ASHRAE Fundamentals of Thermodynamics and Psychrometrics

697 J68f

ASHRAE Continuing Education

Fundamentals of Thernlodynamics & Psychrometries

A Self-Directed Learning Course For Professional Development

Prepared by

Richard R. Johnson, Ph.D. North Carolina State University

American Society of Heating, Refrigerating and Air-Conditioning Engineers Inc. 1791 Tullie Circle NE • Atlanta, GA 30329

Copyright © 1997 by the American Society of Heating, Refrigerating and Air-Conditioning Engineers (ASHRAE). All rights reserved. No part ofthis book may be reproduced without written permission from ASHRAE, except by a reviewer who may quote brief passages or reproduce illustrations in a review with appropriate credit; nor may any part of this book be reproduced, stored in a retrieval system, or transmitted in any form or by any means (electronic, photocopying, recording or other) without written permission from ASHRAE. ASHRAE has compiled this publication with care, but ASHRAE has not investigated, and ASHRAE expressly disclaims any duty to investigate, any product, service, process, procedure, design or the like that may be described herein. The appearance of any technical data or editorial material in this publication does not constitute endorsement, warranty or guaranty by ASHRAE of any product, service, process, procedure, design or the like. ASHRAE does not warrant that the information in this publication is free of errors. The entire risk of the use of any information in this publication is assumed by the user. Comments, criticism and suggestions regarding the subject matter are invited. Any errors or omissions in the data should be brought to the attention of Martin Kraft, Continuing Education Course Editor. ASHRAE Education Department: Anne Spengler, Director of Education Martin Kraft, Continuing Education Course Editor Laura Tracy, Education Coordinator Marietta Henry, Secretary F or course information or to order additional materials, please contact:

Education Department ASHRAElnc. 1791 Tullie Circle NE Atlanta, GA 30329 Telephone: 404/636-8400 Fax: 404/321-5478

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Table of Contents

Chapter 1 • • • • • • •

Introduction to HVAC

Instructions Study Objectives of Chapter 1 1.1 Introduction 1.2 HV AC: Heating, Ventilating and Air-Conditioning 1.3 The Major Processes of Air-Conditioning 1.4 Thermodynamics and Psychrometries The Next Step

• Summary • Bibliography • Skill Development Exercises for Chapter 1

Chapter 2

Systems, Properties, States and Processes

• Instructions • Study Objectives of Chapter 2 • 2.1 Introduction • 2.2 Closed and Open Systems in Thermodynamics • 2.3 Forms of Energy • 2.4 Properties of a System • 2.5 Equilibrium States • 2.6 Processes and Cycles • The Next Step • Summary • Skill Development Exercises for Chapter 2

Chapter 3

Property Diagrams for Pure Substances

• Instructions • Study Objectives of Chapter 3 • • • • •

3.1 The State Postulate 3.2 Properties 3.3 Phases and Phase Changes for Pure Substances 3.4 Property Diagrams The Next Step

• Summary • Skill Development Exercises for Chapter 3 Fundamentals of Thermodynamics and Psychrometries

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Chapter 4 • • • • • •

Thermodynamic Tables and Charts

Instructions Study Objectives of Chapter 4 4.1 Introduction 4.2 Tables for Saturated Liquid and Vapor 4.3 Tables for Superheated Vapor and Compressed Liquid 4.4 Thermodynamic LiquidlVapor Charts

• The Next Step • Summary • Skill Development Exercises for Chapter 4

Chapter 5 • • • • • • •

Ideal Gas Law and Air Tables

Instructions Study Objectives of Chapter 5 5.1 Ideal Gas and the Ideal Gas Law 5.2 The Ideal Gas Law as an Equation of State 5.3 Constant Specific Heat 5.4 Air Tables The Next Step

• Summary • Skill Development Exercises for Chapter 5

Chapter 6

Heat and Work

• Instructions • Study Objectives of Chapter 6 • 6.1 Introduction • • • •

6.2 Heat 6.3 Work 6.4 Flow Work The Next Step


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Chapter 7 • • • •

First Law of Thermodynamics Applied to Closed Systems

Instructions Study Objectives of Chapter 7 7.1 Introduction to Controlled Mass Approach 7.2 The First Law of Thermodynamics for a Closed System

• 7.3 • 7.4

Conservation of Energy First Law Applied in Example Cases


Chapter 8 • • • • •

First Law of Thermodynamics Applied to Open Systems

Instructions Study Objectives of Chapter 8 8.1 Introduction to the Control Volume Approach 8.2 Conservation of Mass 8.3 Conservation of Energy and the First Law for Open Systems

• 8.4 • 8.S

Steady-Flow Processes First Law Applied in Examples of Steady-Flow Processes


Chapter 9 • • • • •

Applications of the First Law of Thermodynamics

Instructions Study Objectives of Chapter 9 9.1 Compressors, Turbines, Pumps and Fans 9.2 Throttling Valves and Metering Devices 9.3 Heat Exchangers, Condensers and Evaporators



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Chapter 10

The Carnot Cycle

• Instructions • Study Objectives of Chapter 10 • 10.1 The Second Law of Thennodynamics Heat Engines, Refrigerators and Heat Pumps • 10.2 Reversible and Irreversible Processes • 10.3 • 10.4 The Camot Cycle and the Reversed Camot Cycle • The Next Step • Summary • Skill Development Exercises for Chapter 10

Chapter 11

Refrigeration Cycles

• • • •

Instructions Study Objectives of Chapter 11 11.1 Refrigeration Equipment 11.2 The Ideal Vapor-Compression Refrigeration Cycle Analysis of the Components and Cycle • 11.3 Perfonnance of Ideal Refrigerators and Heat Pumps • 11.4 The Actual Vapor-Compression Refrigeration Cycle • 11.5 Absorption Refrigeration System ·11.6 • 11.7 Gas Refrigeration System • The Next Step • Summary • Skill Development Exercises for Chapter 11

Chapter 12 • • • • • • • • •

Moist Air as a Mixture of Ideal Gases

Instructions Study Objectives of Chapter 12 12.1 Introduction 12.2 Mixtures ofIdeal Gases Mixture of Dry Air and Water Vapor 12.3 12.4 Moist Air Tables The Next Step Summary Skill Development Exercises for Chapter 12

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Chapter 13 • • • • •

Properties of Moist Air

Instructions Study Objectives of Chapter 13 13.1 Humidity Ratio and Relative Humidity 13.2 Dewpoint Temperature 13.3 Adiabatic Saturation and Wet-Bulb Temperatures


Chapter 14

The Psychrometric Chart

• • • •

Instructions Study Objectives of Chapter 14 14.1 Introduction 14.2 Description ofthe Axes and Lines of the Chart Finding Property Values on the Psychrometric Chart • 14.3 • 14.4 Human Comfort and the Psychrometric Chart • The Next Step • Summary • Skill Development Exercises for Chapter 14

Chapter 15

Air-Conditioning Processes on the Psychrometric Chart

• Instructions • Study Objectives of Chapter 15 • 15.1 Simple Heating or Cooling at Constant Humidity Ratio Heating with Humidification • 15.2 Cooling Coil Dehumidification • 15.3 Evaporative Cooling • 15.4 Adiabatic Mixing • 15.5 • 15.6 Sensible Heat Ratio • Summary • Skill Development Exercises for Chapter 15

Fundamentals of Thermodynamics and Psychrometrics

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Appendices • Appendix A-I • Appendix A-2 • Appendix B-1 • Appendix C-l • Appendix C-2 • Appendix C-3 • Appendix C-4 • Appendix D-l • Appendix E-l

Table of Contents

Dimensions and Units Used in Air-Conditioning Applications Unit Conversion Factors Thermodynamic Properties of Water at Saturation R-22 Properties of Saturated Liquid and Saturated Vapor Pressure-Enthalpy Diagram for R-22 R-134a Properties of Saturated Liquid and Saturated Vapor Pressure-Enthalpy Diagram for R-134a Thermodynamic Properties of Moist Air at 14.696 psia ASHRAE Psychrometric Chart


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Chapter 1 Introduction to HVAC

Contents of Chapter 1 • Instructions • Study Objectives of Chapter 1 • 1.1 • 1.2

Introduction HV AC: Heating, Ventilating and Air-Conditioning

• 1.3

The Major Processes of Air-Conditioning

• 1.4

Thermodynamics and Psychrometrics

• The Next Step • Summary • Bibliography • Skill Development Exercises for Chapter 1

Instructions Read the material of Chapter 1. Re-read the parts of the chapter that are emphasized in the summary and memorize important definitions. At the end of the chapter, complete all of the skill development exercises without consulting the text.

Study Objectives of Chapter I Chapter 1 is intended to introduce the processes of HVAC and provide substance to the assertion that it is important to study thermodynamics to gain a better understanding of HVAC. Chapter 1 concludes with a definition of thermodynamics and psychrometrics. After studying the chapter, you should be able to: • Define air-conditioning; • Name and describe seven major air-conditioning processes; • Provide a description of thermodynamics as the science of energy; and • List HVAC applications where knowledge of basic thermodynamics is essential.



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1.1

Introduction

Modem air-conditioning systems are part of our everyday life. Whether it is the comfort enjoyed while relaxing at home, or the freedom from discomfort that allows concentration on activities at work, we can appreciate air-conditioning for making our lives healthier, more enjoyable and more productive. The benefits of air-conditioning are also apparent in the wide range of high quality products that are available in the marketplace. Many products and manufacturing processes are improved by air-conditioning because many goods can be produced better, faster and more economically in a properly controlled environment. Air-conditioning allows for the creation of a comfortable environment indoors no matter what the weather is like outdoors. Whether it's a sweltering hot day in August or a bitterly cold night in January, air-conditioning can be a boon. However, there is a cost to air-conditioning, which appears in the form of energy consumption. Systems installed in the United States prior to the energy scare of 1973 were generally designed with little attention to energy conservation because energy was perceived to be plentiful and inexpensive. There was less concern then of the potentially negative environmental impact of energy consumption on a wide scale. Times have changed. Today, energy conservation is a key component in air-conditioning design. Efficiency is the watchword. The results of energy conservation efforts have led to some interesting shifts in the air-conditioning marketplace. There are many more system configurations than ever before as designers try to take advantage of any opportunity to save energy, and there are tradeoffs to be made between energy conservation and performance. This means that it is important to know and understand how to best select components for an HVAC system. Besides the actual redesign of components such as compressors, furnaces, refrigerators and heat exchangers for better efficiency, there has also been a dramatic expansion in the logic and sophistication of electronic controllers used in air-conditioning systems. Properly implemented changes in system control have the potential to lead to significant improvements in performance. Today, buildings often have an energy management system. Besides energy conservation, the issues of health and air quality have drawn a lot of recent attention in air-conditioning systems. For instance, there is a current controversy about the question of ventilation rates and air quality on board large passenger airplanes. Energy considerations call for minimizing the ratio of outside air to recirculated air. In contrast, the ventilation need is for plenty offresh outside air to avoid the build-up of contaminant gases such as carbon dioxide. What constitutes the right balance? There are standards and guidelines for such ventilation requirements, but they must be continually reviewed because of new situations and revised health evidence.


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To study the historical record of air-conditioning is to take a fascinating trip through the tremendous technical and scientific record of our society. There are the pioneers such as Robert Boyle, Sadi Carnot, John Dalton, James Watt, Benjamin Franklin, John Gorrie, Lord Kelvin, Ferdinand Carre', Willis Carrier, Thomas Midgley and many others who have brought us to our current state. We can expect that there is further history still to be written. Air-conditioning could not have developed the way it has were it not for all of the accomplishments of science and engineering in this century. Advances in thermodynamics, fluid mechanics, electricity, electronics, construction, materials, medicine, controls and social behavior are the building blocks to better engineered products of air-conditioning. Many of these famous names are connected to the study of thermodynamics. The term thermodynamics was first used by Lord Kelvin in 1849, and the first textbook on thermodynamics was written by William Rankine in 1859. Thermodynamics is the cornerstone to understanding the processes of air-conditioning.

1.2

HVAC: Heating, Ventilating andAir-Conditioning

For a long time, the term air-conditioning was associated with the act of air cooling only. This has changed and the definition of air-conditioning has been broadened to include all of HV AC. HVAC stands for Heating, Ventilating and Air-Conditioning (cooling). This dual meaning of air-conditioning may seem confusing but, when air-conditioning is mentioned by itself in this course, it will stand for both heating and cooling, and when it is used in the context of HVAC, it will mean only cooling. The use of HV AC as a descriptive term continues because of its historical roots. In its modem sense, air-conditioning is the control of temperature, moisture content, cleanliness, air quality and air circulation as required by occupants, a process or a product in the space. This is a definition given by Willis Carrier (1876-1950) in the book, Father ofAirConditioning.! Imbedded in this definition of air-conditioning are many ideas that have been shaped over time about our perceived need for air-conditioning and the changing technology that makes air-conditioning possible. This is a wide ranging definition that gives interpretation to the idea of conditioning the air in a given space for some specific purpose. Air-conditioning is more than an occasional luxury. It is important to our everyday comfort, health and well-being whether we are at home, in the workplace, out shopping or travelling from one place to another. It includes heating, cooling, humidifying, dehumidifying, ventilating and control of air quality year-round. Besides being important for humans and animals, air-conditioning is also essential to many manufacturing processes. For example, precision-engineered products are often made under closely controlled conditions that can only be produced in an air-conditioned environment. Fundamentals ofThermodynamics and Psychrometries


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1.3

The Major Processes ofAir-Conditioning

The seven major processes of air-conditioning are heating, cooling, humidifying, dehumidifying, cleaning, ventilating and air movement.

• Heating is the process of adding thermal energy (heat) to the air in the conditioned space for the purposes of raising or maintaining the temperature of the aIr. • Cooling is the process of removing thermal energy (heat) from the air in the conditioned space for the purposes of lowering or maintaining the temperature of the air. • Humidifying is the process of adding water vapor (moisture) to the air in the conditioned space for the purposes of raising or maintaining the moisture content of the air. • Dehumidifying is the process of removing water vapor (moisture) from the air in the conditioned space for the purposes of lowering or maintaining the moisture content of the air. • Cleaning is the process of removing particulate, chemical and biological contaminants from the air in the conditioned space for the purposes of improving or maintaining the air quality. • Ventilating is the process of exchanging air between the outdoors and the conditioned space for the purposes of diluting the gaseous contaminants in the air and improving or maintaining air quality, composition and freshness. • Air Movement (circulation and mixing) is the process of moving air through conditioned spaces in the building for the purposes of achieving the proper ventilation and facilitating the thermal energy transfer, humidification (or dehumidification) and cleaning processes outlined above.

These are the seven major air-conditioning processes used in HVAC systems. Heating and cooling are the most basic and probably the most recognized. If the space is too hot or too cold, then the occupants are unlikely to be comfortable or productive and there is a need to adjust the temperature by adding or removing thermal energy. Heating has been around for a long time and may utilize a variety of heat sources such as the burning of a fuel, electricity or chemical reaction to supply the heat. Cooling is more complicated. It requires a means for removing energy from the air and usually involves a refrigeration cycle or evaporative cooling technique. It takes energy work to drive the refrigeration cycle.



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The term air is used to mean a mixture of dry air and water vapor. Strictly speaking, the term moist air should be used, but it is common in air-conditioning terminology to use air when referring to moist air. There are several standard measures used to describe the proportions of this mixture such as relative humidity, humidity ratio and moisture content. It is important to understand that there is only a limited range of mixture proportions that are considered to be comfortable by humans. The range is also dependent on temperature. The proportion of water vapor to dry air in the mixture is changed by adding water vapor (humidification) or removing water vapor (dehumidification). The study of the properties and processes of moist air at atmospheric pressures is known as psychrometrics and will be discussed in later chapters. Humidification may be accomplished by allowing the air to blow over water on a wet porous surface, by spraying small droplets into the air stream, by misting with an ultrasonic mister, or by steam injection. On the other hand, dehumidification is mostly commonly accomplished by cooling. Cooling removes the energy from the water vapor to convert it to liquid water. There is a limit to the amount of water vapor that can be held by the air. This limit is known as the saturation condition and it decreases with decreasing temperature. If the temperature of saturated air is decreased, then the vapor condenses into water droplets and drips (or rains) out of the mixture. The result is that the air contains less water vapor. The change of state from water vapor to liquid water (condensation) or from liquid water to water vapor (evaporation) involves transfer of substantial amounts of thermal energy known as latent heat. Because heating and cooling are about the management of thermal energy to control temperature, it is clear that the effects of humidification and dehumidification will be interrelated with heating and cooling in air-conditioning. There are many types of airborne pollutants. Particulates such as pollen, dust and smoke are common items that must be removed from a typical conditioned space. Microscopic organisms such as bacteria, fungi and viruses may be important because of health reasons. Some of these pollutants are just irritating, but others may be deadly. The removal of these pollutants is often achieved by filtering or treating the air stream. Heating and dehumidification are powerful deterrents to many small organisms. Air exchange between outdoors and the conditioned space is either by ventilation (intentional) or by infiltration (unintentional). It is important because the outdoor air is often used to dilute pollutants (such as carbon dioxide) and replenish oxygen to the indoor air. Forced ventilation provides for the best control of air exchange rate and air distribution in a building and is usually mandatory in larger buildings where a minimum air exchange rate is required. On the other hand, natural ventilation is driven by wind pressure, indoor-outdoor temperature differences or the opening of a window, and it is not closely controlled. Infil-

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tration refers to air that leaks in through cracks, incomplete seals and unintentional openings. Most residential homes depend on infiltration for an adequate air exchange rate. However, this has become a concern in many modem houses which are purposefully tightly sealed as an energy conservation measure. Air movement, circulation and mixing are all fluid mechanics processes that involve moving the air between and through the conditioned spaces and to places where the other processes of heating, cooling, humidifying, cleaning and ventilating can occur. Air movement may be forced or free. Forced air movement as produced by fans may be carefully controlled. However, free air movement is often driven by temperature gradients in the conditioned space and occurs naturally. Mixing is an important process that occurs when outdoor air is mixed with indoor air as part of ventilation or when newly conditioned air mixes with air already in the conditioned space.

1.4

Thermodynamics and Psychrometries

To better understand and analyze each of the HVAC processes, it is essential to study and develop an appreciation of thermodynamics and psychrometrics. The terms thermodynamics and psychrometries may sound quite formidable at first, but as we shall see, they are descriptive terms for some common-sense ideas. Studying thermodynamics will provide the tools for quantifying, analyzing and interpreting the very practical matters of HV AC. Studying psychrometrics will provide the understanding needed for controlling humidity. Thermodynamics may be defined as the science of energy. Everybody has some feeling for what is generally meant by energy, but a precise definition takes some development. In the next chapter, some specific definitions and derivations will make it clear how thermodynamics describes the relationships between the different forms of energy and the conversion of one form of energy to another. These relationships occur in almost every aspect of our daily lives. For example, how is it possible to convert the electrical energy ofthe power supply into a cooling effect in an air-conditioned room? How can solar energy be effectively captured and put to good use? How are the heating needs for a house calculated for the middle of winter? These and many other questions may be answered by conducting an energy, or thermodynamic, analysis. You may be familiar with the conservation of energy principle. It states that, during an interaction, energy can change from one form to another, but that the total amount of energy remains constant. Said another way, it means that energy is neither destroyed nor created,. but only changes from one form to another. Forms of energy include internal (such as chemical, thermal, nuclear and electrical), potential and kinetic. To effectively use this principle, Chapter 1 Introduction to HVAC


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it is essential to understand the different forms of energy, the conversion of one energy form to another, the flow of energy across boundaries, and the ability to store energy. Thermodynamics is about the conversion of one form of energy to another. Energy conversions are described in terms of the various properties of substances and the changes of the properties that occur as a result of energy transformations. Thermodynamics is based on experimental observations. The observations are then expressed in terms of some basic laws. For example, the First Law of Thermodynamics is an expression of the conservation of energy principle. The observation that some processes occur in one direction, but not the reverse (for example, the flow of heat from a hot object to a cold one, or the generation of heat by friction) is embodied in the Second Law of Thermodynamics. On the other hand, psychrometrics is the branch of thermodynamics that deals with the thermodynamic properties and processes of moist air. The term humidity is used in connection with the proportion of water vapor in the mixture. It is important to study psychrometrics because, in HVAC, we are interested in controlling humidity as well as temperature. There is a comfort envelope that is framed by temperature and humidity constraints. The study of psychrometrics will be aided by a chart of moist air properties known as the psychrometric chart. We will learn how to plot thermodynamic processes of moist air on the chart. The applications for thermodynamics are numerous. In HVAC, the study of thermodynamics is particularly applied to: • Refrigerators and heat pumps • Heat exchangers • Cooling and heating coils • Radiators • Furnaces • Fans • Pumps • Water and air distribution systems • Building load calculations • Human occupation and activity levels

Both thermodynamic and psychrometric principles are applied to these processes: • Air flow over cooling and heating coils • Humidifiers and dehumidifiers • Outdoor weather conditions • Air in conditioned space (with interest in the comfort envelope) • Breathing by humans and animals



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The following chapters of this course will explain the basics of thermodynamics and psychrometries, and then consider the practical applications to many of the areas listed above.

The Next Step This first chapter provided an introduction to HVAC and the importance ofthermodynamics. In the next chapter, we will begin to build an understanding of thermodynamics by starting with the basics. In Chapter 2, we will introduce the concepts of open and closed systems, forms of energy, equilibrium and property definition. Chapter 2 will also include explanations of the state of a substance, and the process of state change shown on a property diagram.

Summary

Chapter 1 has been an introductory chapter. Following an introduction and a definition of air-conditioning, the seven main processes of HVAC were listed as heating, cooling, humidifying, dehumidifying, cleaning, ventilating and air movement. Each of these processes involves equipment for implementation and deals in different forms of energy. The study of thermodynamics (or the science of energy) is all about the relationship between different forms of energy, and is essential to fully understand the processes ofHVAC. The First Law of Thermodynamics was introduced as a statement of the conservation of energy principle. The Second Law of Thermodynamics was said to embody the idea of irreversibility of some naturally occurring events. Psychrometries was introduced as the science of an air and water vapor mixture. After studying this chapter, you should be able to: • Define air-conditioning; • Name and describe the seven major air-conditioning processes; • Provide a description of thermodynamics as the science of energy; and • List HVAC applications where knowledge of basic thermodynamics is essential.



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Bibliography 1. Carrier, W. 1990. Father ofAir-Conditioning. Louisville, KY: Fetter Printing Co.

Skill Development Exercises for Chapter 1 Complete these questions by writing your answers on the worksheets at the back of this book. Be sure to include your name and address. Send your completed questions to the ASBRAE Education Department.

1-01. List the seven main air-conditioning processes, give a one-sentence description of each process, and give another one-sentence reason why the process is likely to be found in a hospital.

1-02. Define what is meant by air-conditioning.

1-03. Name five pieces of normal BVAC equipment for which the application ofthermodynamic principles is essential to predict performance.

1-04. In your own words, state what is meant by the conservation of energy principle.

1-05. Provide two examples of a process that goes in one direction but will not go in the reverse direction without applying an external energy source.

1-06. What is the difference between dry air and moist air?

Fundamentals ofThernwdynamics and Psychrometries


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Chapter 2 Systems, Properties, States and Processes

Contents of Chapter 2 • Instructions • Study Objectives of Chapter 2 • 2.1

Introduction

• 2.2

Closed and Open Systems in Thermodynamics

• 2.3

Forms of Energy

• 2.4

Properties of a System

• 2.5

Equilibrium States

• 2.6

Processes and Cycles



Study Objectives of Chapter 2 The building blocks of thermodynamics (such as system, properties, states, processes and cycles) will be introduced in this chapter. Many new terms will be introduced and defined. The definitions may seem far removed from the applications of interest, but this development of a terminology and logic is essential to understanding how to apply the laws of thermodynamics.



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After studying Chapter 2 you will be able to: • Define and describe open and closed systems; • Define and describe the different forms of energy; • Define and describe properties and states; and • Define and describe processes and cycles.



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2.1

Introduction

Thermodynamics is about the conversion of one form of energy to another. Energy co.nversions, in tum, are described in terms of the various properties of substances and the changes of the properties that occur as a result of energy transformations. Thermodynamics is based on experimental observations. The observations are then expressed in terms of basic laws. For example, the First Law of Thermodynamics is an expression of the conservation of energy principle. The observation that some processes occur in one direction, but not in the reverse direction (for example, the flow of heat from a hot object to a cold one, or the generation of heat by friction) is embodied in the Second Law of Thermodynamics. However, before we can apply these laws as analysis tools to solve practical applications, we must develop a methodology of analysis. This methodology starts with the definitions of systems, properties, states, processes and cycles. The study of thermodynamics is framed around a system identified for analysis, and the interactions between that system and its surroundings. The condition, or equilibrium state, of a system is defined in terms of the values of the system's properties when at that condition. Properties are any characteristics of the system. The dynamics part of thermodynamics suggests the idea of change, and so changes of state are described in terms of processes and paths. Each of these ideas will be more fully explained.

2.2

Closed and Open Systems in Thermodynamics

In thermodynamics, the first step in finding a solution is identifying the system to be analyzed. The term thermodynamic system has a special meaning in this context as any quantity of matter, or any region in space, selected for study. The system is contained within an imagined system boundary, and everything outside the boundary is known as the surroundings. Heat and work (to be defined later) can cross the boundary in a form of interaction between the system and its surroundings, but it depends on the type of system as to whether mass can cross the boundary. There are two types of systems: closed and open. A closed system is one in which a quantity of matter is identified for study. A closed system definition is often called the control mass approach. A quantity of matter means that the mass to be studied will remain the same throughout the analysis. The matter in the system may change form, phase or location in space, but the system is identified as that matter that was originally targeted for study. There is an imaginary boundary drawn to contain the matter. The location of the matter, and therefore the shape of the boundary, may change with time; but to remain a closed system, no mass enters or leaves the system through the


Chapter:1 Systems, Properties, States and Processes

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boundary. Everything outside the boundary is known as the surroundings. Although there is no mass exchange across the boundary in closed systems, there may still be work and heat crossing the boundary as the system interacts with its surroundings. The other type of system is the open system. An open system is one in which a region in space is identified for study rather than a quantity of matter. An open system definition is often called the control volume approach. In an open system, mass can cross the imaginary boundary between the system and its surroundings. This is important because, as we shall see later, the mass crossing the boundary can also be a vehicle for carrying energy across the boundary. The region in space does not have to be fixed in time, although it is often chosen that way for convenience. In a way, the closed system is a special case of an open system, for which there is no mass flow across the boundary. An example of a closed system is shown in Figure 2-1. The object chosen for demonstra-

tion is a piston and cylinder combination with air in the cylinder. The air is trapped in the cylinder and there is no way for the air to escape, or for new air to come into the cylinder. The system is chosen to be the matter (mass of air) that is trapped in the cylinder. The system boundary is shown by the dotted line. The surroundings are then the piston itself, the cylinder wall, the cylinder head and everything else outside the dotted line. As weights are added to the piston and the piston moves to reduce the volume in the cylinder, the system (air in the cylinder) is compressed and changes shape, but it still maintains the same air that was originally targeted as the system.

Piston

I

Weight

I

Piston

Air System

, - - - -- -- -- --I

Ai r 1 1 :I ___________ System 1 J

1

1

After Adding Weight

Before Adding Weight Figure 2-1. Closed System


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Open systems are shown in Figures 2-2 and 2-3. Figure 2-2 shows a piston and cylinder arrangement like before, only now there is an open valve connecting the cylinder head to a storage tank. Once again, the system is identified as the matter (mass of air) that is in the cylinder at any instant in time. However, with the open system, the amount of air in the cylinder will change when a weight is placed on the piston and the piston moves. Some of the mass initially in the system is pushed across the system boundary and into the storage tank that is part of the surroundings. There is mass flow across the boundary. Although the open system analysis is also called a control volume analysis, the volume does not have to remain fixed in time or space. As can be seen in this case, the control volume changes shape and size during the change. Before leaving Figure 2-2, it is worthwhile to note that the case of the cylinder and the storage tank could have been treated as a closed system if the matter in the system had included both the air in the cylinder and the air in the storage tank. It may also be noted that a different choice of open system could have been to draw a boundary to include the piston and the weights in addition to the air. Which boundary is better depends on what is known, what is sought and what processes are occurring during the change.

Piston

------------

I

I

Weight Piston

Air System

r- - - - - - - - - - - -,

1 1 Air 1 I 1 I System 1 1 ______ - - - - _ ...I

------------

1

Storage Tank

Storage Tank

After Adding Weight

Before Adding Weight Figure 2-2. Open System



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Figure 2-3 depicts another example of an open system; in this case, a reciprocating compressor. The system boundary is drawn to include both the compressor and the air inside the compressor. Mass flows in across the boundary at the inlet, and out across the boundary at the outlet. In this open system, the control volume remains the same in time and space.

There are many choices when it comes to selecting the system and system boundary. The skill is in choosing a system that allows the most convenient analysis and calculation of the unknowns. We will return to the topic of system selection once we have introduced the First Law of Thermodynamics. The conceptual model of system, system boundary and surroundings will be used later to apply the laws of thermodynamics in a way that will permit calculation of practical and useful results. There is only one set of laws, but the thermodynamic relations (equations) that are applicable to closed and open systems are different, making it important to be able to recognize the type of system before starting the analysis. Before applying laws (such as the conservation of energy), we must be able to describe the condition of the system, and define what we mean by energy.

Inlet

-i---------------+

Outlet

-

.. _-------------Figure 2-3. Open System Reciprocating Compressor

Chapter 2 Systems. Properties. States and Processes


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2.3

Forms ofEnergy

The different forms of energy (such as thermal, mechanical, kinetic, potential, electric, magnetic, chemical and nuclear) add together to give the total energy of a system, E. The total energy on a unit mass basis, e, is defined as:

e=Elm

In thermodynamic analysis, it is common to group the forms of energy into two broad categories: macroscopic and microscopic. The macroscopic forms of energy are those that a system possesses as a whole with respect to some outside reference frame. Kinetic and potential energies are examples of macroscopic forms of energy because, as we shall see later, they are defined in terms of an outside reference frame. The microscopic forms of energy are those related to the molecular structure of a system and the degree of molecular activity. Microscopic forms are independent of an outside reference frame. The sum of all microscopic forms of energy is known as the internal energy of the system, U. On a unit mass basis, the specific internal energy is:

u=Ulm

Internal energy is measured at the molecular level. The individual molecules of a system are in continuous motion and the motion contributes to translational, rotational and vibrational molecular energy. The portion of the internal energy associated with this molecular motion is known as sensible energy. The degree of molecular activity increases with temperature; therefore, the sensible energy is higher at higher temperatures. The internal energy is also related to the intermolecular forces between molecules of the system. The forces that bind molecules are strongest in solids and weakest in gases. If sufficient energy is added to a solid or liquid, it will change phase and become a gas. The energy added to make the change is known as latent energy. Latent energy is part of internal energy. Sensible and latent energy changes involve no chemical change in the substance of the system. Chemical energy is part of internal energy. Chemical energy is associated with the chemical bonds between different substances. In cases of chemical change (such as occur in combustion or chemical reaction), there is breaking of some bonds and formation of new bonds, resulting in a change in chemical energy of the system.



2: 8

Within the nucleus of an atom are strong binding forces that also contribute to the internal energy of a system. These nuclear bonds, and therefore the nuclear energy, are changed only during fission or fusion reactions. Because thermodynamics is about change in energy forms and because HVAC&R does not commonly include nuclear fission or fusion, nuclear energy will not appear in the energy change relations. The macroscopic energy that a system possesses by virtue of its elevation in a gravitational field is called the potential energy. The potential energy, PE, and the potential energy on a unit mass basis,pe, are:

PE=mgz pe=gz where g is the gravitational acceleration (g = 32.2 ftls 2 on earth at sea level), m is the mass in Ibm' and z is the elevation of the center of gravity of a system relative to some arbitrarily chosen reference frame, in ft. There is often a unit conversion required to express the potential energy in the common unit offt·lbj known as gc' The parameter gc is a unit conversion constant, whose value depends on what units are used for mass. When using the units of mass as Ibm' the value is gc = 32.2 (Ibm ·ftls2)/lbj . The use of g and gc with apparently the same numerical value may seem odd, but it is a result of the fact that one lbj will accelerate one Ibm at 32.2 ftls2. To convert ft'lbj to Btu, the conversion factor is 1 Btu = 778 ft·lbj . Thermodynamics is the study of energy change rather than the study in absolute values of energy, and the use of an arbitrary reference, or datum, is acceptable. An example can be seen in analyzing the increase in potential energy when lifting a weight from a chair on to the top of a table. It is only the height difference between the chair and tabletop that matters, and not how high the table and chair are above the floor. In thermodynamics, it is change of energy that is important and the reference level of absolute total energy is chosen at some convenient datum level. As an example of the change of potential energy and the units of energy, consider a mass of214 Ibm lifted 2 ft from the chair to the table:

PE increase

mgz (214 Ibm )(32.2

ftls 2)(2

ft)/ {32.2 (Ibm ·ftls2)/lbj

}

= 428 ft·lbj = 0.55 Btu pe Increase

= 2 ft·lbj/lb m = 0.0026 Btullbm


Fundamentals ofThermot/ynamics and Psychrometries

2: 9

The macroscopic energy that a system possesses by virtue of its motion relative to some reference frame is known as the kinetic energy. When all parts of the system move with the same velocity, V (ft/s), then the kinetic energy, KE, and the kinetic energy on a unit mass basis, ke, are:

The units of kinetic energy are handled similarly to those of potential energy and include the unit conversion constant gc' and often the energy conversion factor I Btu = 778 ft·lb!. The magnetic, electric, nuclear and surface tension effects are important in some cases, but are usually not significant in HVAC&R problems and will not be dealt with here. The total energy, E, and the total energy per unit mass, e, may be written as the sum ofthe internal, potential and kinetic energies:

E = U + PE + KE

= U + mgz + m V2 /2 2

e = u + pe + ke = u + gz + V /2

You must be careful about units when doing a calculation involving potential, kinetic and internal energy so as to be sure that they are each in compatible units.

2.4

Properties ofa System

A property of a system is any observable characteristic of the system. Common properties are mass m, temperature t, pressure p, volume V, specific volume v, density p, specific heats Cp and C", internal energy U, enthalpy H, and entropy S. Some of the terms you will immediately recognize, but others we will have to discuss in more detail in the next chapter on properties. Properties that are independent of the size of the system (such as pressure, temperature and density) are known as intensive properties. Properties that vary with the size, or extent of the system (such as mass, volume and total energy) are known as extensive properties. When extensive properties are expressed per unit mass, they become intensive properties and are known as specific properties, such as specific total energy (e = Elm), specific internal energy (u = Ulm), and specific enthalpy (h = Him).



2: 10

Common properties will be more fully defined and described in the next chapter. Properties playa key role in conducting a thermodynamic analysis because of the following postulate: The state of a simple compressible system is completely specified by two independent, intensive properties. To fully understand the impact of this postulate, we must further define what is meant by a thermodynamic state, and then elaborate on properties and independent properties.

2.5

Equilibrium States

Consider a system that is not undergoing any change. It is possible to calculate all the properties of the system. This set of properties then completely determines the condition, or state, of the system. Said another way, the thermodynamic state is the condition of a system in which all the properties have a fixed value. If the value of anyone property changes, then the system will be in a different state. The previous paragraph started with the reference to a system not undergoing change. In thermodynamics, this is known as an equilibrium state. Equilibrium conveys a sense of balance. For a system in an equilibrium state, there are no unbalanced driving forces within the system. For thermal equilibrium, there are no internal temperature gradients that will cause heat flow within the system. For mechanical equilibrium, there is no change of pressure in time to cause mass flow. The static pressure distribution does not violate equilibrium conditions because it does not induce an internal mass flow. For phase equilibrium, there is no relative change in the mass of each phase present in the system. For chemical equilibrium, there are no chemical reactions occurring. The equilibrium state of a system is entirely specified by giving the property values of that state, or conversely, the set of values of the properties is entirely set by identifying the state. Properties are independent of the path followed by the system in coming to a given state. In fact, a property can be defined as any quantity that depends on the state of the system and is independent of the prior history by which the system arrived at the given state.



2: 11

2.6

Processes and Cycles

A change in equilibrium state, as defined by a change in properties, is known as a process. The series of states passed through by the system during a process is known as the path. The description of a process involves specifying the initial and final equilibrium states, the path, and any interactions that occur between the system and the surroundings at the system boundaries during the process. When a system is undergoing change, it is by definition not in equilibrium. However, when the process proceeds in such a manner that the system remains infinitesimally close to equilibrium state at all times, then the process is called a quasi-static, or quasi-equilibrium, process. A quasi-equilibrium process is an idealization of real processes, but it can be a very powerful tool in helping to analyze systems. Most of the systems in HVAC&R that we want to analyze are reasonably described by this idealization, and are adequately analyzed in this way. A quasi-equilibrium process can be thought of as a process that proceeds slowly enough that the system can adjust internally so that the properties in one part of the system do not change significantly faster than the properties in another part of the system.

Figure 2-4 depicts a quasi-equilibrium process of air being compressed in a cylinder. The piston and cylinder are shown horizontally at the bottom of the diagram. The closed system is the air trapped in the cylinder at any Pressure ®Final time. The piston is ________ State , initially in Position ,,, 1. With the piston in ,,, Position 1, the propp --------l.-----------~---...... G) Initial erties of the system , State (air in cylinder) are ,,I mass m l , pressure ,,, PI' temperature tl' ' - - - - - - - I . - : - 7 ' " - - - - - - - - - ! - : - 7 " " " - -.. Volume V V2 volume VI' specific ,,: V1 volume vI' internal ,-------, energy VI' specific ,, A',r ,, internal energy up , System , ,, ,, entropy S I and spe,--------, cific entropy S I' These property val® CD ues determine the Figure 2-4. Change of State for a Closed System initial state as State 1. The piston is ~

t


t


2: 12

slowly moved to the left. After some time, it comes to the final Position 2. The properties have changed. We can see that the volume has changed to a new, smaller volume V2• Because this is a closed system and the mass has not changed (m2 = ml ), the specific volume will have changed to v 2• The pressure may have increased to P2.1t is not known if the other properties have changed values, but it is clear the system has gone to State 2 and so the properties are labeled as (2' U2, u2, S2 and S2' Also shown in Figure 2-4 is the property diagram of pressure plotted against volume. The Initial and Final States (1 and 2) are shown as points on the p-V diagram. The process path is shown as the path taken in going from State 1 to State 2. The pressure-specific volume (p-v) diagram for an open system (axial flow fan) is shown in Figure 2-5. The p-v diagram looks very similar to that for the closed system described earlier, but the States 1 and 2 here refer to the inlet and outlet states of the air blowing through the fan. Many of the practical situations we will want to analyze take the form of the open system shown in Figure 2-5. However, there are often multiple inlets and outlets. We will deal with them when we come to applying the mass and energy conservation laws.

Pressure

p

®

~ I I I I I I I

- - - - +- - -- -- - -- - - ---- - ::--=-:-::-=-----0 I I I I I I I

:V1

Specific Volume V

I I

® Figure 2-5. Change of State for an Open System


Fundamentals ofThermotiynamics and Psychrometrics

2: 13

There are many different possible process paths depending on the nature of the interaction between the system and its surroundings and the physical make-up of the system. For example, if a container of fixed volume is heated as shown in Figure 2-6, then the temperature and pressure would increase, but the volume would remain constant. Or, if the constant weight piston of Figure 2-7 is allowed to drop as the air is let out of the cylinder, then the pressure would remain constant, even though the volume changes. There are sets of such processes in which one of the properties remains constant that are very useful for approximating actual situations. They are: • Isobaric process:

Constant pressure process, p

• Isothermal process:

Constant temperature process, t = constant

• Isometric process:

Constant volume process, V = constant

• Isochoric process:

Constant specific volume process, v = constant

• Isentropic process:

Constant entropy process, S = constant

• Adiabatic process:

No heat exchange between system and surroundings

=

constant

The last of the process descriptions given above, the adiabatic process, is different from the others because it does not refer to a property, but instead refers to heat exchange between the system and its surroundings. Heat is not a property, and we will provide a very detailed description of what is meant by heat in a later chapter.

,------1

.------1

1

I I

I 1

I

1

I I

V2

1

P2

I

1

1 I t2 lI ______ JI

I I I ______ JI 1

~Heating

State 1 L

State 2

I

Figure 2-6. Constant Volume Process


Chapter 2 Systems. Properties. States and Processes

2: 14

I weight I piston ,------

I r------ .. weight piston

I I V1 I I I P1 I I I t1 I L I _______

I I I I I I I

I I I I I I I

-------.1

Closed Valve

!

State 1

Open Valve

I weight I piston ,-----. V2

I I I I I I

P2= P1 t 2

I I I I I I

-------..1

18 Closed Valve

State 2

Figure 2-7. Constant Pressure Process

A system is said to have undergone a cycle if, after having been through a series of processes, the final state is identical to the starting state. The initial and final states for a cycle are identical, which means that the initial and final properties are also identical. Examples of two-process and four-process cycles are shown on the p-V diagrams in Figure 2-8. The four-process cycle is an idealization of the thermodynamics that occur in a gasoline engine.



2: 15

Pressure p

Volume V Two-Process Cycle

Pressure p

®

0---+® ®---+@) @)---+(D 0---+0

Isentropic Compression Constant Volume Isentropic Expansion Constant Volume

Volume V Four-Process Cycle Figure 2-8. Property Diagrams for Two- and Four-Process Cycles



2: 16

The Next Step In the next chapter, we will further develop the definition and description of the common properties in the thermodynamics of HVAC&R. The properties to be considered are pressure, temperature, specific volume and density, internal energy, enthalpy and entropy. Also, the idea of a pure substance and the phases of a pure substance will be introduced.

Summary

The starting point of a thermodynamic analysis is the identification of a system. The definition of a thermodynamic system and the ideas of system boundaries and the surroundings were introduced for both open and closed systems. Total system energy was introduced as the sum of the macroscopic and microscopic forms of energy. Macroscopic forms (such as potential and kinetic energies) and microscopic forms of internal energy (such as thermal, phase and chemical energies) add together to make the total energy. A property of a system was described as an observable characteristic of the system. There are intensive and extensive properties. Properties that are independent of the size of the system (such as pressure, temperature and density) are known as intensive properties. Properties that vary with the size or extent of the system (such as mass, volume and total energy) are known as extensive properties. When extensive properties are expressed per unit mass, they become intensive properties and are known as specific properties. A set of property values determines the condition, or state, of a system in equilibrium. A process was defined in terms of the path by which a system changes from one state to a different state. Several idealized processes were described. A cycle is a series of processes linked together so that the initial state is identical to the final state. After studying Chapter 2, you should be able to: • Define and describe open and closed systems; • Define and describe the different forms of energy; • Define and describe properties and states; and • Defme and describe processes and cycles.



2: 17

Skill Development Exercises for Chapter 2 Complete these questions by writing your answers on the worksheets at the back of this book. Be sure to include your name and address. Send your completed questions to the ASHRAE Education Department.

2-01. Using examples, explain the main difference between open and closed systems. 2-02. Automobile engines are usually cooled by antifreeze that is circulated between the engine and the radiator by a centrifugal pump. Air is blown over the radiator to cool the antifreeze. Sketch the following systems and state whether they are open or closed: • Antifreeze in the pump • Antifreeze in the radiator • Antifreeze in the pump, radiator and engine water jacket • Antifreeze in the radiator and air in the fan

2-03. Describe the differences between thermal, phase, chemical and nuclear energies. 2-04.

Give an expression for the specific total energy of a system in terms of the internal, potential and kinetic energies.

2-05.

Sketch on a p- V diagram what happens when trapped air expands in a closed cylinder with a piston. Be sure to label the initial and final states, the path and the major properties.

2-06.

Sketch on a p- V diagram a four-process cycle that has process 1 to 2 as an isobaric decrease in volume, 2 to 3 as an isometric increase in pressure, 3 to 4 as an isobaric increase in volume, and 4 to 1 as an isometric decrease in pressure.

2-07.

Sketch on a t-p diagram the following four-process cycle: 1 to 2 isothermal; 2 to 3 isobaric; 3 to 4 isothermal; and 4 to 1 isobaric.



3: 1

Chapter 3 Property Diagrams for Pure Substances

Contents of Chapter 3 • Instructions • Study Objectives of Chapter 3 • 3.1 The State Postulate • 3.2 Properties • 3.3 Phases and Phase Changes for Pure Substances • 3.4 Property Diagrams • The Next Step • Summary • Skill Development Exercises for Chapter 3


Study Objectives of Chapter 3 Chapter 3 begins with a description of the state postulate and then defines the properties first introduced in the previous chapter. There are descriptions of pressure, specific volume, temperature, internal energy, enthalpy and entropy. The solid, liquid and vapor phases of a pure substance are introduced, including the ideas of phase equilibrium, independent properties, and saturated conditions for liquids and vapors. States and state changes are displayed on property diagrams. After studying Chapter 3, you should be able to: • Describe those properties common to analysis in thermodynamics; • Define a pure substance; • Describe the phases and phase change characteristics of pure substances; and • Explain and give examples of the value of property diagrams in describing the state and state change of a pure substance. Fundamentals ofThernwdynamics and Psychrometries

Chapter 3 Property Diagramsfor Pure Substances

3:2

3.1

The State Postulate

A property of a system is any observable characteristic of the system. Common properties are mass m, temperature t, pressure p, volume V, specific volume v, density p, specific heats Cp and C", internal energy U, enthalpy H, and entropy S. Properties that are independent of the size of the system (such as pressure, temperature and density) are known as intensive properties. Properties that vary with the size or extent of the system (such as mass, volume and total energy) are known as extensive properties. When extensive properties are expressed per unit mass, they become intensive properties and are known as specific properties, such as specific total energy (e = Elm), specific internal energy (u = Ulm), and specific enthalpy (h = Him). Consider a system that is not undergoing any change. It is possible to calculate all the properties of the system. This set of properties then completely determines the condition, or state, of the system. Said another way, the thermodynamic state is the condition of a system in which all the properties have a fixed value. If the value of anyone property changes, then the system will be in a different state. The equilibrium state of a system is entirely specified by giving the property values of that state or, conversely, the property values are entirely set by identifying the state. Properties are independent of the path followed by the system in coming to a given state. In fact, a property can be defined as any quantity that depends on the state of the system and is independent of the prior history by which the system arrived at the given state. There are many different properties of a system, but not all of the properties are independent in value of one another. For example, consider air at a given pressure p and absolute temperature T. From the ideal gas law (to be introduced later), the specific volume v is given by v = RT/p. R is the gas constant for air. Because of this interdependence between some properties, it raises the question of how many properties must be specified before the state of a system is completely specified. Fortunately, for most of the applications we want to consider, the question is reasonably answered by the following state postulate: The state of a simple compressible system is completely specified by two independent, intensive properties.

There are several terms in this postulate that require careful attention. A simple compressible system is one in which there are no electrical, chemical, magnetic, gravitational or surface tension effects. Often these effects, which are produced by external fields, are not significant to the calculation. However, if they are, then an additional property must be specified for each effect. For example, if gravitational effects are important, it will be necessary to specify an elevation and the two thermodynamic properties to specify the state. Chapter 3 Property Diagramsfor Pure Substances


3: 3

An intensive property has already been defined as a property that is independent of the size of the system (such as pressure, temperature and density) or a property that is expressed per unit mass, such as specific total energy (e = Elm), specific internal energy (u = Vim) and specific enthalpy (h = Him). Intensive properties are independent if one property can be varied while the other is held constant. We will examine more closely the question of independence when we discuss phase change in a later section.

3.2

Properties

Pressure. In solid mechanics, force per unit area is known as stress. If the direction of the force is perpendicular to the plane of the area it acts on, it is known as a normal stress. If the force is tangential to the plane ofthe area, it is known as shear stress. In fluids (liquids and gases), there are similar stresses. The normal stress in a fluid is called pressure, and the tangential stress is called shear stress. Pressure is the normal force per unit area in a fluid and is positive when the stress is compressive. For a fluid at rest, the pressure at a given point is equal in all directions. However, the pressure increases with depth because of the weight of the fluid. This is known as the change in hydrostatic pressure. The variation of pressure with depth, while being equal in all directions, is shown in Figure 3.1. When the fluid is in motion, the pressure may vary from point to point anywhere in the fluid .

........ - . . - - - - - - -........- - - - - " - - " ' - - - - - - -.....~----- Water Level

.

-. ..-

t

j ---........--1 Figure 3-1. Hydrostatic Pressure Variation With Depth



3:4

The units of pressure are force per unit area. In the inch-pound (IP) system of units, the pressure is usually given as pound-force per square inch (lb/in2, or psi). In the SI system of units, the unit of pressure is a Pascal, where 1 Pa = 1 N/m2• Because a Pascal is a small unit, pressure is often expressed in kilopascals (kPa = 103 Pa), or megapascals (MPa = 106 Pa). Typical units and unit conversions are given in Appendix A. The units of pressure are'often quoted in terms of the equivalent height of fluid that may be supported by that pressure. In fluids at rest, the hydrostatic pressure varies as the depth according to the relation!J.p = pg!J.z where !J.p is the pressure change corresponding to an elevation change of!J.z. The fluid density is p. For example, consider standard atmospheric pressure of29.921 in. mercury (Hg). The ratio ofthe density of mercury to that of water is 13.6 (known as specific gravity, SG). The density of water is 62.4 Ibm Ift3. A unit conversion factor of 32.2 Ibm·ftI(lbj ·S2) is required. The pressure difference between atmospheric and zero pressure is then: !J.p

=

pg!J.z

=

(13.6 )(62.4lb)ft3)(32.2 ftls 2)(29.921 in.)(ftl12 in.)/{32.2lbm·ftI(lbjs2)}

=

2,1161bj lft2

=

14.7 psi

Similarly, the equivalent height of fluid can be calculated for a given pressure difference. Expressing 14.7 psi in feet of water yields: !J.z

=

!J.pl(pg)

=

(14.7 lbj lin. 2)(144 in. 2/ft2)(32.2 Ibm·ftI(lbj ·S2) )1 {(62.4 Ib)ft3)(32.2 ftls2)}

= 33.9 ft water Absolute pressure is usually used in thermodynamics. However, the pressure is often measured by instruments that give the pressure relative to atmospheric pressure. These instruments read zero when the pressure is atmospheric. To distinguish such pressure from absolute pressure, the measurement is known as gage pressure, and the units are Ibj lin. 2 gage, or psig. Sometimes absolute pressure is given as psia to emphasize the absolute. Figure 3-2 depicts the different terms of pressure measurement to provide a picture of the relationships between absolute, gage and vacuum measures of pressure. Pressure measurements below atmospheric pressure are called vacuum measurements. A standard atmosphere is taken as: 1 atm = 14.696 psi = zero psig = 101.325 kPa = 29.921 in. Hg



3: 5

pressure

P gage P atmos

--------- ------------- fo-------

P absolute ---

P vacuum P atmos P absolute

Figure 3-2. Terms of Pressure Measurement Temperature. Temperature is not an easy property to define. We have a feeling for temperature because we are used to expressing whether something is hot or cold. It is perhaps easier to define the equality of temperature as: Two bodies have equal temperatures when no change in any observable property occurs when they are in thermal communication. This is the same as saying that if two bodies are in thermal equilibrium with a third body, then they are also in thermal equilibrium with each other. In the case where a thermometer is used to measure the temperature, this may be further simplified to say that two bodies are in thermal equilibrium if both have the same temperature reading even if they are not in contact. This principle is the basis for establishing a temperature scale. By examining materials that have temperature-dependent characteristics, we can assign an arbitrary temperature scale to the observed effect. The mercury thermometer is such a device. Mercury expands with temperature. By observing the densities of mercury at the freezing and boiling points of water at atmospheric pressure (known as the ice point and steam point, respectively), it is possible to construct a temperature scale by assigning a temperature change proportional to the density change. Two such scales are the Fahrenheit and Celsius scales. For the Fahrenheit scale, the ice and steam points were assigned values of 32°F and 212°F, respectively. On the Celsius scale, the ice and steam points were assigned values ofO°C and 100°C. There has been a slight revision to the Celsius scale by identifying the triple point of water to be O.Ol°C rather than the ice point. The triple point is the state in which all three phases of water coexist in equilibrium with each other.



3:6

A more useful scale for thermodynamics is the absolute temperature scale. For the absolute scale, the temperature is always positive. It is based on the concept that for an ideal gas at constant volume, the temperature is proportional to the pressure. This means that as the pressure approaches zero, the temperature also approaches zero. On the absolute scale, temperatures are measured in degrees Rankine, R, for the IP unit system, and degrees Kelvin, K, for the 81 unit system. Temperatures on the absolute scale are referred to with an upper case T. The relationships between these scales can be summarized as:

t (OF)

1.8 t(OC) + 32

and

t(OC)

=

[t(OF) - 32]/1.8 t(OC) + 273.15

T(R)

=

t(OF) +459.17

and

T(K)

=

T(R)

=

1.8 T(K)

and

T(K)

= T(R)I1.8

and

~T(K) =

~T(R) = ~t(OF)

M(OC)

Temperature is an intensive property.

Specific volume. The specific volume of a substance v is the volume per unit mass. The specific volume is an intensive property and has the unit of ft3/lbm . The inverse of the specific volume is the density p (Ibm Ift3 ). In fluid mechanics, it is common to use density, but in thermodynamics, it is just as common to use specific volume.

Internal energy. In thermodynamic analysis, it is common to group the forms of energy into two broad categories: macroscopic and microscopic. The microscopic forms of energy are those related to the molecular structure of a system and the degree of molecular activity. Microscopic forms are independent of an outside reference frame. The sum of all microscopic forms of energy is known as the internal energy of the system, U. On a unit mass basis, the specific internal energy is:

u=Ulm Internal energy is measured at the molecular level. The individual molecules of a system are in continuous motion and this motion contributes to translational, rotational and vibrational molecular energy. The portion of the internal energy associated with this molecular motion is known as sensible energy. The degree of molecular activity increases with temperature; therefore, the sensible energy is higher at higher temperatures. The internal energy is also related to the intermolecular forces between molecules of the system. The forces that bind molecules are strongest in solids and weakest in gases. If sufficient energy is added to a solid or liquid, it will change phase and become a gas. The energy added to make the change is known as latent heat. Latent heat is part of internal energy. Changes in sensible and latent energy involve no chemical change.



3: 7

Chemical energy is part of internal energy. Chemical energy is associated with the chemical bonds between different substances. In cases of chemical change (such as occur in combustion or chemical reactions), there is breaking of some bonds and formation of new bonds, resulting in a change in chemical energy of the system. Within the nucleus of an atom are strong binding forces that also contribute to the internal energy of a system. These nuclear bonds, and therefore the nuclear energy, are changed only during fission or fusion reactions. Because thermodynamics is about change in energy forms, and because HV AC&R does not commonly include nuclear fission or fusion, nuclear energy will not appear in the energy change relations.

Enthalpy. Enthalpy is the name given to the combination of internal energy, pressure and volume. Because it is a combination of properties, enthalpy is a property itself. It is a particularly useful combination, as will be shown when applying the First Law of Thermodynamics to open systems. The enthalpy H, and the specific enthalpy h, are given by the following two expressions: H=U+pV h=u+ pv

The units of specific enthalpy are the same as those of internal energy. If common units of psi and ft3/lb m are used for p and v, then the product pv has to be converted to produce Btu/Ibm as shown by the following: h

=

u (Btu/Ibm) + p (lbj /in. 2) v (ft3/Ib n)(144 in. 2/ft2)(Btu/(778 ft·lbj

))

Enthalpy will become a very convenient property when we apply the First Law of Thermodynamics to open systems. In open systems, enthalpy is a useful descriptor for the internal energy and flow work done by a flow of mass across the system boundary.

Entropy. Entropy is another new property that is not easily measurable or obvious. It is a property that is linked with the Second Law of Thermodynamics. We will not derive the basis for entropy, but we will make use of the tabulated values of entropy to describe certain types of ideal processes. The Second Law of Thermodynamics deals with the fact that many processes proceed in only one direction. For example, hot coffee left out in a cool room will cool to the temperature of the room, but the coffee will not, by itself, become hotter. In another example, when two pieces of wood are rubbed together, there can be enough frictional heating to start a fire, but a hot spot, by itself, will not cause the two pieces of wood to start rubbing against each other. These processes are known as irreversible.



3: 8

In contrast, reversible processes can only be imagined (such as frictionless pistons or heat transfer between objects at the same temperature). Although a frictionless piston does not exist, it is useful to consider it as an ideal, and then compare real behavior to the ideal. Similarly, in thermodynamics, it will prove to be very useful to consider ideal reversible processes, and then compare actual, irreversible processes to the reversible idealizations. Entropy is a convenient property for tracking ideal processes, and also for identifying the degree of irreversibility of real processes. The symbols for entropy and specific entropy are Sand s. The units of entropy are energy over absolute temperature. The units of S are BtuIR, and of s are Btu/(lbm R).

3.3

Phases and Phase Changes for Pure Substances

A pure substance is characterized by the requirements that it be homogeneous and of invariable chemical composition. A pure substance can be a single chemical element, a compound, or a mixture of elements and compounds. Water is a pure substance because it always has the composition of H20, although it may have different phases: solid (ice), liquid (water) or gaseous (steam or water vapor). Air in the gas phase is considered a pure substance even though it is made up of several gases. But air will not be considered as a pure substance if it mixed with liquid air because the composition of the gaseous air and liquid air is different when they are in equilibrium with each other. Similarly, the air of combustion is not a pure substance because it is chemically reacting with the burning fuel. We will start our thermodynamics with the discussion of pure substances. The basic laws and principles of thermodynamics do apply to other than pure substances, but it takes more advanced study. The three principle phases of a pure substance are solid, liquid and gas. As we study phase change in thermodynamics, we do not deal with issues of molecular structure, but some description of molecular structure can provide insight into the nature of the three phases. Solids have a crystalline structure. In solids, the molecular bonds are strong and the molecules are fixed in a three-dimensional lattice. Although the molecules do not change location with respect to each other in the lattice, they continually oscillate about their equilibrium position. The oscillation intensity depends on the temperature. If the temperature is raised high enough, the molecules develop enough oscillation energy to overcome the molecular bonds that keep them fixed in the lattice. With additional energy, the solid melts and becomes a liquid. There are still strong intermolecular bonds and neighboring molecules exert a significant influence over each other, but the molecules can move relative to each other. At a higher temperature, the molecules abChapter 3 Property Diagramsfor Pure Substances

Fundamentals ofTI,ermot/ynamics and Psychrometries

3: 9

sorb energy, the distances between molecules increase and the molecular order breaks down. The liquid evaporates and changes phase to become a vapor (gas phase). In the gas phase, molecules are far apart and move around at random, colliding into each other and the container walls. Under certain circumstances, a solid may sublimate and go directly from the solid to vapor phase when heated. The phases and phase change processes in pure substances are important in thermodynamics. To illustrate the basics of phases and phase changes of a pure substance, we will describe the behavior of water changing from liquid to vapor. Although the example is water, it should remembered that the same basic explanation applies to all pure substances, and there are similar changes between solid-to-liquid and solid-to-vapor. Consider as a system the water in a piston-cylinder arrangement shown in Figure 3-3. Imagine that the piston has no weight so that, although the piston can travel up and down, the pressure in the system remains constant at 14.7 psia (atmospheric pressure) throughout the series of steps described below. The water is initially all liquid at 60 oP. In this state, the water is known as a sub-cooled, or compressed, liquid.

Piston

Vapor Piston

Piston

Liquid

Liquid

Liquid

State 1 compressed liquid t<1sat

State 2 saturated liquid 1 lsat

State 3 saturated (liquid + vapor) t lsat

=

Piston

=

Piston

Piston

Vapor

Vapor

Vapor

State 4 saturated vapor t 1sat

State 5 superheated vapor t> lsat

State 6 gas 1»1sat

=

Figure 3-3. Stages in the Constant Pressure Change of Phase of Water



3: 10

The water in the cylinder is then heated so that the temperature rises. The water expands slightly as the specific volume increases, causing the piston to rise slightly (State 2). The pressure remains constant at 14.7 psia because of the weightless piston and no constraint on piston movement. When the temperature first reaches 212°F, the water is still all liquid and remains liquid unless further heat is added. If further heat is added, then the liquid progressively changes phase from liquid to water vapor or steam (States 3 and 4). The temperature and pressure throughout the whole phase change process remain constant at 212°F and 14.7 psia respectively, but the specific volume increases from 0.0167 ft 3/lb to 27.82 ft3/lb m • The conditions during the phase change are called saturated conditions and the temperature and pressure are known as saturated pressure Psat and saturated temperature tsat • lll

Ifheating continues after the liquid is all converted to vapor, the temperature will rise above tsat" In this case, imagine the temperature to be 250°F and the pressure still 14.7 psia (State 5). The vapor in the system is described as superheated vapor when t> tsat" If the temperature is much greater than the saturated temperature, t >> tsat ' then the vapor is said to behave as a gas (State 6). We will explain this in a later chapter. When the temperature first reached 212°F and the water was all liquid, but before the phase change started to take place, the state of the water is described as saturated liquid (t = 212 of, P = 14.7 psia, and all liquid). When the liquid had just been fully converted to vapor, the state is known as saturated vapor (t = 212°F, P = 14.7 psia, all vapor). Between the saturated liquid and saturated vapor states, both liquid and vapor are in equilibrium with each other. The temperature and pressure are 212 OF and 14.7 psia respectively, but to determine where we are in the phase change, we introduce a new property called the quality. The quality, x, of a pure substance only has meaning during a phase change. The quality during the phase change from liquid to vapor is the ratio of mass of vapor to total mass in the system: x

= mvapor 1m

(1- x)

= mliquid 1m

That means that x is zero when there is only saturated liquid, and 1 when there is only saturated vapor. The quality can easily be related to the specific volumes of saturated liquid and vapor. Let ~ be the volume and Vj be the specific volume of the saturated liquid. Let Vg be the volume and vg be the specific volume of the saturated vapor. The volume of vapor is V g = vg m vapor . The volume of liquid is Vj = Vj mi·IqUi·d'


Fundamentals ofTI,ermot/ynamics and Psychrometries

3: 11

The mass is made up of the sum of the liquid and vapor masses, m = m vapor + m."qUid. :

= mliquid V j + mvapor V g =(l-x)mVj +xmVg Divide through by m and get: V

= (l-x)vj +XVg

This is a useful relationship because it determines the specific volume for a mixture of the two phases. Often the difference between the specific volumes of saturated vapor and liquid is given the designation Vjg = (Vg - Vj)' and the above expression can be rewritten as:

Again, this refers to the property called quality, which has meaning for cases where there is a mixture of two phases of a pure substance. During phase change, we have seen that the temperature and pressure are not independent properties. That means that if the mixture is saturated, then setting either the pressure or temperature will fix the other. It is under these saturated mixture conditions that quality can serve as an independent intensive property. The above discussion used water as an example of a pure substance. All pure substances exhibit similar behavior, but at different values of temperature and pressure. Also, there is not only the phase change between liquid and vapor to be considered. There are also phase changes between solid and liquid, and solid and vapor. These issues will be elaborated in the next section, with the help of property diagrams.


Chapter 3 Properly Diagrams for Pure Substances

3: 12

3.4

Property Diagrams

A two-dimensional property diagram is a plot of one property against another. Because the state postulate asserts that it takes only two independent, intensive properties to determine the state of a pure substance, such diagrams are very useful for visualizing states and changes of state. We will start in this section with a temperature versus specific volume (T-v) diagram to illustrate the constant pressure process that was described earlier for water in a cylinder that was heated and changed from compressed liquid to superheated vapor. The process is shown in Figure 3-4. The States 1,2,3,4,5 and 6 correspond to those first depicted in Figure 3-3. State 2 is saturated liquid, State 4 is saturated vapor, and in between are saturated mixtures of vapor and liquid where the state is determined by the quality. The question arises as to what happens if we follow the same process, but at a higher pressure (we can add a weight to the piston in Figure 3-3). The answer is shown in a more general T-v diagram in Figure 3-5. The constant pressure changes of state are shown as dashed lines for two different pressures PI and P 2 ' for a pure substance being heated from compressed liquid to superheated vapor. The conditions that represent the points of saturated liquid can be joined to draw the saturated liquid line that separates the compressed liquid region from the saturated (liquid + vapor) region. Similarly, the conditions that repre-

Temperature

tOF

CD

Gas

400 Saturated Liquid

® 200

Saturated Mixture

Specific Volume V

Figure 3-4. Constant Pressure Boiling of Water on a T-v Diagram



3: 13

sent the points of saturated vapor may be joined to draw the saturated vapor line. The saturated liquid and vapor lines converge at the top of the curve at the critical point. The critical point has special significance because it characterizes the pure substance. It also represents a temperature/pressure combination above which the transition from liquid to vapor happens without a clear distinction between saturated liquid and vapor.

Temperature

tOF

Critical Point

,s.0 ~"

(I:)~ ~0

Compressed Liquid Region

/

/ /

~

Superheated Vapor Region

/ /

(l)

/.s

li-----------'

/ /

/ / /

::;;, .~

/ / I -'

j

/1 / I I I .a / I

e

c3

Saturated (Liquid & Vapor) Region

Specific Volume V

Figure 3-5. The T-v Diagram

We will become very familiar with diagrams such as the T- v diagram in Figure 3-5, and the p-v diagram in Figure 3-6. In the p-v diagram, lines of constant temperature are shown as dashed lines for the temperatures TJ and T2 • The saturated liquid and saturated vapor lines, and the critical point, look very similar to those of the T- v diagram. Again as a reminder, the real power of these diagrams is that the state of a system is easily represented as a point on the diagram. There is more to property diagrams than the two shown in Figures 3-5 and 3-6. We might plot other property combinations. It will later be useful to plot temperature against entropy and temperature against enthalpy. There is also the extension of the diagram to include phase changes between solid, liquid and vapor.



3: 14

pressure

p

I I I

Compressep Liquid I Region I

Critical Point

I I I I I

Superheated Vapor Region

' "'"

-

I_~ ____ - - - - - -

~

.S!"

N .a

ciJ

,

"

Saturated (Liquid & Vapor) Region

'. ' ,

~

-..J 1:)

,

~~

'lJe/"qfl '

';/"e .,.

, ,

~"

Grr~/61.

'

Q' U

fel1]~

q..oo 1"1..'IfJe

'<

e/"qfll. ~e

l:

1

Specific Volume V

Figure 3-6. The p-v Diagram

The p- T, or phase, diagram is shown in Figure 3-7, where three lines all converge at one point. This point is known as the triple point because the solid, liquid and vapor phases all co-exist in equilibrium with each other at this state. The three lines that separate the solid, liquid and vapor regions on the diagram are known as the fusion line, vaporization line and sublimation line. The vaporization line is also known as the condensation line, depending on whether the change is from liquid to vapor, or vapor to liquid. The vaporization line represents an edge view of the saturated liquid and vapor lines that were presented in the Tv and p-v diagrams. The fusion, or melting, line separates the solid and liquid regions. The sublimation line separates the solid and vapor regions. To fully illustrate the relationship between the properties of pressure, temperature and specific volume, it is possible to construct the p-v-T surface in a three-dimensional picture. The picture helps to visualize the various phases.



3: 15

pressure

p Liquid

Phase

Critical Point

Solid

Phase

Vapor

Phase

Temperature T

Figure 3-7. Phase Diagram of a Pure Substance



3: 16

The Next Step In the next chapter, we will discuss the properties of the common pure substance, water, in more detail. We will examine the thermodynamic tables and charts that give values for the properties over a wide range of conditions.

Summary

The state postulate established that the state of a simple compressible system is completely specified by two independent, intensive properties. Descriptions were provided for pressure, temperature, volume, internal energy, enthalpy and entropy, including explanations of the common units of these properties. Enthalpy was introduced as a property equal to the sum of internal energy and the pressure-volume product. Enthalpy will be very useful when applying the First Law of Thermodynamics to an open system. Entropy was introduced as a property of particular significance when dealing with the Second Law of Thermodynamics, and constructing ideal, reversible processes and cycles. The solid, liquid and vapor phases of a pure substance were introduced, including the ideas of phase equilibrium, independent properties and saturated conditions for liquids and vapors. The quality is introduced as an intensive, independent property relating the mass of vapor to total mass of a system when saturated liquid and vapor exist simultaneously in equilibrium with each other. An expression was developed relating the specific volume of a saturated mixture of liquid and vapor to the quality and the specific volumes of the liquid and vapor. States and state changes are displayed on the T-v and p-v property diagrams. The diagrams are used to introduce the ideas of the saturated liquid line, saturated vapor line and the critical point. After studying Chapter 3, you should be able to: • Describe those properties common to analysis in thermodynamics; • Define a pure substance; • Describe the phases and phase change characteristics of pure substances; and • Explain and give examples of the value of property diagrams in describing the state and state change of a pure substance.



3: 17


3-01.

Give the state postulate and explain what a simple compressive substance is.

3-02. Explain the differences between absolute, gage and vacuum pressures. 3-03. Give the typical units for the following properties, and indicate which properties are intensive and which are extensive: specific enthalpy, total volume, pressure, temperature, specific entropy, mass and quality. 3-04.

Give the temperature of212 cP in units of degrees Rankine, degrees Kelvin and cC.

3-05. Sketch a T-v diagram for a pure substance and show the saturated liquid line, the saturated vapor line, the critical point and a constant pressure process line from compressed liquid to superheated vapor. Also include a constant temperature line. 3-06. Sketch a p-v diagram for a pure substance and show the saturated liquid line, the saturated vapor line, the critical point and a constant temperature process line from compressed liquid to superheated vapor. Also include a constant pressure line. 3-07. A saturated mixture of water is at 230 cP and 103 psia. The specific volume of the liquid is vj = 0.0178 ft3/lb11/ and the specific volume ofthe vapor is vg = 4.312 ft3/lb11/ . If the quality is x = 0.5, what is the specific volume of the mixture? 3-08. A fixed volume container initially contains a pure substance in superheated form. The container is cooled until there is a mixture ofliquid and vapor inside. Using T-v and p-v diagrams, describe the changes taking place in this constant volume process as the contents are cooled.



4: 1

Chapter 4 Thermodynamic Tables and Charts


Introduction

• 4.2

Tables for Saturated Liquid and Vapor

• 4.3

Tables for Superheated Vapor and Compressed Liquid

• 4.4

Thermodynamic LiquidN apor Charts



Study Objectives of Chapter 4 In Chapter 4, you will develop the skill of looking up properties in tables and charts for water (steam) and refrigerants. You willieam how to select the right table, interpolate between entries in the table, and understand the properties normally given in such tables. There are tables for the different phases of a pure substance, and the table format is different for each phase. After studying Chapter 4, you should be able to: • Understand the format and content of the different tables and charts; • Find the values of common properties in the tables; • Interpolate to determine values between the entries in the tables; and • Determine the property values of liquid-vapor mixtures.



4:2

4.1

Introduction

Tables list values of the thermodynamic properties of a pure substance. Charts are property diagrams that have lines of constant property value mapped in on the diagram. The tables and charts for water and refrigerant R-22 will be used in this chapter as the basis for studying how to use tables and charts. Water tables are sometimes referred to as steam tables because of the early development of the science of thermodynamics for steam boilers and steam engines. The tables and charts for water, refrigerants R-22 and R-134a, and moist air are given in Appendices B, C and D at the end of this book. Extracts from these tables will be used extensively in this chapter to illustrate the different operations involving these tables and charts. The tables and charts of Appendices B and C, and thermodynamic tables and charts for many other substances (including many well known refrigerants), are found in the ASHRAE Handbook-Fundamentals, Chapters 6 and 17. R-22 is a cornmon refrigerant used in many appliances, such as horne refrigerators, airconditioners and chillers. Its technical composition is chlorodifluoromethane, and it is one of the CFCs (chlorofluorocarbons). The CFCs have been the subject of extensive discussion in recent years because of the effect that they may have on the ozone layer, high in the earth's atmosphere. The primary refrigerant of concern has been R-12, the CFC that first achieved widespread use and, until recently, was found in applications such as automobile air-conditioning. Production ofR-12 has been discontinued. R-22 is not as potent as R-12 with respect to the ozone question, but even so, there is a proposed agenda to phase out R-22. The CFCs are not naturally occurring substances. As a class of refrigerants, they have been spectacularly successful because of their attractive thermodynamic properties. The concern about the ozone layer has driven a search for non-CFC alternatives. The other older refrigerants (such as ammonia) have important applications, but do not have the full range of attractive properties that were the hallmark of the CFCs. The search has produced some new non-CFC alternatives (such as R-134a) that are very promising. However, of the new alternatives, it is not yet clear as to which ones will be widely adopted by industry as replacements for R-22. In the meantime, R-22 is used in many existing pieces of equipment. Chapter 4 will deal only with property values in the liquid/vapor part of the property diagrams for water and R-22. However, it should be understood that there are also similar tables and charts for the liquid/solid and solid/vapor regions of the property diagram. Except for a short discussion about ice, the tables for solid/liquid and solid/vapor will be left for further study.

Chapter" Thermodynamic Tables and Charts


4: 3

Recalling the introduction of property diagrams in the previous chapter, there are basically three regions of interest for a pure substance undergoing a change from liquid to vapor. These are the compressed liquid, the saturated mixture of liquid and vapor, and the superheated vapor regions as shown in the t-v diagram of Figure 4-1. The regions are separated by the saturated liquid and saturated vapor lines. Each of the three regions is assigned a different table because of the special nature of the three regions. It would seem logical to draw a rectangular grid on the property diagram in Figure 4-1 and then tabulate the values of all properties at each of the grid line intersections. This is what is done for the compressed liquid and superheated vapor tables, except that the grid is drawn on the p-T diagram. However, the technique is different for the mixed liquid/vapor region because the grid is not necessary. The values everywhere in the mixture region can be determined by knowing only the values along the saturated lines. The discussion of tables and charts will be presented in the following order: saturated liquid/vapor tables, superheated vapor and compressed liquid tables, and charts. We will introduce the property values of most interest, and explain how the tables can be used to obtain values for many properties when the state is known. It takes any two independent, intensive, properties to determine the state.

Temperature Critical Point Compressed Liquid Region e«'"

~~"

,

T

I

" .scv ~

, .1:! •••••••••••••• I

I

I

"

I

I I

I I

,

;:,

~

, I

et!O, «~,

I,. .S! I

I

, Superheated Vapor Region

'

"0

cv

.....

EE

.a ~

Saturated (Liquid + Vapor) Region

Specific Volume V

Figure 4-1. Typical t-v Diagram



4:4

4.2

Tables for Saturated Liquid and Vapor

Along either the saturated liquid or saturated vapor lines, and for the region in between, the pressure and temperature are not independent from one another. That means that if we specify the pressure, then the temperature is automatically known. An example was given earlier when we examined boiling water. If the pressure is 14.7 psia, then the water would boil (change phase from liquid to vapor) at a temperature of212°P. The 212°P temperature applies for saturated liquid, saturated vapor and any mixture of saturated liquid and vapor in equilibrium with each other at the 14.7 psia. A similar argument applies for other pressures, and there is a unique temperature corresponding to each pressure at which water will boil. These are known as the saturated valuespsat and t sat. The relationship betweenpsat and t sat can be expressed in mathematical terms by saying that, for a given pure substance, saturated pressure is a function of temperature only: Psat

=/

(tsat)

The question here is what is the function /? Experiments show the function is not a simple function. Mathematical equations for the function/are available, such as those given in the ASHRAE Handbook-Fundamentals (Chapter 6), but the equations are complicated. Therefore, instead of giving the function in mathematical form, the relationship between pressure and temperature is also given in tabulated form. Selected values are taken from the water table of Appendix B and from the R-22 table of Appendix C, and presented in Table 4-1.

Table 4-1. Saturated Pressure and Temperature for Water and R-22 Saturated Water tsat (OF) Psat (psia) 32 62 92 122 152 182 212 242

0.08865 0.27519 0.74394 1.79117 3.91101 7.8589 14.7096 25.9028

Saturated R-22 (OF) Psat (psia) -50.00 11.696 0.00 38.726 50.00 98.799 100.00 210.69 396.32 150.00 200.00 686.11 723.74c 205.06c tsat

The values given in Table 4-1 highlight some features that are common to all saturated property tables. There is usually an entry value and then a value to be found. In this case, the entry value is the temperature and the value to be found is the pressure. The saturated table is sometimes given the other way around, where the temperature is listed for set values of pressure. Whichever way, the basic relationship is the same, and it is only a matter of convenience in which order they appear. Later we will examine the issue of when you wish to find a value that does not correspond exactly to one of the entry values. Chapter" Thermodynamic Tables and Charts


4: 5

The values in the table are usually given to at least five-digit accuracy. Because we are relying on the values to describe the functional relationship, they must be accurate. Although the temperature in the water table appears to be only given to a degree, it is also understood to be accurate to five-digit accuracy. The critical value is the last entry in the saturated R-22 table. The critical values for R-22 are 205.06°F and 723.74 psia. The critical values are the property values where the saturated liquid and saturated vapor lines join. For water, the critical values are 705.44°F and 3,204 psia. The values for saturated water given in Table 4-1 include the 212°F, 14.7 psia combination that we have been using as an example. What happens if we need a value between two table entries? Actually, as can be seen from Appendices B and C, the values in the table are given at a much smaller interval, and usually include a full range of useful values. Even so, the question still remains as to how to compute values that are between entries. Usually it is assumed that linear interpolation between entries in the table provides an accurate enough value of the property. Linear interpolation means that iftwo data entries were plotted on a graph, then a straight line joining the two points would reasonably describe the values between the two points. The closer together the data entries, the better the linear approximation. As an example of linear interpolation, consider the values in Table 4-1 and the need to know the saturation pressure when the saturation temperature is 200.5°F. The 200.5°F lies between the temperature entries 182°F and 212°F and between pressure values of 7.8589 psia and 14.7096 psia. These values are shown in Table 4-2 and in the graph in Figure 4-2 along with a notation of PJ'tl and P2' t2 for the known points and p, t for the desired pair.

Table 4-2. Steps to Interpolate Between Two Pressures Saturated Water

tsat COF) tl = 182 t= 200.5

t2 = 212


Psat (psia) PI-7.8589 p = pI + {(t - 11)/(t2 -tl)x(P2 - pI) = 7.8589 + {(200.5 - 182)/(212 -182)} x (14.7096 - 7.8589) = 12.0835 P2 -14.7096


4: 6

pressure P2 = 14.7096 psia

p=?

P1

=7.8589 psia

t1

=182°F

t

=200.5°F

t2

=212°F

Temperature

Figure 4-2. Linear Interpolation Between Two Temperatures in Saturation Table

From the graph, it can be seen that there are similar triangles in which:

This can be solved for an unknown p as in Table 4-2 as:

or for an unknown t as:

These are two useful expressions that will be applied over and over again as we find values in tables. The linear interpolation technique will be applied to many different variables, not only pressure and temperature.



4: 7

The value obtained for PSal at tsal = 200.5°F, when using values in Table 4-1, is 12.0835 psia. If instead the values in the table given in Appendix B are used, then the linear interpolation gives a more accurate value ofpsal as: p

= P1 + {(t - t1) / (t2 - t1)} X (P2 - PI) = 11.5374 + {(200.5 - 200) / (201- 200)} X (11.7779 -11.5374) = 11.6577 psia

These numbers suggest an error of e% = {(12.0835 - 11.6577)/11.6577) x 100 = 3.6% in the linear interpolation from Table 4-1. This is just an example to demonstrate that the closer the entries are in the table, the more accurate the linear interpolation. If the error is deemed too large, the alternative is to use one of the mathematical equations. For water, find the saturated temperature for a saturated pressure of 5 psia. In Table 4-3, it can be seen that tsal = 160.28°F atPsal = 5 psia.

Table 4-3. Steps to Interpolate Between Two Temperatures Saturated Water tsat (OF) . t} = 152 t = t} + {(P - P/)!(P2 - Pi)} X (t2 - t]) = 152 + {(5 - 3.91101)/(7.8589 - 3.91101)} x (182 - 152) = 160.28 t2 = 182

Psat (psia) p}- 3.91101 p-5

P2 -7.8589

The interpolation relations for P and t are two useful expressions that will be applied over and over again as we look up values in tables. The linear interpolation technique will be applied to many different variables, not only pressure and temperature. The tabulated pressure and temperature values are exactly the same for both the saturated liquid and saturated vapor lines. What then is the difference between the saturated liquid and saturated vapor lines? The difference is that the saturated liquid line represents a quality x = 0 (all liquid) and the saturated vapor line represents a quality of x = 1 (all vapor). Quality, x, is the mass proportion of vapor in a mixture of liquid and vapor. All properties, other than pressure and temperature, have different values for saturated liquid and saturated vapor. It is easy to picture that there is a big difference between specific volume; the liquid is heavy while the vapor is light. To draw the distinction between the specific volume of a saturated liquid and the specific volume of a saturated vapor, we use the subscripts offand g respectively (they are the first letters of the German words for liquid and vapor). There-

Fundamentals ofThermodynamics and Psychrometrics


4: 8

fore, the variable Vj is reserved for the specific volume of a substance that is atpsat ,{sat and has a quality x = 0, and vg is reserved for the specific volume of a substance that is at p t' { and has a quality of x = 1. For a quality of x, the specific volume v of a saturated mixture is given in terms ofvj' Vg and specific volume difference, Vjg = (Vg - Vj j, (see explanation in the previous chapter) as: ~

~

v=vj+XV jg

Values of v j ' Vg and Vjg are given in the saturation table for water as shown in the small subset of Appendix B reproduced in Figure 4-3. Figure 4-3 has column headings that include temperature {sal' pressure Psal' specific volume v j ' Vg and Vjg ' specific enthalpy hj' hg' and h jg , and specific entropy Sj' Sg and Sjg' Enthalpy is a property that will become very useful when applying the the First Law of Thermodynamics to open systems. The enthalpy for any saturated condition can be calculated from: h

=

h j +xhjg

Table 3 Tbermodynamic: Properties of Water at Saturation, Temp. I, of 189 190 191 192 193 194 195 196 197 198 199

Absolute Pressure p psi in.Hg 9.1510 . 9.3493 9.5512 9.7567 9.9659 10.1788 10.3955 10.6160 10.8404 11.0687 11.3010

18.6316 19.0353 ·19.4464 19.8648 20.2907 20.7242 21.1653 21.6143 22.0712 22.5361 23.0091

Specific SaL Uquid

"l 0.01656 0.01657 0.01658 0.01658 0.01659 0.01659 0.01660 0.01661 0.01661 0.01662 0.01663

Volume, Eftp.

ft 3 /1b Sat. Vapor

Enthllpy, Btu/lb Sat. Sit. Uquid Eftp. Vapor

hI

"I,

"t

41.7jO 40.901 40.092 39.301 38.528 37.774 37.035 36.314 35.611 34.923 34.251

41.746 40.918 40.108 39.317 38.544 37.790 37.052 36.331 35.628 34.940 34.268

hl'

. m.07 984.32 158.07 983.71 159.08 983.10 160.08 982.48 161.09 981.87 162.09 981.25 163.10 980.63 164.10 980.02 165.11 979.40 166.11 978.78 167.12 978.16

Entropy, Btu/lb· of SaL Sat. Uquld Eftp. Vapor

h,

I(

1141.39 1141.78 1142.18 1142.57 1142.95 1143.34 1143.73 1144.12 1144.51 1144.89 1145.28

0.2772 0.2787 0.2803 0.2818 0.2834 0.2849 0.2864 0.2880 0.2895 0.2910 0.2926

'l'

't

1.5174 1.7946 1.5141 1.7929 1.5109 1.7911 1.5076 1.7894 1.5043 1.7877 1.5011 1.7860 1.4979 1.7843 1.4946 1.7826 1.4914 1.7809 1.4882 1.7792 1.4850 1.7776

Figure 4-3. Thermodynamic Properties of Saturated Water

Enthalpy is an energy-type property with zero value at some chosen reference state. The reference state can be arbitrarily chosen, so the table is often accompanied by a statement of how the reference state was chosen. For the water table in Appendix B, the reference state is set such that the enthalpy is zero at the triple point of water (32.018°F). In most of the thermodynamic expressions that we will be using, it is the enthalpy difference that is important and not the absolute value of enthalpy, so the choice of reference state is not critical. However, the reference state is critical in topics such as combustion where there is a chemical reaction between components, and enthalpy values have to be brought to a common reference state. The enthalpy difference between saturated vapor and saturated liquid, hjg = (hg - hj ), is also known as the heat of vaporization or the latent heat.



4: 9

The tables offer a way of determining the quality if two independent properties are known. For example, assume that saturated water at 212°F has an enthalpy of h = 800 BtU/Ibm' From the table in Appendix B, the saturation pressure is 14.7096 psia. From the same table, hj = 180.20 and hjg = 970.03 Btu/Ibm' Solving the enthalpy equation given earlier for x gives:

x

= (h-hj )/ hjg = (800 -180.20) / 970.03 =

0.6389 (or 63.89%)

The column headings from the saturation table for R-22, as given in Appendix C, are reproduced in Figure 4-4. The specific volume for liquid is replaced by the density (density is the inverse of specific volume). The columns for specific volume, enthalpy and entropy do not list the values of Vjg ' hjg or Sjg' but these values can be derived easily from the differences between listed values. There are additional columns. Columns that will be of interest in a later chapter are those of coefficient of specific heat at constant pressure, Cp , and the ratio of specific heats, k = Cp / Cv .

Refrigerant 11 (cblorodiDuoromethane)· Properties of Saturated Liquid and Saturated Vapor Delsity, \bIuIt, EaIUlpy, EaIlOpJ, Specific Heat ep' VelocIty .r Solid, VIscosity, 1\tnuI eold, Sarface TeIIp,* Prasart, IbIrtl rl31U1 BIIIIII IIIIIII'·F Ila/ll,.°F eple. WI 1II./r... Itl/l.. fl,·F Teuioa, Temp, 9~~~~_~~~~~~~~~~~~9 -lSO.OO 107.37 -63.169 76.604 -0.2191. 0."952 0.1018 1.291. 395. - -2S0.00 106.41 -240.00 -S6.~ n.629 -0.18786 0.42332 0.1033 1.2860 403. - -:MO.OO -230,00 . 105.48 - -51.569 78.669 -0.16605 0.40101 0.1048 1.2807 411. 36.7S - 230.00 -220,00 0.002 IU..58 1680S. -47.705 79.724 -0.14958 0.38211 0.1064 1.27504 419. 35.70 -220.00 -lIO.OO O.~ 103.70 6982.6 - ".426 80.796 -0.13616 0.36538 0.1080 1.2703 .27. 34.67 - lIO.OO

-

Figure 4-4. Thermodynamic Properties of Saturated R-22

4.3

Tablesfor Superheated Vapor and Compressed Liquid

Superheated vapor tables cover the region to the right ofthe saturated vapor line on the t-v diagram shown in Figure 4-1. The term superheated comes from the fact that, for a given pressure, the temperature is greater than the saturation temperature. In an example of water at atmospheric pressure, the vapor is superheated if the temperature is greater than the saturation value of 212°F. In the superheated region, the pressure and temperature are independent properties. Therefore, the superheated vapor table actually consists of a set of many tables, each for a different pressure, in which the entries are made according to temperature.



4: 10

The compressed liquid tables cover the region to the left of the saturated liquid line on the t-v diagram in Figure 4-1. The term compressed comes from the fact that, for a given temperature, the pressure is greater than the saturation pressure for that temperature. Pressure and temperature are independent properties in the compressed liquid region. The compressed liquid table is a set of tables, each for a different pressure, in which the entries are made according to temperature. The interval in pressures between tables is usually quite large because liquid properties vary less with pressure than do vapor properties. Liquids are often treated as incompressible.

4.4

Thermodynamic Liquid/Vapor Charts

A chart for R-22 is given in Appendix C. A simplified version of that chart is given in Figure 4-5. Figure 4.5 is simplified only in the sense that there are a limited number of constant property lines shown, so that it is easier to explain the chart's complexity. Reference in the next few paragraphs will be to Figure 4-5, but it is the chart in Appendix C that should be used to obtain values. The chart in Figure 4-5 is a plot of pressure against enthalpy. In this case, the pressure is given a log scale. Lines of constant pressure run horizontally across the chart. Lines of constant enthalpy run vertically up and down the chart. The saturated liquid and saturated vapor lines outline a fairly characteristic dome shape to the central part of the chart. The saturated liquid and saturated vapor lines are lines of constant quality (respectively x = 0 and x = 1). Other lines of constant quality can be drawn as shown for x = 0.02 and x = O.S. Because quality is only defined for equilibrium mixtures of saturated liquid and vapor, these lines of constant quality do not extend outside the region bounded by the saturated liquid and vapor lines. The lines of constant temperature on the chart in Figure 4-5 have three distinct parts. Por example, the line for Soop falls steeply in the compressed liquid region until it reaches the saturated liquid line. It then extends horizontally across from the liquid to the vapor line, following the constant pressure line ofPSOI = 15S.40 psia for tsol = SooP. After reaching the saturated vapor line, the line for Soop once again falls very steeply. The line for 200 0 P follows a similar path, but because it is close to the critical temperature of 205.06°P, it seems to barely enter the dome near the critical point. Lines of constant density (inverse of specific volume) are shown for values of 1.0 Ibm /ft3, 15 Ibm /ft3 and 65 lb m /ft3. Lines of constant entropy are shown for 0.3 Btullbm ·R and 0.16 Btullbm .R.

Chapter" Thermodynamic Tables and Charts


4: 11

Enthalpy (Btu/Ibm) 2000.25

0

50

25

125

150

~

u.

1000

100

oensi\.'i :;

:£

0

0 GO II

\'oJtt' 1000

-

400

200 2000

175

'\5

'b /ft::::.. ~r-.;

400

~O'<

~~

~ II) -9: 100

:i? II)

/

Density

~

:J

='I Ib

0..

.Ift3

100 -;-

....

:J

II) II) Q)

II) II)

~

e..

40

40

10

10

....

e..

u.

o

o o

4

4

~ 1 -25

~

__

~L-

0

_ _ _ _ _ _ _ _ _ _ _ _ _ _~_ _ _ _2 -_ _ _ _~_ _ _ _- L_ _ _ _ _ _ _ _ _ _ _ _ _ _ J

25

50

75

100

125

150

175

1

200

Enthalpy (Btu/Ibm)

Figure 4-5. Pressure-Enthalpy Diagram for R-12

There are many alternative charts because it is choice of convenience as to which properties are selected for the axes of the chart. The charts plotting enthalpy against either pressure or temperature are useful in vapor-compression refrigeration because the constant enthalpy and constant pressure processes are easily shown. Previously, when introducing property diagrams, we found it convenient to describe the temperature-volume (t-v) and pressurevolume (P-v) charts. Another combination that is partiCUlarly convenient when studying compressors and turbines is the temperature-entropy, or t-s, chart. The Mollier chart, widely used in steam turbine work, is a plot of enthalpy against entropy (h-s) for water. Charts are difficult to read on the scale that is drawn in this text. Much larger versions of these charts are usually used to determine specific property values. A more common use of property charts is to provide a visual map of a sequence of processes, with a clear indication of the path and end states. Another purpose is to use the chart to determine which table is appropriate to use: compressed liquid, saturated or superheated vapor.



4: 12

The Next Step In this chapter, the emphasis was on finding the property values for substances in a range that included the phase change from liquid to vapor. In Chapter 5, we will consider what to do for ideal gases and for substances that act like ideal gases in the pressure and temperature range of interest. Air at moderate temperatures and pressures is taken as an example of a real gas approximated as an ideal gas.

Summary

In Chapter 4, we developed the understanding and skill of finding properties in the saturated tables for water and R-12. The saturated tables are shown to provide a relationship between P sat and tsat • Values for properties in the mixed liquid and vapor region can be obtained from the saturated values. The tables included values for pressure, temperature, specific volume, specific enthalpy and specific entropy. The linear interpolation technique is used to determine values between the table entries. A chart ofR-22 is used as an example to demonstrate the use of charts. After studying Chapter 4, you should be able to: • Understand the format and content of the different tables and charts; • Find the value of common properties in the tables; • Interpolate to determine values between the entries in the tables; and • Determine the property values of liquid-vapor mixtures.



4: 13


4-01. The saturated temperature of water is 100°F. What is the pressure? What are the liquid and vapor values of specific volume, specific enthalpy and specific entropy? (Use the table in Appendix B.)

4-02. The saturated temperature of water is 100°F and the quality is 50%. What is the value of specific volume and specific enthalpy? (Use the table in Appendix B.)

4-03. The saturation pressure for R-22 is 50 psia. What is the saturation temperature? (Use the table in Appendix C)

4-04.

The saturation temperature for water is 231.5°F and the quality is 0.2. What is the pressure and the specific enthalpy? (Use the table in Appendix B.)

4-05. The enthalpy for R-22 is 50 Btu/Ibm and the pressure is 40 psia. Is the refrigerant in the compressed liquid, saturated or superheated vapor state? What is the quality? (U se the chart in Appendix C)

4-06.

The enthalpy for R-22 is 50 Btu/Ibm and the temperature is 100°F. Is the refrigerant in the compressed liquid, saturated or superheated vapor state? What is the pressure and the quality? (Use the table in Appendix C)

4-07.

The enthalpy for R-22 is 150 Btu/Ibm and the pressure is 40 psia. Is the refrigerant in the compressed liquid, saturated or superheated vapor state? What is the value of the density? (Use the chart in Appendix C)



5: 1

Chapter 5 Ideal Gas Law and Air Tables


Ideal Gas and the Ideal Gas Law

• 5.2

The Ideal Gas Law as an Equation of State

• 5.3

Constant Specific Heat

• 5.4

Air Tables


Instructions Read the material of Chapter 5. Re-read the parts of the chapter that are emphasized in the summary and memorize important definitions. At the end of the chapter, complete all ofthe skill development exercises without consulting the text.

Study Objectives of Chapter 5 The ideal gas law expresses the relationship between pressure, temperature and specific volume for an idealized model of a gas. The study objective for Chapter 5 is to understand how the ideal gas law may be used to determine the state properties of pure substances that can be reasonably assumed to act as an ideal gas. Air at room temperature and pressure can often be treated as an ideal gas. There is an introduction to the use of air tables. After studying Chapter 5, you should be able to: • State the ideal gas law; • Express the limitations to the use of the ideal gas equation; • Describe the use of the ideal gas law in determining state properties; • Describe the coefficients of specific heat as properties; and • Determine the enthalpy for an ideal gas with constant specific heat.



5:2

5.1

Ideal Gas and the Ideal Gas Law

An ideal gas does not exist in reality. The theoretical construction of an ideal gas is one where the molecules that make up the gas are so far apart from each other that they have no influence on each other. It is impossible for molecules to have no influence on each other, but it is possible for a substance oflow enough density to behave approximately as an ideal gas. In fact, many common substances that we describe as gases (such as oxygen, nitrogen, carbon dioxide and air) have low enough densities at atmospheric pressures and temperatures to be reasonably described as ideal gases. This really means they behave as ideal gases to an acceptable limit of error. This does not imply that they will behave as ideal gases at all pressures and temperatures. To the contrary, to be reasonably treated as an ideal gas, the state of the substance must be such that the density is small enough so that individual molecules are far enough apart from each other and that they seldom run into each other. The density is the inverse of the specific volume. Therefore, an ideal gas is a substance in which the specific volume is large enough, and the individual molecules are far enough apart, that the molecules have negligible effect on each other. As described earlier, the specific volume varies depending on the state of the substance. For all pure substances, there are solid, liquid and vapor phases. A substance will not behave as an ideal gas in the solid, liquid or much of the vapor phase areas of the phase diagram. Only in the vapor phase range at low enough pressures and high enough temperatures will the vapor behave as an ideal gas. The vapor behaves less and less like an ideal gas the closer the state is to the critical point or the saturated vapor line. For the theoretical ideal gas, the pressure varies proportionally to the absolute temperature and inversely proportionally to specific volume, leading to the ideal gas law which states:

p=R(Tlv) pv=RT The R in this equation is the constant of proportionality in the relationship and is known as the gas constant. It is important that the temperature in the equation is in absolute units (Rankine or Kelvin). To draw a clear distinction of places where the temperature must be in absolute units, we will use the upper case T. The value of the gas constantR is different for every gas and may be determined by dividing the universal gas constant, Ru ' by the molar mass, M, of the substance (often called molecular weight):

Chapter 5 Ideal Gas Law andAir Tables


5:3

The units of molar mass are lbmoeI lIb m . The value of the universal gas constant is:

Ru

= 10.73 (psia.fe)/(lb mole .R) = 1.98 Btu I (lb mo1e • R) = 8.314 kJ I (k mo1e .K)

The values for molar mass and gas constants for some common gases are given in Table 5-1. The ideal gas law may be expressed in several different forms. For example, the specific volume may be replaced by the total volume divided by the mass of the system. Another form is to write the gas equation for a given number, N, oflbmo'es of gas rather than a mass in lbm . The various forms are summarized below:

pv=RT pV=mRT pV = NRuT

Table 5-1. Gas Constants for Certain Substances ;

Gas Carbon dioxide , Hydrogen , Nitrogen Oxygen Air , R-22 Water

,

Molar Mass (lb",llb mo1e) 44.0995 2.01594 28.0134 3,1.9988 28.9645 102.92 18.01528


,

,

Gas 'Constant (psiaxft~/(lbmXR) , 0.2438 5.3224 0.3830 0.3353 0.3704 0.1043 , 0.5956


5:4

5.2

The Ideal Gas Law as an Equation ofState

In the previous chapter, we said that the functional relationships between properties such as pressure, temperature and specific volume were mathematically complex enough that these values are tabulated for water and other refrigerants. The tables are cumbersome but necessary. However, in areas where the substance behaves as an ideal gas, the ideal gas equation provides a more convenient alternative. The simple relationship among pressure, temperature and specific volume in the ideal gas law suggests that given any two properties, we can calculate the third. The ideal gas law is an example of an equation of state for substances that obey the ideal gas relationship between pressure, temperature and specific volume. In HVAC, it is fortunate that air, and the water vapor in air, can be treated as ideal gases for typical atmospheric conditions. There will be a more careful evaluation of this statement in the chapters on psychrometrics, but for now, consider some examples applying the ideal gas equation of state to air. How would you determine the mass of air in a room of dimensions 10 ft x 20 ft x 8 ft, where the pressure is 14.7 psia and the temperature is 75°P?

m= pV I RT

= (14.7 psia)(10 x 20 x 8 ft3) I {( 0.3704

(psia. ft3))1 (Ibm· R)(75 +460 R)}

= (118.7 Ibm)

What pressure is required if air at 32°P is to have a density ofO.15 Ibm/ft3 (density, p, is the inverse of the specific volume)?

p= RTlv =pRT

= (0.15 Ibm I ft 3){0.3704(psia. ft3) I (Ibm· R)}(32 +460 R) = 27.3 psia The relationship between two states of an ideal gas can be seen by equating the gas constant when the gas is at State 1 to the same gas constant when the gas is at State 2: p\v\ II;

= P2V 2 IT;



5: 5

This relationship holds as long as the gas can be treated as an ideal gas. It does not depend on the path by which the substance goes from State 1 to State 2. Consider a constant volume process for which the initial pressure and temperature are PI = 14.7 psia and tl = 72°F. What is the final pressure if the substance is air, can be treated as an ideal gas and is cooled to 32°F? V2

= VI; PI = 14.7 psia; ~ = (72 +460) = 532 R;

Tz = (36+460) = 496 R

From the ideal gas law:

P2 = (VI / v2)(Tz / ~)(PI) =(496/532)(14.7) = 13.71 psia It is again worth noting that when working with the ideal gas law, it is important to use

absolute temperatures. In the example above, if absolute temperatures were not used, then the expression would have incorrectly predicted a final pressure of one-half of the initial pressure.

5.3

Constant Specific Heat

The ideal gas equation of state relates pressure, temperature and specific volume, but what about the other properties that were listed in the water and refrigerant tables given earlier? How is the enthalpy, or the entropy, determined for an ideal gas. To answer that, we will introduce another new property, specific heat, and we will make the observation that it is the change of enthalpy that interests us more than the absolute value of enthalpy. The term coefficient ofspecific heat is probably known to you from its application in physics. The coefficient of specific heat of unity for liquid water means that it takes one Btu to heat one lbIII of water by one degree Fahrenheit. For gases, we want to apply a similar idea, but it takes a more careful definition because there are two coefficients of specific heat. One is the coefficient of specific heat at constant volume, cv' and the other is the coefficient of specific heat at constant pressure, cp' Contrary to their names, they are not meant to be used only in cases of constant volume or constant pressure. Rather, they are defined in terms oftheir connection to either internal energy, U, or to enthalpy, h.



5:6

For an ideal gas, the coefficient of specific heat at constant volume is: Change of internal energy = cl' times change in temperature

and, for an ideal gas, the coefficient of specific heat at constant pressure is: Change of enthalpy = cp times change in temperature

Strictly speaking, the expressions above apply for only constant values of specific heat. It is possible for the specific heat to vary with temperature and, in those cases, the changes have to be taken over a small temperature range at a time. The distinction between these two specific heats will be elaborated on later when we have developed the idea of work in thermodynamics. For ideal gases, the specific heats do not depend on pressure. Therefore, there is an important deduction that says: For ideal gases, the internal energy and enthalpy are functions of temperature only_

F or the case of constant specific heat, it is straightforward to construct a table of internal energy or enthalpy values as a function of temperature by applying the relationships given above. The coefficients of specific heat can be related to each other. Because enthalpy is related to to internal energy by h = (u + pV), and because the ideal gas law is pv = RT, we can write:

(~-~)=(U2 -UI )+(P2 V 2 -

P1V I )

cAt2 - tl ) = Cv (t2 - t l ) + R(t2 - t l )

Dividing through by the temperature difference gives the relationship between the specific heats and the gas constant as:

cp =cl' +R



5: 7

Another useful property is the ratio of specific heats, k.

Typical values of the coefficients of specific heat for air treated as an ideal gas at 80°F are: C = 0.240 Btullb .OF and C = 0.171 Btullb .oF, which result in a k value of 1.4. p m v III

To determine the values of entropy for an ideal gas (comparable to the values found in the tables for water and refrigerants given earlier), we must have a better understanding of entropy. We are not going to discuss entropy here, because it first takes the introduction of the Second Law of Thermodynamics, but below are the expressions that are used to determine entropy changes for an ideal gas in the cases of constant specific heat:

(S2 -SI) = cl' In(~ 17;)+ R In(v2 Iv l ) (S2 -

SI) = cp

In(~ 17;) + R In(p2 I PI)

For ideal gases, enthalpy is a function of temperature only, but entropy depends on both pressure and temperature.

5.4

Air Tables

The thermodynamic properties for air can be tabulated the same way as they are done for water and refrigerants. The tabulations take into account the way specific heat changes with temperature. The values are usually listed as a function of temperature for a given pressure. An example of a moist air table is given in Appendix D for a pressure of 14.696 psia. The moist air table includes dry air as one of the cases. The moist air table will be more fully explained during the discussion of psychrometrics.



5: 8

The Next Step Now that we have defined a number of useful properties, and learned how to find numbers for those properties, we are ready to return to the discussion about different forms of energy. In the next chapter, we will introduce the ideas of work and heat in preparation for describing the First Law of Thermodynamics.

Summary The ideal gas law expresses the relationship among pressure, temperature and specific volume for an idealized model of a gas. The relationship is given by pv = RT. The use of the equation is limited to applications where the behavior of the substance can be reasonably approximated by that of an ideal gas. This is usually in the low pressure/high temperature range of the superheated vapor region. Vapors do not behave like an ideal gas near the saturated vapor line, or near the critical point. The use of the ideal gas law as an equation of state permits the determination of the remaining value of the properties pressure, temperature or specific volume, if any two of these properties are known. The properties of specific heat at constant volume and specific heat at constant pressure are introduced as a means for determining internal energy and enthalpy for an ideal gas. For many applications, air at room conditions may be treated as an ideal gas with constant coefficients of specific heat. After studying Chapter 5, you should be able to: • State the ideal gas law; • Express the limitations to the use of the ideal gas equation; • Describe the use of the ideal gas law in determining state properties; • Describe the coefficients of specific heat as properties; and • Determine the enthalpy for an ideal gas with constant specific heat.



5: 9


5-01.

Air is often treated as an ideal gas. For the conditions of p = 14.696 psia and t = 72°F, find the specific volume for dry air, va' in Appendix D, and compare the value read in the table to the value calculated using the Ideal Gas Law. Use T= t + 459.67 to convert from of to R. Use the value of gas constant given in Table 5-1. Express the answer as a percent error by treating the tabulated value as being correct.

5-02.

Water vapor is sometimes treated as an ideal gas. For the condition ofp = 67.0341 psia, find the specific volume for saturated vapor, vg , in Appendix B, and compare the value read in the table to the value caclulated using the Ideal Gas Law. Use T= t + 459.67 to convert from of to R. Use the value of gas constant given in Table 5-1. Express the answer as a percent error by treating the tabulated value as being correct.

5-03.

Repeat Exercise 5-02 except for conditions of saturated vapor at p = 0.3628 psia. Compare the error found with the error determined in Exercise 5-02 and comment on the difference.

5-04. A room of dimensions 10ft x 20 ft x 8 ft at atmospheric pressure contains 120 Ibm of air. Assume an ideal gas and estimate the temperature in the room in of. 5-05. Consider the constant pressure heating of air from 70°F to 200°F. If the pressure is 14.7 psia, what is the ratio of specific volumes before and after the heating? 5-06. Make a table listing the values of internal energy, enthalpy and entropy for air at 14.696 psia, with temperatures ranging from O°F to lOO°F in steps of lOoF, using the coefficients of specific heat suggested for air at 80°F in the text. Use reference values of zero internal energy, zero enthalpy and zero entropy at O°F. 5-07. Find the values of enthalpy for dry air in the table in Appendix D, at the same temperature entries used in the table created in Exercise 5-06 and compare calculated and look-up values. Calculate percent error assuming the tabulated values to be correct.

Fundamentals ofThermotlynamics and Psychrometries


6: 1

Chapter 6 Heat and Work


Introduction

• 6.2

Heat

• 6.3

Work

• 6.4

Flow Work


Instructions Read the material of Chapter 6. Re-read the parts of the chapter that are emphasized in the summary and memorize important definitions. At the end ofthe chapter, complete all ofthe skill development exercises without consulting the text.

Study Objectives of Chapter 6 There are three ways in which energy can cross a system boundary. It can cross the boundary as heat, as work, or (in the case of an open system) as an item carried by mass flow. The energy forms of heat and work are the topics of study in this chapter. Heat is an expression of the flow of thermal energy. Work may appear as moving boundary work, shaft work, electrical work, gravitational work, acceleration work or flow work. After studying Chapter 6, you should be able to: • Describe the concept of heat as an energy form crossing a system boundary; • Describe the concept of work as an energy form crossing a system boundary; • Describe the different types of mechanical work; and • Understand the termflow work and relate it to enthalpy.



6:2

6.1

Introduction

There are three important ways in which energy can be exchanged between a system and its surroundings that warrant special description in thermodynamics. Energy can cross a system boundary as heat or work, or it can be transported by a flow of mass. Earlier, we introduced the concept of internal energy as a property of a system that could be used to help determine the state of the system. Heat and work are different. They are forms of energy that are not properties of the system and, therefore, do not determine the state of the system. Heat and work can cause the state of a system to be changed (as we shall see later), but they cannot define the state of the system. Heat and work are associated with the mechanisms of energy exchange between the system and its surroundings. They are known as transient forms of energy because they only have meaning as the action of energy exchange is taking place. Another way of stating the difference is to say that although a system may contain a certain amount of energy at a given state (internal energy, potential energy or kinetic energy), the system never contains heat or work. The system gains or loses energy by heat and work as it changes from one state to another. Besides the exchange of energy by work or heat, energy can also be carried in or out of an open system when there is mass flow across the system's boundary. Mass flow at the boundary, and its resulting effect on the energy balance of the system, will be discussed when applying the First Law of Thermodynamics to open systems. Heat and work are both forms of energy, but they differ in their driving mechanisms. Heat exchange occurs because of a difference in temperature between the system and its surroundings. In contrast, work has several manifestations such as boundary work, shaft work, electrical work, gravitational work, acceleration work and flow work, which generally occur because of a difference in forces.

Figure 6.1 shows a closed system with energy exchange with the surroundings in the Chapter 6 Heat and Work

Surroundings

Work, W

r--------------System

Heat, Q Figure 6-1. Closed System With Sign Convention for Heat and Work


6: 3

form of heat, Q, and work, W. Heat and work can cross the boundary in either direction, in or out, but the sign conventions for the two are different. Heat is positive when it is directed into the system. Work is positive when it is directed out of the system, or when the system is said to do work on the surroundings. The sign convention originates with the idea of a heat engine in which heat is supplied to the system so that it can do work as an output. An automobile is an example of a heat engine because the energy from burning fuel is converted into shaft work to tum the wheels. Heat and work are the topics of the next two sections.

6.2

Heat

Consider a hot potato that has just been taken from the oven. With time, the potato will cool as energy is transferred from it to the surrounding air. The temperature difference between the potato and the air causes the exchange of energy, and when the temperature between the potato and the air becomes negligibly small, the exchange of energy will cease. The direction of the energy exchange is always from the higher temperature body to the one at a lower temperature. Heat is the name given to the transitory form of energy as it crosses a boundary separating systems (or a system and its surroundings) by virtue of a difference in temperature. For the potato example, heat is the energy that passes from the potato to the air. This means that some of the potato's internal energy is converted to heat as it flows across the boundary, and then is converted back to become part of the air's internal energy. We do not speak of either the potato or the air as containing heat. In this sense, the use of the term heat is quite specific in thermodynamics and does not include many everyday references that actually more correctly identify internal energy rather than heat. The expressions heat content offuel and body heat are examples of situations where it is really the internal energy that is referred to. Why make such a strong distinction between heat and internal energy? It is important because internal energy is a property and heat is not. Whereas properties are used to identify states, heat)~~ only help determine the process between states. Heat has the units of energy and is usually given the symbol Q. Because heat is a transitory energy form that often occurs as a system goes from one state to another, the states are added to the symbol designation as JQ2, meaning the heat in going from State 1 to State 2. The heat symbol Q is never given a single state designation, such as QJ' to emphasize that heat is not a property associated with any particular state. The direction of flow of energy across the boundary is important. The sign convention is that a positive value of Q means flow of energy from the surroundings into the system. A negative value of Q means a flow of energy out of the system. Fundamentals ofTilermotiynamics and Psychrometries


6:4

This sign convention originates from the idea of a heat engine where heat is added to the system to make the system do work, as with an automobile. Sometimes it is inconvenient to keep up with the positives and negatives, and the alternative is to subscript Q with the labels in or out to indicate the direction of energy flow, and then remember to place the appropriate sign when doing an energy balance. We shall see many examples of energy balances when we discuss the First Law of Thermodynamics in the next chapters. The energy flow as heat into the system is referred to as heat addition, while the energy flow as heat out of the system is known as heat rejection. There are two other ways in which the heat term will commonly appear in thermodynamics equations. There is the heat per unit mass, known as q (Btu/Ibm)' and the heat per unit time, known as

Q(Btu/h).

Consider again the potato taken from the oven (as shown in Figure 6-2). The potato is initially at 260°F and the surrounding air is at 76°F. From our discussions about internal energy and the coefficient of specific heat for solids in Chapter 5, we can estimate the energy content of the potato at any time. Assume a reference datum for internal energy as zero at 32°F (arbitrary choice), amass ofm = 0.5 Ibm ,and a coefficient of specific heat of cp = 1 Btu/Ibm .0F, then the internal energy when the potato is first taken from the oven (State 1), and the internal energy after a long time when the potato has come to equilibrium with the room air (State 2), are respectively:

U\ =mcAt\ -tref )

= (0.5 lbm)(l Btu /lb m •o F)(260 = 114 Btu

32° F)

U 2 =mcAt2 -tref )

= (0.5 lbm)(l Btu /lb m·o F)(76 - 32° F) =22 Btu

The heat crossing the boundary, JQ2 , is: \Q2

=(U2 -U\) = -92 Btu



6: 5

,'" _.... ~ \\

>. \. ..I

Alternative = 92 Btu

.. Qout

State 1

Process 1- 2

State 2

Figure 6-2. Potato (Closed System) Cooling From 260°F to 76°F

Note that this is a special form of the First Law of Thermodynamics in which the work is zero because there is no change in volume of the potato. Descriptions of work appear in the next section, and the First Law is in Chapter 7. The minus sign means that the heat is being rejected. The arrow is shown in Figure 6-2 flowing into the potato, but the negative value means that it really is in the opposite direction. Alternatively, this could be described as 92 Btu rejected from the system, and be expressed as:

IQ2 oul

= 92 Btu

The arrow pointing out of the system with the positive value of 92 Btu is shown as an alternative (dashed) in Figure 6-2. A process in which there is no heat is known as an adiabatic process. There are two circumstances under which a process may be adiabatic. The system may be well insulated, resulting in negligible energy flow even if there is a temperature difference, or there may be no temperature difference between the system and its surroundings. The temperature in an adiabatic process may, or may not, remain constant, depending on other energy flows taking place, such as work. Therefore, an adiabatic process is not the same as an isothermal process. In an isothermal process, the temperature of the system does not change while going from one state to the next, but there mayor may not be heat transferred.



6:6

6.3

Work

Work is the name given to the transitory fonn of energy as it crosses a boundary separating systems (or a system and its surroundings) by virtue of a difference in pressure or force of any kind. Like heat, work is not a property, and is only defined while the energy interchange is occurring across the system boundary. Work has units of energy and is assigned the symbol W. When the work occurs while going from State 1 to State 2, then the work is specified as 1 W2 • The work per unit mass is w Btu/Ibm' The rate of doing work, W Btu/h, is also known as power, P. Power appears in many different units, and may be given as Btu/h, ft-Ib/s, horsepower hp, or kW. There are many different fonns of mechanical work, each in some way connected to a force being moved over some distance. From physics, work is defined as the product of force and displacement (in the direction of the force). We will use that basic definition and develop the expressions for some common fonns of work.

Moving boundary work. Consider a piston-cylinder arrangement as shown in Figure 6-3. The pressure in the cylinder is p, the piston area is A, and the volume of the air in the cylinder is V. The piston is weightless and frictionless and it takes a force of F = pA to keep the piston in place. If the piston (and therefore the system boundary) moves a small distance f:ls, and the system volume increases by a small amount from V to (V + ~V), then the small amount of work done by the force F moving through distance f:ls is:

,-

, __...>1

~W

= F·f:ls

------------~--------

= (pA)f:ls

=p(Af:ls) =p~V

Pressure p

Volume V Figure 6-3. Boundary Work by a Piston in a Cylinder



6: 7

The work is given as the product of the pressure and the small change of volume. The small change in volume is emphasized here because to get this result, we actually assumed that the pressure remained constant while the volume was changing. Constant pressure represents only a special case of the more general situation where pressure can vary in a variety of different ways with volume change. A case of air expanding in a cylinder is shown in the p- V diagram of Figure 6-4, where the air is shown starting at State 1 and expanding to a lower pressure at State 2. The path between States 1 and 2 is drawn as an arbitrarily shaped curve. The expression developed above for work as p!1Vapplies for the thin slice shown in Figure 6-4 and is the area of that slice on the p- V diagram. To find the work done in going from State 1 to State 2, it would be necessary to add up the areas of many thin slices from the p- V diagram: I

~

= Area under the p - V diagram =LP!1V

The symbol sigma, ~, is used to indicate the sum of all the small areas p!1V. The boundary work is the area under the p- V curve. In evaluating moving boundary work of a closed system, it is very useful to sketch the p- V diagram to obtain a visual picture of the work done. The results to the example of a piston in a cylinder can be stated more carefully and generally. Moving boundary work for the quasi-equilibrium and reversible movement of the boundary of a closed system can be estimated from the area under the p- V diagram. Quasiequilibrium change was described in Chapter 2 as a change that proceeds in such a manner that the system remains infinitesimally close to an equilibrium state at all times. A quasiequilibrium process is an idealization of a real process. Reversibilty is a concept yet to be introduced and will be an important part of a later chapter. A process is said to be reversible ifboth the system and its surroundings can be restored to their original condition. Effects such as friction, non-quasi-equlibrium or heat transfer through a finite temperature difference render a process irreversible. Therefore, reversibility is also an idealization of a real process because no process can actually meet all of the requirements. The sign convention for work is that it has a positive value when work is done by the system on the surroundings. That is the case for the expansion shown in Figure 6-4. If the piston moved in the opposite direction so as to decrease the volume (as in compression), then work would be done on the system, and the sign would be negative. This is the opposite sign convention to that applied to heat.



6: 8

Pressure

P

I I I I I I ____ _ - __ 1I I

- - - ~ - - - - -~ - - - - - - - -=--....-....-......-.~ ®

~

:

~

1

I

V2

AV=AAS

Volume

AS ~

f

1

l+-

I I

~

l}

I I I I

.

Figure 6-4. Boundary Work as the Area Under the p-V Curve

There are many different paths by which the pressure can change as a system goes from State 1 to State 2. The work done by the moving boundary (area under the p- V curve) for a closed system is given for a list of specific processes below and shown in the p-V diagrams in Figure 6-5: Constant pressure process,p] = P2 ' any substance:

IT¥; = PI(V2 - v;) = P2(V2 - v;) Constant volume process, ~

=

V2 ' any substance:

IT¥; = zero


Fundamentals ofThermotiynamics and Psychrometries

6: 9

Isothermal process, T] = T2 ' ideal gas only P V = constant: 1

~

= Pl~ In(V2 I~) = P2 V2 In(~ I~)

Isentropic process, s] I

~ = (P2 ~ -

= S2

PI

,P V" = constant, ideal gas only, P V = mRT:

~) 1(1- k) = mR( 1; -

Polytropic process, P vn = constant, n 1

7;) 1(1 - k)

* 1, ideal gas only, P V = mRT:

~ = (P2~ - pl~)1 (1- n) = mR(1; - 7;)1 (1- n)

Pressure

Pressure Constant Pressure

Constant Pressure

CD ® Volume

Pressure

Volume

Pressure

CD

Isothermal, Ideal Gas

Isentropic, Ideal Gas

pV = constant

V2

Volume

V2

Volume

Figure 6-5. Work as Area Under the p-V Curve for Closed Systems and Different Processes



6: 10

This may seem a formidable list, but you do not need to memorize all these expressions. They are given to emphasize that the work is a path-dependent function and that the moving boundary work may be calculated when needed. The above expressions apply to closed systems. The derivations of the above expressions are not shown, but it can be seen that they depend on the type of process and the substance. In the case of an isentropic process with ideal gas, it can be shown that the process is given as p V" = constant, where k is the ratio of specific heats. Often, it is experimentally seen that the compression or expansion follows a process that is not quite isentropic, but where the process can be described with an expression p V" = constant. This is called a polytropic process, and n is the polytropic exponent.

Example 6-1. A frictionless piston-cylinder device contains 10 Ibm of saturated liquid water at 212°F. The fixed weight piston is not attached to a shaft. Heat is added to the water until it is all saturated vapor. Calculate the work done by the system. Solution. The piston-cylinder and p-V diagram are shown in Figure 6-6. Because of the fixed weight piston with no shaft, the pressure remains constant. The pressure may be found for saturated water at 212°F from the saturated water tables in Appendix B-1 as 14.7096 psia. From the same table, the specific volume of saturated water is vj = v J = 0.01671 ft3/lb m and the specific volume for saturated vapor is Vg = v2 = 26.780 ft 3/lb m• The volume is the mass times the specific volume. The rectangular area under the curve gives the work as:

I~

=

Pl m(v2 -VI)

= (14.7096 psia)(10 lb m )[ (26.780- 0.01671) ft3 /lb m ][ (1 Btu) I (5.404 psia· ft3)] = 728.3 Btu The work is positive because the system does work on the surroundings.



6: 11

Pressure

p=

r- - - ---- --

14.7096 pSia

Saturated /VaporLine

,-f,-fl-Vapor

ttttt Liquid

Vg

Vt

ft3

=0.0167116

ft3 =26. 780 16

Specific Volume V

Figure 6-6. p-v Diagram for Constant Pressure Process

ft3

Example 6-2. A piston-cylinder device initially contains 1.0 of air at 20 psia in such a way that the temperature and 76°P. The air is compressed to 0.1 remains constant. Calculate the work done by the system.

ft3

Solution. The piston-cylinder andp-v diagram are shown in Figure 6-7. Assume that air can be treated as an ideal gas, then the relationship PV = mRT = constant applies. The area under the curve gives the work as:

IUS = PIV;

In(~ IV;)

= (20 psia)( 1.0 ft3) In( 0.1 11.0)[(1 Btu) I (5.404 psia· ft3)] = -8.52 Btu

The negative sign indicates that, for compression, work is done on the system.



6: 12

P2 = 200 psi a

P1

= 20 psia

--- -

V

IJ

Figure 6-7. p-VDiagram for Constant Temperature Process of Ideal Gas

In the situation given in Example 6-2, it is also possible to calculate the pressure at the end of the process and the mass in the system. Because it is an ideal gas at constant temperature:

P2 = PI (V; IV;) = (20 psia)[ (1.0 fe I (0.1 fe))] = 200 psia m = (pIV;)(R~)

= (20 psia)(1.0 fe)1 {[0.3704 (psia.fe)/(lb m. R)][(76 + 460)R]} = 0.101 Ibm



6: 13

The moving boundary work for a closed system process is represented by the area under the process curve on the pressure-volume diagram. The sign of the work depends on whether the volume increases or decreases. A cycle is formed by a sequence of processes such that the final state of the final process is the same as the initial state of the initial process. This means that the moving boundary work of the cycle is the net area enclosed within the process curves on the p-V diagram, as shown in Figure 6-8. Shaft work. Shaft work usually refers to the energy transmission by a rotating shaft. If the torque on the shaft is T, and the shaft rotates through an angle q, then the shaft work is:

W=Te s

The angle of rotation is often re-expressed in terms of the number of revolutions as 8 = 2 nN and the work is then: Ws = 2nNT

Pressure p

®

®:1 1

1 1

---t-----1 1 1

---r~------------------

1 1 I

leD I I

1

I

I

I

"---.......1--------------.....,1--... Volume V Figure 6-8. Boundary Work for a Cycle as the Net Area Under the p-V Curve



6: 14

The rate of doing work is:

Ws = (27tnT) / 60 ft ·lbf

/

s

= (27tnT) / 12.97 Btu / h =(27tnT) /33,000 hp =(27tnT) / 44,248 kW where n is the rotational speed in revolutions per minute and T is the torque in ft.lbr Often the shaft work is not given in terms of the torque and speed, but is specified either as a desired output of an engine or the needed power input to run a compressor or chiller. Electrical work. Electrical work appears in two forms in thermodynamics. The first form results from the electrons flowing through a resistance element and doing work at a rate of:

We = VI = 12 R =V2 / R The electrical current is I, the voltage across the element is V, and the resistance of the element is R. If I is in amps, V is in volts and R is in ohms, then the units of W are watts. The other common way of encountering electrical work is through an electrical motor, generator or alternator. The conversion between electrical work and mechanical work is done internally to the device and the result often shows up as shaft work.

6.4

Flow Work

F or open systems, there is also the possibility of mass flow across the boundary. That flow carries a certain amount of internal energy with it as it crosses the boundary, but in addition, the mass flow has to do work at the boundary to either enter or leave the system. A flow stream entering a system has to push its way into the system. This is known as the flow work. The rate of flow work W can be expressed very much as boundary work because it amounts to the boundary being pushed by a force F= pA at a velocity V The pressure,p, is constant at the location where the flow enters the system. The flow work is usually expressed as specific work (work per unit mass flow) so that the work can be identified with the flow



6: 15

stream entering or leaving the system. If the mass flow rate across the boundary is ~ VA/v, then the specific flow work is: Wflow

=

WI m

=

(pA)(V)I (VA I v)

=

=pv The flow work is pv. However, in most cases, the flow work is not identified in a category by itself. It is usually taken care of by considering the enthalpy, instead of the internal energy, that is carried into, or out of, the system by the mass flow. The enthalpy of the system is defined as:

h

= u+ pv

The enthalpy is the sum of the internal energy and the flow work. The units of internal 3 energy and enthalpy are usually given as Btu/Ibm. When p is in psia and v is in ft /lb m, it is necessary to apply the conversion factor 1 Btu = 5.404 psia·ft3/lb mto the pv term. We will see another derivation of the flow work and the enthalpy when we consider the First Law of Thermodynamics for open systems in Chapter 8.



6: 16

The Next Step Now the pieces are in place to discuss the First Law of Thermodynamics. In the next chapter, we will analyze the First Law as applied to closed systems.

Summary Heat and work are transitory forms of energy that exist when energy is crossing a system boundary. Heat is the name given to the transitory form of energy as it crosses a boundary separating systems (or a system and its surroundings) by virtue of a difference in temperature. Work is the name given to the transitory form of energy as it crosses a boundary separating systems (or a system and its surroundings) by virtue of a difference in pressure or force of any kind. Heat and work have many similarities: • Heat and work are not properties; they are associated with a process and not a state. • Heat and work are recognized only at the boundaries of the system as they cross that boundary. • Systems contain energy, but they do not contain heat or work. • Heat and work are path functions because their values depend on the process path as well as the end-states. • Heat and work have direction. The sign convention for heat is that energy addition is positive and the sign convention for work is that work done by the system is positive. Boundary, gravitational and acceleration work may be expressed as: Boundary work for a closed system, ] W2 = Area under the P-V diagram Constant pressure process,p] = P2' any substance:

Constant volume process, V] = V2 ' any substance:

lU;; = zero Isothermal process, T] = T2 ' ideal gas only P V = constant:

1U;; = PI V; In(~ / V;) = P2 ~ In(~ / V;)



6: 17

Isentropic process, s 1 = I

S2

,p JIk = constant, ideal gas only, P V = mRT:

W; = (P2 V2 - PI ~) / (1 - k) = mR(1; - I;) / (1- k)

Polytropic process, P vn = constant, nil, ideal gas only, P V = mRT:

Also important are shaft work and electrical work. For open systems where there is mass flow into or out of the system, there is work associated with the flow called flow work. Flow work is the product of the pressure and the specific volume and is accounted for in open systems by adding the flow work to the internal energy that is carried across the boundary by the flow. The combination of the internal energy and the flow work is the enthalpy: h = u+ pv

After studying Chapter 6, you should be able to: • Describe the concept of heat as an energy form crossing a system boundary; • Describe the concept of work as an energy form crossing a system boundary; • Describe the different types of mechanical work; and • Understand the termflow work and relate it to enthalpy.



6: 18


6-01.

Provide a description of heat, and describe how it is different from internal energy.

6-02. A 0.35 Ibm potato at 70°F is placed in an oven at 350°F. After 45 minutes, the potato temperature is 350°F. Assume the coefficient of specific heat for the potato is cp = 1.0 Btu/lbm·oF. If the reference datum for internal energy is U = 0 at t = OaF, then estimate the internal energy content of the potato at the beginning and end ofthe 45 minutes. What has been the heat during the 45 minutes and is it positive or negative? What was the average rate of heat transfer?

6-03.

Provide a description of work in general, and then give short (one- or two-sentence) descriptions of boundary work and shaft work.

6-04. Provide a description of flow work, and say how it is included in enthalpy.

6-05.

A frictionless piston-cylinder device contains 2 Ibm ofwater at 212°F and quality x = 0.4. The fixed weight piston is not attached to a shaft. Heat is added to the water until it is all saturated vapor. Calculate the work done by the system.

6-06.

A piston-cylinder device initially contains 0.2 ft3 of air at 80 psia and 76°F. The air is expanded to 0.9 ft3 in such a way that the temperature remains constant. Calculate the work done by the system, the final pressure and the mass of air.

6-07. In the isentropic compression of air in a cylinder, the volume is reduced from 1.5 ft3 to 0.5 ft3. The ratio of specific heats for air is k = 1.4. The initial pressure is 14.7 psia and the initial temperature is 45°F. Assume the air acts as an ideal gas. Estimate the final pressure, the final temperature and the work done.



7: 1

Chapter 7 First Law of Thermodynamics Applied to Closed Systems


Introduction to Controlled Mass Approach

• 7.2

The First Law of Thermodynamics for a Closed System

• 7.3

Conservation of Energy

• 7.4

First Law Applied in Example Cases



Study Objectives of Chapter 7 The objective of Chapter 7 is to develop an understanding of the First Law of Thermodynamics as applied to closed systems. So far, we have developed the ideas and definitions of properties, states, change of states and cycles, and introduced the concepts of heat and work. It is now time to consider the relationship between work, heat and change of energy as we change from one state to another. The relationship is expressed as the principle of energy conservation (or the First Law of Thermodynamics). Initially, the First Law is applied to a closed system (a system for which no mass crosses the system boundary) because the resulting equations are simpler than for an open system.


Chapter 7 First Law and Closed Systems

7:2

However, the principle to be applied is the same for closed and open systems. After studying Chapter 7, you should be able to: • Understand the First Law as an expression of the conservation of energy principle; • Write the First Law in equation form by applying an energy balance to a closed system; and • Apply the First Law to simple cases of closed systems.



7: 3

7.1

Introduction to Controlled Mass Approach

In an earlier chapter, it was stated that the first step in finding a solution to a thermodynamics problem is to identify the system to be analyzed. A closed system is one in which a quantity of matter is identified for study. A closed system definition is often called the control mass approach. A quantity of matter means that the mass to be studied will remain the same mass throughout the analysis. The matter in the system may change form, phase or location in space, but the system is identified as that matter that originally was targeted for study. There is an imaginary boundary drawn to contain the matter. The location of the matter and, therefore the shape of the boundary, may change with time, but to remain a closed system, no mass enters or leaves the system through the boundary. Everything outside the boundary is known as the surroundings. Although there is no mass exchange across the boundary in closed systems, there may still be work and heat crossing the boundary as the system interacts with its surroundings. An example of closed system is shown in Figure 7-1. The object chosen for demonstration is a piston-and-cylinder combination with air in the cylinder. The air is trapped in the cylinder and there is no way for the air to escape, or for new air to come into the cylinder. The system is chosen to be the matter (mass of air) that is trapped in the cylinder. The system boundary is shown by the dotted line. The surroundings are then the piston, the cylinder

Piston

r----------------'

I

Weight

I

Piston

Closed System

r----------------j I I I I

Closed System

I I I I

~-----------------

Before Adding Weight

After Adding Weight

Figure 7-1. Closed System



7:4

wall, the cylinder head and everything else outside the dotted line. As weights are added to the piston and the piston moves to reduce the volume in the cylinder, the system (air in the cylinder) is compressed and changes shape, but still maintains the same air that was originally targeted as the system. A closed system analysis is one in which the mass does not change. The analysis is known as the controlled mass approach, and it is characterized by the absence of mass entering or leaving the system, and no change of mass inside the system.

7.2

The First Law of Thermodynamics for a Closed System

F or a closed system undergoing a small change of state, the increase of total energy of the system is equal to the difference between the heat added to the system and the amount of work done by the system on the surroundings: 11£=oQ-oW

Here 11£ is the increase in total energy of the system due to heat transfer into the system oQ and work going out of the system 0 W. The 11 indicates a small change inside the system while the 0 indicates a small flow across the system boundary. 11 and 0 are both said as delta. A negative change in energy would indicate a decrease of energy while negative flow of heat or work would mean they were in the opposite direction to the above conventions.

7.3

Conservation ofEnergy

The First Law of Thermodynamics states that energy can neither be created nor destroyed, it can only change form. In terms of the universe, this may be difficult to see, but if the universe is divided into two parts (a closed system and its surroundings), then it is possible to consider the implications of the First Law by examining the energy associated with the system and the exchange of energy between the system and its surroundings. Because the system is a well defined mass, m, we can talk about the total energy of the system E as being made up of the sum of the microscopic energy (internal energy U) and the macroscopic energy (potential energy PE and kinetic energy KE):

E=U+PE+KE

=m(u+ pe+ke)


Fundamentals ofTI,ermodynamics and Psychrometries

7: 5

There are two ways that energy can cross the system boundary when there is no mass exchange between the system and its surroundings. Energy can cross the boundary as heat (Q, positive when entering system) and work (W, positive when done by the system on the surroundings). If energy can neither be created nor destroyed, then an energy account for system states that: Net energy transfer to (or from) the system as heat or work = Net increase (or decrease) in the total energy of the system or

where I

Q2

= net heat transfer across the system boundary in going from State 1 to State 2 = L 1Q2 in

I

-

L

1

Q2 out

~ = net heat transfer across the system boundary in going from State 1 to State 2

!:!ill = net change in total energy of system between states 1 and 2

= E2 -E) =!::,U+ME+11KE

=m(u2 -ul )+mg(Z2 -z))+m(V;2 -Vn12 and L = the sum of all components

Therefore the First Law of Thermodynamics (or conservation of energy) for a closed system undergoing a process from State 1 to State 2 may be stated as:

For negligible change in the kinetic and potential energy terms, the equation is:



7:6

7.4

First Law Applied in Example Cases

To develop an understanding of how the First Law of Thermodynamics may be applied to closed systems, we will consider a series of simplified applications in which it will be assumed that the changes in potential and kinetic energy are negligible. These applications are based on two configurations: a rigid tank in which the volume remains unchanged during the process; and a piston-cylinder configuration in which the volume may change as the piston moves. For each configuration, there will be examples of the working fluid being an ideal gas (such as air) or a liquid-vapor combination (such as steam or a refrigerant). The applications will include actual numbers so that the values of properties can be calculated, or read from tables and charts as appropriate, but the solution technique will emphasize the basic solution in symbols before substituting in for the numbers. For each application, there will be a sketch of the configuration and a property diagram (usually p-v) to show the process. Application 1. A rigid tank with a volume of 3 ft3 is initially filled with air at 352.07 psia and 140°F. The air is cooled to 20°F. Treat the air as an ideal gas. Apply the Ideal Gas Law to determine the mass of air in the tank. Apply the continuity equation to determine the final pressure in the tank. Assuming the air in the tank constitutes a closed system, apply the First Law of Thermodynamics to determine the heat transferred during the process. Solution. The air in the tank is taken as the system. The process is assumed to start when the tank is already full of air at p = 352.07 psi a and t1 = 140°F (State 1). The process ends when the air in the tank is cooled to t2 = 20°F,P2 =? Because no mass enters or leaves the tank during the process, this is a closed system. Because it is a rigid tank, the volume remains constant. The configuration is shown in Figure 7-2 along with the p-v diagram showing the constant volume process from State 1 to State 2.

The mass in the tank may be calculated by applying the Ideal Gas Law to the air at State 1:

m= Pl~ / RT; ForP1 = 352.07 psia; V1= 3 ft3; R = 0.3704 (psia·ft3)/(lbm·R) (see Table 5-1); and T1 = (140 + 460)R, the mass is:

m = (352.07 psia)(3 fe) / ([ 0.3704(psia. ft3) / (Ibm' R)]( 600 R)}

=4.75 Ibm The mass of a closed system remains unchanged. The Ideal Gas Law is applied to State 2 to determine the unknown pressure:

P2

= mRT; /~



7: 7

For m = 4.75 Ibm; V2 = VI = 3 ft3 (rigid tank); and T2 = (20 + 460)R, the pressure is:

pz = (4.75 lb n,)[ 0.3704(psia. ft3) / (Ibm' R)](480 R) / (3 ft3) = 281.7 psia The calculation to find the final pressure could also have been done without finding the mass. Consider equating the mass from the Ideal Gas Law applied at States 1 and 2, equating volumes, and canceling the gas constant to get: P2

= (1; / I;)PI = (480/600)(352.07) = 281.7 psia

The First Law of Thermodynamics applied to the process from State 1 to State 2 is:

CQ2 - I Tfi)=m(u2 -u1) There is no change of either potential energy (Z2 - zJ = 0) or kinetic energy (velocity V =0) in this application, and the equation may be solved for the heat transferred as: I

Q2 =1

Tfi + m(u2 -

U1)

r----------, I I I I I I

AIR

I I I I

:

V=3ft'l

:

I I I

t Pressure

p

CD]

PI

= 140°F = 352.07 psia

®

I I I

-----------..1

Specific Volume, v

Figure 7-2. Rigid Tank and p-v Diagram for Application 1



7:8

There is no shaft work. The boundary work is also zero because there is no change in volume. Recall that the boundary work could be expressed as the area under the curve on the p-V diagram. Because the curve is a vertical straight line, as shown in Figure 7-2, there is no area under the curve and the boundary work is zero:

°

I

U;; = (constant volume process)

I

Q2

= m(U2 - ul )

The change of specific internal energy for an ideal gas can be expressed as the product of the specific heat at constant volume and the change in temperature, so that:

IQ2

= mc.(t2 -tl)

Form =4.75Ibm ; cv =0.171 Btu/lbm .oF;t2 =20°F;andtl =140°F,theheattransferis:

IQ2 =(4.75Ib m)(O.l71 Btu/Ib m·oF)(20-140)o = -97.5 Btu The negative sign indicates that the direction of heat flow was out of the system, which agrees with our sense that cooling the air means to remove heat. Application 2. A rigid tank with a volume of 3 ft3 is initially filled with R-22 in a saturated vapor state at 140°F. The refrigerant is cooled to 20°F. Use the refrigerant tables (in Appendix C-J) to determine the mass of refrigerant in the tank. Determine the final pressure. Assuming the refrigerant in the tank to constitute a closed system, apply the First Law of Thermodynamics and determine the heat transferred during the process. Solution. The refrigerant in the tank is taken as the system. The process is assumed to start when the tank is already full atpI = 352.07 psia and tl = 140°F (State 1). The process ends when the refrigerant in the tank is cooled to t2 = 20°F, P2 = ? Because no mass enters or leaves the tank during the process, this is a closed system. Because it is a rigid tank, the volume remains constant. The configuration is shown in Figure 7-3 along with the p-v diagram showing the constant volume process from State 1 to State 2.

From the refrigerant table in Appendix C-J, the properties of saturated R-22 at 140°F are: PI = 352.07 psia, VI = Vg = 0.1434 ft3/lb m ,and enthalpy hI = hg = 112.784 Btu/Ibm' The mass IS:

m = V;

IVI

=(3 ft3)/(O.l434 ft3 /Ibm) = 20.9 Ibm



7: 9

r----------

I I

: REFRIGERANT I R-12 I I I I I

t

PI

= 140°F

= 352.07 psia

Pressure p

V=3ft3

® Specific Volume, v

Figure 7-3. Rigid Tank and p-v Diagram for Application 2

From Figure 7-3, it can be seen that State 2 is a saturated mixture of liquid and vapor. At 20°F, the properties as read from Appendix C-J are: P2

=Pg = 57.083 psia; vg2 = 0.9343 ft3/lb m ; density ofliquid p = 81.411bm/ft3 (note

that density is the inverse of the specific volume and vfl = 0.0122 ft3/Ib m ); enthalpy of saturated vapor hg2 = 106.434 Btu/Ibm; and enthalpy of saturated liquid hfl = 16.090 Btu/Ibm

The quality ofthe mixture (ratio of mass vapor to total mass) can be found by recognizing that for the constant volume process v2 = VI = vfl +X/Vg2 - vfl ). Therefore, the quality is: x2

=(V2

-v j2 )/(vg2 -v j2 )

= (0.1434 = 0.1423

0.0122) I (0.9343 - 0.0122)

With this quality, the specific enthalpy at State 2 is:

~

=hj2 +x2 (hg2 -hj2) = 16.090 + 0.1423(106.434 -16.090) = 28.946 Btu I Ibm



7: 10

The First Law of Thermodynamics applied to the process from State 1 to State 2 is:

CQ2 -, n;) = m(u2 -u,) The changes in kinetic and potential energy are zero. There is no boundary work because it is a constant volume process. The heat transfer is:

,Q2

= m(u2 -u,)

The tabulated values for internal energy are not given in Appendix C-l, but they can be calculated from the enthalpy by remembering the relationships:

u,

= ~ - p,v, = 112.784 -(352.07 Ib j I in.2 )(144 in? Ife)( 0.1434 fe /lb m)(Btul 778 ft ·lb j

)

= 103.4 Btu I Ibm

u2

= ~ - P2 V 2 2

=28.946-(57.803Ib j lin?)(144 in.2 /ft )(0.1434 ft3 Ilb m )(Btu/778 ft.lb j

)

= 27.4 Btullb m

The heat transfer is:

,Q2

= m(u2 -u,) = 20.9 Ib m(27.4-103.4) Btullb m = 1,588.4 Btu

As in the previous application, the negative sign is an indication of the heat removed from the system. The large amount of heat transfer is an indication of the heat released during change of phase from vapor to liquid (condensation).

Application 3. A piston-cylinder device contains 10 Ibm of air which is maintained at a constant pressure of 150 psia by a floating piston of constant weight. as shown in Figure 7-4. A resistance heater within the cylinder is turned on for five minutes and supplies heat at a rate of 1.5 kW. During those five minutes, a heat loss of 200 Btu occurs through heat transfer to the surroundings. If the initial temperature is 20°F, determine the temperature after the five minutes.



7: 11

Solution. The air and the resistance wire are selected as the system. Even though the piston can move, and the volume can change, there is no mass entering or leaving the closed

system. The First Law of Thermodynamics applied to the process from the initial State 1 to final State 2 is:

The changes in kinetic and potential energy are assumed to be negligibly small. Foran ideal gas, the change in internal energy is expressed as (u 2 - u}) = cv (t2 - t}). There are two forms of work in this problem. The boundary work for the constant pressure process is mp(v2 - v). This may also be seen from the area under the curve on the p-v diagram shown in Figure 7-4. Because the air is treated as an ideal gas, the term pv may be replaced by Rt, so that the boundary work is mR(t2 - t}). The electrical work is negative because it is work done on the system and has a magnitude of the power (1.5 kW) times the time it is on (5 minutes); }W2electrical = - (1.5 kW)(3412 BtuI(hr·kW)(5 minutes)(hr/60 minutes) = 426.5 Btu. The heat transfer ]Q2 is negative 200 Btu because it is a heat loss from the system. The First Law equation now reads:

(IQ2 -I W;) = m(u2 -ul )

[-200-(-426.5)-mR(t2 -tl )]=mc (t 2 -tl ) V

Constant Pressure Process

-----------1 : I I I

AIR P = 150 psi a

I I

Pressure

p

®

CD

... I----~I

I I

: Electrically ~ : I Heated - - - - - ______ 1 Specific Volume, v

Figure 7-4. Piston-Cylinder Arrangement andp-v Diagram for Application 3



7: 12

Solving for t2 :

t2

= t, + (226.5) 1meR + c v )

Since cp = (R + c) = 0.240 Btu/(lbm·oF), the final temperature is:

t2

= 20+(226.5)1 [10(0.24)] = 114.4°F

It may be noted here that the First Law was used to find one of the properties rather than the heat transfer as in the previous two examples. It all depends on which information and which is to be determined that dictates what is to be solved for in the First Law equation. What is important is that it takes knowledge of the First Law along with knowledge of the equation of state (as given by Ideal Gas Law, or as a set of values in a table or chart) and knowledge of the type of process (constant volume, constant pressure, etc.) to solve the problem. Application 4. Air is compressed polytropicaUy in a piston-cylinder device, pv" = constant, where n = 1.2. The initial temperature and pressure are 20°F and 15 psia. The volume ratio between the beginning and end of the stroke (compression ratio) is 10. Determine the work and heat transfer per unit mass during the compression stroke. Solution. The air is selected as the system. Even though the piston can move and the volume can change, there is no mass entering or leaving the closed system. The piston-cylinder arrangement is shown with the p-v diagram in Figure 7-5.

Polytropic Process

-----------1 I I I I I I

I I

AIR

I I

I I

I I

!....---- ______ I

Pressure

p

P,= 15 psia

1,

= 20°F

Figure 7-5. Piston-Cylinder Arrangement and p-v Diagram for Application 4



7: 13

It was stated in a previous chapter that for a polytropic process pV' = constant. Therefore:

= PI (VI /vJn = 15(10)1.2 = 237.7 psia

P2

For an ideal gas, the temperature ratio can be expressed in terms ofthe volume and pressure ratios as:

I; = 7; (v, / vI)(Ps / PI)

= (20+460)(0.10)(237.7115) =

760.7 R

= 300.7°F

The work of a polytropic process was given in a previous chapter as: Polytropic process, P Jill = constant, n:f:. 1, ideal gas only, P V = mRT IW2

=

(P2 V2 - PI VI)/(l-n)= R(I; -7;)/(l-n)

=

[0.411 Btu/ (lbm·oF) ](300.7 - 20)oF / (1-1.2)

= -576.8 Btu / Ibm The negative sign means that work has to be done on the air during compression. From the First Law of Thermodynamics:

CQ2 -1 n;)=m(u2 -UI ) For negligible change in kinetic or potential energy of an ideal gas, this equation becomes:

Iq2

=1

W2+C,,(t2 -t I)

= -576.8 + [0.171 Btu/ (lb m·oF)](300.7 = -528.8 Btu / Ibm

20)° F

The negative sign means that heat is rejected from the air to the surroundings during compression.



7: 14

The Next Step This chapter introduced the First Law of Thermodynamics as applied to closed systems. For closed systems, there is no mass flow across the system boundary and, therefore, no need to keep track of where the mass is going and what energy it carries with it as it crosses a system boundary. In the next chapter, we will introduce the idea of a control volume approach for the First Law applied to open systems so as to account for mass flow. It will include a statement of the mass and energy conservation principles, and will show how the enthalpy is a convenient property to include both internal energy and flow work. Because many thermodynamic problems are taken as steady state, the rate form of the First Law equation will be introduced in the next chapter.

Summary The First Law of Thermodynamics has been introduced as an energy conservation principle. The equation describing the First Law for closed systems is given as:

CQ2

-I

Tfi) = m(u2 -u1)+ mg(z2 -ZI)+ m(V;2 - V/)/2

[Heat Transfer In] - [Work Out] = [Increase of Energy of the System] where, 1

Q2 = net heat transfer across the system boundary in going from State 1 to State 2

=L 1

IQ2in

-I

IQ20ut

Tfi = net heat transfer across the system boundary in going from State 1 to State 2

/)ill

= net change in total energy of system between States 1 and 2 =E2 -EI =/).U+ME+M(E

=m(u2-ul)+mg(z2 -zl)+m(V; -Vn12 and L = the sum of all components



7: 15

From the sample applications, it can be seen that the First Law may be solved for heat transfer, work or one of the state properties, but that it is always used in conjunction with knowledge about the equation of state for the working fluid, and knowledge of the type of process from State 1 to State 2. After studying Chapter 7, you should be able to: • Understand the First Law as an expression of the conservation of energy principle . • Be able to write the First Law in equation form by applying an energy balance to a closed system. • Apply the First Law to simple cases of closed systems.



7: 16


7-01.

Give a one-page description of the heat, work and energy change terms in the First Law equation for a closed system and describe how they relate to each other (a word description of the conservation of energy principle).

7-02.

A rigid tank with a volume of3 ft3 is initially filled with air at 15 psi a and 20°F. The air is heated to 200°F. Treat the air as an ideal gas. Apply the Ideal Gas Law to determine the mass of air in the tank. Apply the continuity equation to determine the final pressure in the tank. Assuming the air in the tank to constitute a closed system, apply the First Law to determine the heat transferred during the process.

7-03. A rigid tank with a volume of 1.5 ft3 is initially filled with R-22 in a saturated vapor state at 140°F. The refrigerant is cooled until the pressure is 57.083 psia. Sketch the process on a p-v diagram. Use the refrigerant tables (in Appendix C-l) to determine the mass of refrigerant in the tank. Determine the final temperature. Assuming the refrigerant in the tank to constitute a closed system, apply the First Law and determine the heat transferred during the process. 7-04.

A piston-cylinder device contains 10 Ibm of air which is maintained at a constant pressure of 150 psia by a floating piston of constant weight. How much heat must be added to raise the temperature from 20°F to 300°F?

7-05. A piston-cylinder contains 1 Ibm of saturated liquid R-22 at 20°F. The refrigerant is maintained at constant pressure by a floating piston. Determine the work and heat if the refrigerant is heated until the quality is 50%. Sketch the process on a p-v diagram. 7-06. Describe experiments in which the First Law may be used to determine the specific heats of an ideal gas of unknown properties placed in a rigid tank, and in a cylinder in which the pressure is kept constant with a floating, frictionless piston.



8: 1

Chapter 8 First Law of Thermodynamics Applied to Open Systems


Introduction to the Control Volume Approach

• 8.2 .8.3

Conservation of Mass Conservation of Energy and the First Law for Open Systems

• 8.4

Steady-Flow Processes

• 8.5

First Law Applied in Examples of Steady-Flow Processes



Study Objectives of Chapter 8 The objective of Chapter 8 is to understand the First Law of Thermodynamics as applied to open systems. To develop that understanding, it will be necessary to first introduce the ideas of a control volume approach and the way to write the equation for the conservation of mass.


Chapter 8 First Law and Open Systems

8:2

The concept of flow work and enthalpy will be used in the First Law equation, especially for steady flow problems. After studying Chapter 8, you will be able to: • Apply the conservation of mass principle to an open system; • Understand the First Law as an expression of the conservation of energy principle; • Write the First Law in equation form by applying an energy balance to an open system; and • Use the principles of conservation of mass and energy to solve steady flow problems of open systems.



8: 3

8.1

Introduction to the Control Volume Approach

In the last chapter on closed systems, the system boundary was defined as that boundary that separated the designated mass from its surroundings. For an open system (where mass can flow across the system boundary), it is convenient to define the boundary in terms of a surface that separates a control volume from its surroundings. The control volume may vary in time and space. Mass can enter or leave the control volume. An open system is defined as the mass contained in a control volume at any instant in time. From one instant to another, the mass of the open system may change. The boundary between the control volume and the surroundings is known as the control surface. The selection of an appropriate control volume depends on the problem to be solved. Ifit is of interest to determine the conditions in a ventilated room, then it would be appropriate to choose the room's interior as the control volume and examine the flow of air into and out of the room, heat conducted through the walls, and heat and moisture generated inside the room. If it is desired to determine the overall performance of an engine, it may be useful to select a control volume that has air and fuel entering the control volume and exhaust gas leaving the control volume. Alternatively, ifthe interest is the thermodynamic performance of one of the engine cylinders, then it would be appropriate to select a control volume as the inside of that cylinder. The control surface that defines the control volume can either be a real physical boundary (such as the cylinder wall) or an imaginary surface (such as a plane across the intake port to the cylinder). The choices of control volume are infinite, but as we shall see, some choices are more useful than others. The control volume approach of defining an open system will prove convenient for applying the First Law of Thermodynamics to open systems.

8.2

Conservation ofMass

Mass is a conserved property. For a closed system, the conservation of mass is automatically satisfied because the system is defined in terms of a set mass. However, for an open system, it is necessary to assert the conservation of mass principle by keeping track of changes in mass. Consider an arbitrarily selected control volume, defined by a control surface separating the volume from its surroundings. The mass inside the control volume at any instant of time is m ev ' Consider a change of mass inside the control volume (~mev) over a period of time. If the amount of mass entering the control volume during that same time period is min and the



8:4

mass leaving is mout ' then the principle of conservation of mass applied to the system is: (mass entering CV)

- (mass leaving CV) = (net increase in mass inside CV)

or,

This is not much different than tracking the money in your checking account by monitoring deposits, withdrawals and the change in the account balance. The special case where the mass leaving is just balanced by the mass entering an open system (leading to no change of the mass inside the system) is still different from a closed system. For an open system to be treated as closed, there must be no mass entering or leaving the control volume.

8.3

Conservation ofEnergy and the First Law for Open Systems

The conservation of energy principle (or First Law of Thermodynamics) was introduced in a previous chapter and applied to a closed system (a system of fixed mass). For a closed system, it was explained how the change of energy of the system was related to the heat and work that crosses the system boundary. For a closed system, the balance was:

(Q - W) = !1E (closed system) However, for an open system, there are additional terms to be accounted for in the balance. The mass flow in or out of the control volume carries with it internal, kinetic and potential energies. Furthermore, for an open system, the work term also needs to include the flow work term first described in Chapter 6. The differences between a closed system and an open system in the ways that energy can be affected by flows across the boundary are shown in Figure 8-1. The energy balance applied to a control volume is written as: Energy entering CV + Energy of mass - Energy of mass as heat or work

entering CV

leaving CV

=

Increase of energy inside the CV

or,



8: 5

t

Boundary Work

-----------, I 1 1

I

--~I~

1

Closed System

1

1

Electrical 1 1 Work :__________ I:

Boundary Work

-----------,

--~~

Electrical Work Mass ----. Flow In

I

a) flow in carrying internal energy, ke and pe b) flow work

Heat

1 1 1

1 1 1 1 1 1

CV

'------1---: I Heat

----.

Mass

FlowOu!

a) flow out carrying internal energy, ke and pe b) flow work

Figure 8-1. Closed and Open Systems

The signs on Qand W reflect the sign convention that heat transfer is positive for heat added to the control volume, and work is positive when work is done by the system on the surroundings. As before, the heat transfer is the energy crossing the system boundary by virtue of a temperature difference between the system and its surroundings. The work term includes boundary work, shaft work and flow work. The energy of the mass entering or leaving the control volume may be regarded as the product of the mass entering or leaving and the energy content per unit of mass entering or leaving (specific internal, kinetic and potential energy). The change of energy in the control volume is the difference between the final and initial energy content of the control volume (internal, kinetic and potential energy).

8.4

Steady-Flow Processes

Many applications of the First Law of Thermodynamics for open systems are for situations that can be reasonably described as steady-flow processes. We will examine the First Law for steady-flow problems in some detail before going back to the more general case, because there are several concepts that are more easily introduced under the conditions of steady flow. Steady means unchanging in time. A steady-flow process may be described as one in which a fluid steadily flows through a control volume such that properties within the control volume, and the properties of the flow streams entering and leaving the control volume, remain constant at all times. Properties may vary from point to point in the control



8:6

volume, but do not change in time. Because properties remain constant, there is no change in the mass, energy or volume of the control volume. This simplifies the conservation of mass and energy equations by setting f)"m ev ' Mev and the boundary work to zero. For steady-flow processes, the flows of mass and energy are best expressed as rates. Consider the open system pictured in Figure 8-2. Mass enters the control volume at State 1 at a steady rate of in! .Mass leaves the control volume at State 2 at a steady rate of in2' (The dot above the variable indicates the rate of change with time of that variable.) This is a singlestream, steady-flow process. Because /).mev and Mev are zero, the conservation of mass and energy equations may be written as: (single-stream, steady flow)

(single-stream, steady flow)

where:

m

is the mass flow rate (lbm /h)

Q

is the heat transfer rate (Btu/h)

TV

is the rate of work (same as power) (Btulh, kW, ft·lbj/s, or hp)

E

is the rate of energy carried across the boundary by the flow of mass (BtuIh)

The energy carried across the boundary by the flow of mass may be written as the product of the mass and the specific energy of that mass:

E =m(u+ pe+ke) = m[u+ gz+ V2 /2]

The boundary work for a steady-flow process is zero, therefore the rate of work is a combination of shaft work and flow work:

TV = ~haft + TVjlOW



8: 7

Shaft Work

6 -+-.....,....... Velocity V2 CV

Velocity

----~--------------------------~------------- datum

Figure 8-2. Single Stream, Steady-Flow Open System

The flow work was shown in Chapter 6 to be Wflow = ~ pv . Applying the expressions for energy and work to both the incoming mass flow at State 1 (negative flow work) and the outgoing flow at State 2 (positive flow work), then the conservation of energy equation may be written as:

or, reorganizing terms:

In this equation, we can see how the flow work term (pv) and the internal energy term (u) come together to give the enthalpy h = u + pv. After moving some terms to the right hand side of the equation, the first law equation for a single-stream, steady-flow open system is:

Q- ~ = m(~ + pe2 + ke 2 ) - m(~ + pel + kel ) or, (rate of heat transfer into CV) - (rate of shaft work by CV) = (rate of flow of enthalpy, ke, pe, out of CV) - (rate of flow of enthalpy, ke, pe, into CV)



8: 8

Dividing through by

m, and setting q = QI m and Ws = WI m, the equation is often written

per unit of mass flow as: q-ws = (~-~)+(pe2 - pel) + (ke 2 -kel )

q-ws =(~ -~)+g(Z2 -ZI)+(V;2 -V;2)/2 These expressions are very much like those for a closed system except that the enthalpy is used in place of the internal energy, and the work is the shaft work. For many problems, the change in potential energy and the change in kinetic energy are negligible and the First Law equation may be simplified to:

Q-~ =m(~ -~) (single-stream, steady-flow, negligible change in pe and ke) q-ws =(~ -~) (single-stream, steady-flow, negligible change inpe and ke) The result can easily be expanded to steady-state applications with more than one flow by considering many flow streams in and out. The conservation of mass and energy equations are then: (steady-flow)

(steady-flow)

where L is the summation symbol to indicate that, for more than one flow stream, the contributions of the different streams must be added together. In later equations, we will omit the summation sign out of simplicity, but we will understand that it is there when needed.

8.S

First Law Applied in Examples ofSteady-Flow Processes

Many HV AC applications (such as pipe flow, duct flow, nozzles, diffusers, turbines, compressors, fans, throttling valves, heat exchangers, chillers, and heating or cooling systems) can reasonably be assumed to be undergoing steady-flow processes. Many of these devices



8: 9

will be discussed in detail in Chapter 9. However, to introduce the application of the control volume approach in applying the First Law of Thermodynamics to steady-flow, open systems, we will look at a few simple configurations here. As before, we will select some applications that use air (ideal gas), and some applications that use refrigerant or steam (property tables). Application 1. The inlet temperature to a centrifugal flow, air compressor is 70°F. The outlet temperature is 120°F. The steady mass flow rate is 0.04 Ibm/s. If the compressor is adiabatic, estimate the rate of doing work. Neglect the kinetic and potential energies of the flow. Solution. The compressor is pictured in Figure 8-3. The control volume is the inside of the compressor. The single-stream, steady-flow form of the First Law equation is:

Q-W: =m(~ -~) The heat transfer is zero. The enthalpy change for an ideal gas with constant coefficients of specific heat is (h2 - h J ) = Cp (t2 - tJ ). Therefore, solving for that work yields:

w: = -mCAt

2 -

tl )

= -(0.04)(lbm/ s)(3600 s/ hr)(0.24)(Btu / lb m·o F)(120 -70)° F =

-1728 Btu / h

=-0.51 kW

Work done

Air Compressor

on compressor

Figure 8-3. Air Compressor of Application 1



8: 10

Application 2. The inlet condition to a centrifugal flow, R-134a compressor is saturated vapor at 20°F. The steady mass flow rate is 0.04 Ibm Is. If the heat lost from the compressor is 1000 BtuIh and the work to run the compressor is 1.0 kW, estimate the specific enthalpy of the refrigerant leaving the compressor. Neglect the kinetic and potential energies of the flow. Solution. The compressor is pictured in Figure 8-4. The First Law equation is:

The heat transfer is given as 1000 BtuIh. The work is given as -1.0 kW = -3,412 BtuIh. The flow rate is 0.041bm Is = 144lbm Ih. The enthalpy at inlet is found from the saturated vapor tables for R-134a in Appendix C-3 to be h J = 105.907 Btu/Ibm' Substituting into the First Law equation and solving for h2 gIves:

~ = ~ +(Q-~)Im

= 105.907 + [-1000-(-3412)]/144 = 122.7 Btullbm

r

Saturated Vapor t1 = 20°F

r---_ --- ---Work

---- ----

Refrigerant 134a Compressor

Figure 8-4. R-134a Compressor of Application 2



8: 11

Application 3. A nozzle is a device for increasing flow from low velocity to high velocity. The entry conditions of a steady flow of air through a nozzle are 20 ftI s and 70°F. If the flow is adiabatic and the outlet temperature is observed to be 69°F, estimate the velocity at exit. Solution. The nozzle and a chosen control volume are shown in Figure 8-5. This is a singlestream, steady-flow problem. The difference in potential energy of the flow in and out is negligible, but the change in kinetic energy is the function of the nozzle. The First Law expressed per unit mass is:

q-ws

=(h,. _~)+(V;2 -V;2)/2

There is no shaft work. The heat transfer is zero because the flow is adiabatic. The difference in potential energy of the flow in and out is negligible. For an ideal gas, the enthalpy change can be written as (h2 - h]) = Cp (t2 - t]). Therefore:

0= CAt2-t,)+(V;2 -V;2)/2 Solving for the outlet velocity:

V2 = {V;2 -CAt2_t,)(2)}"2

= {202( ft2 / S2)_ 0.24(Btu/lb m.oF)(778 ft .lb j = 108 ft/ s

/Btu)(69 -70)OF(2)(32.2)(lb m. ft /lb js2)} 112

Note that it is important to develop the correct units for the kinetic energy term. The relationship between lbjand Ibm is taken from the expression that 1 lbj will accelerate 1 Ibm at

32.2 ftls2.

t1=700~ velocity 20 ft/s

Figure 8-5. Nozzle of Application 3



8: 12

Application 4. A mixing tank has two steady inflows and one steady outflow. The conditions of the first inflow is superheated steam at t1 = 400°F, h1 = 1,239 Btu/Ibm' and flow rate

m1 = 20 lbmlh. The second inflow is compressed liquid at t2 = 70°F, h2 = 38 Btullb . If the rate of heat removal from the tank is 15,000 BtuIh, then estimate the inflow ofwate;needed to result in the outflow being saturated liquid water at 212 OF. Neglect changes in kinetic and potential energy. Solution. The tank is pictured in Figure 8-6. There is more than single flow, therefore we use the conservation of mass and energy conservation equations as: (steady-flow)

(steady-flow)

For this case, the equations become:

tf I

Superheated Vapor t1 = 400°F h1 = 1239 Btu/lb 1 = 20 Ib/h

m

v-~He.at Removal

--------------., I

1

Compressed Liquid

H

-

I

t2 =70°F h2

:

=38 Btu/lb

:

I

Mixing / :

Tank

:

+

I I '-1______________ 1

I

Ii

.

Saturated Liquid

ts = 212°F

Figure 8-6. Mixing Tank of Application 4



8: 13

For first inflow hi = 1,239 Btullbm ,and flow rate mJ = 20 lbmIh. For second inflow, h2 = 38 Btu/Ibm. For outflow, the enthalpy can found in Appendix B-1 as h3 = 180.2 Btu/Ibm. There is no shaft work shown for the tank.

-15,000 = ~(180.2)-(20)(1,239)-~(38) Solving these two equations for ~ and ~ results in: ~ = 43.4 Ibm ~

Ih

= 63.4 Ibm Ih

The Next Step In the next chapter, we will take a closer look at some of the devices commonly used in HVAC applications and see how the First Law of Thermodynamics can help when estimating the performance of these devices. The applications and devices include pipe flow, duct flow, nozzles, diffusers, turbines, compressors, fans, throttling valves, metering valves, heat exchangers, chillers, and heating or cooling systems. Cycle analysis will not be part of Chapter 9, but will appear in later chapters.

Summary The control volume approach is introduced for open systems and the conservation of mass and First Law of Thermodynamics are written as: (Mass entering CV)

- (mass leaving CV) = (net increase in mass inside CV)

or,



8: 14

Energy entering CV + Energy of mass - Energy of mass leaving CV as heat or work entering CV

=

Increase of energy inside the CV

or,

Particular emphasis is placed on the rate form of the steady-flow mass and energy conservation equations. The equations for many flows be summarized as follows: (steady-flow)

(steady-flow)

For single-flow, steady-flow the First Law equation becomes: Rate of heat transfer into CV

Rate of shaft work byCV

Rate of flow of enthalpy, ke, pe, outofCV

Rate of flow of enthalpy, ke, pe, into CV

or,

The equation is often written per unit of mass flow as:

q-ws =(~ -~)+g(Z2 -ZI)+(V;2 -V/)/2

(single-stream, steady-flow)

For many problems, the change in potential energy and the change in kinetic energy are negligible and the First Law equation may be simplified to:

Q- Tf: = m(~ -~)

q-

Ws

= (~ -~)


(single-stream, steady-flow, negligible change inpe and ke)

(single-stream, steady-flow, negligible change in pe and ke)


8: 15

After studying Chapter 8, you are now able to: • Apply the conservation of mass principle to an open system. • Understand the First Law as an expression of the conservation of energy principle. • Write the First Law in equation form by applying an energy balance to an open system. • Use the principles of conservation of mass and energy to solve steady flow problems of open systems.

Fundamentals ofTI,ernwdynamics and Psychrometries


8: 16


8-01.

The inlet temperature to a centrifugal flow, 0.8 kWair compressor is 20 o P. The steady mass flow rate is 0.04 Ibm Is. If the compressor is assumed adiabatic and the kinetic and potential energies are assumed negligible, then estimate the outlet temperature of the compressor.

8-02.

The inlet condition to a centrifugal flow, R-134a compressor is saturated vapor at 20 o P. The steady mass flow rate is 0.08 Ibm Is. If the heat lost from the compressor is 800 BtuIh and the work to run the compressor is 3.0 kW, estimate the specific enthalpy of the refrigerant leaving the compressor. Neglect the kinetic and potential energies of the flow.

8-03.

Air enters an adiabatic turbine at 300 0 P and with negligible velocity. The rate of doing work is 2.0 kW. The outlet temperature is 20 o P. If the outlet velocity is 300 ft/s, then calculate the mass flow rate of air.

8-04.

A mixing tank has two steady inflows and one steady outflow. The conditions of the first inflow are superheated steam at t1 = 350 oP, h1 = 1,140 Btu/Ibm' and flow rate

m = 30 Ibm Ih. The second inflow is compressed liquid at t2 = 70 oP, h2 = 38 Btu/Ibm' 1

If the rate of heat removal from the tank is 25,000 BtuIh, then estimate the inflow of water needed to result in the outflow being saturated liquid water at 212 0p. Neglect changes in kinetic and potential energy. 8-05.

Two air streams are mixed together in an adiabatic mixer to produce an output stream of air at 95°P. If the two incoming streams are at 1200 P and 35°P respectively, find the ratio of volume flow rates of the hot to cold air to produce the 95°P air.



9: 1

Chapter 9 Applications of the First Law of Thermodynamics

Contents of Chapter 9 • Instructions • Study Objectives of Chapter 9

.9.1

Compressors, Turbines, Pumps and Fans

• 9.2

Throttling Valves and Metering Devices

• 9.3

Heat Exchangers, Condensers and Evaporators



Study Objectives of Chapter 9 The First Law of Thermodynamics is such an important and useful tool in analyzing problems that this chapter is devoted to the application of the law to many of the mechanical/ thermal devices commonly found in HVAC systems. The objective of Chapter 9 is to provide background to the functioning ofthese devices and insight into ways in which the First Law may be employed to conduct practical analyses of these devices. After completing Chapter 9, you should be able to: • Understand the functioning of many components of HVAC&R systems; and • Understand how to apply the First Law of Thermodynamics in a variety of cases.


Chapter 9 Applications ofthe First Law

9:2

9.1

Compressors, Turbines, Pumps and Fans

Compressors, pumps and fans are devices used to increase the pressure of a fluid. Although all three may work in similar ways, the purpose behind each is somewhat different. Compressors are used to achieve high pressures in gases and vapors. In contrast, the important function of fans is to move air around, and the pressure rise is quite small. Pumps handle liquids, not gases, and are used to raise the pressure and move liquids such as water and oil. All three devices are driven by an external source, usually by a rotating shaft. Therefore, the work for these devices is negative. Compressors are used to raise the pressure of gases and vapors by a significant amount. For example, air compressors are employed to provide the pressurized air often used in workshops to power tools or to operate pneumatic control systems. Refrigerant vapor compressors are an essential part of the refrigeration cycle (as will be discussed in Chapter 11). Compressors are often described in terms of the way they work. There are reciprocating, rotary (vane, scroll and screw), and centrifugal compressors. In reciprocating compressors, the gas is compressed by a piston in a cylinder. The piston goes up and down in a reciprocating motion as the gas is alternatively drawn into the cylinder on the suction stroke and compressed in the cylinder on the compression stroke. On the compression stroke, the gas is trapped in the cylinder with both valves closed, and the pressure and temperature increase as the volume decreases. For the compression stroke, with the air trapped in the cylinder, it seems reasonable to conduct a closed system analysis of the air in the cylinder. On the other hand, if we consider the whole compressor over a longer period of time, because there are often hundreds of compression strokes a minute, it is clear that there is a general flow of fresh gas into the compressor and a pseudo-steady delivery of gas to a high pressure receiver. This view lends itself to an open system analysis, which is often assumed to be steady-flow. Centrifugal and rotary compressors are also usually treated as open systems. Turbines are the opposite of compressors and are intended to convert the energy content of high energy fluids into work. As the fluid passes through the turbine, work is done on the blades attached to a rotating shaft. Common applications are found in steam power stations where high-pressure and -temperature steam is expanded, producing work to drive an electric generator, or in hydroelectric turbines in which the energy of high-head water or water at· high velocity is converted into work. Turbines used in refrigeration systems are often called expanders.



9: 3

F or compressors, fans, pumps and turbines, there are some generalizations of the energy terms in the steady-state First Law equation: • The shaft work term is important in all of these devices and

Tt: =I:- 0 . For com-

pressors, fans and pumps, work is done on the system to make it function; whereas work is the output of turbines. • The heat transfer is often assumed to be negligible compared to the work,

Q= 0,

except for cases where there is a specific attempt to cool the compressor, or when the process is isothermal or polytropic. This is not a valid assumption for small compressors of any type. • The change of kinetic energy of the flow stream entering and leaving the device is usually assumed to be negligible compared to the work and change of enthalpy, !l.ke = O. However, the changes in kinetic energy are important when analyzing the internal processes of centrifugal compressors and pumps. • The change of potential energy is usually assumed to be negligible compared to the work and change of enthalpy, !l.pe = 0, except in the case of water turbines that are driven by gravitational head.

Application 1. Air is compressed reversibly and adiabatically (isentropically) in a compressor at a rate ofO.04Ibm /s from 83.28 psia to 400 psia. The inlet temperature is 40°F. Estimate the rate of work (power) required to keep the compressor running. Solution. The compressor is pictured in Figure 9-1 along with a p-v diagram. The region within the compressor is chosen as the control volume and it is treated as an open system. The work for an open system is not given as the area under the p-v curve as was used earlier for a closed system; rather, the work is estimated from the First Law equation. In this case, the changes in potential and kinetic energies are negligible and the equation for a singlestream can be used:

For the adiabatic process, the heat transfer rate is zero. For an ideal gas, the change in enthalpy can be written as (h2 - hJ ) = Cp (t2 - tJ ). Therefore, it is necessary to find the outlet temperature. This can be done by using a combination of the ideal gas equation, pv = RT, and the isentropic process, pv" = constant: ideal gas PJ vJ

/

TJ = P2

vi T2

' and isentropic PJvJ k = P2 v2 k



9:4

t1 = 40°F P1 = 83.28 psi a Pressure p I I I I I I

---_L
~-- ..........

-- ... -Specific Volume, v

Figure 9-1. Air Compressor of Application 1

Combining we get an important result for any ideal gas undergoing an isentropic process:

1; 17; = ( v2 I VI )

I-k

= (P2 I PI )(k-I)lk

Therefore:

1; = 7;(P2 I PIYk-I)lk = (40 +460)(400 I 83.28i1.4- I)/1.4 =782.9 R = 322.9°F

Then the change in enthalpy is:

(~-~) = CAt2 -tI ) = (0.24 Btu Ilb m ·oF)(322.9 -40)(° F) = 67.9 Btullbm

The rate of doing work is:

~ =-m(~ -~) = -(0.04 Ibm I s)(3,600 s/h)(69.22)(Btullb m ) = -9,967 Btu/h =-2.92 kW



9: 5

Note that in applying the First Law for an open system, we need the difference in enthalpy as (h2 - h) = Cp (t2 - t 1 ) rather than change of internal energy (u2 - u1 ) = C y (t2 - t) needed in the closed system problems. Although the coefficients C y and Cp are the specific heats at constant volume and constant pressure, the appropriate application for each depends more on whether we are seeking change of internal energy or enthalpy than whether the process is constant volume or constant pressure. In the problem just solved, both the pressure and specific volume change through the compressor. The relationships developed between the absolute temperature ratio, pressure ratio and volume ratio for an ideal gas undergoing an isentropic process are very useful:

1; /7;

= ( V2 / VI ) I-k = ( P2 / PI )(k-I)lk

Similarly the expressions can be developed for an ideal gas undergoing a polytropic process:

Application 2. Refrigerant R-22 is compressed isentropically in a compressor from 83.28 psia to 400 psia at a rate of 0.04 lbm/s. The refrigerant is a saturated vapor at the inlet. Estimate the power required to keep the compressor going. Solution. The compressor and the process on the p-h diagram are shown in Figure 9-2. The kinetic and potential energies are negligible. The single-stream, steady-flow First Law equation is:

Saturated Vapor p, =83.28 psia

I-I .................

® Pressure p

I I I I I I

:....................... -Enthalpy. h

Figure 9-2. R-22 Compressor of Application 2



9:6

For reversible isentropic flow, the heat transfer is zero. From the saturated vapor tables in Appendix C-1 (or the p-h diagram in Appendix C-2), the enthalpy at state 1 is h] = 108.2 Btu/Ibm' To find the enthalpy at State 2, we will use the p-h diagram in Appendix C-2. First locate State 1 as saturated vapor at 83.28 psia, 40°F. Then follow the constant entropy line (s] = S2 = 0.22 Btu/lbm·oF) until it intersects P2 = 400 psia. Read from the diagram that h2 = 125 Btu/Ibm' t2 = 195°F. The rate of work is then:

~=m(~-~) = -(0.04 Ibm / s)(3,600 s/ h)(125 -108.2) Btu/ Ibm

= -2,419 Btu/ h =-0.71 kW

9.2

Throttling Valves and Metering Devices

A throttling valve is a flow-restricting device that causes a pressure drop without any work. As we shall see in Chapter 11, the pressure in the refrigerant line of a vapor-compression refrigeration cycle needs to drop from the pressure of the condenser to the pressure of the evaporator. Common devices to achieve this are the capillary tube (a small tube with a large frictional pressure drop) and a thermostatic expansion valve. These are sometimes called metering devices because they also regulate the mass flow of refrigerant. The process in a throttling valve happens too quickly to allow any heat transfer, there is no shaft work, and the change in kinetic and potential energies are usually negligible. Considering the First Law of Thermodynamics, this means that the enthalpy does not change between the inlet and outlet of a throttling valve. Fluid passing through a throttling valve undergoes a constant enthalpy, or isenthalpic, process.

Application 3. Refrigerant R-134a enters a capillary tube of a refrigerator as saturated liquid at 180 psia and is throttled to a pressure of 33.11 psia. Determine the quality of the refrigerant at the final state and the temperature drop during this process. Solution. A schematic of the capillary tube is shown in Figure 9-3. At State 1, the conditions of saturated liquid may be found from Appendix C-3. It is necessary to interpolate between the entries at 173.11 psia and 185.84 psia as shown in Table 9-1. Throttling is a constant enthalpy device, therefore: ~

= ~ = 51.338 Btullb m



9: 7

Table 9-1 Pressure, psia 173.11 180.00

Temperature, of 115.0 t= q5.0+ ({I 80-173.1 1)/(185.84 - 173.11)} x (120 - 115)

Enthalpy (sat. liquid), Btullb 50.343 h = 50.343 + {(l80-I73.ll)1l85.84 -I73.ll)} x (52.181 - 50.343)

= 51.338

= 117.71

52.181

185.84

120.0

1

I

I

I

I

Saturated Liquid p1 180 psia

=

P2=33.11 psia

Figure 9-3. Capillary Tube of Application 3

The temperature at the outlet is found in Appendix C-3. It is the saturated temperature at p 2 = 33.11 psia. We know that the condition of the refrigerant is some mixture of vapor and liquid because the enthalpy lies between the saturated liquid and saturated vapor values (see Table 9-2).

Table 9-2 Temperature, . of 20.0

Pressure, psia 33.11

The temperature at the outlet is T2 the outlet is therefore: M

=

Entbalpy (sat. Iiq..id), Enthalpy (sat. vapor), Btullb Btullb 105.907 18.318

20.00 oP, and the temperature drop between the inlet to

= 117.71-20.00 = 97.71°P



9: 8

The quality at the outlet is: X2

= (~ - hj ) 1(hg - hj ) = (51.338 -18.318)1 (105.907 -18.318) = 0377 =37.7%

9.3

Heat Exchangers, Condensers and Evaporators

The purpose of a heat exchanger is to transfer heat from one flow stream to another. Heat exchangers come in many different shapes, sizes and flow configurations, but the essence of the heat exchanger may be captured by examining a pipe within a pipe arrangement, as shown in Figure 9-4. Flow stream A enters at State 1 and exits at State 2. Flow stream B enters at State 3 and exits at State 4. The flow arrangement in Figure 9-4 is known as counterflow because streams A and B flow in opposite directions. Ifflow streams A and B were in the same direction, it would be known as parallel flow; and if the flow streams were perpendicular to each other, it would be cross-flow. In most heat exchangers, the flow patterns do not neatly fit into one of these categories. However, the simple counterflow model can be used to understand the basic analysis tools for heat exchangers. The changes in kinetic and potential energy are usually negligible for heat exchangers, and there is no shaft work. Therefore, the important terms in the First Law equation are the heat transfer and the enthalpy change. There are basically three possible choices of control volume for a heat exchanger: flow stream A, flow stream B, or the total heat exchanger. The First Law applied separately to flow streams A and B results in:

QA =mA~ -~) QB = mB(h4-~) Heat exchangers are intended to exchange heat between streams but not to lose heat to the surroundings, and they are usually insulated. Therefore, when applying the First Law to the total heat exchanger, the net heat transfer is zero. The heat given up by the hot stream Q is just equal to the heat taken up by the cold stream:

Q= QA =-QB



9: 9

The flow streams may be liquid, gas, vapor or involve change of phase. Shown in Figure 9-4 are the temperature profiles from one end of the exchanger to the other for the cases of both flow streams being liquid or ideal gas in a pipe-in-pipe, counterflow heat exchanger. The change of enthalpy can be written as the product of the coefficient of specific heat and the temperature change from inlet to exit. In addition to being consistent with the First Law of Thermodynamics, the heat transfer between streams depends on the thermal resistance of the barrier separating the two streams and the effective temperature difference between the flows. The temperature difference changes from one end ofthe heat exchanger to the other, but it can be shown that the effective temperature difference is well represented by the Log

~

1

I~

l::t I~

FIUidA-U

I

Fluid B

X=O

X=L Fluid B

X=L

X=o Figure 9-4. Pipe-in-Pipe Counterflow Heat Exchanger

Fundamentals ofTilermodynamics and Psycllrometrics

Cllapter 9 Applications oftile First Law

9: 10

Mean Temperature Difference (LMTD). If the one end of the heat exchanger is identified as x = 0 and the other end as x = L, then the LMTD is:

LMTD = (!1tX=L - Mx=o) I In{!1tx=L I !1tx=o}

= {(t3 -t2)-(t4 -t1 )}lln{(t3-t2)/(t4 -t1)}

You are not expected to remember this expression, but it and the overall heat transfer coefficient, U (Btu/h·ft2 .oF), are important parts of the heat transfer expression:

Q=UA(LMTD)

The area, A, is the heat transfer area for the heat exchanger. The equations for a heat exchanger may be summarized as:

Q=mA~-~) Q=mB(~ -h4 ) Q=UA(LMTD)

Application 4. Consider a pipe-in-pipe, counterflow, water-to-air heat exchanger as shown in Figure 9-4. The water flow rate (stream B) of 0.2 Ibm Is enters at 180°F and leaves at lOO°F. The air flow rate (stream A) of 1.2 Ibm Is enters at 55°F. The coefficient of specific heat of water is Cp = 1.0 Btullbm.oF, and for air is CP = 0.24 Btullbm.oF: • Determine the outlet temperature of the air • Determine the LMTD from the expression:

LMTD = (!1tx=L - !1tx=o)/ln{Mx=L IMx=o} • Determine the UA product for the heat exchanger

Solution. Applying the First Law to the water flow stream: Qwater

= mwaterCp(t4 -t3 ) = (0.2 Ibm I s)(3,600 s/h)(l.O Btullb m ·oF)(lOO-180)oF = -57,600 Btu/h



9: 11

Applying the First Law to the air, Qair

t2

= mair CAt2 - tl )

and remembering that Qair

= -Qwater

= tl + Qair 1(mairCp ) =55+(57,600)1 {L2lbm 1 s (3,600 s/h)(0.24 Btu Ilb m ·o F)} = 110.5 of

LMTD = (J:::,.tX=L - Mx=o) 1In{J:::,.tx=L 1J:::,.tx=o}

= {(180-110.5) = 56.4°F

(100- 55)} Iln{ (180-110.5) 1(100 - 55)}

VA = Q/(LMTD)

= 57,600156.4 = 1,021 Btu/h·oF Condensers and evaporators are special forms of heat exchangers where one of the flow streams changes phase. The change of phase takes place at constant temperature. The temperature profiles for condensers and evaporators are shown in Figure 9-5.

t3

tc

te

t2

4 Ie

t1

x=o

Condenser

x=L

x=o

Evaporator

x=L

Figure 9-5. Temperature Profiles for Condensers and Evaporators



9: 12

Application 5. Refrigerant R-22 enters a condenser at a rate of 0.02 Ibm /s with an enthalpy of 120 Btu/Ibm and exits as a saturated liquid at 100°F. Airflows over the condenser at a rate ofO.2lbm Is. If the air enters at 72°F, what is the air exit temperature? Solution. The condenser is shown in Figure 9-6. The First Law applied to the total heat exchanger may be written as:

mR-22(~ -hr)=mair CAt3-t4)

From Appendix C-l, h2 = 39.538 Btu/Ibm. For Cp = 0.24 BtU/lbm·oF, the result is:

t4

= t3 + mR-22 (~ - hr) / mairCp = 72+0.02(120-39.538)/ {(0.2)(0.24)} = 105.5°F

1

1

Refrigerant

R-22

1

Air

1

Air

Air

Figure 9-6. Condenser of Application 5



9: 13'

The Next Step In the next chapter, we will consider the Second Law of Thermodynamics. Key to understanding the Second Law are the concepts of reversibility and irreversibility. These concepts will be introduced along with models for ideal processes. The Carnot cycle will be described and the Carnot heat engine and Carnot refrigerator analyzed.

Summary

The First Law of Thermodynamics is very useful for analyzing the performance of the many devices commonly found in HVAC&R systems. The devices considered here included compressors, turbines, pumps, fans, throttling valves, heat exchangers, condensers and evaporators. Some useful expressions were developed. For an ideal gas undergoing an isentropic process (pv" = constant), the properties at State 1 and State 2 are related by the following expressions:

1; /7; = ( v2 /

VI )

I-k

(

= P2 / PI

)(k-I)/k

Similarly, the expressions for an ideal gas undergoing a polytropic process (pv" = constant) are:

1; -7;

= ( v2 / VI )

I-n

= ( P2 / PI )(n-I)/1I

For heat exchangers, the main equations are:

Q=mAhz -~) Q=mB(~-h4) Q=UA(LMTD)

where

Qis the rate at which heat is transferred from one stream to the other, U is the overall

heat transfer coefficient between streams, A is the area of heat transfer, and LMTD is the Log Mean Temperature Difference between the temperatures in the two flow streams. After completing Chapter 9, you should be able to: • Understand the functioning of many components ofHVAC&R systems. • Understand how to apply the First Law of Thermodynamics in a variety of cases.



9: 14


9-01. Determine the outlet temperature and mass flow rate of a 0.7 k W air compressor that compresses air from 15 psia to 100 psia. The inlet temperature is 72°P. Assume isentropic compression of an ideal gas, k= 1.4, Cp = 0.24 Btu/lbm·oP. 9-02. Determine the outlet temperature and mass flow rate of a 0.7 k W air compressor that compresses air from 15 psia to 150 psia. The inlet temperature is 72°P. Assume polytropic compression of an ideal gas, n = 1.3, Cp = 0.24 Btu/lbm·oP and a heat loss rate of 0.1 KW. 9-03. The inlet condition to a R-22 compressor is saturated vapor at -40oP. If the specific work for the isentropic compression is 30 Btu/Ibm' then calculate the specific enthalpy of the refrigerant at the compressor exit. Sketch the process on the p-h diagram and use Appendix C-2 to estimate the delivery pressure. (The diagram of Appendix C-2 is not easy to read, so an approximate value is acceptable.) 9-04. The inlet to an adiabatic steam turbine is superheated steam at 1,000oP, 400 psia and enthalpy of 1,526 Btu/Ibm' The steam leaves the turbine at 150oP, with a quality of 0.90. Determine the specific work in Btu/Ibm of the turbine. 9-05. The inlet condition to a throttling valve in a R-134a refrigerator is saturated liquid at 110oP. Determine the outlet condition of the refrigerant (quality) if the outlet pressure is 21.162 psia. 9-06. Consider a pipe-in-pipe, counterflow, water-to-air heat exchanger. The water flow rate ofO.1lbm Is enters at 200 0 P and leaves at 1000 P. The air flow rate ofO.9lbm Is enters at 72°P. The coefficient of specific heat of water is Cp = 1.0 Btu/lbm.op, and for air C = 0.24 Btu/lbm·oP. Por both water and air, the change in enthalpy can be expressed as the coefficient of specific heat times the change in temperature. Complete the following: • Sketch the heat exchanger and the temperature profiles of the two streams through the length of the heat exchanger. • Determine the outlet temperature of the air. • Determine the LMTD from the expression: (MX=L - t:.tx=o)/ln{t:.tx=LIMx=o} • Determine theUA product for the heat exchanger. LMTD

=



10: t

Chapter 10 The Carnot Cycle


The Second Law of Thermodynamics

• 10.2

Heat Engines, Refrigerators and Heat Pumps

• 10.3

Reversible and Irreversible Processes

• 10.4

The Carnot Cycle and the Reversed Carnot Cycle



Study Objectives of Chapter 10 In Chapter 10, we will introduce the Second Law of Thermodynamics, reversibility, irreversibility, ideal processes and the Carnot cycle. The Camot cycle is the ultimate in ideal cycles. By applying the First Law to the Carnot cycle, we can deduce some basic expectations and limits on performance of any cycle. The reversed Carnot cycle is introduced because of its relevance to refrigeration. After studying Chapter 10, you should be able to: • Describe the concepts of reversibility and irreversibility; • Describe each of the processes that constitute a Carnot cycle; and • Analyze the performance of Carnot heat engines and Carnot refrigerators.



10:2

10.1

The Second Law of Thermodynamics

Whereas the First Law of Thermodynamics is a quantitative conservation of energy law about a system undergoing a process, the Second Law of Thermodynamics is a statement about the quality of energy, the preferred direction of energy change, and the differences between what is possible and not possible. There are several ways to state the Second Law of Thermodynamics, but rather than start with a statement of the law, we will make the case that there is a need for a Second Law if we are to describe observed behavior. The First Law of Thermodynamics always applies, but the First Law by itself is not enough. For example, consider a hot potato sitting on a table in a cool room. The First Law does not assert the direction of heat flow. We know by observation that the temperature of the potato will decrease with time, and the heat transfer is from the hot potato to its cool surroundings. The First Law states that the energy lost from the potato is just equal to the gain in energy by the air. Consider the reverse process in which heat flows from the cool room to the hot potato. We know that this is not possible, but that would not be clear from the First Law because the First Law would not be violated by a balance that states that the heat gain of the potato is equal to the energy lost by the air. There are many examples where a process will proceed in one direction, but not, by itself, proceed in the reverse direction. The simple action of rubbing two sticks together to produce heat is not reversed if heat is applied to the sticks. Heat applied to a wire does not make electricity the way that electricity passing along the wire produces heat. Combustion of gasoline and air together releases heat, but heat applied to the products of combustion will not produce gasoline. The First Law places no restriction on the direction of the process. Satisfying the First Law is necessary, but it is not a sufficient condition to ensure that the process will occur. The Second Law is introduced to overcome the inadequacy of the First Law. It will be shown that processes that are not possible violate the Second Law. A process will not occur unless both the First and Second Laws of Thermodynamics are satisfied. The use of the Second Law is not limited to identifying the direction of a process. It is also used in the concept of quality of energy, and the degradation of that quality during a process. The formulation of the Second Law will be helped by elaboration on the concepts of thermal energy reservoirs, heat engines, irreversibility, entropy and the Camot cycle.

.



10: 3

10.2 Heat Engines, Refrigerators and Heat Pumps The conversion of work into heat is as straightforward as rubbing two sticks together. However, the conversion of heat into work is not so easy, and requires a device called a heat engine. Without looking at the physical ways in which a heat engine may be constructed, consider the situation in Figure 10-1 where heat is transferred from a high-temperature energy source to a heat engine, and heat is separately transferred from the heat engine to a low-temperature energy sink. The First Law applied to the heat engine shows that the net work output of the heat engine is equal to the difference in magnitude between the heat flow from the high temperature source and the heat flow to the low-temperature sink:

F or convenience, the sign convention used here regards QHand QL as positive in magnitude, and the direction of heat flow is accounted for by the sign in the equation.

QH is the magnitude of the heat transfer between the high-temperature reservoir and the heat engine.

QL is the magnitude of the heat transfer between the heat engine and the lowtemperature reservoir. Wnet, out is the magnitude of the work of the heat engine. High-Temperature Heat Source, T H

The balance is in accordance with our sense that heat flows from high to low temperature.

Heat Engine

---.~ Wnet, out

Low-Temperature Heat Sink, T L

Figure 10-1. Heat Engine Operating Between High- and Low-Temperature Reservoirs



10:4

A typical cycle of a heat engine may be found by examining a steam power plant cycle as shown in Figure 10-2. There are four main components to the cycle: the steam generator, the steam turbine, the condenser and the pump. Liquid water enters the pump at low pressure and is pumped to the high pressure, requiring some work. The water enters the steam generator as a liquid at high pressure. Heat is added in the steam generator to convert the liquid to superheated vapor. The heat comes from a high-temperature heat source, such as the burning of fuel in a furnace. The superheated vapor is expanded through a steam turbine, where it does work, and then enters the condenser as a low pressure vapor. Heat is rejected from the condenser as the vapor is converted to liquid. The heat rejection takes place to a surrounding low-temperature heat sink, such as a lake, river or the atmosphere. For the cycle to work, the temperature in the steam generator must be less than the temperature in the furnace, and the temperature in the condenser must be greater than the lowtemperature surroundings. The thermal efficiency of a heat engine is measured as the fraction of heat input that is converted to work output. For the heat engines shown in Figures 10-1 and 10-2, the thermal efficiency is:

11th

= (Net work output) I (heat input) =U:et,out I QH =(QH -QL)I QH = l-QL I QH

Steam Generator

Wpump,in

Wturbine, out

Condenser

Figure 10-2. Typical Heat Engine: Steam Power Plant Cycle



10:5

The efficiency of a heat engine cannot be greater than 100% because both QH and QL are positive. Furthermore, the Second Law of Thermodynamics asserts that QL cannot be zero and it is impossible for a heat engine to have a thermal efficiency of 100%. The KelvinPlanck statement of the Second Law of Thermodynamics is: It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce an equivalent amount of work.

This means that the rejection of some heat to the surroundings is an essential ingredient to any device that converts heat into work. This limitation is not imposed by friction or dissipative effects, but is fundamental to both ideal and actual cycles. The flow of heat naturally occurs from high to low temperature. However, to transfer heat from a low-temperature reservoir to a high-temperature reservoir takes a device known as a refrigerator. Consider the refrigerator shown in Figure 10-3 operating between low and high-temperature reservoirs. As before, QHis the magnitude of the heat transfer between the refrigerator and the high-temperature reservoir, and QL is the heat transfer between the lowtemperature reservoir and the refrigerator. QH and QL are positive quantities, but in the opposite direction from those of the heat engine. The work, Wnet,in is also in the opposite direction to a heat engine and is regarded as the work required to make the refrigerator function. The First Law applied to the refrigerator results in: w"et,in = QH - QL

High-Temperature Reservoir, TH

Wnet, in ---~~

Refrigerator

Low-Temperature Reservoir, TL

Figure 10-3. Refrigerator Transferring Heat From Low- to High-Temperature Reservoir



10: 6

A typical example of a refrigeration device is the vapor-compression cycle shown in Figure 10-4. There are four main components to the cycle: the refrigerant compressor, condenser, metering valve and evaporator. Refrigerant vapor is compressed in the compressor and enters the condenser as a superheated vapor at high pressure and temperature. The condensation temperature is higher than that of the high-temperature surroundings, and heat is rejected from the condensing refrigerant to the surroundings. The refrigerant liquid then passes to an expansion valve where the flow is throttled from high to low pressure. The liquid evaporates in the evaporator at a temperature that is less than that of the surroundings, thus absorbing the heat removed from the refrigerated space.

High-Temperature Surroundings

Condenser

Expansion Valve

--... Compressor Wcompressor, in

Evaporator

Refrigerated Space Figure 10-4. Refrigeration Device: Vapor Compression Cycle



10:7

The performance measure for a refrigerator is the coefficient of performance (COPR ). The COPR is the ratio of the cooling effect to the work required to make the cycle function: COPR

= (Cooling Effect) I (Work Required) =QL I w"et in = QL I(QH-QJ = 1I (QH I QL -1)

The coefficient of performance is positive and may be greater than unity. In other words, the heat removed from the refrigerated space may be greater than the work done on the compressor. Although this may at first sound like getting more than you pay for, it does not violate the First Law. It should not be viewed as work being converted into heat, but rather as heat being pumped from low to high temperature by consuming work. The term heat pump is commonly used when the effect of interest is heating at the condenser. The coefficient of performance of a heat pump is: COPHP

= (Heating Effect) I (Work Required) =QH I w"et in =QH1(QH-QJ

= 1I (1- QL I QH) The coefficients of performance are related to each other as: COPHP

= COPR + 1

As in the discussion of heat engines, the discussion of refrigerators leads to an alternative statement of the Second Law of Thermodynamics, the Clausius statement: It is impossible to construct a device that operates in a cycle and produces no other effect than the transfer of heat from a lower temperature body to a higher temperature body.

The Clausius statement says that external work is required to drive the cycle ifheat is to be transferred from low to high temperature. Both the Clausius and Kelvin-Planck statements are negative statements and therefore do not have a direct proof; but both statements are such that violating either is the same as violating both.

Fundamentals ofTilermodynamics and Psycltrometrics

Cllapter 10 Tile Carnot Cycle

10:8

10.3 Reversible and Irreversible Processes The Second Law of Thermodynamics asserts that no heat engine can have an efficiency of 100%. What then is the highest possible efficiency? To answer this question, we must first examine the processes that make up a cycle and, in particular, examine the concept of reversibility. At the beginning of the chapter when introducing the Second Law of Thermodynamics, we cited several cases where a process could be expected to occur in one direction only. A hot cup of coffee sitting on a table will cool with time and not get hotter. A block of material will slide down, but not up, an incline. Electricity passing through a conductor will produce heat, but heat applied to the conductor will not make electricity. The burning of gasoline releases heat and produces exhaust gases, but heating the exhaust gases does not produce gasoline. These, and all other naturally occurring processes, are irreversible. Only idealized processes are reversible. A reversible process is defined as a process that can be reversed without leaving a trace on either the system or its surroundings. Reversible processes are the idealizations of actual processes. Work producing devices (such as heat engines) deliver the most work if the processes are reversible. Work consuming devices (such as refrigerators) use the least amount of work if the processes are reversible. Why are reversible processes of interest even though they are not achievable? Reversible processes are easier to analyze than actual processes and it is therefore possible to establish the theoretical limits to performance. Furthermore, a Second Law efficiency can then be established to compare how irreversible is an irreversible process. Some processes are more efficient than others. Phenomena that contribute to the irreversibility of a process are called irreversibilities. The main irreversibilities that we will explore further are friction, unrestrained expansion, mixing and heat transfer across a finite temperature difference. Friction is a well-known irreversibility. The friction force between two solid bodies always acts so as to oppose motion. That means that work is done to overcome friction in both directions and the process is not reversible because, when the body is returned to its original position, there has been some net heat transferred to the surroundings. Friction is also part of fluid flow. The pressure drop along the direction of flow in a pipe because of the viscosity always acts so as to oppose the flow. It may be possible to reduce the effects of friction, but it is never possible to eliminate friction altogether. Non-quasi-equilibrium expansion or compression is an irreversibility. Earlier, we defined a quasi-equilibrium process as one in which the system remains infinitesimally close to a state of equilibrium as the change of state progresses. Quasi-equilibrium can never be fully achieved, but it may be approached in expansion or compression processes if the process takes place slowly.



10: 9

Heat transfer can only occur when there is a temperature difference and, by observation, we see that heat transfer always occurs in the direction from high to low temperature. Conceptually, the only reversible heat transfer is between bodies at the same temperature, because then the direction of heat flow would be reversible. But if there is no temperature difference, then there is no driving force for heat transfer to take place. Therefore, heat transfer is fundamentally an irreversible process that can only approach reversibility if it takes place isothermally between bodies of the same temperature.

10.4 The Carnot Cycle and the Reversed Carnot Cycle The Second Law of Thermodynamics asserts that no heat engine can have an efficiency of 100%. What then is the theoretical best efficiency? The best efficiency may be expected from a heat engine cycle that includes only ideal, or reversible, processes. The ideal heat engine cycle is called the Carnot cycle. The Carnot cycle is made up of four reversible processes as shown in the schematic and property diagram of Figure 10-5: • Process 1-2. Reversible Isothermal Heat Addition at Temperature TH' Heat from the high-temperature reservoir is added to the heat engine at the isothermal temperature. Both the reservoir and inside the heat engine are at temperatureTw The isothermal condition is necessary for reversibility. The amount of heat transferred is QH' • Process 2-3. Reversible, Adiabatic Expansion. The temperature drops from TH to TL as the working fluid expands and the device does work. The process is assumed adiabatic because heat transfer to the surroundings would be considered a loss. To be reversible, the expansion must be a quasi-equilibrium process at all times. Earlier, we stated that a reversible, adiabatic process is the same as an isentropic process. • Process 3-4. Reversible Isothermal Heat Rejection at Temperature TL • Heat is rejected from the process to the surrounding low-temperature reservoir at constant temperature TL . The amount of heat transferred is QL' • Process 4-1. Reversible, Adiabatic Compression. The temperature is raised from TL to TH as the working fluid is compressed by the work of compression. The reversible adiabatic process is an isentropic process.

The net work of the cycle is the difference in magnitudes of the work of Process 2 to 3 and Process 4 to 1.



10: 10

Temperature

1 2 TH -----~----~----~

----. ......._ _---1

TL -----~----4-----1 4 3

Wnet, out

Entropy S

Figure 10-5. The Carnot Cycle

F or a totally reversible heat engine, the ratio of the heat transfers at the high- and lowtemperature reservoirs may be used to define an absolute thermodynamic temperature scale such that QHIQL = THITL. It is emphasized that this relation is not generally true for all heat engines, but is applicable only for a reversible heat engine such as the Carnot cycle. The thermal efficiency of a Carnot heat engine is then:

11 th,Camot

= 1- QL I QH =I-TL ITH

This is theoretically the best efficiency of any heat engine operating between these two absolute temperatures. Any real effect would cause the efficiency to be less than the Carnot efficiency. For example, consider an automobile engine. The gasoline burns at a maximum possible temperature of 1,540°F, and the heat is rejected from the engine to the atmosphere at 70°F: nth,Camot

= 1- TL I TH

= 1- (70 + 460) 1(1,540 + 460) = 0.735 = 73.5%



10: 11

In reality, the actual thermal efficiency is much less than this because not all of the heat is added at the highest temperature, the heat transfer is not reversible because it involves finite temperature differences, and there is friction. The reverse Carnot cycle is of interest because it forms the theoretical best model for the refrigeration cycle. The Carnot refrigerator is shown in Figure 10-6. There are four components to the cycle with the following processes: • Process 1-2. Reversible Isothermal Heat Addition at Temperature TL (cooling effect on surroundings). • Process 2-3. Reversible, Adiabatic Compression (isentropic). • Process 3-4. Reversible Isothermal Heat Rejection at Temperature TH (heating effect on surroundings). • Process 4-1. Reversible, Adiabatic Expansion (isentropic).

Temperature

4

3

TH -----~----.-----~

----.

TL -----1-----~----4 1 2

Wnet, in

Entropy S

Figure 10-6. The Reversed Carnot Cycle



10: 12

The coefficients of performance of the Reversed Camot Cycle are:

COPR,Camot COPHp,Camot

= 11 (QH 1QL -1) = 11(TH 1~ - 1) = 11 (1- QL 1QH) = 11 (1- TL 1TH)

For example, consider a household refrigerator. If the freezer temperature is 10°F and the room temperature is 72°F, then the best possible COP is:

COPR,Camot

= 11 (TH 1TL -1) = 11{(72 + 460) 1(10 + 460) = 7.58

1}

This means that the heat removal from the refrigerated space is 7.58 times the amount of electricity used to drive the refrigerator. An actual refrigerator has a COP of around two. For example, consider an average house air-conditioning unit that has to provide 50,000 Btu/h of cooling at 45°F. The air-conditioner rejects heat to the outdoors at 95°F. Compare the work required if it was a Camot refrigerator to the work required if it was an actual airconditioner with a COP of2.4:

COPR,camot

= 1I(TH 1TL -1) = 11 {(95+ 460)1 (45 + 460)} = 10.1

W;n,camot = ( Cooling Effect) 1COPR,camot

= (50,000 Btu/h)(1 kW 13,412 Btu 1h) 110.1 = 1.45 kW

W;n,Actual

= (Cooling Effect) 1COPR,Actual = (50,000 Btu 1h)(1 kW 13,412 Btu 1h) 12.4 = 6.11 kW

The Camot cycle provides a convenient model of the ideal heat engine. The Second Law efficiency of any other cycle is taken as the ratio of the thermal efficiency of the cycle divided by the thermal efficiency of a Camot cycle operating between the same maximum



10: 13

and minimum temperatures. A similar idea applies to the Second Law efficiency of a refrigerator: Second Law efficiency of a heat engine = 11 Ih /11 Ih Carnol Second Law efficiency of a refrigerator = COPR / COPR Carnol

The Second Law efficiency does not estimate the amount of work delivered or required, but instead it allows comparisons between different cycles about how effectively the quality of energy is being utilized. The clear message from the analysis of the Carnot cycle is that, to achieve the highest heat engine efficiency, the high temperature should be as high as possible and the low temperature as low as possible. To achieve the highest COP for a refrigerator, the high temperature should be as low as possible while the low temperature should be as high as posssible.

The Next Step This chapter discussed the somewhat abstract conceptual models for reversible Carnot heat engines and refrigerators. In the next chapter, we will examine the vapor-compression refrigeration cycle and the heat-driven absorption cycle and discuss the more practical aspects of performance of these important cycles used in refrigerators and chillers.

Summary

The Second Law of Thermodynamics was introduced as a needed addition to the First Law to account for the observation that processes are in general irreversible. The Kelvin-Planck Statement of the Second Law of Thermodynamics for heat engines: It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce an equivalent amount of work. This statement may be interpreted to say that the efficiency of even the ideal heat engine must be less than 100%.



10: 14

The Clausius Statement of the Second Law of Thermodynamics for refrigerators: It is impossible to construct a device that operates in a cycle and produces no other effect than the transfer of heat from a lower temperature body to a higher temperature body.

This statement may be interpreted to say that work is required to transfer heat in the direction of increasing temperature. For a process to be possible, it must satisfy both the First and Second Laws of Thermodynamics. Thermal efficiency is a measure of performance for a heat engine and coefficients of performance are the measure of performance of refrigerators and heat pumps: 11th

= (Net work output) / (heat input)

=T¥..et ,out / QH = 1-QL / QH

COPR

= (Cooling Effect) / (Work Required) =QL /T¥..et in

COPHP

= (Heating Effect)/ (Work Required) =

QH /T¥..et,in

The heat transfer quantities Q and the work quantities W in these relationships may also be replaced with the heat transfer rate

Q and

the power

ffs respectively when considering a

flow problem. The ideal heat engine cycle is the Carnot cycle and the ideal refrigeration cycle is the reversed Carnot cycle. The Carnot cycle is made up of the following four reversible processes: • 1-2. Reversible, Isothermal Heat Addition • 2-3. Reversible, Adiabatic Expansion (isentropic) • 3-4. Reversible, Isothermal Heat Rejection • 4-1. Reversible, Adiabatic Compression (isentropic)



10: 15

The thermal efficiency and coefficients of performance of the Carnot heat engine and reversed Carnot refrigerator are respectively: 11th,Carnot = 1- ~ / TH

= 1/ (TH / TL -1) COPHP,carnot = 1/ (1- TL / TH)

COPR,Carnot

The Carnot cycle efficiency and Carnot COP represent the absolute upper limits on the thermal efficiency and COP of any cycle. A Second Law efficiency is defined as the ratio of cycle efficiency to Carnot cycle efficiency, or COP to Carnot COP. After studying Chapter 10, you should be able to: • Describe the concepts of reversibility and irreversibility; • Describe each of the processes that constitute a Carnot cycle; and • Analyze the performance of Carnot heat engines and Carnot refrigerators.



10: 16


10-1. A refrigerator with a coefficient of performance of 2.8 provides 120,000 Btu of cooling. Calculate the net work required to run the refrigerator and the heat to be rejected to the surroundings. 10.2.

Give the Clausius statement of the Second Law of Thermodynamics and briefly describe in your own words what the statement means.

10.3.

Describe the four processes of the reversed Carnot cycle and show the cycle on a T-s diagram.

10.4.

A Carnot heat engine operates between a source at 2000 R and a sink at 440 R. If the heat engine is supplied heat at a rate of 800 Btu/min, determine the thermal efficiency and the power output of the engine.

10.5.

A heat engine is operating on a Carnot cycle and has a thermal efficiency of 55%. The waste heat from the engine is rejected to a nearby lake at 50 0 P at a rate of 600 Btu/min. Determine the power output of the engine and the temperature ofthe heat source.

10.6. A refrigerator is required to remove heat from a cooled space at a rate of 200 Btu/ min to maintain a temperature of20oP. If the ambient air surrounding the refrigerator is at 72°P, determine the minimum power input required for the refrigerator. 10.7.

An inventor claims to have developed a cooling system that removes heat from a cooled region at lOoP and transfers it to surrounding air at 75°P while maintaining a COP of7.5. How do you assess the inventor's claim?

10.8.

An air-conditioning system is used to maintain a house at 72°P when the temperature outside is 95°P. The house is gaining heat through the walls and windows at a rate of 50,000 Btu/h, and the heat generation rate within the house by lights and appliances is 5,000 Btu/h. Determine the minimum power input required by the airconditioning system.

Cllapter 10 Tile Carnot Cycle

Fundamentals ofTllermot/ynamics and Psycllrometries

11: 1

Chapter 11 Refrigeration Cycles


Refrigeration Equipment

• 11.2

The Ideal Vapor-Compression Refrigeration Cycle

• 11.3

Analysis of the Components and Cycle

• 11.4

Performance of Ideal Refrigerators and Heat Pumps

• 11.5

The Actual Vapor-Compression Refrigeration Cycle

• 11.6

Absorption Refrigeration System

• 11.7

Gas Refrigeration System



Study Objectives of Chapter 11 Chillers and refrigerators are important in many HVAC&R applications. In Chapter 11, we will present the processes of several refrigeration cycles and describe in detail the application ofthe First Law of Thermodynamics to cycle analysis. The objectives are to understand the working and performance evaluations ofthe vapor-compression refrigeration cycle, the absorption cycle, and the gas refrigeration cycle.



11: 2

After studying Chapter 11, you should be able to: • Describe the components and processes of a vapor-compression refrigeration cycle; • Apply the First Law to cycle analysis of a vapor-compression refrigeration cycle; • Evaluate the performance measures of refrigerators and heat pumps; • Evaluate the effect of non-ideal processes on the cycle; • Describe the absorption refrigeration cycle; and • Describe the gas refrigeration system.

CIlapter 11 Refrigeration Cycles

Fundamentals ofTIlermodynamics and Psychrometries

11: 3

11.1 Refrigeration Equipment Refrigeration is the thermodynamic process for producing a cooling effect. There are different forms of refrigeration, but the most common is the vapor-compression cycle. It derives its name from the use of a compressor as the primary driver of the cycle. The refrigeration cycle has four mechanical components through which the refrigerant is circulated in a closed loop. The refrigerant, or working fluid, is always separated from the air or water to be chilled. The refrigerant enters the compressor as a low pressure vapor and is compressed to a high pressure by the work of the compressor. From the compressor, the high pressure vapor flows to the condenser. The condenser is a heat exchanger where heat is rejected from the refrigerant so that it is converted to a high pressure liquid. Following the condenser is the thermal expansion valve, which lowers the pressure of the liquid refrigerant and is often used to control the mass flow of refrigerant. After expansion, the low pressure refrigerant enters the evaporator, which is a heat exchanger where the liquid refrigerant changes to vapor. To make the change, the refrigerant requires heat (like a low temperature boiler). The removal of this heat from the surroundings is the refrigeration, or cooling, effect desired. Exiting from the evaporator, the refrigerant is again at the low pressure vapor state ready to reenter the compressor for a repeat of the cycle. The four components in the vapor compression cycle are shown schematically in Figure 11-1.

1 q,

Condenser

Expansion Valve

W Evaporator

1

q,

Figure 11-1. Components of the Vapor-Compression Refrigeration Cycle



11: 4

The compressor is usually driven by an electric motor or other engine. There are several different types of compressors in common use that are described by their means of compression. Reciprocating, rotating and helical rotating (screw) compressors are known as positive displacement machines because they work by trapping a volume of vapor at inlet and then reducing that volume and raising the pressure until delivery. In contrast, centrifugal compressors are characterized by continuous exchange of angular momentum between a rotating impeller and a steadily flowing vapor. High rotational speed and steady flow contribute to centrifugal compressors having a high volumetric efficiency for their physical SIze. The evaporator is the key component when considering cooling. The heat that is removed from the air or water to evaporate the refrigerant in the evaporator is the cooling effect desired in a refrigerator. For air-conditioning purposes, this heat removal must be done at a low enough temperature to chill the water or air to the desired low temperature. Often the temperature at the evaporator is selected to be cool enough to also remove some moisture from the air and provide humidity control. While the evaporator cools, the condenser heats because, during condensation, energy is released from the refrigerant. This heat rejection takes place at a temperature above that of the surroundings. Sometimes a cooling tower is used to help with the heat rejection. Heat flow from high temperature to low temperature occurs naturally. With a refrigerator, the heat flow is in the other direction. By doing work in the compressor, the heat is made to flow from the cold side at the evaporator to the hot side at the condenser. Therefore, a refrigerator is sometimes called a heat pump. This is especially true if the interest is to supply a heating effect at the condenser rather than the cooling effect at the evaporator. The efficiency with which a refrigerator or heat pump utilizes electricity to produce either cooling or heating is known as the coefficient of performance. The cooling coefficient of performance, COPcool ' is the cooling effect at the evaporator divided by the work done at the compressor. The heating coefficient of performance, COPheat , is the heating effect at the condenser divided by the work done at the compressor. The coefficients of performance are dimensionless and typically larger than one. A dimensional version of the COP is the Energy Efficiency Ratio, EER, which is the cooling or heating effect in Btu/h divided by the work of the compressor in watts. The refrigerator may either cool the circulated air directly (as in most residential applications), or it may chill water for distribution to cooling coils located close to the cooled space (as in most commercial applications). When used to chill water, the refrigerator is called a chiller. Chiller capacity is measured in refrigeration tons. One ton of refrigeration is equivalent to a rate of cooling of 12,000 Btu/h. Furthermore, one ton of refrigeration will freeze



11:5

one physical ton of water to ice per day. This leads to an efficiency term that is commonly used for chillers which is the work of the compressor expressed in k W over the refrigeration effect given in tons, or kW/ton. This measure is the inverse of the coefficient of performance in different units.

11.2

The Ideal Vapor-Compression Refrigeration Cycle

The most efficient of all possible refrigeration cycles is the reversed Camot cycle, as described in the previous chapter. The cycle is made up offour reversible processes. Consider the reversed Camot cycle shown in Figure 11-2. The T-s property diagram shows the cycle within the saturated liquid/vapor region of the refrigerant. In this setting, it is reasonable to imagine actual evaporators and condensers operating isothermally because of the phase change. However, actual compressors and turbines cannot be found to closely approximate the isentropic processes when they have to handle a liquid/vapor mixture. An alternative is to place the cycle outside the mixed liquid/vapor region, but then it is the isothermal heat transfer conditions that are difficult to approximate over the full range of the condenser. The result of this discussion is that even though the reversed Camot cycle is the most efficient of all cycles, it cannot be reasonably approximated by physical equipment, and it does not serve as a realistic model for refrigeration cycles. A more appropriate model is the ideal vapor-compression cycle.

Figure 11-2. The Reversed Carnot Cycle on a T-s Property Diagram



11: 6

The ideal vapor-compression cycle is not as efficient as the reversed Carnot cycle but it does match well with some idealized physical pieces of equipment. The ideal vapor-compression cycle has four components and the T-s property diagram for the four processes is shown in Figure 11-3. The processes of an ideal vapor-compression cycle are: • 1-2 - Isentropic compression of the refrigerant vapor in the compressor.

• 2-3 - Constant pressure heat rejection as the refrigerant goes from superheated vapor to saturated liquid through the condenser. • 3-4 - Constant enthalpy throttling in an expansion device. • 4-1 - Constant pressure heat absorption as the refrigerant is evaporated in the

evaporator. Saturated vapor at a low pressure enters the compressor at State 1 and is compressed isentropically (reversible, adiabatic) in the compressor to State 2. State 2 is a superheated refrigerant vapor that is at the highest temperature in the cycle. The constant pressure process from State 2 to State 3 takes place in the condenser, although it includes more than just condensing. The first part of the condenser is actually a desuperheater as the vapor is converted from its superheated state at a very high temperature to a saturated vapor condition at the same pressure, but at the saturated vapor temperature.

Temperature T Condenser

Evaporator

Entropy,s

Figure 11-3. T-s Property Diagram for an Ideal Vapor-Compression Cycle



11:7

The second part of the condenser is where the condensation process occurs. Condensation occurs at constant pressure and constant temperature because of the phase change. The condensation temperature must be greater than the surroundings so that heat is rejected from the condenser. At exit from the condenser, the refrigerant is a saturated liquid. The saturated liquid at State 3 is throttled from the high condenser pressure to the low pressure evaporator by passing through an expansion valve or capillary tube. The valve is both a pressure reduction device and a metering device to control flow. The valve is an isenthalpic device and the refrigerant emerges at State 4 as a liquid/vapor mixture at a temperature that is below the temperature of the refrigerated space. Process 4-1 occurs in the evaporator as the refrigerant liquid is converted to vapor. The heat needed to produce the phase change is removed from the surrounding refrigerated space, thus providing the desired cooling effect. The T-s diagram in Figure 11-3 is a logical property diagram for this cycle because it shows the isentropic assumption of the compression process, and the isothermal nature of the evaporation, and part of the condensation, processes. Another logical property diagram is the p-h diagram in Figure 11-4 because it shows the constant pressure processes of the condenser and evaporator, and the constant enthalpy process of the expansion valve. It is the data for the p-h diagram that is given for R-22 and R-134a in Appendices C-2 and C-4.

Pressure

p

Condenser

Evaporator

Enthalpy, h

Figure 11-4. p-h Property Diagram for an Ideal Vapor-Compression Cycle



11: 8

11.3 Analysis ofthe Components and Cycle First Law analysis can be applied to each of the four components of the ideal vapor-compression cycle, and then to the complete cycle. The kinetic and potential energy terms are negligible. Therefore, the single-stream, steady-flow form of the First Law can be used on the components. The sign convention used here is that heat transfer and work are given as the magnitude (always positive) and the direction is taken care of by the sign in the equation. The components are shown in Figure 11-3. Compressor: reversible, adiabatic (isentropic S2

= Sl):

Condenser: no work, constant p (p 3 = P2): Expansion valve: no work, no heat transfer (isenthalpic): Evaporator: no work, constant p (p 4 For the cycle:

wcompressor

=~ - ~

=~ - ~ h4 = ~

qH

= PI): Wcompressor

= qH -

qL

11.4 Performance ofIdeal Refrigerators and Heat Pumps The performance measure for a refrigerator is the coefficient of performance, which is defined as the cooling effect divided by the work of the compressor. The COPcoo1 for an ideal vapor compression cycle is:

If the interest in the cycle is not the cooling effect, but the heating effect, then the vaporcompression cycle is known as a heat pump. Heat pumps have become quite popular in residential houses because the same equipment may be used as an air-conditioner in summer and as a heat pump in winter. The coefficient of performance of a heat pump is defined as the heating effect divided by the compressor work. For the ideal vapor-compression cycle, the COPheal is greater than one, making the heat pump a more efficient use of electricity for residential heating than just a resistance heater.



11: 9

Example 11.1. Consider a 5-ton refrigeration system that operates on an ideal vapor-compression cycle with R-22 as the working fluid. The refrigerant enters the compressor as a saturated vapor at 83.28 psia and is compressed to 396.32 psia. Find the coefficient of performance of the refrigerator and the power of the compressor. Solution. The property diagrams are shown schematically in Figures 11-3 and 11-4. Starting with state 1, the conditions at all states may be determined from Appendix C-l and Appendix C-2 as: • State 1: Given saturated vapor at 83.28 psia:

Appendix C-l: tl = 40°F, hI = 108.191 Btu/Ibm' SI = 0.22 Btu/lbm·oF • State 2: GivenP2 = 396.32 psia, S2 = SI = 0.22Btu/lbm·oF

Appendix C-2: t2 = 195°F, h2 = 125 Btu/Ibm • State 3: Given saturated liquid atp3

=

P2 = 396.32 psia

Appendix C-l: t3 = 150°F, h3 = 56.37 Btu/Ibm • State 4: GivenP4

=

PI = 83.28 psia, and h4 = h3 = 56.37 Btu/Ibm

Appendix C-l: x4= (h 4- hJ)/(hg - hJ) = (56.37 - 21.688)/(108.191 - 21.688) = 0.40 The coefficient of performance is:

CO~oOI

= q L 1wcompressor = (~ - h4 ) 1(~ -~) = (108.191- 56.37) 1(125 -1 08.191) =3.08

The power of the compressor is:

U:ompressor

= QL 1CO~ool = (5 tons)(12,000 Btu/h.ton)(kW 13,412 Btu 1 h)1 3.08 =5.71 kW

If this had been a Carnot cycle operating between the evaporator temperature of 40°F and the heat rejection temperature of 150°F, then the Carnot coefficient of performance would have been: COPCamo / = 11 (TH 1 TL -1) = 11 {(150 +460) 1 (40 +460)-1} = 4.55



11: 10

It should be noted that the Carnot coefficient of performance is usually calculated for the surrounding temperatures rather than the refrigerant temperatures as done here. If surrounding temperatures had been used, the Carnot COP would have been higher. Remember that the Carnot COP is the unattainable ideal.

11.5

The Actual Vapor-Compression Refrigeration Cycle

There are several reasons why the actual performance of a refrigerator is less than the ideal. Besides friction, there are issues of heat loss and pressure drop that are not accounted for in the ideal cycle. Shown in Figure 11-5 is the ideal cycle (solid line) and the more realistic actual cycle (dotted line) as they appear on a T-s and p-h diagrams. The ideal process in the compressor is from State 1 to State 2s. The term 2s for the end of the compression process is to indicate that the process was isentropic. The actual compression process from State 1 to State 2a is likely to deviate from isentropic because of two effects. Heat loss from the compressor decreases the entropy. Irreversible effects, such as friction, increase entropy. Friction always increases the work required to compress the vapor from PI to P r If there is no heat loss and the increase in entropy is only due to irreversibilities, then the ratio of the ideal work to actual work is known as the isentropic efficiency of the compressor: 11 isentropic

= wideal / Wactual = (~s - ~) / (~a - ~)

p

T

s

h

Figure 11-5. Property Diagrams for Vapor-Compression Cycles With an Adiabatic Compressor



11: 11

The ideal processes from State 2 to State 3 and from State 4 back to State 1 are constant pressure processes. For the actual processes, there is a pressure loss along the length of each heat exchanger, resulting in the actual States 3a and 4a. Example 11.2. Consider the same problem as in Example 11.1 but with an isentropic efficiency for the adiabatic compressor of 80%, a pressure drop in the condenser of 4.52 psi, and a pressure drop in the evaporator of7.511 psi. The 5-ton refrigeration system operates on an actual vapor-compression cycle with R-22 as the working fluid. The refrigerant enters the compressor as a saturated vapor at 83.28 psia and is compressed to 396.32 psia. Find the coefficient of performance of the refrigerator and the power of the compressor. Solution. The property diagrams are shown schematically in Figure 11-5. Starting with State 1, the conditions at all states may be determined from Appendix C-1 and Appendix C-2 as:

• State 1: Given saturated vapor at 83.28 psia:

= 40°F, h] = 108.191 Btu/Ibm's] = 0.22 Btu/lbm·oF • State 2: GivenP2 = 396.32 psia, S2s = S] = 0.22 Btu/lbn/F Appendix C-2: t2s = 190°F, h2s = 125 Btu/Ibm Appendix C-1: t]

Isentropic efficiency:

TJisentropic

=

wideal

1Wactual

= (~s -~) 1(~a -~)

Solving for the actual enthalpy:

~a = ~ +(~s -~)/TJisentroPic

= 108.191 +(125-108.191)1 0.8 = 129.202 Btu/lb m • State 3: Given saturated liquid at P3a = (P2 - 4.522) = (396.32 - 4.52) = 391.79 psia Interpolating between values in Appendix C-1:

t3a

= 145.00+(150.00-145.00)(391.79 = 144°F

373.71) 1(396.32 - 373.71)

~a = 54.553 + (56.370- 54.553)(391.79 - 373.71) 1(396.32 - 373.71)

= 56.00 Btu/lb m



11: 12

• State 4: P4a = Pi + 7.511 = (83.28 + 7.511) = 90.791 psi a, and h4a = h3a = 56.00 Btullbm From Appendix C-1: t4a = 45.00°F

x4a = (h4a - hj ) 1(hg - hj

)

= (56.00 - 23.111) 1(108.600 - 23.111)

= 0.385 The coefficient of performance is:

CO~ool = q L 1wcompressor = (~ - h4a ) 1(~a -~)

=(108.191-56.00)1 (129.202 -108.191) =2.48 The power of the compressor is:

~ompressor = QL 1CO~ool = (5 tons)(12,000 Btu/h.ton)(kW 13,412 Btu 1 h) 12.48

=7.09 kW The power required for the actual cycle is greater than for an ideal cycle, and the coefficient of performance is smaller.

Example 11.3. Re-work the ideal cycle of Example 11.1, except that there is heat removal in the compressor such that the compressor outlet temperature is 175°F. The 5-ton refrigeration system operates on a vapor-compression cycle with R-22 as the working fluid. The refrigerant enters the compressor as a saturated vapor at 83.28 psia and is compressed to 396.32 psia. Find the coefficient of performance of the refrigerator and the power of the compressor. Solution. The property diagrams are shown schematically in Figure 11-6. States 1, 3 and 4 are unchanged from Example 11.1. The only difference is in State 2. The conditions at State 2a may be determined from Appendix C-2 as:

• State 2: GivenP2a = 396.32 psia, t2a = 175°F Appendix C-2: h2 = 124 Btu/Ibm


Fundamentals ofThermot/ynamics and Psychrometrics

11: 13

The chart in Appendix C-2 is not easy to read exactly. The value has to be visually interpolated to lie between the lines of temperature of 160°F and 200°F. The coefficient of performance is:

CO~ool = qL / wcompressor = (~ - h4 ) / (~a -~) = (108.191- 56.37) / (124 -108.191) = 3.28 The power of the compressor is:

Ft:ompressor

= QL / CO~ool = (5 tons)(12,000 Btu/h·ton)(kW / 3,412 Btu/h)/ 3.28 =5.36kW

It is a practical matter that the COP can be improved through heat removal from the compressor. It is standard practice and is also important in multi-staging of compressors. From a theoretical viewpoint, the process is not ideal, but that is counteracted by the fact that the heat removed from the compressor is rejected at a lower temperature than the peak superheat temperature that occurs after isentropic compression, leading to a higher COP.

..---

.....

~

25

Figure 11-6. Property Diagrams for Vapor-Compression Cycle With Heat Removal from the Compressor



11: 14

11.6 Absorption Refrigeration System There is another form of refrigeration known as the absorption cycle. This cycle is heat driven rather than work driven. It involves a mixture of two components that have some attraction for each other. Common mixtures are ammonia with water, and lithium bromide with water. For discussion purposes, we will use the water/lithium bromide mixture shown in Figure 11-7 to describe how the cycle works. Lithium bromide is a salt that dissolves in water. The water and lithium bromide are in solution in the generator as heat qgenerator is applied. The heat provides energy to break the bonds of mixing and the heat of vaporization to boil off some of the water. The lithium bromide-rich solution left behind goes to the absorber while the water vapor goes to a condenser where it is converted to liquid at the high pressure. The water is considered the refrigerant and is passed through a metering valve to enter the evaporator at a low pressure. The water evaporates in the evaporator, providing the cooling effect qcool on the surroundings as described before for refrigeration. The vapor leaving the evaporator is at low pressure but needs to be returned to the generator as a high pressure liquid. High Pressure Water Vapor

~ndenser Condenser

Generator

~en.ra1Dr

High Pressure Water Liquid Metering Device

Throttle Valve Weak Solution (Lithium Bromide)

Strong Solution (Lithium Bromide)

Evaporator

Absorber Pump

~absorber

Low Pressure Water Liquid

Low Pressure Water Vapor

~OOI

Figure 11-7. Components of the Absorption Refrigeration System



11: 15

In the vapor-compression cycle, this is achieved with a compressor; whereas in absorption refrigeration, this is achieved by mixing the low pressure water vapor with the lithium bromide solution in the absorber. The lithium bromide has a strong attraction for the water vapor and will convert it to liquid upon mixing. (You can understand this effect if you set out some dry salt on the counter on a humid day; you will see that the salt is able to pull the moisture out of the air to convert it into a wet drop on the counter.) During this process, the condensing water vapor will give up heat in the absorber which must be rejected from the absorber. The low-pressure mixture of water and lithium bromide is then pumped up to the generator pressure to start the cycle over. The coefficient of performance is given as the ratio of the cooling effect to the heat used to drive the cycle, qcooI / qgenerator . At face value, the COP for the absorption cycle is less than that of the vaporcompression cycle, but it uses heat instead of work to drive the cycle and therefore avoids a heat engine to first convert the heat into work.

11. 7

Gas Refrigeration System

Another refrigeration cycle is shown in Figure 11-8. The gas refrigeration system operates on the principle of compressing a gas, removing heat, and then expanding the gas to a low temperature. A gas, like air, enters the compressor and is compressed isentropically to a high pressure and temperature. Heat is removed from the gas in a heat exchanger that follows a line of constant pressure. The air enters the turbine at high pressure but at moderate temperature. Through the turbine, the gas is expanded to a low pressure and low tempera-

T

5

Figure 11-8. Components of the Gas Refrigeration System



11: 16

ture as energy is removed in the form of work. It is the low temperature gas that provides the cooling effect in the second heat exchanger. The open cycle form of gas refrigeration is commonly used on aircraft because of the light weight. The more efficient closed cycle gas refrigeration cycle includes regeneration whereby some of the warm gas leaving the high pressure heat exchanger preheats the cold gas entering the compressor.

The Next Step The past chapters have used the thermodynamic properties of single pure substances. The next step is to explore the thermodynamics of mixtures. In particular, we will be interested in the mixture of air and moisture vapor, but to build up to that understanding we must first understand the rules for a mixture of ideal gases. The topic of the thermodynamics of moist air is known as psychrometrics.

Summary

The ideal vapor-compression cycle is not as efficient as the reversed Carnot cycle, but it does match well with some idealized physical pieces of equipment. The ideal vapor-compression cycle has four components: compressor, condenser, expansion valve and evaporator. The processes of an ideal vapor-compression cycle are: • 1-2 - Isentropic compression of the refrigerant vapor in the compressor. • 2-3 - Constant pressure heat rejection as the refrigerant goes from superheated vapor to saturated liquid through the condenser. • 3-4 - Constant enthalpy throttling in an expansion device. • 4-1 - Constant pressure heat absorption as the refrigerant is evaporated in the evaporator. The performance measure for a refrigerator is the coefficient of performance, defined as the cooling effect divided by the work of the compressor. The COPcool for an ideal vapor compression cycle is:

CO~ool

=

qL / wcompressor

=

(~ - h4 ) / (~ -~)

If the interest in the cycle is not the cooling effect, but the heating effect, then the vaporcompression cycle is known as a heat pump. Heat pumps have become quite popular in residential houses because the same equipment may be used as an air-conditioner in sum-



11: 17

mer and a heat pump in winter. The coefficient of performance of a heat pump is defined as the heating effect divided by the compressor work. For the ideal vapor-compression cycle, the COPheat is greater than one, making the heat pump a more efficient use of electricity for residential heating than just a resistance heater: COp"eat

= qH / wcompressor = (~ -~)/ (~ -~)

The actual vapor-compression cycle includes irreversible effects such as friction and pressure drop. The isentropic efficiency of an adiabatic compressor is: 11 isentropic

= wideal / Wactual = (~s - ~) / (~a - ~)

The performance of a vapor-compression refrigerator can be improved through heat removal during the compression process. The heat removal reduces the amount of work required in the compressor. The absorption refrigeration cycle is a heat-driven cycle that depends on a mixture of two substances that have an affinity for each other. The condenser, expansion valve and evaporator are similar to the vapor-compression cycle, but the compressor is replaced by a generator-absorber combination. The COP for the absorption cycle is typically much lower than for a vapor-compression cycle, but is useful in places where there is a heat source to drive the cycle. The gas refrigeration cycle operates on the principle of compressing a gas, removing heat and then expanding the gas to a low temperature. The refrigerant is the gas. Open cycle gas refrigeration is commonly used on aircraft because of the light weight. The more efficient closed loop gas refrigeration cycle includes internal regeneration within the cycle. After studying Chapter 11, you should be able to: • Describe the components and processes of a vapor-compression refrigeration cycle. • Apply the First Law to cycle analysis of a vapor-compression refrigeration cycle. • Evaluate the performance measures of refrigerators and heat pumps. • Evaluate the effect of non-ideal processes on the cycle. • Describe the absorption refrigeration cycle. • Describe the gas refrigeration system.



11: 18


11-01. Sketch the basic elements of the vapor-compression cycle: compressor, condenser, expansion valve and evaporator and describe the basic function of each.

11-02. Sketch the T-s and p-h property diagrams of an ideal vapor-compression cycle and describe the thermodynamic processes of the cycle.

11-03. A 3-ton refrigeration system operates on an ideal vapor-compression cycle with R-134a as the working fluid. The refrigerant enters the compressor as a saturated vapor at 8.577 psia and is compressed to 199.25 psia. Find the coefficient of performance of the refrigerator and the power of the compressor. Compare the coefficient of performance with that of a Carnot cycle operating between -35°F and 125°F.

11-04. A 3-ton refrigeration system operates on an actual vapor-compression cycle with R-134a as the working fluid. The isentropic efficiency of the adiabatic compressor is 85%. The refrigerant enters the compressor as a saturated vapor at 8.577 psia and is compressed to 199.25 psia. The pressure drop in the condenser is 13 Al psi. The pressure drop in the evaporator is 2.72 psi. Find the coefficient of performance of the refrigerator and the power of the compressor.

11-05. A 5 kW compressor drives an ideal vapor-compression cycle heat pump with R-22 as the working fluid. The refrigerant enters the compressor as a saturated vapor at 83.28 psia and is compressed to 396.32 psia. Find the heating capability ofthe heat pump and the coefficient of performance.



11: 19

11-06. An ice-making machine operates on a vapor-compression cycle using R-22. The refrigerant enters the compressor as a saturated vapor at 10°F and leaves the compressor with an enthalpy of 120 Btu/Ibm. The condenser pressure is 210.69 psia. Water enters the ice-machine at 55°F and leaves as ice at 25°F. For an ice production rate of 10 Ibm Ih, estimate the power input to the refrigerator. The heat removed to convert 55°F water to ice at 25°F is 169 Btu/lbm .

11-07. Sketch and describe the operation of an absorption refrigeration cycle that uses a mixture of ammonia and water. Ammonia serves as the refrigerant.

11-08. Sketch and describe the operation of an open cycle gas refrigeration system.



12: 1

Chapter 12 Moist Air as a Mixture of Ideal Gases


Introduction

• 12.2

Mixtures ofldeal Gases

• 12.3

Mixture of Dry Air and Water Vapor

• 12.4

Moist Air Tables



Study Objectives of Chapter 12 The objective of Chapter 12 is to develop a thermodynamic understanding of the behavior of mixtures of ideal gases. The topic is important because it serves as the basis upon which we will idealize the behavior of the mixture of dry air and water vapor as found in many HV AC applications. The mixture properties will be described in terms of the mass and mole fractions of the components. The conceptual models of Dalton and Amagat will be introduced and used to develop the idea of partial pressure. Moist air will be considered as a example of a mixture of ideal gases. The property expressions and moist air tables will be explained.


Chapter 12 MoistAir

12:2

After studying Chapter 12, you should be able to: • Understand the Dalton and Amagat models as they apply to the basic description of a mixture of ideal gases, including the concepts of partial pressure and partial volume; • Calculate mixture properties given the constituent properties and the composition of the mixture; • Describe why the model of a mixture of ideal gases is applicable to moist air; and • Calculate the properties of moist air as a function of temperature.

Chapter 12 MoistAir


12: 3

12.1 Introduction Understanding the thermodynamics of mixtures is very important for many practical applications. Air is a mixture of many gases. Until now, we have been able to treat air as a pure substance, even though it is a mixture, because we have not needed to consider the change in the proportions of oxygen, nitrogen, carbon dioxide and other gases that make up air. However, what happens in thermodynamic calculations in which the moisture content of the moist air changes from one state to another? Are the components affected differently, and ifthey are, how is each component to be treated? Air and water are known as a nonreacting gas mixture. As a different example, consider the reactive process of combustion. During combustion, a chemical reaction occurs between the fuel and air. How can thermodynamics be applied to find the heat released and the makeup of the resultant exhaust gases? Combustion is an example of a reactive mixture. In this chapter, we will restrict our studies to the non-reactive mixture of ideal gases only. The general study of the thermodynamics of mixtures can be quite complicated, but we can gain insight by looking at the basics and studying a mixture of only ideal gases. The ideal gas model will serve well to describe the thermodynamics of a mixture of air and moisture, because at the typical room temperatures, and because of the small amount of moisture in the air, both the air and water vapor can reasonably be approximated as ideal gases.

12.2 Mixtures ofIdeal Gases Ideal gases are special. The kinetic model for an ideal gas is such that the individual molecules that make up the gas are dispersed so far apart from each other that the number of direct collisions between molecules is negligible as the molecules move around in some common space. Descriptively, we can say that any particular molecule in an ideal gas does not know that it shares common space with any of the other molecules and it behaves as though it were alone in that space. Consider the molecules of an ideal gas contained within a container of volume V. The molecules will collide with the walls of the container. The frequency and intensity of the collisions depend on the type of molecule, number of molecules in the container and the kinetic energy of each molecule. The kinetic energy of the molecules is a function of the gas temperature such that the higher the temperature, the more active the molecules. The pressure in the vessel is a manifestation of the collisions with the container wall.


Chapter 12 MoistAir

12:4

Keeping in mind that an ideal gas is one in which the molecules are spread far enough apart so that individual molecules do not collide with each other, consider the following two ways of how a mixture of ideal gases of volume V, pressure p, and temperature Tmay have arisen. (Whenever there is no subscript on the symbol, then the symbol represents a property of the mixture.) The Dalton model is shown in Figure 12-1. Here, a volume V of ideal gas A at pressure PA and temperature T is mixed with a volume V of ideal gas B at pressure PB and temperature T and placed in a container of the same volume Vat the same temperature T. Each molecule of gas in the mixture, A or B, behaves as though it alone occupies the volume V. The result is that the pressure of the mixture, or total pressure, is the sum of the pressures PA and PB because there is a simple additive effect of the molecule collisions with the container walls. The pressuresPA andpB are known as the partial pressures. Dalton's law of additive pressures is: The pressure of an ideal gas mixture is equal to the sum of the pressures each gas would exert if it existed alone at the mixture temperature and volume.

Gas A Pressure Volume Mass Moles

PA

V rnA NA

+

GasB Pressure Volume Mass Moles

Pe

V me NB

-----+-

Mixture Pressure (PA+Pe) Volume V (rnA+rn e) Mass Moles (NA+N B)

Figure 12-1. Dalton's Law of Additive Pressures

The Amagat model is shown in Figure 12-2. Here, a volume VA of ideal gas A at pressure P and temperature T is mixed with a volume VB of ideal gas B at the same pressure P and temperature T and placed in a container of volume Vat the same pressure P and temperature T. Each molecule of gas in the mixture, A or B, behaves as though it alone contributes to pressure p. The result is that the volume of the mixture, or total volume, is the sum of the volumes VA and VB because there is a simple additive effect of the space occupied. The volumes VA and VB are known as the partial volumes. Amagat's law of additive volumes is: The volume of an ideal gas mixture is equal to the sum of the volumes each gas would occupy if it existed alone at the mixture temperature and pressure.

Chapter 12 MoistAir


12: 5

Gas A Pressure Volume Mass Moles

P VA rnA

NA

+

GasB Pressure Volume Mass Moles

P VB mB NB

----...

Mixture Pressure p Volume lYA+VB) Mass (rnA+rn B) Moles (NA+Ne)

Figure 12-2. Amagat's Law of Additive Volumes

This means that when looking at a mixture of ideal gases, regardless of how the mixture was made, the mixture can be described as a volume V in which the pressure p is made up of the sum of partial pressures, or as a mixture at pressure p for which the volume V is made up of the sum of the partial volumes. The description using the partial pressures will be more convenient when we come to considering the mixture of air and water vapor. There are some other useful measures that can be developed from a mixture of ideal gases by considering the models given in Figures 12-1 and 12-2. The properties of a gas mixture depend on the properties of the individual components and the composition of the mixture. There are two ways of specifying the composition: • If the composition is given as the number of moles of each component in the mixture, it is known as a molar analysis. • If the composition is given as the mass of each component, it is known as a gravimetric analysis.

For a mixture of ideal gases, the mass is:

The mass fractions are: mfA = mAl m and

mfB =mB1m For a mixture of ideal gases, the number of moles is:


Chapter 12 MoistAir

12:6

(Note that for reacting mixtures, this would be different because one mole reacted with another mole produces one mole of product, but here we are only considering nomeacting ideal gases.) The mole fractions are therefore:

YA=NAIN and YB = NB IN The mole and mass fractions are useful in determining the properties of a mixture. For example, consider the specific enthalpy of the mixture:

h = mfAhA + mfBhB h

= y}iA + YBhB

The overbar indicates per unit mole. The molar mass of a substance is the mass divided by the number of moles. Therefore, the molecular weight of the mixture is:

M=mIN =mAI N+mBI N =NAMAIN+NBMBIN =YAMA+YBMB F or ideal gases, there is a relationship between the partial pressure and the mole fraction. The ideal gas law applied to each of the three boxes in the Dalton model shown in Figure 12-1 is: For gas A:

=NA Ru T

For gas B:

=NR T B u

For mixture:

=NRu T

The universal gas constant, Ru = 1986 Btu/(lbmo/R) = 10.73 psia.ft3/(lbmo/R) Dividing the equations for gas A and gas B by the equation for the mixture results in:

PAl p=NAI N=YA PBlp=NBIN=YB A similar argument for the Amagat model in Figure 12-2 leads to:

VA I V = N A I N =Y A VB IV = NB IN = YB

Chapter 12 MoistAir


12: 7

The partial pressure only changes if there is a change of composition or a change in total pressure. In words, the result may be expressed as: Partial pressure = mole fraction x total pressure = volume fraction x total pressure The above expressions are all derived for just two ideal gases in a mixture. The results can be easily extended to mixtures of any number of ideal gases. On the other hand, real gases do not always behave as ideal gases. The conceptual models derived above are applicable to real gases, but adjustments must be made to the detailed results.

12.3 Mixture ofDry Air and Water Vapor For the following discussion, we need to be specific about the use of the term air. Moist air as used here is to be considered as a mixture of dry air (itself a mixture of nitrogen, oxygen and some other gases) and water vapor (sometimes referred to as moisture). Dry air may be accurately treated as an ideal gas with constant specific heat at atmospheric pressure and at the temperatures commonly encountered in HVAC applications. The value of the specific heat for dry air is Cpa = 0.240 BtuI(lbn/F). Therefore, the enthalpy (with a reference value of 0 at O°F) and change in enthalpy for dry air may be calculated as:

ha

=

!::.ha

Cpa t = 0.240 t Btullbdryatr.

= Cpa !::. t = 0.240 /).t Btullbdryatr.

with temperature t in OF with temperature change in OF or R

Water vapor may also be accurately treated as an ideal gas under the right conditions. For common HV AC conditions, the total pressure of the dry air and water vapor mixture is atmospheric and the temperature is in the range ofO°F to 125°F. The saturation pressure of water vapor at 125°F is only 2 psia (see Appendix B-1). At and below this low pressure, water vapor may be treated as an ideal gas with negligible error (under 0.2%) even when it is in a saturated condition. If water vapor is treated as an ideal gas, then the enthalpy depends only on temperature, and the enthalpy can be taken to be equal to the saturated vapor value at the appropriate temperature. Saturated values for water vapor are given in Appendix B-1. Over the range of temperatures commonly encountered in HV AC, the enthalpy of saturated water may be reasonably calculated as: hg =

1061.5 + 0.435 t

Btullbvapor

Therefore, both the dry air and water vapor components of moist air behave as ideal gases and may be analyzed as though each alone occupied the space. Therefore, the pressure of


Chapter 12 MoistAir

12: 8

the air (usually atmospheric) is the sum of the partial pressures of dry air and water vapor:

The subscripts a and v refer to dry air and water vapor, respectively. The partial pressure of water vapor is often also referred to as the vapor pressure. The vapor pressure is the pressure the water vapor would exert if it alone occupied the space at the temperature and volume of the mixture. There is an upper limit as to how much water vapor can be present in moist air. That limit depends on the temperature. Consider a mixture of given composition and at a specified total pressure. As the mixture is cooled, the temperature will fall until it reaches a value when the saturation pressure of the water vapor (Pg) is just equal to the partial pressure of the vapor in the mixture (p J. Any cooling below this saturated temperature will cause some of the water vapor to condense to saturated liquid. At the saturation temperature, the moist air is said to be saturated and incapable of holding any more water vapor.

12.4 MoistAir Tables The properties of moist air are tabulated in the table on thermodynamic properties of moist air at standard pressure in Appendix D-1. A sample of that table is shown in Figure 12-3. As can be seen from the column headings, the properties for all but the condensed water are given per Ibm of dry air, and not per Ibm of mixture. As we shall see later, this is a convenient form for the application of the equations of First Law of Thermodynamics. The entries in the table are listed according to temperature. The humidity ratio is defined as the mass ratio of water vapor to dry air in the mixture. The value given in the table is the value at the saturation condition. For example, that means at 70°F, the humidity ratio could be anywhere between 0 and a maximum ofO.015832lbjlbda (Ibm water vapor to Ibm dry air). The specific volume, enthalpy and entropy are all given at both the conditions of zero humidity ratio (dry air) and maximum humidity ratio (saturated air) with the possibility of linearly interpolating between those extremes according to humidity ratio. The vapor pressure given under the heading of condensed water is actually the saturated pressure of water at the specified temperature. The units of pressure are given as inches of mercury because it is easy to compare with the standard atmospheric pressure in inches of mercury. For calculation purposes, it is just as convenient to find the saturation pressure in psia from the tables for water (see Appendix B-1).

Chapter 12 MoistAir


12: 9

Condensed Water Temp. Humidity t, Ratio, of Ib.,JIbda Ws 59 60 61 62 63 64 65

66 67 68 69 70 71 72 73 74 75 76 77

78 79

Enthalpy, Btullb dry air

Volume, ft311b dry air

Entropy, Btul(lb dry air) • of hga

Vapor Enthalpy, Entropy, Press., Btullb Btullb' of in.Hg

vG

vllg•

va g

hG

0.010692 0.011087 0.011496 0.011919 0.012355 0.012805 0.013270 0.013750 0.014246 0.014758 0.015286

13.071 13.096 13.122 13.147 13.172 13.198 13.223 13.248 13.273 13.299 13.324

0.224 0.233 0.242 0.251 0.261 0.271 0.281 0.292 0.303 0.315 0.326

13.295 13.329 13.364 13.398 13.433 13.468 13.504 13.540 13.577 13.613 13.650

14.174 14.415 14.655 14.895 15.135 15.376 15.616 15.856 16.097 16.337 16.577

11.618 12.052 12.502 12.966 13.446 13.942 14.454 14.983 15.530 16.094 16.677

25.792 26.467 27.157 27.862 28.582 29.318 30.071 30.840 31.626 32.431 33.254

0.02901 0.02947 0.02994 0.03040 0.03086 0.03132 0.03178 0.03223 0.03269 0.03315 0.03360

0.02358 0.02442 0.02528 0.02617 0.02709 0.02804 0.02902 0.03003 0.03107 0.03214 0.03325

0.05259 0.05389 0.05522 0.05657 0.05795 0.05936 0.06080 0.06226 0.06376 0.06529 0.06685

27.11 28.11 29.12 30.11 31.11 32.11 33.11 34.11 35.11 36.11 37.11

0.015832 0.016395 0.016976 0.017575 0.018194 0.018833 0.019491 0.020170 0.020871 0.021594

13.349 13.375 13.400 13.425 13.450 13.476 13.501 13.526 13.551 13.577

0.339 0.351 0.365 0.378 0.392 0.407 0.422 0.437 0.453 0.470

13.688 13.726 13.764 13.803 13.843 13.882 13.923 13.963 14.005 14.046

16.818 17.058 17.299 17.539 17.779 18.020 18.260 18.500 18.741 18.981

17.279 17.901 18.543 19.204 19.889 20.595 21.323 22.075 22.851 23.652

34.097 34.959 35.841 36.743 37.668 38.615 39.583 40.576 41.592 42.633

0.Q3406 0.03451 0.03496 0.03541 0.03586 0.03631 0.03676 0.03721 0.03766 0.03811

0.03438 0.03556 0.03677 0.03801 0.03930 0.04062 0.04199 0.04339 0.04484 0.04633

0.06844 0.07007 0.07173 0.07343 0.07516 0.07694 0.07875 0.08060 0.08250 0.08444

38.11 39.11 40.11 41.11 42.11 43.11 44.10 45.10 46.10 47.10

haga

SG

Saga

sa g

hw

Temp.,

p/

~

0.0536 0.0555 0.0575 0.0594 0.0613 0.0632 0.0651 0.0670 0.0689 0.0708 0.0727

0.50363 0.52193 0.54082 0.56032 0.58041 0.60113 0.62252 0.64454 0.66725 0.69065 0.71479

59 60 61 62 63 64 65 67 68 69

0.0746 0.0765 0.0783 0.0802 0.0821 0.0840 0.0858 0.0877 0.0896 0.0914

0.73966 0.76567 0.79167 0.81882 0.84684 0.87567 0.90533 0.93589 0.96733 0.99970

70 71 72 73 74 75 76 77 78 79

S..

66

Figure 12.3 Example of Moist Air Table

Example 12.1 - Consider moist, atmospheric air at 70 0 P and determine from Appendix D-1 the humidity ratio and enthalpy in the following cases:

• Saturated air • Humidity ratio ofO.01lbw Ilbda •

Solution -

• Saturated air: Humidity ratio = 0.015832 lb w Ilbda , enthalpy = 34.097 Btu/lb a • Interpolation Table 12-1 Humidity Ratio, IbJlb da 0.0 0.01

0.015832


Enthalpy, Btullba 16.818 h -16.818 + (0.01/0.015832)(17.279) = 27.73 34.097

Chapter 12 MoistAir

12: 10

The Next Step Having formulated the basics about treating moist air as a mixture of ideal gases, we are now ready to see how these basics can apply to the terms and calculations of processes such as cooling, dehumidification and comfort that involve moist air. In Chapter 13, we will define the terms for relative humidity, dewpoint and wet-bulb temperature, and develop expressions for the application of the First Law of Thermodynamics to moist air.

Summary

In this chapter, we laid the groundwork for the study of moist air or psychrometrics. The basis of the analysis was that moist air could be treated as a mixture of ideal gases. The Dalton and Amagat laws for ideal gases are: • Dalton's law of additive pressures: The pressure of an ideal gas mixture is equal to the sum of the pressures each gas would exert if it existed alone at the mixture temperature and volume. • Amagat's law of additive volumes: The volume of an ideal gas mixture is equal to the sum of the volumes each gas would occupy if it existed alone at the mixture temperature and pressure.

The Dalton and Amagat models applied to the mixture of two ideal gases, gas A and gas B, led to the following expressions of partial pressure ratio, volume fraction and mole fraction:

P A / P = VA / V = N A / N = Y A PB / P = VB / V = N B / N = Y B The mass and mole fractions are useful in determining the properties of a mixture. For example, the specific enthalpy of the mixture of ideal gases is:

The molecular weight of the mixture is:

M=YAMA +YBMB

Chapter 12 MoistAir


12: 11

The enthalpy of dry air at pressures and temperatures commonly encountered in HVAC applications may be accurately calculated as: ha

= Cpa

llha

=

t

=

with temperature t in OF

0.24 t Btullbdryair

Cpa II t = 0.24 II t Btullbdrymr.

with temperature change in of or R

The enthalpy of water vapor at partial pressures and temperatures commonly encountered in HVAC applications may be accurately calculated as:

hg

=

1061.5 + 0.435 t Btullbvapor

with t in of, and

-20°F < t < 125°F

Dry air and water vapor may be treated as ideal gases when combined as moist air at the temperatures and pressures commonly encountered in HVAC applications. For moist air, the total pressure is the sum of the dry air and vapor pressures:

p

=

Pa + Pv

The partial pressure of water vapor is often also referred to as the vapor pressure. It is the pressure the water vapor would exert if it alone occupied the space at the temperature and volume of the mixture. There is an upper limit as to how much water vapor can be present in moist air. That limit depends on the temperature and is known as the saturated air condition. It occurs when the vapor pressure is equal to the saturated vapor pressure at that temperature, p v = Pg • Moist air tables shown in Appendix D-llist the property values of moist air by temperature for both dry air (zero humidity ratio) and saturated conditions. The tabulated values are given per unit mass of dry air. After studying Chapter 12, you are able to: • Understand the Dalton and Amagat models as they apply to the basic description of a mixture of ideal gases, including the concepts of partial pressure and partial volume. • Calculate mixture properties given the constituent properties and the composition of the mixture. • Describe why the model of a mixture of ideal gases is applicable to moist air. • Calculate the properties of moist air as a function of temperature.


Chapter 12 MoistAir

12: 12

Skill Development Exercisesfor Chapter 12 Complete these questions by writing your answers on the worksheets at the back of this book. Be sure to include your name and address. Send your completed questions to the ASHRAE Education Department.

12-01. State Dalton's law of additive pressures, and then sketch and explain how this model may result from the addition of two gases of equal volume and temperature.

12-02. Calculate the mass fractions and mole fractions when 2 Ibm nitrogen (molecular weight MN2 = 28.013 Ibmllbmo/J is mixed with 6 Ibm hydrogen (molecular weight MH2 = 2.016Ib /lb J· What is the molecular weight of the mixture? What are the m mo/ partial volume fractions and the partial pressure fractions?

12-03. 2lbm dry air (molecular weight MAlr, = 28.97 lbm lIbmoeI) is mixed with 0.1 lbm water vapor (molecular weight MH20 = 18.015Ibm/lbmo/J. What is the molecular weight of the mixture? What is the partial pressure of the water vapor if the total pressure is atmospheric (14.97 psi a)? Compare the partial pressure with the saturated pressure at 100°F and say whether this mixture is possible at this temperature.

12-04. Find the humidity ratio, specific volume (per Ibm of dry air) and specific enthalpy (per Ibm of dry air) of saturated moist air at standard atmospheric pressure and 120°F.

12-05. Find the specific enthalpy (per Ibm of dry air) of moist air at standard atmospheric pressure and 120°F if the humidity ratio is 0.05.

12-06. 2lbm dry air is mixed with 0.11bm water vapor at atmospheric pressure and a temperature of 120°F. Calculate the enthalpy of the mixture (per Ibm of dry air) from: the moist air tables; and the expressions given for the enthalpy of dry air and water vapor in the text.

Chapter 12 MoistAir


13: 1

Chapter 13 Properties of Moist Air


Humidity Ratio and Relative Humidity

• 13.2

Dewpoint Temperature

• 13.3

Adiabatic Saturation and Wet-Bulb Temperatures



Study Objectives of Chapter 13 The topics of Chapter 13 are the terms and parameters useful in common air-conditioning applications involving moist air. Because ofthe moderate temperatures and low vapor pressure of the water vapor in moist air, the water vapor may reasonably be idealized as behaving as an ideal gas. The objectives of Chapter 13 are to develop an understanding of the following terms: humidity ratio, relative humidity, dewpoint temperature, adiabatic saturation temperature and wet-bulb temperature. After studying Chapter 13, you should be able to: • Understand and calculate humidity ratio and relative humidity; • Understand the basis for determining the dewpoint, adiabatic saturation and wetbulb temperatures; and • Apply the First law of Thermodynamics to problems involving moist air.


Chapter 13 Properties ofMoist Air

13:2

13.1 Humidity Ratio and Relative Humidity The amount of water vapor in the moist air can be specified in various ways. One way already introduced in the previous chapter is the humidity ratio. The humidity ratio is the mass of water vapor per unit mass of dry air. The humidity ratio is sometimes called specific humidity and is denoted by W (not to be confused with work which is also denoted W). The subscripts v and a refer to water vapor and dry air, respectively: Ibm vapor /lb m dry air With the assumption that the dry air and water vapor can be treated as ideal gases, a relationship can be developed between the humidity ratio and the vapor pressure by using the ideal gas equation, P V = mRT. W=ml'lma

=(PI'V I RI'T) I (PaV I RaT) =(Ra I Rv)(Pv I Pa) =

0.622 (PI' I Pa) Ibm vapor per Ibm dry air

The gas constants are Rv = 0.1102 Btu/(lbmvapor·R) and Ra = 0.06855 BtuI(lbmair·R) and therefore Ra IR v = 0.6221bm vaporperlbm dry air. Therefore, the units of humidity ratio are also Ibm vapor per Ibm dry air. The total pressure is the sum of the partial pressures, P = P a + PI'. Solving for Paand substituting into the equation for humidity ratio results in the following expression: W = 0.622 PI' I (p - PI') Ibm vapor per Ibm dry air

The total pressure P is often atmospheric pressure when dealing with many of the problems in air-conditioning. We will see later why this is such a useful relationship. Another measure of the moisture content of moist air is the relative humidity,~. Relative humidity is the ratio of the mass of vapor in the moist air compared to the maximum moisture that moist air can hold at that temperature. To explain relative humidity, first consider dry air. Dry air has a humidity ratio of zero. Imagine that water vapor is added to the dry air, thereby increasing the humidity ratio. The more moisture that is added, the higher the humidity ratio, until the air can no longer hold any more moisture. The moist air is then saturated. Any further moisture added to saturated air will cause condensation of the vapor to liquid water. The maximum amount of water vapor in the moist air is at the saturated condition and can be found by setting PI' equal to the saturation pressure Pg in the expression for



13: 3

humidity ratio and in the ideal gas equation. Therefore, the relative humidity is: ~

=ml' /mg = (Pl'v / RvT)/(pgV / RT) = PI' / Pg

where P g = PSal at temperature t. Relative humidity is dimensionless. Combining the relationship for relative humidity with the relationship for humidity ratio by eliminatingp" yields the following two useful expressions:

~ = W(p/ Pg)/(O.622+W) W

= O.622~ P g / (p-~ pg )

The relative humidity ranges from zero for dry air to a value of one for the saturated condition. The relative humidity depends on the temperature because the maximum moisture that the air can hold depends on its temperature. In contrast, the humidity ratio depends on composition and not on temperature except at the saturated condition. Therefore, the relative humidity may change with temperature even when the humidity ratio is constant. The relative humidity ratio is important because the comfort level depends more on relative humidity than on humidity ratio. The humidity ratio is the quantity that is most useful when doing analysis and applying the First Law of Thermodynamics. The total enthalpy of moist air includes both the enthalpies of the dry air and of the water vapor:

H= ha +Hv = maha +mvhv The specific enthalpy of moist air is commonly written per unit mass of dry air because the amount of dry air is usually fixed while the amount of moisture varies. Dividing through by the mass of dry air gives the specific enthalpy as:

h = ha +(mv / ma)h"

=ha + Whv

Btu/Ibm dry air

(hI' is in Btu/Ibm vapor)

Because it is treated as an ideal gas, hI' can be approximated by hg at the same temperature.



13:4

Example 1. The conditions on a hot summer day are given as 95°F and a relative humidity of 65%. Calculate the humidity ratio. Solution. From Appendix B-1, the saturated pressure at 95°F is p g

=

0.81625 psia.

With ~ = 0.65, the humidity ratio is: W

= 0.622~ Pg I(p-~ pg) = 0.622(0.65)(0.81625) 1{ 14.696 - (0.65)(0.81625)} = 0.0238 Ibm vapor perlbmdry air

Note that the units of Ibm vapor per Ibm dry air are contained in the term 0.622 because it came from the ratio of gas constants of vapor and dry air.

Example 2. If the temperature of Example 1 drops from 95°F down to 85°F while the humidity ratio remains at 0.0238 Ibm vapor per Ibm dry air, estimate the relative humidity at the new temperature. Solution. From Appendix B-1, the saturated pressure at 85°F is p g

=

0.59647 psia.

With W = 0.0238 Ibm vapor per Ibm dry air:

~ = W(pl pg)/(0.622+W)

= 0.0238 (14.696/0.59647)/(0.622 + 0.0238) = 0.908 The relative humidity is 90.8% and the day will feel closer to rain.

13.2 Dewpoint Temperature Consider what happens when moist air at constant total pressure is steadily cooled: the temperature will decrease. Initially, the humidity ratio remains constant because there is no change in composition of the moist air. However, the relative humidity will increase, as shown in an earlier example. When the relative humidity becomes 100%, the air reaches its saturated condition. The temperature at which this occurs is the dewpoint temperature. Any further cooling means that the air cannot hold all the vapor and the vapor condenses as a liquid.



13: 5

This process is shown on the t-s diagram in Figure 13-1. The diagram shows only the water vapor behavior and not the dry air behavior. State 1 represents the state of the water vapor at vapor pressure of P. and a relative humidity of less than one. As the moist air is cooled at constant pressure, it follows the line of constant p. until it meets the saturated vapor curve. State 2 is the saturated condition at the intersection of the constant pressure line and the saturation curve. The conditions at State 2 are p g = p.' ~ = 1, and the temperature is the dewpoint temperature:

The dewpoint temperature is defined as the temperature at which condensation begins if the moist air is cooled at constant vapor pressure. If the moist air is cooled below the dewpoint temperature, then there is condensation of some of the vapor to liquid. This can be traced on the t-s diagram in Figure 13-1. The liquid that forms will be at the condition on the saturated liquid curve between States 2f and 3f. The vapor that remains in the moist air will follow the saturated vapor curve from State 2 to State 3g. For the process from State 2 to State 3g, the relative humidity remains constant at 100%, but the humidity ratio steadily decreases.

Temperature t

Entropy,s

Figure 13-1. Dewpoint Temperature on the t-s Diagram for Water Vapor



13:6

The dewpoint temperature can be measured by chilling a surface until dew is first observed and measuring the temperature of the surface. The dewpoint temperature can then be used to determine the vapor pressure. With the vapor pressure, it is then possible to calculate the relative humidity from ~ = p v /pg. Although this is a straightforward procedure, it is cumbersome and not that convenient.

13.3 Adiabatic Saturation and Wet-Bulb Temperatures The relative humidity and humidity ratio are both useful parameters in calculations, but there are no direct measures for these quantities. The dewpoint temperature measurement can lead to the relative humidity as explained earlier, but it is not a convenient measurement. The most common way of determining the relative humidity and humidity ratio is to measure something known as the wet-bulb temperature and use that in a calculation. However, before describing the wet-bulb temperature and how it is used, it is informative to first consider another temperature known as the adiabatic saturation temperature that results from the adiabatic saturation process. Consider the apparatus shown in Figure 13-2. A steady flow of moist air, with unknown humidity ratio WJ and relative humidity ~ J ' enters the long insulated channel at State 1. The channel contains a pool of water. As the· air flows through the channel, some of the water evaporates and mixes with the stream of moist air. As a result, the humidity ratio changes along the length of the channel. If the channel is long enough, the moist air will emerge from the channel at State 2 in a saturated condition, ~2 = 100%. The process for the water vapor change in the moist air is shown as Process 1 to 2 on the t-s diagram of Figure 13-3. The temperature of the moist air changes along the length of the channel because the air has

. . . 92.,.CI=====~~~=====~~~=~I ~ ~aturated

Moist Air

Humidity Ratio w1 Temperature ~ Relative Humidity cj>1 Enthalpy h1

[l

I~

[f

Air

Humidity Ratio Temperature Relative Humidity Enthalpy

w2

t2 cj>2

h2

Saturated Water

Temperature Enthalpy

~

hf

Figure 13-2. The Adiabatic Saturator


Fundamentals ofThermody"amics and Psychrometries

13: 7

Temperature t

Adiabatic Saturation Temperature

Entropy,s

Figure 13-3. The Adiabatic Saturation Process to provide the heat of evaporation of the moisture added to the air stream. The temperature at State 2 is known as the adiabatic saturation temperature because it comes at the end ofthe adiabatic (insulated channel) saturation (water vapor picked up along the channel) process. The adiabatic saturation temperature is different from the dewpoint temperature, as can be seen from Figure 13-3. The moist air composition and humidity ratio change along the channel and, therefore, the saturation process is not a constant vapor pressure process as described for dewpoint temperature. The adiabatic saturation temperature can be related to the unknown inlet humidity ratio and relative humidity by applying the conservation of mass and steady-flow form of the First Law of Thermodynamics to the adiabatic saturation process. The subscript a refers to dry air, v refers to vapor, g refers to saturated vapor, and Ito saturated liquid water. It is helpful to think of the moist air to be made up of two streams: the dry air stream and the water vapor stream. For the conservation of mass, this means two equations, one for each stream. But for the First Law of Thermodynamics, there is only one energy balance with both streams in the same equation.



13: 8

Conservation of mass (air stream):

Conservation of mass (water vapor stream):

The humidity ratios are ~ = mvl / rna and ~ = m1'2 / rna . Solving for mvl and m1'2 and substituting into the equation above, the conservation of mass of the water vapor stream becomes:

Reorganizing terms and solving for ml3

ml3

= (~ - ~)ma

First Law of Thermodynamics (no work and no heat transfer):

Substitute for mvl and mv2 as before. Then using the expression for ml3 from the mass balance in the energy balance, and dividing through by rna leads to: hal + ~hvl +(~ - ~)hl3

Solving for humidity ratio

= ~ + ~h1'2

~:

In this equation, the air and the water vapor are treated as ideal gases. The enthalpy change of the air can be replaced by ha2 - haJ = C/t2 - tl ). Also the enthalpy of the vapor can be found as the saturated vapor value at the appropriate temperature because, for an ideal gas, the enthalpy depends only on temperature. Furthermore, the difference between the vapor and liquid enthalpy can be written as h1'2 - hf3 = hlg2 for the liquid water at t 2. With all these substitutions, the energy equation may be written as: = =

{Cp (t2- tJ ) + W2 hlg2 }/ (h gl - hj2)

O.622pg2 . /(p - p g2)



13: 9

After a long derivation, this is a result that can be used to determine the humidity ratio by measurement of two temperatures and the total pressure. The temperatures are the moist air temperature t] and the adiabatic saturation temperature t2. Knowing t2 , the values of hjg2 , hp and Pg2 can be found in the water tables. Knowing t] , the value of hg] can be found in the water tables. The adiabatic saturator provides the measurements to calculate the humidity ratio, but it is a cumbersome piece of equipment. Fortunately, there is another device - the wet-bulb thermometer - that gives temperatures very similar to the adiabatic saturation temperatures even though the way it functions is quite different. The wet-bulb temperature is measured by a thermometer whose bulb is covered with a water saturated cotton wick. As the unsaturated air blows over the wet wick, some of the water in the wick evaporates, thus cooling the thermometer to a lower temperature. The temperature that results when there is a balance between the heat needed for evaporation and the heat transfer to the air stream is the wetbulb temperature. In general, the wet-bulb temperature and the adiabatic saturation temperature are not the same but, for moist air at atmospheric pressure, they have almost identical values. Therefore, the wet-bulb temperature can be used in place of the adiabatic saturation temperature in the equations developed earlier. Sometimes, the usual temperature measured by just a plain unwrapped thermometer is called the dry-bulb temperature. Wet-bulb temperatures are sometimes measured with a sling psychrometer. The psychrometer has a wick-covered thermometer that is kept wet and is mounted on a handle so that it can be made to rapidly swing through the air. This brings the wet-bulb temperature to quick equilibrium. The sling psychrometer is handy for a single measurement but, for continuous measurement of wet-bulb temperature, it is more common to use a wick-covered thermocouple. Example 13.3. On a 82°F day, the wet-bulb temperature is measured as 65°F. Calculate the humidity ratio and the relative humidity. Solution. The expression for humidity ratio is:

w;. = {cAt2 -

tl ) + ~hjg2} / (h gl - hj2 )

Ibm vapor / Ibm dry air

From the water tables in Appendix B-1, the property values at t2 = 65°F and t] = 82°F are:

hjg2

= 1056.5 Btullbmvapor,

hj2

= 33.07 Btullbmvapor

andpg 2

= 0.30574 psia

hgl = 1096.95 Btu / Ibm vapor



13: 10

The saturated humidity ratio at the wet-bulb temperature is:

~ = 0.622 Pg2 / (p - P g2 ) = (0.622 Ibm vapor Ilb m dry air )(0.30574 psia)/ {(14.696- 0.30574) psia} = 0.0132 Ibm vapor / Ibm dry air Therefore, the humidity ratio of the moist air is:

~ = {CAt2 -tl)+~hfg2}/(hgl-hf2) = {0.24(65- 82) + 0.0132(1056.5)} Btullb m dry air / {(1096.5- 33.07) Btullbm vapor} = 0.0087 Ibm vapor / Ibm dry air The relative humidity is: cj>

= W(p/ Pg)/(0.622+W)

= 0.0087(14.696/0.30574) / (0.622 + 0.0087) =0.663 = 66.3%

Example 13.4. Atmospheric air at 95°F and 50% relative humidity is cooled to 80°F or 60°F. In each case, calculate the final relative humidity. Solution.

W = 0.622 cj> Pg /(p-cj> p g) Ibm vapor Ilb m dry air cj>

= W(p/ Pg)/(0.622+W)

At 95°F, the saturated pressure is found in Appendix B-1 as Pg = 0.81625 psia. Therefore, the humidity ratio is:

~ =0.622 cj>Pg/(p-cj>Pg) = 0.622(0.5)(0.81625) / {14.696 - (0.5)(0.81625)} = 0.0178 Ibm vapor Ilb m dry air



13: 11

The humidity ratio will remain unchanged if there is no condensation. To check if there is condensation, first find the dewpoint temperature. If the final temperature is above the dewpoint temperature, then W2 = WI ; if the final temperature is less than the dewpoint temperature, then ~ 2 = 1:

= t sat @ Pv ~ = Pv / Pg

tdp

Pv =~ Pg = 0.5(0.81625)

= 0.40813 psia From the water table in Appendix B-1, the dewpoint temperature may be interpolated as shown in Table 13-1:

Table 13-1

= 73 + (74 - 73)(0.40813 - 0.40217)1

Pressure, psia 0.40217 0.40813

(0.41592 - 0.40217) t= 73.44 74

0.41592

Temperature, of 73 t

• Cooled to t2 = 80°F. Because t 2 > tdp' the humidity ratio is unchanged and W2 = WI = 0.0178 Ibm vapor/ Ibm dry air. From the saturated water tables in Appendix B-1, the saturation pressure at 80°F is 0.50736 psia. The relative humidity can be calculated as:

~ =W(p/ Pg)/(0.622+W) = 0.0178(14.696/0.50763) / (0.622 + 0.0178)

= 0.805 = 80.5% • Cooled to t2 = 60°F. Because t2 < t dp ' the final condition is saturated. Therefore: ~2

= 100%



13: 12

The Next Step In the next chapter, we will introduce the psychrometric chart. The chart contains the values of the parameters introduced in this chapter as well as the other moist air data in a very compact and easy-to-use format.

Summary For moist air, the humidity ratio is the mass of water vapor per unit mass of dry air: Ibm vapor / Ibm dry air

W = 0.622 Pv / (p - Pv) Another measure of the moisture content of moist air is the relative humidity, ~. Relative humidity is the ratio of the mass of vapor in the moist air compared to the maximum moisture that the air can hold at that temperature. The relative humidity is: ~

=mv /mg =

Pv / Pg

where Pg = PSal at temperature t. Relative humidity is dimensionless. Combining the relationship for relative humidity with the relationship for humidity ratio yields the following two useful expressions:

~ = W(p/ Pg)/(0.622+W) W=

0.622~ P g /(p-~ p g ) Ibm vapor/ Ibm dry air

The relative humidity ranges from zero for dry air to a value of one for the saturated condition. Relative humidity depends on the temperature as well as the amount of vapor in the

au. The specific enthalpy of moist air is commonly written per unit mass of dry air:

Cltapter 13 Properties ofMoist Air

Fundamentals ofTltermodynamics and Psycltrometrics

13: 13

The dewpoint temperature is defined as the temperature at which condensation begins if the moist air is cooled at constant vapor pressure:

The relative humidity and humidity ratio are both useful parameters in calculations, but there are no direct measures for these quantities. The dewpoint temperature measurement can lead to the relative humidity as explained earlier, but it is not a convenient measurement. The most common way of determining the relative humidity and humidity ratio is to measure the wet-bulb temperature and use that in a calculation. The wet-bulb temperature temperature is measured by a thermometer whose bulb is covered with a water-saturated cotton wick. The calculation is based on the First Law of Thermodynamics applied to an adiabatic saturation process. For moist air at atmospheric pressure, the wet-bulb temperature replaces the adiabatic saturation temperature in the expression:

W; = {cAt2 - t l ) + ~hjg2} / (h gl - hj2 )

Ibm vapor / Ibm dry air

where: W2

=

0.622 Pg2 /(p - Pg2 )

t2

= the wet-bulb temperature

tJ

= the dry-bulb temperature

hjg2 ,hj2 and Pg2 are found in the saturated water tables at t2 hg]

is found in the saturated water tables at t]

Cp

=

the coefficient of specific heat for dry air, Cp = 0.24 Btu/(lbmdry air·R)


Chapter 13 Properties ofMoistAir

13: 14


13-01. Give a short description of humidity ratio and relative humidity. 13-02. Using the definitions of humidity ratio, relative humidity and the Ideal Gas Law, show that the relationship between humidity ratio and relative humidity can be written as:

~ = W(p/ Pg)/(0.622+W) W = 0.622~ Pg / (p-~ pg ) Ibm vapor/ Ibm dry air 13-03. Consider moist air at 72°P and atmospheric pressure. Answer the following questions: • What is the humidity ratio and relative humidity if the air is saturated? • What is the humidity ratio if the relative humidity is 60%? • What is the relative humidity ifthe humidity ratio is 0.015 Ibm vapor/ Ibm dry air?

13-04. Explain the term dewpoint temperature. Calculate the dewpoint temperature for 72°P atmospheric air with a relative humidity of 60%. 13-05. On a 72°P day, the wet-bulb temperature is measured as 55°P. Calculate the humidity ratio and the relative humidity. 13-06. Atmospheric air at 90 0 P and 60% relative humidity is cooled to 80 0 P or 60 o P. In each case, calculate the final relative humidity.

13-07. Moist air at 95°P, atmospheric pressure and relative humidity of 60% is cooled at constant pressure to saturation conditions at a rate of 2,000 Ibm dry air per hour. Estimate the rate of heat removal.



14: 1

Chapter 14 The Psychrometric Chart


Introduction

• 14.2

Description of the Axes and Lines of the Chart

• 14.3

Finding Property Values on the Psychrometric Chart

• 14.4

Human Comfort and the Psychrometric Chart


Instructions Read the material of Chapter 14. Re-read the parts ofthe chapter that are emphasized in the summary and memorize important definitions. At the end of the chapter, complete all of the skill development exercises without consulting the text.

Study Objectives of Chapter 14 The psychrometric chart is a comprehensive property chart for moist air. For a given total pressure, the chart covers a range of dry- and wet-bulb temperatures commonly found in HV AC applications. The objective of Chapter 14 is to understand the layout and how to find property values on the chart. After studying Chapter 14, you should be able to: • Fully describe the psychrometric chart; • Look up states on the chart; and • Identify the ASHRAE comfort zone on the chart.



14:2

14.1 Introduction In the previous chapter, we introduced humidity ratio, relative humidity, moist air enthalpy, dry-bulb temperature, wet-bulb temperature and dewpoint temperature. It was also shown that given any two of these properties, the other properties could be calculated by using the expressions that relate different properties, the moist air tables and the saturated water tables. However, the calculations are cumbersome and repetitive. They are more conveniently done by using a plot of the properties on the psychrometric chart. The chart is used extensively in air-conditioning calculations because it is good for finding properties and because many of the processes of air-conditioning are easy to map out on it. The ASHRAE psychrometric chart is shown in Appendix E-l. (It is suggested that you make a few photocopies of the chart that you can use for practice and for developing a familiarity with the many aspects of the chart). The chart includes a wealth of infonnation, but the first thing is to learn the basics of the chart. The properties represented on the chart are: • Dry-bulb temperature, tdb: The temperature ofthe moist air. • Wet-bulb temperature, twb: The equilibrium temperature measured when a thermometer covered in a wet wick is exposed to moist air at tdb' • Dewpoint temperature, tdp : The saturation temperature at a saturation pressure equal to the vapor pressure in the moist air (tsat at Pv = Pg ). In more practical tenns, it is the temperature at which condensation first appears if moist air is cooled at constant vapor pressure. • Humidity ratio, W: The mass ratio of moisture to dry air in the moist air (mass basis). • Relative humidity, ~: The ratio of moisture in the moist air to the amount of moisture in the moist air if it was saturated at the same temperature. • Specific volume, v: The specific volume of the moist air (mixture of dry air and water vapor) in ft3 per Ibm dry air. • Specific enthalpy, h: The specific enthalpy of moist air in Btu per Ibm dry air. The enthalpy includes the contributions of both dry air and water vapor, h = ha + Whg There is a second drawing on the same chart in the left-hand top comer that looks like protractor and includes the ratios of sensible heat to total heat and enthalpy change to humidity ratio change. Unlike the main chart which is used for properties and processes, the protractor only applies to processes, and will be explained further in the next chapter.



14: 3

14.2 Description ofthe Axes and Lines ofthe Chart The basic features of the chart are shown in Figure 14-1. Along the horizontal axis is the dry-bulb temperature. Along the vertical axis on the right side of the chart is the humidity ratio. On the left boundary of the chart, there are two lines, neither of them a vertical axis. There is a straight sloped line that serves as the axis for values of enthalpy. There is also a sloped curve that is the saturation line. The saturated moist air states are located along the saturation line. The saturation line corresponds to a line of constant relative humidity of 100%. The lines of constant property are shown in Figure 14-2. The dry-bulb temperature and humidity ratio form the basis of the chart, with lines of constant humidity running horizontally and the lines of constant dry-bulb temperature running up and down. Lines of constant huinidity ratio are lines of constant dewpoint temperature. The lines of constant enthalpy and constant wet-bulb temperature are straight and slope down and to the right on the chart. They have very nearly the same slope, but not quite. The lines of constant wet-bulb temperature are not drawn as boldly as the lines of constant enthalpy so as to distinguish them from each other. To read a value of enthalpy that lies between two enthalpy lines, draw a

Dew-point temperature

-<0%l"~ 6r:::

C' CD 3

~

~

16

~

~~

~~

~&

~@

Dry-Bulb Temperature

Figure 14-1. Basic Features of the Psychrometric Chart



14:4

Constant Humidity Ratio, W

Figure 14-2. Lines of Constant Property Value

Dew-point temperature, 70°F

70°F

Figure 14-3. Dry-Bulb, Wet-Bulb and Dewpoint Temperatures and the Saturation Line



14:5

line parallel to the enthalpy line and read the result where this intersects the enthalpy axis. The lines for constant specific volume are also straight lines that slope down to the right of the chart, but they are steeper than enthalpy or wet-bulb temperature. Lines of constant relative humidity are curved up and to the right of the chart and are shaped similarly to the saturated line. At any point on the saturation line, the wet-bulb temperature, dry-bulb temperature and dewpoint temperature are the same as shown in Figure 14-3. For non-saturated conditions, the line of constant dewpoint temperature is a horizontal straight line. The lines of constant property will be shown to be important in the next chapter when we look at applications of the psychrometric chart to air-conditioning processes. Often, processes such as simple heating and cooling involving no humidification are equivalent to lines of constant humidity ratio. The chart is good for visualizing the processes of airconditioning. The general direction of increasing property values is shown in Figure 14-4.

Figure 14-4. Direction of Increasing Property Values



14:6

14.3

Finding Property Values on the Psychrometric Chart

The methodology using two known property values to find any other property value for moist air on the psychrometric chart is best shown in the following series of examples. In each case, two of the seven properties will be known. The seven properties are: • Dry-bulb temperature, tdb • Wet-bulb temperature, twb • Dewpoint temperature, tdp • Humidity ratio, W • Relative humidity, ~ • Specific volume, v • Specific enthalpy, h

Example 14.1. Atmospheric moist air has a dry-bulb temperature of SO°F and a humidity ratio of 0.012 Ibm moisture per Ibm dry air. Find the other five properties.

Solution. Looking at Figure 14.4 and consulting the psychrometric chart in Appendix E-1 at tdb = SO°F and W = 0.012, the other properties are: twb tdp

=

6S0F (read from the faint dotted line for constant wet-bulb temperature)

=

~

=

v

=

62°F (read from extending a horizontal line until it intersects the saturation line) 55% (interpolated between the lines of 50% and 60%) 13.S7 ft3 per Ibm dry air (interpolated between lines of 13.S and 13.9) 32.2 Btu per Ibm dry air (read from the intersection of the enthalpy axis and a straight line drawn through the state point with a slope like that of the lines of constant enthalpy)

h

Example 14.2. Atmospheric moist air has a dry-bulb temperature of SO°F and a wet-bulb temperature of 64°F. Find the other five properties.

Solution. Consulting the psychrometric chart in Appendix E-1 at tdb = SO°F and twb = 64°F, the other properties are:

tdp

=

W

=

~

=

v

54.6°F (read from extending a horizontal line until it intersects the saturation line) 0.00921bm moisture per lbm dry air 42% (interpolated between the lines of 40% and 50%) 13.S0 ft3 per Ibm dry air



14: 7

h

=

29.1 Btu per Ibm dry air (read from the intersection ofthe enthalpy axis and a straight line drawn through the state point with a slope like that of the lines of constant enthalpy)

Example 14.3. Atmospheric moist air has a relative humidity of 40% and a specific enthalpyof 38.6 Btu per Ibm dry air. Find the other five properties. Solution. Consulting the psychrometric chart in Appendix E-1 at ~ = 40 % and h = 29.1 Btu per Ibm dry air, the other properties are: tdb

twb

=

tdp

=

W = V

=

95°F 75.1°F 66.8°F (read from extending a horizontal line until it intersects the saturation line) 0.01421bm moisture per lbm dry air 14.3

ft3

perlbmdry air

The accuracy is limited by the ability to read and interpolate on the chart. Fortunately, there are only linear scales and it is not necessary to interpolate on a log scale, as may be found in heat transfer or fluid mechanics. The only combination that is not easy to pinpoint with accuracy on the chart is when wet-bulb temperature and enthalpy are known. These lines are almost parallel, making the task of finding the intersection point difficult. Fortunately, the known wet-bulb temperature and enthalpy combination are not commonly encountered. So far, we have only described the psychrometric chart for standard atmospheric pressure. There are other psychrometric charts, with similar information, for moist air at pressures slightly above and below that of standard sea-level pressure. Also, many manufacturers of air-conditioning equipment have developed their own psychrometric charts. These vary in the details about scales, units and line drawing techniques, but they are basically the same as the ASHRAE psychrometric chart. The values represented on the charts may now be conveniently, and more accurately, determined by using ASHRAE software that has programmed the calculations in just the same way that the chart was constructed. However, there is still a big value to the charts, as will be seen in the next chapter, because they help us visualize the processes of air-conditioning. Plotting out the change of state on the chart gives a perspective that is very useful in design. As time goes on, we can anticipate that even more of the visual aspects will be programmed for the computer.



14: 8

14.4 Human Comfort and the Psychrometric Chart The psychrometric chart covers a range of temperatures and humidities that include those most likely to be encountered under ordinary conditions. However, there is only a smaller range of conditions for which people say they feel comfortable. Comfort means different things to different people, but in the context of environmental conditions of temperature and humidity, surveys have tried to establish a consensus on the range of values considered comfortable. The ASHRAE envelope is such a range that is said to satisfy better than a 10% dissatisfaction criterion. The ASHRAE comfort envelope is specified in ASHRAE Standard 55-1992. It includes two overlapping regions: one appropriate for people in light summer clothing and the other appropriate for people in the slightly heavier indoor clothing that is worn in winter. The comfort envelope is shown in Figure 14-5. The envelope is given as an area on a chart in which the horizontal axis is the operative temperature (aT) and the vertical axis is either the humidity ratio or the dewpoint temperature. Because the heat balance on the human body includes both convection and radiation heat transfer, the operative temperature is the average of the mean radiant temperature and the dry bulb temperature. 70 On the chart, there are lines of constant relative humidity. The boundaries to the comfort zone are lines of constant relative humidity and lines of constant effective temperature (ET*). The effective temperature is an empirical index that includes the effects of both operative temperature and evaporative cooling. It is the temperature of an environment at 50% relative humidity that results in the same total heat loss from the skin as in the actual environment. The effective temperature depends on the clothing worn and on

C/,apter 14 The Psychrometric Chart

65

~

60

~

:::l

~ Q)

55

c. E ~ 50

E

'0 c.

45

o

40~----+---~~~~~~~r--+----~ 5

~

35 30

25 20 68Er"

10

o 60

65

70

75

80

85

90

Operative Temperature, OF

Figure 14-5. The ASHRAE Comfort Zone

Fundamentals ofT/,ermodynamics and Psychrometries

14: 9

the skin wettedness as well as on the surrounding conditions. This means that the effective temperatures for two different people in the same space could be different. The comfort region shown in Figure 14-5 is appropriate for people engaged in light office work «1.2 met). The range is based on a 10% dissatisfaction criterion. The comfort zone may also be shown on the psychrometric chart as in Figure 14-6 if it is assumed that the surrounding radiative temperature is the same as the air temperature and the operative temperature is then just the dry-bulb temperature. Air-conditioning systems often try to maintain the environmental conditions depicted within the comfort zone if there are human occupants in the conditioned space. Besides effective temperature, several other factors contribute to comfort or discomfort. p or example, air velocity that is too high or asymmetric radiation due to certain heat sources in the room can cause discomfort. A vertical temperature gradient can also be a problem, and the temperature at head level should never be more that sop different from that at feet level. Another comfort factor is floor temperature. In environments where shoes are worn, the floor and air temperatures should be the same. However, when barefoot, the preferred temperature range depends on the floorcovering; for bare concrete, it is from 79°P to 83 0p.

Figure 14..;6. The ASHRAE Comfort Zone on the Psychrometric Chart



14: 10

The Next Step Whereas this chapter described the structure of the psychrometric chart and how to find property values on the chart, the next chapter will discuss using the chart for common airconditioning processes. The processes to be considered are heating, cooling, humidification, dehumidification and mixing.

Summary The psychrometric chart is a very useful tool for determining the values of moist air properties. There are seven properties plotted on the chart: • Dry-bulb temperature,

tdb

• Wet-bulb temperature, twb • Dewpoint temperature, tdp • Humidity ratio, W • Relative humidity,

~

• Specific volume, v • Specific enthalpy, h

The basic features of the chart include the dry-bulb temperature on the horizontal axis and the humidity ratio on the vertical axis. On the left boundary of the chart, there are two lines. There is a straight sloped line that serves as the axis for values of enthalpy. There is also a sloped curve that is the saturation line. The saturated moist air states are located along the saturation line. At any point on the saturation line, the wet-bulb temperature, dry-bulb temperature and the dewpoint temperature are the same. A basic rectangular grid is formed with the lines of constant humidity running horizontally and the lines of constant dry-bulb temperature running vertically. The lines of constant enthalpy and constant wet-bulb temperature are straight and slope down and to the right on the chart. They have very nearly the same slope, but not quite. The lines for constant specific volume are also straight lines that slope down to the right of the chart, but they are steeper than enthalpy or wet-bulb temperature. Lines of constant relative humidity are curved up and to the right of the chart and are shaped similarly to the saturated line. The psychrometric chart covers a range of temperatures and humidities that include those most likely to be encountered under ordinary conditions. However, there is only a smaller



14: 11

range of conditions for which people say they feel comfortable. Comfort means different things to different people, but in the context of environmental conditions of temperature and humidity, surveys have tried to establish a consensus on the range of values considered comfortable. The ASHRAE envelope is such a range that is said to satisfy better than a 10% dissatisfaction criterion. After studying Chapter 14, you should be able to: • Fully describe the psychrometric chart. • Look up states on the chart. • Identify the ASHRAE comfort zone on the chart.

Fundamentals ofTllermodynamics and Psycllrometrics

Cllapter 14 Tile Psycllrometric Cllart

14: 12


14-01. Sketch the outline of the psychrometric chart showing typical lines of constant value of dry-bulb temperature, wet-bulb temperature, humidity ratio, relative humidity, specific enthalpy and specific volume.

14-02. List the seven properties that are represented on the psychrometric chart.

The following five problems are to be solved by looking up the properties of moist air on the psychrometric chart for standard atmospheric pressure (sea-level). Given two properties, you must look up the other five properties.

14-03. Dry-bulb temperature of 90°F and a wet-bulb temperature of 75°F.

14-04. Humidity ratio of 0.008 Ibm moisture per Ibm dry air and relative humidity of 80%.

14-05. Humidity ratio of 0.020 Ibm moisture per Ibm dry air and dry-bulb temperature of 80°F.

14-06. Enthalpy of25 Btu per Ibm dry air and relative humidity of20%.

14-07. Saturated moist air at a dry-bulb temperature of 72°F.


Fundamentals ofTI,ermodynamics and Psychrometries

15: 1

Chapter 15 Air-Conditioning Processes on the Psychrometric Chart

Contents of Chapter 15 • Instructions • Study Objectives of Chapter 15 • 15.1 Simple Heating or Cooling at Constant Humidity Ratio • 15.2

Heating with Humidification

• 15.3

Cooling Coil Dehumidification

• 15.4

Evaporative Cooling

• 15.5

Adiabatic Mixing

• 15.6

Sensible Heat Ratio


Instructions Read the material of Chapter 15. Re-read the parts of the chapter that are emphasized in the summary and memorize important definitions. At the end ofthe chapter, complete all ofthe skill development exercises without consulting the text.

Study Objectives of Chapter 15 The objectives of Chapter 15 are to develop an understanding of how to use the psychrometric chart to analyze the processes of air-conditioning. After studying Chapter 15, you should be able to use the chart and the First Law of Thermodynamics to solve problems involving: • Simple heating and cooling processes; • Heating processes with humidification; • Cooling processes with dehumidification; • Evaporative cooling processes; and • Processes with a constant sensible heat ratio.


Chapter 15 Air-Conditioning Processes

15:2

15.1

Simple Heating or Cooling at Constant Humidity Ratio

Simple heating or cooling processes do not involve any moisture addition or removal, or any flow mixing. They appear on the psychrometric chart as straight horizontal lines. Simple heating (as found with an electric heating system, a heat pump or a heating coil with no moisture addition) is shown as a constant humidity ratio process 1 to 2 on the chart in Figure 15-1. Simple cooling is similarly a straight horizontal line, but towards decreasing temperature to the left rather than to the right. The humidity ratio does not change with simple heating or cooling but the relative humidity does. The relative humidity decreases with simple heating and increases with simple cooling. The First Law of Thermodynamics applied to simple heating or cooling is:

The enthalpy of the moist air includes both that of dry air and water vapor, h The sign of the value of heat transfer

=

ha + Wh g

'

Q depends on whether there is heating (positive) or

cooling (negative).

CD. t1

:

"1 : AIR c::> :

··

w1 h1

.

1 .'

.. • Cooling'. "

,""

._.I

I

I

I

~.

.

'Q"

~

2 .'

Heati~p.'':

tl

Figure 15-1. Simple Heating or Cooling



15: 3

15.2 Heating with Humidification As observed earlier, the relative humidity decreases during heating. The decrease is quite noticeable in winter, when outdoor air with a low relative humidity is heated and becomes uncomfortably dry in the conditioned space. The situation can be remedied by adding moisture. The process of adding moisture is known as humidification. Consider the case of a heater followed by a humidifier as shown in Figure 15-2. Through the heating section from State 1 to State 2, the dry-bulb temperature increases while the humidity ratio remains the same. Through the humidification system from State 2 to State 3, the humidity ratio increases, but the dry-bulb temperature may increase or decrease depending on the way in which humidification is accomplished. If the moisture is added by evaporation from a pool of water (as in the adiabatic saturator described in an earlier chapter or by water spray), then the dry-bulb temperature will decrease slightly because the air stream needs to provide the energy for evaporation. However, ifthe moisture is introduced as steam as shown in Figure 15-2, then the dry-bulb temperature could be expected to increase slightly because of the high energy content of the steam.

1,

til, AIR

•

c::> :

1a

tIIs

: C::>AIR

w, h,

Heater

t

Steam Humidifier

Figure 15-2. Heating With Humidification

Example 15.1. 2,000 cfm of outdoor air at 35°F and 30% relative humidity is to be heated and humidified to a final condition of 80°F and 50% relative humidity with a heater followed by a steam humidifier. The temperature after the heater is 74°F. Find the following:

• The heat added in the heater section; • The mass flow rate of steam; and • The enthalpy of the steam.



15:4

Solution. The process can be mapped onto the psychrometric chart as shown in Figure 15-2, and the appropriate properties determined at States 1, 2 and 3 as needed. The First Law of Thermodynamics applied to the heater is:

The mass flow rate can be calculated by knowing the volume flow rate and reading from the chart that the specific volume is VI = 12.48 ft3 per Ibm dry air.

ma = V IVI = (2,000 ft3 I min) I (12.48 ft3 /lb mdry air) = 160.3 Ibm dry air per minute From the chart, the enthalpies are hI = 9.2, hi = 19.0 Btu per Ibm dry air. Therefore, the heat added in the heater is:

Q=ma(~ -~)

= (160.3 Ibm dry air per minute)(19.0- 9.2) Btu per Ibm dry air = 1,570 Btu I min =

94,231 Btu I h

The mass flow of steam can be found by considering the conservation of mass in the humidifier:

= msteam + mV3 msteam = mV3 - m v2 m v2

From the chart, W2 = WI

msteam

=

0.0008 and W3

=

0.011 Ibm vapor per Ibm dry air:

= (0.011- 0.0008)(lb mvapor I Ibm dry air)(160J Ibm dry air I minute) = 1.64 Ib steam per minute



15: 5

The enthalpy of the steam can be calculated by applying the First Law of Thermodynamics to the adiabatic humidifier:

From the chart, the enthalpies are h2 = 19.0 and h3 = 31.4 Btu per Ibm dry air: hsteam

= (160.3 / 1.64)(31.4 -19.0)

= 1,212 Btu per Ibm steam 15.3 Cooling Coil Dehumidification Consider the cooling of moist air shown on the psychrometric chart in Figure 15-3. Initially, there is simple cooling at a constant humidity ratio from State 1 to State 2 as the relative humidity increases. When the air becomes saturated at a relative humidity of 100%, any further cooling has to follow the saturated line from State 2 to State 3 as water vapor is condensed to liquid and the humidity ratio decreases. Applications usually call for a final moist air condition that is in the comfort zone and not on the saturation line. Therefore, the overall cooling path includes a final process during which the relative humidity is reduced from 100% to the desired leveL The final process may be one of simple heating or by mixing the saturated air with warmer, non-saturated air. The overall process may be thought of as one in which there is both temperature control through heat removal and relative humidity control through heating.

3

..,

~

'Q"

.1 AIR c::> : I

•

WI

I I

hI

+

.-_----I w" = . •

'Q-":AF--....-r-...... I

Condensate Drain

fa

t2

•

4

W3

tl

Figure 15-3. Cooling Coil With Dehumidification



15:6

Example 15.2. Outdoor air at 95°F and 60% relative humidity is to be cooled to a final condition of 80°F and 50% relative humidity using a cooling coil and a simple heater. Find the intermediate temperature between the cooling coil and the heater, the specific heat removed in the cooling coil, and the specific heat added in the heater. Solution. The solution is best found by drawing all four processes on the psychrometric chart as shown in Figure 15-3. Start with the initial condition and draw a horizontal line across to the saturation line to describe the cooling process 1 to 2. Next draw a horizontal straight line across from the desired final condition to the saturation line. This is the final heating process from State 3 to State 4. The intermediate temperature can be read as t3 = 60°F. It is worth noting that this is also the dewpoint temperature of the final condition.

From the chart, the enthalpies are: hI = 46.3, h3 = 26.4, and h4 = 31.4 Btu per Ibm dry air. The heat transfer may be found by applying the First Law of Thermodynamics to the cooling coil and heater: qcoil

=~-~

= 26.4-46.3 = -19.9 Btu per Ibm dry air (negative means cooling) qheater

= h4 -

~

= 31.4-26.4

= 5.0 Btu per Ibm dry air (positive means heating) In actual cooling coils, the process does not usually follow the idealized process path of constant humidity ratio from 1 to 2, along the saturation line from 2 to 3, and simple heating from 3 to 4. The more practical path is similar to the dotted line shown in Figure 15-3. In the air stream, only the air closest to the coils or fins is cooled to saturation. The state leaving the coil is not on the saturation line because it is a mixture of the parts of the air stream that reach saturation and parts that do not. A reasonable rule of thumb is to assume 90% relative humidity. The air stream is then heated or mixed with another stream to achieve the desired final state.


Fundamentals ofTllermodynamics and Psychrometries

15: 7

15.4 Evaporative Cooling Evaporative cooling is the process that occurs when water is sprayed into an air stream or when air blows over a wet surface. The energy required to evaporate the water must come from somewhere, and if the process is adiabatic, then that energy is taken out of the air stream. It is the same process that was described earlier as happening in an adiabatic saturator. It is a practical cooling method if the air is initially quite dry (low relative humidity) as is often the case in semi-arid areas. It can be used for space cooling provided the final state is not too humid. Conservation of mass applied to the evaporative cooling process is:

mv1 = mw +mv2 mw =mv2 -mv1

Therefore the mass ratio is:

The First Law of Thermodynamics applied to the adiabatic evaporative cooling process is: ma~

= mwhw + ma~ hw = (ma I mw)(~ -~) =(~ -~)/(n;

-ffn

=l:!.hll:!.W That means that evaporative cooling is a process in which the ratio of the enthalpy change to humidity change is constant and equal to the enthalpy of the water being sprayed into the air stream. On the psychrometric chart, the evaporative cooling process is represented by a straight line with its slope determined from the protractor as shown in Figure 15-4. The protractor can be seen to have I:!.hll:!. W printed as a scale around its perimeter. The line drawn by connecting the center of the protractor to the point on the perimeter corresponding to the value of hw has the same slope as a line drawn on the psychrometric chart itself that goes from State 1 to State 2. The slope of the line is fixed by the enthalpy ofthe added water, and the actual location of State 2 along that line is determined by the mass flow rate of added water. There is an upper limit to the evaporative cooling effect when the outlet condition reaches the adiabatic saturation temperature.



15: 8

t

Water spray hwaler

~ ... M1 = hwater

tlw

'Q'lf ,

'0'

_____ ___

2~':::" ~ •

,

#'

o.

I

fiII'It

:

-:'.....

,',

!It

,. ;1'"-..""""-tj£fl11e/

,. • •

W 2

•

w,

~i

0

Figure 15-4. Evaporative Cooling With Water Spray

15.5 Adiabatic Mixing Mixing of two air streams is a common situation in HVAC systems. The air-conditioning of buildings requires a certain amount of mixing of conditioned air with fresh outdoor ventilated air. The physical means of mixing may be brought about in many different ways, but the case of two air streams merging into one is shown in Figure 15-5. There are three states involved: the flow streams are at States I and 2 respectively, and the mixture state is State 3. It is common to assume that mixing is an adiabatic process. The conservation of mass equation can be applied to the dry air and to the water vapor:

mal + ma2 = ma3 W;mal + ~ma2 = W; ma3 The First Law of Thermodynamics applied to the mixing process is:

After substituting for

ma3 from the first equation into each of the other two equations and

then dividing through by

ma2 and solving for mal I ma2 the result is:

mal Ima2 =(~ -W;)/(W; -W;) =(~ -~)/(~ -~)



15:9

The geometrical interpretation of this result is shown on the psychrometric chart in Figure 15-5. State 3 lies on a straight line connecting States 1 and 2, such that the relative distances between the mixed state and each ofthe original unmixed states are proportional to the ratio of the air stream flow rates. Although not exact because there is a slight change in specific heat with humidity ratio, the flow ratio can be reasonably estimated from the rule of proportionalities applied to the dry-bulb temperature differences:

= (t2 -t3)1 (t3 -tl )

mall ma2

Example 15.3. Saturated air leaving the cooling section of an air-conditioning system at 55°F at a rate of 200 Ibm per minute is mixed with outside air at 95°F and 40% relative humidity at a rate of 100 Ibm per minute. Determine the temperature and relative humidity of the mixture.

Solution. The mixed State 3 is most easily located by drawing a straight line between air stream States 1 and 2 and using the dry-bulb temperature calculated from proportionality to locate State 3 on the straight line. From proportionality: mal Ima2 =(t2 -t3)/(t2 -tl ) 2001100 = (95 - t3) 1(t3 - 55) Solving for the temperature results in a mixed temperature of t3 relative humidity is ~ = 63%.

75°F. From the chart, the

=

I

\').

> ,, ,

~').~\ ~').

'(\').

,

..

~

').

·· .. .. .. ·· ..

1I

-- •• ~ ••• ~ •••••••• Wi I

•

Figure 15-5. Adiabatic Mixing of Two Streams of Moist Air



15: 10

15.6 Sensible Heat Ratio In air-conditioning applications, the object is to maintain the temperature and humidity of the conditioned space within certain comfort limits. To maintain steady conditions means removing the heat load due to people, appliances, conduction through walls, solar gain and other various heat and moisture sources. Because both temperature and humidity are important, it is sometimes useful to think of the total load as being made up of two components: the sensible load and the latent load. The sensible load is the heat to be removed to maintain the dry-bulb temperature without changing the humidity ratio. The latent load is the heat to be removed to change the humidity ratio without changing the dry-bulb temperature. The ratio of the sensible heat to total heat (sensible plus latent) is known as the sensible heat ratio (SHR). This idea can also be related to the psychrometric chart. Consider the change of state shown in Figure 15-6. The line may be thought of as a line of constant SHR. By looking at the protractor, it can be seen that the slope of the line on the psychrometric chart is the same as a line drawn from the center of the protractor to a point on the perimeter corresponding to a value of the SHR. In practice, the idea of the SHR may be useful when approached from the perspective that typical values for SHR may be known Figure 15-6. Sensible Heat Ratio from the desired design conditions. With the known value of SHR and using the slope from a line on the protractor, a line can be drawn through the desired setpoint on the psychrometric chart. Any air delivered to the conditioned space with conditions along this lirie will remove heat in the right ratio of sensible to latent change. Example 15.5. The sensible heat load ratio for a cooling problem is known to be SHR = +2.0. The setpoint condition for the conditioned space is 72°F and 50% relative humidity. Find the relative humidity of the 60°F cooling air if the correct ratio of sensible to latent heat is to be removed from the conditioned space.



15: 11

Solution. The positive sign on SHR means that the sensible enthalpy change and the total enthalpy change have the same sign. On the protractor, draw a line from the center to the value of SHR = 2.0 on the perimeter. Draw a parallel line on the psychrometric chart that goes through the state point tdb = 72°F, ~ = 50%. Read the relative humidity for a point on this line at a dry-bulb temperature of 60°F. The relative humidity of the supply air should be ~ = 87%.

Summary Simple heating or cooling at constant humidity ratio is a horizontal straight line on the psychrometric chart for which the heat transfer is given as:

Cooling coil operation with dehumidification is idealized as a two-step process. First there is simple cooling at constant humidity ratio from State 1 to State 2 as the relative humidity increases to 100%, followed by a path along the saturation line from State 2 to State 3 as water vapor is condensed to liquid and the humidity ratio decreases. Applications usually call for a final moist air condition that is in the comfort zone and not on the saturation line. Therefore, the overall cooling path includes a final process during which the relative humidity is reduced to the desired level either by heating or by mixing the saturated air with warmer, non-saturated air. In actual cooling coils, a more practical process path does not reach the saturated condition because there is a mixture of the parts of the air stream that reach saturation and the parts that do not. A reasonable rule of thumb is to assume 90% relative humidity. Evaporative cooling is the process that occurs when water is sprayed into an air stream or when air blows over a wet surface. Adiabatic evaporative cooling is characterized on the psychrometric chart by:

hw = (~-~)/(Wz -~)

=I:1h / I:1W This means that evaporative cooling is a process in which the ratio ofthe enthalpy change to humidity change is constant and equal to the enthalpy ofthe water being sprayed into the air stream. On the psychrometric chart, the evaporative cooling process is represented by a straight line with its slope determined from the protractor.



15: 12

Mixing of two air streams is a common situation in HVAC systems. If the flow streams are at States 1 and 2, respectively, and the mixture is at State 3, then for adiabatic mixing, the mass ratio may be expressed as:

mal Ima2

=(W; -U;;)/(U;; -~) =(hz-~)/(~-~)

=(t2 - t3) I (t3 - t l ) The geometrical interpretation of this result on the psychrometric chart is that State 3 lies on a straight line connecting States 1 and 2, such that the relative distances between the mixed state and each of the original unmixed states are proportional to the ratio of the air stream flow rates. In air-conditioning applications, the object is to maintain the temperature and humidity of the conditioned space within certain comfort limits. To maintain that relatively steady state means removing the heat load due to people, appliances, conduction through walls, solar gain and other heat and moisture sources. The total load may be considered as being the sensible load plus the latent load. The sensible load is the heat to be removed to maintain the dry-bulb temperature without changing the humidity ratio. The latent load is the heat to be removed to change the humidity ratio without changing the dry-bulb temperature. The ratio of the sensible heat to total heat (sensible plus latent) is known as the sensible heat ratio (SHR). With a known value ofSHR and using the slope from a line on the protractor, a line can be drawn through the desired setpoint on the psychrometric chart. Any air delivered to the conditioned space with conditions along this line will remove heat in the right ratio of sensible to latent change to satisfy the load.



15: 13 .


15-01. Moist air is cooled from 80°F, 50% relative humidity to 70°F with no moisture added or removed. What is the final relative humidity and the heat removed perlbm of dry air? 15-02. Moist air is cooled from 80°F, 50% relative humidity to 50°F. What is the final relative humidity, humidity ratio and change of enthalpy? 15-03. 1,000 cfm of outdoor air at 40°F and 20% relative humidity is to be heated and humidified to a final condition of 72°F and 50% relative humidity with a heater followed by a steam humidifier. The temperature after the heater is 64°F. Find the following: the heat added in the heater section; the mass flow rate of steam; and the enthalpy of the steam. 15-04. Outdoor air at 95°F and 50% relative humidity is to be cooled to a final condition of 72°F and 50% relative humidity using a cooling coil and a simple heater. Find the intermediate temperature between the cooling coil and the heater, the specific heat removed in the cooling coil, and the specific heat added in the heater. 15-05. Outdoor air at 95°F and 20% relative humidity is cooled in an evaporative cooler by spraying with saturated liquid water at 70°F. What is the final temperature if the final relative humidity is 60%? (Hint: use the protractor to get the slope of the process line.) How much water will it take per Ibm of dry air? (Hint: use the change in humidity ratio.) 15-06. 20 Ibm per minute of air at 40°F and 90% relative humidity is adiabatically mixed with moist air at 80°F, but unknown relative humidity. The final mixture is at 72°F and 50% relative humidity. What is the relative humidity and mass flow rate ofthe second air stream? 15-07. The sensible heat load ratio for a heating problem is known to be SHR = -2.0. The setpoint condition for the conditioned space is 72°F and 50% relative humidity. Find the relative humidity of the 80°F heating air if the correct ratio of sensible to latent heat is to be removed from the conditioned space.



A:1

Appendix A-I: Dimensions and Units Used in Air-Conditioning Applications

Dimension

SI Unit

IP Unit

Acceleration

mls2

ftlsec 2

Area

m2

ft2

Density

kg/m3

lb m Ift3

Energy

N-m, Joule (J)

Btu, ft-Ib

Force

(kg-m)/s2, Newton (N)

pound (Ibj

Length

m, meter (m)

foot (ft)

Mass

kg, kilogram (kg)

pound mass (Ibm)

Power

J/s, Watt (W)

Btu/h

Pressure

N/m2, Pascal (P)

pSI

Specific Heat

J/(kgeOC)

Btu/(IbmeOF)

Time

second (s)

second (sec)

Temperature (absolute)

degree Kelvin (K)

degree Rankine (R)

Temperature

degree Celsius (OC)

degree Fahrenheit eF)

Thermal Conductivity

W/(meOC) W/m2

Btu/(hefteOF)

ft/sec, ftlmin, fpm

Volume

mls m3

Volume Flow Rate

m 3/s

ft3/sec, ft 3/min, cfm

Thermal Flux Density Velocity


)

Btu/(heft2) ft3

Appendix A

A:2

Appendix A-2: Unit Conversion Factors

Dimension

SI Unit

IP Unit

Length

1 ft = 0.305 m

Area

1 m = 3.281 ft 1 m2= 10.76 ft2

Volume

1 m3= 35.32 ft3

1 ft2 = 0.0929 m2 1 ft3 = 0.0929 m3

Mass

1 kg = 2.205 Ibm

1 Ibm = 0.435 kg

Force

1 N = 0.2248 lbf

1 lbf = 4.448 N

Energy

1 kJ = 0.9478 Btu 1 J = 0.7376 ft-Ibf 1 kWh = 3.412 X 103Btu

1 Btu = 778.2 ft-Ibf = 1.055 kJ 1 ft-Ibf = 1.356 J 1 Btu = 3.183 x 10-2kWh

Specific Energy Specific Enthalpy

1 kJ/kg = 0.4298 Btu/Ibm

1 Btu/Ibm = 2.326 kJ/kg

Power

1 W = 3.412 Btu/h 1 kW = 1.341 hp 1 kW = 0.2844 ton refrigeration

Pressure

1 Pa = 1.450 x 10-4 psi 1 atm = 101 kPa

1 Btu/h = 0.318 W 1 hp = 2545 Btu/h = 0.746 kW 1 ton = 12,000 Btu/h = 3.516 kW 1 psi = 6.897 x 103Pa 1 atm = 14.7 psi = 29.921 in. Hg

1°C diff. = 9/5°F diff. tC = [(9/5)y + 32]OF K = °c + 273.15 Velocity 1 mls = 1.969 x 102ftlmin 1 kglm3= 6.243 x 1O-2Ib m/ft3 Mass Density Mass Flow Rate 1 kgls = 2.205 Ibm/sec 1 kgls = 7.937 x 1031b)h Volume Flow Rate 1 m3/s = 2.119 x 103cfin 1 m3/s = 1.585 x 104 gal/min 1 W/(moOC) = Thermal Temperature

Conductivity Heat Transfer Coefficient Specific Heat

Appendix A

1°F diff. = 5/9°C diff. tF = (y - 32)(5/9)OC R = OF + 459.67 1 ftlmin = 1.602 x 10-3mls Ilbm/ft3 = 1.602 x 103kglm3 1 Ibm/sec = 0.4535 kgls Ilbmlh = 1.260 x lO-4kgls 1 cfm = 4.719 x 10-4 m3/s 1 gal/min = 6.309 x 10-5 m3/s 1 Btu/(hoftoOF) =

0.5778 Btu/(hoftoOF) 1 W/(m2oOC) = 0.1761 Btu/(hoft2 0F)

1.731 W/(mo°C) 1 Btu/(hoft2 0F) = 5.679 W/(m2o°C)

1 J/(kgoOC) = 2.389 x 10-4 Btu/(lbmoaF)

1 Btu/(lbmoOF) = 4.186 x 10-5 J/(kgoOC)

0

0


B: 1

Appendix B-1: Thermodynamic Properties of Water at Saturation Specific Volume, ft311b Temp. t, of

Absolute Pressure p

Sat. SolidILiq.

Evap.

Sat. Vapor

Entropy, Btullb·oF

Enthalpy, Btullb Sat. SolidILiq. hi

Evap.

Sat. Vapor

Sat. SolidILiq.

Evap.

Sat. Vapor

h jg

hg

si

Sig

Sg

Temp., of

psi

in.Hg

vi

Vig

Vg

-80 -79 -78 -77 -76 -75 -74 -73 -72 -71

0.000116 0.000125 0.000135 0.000145 0.000157 0.000169 0.000182 0.000196 0.000211 0.000227

0.000236 0.000254 0.000275 0.000296 0.000319 0.000344 0.000371 0.000399 0.000430 0.000463

0.01732 0.01732 0.01732 0.01732 0.01732 0.01733 0.01733 0.01733 0.01733 0.01733

1953234 1814052 1685445 1566663 1456752 1355059 1260977 1173848 1093149 1018381

1953234 1814052 1685445 1566663 1456752 1355059 1260977 1173848 1093149 1018381

-193.50 -193.11 -192.71 -192.31 -191.92 -191.52 -191.12 -190.72 -190.32 -189.92

1219.19 1219.24 1219.28 1219.33 1219.38 1219.42 1219.47 1219.51 1219.55 1219.59

1025.69 1026.13 1026.57 1027.02 1027.46 1027.90 1028.34 1028.79 1029.23 1029.67

-0.4067 -0.4056 -0.4046 -0.4036 -0.4025 -0.4015 -0.4005 -0.3994 -0.3984 -0.3974

3.2112 3.2029 3.1946 3.1964 3.1782 3.1701 3.1619 3.1539 3.1459 3.1379

2.8045 2.7972 2.7900 2.7828 2.7757 2.7685 2.7615 2.7544 2.7475 2.7405

-80 -79 -78 -77 -76 -75 -74 -73 -72 -71

-70 -69 -68 -67 -66 -65 -64 -63 -62 -61

0.000245 0.000263 0.000283 0.000304 0.000326 0.000350 0.000376 0.000404 0.000433 0.000464

0.000498 0.000536 0.000576 0.000619 0.000664 0.000714 0.000766 0.000822 0.000882 0.000945

0.01733 0.01733 0.01733 0.01734 0.01734 0.01734 0.01734 0.01734 0.01734 0.01734

949067 884803 825187 769864 718508 670800 626503 585316 548041 511446

949067 884803 825187 769864 718508 670800 626503 585316 547041 511446

-189.52 -189.11 -188.71 -188.30 -187.90 -187.49 -187.08 -186.67 -186.26 -185.85

1219.63 1219.67 1219.71 1219.74 1219.78 1219.82 1219.85 1219.88 1219.91 1219.95

1030.11 1030.55 1031.00 1031.44 1031.88 1032.32 1032.77 1033.21 1033.65 1034.09

-0.3963 -0.3953 -0.3943 -0.3932 -0.3922 -0.3912 -0.3901 -0.3891 -0.3881 -0.3870

3.1299 3.1220 3.1141 3.1063 3.0985 3.0907 3.0830 3.0753 3.0677 3.0601

2.7336 2.7267 2.7199 2.7131 2.7063 2.6996 2.6929 2.6862 2.6730 2.6730

-70 -69 -68 -67 -66 -65 -64 -63 -62 -61

-60 -59 -58 -57 -56 -55 -54 -53 -52 -51

0.000498 0.000533 0.000571 0.000612 0.000655 0.000701 0.000750 0.000802 0.000857 0.000916

0.001013 0.001086 0.001163 0.001246 0.001333 0.001427 0.001526 0.001632 0.001745 0.001865

0.01734 0.01735 0.01735 0.01735 0.01735 0.01735 0.01735 0.01735 0.01735 0.01736

478317 447495 418803 392068 367172 343970 322336 302157 283335 265773

478317 447495 418803 392068 367172 343970 322336 302157 283335 265773

-185.44 -185.03 -184.61 -184.20 -183.78 -183.37 -182.95 -182.53 -182.11 -181.69

1219.98 1220.01 1220.03 1220.06 1220.09 1220.11 1220.14 1220.16 1220.18 1220.21

1034.54 1034.98 1035.42 1035.86 1036.30 1036.75 1037.19 1037.63 1038.07 1038.52

-0.3860 -0.3850 -0.3839 -0.3829 -0.3819 -0.3808 -0.3798 -0.3788 -0.3778 -0.3767

3.0525 3.0449 3.0374 3.0299 3.0225 3.0151 3.0077 3.0004 2.9931 2.9858

2.6665 2.6600 2.6535 2.6470 2.6406 2.6342 2.6279 2.6216 2.6153 2.6091

-60 -59 -58 -57 -56 -55 -54 -53 -52 -51

-50 -49 -48 -47 -46 -45

0.001992 0.002128 0.002272 0.002425 0.002587 0.002760 0.002943 0.003137 0.003343 0.003562

0.01736 0.01736 0.01736 0.01736 0.01736 0.01736 0.01736 0.01737 0.01737 0.01737

249381 234067 219766 206398 193909 182231 171304 161084 151518 142566

249381 234067 219766 206398 193909 182231 171304 161084 151518 142566

-181.27 -180.85 -180.42 -180.00 -179.57 -179.14 -178.72 -178.79 -177.86 -177.43

1220.23 1220.25 1220.26 1220.28 1220.30 1220.31 1220.33 1220.34 1220.36 1220.37

1038.96 1039.40 1039.84 1040.28 1040.73 1041.17 1041.61 1042.05 1042.50 1042.94

-0.3757 -0.3747 -0.3736 -0.3726 -0.3716 -0.3705 -0.3695 -0.3685 -0.3675 -0.3664

2.9786 2.9714 2.9642 2.9570 2.9499 2.9429 2.9358 2.9288 2.9218 2.9149

2.6029 2.5967 2.5906 2.5844 2.5784 2.5723 2.5663 2.5603 2.5544 2.5485

-50 -49 -48 -47 -46 -45

-43 -42 -41

0.000979 0.001045 0.001116 0.001191 0.001271 0.001355 0.001445 0.001541 0.001642 0.001749

-40 -39 -38 -37 -36 -35 -34 -33 -32 -31

0.001863 0.001984 0.002111 0.002247 0.002390 0.002542 0.002702 0.002872 0.003052 0.003242

0.003793 0.004039 0.004299 0.004574 0.004866 0.005175 0.005502 0.005848 0.006213 0.006600

0.01737 0.01737 0.01737 0.01737 0.01738 0.01738 0.01738 0.01738 0.01738 0.01738

134176 126322 118959 112058 105592 99522 93828 88489 83474 78763

134176 126322 118959 112058 105592 99522 93828 88489 83474 78763

-177.00 -176.57 -176.13 -175.70 -175.26 -174.83 -174.39 -173.95 -173.51 -173.07

1220.38 1220.39 1220.40 1220.40 1220.41 1220.42 1220.42 1220.43 1220.43 1220.43

1043.38 1043.82 1044.27 1044.71 1045.15 1045.59 1046.03 1046.48 1046.92 1047.36

-0.3654 -0.3644 -0.3633 -0.3623 -0.3613 -0.3603 -0.3592 -0.3582 -0.3572 -0.3561

2.9080 2.9011 2.8942 2.8874 2.8806 2.8738 2.8671 2.8604 2.8537 2.8470

2.5426 2.5367 2.5309 2.5251 2.5193 2.5136 2.5078 2.5022 2.4965 2.4909

-40 -39 -38 -37 -36 -35 -34 -33 -32 -31

-30 -29 -28 -27 -26 -25 -24 -23 -22 -21

0.003443 0.003655 0.003879 0.004116 0.004366 0.004630 0.004909 0.005203 0.005514 0.005841

0.007009 0.007441 0.007898 0.008380 0.008890 0.009428 0.009995 0.010594 0.011226 0.011892

0.01738 0.01738 0.01739 0.01739 0.01739 0.01739 0.01739 0.01739 0.01739 0.01740

74341 70187 66282 62613 59161 55915 52861 49986 47281 44733

74341 70187 66282 62613 59161 55915 52861 49986 47281 44733

-172.63 -172.19 -171.74 -171.30 -170.86 -170.41 -169.96 -169.51 -169.07 -168.62

1220.43 1220.43 1220.43 1220.43 1220.43 1220.42 1220.42 1220.41 1220.41 1220.40

1047.80 1048.25 1048.69 1049.13 1049.57 1050.01 1050.46 1050.90 1051.34 1051.78

-0.3551 -0.3541 -0.3531 -0.3520 -0.3510 -0.3500 -0.3489 -0.3479 -0.3469 -0.3459

2.8404 2.8338 2.8272 2.8207 2.8142 2.8077 2.8013 2.7948 2.7884 2.7820

2.4853 2.4797 2.4742 2.4687 2.4632 2.4577 2.4523 2.4469 2.4415 2.4362

-30 -29 -28 -27 -26 -25 -24 -23 -22 -21

-20 -19 -18 -17 -16 -15 -14

0.006186 0.006550 0.006933 0.007337 0.007763 0.008211 0.008683

0.012595 0.013336 0.014117 0.014939 0.015806 0.016718 0.017678

0.01740 0.01740 0.01740 0.01740 0.01740 0.01740 0.01741

42333 40073 37943 35934 34041 32256 30572

42333 40073 37943 35934 34041 32256 30572

-168.16 -167.71 -167.26 -166.81 -166.35 -165.90 -165.44

1220.39 1220.38 1220.37 1220.36 1220.34 1220.33 1220.31

1052.22 1052.67 1053.11 1053.55 1053.99 1054.43 1054.87

-0.3448 -0.3438 -0.3428 -0.3418 -0.3407 -0.3397 -0.3387

2.7757 2.7694 2.7631 2.7568 2.7506 2.7444 2.7382

2.4309 2.4256 2.4203 2.4151 2.4098 2.4046 2.3995

-20 -19 -18 -17 -16 -15 -14

-44

Fundamentals ofTllermot/ynamics and Psycllrometries

-44 -43 -42 -41

AppendixB

8:2 Specific Volume, ft3IIb Temp. t, of

Absolute Pressure p

Sat. SolidILiq.

Entbalpy, BtulIb

Evap.

Sat. Vapor

Sat. SolidILiq. hi

Entropy, BtulIb·OF

Evap.

Sat. Vapor

Sat. SolidILiq.

Evap.

Sat. Vapor

hig

hg

si

Sig

Sg

Temp., of

1220.30 1220.28 1220.26 1220.24 1220.22 1220.20 1220.18 1220.16 1220.13 1220.11 1220.08 1220.05 1220.02

1055.32 1055.76 1056.20 1056.64 1057.08 1057.53 1057.97 1058.41 1058.85 1059.29 1059.73 1060.17 1060.62

-0.3377 -0.3366 -0.3356 -0.3346 -0.3335 -0.3325 -0.3315 -0.3305 -0.3294 -0.3284 -0.3274 -0.3264 -0.3253

2.7320 2.7259 2.7197 2.7136 2.7076 2.7015 2.6955 2.6895 2.6836 2.6776 2.6717 2.6658 2.6599

2.3943 2.3892 2.3841 2.3791 2.3740 2.3690 2.3640 2.3591 2.3541 2.3492 2.3443 2.3394 2.3346

-13 -12 -II -10 -9 -8 -7 -6 -5 -4 -3 -2 -I

1220.00 1061.06 1219.96 1061.50 1219.93 1061.94 12J9.90 1062.38 1219.87 1062.82 1219.83 1063.26 1219.80 1063.70 1219.76 1064.14 1219.72 1064.58 -1219.68 -1065.03

-0.3243 -0.3233 -0.3223 -0.3212 -0.3202 -0.3192 -0.3182 -0.3171 -0.3161 -0.3151

2.6541 2.6482 2.6424 2.6367 2.6309 2.6252 2.6194 2.6138 2.6081 -2.6024

2.3298 2.3249 2.3202 2.3154 2.3107 2.3060 2.3013 2.2966 2.2920 -2.2873

0 I 2 3 4 5 6 7 8 9

psi

in.Hg

vi

Vig

Vg

-13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -I

0.009179 0.009702 0.010252 0.010830 0.011438 0.012077 0.012749 0.013456 0.014197 0.014977 0.015795 0.016654 0.017556

0.018689 0.019753 0.020873 0.022050 0.023288 0.024590 0.025958 0.027396 0.028906 0.030493 0.032159 0.033908 0.035744

0.01741 0.01741 0.01741 0.01741 0.01741 0.01741 0.01742 0.01742 0.01742 0.01742 0.01742 0.01742 0.01742

28983 27483 26067 24730 23467 22274 21147 20081 19074 18121 17220 16367 15561

28983 27483 26067 24730 23467 22274 21147 20081 19074 18121 17220 16367 15561

-164.98 -164.52 -164.06 -163.60 -163.14 -162.68 -162.21 -162.75 -161.28 -160.82 -160.35 -159.88 -159.41

0 I 2 3 4 5 6 7 8 9

0.018502 0.019495 0.020537 0.021629 0.022774 0.023975 0.025233 0.026552 0.027933 0.029379

0.037671 0.039693 0.041813 0.044037 0.046369 0.048813 0.051375 0.054059 0.056872 0.059817

0.01743 0.01743 0.01743 0.01743 0.01743 0.01743 0.01743 0.01744 0.01744 0.01744

14797 14073 13388 12740 12125 11543 10991 10468 9971 9500

14797 14073 13388 12740 12125 11543 10991 10468 9971 9500

-158.94 -158.47 -157.99 -157.52 -157.05 -156.57 -156.09 -155.62 -155.14 -154.66

10 II 12 13 14 15 16 17 18 19

0.030894 0.032480 0.034140 0.035878 0.037696 0.039597 0.041586 0.043666 0.045841 0.048113

0.062901 0.066131 0.069511 0.073047 0.076748 0.080621 0.084671 0.088905 0.093332 0.097960

0.01744 0.01744 0.01744 0.01745 0.01745 0.01745 0.01745 0.01745 0.01745 0.01745

9054 8630 8228 7846 7483 7139 6811 6501 6205 5924

9054 8630 8228 7846 7483 7139 6811 6501 6205 5924

-154.18 -153.70 -153.21 -152.73 -152.24 -151.76 -151.27 -150.78 -150.30 -149.81

1219.64 1219.60 1219.56 1219.52 1219.47 1219.43 1219.38 1219.33 1219.28 1219.23

1065.47 1065.91 1066.35 1066.79 1067.23 1067.67 1068.11 1068.55 1068.99 1069.43

-0.3141 -0.3130 -0.3120 -0.3110 -0.3100 -0.3089 -0.3079 -0.3069 -0.3059 -0.3049

2.5968 2.5912 2.5856 2.5801 2.5745 2.5690 2.5635 2.5580 2.5526 2.5471

2.2827 2.2782 2.2736 2.2691 2.2645 2.2600 2.2556 2.2511 2.2467 2.2423

10 II 12 13 14 15 16 17 18 19

20 21 22 23 24 25 26 27 28 29

0.050489 0.052970 0.055563 0.058271 0.061099 0.064051 0.067133 0.070349 0.073706 0.077207

0.102796 0.107849 0.113128 0.118641 0.124398 0.130408 0.136684 0.143233 0.150066 0.157195

0.01746 0.01746 0.01746 0.01746 0.01746 0.01746 0.01747 0.01747 0.01747 0.01747

5657 5404 5162 4932 4714 4506 4308 4119 3940 3769

5657 5404 5162 4932 4714 4506 4308 4119 3940 3769

-149.32 -148.82 -148.33 -147.84 -147.34 -146.85 -146.35 -145.85 -145.35 -144.85

1219.18 1219.13 1219.08 1219.02 1218.97 1218.91 1218.85 1218.80 1218.74 1218.68

1069.87 1070.31 1070.75 1071.19 1071.63 1072.07 1072.50 1072.94 1073.38 1073.82

-0.3038 -0.3028 -0.3018 -0.3008 -0.2997 -0.2987 -0.2977 -0.2967 -0.2956 -0.2946

2.5417 2.5363 2.5309 2.5256 2.5203 2.5149 2.5096 2.5044 2.4991 2.4939

2.2379 2.2335 2.2292 2.2248 2.2205 2.2162 2.2119 2.2077 2.2035 2.1992

20 21 22 23 24 25 26 27 28 29

30 31 32 32* 33 34 35 36 37 38 39

0.080860 0.084669 0.088640 0.08865 0.09229 0.09607 0.09998 0.10403 0.10822 0.11257 0.11707

0.164632 0.172387 0.180474 0.18049 0.18791 0.19559 0.20355 0.21180 0.22035 0.22919 0.23835

0.01747 0.01747 0.01747 0.01602 0.01602 0.01602 0.01602 0.01602 0.01602 0.01602 0.01602

3606 3450 3302 3302.07 3178.15 3059.47 2945.66 2836.60 2732.13 2631.88 2535.86

3606 3450 3302 3302.09 3178.16 3059.49 2945.68 2836.61 2732.15 2631.89 2535.88

-144.35 -143.85 -143.35 -0.02 0.99 2.00 3.00 4.01 5.02 6.02 7.03

1218.61 1218.55 1218.49 1075.15 1074.59 1074.02 1073.45 1072.88 1072.32 1071.75 1071.18

1074.26 1074.70 1075.14 1075.14 1075.58 1076.01 1076.45 1076.89 1077.33 1077.77 1078.21

-0.2936 -0.2926 -0.2915 0.0000 0.0020 0.0041 0.0061 0.0081 0.0102 0.0122 0.0142

2.4886 2.4834 2.4783 2.1867 2.1811 2.1756 2.1700 2.1645 2.1590 2.1535 2.1481

2.1951 2.1909 2.1867 2.1867 2.1832 2.1796 2.1761 2.1726 2.1692 2.1657 2.1623

30 31 32 32 33 34 35 36 37 38 39

40 41 42 43 44 45 46 47 48 49 50 51 52

0.12172 0.12654 0.13153 0.13669 0.14203 0.14755 0.15326 0.15917 0.16527 0.17158 0.17811 0.18484 0.19181

0.24783 0.25765 0.26780 0.27831 0.28918 0.30042 0.31205 0.32407 0.33650 0.34935 0.36263 0.37635 0.39053

0.01602 0.01602 0.01602 0.01602 0.01602 0.01602 0.01602 0.01602 0.01602 0.01602 0.01602 0.01602 0.01603

2443.67 . 2355.22 2270.42 2189.02 2110.92 2035.91 1963.85 1894.71 1828.28 1764.44 1703.18 1644.25 1587.64

2443.69 2355.24 2270.43 2189.04 2110.94 2035.92 1963.87 1894.73 1828.30 1764.46 1703.20 1644.26 1587.65

8.03 9.04 10.04 11.04 12.05 13.05 14.05 15.06 16.06 17.06 18.06 19.06 20.07

1070.62 1070.05 1069.48 1068.92 1068.35 1067.79 1067.22 1066.66 1066.09 1065.53 1064.96 1064.40 1063.83

1078.65 1079.09 1079.52 1079.96 1080.40 1080.84 1081.28 1081.71 1082.15 1082.59 1083.03 1083.46 1083.90

0.0162 0.0182 0.0202 0.0222 0.0242 0.0262 0.0282 0.0302 0.0321 0.0341 0.0361 0.0381 0.0400

2.1426 2.1372 2.1318 2.1265 2.1211 2.1158 2.1105 2.1052 2.1000 2.0947 2.0895 2.0843 2.0791

2.1589 2.1554 2.1521 2.1487 2.1454 2.1420 2.1387 2.1354 2.1321 2.1288 2.1256 2.1224 2.1191

40 4J 42 43 44 45 46 47 48 49 50 51 52

*Extrapolated to represent metastable equilibrium with undercooled liquid.

AppendixB

Fundamentals ofTilermodynamics and Psycilrometrics

B: 3 Specific Volume, ft31lb Temp. t, of

Absolute Pressure p

Entropy, Btullb·oF

Enthalpy, Btullb

Sat. SolidILiq.

Evap.

Sat. Vapor

Sat. SolidILiq.

Evap.

Sat. Vapor

Sat. SolidILiq.

Evap.

Sat. Vapor

hi

hig

hg

si

Sig

Sg

Temp., of

psi

in.Hg

vi

Vig

Vg

53 54 55 56 57 58 59

0.19900 0.20643 0.21410 0.22202 0.23020 0.23864 0.24735

0.40516 0.42029 0.43591 0.45204 0.46869 0.48588 0.50362

0.01603 0.01603 0.01603 0.01603 0.01603 0.01603 0.01603

1533.22 1480.89 1430.61 1382.19 1335.65 1290.85 1247.76

1533.24 1480.91 1430.62 1382.21 1335.67 1290.87 1247.78

21.07 22.07 23.07 24.07 25.07 26.07 27.07

1063.27 1062.71 1062.14 1061.58 1061.01 1060.45 1059.89

1084.34 1084.77 1085.21 1085.65 1086.08 1086.52 1086.96

0.0420 0.0439 0.0459 0.0478 0.0497 0.0517 0.0536

2.0740 2.0689 2.0637 2.0586 2.0536 2.0485 2.0435

2. \159 2.1128 2.1096 2,1064 2.1033 2.0002 2.0971

53 54 55 56 57 58 59

60 61 62 63 64 65 66 67 68 69

0.25635 0.26562 0.27519 0.28506 0.29524 0.30574 0.31656 0.32772 0.33921 0.35107

0.52192 0.54081 0.56029 0.58039 0.60\12 0.62249 0.64452 0.66724 0.69065 0.71478

0.01604 0.01604 0.01604 0.01604 0.01604 0.01604 0.01604 0.01605 0.01605 0.01605

1206.30 1166.38 1127.93 1090.94 1055.32 1020.98 987.95 956.11 925.44 895.86

1206.32 1166.40 1127.95 1090.96 1055.33 1021.00 987.97 956.12 925.45 895.87

28.07 29.07 30.07 31.07 32.07 33.07 34.07 35.07 36.07 37.07

1059.32 1058.76 1058.19 1057.63 1057.07 1056.50 1055.94 1055.37 1054.81 1054.24

1087.39 1087.83 1088.27 1088.70 1089.14 1089.57 1090.01 1090.44 1090.88 1091.31

0.0555 0.0575 0.0594 0.0613 0.0632 0.0651 0.0670 0.0689 0.0708 0.0727

2.0385 2.0334 2.0285 2.0235 2.0186 2.0136 2.0087 2.0039 1.9990 1.9941

2.0940 2.0909 2.0878 2.0848 2.0818 2.0787 2.0758 2.0728 2.0698 2.0668

60 61 62 63 64 65 66 67 68 69

70 71 72 73 74 75 76 77 78 79

0.36328 0.37586 0.38882 0.40217 0.41592 0.43008 0.44465 0.45966 0.47510 0.49100

0.73964 0.76526 0.79164 0.81883 0.84682 0.87564 0.90532 0.93587 0.96732 0.99968

0.01605 0.01605 0.01606 0.01606 0.01606 0.01606 0.01606 0.01607 0.01607 0.01607

867.34 839.87 813.37 787.85 763.19 739.42 716.51 694.38 673.05 652.44

867.36 839.88 813.39 787.87 763.21 739.44 726.53 794.40 673.06 652.46

38.07 39.07 40.07 41.07 42.06 43.06 44.06 45.06 46.06 47.06

1053.68 1053.11 1052.55 1051.98 1051.42 1050.85 1050.29 1049.72 1049.16 1048.59

1091.75 1092.18 1092.61 1093.05 1093.48 1093.92 1094.35 1094.78 1095.22 1095.65

0.0746 0.0765 0.0783 0.0802 0.0821 0.0840 0.0858 0.0877 0.0896 0.0914

1.9893 1.9845 1.9797 1.9749 1.9702 1.9654 1.9607 1.9560 1.9513 1.9466

2.0639 2.0610 2.0580 2.0552 2.0523 2.0494 2.0465 2.0437 2.0409 2.0380

70 71 72 73 74 75 76 77 78 79

80 81 82 83 84 85 86 87 88 89

0.50736 0.52419 0.54150 0.55931 0.57763 0.59647 0.61584 0.63575 0.65622 0.67726

1.03298 1.06725 1.10250 1.13877 1.17606 1.21442 1.25385 1.29440 1.33608 1.37892

0.01607 0.01608 0.01608 0.01608 0.01608 0.01609 0.01609 0.01609 0.01609 0.01610

632.54 613.35 594.82 576.90 559.63 542.93 526.80 511.21 496.14 481.60

632.56 613.37 594.84 576.92 559.65 542.94 526.81 511.22 496.15 481.61

48.06 49.06 50.05 51.05 52.05 53.05 54.05 55.05 56.05 57.04

1048.03 1047.46 1046.89 1046.33 1045.76 1045.19 1044.63 1044.06 1043.49 1042.92

1096.08 1096.51 1096.95 1097.38 1097.81 1098.24 1098.67 1099.11 1099.54 1099.97

0.0933 0.0951 0.0970 0.0988 0.1006 0.1025 0.1043 0.1061 0.1080 0.1098

1.9420 1.9373 1.9327 1.9281 1.9235 1.9189 1.9144 1.9098 1.9053 1.9008

2.0352 2.0324 2.0297 2.0269 2.0242 2.0214 2.0187 2.0160 2.0133 2.0106

80 81 82 83 84 85 86 87 88 89

90 91 92 93 94 95 96 97 98 99

0.69889 0.72111 0.74394 0.76740 0.79150 0.81625 0.84166 0.86776 0.89456 0.92207

1.42295 1.46820 1.51468 1.56244 1.61151 1.66189 1.71364 1.76678 1.82134 1.87736

0.01610 0.01610 0.01611 0.01611 0.01611 0.01612 0.01612 0.01612 0.01612 0.01613

467.52 453.91 440.76 428.04 415.74 403.84 392.33 381.20 370.42 359.99

467.53 453.93 440.78 428.06 415.76 403.86 392.34 381.21 370.44 360.01

58.04 59.04 60.04 61.04 62.04 63.03 64.03 65.03 66.03 67.03

1042.36 1041.79 1041.22 1040.65 1040.08 1039.51 1038.95 1038.38 1037.81 1037.24

1100.40 1100.83 1101.26 1101.69 1102.12 1102.55 1102.98 1103.41 1103.84 1104.26

0.1 \16 0.1134 0.1152 0.1170 0.1188 0.1206 0.1224 0.1242 0.1260 0.1278

1.8963 1.8918 1.8874 1.8829 1.8785 1.8741 1.8697 1.8653 1.8610 1.8566

2.0079 2.0053 2.0026 2.0000 1.9973 1.9947 1.9921 1.9895 1.9870 1.9844

90 91 92 93 94 95 96 97 98 99

100 101 102 103 104 105 106 107 108 109

0.95031 0.97930 1.00904 1.03956 1.07088 1.10301 1.13597 1.16977 1.20444 1.23999

1.93485 1.99387 2.05443 2.11667 2.18034 2.24575 2.31285 2.38168 2.45226 2.52464

0.01613 0.01613 0.01614 0.01614 0.01614 0.01615 0.01615 0.01616 0.01616 0.01616

349.91 340.14 330.69 321.53 312.67 304.08 295.76 287.71 279.91 272.34

349.92 340.15 330.71 321.55 312.69 304.10 295.77 287.73 279.92 272.36

68.03 69.03 70.02 71.02 72.02 73.02 74.02 75.01 76.01 77.01

1036.67 1036.10 1035.53 1034.95 1034.38 1033.81 1033.24 1032.67 1032.10 1031.52

1104.69 1105.12 1105.55 1105.98 1106.40 1106.83 1107.26 1107.68 1108.11 1108.54

0.1296 0.1314 0.1332 0.1349 0.1367 0.1385 0.1402 0.1420 0.1438 0.1455

1.8523 1.8479 1.8436 1.8393 1.8351 1.8308 1.8266 1.8223 1.8181 1.8139

1.9819 1.9793 1.9768 1.9743 1.9718 1.9693 1.9668 1.9643 1.9619 1.9594

100 101 102 103 104 105 106 107 108 109

110 III 112 113 114 \15 116 117 118 119 120

1.27644 1.31381 1.35212 1.39138 1.43162 1.47286 1.51512 1.55842 1.60277 1.64820 1.69474

2.59885 2.67494 2.75293 2.83288 2.91481 2.99878 3.08481 3.17296 3.26327 3.35577 3.45052

0.01617 0.01617 0.01617 0.01618 0.01618 0.01619 0.01619 0.01619 0.01620 0.01620 0.01620

265.02 257.91 251.02 244.36 237.89 231.62 225.53 219.63 213.91 208.36 202.98

265.03 257.93 251.04 244.38 237.90 231.63 225.55 219.65 213.93 208.37 202.99

78.01 79.01 80.01 81.01 82.00 83.00 84.00 85.00 86.00 87.00 88.00

1030.95 1030.38 1029.80 1029.23 1028.66 1028.08 1027.51 1026.93 1026.36 1025.78 1025.20

1108.96 1109.39 1109.81 1110.24 \110.66 1111.09 1111.51 \111.93 1112.36 1112.78 1113.20

0.1473 0.1490 0.1508 0.1525 0.1543 0.1560 0.1577 0.1595 0.1612 0.1629 0.1647

1.8097 1.8055 1.8014 1.7972 1.7931 1.7890 1.7849 1.7808 1.7767 1.7726 1.7686

1.9570 1.9546 1.9521 1.9497 1.9474 1.9450 1.9426 1.9402 1.9379 1.9356 1.9332

\10 111 112 113 114 115 116 117 118 119 120


AppendixB

8:4 Specific Volume, ft31lb Temp. t, of

Absolute Pressure p

Entropy, Btullb·oF

Enthalpy, Btullb

Sat. SolidILiq.

Evap.

Sat. Vapor

Sat. SolidILiq.

Evap.

Sat. Vapor

Sat. SolidILiq.

Evap.

Sat. Vapor

si

Sig

Sg

Temp., of

psi

in.Hg

Vi

Vig

Vg

hi

h ig

hg

121 122 123 124 125 126 127 128 129

1.74240 1.79117 1.84117 1.89233 1.94470 1.99831 2.05318 2.10934 2.16680

3.54755 3.64691 3.74863 3.85282 3.95945 4.06860 4.18032 4.29465 4.41165

0.01621 0.01621 0.01622 0.01622 0.01623 0.01623 0.01623 0.01624 0.01624

197.76 192.69 187.78 182.98 178.34 173.85 169.47 165.23 161.11

197.76 192.69 187.78 182.99 178.36 173.86 169.49 165.25 161.12

89.00 90.00 90.99 91.99 92.99 93.99 94.99 95.99 96.99

1024.63 1024.05 1023.47 1022.90 1022.32 1021.74 1021.16 1020.58 1020.00

1113.62 1114.05 1114.47 1114.89 1115.31 1115.73 1I16.15 1116.57 II 16.99

0.1664 0.1681 0.1698 0.1715 0.1732 0.1749 0.1766 0.1783 0.1800

1.7645 1.7605 1.7565 1.7525 1.7485 1.7445 1.7406 1.7366 1.7327

1.9309 1.9286 1.9263 1.9240 1.9217 1.9195 1.9172 1.9150 1.9127

121 122 123 124 125 126 127 128 129

130 131 132 133 134 135 136 137 138 139

2.22560 2.28576 2.34730 2.41025 2.47463 2.54048 2.60782 2.67667 2.74707 2.81903

4.53136 4.65384 4.77914 4.90730 5.03839 5.17246 5.30956 5.44975 5.59308 5.73961

0.01625 0.01625 0.01626 0.01626 0.01627 0.01627 0.01627 0.01628 0.01628 0.01629

157.11 153.22 149.44 145.77 142.21 138.74 135.37 132.10 128.92 125.83

157.12 153.23 149.46 145.78 142.23 138.76 135.39 132.12 128.94 125.85

97.99 98.99 99.99 100.99 101.99 102.99 103.98 104.98 105.98 106.98

1019.42 1018.84 1018.26 1017.68 1017.10 1016.52 1015.93 1015.35 1014.77 1014.18

1117.41 II 17.83 II 18.25 1118.67 1119.08 1119.50 11I9.92 1120.34 1120.75 1121.17

0.1817 0.1834 0.1851 0.1868 0.1885 0.1902 0.1919 0.1935 0.1952 0.1969

1.7288 1.7249 1.7210 1.7171 1.7132 1.7093 1.7055 1.7017 1.6978 1.6940

1.9105 1.9083 1.9061 1.9039 1.9017 1.8995 1.8974 1.8952 1.8930 1.8909

130 131 132 133 134 135 136 137 138 139

140 141 142 143 144 145 146 147 148 149

2.89260 2.96780 3.04465 3.12320 3.20345 3.28546 3.36924 3.45483 3.54226 3.63156

5.88939 6.04250 6.19897 6.35888 6.52229 6.68926 6.85984 7.03410 7.21211 7.39393

0.01629 0.01630 0.01630 0.01631 0.01631 0.01632 0.01632 0.01633 0.01633 0.01634

122.82 119.90 117.05 114.29 111.60 108.99 106.44 103.96 101.55 99.21

122.84 II 9.92 II 7.07 114.31 II 1.62 109.00 106.45 103.98 101.57 99.22

107.98 108.98 109.98 110.98 111.98 112.98 113.98 114.98 115.98 116.98

1013.60 1013.01 1012.43 1011.84 1011.26 1010.67 1010.09 1009.50 1008.91 1008.32

1121.58 1122.00 1122.41 1122.83 1123.24 1123.66 1124.07 1124.48 1124.89 1125.31

0.1985 0.2002 0.2019 0.2035 0.2052 0.2068 0.2085 0.2101 0.2118 0.2134

1.6902 1.6864 1.6827 1.6789 1.6752 1.6714 1.6677 1.6640 1.6603 1.6566

1.8888 1.8867 1.8845 1.8824 1.8803 1.8783 1.8762 1.8741 1.8721 1.8700

140 141 142 143 144 145 146 147 148 149

150 151 152 153 154 155 156 157 158 159

3.72277 3.81591 3.91101 4.00812 4.10727 4.20848 4.31180 4.41725 4.52488 4.63472

7.57962 7.76925 7.96289 8.16061 8.36247 8.56854 8.77890 8.99360 9.21274 9.43637

0.01634 0.01635 0.01635 0.01636 0.01636 0.01637 0.01637 0.01638 0.01638 0.01639

96.93 94.70 92.54 90.44 88.39 86.40 84.45 82.56 80.72 78.92

96.94 94.72 92.56 90.46 88.41 86.41 84.47 82.58 80.73 78.94

117.98 118.99 119.99 120.99 121.99 122.99 123.99 124.99 125.99 126.99

1007.73 1007.14 1006.55 1005.96 1005.37 1004.78 1004.19 1003.60 1003.00 1002.41

1125.72 1126.13 1126.54 1126.95 1127.36 1127.77 1128.18 1128.59 1128.99 1129.40

0.2151 0.2167 0.2184 0.2200 0.2216 0.2233 0.2249 0.2265 0.2281 0.2297

1.6529 1.6492 1.6455 1.6419 1.6383 1.6346 1.6310 1.6274 1.6238 1.6202

1.8680 1.8659 1.8639 1.8619 1.8599 1.8579 1.8559 1.8539 1.8519 1.8500

150 151 152 153 154 155 156 157 158 159

160 161 162 163 164 165 166 167 168 169

4.7468 4.8612 4.9778 5.0969 5.2183 5.3422 5.4685 5.5974 5.7287 5.8627

9.6646 9.8974 10.1350 10.3774 10.6246 10.8768 11.1340 11.3963 11.6638 11.9366

0.01639 0.01640 0.01640 0.01641 0.01642 0.01642 0.01643 0.01643 0.01644 0.01644

77.175 75.471 73.812 72.196 70.619 69.084 67.587 66.130 64.707 63.320

77.192 75.488 73.829 72.213 70.636 69.101 67.604 66.146 64.723 63.336

127.99 128.99 130.00 131.00 132.00 133.00 134.00 135.00 136.01 137.01

1001.82 1001.22 1000.63 1000.03 999.43 998.84 998.24 997.64 997.04 996.44

1129.81 1130.22 1130.62 1131.03 1131.43 1131.84 1132.24 1132.64 1133.05 1133.45

0.2314 0.2330 0.2346 0.2362 0.2378 0.2394 0.2410 0.2426 0.2442 0.2458

1.6167 1.6131 1.6095 1.6060 1.6025 1.5989 1.5954 1.5919 1.5884 1.5850

1.8480 1.8461 1.8441 1.8422 1.8403 1.8383 1.8364 1.8345 1.8326 1.8308

160 161 162 163 164 165 166 167 168 169

170 171 172 173 174 175 176 177 178 179

5.9993 6.1386 6.2806 6.4253 6.5729 6.7232 6.8765 7.0327 7.1918 7.3539

12.2148 12.4983 12.7874 13.0821 13.3825 13.6886 14.0006 14.3186 14.6426 14.9727

0.01645 0.01646 0.01646 0.01647 0.01647 0.01648 0.01648 0.01649 0.01650 0.01650

61.969 60.649 59.363 58.112 56.887 55.694 54.532 53.397 52.290 51.210

61.986 60.666 59.380 58.128 56.904 55.711 54.549 53.414 52.307 51.226

138.01 139.01 140.oI 141.02 142.02 143.02 144.02 145.03 146.03 147.03

995.84 995.24 994.64 994.04 993.44 992.83 992.23 991.63 991.02 990.42

1133.85 1134.25 1134.66 1135.06 1135.46 1135.86 1136.26 1136.65 1137.05 1137.45

0.2474 0.2490 0.2506 0.2521 0.2537 0.2553 0.2569 0.2585 0.2600 0.2616

1.5815 1.5780 1.5746 1.5711 1.5677 1.5643 1.5609 1.5575 1.5541 1.5507

1.8289 1.8270 1.8251 1.8233 1.8214 1.8196 1.8178 1.8159 1.8141 1.8123

170 171 172 173 174 175 176 177 178 179

180 181 182 183 184 185 186 187 188 189

7.5191 7.6874 7.8589 8.0335 8.2114 8.3926 8.5770 8.7649 8.9562 9.1510

15.3091 15.6518 16.0008 16.3564 16.7185 17.0874 17.4630 17.8455 18.2350 18.6316

0.01651 0.01651 0.01652 0.01653 0.01653 0.01654 0.01654 0.01655 0.01656 0.01656

50.155 49.126 48.122 47.142 46.185 45.251 44.339 43.448 42.579 41.730

50.171 49.143 48.138 47.158 46.202 45.267 44.356 43.465 42.595 41.746

148.04 149.04 150.04 151.05 152.05 153.05 154.06 155.06 156.07 157.07

989.81 989.20 988.60 987.99 987.38 986.77 986.16 985.55 984.94 984.32

1137.85 1138.24 1138.64 1139.03 1139.43 1139.82 1140.22 1140.61 1141.00 1141.39

0.2632 0.2647 0.2663 0.2679 0.2694 0.2710 0.2725 0.2741 0.2756 0.2772

1.5473 1.5440 1.5406 1.5373 1.5339 1.5306 1.5273 1.5240 1.5207 1.5174

1.8105 1.8087 1.8069 1.8051 1.8034 1.8016 1.7998 1.7981 1.7963 1.7946

180 181 182 183 184 185 186 187 188 189

AppendixB

Fundamentals ofTilermodynamics and Psycllrometrics

B: 5 Specific Volume, ft 311b Temp. t, of

Absolute Pressure p psi

in.Hg

Entropy, Btullb·oF

Enthalpy, Btullb

Sat. SolidILiq.

Evap.

Sat. Vapor

Sat. SolidILiq.

Evap.

Sat. Vapor

vi

Vig

Vg

hi

h ig

hg

Sat. SolidILiq.

Evap.

Sat. Vapor

si

Sig

Sg

Temp., of

190 191 192 193 194 195 196 197 198 199

9.3493 9.5512 9.7567 9.9659 10.1788 10.3955 10.6160 10.8404 11.0687 11.3010

19.0353 19.4464 19.8648 20.2907 20.7242 21.1653 21.6143 22.0712 22.5361 23.0091

0.01657 0.01658 0.01658 0.01659 0.01659 0.01660 0.01661 0.01661 0.01662 0.01663

40.901 40.092 39.301 38.528 37.774 37.035 36.314 35.611 34.923 34.251

40.918 40.108 39.317 38.544 37.790 37.052 36.331 35.628 34.940 34.268

158.07 159.08 160.08 161.09 162.09 163.10 164.10 165.11 166.11 167.12

983.71 983.10 982.48 981.87 981.25 980.63 980.02 979.40 978.78 978.16

1141.78 1142.18 1142.57 1142.95 1143.34 1143.73 1144.12 1144.51 1144.89 1145.28

0.2787 0.2803 0.2818 0.2834 0.2849 0.2864 0.2880 0.2895 0.2910 0.2926

1.5141 1.5109 1.5076 1.5043 1.5011 1.4979 1.4946 1.4914 1.4882 1.4850

1.7929 1.7911 1.7894 1.7877 1.7860 1.7843 1.7826 1.7809 1.7792 1.7776

190 191 192 193 194 195 196 197 198 199

200 201 202 203 204 205 206 207 208 209

11.5374 11.7779 12.0225 12.2713 12.5244 12.7819 13.0436 13.3099 13.5806 13.8558

23.4904 23.9800 24.4780 24.9847 25.5000 26.0241 26.5571 27.0991 27.6503 28.2108

0.01663 0.01664 0.01665 0.01665 0.01666 0.01667 0.01667 0.01668 0.01669 0.01669

33.594 32.951 32.324 31.710 31.110 30.523 29.949 29.388 28.839 28.303

33.610 32.968 32.340 31.726 31.127 30.540 29.965 29.404 28.856 28.319

168.13 169.13 170.14 171.14 172.15 173.16 174.16 175.17 176.18 177.18

977.54 976.92 976.29 975.67 975.05 974.42 973.80 973.17 972.54 971.92

1145.66 1146.05 1146.43 1146.81 1147.20 1147.58 1147.96 1148.34 1148.72 1149.10

0.2941 0.2956 0.2971 0.2986 0.3002 0.3017 0.3032 0.3047 0.3062 0.3077

1.4818 1.4786 1.4755 1.4723 1.4691 1.4660 1.4628 1.4597 1.4566 1.4535

1.7759 1.7742 1.7726 1.7709 1.7693 1.7677 1.7660 1.7644 1.7628 1.7612

200 201 202 203 204 205 206 207 208 209

210 212 214 216 218 220 222 224 226 228

14.1357 14.7096 15.3025 15.9152 16.5479 17.2013 17.8759 18.5721 19.2905 20.0316

28.7806 29.9489 31.1563 32.4036 33.6919 35.0218 36.3956 37.8131 39.2758 40.7848

0.01670 0.01671 0.01673 0.01674 0.01676 0.01677 0.01679 0.01680 0.01682 0.01683

27.778 26.763 25.790 24.861 23.970 23.118 22.299 21.516 20.765 20.045

27.795 26.780 25.807 24.878 . 23.987 23.134 22.316 21.533 20.782 20.062

178.19 180.20 182.22 184.24 186.25 188.27 190.29 192.31 194.33 196.35

971.29 970.03 968.76 967.50 966.23 964.95 963.67 962.39 961.11 959.82

1149.48 1150.23 1150.98 1151.73 1152.48 1153.22 1153.96 1154.70 1155.43 1156.16

0.3092 0.3122 0.3152 0.3182 0.3212 0.3241 0.3271 0.3301 0.3330 0.3359

1.4503 1.4442 1.4380 1.4319 1.4258 1.4197 1.4136 1.4076 1.4016 1.3957

1.7596 1.7564 1.7532 1.7501 1.7469 1.7438 1.7407 1.7377 1.7347 1.7316

210 212 214 216 218 220 222 224 226 228

230 232 234 236 238 240 242 244 246 248

20.7961 21.5843 22.3970 23.2345 24.0977 24.9869 25.9028 26.8461 27.8172 28.8169

42.3412 43.9461 45.6006 47.3060 49.0633 50.8738 52.7386 54.6591 56.6364 58.6717

0.01684 0.01686 0.01688 0.01689 0.01691 0.01692 0.01694 0.01695 0.01697 0.01698

19.355 18.692 18.056 17.446 16.860 16.298 15.757 15.238 14.739 14.259

19.372 18.709 18.073 17.463 16.877 16.314 15.774 15.255 14.756 14.276

198.37 200.39 202.41 204.44 206.46 208.49 210.51 212.54 214.57 216.60

958.52 957.22 955.92 954.62 953.31 952.00 950.68 949.35 948.03 946.70

1156.89 1157.62 1158.34 1159.06 1159.77 1160.48 1161.19 1161.90 1162.60 1163.29

0.3389 0.3418 0.3447 0.3476 0.3505 0.3534 0.3563 0.3592 0.3621 0.3649

1.3898 1.3839 1.3780 1.3722 1.3664 1.3606 1.3548 1.3491 1.3434 1.3377

1.7287 1.7257 1.7227 1.7198 1.7169 1.7140 1.7111 1.7083 1.7055 1.7026

230 232 234 236 238 240 242 244 246 248

250 252 254 256 258 260 262 264 266 268

29.8457 30.9043 31.9934 33.1135 34.2653 35.4496 36.6669 37.9180 39.2035 40.5241

60.7664 62.9218 65.1391 67.4197 69.7649 72.1760 74.6545 77.2017 79.8190 82.5078

0.01700 0.01702 0.01703 0.01705 0.01707 0.01708 0.01710 0.01712 0.01714 0.01715

13.798 13.355 12.928 12.526 12.123 11.742 11.376 11.024 10.684 10.357

13.815 13.372 12.945 12.147 12.140 11.759 11.393 11.041 10.701 10.374

218.63 220.66 222.69 226.73 226.76 228.79 230.83 232.87 234.90 236.94

945.36 944.02 942.68 939.99 939.97 938.61 937.25 935.88 934.50 933.12

1163.99 1164.68 1165.37 1166.72 1166.73 1167.40 1168.08 1168.74 1169.41 1170.Q7

0.3678 0.3706 0.3735 0.3764 0.3792 0.3820 0.3848 0.3876 0.3904 0.3932

1.3321 1.3264 1.3208 1.3153 1.3097 1.3042 1.2987 1.2932 1.2877 1.2823

1.6998 1.6971 1.6943 1.6691 1.6889 1.6862 1.6835 1.6808 1.6781 1.6755

250 252 254 256 258 260 262 264 266 268

270 272 274 276 278 280 282 284 286 288

41.8806 43.2736 44.7040 46.1723 47.6794 49.2260 50.8128 52.4406 54.1103 55.8225

85.2697 88.1059 91.0181 94.0076 97.0761 100.2250 103.4558 106.7701 110.1695 113.6556

0.01717 0.01719 0.01721 0.01722 0.01724 0.01726 0.01728 0.01730 0.01731 0.01733

10.042 9.737 9.445 9.162 8.890 8.627 8.373 8.128 7.892 7.664

10.059 9.755 9.462 9.179 8.907 8.644 8.390 8.146 7.910 7.681

238.98 241.03 243.07 245.11 247.16 249.20 251.25 253.30 255.35 257.40

931.74 930.35 928.95 927.55 926.15 924.74 923.32 921.90 920.47 919.03

1170.72 1171.38 1172.02 1172.67 1173.31 1173.94 1174.57 1175.20 1175.82 1176.44

0.3960 0.3988 0.4016 0.4044 0.4071 0.4099 0.4127 0.4154 0.4182 0.4209

1.2769 1.2715 1.2661 1.2608 1.2554 1.2501 1.2448 1.2396 1.2343 1.2291

1.6729 1.6703 1.6677 1.6651 1.6626 1.6600 1.6575 1.6550 1.6525 1.6500

270 272 274 276 278 280 282 284 286 288

290 292 294 296 298 300

57.5780 59.3777 61.2224 63.1128 65.0498 67.0341

117.2299 120.8941 124.6498 128.4987 132.4425 136.4827

0.01735 0.01737 0.01739 0.01741 0.01743 0.01745

7.444 7.231 7.026 6.827 6.635 6.450

7.461 7.248 7.043 6.844 6.652 6.467

259.45 261.51 263.56 265.62 267.68 269.74

917.59 916.15 914.69 913.24 911.77 910.30

1177.05 1177.66 1178.26 1178.86 1179.45 1180.04

0.4236 0.4264 0.4291 0.4318 0.4345 0.4372

1.2239 1.2187 1.2136 1.2084 1.2033 1.1982

1.6476 1.6451 1.6427 1.6402 1.6378 1.6354

290 292 294 296 298 300


AppendixB

c:

1

Appendix C-l: R-22 Properties of Saturated Liquid and Saturated Vapor Density, Volume, Enthalpy, Entropy, Specific Heat cp ' Velocity of Viscosity, Thermal Cond, Surface •• 1 Btullb Sound, fils _ _ _ Ib.,lft·oF Btulh·ft·oF Temp,· Pressure, Iblft3 .,lIb _ _Btullb·OF _ _ _ _ _ _Btullb·oF _ _ _ _ cplc, -:-:-_ __ "'-__- - _ _ _ _ _ _ Tension, Temp,. of psia Liquid Vapor Liquid Vapor Liquid Vapor Liquid Vapor Vapor Liquid Vapor Liquid Vapor Liquid Vapor dyne/cm OF -250.00 -240.00 -230.00 -220.00 -210.00

0.002 0.004

107.37 -63.169 106.41 -56.462 105.48 -51.569 104.58 16805. -47.705 103.70 6982.6 -44.426

76.604 77.629 78.669 79.724 80.796

-0.21914 -0.18786 -0.16605 -0.14958 -0.13616

0.44952 0.42332 0.40101 0.38211 0.36538

0.1018 0.1033 0.1048 0.1064 0.1080

1.2914 1.2860 1.2807 1.2754 1.2703

395. 403. 411. 419. 427.

-250.00 -240.00 36.75 -230.00 35.70 -220.00 34.67 -210.00

-200.00 -190.00 -180.00 -170.00 -160.00

0.010 0.022 0.044 0.084 0.151

102.81 101.92 101.03 100.12 99.22

3151.5 1527.4 787.79 429.22 245.51

-41.474 -38.706 -36.038 -33.424 -30.839

81.882 82.984 84.100 85.230 86.373

-0.12457 -0.11411 -0.10439 -0.09521 -0.08644

0.35048 0.33715 0.32518 0.31441 0.30470

0.1096 0.1113 0.1130 0.1147 0.1165

1.2653 1.2604 1.2558 1.2515 1.2474

435. 442. 449. 456. 463.

33.63 32.61 31.59 30.58 29.57

-200.00 -190.00 -180.00 -170.00 -160.00

-150.00 -140.00 -130.00 -120.00 -110.00

0.262 0.435 0.6% 1.080 1.626

98.30 97.38 %.46 95.53 94.60

146.65 91.059 58.544 38.833 26.494

-28.269 -25.708 -23.150 -20.594 -18.D38

87.528 88.692 89.864 91.040 92.218

-0.07800 -0.06986 -0.06198 -0.05435 -0.04694

0.29594 0.28801 0.28082 0.27430 0.2555 0.26838 0.2555

0.1183 0.1201 0.1221 0.1241 0.1262

1.2437 1.2403 1.2374 1.2349 3483. 1.2329 3384.

470. 476. 482. 488. 494.

0.0765

28.57 27.57 26.59 25.61 24.64

-150.00 -140.00 -130.00 -120.00 -110.00

-100.00 -90.00 -80.00 -70.00 -60.00

2.384 3.413 4.778 6.555 8.830

93.66 92.71 91.75 90.79 89.81

18.540 -15.481 13.275 -12.921 9.7044 -10.355 7.2285 -7.783 5.4766 -5.201

93.397 94.572 95.741 %.901 98.049

-0.03973 -0.03271 -0.02587 -0.01919 -0.01266

0.26298 0.25807 0.25357 0.24945 0.24567

0.2557 0.2561 0.2567 0.2574 0.2584

0.1285 0.1308 0.1334 0.1361 0.1389

1.2315 1.2307 1.2305 1.2310 1.2323

3290. 3198. 3110. 3023. 2937.

500. 505. 510. 514. 519.

0.0749 0.0734 0.0718 0.0703 0.0688

0.00292 0.00315 0.00338 0.00360

23.67 -100.00 22.71 -90.00 21.76 -80.00 20.82 -70.00 19.89 -60.00

-50.00 -45.00 -41.44b -40.00 -35.00 -30.00

11.6% 13.383 14.6% 15.255 17.329 19.617

88.83 88.33 87.97 87.82 87.32 86.81

4.2138 -2.608 99.182 -0.00627 0.24220 0.25% 3.7160 -1.306 99.742 -0.00312 0.24056 0.2604 3.4048 -0.377 100.138 -0.00090 0.23944 0.2609 3.2880 0.000 100.296 0.00000 0.23899 0.2611 2.9185 1.310 100.847 0.00309 0.23748 0.2620 2.5984 2.624 101.391 0.00616 0.23602 0.2629

0.1420 0.1436 0.1448 0.1453 0.1471 0.1489

1.2344 1.2358 1.2369 1.2374 1.2393 1.2414

2852. 2810. 2780. 2768. 2725. 2683.

522. 524. 525. 526. 527. 529.

0.0673 0.0665 0.0660 0.0658 0.0651 0.0643

0.00382 0.00393 0.00401 0.00404 0.00414 0.00425

18.% 18.50 18.18 18.05 17.59 17.14

-50.00 -45.00 -41.44 -40.00 -35.00 -30.00

-25.00 22.136 -20.00 24.899 -15.00 27.924 -10.00 31.226 -5.00 34.821

86.29 85.77 85.25 84.72 84.18

2.3202 2.0774 1.8650 1.6784 1.5142

3.944 5.268 6.598 7.934 9.276

101.928 102.461 102.986 103.503 104.013

0.00920 0.01222 0.01521 0.01818 0.02113

0.23462 0.2638 0.1507 1.2437 0.23327 0.2648 0.1527 1.2463 0.23197 0.2659 0.1547 1.2493 0.23071 0.2671 0.1567 1.2525 0.22949 0.2684 0.1589 1.2560

2641. 2599. 2557. 2515. 2473.

530. 531. 532. 533. 534.

0.0636 0.0629 0.0622 0.0614 0.0607

0.00435 0.00445 0.00456 0.00466 0.00476

16.69 16.24 15.79

-25.00 -20.00 -15.00 -10.00 -5.00

0.00 5.00 10.00 15.00 20.00

38.726 42.960 47.538 52.480 57.803

83.64 83.09 82.54 81.98 81.41

1.3691 1.2406 1.1265 1.0250 0.9343

10.624 11.979 13.342 14.712 16.090

104.515 105.009 105.493 105.%8 106.434

0.02406 0.02697 0.02987 0.03275 0.03561

0.22832 0.22718 0.22607 0.22500 0.22395

0.2697 0.2710 0.2725 0.2740 0.2756

0.1611 0.1634 0.1658 0.1683 0.1709

1.2599 1.2641 1.2687 1.2737 1.2792

2431. 2389. 2346. 2304. 2262.

535. 535. 535. 536. 536.

0.615 0.597 0.580 0.563 0.546

0.0268 0.0271 0.0274 0.0276 0.0279

0.0600 0.0593 0.0586 0.0579 0.0572

0.00486 0.004% 0.00506 0.00516 0.00526

0.00 5.00 10.00 15.00 20.00

25.00 30.00 35.00 40.00 45.00

63.526 69.667 76.245 83.280 90.791

80.84 80.26 79.67 79.07 78.46

0.8532 0.7804 0.7150 0.6561 0.6029

17.476 18.871 20.275 21.688 23.111

106.891 107.336 107.769 108.191 108.600

0.03846 0.04129 0.04411 0.04692 0.04972

0.22294 0.22195 0.22098 0.22004 0.21912

0.2773 0.2791 0.2809 0.2829 0.2849

0.1737 0.1765 0.1794 0.1825 0.1857

1.2851 1.2915 1.2984 1.3059 1.3141

2219. 2177. 2134. 2091. 2048.

536. 536. 535. 535. 534.

0.530 0.515 0.499 0.484 0.470

0.0282 0.0284 0.0287 0.0290 0.0292

0.0566 0.0559 0.0552 0.0545 0.0538

0.00536 0.00546 0.00555 0.00565 0.00575

25.00 30.00 35.00 40.00 45.00

50.00 98.799 55.00 107.32 60.00 116.38 65.00 126.00

77.84 77.22 76.58 75.93

0.5548 0.5111 0.4715 0.4355

24.544 108.997 25.988 109.379 27.443 109.748 28.909 110.103

0.05251 0.05529 0.05806 0.06082

0.21821 0.21732 0.21644 0.21557

0.2870 0.2893 0.2916 0.2941

0.1891 0.1927 0.1964 0.2003

1.3229 1.3324 1.3428 1.3540

2005. 1%2. 1919. 1876.

533. 532. 531. 530.

0.456 0.442 0.429 0.416

0.0295 0.0298 0.0301 0.0303

0.0532 0.0525 0.0518 0.0512

0.00584 0.00594 0.00604 0.00613

50.00 55.00 60.00 65.00

70.00 75.00 80.00 85.00 90.00 95.00

136.19 146.98 158.40 170.45 183.17 1%.57

75.27 74.60 73.92 73.22 72.51 71.79

0.4026 0.3726 0.3451 0.3199 0.2%8 0.2756

30.387 31.877 33.381 34.898 36.430 37.977

110.441 110.761 111.066 111.350 111.616 111.859

0.06358 0.06633 0.06907 0.07182 0.07456 0.07730

0.21472 0.2%7 0.2045 0.21387 0.2994 0.2089 0.21302 0.3024 0.2135 0.21218 0.3055 0.2185 0.21134 0.3088 0.2238 0.21050 0.3123 0.2295

1.3663 1.37% 1.3941 1.4100 1.4275 1.4467

1832. 1788. 1744. 1700. 1655. 1611.

528. 527. 525. 523. 520. 518.

0.404 0.392 0.380 0.369 0.358 0.348

0.0505 0.0499 0.0492 0.0486 0.0479 0.0473

0.00623 0.00632 0.00642 0.00652 0.00661 0.00671

70.00 75.00 80.00 85.00 90.00 95.00

100.00 105.00 110.00 115.00 120.00

210.69 225.53 241.14 257.52 274.71

71.05 70.29 69.51 68.71 67.89

0.2560 0.2379 0.2212 0.2058 0.1914

39.538 41.119 42.717 44.334 45.972

112.081 112.278 112.448 112.591 112.704

0.08003 0.08277 0.08552 0.08827 0.09103

0.20%5 0.20879 0.20793 0.20705 0.20615

0.3162 0.3203 0.3248 0.3298 0.3353

0.2356 0.2422 0.2495 0.2573 0.2660

1.4678 1.4912 1.5173 1.5464 1.5791

1566. 1520. 1474. 1428. 1382.

515. 512. 509. 506. 502.

0.338

0.0466 0.0460 0.0454 0.0447 0.0441

0.00680 0.00690 0.00699 0.00709 0.00719

100.00 105.00 110.00 115.00 120.00

125.00 130.00 135.00 140.00 145.00

292.73 311.61 331.38 352.07 373.71

67.05 66.17 65.27 64.33 63.35

0.1781 0.1657 0.1542 0.1434 0.1332

47.633 49.319 51.032 52775 54.553

112.783 112.825 112.826 112.784 112.692

0.09379 0.09657 0.09937 0.10220 0.10504

0.20522 0.20427 0.20329 0.20227 0.20119

0.3413 0.3482 0.3559 0.3648 0.3752

0.2756 0.2864 0.2985 0.3123 0.3282

1.6160 1.6581 1.7063 1.7621 1.8275

1334. 1287. 1238. 1189. 1139.

498. 494. 489. 485. 479.

125.00 130.00 135.00 140.00 145.00

150.00 160.00 170.00 180.00 190.00

3%.32 444.65 497.35 554.82 617.53

62.33 60.12 57.59 54.57 50.62

0.1237 0.1063 0.0907 0.0763 0.0625

56.370 60.145 64.175 68.597 73.742

112.541 112.035 111.165 109.753 107.398

0.10793 0.11383 0.12001 0.12668 0.13432

0.20006 0.19757 0.19464 0.19102 0.18613

0.3873 0.4198 0.4711 0.5657 0.7952

0.3468 0.3957 0.4716 0.6073 0.9222

1.9050 1088. 2.1126 983. 2.4409 873. 3.0349 752. 4.4150 616.

474. 462. 448. 433. 415.

150.00 160.00 170.00 180.00 190.00

200.00 686.11 205.06c 723.74

44.44 32.70

0.0478 80.558 0.0306 91.052

102.809 91.052

0.14432 0.17805 0.15989 0.15989

o.

o.

"temperatures are on the ITS-90 scale

b = boiling point


0.00

200.00 205.06

c =critical point

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Appendix C-3: R-134a Properties of Saturated Liquid and Saturated Vapor Temp,. Pressure, OF

psia

Density, Volume, Enthalpy, Entropy, Specific Heat cp' Velocity or Viscosity, Thennal Cond, Surface o Ib/rt 3 rt 311b _ _B_t_uII_b_ _ _ _B_t_uII_b_o_F_ _ _ _B_tull_b_oo_F_ c/c, _S_o_un_d_,rtI_s_ _ _I_b",..If_t_oO_F_ _ _ _ B_tu/h_o_ft_oo_F_ Tension, Temp,. Liquid

Vapor Liquid

Vapor

Liquid

dyne/cm

OF

32.989 31.902 29.093 26.238

80.235 80.783 82.190 83.618

0.09154 0.27880 0.08801 0.27588 0.07908 0.26903 0.07029 0.26294

0.2740 0.1397 1.1628 0.2776 0.1409 1.1615 0.2837 0.1438 1.1583 0.2870 0.1467 1.1554

3723. 3670. 3552. 3448.

416. 418. 424. 430.

5.289 4.913 4.117 3.497

0.0160 0.0163 0.0168 0.0173

28.15 27.76 26.78 25.81

-153.94 -150.00 -140.00 -130.00

23359 20.467 17.569 14.665 11.755

85.066 86.531 88.011 89.504 91.005

0.06169 0.05330 0.04513 0.03717 0.02940

0.25752 0.25270 0.24842 0.24462 0.24125

0.2886 0.2895 0.2900 0.2906 0.2913

1.1528 1.1507 1.1489 1.1475 1.1466

3352. 3261. 3173. 3087. 3001.

436. 441. 446. 451. 456.

3.006 2.613 2.292 2.028 1.808

0.0179 0.0184 0.0190 0.0195 0.0200

0.0721 0.0706

24.86 23.92 22.99 22.08 21.17

-120.00 -110.00 -100.00 -90.00 -80.00

16.680 10.297 14.138 8.837 12.045 7.374 10.310 5.907 8.8656 4.437

91.759 92.514 93.270 94.026 94.783

0.02559 0.02182 0.01809 0.01440 0.01075

0.23972 0.23827 0.23691 0.23563 0.23443

0.2917 0.1641 1.1463 0.2922 0.1658 1.1462 0.2928 0.1676 1.1461 0.2935 0.1694 1.1462 0.2943 0.1712 1.1465

2959. 2916. 2874. 2832. 2790.

458. 460. 462. 464. 466.

1.712 1.623 1.542 1.466 1.396

0.0203 0.0206 0.0209 0.0212 0.0214

0.0699 0.0691 0.0684 0.0677 0.00405 0.0669 0.00423

20.73 20.28 19.84 19.40 18.96

-75.00 -70.00 -{;5.00 -{;D.OO -55.00

2.963 1.484 0.000 1.489 2.984

95.539 96.295 97.050 97.804 98.556

0.00713 0.00355 0.00000 0.00352 0.00701

0.23331 0.23225 0.23125 0.23032 0.22945

0.2951 0.2960 0.2970 0.2981 0.2992

0.1731 0.1750 0.1769 0.1789 0.1810

1.1468 1.1473 1.1479 1.1487 1.1497

2747. 2705. 2663. 2621. 2579.

468. 470. 471. 473. 474.

1.331 1.271 1.215 1.163 1.113

0.0217 0.0220 0.0223 0.0225 0.0228

0.0662 0.00440 0.0654 0.00457 0.0647 0.00473 0.0639 0.00489 0.0632 0.00505

18.52 18.09 17.66 17.23 16.81

-50.00 -45.00 -40.00 -35.00 -30.00

86.82 86.32 85.81 85.80 85.29 84.77

3.9014 4.484 3.4452 5.991 3.0519 7.505 3.0462 7.529 2.7116 9.026 2.4161 10.554

99.306 100.054 100.799 100.811 101.542 102.280

0.01048 0.01392 0.01733 0.01739 0.02073 0.02409

0.22863 0.22786 0.22714 0.22713 0.22647 0.22584

0.3004 0.3017 0.3031 0.3031 0.3045 0.3060

0.1831 0.1852 0.1874 0.1874 0.1897 0.1920

1.1508 1.1521 1.\535 1.\535 1.\552 1.\570

2536. 2494. 2452. 2451. 2410. 2367.

476. 477. 478. 478. 479. 480.

1.068 1.024 0.984 0.983 0.946 0.910

0.0231 0.0234 0.0237 0.0237 0.0240 0.0243

0.0625 0.0617 0.0610 0.0610 0.0602 0.0595

0.00521 0.00536 0.00550 0.00551

0.00565 0.00579

16.38 15.96 15.54 15.54 15.13 14.71

-25.00 -20.00 -15.00 -14.92 -10.00 -5.00

21.162 23.767 26.617 29.726 33.110

84.25 83.72 83.18 82.64 82.10

2.1587 1.9337 1.7365 1.5630 1.4101

103.015 103.745 104.471 105.192 105.907

0.02744 0.03077 0.03408 0.03737 0.04065

0.22525 0.22470 0.22418 0.22370 0.22325

0.3075 0.3091 0.3108 0.3126 0.3144

0.1943 0.1968 0.1993 0.2018 0.2045

1.1590 1.\613 1.\ 637 1.1664 1.1694

2325. 2283. 2240. 2198. 2155.

481. 481. 482. 482. 482.

0.876 0.843 0.813 0.784 0.756

0.0245 0.0248 0.0251 0.0254 0.0257

0.0588 0.00593 0.0580 0.00607 0.0573 0.00621 0.0565 0.00635 0.0558 0.00649

14.30 13.89 13.48 13.08 12.67

0.00 5.00 10.00 15.00 20.00

25.00 30.00 35.00 40.00 45.00

36.785 40.768 45.075 49.724 54.732

81.55 80.99 80.42 79.85 79.26

1.0484 23.085 108.016 0.9534 24.694 108.705 0.8685 26.314 109.386

0.04391 0.22283 0.04715 0.22244 0.05038 0.22207 0.05359 0.22172 0.05679 0.22140

0.3162 0.3182 0.3202 0.3223 0.3244

0.2072 0.2100 0.2129 0.2159 0.2190

1.1726 1.1761 1.\799 1.1841 1.1886

2113. 2070. 2027. 1985. 1942.

482. 482. 482. 482. 481.

0.730 0.705 0.681 0.658 0.636

0.0260 0.0263 0.0267 0.0270 0.0273

0.0550 0.00662 0.0543 0.00676 0.0536 0.00690 0.0528 0.00704 0.0521 0.00718

12.27 11.87 11.48 11.08 10.69

25.00 30:00 35:00 40.00 45.00

50.00 55.00 60.00 65.00 70.00

60.116 65.895 72.087 78.712 85.787

78.67 78.07 77.46 76.84 76.21

0.7925 0.7243 0.6630 0.6077 0.5577

27.944 29.586 31.239 32.905 34.583

110.058 11 0.722 111.376 112.019 112.652

0.05998 0.06316 0.06633 0.06949 0.07264

0.22110 0.22081 0.22054 0.22028 0.22003

0.3267 0.2222 1.\935 0.3290 0.2255 1.1988 0.3314 0.2289 1.2046 0.3339 0.2325 1.2109 0.3366 0.2363 1.2178

1899. 1856. 1813. 1770. 1726.

481. 480. 479. 477. 476.

0.615 0.595 0.576 0.557 0.539

0.0276 0.0280 0.0283 0.0286 0.0290

0.0513 0.00732 0.0506 0.00746 0.0499 0.00761 0.0491 0.00776 0.0484 0.00791

10.30 9.91 9.53 9.15 8.77

50.00 55.00 60.00 65.00 70.00

75.00 80.00 85.00 90.00 95.00

93.333 101.37 109.92 1\9.00 128.63

75.57 74.91 74.25 73.57 72.87

0.5125 0.4715 0.4343 0.4004 0.3694

36.274 37.978 39.697 41.430 43.179

113.272 113.880 114.475 115.055 115.619

0.07578 0.07892 0.08205 0.08518 0.08830

0.21979 0.21957 0.21934 0.21912 0.21890

0.3393 0.3422 0.3453 0.3485 0.3519

1683. 1640. 1596. 1552. 1509.

474. 472. 470. 468. 466.

0.522 0.505 0.489 0.473 0.458

0.0294 0.0297 0.0301 0.0305 0.0309

0.0476 0.00806 0.0469 0.00822 0.0462 0.00838 0.0454 0.00855 0.0447 0.00872

8.39 8.02 7.65 7.28 6.91

75.00 80.00 85.00 90.00 95.00

100.00 105.00 110.00 1\5.00 120.00

138.83 149.63 161.05 173.11 185.84

72.16 71.43 70.68 69.91 69.12

0.3411 0.3153 0.2915 0.2697 0.2497

44.943 46.725 48.524 50.343 52.181

116.166 116.694 117.203 117.690 118.153

0.09142 0.21868 0.09454 0.21845 0.09766 0.21822 0.10078 0.21797 0.10391 0.21772

0.3555 0.2633 1.2748 1465. 0.3594 0.2689 1.2880 1421. 0.3635 0.2748 1.3026 1376. 0.3680 0.2812 1.3189 1332. 0.3728 0.2881 1.3372 1288.

463. 460. 457. 454. 450.

0.443 0.428 0.414 0.400 0.387

0.0313 0.0318 0.0322 0.0327 0.0332

0.0439 0.0432 0.0425 0.0417 0.04 10

0.00890 0.00908 0.00926 0.00946 0.00965

6.55 5.84 5.49 5.15

100.00 105.00 110.00 115.00 120.00

125.00 130.00 135.00 140.00 145.00

199.25 213.38 228.25 243.88 260.31

68.31 67.47 66.60 65.70 64.77

0.2312 0.2141 0.1983 0.1836 0.1700

54.040 55.923 57.830 59.764 61.727

118.591 119.000 119.377 119.720 120.024

0.10704 0.11018 0.11333 0.11650 0.11968

0.3781 0.3839 0.3903 0.3974 0.4053

1243. 1198. 1153. 110S. 1062.

446. 442. 437. 432. 427.

0.374 0.361 0.348 0.335 0.323

0.0338 0.0343 '0.0349 0.0356 0.0363

0.0403 0.0395 0.0388 0.0380 0.0373

0.00986 0.01007 0.01029 0.01052 0.01075

4.80 4.47 4.\3 3.81 3.48

125.00 130.00 135.00 140.00 145.00

150.00 155.00 160.00 165.00 170.00

277.57 295.69 314.69 334.62 355.51

63.80 62.78 61.72 60.60 59.42

0.1574 0.1455 0.1345 0.1241 0.1144

63.722 65.752 67.823 69.939 72.106

120.284 120.495 120.650 120.739 120.753

0.12288 0.21566 0.12611 0.21517 0.12938 0.21463 0.13268 0.21400 0.13603 0.21329

0.4144 0.3486 1.5143 1017. 0.4247 0.3640 1.5630 97 I. 0.4368 0.3821 1.6213 924. 0.4511 0.4036 1.6921 877. 0.4683 0.4299 1.7798 829.

421. 416. 409. 403. 396.

0.311 0.298 0.286 0.274 0.262

0.0370 0.0378 0.0387 0.0397 0.0407

0.0366 0.0358 0.0351 0.0343 0.0336

om I 00 0.01125 om151 0.01178 0.01206

3.17 2.86 2.55 2.26 1.97

150.00 155.00 160.00 165.00 170.00

175.00 180.00 185.00 190.00 195.00

377.40 400.34 424.37 449.55 475.95

58.16 56.80 55.33 53.70 51.86

0.1052 0.0965 0.0881 0.0801 0.0723

74.335 76.636 79.027 81.534 84.196

120.677 120.493 120.175 119.684 118.963

0.13945 0.14295 0.14655 0.15029 0.15423

0.21247 0.21151 0.21037 0.2090 I 0.20733

0.4896 0.4627 1.8908 0.5168 0.5048 20354 0.5527 0.5612 2.2309 0.6031 0.6406 2.5087 0.6794 0.7612 29338

781. 731. 680. 627. 572.

388. 380. 372. 363. 353.

0.249 0.237 0.224 0.211 0.197

0.0420 0.0434 0.0450 0.0469 0.0493

0.0329 0.0321 0.0314

0.01235 0.01265 0.01296

1.69 1.41 1.15 0.90 0.67

175.00 180.00 185.00 190.00 195.00

200.00 205.00 210.00 213.85c

503.64 532.72 563.34 588.27

49.70 47.00 43.03 32.04

0.0646 87.088 0.0566 90.368 0.0474 94.548 0.0312103.775

117.906 116.289 113.411 103.775

0.15847 0.16326 0.16933 0.18128

0.20519 0.20226 0.19750 0.18128

0.8100 0.9673 3.6640 1.0906 1.4042 5.2174

514. 450. 378. O.

343. 331. 316.

0.182 0.164 0.142

0.0523 0.0566 0.0639

0.45 0.25 0.09 0.00

200.00 205.00 210.00 213.85

-153.94a -150.00 -140.00 -130.00

0.057 0.072 0.130 0.222

99.34 98.95 97.98 97.01

564.85 449.29 259.15 155.69

-120.00 -110.00 -100.00 -90.00 -80.00

0.367 0.586 0.906 1.363 1.997

96.05 95.09 94.13 93.17 92.21

97.027 62.509 41.496 28.303 19.783

-75.00 -70.00 -{;5.00 -{;D.OO -55.00

2.396 2.859 3.393 4.006 4.707

91.73 91.25 90.77 90.28 89.80

-50.00 -45.00 -40.00 -35.00 -30.00

5.505 6.409 7.429 8.577 9.862

89.31 88.82 88.32 87.83 87.33

7.6569 6.6405 5.7819 5.0533 4.4325

-25.00 -20.00 -15.00 -14.92b -10.00 -5.00

11.297 12.895 14.667 14.696 16.626 18.787

0.00 5.00 10.00 15.00 20.00

12.090 13.634 15.187 16.748 18.318

1.2749 19.897 106.617 \.1550 21.486 107.320

"temperatures are OR the ITS-90 scale

Vapor

0.21744 0.21715 0.21683 0.21648 0.21609

Liquid

a =triple point


Vapor Vapor Liquid Vapor

0.1497 0.1527 0.1559 0.1591 0.1624

0.2402 0.2444 0.2487 0.2533 0.2582

0.2957 0.3040 0.3133 0.3236 0.3353

1.2252 1.2334 1.2424 1.2522 1.2630

1.3577 1.3810 1.4075 1.4379 1.4731

Liquid Vapor

O. b = boiling point

Liquid

Vapor

6.20

c =critical point

AppendixC

20

60

100

140

180

220

260

IJ I

3000 2000

1000~

1000

~

g

~

-

as 'u; c.

-

~ I

..

200 ~

200

~

100

100

w a:

40

::l

en en w

a:

D.

20

20 10

~

~

= ~ ~ = ......

=-~ ..... ~ C ..... ~

(JQ

~

4

9

4

~

11

2

2 1

260

ENTHALPY

(Btu/Ibm)

~

I-l ~ ~ ~

D: 1

Appendix D-l: Thermodynamic Properties of Moist Air at 14.696 psia Condensed Water Temp. Humidity Ratio, I, OF Ib,.JIbda Ws


Volume, air

ft 31lb dry

va

va g

hga

Entropy, Btul(Ib dry air) • OF Sg a Saga Sa

Vapor Enthalpy, Entropy, Press., Btullb Btullb· OF in.Hg pg" h., S.,

Temp., OF

9.553 9.579 9.604 9.629 9.655 9.680 9.705 9.731 9.756 9.782

ha -19.221 -18.980 -18.740 -18.500 -18.259 -18.019 -17.778 -17.538 -17.298 -17.057

ha/ 0.005 0.005 0.006 0.007 0.007 0.007 0.008 0.009 0.010 0.010

-19.215 -18.975 -18.734 -18.493 -18.252 -18.0ll -17.770 -17.529 -17.288 -17.047

-
0.00001 0.00002 0.00002 0.00002 0.00002 0.00002 0.00002 0.00002 0.00003 0.00003

-
-193.45 -193.06 -192.66 -192.27 -191.87 -191.47 -191.07 -190.68 -190.27 -189.87

-
0.000236 -80 0.000255 -79 0.000275 -78 0.000296 -77 0.000319 -76 0.000344 -75 0.000371 -74 0.000400 -73 0.000430 -72 0.000463 -71

-80 -79 -78 -77 -76 -75 -74 -73 -72 -71

0.0000049 0.0000053 0.0000057 0.0000062 0.0000067 0.0000072 0.0000078 0.0000084 0.0000090 0.0000097

9.553 9.579 9.604 9.629 9.655 9.680 9.705 9.731 9.756 9.781

va/ 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

-70 -69 -68 -67 -66 -65 -63 -62 -61

0.0000104 0.0000 II 2 0.0000120 0.0000129 0.0000139 0.0000149 0.0000160 0.0000172 0.0000184 0.0000198

9.807 9.832 9.857 9.883 9.908 9.933 9.959 9.984 10.009 10.035

0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

9.807 9.832 9.858 9.883 9.908 9.934 9.959 9.984 10.010 10.035

-16.806 -16.577 -16.336 -16.096 -15.856 -15.616 -15.375 -15.117 -14.895 -14.654

0.01l 0.012 0.013 0.013 0.015 0.015 0.017 0.018 0.019 0.021

-16.817 -16.565 -16.324 -16.083 -15.841 -15.600 -15.359 -15.135 -14.876 -14.634

-
0.00003 0.00003 0.00003 0.00004 0.00004 0.00004 0.00005 0.00005 0.00005 0.00006

-
-189.47 -189.07 -188.66 -188.26 -187.85 -187.44 -187.04 -186.63 -186.22 -185.81

-
0.000498 0.000536 0.000576 0.000619 0.000665 0.000714 0.000766 0.000822 0.000882 0.000945

-60 -59 -58 -57 -56 -55 -54 -53 -52 -51

0.0000212 0.0000227 0.0000243 0.0000260 0.0000279 0.0000298 0.0000319 0.0000341 0.0000365 0.0000390

10.060 10.085 10.111 10.136 10.161 10.187 10.212 10.237 10.263 10.288

0.000 0.000 0.000 0.000 0.000 0.000 0.001 0.001 0.001 0.001

10.060 10.086 10.111 10.137 10.162 10.187 10.213 10.238 10.263 10.289

-14.414 -14.174 -13.933 -13.693 -13.453 -13.213 -12.972 -12.732 -12.492 -12.251

0.022 0.024 0.025 0.027 0.029 0.031 0.033 0.035 0.038 0.041

-14.392 -14.150 -13.908 -13.666 -13.424 -13.182 -12.939 -12.697 -12.454 -12.211

-
0.00006 0.00006 0.00007 0.00007 0.00008 0.00008 0.00009 0.00009 0.00010 0.000 II

-0.03354 -0.03294 -0.03233 -0.03173 -
-185.39 -184.98 -184.57 -184.15 -183.74 -183.32 -182.90 -182.48 -182.06 -181.64

-0.3860 -0.3850 -0.3839 -0.3829 -0.3819 -
0.001013 -60 0.001086 -59 0.001163 -58 0.001246 -57 0.001333 -56 0.001427 -55 0.001526 -54 0.001632 -53' 0.001745 -52 0.001865 -51

-50 -49 -48 -47 -46 -45 -44 -43 -42 -41

0.0000416 0.0000445 0.0000475 0.0000507 0.0000541 0.0000577 0.0000615 0.0000656 0.0000699 0.0000744

10.313 10.339 10.364 10.389 10.415 10.440 10.465 10.491 10.516 10.541

0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001

10.314 10.340 10.365 10.390 10.416 10.441 10.466 10.492 10.517 10.543

-12.011 -11.771 -11.531 -11.290 -11.050 -10.810 -10.570 -10.329 -10.089 -9.849

0.043 0.046 0.050 0.053 0.056 0.060 0.064 0.068 0.073 0.078

-11.968 -11.725 -11.481 -11.237 -10.994 -10.750 -10.505 -10.261 -10.016 -9.771

-
0.00011 0.00012 0.00013 0.00014 0.00015 0.00016 0.00017 0.00018 0.00019 0.00020

-
-181.22 -180.80 -180.37 -179.95 -179.52 -179.10 -178.67 -178.24 -177.81 -177.38

-0.3757 -0.3747 -0.3736 -0.3726 -0.3716 -0.3705 -0.3695 -0.3685 -0.3675 -0.3664

0.001992 -50 0.002128 -49 0.002272 -48 0.002425 -47 0.002587 -46 0.002760 -45 0.002943 -44 0.003137 -43 0.003343 -42 0.003562 -41

-40 -39 -38 -37 -36 -35 -34 -33 -32 -31

0.0000793 0.0000844 0.0000898 0.0000956 0.0001017 0.0001081 0.0001150 0.0001222 0.0001298 0.0001379

10.567 10.592 10.617 10.643 10.668 10.693 10.719 10.744 10.769 10.795

0.001 0.001 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002

10.568 10.593 10.619 10.644 10.670 10.695 10.721 10.746 10.772 10.797

-9.609 -9.368 -9.128 -8.888 -8.648 -8.407 -8.167 -7.927 -7.687 -7.447

0.083 0.088 0.094 0.100 0.106 0.113 0.120 0.128 0.136 0.145

-9.526 -9.280 -9.034 -8.788 -8.541 -8.294 -8.047 -7.799 -7.551 -7.302

-
0.00021 0.00022 0.00024 0.00025 0.00027 0.00028 0.00030 0.00032 0.00034 0.00036

-
-176.95 -176.52 -176.08 -175.65 -175.21 -174.78 -174.34 -173.90 -173.46 -173.02

-0.3654 -0.3644 -0.3633 -0.3623 -0.3613 -0.3603 -0.3529 -0.3582 -0.3572 -0.3561

0.003793 0.004039 0.004299 0.004575 0.004866 0.005175 0.005502 0.005848 0.006214 0.006601

-40 -39 -38 -37 -36 -35 -34 -33 -32 -31

-30 -29 -28 -27 -26 -25 -24 -23 -22 -21

0.0001465 0.0001555 0.0001650 0.0001751 0.0001858 0.0001970 0.0002088 0.0002214 0.0002346 0.0002485

10.820 10.845 10.871 10.896 10.921 10.947 10.972 10.997 11.022 11.048

0.003 0.003 0.003 0.003 0.003 0.003 0.004 0.004 0.004 0.004

10.822 10.848 10.873 10.899 10.924 10.950 10.976 11.001 11.027 11.052

-7.206 -6.966 -6.726 -6.486 -6.245 -6.005 -5.765 -5.525 -5.284 -5.044

0.154 0.163 0.173 0.184 0.195 0.207 0.220 0.233 0.247 0.261

-7.053 -6.803 -6.553 -6.302 -6.051 -5.798 -5.545 -5.292 -5.038 -4.783

-0.01621 -
0.00038 0.00040 0.00043 0.00045 0.00048 0.00051 0.00054 0.00057 0.00060 0.00063

-
-172.58 -172.14 -171.70 -171.25 -170.81 -170.36 -169.92 -169.47 -169.02 -168.57

-0.3551 -0.3541 -0.3531 -0.3520 -0.3510 -0.3500 -0.3489 -0.3479 -0.3469 -0.3459

0.007009 0.007442 0.007898 0.008381 0.008890 0.009428 0.009995 0.010594 0.011226 0.011893

-30 -29 -28 -27 -26 -25 -24 -23 -22 -21

-20 -19 -18

0.0002632 0.0002786 0.0002950 0.0003121 0.0003303 0.0003493 0.0003694 0.0003905 0.0004128 0.0004362

11.073 11.098 11.124 11.149 11.174 11.200 11.225 11.250 11.276 11.301

0.005 0.005 0.005 0.006 0.006 0.006 0.007 0.007 0.007 0.008

11.078 11.103 11.129 11.155 11.180 11.206 11.232 11.257 11.283 11.309

-4.804 -4.564 -4.324 -4.084 -3.843 -3.603 -3.363 -3.123 -2.882 -2.642

0.277 0.293 0.311 0.329 0.348 0.368 0.390 0.412 0.436 0.460

-4.527 -4.271 -4.013 -3.754 -3.495 -3.235 -2.973 -2.710 -2.447 -2.182

-
0.00067 0.00071 0.00075 0.00079 0.00083 0.00088 0.00093 0.00098 0.00103 0.00109

-0.01002 -0.00943 -
-168.12 -167.67 -167.21 -166.76 -166.30 -165.85 -165.39 -164.93 -164.47 -164.01

-0.3448 -0.3438 -0.3428 -0.3418 -0.3407 -0.3397 -0.3387 -0.3377 -
0.012595 0.013336 0.014117 0.014939 0.015806 0.016718 0.017679 0.018690 0.019754 0.020873

-20 -19 -18 -17 -16 -15 -14 -13 -12 -II

-64

-17 -16 -15 -14 -13 -12 -11

-70 -69 -68 -67 -66 -65 -64

-63 -62 -61

"Note that the values for subSCript g are actually solids (s) until O°F.


AppendixD

0:2 Condensed Water Temp. Humidity t, Ratio, of Ib.,.llbda Ws

Volume,


ft 311b dry air

Vapor Enthalpy, Entropy, Press., Btullb Btullb· OF in.Hg

h ga

Sa

Saga

Sg"

hw

Sw

p ga

Temp., of

0.487 0.514 0.543 0.574 0.606 0.640 0.675 0.712 0.751 0.792

-1.915 -1.647 -1.378 -1.108 -0.835 -0.561 -0.286 -0.008 0.271 0.552

-0.00528 -0.00475 -0.00422 -0.00369 -0.00316 -0.00263 -0.00210 -0.00157 -0.00105 -0.00052

0.00115 0.00121 0.00127 0.00134 0.00141 0.00149 0.00157 0.00165 0.00174 0.00183

-0.00414 -0.00354 -0.00294 -0.00234 -0.00174 -0.00114 -0.00053 0.00008 0.00069 0.00130

-163.55 -163.09 -162.63 -162.17 -161.70 -161.23 -160.77 -160.30 -159.83 -159.36

-0.3346 -0.3335 -0.3325 -0.3315 -0.3305 -0.3294 -0.3284 -0.3274 -0.3264 -0.3253

0.022050 0.023289 0.024591 0.025959 0.027397 0.028907 0.030494 0.032160 0.033909 0.035744

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1

0.0 0.240 0.480 0.721 0.961 1.201 1.441 1.681 1.922 2.162

0.835 0.880 0.928 0.978 1.030 1.085 1.143 1.203 1.266 1.332

0.835 1.121 1.408 1.699 1.991 2.286 2.584 2.884 3.188 3.494

0.00000 0.00052 0.00104 0.00156 0.00208 0.00260 0.00311 0.00363 0.00414 0.00466

0.00192 0.00202 0.00212 0.00223 0.00235 0.00247 0.00259 0.00635 0.00286 0.00300

0.00192 0.00254 0.00317 0.00380 0.00443 0.00506 0.00570 0.00272 0.00700 0.00766

-158.89 -158.42 -157.95 -157.47 -157.00 -156.52 -156.05 -155.57 -155.09 -154.61

-0.3243 -0.3233 -0.3223 -0.3212 -0.3202 -0.3192 -0.3182 -0.3171 -0.3161 -0.3151

0.037671 0.039694 0.041814 0.044037 0.046370 0.048814 0.051375 0.054060 0.056872 0.059819

0 1 2 3 4 5 6 7 8 9

va

vag"

va g

0.0004608 0.0004867 0.0005139 0.0005425 0.0005726 0.0006041 0.0006373 0.0006722 0.0007088 0.0007472

11.326 11.351 11.377 11.402 11.427 11.453 11.478 11.503 11.529 11.554

0.008 0.009 0.009 0.010 0.010 0.011 0.012 0.012 0.013 0.014

11.335 11.360 11.386 11.412 11.438 11.464 11.490 11.516 11.542 11.568

-2.402 -2.162 -1.922 -1.681 -1.441 -1.201 -0.961 -0.721 -0.480 -0.240

0 0.0007875 1 0.0008298 2 0.0008742 3 0.0009207 4 0.0009695 5 0.0010207 6 0.0010743 7 0.0011306 8 0.0011895 9 0.0012512

11.579 11.604 11.630 11.655 11.680 11.706 11.731 11.756 11.782 11.807

0.015 0.015 0.016 0.017 0.018 0.019 0.020 0.021 0.022 0.024

11.594 11.620 11.646 11.672 11.699 11.725 11.751 11.778 11.804 11.831

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1

Entropy, Btul(lb dry air) . of

ha

hag"

10 11 12 13 14 15 16 17 18 19

0.0013158 0.0013835 0.0014544 0.0015286 0.0016062 0.0016874 0.0017724 0.0018613 0.0019543 0.0020515

11.832 11.857 11.883 11.908 11.933 11.959 11.984 12.009 12.035 12.060

0.025 0.026 0.028 0.029 0.031 0.032 0.034 0.036 0.038 0.040

11.857 11.884 11.910 11.937 11.964 11.991 12.018 12.045 12.072 12.099

2.402 2.642 2.882 3.123 3.363 3.603 3.843 4.084 4.324 4.564

1.402 1.474 1.550 1.630 1.714 1.801 1.892 1.988 2.088 2.193

3.804 4.117 4.433 4.753 5.077 5.404 5.736 6.072 6.412 6.757

0.00517 0.00568 0.00619 0.00670 0.00721 0.00771 0.00822 0.00872 0.00923 0.00973

0.00315 0.00330 0.00347 0.00364 0.00381 0.00400 0.00419 0.00439 0.00460 0.00482

0.00832 0.00898 0.00966 0.01033 0.01102 0.01171 0.01241 0.01312 0.01383 0.01455

-154.13 -153.65 -153.17 -152.68 -152.20 -151.71 -151.22 -150.74 -150.25 -149.76

-0.3141 -0.3130 -0.3120 -0.3110 -0.3100 -0.3089 -0.3079 -0.3069 -0.3059 -0.3049

0.062901 0.066131 0.069511 0.073049 0.076751 0.080623 0.084673 0.088907 0.093334 0.097962

10 11 12 13 14 15 16 17 18 19

20 21 22 23 24 25 26 27 28 29

0.0021531 0.0022592 0.0023703 0.0024863 0.0026073 0.0027339 0.0028660 0.0030039 0.0031480 0.0032984

12.085 12.110 12.136 12.161 12.186 12.212 12.237 12.262 12.287 12.313

0.042 0.044 0.046 0.048 0.051 0.054 0.056 0.059 0.062 0.065

12.127 12.154 12.182 12.209 12.237 12.265 12.293 12.321 12.349 12.378

4.804 5.044 5.285 5.525 5.765 6.005 6.246 6.486 6.726 6.966

2.303 2.417 2.537 2.662 2.793 2.930 3.073 3.222 3.378 3.541

7.107 7.462 7.822 8.187 8.558 8.935 9.318 9.708 10.104 10.507

0.01023 0.01073 0.01123 0.01173 0.01223 0.01272 0.01322 0.01371 0.01420 0.01470

0.00505 0.00529 0.00554 0.00580 0.00607 0.00636 0.00665 0.00696 0.00728 0.00761

0.01528 0.01602 0.01677 0.01753 0.01830 0.01908 0.01987 0.02067 0.02148 0.02231

-149.27 -148.78 -148.28 -147.79 -147.30 -146.80 -146.30 -145.81 -145.31 -144.81

-0.3038 -0.3028 -0.30\8 -0.3008 -0.2997 -0.2987 -0.2977 -0.2967 -0.2956 -0.2946

0.102798 0.107849 0.113130 0.118645 0.124396 0.130413 0.136684 0.143233 0.150066 0.157198

20 21 22 23 24 25 26 27 28 29

30 31 32 32* 33 34 35 36 37 38 39

0.0034552 0.0036190 0.0037895 0.003790 0.003947 0.004109 0.004277 0.004452 0.004633 0.004820 0.0050J4

12.338 12.363 12.389 12.389 12.414 12.439 12.464 12.490 12.515 12.540 12.566

0.068 0.072 0.075 0.075 0.079 0.082 0.085 0.089 0.093 0.097 0.101

12.406 12.435 12.464 12.464 12.492 12.521 12.550 12.579 12.608 12.637 12.667

7.206 7.447 7.687 7.687 7.927 8.167 8.408 8.648 8.888 9.128 9.369

3.711 3.888 4.073 4.073 4.243 4.420 4.603 4.793 4.990 5.194 5.405

10.917 11.335 11.760 11.760 12.170 12.587 13.010 13.441 13.878 14.322 14.773

0.01519 0.01568 0.01617 0.01617 0.01665 0.01714 0.01763 0.01811 0.01860 0.01908 0.01956

0.00796 0.00832 0.00870 0.00870 0.00905 0.00940 0.0,0977 0.01016 0.01055 0.01096 0.01139

0.02315 -144.31 -0.2936 0.164631 0.02400 -143.80 -0.2926 0.172390 0.02487 -143.30 -0.2915 0.180479 0.02 0.0000 0.18050 0.02487 1.03 0.0020 0.18791 0.02570 0.02655 2.04 0.0041 0.19559 0.02740 3.05 0.006J 0.20356 0.02827 4.05 0.0081 0.21181 5.06 0.0102 0.22035 0.02915 0.03004 6.06 0.0122 0.22920 7.07 0.0142 0.23835 0.03095

30 31 32 32 33 34 35 36 37 38 39

40 41 42 43

0.005216 0.005424 0.005640 0.005863 0.006094 0.006334 0.006581 0.006838 0.007103 0.007378 0.007661

12.591 12.616 12.641 12.667 12.692 12.717 12.743 12.768 12.793 12.818 12.844

0.105 0.110 0.114 0.119 0.124 0.129 0.134 0.140 0.146 0.152 0.158

12.696 12.726 12.756 12.786 12.816 12.846 12.877 12.908 12.939 12.970 13.001

9.609 9.849 10.089 10.330 10.570 10.810 11.050 11.291 11.531 11.771 12.012

5.624 5.851 6.086 6.330 6.582 6.843 7.114 7.394 7.684 7.984 8.295

15.233 15.700 16.175 16.660 17.152 17.653 18.164 18.685 19.215 19.756 20.306

0.02004 0.02052 0.02100 0.02148 0.02196 0.02244 0.02291 0.02339 0.02386 0.02433 0.02480

0.01183 0.01228 0.01275 0.01324 0.01374 0.01426 0.01479 0.01534 0.01592 0.01651 0.01712

0.03187 0.03281 0.03375 0.03472 0.03570 0.03669 0.03770 0.03873 0.03978 0.04084 0.04192

8.07 9.08 10.08 11.09 12.09 13.09 14.10 15.10 16.10 17.10 18.11

0.0162 0.0182 0.0202 0.0222 0.0242 0.0262 0.0282 0.0302 0.0321 0.0341 0.0361

0.24784 0.25765 0.2678J 0.27831 0.28918 0.30042 0.31206 0.32408 0.33651 0.34937 0.36264

40 41 42 43 44 45 46 47 48 49 50

12.252 8.616 13.033 0.164 12.869 51 0.007955 8.949 12.492 13.065 12.894 0.171 52 0.008259 9.293 12.732 13.097 0.178 12.920 53 0.008573 9.648 12.973 13.129 12.945 0.185 54 0.008897 13.213 10.016 13.162 0.192 12.970 55 0.009233 10.397 13.453 13.195 12.995 0.200 56 0.009580 13.694 10.790 13.228 0.207 13.021 57 0.009938 13.934 11.197 13.262 13.046 0.216 58 0.010309 *Extrapolated to represent metastable equilibrium wilh undercooled liquid. "Note that the values for subscript g are actually solids (s) until O°F.

20.868 21.441 22.025 22.621 23.229 23.850 24.484 25.131

0.02528 0.02575 0.02622 0.02668 0.02715 0.02762 0.02808 0.02855

0.01775 0.01840 0.01907 0.01976 0.02048 0.02122 0.02198 0.02277

0.04302 0.04415 0.04529 0.04645 0.04763 0.04884 0.05006 0.05132

19.11 20.11 21.11 22.11 23.11 24.11 25.11 26.11

0.0381 0.0400 0.0420 0.0439 0.0459 0.0478 0.0497 0.0517

0.37636 0.39054 0.40518 0.42030 0.43592 0.45205 0.46870 0.48589

51 52 53 54 55 56 57 58

44

45 46 47 48 49 50

AppendixD


0:3 Condensed Water Temp. Humidity Ratio, t, of Ib,.JIbda Ws

Vag" 0.224 0.233 0.242 0.251 0.261 0.271 0.281 0.292 0.303 0.315 0.326

Entropy, Btu/(Ib dry air) . of


Volume, dry air

ft 311b

va g 13.295 13.329 13.364 13.398 13.433 13.468 13.504 13.540 13.577 13.613 13.650

ha 14.174 14.415 14.655 14.895 15.135 15.376 15.616 15.856 16.097 16.337 16.577

hag" 11.618 12.052 12.502 12.966 13.446 13.942 14.454 14.983 15.530 16.094 16.677

hga

sa g

Vapor Enthalpy, Entropy, Press., Btullb Btullb' of in.Hg pg" Sw hw

Temp., OF

25.792 26.467 27.157 27.862 28.582 29.318 30.071 30.840 31.626 32.431 33.254

Sa 0.02901 0.02947 0.02994 0.03040 0.03086 0.03132 0.03178 0.03223 0.03269 0.03315 0.03360

sag" 0.02358 0.02442 0.02528 0.02617 0.02709 0.02804 0.02902 0.03003 0.03107 0.03214 0.03325

0.05259 0.05389 0.05522 0.05657 0.05795 0.05936 0.06080 0.06226 0.06376 0.06529 0.06685

27.11 28.11 29.12 30.11 31.11 32.11 33.11 34.11 35.11 36.11 37.11

0.0536 0.0555 0.0575 0.0594 0.0613 0.0632 0.0651 0.0670 0.0689 0.D708 0.0727

0.50363 0.52193 0.54082 0.56032 0.58041 0.60113 0.62252 0.64454 0.66725 0.69065 0.71479

59 60 61 62 63 64 65 66 67 68 69

59 60 61 62 63 64 65 66 67 68 69

0.010692 0.011087 0.011496 0.011919 0.012355 0.012805 0.013270 0.013750 0.014246 0.014758 0.015286

va 13.071 13.096 13.122 13.147 13.172 13.198 13.223 13.248 13.273 13.299 13.324

70 71 72 73 74 75 76 77 78 79

0.015832 0.016395 0.016976 0.017575 0.018194 0.018833 0.019491 0.020170 0.020871 0.021594

13.349 13.375 13.400 13.425 13.450 13.476 13.501 13.526 13.551 13.577

0.339 0.351 0.365 0.378 0.392 0.407 0.422 0.437 0.453 0.470

13.688 13.726 13.764 13.803 13.843 13.882 13.923 13.963 14.005 14.046

16.818 17.058 17.299 17.539 17.779 18.020 18.260 18.500 18.741 18.981

17.279 17.901 18.543 19.204 19.889 20.595 21.323 22.075 22.851 23.652

34.097 34.959 35.841 36.743 37.668 38.615 39.583 40.576 41.592 42.633

0.03406 0.03451 0.03496 0.03541 0.03586 0.03631 0.03676 0.03721 0.03766 0.03811

0.03438 0.03556 0.03677 0.03801 0.03930 0.04062 0.04199 0.04339 0.04484 0.04633

0.06844 0.07007 0.07173 0.07343 0.07516 0.07694 0.07875 0.08060 0.08250 0.08444

38.11 39.11 40.11 41.11 42.11 43.11 44.10 45.10 46.10 47.10

0.0746 0.0765 0.0783 0.0802 0.0821 0.0840 0.0858 0.0877 0.0896 0.0914

0.73966 0.76567 0.79167 0.81882 0.84684 0.87567 0.90533 0.93589 0.96733 0.99970

70 71 72 73 74 75 76 77 78 79

80 81 82 83 84 85 86 87 88 89

0.022340 0.023109 0.023902 0.024720 0.025563 0.026433 0.027329 0.028254 0.029208 0.030189

13.602 13.627 13.653 13.678 13.703 13.728 13.754 13.779 13.804 13.829

0.487 0.505 0.523 0.542 0.561 0.581 0.602 0.624 0.646 0.669

14.089 14.132 14.175 14.220 14.264 14.310 14.356 14.403 14.450 14.498

19.222 19.462 19.702 19.943 20.183 20.424 20.664 20.905 21.145 21.385

24.479 25.332 26.211 27.120 28.055 29.021 30.017 31.045 32.105 33.197

43.701 44.794 45.913 47.062 48.238 49.445 50.681 51.949 53.250 54.582

0.03855 0.03900 0.03944 0.03988 0.04033 0.04077 0.04121 0.04165 0.04209 0.04253

0.04787 0.04945 0.05108 0.05276 0.05448 0.05626 0.05809 0.05998 0.06192 0.06392

0.08642 0.08844 0.09052 0.09264 0.09481 0.09703 0.09930 0.10163 0.10401 0.10645

48.10 49.10 50.10 51.09 52.09 53.09 54.09 55.09 56.09 57.09

0.0933 0.0951 0.0970 0.0988 0.1006 0.1025 0.1043 0.1061 0.1080 0.1098

1.03302 1.06728 1.10252 1.13882 1.17608 1.21445 1.25388 1.29443 1.33613 1.37893

80 81 82 83 84 85 86 87 88 89

90 91 92 93 94 95 96 97 98 99

0.031203 0.032247 0.033323 0.034433 0.035577 0.036757 0.037972 0.039225 0.040516 0.041848

13.855 13.880 13.905 13.930 13.956 13.981 14.006 14.032 14.057 14.082

0.692 0.717 0.742 0.768 0.795 0.823 0.852 0.881 0.912 0.944

14.547 14.597 14.647 14.699 14.751 14.804 14.858 14.913 14.969 15.026

21.626 21.866 22.107 22.347 22.588 22.828 23.069 23.309 23.550 23.790

34.325 35.489 36.687 37.924 39.199 40.515 41.871 43.269 44.711 46.198

55.951 57.355 58.794 60.271 61.787 63.343 64.940 66.578 68.260 69.988

0.04297 0.06810 0.04384 0.04427 0.04471 0.04514 0.04558 0.04601 0.04644 0.04687

0.06598 0.04340 0.07028 0.07253 0.07484 0.07722 0.07968 0.08220 0.08480 0.08747

0.10895 0.11150 0.11412 0.11680 0.11955 0.12237 0.12525 0.12821 0.13124 0.13434

58.08 59.08 60.08 61.08 62.08 63.08 64.07 65.07 66.07 67.07

0.1116 0.1134 0.1152 0.1170 0.1188 0.1206 0.1224 0.1242 0.1260 0.1278

1.42298 1.46824 1.51471 1.56248 1.61l54 1.66196 1.71372 1.76685 1.82141 1.87745

90 91 92 93 94 95 96 97 98 99

100 101 102 103 104 105 106 107 108 109

0.043219 0.044634 0.046090 0.047592 0.049140 0.050737 0.052383 0.054077 0.055826 0.057628

14.107 14.133 14.158 14.183 14.208 14.234 14.259 14.284 14.309 14.335

0.976 1.010 1.045 1.081 1.118 1.156 1.196 1.236 1.279 1.322

15.084 15.143 15.203 15.264 15.326 15.390 15.455 15.521 15.588 15.657

24.031 24.271 24.512 24.752 24.993 25.233 25.474 25.714 25.955 26.195

47.730 49.312 50.940 52.621 54.354 56.142 57.986 59.884 61.844 63.866

71.761 73.583 75.452 77.373 79.346 81.375 83.460 85.599 87.799 90.061

0.04730 0.04773 0.04816 0.04859 0.04901 0.04944 0.04987 0.05029 0.05071 0.05114

0.09022 0.09306 0.09597 0.09897 0.10206 0.10525 0.10852 0.11189 0.1l537 0.11894

0.13752 0.14079 0.14413 0.14756 0.15108 0.15469 0.15839 0.16218 0.16608 0.17008

68.07 69.07 70.06 71.06 72.06 73.06 74.06 75.06 76.05 77.05

0.1296 0.1314 0.1332 0.1349 0.1367 0.1385 0.1402 0.1420 0.1438 0.1455

1.93492 1.99396 2.05447 2.11661 2.18037 2.24581 2.31297 2.38173 2.45232 2.52473

100 101 102 103 104 105 106 107 108 109

110 III 112 113 114 115 116 117 118 119

0.059486 0.061401 0.063378 0.065411 0.067512 0.069676 0.071908 0.074211 0.076586 0.079036

14.360 14.385 14.41l 14.436 14.461 14.486 14.512 14.537 14.562 14.587

1.367 1.414 1.462 1.511 1.562 1.615 1.670 1.726 1.784 1.844

15.727 15.799 15.872 15.947 16.023 16.101 16.181 16.263 16.346 16.432

26.436 26.677 26.917 27.158 27.398 27.639 27.879 28.120 28.361 28.601

65.950 68.099 70.319 72.603 74.964 77.396 79.906 82.497 85.169 87.927

92.386 94.776 97.237 99.760 102.362 105.035 107.786 110.617 1l3.530 1l6.528

0.05156 0.05198 0.05240 0.05282 0.05324 0.05366 0.05408 0.05450 0.05492 0.05533

0.12262 0.12641 0.13032 0.13434 0.13847 0.14274 0.14713 0.15165 0.15631 0.16111

0.17418 0.17839 0.18272 0.18716 0.19172 0.19640 0.20121 0.20615 0.21122 0.21644

78.05 79.05 80.05 81.05 82.04 83.04 84.04 85.04 86.04 87.04

0.1473 0.1490 0.1508 0.1525 0.1543 0.1560 0.1577 0.1595 0.1612 0.1629

2.59891 2.67500 2.75310 2.83291 2.91491 2.99883 3.08488 3.17305 3.26335 3.35586

110 III 112 113 114 115 116 117 118 119

120 121 122 123 124 125 126 127 128 129

0.081560 0.084169 0.086860 0.089633 0.092500 0.095456 0.098504 0.101657 0.104910 0.108270

14.613 14.638 14.663 14.688 14.714 14.739 14.764 14.789 14.815 14.840

1.906 1.971 2.037 2.106 2.176 2.250 2.325 2.404 2.485 2.569

16.519 16.609 16.700 16.794 16.890 16.989 17.090 17.193 17.299 17.409

28.842 29.083 29.323 29.564 29.805 30.045 30.286 30.527 30.767 31.008

90.770 93.709 96.742 99.868 103.102 106.437 109.877 113.438 117.111 120.908

1l9.612 122.792 126.065 129.432 132.907 136.482 140.163 143.965 147.878 151.916

0.05575 0.05616 0.05658 0.05699 0.05740 0.05781 0.05823 0.05864 0.05905 0.21800

0.16605 0.17115 0.17640 0.18181 0.18739 0.19314 0.19907 0.20519 0.21149 0.21810

0.22180 0.22731 0.23298 0.23880 0.24480 0.25096 0.25729 0.26382 0.27054 0.27745

88.04 89.04 90.03 91.03 92.03 93.03 94.03 95.03 96.03 97.03

0.1647 0.1664 0.1681 0.1698 0.1715 0.1732 0.1749 0.1766 0.1783 0.1800

3.45052 3.54764 3.64704 3.74871 3.85298 3.95961 4.06863 4.18046 4.29477 4.41181

120 121 122 123 124 125 126 127 128 129

"Note that the values for subscript g are actually solids (s) until O°F.

Fundamentals ofTllermodynamics and Psycllrometrics

AppendixD

0:4 Condensed Water Temp. Humidity t, Ratio, OF Ib,.llbda Ws


Volume, dry air

ft311b va

V a/

va g

ha

Entropy, Btu/(Ih dry air) . OF

haga

h a g

Sa

Saga

S a g

Vapor Enthalpy, Entropy, Press., Btullb Btullb· OF in.Hg hO'

SO'

pg"

Temp., OF

130 131 132 133 134 135 136 137 138 139

0.111738 0.115322 0.119023 0.122855 0.126804 0.130895 0.135124 0.139494 0.144019 0.148696

14.865 14.891 14.916 14.941 14.966 14.992 15.017 15.042 15.067 15.093

2.655 2.745 2.837 2.934 3.033 3.136 3.242 3.352 3.467 3.585

17.520 17.635 17.753 17.875 17.999 18.127 18.259 18.394 18.534 18.678

31.249 31.489 31.730 31.971 32.212 32.452 32.693 32.934 33.175 33.415

124.828 128.880 133.066 137.403 141.873 146.504 151.294 156.245 161.374 166.677

156.076 160.370 164.796 169.374 174.084 178.957 183.987 189.179 194.548 200.092

0.05986 0.06027 0.06068 0.06109 0.06149 0.06190 0.06230 0.06271 0.06311 0.06351

0.22470 0.23162 0.23876 0.24615 0.25375 0.26161 0.26973 0.27811 0.28707 0.29602

0.28457 0.29190 0.29944 0.30723 0.31524 0.32351 0.33203 0.34082 0.35018 0.35954

98.03 99.02 100.02 101.02 102.02 103.02 104.02 105.02 106.02 107.02

0.1817 0.1834 0.1851 0.1868 0.1885 0.1902 0.1919 0.1935 0.1952 0.1969

4.53148 4.65397 4.77919 4.90755 5.03844 5.17258 5.30973 5.44985 5.59324 5.73970

130 131 132 133 134 135 136 137 138 139

140 141 142 143 144 145 146 147 148 149

0.153538 0.158643 0.163748 0.169122 0.174694 0.180467 0.186460 0.192668 0.199110 0.205792

15.118 15.143 15.168 15.194 15.219 15.244 15.269 15.295 15.320 15.345

3.708 3.835 3.967 4.103 4.245 4.392 4.545 4.704 4.869 5.040

18.825 18.978 19.135 19.297 19.464 19.637 19.815 19.999 20.189 20.385

33.656 33.897 34.138 34.379 34.620 34.860 35.101 35.342 35.583 35.824

172.168 177.857 183.754 189.855 196.183 202.740 209.550 216.607 223.932 231.533

205.824 211.754 217.892 224.233 230.802 237.600 244.651 251.949 259.514 267.356

0.06391 0.06431 0.06471 0.06511 0.06551 0.06591 0.06631 0.06671 0.06710 0.06750

0.30498 0.31456 0.32446 0.33470 0.34530 0.35626 0.36764 0.37941 0.39160 0.40424

0.36890 0.37887 0.38918 0.39981 0.41081 0.42218 0.43395 0.44611 0.45871 0.47174

108.02 109.02 110.02 111.02 112.02 113.02 114.02 115.02 116.02 117.02

0.1985 0.2002 0.2019 0.2035 0.2052 0.2068 0.2085 0.2101 0.2118 0.2134

5.88945 6.04256 6.19918 6.35898 6.52241 6.68932 6.86009 7.03435 7.21239 7.39413

140 141 142 143 144 145 146 147 148 149

150 151 152 153 154 155 156 157 158 159

0.212730 0.219945 0.227429 0.235218 0.243309 0.251738 0.260512 0.269644 0.279166 0.289101

15.370 15.396 15.421 15.446 15.471 15.497 15.522 15.547 15.572 15.598

5.218 5.404 5.596 5.797 6.005 6.223 6.450 6.686 6.933 7.190

20.589 20.799 21.017 21.243 21.477 21.720 21.972 22.233 22.505 22.788

36.064 36.305 36.546 36.787 37.028 37.269 37.510 37.751 37.992 38.233

239.426 247.638 256.158 265.028 274.245 283.849 293.849 304.261 315.120 326.452

275.490 283.943 292.705 301.816 311.273 321.118 331.359 342.012 353.112 364.685

0.06790 0.06829 0.06868 0.06908 0.06947 0.06986 0.07025 0.07065 0.07104 0.07143

0.41735 0.43096 0.44507 0.45973 0.47494 0.49077 0.50723 0.52434 0.54217 0.56074

0.48524 0.49925 0.51375 0.52881 0.54441 0.56064 0.57749 0.59499 0.61320 0.63216

118.02 119.02 120.02 121.02 122.02 123.02 124.02 125.02 126.02 127.02

0.2151 0.2167 0.2184 0.2200 0.2216 0.2233 0.2249 0.2265 0.2281 0.2297

7.57977 7.76958 7.96306 8.16087 8.36256 8.56871 8.77915 8.99378 9.21297 9.43677

150 151 152 153 154 155 156 157 158 159

160 161 162 163 164 165 166 167 168 169

0.29945 0.31027 0.32156 0.33336 0.34572 0.35865 0.37220 0.38639 0.40131 0.41698

15.623 15.648 15.673 15.699 15.724 15.749 15.774 15.800 15.825 15.850

7.459 7.740 8.034 8.341 8.664 9.001 9.355 9.726 10.117 10.527

23.082 23.388 23.707 24.040 24.388 24.750 25.129 25.526 25.942 26.377

38.474 38.715 38.956 39.197 39.438 39.679 39.920 40.161 40.402 40.643

338.263 350.610 363.501 376.979 391.095 405.865 421.352 437.578 454.630 472.554

376.737 389.325 402.457 416.175 430.533 445.544 461.271 477.739 495.032 513.197

0.07181 0.07220 0.07259 0.07298 0.07337 0.07375 0.07414 0.07452 0.07491 0.07529

0.58007 0.60025 0.62128 0.64325 0.66622 0.69022 0.71535 0.74165 0.76925 0.79821

0.65188 0.67245 0.69388 0.71623 0.73959 0.76397 0.78949 0.81617 0.84415 0.87350

128.02 129.02 130.03 131.03 132.03 133.03 134.03 135.03 136.03 137.04

0.2314 0.2330 0.2346 0.2362 0.2378 0.2394 0.2410 0.2426 0.2442 0.2458

9.6648 9.8978 10.1353 10.3776 10.6250 10.8771 11.1343 11.3965 11.6641 11.9370

160 161 162 163 164 165 166 167 168 169

170 171 172 173 174 175 176 177 178 179

0.43343 0.45079 0.46905 0.48829 0.50867 0.53019 0.55294 0.57710 0.60274 0.63002

15.875 15.901 15.926 15.951 15.976 16.002 16.027 16.052 16.078 16.103

10.959 11.414 11.894 12.400 12.937 13.504 14.103 14.741 15.418 16.139

26.834 27.315 27.820 28.352 28.913 29.505 30.130 30.793 31.496 32.242

40.884 41.125 41.366 41.607 41.848 42.089 42.331 42.572 42.813 43.054

491.372 511.231 532.138 554.160 577.489 602.139 628.197 655.876 685.260 716.524

532.256 552.356 573.504 595.767 619.337 644.229 670.528 698.448 728.073 759.579

0.07567 0.07606 0.07644 0.07682 0.07720 0.07758 0.07796 0.07834 0.07872 0.07910

0.82858 0.86058 0.89423 0.92962 0.96707 1.00657 1.04828 1.09253 1.13943 1.18927

0.90425 0.93664 0.97067 1.00644 1.04427 1.08416 1.12624 1.17087 1.21815 1.26837

138.04 139.04 140.04 141.04 142.04 143.05 144.05 145.05 146.05 147.06

0.2474 0.2490 0.2506 0.2521 0.2537 0.2553 0.2569 0.2585 0.2600 0.2616

12.2149 12.4988 12.7880 13.0823 13.3831 13.6894 14.0010 14.3191 14.6430 14.9731

170 171 172 173 174 175 176 177 178 179

180 181 182 183 184 185 186 187 188 189

0.65911 0.69012 0.72331 0.75885 0.79703 0.83817 0.88251 0.93057 0.98272 1.03951

16.128 16.153 16.178 16.204 16.229 16.254 16.280 16.305 16.330 16.355

16.909 17.730 18.609 19.551 20.564 21.656 22.834 24.111 25.498 27.010

33.037 33.883 34.787 35.755 36.793 37.910 39.113 40.416 41.828 43.365

43.295 43.536 43.778 44.019 44.260 44.501 44.742 44.984 45.225 45.466

749.871 785.426 823.487 864.259 908.061 955.261 1006.149 1061.314 1121.174 1186.382

793.166 828.962 867.265 908.278 952.321 999.763 1050.892 1106.298 1166.399 1231.848

0.07947 0.07985 0.08023 0.08060 0.08098 0.08135 0.08172 0.08210 0.08247 0.08284

1.24236 1.29888 1.35932 1.42396 1.49332 1.56797 1.64834 1.73534 1.82963 1.93221

1.32183 1.37873 1.43954 1.50457 1.57430 1.64932 1.73006 1.81744 1.91210 2.01505

148.06 149.06 150.06 151.07 152.07 153.07 154.08 155.08 156.08 157.09

0.2632 0.2647 0.2663 0.2679 0.2694 0.2710 0.2725 0.2741 0.2756 0.2772

15.3097 15.6522 16.0014 16.3569 16.7190 17.0880 17.4634 17.8462 18.2357 18.6323

180 181 182 183 184 185 186 187 188 189

190 191 192 193 194 195 196 197 198 199 200

1.10154 1.16965 1.24471 1.32788 1.42029 1.52396 1.64070 1.77299 1.92472 2.09975 2.30454

16.381 16.406 16.431 16.456 16.481 16.507 16.532 16.557 16.583 16.608 16.633

28.661 30.476 32.477 34.695 37.161 39.928 43.046 46.580 50.636 55.316 60.793

45.042 46.882 48.908 51.151 53.642 56.435 59.578 63.137 67.218 71.923 77.426

45.707 45.949 46.190 46.431 46.673 46.914 47.155 47.397 47.638 47.879 48.121

1257.614 1303.321 1335.834 1381.783 1422.047 1468.238 1517.581 1564.013 1623.758 1670.430 1742.879 1789.793 1877.032 1924.188 2029.0692076.466 2203.464 2251.102 2404.6682452.547 2640.084 2688.205

0.08321 0.08359 0.08396 0.08433 0.08470 0.08506 0.08543 0.08580 0.08617 0.08653 0.08690

2.04412 2.16684 2.30193 2.45144 2.61738 2.80332 3.01244 3.24914 3.52030 3.83275 4.19787

2.12733 2.25043 2.38589 2.53576 2.70208 2.88838 3.09787 3.33494 3.60647 3.91929 4.28477

158.09 159.09 160.10 161.10 162.11 163.11 164.12 165.12 166.13 167.13 168.13

0.2787 19.0358 0.2803 19.4468 0.2818 19.8652 0.2834 20.2913 0.2849 20.7244 0.2864 21.1661 0.2880 21.6152 0.289522.0714 0.2910 22.5367 0.2926 23.0092 0.2941 23.4906

190 191 192 193 194 195 196 197 198 199 200

aNote that the values for subscript g are acrually solids (s) until O°F.

AppendixD


E: 1

ASHRAE Psychrometric Chart II:

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II: Q

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NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY

SEA LEVEL Fundamentals ofThermodynamics and Psychrometries

AppendixE

Skill Development Exercises

Contents • Chapter 1

Introduction to HVAC

• Chapter 2

Systems, Properties, States and Processes

• Chapter 3

Property Diagrams for Pure Substances

• Chapter 4


• Chapter 5


• Chapter 6

Heat and Work

• Chapter 7

First Law of Thermodynamics Applied to Closed Systems

• Chapter 8

First Law of Thermodynamics Applied to Open Systems

• Chapter 9

Applications of the First Law of Thermodynamics

• Chapter 10

The Carnot Cycle

• Chapter 11

Refrigeration Cycles

• Chapter 12

Moist Air as a Mixture of Ideal Gases

• Chapter 13

Properties of Moist Air

• Chapter 14

The Psychrometric Chart

• Chapter 15

Air-Conditioning Processes on the Psychrometric Chart

Instructions After reading each chapter, answer all of the questions pertaining to that chapter on the following worksheets. Remove and send your completed worksheets to ASHRAE. Be sure to include your name and address


Skill Development Exercises

1: 1

Skill Development Exercises for Chapter 1

Complete the following questions by writing your answers on these worksheets. Be sure to include your name and address below. Send the completed worksheets to ASHRAE. Name Company/Organization Address

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1-01. List the seven main air-conditioning processes, give a one-sentence description of each process, and give another one-sentence reason why the process is likely to be found in a hospital.

Question Sheet


1: 2

1-02. Define what is meant by air-conditioning.


Question Sheet

1: 3

1-03. Name five pieces of normal HVAC equipment for which the application of thermodynamic principles is essential to predict performance.

1-04. In your own words, state what is meant by the conservation of energy principle.

Question Sheet


1: 4

1-05. Provide two examples of a process that goes in one direction but will not go in the reverse direction without applying an external energy source.

1-06.

What is the difference between dry air and moist air?

Send your completed worksheets to: ASHRAElnc. Education Department 1791 Tullie Circle NE Atlanta, GA 30329-2305 Phone: 404/636-8400 Fax: 404/321-5478


Question Sheet

2: 1


Complete the following questions by writing your answers on these worksheets. Be sure to include your name and address below. Send the completed worksheets to ASHRAE. Name CompanyIOrganization Address City

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2-01. Using examples, explain the main difference between open and closed systems.

Question Sheet


2:2

2-02. Automobile engines are usually cooled by antifreeze that is circulated between the engine and the radiator by a centrifugal pump. Air is blown over the radiator to cool the antifreeze. Sketch the following systems and state whether they are open or closed: • Antifreeze in the pump • Antifreeze in the radiator • Antifreeze in the pump, radiator and engine water jacket • Antifreeze in the radiator and air in the fan


Question Sheet

2: 3

2-03.

2-04.

Describe the differences between thermal, phase, chemical and nuclear energies.

Give an expression for the specific total energy of a system in terms of the internal, potential and kinetic energies.

Question Sheet


2:4

2-05.

Sketch on a p- V diagram what happens when trapped air expands in a closed cylinder with a piston. Be sure to label the initial and final states, the path and the major properties.

Chapter 2 SystelllS, Properties, States and Processes

Question Sheet

2: 5

2-06.

Sketch on a p- V diagram a four-process cycle that has process 1 to 2 as an isobaric decrease in volume, 2 to 3 as an isometric increase in pressure, 3 to 4 as an isobaric increase in volume, and 4 to 1 as an isometric decrease in pressure.

Question Sheet


2: 6

2-07.

Sketch on a t-p diagram the following four-process cycle: 1 to 2 isothermal; 2 to 3 isobaric; 3 to 4 isothermal; and 4 to 1 isobaric.


Chapter 2 SystelllS, Properties, States and Processes

Question Sheet

3: 1


Complete the following questions by writing your answers on these worksheets. Be sure to include your name and address below. Send the completed worksheets to ASHRAE. Name Company/Organization

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3-01.

State

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Give the state postulate and explain what a simple compressive substance is.

Question Sheet


3:2

3-02. Explain the differences between absolute, gage and vacuum pressures.

3-03.

Give the typical units for the following properties and indicate which properties are intensive and which are extensive: specific enthalpy, total volume, pressure, temperature, specific entropy, mass and quality.

3-04.

Give the temperature of212 cF in units of degrees Rankine, degrees Kelvin and cC.


Question Sheet

3: 3

3-05.

Sketch a T-v diagram for a pure substance and show the saturated liquid line, the saturated vapor line, the critical point and a constant pressure process line from compressed liquid to superheated vapor. Also include a constant temperature line.

3-06.

Sketch a p-v diagram for a pure substance and show the saturated liquid line, the saturated vapor line, the critical point and a constant temperature process line from compressed liquid to superheated vapor. Also include a constant pressure line.

Question Sheet


3:4

3-07.

A saturated mixture of water is at 230°F and 103 psia. The specific volume of the liquid is vj = 0.0178 ft3/lb m and the specific volume of the vapor is vg = 4.312 ft3/lb m. Ifthe quality is x = 0.5, what is the specific volume of the mixture?

3-08.

A fixed volume container initially contains a pure substance in superheated form. The container is cooled until there is a mixture of liquid and vapor inside. Using T-v and p-v diagrams, describe the changes taking place in this constant volume process as the contents are cooled.



Question Sheet

4: 1



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4-01.

The saturated temperature of water is lOO°F. What is the pressure? What are the liquid and vapor values of specific volume, specific enthalpy and specific entropy? (Use the table in Appendix B.)

4-02.

The saturated temperature of water is 100°F and the quality is 50%. What is the value of specific volume and specific enthalpy? (Use the table in Appendix B.)

Question Sheet

Chapter 4


4:2

4-03.

The saturation pressure for R-22 is 50 psia. What is the saturation temperature? (U se the table in Appendix C.)

4-04.

The saturation temperature for water is 231.5°F and the quality is 0.2. What is the pressure and the specific enthalpy? (Use the table in Appendix B.)

Chapter 4


Question Sheet

4: 3

4-05.

The enthalpy for R-22 is 50 Btu/Ibm and the pressure is 40 psia. Is the refrigerant in the compressed liquid, saturated or superheated vapor state? What is the quality? (Use the chart in Appendix C.)

4-06.

The enthalpy for R-22 is 50 Btu/Ibm and the temperature is 100°F. Is the refrigerant in the compressed liquid, saturated or superheated vapor state? What is the pressure and the quality? (Use the table in Appendix C.)

Question Sheet

Chapter 4


4:4

4-07. The enthalpy for R-22 is 150 Btu/Ibm and the pressure is 40 psia. Is the refrigerant in the compressed liquid, saturated or superheated vapor state? What is the value of the density? (Use the chart in Appendix C.)

Send your completed worksheets to: ASHRAEInc. Education Department 1791 Tullie Circle NE Atlanta, GA 30329-2305 Phone: 404/636-8400 Fax: 404/321-5478

Chapter 4


Question Sheet

5: 1


Complete the following questions by writing your answers on these worksheets. Be sure to include your name and address below. Send the completed worksheets to ASHRAE. Name CompanyIOrganization

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5-01.

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Air is often treated as an ideal gas. For the conditions of p = 14.696 psia and t = 72°F, find the specific volume for dry air, va' in Appendix D, and compare the value read in the table to the value caclulated using the Ideal Gas Law. Use T= t + 459.67 to convert from OF to R. Use the value of gas constant given in Table 5-1. Express the answer as a percent error by treating the tabulated value as being correct.

Question Sheet

Chapter 5


5:2

5-02.

Water vapor is sometimes treated as an ideal gas. For the condition ofp = 67.0341 psia, find the specific volume for saturated vapor, vg , in Appendix B, and compare the value read in the table to the value caclulated using the Ideal Gas Law. Use T= t + 459.67 to convert from OF to R. Use the value of gas constant given in Table 5-1. Express the answer as a percent error by treating the tabulated value as being correct.

5-03.

Repeat Exercise 5-02 except for conditions of saturated vapor at p = 0.3628 psia. Compare the error found with the error determined in Exercise 5-02 and comment on the difference.

Cltapter 5 Ideal Gas Law and Air Tables

Question Slteet

5: 3

5-04.

A room of dimensions 10ft x 20 ft x 8 ft at atmospheric pressure contains 120 lb of air. Assume an ideal gas and estimate the temperature in the room in of.

5-05.

Consider the constant pressure heating of air from 70°F to 200°F. If the pressure is 14.7 psia, what is the ratio of specific volumes before and after the heating?

11/

Question Sheet

Chapter 5

I deal Gas Law and Air Tables

5:4

5-06.

Chapter 5

Make a table listing the values of internal energy, enthalpy and entropy for air at 14.696 psia, with temperatures ranging from Oop to lOOoP in steps of lOoP, using the coefficients of specific heat suggested for air at 80 0 P in the text. Use reference values of zero internal energy, zero enthalpy and zero entropy at OOp.


Question Sheet

5: 5

5-07.

Find the values of enthalpy for dry air in the table in Appendix D, at the same temperature entries used in the table created in Exercise 5-06 and compare calculated and look-up values. Calculate percent error assuming the tabulated values to be correct.


Question Sheet

Chapter 5

Ideal Gas Law andAir Tables

6: 1


Complete the following questions by writing your answers on these worksheets. Be sure to include your name and address below. Send the completed worksheets to ASHRAE. Name CompanyIOrganization Address

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Provide a description of heat, and describe how it is different from internal energy.

Question Sheet

Chapter 6

Heat and Work

6:2

6-02. A 0.35 Ibm potato at 70°F is placed in an oven at 350°F. After 45 minutes, the potato temperature is 350°F. Assume the coefficient of specific heat for the potato is cp = 1.0 BtU/lbm·oF. If the reference datum for internal energy is U= 0 at t = OaF, then estimate the internal energy content of the potato at the beginning and end of the 45 minutes. What has been the heat during the 45 minutes and is it positive or negative? What was the average rate of heat transfer?

6-03. Provide a description of work in general, and then give short (one- or two-sentence) descriptions of boundary work and shaft work.

Chapter 6

Heat and Work

Question Sheet

6: 3

6-04.

Provide a description of flow work, and say how it is included in enthalpy.

6-05. A frictionless piston-cylinder device contains 2 Ibm of water at 212°P and quality x = 0.4. The fixed weight piston is not attached to a shaft. Heat is added to the water until it is all saturated vapor. Calculate the work done by the system.

Question Sheet

Chapter 6

Heat and Work

6:4

6-06.

Chapter 6

A piston-cylinder device initially contains 0.2 ft3 of air at 80 psia and 76°F. The air is expanded to 0.9 ft3 in such a way that the temperature remains constant. Calculate the work done by the system, the final pressure and the mass of air.

Heat and Work

Question Sheet

6: 5

6-07.

In the isentropic compression of air in a cylinder, the volume is reduced from 1.5 ft3 to 0.5 ft3. The ratio of specific heats for air is k= 1.4. The initial pressure is 14.7 psia and the initial temperature is 45°F. Assume the air acts as an ideal gas. Estimate the final pressure, the final temperature and the work done.


Question Sheet

Chapter 6

Heat and Work

7: 1



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7-01.

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Give a one-page description of the heat, work and energy change terms in the First Law equation for a closed system and describe how they relate to each other (a word description of the conservation of energy principle).

Question Sileet

Cllapter 7 First Law and Closed Systems

7:2

7-02. A rigid tank with a volume of 3 ft3 is initially filled with air at 15 psi a and 20°F. The air is heated to 200°F. Treat the air as an ideal gas.' Apply the Ideal Gas Law to determine the mass of air in the tank. Apply the continuity equation to determine the final pressure in the tank. Assuming the air in the tank to constitute a closed system, apply the First Law to determine the heat transferred during the process.


Question Sheet

7: 3

7-03.

A rigid tank with a volume of 1.5 ft3 is initially filled with R-22 in a saturated vapor state at 140°F. The refrigerant is cooled until the pressure is 57.083 psia. Sketch the process on ap-v diagram. Use the refrigerant tables (in Appendix C-l) to determine the mass of refrigerant in the tank. Determine the final temperature. Assuming the refrigerant in the tank to constitute a closed system, apply the First Law and determine the heat transferred during the process.

Question Sheet

Chapter 7

First Law and Closed Systems

7:4

7-04.

Chapter 7

A piston-cylinder device contains 10 lb of air which is maintained at a constant pressure of 150 psia by a floating piston of constant weight. How much heat must be added to raise the temperature from 20°F to 300°F?


/11

Question Sheet

7: 5

7-05. A piston-cylinder contains 1 Ibm of saturated liquid R-22 at 20°F. The refrigerant is maintained at constant pressure by a floating piston. Determine the work and heat if the refrigerant is heated until the quality is 50%. Sketch the process on a p-v diagram.

Question Sheet

Chapter 7


7: 6

7-06.

Describe experiments in which the First Law may be used to determine the specific heats of an ideal gas of unknown properties placed in a rigid tank, and in a cylinder in which the pressure is kept constant with a floating, frictionless piston.



Question Sheet

8: 1


Complete the following questions by writing your answers on these worksheets. Be sure to include your name and address below. Send the completed worksheets to ASHRAE. Name CompanyIOrganization Address City

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Telephone__________________________

8-01.

Fax

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State

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The inlet temperature to a centrifugal flow, 0.8 kW air compressor is 20°F. The steady mass flow rate is 0.04 lbn/s. If the compressor is assumed adiabatic and the kinetic and potential energies are assumed negligible, then estimate the outlet temperature of the compressor.

Question Sheet

Chapter 8

First Law and Open Systems

8:2

8-02. The inlet condition to a centrifugal flow, R-134a compressor is saturated vapor at 20°F. The steady mass flow rate is 0.08 Ibm Is. If the heat lost from the compressor is 800 Btulh and the work to run the compressor is 3.0 kW, estimate the specific enthalpy of the refrigerant leaving the compressor. Neglect the kinetic and potential energies of the flow.

8-03. Air enters an adiabatic turbine at 300°F and with negligible velocity. The rate of doing work is 2.0 kW. The outlet temperature is 20°F. If the outlet velocity is 300 fils, then calculate the mass flow rate of air.

Chapter 8


Question Sheet

8: 3

8-04. A mixing tank has two steady inflows and one steady outflow. The conditions of the first inflow are superheated steam at t] = 350°F, h] = 1,140 Btu/Ibm' and flow rate rill = 30 Ibm Ih. The second inflow is compressed liquid at t2 = 70°F, h2 = 38 Btu/Ibm' Ifthe rate of heat removal from the tank is 25,000 Btu/h, then estimate the inflow of water needed to result in the outflow being saturated liquid water at 212°F. Neglect changes in kinetic and potential energy.

Question Sheet

Chapter 8


8:4

8-05.

Two air streams are mixed together in an adiabatic mixer to produce an output stream of air at 95°F. lfthe two incoming streams are at 120°F and 35°F respectively, find the ratio of volume flow rates of the hot to cold air to produce the 95°F air.


Chapter 8


Question Sheet

9: 1



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Telephone_ _ _ _ _ _ _ _ _ _ _ __

9-01.

Fax

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----------------------------

Determine the outlet temperature and mass flow rate of a 0.7 k W air compressor that compresses air from 15 psia to 100 psia. The inlet temperature is 72°F. Assume isentropic compression of an ideal gas, k = 1.4, Cp = 0.24 Btullb111 .oF.

Question Sheet


9:2

9-02.

Determine the outlet temperature and mass flow rate of a 0.7 k W air compressor that compresses air from 15 psia to 150 psia. The inlet temperature is 72°F. Assume polytropic compression of an ideal gas, n = 1.3, Cp = 0.24 Btullbm.of and a heat loss rate of 0.1 KW.

9-03. The inlet condition to a R-22 compressor is saturated vapor at -40°F. If the specific work for the isentropic compression is 30 Btu/Ibm' then calculate the specific enthalpy of the refrigerant at the compressor exit. Sketch the process on the p-h diagram and use Appendix C-2 to estimate the delivery pressure. (The diagram of Appendix C-2 is not easy to read, so an approximate value is acceptable.)


Question Sheet

9: 3

9-04. The inlet to an adiabatic steam turbine is superheated steam at 1,OOO°F, 400 psia and enthalpy of 1,526 Btullb The steam leaves the turbine at 150°F, with a quality of 0.90. Determine the specific work in Btullb111 of the turbine. 111.

9-05.

The inlet condition to a throttling valve in a R -134a refrigerator is saturated liquid at 110°F. Determine the outlet condition of the refrigerant (quality) if the outlet pressure is 21.162 psia.

Question Sheet


9:4

9-06.

Consider a pipe-in-pipe, counterflow, water-to-air heat exchanger. The water flow rate ofO.1lbm /s enters at 200°F and leaves at 100°F. The air flow rate ofO.9lbm /s enters at 72°F. The coefficient of specific heat of water is Cp = 1.0 Btullbm .oF, and for air Cp = 0.24 Btullbm .oF. For both water and air, the change in enthalpy can be expressed as the coefficient of specific heat times the change in temperature. Complete the following: • Sketch the heat exchanger and the temperature profiles of the two streams through the length of the heat exchanger. • Determine the outlet temperature of the air. • Determine the LMTD from the expression:

• Determine the UA product for the heat exchanger.



Question Sheet

10: 1



-------------------------------------------------

Address City

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Telephone______________

10-1.

Fax

State

----

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----------------------------

A refrigerator with a coefficient of performance of 2.8 provides 120,000 Btu of cooling. Calculate the net work required to run the refrigerator and the heat to be rejected to the surroundings.

Question Sheet

Chapter 10

The Carnot Cycle

10:2

10.2.

Give the Clausius statement of the Second Law of Thermodynamics and briefly describe in your own words what the statement means.

10.3.

Describe the four processes of the reversed Carnot cycle and show the cycle on a T-s diagram.

10.4.

A Carnot heat engine operates between a source at 2000 R and a sink at 440 R. lfthe heat engine is supplied heat at a rate of 800 Btu/min, determine the thermal efficiency and the power output of the engine.


Question Sheet

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10.5.

A heat engine is operating on a Camot cycle and has a thermal efficiency of 55%. The waste heat from the engine is rejected to a nearby lake at 50°F at a rate of 600 Btu!min. Determine the power output of the engine and the temperature of the heat source.

10.6.

A refrigerator is required to remove heat from a cooled space at a rate of 200 Btu! min to maintain a temperature of 20°F. If the ambient air surrounding the refrigerator is at 72°F, determine the minimum power input required for the refrigerator.

Question Slteet

Cltapter 10 Tlte Carnot Cycle

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10.7.

An inventor claims to have developed a cooling system that removes heat from a cooled region at 10°F and transfers it to surrounding air at 75°F while maintaining a COP of7.5. How do you assess the inventor's claim?

10.S.

An air-conditioning system is used to maintain a house at 72°F when the temperature outside is 95°F. The house is gaining heat through the walls and windows at a rate of 50,000 Btu/h, and the heat generation rate within the house by lights and appliances is 5,000 Btu/h. Determine the minimum power input required by the airconditioning system.



Question Sheet

11: 1



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11-01. Sketch the basic elements of the vapor-compression cycle: compressor, condenser, expansion valve and evaporator and describe the basic function of each.

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11-02. Sketch the T-s and p-h property diagrams of an ideal vapor-compression cycle and describe the thermodynamic processes of the cycle.

11-03. A 3-ton refrigeration system operates on an ideal vapor-compression cycle with R-134a as the working fluid. The refrigerant enters the compressor as a saturated vapor at 8.577 psia and is compressed to 199.25 psia. Find the coefficient of performance of the refrigerator and the power of the compressor. Compare the coefficient of performance with that ofa Carnot cycle operating between -35°F and 125°F.

Chapter II Refrigeration Cycles

Question Sheet

11:3

11-04. A 3-ton refrigeration system operates on an actual vapor-compression cycle with R-134a as the working fluid. The isentropic efficiency ofthe adiabatic compressor is 85%. The refrigerant enters the compressor as a saturated vapor at 8.577 psi a and is compressed to 199.25 psia. The pressure drop in the condenser is 13.41 psi. The pressure drop in the evaporator is 2.72 psi. Find the coefficient of performance of the refrigerator and the power ofthe compressor.

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11-05. A 5 kW compressor drives an ideal vapor-compression cycle heat pump with R-22 as the working fluid. The refrigerant enters the compressor as a saturated vapor at 83.28 psia and is compressed to 396.32 psia. Find the heating capability of the heat pump and the coefficient of performance.

11-06. An ice-making machine operates on a vapor-compression cycle using R-22. The refrigerant enters the compressor as a saturated vapor at 10°F and leaves the compressor with an enthalpy of 120 Btu/Ibm. The condenser pressure is 210.69 psia. Water enters the ice-machine at 55°F and leaves as ice at 25°F. For an ice production rate of 10 lbn/h, estimate the power input to the refrigerator. The heat removed to convert 55°F water to ice at 25°F is 169 Btullbm .

Chapter II Refrigeration Cycles

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11-07. Sketch and describe the operation of an absorption refrigeration cycle that uses a mixture of ammonia and water. Ammonia serves as the refrigerant.

Question Sheet


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11-08. Sketch and describe the operation of an open cycle gas refrigeration system.



Question Sheet

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Complete the following questions by writing your answers on these worksheets. Be sure to include your name and address below. Send the completed worksheets to ASHRAE.

Name f

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12-01. State Dalton's law of additive pressures, and then sketch and explain how this model may result from the addition of two gases of equal volume and temperature.

Question Sileet

Chapter 12 Moist Air as a Mixture ofIdeal Gases

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12-02. Calculate the mass fractions and mole fractions when 2 Ibm nitrogen (molecular weight MN2 = 28.013 Ibl/,1lb mo,) is mixed with 6 Ibm hydrogen (molecular weight MH2 = 2.0161b m lIb moe I). What is the molecular weight of the mixture? What are the partial volume fractions and the partial pressure fractions?

12-03. 2 lbm dry air (molecular weight M Alr. = 28.97 lbm lIb moeI) is mixed with 0.1 lbm water vapor (molecular weightMH20 = 18.015Ibmllbmo'e). What is the molecular weight of the mixture? What is the partial pressure of the water vapor if the total pressure is atmospheric (14.97 psia)? Compare the partial pressure with the saturated pressure at 100°F and say whether this mixture is possible at this temperature.

Chapter 12 Moist Air as a Mixture ofIdeal Gases

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12-04. Find the humidity ratio, specific volume (per Ibm of dry air) and specific enthalpy (per Ibm of dry air) of saturated moist air at standard atmospheric pressure and 120°F.

12-05. Find the specific enthalpy (per Ibm of dry air) of moist air at standard atmospheric pressure and 120°F if the humidity ratio is 0.05.

Question Sheet

Chapter i2

Moi.~t Air as

a Mixture ofideal Gases

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12-06. 2 lb m dry air is mixed with 0.1 lb m water vapor at atmospheric pressure and temperature of 120°F. Calculate the enthalpy of the mixture (per Ibm of dry air) from: the moist air tables; and the expressions given for the enthalpy of dry air and water vapor in the text.


Cl,apter 12 Moist Air as a Mixture ofIdeal Gases

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13-01. Give a short description of humidity ratio and relative humidity.

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13-02. Using the definitions of humidity ratio, relative humidity and the Ideal Gas Law, show that the relationship between humidity ratio and relative humidity can be written as:

~ = W(pi pg)/{0.622+W) W = 0.622~ Pg I(p-~ pg)

lb vapor/lb dry air III

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13-03. Consider moist air at 72°P and atmospheric pressure. Answer the following questions: • What is the humidity ratio and relative humidity if the air is saturated? • What is the humidity ratio if the relative humidity is 60%? • What is the relative humidity if the humidity ratio is 0.015 Ibm vapor/lb dry air? fII


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13-04. Explain the term dewpoint temperature. Calculate the dewpoint temperature for nop atmospheric air with a relative humidity of 60%.

13-05. On a 72°P day, the wet-bulb temperature is measured as 55°P. Calculate the humidity ratio and the relative humidity.

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13-06. Atmospheric air at 90°F and 60% relative humidity is cooled to 80°F or 60°F. In each case, calculate the final relative humidity.

13-07. Moist air at 95°F, atmospheric pressure and relative humidity of 60% is cooled at constant pressure to saturation conditions at a rate of 2,000 Ibm dry air per hour. Estimate the rate of heat removal.


Chapter 13 Properties ofMoistAir

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14-01. Sketch the outline of the psychrometric chart showing typical lines of constant value of dry-bulb temperature, wet-bulb temperature, humidity ratio, relative humidity, specific enthalpy and specific volume.

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14-02. List the seven properties that are represented on the psychrometric chart.

The following five problems are to be solved by looking up the properties of moist air on the psychrometric chart for standard atmospheric pressure (sea-level). Given two properties, you must look up the other five properties.

14-03. Dry-bulb temperature of90oP and a wet-bulb temperature of75°P.

14-04. Humidity ratio of 0.008 lb m moisture per lb dry air and relative humidity of 80%. I1l

14-05. Humidity ratio of 0.020 lbm moisture per lb111 dry air and dry-bulb temperature of 80 oP.


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14-06. Enthalpy of25 Btu per Ibm dry air and relative humidity of20%.

14-07. Saturated moist air at a dry-bulb temperature of 72°F.


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15-01. Moist air is cooled from 80°F, 50% relative humidity to 70°F with no moisture added or removed. What is the final relative humidity and the heat removed per lb of dry air? 11/

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15-02. Moist air is cooled from 80°F, 50% relative humidity to 50°F. What is the final relative humidity, humidity ratio and change of enthalpy?

15-03. 1,000 cfm of outdoor air at 40°F and 20% relative humidity is to be heated and humidified to a final condition of 72°F and 50% relative humidity with a heater followed by a steam humidifier. The temperature after the heater is 64°F. Find the following: the heat added in the heater section; the mass flow rate of steam; and the enthalpy of the steam.


Question Sheet

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15-04. Outdoor air at 95°F and 50% relative humidity is to be cooled to a final condition of 72°F and 50% relative humidity using a cooling coil and a simple heater. Find the intermediate temperature between the cooling coil and the heater, the specific heat removed in the cooling coil, and the specific heat added in the heater.

15-05. Outdoor air at 95°F and 20% relative humidity is cooled in an evaporative cooler by spraying with saturated liquid water at 70°F. What is the final temperature if the final relative humidity is 60%? (Hint: use the protractor to get the slope of the process line.) How much water will it take per Ibm of dry air? (Hint: use the change in humidity ratio.)

Question Sheet


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15-06. 20 Ibm per minute of air at 40°F and 90% relative humidity is adiabatically mixed with moist air at 80°F, but unknown relative humidity. The final mixture is at 72°F and 50% relative humidity. What is the relative humidity and mass flow rate of the second air stream?

15-07. The sensible heat load ratio for a heating problem is known to be SHR = -2.0. The setpoint condition for the conditioned space is 72°F and 50% relative humidity. Find the relative humidity of the 80°F heating air if the correct ratio of sensible to latent heat is to be removed from the conditioned space.



Question Sheet

ASHRAE Fundamentals of Thermodynamics and Psychrometrics

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