HVACR Principles and Applications 3031452666, 9783031452666

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Nuggenhalli S. Nandagopal

HVACR Principles and Applications

HVACR Principles and Applications

Nuggenhalli S. Nandagopal

HVACR Principles and Applications

Nuggenhalli S. Nandagopal University of Houston-Downtown Houston, TX, USA

ISBN 978-3-031-45266-6 ISBN 978-3-031-45267-3 https://doi.org/10.1007/978-3-031-45267-3

(eBook)

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland Paper in this product is recyclable.

Preface

The field of Heating, Ventilation, Air-conditioning, and Refrigeration (HVACR) is very much practice oriented. HVACR practitioners tend to use practice-oriented design equations without a comprehensive understanding of the theory and concepts behind those equations. The motivation for writing HVACR Principles and Applications is to provide a clear understanding of the concepts and principles used in HVACR design practice. The book starts off with four chapters on support areas of fluid mechanics, heat transfer, thermodynamics, and psychrometrics. Thus, half the book is dedicated to the support areas which form the pillars of design equations and methodologies used by practitioners in the field. The support areas have been written with exclusive focus on applications of those areas in HVACR practice. The example and practice problems in the support area chapters clearly reflect this intention. For example, the support area of heat transfer provides a comprehensive, clear understanding of the calculation of overall heat transfer coefficients (U ) of conduction-convections systems used in the field. After a thorough review and understanding of the support areas, the reader can follow the concepts and equations presented in the ASHRAE Handbooks and other resources with increased levels of confidence. The four remaining chapters after the support areas focus on HVACR design calculations and applications of fundamental principles through the use of many realistic example and practice problems with detailed solutions. The challenge while solving these problems is the requirement of reliable, trusted data. For this, one needs to look no further than ASHRAE Handbooks and ASHRAE Standards, the trusted resources for HVACR professionals around the globe. The problem solutions clearly illustrate how to make extensive use of the aforementioned resources. The presentation of equations, concepts, problems, and solutions in this book make full use of both Inch-Pound (I P) and S I units for the benefit of readers and users around the globe.

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Preface

It is sincerely hoped that readers of this book will be satisfied with the integration and balance between theory and practice in HVACR. Comments and feedback are most welcome. Happy reading in your journey through HVACR! Houston, TX, USA

Nuggenhalli S. Nandagopal

Contents

1

2

Support Areas: Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Fluid Properties and Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Conservation of Mass: Continuity Equation . . . . . . . . . . . . . . . . . 1.3.1 Static Pressure and Velocity Pressure in a Duct . . . . . . . . . 1.3.2 Pitot Tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.3 Hydraulic Diameter of Rectangular Ducts . . . . . . . . . . . . . 1.4 Friction Head Loss and Pressure Drop for Fluid Flow in HVAC Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 Reynolds Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.2 Darcy’s Formula for Calculating Friction Head Loss for Flow Through Ducts . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.3 Duct Sizing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.4 Additional Friction Losses and Pressure Drop Due to Fittings and Components in HVAC Systems . . . . . . . . . 1.5 Fans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.1 Fan Power Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.2 Fan Performance Curves, System Curve, and Operating Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.3 Fan Laws or Affinity Laws . . . . . . . . . . . . . . . . . . . . . . . . 1.5.4 Fan Types and Fan Selection Criteria . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 1 7 11 15 16

Support Areas: Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Relevance of Heat Transfer in HVAC Design and Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Heat Transfer Principles and Modes of Heat Transfer . . . . . . . . . . 2.2.1 Heat Transfer Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . .

59

18 19 20 27 28 37 38 45 47 49 51 52 57

59 59 61 vii

viii

Contents

2.3

3

Conduction Heat Transfer Principles . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Multilayer Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 R-values for Insulation and Building Materials . . . . . . . . . 2.4 Convection Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Newton’s Law of Cooling . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Convection Heat Transfer Resistance . . . . . . . . . . . . . . . . 2.4.3 Typical Convection Heat Transfer Coefficients Used in HVAC-R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.4 Overall Heat Transfer Coefficients in Conduction: Convection Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Heat Exchangers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Heat Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Log Mean Temperature Difference (LMTD) . . . . . . . . . . . 2.5.3 Heat Exchanger Design Equation . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

61 63 64 66 66 67

Support Areas: Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Thermodynamic Properties and Units . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Specific Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Concept of Mole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Universal Gas Constant and Ideal Gas Equation in Molar Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Thermodynamic Processes Involving Ideal Gases . . . . . . . . . . . . . 3.4.1 Isothermal Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Isentropic Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.3 Constant Enthalpy/Throttling Process . . . . . . . . . . . . . . . . 3.4.4 Calculation of Work for Thermodynamic Processes . . . . . . 3.5 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 First Law of Thermodynamics for a Closed System . . . . . . 3.5.2 First Law of Thermodynamics for Open Systems: Energy Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3 First Law of Thermodynamics Applied to Compressors . . . 3.5.4 Isentropic Efficiency of Compressors . . . . . . . . . . . . . . . . 3.6 Thermodynamic Properties of Refrigerants . . . . . . . . . . . . . . . . . . 3.6.1 Thermodynamic Properties of Refrigerants from Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.2 Properties of Liquid–Vapor Mixtures of Refrigerants . . . . . 3.6.3 Properties of Compressed Liquid . . . . . . . . . . . . . . . . . . . 3.6.4 Pressure: Enthalpy (P–h) Phase Diagrams for Refrigerants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

85 85 85 88 89 90

68 68 73 74 77 78 80 81 83

91 93 94 94 96 96 99 99 102 103 105 107 108 108 110 111

Contents

ix

3.7

115

Reversed Carnot Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.1 Performance Measure of Reversed Carnot Cycle: Coefficient of Performance (COP) . . . . . . . . . . . . . . . . . . . 3.8 Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4

5

Support Areas: Psychrometrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Ideal Gas Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Key Definitions for Ideal Gas Mixtures . . . . . . . . . . . . . . . 4.2.2 Laws Related to Ideal Gas Mixtures . . . . . . . . . . . . . . . . . 4.3 Air–Water Vapor Mixture and Psychrometrics . . . . . . . . . . . . . . . 4.3.1 Moist Air Properties and Definitions . . . . . . . . . . . . . . . . . 4.3.2 Relationship Between Humidity Ratio and Relative Humidity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.3 Other Properties of Moist Air . . . . . . . . . . . . . . . . . . . . . . 4.3.4 Use of Psychrometric Chart to Obtain Properties of Moist Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Air-Conditioning Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Cooling and Dehumidification . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Actual Process of Cooling and Dehumidification: Apparatus Dew Point (ADP) and By-pass Factor (BF) . . . . 5.3 Heating and Humidification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Cooling Towers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Range, Approach, and Cooling Efficiency of Cooling Towers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Sensible Heating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Sensible Cooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Other Air Conditioning Processes . . . . . . . . . . . . . . . . . . . . . . . . 5.7.1 Isothermal Humidification . . . . . . . . . . . . . . . . . . . . . . . . 5.7.2 Isothermal Dehumidification . . . . . . . . . . . . . . . . . . . . . . . 5.7.3 Evaporative Cooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7.4 Chemical Dehumidification . . . . . . . . . . . . . . . . . . . . . . . 5.8 Adiabatic Saturation Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9 Adiabatic Mixing of Air Streams . . . . . . . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

116 120 120 125 127 127 127 127 129 129 130 132 134 137 141 142 145 147 147 147 151 154 159 163 165 167 168 168 171 171 173 173 175 179 180 187

x

6

7

8

Contents

Cooling and Heating Load Calculations . . . . . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Comparisons Between Cooling and Heating Load Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Cooling Load Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 Rules of Thumb for Cooling Loads . . . . . . . . . . . . . . . . . . 6.3.2 Cooling Load Components and Sources of Heat Gain . . . . 6.3.3 Indoor and Outdoor Design Parameters . . . . . . . . . . . . . . . 6.3.4 Basics of Cooling Load Calculations . . . . . . . . . . . . . . . . . 6.3.5 Procedure for Estimating Cooling Loads for Facilities . . . . 6.4 Heating Load Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.1 Indoor and Outdoor Design Conditions for Space Heating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.2 Equations for Calculating Heating Loads . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Refrigeration Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Reversed Carnot Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.1 Performance Measure of a Reversed Carnot Cycle: Coefficient of Performance (COP) . . . . . . . . . . . . . . . . . . . 7.3 Absorption Refrigeration Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.1 Analysis of Absorption Refrigeration Cycle . . . . . . . . . . . . 7.4 Vapor Compression Refrigeration Cycle . . . . . . . . . . . . . . . . . . . . 7.4.1 Analysis of Vapor Compression Cycle . . . . . . . . . . . . . . . 7.4.2 Overall Energy Balance for a Vapor Compression Refrigeration Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.3 Condenser in Refrigeration Cycles . . . . . . . . . . . . . . . . . . 7.4.4 Performance Measures for Vapor Compression Cycles . . . . 7.5 Refrigeration and Freezing of Foods . . . . . . . . . . . . . . . . . . . . . . 7.5.1 Thermal Properties of Foods . . . . . . . . . . . . . . . . . . . . . . . 7.6 Refrigerants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.1 Terminologies Associated with Refrigerants . . . . . . . . . . . 7.6.2 Safety Classification of Refrigerants . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HVAC Systems and Equipment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Types of HVAC Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 Direct Expansion-Based Cooling System . . . . . . . . . . . . . . 8.2.2 Chilled-Water-Based Cooling System . . . . . . . . . . . . . . . . 8.2.3 Steam-Based Heating System . . . . . . . . . . . . . . . . . . . . . .

189 189 190 191 191 191 192 193 198 234 235 235 241 244 265 267 267 267 269 272 273 278 278 280 281 282 290 291 295 295 297 297 299 304 305 305 305 306 310 313

Contents

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8.3

315 317

Chilled-Water (CW) Cooling Systems . . . . . . . . . . . . . . . . . . . . . 8.3.1 Water-Cooled and Air-Cooled Chilled-Water Systems . . . . 8.3.2 Classification of Chilled-Water Cooling Systems Based on the Type of Compressor Used . . . . . . . . . . . . . . 8.4 Energy Recovery Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.1 Heat Recovery Ventilators . . . . . . . . . . . . . . . . . . . . . . . . 8.4.2 Energy Recovery Ventilators (ERV): Enthalpy Wheel . . . . 8.5 Acoustics and Noise Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.1 Sound Control and Noise Reduction Methods . . . . . . . . . . 8.6 Vibration and Vibration Control . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 Constant Air Volume (CAV) and Variable Air Volume (VAV) Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7.1 Constant Air Volume Systems . . . . . . . . . . . . . . . . . . . . . 8.7.2 Variable Air Volume Systems . . . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

318 319 319 328 338 339 340 341 342 347 348 349 356

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357

Chapter 1

Support Areas: Fluid Mechanics

1.1

Introduction

Knowledge of the principles of hydraulics and fluid mechanics is required in the design of air handling and air distribution systems in heating, ventilation, air-conditioning, and refrigeration (HVACR) systems. This includes design and sizing of air ducts [1, 3]. Design and sizing of tubes in transporting the refrigerant between different components of refrigerant systems also requires basic knowledge of fluid mechanics.

1.2

Fluid Properties and Units

The important properties that characterize a fluid are pressure (P), density (ρ), specific weight (γ), specific gravity (SG), dynamic viscosity (μ), and kinematic viscosity (ν) [4, 5]. Pressure: The pressure or pressure intensity of a fluid is defined as the force exerted by the fluid per unit area. Pressure increases with increase in depth in a fluid. Units of pressure: Since pressure is force per unit area, it has units of lbf/ft2 (USCS or I-P System), and N/m2 (SI system). However, the more commonly used units in practice are lbf/in2 (psi), and kPa, where N/m2 is also known as pascal (Pa). Another commonly used unit for pressure is ‘bar’ and 1 bar = 105 N/ m2 = 100 kPa. Conversion factors: 1 psi = 6.895 kPa = 0.0689 bar Atmospheric pressure (Patm) is the pressure exerted by local atmospheric air and is also known as barometric pressure, since it is measured by a mercury barometer. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 N. S. Nandagopal, HVACR Principles and Applications, https://doi.org/10.1007/978-3-031-45267-3_1

1

2

1 Support Areas: Fluid Mechanics

Since pressure decreases with increase in elevation, the value of the local atmospheric pressure depends on the elevation of a particular location. Standard atmospheric pressure at sea level: 1 atm = 760 mm Hg = 29.92 in Hg = 14.7 psi = 101.3 kPa = 1.013 bar Pressure decreases with increase in the altitude of a place. The drop in pressure is approximately 10 kPa per 1000 m increase in altitude (up to an altitude of 5000 m). At an altitude of 1500 m, the atmospheric pressure in Denver is about 85 kPa. The atmospheric pressure at the summit of Mt. Everest (approximate altitude of 9000 m) is 32 kPa. It is important to note that the total pressure of ambient, moist air at any location is the same as the atmospheric pressure at that location [1, 3, 5]. The total pressure of moist air is widely used in psychrometric formulas and calculations. Density: The density (ρ) of a fluid is its mass per unit volume. The units for density are lbm/ft3 (USCS) and kg/m3 (SI). The standard density of air commonly used in HVAC calculations is 0.075 lbm/ft3 (at 70 °F and 14.7 psia) and 1.20 kg/m3 (at 20 °C and 101.3 kPa). Standard values of density of moist air at sea-level atmospheric pressure and normal ambient temperature: ρair = 0.075 lbm/ft3 (USCS) ρair = 1.20 kg/m3 (SI) Specific volume (v): The specific volume of a fluid is the volume per unit mass of the fluid. It is the reciprocal of the density of the fluid. Standard values of specific volume of moist air at sea-level atmospheric pressure and normal ambient temperature: vair = 13.33 ft3/lbm (USCS) vair = 0.8264 kg/m3 (SI) The specific volume of moist air is widely used in air conditioning calculations. The density of ambient, moist air can be readily calculated using the ideal gas law [5] (discussed in detail in the thermodynamics chapter). The following formulas and examples illustrate the calculation of specific volume of moist air.

1.2

Fluid Properties and Units

3

ρair =

P Rair T

ð1:1Þ

where P is the absolute pressure (the sum of gage pressure and atmospheric pressure) of air, (for air at atmospheric pressure, the gage pressure is zero) in psi or kPa, and T is the absolute temperature of air in °R (460 + °F) or K (273 + °C). The values of the specific or individual constant are as follows: Rair = 0:37 psia‐ft3 =lbm‐ ° R = 0:2867 kJ=kg  K

ð1:2Þ

Since the specific volume is the reciprocal of density, Eq. 1.1 can be rearranged to obtain the following equation for calculating the specific volume of moist air: vair =

Rair T P

ð1:3Þ

Example 1.1 Validate the standard values of the specific volume of moist air using the ideal gas law. (Solution) USCS / I-P units Covert the normal temperature of air to its absolute value. T = 70 ° F þ 460 ° = 530 ° R Calculate the standard specific volume using Eq. 1.3. 3

psia‐ft 0:37 lbm‐ ° R ð530 ° RÞ R T vair = air = = 13:34 ft3 =lbm 14:7 psia P

SI units Covert the normal temperature of air to its absolute value. T = 20 ° C þ 273 ° = 293 K Calculate the standard specific volume using Eq. 1.3. kJ ð293 KÞ 0:2867 kgK Rair T vair = = 0:8292 m3 =kg = 101:3 kPa P

4

1

Units:

kJ kg

kPa



kNm kg kN m2



Support Areas: Fluid Mechanics

m3 kg

Note: The small discrepancies between the listed values and the calculated values are due to round-off. The volume of a gas, as well as its specific volume, is directly proportional to its absolute temperature (Charles’ Law) and inversely proportional to its absolute pressure [5]. Therefore, using the standard specific volumes as the basis, the specific volume of moist air at any given condition of temperature and pressure can be determined by the following formula: USCS: SI:

v = 13:33 v = 0:8264

ft3 lbm m3 kg

T 530 ° R T 293 K

14:7 psia P

101:3 kPa P

ð1:4aÞ ð1:4bÞ

The pressure as well as temperature of air decreases with increase in altitude. The following formulas can be used for calculating pressures and temperatures at higher altitudes: P = 14:7ð1 - 0:0000069hÞ5:26 t = 59 - 0:0036h

USCS

ð1:4cÞ ð1:4dÞ

USCS

In Eqs. 1.4c and 1.4d, h is the altitude in feet, P is the pressure in psia and t is the temperature in °F. P = 101:3ð1 - 0:000023hÞ5:26 t = 15 - 0:0011h

SI

SI

ð1:4eÞ ð1:4f Þ

In Eqs. 1.4e and 1.4f, h is the altitude in meters, P is the pressure in kPa and t is the temperature in °C. Note: Using standard specific volume in psychrometric and air conditioning calculations for units in Denver (or any place at higher altitudes) introduces an error of approximately 15%. The error gets compounded when the specific volume is multiplied by the mass flow rate of dry air resulting in highly inaccurate results. Hence, there is a separate ASHRAE Psychrometric Chart for an altitude of 5000 ft.

1.2

Fluid Properties and Units

5

Specific weight: The specific weight (γ) of a fluid is the weight per unit volume of the fluid. Weight, W, is the force experienced by a body due to the gravitational acceleration, g. On earth, g = 32.2 ft/sec2 (USCS) and 9.81 m/s2 (SI). Units of specific weight are lbf/ft3 (USCS) and N/m3 (SI). The formulas for specific weight are given here. mg gc

W γ= = V γ=

V

=

m V

ρg g = gc gc

m W mg g = ρg = = V V V

ðUSCSÞ ðSIÞ

ð1:5aÞ ð1:5bÞ

In Eq. 1.5a, gc is the conversion constant that needs to be used to keep units consistent while working with USCS units. gc is derived from the definition of pound force (lbf). 1 lbf is the force required to accelerate 1 lbm at 32.2 ft/sec2. Therefore, gc =

1 lbm × 32:2 secft 2 32:2 lbm‐ft = 1 lbf lbf‐ sec 2

The standard specific weight of water is 62.4 lbf/ft3 (in the range 45–65 °F) and 9810 N/m3 (in the range 5–20 °C). However, since the SI unit of pressure is usually kPa, it is preferable to use the value 9.81 kN/m3 as the standard specific weight of water. Standard Specific Weights of Water. γ w = 62.4 lbf/ft3 (USCS) and γ w = 9.81 kN/m3 (SI) Based on the standard density of air, the calculations for the standard specific weights of air are shown here. USCS units:

γ air =

lbm ft ρair g 0:075 ft3 × 32:2 sec 2 = = 0:075 lbf=ft3 lbm‐ft gc 32:2 lbf‐ 2 sec

SI units: γ air = ρair g = 1:20 Note: N  kg:m s2

kg m 1 kN × 9:81 2 × = 0:0118 kN=m3 3 N 1000 m s

6

1

Support Areas: Fluid Mechanics

Standard Specific Weights of air γ air = 0.075 lbf/ft3 (USCS) and γ air = 0.0118 kN/m3 (SI)

Specific gravity: Specific gravity (SG) is commonly used for liquids and is also known as relative density. The specific gravity (SG) of a liquid is the ratio of the density of the liquid to the standard density of water. It is also the ratio of the specific weight of the liquid to the standard specific weight of water. SG =

ρliquid γ liquid = ρwater,std γ water,std

ð1:6Þ

The specific gravities of commonly used refrigerants are given here. R-22 (freon) = 1.19, R-134a = 1.21, R-410A = 1.06, R-717 (ammonia) = 0.682 Dynamic viscosity: The viscosity of a fluid is a property that indicates the flow resistance of the fluid. A clear understanding of viscosity can be obtained by considering the definition of the absolute or dynamic viscosity (μ) of a fluid, which is defined as the ratio of the applied shear stress to the rate of shear deformation (velocity gradient) of the fluid. μ=

τ

ð1:7Þ

dv dy

where τ is the shear stress parallel to the direction of motion of the fluid. The quantity (dv/dy) is the velocity gradient in the direction perpendicular to the flow direction. The units of dynamic viscosity can be obtained by substituting the units for the physical quantities on the right side of Eq. 1.7. μ

τ dv dy

lbf 2

 ft=ftsec 

μ

ft

τ dv dy

32:2 lbm‐ft lbf‐ sec lbm sec 2 ðsec Þ   2 ft‐ sec ft2 ft N

m   m=s 2

m

Ns  m2

kg‐m s2

m2

ðsÞ



kg ms

Since N/m2 is also known as pascal (Pa), the units of dynamic viscosity in the SI system can be represented as Pa.s. In industrial practice, the units of centipoise (cP) are commonly used for μ, the dynamic viscosity.

1.3

Conservation of Mass: Continuity Equation

7

Poise = dyne.s/cm2 1 cP = 10-2 Poise Conversion factors 1 lbm/ft-sec = 1.4881 kg/m.s = 1.4881 N.s/m2 (Pa.s) 2 1 lbf-sec/ft = 32.2 lbm/ft-sec = 47.88 N.s/m2 (Pa.s) 1 cP = 0.001 N.s/m2, (Pa.s), (kg/m.s) 1 cP = 0.000021 lbf-sec/ft2 = 0.000672 lbm/ft-sec

Kinematic viscosity: Kinematic viscosity (ν), is the ratio of absolute viscosity to the density of the fluid. ν=

μ ρ

ð1:8Þ

The units of kinematic viscosity can be obtained by substituting the units for dynamic viscosity and density in Eq. 1.8: ν

ft2 μ ft‐lbm sec   lbm sec ρ 3

ðUSCSÞ

ft

kg

ν

μ ms m2   s ρ kg3

ðSIÞ

m

A unit of kinematic viscosity that is frequently used in the industry is centistokes (cSt) which is based on the unit Stoke, which is defined as cm2/s. Conversion Factors 1 m2/s = 10.764 ft2/s 1 cSt = 0.000001 m2/s = 0.000011 ft2/s

1.3

Conservation of Mass: Continuity Equation

For steady flow of a fluid, the mass flow is constant. The law of conservation of mass states that the mass flow into a control volume is equal to the mass flow out of the control volume [4, 5]. The typical units for mass flow rates are pound mass per hour (lbm/hr) and kilogram per second (kg/s).

8

1

Support Areas: Fluid Mechanics

Fig. 1.1 Continuity equation

Conversion factor: 1 kg/s = 7937 lbm/hr Consider an air duct with two sections, 1 and 2 as shown in Fig. 1.1. The following nomenclature is used: m_ = mass flow rate of air Q = volume flow rate of air ρ = density of the fluid A = area of cross-section of pipe v = average velocity in the pipe Note: It is common practice to use the symbol ‘Q’ for volume flow rate in fluid mechanics. However, a note of caution that must be remembered is that many sources in HVAC literature use the symbol ‘Q’ is also used for heat transfer rate, cooling load, heating load, etc. In this book, the symbol ‘Q’ is used for volume flow rate and the symbol ‘q’ is used for heat transfer rate, cooling load, heating load, etc. and this practice is strongly recommended by the author. The mass flow into the section is the same as the mass flow out of the section, m_ 1 = m_ 2 = m_ The mass flow rate is the volume flow rate times the density of air and the volume flow rate is the cross-section area of the duct times the average velocity of air in that section. Therefore, m_ = Q1 ρ1 = Q2 ρ2

ð1:9Þ

The specific volume of air, v, is the reciprocal of the density of air. Therefore, m_ =

Q1 Q2 = v1 v2

ð1:9aÞ

1.3

Conservation of Mass: Continuity Equation

m_ = A1 v1 ρ1 = A2 v2 ρ2

9

ð1:10Þ

Equation 1.10 is the most general form of the continuity equation. Due to pressure drop, the pressure at section 2 is less than the pressure at section 1. For gases, the density is directly proportional to pressure and the densities at sections 1 and 2 are not equal because of the pressure drop. Since the density in the flow field is not constant, the flow of gases is described as compressible flow. The density of a liquid is constant and does not vary significantly with pressure. Liquids are essentially incompressible. Since ρ1 = ρ2 for liquids, the volume flow rates at sections 1 and 2 will be equal. Therefore, Q1 = Q2 = Q = A1 v1 = A2 v2 = Av

ð1:11Þ

The volumetric flow rate has units of cubic meters per second (m3/s), and cubic feet per second (ft3/sec). In practice, the units of gallons per minute (gpm) is widely used for the flow of liquids and the units of cubic feet per minute (cfm) for flow of air. Conversion Factors: 1 m3/s = 35.315 ft3/sec (cfs) 1 ft3/sec = 449 gpm 1 m3/s = 2119 cfm

Example 1.2 20,000 cfm of air at 62 °F and 20 in. water gage pressure is flowing in a 24 in. × 18 in. main duct that supplies equal volumes of air to four rooms via four branch ducts. The width-to-height ratio for the branch ducts is the same as that of the main duct. Determine: A. the mass flow rate of air in the main duct (20 in. water gage is equivalent to 0.72 psig and the atmospheric pressure is 14.7 psia). B. the dimensions of the branch ducts if the same velocity is to be maintained in all the ducts. (Solution) A. Calculate the absolute pressure and absolute temperature of air. The absolute pressure is the sum of the gage pressure in the duct and the atmospheric pressure. P = Pabs = Patm þ Pgage = 14:7 psia þ 0:72 psig = 15:42 psia T = 460 ° þ 62 ° F = 522 ° R Calculate the specific volume of the flowing air using Eq. 1.4a.

10

1

v = 13:33 = 13:33

ft3 lbm

ft3 lbm

T 530 ° R

Support Areas: Fluid Mechanics

14:7 psia P

522 ° R 530 ° R

14:7 psia 15:42 psia

= 12:52 ft3 =lbm Calculate the mass flow rate of air by using the continuity equation (Eq. 1.9a). 3

ft Q 20, 000 min = 1598 lbm= min m_ = = ft3 v 12:52 lbm

B. Calculate the cross-section area of the main duct.

A = 24 in: ×

1 ft 12 in:

18 in: ×

1 ft = 3:0 ft2 12 in:

Calculate the velocity of air in the main duct using the continuity equation (Eq. 1.11). 3

Q = Av ) v =

ft Q 20, 000 min = 6667 ft= min = 2 A 3:0 ft

Let W and H be the width and height, respectively, of each branch duct. Calculate the cross-section area of each branch duct in terms of W using the width-to-height ratio of each branch duct is the same as the width-to-height ratio of the main duct. W 24 in: = = 1:333 ) H = 0:75W ) H b = 0:75W b 18 in: H Ab = W b × H b = W b × 0:75W b = 0:75W b 2 Since the volume flow in the main duct is distributed equally between the four branch ducts, the volume flow rate in each branch duct is 3

ft Q 20, 000 min Qb = = = 5000 ft3 = min 4 4

Calculate the dimensions of each branch duct by using the continuity equation (Eq. 1.11) and the same velocity as in the main duct.

1.3

Conservation of Mass: Continuity Equation

11

ft3 Qb 5000 min = 0:75 ft2 ) Qb = Ab vb ) Ab = = ft vb 6667 min 0:75W 2b = 0:75 ft2 ) W b = 1 ft ð 12 in:Þ ) H b = 0:75W b = 0:75 × 1 ft = 0:75 ft ð 9 in:Þ

Therefore, the dimensions of each branch duct are 12 in. × 9 in. Note: Since the volume flow is quartered in each branch, the dimensions of each branch will be halved to maintain the same velocity because the cross-section area is the square of the length dimension.

1.3.1

Static Pressure and Velocity Pressure in a Duct

When air flows through a duct, the total pressure of air at any given section in the duct is the sum of two pressure components, the static pressure, Ps and the velocity pressure, Pv. The static pressure is the pressure exerted by the fluid on the walls of the duct/pipe. The velocity pressure is due to the kinetic energy of the fluid. The static and velocity pressure heads can be explained using Bernoulli’s equation [1, 4, 5] for fluid flow. Bernoulli’s equation is a modification of the mechanical energy equation under the circumstances of frictionless flow without any energy devices in the flow field. The static pressure due to a fluid can be converted to an equivalent head of the fluid column, which is nothing but the height of the fluid column causing the pressure. In this case, the pressure is the weight of the fluid column divided by the area of the surface where the pressure is acting. Pressure can be converted to an equivalent head using the specific weight of the fluid (air in duct flow). Therefore, the equivalent head (static head) for the static pressure is hs =

Ps γ

ð1:12Þ

Bernoulli’s principle states that pressure head plus velocity head plus potential head remains constant in a flow field where no energy is added or removed from the fluid. The pressure head is the pressure energy per unit weight of the fluid, and it is the same as the static head. Velocity head is the kinetic energy per unit weight of the fluid, and it is equivalent to (v2/2g), where v is the velocity of the fluid and g is the acceleration due to gravity. The potential head is potential energy per unit weight of the fluid, and it is equivalent to the elevation of the fluid, Z. Thus, Bernoulli’s equation can be mathematically written as

12

1

Support Areas: Fluid Mechanics

Ps v2 þ þ Z = constant γ 2g

ð1:13Þ

In Eq. 1.13, each term has the units of feet (USCS) or meters (SI). Bernoulli’s equation (USCS or I-P units) can also be written in terms of energy per unit mass [1] as follows. Ps v2 Zg þ = constant þ ρ 2gc gc

ð1:14aÞ

where Ps is the static pressure in lbf/ft2, ρ is the fluid density in lbm/ft3, v is the fluid velocity in ft/sec, gc is the conversion constant 32.2 lbm-ft/lbf-sec2, Z is the fluid elevation in ft, and g is the gravitational acceleration, 32.2 ft/s2. Thus, each term in Eq. 1.14a will have the units ft-lbf/lbm, that is, energy per unit mass. Bernoulli’s equartion (SI units) can be written in terms of energy per unit mass as follows: P s v2 þ þ Zg = constant ρ 2

ð1:14bÞ

where Ps is the static pressure in Pa, ρ is the fluid density in kg/m3, v is the fluid velocity in m/sec, Z is the fluid elevation in m, and g is the gravitational acceleration, 9.81 m/sec2. Each term in Eq. 1.14b will have the units J/kg, that is, energy per unit mass. The details of the units are explained here. kg m2 N × 2 J v P s m2 N  m sec 2 kg    ,   kg 2 kg kg 2 ρ m3 m2 J Thus,  sec 2 kg

kg  m m sec 2 N:m J   kg kg kg

In HVAC applications, the pressure is commonly expressed in terms of equivalent inches of water column. Therefore, the static head in feet of water needs to be multiplied by 12 to obtain equivalent static pressure in terms of inches of water. In Eq. 1.13, the term v2/2g represents the velocity pressure, where v is the velocity in ft/sec and g is the acceleration due to gravity in ft/sec2 resulting in the velocity pressure having equivalent units of ft of fluid. However, the volume flow rate of air or any gas is typically specified in cubic feet per minute (cfm) and using the continuity equation (Eq. 1.11), the velocity will be in feet per min (ft/min or fpm). Any specified pressure can be obtained by multiplying the specific weight of the fluid by the head of the fluid. Consider a fluid column and a water column exerting the same pressure. The equivalent feet of water column can be obtained by using the following formula:

1.3

Conservation of Mass: Continuity Equation

13

γ fluid h γ water fluid

P = γ fluid hfluid = γ water hwater ) hwater =

ð1:15Þ

The standard specific weight of water is γ water = 62.4 lbf/ft3. The following conversion factors are used to obtain the velocity pressure in equivalent inches of water column. v ft × 1 min v2  min 60 sec 2g 2 × 32:2 secft 2

2

×

γ fluid v 12 in ×  γ fluid lbf 1097 ft 62:4 ft3

2

in:water

Therefore, the velocity pressure of any flowing fluid in equivalent inches of water is Pv ðin:water Þ = γ fluid

2

v 1097

ð1:16Þ

where γ fluid is specific weight of the flowing fluid in lbf/ft3 and v is the velocity of the fluid in ft/min. If the flowing fluid is air, as is the case in air conditioning system ducts, the standard specific weight of air can be calculated using Eq. 1.5a. 0:075 lbm 32:2 secft 2 ρair g ft3 = = 0:075 lbf=ft3 γ air = lbm‐ft gc 32:2 lbf‐ sec 2

In USCS units, the numerical values of density and specific weight of a fluid is the same as shown above [5]. Therefore, to obtain velocity pressure of any fluid, in inches of water, Eq. 1.16 can be written as

Pv ðin:waterÞ = ρfluid

2

v 1097

ð1:17Þ

In Eqs. 1.16 and 1.17, the velocity of the fluid is feet per minute (fpm), and it is important to note that the density of the fluid should be in lbm/ft3. Substitute the standard density of air, ρair = 0.075 lbm/ft3, into Eq. 1.16 to obtain an equation for calculating the velocity pressure of air flowing in a duct at standard conditions (14.7 psia and 70 °F). Pv ðin:waterÞ = 0:075

lbm ft3

where v is the velocity of air in ft/min [1, 3].

v 1097

2

=

v 4005

2

ð1:18Þ

14

1

Support Areas: Fluid Mechanics

Conversion Factors: 1 psi = 27.71 in. w c, 1 kPa = 4.02 in. w c, 1 atm = 407.2 in. w c 1 in. w c = 0.0361 psi = 0.2488 kPa = 0.0025 atm

Example 1.3 An air handler is moving 12,000 cfm of air and the connected duct is 4 ft by 5 ft in cross-section. The gage static pressure of air in the duct is 2 psig. Determine (in inches of water), A. the static pressure B. the velocity pressure C. the total pressure (Solution) A. Calculate the static pressure in in. w c by using suitable conversion factor as shown.

Ps = ð2 psiÞ

27:71 in:w c = 55:42 in:w c 1 psi

B. Calculate the velocity of air in feet per min by using the continuity equation (Eq. 1.11). 3

Q = Av ) v =

ft 12, 000 min Q = 600 ft= min = A 4:0 ft × 5:0 ft

Calculate the velocity pressure in in. w c using Eq. 1.18. v Pv ðin:waterÞ = 4005

2

ft 600 min = 4005

2

= 0:0224 in w c

Note: The velocity pressure is very small compared to the static pressure. C. Calculate the total pressure in in. w c. Ptotal = Ps þ Pv = 55:42 in:w c þ 0:0224 in:w c = 55:44 in:w c

1.3

Conservation of Mass: Continuity Equation

15

Fig. 1.2 Pitot Tube with differential manometer

1.3.2

Pitot Tube

A pitot tube is used in measuring fluid velocities in ducts and pipes [4, 5]. As shown in Fig. 1.2, it consists of a bent stagnation tube that is inserted into the flow system. The velocity at the stagnation point is zero. This results in a higher pressure, P0, at the stagnation point. The pressure downstream of the stagnation point is the static pressure, Ps. The velocity of the fluid is proportional to the pressure difference between the stagnation pressure and the static pressure. ΔP = P0 - Ps

ð1:19Þ

The pressure difference can be measured by a differential manometer or by a differential pressure gage. In Fig. 1.2, a differential manometer shows a reading of hm. The equation for the velocity of the fluid can be obtained by applying Bernoulli’s Equation (Eq. 1.13) between the stagnation point and the downstream static point, assuming negligible friction loss between the two points. 0

P0 v 02 P v2 + + z 0 = s + + z s J f 2g J f 2g

v=

2g

ΔP γf

ð1:20Þ

Example 1.4 The velocity of air in a circular duct is determined to be 80 ft/s. The density of air at flow conditions is 0.088 lbm/ft3. Determine the reading of a differential water manometer connected across the pitot tube in inches of water.

16

1

Support Areas: Fluid Mechanics

(Solution) The flowing fluid is air. Calculate the specific weight of air by using the equation for specific weight (Eq. 1.5a, USCS units).

γf =

0:088 lbm 32:2 secft 2 ρf g ft3 = = 0:088 lbf=ft3 sec 2 gc 32:2 lbm‐ft= lbf

Calculate the difference between the stagnation and static pressure by using Eq. 1.20. 2

ft 0:088 lbf 80 sec v2 γ f ft3 = 8:745 lbf=ft2 ΔP = = ft 2g ð2Þ 32:2 sec 2

Using the standard specific weight of water (62.4 lbf/ft3) and using Eq. 1.12, calculate the reading of the differential water manometer. hm =

lbf ΔP 8:745 ft2 = = 0:1401 ft ð lbf γm 62:4 ft3

 1:68 in water - columnÞ

1.3.3

Hydraulic Diameter of Rectangular Ducts

Rectangular ducts are widely used in air handling and air distribution systems because they require much less clearance area compared to circular ducts for a given cross-section area. Fluid flow parameters such as the Reynolds number are commonly defined in terms of diameter of a circular pipe. Flow through non-circular sections are handled by using the ‘Hydraulic Diameter,’ represented by the symbol DH. The diameter, D, in circular pipe formulas is replaced by DH for flow through non-circular sections. The hydraulic diameter is based on the concept of equivalent hydraulic radius (RH) being equal to the cross-section area (A) divided by the wetted perimeter (P). The wetted perimeter is the total length of the duct in contact with the fluid. The following result is obtained by applying this concept to a circular pipe. RH =

π D2 D A = = 4 πD 4 P

The hydraulic diameter can be calculated by using Eq. 1.21.

1.3

Conservation of Mass: Continuity Equation

DH = 4RH = 4

17

Cross SectionArea A =4 Wetted Perimeter P

ð1:21Þ

For a rectangular duct with dimensions of width ‘a’ and height ‘b’, the hydraulic diameter can be obtained by applying Eq. 1.21. DH,Rect: = 4

2 ð a × bÞ A a×b =4 = aþb P 2ða þ bÞ

ð1:22Þ

It is important to note that the optimum duct shape is round or circular [3]. This is because a round duct has less surface area compared to a rectangular duct with the same air-carrying capacity. This results in less friction resistance, lower material costs, and lower operating costs due to less fan power required. Additional advantages of circular ducts are better acoustics and cleaner operation due to the absence of dirt-trapping corners present in rectangular ducts. Example 1.6 illustrates a comparison of perimeters of round and rectangular ducts for the same air-carrying capacity. However, round ducts require more clearance space for installation. Example 1.5 Determine the hydraulic diameter of an air duct with dimensions of 12 in. by 9 in. (Solution) Calculate the hydraulic diameter using Eq. 1.22. DH,Rect: =

2ða × bÞ 2ð12 in: × 9 in:Þ = = 10:29 in: aþb 12 in: þ 9 in:

Example 1.6 The typical width-to-height ratio for a rectangular duct can be taken as 2 to 1. For a capacity of 7000 cubic feet per minute (cfm) of air and specified air velocity of 25 ft/s, compare the dimensions, perimeters, and material requirements of round and rectangular ducts. (Solution) Calculate the required cross-section of the duct using the continuity equation (Eq. 1.11). 3

Q = Av ) A =

ft 7000 min Q = 4:667 ft2 = ft sec v 25 sec × 60min

For calculating the material requirements, assume 0.50 in. ( 0.0417 ft) thickness of sheet metal and 1 ft length of the duct.

18

1

Support Areas: Fluid Mechanics

Round duct: Calculate the diameter of duct using the calculated cross-section area. A = 4:667 ft2 =

π 2 D ) D = 2:438 ft 4

Calculate the perimeter and hence the volume of material required. P = πD = ðπ Þð2:438 ftÞ = 7:659 ft The volume of the material required is perimeter times the length times the thickness. V = P × L × t = 7:659 ft × 1 ft × 0:0417 ft = 0:3156 ft3 Rectangular duct: Calculate the dimensions of the duct using the calculated cross-section area and the given width-to-height ratio (2 to 1). A = 4:667 ft2 = 2H × H = 2H 2 ) H = 1:528 ft ) W = 2H = 2 × 1:528 ft = 3:056 ft Calculate the perimeter and hence the volume of material required. P = 2ðW þ H Þ = 2ð3:056 ft þ 1:528 ftÞ = 9:168 ft The volume of the material required is perimeter times the length times the thickness. V = P × L × t = 9:168 ft × 1 ft × 0:0417 ft = 0:3823 ft3 Thus, the material requirement for a rectangular duct is about 21% greater than that of the round duct.

1.4

Friction Head Loss and Pressure Drop for Fluid Flow in HVAC Systems

Fluid flow is always accompanied by a loss in pressure in the direction of flow. A fluid always flows from a point of higher pressure to a point of lower pressure. Any fluid flowing in a pipe has to overcome the resistance due to friction at the surface of the duct. This results in a loss of energy in the fluid, which can be represented as a

1.4

Friction Head Loss and Pressure Drop for Fluid Flow in HVAC Systems

19

loss in static head corresponding to the pressure drop across the section of the pipe. The pressure drop and the friction head loss can be related by the following equation based on Eq. 1.15 (P = γ fluidhfluid) presented earlier. ΔP = γhf

ð1:23Þ

Useful conversion factors between pressure and static head are presented here for reference. Conversion Factors: 1.0 psi = 6.89 kPa = 2.31 ft of water = 27.71 in. w c = 0.70 m of water 1 in. w c (water column) = 0.0361 psi = 0.2487 kPa

1.4.1

Reynolds Number

The Reynolds number is a dimensionless number used in fluid dynamics. It represents the ratio of inertial forces to viscous forces [5]. The inertial force is responsible for the motion of the fluid due to the momentum it carries and therefore it is proportional to the velocity and density of the fluid. The viscous force tends to resist the flow of the fluid and it is proportional to the viscosity of the fluid. The Reynolds number is represented by Re and can be calculated using the following formula: Re =

Dvρ DH vρ  μ μ

ð1:24Þ

where D is the inside diameter (ID) of the pipe, DH is the equivalent hydraulic diameter of a non-circular section, v is the average velocity of the fluid, ρ is the density of the fluid, and μ is the dynamic viscosity of the fluid. From Eq. 1.8, the ratio of dynamic viscosity to the density of the fluid is the kinematic viscosity of the fluid, ν. Therefore, the formula for the Reynolds number can also be written in terms of kinematic viscosity as shown in Eq. 1.25. Re =

Dvρ Dv DH v = μ = ν μ ρ

ð1:25Þ

Extreme care must be taken to ensure consistency of units while calculating the Reynolds number. All the units in the formula must cancel out leaving a dimensionless result.

20

1

Support Areas: Fluid Mechanics

The Reynolds number is used as a criterion for determining if the flow in a duct is Laminar or Turbulent. Laminar flow occurs in a pipe if the Reynolds number, Re, is less than 2100. Generally, the flow is fully turbulent if Re > 10,000. The flow is in the ‘transient zone’ if 2100 < Re < 10,000.

1.4.2

Darcy’s Formula for Calculating Friction Head Loss for Flow Through Ducts

Intuitively, it can be hypothesized that the friction head loss due to fluid flow in a duct would be directly proportional to the roughness of the pipe surface, the length of the duct, and the velocity of flow and inversely proportional to the hydraulic diameter of the duct. The friction head loss in a duct can be calculated using Darcy’s equation [3–5]. hf = f

L DH

v2 2g



fLv2 2gDH

ð1:26Þ

The following nomenclature is used in the preceding equation. hf = friction head loss (ft, m) f = Darcy friction factor (dimensionless) L = length of the duct (ft, m) DH = hydraulic diameter of the duct or inside diameter of a round duct (ft, m) v = velocity of flow (ft/sec, m/s) g = acceleration due to gravity (32.2 ft/s2, 9.81 m/s2) Note that the term (L/DH) is dimensionless, and the term (v2/2g) is the velocity head equivalent to the velocity pressure discussed earlier and it has the units of ft or m. However, from Eq. 1.18, the velocity pressure for air at standard conditions in inches of water column is Pv ðin:w cÞ =

v 4005

2



v2 2g

Since the friction factor, f, and (L/DH) are dimensionless in Eq. 1.26, the preceding equation for velocity pressure can be combined with Eq. 1.26 to obtain the following equation for the pressure loss in ducts in inches of water: ΔPf ðin:w cÞ  f

L DH

L v2 =f 2g DH

In Eq. 1.27, v is the velocity of air in ft/min (fpm).

v 4005

2

ð1:27Þ

1.4

Friction Head Loss and Pressure Drop for Fluid Flow in HVAC Systems

21

Fig. 1.3 Moody diagram. (Generated by the author, N.S. Nandagopal, P E)

The Darcy friction factor can be determined from the ‘Moody diagram’ shown in Fig. 1.3. The Moody diagram represents the Reynold’s number on the x-axis on a log scale and the Darcy friction factor on the y-axis, also on log scale. The relative roughness, r = ε/D, is represented as parametric curves. To read the Moody diagram, locate the Reynolds number on the x-axis and move along the corresponding ordinate to the appropriate relative roughness curve and locate the intersection of the Reynolds number ordinate and the relative roughness curve. From this intersection point, move horizontally to the left to the y-axis to obtain the Moody friction factor. The roughness of commercial steel is ε = 0.00015 ft  0.05 mm. Alternately, the pressure loss across ducts can be quickly and easily obtained from pressure loss charts [1, 3] for ducts – Fig. 1.4 (USCS units) and Fig. 1.5 (SI units).

Example 1.7 2.25 m3/s of air flows through a 0.50 m × 0.25 m commercial steel duct across 5 m length in a facility. The air is at 20 °C and 101.3 kPa. The properties of air are as follows: ρ = 1.20 kg/m3, ν = 1.6 × 10-5m2/s. Calculate the pressure drop across the duct in inches of water.

22

1

Support Areas: Fluid Mechanics

Fig. 1.4 Pressure loss chart for ducts, USCS / I-P units. (Generated by the author, N.S. Nandagopal, P E)

Fig. 1.5 Pressure loss chart for ducts, SI units. (Generated by the author, N.S. Nandagopal, P E)

1.4

Friction Head Loss and Pressure Drop for Fluid Flow in HVAC Systems

23

(Solution) Calculate the velocity of air using the continuity equation (Eq. 1.11). Q = Av ) v =

Q 2:25 m3 = 18 m=s = A 0:50 m × 0:25 m

Calculate the hydraulic diameter of the duct using Eq. 1.22. DH,Rect: =

2ða × bÞ 2ð0:50 m × 0:25 mÞ = = 0:3333 m aþb 0:50 m þ 0:25 m

Calculate the Reynolds number using Eq. 1.25. Re =

DH v 0:3333 m × 18 ms = 3:75 × 105 = 2 ν 1:6 × 10 - 5 ms

Calculate the relative roughness of the duct. r=

ε 0:05 mm = = 0:00015 mm DH 0:3333 m × 1000 1m

Using the Reynolds number, 3.75 × 10-5, and the relative roughness, 0.00015 as parameters obtain the Darcy friction factor from the Moody diagram (Fig. 1.3) as shown.

From the chart, f = 0.015. Calculate the friction head loss in m of air using Darcy’s equation (Eq. 1.26) for head loss. hf = f

L DH

5m v2 = ð0:015Þ 2g 0:3333 m

18 ms

2

2 × 9:81

= 3:716 m of air Calculate the equivalent head of water column using Eq. 1.15.

m s2

24

1

Support Areas: Fluid Mechanics

kg m3 ð3:716 mÞ kg 1000 3 m = 0:004 m of water

ρair γ hwater = fluid hfluid = h = γ water ρwater air

1:20

 0:1772 in w c Alternate Solution: The pressure drop can be easily and quickly obtained from the pressure loss chart (Fig. 1.5) using the volume flow rate (2.25 m3/s) and the calculated velocity (18 m/s) as shown here.

From the chart, Pressure loss = 0.035 in. w c/m length Multiply the preceding value by the length of the duct to obtain the total pressure loss. Pressure Loss for 5 m =(0.035 in. w c/m)(5 m) = 0.175 in. w c Comment: The result obtained from the friction loss chart compares very well with the result obtained using Darcy’s equation for head loss. Example 1.8 The pressure at a plenum which receives return air is 0.027 psig. The air from the plenum is distributed between two ducts such that the volume flow in the second duct is 2.5 times the volume flow in the first duct. The first duct is 5 in. × 3 in. and is

1.4

Friction Head Loss and Pressure Drop for Fluid Flow in HVAC Systems

25

90 ft long. The second duct is 200 ft long. Both the ducts exhaust to atmosphere and the Darcy friction factor is approximately 0.02 for both the ducts. The entrance losses into the ducts can be ignored. Determine: A. the volume flow rate in the first duct, B. the volume flow rate in the second duct, C. the dimensions of the second duct if the width-to-height ratio is the same as in the first duct, that is, 5:3. (Solution) A. Convert the dimensions of the first duct to feet. a = 5 in ×

1 ft = 0:4167 ft 12 in

b = 3 in ×

1 ft = 0:2500 ft 12 in

Since both the ducts exhaust to the atmosphere, the pressure drop and hence the friction head across each duct will be identical and equal to the plenum gage pressure. Subscript ‘1’ represents the first duct and Subscript ‘2’ represents the second duct. Therefore, ΔP1 = ΔP2 = 0:027 lbf=in:2 ) hf 1 = hf 2 Calculate the friction head loss in each duct using Eq. 1.12 and the standard specific weight of air, γ air = 0.075 lbf/ft3. lbf 144 in 0:027 in 2 × ΔPf ft2 hf = = lbf γ air 0:075 ft3

2

= 51:84 ft air

Calculate the hydraulic diameter of duct 1 using Eq. 1.22. DH1 =

2ða × bÞ 2ð0:4167 ft × 0:2500 ftÞ = 0:3125 ft = 0:4167 ft þ 0:2500 ft aþb

Solve Darcy’s equation (Eq. 1.26) for the velocity of air through duct 1 and substitute the known values to calculate the velocity of air in duct 1. hf =

v1 =

fL1 v21 ) 2gDH1 ft 2 × 32:2 × 0:3125 ft × 51:84 ft 2 2gDH1 hf sec = 0:02 × 90 ft fL1 = 24:07 ft= sec

26

1

Support Areas: Fluid Mechanics

Calculate the volume flow rate of air in duct 1 by using the continuity equation (Eq. 1.11). Q1 = A1 v1 = ð0:4167 ft × 0:2500 ftÞ 24:07

ft sec

= 2:507 ft3 = sec

B. Calculate the volume flow rate of air in duct 2. Q2 = 2:5Q1 = ð2:5Þ 2:507

ft3 sec

= 6:267 ft3 = sec

C. The width-to-height ratio of the second duct is 5:3, which can be used in expressing the height in terms of the width of duct 2. a2 5 = ) 3a2 = 5b2 ) b2 = 0:6a2 : b2 3 Therefore, the cross-section area of the second duct is A2 = a2 × b2 = a2 × 0:6a2 = 0:6a22 Calculate the hydraulic diameter of duct 2 using Eq. 1.22. DH2 =

2ða2 × 0:6a2 Þ 1:2a22 = = 0:75a2 a2 þ 0:6a2 1:6a2

Calculate the velocity of air in duct 2 by using the continuity equation (Eq. 1.11). Q2 = A2 v2 ) v2 =

Q2 = A2

ft3 sec = 10:445 a22 0:6a22

6:267

The friction head loss in duct 2 is the same as the friction loss in duct 1. From Darcy’s equation (Eq. 1.26), hf =

fL2 v22 ) 2gDH2

51:84 ft =

0:02 × 200 ft × 2 × 32:2

10:445 a22

2

ft × 0:75a2 sec 2

) a2 = 0:7051 ft ð8:461 inÞ

1.4

Friction Head Loss and Pressure Drop for Fluid Flow in HVAC Systems

27

Therefore, the dimensions of duct 2 are as follows: a2 = 8:461 in:

and

b2 = 0:6a2 = ð0:6Þð8:461 in:Þ = 5:077 in:

Note: Fig. 1.4 can be used in solving this problem by using the ‘equal area method’ to find the equivalent diameter of a round duct. The cross-section area of the rectangular duct is equated to the same cross-section area of a round duct and solved for the equivalent diameter of the round duct. This equivalent diameter is then used in reading Fig. 1.4.

1.4.3

Duct Sizing

The equal friction method is most widely used in determining the required size of a duct. Using the equal friction method, the ducts are sized such that the pressure loss per 100 ft of duct length is in the range of 0.05 in. to 0.10 in. w c. A pressure loss of 0.10 in. w c/100 ft length is the preferred criteria for designing ducts [3]. Lowering the design pressure drop per 100 ft length from 0.10 to 0.05 in. w c results in approximately 15% increase in duct material costs (capital costs) due to a bigger size of duct required for lower pressure drop. However, because of lower pressure drop, fan power requirement is reduced resulting in about 15–20% savings in energy costs (operating costs). From the preceding discussion, it is clear that sizing a duct is eventually an optimization problem requiring the minimizing of total costs (capital plus operating costs) to determine the optimal duct size. The other methods used in sizing of ducts are velocity reduction method and static regain method. In velocity reduction method, the ducts are sized by specifying velocities for main and branch ducts. In static regain method, ducts are sized based on process of converting velocity pressure to static pressure along the duct as per Bernoulli’s equation. The velocities are successively reduced to create increases in static pressure which compensates partially for the friction losses experienced by the flowing air. Example 1.9 Standard diameters of round ducts available are 16, 18, and 20 in. Select the diameter of a duct required to handle 2000 cfm of air if pressure loss is to be limited to 0.075 in. w c per 100 ft duct length. Subsequently determine the actual velocity and pressure loss in the selected duct. (Solution) Convert the pressure loss to psi per 100 ft duct length. ΔP = 0:075 in:w c ×

1 psi = 0:0027 psi=100 ft 27:71 in w c

28

1

Support Areas: Fluid Mechanics

Using Q = 2000 cfm and ΔP = 0.0027 psi/100 ft as parameters, determine the diameter from Fig. 1.4 as shown.

From the figure, the diameter is slightly higher than 18 in. Therefore, choose the next available higher diameter, that is D = 20 in, in order to keep the pressure loss below 0.075 in. w c per 100 ft length. From the figure, when D = 20 in, the velocity of air is, v = 925 ft/min. The actual pressure loss is ΔP = 0:0022 psi=100 ft: Convert this to inches of water column ΔP = 0:0022 psi ×

1.4.4

27:71 in:w c = 0:0610 in:w c 1 psi

Additional Friction Losses and Pressure Drop Due to Fittings and Components in HVAC Systems

A typical HVAC system includes fittings such as plenums, bends, tees, and transition pieces, and components such as valves, dampers, filters, and terminal devices for air distribution. The pressure loss across fittings and components contribute significantly to the total pressure loss in the system. An accurate estimate of the total pressure loss in the system is necessary to size fans to be used in the system. The

1.4

Friction Head Loss and Pressure Drop for Fluid Flow in HVAC Systems

29

‘velocity head’ method is widely used in calculating head losses across fittings and components, which are also termed as ‘minor losses’. The velocity head is multiplied by the loss coefficient for fittings/components, Kfittings/comp., to obtain the head loss across fittings and components. hf ,fittings=comp: = K fittings=comp:

v2 2g

ð1:28Þ

It is important to recall that the velocity head is the same as the velocity pressure discussed earlier. Combine the preceding equation with Eq. 1.18 to obtain an expression for pressure drop or head loss (in inches of water) across the component. Therefore, the following equation can be used to calculate the pressure drop across fittings/components in air duct systems: ΔPfittings=comp: ðin:w cÞ = K fittings=comp:

v 4005

2

ð1:29Þ

In the preceding equation, v is the velocity of air in feet per minute (fpm).

1.4.4.1

Typical Loss Coefficients for Fittings in HVAC Systems

The preferred source for accurate values of loss coefficients of fittings in air conditioning ducts is ASHRAE Duct Fitting Database [2]. This reference lists loss coefficients for fittings with all possible variations and configurations. The loss coefficients for commonly used fittings are listed here along with dependent parameters. In case of uncertainties, it is best to use the higher values for conservative design purposes. Loss Coefficients for Round Fittings Elbows: The value of the loss coefficient depends on the diameter of the elbow and the ranges are given here. 90° Long Radius, K = 0.10 to 0.30 90° Short Radius, K = 0.20 to 0.40 45° Long Radius, K = 0.10 to 0.20 90° Mitered, K = 0.80 to 1.20 45° Mitered, K = 0.70 to 0.90. Tees: The value of the loss coefficient are different for converging and diverging tees. In a converging tee, the flows from smaller diameter branch and straight pipe converge into a flow to a large diameter. In a diverging tee, the flow from a large diameter pipe splits into a flow into a smaller diameter branch and another flow through the straight pipe. The value of the loss coefficient depends on the area ratios of the branch and straight pipes and the flow ratio between the branch and straight pipes.

30

1

Support Areas: Fluid Mechanics

For a converging tee, with both area and flow ratios being 0.5, K b = K s = 1:34 For a diverging tee, with both area and flow ratios being 0.5, K b = 1:0 and K s = 0:20 Transition pieces: The value of the loss coefficient depends on the area ratio and the angle of the cone. A typical conservative value is K = 0.32. Screens: The value of the loss coefficient depends on the ratio of the area of the screen (AS) to the area of the duct (AD) and the free area of the screen. For example, When AS/AD = 1.0, K = 1.65 for 50% free area. and K = 0.44 for 75% free area When AS/AD = 0.5, K = 6.60 for 50% free area. and K = 1.76 for 75% free area Butterfly damper: The value of the loss coefficient depends on the percent opening of the damper. K 100%open = 0:60, K 70%open = 4, K 50%open = 18

Loss Coefficients for Rectangular Fittings Elbows: The value of the loss coefficient depends on the diameter of the elbow and the ranges are given here. 90° Long Radius, K = 0.18 90° Short Radius, K = 0.22 90° Mitered, K = 1.30 45° Mitered, K = 0.38 90° Mitered (with vanes), K = 0.33. Tees: The value of the loss coefficient are different for converging and diverging tees. The value of the loss coefficient depends on the area ratios of the branch and straight pipes and the flow ratio between the branch and straight pipes. For a converging tee, with both area and flow ratios being 0.5, K b = 1:34 and K s = 2:23

1.4

Friction Head Loss and Pressure Drop for Fluid Flow in HVAC Systems

31

For a diverging tee, with both area and flow ratios being 0.5, K b = 1:35 and K s = 0:40 For a bullhead tee, with both area and flow ratios being 0.5, K branches = 1:01 Transition pieces: The value of the loss coefficient depends on the area ratio and the angle of the cone. A typical conservative value is K = 0.35. Screens: The value of the loss coefficient depends on the ratio of the area of the screen (AS) to the area of the duct (AD) and the free area of the screen. For example, When AS/AD = 1.0, K = 1.65 for 50% free area. and K = 0.44 for 75% free area When AS/AD = 0.5, K = 6.60 for 50% free area. and K = 1.76 for 75% free area Butterfly damper: The value of the loss coefficient depends on the percent opening of the damper. K 100%open = 0:60, K 70%open = 4, K 50%open = 18 Damper with three vanes: K = 0:37

1.4.4.2

Entrance and Exit Losses in HVAC Systems

The fluid will experience a loss in energy as it enters into a duct from a source entity such as a plenum or a fan duct with a much larger cross-section area compared to the duct. This loss in energy is termed as entrance loss and it is calculated based on the velocity head or velocity pressure, that is, v2/2g. The loss coefficient used for duct entrance is typically 0.5 based on a sharp-edged entrance [2,11], although it could be lower in cases where smooth or rounded entrances are used. Equation 1.30 is used for calculating entrance loss into a pipe. ΔPentrance ðin:w cÞ = K entrance

v 4005

2

= 0:5

v 4005

2

ð1:30Þ

32

1

Support Areas: Fluid Mechanics

The velocity head/velocity pressure represents the kinetic energy per unit weight of the fluid. When a fluid leaves a duct and enters a large area such as a room, its velocity falls to zero within the room. This is due to the static nature of the fluid in a large space. Thus, the kinetic energy of the fluid is completely dissipated as it exits a duct into a large space. Thus, the loss coefficient for duct exit into a large area is 1.0 [4, 5], which will result in the velocity head/velocity pressure being completely dissipated. Equation 1.31 is used for calculating exit loss from a duct into a large area. ΔPexit ðin:w cÞ = K exit

v 4005

2

= 1:0

v 4005

2

ð1:31Þ

Similarly, for any fitting the pressure loss in inches of water column can be obtained by the following equation: ΔPfitting ðin:w cÞ = K fitting

v 4005

2

ð1:31aÞ

In the Eqs. 1.30, 1.31, and 1.31a, v is the velocity of air in feet per minute (fpm). If the velocity is in m/s, the equation for pressure loss, first in kPa and then in inches of water column for minor losses can be derived as shown here. Note: Standard specific weight of air, γ air = 0.0118 kN/m3 1 kPa = 4.02 in. w c

ΔPkPa = γ air hf,minor = γ air K minor

v2 kN = K minor 0:0118 3 2g m

ΔPkPa = 0:0006K minor v2 = K minor ΔPin:w c = K minor

v 20:345

v 40:825 2

v2 2 × 9:81

2

m s2

)

ð1:31bÞ ð1:31cÞ

1.4

Friction Head Loss and Pressure Drop for Fluid Flow in HVAC Systems

Table 1.1 Pressure Loss Values for HVAC System Components

1.4.4.3

Component Cooling coil Heater (water) Heater (electrical) Filter Intake damper Noise damper Humidifier Heat exchanger Mixing chamber

Pressure loss (psi) 0.0073 0.0044 0.0080 0.0255 0.0007 0.0022 0.0109 0.0145 0.0116

33 Pressure loss (Pa) 50 30 55 175 5 15 75 100 80

Typical Loss Coefficients for Components in HVAC Systems

Typical, conservative values of pressure loss across components in air conditioning systems are given here in Table 1.1 for reference. More accurate values can be obtained from vendors of components.

1.4.4.4

Equivalent Length of Fittings and Total Effective Length of Duct Segments

In the equivalent length method of determining the head loss across fittings, each fitting is represented by an equivalent length of straight pipe represented by Le [4, 5]. In terms of the equivalent length, the head loss across fittings can be calculated by substituting Le for length in the Darcy equation (Eq. 1.26). hf = f

v2 2g

Le D

ð1:32Þ

The relationship between the loss coefficient and the equivalent length for fittings can be obtained by combining Eqs. 1.28 and 1.32. hfitting = K fitting

v2 v2 =f 2g 2g

Le,fitting D

Therefore, Le,fitting =

K fitting D f

ð1:33Þ

Equation 1.33 provides a method for calculating equivalent length for fittings and valves. For fully turbulent flows, as is the case in most HVAC ducts, the Darcy friction factor can be taken as f = 0.02. Substitute this value for f into Eq. 1.33 to obtain the following equation for equivalent length of fittings:

34

1

Le,fitting =

Support Areas: Fluid Mechanics

K fitting D K fitting D = = 50K fitting D f 0:02

ð1:34Þ

The equivalent lengths for fittings can be obtained from the preceding equation. In addition, many sources list equivalent lengths of fittings with varying values. It is best to use the equivalent lengths from vendor catalogues. The total effective length of a duct section is the sum of the length of the duct and the equivalent lengths of all fittings in that section. Example 1.10 A plan view of an air distribution system is shown in the figure along with relevant information.

Assume entrance and exit loss coefficients as 0.50 and 1.0, respectively. The loss coefficients for the elbow and tee are 0.30 and 1.15, respectively. Determine: A. the velocity in each segment of the duct work B. ΔP1-4 (in. w c) C. ΔP1-5 (in. w c). (Solution) From mass (volume) balance for air at junction 3, Volume flow into junction 3 = volume flow out of junction 3. Therefore, Q1 - 2 = Q2 - 3 = Q3 - 4 þ Q3 - 5 = 2000 cfm þ 3000 cfm = 5000 cfm A. Using the volume flow rates and diameters as parameters, determine the velocities from Fig. 1.4 as shown.

1.4

Friction Head Loss and Pressure Drop for Fluid Flow in HVAC Systems

35

From the solution graph, the velocities are as follows: v1 - 2 = v2 - 3 = 2200 ft= min ðfpmÞ v3 - 4 = 2580 ft= min ðfpmÞ v3 - 5 = 2900 ft= min ðfpmÞ Common Solution Segment for Parts B and C Obtain the pressure drops due to entrance, exit, and across the elbow and across the tee in inches of water column as shown. Obtain the entrance loss using Eq. 1.30. ΔPent = 0:5

v1 - 2 4005

2

= ð0:5Þ

ft 2200 min 4005

2

= 0:1509 in:w c

Obtain the exit losses using Eq. 1.31. ΔPexit,3 - 4 = 1:0

v3 - 4 4005

2

= 1:0

ft 2580 min 4005

2

= 0:4150 in:w c

36

1

ΔPexit,3 - 5 = 1:0

v3 - 5 4005

2

= 1:0

Support Areas: Fluid Mechanics 2

ft 2900 min 4005

= 0:5243 in:w c

Obtain the minor losses due to elbow and due to tee using Eq. 1.31a and appropriate loss coefficient for each fitting. ΔPel = K el

v2 - 3 4005

2

= ð0:30Þ

ft 2200 min 4005

2

= 0:0905 in:w c

To calculate the pressure loss across the tee, use the highest velocity in the tee section, that is, v3-5, for conservative purposes. ΔPtee = K tee

v3 - 5 4005

2

ft 2900 min = 1:15 4005

2

= 0:6030 in:w c

B. Obtain the friction pressure loss in each duct segment from the solution graph and calculate the total pressure loss from point 1 to point 4 as shown. ΔP1 - 4 = ΔPent þ ΔP1 - 2 þ ΔPel þ ΔP2 - 3 þ ΔPtee þ ΔP3 - 4 þ ΔPexit,3‐4 0:0108 psi 27:71 in:w c = 0:1509 in:wc þ × 100 ft × 100 ft psi 0:0108 psi 27:71 in:w c þ0:0905 in:w c þ × 60 ft × 100 ft psi 0:0252 psi 27:71 in:w c þ0:6030 in:w c þ × 30 ft × 100 ft psi þ0:4150 in:w c = 1:9477 in:w c C. Obtain the friction pressure loss in each duct segment from the solution graph and calculate the total pressure loss from point 1 to point 5 as shown. ΔP1 - 5 = ΔPent þ ΔP1 - 2 þ ΔPel þ ΔP2 - 3 þ ΔPtee þ ΔP3 - 5 þ ΔPexit,3‐5 27:71 in:w c 0:0108 psi = 0:1509 in:wc þ × 100 ft × 100 ft psi 27:71 in:w c 0:0108 psi × 60 ft × þ0:0905 in:w c þ 100 ft psi 27:71 in:w c 0:027 psi × 40 ft × þ0:6030 in:w c þ 100 ft psi þ0:5243 in:w c = 2:1468 in:w c

1.5

Fans

37

Note: An intake fan positioned near the plenum should be designed based on the highest possible pressure loss, that is, ΔP1-5 + ΔP3-4 + ΔPexit,3-4. The actual pressure loss to be overcome by the fan is the calculated pressure loss multiplied by a factor of 1.20. The reason for this is explained in the next section on fan power calculations.

1.5

Fans

In a duct system, a fan is used to supply the necessary energy to move the fluid through the duct system at system velocity by overcoming all friction losses including duct, fittings, and component losses. From the fundamental principles of physics, energy or work input provided by a device is the force exerted by the device times the distance traversed by the object. Rate of energy consumed is power consumption. Hence, the fan power required is the force provided by the fan times the distance covered by the air movement divided by time. Distance divided by time is the velocity of air and the force exerted is the pressure difference times the area. However, as per the continuity equation (Eq. 1.11) area times velocity is the volume flow rate of air. From the preceding discussion, the theoretical fan power required is the total pressure difference the fan has to overcome times the volume flow rate of air. The corresponding mathematical equation is

FPtheo = Q × ΔTP

ð1:35Þ

where FPtheo is the theoretical fan power required, Q is the volume flow rate of air, and ΔTP is the total pressure difference to be overcome by the fan. Air Movement and Control Association (AMCA) defines system pressure loss as ‘the sum of the static-pressure losses due to friction, shock, dissipation of velocity pressure at the system discharge, and the static-pressure differences between the entry and discharge openings of an air system.’ As discussed earlier, the total pressure is the sum of the static pressure and the velocity pressure which can be represented by the following equation: TP = SP þ VP

ð1:36Þ

In Eq. 1.36, the velocity pressure, VP, accounts for the dissipation of velocity pressure at system discharge (exit loss).

38

1

Support Areas: Fluid Mechanics

The total static pressure has two components – external static pressure (ESP) and internal static pressure (ISP). SP = ESP þ ISP

ð1:37aÞ

External static pressure is the static pressure created downstream of the air handling unit and it includes all the duct losses from the fan until it reaches the discharge point, and it also includes the negative static pressure on the suction side of the fan. Internal static pressure is the static pressure loss across the filters, coils, louvers, dampers, and all components within the air handling unit. ISP is usually provided by the supplier of the unit. Note: The calculated maximum static pressure required is multiplied by a safety factor of 1.20 to allow for the accumulation of dirt on the filters and coils, and to account for fan system effects. The system effect for a fan is the degradation in the fan performance due to the configuration and connection of the fan to the duct. Therefore, Actual SP = 1:20 × Ideal SP

1.5.1

ð1:37bÞ

Fan Power Calculations

From Eq. 1.35, the theoretical fan power required is FPtheo = Q × ΔTP. In the preceding equation, if the fan capacity Q is in ft3/sec and ΔTP is in lbf/ft2, then the theoretical fan power required will be in ft-lbf/sec. Typically, fan capacities are in cfm and the total pressure loss is expressed in terms of inches of water column. Then, the inches of water column must be converted to feet and multiplied by the specific weight of water, 62.3 lbf/ft3 to obtain the pressure loss in lbf/ft2. If a fan delivers a quantity of air (cfm) with a total pressure difference, ΔTP, (inches of water), then the equation for air horsepower is derived as shown here: Air horsepower (AHP) =



§ · ¨ ¸ § 1 ft · § ft 3 · § 1 hp lbf · ¨ ¸ 'TP in. ¨ ¸¸ ¨¨ 62.3 3 ¸¸ ¸ ¨¨ ft ¹ ¨ 550 ft-lbf u 60 s ¸ © 12 in. ¹ © min ¹ © ¨ s min ¸¹ © cfm u 'TP 6356



AHP

ð1:38Þ

cfm u 'TPin.wc 6356

Note that AHP is the power transferred to air. Fans are not 100% efficient due to energy losses. Hence, the fan has to supply more power than the indicated AHP. The Brake Horsepower (BHP) is the power supplied by the fan and it can be calculated using the following equation:

1.5 Fans

39

BHP =

AHP ηfan

ð1:39Þ

Fans are driven by motors and the motor and drive train efficiency must be included to calculate the actual electrical power consumed by the fan, which is commonly reported in kilowatts (kW). Using the conversion factor, 1 hp. = 0.746 kW, the actual electrical power input to the fan can be calculated using the following equation: Pinput ðkWÞ = 0:746

AHP BHP = 0:746 ηfan × ηmotor ηmotor

ð1:40Þ

SI units: If the air flow rate is specified in m3/s and the difference in total pressure is specified in kPa, then the electrical power input required can be calculated using Eq. 1.41: Pinput ðkWÞ = Units:

Qm3 =s × ΔTPkPa ηfan × ηmotor

ð1:41Þ

m3 m3 kN kN  m kJ × kPa  × 2   kW s s s s m

To allow for the accumulation of dirt on the filters and coils, and for possible changes in the installation of duct work, a safety factor (10–20%) is usually used for the pressure difference used in fan power calculations. Example 1.11 An air conditioning unit delivers 10,000 cfm of air. The connected duct, with dimensions of 3 ft × 2 ft, has a total equivalent length of 275 ft. The resistance to air flow due to the grill, filter, and other components in the air handler unit (AHU) is equivalent to 0.255 psig. Determine: A. the total pressure loss in the system in inches of water and in psi. B. the power input required (in kW) if the fan efficiency is 70% and the motor efficiency is 75%. (Solution) A. Use the ‘equal area method’ to find the equivalent diameter of a round duct. The cross-section area of the rectangular duct is equated to the same cross-section area of a round duct and the resulting equation is solved for the equivalent diameter of the round duct. This equivalent diameter is then used in reading Fig. 1.4. π 2 D ) Deq: = 2:764 ft ð33:2 inÞ Acs = 3 ft × 2 ft = 4 eq: Using Q = 10000 cfm and D = 33 in. as parameters, determine the pressure loss per 100 ft from Fig. 1.4 as shown.

40

1

Support Areas: Fluid Mechanics

From the figure, ΔP/100 ft = 0.00324 psi. Calculate the pressure loss for the total equivalent length as specified. ΔPext: =

0:00324 psi × 275 ft = 0:00891 psi 100 ft

Add both the internal and external static pressure losses to obtain the total pressure loss in the system. ΔTP = ΔPint: þ ΔPext: = 0:255 psi þ 0:00891 psi = 0:26391 psi

ΔTP = ð0:26391 psiÞ

27:71 in:w c = 7:3130 in:w c 1 psi

B. Multiply the preceding value by a safety factor of 1.20 to calculate the design total pressure losses to be overcome by the fan. ΔTPdesign = 1:20 × ΔTPactual = 1:20 × 7:3130 in:w c = 8:7756 in:w c Calculate the air horsepower using Eq. 1.38. AHP =

cfm × ΔTPdesign,in:wc 10, 000 cfm × 8:7756 in:w c = = 13:81 hp 6356 6356

Calculate the power input required in kW using Eq. 1.40. Pinput ðkWÞ = 0:746

AHP 13:81 hp = 19:62 kW = ð0:746Þ 0:70 × 0:75 ηfan × ηmotor

1.5

Fans

41

Example 1.12 The layout for an air conditioning system is shown here. Outside air is drawn through Louvre L1, return air is drawn through Louvre L1 and the combined stream is conditioned in the air handling unit (AHU). All the lengths shown in the figure are equivalent lengths including the length of the duct and the equivalent length of fittings. The pressure losses across the louvres are 0.10 in. w c and the pressure drop across the supply diffusers are 0.17 in. w c. The pressure loss across the AHU is 1.35 in. w c. Determine the maximum air horsepower for which the fan needs to be designed.

(Solution)

42

1

Support Areas: Fluid Mechanics

Determine the entrance losses from louvres L1 and L2 into their respective ducts by using the velocities as shown in the excerpt from Fig. 1.4 and using Eq. 1.30. ΔPent,L1 = 0:5

vL1 ‐I 4005

2

ΔPent,L2 = 0:5

vL2 ‐H 4005

2

ft 1000 min 4005

2

= ð0:5Þ

ft 1400 min 4005

2

= ð0:5Þ

= 0:0312 in:w c

= 0:0611 in:w c

Determine the exit losses from the ducts into the respective supply diffusers S1, S2, and S3 by using the velocities as shown in the excerpt from Fig. 1.4 and using Eq. 1.31. v ΔPexit,S1 = 1:0 E‐S1 4005

2

ΔPexit,S2 = 1:0

vF‐S2 4005

2

ΔPexit,S3 = 1:0

vF‐S3 4005

2

ft 1000 min = 1:0 4005

2

ft 1600 min 4005

2

= 1:0

ft 1000 min 4005

2

= 1:0

= 0:0623 in:w c

= 0:1596 in:w c

= 0:0623 in:w c

Define the flow paths as shown and calculate the pressure loss across each path. Choose the flow path with the maximum pressure loss for calculating the maximum air horsepower required for the fan. Path 1: L1-I-A-B-C-D-E-S1 Path 2: L1-I-A-B-C-D-E-F-S2 Path 3: L1-I-A-B-C-D-E-F-G-S3 Path 4: L2-H-I-A-B-C-D-E-S1 Path 5: L2-H-I-A-B-C-D-E-F-S2 Path 6: L2-H-I-A-B-C-D-E-F-G-S3 For all the paths specified, the flow from point I to point E is common. Determine ΔPI ‐ E and use it in calculating pressure drop across each path. The length of the duct segment I-A is not specified, and it can be reasonably assumed to be 4 ft. The flow rate from point I to point E is 1000 cfm and it is applicable to all duct segments in that range. Also, all the ducts in the path I to E are 12 in. in diameter with pressure loss of 0.0065 in. w c/100 ft length. Therefore, the following results can be obtained from the excerpt from Fig. 1.4 shown earlier in the solution

1.5

Fans

43

ΔPI‐E = ΔPI‐A þ ΔPAHU þ ΔPB‐C þ ΔPD‐E = ΔPI‐A þ ΔPB‐C þ ΔPD‐E þ ΔPAHU =

27:71 in:w c psi ð4 ft þ 6 ft þ 50 ftÞ þ 1:35 in:w c 100 ft

0:0065 psi ×

= 1:458 in:w c

ΔPL1 ‐I =

0:0108 psi × 27:71psiin:w c ð10 ftÞ = 0:0299 in:w c 100 ft

ΔPL2 ‐I =

0:0101 psi × 27:71psiin:w c ð130 ftÞ = 0:3638 in:w c 100 ft

ΔPE‐S1 =

0:0108 psi × 27:71psiin:w c ð30 ftÞ = 0:0898 in:w c 100 ft

ΔPF‐S2 =

0:0180 psi × 27:71psiin:w c ð30 ftÞ = 0:1496 in:w c 100 ft

ΔPG‐S3 =

0:0108 psi × 27:71psiin:w c ð30 ftÞ = 0:0898 in:w c 100 ft

0:0101 psi × 27:71psiin:w c ΔPE‐F = ð40 ftÞ = 0:1119 in:w c 100 ft

ΔPF‐G =

0:0108 psi × 27:71psiin:w c ð40 ftÞ = 0:1197 in:w c 100 ft

Using the preceding results determine the pressure loss across each path. Subscript ‘P’ represents ‘Path’. ΔPSD(=0.17 in. w c) is the pressure loss across the supply diffusers, S1, S2, and S3. The units of each pressure loss term is ‘in. w c’, and they are not shown here for the sake of brevity of equations.

44

1

Support Areas: Fluid Mechanics

ΔPP1 = ΔPent,L1 þ ΔPL1 ‐I þ ΔPI‐E þ ΔPE‐S1 þ ΔPSD,S1 þ ΔPexit,S1 = 0:0312 þ 0:0299 þ 1:458 þ 0:0898 þ 0:17 þ 0:0623 = 1:8412 in:w c ΔPP2 = ΔPent,L1 þ ΔPL1 ‐I þ ΔPI‐E þ ΔPE‐F þ ΔPF‐S2 þ ΔPSD,S2 þ ΔPexit,S2 = 0:0312 þ 0:0299 þ 1:458 þ 0:1119 þ 0:1496 þ 0:17 þ 0:1596 = 2:1102 in:w c ΔPP3 = ΔPent,L1 þ ΔPL1 ‐I þ ΔPI‐E þ ΔPE‐F þ ΔPF‐G þ ΔPG‐S3 þ ΔPSD,S3 þ ΔPexit,S3 = 0:0312 þ 0:0299 þ 1:458 þ 0:1119 þ 0:1197 þ 0:0898 þ 0:17 þ 0:0623 = 2:0728 in:w c

ΔPP4 = ΔPent,L2 þ ΔPL2 ‐I þ ΔPI‐E þ ΔPE‐S1 þ ΔPSD,S1 þ ΔPexit,S1 = 0:0611 þ 0:3638 þ 1:458 þ 0:0898 þ 0:17 þ 0:0623 = 2:2050 in:w c ΔPP5 = ΔPent,L2 þ ΔPL2 ‐I þ ΔPI‐E þ ΔPE‐F þ ΔPF‐S2 þ ΔPSD,S2 þ ΔPexit,S2 = 0:0611 þ 0:3638 þ 1:458 þ 0:1119 þ 0:1496 þ 0:17 þ 0:1596 = 2:4740 in:w c ΔPP6 = ΔPent,L2 þ ΔPL2 ‐I þ ΔPI‐E þ ΔPE‐F þ ΔPF‐G þ ΔPG‐S3 þ ΔPSD,S3 þ ΔPexit,S3 = 0:0611 þ 0:3638 þ 1:458 þ 0:1119 þ 0:1197 þ 0:0898 þ 0:17 þ 0:0623 = 2:3169 in:w c

From the preceding results, it is clear that Path 5 has the highest pressure loss at 2.4740 in. w c. Multiply this by the normal factor of safety of 1.20 to obtain the design total pressure drop in inches of water column. ΔTPdesign,in:wc = ð1:20Þð2:474 in:w cÞ = 2:9688 in:w c Calculate the maximum AHP for which the fan should be designed using Eq. 1.38. AHP =

cfm × ΔTPdesign,in:w c 1000 cfm × 2:9688 in:w c = = 0:4671 hp 6356 6356

1.5

Fans

1.5.2

45

Fan Performance Curves, System Curve, and Operating Point

Fan performance curves are plots of static pressure losses in the system vs. volume throughput of the fan [3, 4]. Fan performance curves are generated by the manufacturer based on tests conducted on the fan. The system curve is a plot of the actual pressure losses for the system (as it is configured) vs. the volume throughput handled by the system. Figure 1.6 illustrates fan performance curves at different speeds (rpm) and the system curve. As shown in Fig. 1.6, the fan performance curve and the system curve will intersect. At the intersection point, the volume flow rate handled by the fan will be the same as the air volume flow rate in the system. This intersection point is the operating point for the fan. From the illustration in Fig. 1.6, point 1 is the operating point for the fan operating at 1500 rpm and reading from the graph the operating point at 1500 rpm occurs at about 8300 cfm and static pressure loss of 5 in. w c. The fan speed can be varied. For example, when the fan speed is increased to 2000 rpm, fan laws (discussed in the next section) can be used to plot a new fan curve. However, the system curve will remain the same and point 2 becomes the new operating point as shown in Fig. 1.6. Operating point 2 occurs at about 10,200 cfm and static pressure loss of 7.7 in. w c. Example 1.13 In an air conditioning system, the resistance to air flow due to the grill, filter, and other components in the air handler unit (AHU) is equivalent to 0.0752 psig. The connected duct, with dimensions of 2 ft × 1 ft, has a total equivalent length of 250 ft.

Fig. 1.6 Fan performance curves, system curve, and operating point

46

1 Support Areas: Fluid Mechanics

Generate the system curve for the parameters specified and determine the operating point based on the fan performance curve at 1000 rpm shown in the figure.

(Solution) Use the ‘equal area method’ to find the equivalent diameter of a round duct. The cross-section area of the rectangular duct is equated to the same cross-section area of a round duct and solved for the equivalent diameter of the round duct. This equivalent diameter is then used in reading Fig. 1.4. Acs = 2 ft × 1 ft =

π 2 D ) Deq: = 1:596 ft ð ≃ 20 in:Þ 4 eq:

Using D = 20 in. and various air flow rates as parameters, determine the pressure loss per 100 ft in each case from Fig. 1.4 as shown.

1.5

Fans

47

The external pressure drop in the system is calculated by multiplying the pressure loss per 100 ft by the equivalent length specified, which is, 250 ft. In essence, multiply the value of pressure loss obtained from the graph by a factor of 2.5 (250 ft/ 100 ft = 2.5). The total actual pressure drop in the system is obtained by adding the external pressure loss to the internal pressure loss due to the grill, filter, and other components in the air handler unit (AHU) which is equivalent to 0.0752 psig. The total actual pressure drop is multiplied by the safety factor of 1.20 to obtain the design pressure loss in psi. The results are tabulated and plotted on the given fan performance curve as shown to obtain the operating point for the fan.

From the figure, the operating point for the fan is 11,000 cfm and 0.21 psig total pressure loss.

1.5.3

Fan Laws or Affinity Laws

There are three basic fan laws which provide relationships between fan air flow rate, static pressure, speed, and fan power required. Fan laws are also known as affinity

48

1 Support Areas: Fluid Mechanics

laws, and they are useful in predicting the outcomes when operating conditions are changed from a known fan performance parameters to a new set of fan performance parameters. The fan laws along with the corresponding mathematical representation are given below. • The flow rate handled by a fan is proportional to the speed of the impeller. For example, the fan capacity can be increased by 20% by increasing the impeller speed (rpm) by 20%. Q/N)

Q2 N 2 = Q1 N 1

ð1:42Þ

Q: volume flow rate (cfm, m3/s), N: fan speed (rpm) • The total static pressure overcome by the fan is proportional to the square of the impeller speed. For example, if the static pressure resistance overcome by the fan increases by 10%, then the impeller speed needs to increase by 4.9% (since 1.0492 = 1.10).

SP / N 2 )

2

SP2 N2 = SP1 N1

ð1:43Þ

SP: static pressure resistance overcome by the fan (usually in inches w c). • The power consumed by the fan is proportional to the cube of the impeller speed of the fan. For example, increasing the impeller speed by 10% will increase the power consumption by 33.1% (1.103 = 1.331).

P / N3 )

P2 N2 = P1 N1

3

ð1:44Þ

P: fan power required (hp or W). The density of air can change with altitude. Since static pressure is directly proportional to the density (P = ρgh) and since the fan power required is directly proportional to the pressure resistance (P = ΔTP × Q), the following equations can be used to account for changes in air density: SP2 ρ2 = SP1 ρ1

ð1:45Þ

P2 ρ2 = P1 ρ1

ð1:46Þ

1.5

Fans

49

Example 1.14 For a fan operating at 1500 rpm, the operating point occurs at 4.10 in. w c pressure resistance and 4200 cfm air flow rate. At this operating point, the power consumed is 5.86 hp. If the air flow rate increases by 10%, determine the new speed, and the new power consumption. Specify the revised operating point. (Solution) Calculate the new flow rate, Q2 = 1.1(Q1) = 1.1(4200 cfm) = 4620 cfm. From Eq. 1.42, Q2 N 2 = ) Q1 N 1 Q2 4620 cfm = ð1500 rpmÞ = 1650 rpm N2 = N1 Q1 4200 cfm From Eq. 1.44, P2 = P1

3

N2 N1

= ð5:86 hpÞ

4620 cfm 4200 cfm

3

= 7:80 hp

From Eq. 1.43, SP2 = ðSP1 Þ

N2 N1

2

= ð4:10 in:w cÞ

4620 cfm 4200 cfm

2

= 4:961 in:w c

Therefore, the new operating point is 4620 cfm and 4:961 in:w c

1.5.4

Fan Types and Fan Selection Criteria

Fans can be categorized into two broad categories – centrifugal (impeller driven) and axial (propeller driven). Centrifugal fans are capable of delivering higher volume flow rates. At the same airflow, a centrifugal fan can overcome much higher pressure losses compared to an axial fan and in the process consumes additional power. Centrifugal fans offer the flexibility of using a belt drive or direct shaft-mounted drive. On the flip side centrifugal fans consume more power and produce more noise compared to axial fans and they are used only if the application requires it’s use. In HVAC systems, one of the main applications for centrifugal fans is in air-handling units, which are required to blow air into ductwork. The types of centrifugal fans include forward curved, backward inclined, and airfoil types.

50

1

Support Areas: Fluid Mechanics

Axial fans are simple in construction with propeller blades arranged around a rotating shaft. This results in lower costs compared to centrifugal fans. Axial fans are suitable in situations where both the airflow and static pressure losses are relatively low. Examples of such situations are cooling towers, outdoor air conditioner condensers, and electronic component cooling systems, all of which have open discharge of air resulting in lesser pressure losses. Axial or propeller-type fans include tube-axial, and vane-axial types. The five important factors to be considered for selecting an appropriate fan in any given situation are as follows: type of fan/fan dive, volume flow rate handled, static pressure to be overcome, fan efficiency, and noise level. Type of fan and fan drive used is dictated by the volume flow rate to be handled. Centrifugal fans are preferred when volume flow rate exceeds 5000 cfm and the static pressure to be overcome is greater than 0.50 in. w c. Direct shaft-mounted drives are preferred for volume flow rates below 2000 cfm. Belt drives are preferred at higher volume flow rates. The volume flow rates can be varied by up to 25% by using adjustable pulleys in belt drive systems. AMCA Standard 205–12 specifies minimum Fan Efficiency Grading (FEG) for fans handling a range of volume flow rate. Fan performance data provided by fan manufacturers include a plot of efficiency vs. volume flow rate. Typically, the efficiency curve attains a peak value at a given volume flow rate and it is desirable to operate the fan within 15% of this peak efficiency value. The plots of Peak total efficiency vs. impeller diameter indicate that efficiency increases with increase in impeller diameter. Hence, it is preferrable to have larger diameter fans operating at lower rotation speeds. Although the initial costs are higher, studies have shown that the payback period is very favourable for selection of larger diameter fans. AMCA standards recommend a range of sound levels for fans operating at different facilities. The sound levels are measured in terms of ‘Sone’ and 1 Sone is the noise of a quiet refrigerator heard five (5) feet away in a standard acoustic setting. For example, the sound range recommended for private homes are 1.5–4.0 Sones and the sound range recommended for hospital wards are 2.5–8.0 Sones. The volume flow rate to be handled by a fan can be determined by the following formula: cfm =

V × ACH 60

ð1:47Þ

where V is the volume of the facility/room, and ACH is the air changes per hour recommended by ASHRAE Standard 62.1 Ventilation and Acceptable Indoor Air Quality (IAQ). Some fan manufacturers use minutes per change (MPC) instead of air changes per hour and the relationship between the two quantities is

Practice Problems

51

MPC =

60 ACH

ð1:48Þ

Example 1.15 An office space has dimensions of 30 ft × 40 ft × 10 ft. Determine the cfm to be handled by a fan if 15 air changes are required per hour. (Solution) From Eq. 1.47, cfm =

V × ACH ð30 ft × 40 ft × 10 ftÞ = 60 min 60 hr

15 hr

= 3000 cfm

Practice Problems Practice Problem 1.1 At an altitude of 5280 ft (1609 m), Denver is known as ‘mile high city’. The atmospheric pressure in Denver is 12.23 psia (84.32 kPa) and the temperature is 60 °F (15.6 °C). Determine the specific volume of atmospheric air in Denver both in USCS and SI units. Practice Problem 1.2 A large room uses a fan to draw in atmospheric air at 20 °C through a 40 cm by 30 cm commercial-steel duct 10 m long. Estimate the air flow rate in m3/hr if the room pressure is 10 Pa vacuum. Practice Problem 1.3 16,000 cfm of air is flowing in a 36 in. duct. Determine the friction head loss across 300 ft of the duct. Practice Problem 1.4 Using the same data from Practice Problem 1.2 and Darcy friction factor of 0.02, determine the air flow rate when entrance and exit losses are included. Compare the flow rates with and without the entrance and exit losses and comment. Practice Problem 1.5 The air conditioning coil shown in the figure processes 5 m3/s of air. The duct dimensions are 60 cm × 40 cm. The total effective lengths of the duct sections are specified in the figure. In addition, the pressure losses across components are as follows: return grill (RG) and filter = 0.15 in. w c, A/C coil = 0.25 in. w c, and supply grill (SG) = 0.04 in. w c. The fan efficiency is 65% and the motor efficiency is 75%. Using a safety factor of 1.25, determine the fan power (kW) required to run the system.

52

1

Support Areas: Fluid Mechanics

Solutions to Practice Problems Practice Problem 1.1 (Solution) USCS units: Convert the temperature of air to its absolute value. T = 60 ° F þ 460 ° = 520 ° R Calculate the specific volume of atmospheric air in Denver in USCS units by substituting all the known values into Eq. 1.4a. v = 13:33

ft3 lbm

T 530 ° R

= 13:33

ft3 lbm

520 ° R 530 ° R

14:7 psia P 14:7 psia 12:23 psia

= 15:72 ft3 =lbm SI units: Convert the temperature of air to its absolute value. T = 15:6 ° C þ 273 ° = 288:6 K Calculate the specific volume of atmospheric air in Denver in SI units by substituting all the known values into Eq. 1.4b. v = 0:8264

m3 kg

T 293 K

= 0:8264

m3 kg

288:6 K 293 K

= 0:9779 m3 =kg

101:3 kPa P 101:3 kPa 84:32 kPa

Solutions to Practice Problems

53

Practice Problem 1.2 (Solution) The pressure drop across the duct is the difference between the atmospheric pressure and the room pressure. This pressure drop is due to friction loss in the duct. Therefore, ΔPf = ð10 PaÞ

1 kPa = 0:01 kPa 1000 Pa

Convert the pressure to inches of water column. ΔPf = ð0:01 kPaÞ

4:02 in:w c = 0:0402 in:w c=10 m length 1 kPa

) ΔPf = 0:00402 in:w c=m length Use the ‘equal area method’ to find the equivalent diameter of a round duct. The cross-section area of the rectangular duct is equated to the same cross-section area of a round duct and solved for the equivalent diameter of the round duct. This equivalent diameter is then used in reading Fig. 1.4. Acs = 40 cm × 30 cm =

π 2 D ) Deq: = 39:09 cm ð390 mmÞ 4 eq:

Using ΔPf = 0.00402 in. w c/m length and D = 390 mm as parameters, determine the air flow rate from Fig. 1.5 as shown.

From the graph, the air flow rate is 0.71 m3/s. Convert this to m3/hr. Q = 0:71

m3 3600 s × = 2556 m3 =hr s hr

54

1

Support Areas: Fluid Mechanics

Practice Problem 1.3 (Solution)

Using Q = 16, 000 cfm and D = 36 in. as parameters, determine the pressure drop per 100 ft from Fig. 1.4 as shown. From the figure, ΔPf = 0.0050 psi/100 ft length. Calculate the friction head loss for 300 ft of the duct using the standard specific weight of air. 2

ΔPf = hf = γ air

0:0050 lbf2 × 144 2in in: ft 100 ft

× 300 ft = 28:8 ft air 0:075 lbf ft3

Practice Problem 1.4 (Solution) From the data given in Practice Problem 1.2, the total pressure drop in the duct will still be 10 Pa. However, now this total pressure drop will include entrance loss into the duct and exit loss from the duct. Divide the total pressure loss in kilo pascal by the standard specific weight of air to obtain the total head loss in meters of air. hf ,total =

1 kPa ΔPtotal ð10 PaÞ 1000 Pa = = 0:8475 m air γ air 0:0118 kN m3

Note : kPa 

kN m2

From the data in Practice Problem 1.2, the dimensions of the rectangular duct are 40 cm × 30 cm, and the length of the duct is 10 m. Calculate the hydraulic diameter of the duct using Eq. 1.22. DH,Rect: =

2ða × bÞ 2ð40 cm × 30 cmÞ = 34:29 cm ð0:3429 mÞ = 40 cm þ 30 cm aþb

Solutions to Practice Problems

55

The friction head loss in the duct is calculated by the Darcy equation (Eq. 1.26). The entrance and exit head losses are obtained by the velocity head method, where v2/2g is the velocity head. hf,total = hf,duct þ hentþexit = f v2 2g

hf,total =

L DH

v2 v2 þ ðK ent þ K exit Þ 2g 2g

)

fL þ K ent þ K exit DH

Solve the preceding equation for v and calculate v by substituting all the known values. Entrance and exit loss coefficients are Kent = 0.5 and Kexit = 1.0, respectively. 2ghf,total

v= f

L DH

=

þ K ent þ K exit

m × 0:8475 m s2 10 m þ 0:5 þ 1:0 0:02 0:3429 m 2 × 9:81

= 2:825 m=s Calculate the volume flow rate of air using the continuity equation (Eq. 1.11). Q = Av = ð0:40 m × 0:30 mÞ 2:825

m = 0:339 m3 =s s

Comparison and comments: From the solution to Practice Problem 1.2, the volume flow rate without considering entrance and exit losses is Qw/ 3 o = 0.71 m /s and the volume flow rate with consideration of entrance and exit losses is Qwith = 0.339 m3/s. There is a 52% decrease in volume flow rate due to entrance and exit losses and they must be considered for accuracy. Because of the high magnitude of typical air velocities in ducts, the entrance and exit losses make a significant contribution to the total pressure loss and cannot be ignored.

Practice Problem 1.5 (Solution) Use the ‘equal area method’ to find the equivalent diameter of a round duct. The cross-section area of the rectangular duct is equated to the same cross-section area of a round duct and solved for the equivalent diameter of the round duct. This equivalent diameter is then used in reading Fig. 1.5. Acs = 60 cm × 40 cm =

π 2 D ) Deq: = 55:28 cm ð553 mmÞ 4 eq:

56

1

Support Areas: Fluid Mechanics

Using Q = 5 m3/s and D = 553 mm as parameters, determine the pressure loss per meter length of the duct from Fig. 1.5 as shown in the excerpt.

From the graph, ΔPduct = 0.024 in. w c/m length. The total equivalent length of all the duct segments is Ltotal = 20 m þ 10 m þ 6 m þ 5 m þ 10 m þ 15 m þ 1 m = 67 m Therefore, the total pressure drop due to the duct and fittings is ΔPduct,total =

0:024 in:w c × 67 m = 1:608 in:w c m

From the excerpt of Fig. 1.5, the velocity of air in the duct is, v = 20 m/s. Determine the entrance loss from the return grill to the duct using Eq. 1.31c. In this case, Kminor = Kent.. ΔPent:,in:w c = K ent:

v 20:345

2

= 0:5

20 ms 20:345

2

= 0:4832 in:w c

Determine the exit loss from the duct to the supply grill using Eq. 1.31c. In this case, Kminor = Kexit. ΔPexit,in:w c = K exit

v 20:345

2

= 1:0

20 ms 20:345

2

= 0:9664 in:w c

References

57

Determine the total pressure loss in the system as shown. Subscript ‘RGF’ represents return grill and filter, and subscript ‘SG’ represents supply grill. The units of each pressure loss term is in. w c, and they are not shown here for the sake of brevity of equations. ΔTPin:wc = ΔPRGF þ ΔPent: þ ΔPduct,total þ ΔPAC coil þ ΔPexit þ ΔPSG = 0:15 þ 0:4832 þ 1:608 þ 0:25 þ 0:9664 þ 0:04 = 3:4976 in:w c Multiply the calculated total pressure drop in the system by the factor of safety (1.25) and convert the result to kPa. ΔTPkPa,act: = ð1:25ÞðΔTPin:wc Þ

1 kPa 4:02 in:w c

= ð1:25Þð3:4976 in:w cÞ

1 kPa 4:02 in:w c

= 1:0875 kPa Calculate the power input required using Eq. 1.41. Qm3 =s × ΔTPkPa 5 ms × 1:0875 kPa = = 11:154 kW ηfan × ηmotor 0:65 × 0:75 3

Pinput ðkWÞ =

References 1. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook Fundamentals, I P edn. ASHRAE, Peachtree (2021) 2. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Duct Fitting Database (DFDB). ASHRAE, Peachtree (2016) 3. Bhatia, A.: HVAC – How to Size and Design Ducts. CED Engineering, https:\crwww. cedengineering.com (2007) 4. Mitchell, J.W.: Fox and McDonald’s Introduction to Fluid Mechanics, 10th edn. Wiley, New York (2020) 5. Nandagopal, N.S.: Fluid and Thermal Sciences – A Practical Approach for Students and Professionals. Springer Nature, Cham (2022)

Chapter 2

Support Areas: Heat Transfer

2.1

Relevance of Heat Transfer in HVAC Design and Calculations

The function of heating, ventilation, and air-conditioning (HVAC) systems is to provide the required cooling or heating for homes and facilities [1]. As such, heat transfer plays a key role in designing strategies for preserving the cooling or heating provided by HVAC systems. Heat transfer principles are used in calculating heat gain by air-conditioned spaces and also in calculating heat losses from heated spaces [1].

2.2

Heat Transfer Principles and Modes of Heat Transfer

Heat transfer is the flow of heat from a high-temperature zone to a low-temperature zone. Heat is a form of energy and therefore heat transfer rates are represented by units of energy flow per unit time. The symbol used for the heat transfer rate is ‘q’ and the units for heat transfer rate are Btu/hr. (USCS / I-P System) and kW (SI). Note
Js. that watt (W) is the rate of energy transfer in SI units, and W
Joules s Conversion factor: 1 kW = 3412 Btu/hr

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 N. S. Nandagopal, HVACR Principles and Applications, https://doi.org/10.1007/978-3-031-45267-3_2

59

60

2

Support Areas: Heat Transfer

Electrical Analogy The equation for the current flow is as follows: Current Flow =

Potential Difference V )I= Resistance R

Similarly, the equation for heat transfer can be written in terms of driving force and resistance to heat flow [1, 2, 4]. The driving force for heat transfer is the temperature difference between the hot and cold regions. The resistance to heat flow is known as ‘thermal resistance’ and it is represented by Rth. Heat Flow =

Temperature Difference Resistance to heat flow ΔT q= Rth

ð2:1Þ

The denominator in Eq. 2.1 indicates the summation of all the thermal resistances since there can be multiple thermal resistances in building entities (such as walls and roofs). The equation for thermal resistance depends on the particular mode of heat transfer. The different modes of heat transfer are discussed in the next section. The units for thermal resistance can be obtained by rearranging Eq. 2.1. Rth =

ΔT °F °C

q Btu=hr kW

It is useful to write Eq. 2.1 in terms of heat flow per unit cross-section area (also known as heat flux) perpendicular to the heat flow direction. Correspondingly, the thermal resistance becomes thermal resistance per unit cross-section area, and it can be written as Rth,ua, where ‘ua’ represents unit cross-section area [1, 4]. q = A

ΔT Rth,ua

ð2:2Þ

The units for thermal resistance per unit cross-section area are as follows: ∘

hrft2  F=Btu ðUSCS=I - P SystemÞ

∘

m2  C=W ðSIÞ

Conversion factor: 1 hr ‐ ft2 ‐ ° F/Btu = 0.1761 m2  ° C/W

2.3

Conduction Heat Transfer Principles

2.2.1

61

Heat Transfer Modes

The principal modes of heat transfer are conduction, convection, and radiation [1, 2, 4 ]. Conduction heat transfer is the transfer of heat through solid media such as a metal pipe wall, building walls, and insulation. Flow of heat through an insulated roof is a typical example of conduction heat transfer. The heat will flow through the insulation and the roof material by means of conduction mechanism. Convection heat transfer is heat transfer between a solid surface and an adjoining fluid in contact with the solid surface. Consider a flat hot plate exposed to ambient air. Heat is transferred from the hot plate surface to the cooler air by means of convection mechanism. Unlike conduction and convection, radiation heat transfer does not require any media (solid or fluid) for heat transfer to occur. The radiation heat from sun reaches earth after traveling through vast, empty space (vacuum). Thus, the only requirement for radiation heat transfer is a temperature difference between the hot and cold entities.

2.3

Conduction Heat Transfer Principles

Fourier’s Law of Heat Conduction Conduction heat transfer is governed by Fourier’s Law of Heat Conduction [1, 2, 4]. According to Fourier’s Law, for one-dimensional heat flow, the rate of heat transfer per unit cross-section area is proportional to the temperature gradient in the direction of heat flow. The heat transfer per unit cross-section area (q/A) is also known as heat flux. Fourier’s Law can be mathematically written as q dT = -k A dx

ð2:3Þ

Since the temperature gradient in the direction of heat flow is always a negative quantity, a negative sign is necessary to keep heat flow as a positive quantity. The variable ‘k’ in Fourier’s Law is called thermal conductivity. Thermal conductivity is a physical property of materials, and it has the following units: ∘

k
Btu=hrft F

ðUSCS=I - P SystemÞ

Conversion Factor: 1 Btu/hr-ft-°F = 1.731 W/m.K

k
W=m:K

ðSIÞ

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Fig. 2.1 Fourier’s law of heat conduction

Typically, metals have relatively high thermal conductivities while insulating materials have low thermal conductivities. Single Layer Heat Conduction Through a Rectangular Slab Consider heat flow across a solid slab of thickness ΔX as shown in Fig. 2.1. The heat flow is normal to the cross-section area of the slab. After separating the variables, Fourier’s Law (Eq. 2.3) can be integrated to obtain the following result: X2

q A

T2

dX = - k X1

dT T1

q ΔT =k A ΔX

ð2:4Þ

ΔT = T1 - T2 and ΔX = X2 - X1. Equation 2.4 is the integrated form of Fourier’s Law for one-dimensional heat flow through a rectangular slab [1, 2, 4]. Equation 2.4 can be written in the form of thermal resistance per unit area as shown here. q ΔT ΔT = ΔX =k A ΔX k

ð2:5Þ

Compare Eq. 2.5 with the basic equation for heat flow per unit area (Eq. 2.2) reproduced here for quick reference. q = A

ΔT Rth,ua

2.3

Conduction Heat Transfer Principles

63

From the comparison of the two equations, the conduction thermal resistance per unit area is as follows: Rcond,ua =

ΔX k

ð2:6Þ

Equation 2.5 can be written in terms of thermal resistance per unit area. q ΔT = A Rcond,ua

2.3.1

ð2:7Þ

Multilayer Conduction

Conduction through multiple layers can be modelled by adding up the thermal resistances of each layer [1, 4]. Consider heat flow through two solid layers, Layer 1 and Layer 2 as shown in Fig. 2.2. At steady state, the overall heat flow across the multiple layers as well as the heat flow across each layer will be constant and identical in value. The heat transfer rate can be calculated by dividing the overall temperature difference by the sum of the resistances of all layers (Eq. 2.2). The heat transfer rate can also be calculated by dividing the temperature difference of a particular layer by the thermal resistance of the same layer [4]. Once again, it is useful to write the equations in terms of heat transfer per unit cross-section area normal to the heat flow direction. q ΔT overall T -T = ΔX11 ΔX3 2 = A Rth,ua k1 þ k2

Fig. 2.2 Conduction through multiple layers

ð2:8Þ

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2

Support Areas: Heat Transfer

q ðΔT Þ1 T 1 - T 2 ðΔT Þ2 T 2 - T 3 = ΔX 1 = = ΔX 2 = R1,ua R2,ua A k1

ð2:9Þ

k2

Example 2.1 The wall of a home consists of a 30 mm thick common brick (k = 0.6929 W/m.K) on the outside followed by a 4-mm layer of plaster (k = 0.4750 W/m.K). During the winter months, the inside temperature is maintained at 21 °C and the outside temperature averages around 7 °C. Calculate: A. The heat loss per unit area of the wall. B. The temperature at the interface of the brick and the plaster. (Solution) A. Calculate the heat loss per unit area of the wall by using Eq. 2.8. Layer 1 is the plaster adjacent to the inside air and Layer 2 is the brick exposed to the outside air. ðT - T Þ q = Δx11 Δx32 = A þ k1

k2

21 ° C - 7 ° C 4 mm 1000 mm m W 0:4750m:K

þ

30 mm 1000 mm m W 0:6929m:K

= 271 W=m2

B. Calculate the temperature at the interface of the brick and the plaster by using Eq. 2.9. W 21 ° C - T q T1 - T2 = ΔX 1 ) 271 2 = 0:004 m 2 ) T 2 = 18:7 ° C A m W k1

2.3.2

0:4750m:K

R-values for Insulation and Building Materials

HVAC design and calculations routinely use R-values for insulating and building materials [1, 4]. The R-value is the same as the conduction resistance per unit crosssection area as specified in Eq. 2.6. R‐value = Rcond,ua =

ΔX k

ð2:10Þ

Units for R-value: Since insulation thickness is specified in inches or centimetres, R-values are reported in terms of conduction resistance per inch thickness (or per centimetre thickness). However, the units of R-value will be the same as the units of heat transfer resistance per unit area, that is, hr-ft2-°F/Btu or m2.°C/W as illustrated

2.3

Conduction Heat Transfer Principles

65

Table 2.1 Typical R values of insulation and building materials

Material Fiber glass (batt) Rock wool (batt) Cellulose (blown) Polyurethane (foam) Polyisocyanurate Common brick Concrete Granite Limestone Wood Plywood Particle board Insulated vinyl siding Asphalt shingles Wood shingles Single pane glass window (0.25 in. thick) Double pane glass window with 0.25 in. air gap Double pane glass window with 0.50 in. air gap

R value per inch thickness hr-ft2-°F/Btu 3.67 3.53 3.63 6.15 7.20 0.20 0.42 0.05 0.08 1.24 1.25 1.06 1.80 0.44 0.97 0.91 1.69 2.04

R value per cm thickness m2. K/W 0.2543 0.2446 0.2516 0.4262 0.4990 0.0139 0.0291 0.0035 0.0055 0.0859 0.0866 0.0735 0.1247 0.0305 0.0672 0.0631 0.1171 0.1414

Note: To obtain the R Value per centimetre thickness in m2. K/W multiply the USCS values in the table by 0.0693

in the following example. Consider 1 in. of fiberglass insulation, which has a thermal conductivity of 0.023 Btu/hr-ft-°F. Substitute these values into Eq. 2.10 and use the conversion factor from inch to feet. R‐value =

ΔX ð1 in:Þ 121 ftin = 3:623 hr‐ft2 ‐ ° F=Btu = Btu k 0:023 hr‐ft‐ °F

The preceding R-value is commonly reported and used for 1 in. thick fiberglass insulation. Table 2.1 lists R-values of commonly used insulation and building materials. The values in the table are indicated values and more accurate data can be obtained from the vendors. Table 1 Building and Insulation Materials: Design Values, Chapter 26 in the ASHRAE Fundamentals Handbook [1] is an extremely important and valuable resource in HVAC design calculations involving heat transfer.

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Example 2.2 3 A typical roof configuration consists of 16 in: asphalt shingles, followed by 12 in: thick plywood sheathing and 3 58 in: fiberglass batting. Using the R-values provided in Table 2.1, determine the total conduction resistance per unit area (combined R-value) for this roof configuration. (Solution) The R-values given in Table 2.1 are per inch thickness of the material. Therefore, to obtain the combined R-value of the roof assembly, multiply the table R-value of each component by its thickness and add up the resulting component values as shown in the following table. Component Asphalt shingles

R-value (per in.), hr-ft2-°F/ Btu 0.44

Plywood

1.25

Fiberglass batting

3.67

Thickness (in) 3 16 1 2 3 58

∑Rcond, ua

2.4

Resistance, R (hr-ft2-°F/ Btu) 0.0825 0.625 13.304 14.01

Convection Heat Transfer

As mentioned earlier, convection heat transfer is heat transfer between a solid surface and a fluid adjacent to the solid surface. In HVAC applications, wall surfaces are typically exposed to outside air on one side and the conditioned air on the other side. Hence it is important to consider convection effects in heat gain/heat loss calculations.

2.4.1

Newton’s Law of Cooling

The heat transfer mechanism in convection heat transfer involves the motion and circulation of fluid particles. The equation governing convection heat transfer is called Newton’s Law of Cooling, which states that the heat transfer per unit area of the exposed surface (also known as heat flux) is proportional to the difference between the surface temperature Ts and the temperature of the fluid far away from the surface, T1, also known as the bulk temperature of the fluid [2, 4]. The proportionality constant in Newton’s Law of Cooling is known as the heat transfer coefficient, h, also known as film coefficient or film conductance.

2.4

Convection Heat Transfer

67

The heat transfer coefficient h has the units Btu/hr-ft2-oF (USCS / I-P System) and W/m2.°C (SI). Equation 2.11 is the mathematical representation of Newton’s Law of Cooling. q = hAs ðT s - T 1 Þ

ð2:11Þ

where As is the area of the surface in contact with the fluid. Conversion factor: 1 W/m2.°C = 0.1761 Btu/hr-ft2-°F

2.4.2

Convection Heat Transfer Resistance

Equation 2.11 can be rearranged to obtain the following equation: q=

Ts - T1 1 hAs




ΔT Rconv

ð2:12Þ

Comparing the terms on the right side of Eq. 2.12, the resistance to convection heat transfer [2, 4] can be written as shown in Eq. 2.13. Rconv =

1 hAs

ð2:13Þ

Equation 2.11 can be written in terms of heat transfer rate per unit surface area. q T -T ΔT conv = s 1 1
R As conv,ua h

ð2:14Þ

Comparing the terms on the right side of Eq. 2.14, the convection heat transfer resistance per unit surface area can be written as shown in Eq. 2.15. Rconv,ua =

1 h

ð2:15Þ

Thus, the resistance to convection heat transfer per unit surface area is the reciprocal of the heat transfer coefficient. Resistance is always inversely proportional to the transfer coefficient. (continued)

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Note: Table 3: Effective Thermal Resistance of Plane Air Spaces, Chapter 26, ASHRAE Fundamentals Handbook [1], is a useful reference in HVAC calculations involving building assemblies (composite wall systems and composite roof systems).

2.4.3

Typical Convection Heat Transfer Coefficients Used in HVAC-R

In HVAC-R applications, convection heat transfer takes place between surfaces in the conditioned space and air and also between exposed surfaces to outside environment and air. The heat transfer coefficient for still air is typically in the range 2 Btu/hr-ft2-°F (11 W/m2.K) to 6 Btu/hr-ft2-°F (34 W/m2.K) and the heat transfer coefficient for moving air is typically in the range 10 Btu/hr-ft2-°F (57 W/m2.K) to 60 Btu/hr-ft2-°F (340 W/m2.K). The following values (in Table 2.2) of heat transfer coefficients (also known as Surface Film Coefficients) and corresponding resistances are listed in Table 10, Chapter 26 of ASHRAE Fundamentals Book [1].

2.4.4

Overall Heat Transfer Coefficients in Conduction: Convection Systems

Many practical situations involve a combination of conduction and convection heat transfer [1, 4]. For example, heat transfer from a heated room to cooler ambient air involves convection heat transfer from room air to inside wall surface, conduction heat transfer through the room wall, and finally convection heat transfer from the outside wall surface to ambient air. Thus, this situation involves three resistances to heat flow. Consider the conduction–convection system shown in Fig. 2.3. Table 2.2 Convection heat transfer coefficients/resistances for air in HVAC systems Indoor air Upward heat flow Downward heat flow Horizontal heat flow Outdoor air Winter Summer

h (Btu/hr-ft2-°F)

R (hr-ft2-°F/Btu)

h (W/m2. K)

R (m2.K/W)

1.63 1.08 1.46

0.61 0.93 0.68

9.26 6.13 8.29

0.11 0.16 0.12

6.00 4.05

0.17 0.25

34 23

0.03 0.04

2.4

Convection Heat Transfer

69

Fig. 2.3 Conduction – convection system

The heat flow is represented by Eq. 2.16 and the three resistances are included in the denominator. q=

ΔT overall T room - T air = Rconv,i þ Rcond þ Rconv,o Rth

ð2:16Þ

Under steady-state conditions, the heat flow is constant through the room (inside) air, through the wall, and through the ambient (outside) air. The cross-section area of the wall perpendicular to the heat flow direction is also constant. Eq. 2.16 can be written in terms of heat flow per unit cross-section area and the corresponding resistances (due to conduction and convection) per unit area (Eqs. 2.6 and 2.15). q ΔT overall ΔT overall = = Rconv,i,ua þ Rcond,ua þ Rconv,o,ua A Rth,ua ΔT overall = 1 ΔX 1 þ þ hi k ho

ð2:17Þ

where hi and ho are the convection heat transfer coefficients for the inside and outside air, respectively. The three heat transfer resistances can be combined and represented by a single resistance, which is the reciprocal of the overall heat transfer coefficient. The overall heat transfer coefficient is commonly represented by U. q ΔT overall ΔT overall = = 1 1 A h1 þ ΔX þ U k ho i

ð2:18Þ

The following equations can be written by comparing terms in the denominator of Eq. 2.18 and by cross multiplying the terms in Eq. 2.18.

70

U=

1 hi

1 = 1 þ ΔX k þ ho

2

Support Areas: Heat Transfer

1 Rth,ua

ð2:19Þ

q = UAΔT overall

ð2:20Þ

The overall heat transfer coefficient is the inverse of the overall resistance to heat flow in conduction–convection systems. Overall heat transfer coefficients are extensively used in HVACR design equations as illustrated in the following example problems. Example 2.3 A residential wall consists of the following components while traversing from outside to inside: 0.35 in. insulated vinyl siding, 0.5 in. plywood, 3.50 in. rock wool insulation, 0.8 in. particle board. The outside temperature is 15 °F and the inside temperature is 70 °F. Using the data available from Tables 2.1 and 2.2, determine, A. B. C. D. E.

the overall heat transfer coefficient the heat loss from the home per unit cross-section of the wall the per cent of total heat transfer resistance due to each entity the temperature at the surface of the vinyl siding the increase in the thickness of rock wool insulation required to maintain the same heat loss as in part B when the outside temperature drops to -10 °F.

(Solution) A. Table 2.1 provides the R-values (per in. thickness) of different components of the wall, which are multiplied by the respective thicknesses to obtain the thermal resistance of each component. Table 2.2 provides the convection resistances for the outside air (winter) and inside air. The results are summarized in the following table. Component Outside air Vinyl siding Plywood Rock wool insulation Particle board Inside air

R-value (per in.), hr-ft2-°F/ Btu (Winter) 1.80 1.25 3.53

Thickness (in)

1.06 (Horizontal flow)

0.80

0.35 0.50 3.50

∑Rth, ua

Resistance, R (hr-ft2-°F/ Btu) 0.17 0.63 0.625 12.355 0.848 0.68 15.308

2.4

Convection Heat Transfer

71

From Eq. 2.19, the overall heat transfer coefficient is the reciprocal of the sum of the thermal resistances per unit area. U=

1 1 = = 0:0653 Btu=hr‐ft2 ‐ ° F Rth,ua 15:308 hr‐ft2 ‐ ° F Btu

B. Calculate the heat loss per unit cross-section area of the wall using Eq. 2.20. q = UAΔT overall )

q Btu = UΔT overall = 0:0653 ð70 ° F - 15 ° FÞ = 3:592 Btu=hr‐ft2 A hr‐ft2 ‐ ° F

C. The per cent heat transfer resistance (%HTR) offered by each entity, ‘i’ can be calculated by using the following equation: %HTR =

Ri 100 Rth,ua

The results are given in the last column of the following table, which is an appended version of the table presented in the solution to part A. Component Outside air Vinyl siding Plywood Rock wool insulation Particle board Inside air

R-value (per in.), hr-ft2-° F/Btu (Winter) 1.80 1.25 3.53

Thickness (in)

1.06 (Horizontal flow)

0.80

0.35 0.50 3.50

∑Rth, ua

Resistance, R (hr-ft2-° F/Btu) 0.17 0.63 0.625 12.355 0.848 0.68 15.308

%HTR 1.11% 4.11% 4.08% 80.68% 5.56% 4.46%

D. At steady state, the heat flow through each layer is the same as the heat flow through the wall assembly. In each case, the heat flow is the temperature difference across the layer divided by thermal resistance of that layer. If Ts is the temperature at the surface of vinyl siding, then for the outside convection layer, q ΔT overall ΔT o,conv: Btu T - 15 ° F = ) 3:592 = s ) T s = 15:6 ° F = 2 0 ‐ F Ro,conv: A Rth,ua hr‐ft2 0:17 hr‐ft Btu E. The heat loss is the same as in part B, but the temperature difference will increase due to the drop in the outside temperature. Hence, there should be a corresponding increase in the overall thermal resistance. Subscript ‘1’ represents

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the old situation and subscript ‘2’ represents the new situation with the drop in outside temperature to -10 °F. Therefore, from Eq. 2.17, 70 ° F - ð- 10 ° FÞ Btu q ΔT overall,2 ) 3:592 = ) = 2 A Rth,ua,2 Rth,ua,2 hr‐ft Rth,ua,2 = 22:27 hr‐ft2 ‐ ° F=Btu The difference in thermal resistance between the new situation and the old situation is exclusively due to the increase in the thickness of rock wool insulation. The data for the old situation and R-value per inch for rock wool insulation can be obtained from the table in the solution to part B. Subscript ‘RW’ represents rock wool insulation. Consequently, Rinc:,RW =

Rth,ua,2 -

Rth,ua,1

hr‐ft2 ‐ ° F hr‐ft2 ‐ ° F - 15:31 Btu Btu = 6:96 hr‐ft2 ‐ ° F=Btu = 22:27

6:96 hr‐ftBtu‐ ° F 2

ΔX inc:,RW =

‐°F 3:53 hr‐ft Btu‐in: 2

= 1:97 in:

Therefore, the increase in thickness of rock wool insulation required after the temperature drop is 1.97 in. Example 2.4 Consider heat transfer through the roof of a building during the summer months when the average ambient temperature is 90 °F and the building interior is maintained at 70 °F. The composition of the roof is a top layer of asphalt shingles (R = 0.440 hr-ft2-°F/Btu), followed by 2.75 in. layer of fiberglass batt insulation (k = 0.026 Btu/hr-ft-°F), followed by 0.20 in. thick metal deck with negligible R-value. Adjacent to the metal deck is 1.5 in. air space with emissivity of 0.50 followed by 1 in. thick sound deadening board (k = 0.031 Btu/hr-ft-°F). The preceding thermal conductivity and R-values are from Table 1: Data for Building and Insulation Materials, Chapter 26, ASHRAE Fundamentals Handbook [1]. Determine A. the U value of the composite roofing. B. the building heat gain through the roof per unit cross-section area of the roof. (Solution) An excerpt of Table 3: Effective Thermal Resistance of Plane Air Spaces, Chapter 26, ASHRAE Fundamentals Handbook [1] is shown here. In this case, the

2.4

Convection Heat Transfer

73

air space is horizontal, and the heat flow is downward. The average temperature is 80 °F and the temperature difference is 20 °F.

From the table excerpt, for an effective emissivity of 0.50, the resistance of the plane air space by interpolation is Rair space = 1.55 hr ‐ ft2 ‐ ° F/Btu. From Table 2.2, the R-value (summer) of outdoor air is ROA = 0.25 hr ‐ ft2 ‐ ° F/Btu and the R-value of indoor air (downward heat flow) is RIA = 0.93 hr ‐ ft2 ‐ ° F/Btu. Subscripts used in the equation for calculation of U value are as follows: ‘AS’ – Asphalt Shingles, ‘FG’ – Fiber Glass insulation, ‘SDB’ – Sound Deadening Board. The units for R-values are hr ‐ ft2 ‐ ° F/Btu and they are not shown for the sake of brevity. Calculate the U value of the composite roofing using Eq. 2.19. A:

U=

1 Rth,ua

1 ΔX FG ΔX SDB ROA þ RAS þ þ Rair sp þ þ RIA k FG kSDB 1 = 2:75 1 ft ft 12 12 0:25 þ 0:44 þ þ 1:55 þ þ 0:93 Btu Btu 0:026 0:031 hr‐ft‐ ° F hr‐ft‐ ° F = 0:0682 Btu=hr‐ft2 ‐ ° F =

B. Calculate the heat loss per unit cross-section area of the roof using Eq. 2.20. q Btu = UΔT overall = 0:0682 ð90 ° F - 70 ° FÞ A hr‐ft2 ‐ ° F = 1:364 Btu=hr‐ft2 Chapter 27 in the ASHRAE Fundamentals Handbook [1] contains useful illustrative examples on HVAC design calculations for wall and roof assemblies.

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Fig. 2.4 Shell and tube heat exchanger schematic. (Source: https://commons.wikimedia.org/wiki/ File:Shell_heat_exchanger_LS.JPG, Public Domain)

2.5

Heat Exchangers

Heat exchangers [1, 3, 4] facilitate heat transfer from a hot fluid to a cold fluid. Condensers are shell and tube heat exchangers [1, 4] with condensing vapor on the shell side and cooling water in the tubes. Condensers are an integral part of refrigeration cycles, and they condense superheated refrigerant vapor resulting from the compression process. Figure 2.4 illustrates the typical layout of a shell and tube heat exchanger.

2.5.1

Heat Balance

The heat balance [3, 4] in a heat exchanger is an application of conservation of energy principle. Ideally, the exchanger is well insulated, resulting in negligible heat transfer to/from the surroundings. With this reasonable assumption, all the energy released by the hot fluid (subscript ‘h’) is absorbed by the cold fluid (subscript ‘c’). The rate of heat exchange between the hot and cold fluids is also known as the ‘Heat Duty’ of the exchanger and it is represented by the symbol q. The heat balance equation is as follows: q = m_ h cph ΔT h = m_ c cpc ΔT c

ð2:21Þ

where m_ h , m_ c : mass flow rates of the hot and cold fluids, respectively cph, cpc: specific heats (heat capacities) of the hot and cold fluids, respectively ΔTh: temperature decrease of the hot fluid ΔTc: temperature increase of the cold fluid

2.5

Heat Exchangers

75

In a condenser, the condensing vapor releases the latent heat of condensation (which is more significant than any sensible heat changes) and the heat balance equation is. heat released by the condensing vapor = heat gained by cooling water q = m_ R ðh2 - h3 Þ = m_ w cpw ΔT w

ð2:22Þ

where subscript ‘R’ refers to the refrigerant and subscript ‘w’ refers to cooling water. Further, q = heat duty of the condenser (Btu/hr, kW) m_ R = mass flow of the refrigerant (lbm/hr, kg/s) h2 = enthalpy of superheated refrigerant vapor, and h3 = enthalpy of saturated/sub-cooled refrigerant liquid h2 and h3 are obtained from refrigerant property tables or charts (explained in detail in Chap. 3: Thermodynamics) m_ w = mass flow rate of cooling water (lbm/hr, kg/s) cpw = heat capacity of water (1 Btu/lbm-°F, 4.187 kJ/kg.°C) ΔTw = temperature difference of cooling water = Tw,out - Tw,in (°F, °C) Example 2.5 The heat duty of a condenser in a refrigeration cycle is 30,000 Btu/hr. Cooling water enters the condenser at 65 °F and leaves at 80 °F. The schematic for a refrigeration cycle is shown in the figure. If the gpm of cooling water required in the condenser per ton of refrigeration, TR, (12,000 Btu/hr) is 3.5, calculate the power input to the compressor in kW.

(Solution) Calculate the mass flow rate of water using the heat duty of the condenser and heat balance (Eq. 2.22). heat rejected in the condenser = heat absorbed by water.

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Therefore, Q_ out = qcond = m_ w cpw ΔT w m_ w =

30, 000 Btu qcond hr = Btu = 2000 lbm=hr cpw ΔT w 1 lbm ° F × 15 ° F

The standard density of water is ρw = 8.34 lbm/gal. Calculate the volume flow rate of cooling water. 1 hr 2000 lbm m_ hr 60 min = 4:0 gpm V_ water = w = ρw 8:34 lbm gal

Determine the tons of refrigeration achieved, Q_ in , using the specified ratio of (the) gpm of water required per ton of refrigeration. gpm water 4:0 gpm = 3:50 ) Q_ in = 1:143 TR = ton refrig: Q_ in Convert TR achieved to Btu/hr. Q_ in = ð1:143 TRÞ

12, 000 Btu hr = 13, 716 Btu=hr TR

Consider the boundary indicated by the red, dashed line for the refrigeration cycle. Energy is added to the cycle from two sources – heat absorbed by the refrigerant from the refrigerated space (occurring in the evaporator of the refrigeration system), and the work input to the compressor. Energy is removed from the refrigeration cycle in the form of the heat rejected by the condenser. Energy balance for the refrigeration cycle results in the following set of equations, which can be solved for the power input to the compressor. energy added in the cycle = energy removed from the cycle _ c,in = Q_ out ) Q_ in þ W _ c,in = Q_ out - Q_ in = 30, 000 Btu - 13, 716 Btu = 16, 284 Btu=hr W hr hr Convert the power input to the compressor to kW. _ c,in = 16, 284 Btu W hr

1 kW = 4:77 kW 3414 Btu hr

2.5

Heat Exchangers

77

Example 2.6 Ambient cold air at 40 °F is heated by direct contact with a finned tube heat exchanger. Hot water enters the tubes at 150 °F and leaves at 110 °F. Due to fin efficiency, the temperature of fin surfaces can be considered to be 80% of average temperature of hot water in the tube. A fraction of ambient air completely bypasses the surfaces of the fins. The air leaving the heat exchanger is to be supplied to a facility at a temperature of 70 °F. Determine the fraction of air that bypasses the surfaces of the fins in the heat exchanger. (Solution) Ideally, the temperature of fin surfaces is the average temperature of hot water flowing in the tube. However, this will not be the case due to the efficiency of the fins. Therefore, the temperature of the fin surfaces can be calculated as shown. T s,actual = ðηfins ÞðT s,ideal Þ = ðηfins Þ

T w,in þ T w,out 150 ° F þ 110 ° F = ð0:80Þ = 104 ° F 2 2

Let m_ a be the total mass flow rate of ambient air and f be the fraction of ambient air that bypasses the surfaces of the fins in the heat exchanger. ð1 - f Þm_ a is the mass flow of air that is heated by the fin surfaces to the actual surface temperature of the fins. The fraction of air bypassing the fin surfaces can be determined by heat balance due to mixing of bypass air and heated air as shown. Enthalpy of bypass air + enthalpy of heated air = enthalpy of supply air to facility f m_ a cpa ðT a,in - T ref Þ þ ð1 - f Þ m_ a cpa ðT s,actual - T ref Þ = m_ a cpa T a,out - T ref ) f ð40 ° F - 0 ° FÞ þ ð1 - f Þð104 ° F - 0 ° FÞ = 70 ° F - 0 ° F ) f = 0:53ð53%Þ

In preceding equations, Tref is an arbitrary reference temperature used for enthalpy calculations (since Δh = cpΔT ) and Tref is taken as 0 ° F for convenience.

2.5.2

Log Mean Temperature Difference (LMTD)

The driving force for heat transfer in a heat exchanger is the temperature difference between the hot fluid and the cold fluid. In a heat exchanger, the hot fluid continuously loses heat, and the cold fluid continuously gains heat across the length of the exchanger. Hence, the temperature difference between the hot fluid and the cold fluid varies along the length of the exchanger. The log-mean temperature difference (LMTD), represents an average value of the temperature difference between the hot fluid and the cold fluid. The log-mean temperature difference can be calculated by using the following formula written in terms of temperature differences at end ‘1’ and at end ‘2’ of the heat exchanger.

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2

Support Areas: Heat Transfer

1

2 Hot Th1  o Th 2

Fig. 2.5 Flow schematic for hot and cold fluids in counter-flow heat exchanger

Cold Tc1 m  Tc 2 'T1

ΔT LMTD =

ΔT 1 - ΔT 2 ln

ΔT 1 ΔT 2

=

ΔT 2 - ΔT 1 ln

ΔT 2 ΔT 1

Th1  Tc1

'T2

Th 2  Tc 2

ð2:23Þ

The temperature differences at the two ends of a heat exchanger can be quickly determined by using a flow schematic for the hot and cold fluids as shown in Fig. 2.5. In a condenser used in a refrigeration cycle, Th1 = the temperature of the superheated refrigerant vapor entering the condenser Th2 = the temperature of the saturated/sub-cooled refrigerant liquid leaving the condenser Tc1 = the temperature of cooling water leaving the condenser Tc2 = the temperature of cooling water entering the condenser.

2.5.3

Heat Exchanger Design Equation

The heat exchanger design equation [3, 4] is used in determining the heat transfer surface area required for the exchanger. The general form of the heat exchanger design equation is q = UAs FΔT LMTD

ð2:24Þ

where the heat duty, q, is determined from heat balance (Eqs. 2.21 and 2.22). As represents the heat transfer surface area required and it is the outside surface area of the tubes in a Shell and Tube Heat Exchanger. U is the overall heat transfer coefficient to be used in the design of the exchanger. The overall heat transfer coefficient is the reciprocal of the overall resistance to heat transfer. In the most general case, there are five resistances to heat transfer between the hot and cold fluids in a heat exchanger. However, in case of condensers used in refrigeration cycles, there are only three significant resistances to heat transfer from the condensing refrigerant vapor to the cooling water. These include the convection resistance of the condensing vapor, the fouling resistance on the cooling water side of the tubes, and the convection resistance of cooling water. Generally, the convection resistance of the condensing vapor is several orders of magnitude lower than the convection resistance of cooling water. The MTD Correction Factor, F, is used only for multi-pass heat exchangers and it can be determined from MTD Correction Factor graphs. The MTD Correction

2.5

Heat Exchangers

79

Factor is 1 for a single-pass exchanger and also for multi-pass condensers and evaporators frequently encountered in HVAC-R applications. Example 2.7 The mass flow rate of refrigerant R-134a in a vapor compression refrigeration cycle is 430 lbm/hr. The refrigerant enters the condenser as a superheated vapor at state point 2 and leaves the condenser as a saturated liquid at state point 3. The following thermodynamic data are available at state points 2 and 3 for R-134a: State point 2: 180 psia, 140° F, h2 = 124 Btu/lbm State point 3: 180 psia, 118° F, h3 = 51 Btu/lbm The overall design heat transfer coefficient for the condenser is 29 Btu/hr-ft2-°F. Cooling water enters the condenser at 65 °F and exits at 80 °F. Determine A. B. C. D.

the heat duty of the condenser gpm of cooling water required the log mean temperature difference the heat transfer surface area required in the condenser

(Solution) A. The heat duty of the condenser is the heat lost by the condensing R-134a superheated vapor (Eq. 2.22).

q = m_ R ðh2 - h3 Þ = 430

lbm hr

124

Btu Btu = 31, 390Btu=hr - 51 lbm lbm

B. First, obtain the mass flow rate of the cooling water using heat balance (Eq. 2.22). Heat lost by R-134a = heat gained by cooling water = heat duty of the condenser q = m_ w ðhw,out - hw,in Þ ) m_ w =

q hw,out - hw,in

From steam tables, hw,out = hf at 80 ° F = 48:06 Btu=lbm hw,in = hf at 65 ° F = 33:08 Btu=lbm Substitute the preceding results and the calculated heat duty of the condenser into the heat balance equation. m_ w =

31, 390 Btu q hr = Btu Btu = 2095 lbm=hr hw,out - hw,in 48:06 lbm - 33:08 lbm

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Support Areas: Heat Transfer

Calculate the gpm of cooling water required by using its standard density (8.34 lbm/gallon) and suitable conversion factors applied to the preceding result. gpm =

1 hr 2095 lbm hr × 60 min = 4:19 gpm 8:34 lbm gal

Note: The conversion factor 500 lbm – min/hr – gal can be directly used to convert lbm/hr of water to gpm of water. C. A schematic temperature diagram showing the temperatures of R-134a and water in the counterflow condenser is shown here along with the temperature differences at end ‘1’ and at end ‘2’.

Calculate the log mean temperature difference using Eq. 2.23. ΔT LMTD =

ΔT 1 - ΔT 2 ln

ΔT 1 ΔT 2

=

60 ° F - 53 ° F = 56:43 ° F °F ln 60 53 ° F

D. Calculate the heat transfer surface area required in the condenser using the heat exchanger design equation (Eq. 2.24) assuming a single pass condenser (MTD Correction Factor, F = 1.0). q = UAFΔT LMTD ) A =

31, 390 Btu q hr = 19:18 ft2 = UFΔT LMTD ð 1:0 Þ ð 56:43 ° F Þ 29 hr‐ftBtu 2 ‐ °F

Practice Problems Practice Problem 2.1 The cross-section of a partition wall consists of 0.80 in. thick wooden panel exposed to a temperature of 85 °F. Adjacent to the wooden panel is 3.25 in. thick fiberglass

Solutions to Practice Problems

81

insulation followed by 0.75 in. thick particle board exposed to a conditioned space maintained at 70 °F. Only the conduction resistances are significant in this situation. Using the R-values given in Table 2.1, determine. A. the combined R-value of the wall assembly. B. the rate of heat gain by the conditioned space if the wall cross-section is 20 ft × 10 ft. Practice Problem 2.2 A building wall consists of a 50 mm thick brick with thermal conductivity 0.67 W/m. K and an adjacent layer of 10 mm thick wall board with thermal conductivity 0.18 W/m.K. The brick is exposed to ambient air with heat transfer coefficient of 36 W/m2.K and the heat transfer coefficient on the wall side is 11 W/m2.K. During the summer months, the average ambient temperature is 42 °C. Typically, the inside is maintained at 21 °C. Determine A. the R-values of the brick and the wall board for the given thickness of each B. the rate of heat gain in the building per unit area of the wall Practice Problem 2.3 A wall partition inside a building separates an unconditioned corridor space at 27 °C from a conditioned space maintained at 21 °C. The wall assembly consists of 20 mm thick gypsum wall board (R = 0.0793 m2.K/W) exposed to ambient air on both sides and 50 mm of fiberglass batt insulation (R = 2.291 m2.K/W) in between the exposed surfaces. Calculate the heat gained by the conditioned space per unit area of the wall surface after including the convection resistances at the exposed surfaces.

Solutions to Practice Problems Practice Problem 2.1 (Solution) A. The R-values given in Table 2.1 are per inch thickness. Therefore, to obtain the combined R-value of the wall assembly, multiply the table R-value of each component by its thickness and then add up the resulting component values as shown in the following table. Component Wooden panel Fiberglass insulation Particle board

R-value (per in.), hr-ft2-°F/ Btu 1.24 3.67

Thickness (in) 0.80 3.25

Resistance, R (hr-ft2-°F/ Btu) 0.992 11.927

1.06

0.75 ∑Rcond, ua

0.795 13.71

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Support Areas: Heat Transfer

B. Calculate the heat gain through the wall using Eq. 2.8. qwall =

ð20 ft × 10 ftÞð85 ° F - 70 ° FÞ AΔT overall = = 218:82 Btu=hr 2 0 Rcond,ua 13:71 hr‐ft ‐ F Btu

Practice Problem 2.2 (Solution) A. Subscript ‘wb’ represents the wall board. Calculate the R-values of the brick and the wall board using Eq. 2.10.

Rbrick = Rwb =

1m ΔX brick 50 mm × 1000 mm = = 0:0746 m2  K=W W kbrick 0:67 mK 1m ΔX wb 10 mm × 1000 mm = = 0:0556 m2  K=W kwb 0:18 mWK

B. Calculate the heat loss per unit cross-section area of the wall using Eq. 2.17. ΔT overall q ΔT overall = ) = 1 1 A Rth,ua þ Rcond,ua þ ho hi q 42 ° C - 21 ° C = A 1 m2  K 1 m2  K þ 0:0556 þ þ 0:0746 W W W W 36 2 11 2 m K m K = 84:36 W=m2 Note that ΔT ° C = ΔT K. Practice Problem 2.3 (Solution) Subscript ‘GWB’ represents gypsum wall board, and subscript ‘FG’ represents fiberglass insulation. From Table 2.2, for horizontal heat flow across a vertical wall, the convection R-value is R = 0.12 m2.K/W. Calculate the R-value of the wall assembly (including the convection resistances) by adding up all the resistances as shown. Rth,ua = 2ðRconv: Þ þ 2ðRGWB Þ þ RFG = 2 0:12

m2  K m2  K m2  K þ 2 0:0793 þ 2:291 W W W

= 2:690 m2  K=W

References

83

Calculate the heat gained by the conditioned space per unit cross-section area of the partition wall using Eq. 2.17. q ΔT overall 27 ° C - 21 ° C = = 2:230 W=m2 = 2 A Rth,ua 2:690 mWK Note that ΔT ° C = ΔT K.

References 1. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook Fundamentals, I P edn, ASHRAE, Peachtree (2021) 2. Bergman, T.L., Lavine, A.S., Incropera, F.P., DeWitt David, P.: Fundamentals of Heat and Mass Transfer, 8th edn. Wiley, New York (2019) 3. Ezgi, C.: Basic Design Methods of Heat Exchangers. Intech Open (2017) Download from https:// www.intechopen.com/chapters/54521 4. Nandagopal, N.S.: Fluid and Thermal Sciences – A Practical Approach for Students and Professionals. Springer Nature, Cham (2022)

Chapter 3

Support Areas: Thermodynamics

3.1

Introduction

The term ‘thermodynamics’ refers to transformations and interactions of energy systems [2]. Thermodynamic principles are extensively used in the field of heating, ventilation, air conditioning, and refrigeration (HVAC-R) [1]. The process of heating a facility involves knowledge of heat sources (fuels), the process of heat generation (combustion), and the dissemination of the generated heat, all of which involve the use of thermodynamic principles. Air conditioning and refrigeration systems typically use compressors, which are designed based on thermodynamic principles. Vapor compression refrigeration cycles [1, 4] are based on the reversed Carnot cycle, which is a cyclic thermodynamic process. The analysis and calculations associated with heat pumps are also based on thermodynamic principles. Finally, psychrometric formulas and charts are entirely based on thermodynamic principles of ideal gas mixtures and enthalpy calculations both within a phase and during phase transformations, e.g. condensation of moisture from humid air.

3.2

Thermodynamic Properties and Units

Thermodynamic properties [2–4] are defined with respect to the species within a defined system. A thermodynamic system is simply a defined bounded area for study. Thermodynamic systems are broadly classified into open systems and closed systems. Open systems allow the flow of heat and matter across its boundaries while closed systems allow just heat flow. A simple example of a closed thermodynamic system is air contained in a closed tank as shown in Fig. 3.1. The mass of air in the tank is m and it remains constant in this closed tank. The total thermodynamic properties of the air in the tank are represented by upper case variables as shown in Fig. 3.1. The thermodynamic properties are defined as follows. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 N. S. Nandagopal, HVACR Principles and Applications, https://doi.org/10.1007/978-3-031-45267-3_3

85

86

3

Support Areas: Thermodynamics

Fig. 3.1 Closed thermodynamic system and thermodynamic properties

Pressure (P) The pressure of the system is the absolute pressure exerted by the substance within the system. Absolute pressure is the total pressure relative to zero pressure or perfect vacuum. It is the sum of the surrounding atmospheric pressure and the gage pressure. Pabs = Patm þ Pgage

ð3:1Þ

In Chap. 1 on Fluid Mechanics, the static gage pressure was extensively used in the analysis of flow through ducts. However, in thermodynamic calculations it is customary to always use absolute pressure. Conversion factors: 1 psi = 6.895 kPa = 0.0689 bar. Standard atmospheric pressure at sea level: 1 atm = 760 mm Hg = 29.92 in Hg = 14.7 psi = 101.3 kPa = 1.013 bar = 407.2 in w c. Volume (V ) It is the total space occupied by the substance in the closed system. The typical units for volume are cubic feet, ft3, (USCS / I-P System) and cubic meter, m3, (SI). Occasionally, the unit of litre (L) is also used for volume. Conversion factors: 1 m3 = 35.31 ft3 = 1000 L. Temperature (T ) The temperature of the substance within the system is indicative of the energy level of the system. It is customary to use the absolute temperature in thermodynamic calculations. The units for absolute temperature are degree Rankine, °R, (USCS / I-P System) and kelvin, K, (SI). The absolute temperature can be obtained by using the following equations. ° R = 460 ° þ ° F

ð3:2Þ

K = 273 ° þ ° C

ð3:3Þ

Conversion Factors: °R = 0.5556 K, °F = 1.8 °C + 32°

3.2

Thermodynamic Properties and Units

87

Note: Temperature difference will be identical in normal (common) and corresponding absolute scales since the same constant will be added and subtracted. Consider the temperature of water being increased from 20 °C to 50 °C. The temperature difference calculations are shown here. ΔT ° C = 50 ° C - 20 ° C = 30 ° C Calculate the absolute temperatures in kelvin by using Eq. (3.3). T 1 = 273 ° þ 20 ° C = 293 K and T 2 = 273 ° þ 50 ° C = 323 K ΔT K = T 2 - T 1 = 323 K - 293 K = 30 K Thus, ΔT K = ΔT °C and similarly, ΔT°R = ΔT °F. Also, ΔT ° F = 1.8(ΔT ° C). Mass (m) This quantity represents the mass of the substance within the closed system. The units for mass are pound mass, lbm (USCS / I-P System) and kilogram, kg (SI). Conversion factor: 1 lbm = 0.4536 kg. Internal Energy (U ) The molecules of the substance within the system have different types of energy such as kinetic energy, vibrational energy, and rotational energy. The internal energy of a system sum total of the energies of all the molecules within the system. Since molecules have higher levels of energy at higher temperatures, the internal energy of a system is directly proportional to the temperature of the substance. The units of internal energy are British Thermal Unit, Btu (USCS / I-P System), and Kilo Joules, kJ (SI). Conversion Factor: 1 Btu = 1.055 kJ.

Enthalpy (H ) The enthalpy of a substance is the sum of the internal energy of the substance and the product of its pressure and volume. The enthalpy of a substance represents the total heat content of the substance. Mathematically, H = U þ PV

ð3:4Þ

Enthalpy has energy units (Btu and kJ). The product of pressure times the volume of a substance will have units of energy as shown here.

88

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Support Areas: Thermodynamics

USCS / I-P System: PV 

lbf ft2

ft3  ft - lbf

Usually, the system pressure is specified in pound force per square inch (lbf/in2) or psi. 1 ft2 = 144 in2 and 1 Btu = 778 ft. - lbf. Therefore, the equation to ensure consistent units (Btu) for enthalpy in USCS is

H =U þ

144 in2 ðV Þ ft2 778 ft - lbf Btu

P×

ð3:5Þ

where the pressure P is in psi and the volume V is in ft3. Entropy (S) The entropy of a substance represents the degree of disorder of the molecules within the substance. Consider the same species, water (H2O) in three different phases – solid (ice), liquid (water), and gas or vapor (steam). In ice, the water molecules have an organized crystalline structure. After ice melts, the molecules in liquid water are more mobile enabling the water to flow under the application of an external force. Finally, when liquid water vaporizes, the molecules in steam have random motion resulting in greater level of disorder. Therefore, Ssteam > Sliq:water > Sice Entropy also represents the thermal energy per unit of absolute temperature that cannot be converted to useful work, also known as Unavailable Energy. Entropy has the units of energy divided by absolute temperature (Btu/°R in USCS and kJ/K in SI).

3.2.1

Specific Properties

Thermodynamic tables and charts provide thermodynamic properties per unit mass of a substance [2–4]. This makes it convenient for performing thermodynamic calculations in different situations. The properties per unit mass can be multiplied by the mass of the substance in thermodynamic calculations for closed systems and they can be multiplied by mass flow rate of the species for situations involving open systems. The properties per unit mass are known as Specific Properties. Various specific properties and their symbols and units are presented here. Specific properties are represented by the corresponding lower case letter of the upper case letter used in representing total properties. For example, specific volume is represented by v and the total volume is represented by V.

3.3

Ideal Gas Law

89

Specific Volume (v) The specific volume of a substance is the volume per unit mass of the substance, and it has units of ft3/lbm (USCS) and m3/kg (SI). Since density is mass per unit volume, specific volume is the reciprocal of density: v=

V 1 = m ρ

ð3:6Þ

Standard values of specific volume of moist air at sea-level atmospheric pressure and normal ambient temperature: vair = 13.33 ft3/lbm (USCS), vair = 0.8264 kg/m3 (SI). Standard values of density of moist air at sea-level atmospheric pressure and normal ambient temperature: ρair = 0.075 lbm/ft3 (USCS), ρair = 1.20 kg/m3 (SI).

Specific Internal Energy (u) The specific internal energy of a substance is the internal energy per unit mass of the substance, and it has units of Btu/lbm (USCS) and kJ/kg (SI). u=

U m

ð3:7Þ

Specific Enthalpy (h) The specific enthalpy of a substance is the enthalpy per unit mass of the substance, and it has units of Btu/lbm (USCS) and kJ/kg (SI). h=

H U þ PV = = u þ Pv m m

ð3:8Þ

Specific Entropy (s) The specific entropy of a substance is the entropy per unit mass of the substance, and it has units of Btu/lbm – °R (USCS) and kJ/kg.K (SI). s=

3.3

S m

ð3:9Þ

Ideal Gas Law

The thermodynamic state of a gaseous system is usually defined by the pressure, temperature, and volume of the system. The ideal gas law [2, 4] is an equation of state that relates the absolute pressure (P), volume (V ), and the absolute temperature (T ) of a gaseous (or vapor) system.

90

3

Support Areas: Thermodynamics

Consider a gas in a piston–cylinder closed system. The piston is free to move back and forth depending on the state of the gas. When the gas in the cylinder is heated, it expands and pushes the piston. It is a well-known fact that gas expands upon heating, that is, the volume of the gas increases upon heating. Conversely, the volume of a gas decreases upon cooling. This suggests a direct relationship between volume and temperature. When a gas is compressed by pushing the piston, it occupies less volume. However, because of the compression process, the pressure of the gas is increasing. This suggests an inverse relationship between volume and pressure. The ideal gas law combines the effects of increasing the temperature and increasing the pressure of a gas and it is mathematically represented by PV = mRT

ð3:10Þ

where P is the absolute pressure, V is the volume, m is the mass, R is the individual gas constant [2, 4], and T is the absolute temperature. The value of R is different for different gases, and it depends on the molecular weight, M, of the gas. The molecular weight of a substance is also known as molar mass, and it has units of lbm/lbmol (USCS / I-P System) and kg/kmol (SI). The molecular weight of a substance can be calculated by using the molecular formula and atomic weights. The molecular formula for water is H2O, the atomic weights of hydrogen and oxygen are 1 lbm/lbmol and 16 lbm/lbmol, respectively. Therefore, the molecular weight of water is, M H2 O = 2 1

lbm lbm þ 16 = 18 lbm=lbmol lbmol lbmol

Since the chemical or molecular formula of a substance remains the same, the numerical value of the molecular weight remains the same in different systems of units with corresponding changes for units of mass and mols. For example, the molecular weight of air (which is very useful in HVAC-R calculations and formulas) is, M air = 29 lbm=lbmol = 29 kg=kmol

3.3.1

Concept of Mole

A mole (gram-mole or gmol or mol) of a substance has a mass equivalent to its molecular weight and it consists of Avogadro Number (6.022 × 1023) of particles.

3.3

Ideal Gas Law

91

Examples: The atomic weight of carbon is 12 g/mol. Therefore, 1 mol of carbon will have a mass of 12 g and will have 6.022 × 1023 atoms of C. The molecular weight of water is 18 g/mol. Therefore, 1 mol of water will have a mass of 18 g and will have 6.022 × 1023 molecules of H2O. Mole is represented by the symbol N and the mols of a substance can be obtained by dividing the mass of the substance by the molecular weight of the substance. N=

m M

ð3:11Þ

Example for calculation of mols: Calculate the kilogram mols of air in a room with dimensions of 5 m × 4 m × 3 m. The density of air is 1.20 kg/m3. Calculate the mass of air in the room mair = V air × ρair = ð5 m × 4 m × 3 mÞ 1:20

kg m3

= 72 kg

Calculate the mols of air using Eq. (3.11). N air =

3.3.2

mair 72 kg = = 2:483 kmol kg M air 29 kmol

Universal Gas Constant and Ideal Gas Equation in Molar Form

The gas constant, R, in the ideal gas equation (Eq. 3.10) is known as the individual gas constant and has different values for different gases depending on the molecular formula of the gas. However, based on the concept of mols and Avogadro Number, it is clear that equal mols of substances will have the same number of molecules of the substance. Based on this concept, a Universal Gas Constant, represented by the symbol R can be used in the ideal gas equation based on mols as shown here. Manipulate the ideal gas equation by multiplying and dividing the right-hand side of Eq. 3.10 by the molecular weight of the gas.

92

3

PV =

m ðR × M ÞT M

Support Areas: Thermodynamics

ð3:12Þ

From Eq. 3.11, the mass of the gas divided by its molecular weight is the mols of the gas, represented by the symbol N. The product of the individual gas constant and the molecular weight is the universal gas constant [4], R, which has the same value for all gases and hence the term Universal is used. Substitute the preceding results into Eq. (3.12). PV = NRT

ð3:13Þ

Equation (3.13) is the ideal gas equation in molar form. The values of the universal gas constant in different systems of units [4] are presented here. Universal Gas Constant R 1545 ft.-lbf/lbmol-°R 10.73 psi-ft3/lbmol-°R 0.7302 atm-ft3/lbmol-°R 1.987 Btu/lbmol-°R 8.314 kJ/kmol.K 0.08314 bar.m3/kmol.K 83.14 bar.cm3/kmol.K 0.0821 L.atm/mol.K 1.987 cal/mol.K Note: mol  g mole. The individual gas constant can be obtained by dividing the universal gas constant by the molecular weight of the gas. R=

R M

ð3:14Þ

Note: Example 1.1 and Practice Problem 1.1 in Chap. 1, which illustrate the utility of the ideal gas law in calculating the specific volumes (and hence densities) of air under different conditions, are worth reviewing again.

Example 3.1 At an altitude of 1000 m, the atmospheric pressure is 89.88 kPa. Determine the specific volume and density of atmospheric air at a location where the altitude is 1000 m, and the temperature is 20 °C.

3.4

Thermodynamic Processes Involving Ideal Gases

93

(Solution) Convert the temperature of air to its absolute value. T = 20 ° C þ 273 ° = 293 K

Calculate the individual gas constant of air using Eq. 3.14. Rair =

kJ 8:314 kmolK R = = 0:2867 kJ=kg  K kg M air 29 kmol

Calculate the specific volume (volume per unit mass) of air using the ideal gas equation (Eq. 3.10) PV = mRair T ) v=

V R T = air = P m

kJ ð293 KÞ kg  K = 0:9346 m3 =kg 89:88 kPa

0:2867

kJ N ˜ m m3 { { Units: kg ˜ k Pa kg N kg ˜ 2 m

Calculate the density of air using Eq. 3.6 (density is the reciprocal of specific volume). ρ=

1 1 = 1:07 kg=m3 = v 0:9346 m3 kg

The preceding value is lower than the standard density of air, 1.20 kg/m3, at the standard pressure of 101.3 kPa. This is to be expected since air will occupy more volume due to lower pressure at higher altitudes.

3.4

Thermodynamic Processes Involving Ideal Gases

A thermodynamic process is a procedure that causes a change in the state of the system. The different types of thermodynamic processes [2, 4] are described here. The ideal gas law (Eq. 3.10) can be written for any general thermodynamic process occurring between the initial state ‘1’ and the final state ‘2’. The mass, m, remains constant in a closed system and the individual gas constant, R, is constant for the gas undergoing the thermodynamic process.

94

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Support Areas: Thermodynamics

P1 V 1 = mRT 1

ð3:15Þ

P2 V 2 = mRT 2

ð3:16Þ

Divide Eq. (3.16) by Eq. (3.15). Note that m and R cancel out since they remain constant. P2 V 2 T 2 = P1 V 1 T 1

ð3:17Þ

Equation (3.17) is very useful in the analysis of different thermodynamic processes.

3.4.1

Isothermal Process

In an isothermal process, the temperature of the substance remains constant. Hence, T2 = T1 From Eq. 3.17, the following equation can be written for an isothermal process. P2 V 2 = P1 V 1 , PV = constant, V /

1 P

ð3:18Þ

An isothermal expansion process is shown on a P–V diagram in Fig. 3.2.

3.4.2

Isentropic Process

In an isentropic process, the entropy of the substance remains constant, that is, S = constant. Since the entropy does not change, the amount of unavailable energy remains constant during the process. This implies that all the available energy is converted to useful work in an isentropic process. Therefore, an isentropic process has the maximum possible efficiency in converting energy to useful work. The pressure–volume relationship for an isentropic process [2, 4] is P2 V 2 k = P1 V 1 k = PV k = C, a constant where k is the ratio of specific heats, k = cp/cv. cp = specific heat of the gas at constant pressure cv = specific heat of the gas at constant volume.

ð3:19Þ

3.4

Thermodynamic Processes Involving Ideal Gases

95

Fig. 3.2 P–V Diagram for an isothermal expansion process

Fig. 3.3 P–V Diagram for isothermal and isentropic expansion processes

Figure 3.3 shows a comparison of P–V diagrams for isothermal and isentropic expansion processes. The curve for isentropic expansion is steeper than the curve for isothermal expansion. This is because the exponent of V in an isentropic process, k, is always greater than 1. In an isothermal process, the exponent of V is 1. By combining the ideal gas equation (Eqs. 3.10) and (3.19), the following additional P–V–T relationships can be written for an isentropic process. T2 V1 = T1 V2 T2 P2 = T1 P1

k-1

k-1 k

ð3:20Þ ð3:21Þ

96

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Support Areas: Thermodynamics

Note: An isentropic process is also known as a reversible adiabatic process. There is no heat transfer in an adiabatic process, that is, Q = 0 in an isentropic process.

3.4.3

Constant Enthalpy/Throttling Process

During a throttling process [2, 4], the enthalpy remains constant. A common example of a throttling process is flow of a fluid across a pressure-reducing throttling valve. The enthalpy remains constant because the throttling valve has no work or heat interaction with the surroundings. Other examples of throttling process are pressure and flow control valves, and pressure relief valves. The throttling process is also used in liquefaction of gases. Constant enthalpy process (throttling) is a common feature of vapor compression refrigeration cycles (discussed in detail in Chap. 7), where the pressure of the refrigerant is reduced from the high value of the condenser pressure to a low value of the evaporator pressure. Since there is a reduction in pressure, the entropy of the refrigerant will increase due to greater mobility of molecules at lower pressure.

3.4.4

Calculation of Work for Thermodynamic Processes

Work is the product of force and distance. The pressure of a gas is the force exerted by the gas per unit cross-section area. Consider a gas expanding in a piston–cylinder system as shown in Fig. 3.4. When the gas expands, it moves the piston thereby performing work. If Acs is the cross-section area of the piston the force exerted by the gas on piston is, F = P × Acs The differential work performed by the gas in moving the piston through a distance dL is dW = P × Acs × dL However, the product of the area of cross-section and the differential length is the differential increase in the volume of the gas.

3.4

Thermodynamic Processes Involving Ideal Gases

97

Fig. 3.4 Work due to expansion of a gas

dL × Acs = dV Therefore, dW = PdV

ð3:22Þ

To get the total work of expansion performed by the gas in moving the piston from the initial position (‘1’) to the final position (‘2’), the preceding equation is integrated with the corresponding limits for the volume. V2

dW = W =

ð3:23Þ

PdV V1

3.4.4.1

Work for a Constant Temperature (Isothermal) Process

From Eq. 3.18 for an isothermal process, P1 V 1 = P2 V 2 = PV = C, a constant: Therefore, P =

C V

Substitute the preceding result into Eq. 3.23 and integrate. V2

PdV =

W= V1

W = C ln

V2

C

dV V

V1

V V V2 = P1 V 1 ln 2 = P2 V 2 ln 2 V1 V1 V1

From the ideal gas law (Eq. 3.10), for a constant temperature process,

ð3:24Þ

98

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Support Areas: Thermodynamics

P1 V 1 = P2 V 2 = mRT

ð3:25Þ

From Eqs. (3.24 and 3.25), the equations for work in an isothermal process can be written as V2 V V = P2 V 2 ln 2 = mRT ln 2 V1 V1 V1

W = P1 V 1 ln

ð3:26Þ

The volume ratio in Eq. 3.26 can be replaced by pressure ratio, VV 21 = another set of equations for an isothermal process. W = P1 V 1 ln

3.4.4.2

P1 P2 ,

P P1 P = P2 V 2 ln 1 = mRT ln 1 P2 P2 P2

to obtain

ð3:27Þ

Work for a Constant Entropy (Isentropic) Process

For an isentropic process, P2V2k = P1V1k = PVk = C. Solve the preceding equation for the pressure, P. P = Substitute for P in the equation for work (Eq. 3.23). V2

V2

PdV =

W= V1

V1

V - kþ1 C dV = C k kþ1 V

V2 V1

=C

C . Vk

V 21 - k - V 11 - k 1-k

Substitute for C from the P–V relationship. W =C

V 21 - k - V 11 - k = 1-k

P2 V 2 k V 2 1 - k - P1 V 1 k V 1 1 - k 1-k

=

P2 V 2 - P1 V 1 1-k

Substitute for the product PV from the ideal gas law, PV = mRT, to get the equations for work in an isentropic process [2, 4]. W=

P2 V 2 - P1 V 1 mRðT 2 - T 1 Þ = 1-k 1-k

ð3:28Þ

3.5

First Law of Thermodynamics

3.5

99

First Law of Thermodynamics

The first law of thermodynamics is based on the principle of conservation of energy [2, 4]. The principles of the first law of thermodynamics are discussed in detail in Chap. 2, pg. 2.2 of the ASHRAE Fundamentals Handbook [1]. The different forms of energy in a system are typically potential energy, kinetic energy, and internal energy. The potential energy of a system is based on its position relative to the datum, that is, the elevation of the system. The kinetic energy of a system is based on the velocity of the system. The internal energy of a system is based on the energy of the molecules within the system, which in turn is dependent on the temperature of the system. The temperature of the system can change due to addition of heat to the system or due to removal of heat from the system. The principle of conservation of energy states that energy is neither created nor destroyed. However, energy can be transformed from one form to another, without any net loss or net gain in the total energy of the system and the surroundings taken together.

3.5.1

First Law of Thermodynamics for a Closed System

Consider a closed system that is clearly defined by its boundaries as shown in Fig. 3.5. The mass of the substance within the system remains constant since there is no transfer of mass across the boundaries of a closed system. However, energy can be added to the system and energy can also be removed from the system. Thus, the system boundaries are closed for mass transfer but open for energy transfer. As shown in Fig. 3.5, energy is added to the system in the form of heat, Q, and energy is removed from the system in the form of work, W. As a result of the energy transfers, the total energy of the system changes from an initial value of Ei to a final value of Ef. Intuitively, the first law of thermodynamics for a closed system can be stated as: due to energy transformations, the final energy of the system will be the initial energy of the system plus the energy added to the system minus the energy removed from the system. In simple terms, the first law of thermodynamics for a closed system can be stated as: change in system energy will be equal to the net energy added to the system [1, 2, 4]. The preceding statement can be mathematically represented by Eq. (3.29). ΔE = E f - E i = Eadded - E removed

Fig. 3.5 Energy transformations for a closed thermodynamic system

ð3:29Þ

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Note that the change, delta, is always the final value minus the initial value. For the most general case, all forms of energy are included, and Eq. 3.29 can be written as follows: ΔE = ðU þ PE þ KE Þf - ðU þ PE þ KE Þi = Eadded - E removed

ð3:30Þ

Most thermodynamic closed systems are stationary and consequently there are no changes in potential and kinetic energies. For a stationary, closed, thermodynamic system as shown in Fig. 3.5, the equation for the first law simplifies to Eq. 3.31. ΔU = U f - U i = Eadded - E removed = Q - W

ð3:31Þ

In Eq. 3.31, the heat transfer, Q, is positive since heat is added to the system, and the work, W, is negative since work is extracted from the system, as illustrated in Fig. 3.5. If heat is removed from the system, Q, will have a negative sign and if work is performed on the system, then W, will have a positive sign. All confusion about the signs for Q and W can be overcome by drawing a schematic diagram as per the problem statement and then applying the first part of Eq. 3.31.

Example 3.2 1 lbm of R-134a at 60 °F and 15 psia is compressed to one-fifth of its original volume in a closed piston–cylinder arrangement. The relevant thermodynamic properties of R-134a are as follows: Molecular weight, M = 102 lbm/lbmol. Ratio of specific heats, k = 1.12. Specific heat at constant volume, cv = 0.18 Btu/lbm ‐ ° R. Determine the final temperature of R-134a when A. the compression is isentropic. B. there is a 10% heat loss based on the isentropic work required. (Solution) A. Calculate the initial absolute temperature of R-134a using Eq. 3.2.

3.5 First Law of Thermodynamics

101

T 1 = 460 ° þ 60 ° F = 520 ° R: The ratio of the final volume to the initial volume is 1:5. Use the temperature– volume relationship for an isentropic process (Eq. 3.20) to calculate the final temperature as shown. For R-134a, the ratio of specific heats is, k = 1.12.

T2 = T1

V2 V1

1-k

= ð520 ° RÞ

1 5

1 - 1:12

= 631 ° R

B. Calculate the individual gas constant for R-134a using Eq. 3.14. R=

ft‐lbf 1545 lbmol‐ R °R = 15:15 ft‐lbf=lbm‐ ° R = lbm M 102 lbmol

Calculate the isentropic work required using Eq. (3.28). ft‐lbf ð631 ° R - 520 ° RÞ lbm‐ ° R 1 - 1:12 1 Btu = ð -14, 014 ft‐lbf Þ 778 ft‐lbf = -18:01 Btu

mRðT 2 - T 1 Þ W= = 1-k

ð1 lbmÞ 15:15

The negative sign for the work indicates that the compression work is done on the gas. C. Calculate the heat loss based on the isentropic compression work. Q = ð0:10Þð18:01 BtuÞ = 1:801 Btu Draw a schematic for the process with an appropriate representation of arrows for work and heat transfer.

Apply the first law, Net energy added to the system = change in system energy, to the schematic shown.

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ΔU = W - Q = 18:01 - 1:801 = 16:21 Btu The internal energy of the system increases due to the work input to the system. For an ideal gas, ΔU = cv ΔT = cv ðT 2 - T 1 Þ ) T2 = T1 þ

ΔU 16:21 Btu = 520 ° R þ = 610 ° R mcv ð1 lbmÞ 0:18 lbmBtu - °R

Due to heat loss, the final temperature is less than the value of 631 °R for isentropic compression. Isentropic compression is a reversible–adiabatic (no heat transfer) process.

3.5.2

First Law of Thermodynamics for Open Systems: Energy Balance

In an open system, the system boundaries are open to both mass and energy flows. Consider the open system Control Volume (CV) shown in Fig. 3.6. Subscript ‘i’ is used for inlet streams and subscript ‘e’ is used for exit streams. The species going in and out of the control volume are labelled 1,2,. . . .. . . ..n. The fundamental equation for the First Law applied to an open system at steady state [2, 4] is as follows: Rate of energy into the CV = Rate of energy out of the CV

ð3:32Þ

_ is the rate of energy (power) Q_ is the rate of heat transfer into the CV and W extracted from the CV. The total energy of each stream consists of internal energy, kinetic energy, and potential energy. Since the streams flow in and out of the control volume of open systems, the flow energies need to be included for each stream. This Fig. 3.6 Control volume and streams for an open thermodynamic system

3.5

First Law of Thermodynamics

103

implies that specific enthalpies (h = u + Pv) should be used for open systems unlike specific internal energy used for closed systems. Note that the Pv term represents the flow energy per unit mass. The kinetic and potential energies per unit mass are v2/2 and gz, respectively. v is the velocity of stream and z is the elevation of the stream. Apply Eq. (3.32) to the control volume in Fig. 3.6. n

m_ ji hji þ

j=1 n j=1

m_ je hje þ

v2ji þ gzji 2

v2je þ gzje 2

þ Q_ =

n

m_ je hje þ

j=1

-

n j=1

m_ ji hji þ

v2je þ gzje 2

_ þW

v2ji _ þ gzji = Q_ - W 2

ð3:33Þ

Equation 3.33 is known as the Steady Flow Energy Equation (SFEE) [2, 3]. Most thermodynamic systems consist of a single stream of steam, water, or air. In addition, in case the potential and kinetic energy changes are very small relative to enthalpy _ changes. At steady state, the mass flow rate remains constant, that is, m_ i = m_ e = m. Apply the preceding simplifications to Eq. (3.33) to obtain the following equation: _ m_ ðhe - hi Þ = Q_ - W

ð3:34Þ

Equation (3.34) is called Single Stream Steady Flow Energy Equation (SSSFEE) [2, 4] and it can be used on most occasions for devices like turbines, compressors. However, as was the case in closed systems, it is always preferrable to draw the schematic for the system device and represent all the mass and energy streams accurately and then apply the simple energy balance equation, Rate of energy in = Rate of energy out (Eq. 3.32), to the schematic.

3.5.3

First Law of Thermodynamics Applied to Compressors

For isentropic compressors, the changes in potential and kinetic energies are very small compared to the changes in enthalpy [4]. Further, since an isentropic process is a reversible adiabatic process, there is no heat transfer from the compressor. Therefore, only the enthalpy changes and work need to be considered for energy balance for compressors as shown in Fig. 3.7. Note that a compressor requires work input. Apply the First Law for Open System (Energy Balance) to the compressor shown in Fig. 3.7. The variables associated with arrows pointing inwards should be on the lefthand side of the following equation and the variables associated with arrows pointing outwards should be on the right-hand side of the equation.

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Fig. 3.7 Energy balance schematic diagram for a compressor

Rate of energy in = Rate of energy out _ c = mh _ c = m_ ðh2 - h1 Þ _ 2)W _ 1þW mh

ð3:35Þ

The work consumed by the compressor per unit mass of the fluid is _c W = wc = h 2 - h 1 m_

ð3:36Þ

Compressors handle gases and, in most cases, ideal gas behaviour can be assumed. For an ideal gas, the change in enthalpy can be calculated by using the following equation: Δh = cp ΔT ) h2 - h1 = cp ðT 2 - T 1 Þ Combine the preceding equation with Eq. 3.35 to obtain the following equation for compressor power required: _ c = mc _ p ðT 2 - T 1 Þ W

ð3:37Þ

Note The isentropic or ideal compressor power required can also be calculated by using the following equation: _ c,ideal = W

k _ 1Þ ðmRT k-1

P2 P1

k-1 k

-1

ð3:38Þ

From Eq. (3.37), the ideal compressor work required per unit mass is _c W = wc = cp ðT 2 - T 1 Þ m_

ð3:39Þ

3.5

First Law of Thermodynamics

105

Example 3.3 The compressor in an ammonia (R-717) based cooling system receives 0.27 kg/s of the refrigerant as a saturated vapor at 200 kPa and compresses it to a discharge pressure of 1350 kPa. The enthalpies of refrigerant R-717 can be obtained from tables or online resources (for example, https://irc.wisc.edu/properties/) and they are provided here for reference. h1 ð200 kPa, saturated vaporÞ = 1420 kJ=kg s1 = 5:6 kJ=kg  K = s2 h2 ð1350 kPa, s2 = 5:6Þ = 1700 kJ=kg Determine the isentropic power required in kW. (Solution) Calculate the compressor power required using Eq. 3.35. _ c = m_ ðh2 - h1 Þ = 0:27 kg W s

3.5.4

1700

kJ kJ = 75:6 kW - 1420 kg kg

Isentropic Efficiency of Compressors

Compressors are extensively used in gas compression systems and in gas turbines. Figure 3.8 is a temperature–specific entropy diagram (T–s diagram) that illustrates and compares isentropic and non-isentropic compression processes. Isentropic compression occurs along the path 1–2 since entropy remains constant along this vertical line. Isentropic compression requires the least amount of work input to the compressor [4]. However, isentropic (reversible, adiabatic) conditions are difficult to achieve in practice due to heat losses and friction losses. As a result, the actual work used in the compression process is always greater than the isentropic or ideal work used. Non-isentropic compression takes place along the path 1–2′, which results in the increase in the entropy of the fluid. This implies that a portion of the input work is used in increasing the temperature and entropy of the gas. Due to the efficiency of a compressor, the actual work required to run the compressor is greater than the ideal work required to run the compressor. The isentropic efficiency of a compressor is defined as the ratio of the ideal work required to the actual work required. Subscript ‘c’ represents the compressor. ηc =

_ c,ideal W w h - h1 T - T1 = c,ideal = 2 = 2 _ w h h T c,actual 2′ 1 2 ′ - T1 W c,actual

ð3:40Þ

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Support Areas: Thermodynamics

Fig. 3.8 Isentropic and non-isentropic compression

Since equations are available to obtain isentropic work, it is obtained first and then the actual compressor work required is obtained by using the isentropic efficiency. Example 3.4 0.20 kg/s of saturated R-134a vapor is compressed from an initial pressure of 125 kPa to a final pressure of 500 kPa. The relevant thermodynamic properties of R-134a are as follows: Saturation temperature at 125 kPa = -21 °C. Molecular weight, M = 102 lbm/lbmol. Ratio of specific heats, k = 1.20. The isentropic efficiency of the compressor is 80%. Determine A. the compressor power required (kW), B. the final temperature of R-134a. (Solution) A. Calculate the absolute value of the initial temperature of R-134a using Eq. 3.3. K = 273 ° þ ° C ) T 1 = 273 ° - 21 ° C = 252 K

Calculate the individual gas constant for R-134a using Eq. 3.14. R=

kJ 8:314 kmolK R = 0:0815 kJ=kg  K = kg M 102 kmol

Calculate the ideal compressor power required using Eq. 3.38.

3.6

Thermodynamic Properties of Refrigerants

_ c,ideal = W

k _ 1Þ ðmRT k-1

1:20 = 1:2 - 1

P2 P1

kg 0:20 s

k-1 k

107

-1

kJ 0:0815 ð252 KÞ × kg  K

500 kPa 125 kPa

1:20 - 1 1:20

-1

= 6:406 kW Calculate the actual compressor work required using Eq. 3.40 _ c,ideal W ) _ W c,actual _ _ c,actual = W c,ideal = 6:406 kW = 8:008 kW W 0:80 ηc ηc =

B. Calculate the final temperature of R-134a for isentropic compression using Eq. 3.21. T2 P2 = T1 P1 ) T2 = T1

k-1 k

P2 P1

k-1 k

= ð252 KÞ

500 kPa 125 kPa

1:20 - 1 1:20

= 317:5 K

Calculate the actual final temperature of R-134a using Eq. 3.40 T2 - T1 ) T2 ′ - T1 317:5 K - 252 K 0:80 = ) T 2 ′ = 333:9 K ð60:9 ° CÞ T 2 ′ - 252 K ηc =

3.6

Thermodynamic Properties of Refrigerants

The thermodynamic properties of refrigerants [2, 4] can be obtained from different sources: • refrigerant tables, • graphs based on phase diagrams (pressure – enthalpy diagrams), • online calculator programs (https://irc.wisc.edu/properties).

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Support Areas: Thermodynamics

Typically, pressure and/or temperature are the key variables required for obtaining thermodynamic properties of refrigerants. Thermodynamic data in the tables and charts are always listed based on properties per unit mass (specific properties). Thus, one can obtain the specific volume, the specific enthalpy, and specific entropy from refrigerant tables and charts. Chapter 30 in the ASHRAE Fundamentals Handbook [1] contains property tables and P–h diagrams of more than thirty (30) different refrigerants classified under the categories of halocarbon, hydrocarbon, cryogenic, and inorganic refrigerants as well as the thermodynamic properties of ammonia-water and water-lithium bromide absorption solutions.

3.6.1

Thermodynamic Properties of Refrigerants from Tables

Refrigerant property tables are categorized as Saturated Tables and Superheated Tables. Saturated tables provide the properties of refrigerants in the saturation range, that is from saturated liquid state to saturated vapor state. Superheated tables provide the properties of refrigerants in the superheated vapor state. Figure 3.9 is an excerpt of saturated table for refrigerant R22. In saturated tables, subscript ‘f’ represents saturated liquid, and subscript ‘g’ represents saturated vapor. Figure 3.10 is an excerpt of superheated table for refrigerant R22 at a pressure of 30 psia. As per the Gibbs Phase Rule, since there is only a single superheated vapor phase, there should be two degrees of freedom. Hence, both pressure and temperature need to be specified to determine the properties of superheated vapor.

3.6.2

Properties of Liquid–Vapor Mixtures of Refrigerants

The state of the substance between saturated liquid and saturated vapor is a liquid– vapor mixture. Liquid–vapor mixtures are characterized by the mass fraction of vapor in the mixture, which is known as the Quality of mixture and is represented by

Fig. 3.9 Excerpt from R22 saturated table

3.6

Thermodynamic Properties of Refrigerants

109

Fig. 3.10 Excerpt from R22 superheated table

the symbol x. The subscript f represents the liquid, and the subscript g represents the vapor. The mathematical representation of the quality of a liquid–vapor mixture is shown as follows: Quality, x =

mg mass of vapor = mass of L‐V mixture mf þ mg

ð3:41Þ

The quality of a liquid–vapor mixture represents the extent to which vaporization (or condensation) has occurred. The subscript ‘fg’ is used in representing the change in any specific property due to complete vaporization. For example, vfg, represents the increase in specific volume due to complete vaporization of the liquid, that is, vfg = vg - vf. To find the value of any specific property of a liquid mixture, the change in property due to complete vaporization is weighted by the quality of the liquid–vapor mixture and the result is added to the value of the specific property of saturated liquid. The following equations can be used in calculating the properties of liquid–vapor mixtures. v = vf þ xvfg

ð3:42Þ

u = uf þ xufg

ð3:43Þ

h = hf þ xhfg

ð3:44Þ

s = sf þ xsfg

ð3:45Þ

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3

3.6.3

Support Areas: Thermodynamics

Properties of Compressed Liquid

The thermal properties of liquids are strongly dependent on temperature and not pressure. This is because liquids are essentially incompressible. Increasing the pressure of a liquid does not increase its enthalpy significantly. The increase in enthalpy of a liquid due to compression can be calculated by using Eq. 3.46. Δhcomprn: = vðP2 - P1 Þ

ð3:46Þ

Consider liquid water at ambient conditions of 30 °C and 100 kPa. From saturated steam tables, the saturation pressure at 30 °C is 0.0042 MPa (4.2 kPa). At the specified conditions, since P > Psat, water is a compressed liquid at 30 °C and 100 kPa. Also, from the saturated steam tables, the enthalpy of saturated liquid at 30 °C is, hf = 125.73 kJ/kg and the specific volume of saturated liquid at the same temperature is, vf = 0.001 m3/ kg. Calculate the increase in enthalpy due to compression by substituting the known values into Eq. 3.46. m3 ð100 kPa - 4:2 kPaÞ kg = 0:0958 kJ=kg 3 kN m3 kN  m kJ m Note : kPa ×  2 ×   kg m kg kg kg

Δhcomprn: = vðP2 - P1 Þ = 0:001

The increase in enthalpy due to compression is added to the enthalpy of saturated liquid at the given temperature of 30 °C to obtain the enthalpy of the compressed liquid at 30 °C and 100 kPa. hcomp:liq = hf at 30 ° C þ Δhcomprn = 125:73

kJ kJ þ 0:096 = 125:83 kJ=kg kg kg

The enthalpy of the compressed liquid is almost equal to the enthalpy of the saturated liquid at the given temperature. For compressed liquids, we can use the properties of the saturated liquid at the given temperature without any loss in accuracy.

Example 3.5 Determine the state, specific enthalpy, specific entropy, and specific volume of R22 under the following conditions of temperature and pressure using the thermodynamic property tables of R22.

3.6

Thermodynamic Properties of Refrigerants

111

A. T = -20 ° F, P = 24.91 psia, saturated vapor B. T = -20 ° F, P = 24.91 psia, liquid–vapor mixture with 80% vapor by mass C. T = -40 ° F, P = 20 psia (Solution) A. From R 22 saturated table in Fig. 3.9, hg(-20 ° F) = 102.52 Btu/lbm. B. From R 22 saturated table in Fig. 3.9, hg ð- 20 ° FÞ = 102:52 Btu=lbm hf ð- 20 ° FÞ = 5:26 Btu=lbm hfg = hg - hf = 102:52

Btu Btu - 5:26 = 97:26 Btu=lbm lbm lbm

Calculate the enthalpy of the liquid–vapor mixture using Eq. 3.44. h = hf þ xhfg Btu Btu = 5:26 = 83:07 Btu=lbm þ ð0:80Þ 97:26 lbm lbm C. From R 22 saturated table in Fig. 3.9, at T = -40 ° F, Psat = 15.62 psia. Since, given P = 20 psia > Psat, R 22 is a compressed liquid and for a compressed liquid, h = hf ð-40 ° FÞ = 0:000 Btu=lbm

3.6.4

Pressure: Enthalpy (P–h) Phase Diagrams for Refrigerants

P–h phase diagrams [1, 2, 4] are extensively used in analysis and calculations involving refrigeration and air-conditioning (HVAC) systems. The schematic layout of a typical P–h phase diagram is shown in Fig. 3.11. The P–h phase diagram has a dome-shaped phase boundary with saturated liquid on the left phase boundary and saturated vapor on the right phase boundary. The state of the refrigerant is a liquid– vapor mixture between these boundaries inside the dome. To the left of the saturated liquid, state of the refrigerant is compressed (or sub-cooled) liquid and the refrigerant is superheated to the right of the saturated vapor line. Figures 3.12 (USCS / I-P units) and 3.13 (SI units) are pressure–enthalpy diagrams for refrigerant R-134a.

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Support Areas: Thermodynamics

Fig. 3.11 Typical P–h diagram for refrigerants

Fig. 3.12 P–h Diagram for refrigerant R-134a, USCS / I-P units. (Generated by the author, N.S. Nandagopal, P E)

3.6

Thermodynamic Properties of Refrigerants

113

Fig. 3.13 P–h Diagram for refrigerant R-134a, SI units. (Generated by the author, N.S. Nandagopal, P E)

Example 3.6 1.15 lbm/s of R-134a at 20 psia with 20 °F superheat is compressed to a final pressure of 160 psia and the isentropic efficiency of the compressor is 76%. Determine A. the compressor horsepower required, B. the enthalpy and temperature of R-134a leaving the compressor. (Solution) A. From the excerpt of the R-134a P–h diagram shown, the saturation temperature of R-134a at 20 psia is -2 °F. With 20 °F superheat, the temperature of R-134a entering the compressor is T1 = -2 ° F + 20 ° F = 18 ° F. Locate state point 1 (20 psia, 18 °F) on the excerpt of the P–h diagram as shown. At state point 1, the enthalpy of R-134a is h1 = 105 Btu/lbm. Assuming isentropic compression, move along the constant entropy line from state point 1 to state point 2 (160 psia, s2 = s1) as shown.

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Support Areas: Thermodynamics

At state point 2, the enthalpy of R-134a is h2s = 123 Btu/lbm after isentropic compression of the refrigerant. The subscript ‘s’ indicates isentropic compression. Determine the ideal compressor power required using Eq. 3.36. _ c,ideal = m_ ðh2s - h1 Þ W lbm Btu Btu = 1:15 123 - 105 sec lbm lbm = 20:7 Btu= sec ) _ c,ideal = 20:7 Btu W sec

1:414 hp Btu 1 sec

= 29:27 hp

Calculate the actual compressor horsepower required using Eq. 3.40 and the compressor efficiency.

ηc =

_ c,ideal _ W _ c,actual = W c,ideal = 29:27 hp = 38:51 hp )W _ c,actual ηc 0:76 W

B. Calculate the enthalpy of R-134a leaving the compressor using Eq. 3.40 and the compressor efficiency. The units for specific enthalpy is Btu/lbm and it is not shown in the equation for the sake of brevity. ηc =

h2s - h1 123 - 105 ) 0:76 = ) h2 ′ = 128:7 Btu=lbm h2 ′ - 105 h2 ′ - h1

Locate state point 2′ (P2 = 160 psia, h2' = 129 Btu/lbm) on the excerpt of the P–h diagram as shown. State point 2′ represents the state of R-134a leaving the compressor. Determine the temperature at state point 2′ )T2' = 160 ° F.

3.7

Reversed Carnot Cycle

115

Fig. 3.14 (a) Reversed carnot engine (b) Reversed carnot cycle T–s diagram

3.7

Reversed Carnot Cycle

As the name implies, a Reversed Carnot Cycle is the exact opposite of the Carnot cycle. It is the same as the Carnot cycle operating in the reverse direction [2, 4]. Similar to Carnot cycle being the basis for cycles producing net power, the Reversed Carnot Cycle forms the basis for power-consuming refrigeration cycles and heat pumps. The schematic diagram and the T–s diagram for the Reversed Carnot Cycle are shown in Fig. 3.14. A Reversed Carnot Engine takes heat from a low-temperature heat source, such as a refrigerated or air-conditioned space, and transfers the heat to a high-temperature heat sink. Heat cannot spontaneously flow from low temperature to high temperature. Therefore, the Reversed Carnot Engine has to use external work to push the heat from low temperature to high temperature. As shown in Fig. 3.14b, the processes in the Reversed Carnot cycle are as follows: 1–2: Isentropic compression of the working fluid, which is usually a refrigerant. (vertical, constant entropy line in the T–s diagram and an increase in temperature of the working fluid implies an isentropic compression process). This compression process requires external work input. 2–3: Heat rejection at constant temperature (horizontal, constant temperature line in the T–s diagram and decreasing entropy implies isothermal compression process, that is, isothermal condensation). 3–4: Isentropic expansion of the working fluid, producing work (vertical, constant entropy line in the T–s diagram and decreasing temperature due to decrease in pressure implies an isentropic expansion process). 4–1: Heat absorption at constant temperature and isothermal expansion of the working fluid (horizontal, constant temperature line in the T–s diagram and increasing entropy due to evaporation of the working fluid implies an isothermal expansion process).

116

3.7.1

3

Support Areas: Thermodynamics

Performance Measure of Reversed Carnot Cycle: Coefficient of Performance (COP)

The performance measure of any device or process can be defined as a measure of desired effect achieved divided by the cost of achieving it. The desired effect in Reversed Carnot Cycle is the heat absorbed from the low-temperature heat source (also known as the refrigerating effect) and cost of achieving it is the work input to the compressor. The preceding performance measure of Reversed Carnot Cycle is known as Coefficient of Performance (COP). A detailed presentation of COP is given in Chap. 2, of the ASHRAE Fundamentals Handbook [1]. Referring to Fig. 3.14a, the mathematical representation of COP for refrigeration or air conditioning (heat removal) is COPrefg:=AC =

desired effect Q = L W cost of achieving it

ð3:47Þ

Energy balance for the Reversed Carnot Engine [Fig. 3.14a] results in the following equations: Energy In = Energy Out QL þ W = QH ) W = QH - QL Substitute the preceding result into Eq. 3.47. Also, the heat energies can be replaced by the corresponding absolute temperatures. This results in a comprehensive set of equations for Coefficient of Performance for refrigeration and air conditioning. COPrefg:=AC =

QL QL TL = = W QH - QL T H - T L

ð3:48Þ

The Reversed Carnot Cycle can also be used for space heating in contrast to space cooling in refrigeration and air conditioning processes. Space heating is achieved by capturing the heat rejected at the high-temperature heat sink (condenser). When the Reversed Carnot Cycle is used for space heating it becomes a Heat Pump [2, 4]. The desired effect in using a heat pump is the heat rejected at the heat sink, QH. Therefore, the COP for a heat pump will be COPHeat Pump =

QH QH TH = = W QH - QL T H - T L

ð3:49Þ

The COP for a heat pump is always a number greater than one, with higher values indicating better performance.

3.7

Reversed Carnot Cycle

117

The rate of cooling can be represented as tons of refrigeration (TOR) required. One ton of cooling is the rate of cooling achieved when 1 short ton (2000 lbm) of ice at 0 °C melts in a 24-h time period. By using the latent heat of melting (fusion) of ice, it can be shown that 1 ton of refrigeration  12, 000 Btu=h = 3:517 kW

ð3:50Þ

Example 3.7 A Carnot refrigerator (freezer) operates between temperature limits of -18 °C and 25 °C. It is desired to remove heat equivalent to 75 kW from the cold space. Determine: A. the COPrefg./AC. B. the power input required (kW). C. the rate of heat rejection (kJ/h). (Solution) A. Calculate absolute temperatures using Eq. 3.3. TH = 25 ° C + 273 ° = 298 K TL = -18 ° C + 273 ° = 255 K Calculate the Coefficient of Performance for the freezer using the last part of Eq. 3.48 and substitute the known values. COPrefg:=AC =

TL 255 K = = 5:93 T H - T L 298 K - 255 K

B. Calculate the power input required by using the first part of Eq. 3.48. The rate of heat removal from cold space is 75 kW.

COPrefg:=AC =

Q_ L Q_ L 75 kW _ = )W = = 12:65 kW _ COPrefg:=AC 5:93 W

C. Calculate the rate of heat rejection using the middle part of Eq. 3.48.

COPrefg:=AC =

75 kW Q_ L ) ) 5:93 = Q_ H - Q_ L Q_ H - 75 kW

Q_ H = 87:65 kW

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3

Support Areas: Thermodynamics

Note: The same result can be obtained by using the energy balance equation for a Reversed Carnot Engine. _ = 75 kW þ 12:65 kW = 87:65 kW Q_ H = Q_ L þ W Example 3.8 A Carnot heat pump is used in heating 1700 cfm of atmospheric air from 40 °F to 70 °F. The specific heat of air at constant pressure is 0.24 Btu/lbm-°R. Calculate the COP of the heat pump if the power input to the compressor is 5 hp. (Solution) Calculate the individual gas constant for air using Eq. 3.14. 3

psia‐ft 10:73 lbmol‐ R °R = = 0:37 psia‐ft3 =lbm‐ ° R Rair = lbm M air 29 lbmol

Calculate the average temperature of air and its absolute value using Eq. 3.2. Tavg = (40 ° F + 70 ° F)/2 = 55 ° F )Tair = 460 ° + 55 ° F = 515 ° R Calculate the density of air at its average temperature using the ideal gas law (Eq. 3.10). ρ=

m P 14:7 psia = = 0:0771 lbm=ft3 = psia‐ft3 V Rair T air 0:37 lbm‐ ° R ð515 ° RÞ

Calculate the mass flow rate of air using the continuity equation (Eq. 1.9). m_ air = ρair V_ air = 0:0771

lbm ft3

1700

ft3 min

= 131:1 lbm= min

The specific heat of air at constant pressure is 0.24 Btu/lbm-°R. Energy balance at the high-temperature heat sink, results in the following set of equations: Rate of heat rejected at the heat sink = rate of heat absorption by air lbm Btu 0:24 ð70 ° F - 40 ° FÞ Q_ H = m_ air cp,air ΔT air = 131:1 min lbm‐ ° R = 943:9 Btu= min

3.7

Reversed Carnot Cycle

119

Calculate the coefficient of performance for the heat pump using Eq. 3.49. 0:0236 hp Btu 943:9 min Btu 1 min Q_ H = COPHeat Pump = = 4:45 _ 5 hp WC

Example 3.9 The inside temperature of a facility is to be maintained at 65 °F using a heat pump with a COP of 3.15 and a normal heating capacity of 35,000 Btu/h. However, the heating capacity of the heat pump is insufficient on an extremely cold day requiring the use of the supplemental electrical heating system with an efficiency of 80% and power consumption of 1.25 kW. Determine A. the total heat loss from the facility on that day, B. the total energy consumption in kWh by the facility for 10 h of operation. (Solution) Subscript ‘SHS’ represents supplemental heating system and subscript ‘HP’ represents the heat pump. A. Calculate the heating capacity of the supplemental system using its power consumption and efficiency as shown. actual rate of heating produced QSHS = ) PSHS power consumed QSHS = ðηSHS ÞðPSHS Þ = ð0:80Þð1:25 kWÞ = 1:0 kW

ηSHS =

Convert the heating capacity of the supplemental system to Btu/h. QSHS = ð1 kWÞ 3414

Btu=h = 3414 Btu=h kW

The total heat loss from the facility on the given day is the sum of the heating capacities of the heat pump and the supplemental heating system. QTotal = QHP þ QSHS = 35000

Btu Btu þ 3414 = 38, 414 Btu=h h h

B. Calculate the power consumption of the heat pump using Eq. 3.49. Btu 35, 000 Q_ HP h = 11, 111 Btu=h _ = W HP = 3:15 COPHP 1 kW _ HP = ð11, 000 Btu=hÞ )W = 3:222 kW 3414 Btu=h

120

3 Support Areas: Thermodynamics

Calculate the total energy consumption for 10 h of operation as shown. E = ð3:222 kW þ 1:25 kWÞð10 hÞ = 44:72 kWh

3.8

Practice Problems

Practice Problem 3.1 0.73 kg/s of R-134a saturated vapor is compressed from an initial pressure of 1.5 bar to a final pressure of 7.0 bar. Dynamometer tests on the compressor indicate a power consumption of 35 kW. Determine the overall efficiency of the compressor. Practice Problem 3.2 Determine the actual compressor power required and the final temperature of R-134a under the same conditions specified in Example 3.4 using the following sources: A. Pressure–enthalpy diagram for R-134a. B. Online resources for properties of refrigerants. Practice Problem 3.3 Saturated R-134a liquid is throttled from 200 psia to 20 psia. Determine: A. the initial and final enthalpy. B. the change in temperature. C. the change in entropy. Practice Problem 3.4 An air-conditioning system with a capacity of 0.80 tons has a COP of 3.25. Determine the compressor horsepower required.

3.9

Solutions to Practice Problems

Practice Problem 3.1 (Solution) State point 1 is the start of the compression process and state point 2 is the end of the compression process. Convert the pressures to Mpa. P1 = ð1:5 barÞ

0:1 MPa = 0:15 MPa 1 bar

P2 = ð7:0 barÞ

0:1 MPa = 0:70 MPa 1 bar

3.9

Solutions to Practice Problems

121

Locate state points 1 (0.15 bar, saturated vapor) and 2 (7.0 bar, s2 = s1) assuming isentropic compression as shown in the graph. Determine the enthalpies at state.

points 1and 2 from the graph. h1 = 240 Btu=lbm h2 = 275 Btu=lbm Calculate the ideal compressor power required using Eq. 3.36. _c W _ c,ideal = m_ ðh2 - h1 Þ ) = h 2 - h1 ) W m_ _ c,ideal = 0:73 kg 275 Btu - 240 Btu = 25:55 kW W s lbm lbm

Calculate the overall efficiency of the compressor using Eq. 3.40. ηc =

_ c,ideal W 25:55 kW = = 0:73 ð73%Þ _ 35 kW W c,actual

Practice Problem 3.2 (Solution) The conditions specified in Example 3.4 are as follows: 0.20 kg/s saturated R-134a vapor is compressed from an initial pressure of 125 kPa to a final pressure of 500 kPa and the isentropic compressor efficiency is 80%. Subscript ‘1’ represents compressor intake and subscript ‘2’ represents compressor outlet. Convert the given pressures to Mpa.

122

3

Support Areas: Thermodynamics

A. Using the P-h diagram for R-134a. P1 = ð125 kPaÞ

1 Mpa = 0:125 MPa 1000 kPa

P2 = ð500 kPaÞ

1 Mpa = 0:500 MPa 1000 kPa

Locate state points 1 and 2 on the P-h diagram of R-134a as shown. State point 1: P1 = 0.125 MPa, saturated vapor. State point 2: P2 = 0.500 MPa, s2 = s1.

From the graph, h1 = 235 kJ=kg,

h2 = 260 kJ=kg

Calculate the ideal compressor power required using Eq. 3.36. _c W _ c,ideal = m_ ðh2 - h1 Þ ) = h2 - h1 ) W m_ _ c,ideal = 0:20 kg 260 kJ - 235 kJ = 5 kW W s kg kg Calculate the actual compressor work required using Eq. 3.40 ηc =

_ c,ideal W ) _ c,actual W

_ _ c,actual = W c,ideal = 5 kW = 6:25 kW W ηc 0:80

3.9

Solutions to Practice Problems

123

Determine the temperature of R-134a at state points 1 and 2 from the graph as shown. T1 = -22 ° C = 251 K

T2 = 22 ° C = 295 K.

Determine the actual temperature of R-134a leaving the compressor using Eq. 3.40. T2 - T1 ) T2 ′ - T1 295 K - 251 K 0:80 = ) T 2 ′ = 306 K ð33 ° CÞ T 2 ′ - 251 K

ηc =

B. Using online resources for properties of R-134a. From the online calculator program, (https://irc.wisc.edu/properties), At state point 1: P1 = 125 kPa, saturated vapor, h1 = 238 kJ/kg, s1 = 0.9470 kJ/kg  K T 1 = - 21:4 ° C = 251:6 K At state point 2: P2 = 500 kPa, s2 = s1 = 0.9470 kJ/kg  K, h2 = 266 kJ/kg, T2 = 22.7 ° C = 295.7 K Calculate the ideal compressor required using Eq. 3.36. _c W _ c,ideal = m_ ðh2 - h1 Þ ) = h2 - h1 ) W m_ _ c,ideal = 0:20 kg 266 kJ - 238 kJ = 5:60 kW W s kg kg

Calculate the actual compressor work required using Eq. 3.40 _ c,ideal W ) _ W c,actual _ _ c,actual = W c,ideal = 5:6 kW = 7:00 kW W ηc 0:80

ηc =

Determine the actual temperature of R-134a leaving the compressor using Eq. 3.40. T2 - T1 ) T2 ′ - T1 295:7 K - 251:6 K 0:80 = ) T 2 ′ = 306:7 K ð33:7 ° CÞ T 2 ′ - 251:6 K ηc =

124

3

Support Areas: Thermodynamics

Comment: The results obtained from the P–h diagram and online sources are in quite good agreement with each other and the result obtained from the formula _ c,actual = 8:008 kW, T 2 ′ = 60:9 ° C in Example 3.4 shows a large deviation W from those obtained using P–h diagram and online sources. This clearly indicates that refrigerant properties should be obtained from reliable tables, graphs, or online sources. Practice Problem 3.3 (Solution)

Locate state point 1 (saturated liquid at 200 psia) of R-134a on the P–h diagram. Enthalpy remains constant during the throttling process. Draw a vertical line from state points 1 to state point 2 (h2 = h1, and 20 psia). From the graph, at A. State point 1: h1 = 55 Btu/lbm, s1 = 0.11 Btu/lbm ‐ ° R, T1 = 125 ° C. State point 2: h2 = 55 Btu/lbm, s2 = 0.12 Btu/lbm ‐ ° R, T2 = 0 ° C

References

125

B. ΔT = T2 - T1 = 0 ° C - 125 ° C = -125 ° C Btu Btu C. Δs = s2 - s1 = 0:12 lbm‐ ° R - 0:11 lbm‐ ° R = 0:01 Btu=lbm‐ ° R As a check, the following results were obtained from online resource (https:// irc.wisc.edu/properties) A. State point 1: h1 = 54.3 Btu/lbm, s1 = 0.108 Btu/lbm ‐ ° R, T1 = 125 ° C. State point 2: h2 = 54.3 Btu/lbm, s2 = 0.12 Btu/lbm ‐ ° R, T2 = -2.4 ° C B. ΔT = T2 - T1 = -2.4 ° C - 125 ° C = -127.4 ° C Btu Btu C. Δs = s2 - s1 = 0:12 lbm‐ ° R - 0:108 lbm‐ ° R = 0:012 Btu=lbm‐ ° R Comment: The results from the P-h diagram and from online sources are in reasonable agreement with each other. Practice Problem 3.4 (Solution) Calculate the compressor horsepower required using the first part of Eq. (3.48) COPrefg:=AC =

QL ) W

Q_ L _ c= = W COPrefg:=AC

ð0:8 tonsÞ

12000 Btu=h 1 ton 3:25

= 2954 Btu=h

Convert the compressor power required to horsepower. _ c = 2954 Btu=h × W

1 hp = 1:16 hp 2545 Btu=h

References 1. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook Fundamentals, IP Edition. ASHRAE (2021) 2. Cengel, Y., Boles, M.: Thermodynamics – An Engineering Approach, 9th edn. McGraw Hill (2019) 3. Conner, N.: Thermodynamic Properties. Thermal Engineering (2019) Download from https:// www.thermal-engineering.org/what-is-thermodynamic-property-definition/ 4. Nandagopal, N.S.: Fluid and Thermal Sciences – A Practical Approach for Students and Professionals. Springer, Dordrecht (2022)

Chapter 4

Support Areas: Psychrometrics

4.1

Introduction

As per ASHRAE, ‘Psychrometrics deals with thermodynamic properties of moist air and uses these properties to analyze conditions and processes involving moist air.’ Atmospheric air, also known as moist air or humid air, is an ideal gas mixture of dry air and water vapor (moisture). Hence, understanding the principles of psychrometrics requires knowledge of definitions and properties of ideal gas mixtures [2]. Chapter 1 in ASHRAE Fundamentals Handbook [1] has extensive coverage of air–water vapor mixtures and psychrometrics.

4.2

Ideal Gas Mixtures

An ideal gas mixture will have several different gases as components of the mixture. Each component in the mixture obeys the ideal gas law and the mixture as a whole will also obey the ideal gas law [2, 3]. The components are represented by subscript i and, i can take on values ranging from 1 to n, where n is the number of components in the mixture. The mixture properties are represented by variables without subscripts whereas the variables representing component properties will have subscripts representing the component. As an example, the variable P represents the pressure of the mixture as a whole and P2 represents the pressure exerted by component ‘2’ in the mixture. P2 is also known as the partial pressure of component ‘2’. The basic variables used in ideal gas mixtures are illustrated in Fig. 4.1.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 N. S. Nandagopal, HVACR Principles and Applications, https://doi.org/10.1007/978-3-031-45267-3_4

127

128

4

Support Areas: Psychrometrics

Fig. 4.1 Mixture of ideal gases

4.2.1

Key Definitions for Ideal Gas Mixtures

Mass fraction (xi): The mass fraction of component ‘i’ is the mass of the component ‘i’ divided by the total mass of the mixture. xi =

mi m

ð4:1Þ

Mole fraction (yi): The mole fraction of component ‘i’ is the moles of the component ‘i’ divided by the total moles of the mixture. yi =

Ni N

ð4:2Þ

Partial pressure of component i (Pi): The partial pressure of component ‘i’, is defined as the pressure exerted by that component when it occupies the mixture volume, V, at the mixture temperature, T. From the ideal gas law (Eq. 3.10, Chap. 3) Pi =

mi Ri T V

ð4:3Þ

Partial volume of component ‘i’ (Vi): The partial volume of component ‘i’, is defined as the volume occupied by that component at the mixture pressure, P, and at the mixture temperature, T. From the ideal gas law (Eq. 3.10, Chap. 3) Vi =

mi Ri T P

ð4:4Þ

Since temperature is an intensive property, it is uniform throughout the mixture and therefore the temperature of each component in the ideal gas mixture is the same as the temperature of the mixture itself.

4.3

Air–Water Vapor Mixture and Psychrometrics

129

From the molar form of the ideal gas law, PV = NRT (Eq. 3.13, Chap. 3), both pressure and volume are directly proportional to the moles of the gas. Hence, for an ideal gas mixture, the mole fraction, the pressure fraction, and volume fraction of each component are identical. For any component, ‘i’, in an ideal gas mixture, mole fraction = pressure fraction = volume fraction [1–3].

yi =

N i Pi V i = = N P V

4.2.2

Laws Related to Ideal Gas Mixtures

4.2.2.1

Dalton’s Law

ð4:5Þ

Dalton’s Law [1, 3] states that the sum of partial pressures of all the components in an ideal gas mixture will be equal to the total pressure of the mixture. n

Pi = P

ð4:6Þ

i=1

4.2.2.2

Amagat’s Law

Amagat’s Law [1, 3] states that the sum of partial volumes of all the components in an ideal gas mixture will be equal to the total volume of the mixture. n

Vi = V

ð4:7Þ

i=1

4.3

Air–Water Vapor Mixture and Psychrometrics

Psychrometrics involves the following basic assumptions [1]: • Dry air and water vapor are in equilibrium with each other and all of them obey the ideal gas law.

130

4

Support Areas: Psychrometrics

• Moist air, dry air, and water vapor are all at the same temperature. Usually, the reference temperature is the temperature of the ambient atmospheric air. • Moist air, dry air, and water vapor would occupy the same volume of the containment space. • The atmospheric pressure, P, is the total pressure due to the partial pressures of the two components, dry air (denoted by subscript a) and water vapor (denoted by subscript w). According to Dalton’s Law, the total pressure of the gas mixture is the sum of the partial pressures of the components. Therefore, P = Pa þ Pw

ð4:8Þ

In Eq. 4.8, P is the atmospheric pressure, Pa is the partial pressure of dry air and Pw is the partial pressure of water vapor. It is important to study and understand the properties of moist air [1, 3] so that suitable methods can be used to manipulate the properties of moist air as desired. For example, atmospheric air can be hot and humid during summer months at certain locations. For human comfort, it is necessary to cool the hot, moist air and remove the condensed moisture from this air. This process is known as dehumidification and cooling, which is commonly known as air conditioning [1].

4.3.1

Moist Air Properties and Definitions

Dry bulb temperature (TDB): The dry bulb temperature is the ordinary, atmospheric temperature of air measured by a regular thermometer. Wet bulb temperature (TWB): The wet bulb temperature is the temperature measured by a wet bulb thermometer. As the name implies, in a wet bulb thermometer the thermometer bulb is surrounded by a moist wick. Depending on the condition of the surrounding air, moisture from the wick will evaporate into air, using the internal energy of the thermometer itself. This results in a corresponding decrease in the thermometer reading, resulting in the wet bulb temperature. Dew point temperature (TDP): Moist air at equilibrium can hold a maximum amount of water vapor at a given temperature. This condition is known as saturated air, that is, moist air is saturated with water vapor. The saturation limit decreases as the temperature of ambient atmospheric air decreases. Hence, when moist air is cooled, it becomes supersaturated at a lower temperature, causing the excess moisture to condense out of the mixture leaving behind saturated air. The temperature at which condensation begins is called dew point temperature. Note: For moist air, TDP ≤ TWB ≤ TDB

4.3

Air–Water Vapor Mixture and Psychrometrics

131

Humidity ratio (ω): Humidity ratio quantifies the moisture content of air on a mass basis. It is the ratio of the mass of water vapor to the mass of dry air. The mathematical equation for the humidity ratio is ω=

lbm water vapor kg water vapor = lbm dry air kg dry air

ð4:9Þ

Since specific humidity is dimensionless, it will have the same numerical value in all systems of units. Relative humidity (ϕ or rh): Relative humidity is a measure of how close the air is to the saturated state at a given temperature. Air can hold a limited amount of moisture at any given temperature. On nights when there is a rapid drop in temperature of ambient atmospheric air, the air tends to become supersaturated since the amount of moisture air can hold will decrease with decrease in temperature of ambient atmospheric air. Supersaturated air has excess moisture which condenses on exposed surfaces. This type of condensation on surfaces can be noticed after a cool night. Fully saturated air has 100% relative humidity. Mathematically, relative humidity is the ratio of the partial pressure of water vapor in the given mixture to the partial pressure of water vapor if the mixture were to be saturated at the mixture temperature. The saturation partial pressure can be obtained from steam tables, and it is the saturation pressure corresponding to the mixture temperature. The formula for relative humidity is ϕ=

Pw Psat at T DB of mixture

ð4:10Þ

Example 4.1 Atmospheric air at 14.7 psia is at a temperature of 80 °F and has a relative humidity of 70%. What is the partial pressure of dry air in the mixture? (Solution) From steam tables, at 80 °F, the saturation pressure is Psat = 0.51 psia. Calculate the partial pressure of water vapor in moist air by using the definition of relative humidity (Eq. 4.10). Pw = ϕPsat = 0:7 × 0:51 psia = 0:357 psia Since atmospheric pressure is the sum of partial pressures of water vapor and dry air (Eq. 4.8), calculate the partial pressure of dry air as shown. Pa = P - Pw = 14:7 psia - 0:36 psia = 14:34 psia

132

4.3.2

4

Support Areas: Psychrometrics

Relationship Between Humidity Ratio and Relative Humidity

Humidity ratio (ω) is a measure of moisture content in air and relative humidity (ϕ) is an indication of how close the moisture content is to the saturation level. Hence, it follows that there should be a relationship between humidity ratio and relative humidity [3], which is shown here by using basic thermodynamic relationships discussed in Chap. 3. N represents moles, m represents mass, and M represents molecular weight. Recall from Chap. 3 that mass = moles multiplied by the molecular weight. ω=

mw N w M w Nw = = ma NaMa Na

kg 18:0153 kmol kg 28:9647 kmol

= 0:622

Nw Na

In an ideal gas mixture, the pressure ratio between components is the same as the mole ratio between components. Also, the partial pressure of dry air is, Pa = P - Pw. Therefore, ω = 0:622

Nw P Pw = 0:622 w = 0:622 Na Pa P - Pw

ð4:11Þ

From the definition of relative humidity (Eq. 4.10), Pw = ϕPsat. Substitute this result into the preceding equation to obtain the relationship between humidity ratio and relative humidity. ω = 0:622

ϕPsat Pw = 0:622 P - Pw P - ϕPsat

ð4:12Þ

Example 4.2 An air–water vapor mixture at 14 psia and 65 °F has 1% water vapor by mass. Calculate the relative humidity of this mixture. (Solution) Since the mass fraction of water vapor is 0.01, the mixture will have 0.01 lbm water/ lbm air–water vapor mixture. Calculate the humidity ratio using Eq. 4.9 (humidity ratio is the ratio of mass of water vapor to the mass of dry air). ω=

0:01 lbm H2 O 0:01 lbm H2 O = = 0:0111 lbm H2 O=lbm dry air lbm mixture 0:90 lbm dry air

The total atmospheric pressure is P = 14 psia. Substitute all the known values into Eq. 4.12 and solve for the partial pressure of water vapor, Pw, as shown

4.3

Air–Water Vapor Mixture and Psychrometrics

ω = 0:622 0:0111

Pw P - Pw

133

)

lbm H2 O Pw = 0:622 14 psia - Pw lbm d a Pw = 0:2455 psia

)

From steam tables, at 65 °F, the saturation pressure is Psat = 0.315 psia. Calculate the relative humidity of the mixture using Eq. 4.10. ϕ=

0:2455 psia Pw = = 0:78 ð78%Þ Psat at T DB of mixture 0:315 psia

Example 4.3 At a coastal location, the dry bulb temperature of moist air is 30 °C and the dew point is 20 °C. Determine A. the relative humidity of moist air B. the mole fraction of dry air (Solution) A. From steam tables, determine the saturation pressures of water vapor at 30 °C and 20 °C, respectively. Psat ð30 ° CÞ = 4:25 kPa ðabs:Þ Psat ð20 ° CÞ = 2:34 kPa ðabs:Þ The moist air will have water vapor content corresponding to the dew point of the mixture and since air–water vapor mixture is saturated at dew point, the partial pressure of water vapor is the same as the saturation pressure at dew point. Therefore, in the given moist air, Pw = Psat ð20 ° CÞ = 2:34 kPa ðabs:Þ Calculate the relative humidity of the mixture using Eq. 4.10. ϕ=

2:34 kPa Pw = = 0:55 ð55%Þ Psat at T DB of mixture 4:25 kPa

B. Since a coastal location is specified, the location is at sea level, where the total pressure of moist air is standard atmospheric pressure, P = 101.3 kPa. Calculate the partial pressure of dry air using Eq. 4.8. P = Pa þ Pw ) Pa = P - Pw = 101:3 kPa - 2:34 kPa = 98:96 kPa

134

4

Support Areas: Psychrometrics

Since mole fraction = pressure fraction in moist air, calculate the mole fraction of dry air using Eq. 4.5. ya =

4.3.3

N a Pa 98:96 kPa = = = 0:9769 N P 101:3 kPa

Other Properties of Moist Air

Specific volume (v): The specific volume of moist air is the total volume of moist air per unit mass of dry air (units: ft3/lbm d a, m3/kg d a, d a  dry air). From the ideal gas law (Eq. 3.10), v=

V R T Ra T Ra T = a = = ma Pa P - Pw P - ϕPsat

ð4:13Þ

where Ra= individual gas constant for air 3

Ra =

psia‐ft 10:73 lbmol‐ R psia‐ft3 °R = = 0:37 lbm lbm‐ ° R Ma 29 lbmol

Ra =

kJ 8:314 kmolK R kJ = = 0:2867 kg Ma kg K 29 kmol

ðUSCSÞ

ð4:14aÞ

ðSIÞ

ð4:14bÞ

T = absolute temperature of air (°R, K) P = total pressure of moist air, atmospheric pressure (psia, kPa) Pa = partial pressure of dry air (psia, kPa) Pw = partial pressure of water vapor in moist air (psia, kPa) Equations 4.11 and 4.13 can be combined to obtain an additional equation for the specific volume of moist air. v=

Ra T ð1 þ 1:608ωÞ P

where ω is the humidity ratio of moist air nomenclature is the same as in Eq. 4.13.

lbm H2 O kg H2 O lbm d a , kg d a

ð4:15Þ and the remaining

4.3

Air–Water Vapor Mixture and Psychrometrics

135

Specific enthalpy (h): The specific enthalpy of moist air is the total heat content (enthalpy) of dry air and the associated water vapor per unit mass of dry air (Btu/lbm d a, kJ/kg d a). Since latent heat content is much higher compared to sensible heat content, the contribution of water vapor to the specific enthalpy is significant as shown in the following equations for calculating the specific enthalpy of moist air. h = 0:24T þ ωð1061 þ 0:45T Þ

ðUSCSÞ

ð4:16aÞ

where h = specific enthalpy of moist air (Btu/lbm d a) Specific heat of dry air = 0.24 Btu/lbm d a ‐ ° F T = dry bulb temperature of moist air (°F) H2 O ω = humidity ratio of moist air lbm lbm d a Enthalpy of vaporization of water = 1061 Btu/lbm H2O (at Tref = 60 ° F) Specific heat of water vapor = 0.45 Btu/lbm d a ‐ ° F The formula for calculating specific enthalpy of moist air in SI units is h = 1:0T þ ωð2466 þ 1:88T Þ

ðSIÞ

ð4:16bÞ

where h = specific enthalpy of moist air (kJ/kg d a) Specific heat of dry air = 1.0 kJ/kg d a  ° C T = dry bulb temperature of moist air (°C) ω = humidity ratio of moist air

kg H2 O kg d a

Enthalpy of vaporization of water = 2466 kJ/kg H2O (at Tref = 15 ° C) Specific heat of water vapor = 1.88 kJ/kg H2O  ° C Commonly used psychrometric charts are valid only at sea level where the barometric pressure is 760 mm Hg (1 atmosphere). However, the barometric pressure decreases with increase in altitude. Hence, sea-level, 1 atmosphere barometric pressure psychrometric charts are not accurate at higher altitude locations. In such locations, the psychrometric formulas that have been presented here [1, 3] will have to be used.

136

4

Support Areas: Psychrometrics

Example 4.4 At a given location, the atmospheric pressure is 13 psia and the air temperature is 50 °F. The relative humidity is 40%. For the moist air at this location, determine: A. the humidity ratio B. the specific enthalpy C. the specific volume (Solution) A. Determine the saturation pressure at 50 °F from the steam tables. Psat = 0:18 psia at 50 ° F Calculate the humidity ratio substituting all the known values into Eq. 4.12. ω = 0:622

0:40 × 0:18 psia ϕPsat = 0:622 13 psia - 0:40 × 0:18 psia P - ϕPsat = 0:0035 lbm water=lbm d a

B. Calculate the specific enthalpy of moist air using Eq. 4.16a. h = 0:24T þ ωð1061 þ 0:45T Þ Btu = 0:240 ð50 ° FÞ lbm d a‐ ° F þ 0:0035

lbm H2 O lbm d a

1061

Btu lbm H2 O þ0:45

Btu × 50 ° F lbm H2 O‐ ° F

= 15:79 Btu=lbm d a C. Calculate the specific volume of moist air using Eq. 4.13 and the gas constant for air from Eq. 4.14a. The absolute temperature of air is T = 460 ° + 50 ° F = 510 ° R

v=

Ra T P - ϕPsat

psia‐ft3 ð510 ° RÞ lbm d a‐ ° R = ð13 psia - 0:40 × 0:18 psiaÞ 0:37

= 14:60 ft3 =lbm d a

4.3

Air–Water Vapor Mixture and Psychrometrics

4.3.4

137

Use of Psychrometric Chart to Obtain Properties of Moist Air

The properties of moist air can be determined from the Psychrometric Chart [1, 3]. Figure 4.2 illustrates the general layout and the parameters used in the psychrometric chart for a given barometric pressure. The state point of moist air can be located on the chart by using two parameters, for example, dry bulb temperature and relative humidity. Once the state point is located, all the properties of moist air can be determined [1, 3] as shown in Fig. 4.2. The psychrometric chart (sea level) for USCS units is shown Fig. 4.3. The psychrometric chart in SI units is shown in Fig. 4.4. The following example illustrates the use of psychrometric chart to find the properties of moist air. Example 4.5 Air is at 70 °F and 60% relative humidity. Determine: A. the humidity ratio B. the wet bulb temperature

Fig. 4.2 Determining properties of moist air from the psychrometric chart

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Support Areas: Psychrometrics

Fig. 4.3 Sea-level psychrometric chart, USCS units. (Generated by the Author, N. S. Nandagopal, P E)

Fig. 4.4 Sea-level psychrometric chart, SI units. (Generated by the author, N.S. Nandagopal, P E)

4.3

Air–Water Vapor Mixture and Psychrometrics

139

C. the dew point temperature D. the enthalpy per pound mass dry air E. the specific volume (Solution) Locate the state point of moist air at the intersection of 70 °F dry bulb vertical line and 60% relative humidity curve as shown in the following excerpt of the psychrometric chart. Using the state point as a reference, determine all the properties of moist air as shown in the following excerpt of the psychrometric chart.

A. B. C. D. E.

ω = 0.0092 lbm H2O/lbm dry air TWB = 61 °F TDP = 55.5 °F h = 27 Btu/lbm dry air v = 13.55 ft3/lbm dry air

Example 4.6 Calculate the following standard (sea-level) properties of moist air at 70 °F and 60% relative humidity using psychrometric formulas and comment on the results obtained: A. B. C. D.

humidity ratio dew point enthalpy specific volume

(Solution) A. From steam tables determine the saturation pressure of water vapor at 70 °F. Psat ð70 ° FÞ = 0:3633 psia

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Support Areas: Psychrometrics

At sea-level, the total pressure of moist air is the atmospheric pressure, that is, P = 14.7 psia Calculate the humidity ratio using Eq. 4.12. ω = 0:622

ϕðPsat Þ P - ϕPsat

0:60 × 0:3633 psia 14:7 psia - 0:60 × 0:3633 psia = 0:0094 lbm H2 O=lbm dry air = 0:622

From the psychrometric chart (solution to Example 4.5), ω = 0.0092 lbm H2O/lbm dry air. B. Calculate the partial pressure of water vapor in the mixture using the given relative humidity and Eq. 4.10. Pw ) Psat at T DB of mixture Pw = ϕPsat = ð0:60Þð0:3633 psiaÞ = 0:2180 psia ϕ=

At the dew point, the mixture is saturated. Therefore, the dew point is the saturation temperature corresponding to the partial pressure of water vapor in the mixture. From steam tables, T DP = T sat ð0:2180 psiaÞ = 55:5 ° F From the psychrometric chart (solution to Example 4.5), TDP = 55.5 °F C. Calculate the enthalpy of moist air using Eq. 4.16a. h = 0:24T þ ωð1061 þ 0:45T Þ Btu ð70 ° FÞ = 0:240 lbm d a‐ ° F lbm H2 O þ 0:0094 lbm d a

1061

Btu lbm H2 O þ0:45

Btu × 70 ° F lbm H2 O‐ ° F

= 27:07 Btu=lbm d a From the psychrometric chart (solution to Example 4.5), h = 27 Btu/lbm d a

Practice Problems

141

D. Calculate the specific volume of moist air using Eq. 4.13 and the gas constant for air from Eq. 4.14a. The absolute temperature of air is T = 460 ° + 70 ° F = 530 ° R

v=

Ra T P - ϕPsat

psia‐ft3 ð530 ° RÞ lbm d a‐ ° R = ð14:7 psia - 0:60 × 0:3633 psiaÞ 0:37

= 13:54 ft3 =lbm d a From the psychrometric chart (solution to Example 4.5), v = 13.55 ft3/lbm dry air Comment: The results from psychrometric formulas are in very good agreement with the results from the psychrometric chart.

Practice Problems Practice Problem 4.1 In atmospheric air at 100 kPa and 30 °C, the partial pressure of water vapor is 1.3 kPa. Calculate: A. the partial pressure of dry air B. the relative humidity of atmospheric air Practice Problem 4.2 At a certain location, the atmospheric pressure is 90 kPa and air is at 30 °C with 55% relative humidity. Calculate the humidity ratio of moist air at this location. Practice Problem 4.3 Air is at 30 °C, dry bulb, and 20 °C wet bulb. Determine: A. B. C. D. E.

the humidity ratio the relative humidity the dew point temperature the enthalpy per kilogram mass of dry air the specific volume

Practice Problem 4.4 The atmospheric pressure and temperatures at an altitude of h meters can be obtained from Eqs. 1.4e and 1.4f in Chap. 1 (reproduced here for quick reference). P = 101:3ð1 - 0:000023hÞ5:26 t = 15 - 0:0011h h in m, P in kPa, and t in ° C

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4

Support Areas: Psychrometrics

Determine the following properties of moist air at an altitude of 1500 m when the relative humidity is 20%. A. B. C. D.

dew point humidity ratio specific volume specific enthalpy

Solutions to Practice Problems Practice Problem 4.1 (Solution) A. Since atmospheric pressure is the sum of partial pressures of water vapor and dry air (Eq. 4.8), calculate the partial pressure of dry air as shown. Pa = P - Pw = 100 kPa - 1:3 kPa = 98:7 kPa B. From steam tables, at 30 °C, the saturation pressure is Psat = 0.0042 MPa = 4.2 kPa. Calculate the relative humidity of atmospheric air using Eq. 4.10. ϕ=

1:3 kPa Pw = = 0:31 ð31%Þ Psat at T DB of mixture 4:2 kPa

Practice Problem 4.2 (Solution) From steam tables, at 30 °C, the saturation pressure is, Psat = 0:0042 MPa = 4:2 kPa Calculate the humidity ratio using Eq. 4.12. ω = 0:622

0:55 × 4:2 kPa ϕPsat = 0:622 90 kPa - 0:55 × 4:2 kPa P - ϕPsat = 0:0164 kg H2 O=kg dry air

Practice Problem 4.3 (Solution) Locate the state point of moist air at the intersection of 30 °C dry bulb vertical line and 20 °C wet bulb inclined line as shown in the following excerpt of the psychrometric chart. Using the state point as a reference, determine all the properties of moist air as shown in the following excerpt of the psychrometric chart.

Solutions to Practice Problems

A. B. C. D. E.

143

ω = 0.011 kg water/kg dry air RH = 40% TDP = 15 °C h = 58 kJ/kg dry air v = 0.87 m3/kg dry air.

Practice Problem 4.4 (Solution) Calculate the atmospheric pressure and temperature at the specified altitude using the given formulas. P = 101:3ð1 - 0:000023hÞ5:26 = 101:3ð1 - 0:000023 × 1500 mÞ5:26 = 84:22 kPa t = 15 - 0:0011h = 15 - 0:0011 × 1500 = 13:35 ° C A. From steam tables, determine the saturation pressure of water vapor at 13.35 °C. Psat ð13:35 ° CÞ = 1:533 kPa ðabs:Þ Calculate the partial pressure of water vapor in the mixture using the given relative humidity and Eq. 4.10.

144

4

Support Areas: Psychrometrics

Pw ) Psat at T DB of mixture Pw = ϕPsat = ð0:20Þð1:533 kPaÞ = 0:3066 kPa

ϕ=

At dew point, the mixture is saturated. Therefore, the dew point is the saturation temperature corresponding to the partial pressure of water vapor in the mixture. From steam tables, T DP = T sat ð0:3066 kPaÞ = - 9:14 ° C B. Calculate the humidity ratio using Eq. 4.12.

ω = 0:622

ϕPsat P - ϕPsat

0:20 × 1:533 kPa 84:22 kPa - 0:20 × 1:533 kPa = 0:0023 kg H2 O=kg dry air = 0:622

C. Calculate the absolute temperature of air. T = 273 þ ° C = 273 þ 13:35 ° C = 286:35 K

Ra = 0:2867

kJ kg  K

Calculate the specific volume of moist air using Eq. 4.13. v=

Ra T P - ϕPsat

kJ ð286:35 KÞ kg  K = = 0:9783 m3 =kg d a 84:22 kPa - 0:20 × 1:533 kPa 0:2867

Units:

kJ kg

kPa



kNm kg kN m2

 m3 =kg

D. Calculate the specific enthalpy of moist air using Eq. 4.16b. h = 1:0T þ ωð2466 þ 1:88T Þ = 1:0 × 13:35 ° C þ 0:0023 = 19:08 kJ=kg d a

kg H2 O ð2466 þ 1:88 × 13:35 ° CÞ kg d a

References

145

References 1. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook Fundamentals, I P edn, ASHRAE, Peachtree (2021) 2. Bahrami, M.: Gas Vapor Mixtures and HVAC, SFU Notes (2011). Download from, https://www. sfu.ca/~mbahrami/ENSC%20461/Notes/Gas%20Vapor%20Mixture_HVAC.pdf 3. Nandagopal, N.S.: Fluid and Thermal Sciences – A Practical Approach for Students and Professionals. Springer Nature, Cham (2022)

Chapter 5

Air-Conditioning Processes

5.1

Introduction

An air conditioning process is any process that alters the state of moist air. Air conditioning goes beyond the most commonly perceived air conditioning process of cooling and dehumidification (moisture removal) of hot, humid air. Section 10, Chapter 1 in ASHRAE Fundamentals Handbook [1], contains a summary of typical air-conditioning processes with relevant diagrams and examples. The different air conditioning processes are described in this chapter along with illustrative problems and representations on the psychrometric chart. Figure 5.1 illustrates the paths of commonly used air conditioning processes. The mass flow rate of dry air in any air conditioning process remains constant [1, 2]. This is a key concept used in the solution of problems on air conditioning. This is also the reasoning behind using unit mass of dry air as the basis for moist air properties [1–3]. For example, the humidity ratio is lbm moisture/ lbm dry air or kg moisture/kg dry air.

5.2

Cooling and Dehumidification

The most commonly perceived air conditioning process is cooling and dehumidification of hot, moist air [1, 2]. In this process, shown as process 1 and 2 in Fig. 5.1, the temperature of air decreases along with decreases in moisture content and relative humidity.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 N. S. Nandagopal, HVACR Principles and Applications, https://doi.org/10.1007/978-3-031-45267-3_5

147

148

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Air-Conditioning Processes

Fig. 5.1 Commonly used air conditioning processes

Example 5.1 8000 cfm of air at 80 °F dry bulb and 80% relative humidity is conditioned to 70 °F dry bulb and 60% relative humidity. Determine A. the rate of condensate removal in gpm B. the tons of cooling achieved (1 ton of cooling = 12,000 Btu/hr) (Solution) Obtain the properties of moist air at state point 1 (80 °F DB and 80% RH) and at state point 2 (70 °F DB and 60% RH) from the following excerpt of the psychrometric chart as shown.

5.2

Cooling and Dehumidification

149

State point 1 (80 °F DB and 80% RH): ω1 = 0:018 lbm H2 O=lbm d a, h1 = 39 Btu=lbm d a, v1 = 14:0 ft3 =lbm d a

State point 2 (70 °F DB and 60% RH): ω2 = 0:0095 lbm H2 O=lbm d a, h1 = 27 Btu=lbm d a

A. Draw a schematic diagram for the air conditioner as shown.

Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1.

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ft 8000 min V_ = = 571:43 lbm d a= min ν1 14:0 ft3 lbm d a 3

m_ a =

The mass flow rate of dry air is not affected by any air conditioning process and hence it is constant. Mass balance for water around the air conditioner results in the following equations. Mass flow of water in = mass flow of water out m_ a ω1 = m_ a ω2 þ m_ w,condensed Solve the preceding equation for the rate of condensation of water and substitute the known values from the excerpt of the psychrometric chart. m_ w,condensed = m_ a ðω1 - ω2 Þ lbm d a lbm water lbm water = 571:43 0:018 - 0:0095 min lbm d a lbm d a = 4:857 lbm water= min

Convert the mass rate of condensation to gpm using the density of water, ρw = 8.34 lbm/gal. Water condensed in gpm =

water m_ w,condensed 4:857 lbmmin = = 0:5824 gpm lbm ρw 8:34 gal

B. Calculate the rate of cooling achieved by performing an energy balance for the air conditioner. Rate of energy in = Rate of energy out m_ a ha1 = m_ a ha2 þ m_ w hw þ Q_ out Solve the preceding equation for the rate of heat removal and substitute the known values. From steam tables, the enthalpy of the condensate is, hw = hf at 70 ° F = 38.08 Btu/lbm Q_ out = m_ a ðha1 - ha2 Þ - m_ w hw Btu lbm water lbm d a Btu - 4:857 = 571:43 39 - 27 lbm d a min min lbm d a = 6672 Btu= min

38:08

Btu lbm water

5.2

Cooling and Dehumidification

151

Convert the rate of cooling to tons. Btu 60 min Q_ out ðtonsÞ = 6672 × min hr

5.2.1

1 ton = 33:36 tons 12, 000 Btu hr

Actual Process of Cooling and Dehumidification: Apparatus Dew Point (ADP) and By-pass Factor (BF)

The process of cooling and dehumidification considered in Example 5.1 does not follow the straight line path from state point 1 to state point 2. In reality, the air has to be saturated before any condensation can occur. Therefore, the actual cooling and dehumidification process follows the path 1–1A–1B–2 as shown in Fig. 5.2 [1]. Process 1–1A is sensible cooling of air from state point 1 to state point 1A, where the air is saturated. Process 1A–1B is the dehumidification process where moisture condenses from saturated air resulting in a lower humidity ratio. Process 1B–2 is sensible heating of saturated air to a lower, desired value of relative humidity. Figure 5.2 also illustrates the enthalpy changes associated with the cooling and dehumidification process. The total enthalpy decrease during the sensible cooling and condensation process at saturation is 16 Btu/lbm d a, followed by an enthalpy increase of 4 Btu/lbm d a for sensible heating. This results in a net enthalpy decrease of 12 Btu/lbm d a for the cooling and dehumidification process. By-pass Factor and Coil Efficiency A portion of feed airstream passing over cooling coils completely bypasses cooling surfaces and essentially remains at feed condition. This portion of air is referred to as by-pass air and the fraction of air that does not contact the coil surface is called the ‘Bypass Factor (BF).’ The coil efficiency, also known as ‘Contact Factor (CF)’ is a measure of the effectiveness of the coil surfaces in contacting the feed air stream and it is the complement of the bypass factor [2]. Coil Efficiency, CE = 1 - BF

ð5:1Þ

Apparatus Dew Point (ADP) Under ideal conditions of contact of 100% of inlet air with coil surfaces and with negligible heat transfer resistance of coil wall, the inlet air to the coil will cool to a saturated temperature called the Apparatus Dew Point (ADP), TADP (also known as Effective Coil Temperature) [1, 2]. However, due to a fraction of air bypassing the coil surfaces, only the fraction of air equivalent to the contact factor, (1-BF), will cool down to ADP and this fraction of air mixes with by-passed air. It is important to note that the bypassed air is at the original inlet air temperature, Tinlet. The steadystate temperature of the mixture is the temperature of the air leaving the coil Toutlet.

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Fig. 5.2 Actual process of cooling and dehumidification

The outlet temperature of air and hence the bypass factor (BF) can be derived from energy balance at the junction of the bypass air and contact air streams as shown here. The temperature of each stream in Fig. 5.3 is representative of the enthalpy of that stream. Energy balance at the junction, J, of the bypass stream and the stream in contact with the coil surface yields the following set of equations.

Energy into junction J = Energy out of junction J m_ ðBFÞðT inlet Þ þ m_ ð1 - BFÞðT ADP Þ = m_ ðT outlet Þ ) T outlet = ðBFÞðT inlet - T ADP Þ þ T ADP

BF =

T outlet - T ADP T inlet - T ADP

ð5:2Þ

ð5:3Þ

5.2

Cooling and Dehumidification

153

Fig. 5.3 Schematic for bypass factor and apparatus dew point

If the effective coil temperature (TADP), bypass factor (BF), and the inlet air temperature, (Tinlet), are given, then the outlet air temperature (Toutlet) can be calculated using Eq. 5.3. The coil temperature can be graphically determined from the psychrometric chart by connecting the state points of inlet air and outlet air and then extending the connecting line to the saturation curve. The intersection of the connecting line with the saturation curve is the apparatus dew point or coil temperature. This is illustrated in the solution to the following example. Example 5.2 The following information is available about a cooling coil in an air handling unit. Air entering the cooling coil: dry bulb temp = 27 °C, wet bulb temp = 20 °C Air leaving the cooling coil: dry bulb temp = 15 °C, wet bulb temp = 13 °C Determine the bypass factor and the coil efficiency. (Solution) Locate the state points of air entering the coil (state point 1) and leaving the coil (state point 2) on the excerpt of the psychrometric chart as shown. State point 1: TDB = 27 ° C, State point 2: TDB = 15 ° C,

TWB = 20 ° C TWB = 13 ° C

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Air-Conditioning Processes

The line joining state points 1 and 2 is the ‘process line’. Extend the process line to the saturation curve. The intersection of the process line with the saturation curve is the apparatus dew point (TADP). From the graph, TADP = 9 ° C Calculate the bypass factor using Eq. 5.3. BF =

T outlet - T ADP 15 ° C - 9 ° C = = 0:3333 27 ° C - 9 ° C T inlet - T ADP

Calculate the coil efficiency using Eq. 5.1. Coil Efficiency, CE = 1 - BF = 1 - 0:3333 = 0:6667ð ≃ 67%Þ

5.3

Heating and Humidification

At some locations, the state of moist air could be cold and dry (low humidity) during the winter season. Cold, dry air can damage human skin and sensitive electronic devices. Cold, dry air needs to be conditioned by adding moisture and heat. This process, shown as process 7–8 in Fig. 5.1, is the heating and humidification process [1, 2]. In this process, both the moisture content and temperature will increase, and this is typically accomplished by spraying steam into the cold, dry air.

5.3

Heating and Humidification

155

Note: The process line for process 7–8 shown in Fig. 5.1 has a slope less than 1.0. This will result in an increase in dry bulb and wet bulb temperatures, and increases in humidity ratio, and enthalpy. However, there will be a decrease in relative humidity as indicated by the process line 7–8. When the slope of the process line for heating and humidification is greater than 1.0, then all the parameters (dry bulb, wet bulb, humidity ratio, enthalpy, and relative humidity) will show an increase in value. Further discussion of this phenomenon will be done using actual numbers from Examples 5.3 and 5.4. Example 5.3 300 m3/min of air at 5 °C and 40% relative humidity is to be humidified and heated a using steam spray. The desired conditions of air are 20 °C and 30% relative humidity. Steam is supplied at 100 kPa and 150 °C. Determine: A. the rate at which steam must be supplied. B. the kW of additional heat input required, if any. (Solution)

From the excerpt of the psychrometric chart (as shown), determine the properties of air at state points before (state point 1) and after (state point 2) the heating and humidification process. Therefore, the process line will be from state point 1 to state point 2. Note that the relative humidity decreases during the process. State point 1 (5 ° C DB, 40 % rh): v1 = 0.79 m3/kg d a, ω1 = 0.002 kg H2O/kg d a, ha1 = 12 kJ/kg d a State point 2(20 ° C DB, 30 % rh): ω2 = 0.004 kg H2O/kg d a, ha2 = 33 kJ/kg d a Draw the schematic diagram for the humidifier – heater as shown. Q_ in is the rate of additional heat input required.

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Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1. m 300 min V_ = = 380 kg= min ν1 0:79 m3 kg d a 3

m_ a =

Mass balance for water around the humidifier – heater results in the following set of equations. Mass flow of water in = mass flow of water out m_ a ω1 þ m_ s = m_ a ω2 Solve the preceding equation for the mass flow rate of steam and substitute the known values. m_ s = m_ a ðω2 - ω1 Þ kg kg water kg water = 380 0:004 - 0:002 min kg d a kg d a = 0:76 kg steam= min From steam tables, the enthalpy of steam (100 kPa, 150 °C, superheated steam) is, hs(0.10 MPa, 150 ° C) = 2776.6 kJ/kg. Energy balance for the humidifier-heater results in the following set of equations. Rate of energy in = Rate of energy out m_ a ha1 þ m_ s hs þ Q_ in = m_ a ha2 Solve the preceding equation for the rate of heat input and substitute the known values. Q_ in = m_ a ðha2 - ha1 Þ - m_ s hs kg steam kJ kJ kg d a - 12 - 0:76 = 380 33 min kg d a kg d a min = 5970 kJ= min

2776:6

kJ kg steam

5.3

Heating and Humidification

157

Convert the additional heat input required to kW. Q_ in =

kJ 5870 min 60 s min

= 97:83 kW

Example 5.4 300 m3/min of air at 5 °C and 40% relative humidity is to be humidified and heated a using steam spray. The desired conditions of air are 20 °C and 60% relative humidity. Steam is supplied at 100 kPa and 150 °C. Determine: A. the rate at which steam must be supplied. B. the kW of additional heat input required, if any. (Solution) From the excerpt of the psychrometric chart (as shown), determine the properties of air at state points before (state point 1) and after (state point 2) the heating and humidification process. Therefore, the process line will be from state point 1 to state point 2. Note that the relative humidity increases during the process. State point 1 (5 ° C DB, 40 % rh): v1 = 0.79 m3/kg d a, ω1 = 0.002 kg H2O/kg d a, ha1 = 12 kJ/kg d a State point 2(20 ° C DB, 60 % rh): ω2 = 0.009 kg H2O/kg d a, ha2 = 43 kJ/kg d a

Draw the schematic diagram for the humidifier – heater as shown. Q_ in is the rate of additional heat input required.

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Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1. m 300 min V_ = = 380 kg= min ν1 0:79 m3 kg d a 3

m_ a =

Mass balance for water around the humidifier-heater results in the following set of equations. Mass flow of water in = mass flow of water out m_ a ω1 þ m_ s = m_ a ω2 Solve the preceding equation for the mass flow rate of steam and substitute the known values. m_ s = m_ a ðω2 - ω1 Þ kg kg water kg water = 380 0:009 - 0:002 min kg d a kg d a = 2:66 kg steam= min From steam tables, the enthalpy of steam (100 kPa, 150 °C, superheated steam) is, hs(0.10 MPa, 150 ° C) = 2776.6 kJ/kg. Energy balance for the humidifier-heater results in the following set of equations. Rate of energy in = Rate of energy out m_ a ha1 þ m_ s hs þ Q_ in = m_ a ha2 Solve the preceding equation for the rate of heat input and substitute the known values.

5.4

Cooling Towers

159

Q_ in = m_ a ðha2 - ha1 Þ - m_ s hs kg d a kJ kJ kg steam = 380 43 - 12 - 2:66 min kg d a kg d a min = 4394 kJ= min

2776:6

kJ kg steam

Convert the heat input required to kW. Q_ in =

kJ 4394 min 60 s min

= 73:23 kW

Discussion of results from Examples 5.3 and 5.4: Since there is a decrease in relative humidity for the process in Example 5.3, the steam requirement (0.76 kg steam/min) is much lesser than that in Example 5.4 (2.66 kg steam/min), where there is an increase in relative humidity for the process. Steam is a source of heat and consequently the heat input from steam is less in Example 5.3, requiring greater heat input from the additional source, that is, 97.83 kW vs. 73.23 kW in Example 5.4.

5.4

Cooling Towers

In power plants and chemical processing plants, hot water from condensers and heat exchangers needs to be cooled before recirculating back to the exchangers in a closed loop. Typically, this done in a cooling tower. A cooling tower operates adiabatically [2] since it uses the heat from hot water to heat and humidify the air flowing through the tower in a direction opposite to the hot water spray. Due to the supply of enthalpy of vaporization from the hot water to heat and humidify the air, the hot water cools down. Since some quantity of hot water evaporates into the air stream, makeup water must be supplied to maintain constant water circulation rate through the condensers and heat exchangers. The amount of air required to cool the hot water and the quantity of make-up water required can be determined by using mass and energy balances for the cooling tower as shown here. Figure 5.4 is a schematic of a cooling tower with all the relevant variables. Note that ‘1’ represents the top of the tower, ‘2’ represents the bottom of the tower, and water and air flow in opposite directions (counter flow). Mass balance for water around the envelope (red, dashed lines) shown in Fig. 5.4 results in the following set of equations. Mass flow of water in = mass flow of water out m_ w1 þ m_ a ω2 = m_ w2 þ m_ a ω1 m_ w1 - m_ w2 = m_ a ðω1 - ω2 Þ

ð5:4Þ

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Air-Conditioning Processes

Fig. 5.4 Schematic for a cooling tower

The left hand side of the preceding equation also represents the quantity of makeup water required. m_ m = m_ w1 - m_ w2 = m_ a ðω1 - ω2 Þ

ð5:5Þ

m_ w2 = m_ w1 - m_ a ðω1 - ω2 Þ

ð5:5aÞ

Energy balance around the envelope results in the following set of equations. Rate of energy in = Rate of energy out m_ w1 hw1 þ m_ a ha2 = m_ w2 hw2 þ m_ a ha1

ð5:6Þ

Further, Eqs. 5.5a and 5.6 can be combined to obtain an expression for the mass flow rate of dry air required. m_ a =

m_ w1 ðhw1 - hw2 Þ ðha1 - ha2 Þ - hw2 ðω1 - ω2 Þ

ð5:6aÞ

Example 5.5 8000 kg/hr of water at 40 °C is to be cooled to a temperature of 25 °C by using a cooling tower. Air enters the cooling tower at 20 °C dry bulb and 50% relative humidity and leaves the tower at 32 °C dry bulb and 70% relative humidity. Determine:

5.4

Cooling Towers

161

A. the quantity of air required in terms of Standard Cubic Meters per Minute (SCMM). B. the amount of make-up water required in litres per minute. (Solution) A. Use the same nomenclature as in Fig. 5.4. Locate the state points of moist air on the psychrometric chart and obtain the corresponding state properties as shown. State point 1 is air leaving the cooling tower at the top and state point 2 is air entering the cooling tower at the bottom.

State point 1(32 ° C DB, 70 % rh): ω1 = 0.022 kg H2O/kg d a, ha1 = 88 kJ/kg d a State point 2(20 ° C DB, 50 % rh) : v2 = 0.84 m3/kg d a ω2 = 0.0075 kg H2O/kg d a, ha2 = 39 kJ/kg d a Determine the enthalpies of water from the steam tables: hw1 = hf at 40 ° C = 167.53 kJ/kg water(entering water, top of the tower) hw2 = hf at 25 ° C = 104.83 kJ/kg water(leaving water, tower bottom) Substitute all the known values into Eq. 5.6a to obtain the mass flow rate of dry air.

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m_ w1 ðhw1 - hw2 Þ ðha1 - ha2 Þ - hw2 ðω1 - ω2 Þ kg water kJ kJ 8000 167:53 - 104:83 hr kg water kg water = kg water 0:022 kg d a kJ kJ kJ 88 - 39 - 104:83 kg d a kg d a kg water kg water - 0:0075 kg d a = 10564 kg d a=hr

m_ a =

Calculate the volume flow rate of air at inlet conditions using the specific volume of entering air. kg d a V_ air,in = m_ a ν2 = 10, 564 hr

0:84

m3 = 8874 m3 =hr kg d a

The air inlet temperature is 20 °C (293 K), whereas the reference temperature for standard cubic meter is 15 °C (288 K). Volume is directly proportional to absolute temperature. Calculate the Standard Cubic Meter per Minute (SCMM) of air required using the preceding concept. 1 hr m3 V_ air req: = 8874 × hr 60 min

288 K = 145:37 SCMM 293 K

B. Calculate the quantity of make – up water required by substituting the known values into Eq. 5.5. m_ m = m_ a ðω1 - ω2 Þ kg da kg water kg water = 10, 564 0:022 - 0:0075 kg d a hr kg d a = 153 kg water=hr Calculate the amount of make-up water required in litres per minute. Use the standard density of water, 1 kg/L. m_ m = 153

kg water 1 hr 1 × = 2:55 L= min × hr 60 min 1 kg L

5.4

Cooling Towers

5.4.1

163

Range, Approach, and Cooling Efficiency of Cooling Towers

In a cooling tower, feed hot water cannot be cooled to temperature below the wet bulb temperature of the entering air. Based on this concept, range, approach, and cooling efficiency for cooling towers are defined as follows.

Range: The range for a cooling tower is the temperature difference between entering and leaving water. Referring to Fig. 5.4, the range is Tw1 - Tw2. Approach: The approach for a cooling tower is the difference in leaving water temperature and wet bulb temperature of the air entering the water tower. Referring to Fig. 5.4, the approach is Tw2 - Ta2, wb. Cooling Efficiency: The cooling efficiency for a cooling tower can be defined as the ratio of actual cooling of water to the maximum possible cooling of water. As mentioned earlier, water cannot be cooled to temperatures below the wet bulb temperature of the entering air. Hence, referring to Fig. 5.4, the cooling efficiency of a cooling tower can be calculated by using the following equation.

ηc =

T w1 - T w2 100 T w1 - T a2,wb

ð5:7Þ

Example 5.6 Air at 20 °C and 60% relative humidity is used in a cooling tower to cool hot water entering the tower at 40 °C. The tower is designed to operate with an approach of 10 °C, and the design heat exchange rate in the tower is 147 kW. The average density of the circulating water is 0.99 kg/L. Determine the amount of make-up water required in litres per minute if the water losses in the tower due to all causes is 2.5% of the water circulation rate. (Solution) From the excerpt of the psychrometric chart, determine the wet bulb temperature of air entering the cooling tower (20 °C DB and 60% rh) as shown here.

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Air-Conditioning Processes

From the psychrometric chart, Ta2, wb = 15 ° C. The temperature of hot water entering the cooling tower is, Tw1 = 40 ° C (given). The given approach temperature is 10 °C, therefore, using the definition of approach determine the temperature of hot water leaving the tower. T w2 - T a2,wb = 10 ° C ) T w2 = T a2,wb þ 10 ° C = 15 ° C þ 10 ° C = 25 ° C From steam tables, determine the enthalpy of water entering the tower (1) and leaving the tower (2). hw1 = hf ð40 ° CÞ = 167:5 kJ=kg hw2 = hf ð25 ° CÞ = 104:8 kJ=kg If m_ w is the mass flow rate of hot water entering the tower, then due to 2.5% water losses in the tower, the mass flow rate of hot water leaving the tower is 0:975m_ w . From heat balance for the tower, Rate of heat lost by hot water = design heat exchange rate in the tower ) m_ w hw1 - 0:975m_ w hw2 = qtower ) qtower 147 kW m_ w = = hw1 - 0:975hw2 kJ kJ - ð0:975Þ 104:8 167:5 kg kg If m_ m is the mass flow rate of make-up water, then,

= 2:250 kg=s

5.5

Sensible Heating

165

m_ m = 0:025m_ w = ð0:025Þ 2:250

kg 60 s × s min

= 3:375 kg= min

Therefore, volume flow rate of make-up water required is kg 3:375 min m_ = 3:409 L= min V_ m = m = ρm 0:99 kg L

5.5

Sensible Heating

Sensible heating process, shown as process 3–4 in Fig. 5.1, increases the temperature of moist air without any changes to the moisture content [1]. This process is used to heat up cold air. Sensible heating also decreases the relative humidity of air since warmer air can hold more saturation moisture content compared to cold air, making the warmer air relatively drier than colder air at the same moisture content. Steam/hot water coils, electric heaters, furnaces based on gas heating and heat pumps are some of the options available for sensible heating of cold air. Example 5.7 3000 cfm of cold air at 50 °F dry bulb and 40 °F wet bulb is heated to 70 °F dry bulb at constant moisture content. Determine: A. the change in relative humidity. B. the kW rating of the heating coil required. (Solution) Obtain the properties of moist air at state point 1 (50 °F DB and 40 °F WB) and at state point 2 (70 °F DB and ω2 = ω1) from the excerpt of the psychrometric chart as shown.

From the preceding figure, State point 1 (50 ° F DB, 40 % rh): v1 = 12.9 ft3/lbm d a ω1 = 0.0032 lbm H2O/lbm d a, h1 = 15.5 Btu/lbm d a

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5 Air-Conditioning Processes

State point 2 (70 ° F DB, ω2 = ω1): rh2 = 20% ω2 = ω1 = 0.0032 lbm H2O/lbm d a, h2 = 21 Btu/lbm d a A. Δrh = rh2 - rh1 = 20 % - 40 % = - 20%. There is a 20% decrease in relative humidity. B. Draw a schematic diagram for the heater as shown.

Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1. ft 3000 min V_ = = 232:56 lbm d a= min m_ a = ν1 12:9 ft3 lbm d a 3

The mass flow rate of dry air is not affected by any air conditioning process and hence it is constant. Calculate the rate of heating required by performing an energy balance for the heater. Rate of energy in = Rate of energy out m_ a h1 þ Q_ in = m_ a h2 Solve the preceding equation for the rate of heat input required and substitute the known values. lbm d a Q_ in = m_ a ðh2 - h1 Þ = 232:56 min = 1047 Btu= min

20

Btu Btu - 15:5 lbm d a lbm d a

Convert the heat input rate to kW. Q_ in =

Btu 1047 min Btu 57 min kW

= 18:37 kW

5.6

Sensible Cooling

5.6

167

Sensible Cooling

Sensible cooling process, shown as process 5–6 in Fig. 5.1, decreases the temperature of moist air without any changes to the moisture content [1]. This process is used to cool relatively dry, hot air to a comfortable level of relative humidity and temperature. Sensible cooling is achieved by using cooling coils, which typically have refrigerants or chilled water circulating within them. Example 5.8 1800 m3/hr of air is cooled from 28 °C and 30% relative humidity to a final temperature of 20 °C at constant moisture content. Determine A. the cooling load for the cooler in kW B. the percent increase in relative humidity (Solution) Locate the state points of moist air on the excerpt of the psychrometric chart and obtain the corresponding state properties as shown. State point 1 is before cooling and state point 2 is after cooling.

State point 1(28 ° C DB, 30 % rh): v1 = 0.86 m3/kg d a, h1 = 46 kJ/kg d a State point 2 (20 ° C DB, ω2 = ω1): h2 = 38 kJ/kg d a Draw a schematic diagram for the cooler as shown.

168

5

Air-Conditioning Processes

Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1. 1h 1800 mh × 3600 V_ s = 0:5814 kg d a=s m_ a = 1 = 3 ν1 0:86 kgmd a 3

The mass flow rate of dry air is not affected by any air conditioning process and hence it is constant. Calculate the rate of cooling required by performing an energy balance for the cooler. Rate of energy in = Rate of energy out m_ a h1 = m_ a h2 þ Q_ out Solve the preceding equation for the rate of cooling required and substitute the known values. kg d a Q_ out = m_ a ðh1 - h2 Þ = 0:5814 s = 4:65 kW

5.7

46

kJ kJ - 38 kg d a kg d a

Other Air Conditioning Processes

Other, somewhat less common air conditioning processes are [1, 2] isothermal humidification, isothermal dehumidification, evaporative cooling, and chemical dehumidification, which are illustrated in Fig. 5.5.

5.7.1

Isothermal Humidification

In an isothermal humidification process, the relative humidity and the humidity ratio of air is increased at a constant dry bulb temperature [1] (process 1–2 in Fig. 5.5).

5.7

Other Air Conditioning Processes

169

Fig. 5.5 Other air conditioning processes

This is accomplished by introducing water vapor (steam) into the air space. Water vapor is generated by an electric humidifier or by a boiler. In dry conditions, friction can produce static build-up and sparks. Isothermal humidification is typically used in reducing or eliminating such static build-up. Isothermal humidification is also used in increasing moisture levels to prevent damage to sensitive electronic components and print media under dry conditions. Example 5.9 2000 cfm of air at 70 °F and 20% relative humidity is to be humidified to 50% relative humidity at constant temperature using an electric humidifier. Determine A. the increase in humidity ratio B. the minimum power rating of the electric humidifier in kW (Solution) Obtain the properties of moist air at state point 1 (70 °F DB, 20% rh) and at state point 2 (70 °F DB, 50% rh) from the excerpt of the psychrometric chart as shown.

From the preceding figure,

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5

Air-Conditioning Processes

State point 1 (70 ° F DB, 20 % rh): v1 = 13.45 ft3/lbm d a ω1 = 0.0032 lbm H2O/lbm d a, h1 = 20.2 Btu/lbm d a State point 2(70 ° F DB, 50 % rh): ω2 = 0.0075 lbm H2O/lbm d a, h2 = 25 Btu/lbm d a A. Calculate the increase in humidity ratio. Δω = ω2 - ω1 lbm H2 O lbm H2 O = 0:0075 - 0:0032 lbm d a lbm d a = 0:0043 lbm H2 O=lbm d a B. Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1. ft 2000 min V_ 1 = = 148:70 lbm d a= min 3 ν1 13:45 lbmft d a 3

m_ a =

The mass flow rate of dry air is not affected by any air conditioning process and hence it is constant. Calculate the rate of heating required by performing an energy balance for air. Rate of energy in + Rate of heat added by heater = Rate of energy out m_ a h1 þ Q_ in = m_ a h2 Solve the preceding equation for the rate of heat consumed by the electric humidifier and substitute the known values. lbm d a Q_ in = m_ a ðh2 - h1 Þ = 148:70 min = 714 Btu= min

25

Btu Btu - 20:2 lbm d a lbm d a

Convert the heat input rate to kW. Q_ in =

Btu 714 min Btu 57 min kW

= 12:53 kW

5.7

Other Air Conditioning Processes

5.7.2

171

Isothermal Dehumidification

In an isothermal dehumidification process, the relative humidity and the humidity ratio of air is decreased at a constant dry bulb temperature (process 3–4 in Fig. 5.5). Isothermal dehumidification can be achieved primarily by two main technologies – adsorption of water vapor by using desiccants such as silica gel and selective removal of water vapor using Isothermal Membrane Air Dehumidification (IMAD) processes [1]. A recent article from ASHRAE news highlights the possibility of energy savings in homes by using isothermal dehumidification instead of conventional air conditioning systems. May 2022 ASHRAE Journal Industry News. https://www.ashrae.org/news/esociety/isothermal-dehumidification-potentialhome-energy-saver

5.7.3

Evaporative Cooling

In an evaporative cooling process (process 5–6 in Fig. 5.5) hot dry air, typically found in desert climates, is cooled, and simultaneously humidified to achieve a comfortable combination of dry bulb temperature and relative humidity [1, 2]. The hot air gives up its sensible heat to evaporate water from a water source into the dry air and as a result the dry bulb temperature of air decreases with a simultaneous increase in its relative humidity and moisture content. This process is an entirely selfcontained adiabatic process without the use of any external heat source to humidify the air. The decrease in sensible heat of incoming air shows up as an equivalent increase in the latent heat due to increase in water vapor content of air. Hence, the enthalpy and wet bulb temperature of air remains constant during an evaporative cooling process. The increase in water vapor content of dry air can be measured by the increase in relative humidity as well as an increase in the humidity ratio of air. Evaporative cooling can be achieved by either direct evaporative cooling or indirect evaporative cooling. In direct evaporative cooling, the hot air is in direct contact with the water source. Water sources could be a pool of water in a pan or a film of water on substrates such as a sheet or a packed bed. In indirect evaporative cooling the water source first cools the exhaust air from the facility. The cool exhaust air gains heat from the primary hot air by means of an energy exchange device such as an enthalpy wheel. Example 5.10 On a typical hot summer day in Phoenix, Arizona, the condition of air is 90 °F dry bulb and 10% relative humidity. This air is cooled by an evaporative cooling process to a final temperature of 65 °F. Per pound of dry air, determine A. the final relative humidity B. the increase in humidity ratio C. the pounds of source water consumed

172

5

Air-Conditioning Processes

(Solution) Obtain the properties of moist air at state point 1 (90 °F DB, 10% rh) and at state point 2 (65 °F DB, constant enthalpy) from the excerpt of the psychrometric chart as shown. To locate state point 2, move along the constant enthalpy (or wet bulb) line to intersect with 65 °F dry bulb line.

From the preceding figure, State point 1(90 ° F DB, 10 % rh): ω1 = 0.003 lbm H2O/lbm d a, h1 = 25 Btu/lbm d a State point 2 (65 ° F DB, h2 = h1): rh2 = 67% ω2 = 0.0088 lbm H2O/lbm d a, From the preceding results, A. final relative humidity, rh2 = 67% B. increase in humidity ratio, Δω = ω2 - ω1 lbm H2 O lbm H2 O = 0:0088 - 0:003 lbm d a lbm d a = 0:0058 lbm H2 O=lbm d a C. Mass balance for water results in the following equations. ma ω1 þ mH2 O,source = ma ω2 )

5.8

Adiabatic Saturation Process

173

mH2 O,source = ma ðω2 - ω1 Þ = ð1 lbm d aÞ 0:0088

lbm H2 O lbm H2 O - 0:003 lbm d a lbm d a

= 0:0058 lbm H2 O

5.7.4

Chemical Dehumidification

In a chemical dehumidification process [1, 2], the relative humidity as well as the humidity ratio of air is decreased with a simultaneous increase in dry bulb temperature (process 7–8 in Fig. 5.5). Chemical dehumidification is accomplished by circulating the moist air through a hygroscopic material, which absorbs water vapor from moist air. The absorption of water vapor by hygroscopic materials is generally an exothermic chemical reaction. As a result, the heat released during the reaction is transferred to air. Hence, the end result is simultaneous increase in dry bulb temperature and decrease in moisture content of air.

5.8

Adiabatic Saturation Process

An adiabatic saturation process is similar to evaporative cooling. It uses the sensible heat of hot, dry air to evaporate water and hence increase the humidity of dry air [1, 2]. In contrast to isothermal humidification, the adiabatic humidification process will result in a temperature decrease of the incoming air. The sensible heat energy of the incoming air is converted to latent heat energy of the outgoing air due to the vaporization of water from the water source into the dry air. However, the total energy of the air stream remains constant throughout the adiabatic saturation process. Hence, the adiabatic saturation process line follows the constant enthalpy and constant wet bulb temperature lines. The source of water for adiabatic saturation process could be wet solids that need to be dried, or water that is atomized and sprayed into the air stream (air washer), or a wet pad that is typically used in pad type residential humidifier. The entering air cannot be cooled below its wet bulb temperature. Therefore, the maximum possible sensible cooling of the entering hot air is the difference between its dry bulb and wet bulb temperatures. Since the energy released during sensible cooling is used in saturating the air, the maximum possible sensible cooling is also a measure of the maximum possible saturation. The performance of air washers can be measured by saturation efficiency, which is the ratio of actual saturation achieved to the maximum possible saturation that can be achieved. The difference between the entering and leaving dry bulb temperatures of the hot air is a measure of the actual saturation achieved. Hence, the saturation efficiency of an adiabatic air washer

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5 Air-Conditioning Processes

(or any adiabatic saturation process) can be calculated using the following equation. Typical saturation efficiencies for air washers are in the range of 80–90%. ηsat =

T db,in - T db,out 100 T db,in - T wb,in

ð5:8Þ

Example 5.11 5 m3/s of hot, dry air at 38 °C and 20% relative humidity is processed to a final condition of 65% relative humidity in an adiabatic air washer using atomized water spray. Determine A. B. C. D.

the dry bulb and wet bulb temperatures of the humified air. the percent increase in humidity ratio. the quantity of make-up water required in L/s. the saturation efficiency.

(Solution) Locate the state points of moist air on the excerpt of the psychrometric chart and obtain the corresponding state properties as shown. State point 1 is before the adiabatic saturation process and state point 2 is after the adiabatic saturation process. To locate state point 2, move along the constant enthalpy (or wet bulb) line to intersect with 65% relative humidity curve.

State point 1 (38 ° C DB, 20 % rh): Twb1 = 20 ° C v1 = 0.89 m3/kg d a, ω1 = 0.008 kg H2O/kg d a State point 2 (65 % rh, Twb2 = Twb1 = 20 ° C): Tdb2 = 26 ° C ω1 = 0.0135 kg H2O/kg d a A. Tdb2 = 26 ° C, Twb2 = 20 ° C B.

5.9

Adiabatic Mixing of Air Streams

175

ω2 - ω 1 100 ω1 kg H2 O kg H2 O 0:0135 - 0:008 kg d a kg d a = 100 kg H2 O 0:008 kg d a = 72:5%

%inc:ω =

C. Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1. 5 ms V_ m_ a = 1 = = 5:62 kg d a=s 3 ν1 0:89 kgmd a 3

Mass balance for water results in the following equation. m_ a ω1 þ m_ H2 O,added = m_ a ω2 ) m_ H2 O,added = m_ a ðω2 - ω1 Þ kg H2 O kg d a kg H2 O = 5:62 0:0135 - 0:008 kg d a kg d a s = 0:0309 kg H2 O=s Calculate the make-up water required using the standard density of water, 1 kg/L. kg H2 O V_ make‐upH2 O = 0:0309 s

1

L kg H2 O

= 0:0309 L=s

D. Calculate the saturation efficiency using Eq. 5.8.

ηsat =

5.9

T 1,db - T 2,db 38 ° C - 26 ° C 100 = 100 = 0:6667 ð67%Þ 38 ° C - 20 ° C T 1,db - T 1,wb

Adiabatic Mixing of Air Streams

In commercial buildings, it is a common practice to use a mix of return air and outside air for purposes of air quality, ventilation, and energy conservation. This mixing process is adiabatic with no heat loss or heat gain. After mixing, the air is

176

5 Air-Conditioning Processes

conditioned as required. For purposes of conditioning, it is essential to know the properties of the mixed air. This can be determined by using mass and energy balances [1, 2] or by using a graphical method using the psychrometric chart. These methods are illustrated in the following example. Example 5.12 A facility needs 10,000 cfm of conditioned air. Prior to conditioning, outside air and return air are mixed in the proportion of 30% outside air and 70% return air by volume. The return air is at 70 °F dry bulb and 60% relative humidity and outside air is at 85 °F dry bulb and 80% relative humidity. Determine all the properties of the mixed air. (Solution) Locate the state points of return air (state point 1) and outside air (state point 2) on the excerpt of the psychrometric chart shown and determine the properties of moist air as shown. State point 1, RA: (70 ° F DB, 60 % rh): v1 = 13.55 ft3/lbm d a ω1 = 0.0094 lbm H2O/lbm d a, h1 = 27 Btu/lbm d a State point 2, OA: (85 ° F DB, 80 % rh): v2 = 14.2 ft3/lbm d a ω2 = 0.0210 lbm H2O/lbm d a, h2 = 43.5 Btu/lbm d a

5.9

Adiabatic Mixing of Air Streams

177

Draw a schematic for the mixing process as shown in the figure.

The volume flow rate of return air is V_ 1 = 0:7 × 10, 000 = 7000 cfm. The volume flow rate of outside air is V_ 2 = 0:3 × 10, 000 = 3000 cfm. Calculate the mass flow rates of return air and outside air (dry basis) using the specific volumes. ft 7000 min V_ 1 = = 516:60 lbm d a= min 3 ν1 13:55 lbmft d a 3

m_ a1 =

ft 3000 min V_ 2 = = 211:27 lbm d a= min 3 ν2 14:2 lbmft d a 3

m_ a2 =

Add up the mass flow rates of dry air in streams 1 and 2 to obtain the mass flow rate of dry air in the mixed stream (stream 3). lbm d a lbm d a þ 211:27 min min = 727:87 lbm d a= min

m_ a3 = m_ a1 þ m_ a2 = 516:60

Mass balance for water at the junction, J, results in the following equations. Mass flow rate of water into the junction = Mass flow rate of water out of the junction m_ a1 ω1 þ m_ a2 ω2 = m_ a3 ω3 Solve the preceding equation for ω3 and substitute the known values to obtain ω3. ω3 =

m_ a1 ω1 þ m_ a2 ω2 m_ a3 lbm d a 516:60 min

0:0094

lbm H2 O lbm d a

lbm d a min = lbm d a 727:87 min = 0:0128 lbm H2 O=lbm d a þ 211:27

0:0210

lbm H2 O lbm d a

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5

Air-Conditioning Processes

Energy balance at the junction, J, results in the following equations. Energy flow into the junction = Energy flow out of the junction m_ a1 h1 þ m_ a2 h2 = m_ a3 h3 Solve the preceding equation for h3 and substitute the known values to obtain h3. h3 =

m_ a1 h1 þ m_ a2 h2 m_ a3 lbm d a 516:60 min

27

Btu lbm d a

lbm d a min lbm d a 727:87 min

þ 211:27 = = 31:79 Btu=lbm d a

43:5

Btu lbm d a

The intersection of ω3 = 0.0128 lbm H2O/lbm d a, and h3 = 31.80 Btu/lbm d a determines state point 3, which can be located on the excerpt of the psychrometric chart as shown. At state point 3, the properties of moist air are: ∘

∘

T db3 = 74:6 F, T wb3 = 67:5 F, rh3 = 70%, v3 = 13:8 ft3 =lbm d a ∘

ω3 = 0:0128 lbm H2 O=lbm d a, h3 = 31:8 Btu=lbm d a, T DP3 = 64 F Note: The preceding results can be verified using results from on-line psychrometric calculators [3]. Alternate Method of Solution Using the lever rule, calculate the approximate value of the dry bulb temperature at state point 3 as shown. T DB3 =

V_ 1 × T DB1 þ V_ 2 × T DB2 V_ 3

ft3 ft3 ð70 ° FÞ þ 3000 ð85 ° FÞ min min = ft3 10, 000 min = 74:5 ° F 7000

Practice Problems

179

Join state point 1 and state point 2 with a straight line and locate state point 3 on this line corresponding to the vertical dry bulb temperature line of 74.5 °F. Determine all the properties of the mixed air at state point 3 as shown before. The results from the graphical method are consistent with the results obtained from the analytical method.

Practice Problems Practice Problem 5.1 7 m3/s of inlet air at 23 °C dry bulb temperature and 70% rh is to be conditioned to a final state of 18 °C dry bulb by a cooling coil. During this process, the rate of condensate collection is determined to be 0.05 L/s. Determine A. the remaining parameters (humidity ratio, wet bulb temperature, relative humidity, enthalpy, and specific volume) of the final state of air B. the tons of cooling achieved (1 ton of cooling = 3.517 kW) Practice Problem 5.2 Air enters a cooling coil at 80 °F. The coil temperature is 55 °F and 10% of the entering air bypasses the coil. Determine the temperature of the air leaving the coil. Practice Problem 5.3 On a cold winter day outdoor air is at 40 °F and 30% relative humidity. The outdoor air must be conditioned to 70 °F and 60% relative humidity to replenish 3500 cfm of indoor air at the stated conditions. Determine the rate at which moisture and heat must be added to outdoor air to accomplish this task. Practice Problem 5.4 3000 cfm of air at 80 °F dry bulb and 10% relative humidity is humidified to a final condition where the dew point of air is 60 °F. The process is accomplished by spraying saturated steam vapor at 20 psia into the humidification chamber. Determine A. the steam flow rate required in lbm/min. B. the state of the humidified air (humidity ratio, dry bulb and wet bulb temperatures, relative humidity, enthalpy, and specific volume). Practice Problem 5.5 25,000 cfm of air at 60 °F and 50% relative humidity is used in cooling 150 gpm of hot water (density = 8.3 lbm/gallon). Hot water enters the cooling tower at 90 °F, and the air leaves the tower at 80 °F and 70% relative humidity. For the operation of this cooling tower, determine: A. the exit temperature of water B. the range, approach, and cooling efficiency

180

5

Air-Conditioning Processes

Solutions to Practice Problems Practice Problem 5.1 (Solution) A. Calculate the mass rate of condensate removed using the standard density of water (1000 kg/m3) m_ H2 OðlÞ = 0:05

L 1 m3 × s 1000 L

1000 kg = 0:05 kg=s m3

Determine the properties of the inlet air (state point 1) from the excerpt of the psychrometric chart as shown.

From the excerpt of the psychrometric chart, at state point 1 (23 °C and 70% rh) ω1 = 0:0125 kg H2 O=kg d a, v = 0:86 m3 =kg d a, h1 = 54 kJ=kg d a Calculate the mass flow rate of dry air using the specific volume at inlet conditions. 3 7 ms V_ a1 = = 8:14 kg d a=s m_ a = 3 v1 0:86 m kg d a

m_ H2 OðlÞ represents the rate of condensation of water vapor. Mass balance for water around the air conditioner results in the following set of equations. Mass flow rate of water in = mass flow rate of water out m_ a ω1 = m_ a ω2 þ m_ H2 OðlÞ )

Solutions to Practice Problems

181

m_ H2 OðlÞ = m_ a ðω1 - ω2 Þ ) kg H2 O condensed 0:05 s kg d a kg H2 O - ω2 = 8:14 0:0125 kg d a s ω2 = 0:0064 kg H2 O=kg d a

)

On the excerpt of the psychrometric chart, locate state point 2, the final state of air leaving the coil, at the intersection of ω2 = 0.0064 kg H2O/kg d a, and TDB2 = 18 ° C as shown From the excerpt of the psychrometric chart, determine the following remaining properties at state point 2: h2 = 34 kJ=kg d a, v2 = 0:83 m3 =kg d a,

T WB2 = 11 ° C rh2 = 49%

B. If Q_ is the rate of cooling (heat removal) achieved, then energy balance for the unit results in the following equations. Rate of heat flow in = rate of heat flow out m_ a h1 = m_ a h2 þ Q_ ) Q_ = m_ a ðh1 - h2 Þ kg d a kJ kJ = 8:14 54 - 34 s kg d a kg d a = 162:8 kW heat removal rate: Calculate the tons of cooling achieved by using the conversion factor, 1 ton of refrigeration = 3.517 kW (Eq. 3.50). Q_ = 162:8 kW ×

1 ton = 46:29 tons of cooling 3:517 kW

Practice Problem 5.2 (Solution) The coil temperature is the same as the apparatus dew point. Therefore, T ADP = 55 ° F

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5

Air-Conditioning Processes

Calculate the outlet temperature of air using Eq. 5.2. T outlet = ðBFÞðT inlet - T ADP Þ þ T ADP = ð0:10Þð80 ° F - 55 ° FÞ þ 55 ° F = 57:5 ° F Practice Problem 5.3 (Solution) Locate the state points of the outdoor air (state point 1) and the inside air (state point 2) on the excerpt of the psychrometric chart and determine the properties of moist air at the state points as shown. State point 1, OA: (40 ° F DB, 30 % rh): v1 = 12.65 ft3/lbm d a, ω1 = 0.0017 lbm H2O/lbm d a, ha1 = 11 Btu/lbm d a State point 2, IA: (70 ° F DB, 60 % rh): ω2 = 0.0094 lbm H2O/lbm d a, ha2 = 27 Btu/lbm d a

Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1. ft 3500 min V_ = 276:68 lbm d a= min m_ a = 1 = 3 ν1 12:65 lbmft d a 3

Mass balance for water around the humidifier/heater results in the following equations. Mass flow of water in = mass flow of water out m_ a ω1 þ m_ moist:add = m_ a ω2

Solutions to Practice Problems

183

Solve the preceding equation for the mass flow rate of moisture to be added and substitute the known values. m_ moist:add = m_ a ðω2 - ω1 Þ lbm H2 O lbm d a lbm H2 O = 276:68 0:0094 - 0:0017 lbm d a lbm d a min = 2:13 lbm H2 O= min Energy balance around the humidifier/heater results in the following set of equations. Rate of energy in = Rate of energy out m_ a ha1 þ Q_ in = m_ a ha2 Solve the preceding equation for the rate of heat input and substitute the known values. Q_ in = m_ a ðha2 - ha1 Þ lbm d a = 276:68 min = 4427 Btu= min

27

Btu Btu - 11 lbm d a lbm d a

Practice Problem 5.4 (Solution) Locate the state points of initial air (state point 1) and final air (state point 2) on the excerpt of the psychrometric chart and determine the properties of moist air at state points as shown. Note that the enthalpy at state point 2 is obtained after subsequent calculations as shown.

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5

Air-Conditioning Processes

State point 1 (80 ° F DB, 10 % rh): v1 = 13.62 ft3/lbm d a, ω1 = 0.002 lbm H2O/lbm d a, ha1 = 21.5 Btu/lbm d a At state point 2, TDP2 = 60 ° F. Draw a horizontal line at TDP2 = 60 ° F. The moisture content is constant along the horizontal line. Therefore, ω2 = 0:0110 lbm H2 O=lbm d a Draw a schematic diagram for the humidifier as shown.

A. Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1. ft 3000 min V_ 1 = = 220:26 lbm d a= min 3 ν1 13:62 lbmft d a 3

m_ a =

m_ s is the mass flow rate of steam. Mass balance for water vapor around the humidifier/heater results in the following set of equations. Mass flow of water vapor in = Mass flow of water vapor out m_ a ω1 þ m_ s = m_ a ω2 ) m_ s = m_ a ðω2 - ω1 Þ lbm H2 O lbm d a lbm H2 O = 220:26 0:0110 - 0:002 lbm d a lbm d a min = 1:9823 lbm H2 O ðsteamÞ= min B. From steam tables, enthalpy of saturated steam (vapor) at 20 psia is hg = hs = 1156:4 Btu=lbm H2 O Energy balance around the humidifier/heater results in the following set of equations.

Solutions to Practice Problems

185

Rate of energy in = Rate of energy out m_ a ha1 þ m_ s hs = m_ a ha2 ) 220:26

lbm d a min

21:5

Btu lbm d a

lbm d a Btu lbm H2 O ha2 × 1156:4 = 220:26 min min lbm H2 O ) ha2 = 31:91 Btu=lbm d a

þ 1:9823

On the psychrometric chart, draw the constant enthalpy line at 31.91 Btu/lbm d a. The intersection of this constant enthalpy line with the horizontal line at ω2 = 0.0110 lbm H2O/lbm d a defines state point 2 as shown on the excerpt of the psychrometric chart. At state point 2, T db2 = 81:5 ° F, T wb2 = 67 ° F, rh2 = 48%, v2 = 13:9 ft3 =lbm d a ω2 = 0:0110 lbm H2 O=lbm d a, h2 = 31:91 Btu=lbm d a Practice Problem 5.5 (Solution) A. Subscript ‘1’ represents the top of the tower and subscript ‘2’ represents the bottom of the tower. The water flows from the top of the tower and the air flows from the bottom of the tower. Determine the properties of air entering (state point 2) and leaving the tower (state point 1) from the excerpt of the psychrometric chart as shown.

186

5

Air-Conditioning Processes

State point 2 (60 ° F DB, 50 % rh): v2 = 13.2 ft3/lbm d a, ω2 = 0.0054 lbm H2O/lbm d a, ha2 = 20.5 Btu/lbm d a State point 1(80 ° F DB, 70 % rh): ω1 = 0.0155 lbm H2O/lbm d a, ha1 = 36 Btu/lbm d a From steam tables, for water entering the top of the tower: hw1 = hf ð90 ° FÞ = 58:1 Btu=lbm Calculate the mass flow rate of water entering the tower using the density of water. m_ w1 = 150

gal: min

lbm = 1245 lbm= min gal:

8:3

Calculate the mass flow rate of dry air. ft V_ 2 25, 000 min = = 1894 lb d a= min 3 v2 13:2 lbftd a 3

m_ a =

Mass balance for water around the tower results in the following equations. Mass flow rate of water in = mass flow rate of water out m_ w1 þ m_ a ω2 = m_ w2 þ m_ a ω1 ) m_ w2 = m_ w1 - m_ a ðω1 - ω2 Þ Energy balance for the tower results in the following equation. Rate of Energy in = Rate of Energy out m_ w1 hw1 þ m_ a ha2 = m_ w2 hw2 þ m_ a ha1 Combine the preceding equations to obtain the following results. m_ w1 hw1 þ m_ a ha2 = ½m_ w1 - m_ a ðω1 - ω2 Þ]hw2 þ m_ a ha1 Therefore,

References

187

hw2 =

m_ w1 hw1 þ m_ a ðha2 - ha1 Þ m_ w1 - m_ a ðω1 - ω2 Þ Btu lbm H2 O lbm d a 58:1 1245 þ 1894 min lbm H2 O min 20:5

= 1245

Btu Btu - 36 lbm d a lbm d a

lbm d a lbm H2 O - 1894 min min 0:0155

lbm H2 O lbm H2 O - 0:0054 lbm d a lbm d a

= 35:06 Btu=lbm H2 O From steam temperature tables, when hf = 35:06 Btu=lbm H2 O ) T sat: = T w2 = 67 ° F ðexit temperature of waterÞ B. Calculate the range, approach, and cooling efficiency for the tower using the definitions and the corresponding equations Range = T w1 - T w2 = 90 ° F - 67 ° F = 23 ° F Note: From the psychrometric chart, Ta2, wb = 50 ° F Approach = T w2 - T a2,wb = 67 ° F - 50 ° F = 17 ° F Calculate the cooling efficiency of the tower using Eq. 5.7. ηc =

T w1 - T w2 90 ° F - 67 ° F 100 = 58% 100 = 90 ° F - 50 ° F T w1 - T a2,wb

References 1. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook Fundamentals, I P edn, ASHRAE, Peachtree (2021) 2. Nandagopal, N.S.: Fluid and Thermal Sciences – A Practical Approach for Students and Professionals. Springer Nature, Cham (2022) 3. On-line Psychrometric Calculator. Download from http://www.hvac-calculator.net/index.php?v=2

Chapter 6

Cooling and Heating Load Calculations

6.1

Introduction

Accurate cooling and heating and load calculations are important in obtaining the right size and capacity of relevant cooling and heating equipment. Cooling load calculations are based on typical outdoor conditions during summer months while heating load calculations are based on typical outdoor winter conditions. Obviously, the design indoor conditions have to be specified since the cooling and heating loads are highly dependent on the magnitude of the difference between indoor and outdoor conditions. In addition, cooling loads should consider the type of facility, the occupancy, the activity level, and the appliances and equipment used within the facility. For residential and commercial office buildings, the design indoor conditions are specified based on the required thermal comfort for the occupants. For other commercial facilities, the design indoor conditions are specified based on the processes/activities being carried out within the facility. For storage facilities, the design indoor conditions are specified based on the products being stored. The following chapters in ASHRAE Fundamentals Handbooks [2, 3] contain useful and relevant information on calculation of cooling and heating loads. . . . . . .

Chapter 9: Thermal Comfort Chapter 14: Climatic Design Information Chapter 15: Fenestration Chapter 16: Ventilation and Infiltration Chapter 17: Residential Cooling and Heating Load Calculations Chapter 18: Non-Residential Cooling and Heating Load Calculations

The Air Conditioning Contractors of America (ACCA) Manual J Version 8 [1] provides detailed steps required to calculate the heating and cooling loads. It is prudent to keep the following statements from the ACCA manual in mind while estimating the cooling and heating loads for a facility.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 N. S. Nandagopal, HVACR Principles and Applications, https://doi.org/10.1007/978-3-031-45267-3_6

189

190

6

Cooling and Heating Load Calculations

Cooling and heating load calculations should be aggressive, which means that the designer should take full advantage of legitimate opportunities to minimize the size of estimated loads. In this regard, the practice of manipulating the outdoor design temperature, not taking full credit for efficient construction features, ignoring internal and external window shading devices and then applying an arbitrary ‘safety factor’ is indefensible. No additional safety factors are required when load estimates are based on accurate information pertaining to the envelope construction and duct system efficiency. Large errors are possible if there is uncertainty about insulation levels, fenestration performance, envelope tightness or the efficiency of the duct runs installed in the unconditioned space.

6.2

Comparisons Between Cooling and Heating Load Calculations

The data on outdoor conditions play a key role in calculations of cooling loads for facilities [1–3]. However, the outside conditions vary significantly throughout the day due to changing intensity of solar radiation. The peak cooling load occurs sometime during the day, which the system must be able to handle. In addition, all internal sources of heat generation, which can vary due to changes in occupancy, and in usage of lights and appliances, must also be considered in the calculation of cooling loads. For greater accuracy, the thermal storage characteristics of the facility must be considered in cooling load calculations [2, 3]. This results in time lag between the time at which the conditioned space actually gains the heat and the time at which the heat starts to flow or dissipate. This makes the cooling load calculations inherently unsteady and complicated and results in solving transient equations with unsteady boundary conditions [2, 3, 11]. Heating load calculations are carried out to estimate the heat loss from the building in winter so as to arrive at required heat inputs for facilities. During winter months, the peak heating load usually occurs before sunrise and the outdoor conditions do not vary significantly throughout the winter season. Internal heat sources due to occupants and appliances compensate for some of the heat losses from the facility. However, the internal heat gains are not considered with the justification of them not making significant contributions. Due to almost steady outdoor conditions and negligible indoor contributions, the heat load calculations can be carried out assuming steady-state conditions [2, 3, 11]. This is in sharp contrast to the unsteady and complex nature of cooling load calculations.

6.3

Cooling Load Calculations

6.3

191

Cooling Load Calculations

This section considers all aspects of cooling load calculations including all possible heat sources, outdoor climate conditions, and designing the capacity of cooling systems based on the desired indoor conditions.

6.3.1

Rules of Thumb for Cooling Loads

While cooling loads are highly dependent on the particular situation, and outside conditions, general rules of thumb have been derived based on years of experience. These rules of thumb are very useful for initial estimates, and they can also be used as guidelines for checking the validity of the actual cooling calculations [2]. The following are rules of thumb estimates of cooling load requirements in tons of refrigeration (TR) per 1000 ft2 of floor space for different types of facilities. . . . . . .

office buildings with 50% glass 3 to 4.5 TR computer facilities 7 to 10 TR hotel rooms 0.5 to 1.0 TR restaurants 6 to 8 TR retail spaces/shopping malls 3 to 5 TR theaters/auditoriums/cinema halls 0.08 TR per seat

6.3.2

Cooling Load Components and Sources of Heat Gain

The various sources of heat gain in conditioned spaces can be classified into External Sources and Internal Sources, all of which contribute to the cooling load. External Sources of heat gain are outside the envelope of the facility. This includes both conduction and radiation heat gain through the roofs, walls, floor, and windows of the conditioned space [2, 3]. Health standards and codes specify minimum amounts of ventilation of internal air from conditioned spaces [7, 8]. The ventilated air must be replaced by conditioned outside air to maintain constant air circulation within the facility. Thus, ventilation adds to the cooling load of the conditioned space. It is difficult, if not impossible, to maintain complete air tightness, especially in residential facilities. Some amount of outside air creeps into the conditioned space due to cracks and similar openings in the building structure. This is termed as infiltration and infiltration of outside air also adds to the cooling load from external sources [2, 3]. Internal Sources of heat gain in a facility is primarily due to heat generation from occupants, lights, appliances, and other equipment [2, 3].

192

6.3.3

6

Cooling and Heating Load Calculations

Indoor and Outdoor Design Parameters

The main parameters used in the design of cooling systems are dry bulb and relative humidity (or wet bulb temperature) of both the indoor conditioned space and the outdoor environment. Indoor Parameters The indoor conditions are dictated by the thermal comfort required. The indoor design conditions are directly related to human comfort. Comfort standards, primarily ASHRAE Standard 55 [6], specify conditions for acceptable thermal environments for human occupancy. Another comfort standard is ISO Standard 7730 [10]. In summary, the preceding standards specify a ‘comfort zone,’ representing the optimal range and combinations of thermal factors and personal factors with which at least 80% of the building occupants are expected to express satisfaction. The key thermal factors are indoor air temperature, and humidity level, while personal factors include clothing and activity level.

The recommended indoor relative humidity range is 40–55% during the summer months. The cooling load calculations are usually based on indoor conditions of 75 °F DB and 50% relative humidity. Outdoor Parameters for Cooling Loads Chapter 14/Appendix of ASHRAE Fundamentals Handbooks [2, 3] are sources of extensive climate conditions for US, Canada, and other international locations. The following data provided in this table can be used specifically for cooling load calculations: Dry bulb (DB) temperature corresponds to 0.4, 1.0, and 2.0 percentiles of the annual cumulative frequency of occurrence and the mean coincident wet-bulb temperature (MCWB). These conditions are used in determining both the sensible and latent (explained later in this section) loads in the cooling mode. For example, Table 1 in Chapter 14 of ASHRAE Fundamentals Handbook, S I Edition [3] consists of extensive climatic data for Atlanta, GA, USA, followed by a detailed explanation of the significance of each table heading and its meaning. From this table, during the hottest month of July, the 0.40 percentile cumulative frequency DB is 34.4 °C and MCWB is 23.4 °C. What this means is, statistically only 0.4% of cumulative data exceeds the preceding values and therefore it makes sense to use this input to design cooling systems for this particular location. Another useful data in Table 1 is the mean daily range (DR) of the dry bulb temperature, which is the mean of the temperature difference between daily maximum and minimum temperatures for the warmest month (highest average dry-bulb temperature). For comfort cooling, use of the 2.0% percent cumulative frequency of occurrence of the outdoor condition is recommended as a design condition in calculating the maximum heat gain in facilities [2, 3].

6.3

Cooling Load Calculations

193

The 2.0% design condition means that the outside summer DB and MCWB will be exceeded only during 2.0% of the total hours from June to September, that is for about 59 hours out of a total of 2928 hours of these summer months.

6.3.4

Basics of Cooling Load Calculations

The total heat gain in a conditioned space consists of sensible heat gain and latent heat gain. The sensible heat gain is simply the heat transfer caused due to the difference between the outdoor and indoor temperatures. Sensible heat increases the temperature of a substance due to absorption of heat. It is important to note that the sensible heat does not change the state of the substance, for example from liquid state to vapor state. Sensible heat gain sources include [2, 3]: . . . . . .

heat transmitted through floors, ceilings, and walls heat emitted by occupants heat emitted by lights and appliances radiation heat gain through glass and other transparent entities such as skylights heat gained due infiltration of outside air (sensible component) heat gained from outside air due to ventilation (sensible component) Latent heat gain occurs due to addition of moisture to the conditioned space. Latent gain sources include [2, 3]:

. moist air from outside due to infiltration and ventilation (latent components) . moisture due to respiration and activities of occupants . moisture from equipment and appliances In order to maintain a comfortable level of humidity in the conditioned space, the cooling system must condense water vapor added to the conditioned space at a rate equal to the rate of addition. This dehumidification process is energy intensive (1084 Btu/lbm H2O(v) condensed, 2520 kJ/kg H2O(v) condensed).

6.3.4.1

Basic Equations for Cooling Load Calculations

Energy balance across a cooling coil shown in Fig. 6.1, results in the following set of equations: Rate of energy in = rate of energy out m_ a h1 = m_ a h2 þ qT )

194

6

Cooling and Heating Load Calculations

Fig. 6.1 Energy balance across a Cooling Coil

qT = m_ a ðh1 - h2 Þ

ð6:1Þ

where m_ a = mass flow rate of dry air (lbm d a/hr., kg d a/s) h1 = enthalpy of moist air entering the coil (Btu/lbm d a, kJ/kg d a) h2 = enthalpy of moist air leaving the coil (Btu/lbm d a, kJ/kg d a) qT = total rate of heat removal by the coil (Btu/hr., kW) Also, as shown in Fig. 6.1, the total rate of heat removed by the coil is the sum of the rate of sensible heat removed by the coil (qS)and the rate of latent heat removed by the coil (qL). Therefore, qT = q S þ q L

ð6:2Þ

Divide both sides of Eq. 6.2 by qT, the total rate of heat removal by the coil, to obtain the following equation: qS q L þ = 1 ) SHR þ LHR = 1:0 qT qT

ð6:3Þ

qS = Sensible heat ratio ðSHRÞ qT qL = Latent heat ratio ðLHRÞ qT

ð6:4Þ

where

ð6:5Þ

A sensible heat ratio around 0.70 is commonly used in air-conditioning systems [2, 3]. This means that 70% capacity of the cooling coil is used in handling the sensible load and the remaining 30% cooling capacity takes care of the latent load. For air conditioning systems, the volume flow rate of air across the cooling coil is usually specified in cubic feet per minute (cfm). Assuming standard density of air of

6.3

Cooling Load Calculations

195

0.075 lbm/ft3, the mass flow rate of air in lbm/hr. can be calculated using Eq. 1.9, in Chap. 1. m_ a = cfm × ρa = ðcfmÞ

60 min hr

0:075

lbm = 4:5 × cfm lbm=hr ft3

Substitute the preceding result into Eq. 6.1 to obtain the following equation for the total rate of heat removed by the coil: qT = 4:5ðcfmÞðh1 - h2 Þ

ð6:6Þ

From Chap. 3, Thermodynamics, the change in enthalpy for an ideal gas (air, in this case) only in the vapor phase can be calculated by using the following equation: Δh = cp ΔT ) h2 - h1 = cpa ðT 2 - T 1 Þ It is important to keep in mind that the preceding equation accounts only for the rate of sensible heat change of air across the coil and does not include latent heat changes. Using the standard specific heat of air of cpa = 0.24 Btu/lbm ‐ ° F, the preceding equation, and the equation for mass flow rate of air with standard density, the following equation is obtained for calculating the rate of sensible heat removed by the coil (Btu/hr). qs = m_ a cpa ðT 1 - T 2 Þ = ð4:5ÞðcfmÞ 0:24

Btu ðT 1 - T 2 Þ ) lbm‐ ° F

qS = ð1:08ÞðcfmÞðT 1 - T 2 Þ

ð6:7Þ

The condensing or latent heat load on the coil is dictated by the difference in the moisture content between the entering air and the exit air. From mass balance for water across the coil, the rate of condensation of water vapor is m_ H2 O,cond = m_ a ðω1 - ω2 Þ In the preceding equation, ω1and ω2are humidity ratios of the air entering and leaving the coil, respectively. The right hand side of the preceding equation has to be multiplied by the enthalpy of condensation of water vapor (steam) to obtain the latent heat load on the cooling coil. The latent heat of condensation (hfg) is obtained by subtracting the enthalpy of the condensed liquid water (hf) at 50 °F from the enthalpy of water vapor (hg) at 75 °F, which can be obtained from steam tables or on-line resources (https://www.tlv.com/global/TI/calculator/steam-table-temperature.html). Hence, hfg = hg,750 F - hf ,500 F = 1093:6

Btu Btu - 18:1 = 1075:5 Btu=lbm lbm lbm

196

6

Cooling and Heating Load Calculations

Using this value of latent heat of condensation, the preceding equation, and the equation for mass flow rate of air with standard density, the following equation is obtained for calculating the rate of latent heat removed by the coil (Btu/hr) Btu Btu = ð4:5ÞðcfmÞðω1 - ω2 Þ 1075:5 lbm lbm ) qL = ð4840ÞðcfmÞðω1 - ω2 Þ

qL = m_ a ðω1 - ω2 Þ 1075:5

ð6:8Þ

where ω1 and ω2 are the humidity ratios (specific humidities) of the air entering and leaving the coil, respectively (lbm H2O/lbm d a). Note: Eqs. 6.6, 6.7, and 6.8 are valid only under standard conditions of air (14.7 psia and 70 °F). However, standard conditions cannot always be assumed. In any case, the mass flow rate of dry air can be determined accurately by using the specific volume of air from psychrometric charts and the continuity equation as shown here. It must be noted that Eqs. 6.6, 6.7, and 6.8 are widely used in practice. In cooling calculations, the common practice is to obtain the total rate of heat removal using Eq. 6.1, calculate the rate of sensible heat removal using Eq. 6.7 or SI equivalent and then obtain the sensible heat ratio (SHR), which is a useful parameter in cooling load calculations. From the continuity equation (Eq. 1.9), the mass flow rate of dry air in lbm d a/hr. is m_ a = 60ðcfmÞρa =

60ðcfmÞ va

ð6:9Þ

where va is the specific volume of air in ft3/lbm d a, which can be obtained from psychrometric charts. Equations 6.6, 6.7, and 6.8 are valid only under standard density conditions of air, that is, ρair = 0.075 lbm/ft3(at sea level). Consequently, the constants 4.5, 1.08, and 4840 are valid only when ρair = 0.075 lbm/ft3. However, for locations at higher altitudes, the constants need to be modified by the following equation, which is based on the fact that the density of a fluid is directly proportional to the absolute pressure and inversely proportional to the absolute temperature (ideal gas law, ρ = P/RT). C alt = C0

Palt P0

T0 T alt

ð6:10Þ

6.3

Cooling Load Calculations

197

where Calt = modified constant at a given altitude C0 = standard values of the constants at sea level, that is, 4.5, 1.08, and 4840 Palt,Talt = absolute pressure and absolute temperature, respectively, at a given altitude P0,T0 = absolute pressure and absolute temperature, respectively, at sea level The atmospheric or barometric pressure decreases with altitude, and it can be calculated using the following equations: 5:256

Palt = P0 ½1 - Elf eet × 6:875 × 10 - 6 ]

USCS=I - P System

Palt = P0 1 - Elm × 2:256 × 10

- 6 5:256

SI

ð6:11aÞ ð6:11bÞ

The temperature of atmospheric air also decreases with increase in altitude up to a certain point. In this range, the temperature lapse rate is 3.3 °F/1000 ft. (6 °C/ 1000 m). Cooling Load Equations in S I Units (S I Equivalents of Eqs. 6.6, 6.7, 6.8, and 6.9) In S I units the standard density of air at 20 °C and 101.3 kPa is 1.20 kg/m3 and the volume flow rate of air is usually specified in m3/s. Hence, the S I equivalent of Eq. 6.6 is qT = 1:2V_ m3 =s ðh1 - h2 Þ

ð6:12Þ

where qT = total rate of heat removal by the cooling coil (kW) V_ = volume flow rate of air (m3/s) h1, h2 = specific enthalpies of air entering and leaving the coil, respectively (kJ/ kg d a) In S I units, the standard specific heat of air is cpa = 1.025 kJ/kg . ° C and the sensible cooling load under standard air conditions can be calculated as follows qs = m_ a cpa ðT 1 - T 2 Þ = ð1:20Þ V_ m3 =s

1:025

kJ ðT 1 - T 2 Þ ) kg . ° C

qS = 1:23 V_ m3 =s ðT 1 - T 2 Þ

ð6:13Þ

In S I units, the latent heat of condensation (hfg) is obtained by subtracting the enthalpy of the condensed liquid water (hf) at 10 °C from the enthalpy of water vapor (hg) at 24 °C, which can obtained from steam tables or on-line resources (https:// www.tlv.com/global/TI/calculator/steam-table-temperature.html). Hence,

198

6

hfg = hg,24 ° C - hf ,10 ° C = 2544:53

Cooling and Heating Load Calculations

kJ kJ - 42:02 = 2502:51 kJ=kg kg kg

Hence, the equation for calculating the rate of latent heat removed by the coil (kW) is obtained as follows. qL = m_ a ðω1 - ω2 Þ 2502:51

kJ kg

= 1:2V_ m3 =s ðω1 - ω2 Þ 2502:51

kJ kg

)

qL = ð3003Þ V_ m3 =s ðω1 - ω2 Þ

ð6:14Þ

where qL = rate of latent heat removal by the coil (kW) ω1and ω2 are humidity ratios of air entering and leaving the coil, respectively (kg H2O/kg d a) From continuity equation (Eq. 1.9), the mass flow rate of dry air in kg d a/s is V_ m_ a = V_ a ρa = a va

ð6:15Þ

where va is the specific volume of air in m3/kg d a, which can be obtained from psychrometric charts.

6.3.5

Procedure for Estimating Cooling Loads for Facilities

As mentioned earlier, calculating cooling loads for facilities is quite complex due to the unsteady nature of boundary conditions and also due to storage and delayed release of thermal energy. The contemporary practice is to use software/spreadsheets to accurately assess cooling loads for facilities on an hourly basis [11]. Although manual calculations of cooling loads can be quite cumbersome, a simplified approach is to use cooling load temperature difference (CLTD) and cooling load factor (CLF) methods. These methods are widely used by air conditioning engineers as they yield reasonably accurate results and estimations can be carried out manually in a relatively short time. The following sub-sections provide detailed descriptions of heat gain calculations with appropriate references to data and tables in ASHRAE Fundamentals Handbooks [2–5].

6.3

Cooling Load Calculations

6.3.5.1

199

Calculation of Heat Gain through Opaque Walls and Roofs

The heat gained by a conditioned space through opaque walls and roofs is sensible heat gain and it is mainly driven by the outdoor - indoor temperature difference. However, the heat transfer is also affected by solar radiation absorbed and stored by the mass of the opaque surfaces. The stored thermal energy is later released into the conditioned space resulting in a time lag in heat transfer. To account for radiation and time delay effects, the Cooling Load Temperature Difference (CLTD) is used rather than the absolute temperature between outside and inside. CLTD is a theoretical temperature difference that accounts for the combined effects of inside and outside air temperature difference, daily temperature range, solar radiation, and heat storage and release from the entities that absorb radiant heat. CLTD Correction CLTD values in tables in 1997 ASHRAE Fundamentals Handbook, I P Edition [4] are obtained under assumed conditions of maximum outdoor temperature 95 °F, mean temperature of 85 °F, and a daily range of 21 °F (defined as the mean of the difference between the daily maximum and minimum dry bulb temperatures for the month). Since the ambient design conditions are not always coincident with the assumed conditions for calculating CLTD, CLTD values from the tables are modified using the following equations for CLTD corrections: ∘

∘

CLTDcorrection = ð78 F - T i Þ þ ðT M - 85 FÞ USCS=I - P System ∘

∘

CLTDcorrection = ð25:5 C - T i Þ þ ðT M - 29:4 CÞ S I

ð6:16aÞ ð6:16bÞ

where CLTDcorrection. = correction to be applied to CLTD values from tables Ti = design indoor temperature TM = mean outdoor temperature=To - (DR/2) To = design outdoor temperature DR = daily range temperature (from local climatic data) The basic equation for calculating heat gain through conduction through opaque walls, and roofs is qroofs,walls = UAðCLTDcorrected: Þ

ð6:17aÞ

where qroofs,walls = heat gain due to conduction through roofs, and walls U = design heat transfer coefficient, which can be calculated using data from Chapter 26, Table 1 Building and Insulating Materials: Design Values, ASHRAE Fundamentals Handbooks, 2021 [2, 3]. U calculation details are clearly explained in the solutions to example and practice problems. A = surface area of the roof, wall, or glass

200

6

Cooling and Heating Load Calculations

CLTDcorrected. = corrected cooling load temperature difference for roof or wall. CLTD can be obtained from Chapter 28, Table 30 (used in conjunction with Table 31 for roofs) and Table 32 (used in conjunction with Table 33A,B,C for walls) ASHRAE Fundamentals Handbooks, 1997 [4, 5]. CLTD obtained from the tables are corrected using Eq. 6.16a (USCS) or 6.16b (SI). Therefore, CLTDcorrected = CLTDtables + CLTDcorrection

ð6:17bÞ

ASHRAE recommends the use of ‘Radiant Time Series Method’ for non-residential cooling load calculations. Further details on this method can be obtained from Chapter 18 of ASHRAE Fundamentals Handbooks [2, 3].

6.3.5.2

Calculation of Heat Gain Through Floors, and Partitions qfloors,partitions = UAðT s - T i Þ

ð6:18Þ

where qfloors,partitions= rate of sensible heat gain from floors, partitions U= design heat transfer coefficient, which can be obtained from Chapter 26, Table 1 Building and Insulating Materials: Design Values, ASHRAE Fundamentals Handbooks, 2021 [2, 3]. A= effective surface area of floor or partition Ts = temperature of space adjacent to the floor or partition surface on the other side of the conditioned space Ti = temperature of the conditioned space Note: 1. For floors directly in contact with the ground or over an underground basement that is neither ventilated nor conditioned, sensible heat transfer is insignificant and can be neglected for cooling load estimates. 2. If the temperature of the adjacent surface is not known, Ts can be assumed to be the outdoor temperature minus 6 °F (3 °C). 6.3.5.3

Calculation of Heat Gain Through Windows and Transparent Surfaces: Fenestration

Heat transmission through windows and transparent surfaces occurs due to conduction as well as radiation and it can be calculated by the following equation: qFen: = UAðT o - T i Þ þ AðSHGC ÞðE t Þ

ð6:19Þ

6.3

Cooling Load Calculations

201

Note: The preceding equation is the same as Eq. 1 in Chapter 15, ASHRAE Fundamentals Handbooks [2, 3]. However, Eq. 6.19 does not include the fenestration air leakage term presented in the ASHRAE equation. where qFen. = heat gain due to conduction and radiation through transparent surfaces U = design heat transfer coefficient which can be obtained from Chapter 15, Table 4: U-Factors for Various Fenestration Products, ASHRAE Fundamentals Handbooks, 2021 [2, 3]. A = surface area of the fenestration surface (windows, skylights, etc.) To = maximum outdoor temperature Ti = indoor temperature SHGC = solar heat gain coefficient, which can be obtained from Table 10 Chapter 15, ASHRAE Fundamentals Handbooks, 2021 [2, 3] Et = incident total irradiance (Btu/hr-ft2, W/m2), obtainable from Table10, Chapter 17, ASHRAE Fundamentals Handbooks, 2021 [2, 3]. Solar heat gain coefficient (SHGC) is the fraction of solar radiation admitted through a window, door, or skylight into the conditioned space. SHGC can be lowered by using louvers, drapes, and curtains. The reduced value of SHGC due to shading facilitated by aforementioned entities can be calculated by multiplying the normal SHGC of glazed surfaces, also known as clear glass SHGC (SHGCcg), by a factor called ‘Indoor Solar Attenuation Coefficient (IAC)’. The effective value of SHGC can be obtained from the following formula: SHGC = SHGC cg ðIAC Þ

ð6:20Þ

IAC values for specific shading situations can be obtained from Tables 14A to 14G, Chapter 15, ASHRAE Fundamentals Handbooks, 2021 [2, 3].

Alternate Methods of Calculating Fenestration Heat Gain Method 1: Solar Cooling Load Factor (SCL) Method The rate of fenestration heat gain into the conditioned space can be calculated by using the following alternate formula: qFen: = UAðT o - T i Þ þ AðSC ÞðSCLÞ

ð6:21Þ

For most part, Eq. 6.21 has the same nomenclature as Eq. 6.19. The new factors introduced in Eq. 6.21 are as follows: SC = shading coefficient, which can be obtained from Table 11, Chapter 29, ASHRAE Fundamentals Handbooks [4, 5].

202

6

Cooling and Heating Load Calculations

SCL = solar cooling load factor which can be obtained from Table 36, Chapter 28, ASHRAE Fundamentals Handbooks [4, 5]. The tables for obtaining solar cooling load factors should be used after obtaining the relevant zone information from Tables 35A and B, Chapter 28, ASHRAE Fundamentals Handbooks [4, 5]. Method 2: Solar Heat Gain Factor (SHGF) Method The rate of fenestration heat gain into the conditioned space can be calculated by using the following alternate formula involving the solar heat gain factor (SHGF): qFen: = UAðT o - T i Þ þ AðSC ÞðSHGF Þ

ð6:22Þ

Equation 6.22 is identical to Eq. 6.21 except for the fact that the solar cooling load (SCL) factor is replaced by the solar heat gain factor (SHGF). SHGFs can be obtained from Tables 15 through 21, for different latitudes, Chapter 29, ASHRAE Fundamentals Handbooks [4, 5]. Solutions to Examples 6.1 through 6.11 provide comprehensive, step by step illustrations of manual cooling load calculations by considering various sources of heat gain. The solutions also illustrate extensive use of data from ASHRAE Fundamentals Handbooks [2–5] Example 6.1 A single-story office building in Atlanta, GA, USA (Latitude 340) occupies a floor space of 50 ft. × 80 ft. with 10 ft. high walls with north-facing glass façade. One of the 50 ft. edges faces north towards the street and is entirely a glass façade with swinging double glass door each 7 ft. high and 4 ft. wide. The east and west side walls each have four equally spaced fixed glass windows, each 10 ft. wide and 4 ft. high. The south side wall has two equally spaced windows of the same size. The building has a flat roof. The inside of the building is to be maintained at 70 °F DB and 50% relative humidity. The cooling system is to be designed based on 94 °F DB and 74 °F mean coincident wet bulb temperature (MCWB), and DR (daily range) of 16.7 °F. (from Table 1, Chapter 14 on climatic conditions, ASHRAE Fundamentals Handbook 2021, I-P Edition [2]). The following information is available about the walls, roof, and glazing systems of the building: Roof: The roof composition from the top layer to the bottom layer is given along with the relevant thermal data, which can also be obtained from Tables 1, 3, and 10, ASHRAE Fundamentals Handbook, 2021, Chapter 26, ASHRAE Fundamentals Handbook, 2021-I P Edition [2]. Outdoor air film resistance, R = 0.2271 hr-ft2-0F/Btu ½ in. built-up roof membrane, R = 0.3531 hr-ft2-0F/Btu 1 in. mineral fiber board, R = 2.951 hr-ft2-0F/Btu 2 in. mineral wool insulation, k = 0.2498 Btu-in./hr-ft2-0F 4 in. concrete, k = 7.6328 Btu-in./hr-ft2-0F 3.5 in. air space (for dropped roof framing), R = 1.8735 hr-ft2-0F/Btu

6.3

Cooling Load Calculations

203

0.5 in. in gypsum board dropped ceiling, k = 1.11 Btu-in./hr-ft2-0F Indoor air film resistance, R = 0.9083 hr-ft2-0F/Btu Wall: The wall composition from the outside layer to the inside layer is given along with the relevant thermal data, which can also be obtained from Tables 1, 3, and 10, Chapter 26, ASHRAE Fundamentals Handbook, 2021-I P Edition [2]. Outdoor air film resistance, R = 0.2498 hr-ft2-0F/Btu 0.5 in. painted cement plastering on the outside, k = 4.9954 Btu-in./hr-ft2-0F 8 in. concrete block (limestone aggregate with two perlite filled cores), R = 2.100 hr-ft2-0F/Btu 0.5 in. painted cement plastering on the inside, k = 4.9954 Btu-in./hr-ft2-0F Indoor air film resistance, R = 0.6812 hr-ft2-0F/Btu Glazing System: Low emissivity (e = 0.10) clear glass double glazing (1/8 in. glass thickness) system with 0.5 in. air space (ID 21/21a) and fixed, insulated frame. The thermal data is given here, and it can be obtained from Tables 4 and 10, Chapter 15, ASHRAE Fundamentals Handbook, 2021-I P Edition [2]. Overall heat transfer coefficient, U = 0.32 Btu/hr-ft2-0F Total window solar heat gain coefficient (SHGC) at normal incidence, SHGC = 0.58. The door can be treated as part of the glass façade and the heat gain due to opening and closing of doors will be included as part of infiltration. The peak irradiance at 350N, approximately the same latitude as Atlanta, from Table 10, Chapter 17, ASHRAE Fundamentals Handbook, 2021-I P Edition [2] is as follows (all values in Btu/hr-ft2): North = 53.61, South = 149.4, East/West = 237.6 Determine the total maximum possible sensible heat gain in the building due to the roof, walls, and glazing systems. (Solution) Maximum possible sensible heat gain through the roof Calculate the R values for the roof entities for which the thickness and thermal conductivity values are given using Eq. 2.6 (Chap. 2). Rcond,ua = Rins =

ΔX ) k

ΔX ins = kins

2 in = 8:0064 hr‐ft2 ‐ ° F=Btu Btu‐in: 0:2498 hr‐ft2 ‐ ° F

204

6 Cooling and Heating Load Calculations

Similarly, Rconc =

ΔX conc 4 in = = 0:5240 hr‐ft2 ‐ ° F=Btu Btu‐in: kconc 7:6328 hr‐ft 2 ‐°F

Rgyp:bd =

ΔX gyp:bd 0:5 in = = 0:4504 hr‐ft2 ‐ ° F=Btu Btu‐in: kgyp:bd 1:11 hr‐ft 2 ‐°F

Add up the resistances of all the roof entities to obtain the total thermal resistance of the roof per unit area. The units for each thermal resistance term is hr-ft2-0F/Btu and it is not included in the calculation equation for the sake of brevity of expression. Rroof = Roa þ Rmem þ Rfib:bd þ Rins þ Rconc þ Rair:sp þ Rgyp:bd: þ Ria = 0:2271 þ 0:3531 þ 2:951 þ 8:0064 þ 0:5240 þ 1:8735 þ 0:4504 þ 0:9083 = 15:2938 hr‐ft2 ‐ ° F=Btu Calculate the overall heat transfer coefficient (reciprocal of the overall thermal resistance) for the roof using Eq. 2.19 (Chap. 2). U roof =

1 1 = = 0:0654 Btu=hr‐ft2 ‐ ° F 2 Rroof 15:2938 hr‐ftBtu‐ ° F

After determining the mean outdoor temperature, calculate CLTD correction using the given design and climate data and Eq. 6.16a. T M = T o - ðDR=2Þ = 94 ° F -

16:7 ° F = 85:65 ° F 2 ∘

∘

CLTDcorrection = ð78 F - T i Þ þ ðT M - 85 FÞ ∘

∘

∘

∘

= ð78 F - 70 FÞ þ ð85:65 F - 85 FÞ ∘ = 8:65 F Determine table CLTD values for the roof. First, obtain the roof number from Chapter 28, Table 31, ASHRAE Fundamentals Handbook, 1997-I P Edition [4]. Since this is a concrete roof (column 5) with R value within the range 15 to 20 (column 3), roof mass outside the insulation (column 1), with suspended (dropped) ceiling (column 2), the roof number is 5 from column 5. Next, obtain the maximum possible CLTD for roof number 5 from Table 30 in the preceding reference. From the table, CLTDmax = 69 ° F, occurring both at 5 p m (hour 17) and 6 p m (hour 18). Next, apply the CLTD correction to the preceding value from the CLTD table. CLTD max,corrected = CLTD max,table þ CLTDcorrection = 69 ° F þ 8:65 ° F = 77:65 ° F

6.3

Cooling Load Calculations

205

Note: The decimal number results for heat gains have been rounded up for conservative estimates. Calculate the maximum possible sensible heat gain through the roof using Eq. 6.17. qroof = U roof Aroof CLTD max,corrected Btu ð50 ft × 80 ftÞð77:65 ° FÞ hr‐ft2 ‐ ° F = 20, 314 Btu=hr = 0:0654

Maximum possible sensible heat gain through the walls Since all the other R values are given, only the R value for the cement plastering needs to be calculated using Eq. 2.6 (Chap. 2). Rcem:plas: =

ΔX cem:plas: 0:50 in = = 1:0001 hr‐ft2 ‐ ° F=Btu Btu‐in: kcem:plas: 4:9954 hr‐ft 2 ‐°F

Add up the resistances of all the wall entities to obtain the total thermal resistance of the roof per unit area. The units for each thermal resistance term is hr-ft2-0F/Btu and it is not included in the calculation equation for the sake of brevity of expression. Rwall = Roa þ Rcem:pl,o þ Rconc:blk þ Rcem:pl,i þ Ria = 0:2498 þ 1:0001 þ 2:100 þ 1:0001 þ 0:6812 = 5:0312 hr‐ft2 ‐ ° F=Btu Calculate the overall heat transfer coefficient (reciprocal of the overall thermal resistance) for the wall using Eq. 2.19 (Chap. 2). U wall =

1 1 = = 0:1988 Btu=hr‐ft2 ‐ ° F Rwall 5:0312 hr‐ft2 ‐ ° F Btu

Calculate the total net surface area of the walls by subtracting the surface area of the windows (4 windows, each 10 ft. × 4 ft., on the east, west walls and 2 windows on the south walls) from the total surface areas of the east (80 ft. × 10 ft), west, and south walls (50 ft. × 10 ft). Awalls,net,E,W = 2ð80 ft × 10 ft - 4 × 10 ft × 4 ftÞ = 1280 ft2 Awalls,net,S = 50 ft × 10 ft - 2 × 10 ft × 4 ft = 420 ft2 Determine the CLTD for the walls.

206

6

Cooling and Heating Load Calculations

First, obtain the code for the primary wall material from Chapter 28, Table 11, ASHRAE Fundamentals Handbook, 1997-I P Edition [4]. The R value obtained for the concrete block is 2.10 hr-ft2-0F/Btu. Code number C18 has the closest R value of 1.96 hr-ft2-0F. Next, determine the relevant wall number from Chapter 28, Table 33B: Wall Mass Evenly Distributed, ASHRAE Fundamentals Handbook, 1997-I P Edition [4]. For an R value in the range 4.75 to 5.5 hr-ft2-0F/Btu (the calculated R value is 5.0312 hr-ft2-0F/Btu), under the column for C18, the wall number is 16. Next, obtain the maximum possible CLTD for wall number 16 from Chapter 28, Table 32, ASHRAE Fundamentals Handbook, 1997-I P Edition [4], and apply the correction of +8.65 °F to the table values to get the corrected maximum CLTDs. CLTDmax,corrected = 40.65 ° Ffor the east facing wall CLTDmax,corrected = 46.65 ° Ffor the west facing wall CLTDmax,corrected = 35.65 ° Ffor the south facing wall Calculate the maximum possible sensible heat gain through each wall using Eq. 6.17. qE,wall = U E,wall AE,wall CLTD max,corrected,E Btu hr‐ft2 ‐ ° F = 10, 344 Btu=hr

= 0:1988

1280 ft2 ð40:65 ° FÞ

qW,wall = U W,wall AW,wall CLTD max,corrected,W Btu hr‐ft2 ‐ ° F = 11, 871 Btu=hr = 0:1988

1280 ft2 ð46:65 ° FÞ

qS,wall = U S,wall AS,wall CLTD max,corrected,S Btu hr‐ft2 ‐ ° F = 2, 977 Btu=hr

= 0:1988

420 ft2 ð35:65 ° FÞ

Calculate the total heat gain through all the three walls. Each term in the following equation has units of Btu/hr.: qwalls,total = 10, 344 þ 11, 871 þ 2, 977 = 25, 192 Btu=hr Maximum possible sensible heat gain through glazed surfaces The sensible heat gain through glazing systems consists of two components – heat gain through conduction and heat gain due to radiation.

6.3

Cooling Load Calculations

207

Conduction Calculate the total surface area of the glazed surfaces as shown. Surface area of each window = 10 ft × 4 ft = 40 ft2 Total number of windows = 4 (east) + 4 (west) + 2 (south) = 10 Surface area of the north-facing glass façade = 50 ft × 10 ft = 500 ft2 Therefore, the total surface area of the glazed surfaces is Atot,glz = 10 × 40 ft2 þ 500 ft2 = 900 ft2 The heat gained through conduction across windows and glass walls can be obtained from the first part of Eq. 6.19. Btu hr‐ft2 ‐ ° F = 6912 Btu=hr

qcond: = UAglz,tot ðT o - T i Þ = 0:32

900 ft2 ð94 ° F - 70 ° FÞ

Note: The doors are operable and have a higher U value of 0.34 Btu/hrft2-0F. However, the same U value as the fixed windows has been used considering the small difference. Radiation The peak irradiance (provided in the problem statement) depends on the orientation of the glazing surfaces and heat gain due to radiation for each orientation is calculated using the second part of Eq. 6.19 and added up to obtain the total heat gain due to radiation. The second part of Eq. 6.19 is qrad = ðSHGC ÞðAN EtN þ AS E tS þ AE E tE þ AW EtW Þ Btu Btu 50 ft × 10 ft × 53:61 þ 2 × 40 ft2 × 149:4 hr‐ft2 hr‐ft2 = ð0:58Þ Btu Btu þ4 × 40 ft2 × 237:6 þ 4 × 40 ft2 × 237:6 2 hr‐ft hr‐ft2 = 66, 579 Btu=hr Therefore, the total sensible heat gain due to fenestration is qfen = qcond þ qrad = 6912

Btu Btu þ 66579 = 73, 491 Btu=hr hr hr

Calculate the total maximum possible sensible heat gain due to the roof, walls, and glazing systems by adding up all the components. Each term in the equation has units of Btu/hr. qsen,tot = qroof þ qwalls þ qfen = 20314 þ 25192 þ 73491 = 118, 997 Btu=hr

208

6

Cooling and Heating Load Calculations

Example 6.2 Determine the radiation heat gain due to fenestration on the north side of the building in Example 6.1 using A. the solar cooling load (SCL) method and B. the solar heat gain coefficient (SHGF) method Compare the preceding results with that obtained by using the solar heat gain coefficient (SHGC). (Solution) A. Step 1: Obtain the Shading Coefficient – From Table 11, Chapter 29, ASHRAE Fundamentals Handbook, 1997-I P Edition [4], for uncoated double-glazing system (ID 5a), the shading coefficient is, SC = 0.87. Step 2: Obtain the relevant zone information. From Tables 35 B, Chapter 28, ASHRAE Fundamentals Handbook, 1997-I P Edition [4], for a space with 3 concrete walls, with no inside shade, the zone type is C for solar glass. Step 3: Obtain the solar cooling load factor which can be obtained from Table 36, Chapter 28, ASHRAE Fundamentals Handbook, 1997-I P Edition [4]. For Zone type C and north facing glazing, the maximum possible solar cooling load is, SCL = 35 Btu/hr ‐ ft2 Step 4: Calculate the radiation heat gain due to fenestration on the north side of the building using the preceding results and the second part of Eq. 6.21. qrad,N = AN ðSCÞðSCLN Þ = 500 ft2 ð0:87Þ 35

Btu = 15, 225 Btu=hr hr‐ft2

B. Step 1: Determine the shading coefficient. From Step 1 in the solution to part A, the shading coefficient is SC = 0.87. Step 2: Determine the maximum possible SHGF in July for north facing glazed surface by interpolating the values from Table 17 (for 320 latitude, SHGF = 40) and Table 18 (for 400 latitude, SHGF = 38), Chapter 29, ASHRAE Fundamentals Handbook, 1997-I P Edition [4]. For 340 latitude for Atlanta, by interpolation, SHGFN = 39.5 Btu/hr ‐ ft2 Step 3: Calculate the radiation heat gain due to fenestration on the north side of the building using the preceding results and the second part of Eq. 6.22. qrad,N = AN ðSCÞðSHGF N Þ = 500 ft2 ð0:87Þ 39:5

Btu = 17, 182 Btu=hr hr‐ft2

6.3

Cooling Load Calculations

209

Comparison of results from the three methods Calculate the radiation heat gain due to fenestration on the north side of the building using SHGC = 0.58 and the second part of Eq. 6.19. qrad,N = AN ðSHGC ÞðE tN Þ = 500 ft2 ð0:58Þ 53:16

Btu = 15, 417 Btu=hr hr‐ft2

ðSHGC MethodÞ In comparison, the radiation heat gain from SCL method is qrad,N = 15, 225 Btu=hr ðSCL MethodÞ The radiation heat gain from SHGF method is qrad,N = 17, 182 Btu=hr ðSHGF MethodÞ Comment: The results from SHGC and SCL methods are very close to each other (1.3% difference). The SHGF method gives a radiation heat gain value approximately 11% higher than that obtained from the SHGC method. This is due to the fact that the SHGF listed in the tables are for single-pane, clear glass which is modulated by the shading coefficient, which needs to be higher in this case because of the air gap.

6.3.5.4

Calculation of Heat Gain Due to Occupants

The heat gain due to people inside a facility obviously depends on the activity levels of the person. The heat gain due to occupants consists of both sensible and latent heat components. The sensible heat gain from occupants is due to radiation heat emitted by a person’s body while the latent heat gain from a person is primarily due to perspiration which causes moisture addition to the conditioned space. Table 1, Chapter 28, ASHRAE Fundamentals Handbook, 2021-I P Edition [2]. lists typical vales of sensible and latent heats added to various types of conditioned spaces at different activity levels. Some of the useful heat gain values are as follows: Activity Moderate activity levels at home or office Heavy activity levels in a factory Athletic activity in a gymnasium

Sensible heat 250 Btu/hr 580 Btu/hr 710 Btu/hr

Latent heat 200 Btu/hr 670 Btu/hr 1090 Btu/hr

210

6

Cooling and Heating Load Calculations

Note: To convert the listed values to W, multiply by 0.2929 The sensible and latent heat gains due to occupants can be calculated by using the following equations: qS,occu: = ðN Þ qS=P ðCLF Þ qL,occu: = ðN Þ qL=P

ð6:23Þ ð6:24Þ

where qS,occ. = total sensible heat gain due to N occupants in the conditioned space qS/P = sensible heat gain per person CLF = cooling load factor to be used due to the radiant nature of sensible heat from people. The radiation heat gain from people is not converted instantaneously to cooling load. It is first absorbed by the surroundings such as floor, furniture etc. and then later transferred to the space by convection, necessitating the use of CLF. The CLF to be used for sensible heat gain from occupants can be obtained from Table 37, Ch.28, ASHRAE Fundamentals Handbook, 1997-I P Edition [4], after ascertaining the zone from Tables 35 A/B. The cooling load factor is not used for latent heat gain calculation since latent heat gain is due to gain of moisture (no radiation involved). where qL,occ. = total latent heat gain due to N occupants in the conditioned space qL/P = latent heat gain per person Example 6.3 Consider the same office building as in Example 6.1. Assume a maximum of fifty (50) occupants at any given time, vinyl flooring, gypsum partitions, and the following sensible and latent heat gain per person qS=P = 270 Btu=hr and qL=P = 220 Btu=hr Calculate the maximum possible total heat gain due to occupants in the building. (Solution) From Table 35 B, Chapter 28, ASHRAE Fundamentals Handbook, 1997-I P Edition [4], for 3 walls, vinyl flooring, concrete block walls, half to no shading, the zone for people is C. Assume 8 hours occupation in space and for 8 hours after entry into the space, from Table 37, Chapter 28, ASHRAE Fundamentals Handbook, 1997-I P Edition [4], the maximum value of the cooling load factor for sensible heat transfer due to people is CLF = 0.91. Calculate the sensible heat gain for the occupants using Eq. 6.23.

6.3

Cooling Load Calculations

211

qS,occu: = ðN Þ qS=P ðCLF Þ = ð50 personsÞ

Btu hr ð0:91Þ person

270

= 12, 285 Btu=hr Calculate the latent heat gain for the occupants using Eq. 6.24. qL,occu: = ðN Þ qL=P = ð50 personsÞ

6.3.5.5

220 Btu hr = 11, 000 Btu=hr person

Calculation of Heat Gain Due to Lights, Appliances, and Equipment

Heat Gain Due to Lights The heat gain due to lighting depends on the wattage of lighting. The heat gain due to lighting is only sensible heat gain and it can be calculated by the following equation: qlighting = 3:41WF ul F sa

ð6:25Þ

where 3.41 is the conversion factor, 3.41 Btu/hr. per Watt W = lighting wattage Ful = utilization factor for lighting (fraction of total lights used) Fsa = lighting special allowance factor due to lamps and ballasts -1.1 to 1.3 An alternative approach to calculate heat gain due to lighting is to use the data in Table 2, Chapter 18, ASHRAE Fundamentals Handbook, 2021-I P Edition [2], which lists power densities (W/ft2) for different space types as per ASHRAE Standard 90.1–2022 [9]. Examples of typical values in W/ft2are as follows: Open office = 0.98, Hotel Lobby = 1.06, Computer Room = 1.71. Example 6.4 Consider the same office building as in Example 6.1. The office space uses thirtyfive (35) energy saving lights, each with a wattage of 20 W. Calculate the sensible heat gain due to lighting in the office space. Solution Calculate the heat gain due to lighting using Eq. 6.25 assuming Ful = Fsa = 1.0.

212

6

Cooling and Heating Load Calculations

qS,lighting = 3:41WF ul F sa Btu=hr ð35 × 20 WÞð1:0Þð1:0Þ W = 2, 387 Btu=hr = 3:41

Heat Gain due to Appliances/Equipment The heat gain due to appliances and equipment depends on the rated power input (W or Btu/hr) to the appliance. The heat gain due to appliances is only sensible heat gain and it can be calculated by the following equation: qs,appl: = qinput F U F R

ð6:26Þ

FL = FU FR

ð6:27Þ

where qs,appl. = sensible heat gain from appliance qinput = rated power input of the appliance FU = utilization factor FR = radiation factor FL = load factor, ratio of sensible heat gain to rated input for the appliance The following tables in Chapter 18, ASHRAE Fundamentals Handbooks, 2021-I P and S I Editions [2, 3], provide relevant values/information on the preceding variables. Table 5A: Recommended Heat Gain for Unhooded Electrical Appliances During Idle Conditions Table 5B: Recommended Heat Gain for Unhooded Electrical Appliances During Cooking Conditions Table 6: Recommended Heat Gain for Typical Medical Equipment Table 7: Recommended Heat Gain for Typical Laboratory Equipment Tables 8A, B, C, D: Recommended Heat Gains for Desktop Computers, Laptop Computers, Tablets, and Computer Monitors Table 9: Recommended Heat Gain for Printers Table 10: Recommended Heat Gain for Miscellaneous Equipment Example 6.5 Consider the same office building as in Example 6.1. The office space uses 25 laptop computers. The office space also has an office services room with a multipurpose copier/printer/scanner and a break room with a microwave oven, water dispenser, and coffee maker. The peak heat gain with the load factor, FL = 1.0, from the preceding appliances are listed here Each laptop, including monitor = 78 W Copier/Printer/Scanner = 400 W

6.3

Cooling Load Calculations

213

Microwave oven = 770 W Water dispenser = 232 W (sensible) + 46 W (latent) Coffee maker = 780 W (sensible) + 650 W (latent) Determine the total sensible and latent heat gains from the office space (Btu/hr). (Solution) The sensible and latent heat gains from the appliances can be obtained by adding up all the given peak heat gains and converting the result to Btu/hr. as shown here. qS,appl: =

25 × 78 W þ 400 W þ 770 W þ232 W þ 780 W qL,appl: = ð46 W þ 650 WÞ 3:41

6.3.5.6

3:41

Btu=hr = 14, 090 Btu=hr W

Btu=hr = 2, 374 Btu=hr W

Calculation of Heat Gain Due to Infiltration

Infiltration is the movement of outside air into the conditioned space through cracks in the walls and ceilings, gaps in seals for fixed windows, gaps in operable doors and windows in the closed position. There is also infiltration of outside air when doors are temporarily opened to facilitate movement of people and material to and from the building. Since infiltration brings in outside air, the heat gain due to infiltration has both sensible and latent heat components. The total, sensible, and latent heat gains due to infiltration can be calculated by using cfminf. in Eqs. 6.6, 6.7, and 6.8 for USCS units. Hence, the resulting heat gain equations for infiltration in USCS units are qT,inf: = 4:5ðcfminf: Þðho - hi Þ

ð6:28Þ

qS,inf: = ð1:08Þðcfminf: ÞðT o - T i Þ

ð6:29Þ

qL,inf: = ð4840Þðcfminf: Þðωo - ωi Þ

ð6:30Þ

The key variable in Eqs. 6.28, 6.29, and 6.30 is the volume flow rate of infiltration air, cfminf., which can be calculated by using two different methods – air change method, and crack method. Using the air change method, the volume flow rate of infiltration is expressed as a factor of the air changes per hour (ACH) for the facility. The factor multiplying ACH varies from 0.10 for newer buildings with tight construction to 1.1 for older buildings with significant amounts of leakage spaces. Thus, if an average value of 0.60ACH is used, then the volume flow rate of infiltration air, cfminf., can be calculated using the following equation: cfminf: =

V space × 0:60ACH 60

ð6:31Þ

214

6 Cooling and Heating Load Calculations

where Vspace = volume of the conditioned space (ft3) 1.25ACH has units of (1/hr) 60 is a conversion constant to convert hours to minutes with units (60 min/hr) The crack method to calculate the volume flow rate of infiltration is much more complex than the air change method and the equation for calculating cfminf. is cfm inf: = A × C × ΔPn

ð6:32Þ

where A = leakage area of the opening C = flow coefficient dependent on the type of opening ΔP = Po - Pi, the pressure difference between outside and inside, which has three components – wind, stack, and applied mechanical forces n = pressure exponent (0.4 to 1.0) depending on the nature of the flow through the crack S I Equations for Calculating Infiltration Heat Gains In S I units, the volume flow rate of air is represented in m3/s. Accordingly, Eqs. 6.12, 6.13, and 6.14 can be used to calculate the total, sensible, and latent _ heat gains, respectively, due to infiltration, with Vrepresenting the volume flow rate 3 of infiltration in m /s. The equations for calculating infiltration heat gains in S I units are qT,inf = 1:2V_ m3 =s, inf: ðh1 - h2 Þ

ð6:33Þ

qS,inf: = 1:23 V_ m3 =s, inf: ðT 1 - T 2 Þ

ð6:34Þ

qL, inf: = ð3003Þ V_ m3 =s, inf: ðω1 - ω2 Þ

ð6:35Þ

Example 6.6 Consider the same office building as in Example 6.1. Assuming an infiltration rate of 0.12ACH, determine the sensible, latent, and total heat gains due to infiltration. (Solution) Calculate the volume of the office space using the dimensions given in Example 6.1. V space = 50 ft × 80 ft × 10 ft = 40000 ft3 Calculate the volume flow rate of infiltration using Eq. 6.31 with 1.20ACH. cfminf: =

V space × 0:12ACH 40000 ft3 × 0:12 hr - 1 = 80 ft3 = min = 60 60 min hr

6.3

Cooling Load Calculations

215

From the data given in Example 6.1, the inside of the building is to be maintained at 70 °F DB and 50% relative humidity and the outside conditions are 94 °F DB and 74 °F WB. Therefore, To = 94 ° F and Ti = 70 ° F. Calculate the sensible heat gain due to infiltration using Eq. 6.29. qS, inf: = ð1:08Þðcfminf: ÞðT o - T i Þ = ð1:08Þð80 cfmÞð94 ° F - 70 ° FÞ = 2, 074 Btu=hr Locate the indoor (i, 70 °F DB and 50% rh) and outdoor (o, 94 °F DB and 74 °F WB) state points on the psychrometric chart as shown.

From the excerpt of the psychrometric chart, ωo = 0.014 lbm H2O/lbm d a, and ωi = 0.0076 lbm H2O/lbm d a. Note: Any data obtained from the psychrometric chart can also be obtained from on-line resources such as on-line psychrometric calculator [12]. Calculate the latent heat gain due to infiltration using Eq. 6.30. qL, inf: = ð4840Þðcfminf: Þðωo - ωi Þ = ð4840Þð80 cfmÞ 0:014

lbm H2 O lbm H2 O - 0:0076 lbm d a lbm d a

= 2, 478 Btu=hr From the excerpt of the psychrometric chart, ho = 38 Btu/lbm d a, and hi = 25 Btu/lbm d a. Calculate the total heat gain due to infiltration using Eq. 6.28. qT, inf: = 4:5ðcfminf: Þðho - hi Þ Btu Btu = ð4:5Þð80 cfmÞ 38 - 25 lbm d a lbm d a = 4, 680 Btu=hr

216

6

Cooling and Heating Load Calculations

Note: The total heat gain due to infiltration should be the sum of the sensible and latent heat gains due to infiltration. This gives a result of 2074 Btu/hr. + 2478 Btu/ hr. = 4552 Btu/hr., which is less than the result obtained by using the total heat gain equation (Eq. 6.28). The discrepancy is due to the inaccuracies while reading the psychrometric chart.

6.3.5.7

Calculation of Heat Gain Due to Ventilation

Ventilation of air from the conditioned space is required to maintain indoor air quality and a healthy environment for the occupants. ANSI/ASHRAE Standards 62.1 and 62.2 [7, 8] are the recognized standards for ventilation system design and acceptable indoor air quality (IAQ) in commercial and residential buildings, respectively. Both standards have been expanded and revised in 2022 and they specify minimum ventilation rates and other measures in order to minimize adverse health effects for occupants. Acceptable indoor air quality is achieved through careful attention to each of the following fundamental elements: . . . .

contaminant source control proper ventilation humidity management adequate filtration

The recommended ventilation rates for different types of facilities are given below. . . . . . . . . .

offices, conference rooms 20 cfm/person telecommunication and data centers 20 cfm/person reception areas, classrooms, auditorium 15 cfm/person hospital patient rooms 25 cfm/person public Restrooms 50 cfm/person elevators 1 cfm/ft2 lockers, dressing rooms, gymnasiums 0.5 cfm/ft2 retail stores, malls 0.3 cfm/ft2 corridors, utilities 0.05 cfm/ft2

Example 6.7 Consider the same office building as in Example 6.1. Assuming a maximum of fifty (50) occupants at any given time, calculate the ventilation requirement for the building and express the result in terms of air changes per hour (ACH). (Solution) As per ASHRAE Standard 62.1, the recommended ventilation is 20 cfm/person. Calculate the ventilation requirement based on this standard and express the result in terms of air changes per hour using 6.31.

6.3

Cooling Load Calculations

217

Fig. 6.2 Configuration of an air conditioning unit

cfmvent: =

20 cfm ð50 personsÞ = 1000 cfm = V_ vent: person 3

ACH =

ft × 60 hrmin cfmvent: × 60 1000 min = = 1:5 ACH 50 ft × 80 ft × 10 ft V space

Heat Gains due to Ventilation A typical configuration of an air conditioning unit is shown in Fig. 6.2. As shown in Fig. 6.2, ventilation of the conditioned space is accomplished by exhausting or removing part of the return air and replacing it with outside air. From the figure it is clear that the cooling load due to ventilation, where outside air is mixed with return air, is on cooling coil and it is not part of the cooling load of the conditioned space. However, some amount of the mixed air and therefore some amount of outside air bypasses the cooling coil. As mentioned in Chapter 5, Sect. 5.2.1, the fraction of air that bypasses the cooling coil is called ‘Bypass Factor. The bypassed air, a portion of which is hot outside air, directly enters the conditioned space and therefore results in both sensible and latent heat gains for the conditioned space. Therefore, the bypass fraction of the outside air replacing the ventilated or exhaust air, contributes to the cooling load of the conditioned space. The sensible, latent, and total heat gains by the conditioned space due to bypass of outdoor air equivalent to ventilation can be calculated by the following equations which are similar to the equations used for calculating heat gains due to infiltration:

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6

Cooling and Heating Load Calculations

qT,bypass = 4:5 V_ vent: ðBFÞðh1 - h2 Þ

ð6:36Þ

qS,bypass = ð1:08Þ V_ vent: ðBFÞðT 1 - T 2 Þ

ð6:37Þ

qL,bypass = ð4840Þ V_ vent: ðBFÞðω1 - ω2 Þ

ð6:38Þ

where V_ vent: = ventilation rate in cfm BF = bypass factor for the cooling coil Other nomenclature is the same as in Eqs. 6.28, 6.29, and 6.30. The equations in S I units for heat gain due to bypass of outdoor air equivalent to ventilation are qT,bypass = 1:2 V_ m3 =s,vent: ðBFÞðh1 - h2 Þ

ð6:39Þ

qS,bypass = 1:23 V_ m3 =s,vent: ðBFÞðT 1 - T 2 Þ

ð6:40Þ

qL,bypass = ð3003Þ V_ m3 =s,vent: ðBFÞðω1 - ω2 Þ

ð6:41Þ

where V_ vent: = ventilation rate in m3/s BF = bypass factor for the cooling coil Example 6.8 Consider the same office building as in Example 6.1. Using the ventilation rate calculated in Example 6.7 and a bypass factor of 0.10 for the coil, calculate the sensible, latent, and total heat gains in the office space caused due to bypass of outdoor air across the coil. (Solution) Using the values of properties of outside and inside air obtained from the psychrometric chart in the solution to Example 6.6 and Eqs. 6.36, 6.37, and 6.38, respectively, calculate the total, sensible, and latent heat gains due to the bypass fraction of outside air used due to ventilation. qT,bypass = 4:5 V_ vent: ðBFÞðh1 - h2 Þ = ð4:5Þð1000 cfmÞð0:10Þ 38

Btu Btu - 25 lbm d a lbm d a

= 5, 850 Btu=hr qS,bypass = ð1:08Þ V_ vent: ðBFÞðT o - T i Þ = ð1:08Þð1000 cfmÞð0:10Þð94 ° F - 70 ° FÞ = 2, 592 Btu=hr

6.3

Cooling Load Calculations

219

qL,bypass = ð4840Þ V_ vent: ðBFÞðω1 - ω2 Þ = ð4840Þð1000 cfmÞð0:10Þ 0:014

lbm H2 O lbm H2 O - 0:0076 lbm d a lbm d a

= 3, 098 Btu=hr Note: The total heat gain in the space due to the bypass fraction of outside air should be the sum of the sensible and latent heat gains due to the bypass fraction of outside air. This gives a result of 2592 Btu/hr. + 3098 Btu/hr. = 5690 Btu/hr., which is less than the result obtained by using the total heat gain equation (Eq. 6.36). The discrepancy is due to the inaccuracies while reading the psychrometric chart.

6.3.5.8

Miscellaneous Heat Gains by the Conditioned Space

Miscellaneous amounts of heat added to the conditioned space include the heat equivalent of the total power consumed (and dissipated in the form of heat to the supply air) by the supply fan motor located downstream of the cooling coil. Fan power has been discussed in detail in Sect. 1.5, Chap. 1. The sensible heat added by the supply fan can be calculated by using the following equations:

qfan =

2545 Btu=hr ðhpÞ hp

qfan =

ηm ηfan PkW ηm ηfan

USCS

ð6:42Þ

SI

ð6:43Þ

where hp, PkW = rated power of the fan required to move air through the system, which can be calculated using Eq. 1.38 (USCS) or the numerator of Eq. 1.41 (S I) ηm = combined efficiency of the motor and the drive ηfan = efficiency of the fan Equations 1.38 and 1.41, use the volume flow rate of air handled by the fan. The initial guess for the volume flow rate can be obtained either from Eq. 6.7 (USCS / I-P System) or Eq. 6.13 (S I), where qSis the total known sensible heat gain by the space, except the sensible heat added by the fan. The fan heat added can be calculated using this initial guess and later refined after iterating for the fan capacity. Example 6.9 Using all the sensible gain results (from Examples 6.1, 6.3, 6.4, 6.5, 6.6, 6.7, and 6.8) for the office building as an initial guess, calculate the heat gain due to the supply fan with the following parameters: total pressure drop of 1.65 in. w c in the distribution system fan efficiency 70% and motor efficiency 80%.

220

6

Cooling and Heating Load Calculations

Refine the result obtained and calculate A. the final sensible heat gain due to the fan B. the final total sensible heat gain for the office space C. the cfm of air circulation required (Solution) The results of sensible heat gains from the solutions to preceding examples are tabulated here. Source Roof Walls Fenestration Occupants Lighting Appliances Infiltration Ventilation, bypass Total

Sensible heat gain, Btu/hr 20,314 25,192 73,491 12,285 2387 14,090 2074 2592 152,425

Generally, a delta Tair of 16 °F to 20 °F across the coil is used. Assume delta Tair of 18 °F across the coil. Using the preceding result, calculate the initial air circulation rate (iteration 1) using Eq. 6.7. qS1 = ð1:08ÞðcfmÞðΔT air,coil Þ ) Btu 152425 qS hr = 7, 841 cfm cfm1 = = ð1:08ÞðΔT air,coil Þ ð1:08Þ 180 F Calculate the required air horsepower of the fan using Eq. 1.38. AHP =

cfm × ΔTPin:wc ð7841 cfmÞð1:65 in:w cÞ = 2:04 hp = 6356 6356

Calculate the sensible heat added by the fan (iteration 1) using Eq. 6.42 2545 qfan1 =

Btu=hr ðhpÞ hp = ηfan ηm

2545

Btu=hr ð2:04 hpÞ hp 0:7 × 0:8

= 9, 271 Btu=hr Add the preceding value to the total sensible heat gain and refine, iterate the calculations to obtain the final cfm, the final fan heat, and the final sensible heat gain by the office space.

6.3

Cooling Load Calculations

221

qS2 = qS1 þ qfan = 152425 cfm2 = AHP =

Btu Btu þ 9271 = 161, 696 Btu=hr hr hr

161696 Btu qS2 hr = = 8, 318 cfm ð1:08ÞðΔT air,coil Þ ð1:08Þð18 ° FÞ

cfm × ΔTPin:wc ð8318 cfmÞð1:65 in:w cÞ = = 2:16 hp 6356 6356 qfan2 =

2545 Btu=hr ð2:16 hpÞ hp 0:7 × 0:8

= 9816 Btu=hr

This iteration process is continued, and the results are presented in the image of the spreadsheet calculations here.

From the spreadsheet, A. qSfan, final = 9, 857 Btu/hr B. qS space, final = 162, 282 Btu/hr C. cfmfinal = 8, 341 cfm Example 6.10 Use the total sensible heat gain result from Example 6.9, and add up all the latent heat gains for the office space (from Examples 6.3, 6.5, 6.6, 6.7, and 6.8) and calculate A. the total heat gain for the office space B. the sensible heat ratio (SHR) (Solution) Add up the latent heats from various sources as shown in the following table. Source Occupants Appliances Infiltration Ventilation, bypass Total

Latent heat gain, Btu/hr 11,000 2374 2478 3098 18,950

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6

Cooling and Heating Load Calculations

From the preceding table, qL,Space = 18, 950 Btu/hr A. Calculate the total heat gain of the office space using Eq. 6.2. Btu Btu þ 18950 hr hr = 181, 232 Btu=hr

qT,Space = qS,Space þ qL,Space = 162282

B. Calculate the sensible heat ratio (SHR) using Eq. 6.4.

SHRSpace =

6.3.5.9

qS,Space 162282 Btu hr = = 0:8954 qT,Space 181232 Btu hr

Total Cooling Load on the Cooling Coil

The total load on the cooling coil will be higher than the cooling load on the conditioned space. From Fig. 6.2, it is clear that the cooling coil is involved in cooling a mixture of outside air and return air. This will place additional sensible and latent loads on the cooling coil. The return air duct (represented by subscript ‘RD’) also adds additional loads on the cooling coil. This includes sensible heat gains due to the return duct and the return air fan. The equations used in calculating the sensible and latent heat gains due to outside air use the bypass factor, BF. This is because the fraction of outside air in contact with the cooling coil is (1-BF). The total sensible cooling load on the cooling coil is qS,Coil = qS,Space þ qS,OA þ qS,RD

ð6:44Þ

where qS,Space is the total sensible cooling load on the space, and subscripts represent the following: S = sensible heat OA = outside air RD = return air duct The sensible cooling load on the coil due to outside air can be calculated using the following equation(s): qS,OA = ð1:08ÞðcfmOA Þð1 - BFÞðT o - T coil Þ USCS

ð6:45aÞ

qS,OA = 1:23 V_ m3 =s,OA ð1 - BFÞðT o - T coil Þ S I

ð6:45bÞ

6.3

Cooling Load Calculations

223

where Tcoil = effective coil temperature or apparatus dew point, ADP (refer to Sect. 5.2.1, Chap. 5) To = outside air temperature The sensible cooling load on the coil due to the return air duct, including the return air fan can be calculated using the following equation: qS,RD = U RD ARD ðT o - T i Þ þ qS,RAfan

ð6:46Þ

where URD = overall heat transfer coefficient for the return air duct with insulation ARD = exposed surface area of the return air duct To = outside air temperature Ti= inside space temperature qS,RAfan = sensible heat added by the return air fan, which can be calculated using either Eq. 6.42 or Eq. 6.43. The total latent cooling load on the cooling coil is qL,Coil = qL,Space þ qL,OA

ð6:47Þ

where qL,Space is the total latent cooling load on the space and subscript OA represents outside air. The latent cooling load on the coil due to outside air can be calculated using the following equation(s): qL,OA = ð4840ÞðcfmOA Þð1 - BFÞðωo - ωcoil Þ

USCS=I - P System

qL,OA = ð3003Þ V_ m3 =s,OA ð1 - BFÞðωo - ωcoil Þ S I

ð6:48aÞ ð6:48bÞ

where ωcoil = humidity ratio of air leaving the coil ωo = humidity ratio of outside air Example 6.11 Use the data and results from Examples 6.1, 6.9, and 6.10. The length of the return air duct is 70 ft., and the dimensions are 3 ft. × 2 ft. The overall heat transfer coefficient for insulated duct is U = 0.2859 Btu/hr‐ft2‐ ° F. The pressure drop in the return duct system is 0.18 in. w c. Calculate A. the sensible cooling load on the cooling coil due to outside air with ventilation as determined in Example 6.7.

224

6

Cooling and Heating Load Calculations

B. the sensible cooling load on the cooling coil due to the return air duct including the return air fan. C. the latent cooling load on the cooling coil due to outside air with ventilation as determined in Example 6.7. D. the total sensible cooling load on the cooling coil. E. the total latent cooling load on the cooling coil. F. the total cooling load on the cooling coil expressed in terms of tons of refrigeration required. G. the sensible heat ratio for the cooling coil H. the percentage of the total cooling load on the cooling coil from different sources and comment on the results (Solution) A. The calculation of the sensible heat load on the coil due to outside air requires the knowledge of the coil temperature (apparatus dew point, ADP). This in turn requires the use of Eq. 5.3, reproduced here for reference. BF =

T outlet - T ADP T inlet - T ADP

ð5:3Þ

In the preceding equation, BF is the bypass factor, and Tinlet, Toutlet are inlet and outlet temperatures of air passing through the coil. As shown in the following figure, the air entering the coil is a mixture of return air and outside air. Therefore, the properties of the mixed air will have to obtained before using Eq. 5.3 to determine the coil temperature. The procedure for obtaining the properties of mixed air is clearly explained in the solution to Example 5.11 (Sect. 5.1.7, Chap. 5) and the same procedure will be used here, along with the nomenclature shown in the figure.

From the data given in Example 6.1, the properties of the streams shown in the figure are as follows: Stream 1 (return air/air from the space): T1DB = 70 ° F, rh1 = 50%. Calculate the volume flow rate of return air as described here. From the solution to Example 6.11, the ventilation rate is 1000 cfm, and this will be the cfm of the outside air as well. From the solution to Example 6.9, the air circulation rate required for the space is 8341 cfm, of which 1000 cfm is exhaust air due to ventilation. Hence, the volume flow rate of stream 1 is,

6.3

Cooling Load Calculations

225

V_ 1 = 8341 cfm - 1000 cfm = 7, 341 cfm Stream 2 (outside air): From Example 6.1, the outside air conditions are as follows: T2,DB = 94 ° F, T2,WB = 74 ° F, and as mentioned earlier, the required outside air flow rate is, V_ 2 = 1, 000 cfm Using the lever rule, calculate the approximate value of the dry bulb temperature at state point 3 as shown. T 3DB =

V_ 1 × T 1DB þ V_ 2 × T 2DB V_ 3 7341

= = 73 ° F

ft3 ft3 ð70 ° FÞ þ 1000 ð94 ° FÞ min min ft3 8341 min

As shown in the following excerpt of the psychrometric chart, join state point 1 and state point 2 with a straight line and locate state point 3 on this line corresponding to the vertical dry bulb temperature line of 73 °F. Determine all the properties of the mixed air at state point 3 as shown in the psychrometric chart.

From the psychrometric chart, T 3DB = 73 ° F, ω3 = 0:0085 lbm H2 O=lbm d a, h3 = 27 Btu=lbm d a

226

6

Cooling and Heating Load Calculations

In Example 6.8, a bypass factor of 0.10 was given and in Example 6.9, the temperature difference across the coil was assumed to be 18 ° F. Hence, the temperature of the outlet air from the coil is T oulet = T 4 = 73 ° F - 18 ° F = 55 ° F: Substitute all the known values into Eq. 5.3 and determine the apparatus dew point as shown. BF =

T outlet - T ADP 55 ° F - T ADP ) 0:10 = ) T ADP = 53 ° F = T coil T inlet - T ADP 73 ° F - T ADP

Calculate the sensible cooling load on the coil due to outside air using Eq. 6.45a. qS,OA = ð1:08ÞðcfmOA Þð1 - BFÞðT o - T coil Þ ft3 ð1 - 0:10Þð94 ° F - 53 ° FÞ min = 39, 852 Btu=hr

= ð1:08Þ 1000

B. Note that the return fan handles the entire air circulation rate of 8341 cfm, including the exhaust air with 1000 cfm and return air with 7341 cfm. Calculate the required air horsepower of the return fan using Eq. 1.38. AHP =

cfmRA × ΔTPin:wc ð8341 cfmÞð0:18 in:w cÞ = 0:24 hp = 6356 6356

Calculate the sensible heat added by the return air fan using Eq. 6.42 2545 qS,RAfan =

Btu=hr ðhpÞ hp = ηfan ηm

2545

Btu=hr ð0:24 hpÞ hp 0:7 × 0:8

= 1091 Btu=hr Calculate the exposed area of the return duct. ARD = perimeter × length = ½2ð3 ft þ 2 ftÞ]½70 ft] = 700 ft2 Substitute all the known values into Eq. 6.46 to calculate the sensible cooling load on the coil due to the return air duct, including the return air fan.

6.3

Cooling Load Calculations

227

qS,RD = U RD ARD ðT o - T i Þ þ qS,RAfan ∘ ∘ Btu Btu = ð0:2859 Þð700 ft2 Þð94 F - 70 FÞ þ 1091 2 ∘ hr hr-ft - F Btu Btu = 4003 þ 1091 = 5094 Btu=hr hr hr C. From the excerpt of the psychrometric chart shown earlier in the solution, the humidity ratios are as follows: ωo = 0:014 lbm H2 O=lbm d a ωcoil = 0:0086 lbm H2 O=lbm d a Calculate the latent cooling load on the coil due to outside air using Eq. 6.48a qL,OA = ð4840ÞðcfmOA Þð1 - BFÞðωo - ωcoil Þ ft3 = ð4840Þ 1000 ð1 - 0:10Þ min

0:014

lbm H2 O lbm d a 0:0086

= 23, 523 Btu=hr

lbm H2 O lbm d a

D. Calculate the total sensible cooling load on the cooling coil using Eq. 6.44 and previous results, including those from prior example problems (Example 6.9). qS,Coil = qS,Space þ qS,OA þ qS,RD=fan Btu Btu Btu = 162282 þ 39852 þ 5094 hr hr hr = 207, 228 Btu=hr E. Calculate the total latent cooling load on the cooling coil using Eq. 6.47 and previous results, including those from prior example problems (Example 6.9). qL,Coil = qL,Space þ qL,OA Btu Btu = 18, 950 þ 23, 523 hr hr = 42, 473 Btu=hr F. Calculate the total cooling load on the cooling coil by adding up the sensible and latent cooling loads. qT,Coil = qS,Coil þ qL,Coil Btu Btu = 207228 þ 42473 hr hr = 249, 701 Btu=hr Convert the preceding result to tons of refrigeration (TR).

228

6

qT,Coil = qS,Coil þ qL,Coil =

Cooling and Heating Load Calculations

249701 Btu=hr 12000 Btu=hr 1 TR

= 20:81 TR

G. Calculate the sensible heat ratio (SHR) for the cooling coil SHRCoil =

207228 Btu qS,Coil hr = = 0:83 qT,Coil 249701 Btu hr

H. Calculate the percentage of the total cooling load on the cooling coil from different sources as shown in the table below.

Comments: From the preceding table, fenestration and outside air replacement for ventilated space air are the major contributors to the load on the cooling coil. This is to be expected. However, methods to reduce heat gains due to fenestration such as using shading entities (blinds) and using a more reflective glazing system can be explored. Fenestration surface area can also be reduced to the extent possible. Example 6.12 Use the results from the solution to Example 6.11 and determine the following A. the process line for air entering and leaving the cooling coil B. the ratio of the change in enthalpy of air to the change in moisture content of air across the cooling coil C. the sensible heat ratio (SHR) using the psychrometric protractor

6.3

Cooling Load Calculations

229

(Solution) A. The schematic diagram for the air flow across the cooling coil is again shown here for reference.

From the excerpt of the psychrometric chart shown in the solution of Example 6.11, part A, state point 3 is defined by T 3,DB = 73 ° F, and ω3 = 0:0085 lbm H2 O=lbm d a From the same solution, the apparatus dew point is TADP = 53 ° F, which will be located on the saturation curve. Draw a line joining state point 3 and the apparatus dew point. The apparatus dew point will be on the extension of the process line. Again, from the same solution, the dry bulb temperature of the supply air is T4,DB = 55 ° F. Draw a vertical line at 55 ° F DB. The intersection of this line with the line joining 3-TADP, is state point 4. Line 3–4 is the process line for the coil. All the preceding information, including the process line is shown in the excerpt of the psychrometric chart shown here.

230

6

Cooling and Heating Load Calculations

B. Calculate Δh, the change in enthalpy across the cooling coil, using Eq. 6.6 and then substituting all the known values from the solutions to Examples 6.9 (for cfm) and 6.11 (for total cooling load). qT = 4:5ðcfmÞðΔhÞ ) qT = Δh = 4:5ðcfmÞ

Btu hr = 6:64 Btu=lbm d a ft3 Þ ð4:5Þð8341 min 249701

Similarly, calculate Δω, the change in moisture content across the cooling coil, using Eq. 6.8 and then substituting all the known values from the solutions to Examples 6.9 (for cfm) and 6.11 (for latent cooling load). qL = 4840ðcfmÞðΔωÞ ) qL = Δω = 4840ðcfmÞ

Btu hr = 0:0010 lbm H2 O=lbm d a ft3 ð4840Þð8341 Þ min 42473

Using the preceding results, calculate the ratio of the change in enthalpy of air to the change in moisture content of air across the cooling coil. Δh 6:64 = = 6640 Δω 0:0010 C. On the psychrometric protractor, draw a line joining the center of the protractor to the value of 6640 for the preceding ratio. This line is shown on the excerpt from the psychrometric chart shown earlier in the solution and this line passes through the point in the protractor where the sensible heat ratio (SHR) is approximately 0.83. This result is in good agreement with the result of SHR = 0.83 obtained in the solution to Example 6.11. The line joining the protractor center line and the Δh/Δω ratio is parallel to the process line as shown in the excerpt of the psychrometric chart. Note: More accurate results can be obtained with a psychrometric protractor with more divisions and better markings compared to the one used in this solution. Example 6.13 Calculate the rate of condensation of moisture, in gpm, from the cooling coil used in Example 6.11. (Solution) The following relevant information can be obtained solutions to previous examples, with the example numbers indicated in parenthesis.

6.3

Cooling Load Calculations

231

Volume flow rate of inlet air to the coil, V_ 3 = 8, 341 ft3 = min (from solution to Example 6.9) Specific volume of air entering the coil, v3 = 13.6 ft3/lbm d a (from psychrometric chart in solution to Example 6.11) Decrease in humidity ratio across the coil, Δω = 0.0010 lbm H2O/lbm d a(from solution to Example 6.12) Calculate the mass flow rate of dry air entering the coil. ft 8341 min V_ 3 = = 613:31 lbm d a= min 3 v3 13:6 lbmft d a 3

m_ a3 =

From mass balance for water, Mass of water condensed from the cooling coil = decrease in moisture content of air across the coil lbm d a lbm H2 O 0:0010 lbm d a min = 0:6133 lbm H2 O= min:

m_ H2 OðlÞ = m_ a3 Δω = 613:31

Convert the rate of condensation to gpm using the density of water. gpm moisture condensed =

H2 O 0:6133 lbmmin = 0:0735 gpm 8:34 lbm gal

Example 6.14 R-134a and chilled water are being considered as options for coolants for the cooling coil of Example 6.11. The properties of R-134a can be obtained from refrigerant tables/on-line sources (IRC Fluid Property Calculator https://irc.wisc.edu/properties/) and it is provided here for reference. hR134a ð10 psia, x = 0:23Þ = 25:1 Btu=lbm hR134a ð10 psia, sat:vap:Þ = 98:7 Btu=lbm Calculate the rate of circulation of the coolant (lbm/min) required in if the coolant is A. Refrigerant R-134a entering the coil at 10 psia and quality of 0.23 and leaving as a saturated vapor. B. Chilled water entering the coil at 40 °F and leaving the coil at 51 °F

232

6

Cooling and Heating Load Calculations

(Solution) A. From the solution to Example 6.11, the total cooling load on the coil is qT,coil = 249701 Btu=hr = 4162 Btu= min Energy balance across the coil results in the following set of equations. Heat transferred from air to the coil (cooling load) = heat absorbed by the refrigerant qT,coil = m_ R ðhRe - hRi Þ ) qT,coil m_ R = = hRe - hRi

4162 98:7

Btu min

Btu Btu - 25:1 lbm lbm

= 56:55 lbm= min

B. From steam tables or online steam property resources (https://www.tlv.com/ global/TI/calculator/steam-table-temperature.html) hcw,e = hf ð51 ° FÞ = 19:07 Btu=lbm hcw,i = hf ð40 ° FÞ = 8:03 Btu=lbm Energy balance across the coil results in the following set of equations. Heat transferred from air to the coil (cooling load) = heat absorbed by chilled water qT,coil = m_ cw ðhcw,e - hcw,i Þ ) qT,coil = m_ R = hcw,e - hcw,i

4162 19:07

Btu min

Btu Btu - 8:03 lbm lbm

= 377 lbm= min

Example 6.15 An air handler unit (AHU) uses chilled water as a coolant. It receives an air mixture consisting of 0.6 m3/s of outside air at 35 °C and 50% relative humidity and 4.4 m3/s of return air at 20 °C dry bulb and 60% relative humidity. The cooling coil operates at an effective coil temperature of 11 °C. The bypass factor is 10%. Determine A. the condition (DB, enthalpy, rh, and moisture content) of air supplied to the space B. the cooling load of the system in kW and tons of refrigeration C. the SHR for the coil Solution A. State point 1 (35 °C DB, 50% rh) represents outside air and state point 2 (20 °C DB, 60% rh) represents return air. Calculate the dry bulb temperature of the mixed air entering the coil using the lever rule.

6.3

Cooling Load Calculations

T 3,DB =

233

T 1 × V_ 1 þ T 2 × V_ 2 35 × 0:6 þ 20 × 4:4 = = 21:8 ° C ≃ 22 ° C 0:6 þ 4:4 V_ 1 þ V_ 2

A line connecting state points 1 (35 °C DB, 50% rh) and 2 (20 °C DB, 60% rh) is drawn on the psych. Chart. The intersection of this line with 22 °C dry bulb line defines state point 3 of the mixed air entering the coil.

From the excerpt of the psychrometric chart, at state point 3, T DB3 = 22 ° C, v3 = 0:85 m3 =kg d a, h3 = 47:5 kJ=kg d a Locate the effective coil temperature, TADP = 11 ° C, on the saturation curve as shown. Calculate the dry bulb temperature of air leaving the coil (supply air to the space, T4, DB, using Eq. 5.3. T - T ADP T outlet - T ADP ) 0:10 = 4,DB ) T inlet - T ADP T 3,DB - T ADP T 4 - 11 ° C 0:10 = ) T 4,DB = 12:1 ° C 22 ° C - 11 ° C

BF =

State point 4 is located at the intersection of the line joining state point 3 and TADP, and the vertical dry bulb line at 12.1 °C. From the excerpt of the psychrometric chart, at state point 4 (air leaving the coil, supplied to the space), T DB4 = 12 ° C, ω4 = 0:0085 kg H2 O=kg d a, h4 = 33 kJ=kg d a, rh4 = 93%

234

6

Cooling and Heating Load Calculations

B. Calculate the mass flow rate of dry air into the coil using the specific volume at coil inlet. Note that the volume flow rate of the mixed inlet air to the coil is 4:4 m3 =s þ 0:6 m3 =s = 5 m3 =s 5 ms V_ 3 = = 5:8823 kg d a=s 3 v3 0:85 kgmd a 3

m_ a =

Calculate the cooling load of the system using the mass flow rate of dry air and coil inlet and exit enthalpies. qT = m_ a ðh3 - h4 Þ kg d a = 5:8823 s = 85:293 kW

47:5

kJ kJ - 33 kg d a kg d a

Cooling load in tons, 1 ton cooling 3:5168 kW = 24:253 tons of cooling

qT = ð85:293 kWÞ

C. Calculate the sensible cooling load using the coil inlet and exit dry bulb temperatures. qS = m_ a cpa ΔT a,DB = 5:8823

kg d a s

1:0

kJ ð22 ° C - 12 ° CÞ = 58:823 kW kg d a . K

Calculate the sensible heat ratio (SHR) using Eq. 6.4. SHR =

6.4

qS 58:823 kW = = 0:69 qT 85:293 kW

Heating Load Calculations

As mentioned earlier, the heating load calculations can be carried out assuming steady-state conditions. This is because of almost steady outdoor conditions and negligible indoor contributions, which is in sharp contrast to the unsteady and complex nature of cooling load calculations.

6.4

Heating Load Calculations

6.4.1

235

Indoor and Outdoor Design Conditions for Space Heating

The indoor design conditions are based on the comfort level of the occupants including health considerations to avoid extreme dry air conditions. The typical indoor design for calculating heating loads are in the range 60 °F to 70 °F DB and relative humidity in the range 30–60%. Usually, thermal contributions from occupants, lights, and appliances are not included for initial sizing of heating equipment. Keeping the preceding concept in mind, it is acceptable to use the lower end of the recommended temperature and humidity range for indoor design conditions in order to avoid oversizing the heating equipment. Chapter 18, ASHRAE Fundamentals Handbook, 2021, I-P Edition [2] recommends maximum 70 °F DB and minimum 30% relative humidity for heating load calculations. The outdoor design conditions are based on the climatic conditions available from Chapter 14, ASHRAE Fundamentals Handbooks [2, 3]. The coldest month 99% heating DB and humidity ratio (HR) are generally used for heating load calculations.

6.4.2

Equations for Calculating Heating Loads

The equations for calculating sensible heat losses through surfaces, including windows, are similar to the equations used in calculating sensible heat gains through surfaces during the hot summer months. The most important difference is obviously the temperature driving force for heat losses, which is the difference between the warm indoor temperature, Ti, and the cold outdoor temperature, To. Hence, the equation for calculating heat loss through any surface is qs,surface = UAðT i - T o Þ

ð6:49Þ

where qs,surface = hot loss through the surface U = overall heat transfer coefficient for the composite/single surface including air film resistances A = surface area exposed to air Ti = design indoor temperature To = design outdoor temperature

236

6

Cooling and Heating Load Calculations

The heating load due to ventilation and infiltration will consist of both sensible and latent heat components and they can be calculated using the following equations, which are similar to the equations used for calculating cooling loads due to infiltration and ventilation. Sensible heating loads due to infiltration (subscript ‘inf.’) and ventilation (subscript ‘vent.’) qS, inf:,htg: = ð1:08Þðcfminf: ÞðT i - T o Þ qS, inf:,htg: = 1:23 V_ m3 =s,inf: ðT i - T o Þ qS,vent:,htg: = ð1:08Þðcfmvent: ÞðT i - T o Þ qS,vent:,htg: = 1:23 V_ m3 =s,vent: ðT i - T o Þ

USCS

ð6:50aÞ

SI

ð6:50bÞ

USCS

ð6:51aÞ

SI

ð6:51bÞ

Latent heating loads due to infiltration (subscript ‘inf.’) and ventilation (subscript ‘vent.’) qL,inf:,htg = ð4840Þðcfminf: Þðωi - ωo Þ qL,inf:,htg: = ð3003Þ V_ m3 =s,inf: ðωi - ωo Þ qL,vent:,htg = ð4840Þðcfmvent: Þðωi - ωo Þ

USCS

ð6:52aÞ

SI

ð6:52bÞ

USCS

ð6:53aÞ

qL,vent:,htg: = ð3003Þ V_ m3 =s,vent: ðωi - ωo Þ S I

ð6:53bÞ

In Eqs. 6.50a, 6.50b, 6.51a, 6.51b, 6.52a, 6.52b, 6.53a, and 6.53b Ti = design indoor temperature To = design outdoor temperature ωi = design indoor humidity ratio ωo = design outdoor humidity ratio The following example problem illustrates the details of heating load calculations. Example 6.16 A single-story office building in Atlanta, GA, USA (Latitude 340) occupies a floor space of 50 ft. × 80 ft. with 10 ft. high walls/glass façade. One of the 50 ft. edges faces north towards the street and is entirely a glass façade with swinging double glass door each 7 ft. high and 4 ft. wide. The east and west side walls each have four equally spaced fixed glass windows, each 10 ft. wide and 4 ft. high. The south side wall has two equally spaced windows of the same size. The building has a flat roof. The inside of the building is to be maintained at 70 °F DB and 50% relative humidity. The heating system is to be designed based on 26 °F DB and 0.0014 lbm H2O/lbm d a humidity ratio (from Table 1, Chapter 14, ASHRAE Fundamentals Handbook, 2021, I P Edition [2]). The following information is available about the walls, roof, and glazing systems of the building.

6.4

Heating Load Calculations

237

Roof: The roof composition from the top layer to the bottom layer is given along with the relevant thermal data, which can also be obtained from Tables 1, 3, and 10, Chapter 26, ASHRAE Fundamentals Handbook, 2021, I P Edition [2]. Outdoor air film resistance, R = 0.2271 hr-ft2-0F/Btu ½ in. built-up roof membrane, R = 0.3531 hr-ft2-0F/Btu 1 in. mineral fiber board, R = 2.951 hr-ft2-0F/Btu 2 in. mineral wool insulation, k = 0.2498 Btu-in./hr-ft2-0F 4 in. concrete, k = 7.6328 Btu-in./hr-ft2-0F 3.5 in. air space (for dropped roof framing), R = 1.8735 hr-ft2-0F/Btu 0.5 in. in gypsum board dropped ceiling, k = 1.11 Btu-in./hr-ft2-0F Indoor air film resistance, R = 0.9083 hr-ft2-0F/Btu Wall: The wall composition from the outside layer to the inside layer is given along with the relevant thermal data, which can also be obtained from Tables 1, 3, and 10, Chapter 26, ASHRAE Fundamentals Handbook, 2021, I P Edition [2]. Outdoor air film resistance, R = 0.2498 hr-ft2-0F/Btu 0.5 in. painted cement plastering on the outside, k = 4.9954 Btu-in./hr-ft2-0F 8 in. concrete block (limestone aggregate with two perlite filled cores), R = 2.100 hr-ft2-0F/Btu 0.5 in. painted cement plastering on the inside, k = 4.9954 Btu-in./hr-ft2-0F Indoor air film resistance, R = 0.6812 hr-ft2-0F/Btu Glazing System: Low emissivity clear glass double glazing system with 1/2 in. air space (ID 21). The thermal data is given here, and it can be obtained from Tables 4 and 10, Chapter 15, ASHRAE Fundamentals Handbook, 2021, I P Edition [2]. Overall heat transfer coefficient, U = 0.32 Btu/hr-ft2-0F The door can be treated as part of the glass façade and the heat gain due to opening and closing of doors will be included as part of infiltration. Note: The information provided thus far (except for the design indoor and outdoor conditions) is the same as that was provided in Example 6.1 and hence some of the calculations from the solution to Example 6.1 will be used here. The heated space in this example (Example 6.16) is the same as the office building described in Example 6.1, and hence the same data is applicable to the roof, the walls, and the windows. Although there is a small variation in the value of the outdoor air film coefficient for summer and winter months, it is not considered here due to the small magnitude of the difference and also due to the relatively small contribution of the air film resistance. Additionally, using the higher summer R value will give conservative results for winter calculations. Hence, calculation results such as R-values and U values are directly taken from the solution to Example 6.1. A. Determine the individual and total heating loads due to surfaces B. Assuming an infiltration rate of 0.9ACH, determine the sensible, and latent heating loads due to infiltration.

238

6

Cooling and Heating Load Calculations

C. Considering the maximum number of occupants in the building (50 persons) and ASHRAE Standard 62.1 ventilation requirement of 20 cfm/person, the required ventilation rate is 1000 cfm, determine the sensible, and latent heating loads due to ventilation. D. Determine the total heating load required for the building. E. Determine the percentage contribution from each entity towards the heating load. (Solution) A. Heating load due to roof From the solution to Example 6.1, the overall heat transfer coefficient for the roof assembly, including inside and outside air convection resistances is, U roof = 0:0654 Btu=hr‐ft2 ‐ ° F Calculate the sensible heat loss through the roof using Eq. 6.49. qs,roof = U roof Aroof ðT i - T o Þ Btu ð50 ft × 80 ftÞð70 ° F - 26 ° FÞ hr‐ft2 ‐ ° F = 11, 510:4 Btu=hr

= 0:0654

Heating load due to walls From the solution to Example 6.1, the overall heat transfer coefficient for the wall assembly, including inside and outside air convection resistances is, U wall = 0:1988 Btu=hr‐ft2 ‐ ° F Calculate the total net surface area of the three walls (since there is a glass façade in North direction) by subtracting the surface area of the windows (4 windows, each 10 ft. × 4 ft., on the east, west walls and 2 windows on the south walls) from the total surface areas of the east (80 ft. × 10 ft), west, and south walls (80 ft. × 10 ft). Awalls,net,E,W = 2ð80 ft × 10 ft - 4 × 10 ft × 4 ftÞ = 1280 ft2 Awalls,net,S = 50 ft × 10 ft - 2 × 10 ft × 4 ft = 420 ft2 Awalls,total = 1280 ft2 þ 420 ft2 = 1700 ft2

6.4

Heating Load Calculations

239

Calculate the sensible heat loss through the walls using Eq. 6.49. qs,walls = U walls Awalls ðT i - T o Þ Btu hr‐ft2 ‐ ° F = 14, 870:2 Btu=hr

1700 ft2 ð70 ° F - 26 ° FÞ

= 0:1988

Heating load due to glazed surfaces Calculate the total area of the glazed surfaces as shown. Surface area of each window = 10 ft × 4 ft = 40 ft2 Total number of windows = 4 (east) + 4 (west) + 2 (south) = 10 Surface area of the north glass façade = 50 ft × 10 ft = 500 ft2 Therefore, the total surface area for conduction heat loss through the glazed surfaces is Atot,glz = 10 × 40 ft2 þ 500 ft2 = 900 ft2 Calculate the heat loss due to conduction through windows and glass facade using Eq. 6.49. qs,glz = U glz Atot,glz ðT i - T o Þ Btu hr‐ft2 ‐ ° F = 12, 672 Btu=hr

= 0:32

900 ft2 ð70 ° F - 26 ° FÞ

Calculate the total sensible heating load due to all surfaces. qs,total = qs,roof þ qs,wall þ qs,glz Btu Btu Btu = 11510:4 þ 14870:2 þ 12672 hr hr hr = 39, 052:6 Btu=hr B. Calculate the infiltration rate using 0.9 ACH. (Note: The specified infiltration rate equivalent to 0.9ACH is very much on the higher side, indicative of an older building with significant openings and leakages, which ought to be fixed by undertaking required repair works in the interest of energy conservation [9]. With this in mind, the infiltration rate should be within the reasonable range of 0.10ACH to 0.5ACH for modern buildings, which should significantly bring down the contribution of infiltration to the heating load.) ð0:9Þð50 ft × 80 ft × 10 ftÞ = 600 cfm V_ inf = hr × 60 hrmin

240

6

Cooling and Heating Load Calculations

Calculate the sensible heating load due to infiltration using Eq. 6.50a qS, inf:,htg: = ð1:08Þðcfminf: ÞðT i - T o Þ Btu=hr ð600 cfmÞð70 ° F - 26 ° FÞ cfm × ° F = 28, 512 Btu=hr

= 1:08

The outside humidity ratio is ωo = 0.0014 lbm H2O/lbm d a. Locate the state point of the inside air (70 °F DB, 50% rh) on the psychrometric chart as shown and determine the humidity ratio of the inside air.

Calculate the latent heating load due to infiltration using Eq. 6.52a qL, inf:,htg = ð4840Þðcfminf: Þðωi - ωo Þ Btu=hr = 4840 ð600 cfmÞ lbm H2 O cfm‐ lbm d a = 18, 586 Btu=hr

lbm H2 O lbm d a lbm H2 O - 0:0014 lbm d a

0:0078

C. Calculate the sensible heating load due to ventilation using Eq. 6.52a. qS,vent:,htg: = ð1:08Þðcfmvent: ÞðT i - T o Þ Btu=hr ð1000 cfmÞð70 ° F - 26 ° FÞ cfm × ° F = 47, 520 Btu=hr

= 1:08

Calculate the latent heating load due to ventilation using Eq. 6.53a. qL,vent:,htg = ð4840Þðcfmvent: Þðωi - ωo Þ Btu=hr = 4840 ð1000 cfmÞ lbm H2 O cfm‐ lbm d a = 30, 976 Btu=hr

lbm H2 O lbm d a lbm H2 O - 0:0014 lbm d a

0:0078

Practice Problems

241

D. and E The total heating load required for the building and the percentage contribution from each entity towards the total heating load is summarized in the following table. (Note: Please refer to the note in the solution to part B. With a more plausible infiltration rate equivalent to a maximum of 0.40ACH, the contribution of infiltration to the heating load should come down significantly, and should be in the range of 10 to 15% of the total heating load.)

Practice Problems Note: The following problem is an S I version of cooling load calculations with the solution being similar to the methods followed in the solutions for Examples 6.1 through 6.11. Practice Problem 6.1 A cooling system is to be designed for a small school building with indoor design conditions of 22 °C DB and 60% relative humidity and outdoor design conditions of 35 °C DB and 28 °C WB. CLTD correction of -3 °C can be used. The design parameters are as follows: Building parameters: overall dimensions 10 m (east and west) by 15 m (north and south) by 4 m (height). Roof: The roof composition from the top layer to the bottom layer is given along with the relevant thermal data, which can also be obtained from Tables 1, 3, and 10, Chapter 26, ASHRAE Fundamentals Handbook, S I Edition, 2021 [3]. outdoor air film resistance, R = 0.044 m2.K/W 10 mm built-up roof membrane, R = 0.059 m2.K/W 50 mm fiber glass insulation, k = 0.042 W/m.K 300 mm asphalt roofing, k = 0.79 W/m.K

242

6

Cooling and Heating Load Calculations

40 mm air space (for dropped roof framing), R = 0.3 m2.K/W 20 mm fiber board dropped ceiling, k = 0.09 W/m.K Indoor air film resistance, R = 0.16 m2.K/W Wall: The wall consists of 300 mm thick fire clay brick, k = 0.79 W/m.K. In addition, Outdoor air film resistance, R = 0.044 m2.K/W Indoor air film resistance, R = 0.12 m2.K/W (Preceding data from Tables 1, 3, and 10, Chapter 26, ASHRAE Fundamentals Handbook, S I Edition, 2021 [3]). Glazing System: Two 1 m × 1 m glass windows and a 3 m wide × 2.5 m high glass double door in east wall, all with the same glazing system Two 1.5 m × 1.5 m glass windows in west wall Three 2.0 m × 1.5 m glass windows each in north and south walls Glazing system specifications and data – Fixed glass windows with insulated fiber glass frames ID # 17a, Low emissivity clear glass double glazing system, 3 mm thick glass with 13 mm air space. The thermal data is given here, and it can be obtained from Tables 4 and 10, Chapter 15, ASHRAE Fundamentals Handbook, S I Edition, 2021 [3]. Overall heat transfer coefficient, U = 2.12 W/m2.K Total window solar heat gain coefficient (SHGC) at normal incidence, SHGC = 0.58. The peak irradiance at the location, from Table 10, Chapter 17, ASHRAE Fundamentals Handbook, S I Edition, 2021 [3], is as follows (all values in W/m2): North = 183, South = 306, East/West = 719 Infiltration and Ventilation: Infiltration, including that due to opening of doors, is equivalent to 0.5 ACH and ventilation is based on 8 L/s per person (maximum occupancy = 60 and bypass factor of 0.13) People, lights, and appliances: Projected maximum occupancy in the building is 60 persons. The cooling loads due to people are as follows: 75 W sensible heat and 55 W latent heat per person as per Table 1, Chapter 18, ASHRAE Fundamentals Handbook, S I Edition, 2021 [3]. Lighting: 2.5 W/m2 and Appliances: 1200 W sensible heat, 750 W latent heat Fans: Pressure differences to be overcome by supply air fan and return air fan are 1.55 in. w c and 0.45 in. w c, respectively. Both fans have efficiencies of 70% and motor/belt efficiencies of 80%. Return Duct: cross section of 1.1 m × 0.70 m, length = 15 m. The overall heat transfer coefficient for the return duct is 1.7023 W/m2.0C

Practice Problems

243

Determine A. the sensible, latent, and total cooling loads for the building B. the sensible, latent, and total cooling loads for the cooling coil C. the tons of refrigeration required for cooling

Practice Problem 6.2 From ASHRAE Fundamentals Handbook, the following data are available for Atlanta, GA, USA region for summer: DB = 90 °F and MCWB = 73 °F. The conditioned space is to be maintained at a comfort level of 70 °F and 60% relative humidity. The required air circulation rate is 10,000 cfm. A mixture of 20% outside air (OA) and 80% return air (RA) enters the coil and the air leaves the coil at 60 °F DB and 57 °F WB. Determine A. B. C. D.

the total cooling load in Btu/hr. and tons of refrigeration the sensible heat ratio (SHR) the bypass factor the coil efficiency

Note: The following problem is an S I version of heating load calculations with the solution being similar to the methods followed in the solution for Examples 6.16. Practice Problem 6.3 A heating system is to be designed for a small school building with indoor design conditions of 22 °C DB and 60% relative humidity and outdoor design conditions of -3 °C DB and humidity ratio of 0.0013 kg H2O/kg d a. The design parameters are as follows: Building parameters: overall dimensions 10 m (east and west) by 15 m (north and south) by 4 m (height). Roof: The roof composition from the top layer to the bottom layer is given along with the relevant thermal data, which can also be obtained from Tables 1, 3, and 10, Chapter 26, ASHRAE Fundamentals Handbook, 2021, S I Edition [3]. outdoor air film resistance, R = 0.044 m2.K/W 10 mm built-up roof membrane, R = 0.059 m2.K/W 50 mm fiber glass insulation, k = 0.042 W/m.K 300 mm asphalt roofing, k = 0.79 W/m.K 40 mm air space (for dropped roof framing), R = 0.3 m2.K/W 20 mm fiber board dropped ceiling, k = 0.09 W/m.K Indoor air film resistance, R = 0.16 m2.K/W Wall: The wall consists of 300 mm thick fire clay brick, k = 0.79 W/m.K. In addition,

244

6

Cooling and Heating Load Calculations

Outdoor air film resistance, R = 0.044 m2.K/W Indoor air film resistance, R = 0.12 m2.K/W (Preceding data from Tables 1, 3, and 10, Chapter 26, ASHRAE Fundamentals Handbook, 2021, S I Edition [3]). Glazing System: Two 1 m × 1 m glass windows and a 3 m wide × 2.5 m high glass double door in east wall, all with the same glazing system Two 1.5 m × 1.5 m glass windows in west wall Three 2.0 m × 1.5 m glass windows each in north and south walls Glazing system specifications and data – Fixed glass windows with insulated fiber glass frames ID # 17a, Low emissivity clear glass double glazing system, 3 mm thick glass with 13 mm air space. The thermal data is given here, and it can be obtained from Tables 4 and 10, Chapter 15, ASHRAE Fundamentals Handbook, 2021, S I Edition [3]). Overall heat transfer coefficient, U = 2.12 W/m2.K Infiltration and Ventilation: Infiltration, including that due to opening of doors, is equivalent to 0.5 ACH and ventilation is based on 8 L/s per person (maximum occupancy = 60). Note: The information provided thus far (except for the design outdoor conditions) is the same as that was provided in Practice Problem 6.1 and hence some of the calculation results, such as U values and R values, from the solution to Practice Problem 6.1 will be used here. Determine: A. B. C. D. E.

the individual and total heating loads due to surfaces the sensible, and latent heating loads due to infiltration the sensible, and latent heating loads due to ventilation the total heating load required for the school building the percentage contribution from each source towards the total heating load

Solutions to Practice Problems Note: The following solution is similar to the methods followed in the solutions for Examples 6.1 through 6.11. Practice Problem 6.1 (Solution) A. Maximum possible sensible heat gain through the roof Calculate the R values for the roof entities for which the thickness and thermal conductivity values are given using Eq. 2.6 (Chap. 2).

Solutions to Practice Problems

ΔX FG,ins ΔX Rcond,ua = = ) RFG,ins = kFG,ins k

245

50 m 1000 = 1:1905 m2 . K=W W 0:042 m.K

Similarly, 300 m ΔX asphalt = 1000 W = 0:3797 m2 . K=W kasphalt 0:79 m.K 20 m ΔX fiber bd: = 1000 W = 0:2222 m2 . K=W Rfiber bd: = kfiber bd: 0:09 m.K

Rasphalt =

Add up the resistances of all the roof entities (proceeding from the outside to inside) to obtain the total thermal resistance of the roof per unit area. The units for each thermal resistance term is m2.K/W and it is not included in the calculation equation for the sake of brevity of expression. Rroof = Roa þ Rmem þ RFG,ins þ Rasphalt þ Rair:sp þ Rfiber bd: þ Ria = 0:044 þ 0:059 þ 1:1905 þ 0:3797 þ 0:3 þ 0:2222 þ 0:16 = 2:3554 m2 . K=W Calculate the overall heat transfer coefficient (reciprocal of the overall thermal resistance) for the roof using Eq. 2.19 (Chap. 2). U roof =

1 1 = = 0:4246 W=m2 . K 2 Rroof 2:3554 mW.K

Determine table CLTD values for roofs. First, obtain the roof number from Chapter 28, Table 31, ASHRAE Fundamentals Handbook, 1997, S I Edition [5]. With suspended (dropped) ceiling (column 2) with R value within the range 1.8 to 2.6 m2.K/W (column 3), the only feasible roof number is 5 from column 4. Next, obtain the maximum possible CLTD for roof number 5 from Table 30 in the preceding reference. From the table, CLTDmax = 38 ° C, occurring both at 5 p m (hour 17) and 6 p m (hour 18). Next, apply the CLTD correction to the preceding value from the CLTD table. CLTD max,corrected = CLTD max,table þ CLTDcorrection = 38 ° C - 3 ° C = 35 ° C

246

6

Cooling and Heating Load Calculations

Calculate the maximum possible sensible heat gain through the roof using Eq. 6.17. qroof = U roof Aroof CLTD max,corrected W = 0:4246 2 ð10 m × 15 mÞð35 ° CÞ m .K = 2229 W Maximum possible sensible heat gain through the walls Since all the other R values are given, only the R value for the fire clay brick needs to be calculated using Eq. 2.6 (Chap. 2).

Rbrick =

300 m ΔX brick = 1000 W = 0:3797 m2 . K=W kbrick 0:79 m.K

Add up the resistances of all the wall entities to obtain the total thermal resistance of the wall per unit area. The units for each thermal resistance term is m2.K/W and it is not included in the calculation equation for the sake of brevity of expression. Rwall = Roa þ Rbrick þ Ria = 0:044 þ 0:3797 þ 0:12 = 0:5437 m2 . K=W Calculate the overall heat transfer coefficient (reciprocal of the overall thermal resistance) for the wall using Eq. 2.19 (Chap. 2). U wall =

1 1 = = 1:8392 W=m2 . K Rwall 0:5437 m2 .K W

Calculate the total net surface area of the walls by subtracting the surface area of the windows/doors (3 windows, each 2 ft. × 1.5 m, in the north, south walls, 2 windows, 1 m × 1 m and 3 m × 2.5 m door in the east wall, and 2 windows, 1.5 m × 1.5 m in the west wall) from the total surface areas of the north, south walls (15 m × 4 m), and east, west walls (10 m × 4 m). Awalls,net,N,S = 15 m × 4 m - 3 × 2 m × 1:5 m = 51 m2 , each ðN, SÞ Awall,net,W = 10 m × 4 m - 2 × 1:5 m × 1:5 m = 35:5 m2 Awall,net,E = 10 m × 4 m - 2 × 1:0 m × 1:0 m - 3:0 m × 2:5 m = 30:5 m2 Determine the CLTD for the walls.

Solutions to Practice Problems

247

First, obtain the code for the primary wall material from Chapter 28, Table 11, ASHRAE Fundamentals Handbook, 1997, S I Edition [5]. The only feasible code number for brick is C4, which has the closest k value of 0.727 W/m.K. Next, determine the relevant wall number from Chapter 28, Table 33B: Wall Mass Evenly Distributed, ASHRAE Fundamentals Handbook, 1997, S I Edition [5]. For an R value in the range 0.53 to 0.62 m2.K/W (the calculated R value is 0.5437 m2.K/W), under the column for C4, the wall number is 11. Next, obtain the maximum possible CLTD for each wall orientation for wall number 11 from Chapter 28, Table 32, ASHRAE Fundamentals Handbook, 1997, S I Edition [5] and apply the correction of -3 °C to the table values to get the corrected maximum CLTDs. CLTDmax,corrected = 14 ° Cfor the east facing wall CLTDmax,corrected = 19 ° Cfor the west facing wall CLTDmax,corrected = 8 ° Cfor the north facing wall CLTDmax,corrected = 13 ° Cfor the south facing wall Calculate the maximum possible sensible heat gain through each wall using Eq. 6.17. qE,wall = U E,wall AE,wall CLTD max,corrected,E W 30:5 m2 ð14 ° CÞ = 0:5437 2 m .K = 232:16 W qW,wall = U W,wall AW,wall CLTD max,corrected,W W = 0:5437 2 35:5 m2 ð19 ° CÞ m .K = 366:73 W qN,wall = U N,wall AN,wall CLTD max,corrected,N W = 0:5437 2 51 m2 ð8 ° CÞ m .K = 221:83 W qS,wall = U S,wall AS,wall CLTD max,corrected,S W = 0:5437 2 51 m2 ð13 ° CÞ m .K = 360:47 W Calculate the total heat gain through all the four walls. Each term in the following equation has units of W: qwalls,total = 232:16 þ 366:73 þ 221:83 þ 360:47 = 1181 W

248

6

Cooling and Heating Load Calculations

Maximum possible sensible heat gain through glazing systems The sensible heat gain through glazing systems consists of two components – heat gain through conduction and heat gain due to radiation. Conduction Calculate the total area of the glazing surfaces as shown. East windows and doors=2 × 1.0 m × 1.0 m + 3.0 m × 2.5 m = 9.5 m2 West windows=2 × 1.5 m × 1.5 m = 4.5 m2 North windows=3 × 2 m × 1.5 m = 9 m2 South windows=3 × 2 m × 1.5 m = 9 m2 Atot,glz = 9:5 m2 þ 4:5 m2 þ 2ð9Þ m2 = 32 m2 : Therefore, the total conduction heat gain through the glazing system (windows and glass doors) can be obtained from the first part of Eq. 6.19. W m2 . K = 882 W

qcond:,glz = U glz Aglz,tot ðT o - T i Þ = 2:12

32 m2 ð35 ° C - 22 ° CÞ

Radiation through glazed surfaces The peak irradiance depends on the orientation of the glazed surfaces and heat gain due to radiation for each orientation is calculated using the second part of Eq. 6.19 and the peak irradiance provided in the problem statement. The radiation from all directions are added up to obtain the total heat gain due to radiation. qrad,glz = ðSHGC ÞðAN E tN þ AS EtS þ AE E tE þ AW EtW Þ W W W 9 m2 × 183 2 þ 9 m2 × 306 2 þ 9:5 m2 × 719 2 m m m = ð0:58Þ W þ4:5 m2 × 719 2 m = 8391 W Therefore, the total sensible heat gain due to fenestration (glazed surfaces) is qglz = qcond,glz þ qrad,glz = 882 W þ 8391 W = 9273 W Cooling Load Due to Infiltration The total, sensible, and latent heat gains due to infiltration can be calculated using Eqs. 6.12, 6.13, and 6.14 and substituting for the infiltration volume flow rate. Calculate the volume of the building space using the given dimensions. V space = 10 m × 15 m × 4 m = 600 m3

Solutions to Practice Problems

249

Calculate the infiltration volume flow rate per second using the given 0.5 ACH. 0:5 V_ m3 =s, inf = hr =

0:5 hr

1 hr 3600 s

V space 600 m3

1 hr 3600 s

= 0:0833 m3 =s

Calculate the sensible heat gain due to infiltration using Eq. 6.34. qS, inf = 1:23 V_ m3 =s, inf ðT o - T i Þ =

1:23

kW

0:0833

m3 . °C s = 1:3325 kWð= 1333 WÞ

m3 ð35 ° C - 22 ° CÞ s

Calculate the latent heat gain due to infiltration using Eq. 6.35. This requires the humidity ratio of inside and outside air. Locate the state points of inside and outside air on the psychrometric chart as shown. Inside air, state point 1 (22 °C DB, 60% rh) Outside air, state point 2 (35 °C DB, 28 °C WB)

250

6

Cooling and Heating Load Calculations

From the excerpt of the psychrometric chart shown, ω2 = 0:0215 kg H2 O=kg d a and ω1 = 0:0099 kg H2 O=kg d a:

Calculate the latent heat gain due to infiltration using Eq. 6.35.

qL, inf =

3003

kW m3 kg H2 O . s kg d a

V_ m3 =s, inf ðω2 - ω1 Þ

3

kW m 0:0833 s m3 kg H2 O . s kg d a = 2:9017 kWð= 2902 WÞ

=

0:0215

3003

kg H2 O kg d a

- 0:0099

kg H2 O kg d a

Calculate the total heat gain due to infiltration. qT, inf: = qS, inf: þ qL, inf: = 1:3325 kW þ 2:9017 kW = 4:2342 kWð= 4234 WÞ Cooling Load on the Building Space Due to Ventilation Equivalent to Bypassed Air Calculate the required rate of ventilation in m3/s based on the maximum projected occupancy. V_ m3 =s,vent: = 8

L × 60 persons s . person

1 m3 = 0:480 m3 =s 1000 L

Calculate the sensible cooling load on the space due to the bypassed air using Eq. 6.40. qS,bypass = 1:23 V_ m3 =s,vent: ðBFÞðT 1 - T 2 Þ =

1:23

kW

0:480

m3 . °C s = 0:9978 kWð= 998 WÞ

m3 ð0:13Þð35 ° C - 22 ° CÞ s

Solutions to Practice Problems

251

Calculate the latent cooling load on the space due to the bypassed air using Eq. 6.41.

qL,bypass =

3003

kW m kg H2 O . s kg d a 3

V_ m3 =s,vent ðBFÞðω2 - ω1 Þ

kW m3 0:480 = 3003 3 ð0:13Þ s m kg H2 O . s kg d a = 2:1737 kWð= 2174 WÞ

0:0215

kg H2 O kg d a

- 0:0099

kg H2 O kg d a

Calculate the total cooling load on the space due to the bypassed air. qT,bypass = qS,bypass þ qL,bypass = 998 W þ 2174 W = 3172 Wð= 3:172 kWÞ Cooling Loads Due to Occupants From Table 35 B, Chapter 28, ASHRAE Fundamentals Handbook, 1997, S I Edition [5], for 4 walls, vinyl flooring, gypsum partition, half to no shading, the zone for people is C. Assume 8 hours occupation in space and for 8 hours after entry into the space, from Table 37, Chapter 28, ASHRAE Fundamentals Handbook, 1997, S I Edition [5], the maximum value of the cooling load factor for sensible heat transfer due to people is CLF = 0.91. Calculate the sensible heat gain for the occupants using Eq. 6.23. qS,occu: = ðN Þ qS=P ðCLF Þ = ð60 personsÞ

75 W ð0:91Þ person

= 4095 W Calculate the latent heat gain for the occupants using Eq. 6.24.

qL,occu: = ðN Þ qL=P = ð60 personsÞ

55 W = 3300 W person

Calculate the total cooling load on the space due to occupants. qT,occu = qS,occu þ qL,occu = 4095 W þ 3300 W = 7395 W

252

6

Cooling and Heating Load Calculations

Cooling Load Due to Lighting, and Appliances qlights = qS, Appl = 1200 W

2:5 W ð15 m × 10 mÞ = 375 W m2

qL, Appl = 750 W

Cooling Load Due to Supply Fan The sensible heat gains from different sources are tabulated here. Source Roof Walls Fenestration Occupants Lighting Appliances Infiltration Ventilation, bypass Total

Sensible heat gain, W 2229 1181 9273 4095 375 1200 1333 998 20,684 (= 20.684 kW)

Generally, a delta Tair of 9 °C to 11 °C across the coil is used. Assume delta Tair of 10 °C. Using the preceding result, calculate the air circulation rate, iteration 1, using Eq. 6.13. qS1 = 1:23 V_ m3 =s ðΔT air,coil Þ ) qS1 V_ m3 =s,1 = = ð1:23ÞðΔT air,coil Þ

20:684 kW 1:23

= 1:6816 m3 =s

kW ð10 ° CÞ m3 . °C s

Calculate the required power (kW) input to the fan, iteration 1, using Eq. 1.41. V_ m3 =s × ΔTPkPa ηfan × ηmotor 0:2488 kPa m3 1:6816 1:55 in wc × s 1 in wc = 0:7 × 0:8 = 1:158 kWð= 1158 WÞ

Pinput1 =

Solutions to Practice Problems

253

The preceding value is the sensible heat added by the fan (qfan1). Add this value to the total sensible heat gain and refine the calculations for air flow rate, fan heat, and final sensible heat gain by the building space (iteration 2). qS2 = qS1 þ qfan1 = 20:684 W þ 1:158 W = 21:842 W V_ m3 =s,1 =

qS2 = ð1:23ÞðΔT air,coil Þ

21:842 kW 1:23 m3kW s .°C

= 1:7758 m3 =s

ð10 ° CÞ

V_ m3 =s,2 × ΔTPkPa ηfan × ηmotor m3 0:2488 kPa 1:7758 1:55 in wc × s 1 in wc = 0:7 × 0:8 = 1:2229 kWð= 1223 WÞ

qfan2 =

qS3 = qS1 þ qfan2 = 20:684 kW þ 1:223 kW = 21:907 kW This iteration process is continued, and the results are presented in the spreadsheet image here.

From the preceding table, the total sensible cooling load for the building is qS,bldg = 21:9106 kW The latent heat gains for the building from different sources is summarized in the following table.

254

6

Cooling and Heating Load Calculations

From the preceding table, the latent cooling load for the building is Source Occupants Appliances Infiltration Ventilation, bypass

Latent heat gain, kW 3.300 0.75 2.9017 2.1737

qL,bldg = 9:1254 kW The total cooling load for the building is qT,bldg = qS,bldg þ qL,bldg = 21:9106 kW þ 9:1254 kW = 31:036 kW B. The sensible cooling load on the coil is the sum of the sensible load on the building space, the sensible load due to outside air, and the sensible load due to the return duct (Eq. 6.44). qS,Coil = qS,Space þ qS,OA þ qS,RD The calculation of the sensible heat load on the coil due to outside air requires the knowledge of the coil temperature (apparatus dew point, ADP). This in turn requires the use of Eq. 5.3, reproduced here for reference. BF =

T outlet - T ADP T inlet - T ADP

ð5:3Þ

In the preceding equation, BF is the bypass factor, and Tinlet, Toutlet are inlet and outlet temperatures of air passing through the coil. As shown in the following figure, the air entering the coil is a mixture of return air and outside air. Therefore, the properties of the mixed air will have to obtained before using Eq. 5.3 to determine the coil temperature. The procedure for obtaining the properties of mixed air is clearly explained in the solution to Example 5.11 (Sect. 5.1.7, Chap. 5) and the same procedure will be used here, along with the nomenclature shown in the figure.

Solutions to Practice Problems

255

From the data given, the properties of the streams shown in the figure are as follows: Stream 1 (return air/air from the space): T1DB = 22 ° C, rh1 = 60%. Earlier, the ventilation rate was determined to be 0.480 m3/s, and this will be the volume flow rate of the outside air as well. From the iteration table for determining the sensible cooling load for the building, the air circulation rate required for the space is 1.7813 m3/s, of which 0.480 m3/s is exhaust air due to ventilation. Hence, the volume flow rate of stream 1 (return air, RA) is, m3 m3 V_ 1,RA = 1:7813 - 0:480 = 1:3013 m3 =s s s Stream 2 (outside air, OA): T2, DB = 35 ° C, T2, WB = 28 ° C, and as mentioned earlier, V_ 2,OA = 0:480 m3 =s Using the lever rule, calculate the approximate value of the dry bulb temperature at state point 3 (mixed air going to the coil) as shown. V_ 1 × T 1DB þ V_ 2 × T 2DB V_ 3 m3 m3 1:3013 ð22 ° CÞ þ 0:480 ð35 ° CÞ s s = m3 1:7813 s = 25:5 ° C

T 3DB =

A bypass factor of 0.13 was specified and the temperature difference across the coil was assumed to be 10 ° C. Hence, the temperature of the outlet air from the coil is T oulet = T 4 = 25:5 ° C - 10 ° C = 15:5 ° C: Substitute all the known values into Eq. 5.3 and determine the apparatus dew point as shown.

256

6

BF =

Cooling and Heating Load Calculations

T outlet - T ADP 15:5 ° C - T ADP ) 0:13 = ) T ADP = 14 ° C = T coil T inlet - T ADP 25:5 ° C - T ADP

The outside air flow rate is equivalent to the ventilation rate multiplied by the complement of the bypass fraction. Calculate the sensible cooling load due to outside air using Eq. 6.45b. qS,OA = 1:23 V_ m3 =s,OA ð1 - BFÞðT o - T coil Þ =

kW

1:23

m3 . °C s = 10:787 kW

0:480

m3 ð1 - 0:13Þð35 ° C - 14 ° CÞ s

Note that the return fan handles the entire air circulation rate of 1.7813 m3/s, including the exhaust air with 0.4800 m3/s and return air with 1.3013 m3/s. Calculate the required power (kW) input to the return fan using Eq. 1.41. V_ m3 =s,RA × ΔTPkPa ηfan × ηmotor m3 0:2488 kPa 1:7813 0:45 in wc × s 1 in wc = 0:7 × 0:8 = 0:3561 kWð= 356:1 WÞ

Pinput,RF =

The preceding value is the sensible heat added by the return fan,qS, RAfan. Calculate the exposed area of the return duct. ARD = perimeter × length = ½2ð1:1 m þ 0:70 mÞ]½15 m] = 54 m2 Substitute all the known values into Eq. 6.46 to calculate the sensible cooling load on the coil due to the return air duct, including the return air fan. qS,RD = U RD ARD ðT o - T i Þ þ qS,RAfan W 54 m2 ð35 ° C - 22 ° CÞ þ 356:1 W = 1:7023 2 m . °C = 1195 W þ 356:1W = 1551:1 Wð= 1:551 kWÞ qS,Coil = qS,Space þ qS,OA þ qS,RD = 21:911 kW þ 10:787 kW þ 1:551 kW = 34:249 kW

Solutions to Practice Problems

257

The total latent cooling load on the coil must include the latent cooling load due to the portion of the outside air flowing across the coil, which can be calculated using Eq. 6.48b. The difference in humidity ratio across the coil can be determined using the psychrometric chart as shown. As shown in the following excerpt of the psychrometric chart, join state point 1 (RA, 22 °C DB, 60% rh) and state point 2 (OA, 35 °C DB, 28 °C WB) with a straight line and locate state point 3 of mixed air going into the coil on this line corresponding to the vertical dry bulb temperature line of 25.5 °C.

Determine the humidity ratio at state point 3 (mixed air at coil inlet) as shown in the psychrometric chart. ω3 = 0:0135 kg H2 O=kg d a Draw a vertical line at the coil dry bulb temperature, Tcoil = 14 ° C (calculated earlier). Since the air leaving the coil is saturated, the coil humidity ratio can be determined at the intersection of coil dry bulb temperature and the saturation curve. From the psychrometric chart, ωcoil = 0:0099 kg H2 O=kg d a

258

6

Cooling and Heating Load Calculations

Calculate the latent cooling load due to the portion of the outside air flowing across the coil using Eq. 6.48b. qL,OA = ð3003Þ V_ m3 =s,OA ð1 - BFÞðω3 - ωcoil Þ =

3003

kW m kg H2 O . s kg d a 3

0:480

0:0135

m3 ð1 - 0:13Þ × s

kg H2 O kg H2 O - 0:0099 kg d a kg d a

= 4:5146 kW Calculate the total latent cooling load on the cooling coil using Eq. 6.47. qL,Coil = qL,Space þ qL,OA = 9:1254 kW þ 4:5146 kW = 13:6400 kW Calculate the total cooling load on the cooling coil by adding up the sensible and latent cooling loads. qT,Coil = qS,Coil þ qL,Coil = 34:249 kW þ 13:640 kW = 47:889 kW C. Convert the preceding result to tons of refrigeration (TR).

TR =

47:889 kW 3:5168 kW 1 TR

= 13:62 TR

Practice Problem 6.2 (Solution) A. State point 1 (90 °F DB, 73 °F WB,) represents outside air and state point 2 (70 °F DB, 60% rh) represents return air. Calculate the dry bulb temperature of the mixed air entering the coil using the lever rule.

T 3,DB =

T 1 × V_ 1 þ T 2 × V_ 2 90 ° F × 2000 cfm þ 70 ° F × 8000 cfm = = 74 ° F 10000 cfm V_ 1 þ V_ 2

Solutions to Practice Problems

259

A line connecting state points 1 (90 °F DB, 73 °F WB) and 2 (70 °F DB, 60% rh) is drawn on the excerpt of the psychrometric chart as shown. The intersection of this line with 74 °F dry bulb line defines state point 3 of the mixed air entering the coil. The properties of the mixed air at state point 3 are determined from the excerpt of psychrometric chart.

At state point 3, T DB3 = 74 ° F, v3 = 13:65 ft3 =lbm d a, h3 = 29 Btu=lbm d a ω3 = 0:0103 lbm H2 O=lbm d a On the excerpt of the psychrometric chart state point 4 is located at the intersection of TDB4 = 60 ° F, and TWB4 = 57 ° F, the given conditions of air leaving the coil. At state point 4 (air leaving the coil, supplied to the space), T DB4 = 60 ° F, ω4 = 0:009 lbm H2 O=lbm d a, h4 = 24 Btu=lbm d a Calculate the mass flow rate of dry air into the coil using the specific volume at coil inlet, v3. ft 10000 min V_ 3 = = 732:6 lbm d a= min 3 v3 13:65 lbmft d a 3

m_ a =

260

6

Cooling and Heating Load Calculations

Calculate the cooling load of the system using the mass flow rate of dry air and coil inlet and exit enthalpies. qT = m_ a ðh3 - h4 Þ lbm d a 60 min = 732:6 × min hr = 219, 780 Btu=hr

29

Btu Btu - 24 lbm d a lbm d a

Cooling load in tons, 1 ton cooling 12000 Btu=hr = 18:315 tons of cooling

qT = ð219780 Btu=hrÞ

B. Calculate the sensible cooling load using the coil inlet and exit dry bulb temperatures. qS = m_ a cpa ΔT a,DB lbm d a 60 min × min hr = 147, 692 Btu=hr = 732:6

0:24

Btu ð74 ° F - 60 ° FÞ lbm d a‐0 F

Calculate the sensible heat ratio (SHR) using Eq. 6.4. SHR =

qS 147692 Btu=hr = = 0:672 qT 219780 Btu=hr

C. Extend the process line, 3–4 to the saturation curve and determine the effective coil temperature, TADP = 53 ° F, on the saturation curve as shown. Calculate the bypass factor, BF, using Eq. 5.3. BF =

T outlet - T ADP T 4,DB - T ADP 60 ° F - 53 ° F = = = 0:3333 T inlet - T ADP T 3,DB - T ADP 74 ° F - 53 ° F

Calculate the coil efficiency using Eq. 5.1. Coil efficiency = 1 - BF = 1 - 0:3333 = 0:6667ð67%Þ Note: The following solution is similar to the methods followed in the solution for Examples 6.16.

Solutions to Practice Problems

261

Practice Problem 6.3 The heated space in this problem (Practice Problem 6.3) is the same as the school building described in Example 6.1, and hence the same data is applicable to the roof, the walls, and the windows. Although there is a small variation in the value of the outdoor air film coefficient for summer and winter months, it is not considered here due to the small magnitude of the difference and also due to the relatively small contribution of the air film resistance. Additionally, using the higher summer R value will give conservative results for winter calculations. (Solution) A. Heating load due to the roof From the solution to Practice Problem 6.1, the overall heat transfer coefficient for the roof is, Uroof = 0.4246 W/m2 . K. Calculate the sensible heat loss through the roof using Eq. 6.49. qs,roof = U roof Aroof ðT i - T o Þ W ð10 m × 15 mÞð22 ° C - ð - 3 ° CÞÞ = 0:4246 2 m .K = 1592:3 W Heating load due to the walls From the solution to Practice Problem 6.1, the overall heat transfer coefficient for the roof is, Uwall = 1.8392 W/m2 . K. From the solution to Practice Problem 6.1, the net area (wall area minus the window areas) of the walls are Awalls,net,N,S = 15 m × 4 m - 3 × 2 m × 1:5 m = 51 m2 , each ðN, SÞ Awall,net,W = 10 m × 4 m - 2 × 1:5 m × 1:5 m = 35:5 m2 Awall,net,E = 10 m × 4 m - 2 × 1:0 m × 1:0 m - 3:0 m × 2:5 m = 30:5 m2 Therefore, the total net area of the walls is Anet,walls = 2 51 m2 þ 35:5 m2 þ 30:5 m2 = 168 m2 Calculate the sensible heat loss through the walls using Eq. 6.49.

262

6

Cooling and Heating Load Calculations

qs,walls = U walls Anet,walls ðT i - T o Þ W 168 m2 ð22 ° C - ð - 3 ° CÞÞ = 1:8392 2 m .K = 7724:6 W Heating Load due to Glazed Surfaces From the solution to Practice Problem 6.1, the total surface area of the glazed surfaces is, Atot, glz = 32 m2. Calculate the sensible heat loss due to conduction through windows and glass façade (glazed surfaces) using Eq. 6.49. qs,glz = U glz Atot,glz ðT i - T o Þ W 32 m2 ð22 ° C - ð - 3 ° CÞÞ = 2:12 2 m .K = 1696 W Calculate the total sensible heating load due to all surfaces. qs,total = qs,roof þ qs,wall þ qs,glz = 1592:3 W þ 7724:6 W þ 1696 W = 11012:9 W B. Heating Load due to Infiltration Calculate the infiltration rate using 0.5 ACH. ð0:5Þð10 m × 15 m × 4 mÞ = 0:0833 m3 =s V_ inf = s hr × 3600 hr Calculate the sensible heating load due to infiltration using Eq. 6.50b qS, inf:,htg: = 1:23 V_ m3 =s, inf: ðT i - T o Þ =

1:23

kW 3

m . °C s = 2:5625 kW

0:0833

m3 ð22 ° C - ð - 3 ° CÞÞ s

The outside humidity ratio is ωo = 0.0013 kg H2O/kg d a. Locate the state point of the inside air (22 °C DB, 60% rh) on the psychrometric chart as shown and determine the humidity ratio of the inside air.

Solutions to Practice Problems

263

From the psychrometric chart, the inside humidity ratio is ωi = 0:010 kg H2 O=kg d a Calculate the latent heating load due to infiltration using Eq. 6.52b qL, inf:,htg: = ð3003Þ V_ m3 =s, inf: ðωi - ωo Þ kW = 3003 3 m kg H2 O . s kg d a = 2:1763 kW

m3 0:0833 s

lbm H2 O lbm d a lbm H2 O - 0:0013 lbm d a

0:010

C. Heating Load due to Ventilation Calculate the required rate of ventilation in m3/s based on the maximum projected occupancy. V_ m3 =s,vent: = 8

L × 60 persons s . person

1 m3 = 0:480 m3 =s 1000 L

Calculate the sensible heating load due to ventilation using Eq. 6.51b.

264

6

Cooling and Heating Load Calculations

qS,vent:,htg: = 1:23 V_ m3 =s,vent: ðT i - T o Þ =

1:23

kW

m3 . °C s = 14:76 kW

0:480

m3 ð22 ° C - ð - 3 ° CÞÞ s

Calculate the latent heating load due to ventilation using Eq. 6.53b. qL,vent:,htg: = ð3003Þ V_ m3 =s,vent: ðωi - ωo Þ kW = 3003 3 m kg H2 O . s kg d a = 12:54 kW

m3 0:480 s

lbm H2 O lbm d a lbm H2 O - 0:0013 lbm d a

0:010

D. and E. The total heating load required for the building and the percentage contribution from each entity towards the total heating load is summarized in the following table.

References

265

References 1. Air Conditioning Contractors of America Association, Inc: Manual J Residential Load Calculation [ANSI/ACCA 2 Manual J], 8th Edition, ACCA, Inc., USA (2016) 2. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook Fundamentals, I P Edition, ASHRAE, USA (2021). 3. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook Fundamentals, S I Edition, ASHRAE, USA (2021). 4. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook Fundamentals, I P Edition, ASHRAE, USA (1997) 5. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook Fundamentals, S I Edition, ASHRAE, USA (1997) 6. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): Standard Thermal Environmental Condition for Human Occupancy [ASHRAE Standard 55], ASHRAE, USA (2022) 7. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): Ventilation and Acceptable Indoor Air Quality [ANSI/ASHRAE Standard 62.1], ASHRAE, USA (2022) 8. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): Ventilation and Acceptable Indoor Air Quality in Residential Buildings [ANSI/ASHRAE Standard 62.2], ASHRAE, USA (2022) 9. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): Energy Standard for Sites and Buildings Except Low Rise Residential Buildings [ASHRAE Standard 90.1], ASHRAE, USA, (2022) 10. International Organization for Standardization, Ergonomics of the Thermal Environment – Analytical Determination and Interpretation of Thermal Comfort Using Calculation of the PMV and PPD indices and Local Thermal Comfort Criteria [ISO 7730], ISO, Switzerland (2005) 11. Spitler, Jeffrey D: Load Calculations Applications Manual, ASHRAE, USA (2014). 12. On-line Psychrometric Calculator, Download from http://www.hvac-calculator.net/index.php? v=2

Chapter 7

Refrigeration Systems

7.1

Introduction

Refrigeration is the process of moving heat from a low-temperature heat source (the refrigerated or air-conditioned space) to a high-temperature heat sink using a refrigerant and energy input in a closed cycle [1, 2, 8]. The energy input is work input to a compressor in a Vapor Compression Refrigeration Cycle and heat input as in an Absorption Refrigeration Cycle. Refrigeration cycles have extensive practical applications in everyday life. These include air conditioning of spaces during the hot summer months, household refrigerators and freezers to store and preserve foods, vegetables, and fruits, commercial refrigerators and freezers in supermarkets, cold storages to preserve meats, and other foods. Heat pumps used in space heating and in producing hot water are based on the utilization of heat rejected during refrigeration cycles. Refrigeration cycles also have several industrial applications, which include cryogenics, liquefaction of gases, and low-temperature feeds and processes in refineries and chemical plants. In any refrigeration cycle, the refrigerating effect is the heat removed from the cooler space per unit mass of the refrigerant. Thus, the refrigerating effect is the difference in enthalpy of the refrigerant across the evaporator situated in the refrigerated/air-conditioned space. The refrigeration cycle is based on the Reversed Carnot Cycle, which is discussed in the following section.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 N. S. Nandagopal, HVACR Principles and Applications, https://doi.org/10.1007/978-3-031-45267-3_7

267

268

7.2

7

Refrigeration Systems

Reversed Carnot Cycle

As the name implies, a Reversed Carnot Cycle is the exact opposite of the Carnot cycle. It is the same as the Carnot cycle operating in the reverse direction [8]. Similar to Carnot cycle being the basis for cycles producing net power, the Reversed Carnot Cycle forms the basis for power-consuming refrigeration cycles and heat pumps. The schematic diagram and the T–s diagram for the Reversed Carnot Cycle are shown in Fig. 7.1. A Reversed Carnot Engine takes heat from a low-temperature heat source, such as a refrigerated space, and transfers the heat to a high-temperature heat sink [8]. Heat cannot spontaneously flow from low temperature to high temperature. Therefore, the Reversed Carnot Engine has to use external work to push the heat from low temperature to high temperature. As shown in Fig. 7.1b, the processes in the Reversed Carnot cycle are as follows: 1–2: Isentropic compression of the working fluid, which is usually a refrigerant. (vertical, constant entropy line in the T–s diagram and an increase in temperature of the working fluid implies an isentropic compression process). This compression process requires external work input. 2–3: Heat rejection at constant temperature (horizontal, constant temperature line in the T–s diagram and decreasing entropy and volume implies isothermal condensation). 3–4: Isentropic expansion of the working fluid, producing work (vertical, constant entropy line in the T–s diagram and decreasing temperature due to decrease in pressure implies an isentropic expansion process). 4–1: Heat absorption at constant temperature and isothermal expansion of the working fluid (horizontal, constant temperature line in the T–s diagram and increasing entropy due to evaporation of the working fluid implies an isothermal expansion process).

Fig. 7.1 (a) Reversed Carnot engine; (b) Reversed Carnot T–s diagram cycle

7.2

Reversed Carnot Cycle

7.2.1

269

Performance Measure of a Reversed Carnot Cycle: Coefficient of Performance (COP)

The performance measure of any device or process can be defined as a measure of desired effect achieved divided by the cost of achieving it. The desired effect in Reversed Carnot Cycle is the amount or rate of heat absorbed from the low-temperature heat source and the cost of achieving it is the work input to the compressor [8]. The preceding performance measure of Reversed Carnot Cycle is known as Coefficient of Performance (COP). Referring to Fig. 7.1a, the mathematical representation of COP for refrigeration or air conditioning [1, 2, 8] (heat removal from low-temperature heat source) is COPrefg:=AC =

Desired effect Q = L W Cost of achieving it

ð7:1Þ

Energy balance for the Reversed Carnot Engine (Fig. 7.1a) results in the following equations: Energy In = Energy Out QL þ W = QH ) W = QH - QL Substitute the preceding result into Eq. 7.1. Also, the heat energies can be replaced by the corresponding absolute temperatures. This results in a comprehensive set of equations for Coefficient of Performance for refrigeration and air conditioning. COPrefg:=AC =

QL QL TL = = W QH - QL T H - T L

ð7:2Þ

The Reversed Carnot Cycle can also be used for space heating. Space heating is achieved by capturing the heat rejected at the high-temperature heat sink, that is, heat rejected due to condensation of refrigerant vapor. When the Reversed Carnot Cycle is used for space heating it becomes a Heat Pump [8]. The desired effect in using a heat pump is the heat rejected at the heat sink, QH. Therefore, the COP for a heat pump [1, 2, 8] will be COPHeat Pump =

QH QH TH = = W QH - QL T H - T L

ð7:3Þ

The COP for a heat pump is always a number greater than one, with higher values indicating better performance.

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The rate of cooling can be represented as tons of refrigeration (TR) required. One ton of cooling is the rate of cooling achieved when 1 short ton (2000 lbm) of ice at 0 ° C melts in a 24-h time period [8]. By using the latent heat of melting (fusion) of ice (144 Btu/lbm or 334 kJ/kg), it can be shown that 1 ton of refrigeration = 12, 000 Btu=hr = 3:517 kW

ð7:4Þ

Example 7.1 A Carnot refrigerator (freezer) operates between temperature limits of -18 °C and 25 °C. It is desired to remove heat equivalent to 75 kW from the cold space. Determine: A. the COPrefg./AC. B. the power input required (kW). C. the rate of heat rejection (kJ/hr). (Solution) A. Calculate absolute temperatures using Eq. 3.3 (Chap. 3 – Thermodynamics). T H = 25 ° C þ 273 ° = 298 K

T L = -18 ° C þ 273 ° = 255 K

Calculate the Coefficient of Performance for the freezer using the last part of Eq. 7.2 and substitute the known values. COPrefg:=AC =

TL 255 K = = 5:93 T H - T L 298 K - 255 K

B. Calculate the power input required by using the first part of Eq. 7.2. The rate of heat removal from the cold space is given as 75 kW.

COPrefg:=AC =

Q_ L Q_ L 75 kW _ = )W = = 12:65 kW _ COPrefg:=AC 5:93 W

C. Calculate the rate of heat rejection using the middle part of Eq. 7.2. Q_ L ) Q_ H - Q_ L 75 kW ) Q_ H = 87:65 kW 5:93 = Q_ H - 75 kW

COPrefg:=AC =

7.2

Reversed Carnot Cycle

271

Note: The same result can be obtained by using the energy balance equation for a Reversed Carnot Engine as shown here. _ = 75 kW þ 12:65 kW = 87:65 kW Q_ H = Q_ L þ W Example 7.2 A Carnot heat pump is used in heating 1700 cfm of atmospheric air from 40 °F to 70 °F. Calculate the COP if the power input is 5 hp. (Solution) Calculate the individual gas constant for air using Eq. 3.14 (Chap. 3 – Thermodynamics). 3

psia‐ft 10:73 lbmol‐ R °R = = 0:37 psia‐ft3 =lbm‐ ° R Rair = lbm M air 29 lbmol

Calculate the average temperature of air and its corresponding absolute value. T avg = ð40 ° F þ 70 ° FÞ=2 = 55 ° F

) T air = 460 ° þ 55 ° F = 515 ° R

Calculate the density of air at its average temperature using the ideal gas law (Eq. 3.10). ρ=

14:7 psia m P = = 0:0771 lbm=ft3 = psia‐ft3 V Rair T air 0:37 lbm‐ ° R ð515 ° RÞ

Calculate the mass flow rate of air using the continuity equation, ðm_ = ρQÞ, Q being the volume flow rate of the fluid, which in this case can also be represented by V_ air . lbm m_ air = ρair V_ air = 0:0771 3 ft

1700

ft3 min

= 131:1 lbm= min

The specific heat of air at constant pressure is 0.24 Btu/lbm-°R. Energy balance at the high-temperature heat sink results in the following set of equations: Rate of heat rejected at the heat sink = Rate of heat absorption by air lbm Btu 0:24 ð70 ° F - 40 ° FÞ Q_ H = m_ air cp,air ΔT air = 131:1 min lbm‐ ° R = 943:9 Btu= min

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Calculate the coefficient of performance for the heat pump using Eq. 7.3.

COPHeat Pump =

7.3

0:0236 hp Btu 943:9 min Btu 1 min Q_ H = = 4:45 _ 5 hp W

Absorption Refrigeration Cycle

An absorption refrigeration cycle uses heat input from external sources to accomplish the refrigeration process. Thus, an absorption refrigeration cycle uses Thermal Compression in contrast to a vapor compression refrigeration cycle, which uses Mechanical Compression. The schematic diagram of an absorption refrigeration cycle is shown in Fig. 7.2. In an absorption refrigeration cycle, a solvent is used to absorb the refrigerant [1, 2]. Typically, ammonia is used as a refrigerant and water is the solvent. As shown in Fig. 7.2, a low-pressure liquid–vapor mixture of ammonia enters the evaporator, where it picks up the heat from the refrigerated/air-conditioned space. Due to heat addition, the ammonia liquid–vapor mixture is converted into a single, saturated vapor phase. The low-pressure ammonia vapor is absorbed by water in the absorber. The absorption process is exothermic due to the generation of the heat of solution. The heat thus generated is dissipated by circulating cooling water through a cooling coil (not shown in the diagram) in the absorber. The strong ammonia–water solution is pumped to an ammonia vapor generator operating at high pressure. The high pressure in the generator is achieved by heating the ammonia–water solution and the

Fig. 7.2 Absorption refrigeration cycle schematic diagram

7.3

Absorption Refrigeration Cycle

273

vapor phase causes the increase in pressure. Ammonia vaporizes first and it is separated from water in the analyser. The weak ammonia–water solution at the bottom of the generator is recycled back to the absorber via a pressure relief valve (PRV) [1, 2, 6]. The resulting high-pressure ammonia vapor from the analyser is condensed in a condenser, where heat is rejected during the isothermal condensation process. The high-pressure liquid ammonia from the condenser is throttled using a throttling valve resulting in the formation of a low-pressure liquid–vapor mixture of ammonia to continue the refrigeration cycle [1, 2].

7.3.1

Analysis of Absorption Refrigeration Cycle

The maximum coefficient of performance (COPmax) of an absorption refrigeration cycle can be calculated using the following equation which has temperatures as indicated in Fig. 7.2: COPmax =

TL TC - TL

TH - TA TH

ð7:5Þ

where TL = absolute temperature of the refrigerated space TC = absolute temperature at which heat is rejected from the condenser and this is usually, the absolute temperature of the surrounding atmosphere TH = absolute saturation temperature of the heating media (typically condensing steam) supplying heat to the generator TA = absolute temperature of the solution in the absorber Note: All the absolute temperatures must be in °R or K. Example 7.3 In an ammonia–water absorption refrigeration cycle, heat is supplied to the generator by condensing steam at a pressure of 30 psia and the refrigerated space is to be maintained at a temperature of 20 °F. The temperature in the absorber is 80 °F and the ambient atmospheric temperature is 70 °F. Determine the maximum possible COP for this refrigeration cycle. (Solution) Find the saturation temperature of steam at 30 psia from steam tables or online sources (https://www.tlv.com/global/TI/calculator/steam-table-pressure.html). For steam at 30 psia, Tsat = 250 ° F ) TH = 460 ° + 250 ° F = 710 ° R. Similarly, convert all the other given temperatures to their corresponding absolute values:

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T L = 460 ° þ 20 ° F = 480 ° R T C = 460 ° þ 70 ° F = 530 ° R T A = 460 ° þ 80 ° F = 540 ° R Substitute all the known values into Eq. 7.5 to obtain the maximum possible COP: COPmax =

TL TC - TL

TH - TA TH

480 ° R 530 ° R - 480 ° R = 2:299

=

710 ° R - 540 ° R 710 ° R

The COP for an actual absorption refrigeration cycle can be calculated by using the definition of COP [1, 2]. The contribution of the pump to the overall cost of achieving the refrigerating effect is very small due to very small magnitude of pump work or power compared to the rate of heat input in the generator. Therefore, ignoring the power consumed by the pump, the actual COP for the absorption refrigeration cycle shown in Fig. 7.2 is COPact:,abs ref: =

Desired effect Q_ = in Cost of achieving Q_ H

ð7:6Þ

The enthalpies of ammonia at the state points indicated in Fig. 7.2 are explained here. Assuming ammonia vapor to be in the saturated state at state 5, h5 = hg at condenser pressure Similarly, assuming ammonia liquid to be in the saturated state at state 6, h6 = hf at condenser pressure Energy balance for the condenser results in the following set of equations: Rate of energy in = rate of energy out m_ NH 3 h5 = Q_ C þ m_ NH 3 h6 ) Q_ C = m_ NH 3 ðh5 - h6 Þ where Q_ C is the rate of heat rejection in the condenser. Throttling is a constant enthalpy process [1, 2, 8]. Therefore, h1 = h6 = hf at condenser pressure

ð7:7Þ

7.3

Absorption Refrigeration Cycle

275

Assuming ammonia vapor to be in the saturated state at state 2, h2 = hg at evaporator pressure Energy balance for the evaporator results in the following set of equations: Rate of energy in = rate of energy out m_ NH 3 h1 þ Q_ in = m_ NH 3 h2 ) Q_ in = m_ NH 3 ðh2 - h1 Þ

ð7:8Þ

where Q_ in is the rate of refrigeration required, usually specified in terms of tons of refrigeration (TR). Hence, with the knowledge of the amount of refrigeration required and the enthalpies of the refrigerant (obtainable from refrigerant tables or charts or online sources), the mass flow rate of the refrigerant can be obtained. Also, the refrigerating effect can be calculated using the following equation: Refrigerating Effect = qin =

Q_ in = h2 - h1 m_ NH 3

ð7:8aÞ

Overall mass balance either for the generator or the absorber results in the following equation: m_ 4 = m_ 3 = m_ NH 3 þ m_ weak soln:

ð7:9Þ

State points 3 and 4 both represent the strong solution. Energy balance for the generator results in the following set of equations: Rate of energy in = rate of energy out m_ 4 h4 þ Q_ H = m_ weak soln: hweak soln: þ m_ NH 3 h5 ) Q_ H = m_ weak soln: hweak soln: þ m_ NH 3 h5 - m_ 4 h4

ð7:10Þ

where Q_ H is the rate of heat addition required in the generator. Once Q_ H and Q_ in are known, the actual coefficient of performance can be calculated using Eq. 7.6. The enthalpies, hweak soln. and h4 (strong solution) can be obtained from ammonia– water, enthalpy–concentration charts. Example 7.4 An ammonia–water absorption refrigeration cycle is used in producing 10 tons of refrigeration. The evaporator and condenser pressures for ammonia are 30 and 230 psi, respectively. The ratio of mass flow rate of weak solution to the mass flow rate of ammonia in the absorber is 2.2 to 1. The specific enthalpy of the strong ammonia solution leaving the absorber and entering the generator is 10 Btu/lbm and that of weak ammonia solution entering the absorber is 100 Btu/lbm. Determine:

276

A. B. C. D.

7

Refrigeration Systems

the refrigerating effect (Btu/lbm) the mass flow rate of ammonia required (lbm/min) the rate of heat input required in the absorber (Btu/min) the COP for the refrigeration cycle

(Solution) From ammonia refrigerant tables/P–h diagram or online sources (https://irc.wisc. edu/properties/), determine the enthalpies of ammonia at different state points in the refrigeration cycle using the same nomenclature as in Fig. 7.3. h2 = hg ðat 30 psiaÞ = 611 Btu=lbm h5 = hg ðat 230 psiaÞ = 633 Btu=lbm h6 = hf ðat 230 psiaÞ = 161 Btu=lbm h1 = h6 = 161 Btu=lbm A. Calculate the refrigerating effect using Eq. 7.8a. Btu Btu - 161 lbm lbm = 450 Btu=lbm

Refrigerating Effect = h2 - h1 = 611

B. Calculate the mass flow rate of ammonia required using the tons of refrigeration specified and Eq. 7.8. Q_ in = m_ NH 3 ðh2 - h1 Þ ) ð10 TRÞ Q_ in = m_ NH 3 = h2 - h1

12, 000

450

Btu 1 hr × hr 60 min TR

Btu lbm

= 4:44 lbm= min

C. Calculate the mass flow rate of weak solution into the absorber. m_ weak soln: = 2:2 ) m_ NH3 m_ weak soln: = 2:2m_ NH3 = 2:2 4:44

lbm min

= 9:768 lbm= min

Calculate the mass flow rate of strong ammonia solution into the generator using Eq. 7.9.

7.4

Vapor Compression Refrigeration Cycle

277

m_ 4 = m_ 3 = m_ NH 3 þ m_ weak soln: lbm lbm = 4:44 þ 9:768 min min = 14:208 lbm= min The enthalpies of weak and strong ammonia solutions are given. hweak soln: = 100 Btu=lbm and h4 ðstrong soln:Þ = 10 Btu=lbm Calculate the heat input to the generator using Eq. 7.10. Q_ H = m_ weak soln: hweak soln: þ m_ NH 3 h5 - m_ 4 h4 = 9:768

lbm min

100

Btu lbm þ 4:44 lbm min - 14:208

633 lbm min

Btu lbm 10

Btu lbm

= 3645 Btu= min D. Calculate the coefficient of performance using Eq. 7.6. 1 hr 12, 000Btu hr × 60 min TR Q_ in ð10 TRÞ COPact:,abs ref: = = Btu 3645 min Q_ H

= 0:55

Typically, the coefficient of performance for absorption refrigeration systems is less than 1.0. However, the viability of such systems is made possible by using waste heat as the heat source. Since waste heat is utilized, the cost of achieving the refrigeration drops down significantly. For example, absorption refrigeration systems become feasible in combined heat power (CHP) or cogeneration power plants, where waste heat from steam after the first stage of expansion in a high-pressure turbine can be used as a heat source for the ammonia generator in the absorption refrigeration cycle.

7.4

Vapor Compression Refrigeration Cycle

Vapor compression cycles are widely used in refrigeration and air conditioning applications, where heat is transferred from a low-temperature environment to surroundings at higher temperature. The objective is to keep the low-temperature environment at a constant, cooler temperature. Vapor compression cycle [1, 2, 8] is a modified version of the reversed Carnot cycle. The modifications result in making the reversed Carnot cycle practically feasible. The explanation for this will follow Fig. 7.3, which compares reversed Carnot cycle and vapor compression cycles.

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Refrigeration Systems

Fig. 7.3 Comparison of reversed Carnot and vapor compression cycles

Fig. 7.4 Vapor compression refrigeration cycle (a) schematic diagram (b) T-s diagram (c) P-h diagram

As shown in Fig. 7.3, the compressor in Reversed Carnot Cycle handles a liquid– vapor mixture at state point 1. This can result in damage to compressor blades (in a rotary compressor) due to the high-velocity impact and collapse of liquid–vapor bubbles on the blade surface. The Vapor Compression Cycle overcomes this problem by compressing ‘dry’ saturated vapor. The term ‘dry’ here refers to the absence of liquid or condensate. The saturated vapor at state point 1′ becomes superheated vapor at state point 2′ due to the addition of compressor work (energy). The other modification is the isentropic expansion process, 3–4, in the Reversed Carnot Cycle is replaced by constant enthalpy expansion process (3′–4′) through a throttling valve, which is a much simpler device compared to an isentropic turbine. Since dry, saturated vapor is compressed, the terminology ‘Vapor Compression Cycle’ is used. Figure 7.4 illustrates the schematic, the temperature–entropy (T–s), and the pressure–enthalpy (P–h) diagrams for a typical vapor compression cycle [1, 2, 8].

7.4.1

Analysis of Vapor Compression Cycle

In Fig. 7.4, m_ R is the mass flow rate of the refrigerant, such as R-134a, circulating in the cycle. Usually, the tons of refrigeration or air conditioning (TR) required is specified. This can be converted to the rate of heat absorption from the refrigerated/ air-conditioned space in terms of Btu/hr or kW using the conversion factors presented earlier in Eq. 7.4 and reproduced here for reference.

7.4

Vapor Compression Refrigeration Cycle

1 ton of refrigeration = 12,000 Btu=hr = 3:517 kW

279

ð7:4Þ

The heat absorbed in the evaporator is used in evaporating the refrigerant from state point 4 (liquid–vapor mixture) to state point 1 (saturated vapor). The heat absorbed in the evaporator per unit mass of refrigerant can be obtained by energy balance around the evaporator. Energy in = Energy out h4 þ qin = h1 ) qin = h1 - h4

ð7:11Þ

Rate of heat absorption = Q_ in = m_ R qin

ð7:12Þ

Equation 7.11 represents the refrigerating effect, that is, the heat removed per unit mass of refrigerant. The preceding equations can be combined to obtain an equation for calculating the mass flow rate of the refrigerant. m_ R =

Q_ in Q_ in = qin h1 - h4

ð7:13Þ

The energy supplied by the compressor results in the refrigerant changing state from saturated vapor (state point 1) to superheated vapor (state point 2). The work input to the compressor per unit mass of refrigerant can be obtained by energy balance around the compressor: Energy in = Energy out h1 þ wc,ideal = h2 ) wc,ideal = h2 - h1 _ c,ideal = m_ R wc,ideal Isentropic power input to thecompressor = W

ð7:14Þ ð7:15Þ

Combining the preceding equations, _ c,ideal = m_ R wc,ideal = m_ R ðh2 - h1 Þ W

ð7:16Þ

The actual power input to the compressor can be obtained by dividing the preceding equation by the isentropic efficiency of the compressor. h2 represents the enthalpy of the refrigerant after isentropic compression. _ _ c,actual = W c,ideal = m_ R ðh2 - h1 Þ = m_ R wc,actual = m_ R ðh2a - h1 Þ W ηc ηc

ð7:17Þ

where h2a represents the actual enthalpy of the refrigerant leaving the compressor.

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Refrigeration Systems

When the superheated refrigerant (state point 2) condenses to saturated liquid (state point 3) heat is rejected in the condenser. The heat rejected in the condenser per unit mass of refrigerant can be obtained by energy balance around the condenser. Energy in = Energy out h2 = h3 þ qout ) qout = h2 - h3

ð7:18Þ

Rate of heat rejection = Q_ out = m_ R qout

ð7:19Þ

Combining the preceding equations, Q_ out = m_ R qout = m_ R ðh2 - h3 Þ

ð7:20Þ

The enthalpies at each state point in the refrigeration cycle can be determined as follows: h1 = hg at evaporator pressure h2 (s2 = s1) at condenser pressure h3 = hf at condenser pressure. Process 3–4 is a constant enthalpy process)h4 = h3. ASHRAE Refrigeration Handbooks [3, 4] provide detailed coverage of equipment, piping, and safety devices used in both absorption and vapor compression refrigeration systems. ASHRAE Standard 15 [6] contains information related to safety of equipment and piping in refrigeration systems as well as safe operating procedures for refrigeration systems.

7.4.2

Overall Energy Balance for a Vapor Compression Refrigeration Cycle

Overall energy balance for the vapor compression cycle shown in Fig. 7.5 results in the following set of equations [1, 2, 8]: Rate of energy in = rate of energy out _ c,actual = Q_ out ) Q_ out = Q_ in þ W _ c,actual Q_ in þ W

ð7:21Þ

7.4

Vapor Compression Refrigeration Cycle

281

Fig. 7.5 Overall energy balance for a vapor compression refrigeration cycle

7.4.3

Condenser in Refrigeration Cycles

The condenser in a refrigeration cycle performs the important function of condensing the refrigerant vapor [1, 2, 8]. The condenser is usually a shell and tube heat exchanger with refrigerant vapor on the shell side being cooled and condensed by cooling water circulating in the tubes of the condenser. Air-cooled condensers can also be used in a refrigeration system. The pros and cons of shell and tube (water cooled) condensers and air-cooled condensers are discussed in Chap. 8. Heat exchangers are covered extensively in Chap. 2. Example 2.6 provides extensive, detailed coverage of different aspects of a condenser specifically used in a vapor compression refrigeration cycle. The heat rejected in the condenser is absorbed by cooling water circulating in the tubes of the condenser. Therefore, heat balance in the condenser results in the following set of equations: Q_ out = m_ w cpw ΔT w ) m_ w =

Q_ out cpw ΔT w

ð7:22Þ

where m_ w is the mass flow rate of cooling water required in the condenser. Usually, m_ w is in lbm/hr and the density of water is 8.34 lbm/gallon. The gpm (gallons per minute) of cooling water required in the condenser can be calculated using the following equation: gpmw =

1 hr m_ w lbm m_ w lbm hr × 60 min hr = min 8:34 lbm 500 lbm‐ gal gal‐hr

ð7:23Þ

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Refrigeration Systems

Btu Further, in Eq. 7.22, the specific heat of water is cpw = 1:0 lbm‐ ° F. Equations 7.22 and 7.23 can be combined with the preceding value of specific heat of water to obtain the following equation for gpm of cooling water in the condenser:

gpmw =

1 hr m_ w lbm Q_ out Btu hr × 60 min hr ) = Btu‐lbm‐ min 8:34 lbm 500 gal lbm‐gal‐hr‐ ° F × ΔT ° F

gpmw =

7.4.4

_

Qout Btu‐ min 500 gal‐hr‐ ° F × ΔT

ð7:24Þ

°F

Performance Measures for Vapor Compression Cycles

Vapor compression cycles can be used for cooling (refrigeration and air conditioning) or for heating [1, 2, 8] (heat pump). Depending on the purpose, the performance measures, coefficients of performance, are defined as follows: qin Q_ in h - h4 = = 1 _ w h c,actual 2a - h1 W c,actual

COPref:=AC = COPheat pump =

qout Q_ out h - h3 = = 2 _ c,actual wc,actual h2a - h1 W

ð7:25Þ ð7:26Þ

It can be shown that for a vapor compression cycle, COPheating = 1 þ COPcooling

ð7:27Þ

The energy efficiency ratio, EER, of a refrigeration cycle is the ratio of heat re moval in Btu/hr to the power input to the compressor in Watts [9]. EER =

Q_ in,Btu=hr = COPref:=AC _ C,W W

3:41

Btu W‐hr

ð7:28Þ

The seasonal energy efficiency ratio, SEER, is the energy efficiency ratio calculated over a period of an entire season [9]. SEER =

Btu removed W‐hr consumed

ð7:29Þ entire season

7.4

Vapor Compression Refrigeration Cycle

283

For heating systems, SEER is known as heating seasonal performance factor, HSPF. Example 7.5 The isentropic power required for a compressor in a vapor compression refrigeration cycle is 24.8 hp. However, there is a 10% energy loss in the compressor due to friction, heat loss, and other irreversibilities. The capacity of the refrigeration system is 15.5 tons and the temperature difference of cooling water across the condenser is 12 °F. Determine: A. B. C. D.

the COPref. the heat rejected in the condenser (Btu/hr), the gpm of cooling water required in the condenser, the COPheating.

(Solution) A. Determine the actual compressor power required by including the energy losses in the compressor. _ c,actual = 24:8 hp þ 0:10W _ c,actual ) W _ c,actual = 27:56 hp W Calculate the coefficient of performance for refrigeration using Eq. 7.25. COPref:=AC =

Q_ in

_ c,actual W

=

12, 000 Btu=hr 1 TR 2545 Btu=hr hp × 1 hp

15:5 TR × 27:56

= 2:65

B. Calculate the rate of heat rejection in the condenser using Eq. 7.21. _ c,actual = 15:5 × 12, 000 Btu þ 27:56 hp × 2545 Btu Q_ out = Q_ in þ W hr hr - hp = 256, 140 Btu=hr C. Calculate the gpm of cooling water required using Eq. 7.24.

gpmw =

256, 140 Btu Q_ out hr = = 42:69 gpm Btu‐ min Btu‐ min 500 gal‐hr‐ × ΔT ° F 500 × 12 °F °F gal‐hr‐ ° F

D. Calculate the coefficient of performance for heating using Eq. 7.26.

COPheating =

256, 140 Btu Q_ out hr = = 3:65 _ c,actual 27:56 hp × 2545Btu hr W 1 hp

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Example 7.6 It is necessary to remove heat at the rate of 40 tons to maintain a cold storage at the desired temperature. This is accomplished by using a vapor compression cycle using R-134a as the refrigerant. The evaporator pressure is 300 kPa and the condenser pressure is 1.5 MPa. Determine: A. the enthalpy of the refrigerant at each state point. B. the refrigerating effect (kJ/kg). C. the power input required for the compressor (kW) if the compressor efficiency is 80%. D. the COPrefg./AC. E. the energy efficiency ratio, EER. (Solution) A. The enthalpies of R-134a at the different state points can be determined from R-134a tables/P–h diagram or online sources (https://irc.wisc.edu/properties/). An excerpt of the P–h diagram for R-134a, SI units (Fig. 3.13 from Chap. 3, Thermodynamics), with the enthalpies at the state points is shown here for reference. Note that 300 kPa = 0.30 Mpa.

h1 = hg, 300 kPa = 250 kJ/kg h2 = h1:5 MPa,s2 = s1 = 285 kJ=kg h3 = hf, 1.5 MPa = 128 kJ/kg h4 = h3 = 128 kJ/kg

7.4

Vapor Compression Refrigeration Cycle

285

B. Calculate the heat absorption per unit mass of refrigerant, that is, the refrigerating effect, using Eq. 7.11. qin = h1 - h4 = 250

kJ kJ - 128 = 122 kJ=kg kg kg

C. Using the information of 40 tons of cooling required and Eq. 7.13, calculate the mass flow rate of refrigerant. Note that W = J/s.

m_ R =

kW Q_ in 40 TR × 3:517 TR = = 1:153 kg=s kJ qin 122 kg

Calculate the actual power input required for the compressor by using Eq. 7.17. 1:153 kgs _ c,actual = m_ R ðh2 - h1 Þ = W ηc

kJ kJ 285 kg - 250 kg

0:80

= 50:44 kW

D. Calculate the Coefficient of Performance for refrigeration using Eq. 7.25.

COPrefg: =

ð40 × 3:517Þ kW Q_ in = = 2:79 _ 50:44 kW W c,actual

E. Calculate the energy efficiency ratio, EER, using Eq. 7.28. EER = COPref:=AC = ð2:79Þ 3:41

3:41

Btu W‐hr

Btu = 9:514 Btu=W‐hr W‐hr

Example 7.7 The following data are obtained from an actual vapor compression refrigeration cycle used in cooling a small building. The refrigerant used is R-134a. The unit operates for an average of 8 h per day and the cooling season stretches over a period of 110 days during a year. Compressor suction and discharge pressures: 40 psia and 200 psia, respectively. The liquid refrigerant leaving the condenser is sub-cooled to 120 °F. Refrigeration required: 8 tons. Power input to the compressor: 9.28 kW. Seasonal energy efficiency ratio (SEER) of 9.5 Btu/W.h.

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Determine the following: A. B. C. D. E. F.

the refrigerant circulation rate (lbm/min). the COP. the isentropic efficiency of the compressor. the enthalpy of R-134a at compressor exit. the energy efficiency ratio (EER). the annual energy consumption of the unit in kWh/year.

(Solution) Assuming isentropic compression, obtain the enthalpies at different state points in the cycle using R-134a tables/P–h diagram or online sources (https://irc.wisc.edu/ properties/). An excerpt of the P–h diagram for R-134a, USCS units (Fig. 3.13 from Chap. 3, Thermodynamics), with the enthalpies at the state points is shown here for reference.

From the excerpt of the R-134a P–h diagram, h1 = hg, 40 psia = 105 Btu/lbm h2 = h200 psia,s2 = s1 = 120 Btu=lbm h3 = hf, 200 psia = 50 Btu/lbm (sub ‐ cooled to 120 ° F) h4 = h3 = 50 Btu/lbm

7.4

Vapor Compression Refrigeration Cycle

287

A. Using the information of 8 tons of cooling required and Eq. 7.13, calculate the mass flow rate of the refrigerant required.

8 TR × m_ R =

Q_ in = h1 - h4

Btu 1 hr × hr 60 min TR

12, 000

105

Btu Btu - 50 lbm lbm

= 29:09 lbm= min

B. Calculate the coefficient of performance for refrigeration using Eq. 7.25.

COPrefg: =

Q_ in

_ c,actual W

=

kW 8 TR × 3:517 TR = 3:03 9:28 kW

C. Calculate the ideal (isentropic) power required using Eq. 7.16. _ c,ideal = m_ R ðh2 - h1 Þ W lbm = 29:09 min = 7:68 kW

120

Btu Btu - 105 lbm lbm

0:0176 kW 1 Btu= min

The isentropic efficiency of the compressor is the ratio of the ideal power required to the actual power required (Eq. 7.17). ηc =

_ c,ideal W 7:68 kW = = 0:83ð83%Þ _ W c,actual 9:28 kW

D. Calculate the actual compressor work per unit mass of refrigerant using a part of Eq. 7.17 and then calculate the actual enthalpy of the refrigerant leaving the compressor using another part of the same equation. _ c,actual = m_ R wc,actual ) W 56:91 Btu= min _ c,actual 9:28 kW × W 1 kW = wc,actual = = 18:15 Btu=lbm lbm m_ R 29:09 min wc,actual = h2,actual - h1 ) h2,actual = wc,actual þ h1 = 18:15

Btu Btu þ 105 = 123:13 Btu= min min min

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7

Refrigeration Systems

E. Calculate the energy efficiency ratio (EER) using the first part of Eq. 7.28 as shown. 12, 000 Q_ in,Btu=hr 8 TR × TR hr = = 10:34 Btu=W . h EER = W _ C,W 9:28 kW × 1000 W kW Btu

F. Calculate the annual energy consumption of the unit as shown. Annual energy cons: =

Heat removed per year SEER Btu 12, 000 hr × 8 hr × 110 days 8 TR × TR day yr

= 9:5 = 8893 kWh=yr

Btu 1000 W × W.h kW

Example 7.8 Use the data (including the P–h diagram) and results from the solution to Example 7.7 and the following additional information: Cooling water enters the condenser at 60 °F and leaves the condenser at 72 °F and the overall heat transfer coefficient for the condenser is 23 Btu/hr-ft2-°F. Determine: A. the condenser heat duty B. the heat transfer area required for the condenser C. the number of tubes required in the condenser if the tube OD is 2 in., and the tube length is 5 ft. D. the gpm of cooling water required Hint: Follow the same methodology used in the solution of Example 2.6 in the Heat Exchanger chapter (Chap. 2). A. The condenser heat duty is the rate of heat rejected in the condenser. From the solution to Example 7.7, Q_ in = 8 TR ×

12, 000 Btu hr = 96, 000 Btu=hr 1 TR

_ c,actual = 9:28 kW × W

3414:43 Btu hr = 31, 685 Btu=hr 1 kW

7.4

Vapor Compression Refrigeration Cycle

289

Calculate the heat rejected in the condenser using Eq. 7.21. _ c,actual Q_ out = Q_ in þ W Btu Btu = 96, 000 þ 31, 685 = 127, 685 Btu=hr hr hr Therefore, qcond = Q_ out = 127, 685 Btu=hr B. R-134a vapor enters the condenser as a superheated vapor at a condenser pressure of 200 psia (data given in Example 7.7) and at an actual enthalpy (h2, actual) of 123.13 Btu/lbm as calculated in the solution to Example 7.7. The temperature at which the superheated vapor enters the condenser corresponds to h = 123:13 Btu=lbm, and P = 200 psia This can be determined from the P–h diagram or from R-134a Tables or online sources (https://irc.wisc.edu/properties/). Accordingly, T R134,in = T ð200 psia, h = 123:3 Þ = 141 ° F: From the data given in Example 7.7, R-134a leaves the condenser as a sub-cooled liquid at 120 °F. Therefore, TR134, out = 120 ° F. Cooling water enters the condenser at 60 °F and leaves the condenser at 72 °F. Draw the schematic for calculating the LMTD (log mean temperature difference) for the condenser as shown.

Calculate LMTD using Eq. 2.23 from the heat transfer chapter (Chap. 2). ΔT LMTD =

ΔT 1 - ΔT 2 ln

ΔT 1 ΔT 2

=

69 ° F - 60 ° F = 64:4 ° F °F ln 69 60 ° F

290

7

Refrigeration Systems

Calculate the heat transfer surface area required in the condenser using the heat exchanger design equation (Eq. 2.24, heat transfer chapter, Chap. 2) assuming a single pass condenser (MTD Correction Factor, F = 1.0). qcond = UAs FΔT LMTD ) qcond = As = UFΔT LMTD

127, 685

Btu hr

Btu 23 ð1:0Þð64:4 ° FÞ hr‐ft2 ‐ ° F

= 86:2 ft2

C. The heat transfer area is the outside surface area of the tubes. Calculate the number of tubes required in the condenser as shown.

n=

86:2 ft2 As = 33 tubes = 2 πDL ðπ Þ 12 ft ð5 ftÞ

D. Calculate the gpm of cooling water required using Eq. 7.24. Q_ out Btu‐ min 500 × ΔT ° F gal‐hr‐ ° F Btu 127, 685 hr = = 21:28 gpm Btu‐ min 500 × 12 ° F gal‐hr‐ ° F

gpmw =

7.5

Refrigeration and Freezing of Foods

Refrigeration and freezing of foods involve three separate steps [3, 4] as shown in Fig. 7.6. Step1: the raw food item is first cooled from its initial temperature to the freezing point of the food. The cooling/refrigeration load for this step is Q1 = mcpaf ðT initial - T FP Þ

ð7:30Þ

where m is the mass of the food item, and cpaf is the specific heat of the food item above freezing point. Step 2: the unfrozen food item (with water content being in the liquid phase) is converted to frozen food at the freezing point. This step is a phase change of water

7.5

Refrigeration and Freezing of Foods

291

Fig. 7.6 Steps involved in producing frozen foods

in the food item from liquid to solid phase involving the latent heat of fusion or freezing. The cooling/refrigeration load for this step is Q2 = mhfs

ð7:31Þ

where m is the mass of the food item, and hfs is the latent heat fusion of the food item at freezing point. Step 3: the frozen food item at freezing point is cooled to its final temperature below the freezing point of the food. The cooling/refrigeration load for this step is Q1 = mcpbf ðT FP - T final Þ

ð7:32Þ

where m is the mass of the food item, and cpbf is the specific heat of the food item below the freezing point. The total cooling/refrigeration load required for producing the frozen food item [3, 4] is obtained by adding up Eqs. 7.30, 7.31, and 7.32. Qtotal,FF = Q1 þ Q2 þ Q3 = mcpaf ðT initial - T FP Þ þ mhfs þ mcpbf ðT FP - T final Þ ) Qtotal,FF = m cpaf ðT initial - T FP Þ þ hfs þ cpbf ðT FP - T final Þ

7.5.1

ð7:33Þ

Thermal Properties of Foods

Chapter 19 of ASHRAE Refrigeration Handbook, 2022, I P and SI Editions [3, 4] provides extensive coverage of thermal properties of foods including water content, initial freezing point, and latent and specific heats. In particular, Table 3 in Chapter 19 of ASHRAE Refrigeration Handbook, 2022, I P and SI Editions [3, 4]

292

7 Refrigeration Systems

Table 7.1 Excerpts from Table 3, Chapter 19 [5] – fruits and vegetables Fruit/ vegetable Broccoli Potatoes Spinach Tomatoes Apple Orange Strawberry

Moisture content, % 90.69 78.96 91.58 93.76 83.93 82.30 91.57

Initial freezing point, °C,a -0.6 -0.6 -0.3 -0.5 -1.1 -0.8 -0.8

Sp.heat, above freezing kJ/kg.Kb 4.01 3.67 4.02 4.08 3.81 3.81 4.00

Sp.heat, below freezing kJ/kg.Kb 1.82 1.93 1.75 1.79 1.98 1.96 1.84

Latent heat kJ/kgc 303 264 306 313 280 275 306

Table 7.2 Excerpts from Table 3, Chapter 19 [5] – dairy, juices, and meats Fruit/ vegetable Swiss cheese Whole milk Vanilla ice cream Orange juice Chicken Ham Sirloin beef Pink Salmon

Moisture content, % 37.21

Initial freezing point, °C,a -10.0

Sp.heat, above freezing kJ/kg.Kb 2.78

Sp.heat, below freezing kJ/kg.K 2.88

Latent heat kJ/kgc 124

87.69

-0.6

3.89

1.81

293

61.00

-5.6

3.22

2.74

204

89.01

-0.4

3.90

1.76

297

65.99 68.26 71.70

-2.8 -2.2 -1.7

4.34 3.47 3.53

3.32 2.22 2.11

220 228 239

76.35

-2.2

3.68

2.17

255

Notes for Tables 7.1 and 7.2: a °F = 1.8(°C) + 32 b Btu

lbm‐ ° F

c Btu

lbm

= 0:2390

= 0:4299

kJ kg.K

kJ kg

provides the unfrozen composition data (percent proteins, fats, and carbohydrates) in addition to the preceding data (that is, moisture contents and specific/latent heats). Table 3, Chapter 19 [3, 4] provides the thermal data for foods under the categories of fruits, vegetables, dairy products, meats, candy, and juices/beverages. Excerpts from Table 3 Chapter 19 of ASHRAE Refrigeration Handbook, 2018, SI Edition [5] are provided in Tables 7.1 and 7.2.

7.5

Refrigeration and Freezing of Foods

293

The following examples illustrate calculations related to refrigeration and freezing of foods. Example 7.9 2200 lbm of unprocessed chicken, delivered to a cold storage unit at 70 °F is to be frozen to a final temperature of 8 °F for storage. The thermal properties for chicken can be obtained from Table 7.2, but they are provided here for reference. Freezing point = 27 °F Specific heat above freezing point = 1.037 Btu/lbm-°F Latent heat of fusion = 94.6 Btu/lbm Specific heat below freezing point = 0.7935 Btu/lbm-°F Determine the percent cooling required in each step of the cooling and freezing process and express the total cooling required in terms of Btu/lbm. (Solution) Determine the cooling required before freezing using Eq. 7.30. Q1 = mcpaf ðT initial - T FP Þ = ð2200 lbmÞ 1:037 = 98, 100 Btu

Btu ð70 ° F - 27 ° FÞ lbm‐0 F

Determine the cooling required for freezing using Eq. 7.31. Q2 = mhfs = ð2200 lbmÞ 94:6

Btu = 208, 120 Btu lbm

Determine the cooling required to reach the final temperature after freezing using Eq. 7.32. Q3 = mcpbf ðT FP - T final Þ = ð2200 lbmÞ 0:7935 = 33, 168 Btu

Btu ð27 ° F - 8 ° FÞ lbm‐ ° F

Determine the total cooling required using Eq. 7.33. Qtotal = Q1 þ Q2 þ Q3 = 98, 100 Btu þ 208, 120 Btu þ 33, 168 Btu = 339, 388 Btu

294

7 Refrigeration Systems

Determine the percent cooling required for each step using the preceding results. %cooling before freezing =

%cooling for freezing =

%cooling after freezing =

Q1 98, 100 Btu 100 = 29% 100 = 339, 388 Btu Qtotal

Q2 208, 120 Btu 100 = 61% 100 = 339, 388 Btu Qtotal

Q3 33, 168 Btu 100 = 10% 100 = 339, 388 Btu Qtotal

Total cooling required per unit mass: Qtotal 339, 388 Btu = = 154:27 Btu=lbm m 2200 lbm Example 7.10 It is estimated that 80 MJ of cooling is required in a food processing plant producing frozen tomatoes at a temperature of -7 °C from raw tomatoes delivered at 20 °C. Determine the mass (kg) of tomatoes processed. (Solution) Obtain the following food properties for tomatoes from Table 7.1. Freezing point = -0.5 °C Specific heat above freezing point = 4.08 kJ/kg.K Latent heat of fusion = 313 kJ/kg Specific heat below freezing point = 1.79 kJ/kg.K Substitute all the known values into Eq. 7.33 and solve for the mass of the tomatoes processed. Qtotal,FF = m cpaf ðT initial - T FP Þ þ hfs þ cpbf ðT FP - T final Þ ) 4:08 80, 000 kJ = m

kJ kJ ð20 ° C - ð - 0:5 ° CÞÞ þ 313 kg . K kg þ 1:79

kJ ð - 0:5 ° C - ð - 7 ° CÞÞ kg . K

) m = 195:95 kg Note: ΔT ° C = ΔT K and 80 MJ = 80, 000 kJ

7.6

Refrigerants

7.6

295

Refrigerants

Refrigerants play the crucial role of absorbing heat from the refrigerant space. Refrigerants vary in characteristics from simple, naturally occurring compounds to complex organic compounds. Ammonia, Carbon dioxide, Water and Air are natural refrigerants. R-134a, a hydrofluoro carbon, is an example of a complex organic refrigerant that is commonly used. The impact of refrigerants on the environment has been extensively studied in recent years following the negative effects of earlier refrigerants on the ozone layer. The factors that need to be considered in the selection of a refrigerant are . . . . . . .

environmental impact cost operating efficiency chemical stability compatibility with equipment materials of construction safety (flammability and toxicity) and end-of-life safe disposal.

The evolution of organic refrigerants has progressed from earlier chlorofluoro carbons (CFCs, 1950s to 1980s) to hydrochlorofluoro carbons (HCFCs, 1980s to 1990s) to contemporary hydrofluoro carbons (HFCs 1990s to 2010) to emerging, future generation refrigerants such as hydrofluoro olefins (HFOs 2010 and beyond). Chapter 29 in ASHRAE Fundamentals Handbooks [1, 2] has extensive coverage of refrigerants, their characteristics/properties, and their safety/toxicity attributes.

7.6.1

Terminologies Associated with Refrigerants

Ozone Depletion Potential (ODP) The ozone depletion potential (ODP) of a refrigerant is a comparative measure of its potential to cause damage to the ozone layer. The comparison is with respect to refrigerant R-11(trichlorofluoromethane, CCl3F, a CFC) which is assigned an ODP 1.0. As an example of ODP values, refrigerant R-22 (chlorodifluoromethane, CHClF2, an HCFC) has an ODP of 0.05. R-11 has a higher ODP because of the presence of three chlorine atoms, which cause more damage to the ozone layer. R-22 has a much lower ODP value because it has only one chlorine atom along with two fluorine atoms. Fluorine atoms cause little or no damage to the ozone layer. Global Warming Potential (GWP) The global warming potential (GWP) of a refrigerant is a comparative measure of its potential to cause global warming. The comparison is with respect to a reference gas, that is, 1 kg of carbon dioxide. Thus, carbon dioxide is assigned a base GWP of 1 and other gases are assigned GWP values with respect to this standard. For example,

296

7 Refrigeration Systems

R-22 has a GWP of 1780, which means 1 kg of R-22 gas has the potential to cause 1780 times the global warming of 1 kg of carbon dioxide over the same period of time. GWP100 measures global warming effects over a time period of 100 years. Chlorofluoro Carbons (CFCs) CFCs contain chlorine and fluorine atoms in addition to carbon atoms. Examples of CFCs are R-11 (trichlorofluoromethane, CCl3F) and R-12 (dichlorodifluoromethane, CCl2F2). CFCs have high ozone depletion potential (ODP) and high global warming potential (GWP). As a result, the manufacture of these refrigerants ended for the most part on January 1, 1996. Hydrochlorofluoro Carbons (HCFCs) Besides carbon, HCFCs contain hydrogen atoms in addition to chlorine and fluorine atoms. Examples of HCFCs are R-22 (chlorodifluoromethane, CHClF2, popularly known as ‘Freon’) and R-123 (2,2-Dichloro-1,1,1-trifluoroethane, C2HCl2F3) HCFCs have lower ozone depletion potential (ODP) and moderate global warming potential (GWP). Since they still contain chlorine atoms, they have been viewed only as temporary replacements for the CFCs. Production caps have been mandated for HCFCs in almost all countries and production is prohibited after 2020 in developed countries and after 2030 in developing countries. Hydrofluoro Carbons (HFCs) HFCs have only carbon, hydrogen, and fluorine atoms. Since they do not have any chlorine atoms, they have zero ozone depletion potential (ODP). HFCs have moderate to high global warming potential (GWP). Examples of HFCs are R-134a (1,1,1,2-tetrafluoroethane, C2H2F4) and R-32 (difluoro methylene, CH2F2). R-134a is being used extensively at present. However, low GWP HFC refrigerants such as R-407A are finding increasing use. Hydrofluoro Olefins (HFOs) HFOs are unsaturated organic compounds (alkenes) consisting of carbon, hydrogen, and fluorine. HFOs have double bonds between carbon atoms unlike CFCs and HCFCs which are saturated organic compounds. HFO refrigerants have the twin advantages of zero ozone depletion potential (ODP) and low global warming potential (GWP) and are certainly a more environmentally friendly alternative to CFC, HCFC, and HFC refrigerants. Examples of HFOs are R-1234yf (2,3,3,3tetrafluoro propene, C3H2F4) and R-1234ze (1,3,3,3-tetrafluoro propene, C3H2F4). Since HFOs have zero ODP and low GWP, they represent emerging and future generation of refrigerants. Ammonia, carbon dioxide, water, and air are natural refrigerants. Table 3B, Chapter 29 in ASHRAE Fundamentals Handbooks [1, 2] provides a summary of the environmental properties of refrigerants. Excerpts from the table are included in Table 7.3.

Practice Problems

297

Table 7.3 Ozone depletion potential (ODP), global warming potential (GWP), and safety designations of selected refrigerants Refrigerant R-11(CFC) R-12 (CFC) R-22 (HCFC, Freon 22) R-123 (HCFC, Freon 123) R-32 (HFC) R-134a (HFC) R-1234yf (HFO) R-1234ze (HFO) R-717 (ammonia)

ODP 1 0.73 0.034 0.01 0.00 0.00 0.00 0.00 0.00

GWP 4660 10,800 1760 79 677 1300 TOA,i, and TEA,i > TEA,e. The actual heat transfer accomplished in the device can be calculated in terms of the heat gained by the cold outside air or in terms of equivalent heat lost by exhaust air using the following equation. The additional subscript, ‘W’ represents winter operation.

8.4

Energy Recovery Systems

323

qHRV,W = qOA,W = qEA,W )

ð8:5Þ

qHRV,W = m_ OA cpa ðT OA,e - T OA,i Þ = m_ EA cpa ðT EA,i - T EA,e Þ

ð8:6Þ

The maximum possible temperature difference is TEA,i - TOA,i, and Cminrepresents the lower value of m_ OA cpa and m_ EA cpa (the products are known as heat transfer capacity rates, COA = m_ OA cpa and C EA = m_ EA cpa ). Therefore, the maximum possible heat transfer in winter is q max,W = C min ðT EA,i - T OA,i Þ

ð8:7Þ

The effectiveness of a heat recovery ventilator is defined as the ratio of actual heat transfer to the maximum possible heat transfer. εHRV,W =

qHRV,W C ðT - T OA,i Þ - T EA,e Þ C ðT = OA OA,e = EA EA,i q max,W Cmin ðT EA,i - T OA,i Þ C min ðT EA,i - T OA,i Þ

ð8:8Þ

Example 8.4 A run-around coil is used in preheating 6000 cfm of winter outside air at 45 °F DB and 40% rh to a temperature of 63 °F DB. The circulation rate of water in the coil is 13 gpm. The source of heat is an equivalent mass flow of exhaust air from the facility, which is maintained at 70 °F DB and 60% rh. Determine: A. the exit temperature of the exhaust air, B. the temperature rise of the circulating water, C. the effectiveness of the HRV. (Solution) A. Since mass flow rates of the outside air and exhaust air are equal, Eq. 8.6 can be simplified, and the exit temperature of the exhaust air can be calculated as shown.

qHRV ,W = m& OAc pa (TOA,e - TOA,i ) = m& EAc pa (TEA,i - TEA,e ) = 630 F - 450 F = 700 F - TEA,e = TEA,e = 520 F

B. Calculate the mass flow rate of the outside air (dry air) by using its specific volume. Obtain the specific volume of the outside air from the psychrometric chart as shown. Locate the state point of the outside air, OA (45 °F DB and 40% rh) on the excerpt of the psychrometric chart. From the excerpt of psychrometric chart shown, vOA = 12:8 ft3 =lbm d a

324

8

HVAC Systems and Equipment

ft 6000 min V_ = 468:75 lbm d a= min m_ OA = OA = 3 vOA 12:8 lbmft d a 3

The mass flow rates of the outside air and exhaust air are equal. Therefore, m_ EA = m_ OA = 468:75 lbm d a= min Energy balance for the water-circulating coil results in the following set of equations. Heat lost by the exhaust air from facility = heat gained by circulating water qHRV,W = m_ EA cpa ðT EA,i - T EA,e Þ = m_ w cpw ΔT w ) m_ EA cpa ðT EA,i - T EA,e Þ ΔT w = m_ w cpw lbm d a Btu 0:24 700 F - 520 F min lbm d a‐0 F = 18:70 F gal lbm H2 O Btu 13 × 8:34 1:0 min gal lbm H2 O‐0 F

468:75 =

C. Since the mass flow rates of the outside air (OA) and exhaust air (EA) are equal,

8.4

Energy Recovery Systems

325

m_ EA cpa = m_ OA cpa ) lbm d a min

CEA = C OA = C min = 468:75

0:24

Btu lbm d a‐0 F

= 112:5 Btu= min ‐0 F Calculate the effectiveness of the heat recovery ventilator using Eq. 8.8.

e HRV ,W =

CEA (TEA,i − TEA,e )

Cmin (TEA,i − TOA,i )

=

700 F − 520 F = 0.72 ( 72% ) 700 F − 450 F

Example 8.5 A plate exchanger, used as an HRV, has an effectiveness of 50%. 0.24 m3/s of fresh outside air at 32 °C DB and 80% rh enters the HRV on one side and on the other side 0.20 m3/s of return / exhaust air from the facility at 24 °C and 65% rh enters the HRV. Standard heat capacity of air can be assumed for both the streams. Determine: A. the conditions (DB and rh) of the air entering the air handling unit, B. the heat exchange rate achieved in the HRV, C. the conditions (DB and rh) of the exhaust air. (Solution) A. Locate the state points of outside air (OA, 32 °C DB, 80% rh) and return exhaust air (EA, 24 °C DB, 65% rh) on the psychrometric chart and determine the specific volume of each entity as shown. From the excerpt of the psychrometric chart shown here, vOA = 0:898 m3 =kg d a, and vEA = 0:857 m3 =kg d a Calculate the mass flow rates of the outside air and exhaust air using the preceding values of specific volumes and the given volumetric flow rates. 0:24 ms V_ m_ OA = OA = = 0:2673 kg d a=s 3 vOA 0:898 kgmd a 3

0:20 ms V_ EA = = 0:2334 kg d a=s 3 vEA 0:857 kgmd a 3

m_ EA =

326

8

HVAC Systems and Equipment

Calculate CEA and COA as shown. COA = m_ OA cpa = 0:2673

kg s

1:0

kJ = 0:2673 kW= ° C kg: ° C

CEA = m_ EA cpa = 0:2334

kg s

1:0

kJ = 0:2334 kW= ° C kg: ° C

Therefore, Cmin = C EA = 0:2334 kW= ° C

8.4

Energy Recovery Systems

327

Calculate the temperature of outside air leaving the heat recovery unit (HRV) and entering the air handling unit (AHU) using Eq. 8.4 for summer operation. C OA ðT OA,i - T OA,e Þ ) Cmin ðT OA,i - T EA,i Þ kW 0:2673 ð32 ° C - T OA,e Þ °C ) T OA,e = 28:5 ° C 0:50 = kW 0:2334 ð32 ° C - 24 ° CÞ °C

εHRV,S =

The moisture content of the outside air does not change during this cooling process in an HRV. Draw a vertical line at 28.5 °C DB to intersect with the constant humidity ratio (ω) horizontal line of the outside air. This intersection point is the exit state point of the outside air as well as the state point of the air entering the AHU. From the excerpt of the psychrometric chart shown earlier in the solution, the relative humidity at this intersection point is 96%. Therefore, the conditions (DB and rh) of the air entering the air handling unit are TDB = 28.5 ° C, rh = 96%. B. Calculate the actual heat exchange rate achieved in the HRV using Eq. 8.2 qHRV,S = m_ OA cpa ðT OA,i - T OA,e Þ = C OA ðT OA,i - T OA,e Þ kW ð32 ° C - 28:5 ° CÞ °C = 0:9356 kW ð936 WÞ

= 0:2673

C. Calculate the exit temperature of the exhaust air using Eq. 8.2. qHRV,S = m_ EA cpa ðT EA,e - T EA,i Þ = CEA ðT EA,e - T EA,i Þ ) 0:9356 kW = 0:2334

kW ðT EA,e - 24 ° CÞ ) T EA,e = 28 ° C °C

The moisture content of the return air does not change during this heating process in an HRV. Draw a vertical line at 28 °C DB to intersect with the constant humidity ratio (ω) horizontal line of the exhaust air. This intersection point is the state point of the exhaust air. From the excerpt of the psychrometric chart shown earlier in the solution, the relative humidity at this intersection point is 52%. Therefore, the conditions (DB and rh) of the exhaust air are T DB = 28 ° C, rh = 52%:

328

8.4.2

8

HVAC Systems and Equipment

Energy Recovery Ventilators (ERV): Enthalpy Wheel

An energy recovery ventilator affects the moisture content as well as temperature of the streams flowing through the energy recovery device [6, 7]. An example of an ERV is an Enthalpy Wheel. An enthalpy wheel is similar to a sensible wheel and the core material matrix in the wheel is coated with a desiccant (moisture-absorbing material) such as silica gel. A schematic diagram of an enthalpy wheel is shown in Fig. 8.10.

8.4.2.1

Energy Recovery Ventilator Analysis

In the analysis of ERV, calculations must include both sensible and latent heat transfer rates [6, 7]. Subscript ‘Sen’ represents sensible heat transfer rate, subscript ‘L’ represents latent heat transfer rate, and subscript ‘T’ represents the total heat transfer rate achieved in the ERV. The symbol ‘ω’ represents the humidity ratio as per the normal convention. Subscript ‘i’ represents inlet conditions of the streams and subscript ‘e’ represents exit conditions of the streams. During summer operation, hot, humid outside air is cooled and dehumidified by the exhaust, return air from the facility. As a result, cooler, less humid air is supplied to the air handling unit (AHU), thereby reducing the cooling load on the AHU. The schematic diagram of the enthalpy wheel for summer operation is shown in Fig. 8.11. The total heat transfer during summer operation is

Fig. 8.10 Schematic diagram of an enthalpy wheel

8.4

Energy Recovery Systems

329

Fig. 8.11 Enthalpy wheel – summer operation

qT,S = m_ OA ðhOA,i - hOA,e Þ = m_ EA ðhEA,e - hEA,i Þ

ð8:9Þ

The sensible heat transfer during summer operation is qSen,S = m_ OA cpa ðT OA,i - T OA,e Þ = m_ EA cpa ðT EA,e - T EA,i Þ

ð8:10Þ

The latent heat transfer during summer operation is qL,S = m_ OA hfg,OA ðωOA,i - ωOA,e Þ = m_ EA hfg,EA ðωEA,e - ωEA,i Þ

ð8:11Þ

In Eq. 8.11, hfg is the enthalpy of condensation/enthalpy of vaporization at the dew point of the air stream. The total heat transfer is the sum of the sensible and latent heat transfer. For summer operation, qT,S = qSen,S þ qL,S

i,

ð8:12Þ

For summer operation, the maximum possible enthalpy difference is hOA, i - hEA, and m_ min represents the lower value of m_ OA and m_ EA . Therefore, the maximum possible total heat transfer in summer is qT max ,S = m_ min ðhOA,i - hEA,i Þ

ð8:13Þ

The total effectiveness of an ERV is defined as the ratio of actual heat transfer to the maximum possible heat transfer. Therefore, for summer operation,

330

8

HVAC Systems and Equipment

Fig. 8.12 Enthalpy wheel – winter operation

εT,ERV,S =

qT,S - hOA,e Þ - hEA,i Þ m_ ðh m_ ðh = OA OA,i = EA EA,e qT max ,S m_ min ðhOA,i - hEA,i Þ m_ min ðhOA,i - hEA,i Þ

ð8:14Þ

If needed, the sensible effectiveness and the latent effectiveness of an ERV can be calculated in a similar manner by using the relevant equations. However, since an ERV also affects the moisture content of the air streams, the moisture effectiveness of an ERV can be defined as the ratio of the actual rate of condensation to the maximum possible rate of condensation. The maximum possible difference in moisture content for summer operation is ωOA, i - ωEA, i. Therefore, the formula for calculating the moisture effectiveness for summer operations is εmoist:,S =

m_ cond

m_ cond, max

=

- ωEA,i Þ m_ OA ðωOA,i - ωOA,e Þ m_ ðω = EA EA,e m_ min ðωOA,i - ωEA,i Þ m_ min ðωOA,i - ωEA,i Þ

ð8:15Þ

The schematic diagram of the enthalpy wheel for winter operation, assuming cold, dry outside air is shown in Fig. 8.12. Equations 8.9 through 8.15, applicable for summer operations can be modified and written as follows for winter operations. Subscript W represents winter operations. The total heat transfer during winter operation is qT,W = m_ OA ðhOA,e - hOA,i Þ = m_ EA ðhEA,i - hEA,e Þ

ð8:16Þ

The sensible heat transfer during winter operation is qSen,W = m_ OA cpa ðT OA,e - T OA,i Þ = m_ EA cpa ðT EA,i - T EA,e Þ

ð8:17Þ

8.4

Energy Recovery Systems

331

The latent heat transfer during winter operation is qL,W = m_ OA hfg,OA ðωOA,e - ωOA,i Þ = m_ EA hfg,EA ðωEA,i - ωEA,e Þ

ð8:18Þ

In Eq. 8.18, hfg is the enthalpy of condensation/enthalpy of vaporization at the dew point of the air stream. The total heat transfer is the sum of the sensible and latent heat transfer. For winter operation, qT,W = qSen,W þ qL,W

ð8:19Þ

For winter operation, the maximum possible enthalpy difference is hEA, i - hOA, i, and m_ min represents the lower value of m_ OA and m_ EA . Therefore, the maximum possible total heat transfer in winter is qT max ,W = m_ min ðhEA,i - hOA,i Þ

ð8:20Þ

The total effectiveness of an ERV is defined as the ratio of actual heat transfer to the maximum possible heat transfer. Therefore, for winter operation, εT,ERV,W =

qT,W - hOA,i Þ m_ ðh m_ ðh - hEA,o Þ = OA OA,e = EA EA,i qT max ,W m_ min ðhEA,i - hOA,i Þ m_ min ðhEA,i - hOA,i Þ

ð8:21Þ

If needed, the sensible effectiveness and the latent effectiveness of an ERV can be calculated in a similar manner by using the relevant equations. The maximum possible difference in moisture content for winter operations is ωEA, i - ωOA, i. Therefore, the formula for calculating the moisture effectiveness for winter operations is εmoist:,W =

m_ cond

m_ cond, max

=

m_ OA ðωOA,e - ωOA,i Þ m_ ðω - ωEA,e Þ = EA EA,i m_ min ðωEA,i - ωOA,i Þ m_ min ðωEA,i - ωOA,i Þ

ð8:22Þ

Example 8.6 During summer, an enthalpy wheel is used in pre-cooling outside air at 92 °F DB and 75% rh using 10,000 cfm of ventilation air from the facility at 70 °F DB and 60% rh. The total effectiveness of the enthalpy wheel is 80% and volume flow of outside air used is the same as the volume flow of the ventilation air. 20% of the total heat transfer achieved in the wheel is sensible heat transfer. The COP for the cooling system is 3.15. The cost of energy is $0.091/kWh. The projected annual usage of the cooling system is 2600 hours. Determine: A. the tons of cooling (TR) achieved by the enthalpy wheel, B. the moisture effectiveness of the enthalpy wheel,

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HVAC Systems and Equipment

C. the reduction in compressor power (kW) achieved due to reduction in system cooling load by using the enthalpy wheel, D. the annual savings in power costs. (Solution) Locate the state points of outside air inlet (OAi, 92 °F DB, 75% rh) and inlet exhaust air (EAi, 70 °F DB, 60% rh) on the psychrometric chart as shown. From the excerpt of the chart shown

hOA,i = 49 Btu=lbm d a, vOA,i = 14:45 ft3 =lbm d a hEA,i = 27 Btu=lbm d a,

vEA,i = 13:55 ft3 =lbm d a

Calculate the mass flow rates of the outside air and exhaust air using their respective specific volumes. ft 10000 min V_ OA,i = = 692:04 lbm d a= min 3 vOA,i 14:45 lbmft d a 3

m_ OA =

ft 10000 min V_ = 738:00 lbm d a= min m_ EA = EA,i = 3 vEA,i 13:55 lbmft d a 3

Therefore, m_ min = m_ OA = 692:04 lbm d a= min

8.4

Energy Recovery Systems

333

A. Calculate the total heat transfer (cooling) achieved by the wheel using the given total effectiveness of the wheel and Eq. 8.14. εT,ERV,S =

qT,S - hOA,e Þ m_ ðh = OA OA,i ) qT max ,S m_ min ðhOA,i - hEA,i Þ

qT,S = ðεT,ERV,S Þ qT max ,S = ðεT,ERV,S Þðm_ min ðhOA,i - hEA,i ÞÞ lbm d a min = 12, 180 Btu= min = ð0:80Þ 692:04

49

Btu Btu - 27 lbm d a lbm d a

Convert the cooling achieved to TR.

qT,S = ð12180 Btu= min Þ

1 TR 200 Btu= min

= 60:9 TR

B. Calculate the exit enthalpy of the outside air using Eq. 8.9. qT,S = m_ OA ðhOA,i - hOA,e Þ ) Btu lbm d a = 692:04 min min ) hOA,e = 31:40 Btu=lbm d a

12180

49

Btu - hOA,e lbm d a

Since 20% of the total heat transfer achieved is sensible heat transfer, calculate the sensible heat transfer achieved during summer operation of the energy wheel. qSen,S = 0:2qT,S = 0:2 12180

Btu = 2436 Btu= min min

Calculate the exit temperature of outside air using Eq. 8.10. qSen,S = m_ OA cpa ðT OA,i - T OA,e Þ ) 2436

Btu lbm d a = 692:04 min min

0:24

∘ Btu ð92 F - T OA,e Þ ∘ lbm d a_ F

∘

) T OA,e = 77:3 F On the excerpt of the psychrometric chart, locate the state point of the exit outside air, OA,e, at the intersection of. hOA,e = 31.40 Btu/lbm d a and TOA,e = 77.3 ° F.

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HVAC Systems and Equipment

Determine the following humidity ratios (lbm H2O/lbm d a) from excerpt of the psychrometric chart as shown. ωOA,i = 0:0248, ωOA,e = 0:0116, ωEA,i = 0:0097 Determine the moisture effectiveness for summer operation using Eq. 8.15. Note that, m_ OA = m_ min .

e moist ., S =

m& OA (wOA,i - wOA,e )

m& min (wOA,i - wEA,i ) lbm H 2 O lbm H 2 O - 0.0116 lbm d a lbm d a = 0.87 ( 87% ) = lbm H 2 O lbm H 2 O 0.0248 - 0.0097 lbm d a lbm d a 0.0248

C. Using the definition of COP and its given value, calculate the kW of compressor power savings achieved due to the use of the enthalpy wheel.

COP =

kW qcooling,wheel = qT ,s 60:9 TR × 3:517 1 TR _ c = 68 kW ) 3:15 = )W _c _c W W

D. Calculate the annual savings in power costs due to the use of the enthalpy wheel using the given information.

Annual savings = ð68 kWÞ

2600 h yr

$0:091 = $16, 089 kWh

Example 8.7 An enthalpy wheel used in summer conditions processes 2.8 m3/s of outside air at 32 °C DB and 80% rh to an exit condition of 27 °C DB and 70% rh. 2.5 m3/s of exhaust air from the facility maintained at 21 °C DB and 60% rh enters the wheel. Determine: A. the total effectiveness of the enthalpy wheel, B. the sensible effectiveness of the enthalpy wheel, C. the temperature, enthalpy, and humidity ratio of the exhaust air leaving the enthalpy wheel, D. the latent effectiveness of the enthalpy wheel, E. the tons of cooling achieved by the enthalpy wheel.

8.4

Energy Recovery Systems

335

(Solution) Locate the state points of inlet outside air (OAi, 32 °C DB, 80% rh), exit outside air (OAe, 27 °C DB, 70% rh) and inlet exhaust air (EAi, 21 °C DB, 60% rh) on the excerpt of the psychrometric chart as shown.

From the excerpt of the chart shown, hOA,i = 95 kJ=kg d a, vOA,i = 0:897 m3 =kg d a ωOA,i = 0:0243 kg H2 O=kg d a hEA,i = 45 kJ=kg d a, vEA,i = 0:845 m3 =kg d a ωEA,i = 0:0095 kg H2 O=kg d a hOA,e = 67 kJ=kg d a, ωOA,e = 0:0157 kg H2 O=kg d a Calculate the mass flow rates of the outside air and exhaust air using their respective specific volumes. 3 2:8 ms V_ OA,i m_ OA = = = 3:121 kg d a=s 3 vOA,i 0:897 kgmd a

2:5 ms V_ EA,i = = 2:959 kg d a=s 3 vEA,i 0:845 kgmd a 3

m_ EA =

Since m_ EA < m_ OA , m_ EA = m_ min = 2:959 kg d a=s

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HVAC Systems and Equipment

A. Calculate the total effectiveness of the enthalpy wheel using Eq. 8.14. m_ OA ðhOA,i - hOA,e Þ m_ min ðhOA,i - hEA,i Þ kg d a kJ kJ 3:121 95 - 67 s kg d a kg d a = kJ kg d a kJ 2:959 95 - 45 kg d a s kg d a

εT,ERV,S =

= 0:59ð59%Þ

B. Calculate the sensible effectiveness of the enthalpy wheel using an equation similar Eq. 8.14.

e Sen, ERV, S =

m& OA c pa (TOA,i - TOA,e )

m& min c pa (TOA,i - TEA,i )

kg d a ) 0 ( 0 | 3.121 | ( 32 C - 27 C ) s ) ( = = 0.48 ( 48% ) kg d a ) 0 ( 0 2.959 32 C 21 C ( ) | | s ) (

C. Calculate the total heat transferred by the enthalpy wheel using the first part of Eq. 8.9. qT,S = m_ OA ðhOA,i - hOA,e Þ = 3:121

kg d a s

95

kJ kJ = 87:388 kW - 67 kg d a kg d a

Calculate the enthalpy of the exit exhaust air using the second part (total heat balance) of Eq. 8.9. qT,S = m_ EA ðhEA,e - hEA,i Þ ) kg d a s hEA,e = 74:5 kJ=kg d a

87:388 kW = 2:959

hEA,e - 45

kJ kg d a

)

8.4

Energy Recovery Systems

337

Calculate the sensible heat transferred by the enthalpy wheel using the first part of Eq. 8.10. qSen,S = m_ OA cpa ðT OA,i - T OA,e Þ = 3:121

kg d a s

1

∘ ∘ kJ ð32 C - 27 CÞ = 15:605 kW kg:∘ C

Calculate the temperature of the exit exhaust air using the second part (sensible heat balance) of Eq. 8.9. qSen,S = m_ EA cpa ðT EA,e - T EA,i Þ ) 15:605 kW = 2:959

kg d a s

1

kJ kg:∘ C

∘

T EA,e - 21 C )

∘

T EA,e = 26:3 C On the excerpt of the psychrometric locate the state point of exit exhaust air, EAe, at the intersection of its enthalpy and temperature EAe : hEA,e = 74:5 kJ=kg d a, T EA,e = 26:3 ° C From the excerpt of the psychrometric chart, ωEA,e = 0:0193 kg H2 O=kg d a Therefore, the conditions of the exit exhaust air are: T EA,e = 26:3 ° C, hEA,e = 74:5 kJ=kg d a ωEA,e = 0:0193 kg H2 O=kg d a D. m_ EA < m_ OA , m_ EA = m_ min = 2:959 kg d a=s From the excerpt of the psychrometric chart, the dew point of the inlet outside air is. TDP, OA, i = 28 ° C. From steam tables or online resources. https://www.tlv.com/global/TI/calculator/steam-table-temperature.html hfg,OA = 2434:6 kJ=kg at 28 ° C From the excerpt of the psychrometric chart, the dew point of the inlet exhaust air is TDP, EA, i = 13 ° C. From steam tables or online resources. https://www.tlv.com/global/TI/calculator/steam-table-temperature.html

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HVAC Systems and Equipment

hfg, min = 2470:1 kJ=kg at 13 ° C Calculate the latent effectiveness of the enthalpy wheel using an equation similar Eq. 8.14.

εL,ERV,S =

m_ OA hfg,OA ðωOA,i - ωOA,e Þ m_ min hfg, min ðωOA,i - ωEA,i Þ kg d a 3:121 s

kJ 2434:6 kg H2 O

= 2:959

kg d a s

2470:1

kJ kg H2 O

0:0243

kg H2 O kg d a

kg H2 O kg d a kg H2 O 0:0243 kg d a - 0:0157

- 0:0095

kg H2 O kg d a

= 0:60ð60%Þ E. In the solution to part C, the total cooling achieved by the enthalpy wheel was calculated. Convert the total cooling achieved to tons of refrigeration (TR). qT,S = 87:388 kW ×

8.5

1 TR = 24:85 TR 3:517 kW

Acoustics and Noise Control

Sound is caused by a disturbance in an elastic medium (solid, liquid, or gas) that can be detected by the human ear. The disturbance creates a pressure wave that acts on the inner ear, resulting in sound that we hear . The amplitude of the sound wave represents the loudness and is measured in decibels (dB). Decibel is proportional to a logarithmic ratio of actual sound pressure to the reference pressure of 20 μPa (0.00002 Pa). Decibel can also be considered in terms of the logarithmic ratio of the actual sound power in watts (W) to the reference sound power of 10 -12 W. Sound levels in decibels can be calculated by using the following defining equations, where P is pressure in pascals in Eq. 8.23 and W is the power in watts in Eq. 8.24. dBPr: = LP = 20 log

P 0:00002

ð8:23Þ

8.5

Acoustics and Noise Control

dBPower = LW = 10 log

339

W 10 - 12

ð8:24Þ

For points of reference, the sound level (with reference to 10 -12 W) of a human voice is in the range of 30 dB (whisper) to 70 dB (conversation) and the sound level of a jet at take-off is about 160 dB. The sound level of a large HVAC fan running is about 110 dB. The sources of noise in HVAC systems are fans, air turbulence, equipment vibrations, pumps, chillers, and diffusers. Different sound rating methods are used in acoustic evaluation of HVAC systems and their components. The dBA rating is used in assessing equipment such as cooling towers, and chillers. The Noise Criteria (NC) rating is used in assessing air terminals and diffusers, and the Room Criteria (RC) is commonly used for ducts.

8.5.1

Sound Control and Noise Reduction Methods

There are a number of different methods available to mitigate noise and sound in HVAC systems. Some of them are listed here. . Reduce the noise at the source by reducing/absorbing noise-generating vibration of equipment. Such noise, when reduced at the source, will not propagate along the ductwork. . Line the interior of the ductwork with a sound-absorbing material such as fiberglass or foam. The lining material must have high sound absorption and must not be flammable. . Wrap ducts and pipes with sound-proofing material. The add-on materials used in reducing noise are known as insertions. The reduction in sound achieved by jacketing, lining, and/or lagging ducts is known as insertion loss. Note: Tables 17 to 25, Chapter 49 of the ASHRAE Applications Handbooks [6, 7] provide the values of insertion losses for different types of ducts over a range of frequencies. The approximate design sound levels recommended for HVAC-related background noise for most facilities is in the range 35 dbA to 45 dbA for most facilities and much lower for places like theaters and music studios. More details can be obtained from Table 1, Chapter 49, ASHRAE Applications Handbooks [3, 4].

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8

HVAC Systems and Equipment

Example 8.8 A supply fan generates the following sound levels at the given frequency bands in a 1.5 ft. × 1.0 ft. rectangular duct. A 10 ft. segment of this duct is lined with 1 in. fiberglass sound-proofing insulation. Determine the sound levels at the end of the lined segment. Frequency, Hz LW (dBW)

125 98

250 96

500 93

1000 89

2000 84

4000 81

(Solution) Obtain the insertion losses at the given frequency bands from Table 17: Insertion Loss for Rectangular Sheet Metal Ducts with 1 in. Fiberglass Lining in Chapter 49, pg. 49.22, ASHRAE Applications Handbook, 2019, I P Edition [5], for an 18 in × 12 in duct. Calculate the sound levels after 10 ft. of the lined segment using the following equation. Sound Level, LW,outlet = LW,inlet -

insertion loss ð10 ftÞ ft

The following table summarizes the values of the insertion losses obtained from the ASHRAE Table mentioned above along with the sound levels at the outlet of the lined segment. All sound levels are in decibels watts sound power. Frequency, Hz LW (dBW), fan outlet Insertion Loss (dBW/ft) [5] LW (dBW), outlet

8.6

125 98 0.3 95

250 96 0.7 89

500 93 1.7 76

1000 89 3.7 52

2000 84 3.5 49

4000 81 2.5 56

Vibration and Vibration Control

Mechanical vibration and vibration-induced noise do occur during the operation of HVACR systems. Vibration is caused by reciprocating or rotary motion of components within mechanical equipment. HVACR equipment such as motors, compressors, pumps, fans, chillers, and even packaged HVACR equipment are sources of vibration. The most commonly used strategy in controlling or reducing the undesired effects of vibration is Vibration Isolation. This strategy decouples the source of vibration, that is, the vibrating machine from the structure it is mounted on. For example, consider an air compressor mounted on a concrete floor. Vibration of the compressor

8.7

Constant Air Volume (CAV) and Variable Air Volume (VAV) Systems

341

causes the vibration effects to travel along the floor creating vibration and noise in the adjacent areas. Putting rubber pads under the compressor support legs will isolate the machine from the floor. This results in the absorption and mitigation of the physical action of the machine vibrating against the floor and consequently the vibration source is isolated from the rest of the system. Thus, vibration isolation is physically decoupling a vibrating unit from another substrate. In addition to padding, other types of isolation methods, include pneumatic shock absorbers, floor springs, and hangers. Vibration isolators must be selected in a such a way that in addition to providing the required isolation, it must be ensured that the natural frequency of the isolated unit is not close to the natural frequency of the system or the base in order to prevent detrimental resonance effects. Table 47, Chapter 49 of the ASHRAE Applications Handbook [3, 4], provides selection guidelines for vibration isolation. This table includes different types of bases such as structural steel, railings, concrete base, and no base, where the isolators are directly attached to the equipment. The proper isolation system can be selected from different types such as rubber/fiberglass pads, rubber mounts, floor-mounted spring isolators/restrained spring isolators, and spring hangers. Example 8.9 What is the recommended method of vibration isolation for a supply fan with a capacity of 700 cfm where the ducted system is suspended from a ceiling with a span of 20 ft. (Solution) The following excerpt can be found in Table 47: Selection Guide for Vibration isolation, Chapter 49, pg. 49.47, ASHRAE Applications Handbook [5]. Equipment type Ducted rotating equipment, small fans and fan powered boxes

Capacity >601 cfm

Base type A

isolator type 3

Min. deflection, in. 0.75 in.

From the notes for this table, base type A is ‘no base, isolators directly attached to equipment’, which is applicable in this case since the ducted system is suspended from the ceiling. The isolator type recommended is Type 3. Again, from the table notes, type 3 isolators are ‘spring floor isolator or hanger’. In this case, a spring hanger would be appropriate for a ducted system suspended from the ceiling.

8.7

Constant Air Volume (CAV) and Variable Air Volume (VAV) Systems

There are two strategies available to meet fluctuating cooling and heating loads of facilities and conditioned space. The first strategy is to supply a constant volume flow rate of conditioned air to the facility from the cooling/heating coil (AHU) and

342

8 HVAC Systems and Equipment

the second strategy is to vary the volume flow rate of the conditioned supply air to the facility. The details and comparisons of the two strategies are presented in the following sub-sections.

8.7.1

Constant Air Volume Systems

In constant air volume systems (CAV), a constant flow of supply air from the cooling/heating coil is introduced into the conditioned space in order to meet the desired cooling or heating load. Fluctuating loads in a given conditioned zone are accommodated by varying the flow rates of either the coolant (chilled water or refrigerant) or the heating fluid (hot water or steam), which will result in required changes to the temperature and condition of the supply air. CAV systems are more suited single-zone applications where the load in a given zone remains fairly constant. Examples of single-zone applications are movie theatres and performance auditoriums. CAV systems become less energy efficient when used in multi-zone applications. Consider a CAV system being used in a two-zone application. The thermostat in zone A is set at 65 °F, while zone B is required to be maintained at 70 ° F during the summer months. The supply temperature of the air leaving the cooling coil must be designed to meet the needs of the zone with the lowest thermostat setting, which in this case is zone A. In order to meet the needs of zone B with a higher set temperature, the supply air will have to be reheated in the duct. This reduces the energy efficiency of the system due to two reasons – extra energy is used in reheating the air and the energy used in cooling the supply air to less than the required value for zone B is wasted. The net result is that air that is overcooled has to be reheated again, resulting in a double dose of energy wastage. The following examples illustrate calculations for a constant air volume (CAV) system. Example 8.9 The total design cooling load for a given space is 40,000 Btu/hr. with a sensible heat ratio (SHR) of 0.50. The inlet air to the air handling unit (AHU) is at 75 °F DB and 70% relative humidity and the supply air is at 55 °F DB. The AHU is designed for constant air volume (CAV) flow. The cooling is achieved by chilled water with a temperature difference of 14 °F across the AHU. Subsequently, the cooling load for the space drops down to 35,000 Btu/hr. with the same SHR. Bypass effects can be ignored. Determine A. the air circulation rate (cfm at AHU inlet) for the design cooling load, B. the chilled-water circulation rate required in gpm for the design cooling load, C. the conditions of the supply air (DB, rh, enthalpy, and humidity ratio) for the design cooling load. D. the conditions of the supply air (DB, rh, enthalpy, and humidity ratio) for the revised cooling load.

8.7

Constant Air Volume (CAV) and Variable Air Volume (VAV) Systems

343

(Solution) A. Calculate the sensible cooling load for the design condition using Eq. 6.4. SHR =

qS qS ) 0:5 = ) qS = 20, 000 Btu=hr qT 40000 Btu hr

Subscript ‘1’ represents AHU inlet and subscript ‘2’ represents supply air at AHU outlet. The conditions of air at AHU inlet are such that standard density of air can be used. Equation 6.7 uses the standard conditions of air. Calculate the cfm of air at AHU inlet using Eq. 6.7. qS = ð1:08ÞðcfmÞðT 1 - T 2 Þ ) Btu=hr Btu 20000 = 1:08 ðcfmÞð75 ° F - 55 ° FÞ ) hr cfm‐ ° F V_ 1 = 926 cfm B. Energy balance for the chilled-water coil results in the following set of equations. Heat removed from air (cooling load) = heat gained by chilled water) qT = m_ w cpw ΔT w Calculate the gpm of chilled water required using Eq. 7.24.

gpmw =

qT

Btu‐ min 500 gal‐hr‐ ° F × ΔT ° F

=

40000 Btu hr = 5:71 gpm Btu‐ min 500 gal‐hr‐ × 14 ° F °F

C. On the psychrometric chart, locate state point 1 (AHU inlet) at the intersection of 75 °F DB and 70% rh as shown in the following excerpt of the chart.

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8

HVAC Systems and Equipment

From the chart excerpt, at state point 1, h1 = 32:5 Btu=lbm d a, v1 = 13:8 ft3 =lbm d a Calculate the mass flow rate of air into the AHU. ft 926 min V_ 1 = = 67:10 lbm d a= min 3 v1 13:8 lbmft d a 3

m_ a =

Calculate the enthalpy of air leaving the AHU (at state point 2) using Eq. 6.1. qT = m_ a ðh1 - h2 Þ ) lbm d a 60 min Btu 40000 = 67:10 × hr min hr ) h2 = 22:6 Btu=lbm d a

32:5

Btu - h2 lbm d a

Locate state point 2 on the psychrometric chart at the intersection of T 2,DB = 55 ° F and h2 = 22:6 Btu=lbm d a From the chart excerpt, at state point 2, T 2,DB = 55 ° F, h2 = 22:6 Btu=lbm d, rh2 = 95% ω2 = 0:0088 lbm H2 O=lbm d a D. Since this is a constant air volume (CAV) system, the air flow into the AHU remains constant at the value calculated in part C. Therefore, from the solution to part C, V_ 1 = 926 cfm, and, m_ a = 67:10 lbm d a= min The inlet temperature to the AHU also remains at the same value as in part C, that is, T1, DB = 75 ° F. However, the sensible cooling changes due to change in the total cooling load to 35,000 Btu/hr. Let 2′ denote the state due to the revised cooling load. Calculate the sensible cooling load for the revised condition using Eq. 6.4.

8.7

Constant Air Volume (CAV) and Variable Air Volume (VAV) Systems

SHR =

345

qS qS ) 0:5 = ) qS = 17, 500 Btu=hr qT 35000 Btu hr

Calculate the temperature of air, T2', at AHU exit using Eq. 6.7. qS = ð1:08ÞðcfmÞðT 1 - T 2 ′ Þ ) Btu=hr Btu 17500 = 1:08 ð926 cfmÞð75 ° F - T 2 ′ Þ ) hr cfm‐ ° F T 2 ′ = 57:5 ° F At the revised cooling load, calculate the enthalpy of air leaving the AHU (at state point 2′) using Eq. 6.1. qT = m_ a ðh1 - h2 ′ Þ ) Btu lbm d a 60 min 35000 = 67:10 × hr min hr ) h2 = 23:8 Btu=lbm d a

32:5

Btu - h2 ′ lbm d a

Locate state point 2′ on the psychrometric chart at the intersection of T 2 ′ ,DB = 57:5 ° F and h2 ′ = 23:8 Btu=lbm d a From the chart excerpt, at state point 2, T 2 ′ ,DB = 57:5 ° F, h2 ′ = 23:8 Btu=lbm d, rh2 ′ = 91% ω2 ′ = 0:0092 lbm H2 O=lbm d a Example 8.10 A constant air volume (CAV) system is used in cooling multiple zones. The supply air is at 15 °C and 95% relative humidity, necessitating a reheat to 18 °C to satisfy the set temperature of one of the zones. The zone requirement is 2.4 m3/s of air at supply air conditions. Determine A. the rate of energy required for the reheat process, B. the additional compressor power required for cooling the supply air by additional, undesired 3 °C if the actual COP of the cooling system is 3.8. C. the additional daily costs incurred by the CAV system if it operates for an average of 10 hours a day and the cost of power is $1.11 per kWh. (Solution) A. Locate the state point of supply air (SA, 15 °C, 95% rh) on the psychrometric chart as shown here in the excerpt of the chart.

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HVAC Systems and Equipment

From the excerpt of the chart, the specific volume of the supply air is, vSA = 0:83 m3 =kg da Calculate the mass flow rate of the supply air, 2:4 ms V_ m_ SA = SA = = 2:89 kg d a=s 3 vSA 0:83 kgmd a 3

Calculate the sensible heat rate required to reheat the supply air from 15 °C to 18 °C. qs = m_ a cpa ðT 1 - T 2 Þ = 2:89

kg d a s

1:0

kJ ð18 ° C - 15 ° CÞ = 8:67 kW kg . ° C

B. The additional sensible cooling rate used during unnecessary cooling is also equivalent to the additional sensible heat rate required for reheating. Calculate the additional compressor power required for unnecessary cooling using the definition of COP for cooling (Eq. 7.25).

8.7

Constant Air Volume (CAV) and Variable Air Volume (VAV) Systems

347

Q_ in ) _ W c,actual 8:67 kW _ c,actual = 2:28 kW )W 3:8 = _ c,actual W

COPref:=AC =

C. Calculate the additional daily costs incurred by the CAV system by adding up the reheat rate required, and the additional compressor power used in unnecessary cooling. Additional daily cost = ð8:67 kW þ 2:28 kWÞ

8.7.2

10 h day

$1:11 = $121:54=day kWh

Variable Air Volume Systems

In contrast to constant air volume systems, variable air volume systems have the flexibility to adjust the supply air volume to meet the changing demands of conditioned space(s). This is accomplished through the use of appropriate feed-back control strategy (as illustrated in Figure1, Chap. 7, ASHRAE Fundamentals Handbooks [1, 2]) and operation of supply air damper in a VAV Box/Terminal Unit (as illustrated in Figures 24, 27, and 28, Chapter 48, ASHRAE Applications Handbooks [3, 4]). Thus, in VAV systems, the supply air temperature is maintained at a constant value and the flow of the supply air is adjusted to match the thermostat settings of a particular space or zone. Since air flow is adjusted to maintain the temperature at a desired value, the need for reheating is eliminated, making the system more energy efficient. Please refer to the solution to Example 8.10, where the additional reheat energy cost calculations are illustrated. VAV systems are ideally suited for multizone applications such as malls, hotels, and office buildings. Example 8.11 For various reasons, the sensible cooling load for a space served by a VAV system varies from a maximum value of 12,000 Btu/hr. to a minimum value of 10,000 Btu/ hr. The supply air temperature from the air handling unit is 52 °F and air enters the AHU at 70 °F. Determine the minimum and maximum flow (cfm) settings to be used in the VAV box of the system. (Solution) Subscript ‘1’ refers to AHU inlet and subscript ‘2’ refers to AHU exit, that is, supply air.

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8

HVAC Systems and Equipment

Calculate the minimum and maximum flow (cfm) settings to be used in the VAV box of the system using Eq. 6.7. qS, min = ð1:08Þðcfmmin ÞðT 1 - T 2 Þ ) Btu=hr Btu ðcfmmin Þð70 ° F - 52 ° FÞ ) = 1:08 hr cfm‐ ° F cfmmin = 514:4 cfm

10000

qS, max = ð1:08Þðcfmmax ÞðT 1 - T 2 Þ ) Btu=hr Btu ðcfmmax Þð70 ° F - 52 ° FÞ ) = 1:08 hr cfm‐ ° F cfmmax = 617:3 cfm

12000

Practice Problems Practice Problem 8.1 A sensible heat recovery run-around coil is used in preheating 10,000 cfm of outside winter air at 40 °F DB and 50% rh to a temperature of 60 °F DB. It is estimated that the sensible heat exchange effectiveness of the run around coil is 70%. If an equivalent mass flow rate of air from the facility is to be ventilated, determine A. the temperature of the air in the facility, B. the actual heat transfer achieved by the system, C. the gpm of the circulating water required if the temperature increase of the water is limited to 10 °F, D. cfm of return / exhaust air if the facility is maintained at 60% rh. Practice Problem 8.2 An enthalpy wheel with a moisture effectiveness of 70% is used in processing 2000 cfm of cold, dry winter air at 40 °F and 30% rh to a pre-heated air temperature of 55 ° F. 2000 cfm of return / exhaust air at 68 °F and 50% rh enters the wheel. Determine A. the humidity ratio and the relative humidity of pre-heated air supplied to the air handling unit, B. the kW of heating achieved by the wheel, C. the conditions of the exhaust air leaving the wheel. Practice Problem 8.3 0.8 m3/s of winter air at 6 °C DB and 3 °C WB is preheated using an equivalent volume flow of facility return / exhaust air at 20 °C and 50% rh. The heat exchange is accomplished in an ERV (enthalpy wheel). The total and sensible effectiveness of the ERV are equal at 85%. Determine

Solutions to Practice Problems

349

A. the conditions (DB temperature, enthalpy, relative humidity, and humidity ratio) of both the air streams exiting the ERV. B. the moisture effectiveness of the ERV, C. the annual savings in energy costs due to the use of ERV if power costs are $1.03 per kWH and the annual usage of the heating system is estimated to be 1000 hours.

Solutions to Practice Problems Practice Problem 8.1 (Solution) Determine the specific volume of inlet outside air by locating its state point (OA, 40 ° F DB, 50% rh) on the psychrometric chart as shown in the excerpt of the chart below.

vOA = 12:7 ft3 =lbm d a ft 10000 min V_ OA = = 787:40 lbm d a= min 3 vOA 12:7 lbmft d a 3

m_ OA =

The mass flow rates of the outside air and exhaust air from the facility are equal. Therefore, m_ EA = m_ OA = 787:40 lbm d a= min

350

8

HVAC Systems and Equipment

A. Since the mass flow rates of the outside air (OA) and exhaust air (EA) are equal, m_ EA cpa = m_ OA cpa ) CEA = C OA = C min = 787:40

lbm d a min

0:24

Btu lbm d a‐0 F

= 188:98 Btu= min ‐0 F Calculate the temperature of air from the facility entering the HRV using Eq. 8.8.

e HRV ,W = 0.70 =

COA (TOA,e - TOA,i )

Cmin (TEA,i - TOA,i )

=

600 F - 400 F = TEA,i = 68.57 0 F TEA,i - 400 F

B. Calculate the actual heat transfer achieved by the system using Eq. 8.6. qHRV,W = m_ OA cpa ðT OA,e - T OA,i Þ = C OA ðT OA,e - T OA,i Þ Btu ð60 ° F - 40 ° FÞ = 188:98 min‐ ° F = 3779:6 Btu= min C. Energy balance for the run-around coil where it transfers heat to cold outside air results in the following set of equations. qHRV, W= heat gained by the outside air = heat lost by the circulating water = qHRV,W = m_ w cpw ΔT w ) q m_ w = HRV,W = cpw ΔT w

3779:6

Btu min

Btu 1:0 ð10 ° FÞ lbm H2 O‐ ° F

= 377:96 lbm H2 O= min

Convert the mass flow rate of water to gpm using the density of water. 377:96 lbm m_ min = 45:32 gpm V_ w = w = ρw 8:34 lbm gal

Solutions to Practice Problems

351

D. Determine the specific volume of return / exhaust air by locating its state point (EA, 69 °F DB, 60% rh) on the psychrometric chart as shown. From the excerpt of the psychrometric chart shown, vEA = 13:52 ft3 =lbm d a lbm d a V_ EA = m_ EA vEA = 787:40 min

13:52

ft3 lbm d a

= 10, 646 ft3 = minðcfmÞ Practice Problem 8.2 (Solution) Locate the state points of inlet outside air (OAi, 40 °F DB, 30% rh), and inlet exhaust air (EAi, 68 °F DB, 50% rh) on the psychrometric chart as shown.

From the excerpt of the chart shown, hOA,i = 11:3 Btu=lbm d a, vOA,i = 12:65 ft3 =lbm d a ωOA,i = 0:0016 lbm H2 O=lbm d a hEA,i = 24:3 Btu=lbm d a, vEA,i = 13:46 ft3 =lbm d a ωEA,i = 0:0073 lbm H2 O=lbm d a Calculate the mass flow rates of the outside air and exhaust air using specific volumes. ft 2000 min V_ OA,i = = 158:10 lbm d a= min 3 vOA,i 12:65 lbmft d a 3

m_ OA =

ft 2000 min V_ = 148:70 lbm d a= min m_ EA = EA,i = 3 vEA,i 13:45 lbmft d a 3

Since m_ EA < m_ OA , m_ EA = m_ min = 148:70 lbm d a= min

352

8

HVAC Systems and Equipment

A. Using the given moisture effectiveness for winter operation, determine the humidity ratio of the exit outside air (supply air to AHU) using Eq. 8.22. m_ OA ðωOA,e - ωOA,i Þ ) m_ min ðωEA,i - ωOA,i Þ lbm d a lbm H2 O ωOA,e - 0:0016 158:10 lbm d a min 0:70 = lbm d a lbm H2 O lbm H2 O 148:70 0:0073 - 0:0016 min lbm d a lbm d a ) ωOA,e = 0:0053 lbm H2 O=lbm d a εmoist:,W =

On the psychrometric chart, locate the state point of the exit outside air (supply air to AHU) at the intersection of T OA,e = 55 ° F, and ωOA,e = 0:0053 lbm H2 O=lbm d a At the state point of the exit outside air (supply air to AHU), the relative humidity is rhOA,e = 60% B. At the state point of the exit outside air (supply air to AHU), the enthalpy is hOA,e = 19:0 Btu=lbm d a Calculate the total heat transfer achieved using Eq. 8.16. qT,W = m_ OA ðhOA,e - hOA,i Þ lbm d a min = 1217:4 Btu= min = 158:10

19

Btu Btu - 11:3 lbm d a lbm d a

Convert the total heat transfer achieved to kW of heating.

qT,W = ð1217:4 Btu= min Þ

0:0176 kW = 21:43 kW 1 Btu= min

C. Calculate the enthalpy of the exit exhaust air using the total heat transfer achieved and Eq. 8.16. qT,W = m_ EA ðhEA,i - hEA,e Þ ) Btu lbm d a = 148:70 min min hEA,e = 16:11 Btu=lbm d a

1217:4

24:30

Btu - hEA,e ) lbm d a

Solutions to Practice Problems

353

Using the given moisture effectiveness for winter operation, determine the humidity ratio of the exit exhaust air using Eq. 8.22. Note that m_ EA = m_ min .

e moist .,W =

m& EA (wEA,i - wEA,e )

m& min (wEA,i - wOA,i )

=

lbm H 2 O ( ) - wEA,e | | 0.0073 lbm d a ( ) 0.70 = lbm H 2 O lbm H 2 O ) ( - 0.0016 | 0.0073 | lbm d a lbm d a ) ( wEA,e = 0.0033 lbm H 2 O / lbm d a

On the psychrometric chart, locate the state point of the exit exhaust air at the intersection of hEA,e = 16:11 Btu=lbm d a, and ωEA,e = 0:0033 lbm H2 O=lbm d a From the psychrometric chart, the conditions of the exit exhaust air, EAe (leaving the enthalpy wheel) are: T DB,EA,e = 520 F, T WB,EA,e = 420 F, hEA,e = 16:11 Btu=lbm d a ωEA,e = 0:0033 lbm H2 O=lbm d a,

rhEA,e = 40%

Practice Problem 8.3 (Solution) Locate the state points of inlet outside air (OAi, 6 °C DB, 3 °C WB), and inlet exhaust air (EAi, 20 °C DB, 50% rh) on the psychrometric chart as shown.

From the excerpt of the chart shown,

354

8

HVAC Systems and Equipment

hOA,i = 15 kJ=kg d a, vOA,i = 0:795 m3 =kg d a ωOA,i = 0:0035 kg H2 O=kg d a hEA,i = 39 kJ=kg d a, vEA,i = 0:84 m3 =kg d a ωEA,i = 0:0076 kg H2 O=kg d a Calculate the mass flow rates of the outside air and exhaust air using specific volumes. 0:8 ms V_ m_ OA = OA,i = = 1:006 kg d a=s 3 vOA,i 0:795 kgmd a 3

0:8 ms V_ EA,i = = 0:9524 kg d a=s 3 vEA,i 0:84 kgmd a 3

m_ EA =

Since m_ EA < m_ OA , m_ EA = m_ min = 0:9524 kg d a=s A. Using the given total effectiveness of the enthalpy wheel and Eq. 8.21, calculate the enthalpy of the exit outside air, that is, hOA, e. m_ OA ðhOA,e - hOA,i Þ ) m_ min ðhEA,i - hOA,i Þ kg d a kJ 1:006 hOA,e - 15 s kg d a 0:85 = kJ kJ kg d a 39 0:9524 - 15 s kg d a kg d a hOA,e = 34:3 kJ=kg d a εT,ERV,W =

)

Using the given sensible effectiveness of the enthalpy wheel and an equation similar to Eq. 8.21, calculate the dry bulb temperature of the exit outside air, that is, TOA,e. εSen,ERV ,W =

m_ OA cpa ðT OA,e - T OA,i Þ ) m_ min cpa ðT EA,i - T OA,i Þ

kg d a s kg d a 0:9524 s 1:006

0:85 =

∘

T OA,e = 17:3 C

∘ kJ ðT OA,e - 6 CÞ kg . ∘ C ) ∘ ∘ kJ ð20 C - 6 CÞ 1:0 ∘ kg . C

1:0

Solutions to Practice Problems

355

On the excerpt of the psychrometric chart locate the state point of exit outside air at the intersection of its enthalpy and dry bulb temperature as shown. OAe : hOA,e = 34:3 kJ=kg d a, T OA,e = 17:3 ° C From the psychrometric chart, the conditions of the exit outside air, OAe (leaving the ERV) are: T DB,OA,e = 17:3 ° C, hOA,e = 34:3 kJ=kj d a ωOA,e = 0:0062 kg H2 O=kg d a, rhOA,e = 52% Calculate the total heat transfer achieved during winter operation using the first part of Eq. 8.16. qT,W = m_ OA ðhOA,e - hOA,i Þ = 1:006

kg d a s

34:3

kJ kJ = 19:42 kW - 15 kg d a kg d a

Calculate the enthalpy of the exit exhaust air using the second part of Eq. 8.16. qT,W = m_ EA ðhEA,i - hEA,e Þ ) kg d a s hEA,e = 18:6 kJ=kg d a

19:42 kW = 0:9524

39

kJ - hEA,e kg d a

)

Calculate the dry bulb temperature of the exit exhaust air using Eq. 8.17 qSen,W = m_ OA cpa ðT OA,e - T OA,i Þ = m_ EA cpa ðT EA,i - T EA,e Þ ) 1:006

kg d a s

1:0

∘ ∘ kJ ð17:3 C - 6 CÞ kg . ∘ C

= 0:9524

kg d a s

1:0

∘ kJ ð20 C - T EA,e Þ kg . ∘ C

∘

) T EA,e = 8:1 C On the psychrometric locate the state point of exit outside air at the intersection of its enthalpy and dry bulb temperature. EAe : hEA,e = 18:6 kJ=kg d a,

T EA,e = 8:1 ° C

356

8

HVAC Systems and Equipment

From the psychrometric chart, the conditions of the exit exhaust air, EA,e (leaving the ERV) are: T DB,EA,e = 8:1 ° C, hEA,e = 18:60 kJ=kj d a ωEA,e = 0:0042 kg H2 O=kg d a, rhEA,e = 62% B. Calculate the moisture effectiveness of the ERV using Eq. 8.22. εmoist:,W =

m_ OA ðωOA,e - ωOA,i Þ m_ min ðωEA,i - ωOA,i Þ

kg d a s = kg d a 0:9524 s = 0:7ð70%Þ 1:006

kg H2 O kg H2 O - 0:0035 kg d a kg d a kg H2 O kg H2 O 0:0076 - 0:0035 kg d a kg d a

0:0062

C. In the solution to part A, the total heat transfer achieved by the ERV was determined to be 19.42 kW. This is the equivalent heat input saved by the usage of the ERV. Therefore, the annual savings due to the usage of ERV is

Annual Savings = ð19:42 kWÞ

1000 h yr

$1:03 = $20, 003=yr kWh

References 1. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook – Fundamentals, I P Edition, ASHRAE, USA (2021) 2. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook - Fundamentals, S I Edition, ASHRAE, USA (2021) 3. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook – HVAC Applications, I P Edition, ASHRAE, USA (2023) 4. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook - HVAC Applications, S I Edition, ASHRAE, USA (2023) 5. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook – HVAC Applications, I P Edition, ASHRAE, USA (2019) 6. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook – HVAC Systems and Equipment, I P Edition, ASHRAE, USA (2020) 7. American Society of Heating Refrigeration and Air-Conditioning Engineers (ASHRAE): ASHRAE Handbook - HVAC Systems and Equipment, S I Edition, ASHRAE, USA (2021) 8. Nandagopal, N.S.: Fluid and Thermal Sciences – a Practical Approach for Students and Professionals. Springer Nature, Switzerland (2022)

Index

A Absorber, 272, 273, 275, 276, 341 Absorption refrigeration systems, 277 Acoustics, 17, 50, 338–340 Actual compressor power, 120, 283 Adiabatic saturation, 173–175 Affinity laws, 47–49 Air change method, 201, 214 Air changes per hour (ACH), 50, 201, 216, 239, 242, 244, 249, 262 Air-conditioning processes, 116, 147, 148, 150, 166, 168–173 Air cooled chilled water systems, 317–318 Air-cooled condensers, 281 Air horse power (AHP), 38, 44 Air Movement and Control Association (AMCA), 37, 50 Air-water vapor mixtures, 129–144 Analyser, 273 Apparatus dew point (ADP), 151–154, 181, 223, 224, 226, 229, 254, 255, 309 Approach, 163–165, 179, 187, 198, 211 ASHRAE, 4, 29, 50, 65, 68, 72, 73, 99, 108, 116, 127, 147, 171, 189, 192, 199–204, 206, 208–212, 216, 235–238, 241–245, 247, 251, 280, 291, 295–297, 305, 306, 310, 313, 315, 317–319, 339–341, 347 ASHRAE Standard 34, 297 Axial fans, 49, 50

B Brake horsepower (BHP), 38 By-pass factor (BF), 151–154, 222, 260

C Carbon, 91, 295, 296 CAV systems, 342, 345, 347 Centrifugal compressors, 318 Centrifugal fan, 49, 50 Chemical dehumidification, 168, 173 Chilled-water cooling systems, 311, 315, 316, 318 Chlorine, 295, 296 Chlorofluoro carbons (CFC), 295–297 Climatic design information, 189 Coefficient of performance (COP), 116–120, 269–277, 283, 285–287, 302, 303, 331, 334, 345, 346 Coil cooling load, 195, 222–234, 258 Coil efficiency, 151, 153, 154, 243, 260 Component friction losses, 28–37 Compressors, 75, 76, 85, 103–107, 113, 114, 116, 118, 120–123, 125, 267, 269, 278, 279, 282–287, 298, 302, 303, 306, 316–318, 332, 334, 340, 341, 345–347 Condenser, 50, 75, 76, 78–80, 96, 116, 159, 273, 275, 280–285, 288–290, 302 Condenser heat duty, 288 Conduction, 61–66, 68–70, 81, 191, 199, 200, 239, 248, 262 Conduction–convection systems, 68, 69 Constant air volume (CAV), 342 Constant enthalpy process, 96, 274, 280 Continuity equation, 7–18, 23, 26, 37, 55, 118, 196, 198, 271, 301 Convection, 61, 66–71, 78, 81, 82, 210, 238 Convection heat transfer coefficient, 68, 69

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 N. S. Nandagopal, HVACR Principles and Applications, https://doi.org/10.1007/978-3-031-45267-3

357

358

Index

Conversion factors, 1, 7–9, 13, 14, 19, 39, 59–61, 65, 67, 80, 86, 87, 181, 211, 278 Cooling, 8, 59, 90, 130, 148, 189, 270, 305 Cooling above freezing, 299 Cooling and dehumidification, 147–154 Cooling below freezing, 299 Cooling efficiency, 163–165, 179, 187 Cooling load, 8, 167, 189, 298, 312 Cooling load calculations, 189–234, 241 Cooling load factor (CLF), 198, 201, 202, 208, 210, 251 Cooling load temperature difference (CLTD), 198–200, 204–206, 241, 245–247 Cooling towers, 50, 159–165, 179, 317, 318, 339 Cooling water, 74–76, 78–80, 272, 281–283, 288–290, 311, 317 COPcooling, 298, 303 COPheating, 282, 283 COPheat pump, 116, 119, 269, 282, 298, 302 COPrefg./AC, 117, 270, 284 Crack method, 201, 214

178, 181, 183, 184, 186, 193, 232, 269, 271, 274, 275, 279, 280, 300, 307, 311, 314, 315, 324, 343, 350 Energy balance for compressors, 103 Energy efficiency ratio (EER), 282, 284–286, 288 Energy recovery systems, 319–338 Energy recovery ventilators (ERV), 319, 328–338, 348, 349, 355, 356 Enthalpy, 75, 85, 135, 150, 194, 267, 306 Enthalpy–concentration diagram, 275 Enthalpy wheels, 171, 328–338, 348, 353, 354 Entrance loss, 25, 31, 35, 42, 54, 56 Entropy, 88, 89, 94, 96, 98, 105, 113, 115, 120, 268, 278 Equal friction method, 27 Equivalent lengths, 33–37, 39–41, 45, 47, 56 Evaporative condensers, 318 Evaporative cooling, 168, 171–173 Evaporator, 76, 79, 96, 267, 272, 275, 279, 280, 284, 306, 307, 311, 316, 318 Exit loss, 31–32, 34, 35, 37, 42, 51, 54–56

D Dampers, 28, 30, 31, 33, 38, 347 Darcy’s formula, 20–27 Dehumidification, 130, 151, 171, 193 Density, 1–3, 6–9, 12, 13, 15, 19, 48, 76, 80, 89, 91–93, 118, 150, 162, 163, 175, 179, 180, 186, 195, 196, 211, 231, 271, 281, 301, 310, 350 Design conditions, 192, 193, 199, 235, 241, 243, 343 Design indoor conditions, 189 Design outdoor conditions, 237, 244 Dew point, 130, 133, 139–142, 144, 179, 329, 331, 337 Direct expansion cooling systems, 306–309 Dry air, 4, 127, 129–135, 139–143, 147, 149, 150, 154, 156, 158, 160, 161, 166, 168, 170, 171, 173–175, 177, 180, 182, 184, 186, 194, 196, 198, 231, 234, 235, 259, 260, 308, 315, 323 Dry bulb (DB), 139, 141, 142, 148, 153, 160, 165, 171, 176, 179, 192, 232, 233, 309 Dry bulb (DB) temperature, 192 Duct sizing, 27–28

F Fan laws, 45, 47–49 Fan performance, 38, 45, 50 Fan performance curve, 45–47 Fan power, 17, 27, 37–44, 47, 48, 51, 219 Fans, 28, 31, 37–51, 219, 220, 222–224, 226, 242, 252, 253, 256, 317, 339–341 Fan selection, 49–51 Fan types, 49–51 Feed-back control, 347 Fenestration, 189, 190, 200–209, 220, 228, 252 Fiberglass lining, 340 Film coefficient, 66, 68, 237, 261 Film conductance, 66 First law applied to compressors, 103–105 First law of thermodynamics, 99–107 Fitting friction losses, 28–37 Floor springs, 341 Fluorine, 295, 296 Freezing of foods, 290–294 Freezing point, 290–294, 299 Friction factor, 20, 21, 23, 25, 33, 51 Friction head loss, 18–37, 51, 54, 55

E Effective coil temperature, 151, 153, 223, 232, 233, 260 Energy balance, 76, 102–103, 116, 118, 150, 152, 156, 158–160, 166, 168, 170, 176,

G Gallons per minute (Gpm), 9, 75, 76, 79, 80, 148, 150, 179, 230, 231, 281–283, 288, 290, 306, 323, 342, 343, 348, 350 Gas constants, 90–93, 101, 106, 118, 134, 136, 141, 271, 301

Index Generator, 272–277 Glazed surfaces, 201, 206–208, 239, 248, 262 Glazing, 202, 203, 206–208, 228, 236, 237, 242, 244, 248 Global warming potential (GWP), 295–298

H Heat exchanger design, 78, 80, 290 Heat gain due to appliances, 211–213 Heat gain due to conduction, 199, 201 Heat gain due to equipment, 211–213 Heat gain due to fenestration, 207–209, 248 Heat gain due to infiltration, 201–216, 249, 250 Heat gain due to lights, 211 Heat gain due to occupants, 209–211 Heat gain due to opaque surfaces, 199–200 Heat gain due to radiation, 206, 207, 248 Heat gain due to ventilation, 216–219 Heat gain sources, 193 Heat loss due to infiltration, 262 Heat loss through surfaces, 235, 239, 262 Heat pumps, 85, 115, 116, 118, 119, 165, 267–269, 271, 272, 282, 301, 302 Heat recovery ventilators (HRV), 319–327, 350 Heat recovery wheels, 321 Heat transfer area, 288, 290 Heat transfer coefficient, 66–68, 79, 81, 199–201 Heat transfer modes, 61 Heat transfer rates, 8, 59, 63, 67, 328 Heat transfer resistance (HTR), 64, 67–68, 70, 71, 151 Heating, 8, 59, 85, 165, 189, 267, 305 Heating and humidification, 154–159 Heating load, 8, 301, 313, 314, 342 Heating load calculations, 189, 190, 234–241, 243 Humidification, 173, 179 Humidity ratio, 131–137, 139–142, 144, 147, 151, 155, 168–174, 179, 195, 196, 198, 223, 227, 231, 235, 236, 240, 243, 249, 257, 262, 263, 309, 327, 328, 334, 342, 348, 349, 352, 353 HVAC equipment, 305–356 HVAC system components, 33 HVAC systems, 18–37, 59, 68, 111, 339 HVAC systems and equipment, 305, 317–319 Hydraulic diameter, 16–20, 23, 25, 26, 54 Hydrochlorofluoro carbons (HCFC), 295–297 Hydrofluoro carbons (HFCs), 295, 296 Hydrofluoro olefins (HFO), 295–297 Hydrogen, 90, 296

359 I Ideal compressor power, 104, 106, 114, 122 Ideal gas law, 2, 3, 89–93, 97, 98, 118, 127–129, 134, 271, 301 Ideal gas mixtures, 85, 127–129, 132 Inches water column, 45 Indoor air quality (IAQ), 50, 216 Indoor conditions, 191, 192 Indoor solar attenuation coefficient (IAC), 201 Infiltration, 189, 191, 193, 201, 203, 214, 216, 217, 220, 221, 236, 237, 239, 240, 242, 244, 248, 249, 252, 254, 262, 263 Insertion losses, 339, 340 Internal energy, 87, 89, 99, 102, 103, 130 In. w c, 14, 27, 34, 42–44, 51, 57, 219, 223, 242 Isentropic compression, 101, 102, 105, 107, 113–115, 121, 268, 279, 286 Isentropic efficiency, 105–107, 113, 279, 286 Isentropic efficiency of compressors, 105–107 Isentropic process, 94–96, 98, 101, 103 Isothermal dehumidification, 168, 171 Isothermal humidification, 168–170, 173 Isothermal process, 94, 95, 97, 98

L Latent cooling, 223, 224, 227, 230, 251, 254, 257, 258 Latent cooling load, 223, 224, 227, 230, 251, 254, 257, 258 Latent heat, 75, 117, 135, 171, 173, 194, 195, 197, 201, 209, 236, 237, 240, 242, 270, 291, 292, 294, 329 Latent heat gain, 193, 201, 209–211, 213–219, 221, 222, 248–251, 253, 254 Latent heat ratio (LHR), 194 Latent load, 194, 222 Lighting special allowance factor, 211 Liquid vapor mixture, 306 Load factor, 202, 212 Log mean temperature difference (LMTD), 77, 79, 80, 289 Loss coefficients, 29–34, 36, 55

M MA, 314 Mass fraction, 108, 128, 132 Mean coincident wet bulb temperature (MCWB), 192, 193, 202, 243 Mean daily range (DR), 192, 202

360 Mixed air, 176, 179, 217, 224, 225, 232, 233, 254, 255, 257–259, 306, 310, 311, 313, 314, 316 Mixing of air streams, 175–179 Moist air, 2–4, 89, 127, 130–131, 133–142, 144, 147–149, 154, 156, 158, 161, 165–170, 172–176, 178, 182–184, 193, 194 Moisture effectiveness, 330, 331, 334, 348, 349, 352, 353, 356 Molar volume, 129 Mole, 90, 132 Mole fraction, 128, 129, 133, 134 Moody diagram, 21, 23 Multilayer heat conduction, 63–64 Multiple-zone, 345 Multiple-zone applications, 345

N Newton’s law of cooling, 66–67 Noise control, 338–340 Noise mitigation in HVAC systems, 339

O Online sources for refrigerant properties, 124 Operating point, 45–47, 49 Optimum duct shape, 17 Outdoor conditions, 189, 190, 192, 234 Overall heat transfer coefficient, 69–71, 78, 203–205, 223, 235, 237, 238, 242, 244–246, 261, 288 Overall resistance, 70, 78 Ozone depletion potential (ODP), 295–297

P Partial pressure, 127–134, 140–144 Partial volume, 128, 129 Percentile cumulative frequency, 192 P–h diagram, 108 Pitot tube, 15–16 Plate and frame heat exchangers, 320 Pressure drop, 9, 18–37, 41, 42, 44, 47, 53, 54, 56, 57, 219, 223 Pressure–enthalpy diagram, 107 Pressure loss charts, 21, 22, 24 Psychrometric charts, 4, 135, 137–142, 147, 148, 150, 153, 155, 157, 161, 163–165, 167, 169, 172, 174, 176, 178, 180–183, 185, 187, 196, 198, 215, 216, 218, 219, 225, 227, 229–231, 233, 240, 249, 250,

Index 257, 259, 262, 263, 308, 309, 311, 312, 314, 323, 325, 327, 332–335, 337, 343–345, 349, 351–353, 355, 356 Psychrometric formulas, 2, 85, 135, 139, 141 Psychrometric properties, 137–144 Psychrometric protractor, 228, 230

R R-22, 6, 111, 295–297 R-134a, 6, 79, 80, 100, 101, 106, 107, 111–114, 120–124, 231, 278, 284–286, 289, 295–298, 300, 306, 307 Radiation, 61, 190, 191, 193, 199–201, 207–210, 212, 248 Range, 5, 27, 29, 30, 42, 50, 68, 108, 163–165, 174, 179, 187, 192, 197, 199, 202, 204, 206, 235, 245, 247, 317, 339 Ratio of specific heats, 94, 100, 101, 106 Refrigerant property tables, 75, 108 Refrigerants, 1, 74, 167, 231, 267, 305 Refrigerant safety designations, 297 Refrigerating effect, 116, 267, 274–276, 279, 284, 285 Refrigeration cycles, 74–76, 78, 79, 85, 96, 115, 267, 268, 272–290, 302, 310 Refrigeration of foods, 290–294 Refrigeration systems, 76, 85, 280, 281, 283, 307, 310 Relative humidity, 131–134, 136, 137, 139–143, 147, 148, 151, 155, 157, 159, 160, 163, 165–169, 171–174, 176, 179, 192, 202, 215, 232, 235, 236, 241, 243, 327, 342, 345, 348, 349, 352 Return air (RA), 24, 41, 175–177, 217, 222–224, 226, 232, 242, 243, 254–258, 319–321, 323, 325, 327, 328, 348, 351 Reversed Carnot Cycle, 85, 115–120, 267–272, 277, 278 Reynolds number, 16, 19–21, 23 rh, 16, 143, 148, 149 Rubber pads, 341 Run around coils, 320, 323, 348, 350 R-values, 64–66, 70–73, 81, 82 R-values of building materials, 64–66 R-values of insulation, 65

S SA, 345 Saturated air, 130, 131, 151 Saturated liquid, 79, 108–111, 124, 280

Index Saturated vapor, 105, 108, 111, 120, 121, 231, 272, 278, 279, 306, 313 Saturation efficiency, 173–175 Screw compressors, 318 Scroll compressors, 318 Seasonal energy efficiency ratio (SEER), 282, 283, 285 Sensible cooling, 151, 167–168, 173, 344, 346 Sensible cooling load, 197, 222–224, 226, 227, 234, 250, 253–256, 260, 343, 344, 347 Sensible effectiveness, 330, 331, 334, 336, 348, 354 Sensible heat gain, 193, 199, 200, 203, 205–207, 209–212, 215, 219–222, 235, 244, 246–249, 251–253 Sensible heat ratio (SHR), 194, 196, 221, 222, 224, 228, 230, 232, 234, 243, 260, 342 Sensible heating, 151, 165–166, 236, 239, 240, 262, 263 Sensible load, 194, 254 Sensible wheels, 320, 321, 328 Set temperature, 342, 345 Shading coefficient (SC), 201, 208 SI and USCS units, 1, 2, 5, 12, 51, 59, 86, 87, 89, 90 Single-zone, 342 Single-zone applications, 342 Solar cooling load (SCL), 201, 202, 208, 209 Solar heat gain coefficient (SHGH), 201, 203, 208, 209, 242 Solar heat gain factor (SHGF), 202, 208, 209 Solvent, 272 Sound-absorbing materials, 339 Space cooling load, 217, 223, 250–251, 342, 347 Specific enthalpy, 89, 103, 108, 110, 114, 135, 136, 142, 144, 197, 275 Specific entropy, 89, 105, 108, 110 Specific heat, 74, 94, 118, 135, 195, 197, 271, 282, 290, 291, 293, 294, 299, 310 Specific heats of gases, 94 Specific humidity, 131, 196 Specific properties, 88–89, 108, 109 Specific volume, 2–4, 8, 9, 51, 52, 88, 89, 92, 93, 108–110, 134, 136, 139, 141, 142, 144, 149, 156, 158, 162, 166, 168, 170, 175, 177, 179, 180, 182, 184, 196, 198, 231, 234, 259, 323, 325, 332, 335, 346, 349, 351, 354 Spring hangers, 341 Stagnation pressure, 15 Standard air, 197 Standard density of air, 2, 5, 13, 93, 194, 197, 343

361 State points, 79, 113, 114, 120–125, 137, 139, 142, 148, 149, 151, 153–158, 161, 165–170, 172, 174–176, 178–186, 215, 225, 229, 232, 233, 240, 249, 255, 257–259, 262, 274–276, 278–280, 284, 286, 298, 300, 302, 308, 309, 311, 314, 315, 323, 325, 327, 332, 333, 335, 337, 343–345, 349, 351–353, 355 Static pressure, 11–16, 27, 37, 38, 40, 45, 47, 48, 50 Steam-based heating systems, 305, 313–315 Strong solution, 275 Summer operation of ERV, 328, 329 Summer operation of HRV, 327 Superheated vapor, 79, 108, 278, 279, 289 Supply air (SA), 77, 219, 229, 233, 242, 309–313, 342, 343, 345–347, 352 System curve, 45–47

T Thermal comfort, 189, 192 Thermal properties of food, 291–294 Thermodynamic processes, 85, 93–98 Thermodynamic properties, 85, 100, 106, 127 Thermodynamic properties of refrigerants, 107–114 Thermostat, 342, 347 Throttling process, 96, 124 Throttling valve, 96, 273, 278 Tons of refrigeration (TR), 75, 76, 117, 191, 224, 227, 232, 243, 258, 270, 275, 276, 278, 306, 312, 317, 318, 331, 333, 338 Total effectiveness, 329, 331, 333, 334, 336, 354

U Unit, 1, 59, 86, 131, 147, 197, 267, 306 Universal gas constant, 91–93 Utilization factor, 211, 212

V Vapor compression refrigeration systems, 280, 306 Variable air volume (VAV) systems, 341–348 Velocity head, 11, 20, 29, 31, 32, 55 Velocity pressure, 11–14, 20, 27, 29, 31, 32, 37 Ventilation, 1, 50, 59, 85, 175, 189, 191, 193, 216–218, 220, 221, 223, 224, 236, 238, 240, 242, 244, 250, 252, 254–256, 263, 264, 319, 331 Vibration control in HVAC equipment, 340

362 Vibration control in HVAC systems, 340 Vibration in HVAC equipment, 340 Vibration in HVAC systems, 339, 340 Vibration isolation, 340, 341 Vibration isolation mechanisms, 340, 341 Viscosity, 1, 6, 7, 19

W Water-cooled chilled water systems, 317 Water-cooled condensers, 281

Index WB, 165, 215, 241, 243, 249, 257–259, 306, 310, 311, 313, 348, 353 Weak solution, 275, 276 Wet bulb temperature (TWB), 130, 137, 139, 163, 171, 173, 179, 192 Winter operation of ERV, 319, 331 Winter operation of HRV, 322 Work, 34, 37, 39, 76, 88, 94, 96–107, 115, 116, 122, 123, 267–269, 274, 278, 279, 287, 300, 310