDFHFRQGXFWDQFHRUWKH¿OP    FRHI¿FLHQW 7KHYDOXHRIhGHSHQGVXSRQWKHIROORZLQJfactors   i  6XUIDFHFRQGLWLRQV²URXJKQHVVDQGFOHDQOLQHVV  ii  *HRPHWU\DQGRULHQWDWLRQRIWKHVXUIDFH²SODWHWXEHDQGF\OLQGHUSODFHGYHUWLFDOO\RU KRUL]RQWDOO\  iii  7KHUPRSK\VLFDO SURSHUWLHV RI ÀXLG²GHQVLW\ YLVFRVLW\ VSHFL¿F KHDW FRHI¿FLHQW RI H[SDQVLRQDQGWKHUPDOFRQGXFWLYLW\  iv  1DWXUHRIÀXLGÀRZ²ODPLQDURUWXUEXOHQW  v  %RXQGDU\OD\HUFRQ¿JXUDWLRQ  vi  3UHYDLOLQJWKHUPDOFRQGLWLRQV b  )URPHQHUJ\EDODQFHZHKDYH ⎛ dt ⎞ Q = − kA ⎜ ⎟ = hA (ts − t∞ )  ⎝ dy ⎠ y = o 

RU

h=−

k ⎛ dt ⎞ (ts − t∞ ) ⎜⎝ dy ⎟⎠ y = o

)URPWKHJLYHQERXQGDU\OD\HUWHPSHUDWXUHSUR¿OH t = A − By + Cy 2 ;   DQG ?   4

dt = − B + 2 Cy dy ⎛ dt ⎞ =−B ⎜⎝ dy ⎟⎠ y=o h=

−k Bk ( − B) = ( $QV.) (ts − t∞ ) (ts − t∞ )

In a nucleate boiling of saturated water at 101.32 kN/m2, vapour bubbles rise at an average velocity of 2.8 m/s. The average diameter of the bubbles is 3.2 mm. 8VLQJWKHIROORZLQJHPSLULFDOUHODWLRQIRUÀXLGÀRZRYHUDVLQJOHVSKHUH hD = (1.2 + 0.53 Re0.54 ) Pr 0.3 k

⎛ μ∞ ⎞ ⎜⎝ μ ⎟⎠ r

0.25

,

valid over the range 1 < Re < 2 × 105; DQGDOOSURSHUWLHVH[FHSWPs are evaluated at the free stream temperature, determine : i  7KHDYHUDJHKHDWWUDQVIHUFRHI¿FLHQW ii The drop in vapour temperature as the bubble moves up through a distance of 0.25 m.                        08 6ROXWLRQGiven pv N1P8f PVD PP'LVWDQFHE\ZKLFKWKHEXEEOH PRYHVXS P i  7KHDYHUDJHKHDWWUDQVIHUFRHI¿FLHQWK )RUVDWXUDWHGZDWHUDWN1PWKHIUHHVWUHDPWHPSHUDWXUHLVƒ&DQGWKHFRUUHVSRQGLQJ WKHUPRSK\VLFDOSURSHUWLHVDUH  Uf NJPkf :Pƒ&vf î±PV  Pf Ps UQ î±1VP Pr cp -NJ.

Chapter : 13

808 

U ∞ D 2.8 × (3.2 × 10−3 ) = = 30168 ν∞ 0.297 × 10 −6 6LQFH5HîWKHJLYHQUHODWLRQDSSOLHV 6XEVWLWXWLQJWKHDSSURSULDWHYDOXHVZHJHW hD = ⎡⎣1.2 + 0.53 × (30168)0.54 ⎤⎦ × (1.75)0.3 × 1 = 165.9  k∞ 5H\QROGVQXPEHU Re =

k∞ 0.682 = 165.9 × =  î  : /P  ƒ& ( $QV.) D 3.2 × 10 −3 ii  7KHGURSLQYDSRXUWHPSHUDWXUHDVWKHEXEEOHPRYHVXSWKURXJKDGLVWDQFHRIP %\DQHQHUJ\EDODQFHDQGDVVXPLQJQRSKDVHFKDQJHZHKDYH hATv±TVDW W Pc3'T UV cp 'T Distance 0.25 τ= = = 0.0893 s ZKHUH Velocity 2.8 h = 165.9 ×

?



A SR SD Sîî±  î±P π π V (volume) = D3 = × (3.2 × 10−3 )3 = 1.716 × 10−8 m3  6 6 $VVXPLQJWKHEXEEOHVFRQWDLQQRFRQGHQVDEOHJDV 2 ⎤ 2σ ⎡ Rv Tsat >(TQ @ Tv − Tsat =  ⎢ ⎥ r ⎣⎢ ρv h fg ⎦⎥ ⎡ ⎤ 462 × 3732 = 0.0204°C ⎢ 3 3⎥ −3 (3.2 × 10 / 2) ⎣ (101.32 × 10 ) × 2257 × 10 ⎦ ?9DSRXUEXEEOHWHPSHUDWXUHULVH hA (Tv − Tsat ) τ ΔT =  ρV c p =



=



2 × 0.058

35.36 × 103 × 32.17 × 10 −6 × 0.0204 × 0.0893 958.4 × 1.716 × 10 −8 × 4220

= .°& ( $QV.)

  4a ,VLWEHWWHUWRDUUDQJHIRUÀRZLQDKHDWH[FKDQJHUWREHSDUDOOHORUFRXQWHU ÀRZ" b  ,QDFRXQWHUÀRZKHDWH[FKDQJHURLOFp = 2.8 kJ/kg K) at the rate of 1350 kg/h is cooled from 90°C to 20°C by water that enters at 15°C at the rate of 1250 kg/h. i  'HWHUPLQHWKHKHDWH[FKDQJHUDUHDIRUDQRYHUDOOKHDWWUDQVIHUFRHI¿FLHQWRI 3800 W/m2-h-K. ii Derive a relationship between oil and water temperatures at any section of WKHKHDWH[FKDQJHU                    0'8 6ROXWLRQ   

a  :KHQDFKRLFHKDVWREHPDGHEHWZHHQSDUDOOHOÀRZDQGFRXQWHUÀRZDUUDQJH PHQWWKHFRXQWHUÀRZGHVLJQLVXVXDOO\preferredIRUWKHIROORZLQJreasons i  7KHH[FKDQJHRIKHDWPD\UDLVHWKHWHPSHUDWXUHRIWKHFROGÀXLGWRmore nearly the initial WHPSHUDWXUHRIWKHKRWÀXLG ii  /07'/RJPHDQWHPSHUDWXUHGLIIHUHQFH LVKLJKHUDQGDFFRUGLQJO\PRUHKHDWFDQEHWUDQV IHUUHG&RQYHUVHO\Dsmaller surface area is required for the same rate of heat transfer b  Given cph N-NJ.mh NJKth ƒ&th ƒ&tc  ƒ&m c  NJKU :PK.

Universities’ Exams. Questions (Latest)...

809

i  +HDWH[FKDQJHDUHDA   +HDWORVWE\KRWÀXLG mhîcphîth±th   îî±  RU tc

+HDWJDLQHGE\FROGÀXLG mcîcpcîtc±tc îîtc± ƒ& θ − θ2 LMTD, θ m = 1 ln (θ1 / θ 2 )





 

(th1 − tc 2 ) − (th 2 − tc1 ) ln [(th1 − tc 2 ) / (th 2 − tc1 ) ]

=

 





(90 − 65.6) − (20 − 15) 19.4 = = 12.24°C ln [ (90 − 65.6) / (20 − 15) ] ln (24.4 / 5)

Q UATm mh × c ph (th1 − th 2 ) Q = +HDWH[FKDQJHUDUHD A = U θm U θm

+HDWH[FKDQJHG 

RU

1350 × 2.8 (90 − 20) = . P  ( $QV.) 3800 × 12.24 ii  $WDQ\VHFWLRQRIWKHKHDWH[FKDQJHUZHKDYH mhîcphî±th  mcîcpc±tc  îî±th  î±tc RU ±th  ±tc RU th ±±tc  tc±$QV   4 A feed water heater with steam outside and water inside the tubes is required to heat water from a temperature ‘t1’ to temperature ‘t2’. The overall heat m ⎡ ⎤ kJ/m 2 -s-°C, where ‘a’ is WUDQVIHUFRHI¿FLHQWLVDQWLFLSDWHGWREH ⎢ ⎣ (m + Ba ) ⎥⎦ 

=

WKHWRWDOFURVVVHFWLRQDODUHDRIZDWHUÀRZLQP2µP¶LVWKHPDVVÀRZUDWHRI water per tube in kg/s and ‘B’ is a dimensional constant. Derive the following relation between length ‘l’ and diameter ‘d’ of the tube : m⎞ 4l ⎛ t − t1 ⎞ ⎛ = c ⎜ B + ⎟ ln ⎜ s d a⎠ ⎝ ⎝ ts − t2 ⎟⎠  ZKHUHµF¶LVWKHVSHFL¿FKHDWRIZDWHUDQGWs is the temperature of condensing steam. 378



6ROXWLRQ $QHQHUJ\EDODQFHRQWKHFRRODQWZDWHU JLYHVWKHKHDWH[FKDQJHUDWHDV Q Nîmîcît±t ZKHUH N 1RRIWXEHVDQG m 0DVVÀRZUDWHRIZDWHUSHUWXEH 7KHRYHUDOOKHDWH[FKDQJHUDWHLVDOVRJLYHQE\   $OVR

⎡ θ − θ2 ⎤ Q = UA θ m = UA ⎢ 1 ⎥ ⎣ ln (θ1 / θ 2 ) ⎦ ⎡ (t − t1 ) − (ts − t2 ) ⎤ = UA ⎢ s ⎥ ⎣ ln {(ts − t1 ) / (ts − t2 )} ⎦ A NîSd l

810 ? 

Chapter : 13 (t2 − t1 ) ⎡ ⎤ ⎛ m ⎞ N × m × c × (t2 − t1 ) = ⎜ × (N × π d l) × ⎢ ⎥ l {( t − t ) / ( t − t )} ⎝ m + Ba ⎟⎠ s 1 2 ⎦ ⎣n s ⎛ t − t1 ⎞ π d l = c (m + Ba ) ln ⎜ s ⎝ ts − t2 ⎟⎠ πd 2 ZHJHW 4 m⎞ 4l ⎛ t − t1 ⎞ ⎛ = c ⎜ B + ⎟ ln ⎜ s 5HTXLUHGH[SUHVVLRQ$QV d a⎠ ⎝ ⎝ ts − t2 ⎟⎠

'LYLGLQJERWKVLGHVE\WKHÀRZDUHD a =    4

 $ERLOHUIXUQDFHLVODLGIURP¿UHFOD\EULFNZLWKRXWVLGHODJJLQJIURPSODWHVWHHO WKHGLVWDQFHEHWZHHQWKHWZRLVTXLWHVPDOOFRPSDUHGZLWKWKHVL]HRIWKHIXUQDFH The brick setting is at an average temperature of 380 K while the steel lagging is at 300 K. The emissivity values are : ε (brick ) = 0.84 and ε ( steel ) = 0.64.

Determine: i  7KHUDGLDQWÀX[ ii The reduction in heat loss if a steel screen having an emissivity value of 0.62 on both sides is placed between the brick and steel setting. Also calculate the desired emissivity of screen if the radiation loss is to be limited to 90 W/m2. 18 6ROXWLRQ Given T .T .HEULFN  HVWHHO   i  7KHUDGLDQWÀX[Q 7KHUDWHRIKHDWLQWHUFKDQJHEHWZHHQEULFNVHWWLQJVXI¿[ DQGVWHHOODJJLQJVXI¿[ LV JLYHQE\ Q12 = ( Fg )1− 2 A1 σ T14 − T24 

(



7KHJUD\IDFWRU ( Fg )1− 2 =

)

1 1 ⎛ 1 − ε1 ⎞ ⎛ 1 − ε 2 ⎞ A1 +⎜ ⎜⎝ ε ⎟⎠ + F ⎝ ε 2 ⎟⎠ A2 1 1− 2

)RULQ¿QLWHORQJSDUDOOHOSODWHVZKLFKVHHHDFKRWKHUDQGQRWKLQJHOVH F± A A ? 1 1 ( Fg )1− 2 = =  1 1 ⎛1 ⎞ ⎛ 1 ⎞ + −1 ⎜⎝ ε − 1⎟⎠ + 1 + ⎜⎝ ε − 1⎟⎠ ε ε 1 2 1 2 

?

=

1 1 1 + −1 0.84 0.64

= 0.57

⎡⎛ 380 ⎞ 4 ⎛ 300 ⎞ 4 ⎤ Q12 = 0.57 × 1 × 5.67 ⎢⎜ −⎜ ⎥ ⎝ 100 ⎟⎠ ⎝ 100 ⎟⎠ ⎥ ⎣⎢ ⎦

  :P$QV ii  7KHUHGXFWLRQLQKHDWORVV :KHQDVFUHHQVXI¿[ LVLQVHUWHGEHWZHHQWKHEULFNVXI¿[ DQGVWHHOODJJLQJVXI¿[  1 Fg = 1− 2 ⎡⎛ 1 − ε1 ⎞ 1 1 ⎛ 1 − ε3 ⎞ A1 ⎤ ⎡⎛ 1 − ε 3 ⎞ ⎛ 1 − ε 2 ⎞ A3 ⎤ +⎜ +⎜ ⎢⎜⎝ ε ⎟⎠ + F ⎟⎠ A ⎥ + ⎢⎜⎝ ε ⎟⎠ + F ε ⎝ ε 2 ⎟⎠ A2 ⎥⎦ ⎝ 1 1− 2 3 3⎦ 3 3−2 ⎣ ⎣

( )

Universities’ Exams. Questions (Latest)...

811

)RUJLYHQDUUDQJHPHQW A A ADQGF± F± 

(F )

?

g 1− 2



=

1 1 1 2 + + −2 ε1 ε 2 ε3

=

1 = 0.25 1 1 2 + + −2 0.84 0.64 0.62

⎡⎛ 380 ⎞ 4 ⎛ 300 ⎞ 4 ⎤  Q12 = 0.25 × 1 × 5.67 ⎢⎜ ⎟ − ⎜⎝ 100 ⎟⎠ ⎥ = . : /P ( $QV) ⎢⎣⎝ 100 ⎠ ⎥⎦ +HQFHWKHUHGXFWLRQLQKHDWORVVGXHWRSODFHPHQWRIVWHHOVFUHHQ ?

 ± :P$QV 



(PLVVLYLW\RIVFUHHQIRUUHGXFLQJORVVWR:PH ,IUDGLDQWKHDWORVVLVWREHOLPLWHGWR:PWKHQ ⎡⎛ 380 ⎞ 4 ⎛ 300 ⎞ 4 ⎤ 90 = ( Fg )1− 2 × 1 × 5.67 ⎢⎜ −⎜ ⎥ ⎝ 100 ⎟⎠ ⎝ 100 ⎟⎠ ⎥ ⎣⎢ ⎦

 

 Fg ±î

?

Fg ±  1 1 1 2 = + + −2 ( Fg )1− 2 ε1 ε 2 ε3

$OVR

1 1 1 2 = + + −2 0.1245 0.84 0.64 ε3

RU

8.032 = 1.1905 + 1.5625 +



2 −2 ε3

'HVLUHGHPLVVLYLW\RIVFUHHQH $QV   4

 $PPRXWVLGHGLDPHWHUSLSHFDUULHVDFU\RJHQLFÀXLGDW.$QRWKHUSLSH RIPPRXWVLGHGLDPHWHUDQGDW.VXUURXQGVLWFRD[LDOO\DQGWKHVSDFH between the pipes is completely evacuated. i  'HWHUPLQHWKHUDGLDQWKHDWÀRZIRUPOHQJWKRISLSHLIWKHVXUIDFHHPLVsivity for both surfaces is 0.25. LL  &DOFXODWHWKHSHUFHQWDJHUHGXFWLRQLQKHDWÀRZLIDVKLHOGRIPPGLDPHWHU and 0.06 surface emissivity is placed between the pipes.      8378



6ROXWLRQ Given : d PP PT .d PP PT .l  PH H ds PP PHs  i  5DGLDQWKHDWÀRZ 7KHUDWHRIKHDWLQWHUFKDQJHEHWZHHQWKHLQQHUVXI¿[ DQGRXWHUVXI¿[ SLSHVLVJLYHQE\ 

Q12 = ( Fg )1− 2 A1 σ (T14 − T24 )

812

Chapter : 13

7KHJUD\ERG\IDFWRU 

( Fg )1− 2 =

1 1 1 ⎛ ⎞ ⎛ 1 ⎞ A +⎜ − 1⎟ 1 ⎜⎝ ε − 1⎟⎠ + F ⎝ ε2 ⎠ A2 1 1− 2

)RUWKHJLYHQFRQ¿JXUDWLRQF±  1 ( Fg )1− 2 =  1 ⎛ 1 ⎞ A + − 1⎟ 1 ε1 ⎜⎝ ε 2 ⎠ A2 1RZ

Q12 =

A1σ (T14 − T24 ) 1 ⎛ 1 ⎞ A + − 1⎟ 1 ε1 ⎜⎝ ε 2 ⎠ A2

)RUPOHQJWKRISLSHWKHSLSHDUHDVDUH A Sdl Sîî P A Sdl Sîî P 

A1 π d1 l d 0.012 = = 1 = = 0.8 A2 π d 2 l d 2 0.015

,QVHUWLQJWKHDSSURSULDWHYDOXHVLQWKHDERYHHTQZHJHW



Q12

⎡⎛ 90 ⎞ 4 ⎛ 290 ⎞ 4 ⎤ 0.1319 × 5.67 ⎢⎜ ⎟ − ⎜⎝ 100 ⎟⎠ ⎥ ⎢⎝ 100 ⎠ ⎣ ⎦⎥ = 1 ⎛ 1 ⎞ + − 1⎟ × 0.8 0.25 ⎜⎝ 0.25 ⎠

52.4 = ±  : 6.4 7KHQHJDWLYHVLJQLPSOLHVWKDWKHDWÀRZVIURPRXWHUSLSHWRLQQHUSLSH$QV ii  DJHUHGXFWLRQLQKHDWÀRZGXHWRLQVHUWLRQRIDVKLHOG 

=−

:KHQDUDGLDWLRQVKLHOGLVSODFHGEHWZHHQWKHWZRSLSHV ( Fg )1− 2 =

 

1 ⎡⎛ 1 ⎤ ⎛⎛ 1 1 1 ⎛ 1 ⎞ A1 ⎡⎛ 1 ⎞ ⎞ ⎞ A ⎞⎤ +⎜⎜ − 1⎟ s ⎟ ⎥ ⎢⎝⎜ ε − 1⎠⎟ + F + ⎝⎜ ε − 1⎠⎟ A ⎥ + ⎢⎝⎜ ε − 1⎠⎟ + F ⎠ A2 ⎠ ⎦ 1− s s s⎦ ⎣ s s − 2 ⎝ ⎝ ε2 ⎣ 1 +HUHF±V FV±  ( Fg )1− 2 =

1 1 ⎛ 1 1 ⎞ A1 ⎛ 1 ⎞ A + − 1⎟ + +⎜ − 1⎟ s ε1 ⎜⎝ ε s A ε ε ⎝ 2 ⎠ A2 ⎠ s s

2 AsVKLHOGSLSHDUHD   π d s l = π × 0.0135 × 3.5 = 0.1484 m



A1 d 0.013 = 1 = = 0.963 0.0135 As ds



As d 0.0135 = s = = 0.9 A2 d2 0.015

Universities’ Exams. Questions (Latest)... ?

Q12 =

(

A1 σ T14 − T24

813

)

1 ⎛ 1 1 ⎞ A ⎛ 1 ⎞ A +⎜ − 1⎟ 1 + +⎜ − 1⎟ s ε1 ⎝ ε s εs ⎝ ε2 ⎠ A2 ⎠ As



⎡⎛ 90 ⎞ 4 ⎛ 290 ⎞ 4 ⎤ 0.1319 × 5.67 ⎢⎜ ⎟ − ⎜⎝ 100 ⎟⎠ ⎥ ⎢⎣⎝ 100 ⎠ ⎥⎦ = 1 1 ⎛ 1 ⎞ ⎛ 1 ⎞ + − 1⎟ × 0.963 + + − 1⎟ × 0.9 0.25 ⎜⎝ 0.06 0.06 ⎜⎝ 0.25 ⎠ ⎠



=−

52.4 = − 1.362 W 4 + 15.087 + 16.67 + 2.7

+HQFHSHUFHQWDJHreductionLQKHDWÀRZ 

=

8.188 − 1.362 × 100 =  $QV 8.188

C. MULTIPLE-CHOICE QUESTIONS WITH ANSWERS   7KH)RXULHU¶VODZRIKHDWWUDQVIHUE\FRQGXFWLRQ LVH[SUHVVHGDV dt dt  a  Q = kA  b  – kA G[ G[ G[ G[  c  Q = kA  d  – kA . dt dt   7KHKHDWWUDQVIHULVFRQVWDQWZKHQ  a  WHPSHUDWXUHUHPDLQVFRQVWDQWZLWKWLPH  b  WHPSHUDWXUHGHFUHDVHVZLWKWLPH  c  WHPSHUDWXUHLQFUHDVHVZLWKWLPH  d  $OORIWKHDERYH   7KHFRHI¿FLHQWRIWKHUPDOFRQGXFWLYLW\LVGH¿QHG DV  a  4XDQWLW\ RI KHDW WUDQVIHU SHU XQLW DUHD SHU RQHGHJUHHGURSLQWHPSHUDWXUH  b  4XDQWLW\ RI KHDW WUDQVIHU SHU RQH GHJUHH WHPSHUDWXUHGURSSHUXQLWDUHD  c  4XDQWLW\RIKHDWWUDQVIHUSHUXQLWWLPHSHU XQLWDUHD  d  4XDQWLW\RIKHDWWUDQVIHUSHUXQLWWLPHSHU XQLW DUHD SHU RQH GHJUHH WHPSHUDWXUH GURS SHUXQLWOHQJWK   7KHWKHUPDOFRQGXFWLYLW\LVH[SUHVVHGDV  a  :P. b  :P.  c  :KP. d  :KP.   +HDW WUDQVIHU IURP KLJKHU WHPSHUDWXUH WR ORZ WHPSHUDWXUHWDNHVSODFHDFFURGLQJWR  a  )RXULHU¶VODZ  b  )LUVWODZRIWKHUPRG\QDPLFV  c  6HFRQGODZRIWKHUPRG\QDPFLV  d  =HURWKODZRIWKHUPRG\QDPLFV

  &RQGXFWLRQWKURXJKÀDWFRPSRVLWHZDOOLVJLYHQ E\ 

a  Q =



b  Q =



c  Q =

 

t1 – t4  [ [1 [ + 2 + 3 k1 A k2 A k3 A t1 – t4 k1 A k2 A k3 A + + [1 [2 [3 (t1 – t4 ) A k1 k2 k3 + + [1 [2 [3

k1 A k2 A k3 A + + [ [2 [3 d  Q = 1 (t1 – t4 )  ZKHUHQ KHDWWUDQVIHUtttDQGtWHPSHUDWXUHV RQ VXUIDFHV RI FRPSRVLWH ZDOO [ [ [ [ WKLFNQHVVHVRIGLIIHUHQWFRPSRVLWHZDOOOD\HUV

  &RQGXFWLRQWKURXJKKROORZUDGLDORQHGLPHQVLRQDO KHDWWUDQVIHULVH[SUHVVHGDV 

a  Q =

2πl (t1 – t2 ) k  log e r2 / r1



b  Q =

2πl (t1 – t2 )  k (r2 – r1 )



c  Q =

2πl log e (t1 / t2 )  (r2 – r1 ) k



d  Q =

2 π l (t1 – t2 ) k . log e r2 / r1

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Chapter : 13

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Chapter : 13

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Chapter : 13

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Chapter : 13

828   

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838

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Universities’ Exams. Questions (Latest)...

839



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GATE and UPSC Examinations’ Questions (Latest-Selected)

14 C H A P T E R

with Answers/Solutions: A. Conventional Questions with Answers/ Solutions B. Multiple-choice Questions with Answers and “Explanations.” A. CONVENTIONAL QUESTIONS WITH ANSWERS/SOLUTIONS 4(Conduction). A steam pipe (inner diameter = 150 mm and outer diameter = 160 mm) having thermal conductivity 58 W/m K is covered with two layers of insulation of thicknesses 30 mm and 50 mm respectively and thermal conductivities 0.18 W/m K and 0.09 W/m K respectively. The temperature of inner surface of the steam pipe is 320°C and that of outer surface of the insulation layers is 40°C. (i) Determine the quantity of heat lost per metre length of the steam pipe and layer contact temperature. LL ,IWKHFRQGLWLRQRIWKHVWHDPLVGU\DQGVDWXUDWHG¿QGWKHTXDOLW\RIWKHVWHDPFRPLQJ RXWRIRQHPHWUHSLSHDVVXPLQJWKDWWKHTXDQWLW\RIVWHDPÀRZLQJLVNJPLQ [Use the data: At 320°C saturation temperature hf = 1463 kJ/kg, hfg = 1240 kJ/kg, hg = 2703 kJ/kg] &6(0DLQV  6ROXWLRQGiven5HIHUWR)LJ riA PProA PPriB PProB PP riC PProC PPt ƒ&t ƒ& i 4XDQWLW\RIKHDWORVWSHUPHWUHOHQJWKRIVWHDPSLSHDQGOD\HUFRQWDFWWHPSHUDWXUHV 5HIHUWR)LJ RWK A 

 



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842 



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843

4(Heat exchanger). In a large steam power plant, a shell and tube type condenser is used which has the following data: Heat exchange data = 2100 MW

Number of shell passes = 1

Number of tubes = 31500

Number of tube passes = 2

Diameter of each tube = 25 mm Condensation temperature = 50°C 0DVVÀRZUDWHRIFRROLQJZDWHU î4 kg/s +HDWWUDQVIHUFRHI¿FLHQWRQWKHVWHDPVLGH :P2K Inlet water temperature = 20°C +HDWWUDQVIHUFRHI¿FLHQWRQWKHZDWHUVLGH :P2K Using only H-NTU method, calculate: (i) the outlet temperature of cooling water; (ii) the length of tube pass. [Properties of water at 27°C are: cp = 4.18 kJ/kg K, P =î6 N s/m2, k = 0.613 W/mK and Pr = 5.83] Neglect the thermal resistance due to tube wall. &6(0DLQV  6ROXWLRQGiven+HDWH[FKDQJHG 0:1XPEHURIVKHOOSDVVHV 1XPEHURIWXEH SDVVHVNp d PP P&RQGHQVDWLRQWHPSHUDWXUHth  th  ƒ&māc î   NJVhVWHDP :P.hZDWHU :P.tc ƒ& 1RZ

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Chapter : 14

844 

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(150 + 30) = 90°C: 2

U = 0.962 kg/m3, P î5 kg/ms k = 0.031 W/mK, cp = 1.01 kJ/kg K

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GATE and UPSC Examinations’ Questions ...

845

4(Free forced convection). A 2-stroke motor cycle petrol engine cylinder consists of ¿QVRQLWVRXWHUVXUIDFH,IWKHRXWVLGHDQGLQVLGHGLDPHWHUVRIHDFK¿QDUHPPDQGPP UHVSHFWLYHO\WKHDYHUDJH¿QVXUIDFHWHPSHUDWXUHLVƒ&DQGWKHDWPRVSKHULFDLUWHPSHUDWXUHLV ƒ&FDOFXODWHWKHKHDWWUDQVIHUUDWHIURPWKH¿QVIRUWKHIROORZLQJFDVHV L  ZKHQWKHPRWRUF\FOHLVVWDWLRQDU\ LL  ZKHQWKHPRWRUF\FOHLVUXQQLQJDWDVSHHGRINPSK 7KH ¿Q PD\ EH LGHDOL]HG DV D VLQJOH KRUL]RQWDO SODWH RI WKH VDPH DUHD DQG WKH VLJQL¿FDQW OHQJWKPD\EHWDNHQDV/ GZKHUHGLVWKHRXWHUGLDPHWHURIWKH¿Q$VVXPHGDVPP 7KHSURSHUWLHVRIDLUPD\EHWDNHQDVIROORZV N î2:Pƒ&Q îm2V3U  )RUWXUEXOHQWÀRZIRUFHGFRQYHFWLRQ 1X 5H 3U 

  

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0.04922 (40.61 × 10 −6 ) 2

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846

ii :KHQPRWRUF\FOHLVUXQQLQJDWNPK 60 × 1000 = 16.66 m/s  9HORFLW\v  60 × 60 

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di

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Condensate Fig. 14.3

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GATE and UPSC Examinations’ Questions ...

847

Cmin   &o Cmax 0D[LPXPWUDQVIHUQPD[ māw cpwthi±tci

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iii 7KHUDWHRIVWHDPFRQVXPSWLRQ 5DWHRIVWHDPFRQVXPSWLRQ 

Qactual 11.495 = 5.09 × 10−3 kg/s $QV   h fg 2257

4(Heat Exchanger).7KHYDSRXUDWWKHVDWXUDWLRQWHPSHUDWXUHRIDQRLOÀRZLQJDW the rate of 500 kg/min, enters a heat exchanger tube, at 355 K and condenses while it is cooled E\ZDWHUÀRZLQJDWWKHUDWHRINJPLQHQWHULQJWKHFRQFHQWULFWXEHRIDSDUDOOHOÀRZKHDW H[FKDQJHUDW.$VVXPLQJRYHUDOOKHDWWUDQVIHUFRHI¿FLHQWRI:P2 K, latent heat of oil as 600 kJ/kg K, calculate the number of tubes required of 25 mm outer diameter and 2 mm thick with a length of 4.87 m. What will be the number of tube passes, if cooling water velocity should not exceed 2 m/s? Take cp for water as 4.18 kJ/kg K and density of water as 1000 kg/m3. &6(0DLQV  6ROXWLRQGivenmā NJPLQth ± ƒ&māw NJPLQtc ± ƒ&U :P.hfgRLO  N-NJ.do PP Pdi ± PP  PL Pcpw N-NJ.Uw NJP

Chapter : 14

848

Concentric tube

. m0 = 500 kg/min (th1)

Heat exchanger tube

Oil vapour

Oil Water

Water do

Oil di

Water (tc2)

. mw = 3600 kg/min (tc1)

Fig. 14.4

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e NTU 



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  Nt 

UAs U × πdi L × n = Cmin m· w c pw 475 × π × 0.021 × 4.87 × Nt (3600 / 60) × (4.18 × 103 ) 0.341 × 3600 × (4.18 × 103 )   $QV 60 × 475 × π × 0.021 × 4.87

π 2 π di = × (0.021) 2 = 3.464 × 10 −4 m 2 4 4 0D[LPXPDOORZDEOHYHORFLW\v PVJLYHQ 0D[LPXPÀRZUDWHSHUWXEH UAv îî±î NJV 3600  WXEHV ?1XPEHURIWXEHVUHTXLUHGSHUSDVVNp  60 × 0.6928 ,QWHUQDODUHDRIWKHWXEHAi 

GATE and UPSC Examinations’ Questions ...

849

1XPEHURIWXEHSDVVHVnp $VVXPHHDFKWXEHUHTXLUHVnpSDVVHV 7RWDOOHQJWKRIWXEHVUHTXLUHGIRUUHTXLUHGKHDWWUDQVIHU    îL î P ?)RUWXEHVZLWKnpSDVVHVHDFKWRWDOOHQJWKRIKHDWH[FKDQJHUVXUIDFHUHTXLUHG    înpîL înpî  2732.07  SDVVHVSDVVHV RU np  87 × 4.87 N 561 = 6.448  SDVVHV    RUnp  t = Np 87 ?WXEHVZLWKSDVVHVHDFKRUWXEHSDVVHVDUHUHTXLUHGWROLPLWPD[LPXPYHORFLW\WR PV$QV  4(Radiation). A thin radiation shield having equal emissivities on both sides is introduced parallel to and in between two large plates with emissivities 0.8 and 0.5 respectively. Determine the emissivity of the radiation shield to reduce the heat transfer rate by 92% of the original. &6(0DLQV  6ROXWLRQGivenH H 5HGXFWLRQLQKHDWWUDQVIHUGXHWRLQWURGXFWLRQRIUDGLDWLRQ VKLHG RIWKHRULJLQDO Hot (T1)

Cold (T2)

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e3

e1 e1 = 0.8

e2

e2 = 0.5

1

1

2

3

2

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σ (T14 − T24 )  1 1 + −1 ε1 ε 2







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σ (T14 − T24 ) σ (T14 − T24 ) = = 0.444 σ (T14 − T24 ) 1 ⎛ 1 ⎞ (1.25 + 2 − 1) + − 1⎟ ⎜⎝ 0.8 0.5 ⎠

b )RUWKHFDVHZKHQUDGLDWLRQVKLHOGLVLQWURGXFHGEHWZHHQWKHWZRSODWHVWKHWRWDOWKHUPDO UHVLVWDQFH 1 1 ⎡1 ⎤ ⎡1 ⎤ − 1⎥ + ⎢ + − 1⎥  RWK WRWDO  ⎢ + ⎣ ε1 ε3 ⎦ ⎣ ε 2 ε3 ⎦

Chapter : 14

850

1 1 2⎤ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎡ + − 1⎟ + ⎜ + − 1⎟ = ⎢1.25 + ⎥  ⎜ ε3 ⎦ ⎝ 0.8 ε3 ⎠ ⎝ 0.5 ε 3 ⎠ ⎣



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RU

σ (T14 − T24 ) = (1 − 0.92) × 0.444 σ (T14 − T24 )  2⎞ ⎛ ⎜⎝1.25 + ε ⎟⎠ 3 1 1.25 +

2 ε3

  î 

2⎞ ⎛    ⎜1.25 + ⎟ ε3 ⎠ ⎝

RU

  0.0444 +

RU RU

σ (T14 − T24 ) = (1 − 0.92) × Q ( Rth ) total

0.071 ε3

0.071   ±  ε3

RU

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4(Conduction). The outer and inner surfaces of a thick hollow cylinder have areas 1.25 m2 and 0.25 m2 respectively. The thickness of the cylinder is 10 cm and the thermal conductivity of the cylinder is 50 W/mK. Find the radial heat transfer through the cylinder for 100°C temperature difference at the surfaces. Derive the formula used. &6(0DLQV  



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Element dr r

i ii

r  r r±r  P

r2 r1

10 cm

Fig. 14.6

1.25 = 1.59 m 2π × 0.125 &RQVLGHUDWKLQF\OLQGULFDOHOHPHQWDWDGLVWDQFHrIURPWKHFHQWUHRIVKHOODQGWKLFNQHVVdr dt dt = − k × (2π rL)  +HDWWUDQVIHUUDWHQ ±kA dr dr

)URPZKLFK

r  PDQGr PDQGL 

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dr k   − dt 2πrL Q

RU

,QWHJUDWLQJERWKVLGHVZHJHW

∫



r2 r1

dr k  − 2πrL Q

t2

∫ dt

t1

1 k ⎛r ⎞ ln 2   (t1 − t2 ) 2πL ⎝⎜ r1 ⎠⎟ Q

RU

?+HDWWUDQVIHUUDWH Q 

(t1 − t2 )   1 ⎛r ⎞ . ln ⎜ 2 ⎟ 2π kL ⎝ r1 ⎠



IRUPXODGHULYHG

+HQFH UDGLDO KHDW WUDQVIHUUHG WKURXJK F\OLQGHU IRU ƒ& WHPSHUDWXUH GLIIHUHQFH EHWZHHQ VXUIDFHV 100   N:   $QV  Q  1 ln (5) 2π × 50 × 1.59 4(Conduction). Two long slender rods A and B, made of different materials having same diameter of 12 mm and length 1 m, are attached to a surface maintained at a temperature of 100°C. The surfaces of the rods are exposed to ambient still air at 20°C. By traversing along the length of the rods with a temperature sensor, it is found that the surface temperatures of rods A and B are equal at positions 15 cm and 7.5 cm respectively away from the base surface. If material of A is carbon steel with thermal conductivity 60 W/mK, what is the thermal conductivity of rod B? List the assumptions made. Assume that thHDYHUDJHFRQYHFWLRQFRHI¿FLHQWDLULV:P2K. Find the ratio of the rate of heat transfer for rods A and B. &6(0DLQV  6ROXWLRQ5HIHUWR)LJ Givenr PP Pl Pto ƒ&ta ƒ&kA :P.h :P. kB

QA  QB 12 mm A

x

Rods dx

ta = 20°C

t0 = 100°C

Qconv. B 12 mm 1m

Q(x+dx)

Qx Element

Fig. 14.7

Chapter : 14

852

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1RZ

Acs  Sr Sî  î±P



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kAcs d 2 t dx

RU

dx 2 d 2θ dx

2

− h ( Pdx) (t − ta ) = 0

− m2θ    Q  t±taDQGm 

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hP kAcs

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x=l

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hP = k A Acs

5 × 0.0377 60 × (1.131 × 10 −4 )

i

= 5.27

7HPSHUDWXUHDWFPIURPWKHEDVHtLVFDOFXODWHGHTQi DVIROORZV θ    e±m x θ0 t − ta   e±m x e±m î e±mAî to − t a

RU RU

t − 20   e±î  e± 100 − 20

RU

t  ± îe± îe± ƒ&

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t − ta   e±mBî to − t a

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56.3 − 20   e±mBî  100 − 20 emBî   RU mB  mB 

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853

5 × 0.0377 k B × 1.131 × 10 −4 5 × 0.0377 k B × 1.131 × 10 −4

     

kB 

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hP = 10.64 hB Acs

5 × 0.0377 1.131 × 10 −4 × 113.21

 :P.

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,QWKLVFDVHWKHKHDWWUDQVIHUIRUDURG 

 

h Pk Acs × (to − ta )

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QA   QB



 

h Pk A Acs × (to − ta ) h P k B Acs × (to − ta )

=

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4(Fins). A very long AISI 316 stainless steel (K = 14 W/m-K) rod 5 mm in diameter has one end maintained at 100°C. The surface of the rod is exposed to ambient air at 30°C with DYHUDJH FRQYHFWLYH KHDW WUDQVIHU FRHI¿FLHQW RI  :P2-K. Neglecting radiation heat transfer, HVWLPDWHKRZORQJWKHURGPXVWEHWRWUHDWLWDV³LQ¿QLWHO\ORQJ´WR\LHOGDUHDVRQDEOHDFFXUDWH estimation of heat loss. If the rod is made of copper K = 350 W/m-K, will the length be different? How much will it be and why? Compare the heat transfer rates for both the rods. The analysis may EHEDVHGRQ¿QWLSKHDWORVVDORQH                  &6(0DLQV  6ROXWLRQ5HIHUWR)LJ Givend PP Pto ƒ&ta ƒ&h :P. l /HQJWKRIWKH¿QSHUSHQGLFXODU WRVXUIDFHZKLFKKHDWLVWREHUHPRYHG 5 mm p2, ta

d

t0 = 100°C Stainless steel l

Fig. 14.8

   

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Chapter : 14

854

 to  7HPSHUDWXUHDWWKHEDVHRIWKH¿QDQG  ta  7HPSHUDWXUHRIWKHDPELHQWVXUURXQGLQJÀXLG 7KHUDWHRIWKHKHDWÀRZDFURVVWKHEDVHRIWKH¿QLVJLYHQE\ 

Q¿Q 

Ph kAcs to±ta 







>(TQ @

)RUDYHU\ORQJURGWHPSHUDWXUHGLVWULEXWLRQLVJLYHQE\  T  Te±mxT t±ta   To to±ta  +HUHIRUUHDVRQDEO\DFFXUDWHHVWLPDWHRIKHDWORVVZHQHHGWRKDYHURGORQJHQRXJKVRWKDW  e±ml    RU Ql  QaîUHDVRQDEOHHVWLPDWLRQ 1RZ 7KHQ RU ? 

m 

?

1 = 100 0.01 $QV

e±l  RUel   l  P Q¿Q ss 

0.0157 × 50 × 14 × (1.963 × 10 −5 ) × (100 − 30) × 0.99

  :

)RUFRSSHUURG m  7KHQ RU

hP 50 × 0.0157 = = 53.45 kAcs 14 × 1.963 × 10 −5

50 × 0.0157 350 × 1.963 × 10 −5

= 10.7

e±ml   e±l  RUel RUl P Q¿Q cu 

0.0157 × 50 × 350 × (1.963 × 10 −5 ) × (100 − 30) × 0.99

   : ,QWKLVFDVHOHQJWKDQGKHDWWUDQVIHUUDWHDUHPXFKKLJKHUEHFDXVHkcu!!kss

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4(Free convection). A hot plate of 100 cm height and 25 cm wide is exposed to atmospheric air at 25°C. The surface temperature of the plate is 95°C. Find the heat loss from ERWKWKHVXUIDFHVRIWKHSODWH$OVR¿QGWKHFKDQJHLQKHDWORVVLIWKHKHLJKWRIWKHSODWHLVUHGXFHG to 50 cm and the width is increased to 40 cm. Use the following relations: Nu = 0.59 (Gr.Pr)0.25 if Gr.Pr < 109, Nu = 0.10 (Gr.Pr)0.33, if Gr.Pr > 109 The properties of air are: U = 1.06 kg/m3, cp= 1004 J/kg-K, k = 0.029 W/m-K, Y î6 m2/sec &6(0DLQV  6ROXWLRQGivenL H FP PB FP Pta ƒ&ts ƒ& +HDWORVVIURPERWKVXUIDFHVRISODWHQ &DVH,L FP PB P g βH 3 ( Δt )

 ν2 95 + 25 = 60°C 7KHPHDQ¿OPWHPSHUDWXUH  2 1 1 1 = RU   E  = T 273 + 60 333 :HNQRZWKDWGr 

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GrPr 

?

μc p k

=

855

ρνc p

g β Δt L3 ν2

k ×

ρνc p k

=

g β Δt L3ρc p νk

9.81 × (1/ 333) × (95 − 25) × (1)3 × 1.06 × 1004



 



  î

(18.97 × 10 −6 ) × 0.029

GrPr  î!WKHÀRZLVWXUEXOHQW

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? 

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GrPr 



9.81 × (1/ 333) × (95 − 25) × (0.5)3 × 1.06 × 1004 (18.97 × 10 −6 ) × 0.029

  î



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?

Nu k 88.137 × 0.029 = = 5.112 W/m 2 K L 0.5 Q  hA't  îîî î h 

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4  (Heat Exchanger). $ WXEXODU JDV KHDWHU KHDWV DLU ÀRZLQJ DW WKH UDWH RI  kg/s from 20°C to 75°C using saturated steam condensing at 1.3 bar (saturation temperature, tsat ƒ& ,WLVSURSRVHGWRGRXEOHWKHÀRZUDWHRIDLUWRKHDWWKHVDPHIRUWKHVDPHULVHLQ temperature in the same gas heater. One way of doing this is to increase the condensing pressure of VDWXUDWHGVWHDP:KDWVKRXOGEHWKHSUHVVXUHQHHGHGLIWKHRYHUDOOKHDWWUDQVIHUFRHI¿FLHQWUHPDLQV WKHVDPHIRUERWKWKHRSHUDWLQJFRQGLWLRQV"6SHFL¿FKHDWRIDLU N-NJ. For steam, the following pressure and the corresponding saturation temperature are known:        p EDU t6DWƒ&       

 

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cpa  N-NJ.

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Chapter : 14

856

Q  m·acpato±ti



Saturated steam

th1 = th2 = 107°C t1 = 20°C

F

Air

t2 = 75°C

Fig. 14.9

QLVDOVRJLYHQE\ Q  UA Tm 

 ZKHUH 

Tm    

(th1 − tc1 ) − (th 2 − tc 2 ) θ1 − θ 2 = ln (θ1 / θ 2 ) ln [(th1 − tc1 ) / (th 2 − tc 2 )] (107 − 20) − (107 − 75) 55 =  55 ln [(107 − 20) / (107 − 75)] ln (87 / 32)

&DVH,,:KHQ māa  î NJV  Q  māaîcpato±ti  U ATm  îcpato±ti 6LQFHcp to±ti UHPDLQVVDPHDQGDOVRU AUHPDLQVVDPH ?

Q  îQ

RU

Tm   Tm î

RU Tm   î  $VVXPLQJFRQGHQVLQJWHPSHUDWXUHRIVWHDPDV th KDYH (t − 20) − (th − 75) 55 =    h 20 t − t ⎛ ⎞ ⎛ − 20 ⎞ ln ⎜ h ln ⎜ h ⎟ ⎝ th − 75 ⎠ ⎝ th − 75 ⎟⎠ RU

55 ⎛ t − 20 ⎞ ln ⎜ h = 0.5   110 ⎝ th − 75 ⎟⎠

7DNLQJORJRQERWKWKHVLGHVZHKDYH th − 20     th − 75 RU RU

th±   th± î th ±  th±

103.66  ƒ& 0.6488 1RZDSSO\LQJLQWHUSRODWLRQWR¿QGRXWSUHVVXUHpZHKDYH 159.77 − 158.8 p−6    165 − 158.8 7−6 RU

th   RU th 

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857

0.97 p−6 0.97    RU p  EDU $QV  6.2 1 6.2

RU

4(Forced convection).$GHFRUDWLYHSODVWLF¿OPRQDFRSSHUVSKHUHKDYLQJPP diameter is cured in an oven at 75° C. Upon removal from the oven, the copper sphere is subjected to an air stream at a pressure, temperature and velocity of 1 bar, 23qC and 10 m/s respectively. How long it will take for the sphere to cool down to 35° ? State the assumptions made and justify the method of analysis used. )RUFRSSHUWKHGHQVLW\VSHFL¿FKHDWDQGWKHUPDOFRQGXFWLYLW\DUHUHVSHFWLYHO\NJP3, 388 J/kg and 350 W/mK. The following correction for forced cxonvection mat be used. Nud = 2 + [0.4 Red1/2 + 0.06 Red2/3]Pr0.4 )RU DLU NLQHPDWLF YLVFRVLW\ WKHUPDO FRQGXFWLYLW\ DQG 3UDQGWO QXPEHU DW WKH PHDQ ¿OP WHPSHUDWXUHXQGHUFRQVLGHUDWLRQDUHî6 m2/s, 0.025 W/m-K and 0.708 respectively &6(0DLQV  6ROXWLRQGivenD PP Pts ƒ&ta ƒ&U PVtf ƒ& 7LPH UHTXLUHG IRU WHPSHUDWXUH DW VSKHUH WR GURS WRƒ&W

Copper sphere (ts = 75°C)

Air

ts + ta 75 + 23 = = 49°C 2 2 5H\QROGVQXPEHUIRUÀRZ tPHDQ 



Re 

UD 10 × 0.01 = 6439   ν 15.53 × 10 −6

ta = 23°C

3UDQGWOQXPEHUPr 

D = 10 mm

:HKDYH 

Fig. 14.10

1/2 2/3 0.4 Nud  2 + [0.4 Red + 0.06 Red ] Pr



 >î  @î 



 >  @î 

hD h × 0.01 or 48.05 = k 0.025 48.05 × 0.025 = 120.125 RU h  0.01 %LRWQXPEHUIRUWUDQVLHQWKHDWWUDQVIHUIURPVSKHUH hD 120.125 × 0.01 = = 3.432 × 10 −3  Bi  k 350 6LQFHBiHQWLUHVSKHUHFDQEHFRQVLGHUHGDVDlumped bodyIRUKHDWWUDQVIHU $VVXPHVSKHUHLVDWWHPSHUDWXUHt DQGIRUDVPDOOFKDQJHLQWHPSHUDWXUHdW RIVSKHUHZHKDYH  Q  hAt±ta dW 7KLVZLOOEHHTXDOWRUHGXFWLRQLQLQWHUQDOHQHUJ\RIVSKHUH  Q  ±m cp dt ? hAt ±ta dW  ±UQ cp dt  1RZ

RU

Nud 

d (t − ta ) hA . dτ    ± (t − ta ) ρν c p

⎡ dt = d (t − ta ) ⎤ ⎢ ⎥ ⎣since ta is constant ⎦

Chapter : 14

858

,QWHJUDWLQJERWKVLGHVZHJHW tf

∫



ts

d (t − ta ) hA  ± t − ta ρν c p

τ

∫ dτ 0

⎡ t f − ta ⎤ hA ln ⎢ . (τ) ⎥  − ρν c p ⎣ (ts − ta ) ⎦ 120.125 × [4π × 0.0052 ]   − ⎛4 ⎞ 8933 × ⎜ π × 0.0053 ⎟ × 388 ⎝3 ⎠ 0.0377 = − 0.02077   − 1.8148 1 ⎛ 35 − 23 ⎞ × ? 7LPHFRQVWDQWW  − ln ⎜ ⎟ ⎝ 75 − 23 ⎠ 0.02077    V $QV RU

4(Radiation). A 0.5 m diameter disc heater is horizontally placed and enclosed concentrically in a hemispherical shaped surface. The surface of the enclosure having an emissivity of 0.7 is maintained at 500 K. The disc heater, having emissivity of 0.8 is maintained at 1200 K. The diameter of the hemisphere is 2 m and the remaining base area enclosed is open to surroundings at 300 K and may be considered as black with reference to radiation network. Using thermal network method, calculate the heat exchange between heater and surroundings. Neglect convection heat transfer. Assume heater and hemispherical surface are opaque, diffuse and grey. &6(0DLQV  π 6ROXWLRQGivenDGLVF PA  î PH H T . 4 T .T .Dhs PRUR P?A SR Sî SP i +HDWH[FKDQJHEHWZHHQKHDWHUDQGHQFORVXUHQ E − Eb 2  Q  b1 ( Rth ) net 



 

 

Hemispherical surface (ε2 = 0.7)

σ (T14 − T24 ) 1 − ε1 1 − ε2 1 + + ε1 A1 A1 F1 − 2 ε 2 A2

2

5.67 × 10 −8 × (1200 4 − 500 4 ) (1 − 0.8) 1 (1 − 0.7) + + 0.8 × (0.1963) 0.1963 0.7 × (2 × π)



114029.4 ⎛ ⎞  ⎜ ⎝ 1.2735 + 5.094 + 0.06821⎠⎟



 î: N:

ε1 = 0.8 Disc heater

1

d= 0.5 m D (a)

Eb1

(Rth )1–2

(Rth )1 =

1 – ε1 A1 ε1

1

=

1 A1 F1–2 (b)

Fig. 14.11

surrounding (T3 = 300 K)

(Rth )2 1–ε = A ε2 2 2

2 Eb2

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4(Conduction). A spherical thermocouple of 2.5 mm diameter is used to measure the WHPSHUDWXUHRIDLUÀRZLQJLQDSLSH,QLWLDOO\ERWKWKHWKHUPRFRXSOHDQGWKHDLUDUHDWDWHPSHUDWXUH of 30°C. The air is heated to a temperature of 235°C and maintained at this temperature. Find WKH WLPH UHTXLUHG IRU WKH WKHUPRFRXSOH WR UHDFK ƒ&$OVR ¿QG RXW WKH WLPH FRQVWDQW RI WKH thermocouple and comment on the suitability of this thermocouple to measure unsteady state temperature. For thermocouple material take: density = 9000 kg/m3VSHFL¿FKHDW N-NJ. and thermal conductivity = 30 W/mK. &RQYHFWLYHKHDWWUDQVIHUFRHI¿FLHQWEHWZHHQWKHUPRFRXSOHVXUIDFHDQGWKHDLULV:P2K. &6(0DLQV  6ROXWLRQGiven: d  PPr PPti ƒ&tf ƒ&tf ƒ&  U  NJPc N-NJ.k :P.  h  :P. Spherical theromocouple 7LPHUHTXLUHGE\VSKHUHWRUHDFKƒ& 5HIHUWR)LJ hL %LRWQXPEHURIVSKHUHBi  c t¥ = 235°C k t = 30°C V 4 / 3 πr 3 r = )RUDVSKHUHFKDUDFWHULVWLFOHQJWKLc  = A 3 4πr 2 120 × (1.25 / 1000) = 1.67 × 10 −3 ? Bi  30 × 3

i

d = 2.5 mm

Fig. 14.12

6LQFHBi HQWLUHERG\can be considered as lumped. &RQVLGHULQJKHDWH[FKDQJHGEHWZHHQDLUDQGUHVXOWDQWLQFUHDVHLQLQWHUQDOHQHUJ\RIVSKHUH ZHJHW  m cp dt  h Astf±t dW hAs dt . dτ  − ρ Vc t − t∞

RU

,QWHJUDWLQJERWKVLGHVZHJHW tf



∫

ti

RU

RU

h As dt  − t − t∞ ρVc

τ

∫ dτ 0

⎛ t f − t∞ ⎞ ⎛ hA ⎞ ⎡ τ ⎤ ln ⎜   ⎜− s ⎟ τ = − ⎢ ⎟ ρVc ⎥ ⎝ ρ Vc ⎠ t − t ⎝ i ∞⎠ ⎢ ⎥ ⎣ hAs ⎦ ⎛ t f − t∞ ⎞ τ ln ⎜  − τ th ⎝ ti − t∞ ⎟⎠

Chapter : 14

860

⎛ ρVc ⎞  ZKHUHWWKLVFDOOHGthermal time constant ⎜ = ⎝ hAs ⎟⎠ +HUHWWK 

3 ρVc ⎛ 0.0025 ⎞ 0.4 × 10 = 9000 × ⎜ ×  V hAs 120 ⎝ 2 × 3 ⎟⎠

7LPHUHTXLUHGE\WKHWKHUPRFRXSOHWRUHDFKƒ& ⎛ t f − t∞ ⎞  W  ±WWKOQ ⎜ ⎝ t − t ⎟⎠ ∞

i



⎛ 200 − 235 ⎞  22.1 s $QV   ±îOQ ⎜ ⎝ 30 − 235 ⎟⎠

,QRUGHUWRPHDVXUHXQVWHDG\WHPSHUDWXUHWKHWKHUPRFRXSOHVKRXOGWDNHleast amount of time to reach environment temperature, tf Thus the value of time constant should be as small as possible. ,QWKLVFDVHLIWKHV\VWHPWHPSHUDWXUHFKDQJHVIDVWHUWKDQWKHWLPHUHTXLUHGE\WKHUPRFRXSOHWR UHDFKWKDWWHPSHUDWXUHWKHQWKHUPRFRXSOHLVXQVXLWDEOHIRUWHPSHUDWXUHPHDVXUHPHQW7KHUHIRUHLI WHPSHUDWXUHRIDLUFKDQJHVIURPƒ&EHIRUHVLWLVnot a good measuring device+HQFHLQ WKLVFDVHEHFDXVHXQVWHDG\WHPSHUDWXUHFKDQJHVHYHU\VHFRQGZKLOHKHUHWLPHFRQVWDQWWth LVDQODUJH DVV7KHWHUPRFRXSOHLQTXHVWLRQLVnot suitableIRUPHDVXULQJXQVWHDG\WHPSHUDWXUH$QV 4(Heat Exchanger). $FRXQWHUÀRZFRQFHQWULFWXEHVKHDWH[FKDQJHULVGHVLJQHG WRKHDWZDWHUIURPƒ&WRƒ&XVLQJKRWRLOÀRZLQJWKURXJKWKHDQQXOXV7KHRLOWHPSHUDWXUH gets reduced from 160°C to 140°C. The nominal diameter of the inner tube is 20 mm and the FRUUHVSRQGLQJRYHUDOOKHDWWUDQVIHUFRHI¿FLHQWLV:P2-K. The heat transfer rate from the oil is 3000 watts. Determine the length of the exchanger. Because of fouling after some days the outlet WHPSHUDWXUHRIZDWHUUHGXFHWRƒ&IRUWKHVDPHÀRZUDWHVDQGVDPHLQOHWFRQGLWLRQV'HWHUPLQH the outlet temperature of oil, the fouling factor and the new heat transfer rate. Sketch the heat exchanger arrangement.

&6(0DLQV 

6ROXWLRQGiven: tc ƒ&tc ƒ&th ƒth ƒ&di PP Ph  :P.Q : tc2 = 80°C th1 = 160°C Oil

Water 20 mm Water Counter flow heat exchanger Fig. 14.13

/HQJWKRIWKHKHDWH[FKDQJHUL  

θ1 − θ 2  ln (θ1 /θ 2 ) (th1 − tc 2 ) − (th 2 − tc1 )   ln [(th1 − tc 2 ) / (th1 − tc1 )]

T m 



 

(160 − 80) − (140 − 20) ln [(160 − 80) / (140 − 20)]



 

(80 − 120) = 98.65°C ln (80 /120)

tc1 = 20°C Oil th2 = 140°C

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1HZRXWOHWWHPSHUDWXUHRIWKHt c c %HFDXVHRIIRXOLQJRXWOHWWHPSHUDWXUHRIZDWHUUHGXFHVWRƒ& For initial condition: +HDWWUDQVIHUUDWHQ  mw cpw±  mo cpo±  RU mwîcpwî  mo cpoî For new condition:  Q  mw cpw±  mo cpo ±t cc  ZKHUHtccLVQHZRXWOHWWHPSHUDWXUHRIRLO 'LYLGLQJii E\i ZHJHW 160 − tc′2 45    20 60 RU ±t cc  RUtcc ƒ& $QV 1HZKHDWWUDQVIHUUDWHQc 2OGKHDWWUDQVIHUUDWHQ mw cpw±  : 3000 = 50 W/°C RU mwcpw  60 1HZKHDWWUDQVIHUUDWH Q c mw cpw±  î : $QV $OVR1HZKHDWWUDQVIHUUDWHQ c hcATm (160 − 65) − (145 − 20) RU   hcîSîî î ln [(160 − 65) / (145 − 20)] 

  hcîî

( − 30) = h ′ × 6.658 ( − 0.2744)

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RU

hc 

⎡)RXOLQJ RU VFDOLQJ  The phenomenon of rust formation and deposition of fluid ⎤ ⎢impurities in the tubes of a heat exchanger, during its normal operation, is ⎥ ⎢ ⎥ ⎢called IRXOLQJ ⎥ ⎢ ⎥ ⎣The reciprocal of scale heat transfer, hs , is called the fauling f actor , R f . ⎦ 

Rf 

1 2 m k/W hs



Rf 

1 1 − hdirty hclean



 

861

1 1 1 1 − = −  î±P.: h ′ h 337.94 500

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Chapter : 14

862

4(Conduction). 7KHWHPSHUDWXUHRIDJDVÀRZLQJWKURXJKDSLSHZDVPHDVXUHGE\ DPHUFXU\LQJODVVWKHUPRPHWHUGLSSHGLQDQRLO¿OOHGVWHHOWXEHZHOGHGUDGLDOO\WRWKHSLSHOLQH The thermometer indicates a temperature lower than the gas temperature. How large is the error in the temperature measurement if the thermometer reads 85°C and the temperature of the pipe wall is 40°C? The steel tube is 125 mm long and has a 1.5 mm thick wall. The thermal conductivity RIWKLVWXEHPDWHULDOLV:P.DQGWKHORFDOKHDWWUDQVIHUFRHI¿FLHQWEHWZHHQWKHJDVDQGWKH tube is 23.5 W/m2-K. In what way the thermometric error can be reduced? &6(0DLQV 6ROXWLRQ5HIHUWR)LJ  l  PP Pt PP Pk :P.  h  :P.3LSHZDOOWHPSHUDWXUHto ƒ&  7HPSHUDWXUHRIJDVtf ƒ& (UURULQWHPSHUDWXUHPHDVXUHPHQW Thermometer Pipeline t0 Oil

l

Gas

δ

tf

Fig. 14.14

/HW tl  7HPSHUDWXUHUHFRUGHGE\WKHUPRPHWHUDWWKHERWWRPRIWKHZHOO 7R¿QGtl XVHWKHUHODWLRQ tl − t f 1        >(TQ @  to − t f cosh (ml ) ZKHUH

m 

h , ZKHUHGLVWKHWKLFNQHVVRIWKHWXEH kδ

RU

m 

23.5 = 16.726 56 × 0.0015

1RZLQVHUWLQJWKHYDOXHVLQDERYHHTQZHJHW 85 − t f 1 1 = = 0.2436    cosh (16.726 × 0.125) cosh (2.09) 40 − t f RU  RU ?

±tf  ±tf î   ±tf



±  tfRUtf ƒ& (UURU  tf±tl ± ƒ&

$QV 

GATE and UPSC Examinations’ Questions ...

863

7RUHGXFHWKHWHPSHUDWXUHPHDVXUHPHQWHUURUmlVKRXOGEHODUJHQHFHVVLWDWLQJWKHIROORZLQJ i /DUJHYDOXHRIhKHDWWUDQVIHUFRHI¿FLHQW  ii 6PDOOYDOXHRIk7KHUPDOFRQGXFWLYLW\  iii /DUJHDQGWKLQZHOOWKHSRFNHWSURWUXGLQJVPDOOWXEH PD\EHSODFHGREOLTXHO\LQFOLQHG LIQHFHVVDU\WRSURYLGHDODUJHLQVHUWLRQRIWKHUPRPHWHU$QV 4(Conduction). A small hemispherical oven is built of two insulating materials. 7KHLQQHUOD\HULVRI¿UHEULFNPPWKLFNDQGRXWHUOD\HULVRIPDJQHVLDPPWKLFN 7KHLQQHUVXUIDFHRIWKHRYHQLVDWƒDQGWKHKHDWWUDQVIHUFRHI¿FLHQWIRUWKHRXWHUVXUIDFHLV 10 W/m2-K. The room temperature is 20°C. Calculate the heat loss through hemisphere with the LQVLGHUDGLXVLVP7KHWKHUPDOFRQGXFWLYLWLHVRI¿UHEULFNDQGPDJQHVLDDUHDQG 0.05 W/m-K respectively. Also calculate the temperature at contact between insulating materials and at the outer surface. &6(0DLQV 6ROXWLRQGivenriA ProA   P riB ProB   Pti ƒ&tf ƒ& kA :P.kB :P.h :P. +HDWORVVWKURXJKKHPLVSKHUH 5HIHUWR)LJ /HW to  7HPSHUDWXUHRIRXWHUVXUIDFH &RQVLGHULQJWKHHOHPHQWRIUDGLXV r DQGWKLFNQHVVdrZHKDYH +HDWWUDQVIHUWKURXJKWKLVHOHPHQW dt dt = − k × 2πr 2 ×  Q  ±kA dr dr dr k − dt    RU Q 2πr 2 ,QWHJUDWLQJZHJHW 1 ⎡1 1 ⎤ k − ⎥   (ti − to )  ⎢ 2π ⎣ ri ro ⎦ Q RU

Q 

1 ⎛1 1⎞ −  RWK  2 k π ⎜⎝ ri ro ⎟⎠ 7RWDOWKHUPDOUHVLVWDQFH  RWK WRWDO  RWK ARWK BRFRQY

RU





RWK WRWDO 

dr

1 2 πk A

1 ⎤ 1 ⎡1 ⎢ r − r ⎥ + 2 πk oA ⎦ B ⎣ iA

Element

B

t0

ti − to 1 ⎛1 1⎞ − 2k π ⎜⎝ ri ro ⎟⎠

?7KHUPDOUHVLVWDQFHRIKHPLVSKHUH



t¥ = 20°C

r

A

Fire brick (A)

125 mm 40 mm 0.6 m

Fig. 14.15

1 ⎤ 1 ⎡1 ⎢ r − r ⎥ + hA oB ⎦ o ⎣ iB

1 1 ⎤ 1 1 ⎤ ⎡ 1 ⎡ 1 − + − 2π × 0.31 ⎢⎣ 0.6 0.725 ⎥⎦ 2π × 0.05 ⎢⎣ 0.725 0.765 ⎥⎦ +

1 10 × 2π × (0.765) 2

Chapter : 14

864

   N: ?+HDWORVVWKURXJKWKHKHPLVSKHUH t −t (800 − 20) W   N: $QV  Q  i ∞ = ( Rth ) total 0.4 7HPSHUDWXUHDWWKHFRQWDFWtc t −t 1RZ Q  o ∞ Rconv. RU $OVR RU

to  QîRFRQYtf î ƒ& t −t Q  c o ( Rth ) B tc  Q.Rth Bto î  ƒ& $QV

4 (Radiation exchange between surfaces). The net radiation from the surfaces of two parallel plates having equal emissivities of 0.8 and at different temperatures of T1 and T2 (T1 > T2) is to be reduced by 99%. How many numbers of radiation screens having equal emissivities of 0.05 are to be placed between the plates to achieve the reduction in heat exchange. *$7( 6ROXWLRQGiven(PLVVLYLW\RIHDFKSDUDOOHOSODWHVXUIDFHH 1HWUDGLDWLRQWREHUHGXFHG  ± RU(PLVVLYLW\RIHDFKUDGLDWLRQVFUHHQH  5HIHUWR)LJ Parallel Plates T1

T2

e1 = 0.8

e2 = 0.05

e1 = 0.8

n screens/shields 1 – e1 Req 1 AF12 e1

1 – e1 1 e1 A2F21

Fig. 14.16. Equivalent circuit diagram.

1XPEHURIUDGLDWLRQVFUHHQVUHTXLUHGn For plate: 

5VKDSH 

1 =1 AF

⎛ 1 − ε1 ⎞ ⎛ 1 − 0.8 ⎞ 5VXUIDFH  î ⎜ ⎟ = 2 × ⎜ 0.8 ⎟   ε ⎝ ⎠ ⎝ 1 ⎠

For screen: 5VKDSH  (1 + n) 

1 = (1 + n) × 1 = n + 1 AF

⎛ 1 − ε2 ⎞ ⎛ 1 − 0.05 ⎞  n = 2n × ⎜ 5VXUIDFH  n ⎜ ⎝ 0.05 ⎟⎠ ⎝ ε 2 ⎠⎟

GATE and UPSC Examinations’ Questions ...

865

7RWDOUHVLVWDQFHZLWKRXWVFUHHQV   7RWDOUHVLVWDQFHZLWKnVFUHHQV  

   >n n@ n n Qscreens Q without screens

 

ΔT

Rscreens

×

Rwithout screens  ΔT

RU

 

1.5 39 n + 2.5

RU

n 

1.5 = 150 0.01

RU

n 

150 − 2.5 = 3.8   $QV 39

45HFWDQJXODU¿Q $UHFWDQJXODU¿QRIOHQJWKFPZLGWKFPDQGWKLFNQHVV PPLVDWWDFKHGWRDVXUIDFHDWƒ&7KH¿QLVPDGHRIDOXPLQLXPN :/P.) and is H[SRVHGWRDLUDWƒ&7KH¿QHQGLVXQLQVXODWHGDQGFDQORVHKHDWWKURXJKLWVHQGDOVR7KH FRQYHFWLRQKHDWWUDQVIHUFRHI¿FLHQWEHWZHHQWKH¿QVXUIDFHDQGDLULV:P2.. Determine L  7KHWHPSHUDWXUHRIWKH¿QDWFPIURPWKHEDVH LL  7KHUDWHRIKHDWWUDQVIHUIURPWKH¿Q LLL 7KH¿QHI¿FLHQF\ ,(6 6ROXWLRQ Given /HQJWK RI ¿Q l   FP   FP :LGWK RI ¿Q b   FP   P 7KLFNQHVVRI¿Qy PP P7KHUPDOFRQGXFWLYLW\k :P.7HPSHUDWXUHRIVXUIDFH t ƒ&7HPSHUDWXUHRIDLUta ƒ&&RQYHFWLRQKHDWWUDQVIHUFRHI¿FLHQWEHWZHHQWKH¿QVXUIDFH DQGDLU :P. i  7HPSHUDWXUHRIWKH¿QDWFPIURPWKHEDVHt 1RZ +HUH DQG ?    

⎡ P = perimeter of the fin, ⎤ ⎢ A = area of cross section.⎥ ⎣ cs ⎦ P  by    P

m 

hP kAcs

AFV  bîy î î±P m 

15 × 0.604   204 × (6 × 10−4 )

h 15  î±   km 204 × 8.6

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h [sin h{m(l − x)}] km  h cosh (ml )} + [sin h(ml )] km x  lZHKDYH

θ   θo

cosh{m(l − x)} +

>(TQ @

Chapter : 14

866

  

RU 

RU 

RU  RU 

m ( l −l ) e m ( l −l ) + e − m ( l − l ) − e − m (l − l ) ⎤ h ⎡e + ⎢ ⎥ km ⎣ 2 2 θl ⎦   ml − ml ml − ml ⎤ θo ⎡ e +e h e −e + ⎥ km ⎢⎣ 2 2 ⎦

t30 − ta 1   8.6 × 0.3 −8.6 × 0.3 to − t a e +e + 8.5 × 10−3 2

⎡ e8.6 × 0.3 − e−8.6 × 0.3 ⎤ ⎢ ⎥ 2 ⎣ ⎦

t30 − 30 1   300 − 30 ⎛ 13.2 + 0.0757 ⎞ −3 ⎛ 13.2 + 0.0757 ⎞ ⎜ ⎟ + 8.5 × 10 ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ t30 − 30 1     300 − 30 6.638 + 0.0558 t  ±  ƒ&$QV 

ii  7KHUDWHRIKHDWWUDQVIHUIURPWKH¿QQ¿Q   

Q¿Q 

  

 

⎡ tan h(ml ) + h ⎤ ⎢ km ⎥  Ph k Acs (to − ta ) ⎢ ⎥ h ⎢⎣1 + km . tan h (ml ) ⎥⎦

>(TQ @

0.604 × 15 × 204 × (6 × 10−4 ) (300 − 30)  ⎡ e8.6 × 0.3 + e −8.6 × 0.3 ⎤ + 8.5 × 10−3 ⎥ ⎢ 8.6 × 0.3 −8.6 × 0.3 −e ⎢ e ⎥ 8.6 × 1.3 − 8.6 × 0.3 ⎢ ⎡e ⎤⎥ +e ⎢1 + (8.5 × 10−3 ) ⎢ 8.6 × 0.3 ⎥ −8.6 × 0.3 ⎥ −e ⎣e ⎦ ⎦⎥ ⎣⎢



  

⎡ 13.2 + 0.0757 −3 ⎤ ⎢ 13.2 − 0.0757 + 8.5 × 10 ⎥ ⎥   î î ⎢ ⎢1 + 8.5 × 10 −3 ⎧13.2 + 0.0757 ⎫ ⎥ ⎨13.2 − 0.0757 ⎬ ⎥ ⎢ ⎩ ⎭⎦ ⎣

  

⎡ ⎤ 1.02   îî ⎢ ⎥ −3 ⎣1 + 8.5 × 10 (1.0115) ⎦

  

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4(Radiation). Find the shape factor F1–2IRUWKHFRQ¿JXUDWLRQVKRZQLQWKH¿JXUH below: *$7(

GATE and UPSC Examinations’ Questions ... 6m

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6m

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2

2

3

3

m

m

6

6

m

m

1

Hypothetical surface

3

1

X=6m

Fig. 14.17

Fig. 14.18

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F  F ±F ±  AF  AF F 

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A2 6×6 × 0.075   $QV  F   A1 21 6 × 6

4(Heat-exchanger). A heat exchanger tube with inner and outer diameters 2.118 FPDQGFPUHVSHFWLYHO\LVHTXLSSHGZLWKHTXDOO\VSDFHG¿QVRIXQLIRUPWKLFNQHVVSODFHG ORQJLWXGLQDOO\DORQJWKHWXEH7KH¿QVDUHFPORQJLQWKHUDGLDOGLUHFWLRQDQGDUHFP WKLFN%RWKWKHWXEHDQGWKH¿QVDUHPDGHRIVWHHOZLWKN :Pƒ&7KHLQVLGHDQGRXWVLGHKHDW WUDQVIHUFRHI¿FLHQWVDUHDQG:Pƒ&UHVSHFWLYHO\ :KDWLVWKHRYHUDOOKHDWWUDQVIHUFRHI¿FLHQWIRUWKHH[FKDQJHEDVHGRQWKHRXWHUH[SRVHG VXUIDFHDUHD&RPSDUHWKLVUHVXOWZLWKWKDWIRUWKHVDPHWXEHZLWKRXW¿QVVXEMHFWWRWKHVDPHKHDW WUDQVIHUFRHI¿FLHQWV ,(6 6ROXWLRQ5HIHUWR)LJ Givend FPRUr FPD FPRUR FP1RRI¿QVn  l¿QUDGLDO  FPt¿Q FPk :Pƒ& h :Pƒ&h :Pƒ&

Chapter : 14

868

0 .1 6

cm

Fins

2.5 cm d D (a) Tube without fins

(b) Tube with 20 fins Fin A23 Area exposed

A21

A22

(c) Section at any fins

Fig. 14.19.

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ln( R /r ) 1 1 1 + +  %DVHGRQH[SRVHGDUHD    h1 A1 2πk h2 A2 U 0 A2

 

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A  SR±nît¿Q Sî±î FP A  îl¿Q î FP A  nt¿Q î FP

?

A  AAA  FP

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A 1 A 1 1   2 . + 2 ln ( R / r ) +  U0 A1 h1 2πk h2 12.979 × 10 12.979 1 × + 6.65 1130 2π × 45

−2

 1.27 ⎞ 1 ln ⎛ + ⎝ 1.059 ⎠ 255



 



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GATE and UPSC Examinations’ Questions ...

869

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A 1 A 1 1 + 2 ln ( R /r ) +   2 A1 h1 2πk h2 U0



 

2πR 1 2πR 1 . + ln ( R /r ) + 2πr h1 2πk h2



 

R 1 R 1 . + ln ( R /r ) + r h1 k h2



 

1.27 × 10 1.27 1 × + 1.059 1130 45



  î±î±î±



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RU

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1.27 ⎞ 1 ln ⎛ + ⎝ 1.059 ⎠ 255

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4(Heat exchanger).&RROLQJZDWHUDWDVWHDG\UDWHRINJVÀRZVWKURXJKDQ inner tube having inner diameter of 25 mm and length of 10 m of tube-in-tube condenser. The mean inlet temperature of cooling water is 10°C. Saturated steam condenses in the annulus at a uniform rate such that the inner surface temperature of the tube is constant throughout the length of the WXEHDWƒ&7KHDYHUDJHFRQGHQVLQJVLGHKHDWWUDQVIHUFRHI¿FLHQWLV:P2K. Neglect the thickness of the heat exchanger tube. Calculate the effectiveness of the heat exchanger and the exit water temperature. Properperties of water are given below: 6SHFL¿FKHDW -NJ. Density = 990 kg/m3 '\QDPLFYLVFRVLW\ î–3 Pascal.sec Thermol conductivity = 0.57 W/mK You may use the relation Nu = 0.023 Re0.8 Pr0.4 *$7( 6ROXWLRQ5HIHUWR)LJ Given:0DVVRIFRROLQJZDWHUmw NJVd PP P twi ƒ& .two ƒ& .ho :P.cp -NJ. U NJPP î±3DVFDOVHFk :P.1X RePr Steam in

25 mm dia

Water in (mw = 0.5kg/s twi = 10°C)

Condensate out

Water out (two = 40°C)

10 m

Fig. 14.20 Tube-in-tube condenser.

(IIHFWLYHQHVVRIKHDWH[FKDQJHUH 5H\QROGVQXPEHU5H 

ρVd μ

Chapter : 14

870

(0.5 / 990)  PV π × (0.025) 2 4 990 × 1.0289 × 0.025    0.8 × 10 −3

9HORFLW\RIZDWHU9

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0.8 × 10 −3 × 4180   0.57  5H 3U 



    



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1X



hi 1  UA



1 1 + hi Ai ho Ao

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⎞ 1 1 ⎞ ⎛ 1 ⎛ + × ⎝ 4259.95 10000 ⎠ ⎜⎝ π × 0.025 × 10 ⎟⎠

RU

UA

 :.



178



UA 2346.26 =   Cmin 4180 × 0.5

+HQFHHIIHFWLYHQHVVH ±e±178 ±e± $QV 4  (Heat exchanger). Find the condenser surface area required for a surface condenser dealing with 25000 kg of saturated stream per hour at a pressure of 0.5 bar. Temperature of condensing water is 25°C. Cooling water is heated from 15°C to 25°C while passing through the FRQGHQVHU$VVXPHDQRYHUDOOKHDWWUDQVIHUFRHI¿FLHQWRIN:/P2.. The condenser has 2 water passes with tubes of 19 PP OD and 1.2 PP thickness. Find the length and number of tubes per pass. Assume velocity of water is 1 m/s. Assume correction factor for 2 tube pass exchanger as 0.86. At 0.5 bar saturation temperature is 32.55°& and latent heat is 2560 N-NJ6SHFL¿FKHDWRIZDWHULV ,(6 4.18 N-NJ. and density is 1000 NJP3. 6ROXWLRQ5HIHUWR)LJ Given m s NJKtVDWDWEDU  ƒ&7HPSHUDWXUHRIFRQGHQVLQJZDWHU ƒ& tc  ƒ&tc  ƒ&U N:P .1RRISDVVHV do PP Pdi ±   î  PP P9HORFLW\RIZDWHU9 PV &RUUHFWLRQ IDFWRU IRU  WXEH SDVV H[FKDQJHU )   h fg  DW  EDU    N-NJ cpw N-NJ.U NJP &RQGHQVHUVXUIDFHDUHDA .

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GATE and UPSC Examinations’ Questions ...

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fg ƒ&

pw

f

25000 >± @ 3600   îN:

  

 

  

Temp.

th1, th2 = tsat = 32.55°C tc2 = 25°C

f

θ1

Steam

2

32.5°C

1

θ2

Cooli

ng w

ater

25°C

3

tc1 = 15°C (a)

(b)

Fig. 14.21 Surface condenser.

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RU  RU 

c

c

   mā wîî±

17.997 × 103 māw   NJV 4.18 × (25 − 15)

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T  ± ƒ&

  

T  ± ƒ&

? 

T m 

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î  Aîî

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A 

17.997 × 103  P$QV 0.86 × 10 × 8.72

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π māw  ⎡⎢ d 22 × V × ρ⎤⎥ × N p ⎣4 ⎦ π    ⎛ × 0.01662 × 1 × 1000⎞ × N p ⎝4 ⎠ Np 

430.55 × 4 M$QV π × 0.01662 × 1000

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4(Free convection).(VWLPDWHWKHFRHI¿FLHQWRIKHDWWUDQVIHUIURPDYHUWLFDOSODWH 2 PîP to the surrounding air at 25°C. The plate surface temperature is 150°C. Also calculate WKHUDWHRIKHDWWUDQVIHUIURPWKHSODWH)RUDLUDVVXPHWKHNLQHPDWLFYLVFRVLW\DVî–5 m2/s. 7KHSURSHUWLHVRIDLUDW¿OPWHPSHUDWXUHDUHGHQVLW\NJP3VSHFL¿FKHDWN-NJ. WKHUPDOFRQGXFWLYLW\î–2 W/mK, Prandtl no. 0.69. The constants C and n in Nusselt no. equation are 0.15 and 1/3 respectively. *$7( 6ROXWLRQGivenL Pts ƒ&tf ƒ&)LOPWHPSHUDWXUHtf 

150 + 25  ƒ& 2

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4(Forced Convection). A refrigerated truck carrying stuff is speeding on a highway at 90 km/h in a desert area where ambient air temperature is 55°C. The body of the truck may be modelled as a rectangular box measuring 11 m long, 4 m wide and 3 m high. Consider the boundary layer to the four walls to be turbulent and heat transfer only from the four surfaces. The wall surfaces of the truck are maintained at 10°&$VVXPHWKHÀRZWREHSDUDOOHOWRPORQJVLGH7KH thermo- physical properties at the mean temperature at tf 55 + 10 are : = 32.5°C 2 U NJP, cp N-NJ. k î±:P.Q î±PVDQGPr 

GATE and UPSC Examinations’ Questions ...

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Workout the following: i  Heat loss from the four surfaces. ii  Tonnage of refrigeration needed. iii Power required to overcome the resistance acting on four surfaces. 6ROXWLRQ Given  9HORFLW\ U   NPK 

,(6

90 × 1000    PV$PELHQWWHPSHUDWXUH 3600

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ReL 

25 × 11 UL =  î ν 16 × 10 −6

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Nu   Re Pr 

RU 

hL   î    k

RU 

h 

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2.673 × 10 −2 k × 19640 = × 19640  :P. L 11

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Chapter : 14

4(Free convection). An electrically heated sphere of 1.5 FP diameter is cooled in a quiescent medium of air at 315 .. In order to maintain the surface temperature of the sphere at 385 ., estimate the amount of heat to be supplied by the electrical heater. Ambient temperature of air is 20°C. 315 + 385 The temperature of air at, Tf = = 350 K are: 2 Kinematic viscosity, Q î–5 m2/s Prandtl number Pr = 0.697 Thermal conductivity, k = 0.03 :/P. 1 1  î–3 K–1 &RHI¿FLHQWRIWKHUPDOH[SDQVLRQE = = T f 350 Use the relationship Nu = 2 + 0.43 *Uî3U 1/4

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  Nu  

     

L3βg Δt D3βg Δt =  ν2 ν2 (0.015)3 × (2.86 × 10 −3 ) × 9.81 × (385 − 315)   (2.076 × 10 −5 ) 2

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4(Forced convection):DWHUDWƒ&ÀRZVQRUPDOWRWKHD[LVRIDFLUFXODUWXEH with a velocity of 1.5 PV. The diameter of the tube is 25 PP. Calculate the average heat transfer FRHI¿FLHQWLIWKHWXEHVXUIDFHLVPDLQWDLQHGDWDXQLIRUPWHPSHUDWXUHRIƒ&$OVRHVWLPDWHWKH 20 + 80 heat transfer rate per unit length of the tube. Properties of water tf = = 50°C are: 2 Fp = 4.1813 N-NJ.   6SHFL¿FKHDW    Kinematic viscosity,

Q  î–6 m2/s

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k = 0.6395 :/P.

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Pr = 3.68 U = 990 NJ/P3

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Pf 20°C  î– 3 NJPV

  

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1/4

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Fig. 14.22. Arrangement of the furnace.

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r)

th2 = 40°C tc1 = 20°C

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4(Radiation exchange between surfaces). An electrically heated industrial furnace cavity is modelled in the form of a cylinder having diameter 10 FP and length 20 FP. It is opened at one end of surroundings that are at a temperature of 300 K. The electrically heated sides and the bottom of the cavity which are well insulated and may be approximated as black bodies are maintained DWDWHPSHUDWXUHRI.DQG.UHVSHFWLYHO\%\VKRZLQJWKHVNHWFKRIWKHIXUQDFH¿QGWKH power required to maintain the surface at this condition. Take shape factor from the bottom surface to surroundings as 0.06. *$7( 6ROXWLRQGivenD FP PL FP PTamb .T .T T  .6KDSHIDFWRU  7KHsketch of the furnaceLVVKRZQLQ)LJ 3RZHUUHTXLUHG *LYHQ  F±  1RZF±F±F±  i RU F± RU F± DQG F±F±F±  F± $OVR 

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T2 = T3 = 1800 K

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3

2

2000 K 1 D = 10 cm

Fig. 14.24. Sketch of the furnace.

Chapter : 14

878

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4 (Radiation exchange between surfaces). The cross-section of a very long black body enclosure consists of a semicircle with its diameter D at the base. The temperature of semicircle is 1000 K and that of diameter is 500 K. Determie the shape factors for diameter-semicircle combination DQGWKHUDGLDWLRQKHDWWUDQVIHUUDWHSHUXQLWZLGWKLQWHUPVRI' 6WHIDQ%ROW]PDQQFRQVWDQW î–8 :P.. *$7( 6ROXWLRQ5HIHUWR)LJ Given6HPLFLUFOHGLDPHWHU D T .T .Vb î±:P. 6KDSHIDFWRUV

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4 (Forced convention). Air with an average velocity of 10 m/s at 300 K enters a copper tube of 11.2 PP diameter and 2.5 P length. The tube wall is maintained at 372 K by condensing steam at atmospheric pressure. Using LMTD method, determine the temperature of air at outlet of tube. Average properties of air are: k = 0.02624 W/mK, cp = 1.005 kJ/kgK U = 1.174 kg/m3, Q î–5 PV Pr = 0.7 0.668 (d /L) RePr Nu = 3.66 + R < 2300 1 + 0.04 (d /RePr ) e Nu = 0.023 Re0.8 Pr0.4Re > 2300 ZKHUHGDQG/DUHGLDPHWHUDQGOHQJWKRIWXEHUHVSHFWLYHO\$VVXPHKHDWWUDQVIHUFRHI¿FLHQW to be constant and neglect conduction thermal resistance of copper. ,(6 .

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[(126) – (298) ] ⎛ 1 – 1⎞ ⎤ ⎜ ⎟⎥ ⎝ 0.028 ⎠⎦

– 195703.25 = 3828.2 kJ/h (35.71 + 0.444 × 34.71)

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q  εσ (T14 – T24 )

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ie 

 



 

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Chapter : 14

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a 

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c  HH

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 ie

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d  P

dt dx

dt  ⎛ 0.14 × A × ⎞ ⎝ dx ⎠ diatomite

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dx 

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907

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Chapter : 14

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⎛ 627 + 273 ⎞ ∝ 34 ∝ 81 $QV  (PLVVLYHSRZHULVSURSRUWLRQDOWRT ie ∝ ⎜ ⎝ 27 + 273 ⎟⎠ 

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909

 EXPLANATION:  ,UUDGLDWLRQRQDVPDOOWHVWVXUIDFHSODFHGLQVLGHDKROORZEODFNVSKHULFDOFKDPEHU VT î±î :P$QV 4$FURVVIORZW\SHDLUKHDWHUKDVDQDUHDRIP7KHRYHUDOOKHDWWUDQVIHUFRHIILFLHQWLV :P.DQGKHDWFDSDFLW\RIERWKKRWDQGFROGVWUHDPLV:.7KHYDOXHRI178LV  a   b   c   d    $QV c   EXPLANATION: AU    178  , A DUHD P Cmin  

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Outer casing

1000°C T1 120°C k = 3 W/mK 0.3 W/mK 0.3 m

ta = 40°C

0.3 m

Fig. 13

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t1 – t2 1000 – 120 880 Q = = = 800   Δx1 Δx2 0.3 0.3 A 1.1 + + 3 0.3 k1 k2 120 – 40 Q 1 800 , or 800 × = 80, and h = = 10 W/m 2 K $QV   1/h h 80 A

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∫

Chapter : 14

910

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GATE and UPSC Examinations’ Questions ...

911

 $QV b   EXPLANATION: k 1.0 W = 43.6 2 , DQGIRU   )RUFRQVWDQWKHDWIOX[DVSHU%D\OH\ h = 4.364 = 4.364 × D 0.1 m K k 1.0 W FRQVWDQWZDOOVXUIDFHWHPSHUDWXUH h = 3.66 = 3.66 × = 36.6 2 $QV D 0.1 m K 47ZRODUJHSDUDOOHOJUD\SODWHVZLWKDVPDOOJDSH[FKDQJHUDGLDWLRQDWWKHUDWHRI:P ZKHQWKHLUHPLVVLYLWLHVDUHHDFK%\FRDWLQJRQHSODWHLWVHPLVVLYLW\LVUHGXFHGWR 7HPSHUDWXUHVUHPDLQXQFKDQJHG7KHQHZUDWHRIKHDWH[FKDQJHVKDOOEHFRPH  a  :P b  :P c  :P d  :P  $QV b   EXPLANATION: 2 2 –1 1000 × ⎛⎜ – 1⎞⎟ 3 ε1 0.5 ⎝ ⎠ = 3 × 10 = 600 W $QV   Q î = 1 1 5 ⎛1 ⎞ ⎛ 1 ⎞ m2 –1+ –1+1 ⎜ ε – 1⎟ + ⎜ ε – 1⎟ + 1 0.5 0.25 ⎝ 1 ⎠ ⎝ 2 ⎠ 47ZRORQJSDUDOOHOSODWHVRIVDPHHPLVVLYLW\DUHPDLQWDLQHGDWGLIIHUHQWWHPSHUDWXUHV DQGKDYHUDGLDWLRQKHDWH[FKDQJHEHWZHHQWKHP7KHUDGLDWLRQVKLHOGRIHPLVVLYLW\ SODFHGLQWKHPLGGOHZLOOUHGXFHUDGLDWLRQKHDWH[FKDQJHWR 1 1 3 3 b   c   d   a   2 4 10 5  $QV c  EXPLANATION:   5HGXFWLRQLQUDGLDWLRQKHDWH[FKDQJHGXHWRLQWURGXFWLRQRIVKLHOG 2 2 –1 –1 ε1 3 0.5    = = $QV 0.5 0.75 10 ⎛ 1 – ε1 ⎞ ⎛ 1 – ε2 ⎞ × + × + 2 2 2 2⎜ ⎟+2⎜ ε ⎟+2 0.5 0.25 ⎝ ε1 ⎠ ⎝ 2 ⎠ 47KH LQVXODWHG WLS WHPSHUDWXUH RI D UHFWDQJXODU ORQJLWXGLQDO ILQ KDYLQJ DQ H[FHVV RYHU DPELHQW URRWWHPSHUDWXUHRIToLV θo θo θ tanh (ml )  a  ToWDQKml  b   c  o  d  sinh (ml ) cosh (ml ) (ml )  $QV c   EXPLANATION: 

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912

Chapter : 14

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914

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