Fundamentals of Psychrometrics SI, 2nd ed..pdf

Fundamentals of Psychrometrics Second Edition Don Brandt

SI

International System

A Course Book for Self-Directed or Group Learning

Includes Skill Development Exercises for PDH, CEU, or LU Credits

Fundamentals of Psychrometrics Second Edition

Don Brandt

A Course Book Book for Self-Directed or or Group Learning Learning

Atlanta

Fundamentals of Psychrometrics (SI), (SI) , Second Edition A Course Book for Self-Directed or Group Learning ISBN 978-1-939200-32-7 (paperback) ISBN 978-1-939200-33-4 (PDF) SDL Number: 00327 © 2008, 2016 ASHRAE All rights reserved.

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• • •

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Don Brandt is an ASHRAE member from Phoenix, Arizona, who spent 39 years (4 of them part time) with Trane Co., mostly as a Commercial Sales Engineer and Sales Manager. He was involved in thousands of projects over that time period, including many with large industrial customers that had special HVAC and process applications. Brandt also taught the Trane A/C Clinic series many times to young engineers starting out in the industry. Brandt has also been active in ASHRAE at all three levels of organization—Chapter, Regional, and Society—during this same period. He is a charter member and past president of the Anthracite Chapter in Northeast Pennsylvania. He was the Technical, Energy and Government Activities Committee (TEGA) Regional Vice Chair for Region X, the TEGA Vice Chair and Chair in 2002, Region X Director and Regional Chair from 2002–2005, on the Board of Directors from 2002– 2005, and a member of Standards Committee from 2005–2009 as a Standards Project Liaison Subcommittee (SPLS) Liaison. In 2001 he became a member of Professional Development Committee, moving up to Chair in 2013. He is currently a member of the Energy Targets Multidisciplinary Task Group, a Nominating Member for Region X, and Member of the Appeals Board for Standards. Brandt is a 1974 graduate of Penn State with a BS in Electrical Engineering and an active member of the Alumni Association. In retirement, he is an instructor for the successful ASHRAE HVAC Essentials Course, both Levels 1 and 2, that is held both in the United States and internationally. He also teaches a portion of the Association of Energy Engineers (AEE), Arizona Chapter, Certified Energy Manager (CEM) preparation class held on an annual basis.

Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xi Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii Chapter 1: Introduction to Psychrometrics. . . . . . . . . . . . . . . . . . . . . 1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Air Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Volumetric Airflow versus Mass Flow Calculations . . . . . . . . . . . . . . 2 Skill Development Exercises for Chapter 1 . . . . . . . . . . . . . . . . . . . . 4 Chapter 2: Properties of Moist Air . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Temperature. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Humidity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Specific Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Using Appendix A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Terminology and Symbols for Psychrometrics . . . . . . . . . . . . . . . . . 7 Skill Development Exercises for Chapter 2 . . . . . . . . . . . . . . . . . . . . 8 Chapter 3: Introducing the Psychrometric Chart . . . . . . . . . . . . . . . 11 The Modern Age of Psychrometrics. . . . . . . . . . . . . . . . . . . . . . . . 11 Creating the Psychrometric Chart . . . . . . . . . . . . . . . . . . . . . . . . . 11 Finding Seven Psychrometric Quantities . . . . . . . . . . . . . . . . . . . . . 14 Climatic Design Information. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Psychrometric Chart for Extended Temperature and Altitude . . . . . 17 Skill Development Exercises for Chapter 3 . . . . . . . . . . . . . . . . . . . 19 Chapter 4: Air-Conditioning Processes on the Psychrometric Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 The Power of the Psychrometric Chart . . . . . . . . . . . . . . . . . . . . . . 21 Sensible Cooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Sensible Heating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Latent Heat Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Latent Heat Removal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Total Heat Content . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Cooling and Humidifying . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Heating and Humidifying . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 Dehumidification and Heating. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Skill Development Exercises for Chapter 4 . . . . . . . . . . . . . . . . . . . 32

viii

Contents

Chapter 5: HVAC Design and the Psychrometric Chart. . . . . . . . . . 35 Schematic of an Air-Conditioning System . . . . . . . . . . . . . . . . . . . . 35 Mixing Airstreams—Cooling Systems . . . . . . . . . . . . . . . . . . . . . . . 35 Mixing Airstreams—Heating Systems . . . . . . . . . . . . . . . . . . . . . . . 37 Sensible Heat Ratio—Cooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Sensible Heat Ratio—Cooling with Outdoor Air . . . . . . . . . . . . . . 40 Psychrometric Process—Heating. . . . . . . . . . . . . . . . . . . . . . . . . . 41 Skill Development Exercises for Chapter 5. . . . . . . . . . . . . . . . . . . 44 Chapter 6: Psychrometrics in HVAC Equipment . . . . . . . . . . . . . . . 47 The Air-Handling Unit: Heart of the Commercial Air-Conditioning System. . . . . . . . . . . . . 47 Psychrometrics of a Cooling Coil . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Psychrometrics of Fan Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Psychrometrics of a Heating Coil . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Humidification Equipment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Skill Development Exercises for Chapter 6. . . . . . . . . . . . . . . . . . . 54 Chapter 7: Psychrometrics in Zoned HVAC Systems . . . . . . . . . . . . 57 Constant-Volume and Variable-Air-Volume Systems. . . . . . . . . . . . 57 Constant-Volume, Single-Zone System . . . . . . . . . . . . . . . . . . . . . 58 Constant-Volume, Single-Zone System with Reheat. . . . . . . . . . . . 59 Constant-Volume, Single-Zone System with Face and Bypass Dampers on the Cooling Coil. . . . . . . . . . . . 60 Constant-Volume System with Terminal Reheat . . . . . . . . . . . . . . 62 Constant-Volume Multizone and Dual-Duct Systems . . . . . . . . . . . 63 Variable-Air-Volume Systems for Multiple Zones . . . . . . . . . . . . . . 65 Variable-Air-Volume Systems with Heating VAV Boxes. . . . . . . . . . 66 Skill Development Exercises for Chapter 7. . . . . . . . . . . . . . . . . . . 69 Chapter 8: Energy Conservation and Psychrometrics. . . . . . . . . . . . 73 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Heat Recovery Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Energy Recovery Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 Air-Side Economizer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 Water-Side Economizer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 Supply Air Temperature Reset . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 Skill Development Exercises for Chapter 8. . . . . . . . . . . . . . . . . . . 85 Chapter 9: Special Applications and Psychrometric Considerations . . .87 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Cooling Towers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Indoor Swimming Pools. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Cleanrooms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

Fundamentals of Psychrometrics (SI), Second Edition

ix

Direct Evaporative Cooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Indirect Evaporative Cooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 Skill Development Exercises for Chapter 9 . . . . . . . . . . . . . . . . . . . 98

Appendix A: Thermodynamic Properties of Moist Air . . . . . . . . . . 101 Appendix B: Dimensions, Units, and Unit Conversion Factors . . . . 105 Appendix C: Climatic Design Information . . . . . . . . . . . . . . . . . . . . 107 Appendix D: Thermodynamic Properties of Water at Saturation . . 137 Skill Development Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

Preface Psychrometrics is a tool HVAC designers use to determine the amount of moisture in the air and to provide solutions to designers for the ultimate comfort of building occupants. It can be used to size airhandling units; optimize energy performance; identify control sensors for building automation; describe the performance of cooling coils, cooling towers, and humidification equipment; and evaluate heat recovery strategies. Yet the use of psychrometrics and the psychrometric chart is different from designer to designer. Some do not use the psychrometric chart, relying instead on simplified formulas or complex computer simulations. Others use the chart only for special situations, such as indoor swimming pool applications. And some use it as their primary system evaluation tool. This course is meant to cover all of these uses of the psychrometric chart, to have something for all these different backgrounds, and to be an introduction for the young designer who has yet to pick an approach. The course addresses the use of psychrometrics and the psychrometric chart for typical applications and systems and includes some theory. This theory not only sets the basics but also gives students an appreciation of the simplification that the psychrometric chart provides. The psychrometric chart gives a visual description of HVAC design, one that could never be appreciated from formulas alone. This second edition of the psychrometrics self-directed learning (SDL) course was rewritten in an attempt to teach the fundamentals of psychrometrics in about half the time as the previous version. The author has used his 42 years of experience in the HVAC industry as the expertise for the format and content.

Acknowledgments I need to acknowledge a few folks who helped me out on this first publishing adventure of my life. My wife JoAnn was the person who put my scribble into a nice Word document. John Duren, Sales Engineer for MPSW in Phoenix, did various equipment selections that are used in the course. Mick Schwedler and John Murphy from Trane Applications Engineering Department in La Crosse, Wisconsin, were there when I need some technical help and review. And I would like to thank ASHRAE staff for their understanding and help to make this a finished publication.

Introduction to Psychrometrics Study Objectives After completing completing this chapter, you should be able to  understand the basic processes of psychrometrics, psychrometrics,  understand enthalpy enthalpy and volumetric airflow, airflow, and  understand the basic basic formulas of HVAC HVAC design.

Instructions Read the material in Chapter 1. At the end of the chapter, complete the skill development exercises without referring to the text.

Introduction Psychrometrics is an analysis tool that HVAC engineers use to provide solutions to comfort issues. These issues can be related to human comfort or process comfort depending depe nding on the applications. If the air surrounding us were totally dry, our job as HVAC engineers would be very easy and probably boring. But, because all air we work with contains some amount of moisture in the form of water vapor, our jobs get more complex. The four basic processes that an HVAC system can perform on moist air are as follows: • • • •

Cooling—Lowe Cooling—Lowering ring the air temperature temperature Heatin Heating—R g—Rais aising ing the air temper temperatu ature re Humidificat Humidification—Ra ion—Raising ising the moisture moisture content in the air Dehumidific Dehumidification— ation—Loweri Lowering ng the the moistu moisture re content content in the the air air

Note that HVAC processes can also be a combination of the above four basic actions. These combinations include the following: •

Heating Heating and humidificat humidification—I ion—Increas ncreasing ing the the temperat temperature ure and and moisture moisture content of the air at the same time

2

Chapter 1 • • •

Introduction to Psychrometrics Heating Heating and dehumidific dehumidification— ation—Incre Increasing asing the tempera temperature ture and decreasing decreasing the moisture content of the air at the same time Coolin Cooling g and humidifi humidificat cation ion—De —Decre creasi asing ng the temperat temperature ure and increas increasing ing the moisture content of the air at the same time Cooling Cooling and dehumidi dehumidificati fication—Dec on—Decreasi reasing ng the tempera temperature ture and and moisture moisture content of the air at the same time

These These combin combinati ation on proces processes ses provid providee an infini infinite te number number of potent potential ial actions. For example, consider the cooling and dehumidification combination process. We have a solution that drops the temperature 12 degrees and only drops the dew point (defined later) 1 degree Another solution drops the tem perature 5 degrees and the dew point 5 degrees. So, there are thousands of solutions with too many temperature and dew-point combinations to list.

Enthalpy We will use the property of the enthalpy of air throughout this course. Enthalpy is the sum of the internal energy or the total heat content of the air. In other words, warm and humid air can have the same heat content as hot and dry air. So the energy required to cool warm/humid air in the Gulf Region of North America might be close to the energy required to cool hot/dry air in the Southwest deserts. Enthalpy h Enthalpy h is is subdivided into the following: • • •

hda = h s = has =

enthalp enthalpy y of of dry dry air at 0% relati relative ve humidi humidity ty enthalpy enthalpy of saturated saturated air or 100% relative relative humidity humidity enthalp enthalpy y differ differenc ences es betwee between n hda and h h s

Air Density Elevation has an effect on psychrometric analyses. As elevation changes, so does the air density. This means the constants used in equations will change and different different psychrometric psychrometric charts (Chapter 3) are required for proper calculacalculation. For this course we will use sea level as atmospheric pressure for all calculations unless otherwise noted.

Volumetric Airflow versus versus Mass Flow Calculations Calculations For easier understanding, we will use volumetric airflow in cubic meters per second (m3/s) rather than mass flow for our calculations throughout this course. Standard air is defined as air at sea level or 101.325 kPa of barometric pressure and 20.8°C t 20.8°C t db. The density of air at sea level is 1.20 kg/m3. The three equations we will use in our calculations are for sensible heat (q s), the total heat required (q (qt ), and latent heat (q (ql ):

Fundamentals of Psychrometrics (SI), Second Edition q s (W) = 1210 × airflow in m 3 /s × (t 1 – t 2) in °C

3 (1-1)

where the constant 1210 is derived from the following: 1.20 kg/m3 × 1.006 kJ/(kg·K) × 1000 J/kJ = 1210 J/(m3·K) where 1.006 kJ/(kg·K) is the specific heat of air. qt (kW) = 1.20 kg/m 3 × airflow in m 3 /s × (h1 – h 2) in kJ/kg

(1-2)

ql (kW) = 3010 × airflow in m 3 /s × (w1 – w 2) in kg/kg of dry air

(1-3)

where the constant 3010 W is derived from the following: 2500 kJ/kg × 1.2 kg/m3 × 1.006 kJ/(kg·K) = 3010 W where 2500 kJ/kg is the latent heat of water vapor and 1.006 kJ/(kg·K) is the specific heat of air.

4

Chapter 1

Introduction to Psychrometrics

Skill Development Exercises for Chapter 1 Complete these questions by writing your answers on the worksheets at the back of this book.

1-1

How many basic processes of air conditioning can be performed on moist air? a) Two b) Three c) Four

1-2

Which combination process will increase both the temperature and the moisture content? a) Cooling and dehumidification b) Heating and dehumidification c) Heating and humidification

1-3

Enthalpy is the total heat content of the air. a) True b) False

1-4

Change in elevation has no effect on the air density. a) True b) False

Properties of Moist Air Study Objectives After completing this chapter, you should be able to  define some fundamental properties used in psychrometrics and  understand how to use Appendix A.


Introduction This chapter defines the properties that an HVAC engineer uses to do psychrometric analysis. The properties are temperature, humidity, enthalpy, and specific volume.

Temperature The temperatures we are concerned about in HVAC systems are the following: •

•

Dry-bulb temperature: The measure of the surrounding air temperature with a standard thermometer in degrees Celsius (°C) without influence on the thermometer by heat sources or solar heat gain. Wet-bulb temperature: The measure of the moist effect on the evaporation process in the air. Using a standard dry-bulb thermometer, place a cotton sock on the sensing portion. Next, soak the sock in ambienttemperature water and, by rotation, move surrounding air across the sock. The temperature will drop below ambient because of the evaporative or cooling effect on the sensing bulb. It will continue to drop until most of the water is evaporated into the surrounding air. This is the wet-bulb temperature and may also be called wet-bulb depression. Any further drying of the sock will result in the temperature going back up to the ambient dry-bulb temperature.

6

Chapter 2 •

•

Properties of Moist Air Dew-point temperature: The measure of the dry-bulb temperature at the point where water vapor starts to condense to liquid or be removed from the air. This is also referred to as the condensation point , because it is the temperature at which the water turns to liquid from vapor in the airstream. Saturation temperature: The temperature at which the air cannot hold any additional water vapor. At the saturation temperature, the dry-bulb, wet-bulb, and dew-point temperatures are identical.

Humidity Humidity is the moisture in the air. We can talk about it in two ways: • •

Humidity ratio: The mass in grams of water vapor per kilogram of dry air (gw /kgda). Relative humidity: The actual amount of moisture in the air at a given dry bulb temperature versus the maximum amount of moisture in the air at the same dry-bulb temperature. It is expressed in percentage because it is a partial moisture/maximum moisture ratio. At constant moisture content, as soon as the dry-bulb temperature changes, so does the relative humidity.

Enthalpy The energy content of air is defined as the enthalpy of the air or the total heat content of the air. It is expressed in kilojoules (kJ) per kilogram of dry air (kJ/kgda). Again, warm/humid air can have the same enthalpy as hot/dry air, so it takes the same amount of energy to cool either airstream to a comfortable condition.

Specific Volume Specific volume is the cubic metres per kilogram of dry air (m3/kgda). It is the inverse of air density (kg/m3). Note that specific volume changes as the dry-bulb temperature changes, but not nearly as much as it changes with the effect of higher altitude.

Using Appendix A The table in Appendix A lists thermodynamic properties of moist air at 101.325 kPa. For each dry-bulb temperature in °C, we have values for the following: • • • •

Humidity ratio at saturation Specific volume at dry, saturated, and differential conditions Specific enthalpy at dry, saturated, and differential conditions Specific entropy at dry, saturated, and differential conditions (not used in this text)


7

We will use this table as we go through the text.

Terminology and Symbols for Psychrometrics h

=

enthalpy of moist air, kJ/kg of dry air

ha

=

specific enthalpy of dry air, kJ/kg of dry air

hw

=

specific enthalpy of water vapor, kJ/kg of dry air

p

=

total pressure, usually barometric, kPa

pw

=

partial pressure of water vapor, kPa

pa

=

partial pressure of dry air, kPa

q

=

rate at which heat is transferred to a process, W

t db

=

dry-bulb temperature of moist air, °C

t wb

=

wet-bulb temperature of moist air, °C

t dp

=

dew-point temperature of moist air, °C

v

=

specific volume of moist air, m3/kg of dry air

va

=

specific volume of dry air, m3/kg of dry air

vw

=

specific volume of water vapor, m3/kg of dry air

v s, v g =

specific volume of saturated water vapor, m3/kg of dry air

W

=

humidity ratio of moist air, kg (water)/kg (dry air)

W s

=

humidity ratio of moist air at saturation, kg (water)/kg (dry air)



=

relative humidity the ratio of actual moisture amount to maximum moisture amount, % rh

For dimensions and units used in air-conditioning applications and a table of unit conversion factors for converting between Inch-Pound (I-P) and Systéme International (SI) measurement units, see Appendix B.

8

Chapter 2

Properties of Moist Air


2-1

Dry-bulb temperature is measured with a wet sock around the sensing bulb. a) True b) False

2-2

Saturation temperature of air is the point at which the dry-bulb, wet-bulb, and dew-point temperatures are equal. a) True b) False

2-3

Relative humidity does not change as the dry-bulb temperature changes. a) True b) False

2-4

The dry-bulb temperature can be above the dew-point temperature. a) True b) False

2-5

According to Appendix A, what is the specific enthalpy h s of saturated air at 5°C? a) 15.231863 b) 18.63 c) 5.02 d) None of the above

2-6

According to Appendix A, under the same condition cited in Exercise 2-5, what is the specific volume v? a) 0.811 b) 0.794 c) 0.006 d) None of the above

2-7

According to Appendix A, what is the specific enthalpy of dry air hda at 50°C? a) 52.33 b) 225.03 c) 50.31 d) None of the above

Fundamentals of Psychrometrics (SI), Second Edition 2-8

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According to Appendix A, under the same condition cited in Exercise 2-7, what is the specific volume v? a) 0.915 b) 1.24 c) 0.012 d) None of the above

Introducing the Psychrometric Chart Study Objectives After completing completing this chapter, you should be able to  describe how the psychrometric chart was developed,  understand how to read read the psychrometric psychrometric chart, and  use climatic design information information from tables published by ASHRAE. ASHRAE.


The Modern Age of Psychrometrics In this chapter, you will learn about the psychrometric chart and how to use it in HVAC problems. You will see how to construct and then use the psychrometric chart. The psychr psychrome ometri tricc chart chart was develop developed ed by Willis Willis Carrier Carrier in the early early 1900s. It has been refined over time for more accuracy to provide better results. The chart contains seven important psychrometric variables, represented on the chart by the following symbols: t db t wb t dp  h W v

= = = = = = =

dry-bu dry-bulb lb temper temperatu ature re wet-bu wet-bulb lb temper temperatu ature re dew-po dew-point int temper temperatu ature re relati relative ve humidi humidity ty enth enthal alpy py humi humidi dity ty rati ratio o speci specifi ficc volu volume me

Creating the Psychrometric Chart We will now discuss how to construct the psychrometric chart and plot the seven important properties on the chart. Using Figure 3-1, start with the x-axis (the horizontal horizontal line across across the bottom) bottom) and plot dry-bulb dry-bulb temperatures temperatures on a linear scale from low on the right to high on the left.

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Next we go to Appendix A and get the saturated humidity ratio W ratio W s values from 0°C to 50°C. Note the W s values are in kilograms and need to be multi plied by 1000 to derive grams. Then, put the y-axis (the vertical line on the right side) on a linear scale from low on the bottom to high on the top. We develop the saturation line (the heavy dark curve shown in Figure 3-1) by the intersection of the W the W s value and the vertical dry-bulb line. Remember, the dry-bulb, wet-bulb, and dew-point temperatures are equal on the saturation line. Lines of constant humidity ratio are all horizontal. Figure 3-2 shows that the dew-point temperature is where the dry-bulb tem perature intersects the saturation line. A line of constant dew point goes horizontally on the psychrometric chart. Going to Figure 3-3, we can get the value of the enthalpy at saturation (h (h s) and again plot that value on the saturation saturation line for that dry-bulb temperature. temperature. To find the other end of the enthalpy line, we simply take the h the h s value and go to Appendix A and look for a very close value in the h the h da column. Once we find it, that dry-bulb temperature temperature is the intersection intersection point for that enthalpy enthalpy line with the x-axis. For example, 15°C db has a value of 42.11 kJ/kg, so we plot this value at the 15°C saturation temperature. We go to Appendix A and search for a value of 42.11 kJ/kg in the hda column and find it at 42.1°C db. Those two points 80%

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establish the 42.11 kJ/kgda enthalpy line. Lines of constant enthalpy are sloped and are solid dark. Figure 3-3 also shows that lines of constant wet-bulb temperature are almost parallel to lines of constant enthalpy. Lines of constant wet-bulb tem perature are sloped and dotted, as shown in Figure 3-4. Figure 3-4 shows specific volume lines. We plot them by going to Appendix A and, at a given saturation temperature, getting the v s value and inserting it on the saturation line. Then we look at the table for the same value at a higher temperature, and that is the x-axis intersection point for the other end of the specific volume line.

Finding Seven Psychrometric Quantities The psychrometric chart shown in Figure 3-4 is sufficient to provide immediate and complete characteristics of a moist air parcel, even if only a small amount of information is known about that parcel. More specifically, if any two of the seven important psychrometric variables (t db, t dp, t wb, , h, v, W ) of a moist air sample are given (for a specific barometric pressure), then all of the remaining ones can be determined immediately from the chart.

Example 3-1 Problem

Given a sample of air where t db = 21°C and  = 60% rh, determine its dew point.

Solution

Using Figure 3-4, the location point is at the intersection of the conditions stated in the problem. Moving to the left in a straight line indicates that the saturation curve is crossed at a temperature of 13°C. This is the dew-point tem perature.

Example 3-2 Problem

What is the enthalpy of the 21°C and 60% rh parcel of air from Example 3-1?

Solution

Again using Figure 3-4, the location point is the same. Following the line of constant enthalpy up the enthalpy scale reveals that the enthalpy of this point is 45 kJ/kgda.

Example 3-3 Problem

Find the wet-bulb temperature for the point in Example 3-1.

Fundamentals of Psychrometrics (SI),

Figure 3-4

ASHRAE Psychrometric Chart No. 1.

Second Edition

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Solution


Again using Figure 3-4, follow the constant wet-bulb line to the saturation line, then drop straight down to read a temperature of 16°C, the wet-bulb temperature.

Example 3-4 Problem

Use Figure 3-4 to find the specific volume of t db = 21°C and  = 60% rh.

Solution

Finding this point on Figure 3-4 reveals that it is located between the values of 0.84 and 0.85 m3/kg for specific volume. Further inspection of the figure indicates that there are more lines of constant specific volume that are unmarked. It appears that each of these represents an increase of 0.01 m3/kgda. Therefore, it can be determined that the intersection is a specific volume of 0.846 m3/kgda.

Example 3-5 Problem

Using the psychrometric chart in Figure 3-4, find the t db, t dp, t wb, , and humidity ratio W of a parcel of air that has a specific volume of 0.88 m3/kgda and an enthalpy of 60 kJ/kgda.

Solution

t db = 32.3°C, t dp = 15.2°C, t wb = 20.9°C,  = 36% rh, W = 10.8 g/kgda Notice that the dew point and relative humidity both needed interpolation. There are many methods of interpolation. Most engineers simply “eyeball” interpolate by doing a visual scaling between the lines of the chart. There is an art to this that is learned by practice, but results in error by less than ±1% can be achieved.

Climatic Design Information Outdoor weather conditions have a lot to do with the air conditioning and heating processes described in this book. Climatic design information for the United States, Canada, and other countries is provided in Appendix C. For our examples, we will use the “2%” column under the “Cooling DB/ MCWB” heading in Appendix C as our design conditions. This means that only 2% of the total hours, in an average year, are above the listed dry-bulb temperature. Note that columns for 0.4% and 1% of the time are also shown. Using a blank psychrometric chart and Appendix C, plot the outdoor design conditions (t db and mean coincident wet-bulb temperature [0.4%]) for summer in the following cities (label them). You will use these outdoor design points as we go further into this course. • •

Miami, Florida, USA Phoenix, Arizona, USA

Fundamentals of Psychrometrics (SI), Second Edition •

Denver, Colorado, USA

•

Sydney, Australia

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Montreal, Quebec, Canada

•

Dubai, UAE

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London, England

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The plotted points are shown in Figure 3-5.

Psychrometric Chart for Extended Temperature and Altitude The psychrometric chart in Figure 3-4 is for sea level and normal temperatures (0°C db to 50°C db). It is available from ASHRAE as Psychrometric Chart No. 1. Psychrometric charts are available at 750, 1500, and 2250 m elevations (Charts No. 4 and No. 5), at low temperatures of –40°C to 10°C (sea level, Chart No. 2), and at high temperatures of 10°C to 120°C (sea level, Chart No. 3).

Figure 3-5

Climatic design information plotted on the psychrometric chart (detail).

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If you do work at these elevations and temperature ranges, please use the appropriate charts. Also, do not forget to correct the sensible heat formula, enthalpy formula, and humidity ratio formula constants for air density changes using the equations included at the end of Chapter 1.


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Skill Development Exercises for Chapter 3 Complete Complete these questions by writing writing your answers answers on the worksheets worksheets at the back of this book.

3-1

On a psychrometr psychrometric ic chart, chart, the y-axis y-axis is humidity humidity ratio ratio and the the x-axis is: a) Relative Relative humidity humidity b) Dew-point temperature c) Dry-bulb Dry-bulb temperature temperature d) Wet-bulb Wet-bulb temperature temperature

3-2

Using the psychrometric chart in Figure 3-4, determine the relative humidity of an air parcel with W with W = = 6.4 and t t db = 15°C. a) 60% 60% rh b) 70% rh c) 80% 80% rh d) 90% 90% rh

3-3

Using the psychrometric chart in Figure 3-4, determine the dew-point temperature of an air parcel with t with t db = 21°C and  =  = 50% rh. a) 10°C 10°C b) 12°C c) 15°C 15°C d) 19°C 19°C

3-4

Using the psychrometric chart in Figure 3-4, determine the humidity ratio W ratio W of of an air parcel with a saturation saturation temperature temperature of t t db = 10°C. a) 6.5 6.5 b) 7.6 c) 30% 30% d) 10°C 10°C

3-5

Using the psychrom psychrometric etric chart chart in Figure 3-4, determin determinee the specific specific volume v of an air parcel with t with t db = 21°C and W W = = 10. a) 0.82 0.82 b) 0.846 c) 0.86 0.86 d) none none of the above above

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Accordi According ng to the psychro psychromet metric ric chart chart in Figure Figure 3-4, what what is the enthalp enthalpy y of t db = 25°C dry air? a) 22 b) 35 c) 76 d) 25

3-7

According According to the psychromet psychrometric ric chart in in Figure 3-4, 3-4, what is the wet-bul wet-bulb b tem perature of a moist air parcel with t with t db = 21°C and  =  = 50% rh air? a) 21°C 21°C b) 14°C c) 10°C 10°C d) 13°C 13°C

3-8

According According to the psychromet psychrometric ric chart chart in Figure Figure 3-4, what is is the dew point point of t db = 10°C saturated air? a) 10°C 10°C b) 4°C c) 0°C 0°C d) –5°C –5°C

3-9

According According to the psychromet psychrometric ric chart in in Figure 3-4, 3-4, what is the wet-bul wet-bulb b tem perature of t t db = 21°C dry air? a) 0°C 0°C b) –4°C c) 4°C 4°C d) 6.5° 6.5°C C

3-10

Using the psychrometric chart in Figure 3-4, plot the points t points t db = 21°C, h 21°C, h = = 24, and t t db = 21°C, t 21°C, t wb = 14°C, then connect the points with a line. Upon investigation of the line, which of the following following is the best description? description? a) The line line is almost almost vertical. vertical. b) The line has a slope of about 45° (angle). c) The line line almost almost horizont horizontal. al.

Air-Conditioning Processes on the Psychrometric Chart Study Objectives After completing this chapter, you should be able to  understand the air-conditioning processes shown on the psychrometric

chart and  understand the use of the HVAC equations provided.


The Power of the Psychrometric Chart This chapter applies the processes discussed in Chapter 2 and the properties of moist air discussed in Chapter 3 to the psychrometric chart. But before we start, we must first define two processes, sensible heat transfer and latent heat transfer. Sensible heat transfer (q s) is changing only the dry-bulb temperature of the air and can be sensible cooling (lowering the temperature) or sensible heating (raising the temperature). On the psychrometric chart, it is pure horizontal movement, right to left or left to right only. We can use the following equation for sensible heat change at sea level: q s (W) = 1210 × airflow × (t 1 – t 2) in °C

(4-1)

where airflow in cubic metres per second, t 1 is the initial temperature, and t 2 is the final temperature. Latent heat transfer (ql ) is changing only the moisture content of the air or changing only the humidity ratio of the air. It is vertical-only movement on the psychrometric chart, top to bottom or bottom to top only. We can use the following equation for latent heat change at sea level: ql (W) = 3010 × airflow × (W 1 – W 2) in g/kg of dry air

(4-2)

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where airflow in cubic metres per second, W 1 is the initial humidity ratio, and W 2 is the final humidity ratio.

Sensible Cooling We will first show the air conditioning process of sensible cooling. It is a horizontal process on the psychrometric chart, moving from the right to left. For example, our entering temperature (T ) is at 38°C db, = 10% rh, and we cool the air to 16°C db as shown in Figure 4-1. The leaving t 2 is at 16°C db and  = 37% rh. Note the humidity ratio of W = 4.1 did not change. If we apply our example airflow of 2.35 m3/s to the problem, then q s = 1210 × airflow × (t 1 – t 2) = 1210 × 2.35 m3/s × (38°C – 16°C) = 1210 × 2.35 × (22) = 62 550 W cooling

Sensible Heating Next we will review the air-conditioning process of sensible heating. It is also a horizontal process on the psychrometric chart, but from left to right. In this example, our entering temperature (T ) is 21°C db,  = 51% rh, and we heat the air to 43°C db as shown in Figure 4-2. The leaving t 2 is 43°C db and  = 14% rh. Note the humidity ratio of W = 7.9 did not change.

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If we apply our example airflow of 2.35 m3/s to our sensible heat equation, then q s = 1210 × airflow × (t 1 – t 2) = 1210 × 2.35 m3/s × (21°C – 43°C) = 1210 × 2.35 × (22) = 62 550 W heating

Latent Heat Addition The addition of latent heat, or the addition of moisture content to air, is the next area of focus. It is a vertical movement, from bottom to top of the psychrometric chart. For example, the entering conditions of t db = 27°C and  = 18% rh have a W = 4.5 humidity ratio. The leaving conditions of t db = 27°C and  = 55% rh have a humidity ratio of W = 13, and the dry-bulb temperature did not change, as shown in Figure 4-3. The latent heat required with our example of 2.35 m3/s can be calculated as follows: ql = 3010 × airflow × (W 1 – W 2) = 3010 × 2.35 m3/s × (4.5 – 13) = 3010 × 2.35 × (8.5) = 60 120 W

Latent Heat Removal The removal of latent heat, or the lowering of moisture content to air, is the process shown in Figure 4-4. The entering conditions of t db = 24°C and  =

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70% rh have a W = 13 humidity ratio. We remove moisture to the leaving conditions of t db = 24°C and  = 17% rh, which have a humidity ratio of W = 3.5. The latent heat removed with our example of 2.35 m3/s is as follows: ql = 3010 × airflow × (W 1 – W 2) = 3010 × 2.35 m3/s × (13 – 3.5) = 3010 × 2.35 × (9.5) = 67 200 W It should be noted at this time that the processes shown in both Figures 4-3 and 4-4 are nearly impossible to do in the real world of HVAC as stand-alone processes. When we humidify the air, we generally have to add heat to the air, even if not desired (steam humidifier). When we dehumidify the air, we need to cool the air dry-bulb temperature below the entering dew-point temperature to start the moisture removal process, so we end up with cooled and dehumidified air.

Total Heat Content We will now discuss the four air-conditioning processes that are combinations of two simple processes. For these combination processes, we use the enthalpy equation to get the total heat required (qt ) at sea level: qt = 1.2 × airflow × (h1 – h 2) in kJ/kg of dry air

(4-3)

where airflow in cubic metres per second, h1 is the initial enthalpy, and h2 is the final enthalpy. Let us start with the cooling and dehumidifying process, because it is the most common in the HVAC industry. Movement on the psychrometric chart is to the left (sensible) and down (latent) from the initial condition. See Figure 4-5 for the actual movement of the air. Also, note that a gradual slope indicates a more sensible than latent load, but a steeper slope shows a more latent than sensible load. The following example will explain this combination process. The entering conditions to our cooling coil are 27°C db and 18°C wb, with h1 = 50.7. The air is cooled and dehumidified all the way down to 12°C db and 11.5°C wb with h 2 = 33. We can find the total heat required by using our new equation with our example of 2.35 m3/s: qt = 1.2 × airflow × (h1 – h 2) = 1.2 × 2.35 m3/s × (50.7 – 33) = 1.2 × 2.35 × (17.7) = 49.91 kW or 49 910 W We can get the same answer by using the individual sensible and latent heat equations: q s = 1210 × airflow × (t 1 – t 2) = 1210 × 2.35 m3/s × (27°C – 12°C) = 1210 × 2.35 × (15) = 42 650 W

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and ql = 3010 × airflow × (W 1 – W 2) = 3010 × 2.35 m3/s × (9.3 – 8.3) = 3010 × 2.35 × (1) = 7075 W q s + q l = q t = 42 650 + 7075 = 49 725 W Almost the same answer, but done in one less step by using the combination equation. Because we are “eyeballing” the values on our psychrometric chart, it is not uncommon that these two answers are not exact and can be off by a few percent. See Figure 4-6 for the breakdown of the sensible and latent components.

Cooling and Humidifying Cooling and humidifying is most easily explained with the process of evaporative cooling. Because evaporative cooling is a constant wet-bulb or adia batic cooling process, the total heat or enthalpy equation does not work. The cooling of the air at dry-bulb temperature is done by the fact that the water in the liquid form is evaporated to water vapor. See Figure 4-7 for an example,

Second Edition

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Sensible and latent components of determining total heat content using the psychrometric chart.

and note that the movement on the psychrometric chart is to the left (sensible) and upward (latent) to complete the process. Our inlet conditions to the evaporative cooler are 38°C db and  =10% rh, resulting in a 17.5°C wb. We move upward and left on the 17.5°C wet-bulb line the distance our evaporative media will provide. Our outlet conditions will be 18.5°C db, 17.5°C wb, and  =85% rh. Note that we have dropped the dry-bulb temperature from 38°C to 18.5°C by using only water. At our 2.35 m3/s exam ple, we are able to obtain a sensible cooling of q s = 1210 × airflow × (t 1 – t 2) = 1210 × 2.35 m3/s × (38°C – 18.5°C) = 1210 × 2.35 × (19.5) = 55 450 W So, how much water do we use in the evaporative process? We can use a new equation to calculate the water usage in grams of water per hour: gw /h = airflow × 1/v × (W 1 – W 2) × 3600 = 2.35 m3/s × 1/0.84 m3/kgda × (4.5 – 12.5) gw /kgda × 3600 s/h = 2.35 × 1/0.84 × (8) × 3600 = 80 570 g of water/h or 80.57 kg/h Evaporative cooling should always be an option if you are doing a project in the hot/dry climates of the world.

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Heating and Humidifying Heating and humidifying is a combination process that is frequently seen in the HVAC industry when it is desirable to attempt to maintain a space at or above a minimum relative humidity setpoint. The movement on the psychrometric chart is to the right and towards the top, as shown in Figure 4-8. In our example, we have an airstream at 18°C db and  = 20% rh, with 2.35 m /s at sea level. We want to maintain a room at 24°C db and  = 50% rh. We can use our equation from the evaporative cooling example to solve for the pounds of water per hour needed to increase the relative humidity of this airstream. 3

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Heating and humidifying shown on the psychrometric chart.

18°C db,  = 20% rh, h = 25 kJ/kgda 24°C db,  = 50% rh, h = 48 kJ/kgda qt = 1.2 × airflow × (h1 – h 2) = 1.2 × 2.35 m3/s × (25 – 48) = 1.2 × 2.35 × (23) = 64.86 kW or 64 860 W Another way to get this answer is to break the problem into the sensible portion and the latent portion. The sensible portion is easy, 18°C db to 24°C db, or: q s = 1210 × airflow × (t 1 – t 2) = 1210 × 2.35 m3/s × (18°C – 24°C) = 17 060 W For the latent portion, we need to go to Appendix D and look at the thermodynamic properties of water at saturation, or steam, tables. In the left-hand column, find 18°C temperature and follow that to the right until you get to the column labeled “Evap. h g ” under the “Specific Enthalpy” heading and get 2533 kJ/kgw. The steam will reach equilibrium at 18°C db soon after injection into the airstream. Because we know the grams of water per hour of the humidifier, the latent portion is

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Air-Conditioning Processes on the Psychrometric Chart ql = 69,670 gw /h × 2533 kJ/kgw = 69.67 kg/h × 2533 kJ/kg = 176 470 kJ/h 176,470 kJ/h × 0.275 W/kJ = 48 530 W Now we add the sensible and latent portions together: q s + q l = q t = 17 060 + 48 530 = 65 590 W

which is close to 64 860 W. Remember, we are eyeballing all these values from the psychrometric chart, so the actual values may be off ±1 or 2%.

Dehumidification and Heating The last combination process is dehumidification and heating, or dehumidification by desiccant moisture absorption. The desiccant material (contained in a wheel) is either rotated through the airstream or sprayed into the airstream and collected in a pan at the bottom. The other portion of the desiccant cycle is the regeneration process that heats up the desiccant to drive off the moisture to the atmosphere and start the cycle over again. The process is shown in Figure 4-9 and has movement to the right and the bottom of the psychrometric chart. For our example, we have our inlet conditions of 27°C db and  = 27% rh, which gives a dew-point temperature of 6°C and which a mechanical vapor compression refrigeration can easily reach. However, our leaving conditions require a dew-point temperature of –3°C, less than freezing (0°C), so mechanical cooling will not work. Our leaving conditions with desiccant dehumidification are 35°C db and  = 8% rh, for a dew-point temperature of –3°C. Note that these conditions are something you may not see in normal human comfort cooling, but they may be used in an industrial process or candy manufacturing facility. You will also have to contact a manufacturer for an exact selection and the regeneration method they use.


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4-1

Moist air that is heated without humidification has the following change in relative humidity: a) Increase b) Decrease c) Stays the same d) Depends on the type of humidifier

4-2

What is the equation that converts enthalpy changes into capacity (kW)? a) 1210 × airflow × (t 1 – t 2) b) 1.2 × airflow × (h1 – h 2) c) 3300 × airflow × (W 1 – W 2) d) None of the above

4-3

Which of the following is true concerning humidification by steam versus by (cold water) atomization? a) Atomization always maintains a constant relative humidity. b) Steam humidification adds no net energy to the airstream. c) Heat to make steam in the steam humidifier comes from the air entering the humidifier. d) Heat to evaporate water in the atomizer comes from the air entering the humidifier.

4-4

A heating coil can provide for both heating and humidification. a) True b) False

4-5

A cooling coil can provide for both cooling and dehumidification. a) True b) False

4-6

What is the change in enthalpy when dry air is heated from 10°C to 23°C? a) 10 b) 13 c) 16 d) 18

Fundamentals of Psychrometrics (SI), 4-7

Second Edition

33

What is the enthalpy change when saturated air at 10°C is conditioned to be saturated air at 23°C? a) 39 b) 35 c) 13 d) 45

4-8

One day in Phoenix, Arizona, the temperature reaches 40.5°C with 20% rh. Water is sprayed into the air to cool it. What will the temperature of the air be when the relative humidity increases to 50% rh? a) 30°C b) 35°C c) 40°C d) 22°C

4-9

If the air entering a heating coil is dry and 21°C db and the leaving air is 43°C, how many watts of cooling are supplied by the coil at 2.35 m3/s if the fan is located at the coil inlet? a) 58 000 W b) 65 000 W c) 61 000 W d) 62 550 W

4-10

Air enters a cooling coil at 38°C and 40% rh and leaves saturated at a temperature of 7°C. What is the total watts of cooling required if a 2.35 m3/s fan is located at the inlet of the cooling coil? a) 190 200 W b) 174 840 W c) 160 000 W d) 158 960 W

HVAC Design and the Psychrometric Chart Study Objectives After completing this chapter, you should be able to  apply HVAC systems to the psychrometric chart,  define and use sensible heat ratio for cooling, and  show the heating and humidification process on the psychrometric chart.


Schematic of an Air-Conditioning System To understand what an air-conditioning system is or what components it has, it is best to look first at the room, or space, that it is to serve. This space is to be occupied and maintained at some psychrometric condition (t db and ). This state is called the room design condition. This condition of temperature and humidity is being constantly defeated by heat flowing through the building envelope, coming in or going out. Furthermore, it is being changed by the activities happening inside. Occupants are providing heat and moisture to the space. There are machines and lights that transfer heat to the space as a byproduct of their operation. There may be things that are cooling the room, and there are things that are adding humidity to the room. These tend to change the interior room conditions. It is the purpose of the air-conditioning system to offset these changes by conditioning the room air to maintain the room at the desired condition. To do this, some air is taken out of the room, conditioned, and returned back to the space. This is done as depicted in the layout of a typical air-conditioning system shown in Figure 5-1.

Mixing Airstreams—Cooling Systems The mixing of two airstreams is common in HVAC systems to ensure the proper ventilation amount in the occupied space. This involves mixing an

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Chapter 5

Figure 5-1

HVAC Design and the Psychrometric Chart

Schematic of a general air-conditioning system.

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Summer design conditions shown on the psychrometric chart.

amount (fixed or variable) of outdoor air with a different amount of return or room air to meet the ventilation code in your local area. First, we will look at summer design conditions for a cooling application. Assume our room design is t db = 24°C and  = 50% rh, with an air outdoor design temperature of t db = 35°C and t wb = 18°C. See the psychrometric chart in Figure 5-2 for the plotted conditions.


37

Next, we draw a straight line between these two plotted points on the psychrometric chart. Our mixed condition will always be on this straight line. We can locate the exact location by using the following formula: t ma = [airflowoa (t oa) + airflowra (t ra)]/airflow sa where t ma

(5-1)

= dry-bulb temperature of mixed air

airflowoa = volume of outdoor air t oa

= temperature of outdoor air

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= volume of return air

t ra

= temperature of return air

airflow sa

= volume of supply air

Assume our HVAC system has a supply air volume (m 3/s sa) of 11 m3/s and a ventilation or outdoor air volume (m3/soa) of 2.2 m3/s. This means the return or room air volume (m3/sra) is the difference between the supply air volume and the return air volume, or m3/s sa = m3/soa + m3/sra

(5-2)

In our example, then, 11 m3/s sa = 2.2 m3/soa + 8.8 m3/sra Then, our mixed-air dry bulb temperature is t ma = [2.2 m3/soa (35°C) + 8.8 m3/sra (24°C)]/11 m3/s sa = [77 + 211.2]/11 = 26.2°C t db Now go back to the psychrometric chart in Figure 5-2 and plot the mixedair condition on this straight line at the intersection with the 26.2°C dry-bulb line marked MA. So our mixed-air conditions for these two airstreams are t db = 26.2°C and t wb = 18°C. This is an important item to know because the cooling coil will be sized using this condition as the entering air to this heat exchanger.

Mixing Airstreams—Heating Systems Now we will look at the same example in the winter heating mode. Assume a room design of t db = 21°C and  = 40% rh with air outdoor design temperature of t db = 0°C and  = 50% rh. See Figure 5-2 for the plotted conditions. We will again plot both points on our psychrometric chart, as shown in Figure 5-3, and connect these points with a new straight line. Using the same formula from the Mixing Airstreams—Cooling Systems section with different temperatures and the same volume, we get:

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Chapter 5

t ma = [m3/soa (t oa) + m3/sra (t ra)]/m3/s sa = [2.2 m3/s (0°C) + 8.8 m 3/s (21°C)]/11 m3/s = [0.22 + 184.8]/11 = 16.8°C Go back to the psychrometric chart in Figure 5-3 and plot this heating mixed-air condition on the straight line at the intersection with the 16.8°C dry bulb temperature line. This results in a mixing of these two airstreams at t db = 16.8°C and t wb = 10.5°C. This will be the entering air condition for the heating coil used in our system. One special note on heating mixed airstreams is that we need only the dry bulb temperature to select our heating coil or heat exchanger. The wet-bulb temperature becomes important only if humidification is needed in the HVAC system serving this area.

Sensible Heat Ratio—Cooling Sensible heat ratio (SHR) is a very important concept in HVAC psychrometric analysis. With the proper use of SHR, we will ensure that both the room dry-bulb temperature and room relative humidity are met in our design. It will ensure our room supply air dry-bulb temperature and relative humidity are cold and dry enough to achieve the room design conditions. Failure to do a proper SHR analysis could result in not meeting one or both of the room design parameters. Here’s an example to explain the concept.

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Winter design conditions shown on the psychrometric chart.

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For our example, we will use a space with a sensible heat gain of 80 000 W and a latent heat gain of 20 000 W, for a total load of 100 000 W. Our room SHR is 80 000 SHR = ------------------ = 0.8 100 000 We plot the room conditions of t db = 24°C and  = 50% rh on a psychrometric chart as shown in Figure 5-4. Then we draw a line from the center score mark to the value of 0.8 on the left side of the half circle. This is now our SHR slope line. Next we transfer it from the upper left corner to the room conditions on the chart. Please make sure the slope of this line is exactly the same as you plotted it. Note that any air condition along the SHR line will meet our room design conditions of t db = 24°C and  = 50% rh. These air conditions are the leaving air temperature off the cooling coil in the air handler. The only thing that changes on these varying leaving air temperatures is the volume. To solve the problem, we go to the sensible heat equation discussed in Chapter 4: q s = 1210 × airflow × (t 1 – t 2) We plot the intersection of the SHR line and a condition around 90% rh to the left of the room condition in Figure 5-4. We have selected a leaving air tem perature of t db = 12°C and  = 87% rh as the desired cooling coil leaving air temperature. It is best practice to contact a cooling coil manufacturer (or run

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Drawing the SHR slope line on the psychrometric chart.

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Chapter 5


their cooling coil selection software) and confirm that they can provide a coil that would perform to these conditions with a volume of q s = 1210 × airflow × (t 1 – t 2) 80 000 = 1210 × airflow × (24°C – 12°C) 80 000 airflow = ------------------------ = 5.5 m3/s 1210  12 So, our cooling coil needs to provide a leaving air temperature of t db = 12°C and  = 87% rh at our entering air temperature of t db = 24°C and  = 50% rh, with 5.5 m3/s of airflow through the coil. We could also have picked a leaving air temperature of t db = 15°C and  = 76% rh, resulting in an volume of q s = 1210 × airflow × (t 1 – t 2) 80 000 = 1210 × airflow × (24°C – 15°C) 80 000 – airflow = ------------------------ = 7.3 m3/s 1210  15 We may or may not have been able to find a cooling coil to perform this duty, because the leaving relative humidity is not close to 90%. Coils that dehumidify typically have a leaving relative humidity close to 90%. By providing this supply air quantity (airflow), supply air dry-bulb tem perature, and relative humidity, we will ensure that our room design conditions are satisfied.

Sensible Heat Ratio—Cooling with Outdoor Air This section covers the psychrometric process that is added to the SHR process to account for the outdoor air (ventilation air) in our HVAC system. We will use the same outdoor air design conditions of t db = 35°C and t wb = 24°C and 20% outdoor air from our example of the Mixing Airstreams—Cooling Systems section. We plot all the conditions on our psychrometric chart as shown in Figure 5-5. The mixed-air conditions are t db = 26.2°C and t wb = 18°C from our calculation of supply air as 5.5 m 3/s with the values from the previous example of outdoor air as 1.1 m3/s and return air as 4.4 m3/s. t ma = [airflowoa (t oa) + airflowra (t ra)]/airflow sa = [1.1 m3/s (35°C) + 4.4 m3/s (24°C)]/5.5 m3/s = [38.5 + 105.6]/5.5 m3/s = [144.1]/5.5 m3/s = 26.2°C t db Again, go to the t db = 26.2°C scale on the psychrometric chart and go up until you intersect the mixed-air line. That is our entering air condition to the cooling coil, t db = 26.2°C and t wb = 18°C. We still need to cool the air down to


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SHR for cooling with outdoor air shown on the psychrometric chart.

t wb = 12°C and  = 87% rh to meet our room conditions. To calculate the total cooling coil load, use the follow total heat equation: qt = 1.2 × airflow × (h1 – h 2) = 1.2 × 5.5 m3/s × (53 kJ/kg – 32 kJ/kg) = 138.6 kW or 138 600 W Note the increased cooling requirement due to the addition of outdoor air into the HVAC system. The room total load was 100,000 W and the outdoor air is an additional 38 600 W total. We can also calculate the room-only coil load with the same total heat equation: qt = 1.2 × airflow × (h1 – h 2) = 1.2 × 5.5 m3/s × (47 kJ/kg – 32 kJ/kg) = 1.2 × 5.5 m3/s × (15 kJ/kg) = 99 kW or 99 000 W Notice that this is not exactly the same as the 100 000 W total heat gain, but it is very close and within acceptable tolerance for HVAC calculations.

Psychrometric Process—Heating This section uses the same HVAC system we’ve been discussing to show how to handle the heating requirements of our space. The air handler has the same 5.5 m3/s. Assume our space has a heat loss of 90 000 W and all this load is sensible load. Our sensible heat is as follows:

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The heating process shown on the psychrometric chart.

q s = 1210 × airflow × (t 1 – t 2) = 1210 × 5.5 m3/s × (t 1 – t 2) = 90 000 W The heating room design t db = 21°C and  = 40% rh. So, our t 1 = 21°C and q s = 90 000 W 90 000 = 1210 × 5.5 m3/s × (21 – t 2) 90 000 = 6655 × (21 – t 2) 90 000 ----------------- = 21 – t 2  6655 13.5°C = (21 – t 2) t 2 = 34.5°C db So, if we supply 34.5°C warm air to our space on the coldest winter day, we will keep the space at t db = 21°C. See Figure 5-6 for how to show the heating process on a psychrometric chart. Now we add the need for humidification in the winter to our space. Assume, for example, that we need to add 15 000 W of latent heating in the form of moisture or water vapor. Our outdoor design is t db = 0°C and  = 50% rh. We add 20% outdoor air into our HVAC system and our new entering air conditions are t db = 16.8°C and  = 44% rh. The new total heating required is qt = 1.2 × airflow × (h1 – h 2) = 1.2 × 5.5 m3/s × (52 kJ/kg – 29 kJ/kg)


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Humidification need shown on the psychrometric chart.

= 1.2 × 5.5 m3/s × (23 kJ/kg) = 151.8 kW or 151 800 W An interesting part of this analysis is that the leaving air temperature from the heating coil has been increased to approximately t db = 35.3°C to account for the temperature drop the humidified air will cause, as shown in Figure 5-7. The amount of water vapor that must be added to the airstream is calculated based on an entering air of t db = 16.8°C and  = 44% rh with a humidity ratio of W 1 = 6.5 gw /kgda and leaving conditions of t db = 34.5°C and  = 20% rh with a humidity ratio of W 2 = 7.5 gw /kgda. Use the following formula: g/h = airflow × 1/Sv × (W 1 – W 2) × 3600 s/h = 5.5 m3/s × 1/0.88 m3/kg × (6.5 g/kg – 7.5 g/kg) × 3600 s/h = 22 500 g/h of water We will discuss humidification more in the next chapter as we differentiate between steam and water spray humidification.

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Chapter 5



5-1

The definition of sensible heat ratio (SHR) is the: a) Ratio of sensible to latent load b) Ratio of latent to sensible load c) Ratio of total load to sensible load d) Ratio of sensible load to total load

5-2

If the sensible load on a building is equal to the latent load, the value of SHR is: a) 2 b) 1 c) 0.5 d) –2

5-3

The psychrometric condition for supply air that will satisfy the requirements of a room depends on: a) The amount of outdoor air needed b) The desired room condition c) Room SHR d) All of the above e) Answers b and c only

5-4

Why is it possible to satisfy a room with a variety of “assumptions” about the temperature change across a coil (heating or cooling)? a) Because there is a corresponding airflow with every t . b) Because the heat/cool load calculation is never accurate. c) Because the comfort zone is large. d) Because there is a wide variety of methods for heating and cooling.

5-5

Which condition below is not possible to show on a psychrometric chart? a) t db = 24°C, h = 54 kJ/kg b) t db = 32°C, t wb = 25°C c) t wb = 25°C, h = 84 d) t db = 24°C,  = 50%

Fundamentals of Psychrometrics (SI), 5-6

Second Edition

45

In a system, 1 m3/s of air at 15³C and 30% rh is mixed with 4 m3/s air at 27°C and 80% rh. Find the mixed-air temperature using the mixing equation. a) 18°C b) 17.5°C c) 16°C d) 25°C

5-7

In Exercise 5-6, what is the mixed-air relative humidity? a) 51% rh b) 40% rh c) 60% rh d) None of these

5-8

In a system, 1 m3/s of air at 4°C and 90% rh is adiabatically mixed with moist air at 26°C but unknown relative humidity. The final mixture is at 22°C and 50% rh. What is the relative humidity and airflow rate of the second airstream? a) 42% rh, 5 m3/s b) 42% rh, 3 m3/s c) 60% rh, 5 m3/s d) 35% rh, 6 m3/s

5-9

If the sensible load is 600 000 W and the latent load is 300 000 W, what is the SHR? a) 2.0 b) 1.0 c) 0.66 d) 0.76

5-10

If the room design is t db = 24°C and  = 50% rh and we mix in 25% outdoor air at t db = 48°C and  = 10% rh, what is the mixed-air dry-bulb temperature? a) 45°C b) 42°C c) 30°C d) Not possible

5-11

From Exercise 5-10, what is the mixed-air relative humidity? a) 33% rh b) 15% rh c) 21% rh d) 28% rh

Psychrometrics in HVAC Equipment Study Objectives After completing this chapter, you should be able to  show the components of an air-handling unit and their psychrometric pro-

cesses and  explain two types of humidification.


The Air-Handling Unit: Heart of the Commercial Air-Conditioning System In Chapter 5, psychrometrics was used to determine the technical characteristics of the air-conditioning system required to perform a specific function. Psychrometrics was used to convert this information into the necessary volume and supply air conditions for both heating and cooling. These conditions not only determined the capacity of the unit in watts but also specified the amount of dehumidification and humidification by determining the entering and leaving dry-bulb and wet-bulb temperatures for both the heating and cooling coils. A manufacturer will usually package all (or most) of the components of an HVAC system into one large enclosure called an air-handling unit ( AHU ). AHUs (Figure 6-1) are almost custom-made for every design because the com ponents are selected from an extensive list of available sizes and capabilities to match the specific application. So that the AHU manufacturer can deliver the proper unit for the application, the design engineer must provide all of the necessary information.

Psychrometrics of a Cooling Coil Let’s start this discussion on what actually happens in a dehumidifying cooling coil as the air goes through it. The entering side of the coil is warmer than the leaving side of the coil. Therefore, the first few rows of the cooling

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Psychrometrics in HVAC Equipment

Chapter 6

Figure 6-1

Flow path through a simple AHU.

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Cooling coil line shown on the psychrometric chart.

coil are doing sensible cooling only, as can be seen in Figure 6-2, the cooling coil line. Note, it is horizontal and moving to the left side of the psychrometric chart. As the air moves further into the coil, the dehumidification process is starting as the cooling coil starts curving downward and to the left. The maximum dehumidification occurs just before the air exits the coil and generally leaves the coil around  = 90% rh. Again, refer to Figure 6-2 to see the final curve showing the completed dehumidification and cooling process.


49

Assume a fan has an airflow of 2.0 m 3/s, entering conditions of t db = 28°C and t wb = 20°C (with outdoor air mixed), and desired leaving conditions of t db = 11°C and approximately  = 90% rh. Now we can calculate the total cooling capacity of the cooing coil with the total heat equation discussed in Chapter 4: qt = 1.2 × airflow × (h1 – h 2) = 1.2 × 2.0 m3/s × (58 – 32 kJ/kg) = 1.2 × 2.0 m3/s × (26 kJ/kg) = 62.4 kW or 62 400 W We can now plot the cooling coil performance on the psychrometric chart as shown in Figure 6-2. We can see the total heat is broken down into a sensible component and a latent component, as also shown in Figure 6-2. Again, the HVAC engineer must provide the above information to the manufacturer of the cooling coils so they can provide an actual selection of either a chilled-water coil or a direct expansion (DX) refrigerant coil. Their output would include the size, height, width, number of coil rows, pressure drop (air pressure and water pressure, if a chilled-water coil), chilled-water temperature rise (you must supply the entering chilled-water temperature), and the actual leaving air conditions.

Psychrometrics of Fan Heat As a fan moves air through an HVAC system, the fan input energy is converted to heat as a result of the heat of compression. All the fan input energy ends up as heat as the fan increases the air pressure to provide air motion. Say, for example, a fan requires 7.46 kW to move 4.72 m3/s against of 750 Pa total pressure. 7.46 kW × 1000 = 7460 W Because the fan is moving 4.72 m3/s, we can use the sensible heat equation discussed in Chapter 4 to calculate the actual temperatures: q s = 1210 × airflow × (t 1 – t 2) 7460 W = 1210 × 4.72 m3/s × (t 1 – t 2) (t 1 – t 2) = 1.3°C temperature rise So, we have the addition of 1.3°C fan heat to account for in our psychrometric analysis. Fan heat is the addition of sensible heat, horizontal moving to the right on the psychrometric chart, either before the cooling coil (blow-through fan) or after the cooling coil (draw-through fan). Be careful with draw-through fans, because with these fans the fan leaving air temperature is higher than the cooling coil leaving air temperature. It is an additional load that must be accounted for in cooling heat gain calculations.

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Chapter 6


ASHRAE Handbook—HVAC Systems and Equipment (2012) gives a general estimate of fan heat as approximately 0.8°C per 750 Pa of total fan pressure. In this example we calculated a little over 1°C, while this general estimate would have given us 0.8°C. Therefore, it is better to perform the calculations.

Psychrometrics of a Heating Coil The process of heating air is a sensible-heating-only psychrometric problem, which means that the point moves from left to right horizontally across the psychrometric chart. This example again uses 2.0 m3/s airflow and the entering conditions to the hot water coil of t db = 15°C and approximately  = 30% rh (with outdoor air mixed) and a leaving condition of t db = 34°C. See Figure 6-3 for the process of heating and use the following equation: q s = 1210 × airflow × (t 1 – t 2) = 1210 × 2.0 m3/s × (15°C – 34°C) = 1210 × 2.0 × (19°C) = 45 980 W Even if you forget that the process is sensible heating only, using the total heat equation discussed in Chapter 4 will give you almost the same result as shown below: qt = 1.2 × airflow × (h1 – h 2) = 1.2 × 2.0 m3/s × (43 – 24 kJ/kg) = 1.2 × 2.0 m3/s × (19 kJ/kg) = 45.6 kW or 45 600 W

Humidification Equipment Humidification equipment can be divided into two groups: 1) isothermal or constant temperature and 2) adiabatic or moisture evaporating (see Figure 6-4). Isothermal humidification generally involves steam humidifiers, with many ways of steam generation, and addition of moisture to an airstream at a constant temperature. Adiabatic humidification is moisture evaporating and involves sprayed, atomized, media, or ultrasonic humidifiers. These humidifiers lower the air temperature as they add moisture to the airstream and are the same as evaporative coolers. In all methods of humidification, the following formula can be used to calculate the amount of water that must be added to the airstream in pounds of water per hour: g/h = airflow in m3/s ×

1 ---- × (w1 – w2) × 3600 s/h va


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Heating coil line shown on the psychrometric chart.

Figure 6-4

Steam (constant-temperature) and spray (adiabatic) humidifiers.

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Figure 6-5

Isothermal humidification shown on the psychrometric chart.

Isothermal humidification is shown in Figure 6-5 and has entering air conditions of t db = 32°C and  = 20% rh and leaving air conditions of t db = 32°C and  = 40% rh. The humidity ratio increases from 7 to 13. Note that the dry bulb temperature stays the same even though we have added moisture to the airstream. Adiabatic humidification is shown in Figure 6-6 and has entering air conditions of t db = 32°C and  = 20% rh and leaving air conditions of t db = 28°C and  = 36% rh. The humidity ratio is increased from 7 to 8.5. Note that the dry bulb temperature decreases in the process of adding moisture to the air. Be careful to not humidify the airstream at greater than  = 90% rh. The dew-point temperature of the interior surface of the ductwork is very important, and failure to observe this rule will result in condensation in the ductwork that will eventually leak out and cause a problem in the building. Please work closely with a humidifier supplier to make sure you are following all the application rules of that product and system.

Reference ASHRAE. 2013. Chapter 1. In ASHRAE handbook—Fundamentals. Atlanta: ASHRAE.


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6-1

Which type of humidification requires the change to not exceed the temperature rise capacity of a heating coil? a) Water spray b) Steam c) Both the same d) Neither has an impact

6-2

From the discussion of the psychrometrics of cooling coils, which “rule of thumb” will best select the cooling coil conditions? a) Temperature drop across a cooling coil should be about 10°C. b) Relative humidity off the coil should be 90%. c) Volume of air across a cooling coil should be kept to a minimum. d) Coil temperatures should be selected to be as low as possible.

6-3

Which of the following statements best describe why cooling coils cannot accommodate large latent loads with small sensible loads? a) Cooling coils rust if too much condensate forms. b) Cooling coils will freeze up if the coil temperature gets too low. c) Cooling coils tend to dehumidify first, then drop the air temperature. d) Condensation requires a drop in air temperature to the dew point.

6-4

Consider a room heating load with a 200 000 W sensible loss and 40 000 W latent loss, with room design conditions of t db = 22°C and approximately  = 40% rh. The air handler has an adiabatic humidifier downstream from a heating coil without any outdoor air. If the leaving air temperature is t db = 38°C after the humidifier, what is the airflow required to satisfy the load? a) 12 b) 10.3 c) 8 d) None of these


55

What is the leaving air temperature t db from the heating coil for the conditions listed in Exercise 6-4? a) 37°C b) 40°C c) 38°C d) None of these

6-6

What is the leaving relative humidity  from the heating coil for the conditions listed in Exercise 6-4? a) 15% rh b) 12% rh c) 20% rh d) 24% rh

6-7

What is the leaving relative humidity  from the adiabatic humidifier for the conditions listed in Exercise 6-4? a) 15% rh b) 25% rh c) 19% rh d) 28% rh

6-8

Using the air handler in Exercise 6-4 and 10.3 m3/s, adding a cooling coil to satisfy a room sensible heat gain of 146 kW and a room latent heat gain of 15 000 W, and room conditions of t db = 24°C and  = 40% rh and without outdoor air, what is the required leaving air temperature t db and  from the cooling coil? a) 12°C t db ,  = 90% rh b) 13°C t db ,  = 80% rh c) 12°C t db ,  = 75% rh

6-9

What is the room sensible heat ratio for the conditions listed in Exercise 6-8? a) 0.89 b) 0.95 c) 0.91 d) 1.0

6-10

Would you attempt to add humidity to the leaving airstream for the conditions listed in Exercise 6-8 in the cooling mode with an adiabatic humidifier? a) Yes b) No c) Not sure

Psychrometrics in Zoned HVAC Systems Study Objectives After completing this chapter, you should be able to  understand the most commonly used HVAC systems,  understand the psychrometric analysis of these HVAC systems, and  explain why we use zoned systems for HVAC.


Constant-Volume and Variable-Air-Volume Systems This chapter covers the major types of HVAC air systems. They can be divided into two major types: 1) constant-volume, variable air temperature and 2) variable-air-volume, constant air temperature. Constant-volume systems deliver the same volume, or airflow, at all load conditions and change the supply air dry-bulb temperature as the load changes. The load changes as the time of day changes, as the time of year changes, as the occupancy changes, as the internal loads change, and as the solar load changes. The room thermostat senses these changes in load and adjusts the supply air temperature accordingly to maintain the room at a constant temperature. In a chilled-water cooling, constant-volume air handler, the chilled-water coil has a control valve, controlled by the room thermostat to vary the supply air temperature. This type of system can provide acceptable comfort because there is an infinite number of chilled-water valve positions to match the large number of load conditions. In a direct expansion (DX) system, which has refrigerant in direct contact with the cooling coil tubes, there are typically steps or stages of cooling capacity. This causes the supply air temperature to be delivered at a set temperature between the design temperature and a few part-load temperatures. The room thermostat must have multiple stages so it can bring on additional capacity if the room temperature rises or remove capacity if the room temperature drops.

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This causes a compromise in room temperature control, as the setpoint is almost never met; the room temperature actually fluctuates around the setpoint, approximately ±0.5°C or ±1°C. In most cases, this fluctuation in room tem peratures can still provide an acceptable comfort level, but not one as good as a chilled-water system. Note that some newer styles of DX systems have variable-refrigerant-flow capacity, which can provide a greater level of comfort in the room. These systems can better match the refrigerant flow required to handle the room’s load variations, thus providing better comfort than stepped DX systems. Variable-air-volume (VAV) systems deliver a variable amount of volume, or airflow, at all load conditions at a constant supply air dry-bulb temperature. Again, the load changes with the five variables mentioned above. The room thermostat senses these load changes and adjusts the quantity of air delivered to the space so the room air temperature remains constant. Note that all VAV systems do a very good job of part-load room humidity control, as the main air handler cools and dehumidifies the air at all load conditions to a constant dew point temperature. In most cases, with modern air-handling systems, the air volume (airflow) is changed with a variable-frequency drive (VFD) wired to the electric fan motor. Because input frequency determines the speed of rotation in induction electric motors as the load changes, the VFD frequency output matches the airflow required by adjusting the fan speed. VAV air handlers can have chilled-water or DX cooling coils installed in them. With chilled water, the chilled-water control valve is infinitely variable and can be controlled to maintain a constant supply air temperature as the air volume goes up and down in response to load changes. With DX cooling coils, the supply air temperature changes in stages, up and down, as the load changes and the airflow increases or decreases. Again, even with stages, you can still obtain acceptable room comfort. One final point: you can use many small fan-coils to provide a lot of zone control in a large building, but each fan-coil is a constant-volume, single-zone subsystem.

Constant-Volume, Single-Zone System Constant-volume, single-zone systems are the most commonly used in HVAC and come in many forms. They include single-zone air-handling room fan-coils, packaged DX single-zone rooftop systems, and residential split systems, to name a few. They have one common trait: the unit is controlled by one room thermostat only. Therefore, they can be applied to only one zone and provide room comfort to only one zone. A zone, by definition, has like occupancy and like thermal characteristics but does not have a defined size. For example, a large interior cubical office area could be a zone as large as 185 m2 if the density and usage are uniform throughout. It could also be a zone as small as 6 m2 with a row of exterior, individual offices lining an outside wall.


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Psychrometrics of single-zone system at full load.

The psychrometric chart of a single-zone system is determined by the room’s sensible and latent loads, the room’s sensible heat ratio (SHR), and either the quantity of outdoor air needed or the code-required ventilation rate. It is shown at full load in Figure 7-1. In all examples in this chapter, we will use room design conditions of t db = 24°C and  = 50% rh and 20% outdoor air for ventilation. Next, look at the psychrometric chart in Figure 7-2 for part-load operation. The part-load supply air temperature is higher than the design temperature because the room thermostat is calling for less cooling. Depending on the partload SHR, the room relative humidity may not be met at this part-load condition. So you must determine if this deviation above the design  = 50% rh is acceptable. Also note that the mixed-air condition will usually change to a lower value, as the outdoor air will typically be lower at part-load conditions.

Constant-Volume, Single-Zone System with Reheat Constant-volume, single-zone systems with reheat are used when we need to control the room relative humidity at all load conditions. The control is simple; the cooling coil supply air dry-bulb temperature is set to a constant leaving tem perature. Therefore, the supply air is dehumidified regardless of the room partload conditions. The supply air is reheated purely to satisfy the room thermostat. It should be noted that this type of system is an expensive one to operate, as we pay to cool and dehumidify the supply air and then pay again to warm up, or reheat, the same airstream. If you are forced to provide this design in a sys-

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Psychrometrics of single-zone system at part load.

tem, you should look for ways to use recovered energy to provide the reheat. Recovered energy sources could be condenser heat from the cooling system, such as a heat recovery chiller or condenser reheat plus sensible heat recovery from exhaust or return airstreams, among others. The psychrometric chart for this system is shown in Figure 7-3 at part-load conditions. Note that even with a SHR of less than the design SHR, we can meet the room relative humidity condition. The psychrometrics of this system at full load are exactly the same as shown in Figure 7-1 because, at full load, we are not doing any reheat.

Constant-Volume, Single-Zone System with Face and Bypass Dampers on the Cooling Coil Constant-volume, single-zone systems with face and bypass dampers on the cooling coil are not as common as they were in the past, but we will analyze this type in our psychrometric training process nonetheless. This system is basically a single-zone air handler with a cooling coil plus an extra damper section; see Figure 7-4 for all the components. The dampers are used to adjust the amount of supply air that goes through the cooling coil or the amount of mixed air that bypasses the cooling coil, thus the name face and bypass. The damper is controlled by the room thermostat as it maintains a constant room dry-bulb temperature. As the room temperature drops, the dampers are adjusted to open the bypass section and close down on the face section, raising the air handler supply dry-bulb temperature. The opposite is true if the room


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Psychrometrics of constant-volume, single-zone system with reheat at part load.

Bypass

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Figure 7-4

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Components of constant-volume, single-zone system with face and bypass dampers on the cooling coil.

temperature rises: the damper adjusts for less bypass air and more cooling coil air to lower the air handler supply temperature. The psychrometrics of this system at part load are show in Figure 7-5. Note that the cooling coil supply air temperature decreases as we decrease the airflow through the cooling coil at part load. Therefore, this system does a much better job of maintaining the room relative humidity at part load than a constant-volume, variable-temperature, single-zone system. The psychrometrics of the face and bypass system at full load are the same as shown in Figure 7-1 because no air is being bypassed at full load. The room thermostat is calling for full cooling, so 100% of the supply air is being cooled in the cooling coil.

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Psychrometrics of constant-volume, single-zone system with face and bypass dampers on the cooling coil at part load.

Constant-Volume System with Terminal Reheat Now we shift our focus to HVAC air systems that are designed to serve many zones from one air handler. The constant-volume with terminal reheat system was the primary commercial office space system from the 1940s to the mid-1970s. The system is fairly simple: a single-zone air handler supplies air ducted throughout the building, and then reheat coils are put in each duct runout to serve any individual zones. The air handler ensures a constant leaving supply air temperature all year round, and each reheat coil tempers the air to meet the room temperature desired in each zone. A few things to note: zones can have different temperature setpoints, zone reheats can be at different stages of tempering as zone loads change, and part-load humidity control by zone is very good because the main supply air is constantly being dehumidified. However, this system is very expensive to operate because it is both cooling and reheating throughout the day, month, and year. Also, most energy codes, such as ANSI/ASHRAE/IES Standard 90.1 (ASHRAE 2013) and California’s Title 24 (CBSC 2013), restrict the use of this system for obvious reasons. It was a popular way to get zone control in buildings when energy was cheap and before VAV was invented. The psychrometrics of this system at full load are the same as those shown in Figure 7-1. The psychrometrics of this system at part load are the same as those shown in Figure 7-2, with the exception that each zone has a separate psychrometric


Figure 7-6

63

Components of constant-volume, multizone system (ASHRAE 2016, Figure 12).

chart, because the amount of the reheat will vary by zone and the SHR can be slightly different by zone. However, a word of caution: the worst zone, from a lowest-SHR standpoint, sets the air handler supply air dry-bulb temperature for the entire system.

Constant-Volume Multizone and Dual-Duct Systems Constant-volume multizone and dual-duct systems are designed to provide comfort to multiple zones by mixing cool air with warm air so that the discharge supply air temperature is satisfied by the zone thermostat. The only difference between these two systems is where the mixing of the hot and cool air occurs. In a multizone system, the blow-through air handler has the mixing dampers mounted on the front or top of the air-handling unit (AHU). The dampers are on a common shaft, but offset by 90°, so when the hot deck is full open, the cold deck is full closed and vice versa. Control of the air handler is simply a constant deck temperature for each, say t db = 12°C for the cold deck and t db = 40°C for the hot deck. A single damper activator is then controlled by a room thermostat to position the dampers to meet the room temperature setpoint. If the zone is too cool, this actuator opens more to the hot deck to warm the air and increase the supply air temperature, thus warming the room. If the zone is too warm, this actuator repositions to open to the cold deck (closes down on the hot deck) to cool the air and lower the supply air temperature. Supply ductwork is run out from this central air handler to each zone. Small units may have as few as three zones and large units as many as 18 to 20 zones. Figure 7-6 shows a constant-volume, multizone system. In the dual-duct system, the blow-through air handler again has a hot deck and a cold deck on the discharge side of the supply fan, but no mixing dampers. Two sets of supply ductwork are run around the building in parallel with each other. At any location that a zone is required, a dual-duct mixing box is installed and dual taps are run to the cool deck duct and the hot deck duct. The mixing box has two dampers on a common shaft offset by 90° rotation with a single actuator. The room thermostat is connected to this actuator to provide zone comfort. The air handler again has controls to maintain constant leaving

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cold-deck and hot-deck temperatures. The operation of this system is identical to that of the multizone system. Figure 7-7 shows a constant-volume, dual-duct system. Both of these systems are very expensive to operate because both decks are kept at a constant leaving air temperature and achieve comfort by mixing both airstreams. Their use is restricted by most energy codes due to the high energy usage required for proper operation. The psychrometrics of the multizone and dual-duct systems are shown in Figure 7-8. Note the mixing line from the cold-deck discharge at t db = 12°C

Figure 7-7

Components of constant-volume, dual-duct system (ASHRAE 2016, Figure 13). 80%

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Psychrometrics of constant-volume multizone and dual-duct systems.


65

and  = 87% rh to the hot-deck discharge at t db = 40°C and  = 22% rh. The actual discharge temperature can be any one of an infinite number of points along this line, and each zone can be at a different discharge point. Also note that we can heat and cool with this mixing of airstreams at each zone. Exercise caution to make sure that the cold-deck dry-bulb temperature is low enough to satisfy the SHR for the worst zone and that the hot-deck tem perature is warm enough to satisfy the heat loss of the worst zone. Figure 7-9 shows the psychrometrics of multizone and dual-duct systems in winter heating mode. In this example, the room is at t db = 21°C and  = 40% rh, with an outdoor design of t db = 0°C and  = 50% rh. The mixed condition is t db = 16°C and  = 45% rh. The cooling coil line is sensible cooling only, or horizontal, and ends at t db = 12°C and  = 57% rh. The heating coil line is sensible heating only and ends at t db = 41°C and  = 12% rh. Note that because both the cooling and heating are sensible only (no latent), the hot-deck and cold-deck mixing line is the sum of the cooling coil line and the heating coil line. Any zone will require supply air that is mixed along this line.

Variable-Air-Volume Systems for Multiple Zones In the early 1970s, the high energy usage of most of the constant-volume systems forced designs and owners to look for systems with lower operational costs. Variable-air-volume (VAV) systems made their debut and are still very popular today, as they provide great fan horsepower savings for most of the year.

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Psychrometrics of constant-volume multizone and dual-duct systems in winter heating mode.

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An air handler with a single supply duct and a VFD on the supply fan provides a constant discharge air temperature of t db = 12°C to the building. As zones are required, a VAV box is tapped into this main supply trunk. The VAV box is a single-damper device that modulates the airflow to the zone in response to the room thermostat. Said another way, it is an air-throttling device that provides comfort to the zone. At full load in the zone, the VAV box is wide open in response to the room thermostat. At all part-load conditions, the VAV box has its damper closed some amount in response to lower demand for cooling in the zone. The psychrometrics for VAV systems are the same as shown in Figure 7-1 for each zone on the AHU. Again, make sure the air handler leaving supply air dry-bulb temperature is low enough to satisfy the worst zone’s SHR. Now at part load we simply go to the sensible heat equation, q s = 1210 × airflow × (t 1 – t 2), to determine how we handle the VAV operation. Because the supply air temperature is constant year round, we reduce the sensible heat by simply reducing the airflow delivered to the zone. If we want half of the sensible load, we only supply half the airflow to the zone. The room thermostat controls the actuator on the damper shaft to keep the zone comfortable. So, at most part-load conditions, the psychrometrics do not change—only the airflow changes to satisfy the reduced load, and Figure 7-1 is still valid. This assumes the room SHR stays close to the full- load SHR at part load. This style of VAV box is a cooling-only box and can typically only be used for the building interior or zones that are in cooling year round.

Variable-Air-Volume Systems with Heating VAV Boxes Most exterior zones require heating for a portion of the year. This section covers two different styles of heating VAV boxes: VAV reheat boxes and fan powered VAV boxes that can have reheat as needed. Note that the central air handler does not change with this design—we have cooling-only boxes on the interior and heating boxes on the exterior. To build a VAV reheat box, we simply take a cooling-only box and put a reheat coil on the discharge of it. The coil can be hot water, steam, or electric duct heater in design. The psychrometrics of a VAV reheat box at full load are the same as shown in Figure 7-1. As the demand for cooling drops, we use the same part-load psychrometric chart as shown in Figure 7-1 but at lower airflow. But at some preset minimum airflow, say 35% of full airflow, we energize the reheat coil. The controls modulate the amount of reheat or temperature rise in response to the room thermostat. The psychrometrics are shown in Figure 7-10, with the maximum reheat to a dry-bulb temperature of 35°C. Remember, only reheat to a discharge air temperature into the zone that will satisfy the room thermostat.


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Figure 7-10

Psychrometrics of a VAV reheat box at part load.

To construct a fan-powered VAV box that can also have reheat as needed, add a sheet metal plenum on the side of a cooling-only box with a small directdrive centrifugal fan. The fan can draw plenum air through an air filter and discharge it into the cooling box downstream of the cooling control damper. A back-draft damper is required on the discharge of the centrifugal fan. This is considered a parallel fan-powered box. The control is similar to that of a VAV reheat box. At full cooling, we have 100% of the cooling air going to the zone (no fan operation). At part load, we throttle down the supply air to a lower amount (no fan operation). At a preset minimum airflow, say 35% of the full-load airflow, we fix the cooling damper to that position and start the centrifugal fan. It draws air from the ceiling plenum and mixes it with the reduced flow t db = 12°C to discharge warmer air into the zone and meet the room thermostat setpoint. This ceiling plenum air can be 1.5°C to 3°C higher than the room temperature as long as the building is occu pied, because it has the heat of the lights added to it. The psychrometrics of a fan-powered VAV box that can have reheat as needed are shown in Figure 7-11. Note that you are mixing supply air at t db = 12°C and  = 87% rh with plenum air at t db = 26°C and  = 42% rh along the mixing line. The location will be determined by the airflow of supply air and the airflow of the plenum air provided by the small centrifugal fan. If needed, another reheat coil could be mounted on the box discharge section to provide additional heating capacity for wintertime zone heat losses. The coil is shown as additional reheat (“Reheat if Needed”) in Figure 7-11.

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Psychrometrics of a fan-powered VAV box with reheat at part load.

References ASHRAE. 2013. ANSI/ASHRAE/IES Standard 90.1, Energy standard for building except low-rise residential buildings. Atlanta: ASHRAE. ASHRAE. 2016. Chapter 4, Air handling and distribution. In ASHRAE handbook—HVAC systems and equipment . Atlanta: ASHRAE. CBSC. 2013. California building standards code. Title 24 of California Code of Regulations. Sacramento, CA: California Building Standards Commission.


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For all of the Skill Development Exercises for Chapter 7, consider three zones in a small office building that we are going to heat and cool. The cooling and heating loads are as follows: Zone

Sensible Cooling, W

LatentCooling, W

HeatingSensible, W

1

12 000

1700

6000

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2000

8000

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10 000

Assume room design conditions of the following: Coolingt db = 24°C and  = 50% rh Heating t db = 21°C and  = 40% rh Use a sea-level psychrometric chart.

7-1

What is the sensible heat ratio for all three zones in order 1, 2, 3? (Round to two decimal places.) a) 0.87, 0.89, 0.86 b) 0.88, 0.9, 0.91 c) 0.87, 0.89, 0.88

7-2

If we provide 25% outdoor air for code-required ventilation to all three zones, what is the mixed air condition in the summer if the outdoor air is t db = 38°C and  = 25% rh? a) t db = 34.5°C and  = 30% rh b) t db = 29°C and  = 36% rh c) t db = 27.5°C and  = 42% rh

7-3

For Zone 1 only, if we use individual fan-coils for each zone, what is the required supply airflow? a) airflow = 0.9 m3/s b) airflow = 0.99 m3/s c) airflow = 1.03 m3/s

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For Zone 1 only, what are the leaving air conditions from the cooling coil assuming we use 25% outdoor air from Exercise 7-2 and the correct supply airflow? a) t db = 12°C and  = 90% rh b) t db = 14°C and  = 88% rh c) t db = 16°C and  = 80% rh

7-5

For Zone 1 only, what is the total cooling capacity, q t , of the cooling coil with the correct airflow and leaving air conditions? a) 18 000 W b) 22 000 W c) 19 008 W

7-6

If all three zones were put on a central air handler with a constant-volume terminal reheat system, what would the airflow of all three zones be, in order 1, 2, 3? (Same outdoor design and percent outdoor air.) a) 0.99, 1.3, 1.6 b) 0.9, 1.4, 1.7 c) 0.99, 1.32, 1.65

7-7

If all three zones were put on a central air handler with a variable-air-volume reheat VAV box and 25% outdoor air, what are the required leaving air conditions from this air handler? a) t db = 13.5°C and  = 91% rh b) t db = 14.5°C and  = 88% rh c) t db = 16°C and  = 82% rh

7-8

With the system in Exercise 7-7, what are the new required airflows by zone in order 1, 2, 3 with the new leaving conditions? a) 0.94, 1.26, 1.57 m3/s b) 0.99, 1.26, 1.6 m3/s c) 1.03, 1.3, 1.55 m3/s

7-9

What is the reheat required by zone in order 1, 2, 3 to meet the total reheat load plus the winter heat loss load? (Use t db = 21°C for room condition and 25% outdoor air.) a) 14 000, 18 000, 25 000 W b) 14 530, 19 930, 24 240 W c) 16 000, 21 000, 26 000 W


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From Exercise 7-7, with the correct leaving conditions and airflow, what is the total cooling capacity of the central air-handler cooling coil? a) 73 280 W b) 70 120 W c) 76 910 W

7-11

If the system in Exercise 7-7 were a constant-volume, dual-duct system, what would be the heat capacity of the hot-deck coil used in the central air handler? (Room at t db = 24°C.) a) 75 000 W b) 65 000 W c) 55 100 W d) 60 500 W

Energy Conservation and Psychrometrics Study Objectives After completing this chapter, you should be able to  understand energy saving systems and strategies and their effects on the

psychrometric analysis and  understand why lower energy costs result from the use of these systems and strategies.


Introduction This chapter covers energy conservation principles and strategies and how they affect the HVAC system design. There are many devices and strategies that can conserve energy, but this text focuses on only the most commonly used in HVAC: heat recovery devices, energy recovery devices, air-side economizers, water-side economizers, and supply air temperature reset. We will examine the psychrometric processes and the energy-saving effects of these five systems.

Heat Recovery Devices Heat recovery is the exchange of dry-bulb air temperature only between two airstreams. In an HVAC system, this is typically between the outdoor air used for ventilation and the common building exhaust airstream. The greater the temperature difference that exists between the two airstreams, the more we can affect the HVAC performance. This is also called sensible heat recovery, because we only change the dry-bulb temperature. The four most common types of heat recovery devices are heat wheels, airto-air heat exchangers, heat pipes, and glycol run-around loops. Figure 8-1 shows three of these devices, and Figure 8-2 shows a schematic of a coil runaround loop.

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Energy Conservation and Psychrometrics

Figure 8-1

Heat and energy recovery devices.

Figure 8-2

Run-around loop (ASHRAE 2012, Figure 14).

A heat wheel is a large-diameter, deep wheel consisting of a honeycomb styling of metal pockets. Half of the wheel is located in the exhaust airstream and the other half is located in the incoming outdoor airstream. In the winter, as the wheel rotates slowly, the cold outdoor air is preheated by the warmer exhaust airstream that is being dumped outdoors. The individual pockets change temperature rapidly as the wheel rotates from one airstream to the other and back again. In the summer, the wheel also precools the hot outdoor air with indoor room-temperature exhaust air from the building. Note that this device must be ducted and positioned in such a way that the airstreams are next to each other somewhere in the system. Also, a small amount of cross-contamination occurs between the airstreams, so care must be taken depending on the application. For example, exhaust air from an office building (toilet, break room, janitorial


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closets, etc.) is generally acceptable to use, but exhaust from hospital isolation rooms is never acceptable. One final note on rotary heat wheels is that the loss of air from the supply side (outdoor air) to the exhaust air side can be as high as 10% of the total airflow. This is the cause of the cross-contamination, but you must also increase the airflow higher than the design required amount to cover this loss or leakage. An air-to-air heat exchanger uses parallel plates of metal (or other material) to separate the exhaust air from the outdoor air. The plates are packed tightly next to each other in the heat exchanger, and the heat transfer goes across each plate. So every other plate has indoor or outdoor air flowing through it in opposite directions. This heat exchanger design does not have any cross-contamination between the two airstreams, and both airstreams must be side by side somewhere in the system. Heat pipes look like one big chilled-water coil or heat exchanger. The difference is that each tube going across the coil is a separate chamber filled with a very small refrigerant charge. Each half of the coil, split side to side, sits in one of the airstreams. In winter, the outdoor air side of the heat pipe condenses the refrigerant in the tube and rejects the heat to the outdoor air, warming it. The other side sits in the warm exhaust air, which vaporizes the refrigerant, absorbing the heat from the warm airstream. By natural pressure difference, the warm refrigerant vapor migrates to the colder side, where it condenses. The coil is tilted slightly so the liquid refrigerant flows back to the warm side on the bottom of each small tube. When the season changes to summer, the tilt must be reversed so the heat pipe can work in the reverse and cool the warm entering outdoor air. A glycol run-around loop is two large coils or heat exchangers placed in the two airstreams that are connected by two pipes and one pump to move the glycol-water solution from one coil to the other. They transfer heat from the exhaust airstream to the outdoor airstream by warming up and then cooling down the pumped glycol-water solution. The advantage of this system is that the airstreams can be located great distances from each other. There is no cross-contamination with this device.

Psychrometric Effects and Savings of Heat Recovery Figure 8-3 shows winter temperature-change-only heat recovery. The exhaust airstream is at t db = 21°C and  = 40% rh and 0.7 m3/s. The outdoor airstream is at t db = 0°C and  = 50% rh and 0.94 m3/s. Note: Typically more outdoor air is brought in than exhausted to positively pressurize the building to keep the indoor environment clean and dust free. However, we must first introduce the concept of heat exchanger effectiveness. Counterflow air-to-air heat exchangers can achieve close to 100% effectiveness. But, the range of 50% to 70% effectiveness for cost and air pressure drop considerations is generally selected. For this example, we will use 60% effectiveness.

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If the exhaust airstream were cooled from t db = 21°C to t db = 0°C the maximum amount of sensible heat transfer would be q s = q max = 1210 × airflow × (t 1 – t 2) = 1210 × 0.7 m3/s × (21 – 0) qmax = 17 780 W Because the effectiveness is 60%, the transferred heat is qmax × effectiveness = q transferral 17 780 W × (0.6) = 10 668 W Then the outdoor air is warmed to q s = 10 668 W = 1210 × 0.94 m3/s × (0 – t 2) difference 10 668 ----------------- = 9.4 C =  0 – t 2  1137 t 2 = 8.4°C Likewise, the exhaust air is cooled to q s = 10 668 W = 1210 × 0.7 m3/s× (21– t 2) 10 668 ---------------- = 12.6 C =  21 – t 2  847 t 2 = 8.4°C


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Heat recovery in the summer.

So 10 668 W of energy has been conserved by preheating the outdoor air by 9.4°C and therefore lowering the operating cost in the process. Next, we will look at the process in the summer with outdoor design conditions of t db = 35°C and  = 40% rh and room conditions of t db = 24°C and  = 50% rh, as shown in Figure 8-4. q s = q max = 1210 × 0.7 m3/s × (35 – 24) = 1210 × 0.7 m3/s × (11) qmax = 9317 W Using the same effectiveness of 60%, q s = q max × (0.60) = q transferral qtransferral = 9317 W × (0.60) = 5590 W So we cool the outdoor air sensibly by q s = 5590 W = 1210 × 0.94 m3/s × (35 – t 2) 5590 ------------ = 4.9C =  35 – t 2  1137 t 2 = 30.1°C

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Energy Conservation and Psychrometrics And we warm the exhaust air by q s = 5590 W = 1210 × 0.7 m3/s × (24 – t 2) 5590 ------------ = 6.5 C =  24 – t 2  8471 t 2 = 30.5°C

An air-to-air heat exchanger was used in this example, but the same procedure applies to the other three heat recovery devices—only the effectiveness will change.

Condensation and Frost Formation We must consider two other items in the use of heat recovery devices, condensation and frost formation. Condensation can occur on a heat exchanger if the exhaust air dew-point temperature is reached. For example, in Figure 8-3, if the exhaust air temperature were lowered below t db = 7°C, then condensation would occur on a small portion of the heat transfer surface. Be sure to specify these devices with a condensate drain to properly collect this water. Frost can form on the leaving side of the exhaust air in the heat recovery device when the outdoor air gets to temperatures of –12°C or less. In the winter, because the exhaust air dew-point temperature is almost always above 2°C, this is a design consideration in Climate Zones 4 to 8 (ASHRAE 2013). Frost formation in the heat recovery device decreases the exhaust airflow and reduces the device’s effectiveness. The methods used to prevent frost formation on heat recovery devices are as follows: • •

•

•

Preheat the outdoor airstream to some preset temperature entering the device (e.g., –12°C) (all types of heat recovery devices). Install a set of bypass dampers around the device to bypass a portion of the outdoor airstream so the heat exchanger does not get so cold (heat wheel, air-to-air heat exchanger, heat pipe). Install a three-way control valve in the glycol piping to control the glycol solution inlet temperature on the exhaust coil to somewhere around –1°C, thus preventing frost formation (run-around loop). Increase the rotational speed of the heat wheel so the outdoor air does not cool the heat transfer part of the wheel to below around –1°C (heat wheel).

Energy Recovery Devices Energy recovery is the transfer of sensible heat and latent heat from the exhaust airstream to the outdoor airstream. At first glance, the wheel looks identical to a heat recovery wheel that transfers temperature only. However, the heat transfer material in an energy recovery device is coated with a desiccant material that absorbs water vapor and then rejects the water vapor to the other


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airstream. So in winter, the moisture in the warmer building air is transferred to the very dry outdoor air, thus helping maintain the indoor relative humidity. And in the summer, the cool and dryer indoor air that is exhausted absorbs some of the moisture in the hot, humid outdoor air. Energy recovery devices come in two types: 1) rotary energy wheels and 2) plate air-to-air heat exchangers with moisture transfer plates that are not solid metal in construction. Following are performance examples of an energy recovery rotary wheel. Summer performance with a total energy effectiveness of 0.87 is shown in Figure 8-5 and is as follows: Outdoor air conditions:

t db = 35°C, t wb = 24°C, outdoor air = 1.08 m3/s

Supply air conditions:

t db = 28°C, t wb = 20°C, supply air = 0.94 m3/s

Return air conditions:

t db = 24°C, t wb = 16°C, return air = 0.7 m 3/s

Exhaust air conditions:

t db = 33°C, t wb = 24°C, exhaust air = 0.84 m3/s

Note the leakage airflow is 0.14 m3/s and the outdoor air total cooling load reduction is qt = 1.2 × airflow × (h1 – h 2) = 1.2 × 0.94 m3/s × (69 – 54) = 1.2 × 0.94 m3/s × (15) = 16.92 kW or 16 920 W

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The cooling system only needs to cool the outdoor air from t db = 28°C and t wb = 20°C to t db = 24°C and  = 50% rh instead of from t db = 35°C and t wb = 24°C, which reduces the cooling energy costs. Winter performance with the same effectiveness is shown in Figure 8-6 and is as follows: Outdoor air conditions:

t db = 0°C, t wb = –3°C, outdoor air = 1.08 m3/s

Supply air conditions:

t db = 13°C, t wb = 9°C, supply air = 0.94 m 3/s

Return air conditions:

t db = 21°C, t wb = 13°C, return air = 0.94 m3/s

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t db = 3°C, t wb = 0°C, exhaust air = 0.84 m3/s

The outdoor air heating load is reduced by qt = 1.2 × airflow × (h1 – h 2) = 1.2 × 0.94 m3/s × (41 – 6) = 1.2 × 0.94 m3/s × (35) = 39.48 kW or 39 480 W The outdoor air now has only to be heated from t db = 13°C to t db = 21°C instead of from t db = 0°C, which reduces the cooling energy costs, as with the winter operation.

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Note that in both cases of heating and cooling to the outdoor air, the humidity ratio changes so you are not only transferring sensible heat, but also latent heat. This latent heat helps maintain the indoor room relative humidity at the design condition.

Air-Side Economizer An air-side economizer is an HVAC system option that allows cooling without the use of mechanical refrigeration, thus making the cooling energy equal to zero. For instance, to satisfy the cooling load, we need the supply air t db = 12°C. Then, any time of the year that the outdoor air temperature is t db = 12°C or less, we can shut the mechanized refrigeration off and open our outdoor air dampers to 100%, pulling in 12°C air. Now this 12°C outdoor air provides all the cooling to the building. This is called full economizer mode and should be considered any place where there is a cooling need in the fall, winter, and spring. You can also implement partial air-side economizing in your HVAC system. For example, the outdoor air temperature is t db = 18°C and your room is at t db = 24°C. Instead of cooling the air from 24°C down to 12°C for the supply air, you only have to cool the air from 18°C to 12°C with mechanical refrigeration. In this example, the refrigeration load is reduced approximately 50%. So partial economizing can be used any time the outdoor air is less than the room cooling condition. A word of caution when using partial air-side economizing on constantvolume, variable-temperature systems and it is very humid or raining outdoors is that the humid outdoor air will cause the room relative humidity to go above the room design condition of  = 50% rh. In this instance, use the outdoor air enthalpy instead of the dry-bulb temperature to initiate partial economizing. The outdoor enthalpy should be at least 8 kJ/kgda less than the room enthalpy condition before you allow partial economizing. This is less of a concern on VAV systems since they control to a constant leaving cooling air temperature at all load conditions. Note that with the air-side economizer option the air-handling system must be capable of bringing in up to 100% outdoor air. This means the outdoor air weather louvers, outdoor air duct, and outdoor air dampers must all be sized and selected for the full airflow of the air handler. Also, because you are bringing up to 100% outdoor air into the building, you must provide a way to relieve, or exhaust, this additional air and provide building pressure control to prevent overpressurization of the building. Failure to do this will result in the exterior doors staying partially open, not fully closing.

Water-Side Economizer The water-side economizer system was developed to provide cooling via the chilled-water system to air-handling systems that do not have any outdoor

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air connection or have minimum outdoor air capability. For this energy-saving option, you must have a chilled-water cooling delivery system, a cooling tower, and a heat exchanger piped between the chilled- and condenser-water systems. Let’s look at the performance of a cooling tower from the psychrometric side. A cooling tower produces cool water dependent on the ambient wet-bulb temperature only. For example, the ambient is t db = 35°C and t wb = 24°C, as in our previous examples. The cooling tower has a 3°C approach temperature at full load or heat rejection. This means the cooling tower can produce 27°C leaving cooling tower water, or t wb = 24°C + 3°C = 27°C. In the process of cooling the water down to 27°C, a portion of the recirculating water is evaporated by slightly cooling the air (lower t db) and greatly adding moisture to the air. It is not uncommon that the air leaving a cooling tower is between 90% and 95% relative humidity. As the outdoor air cools in the fall, winter, and spring, so does the outdoor wet-bulb temperature. Also, the building sensible load decreases in these non peak cooling seasons. For example, the ambient temperature is t db = 4°C and t wb = 1°C. We also have the same approach temperature at part load of 3°C (if the tower is at full load and the same ambient conditions, the approach would be around 6°C). So this tower will make 7°C leaving condenser water. If our heat exchanger has a 1°C approach temperature, then we can make 8°C chilled water to be distributed throughout the building to provide cooling where needed. See Figure 8-7 for a system schematic of a water-side economizer. This water-side economizer is most commonly used on systems that have chilled-water fan-coils or small air handler type systems. Typically, these sys-

Figure 8-7

Water-side economizer schematic.


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tems have no outdoor air or a small amount of outdoor air, and it could be difficult or impossible to duct 100% outdoor air capability to them. The heat exchanger between the condenser water system and the chilledwater system must be cleanable. So, because both can have the condenser water debris easily removed, plate-and-frame and shell-and-tube (tube-side condenser water) are the two heat exchangers most commonly used.

Supply Air Temperature Reset Supply air temperature reset works because in almost all comfort cooling system applications, the sensible heat gain decreases in the fall, winter, and spring. So, if the sensible heat gain to a zone is half the summer peak gain by the sensible heat equation discussed in Chapter 4, q s = 1210 × airflow × (t 1 – t 2), and if our airflow is constant, then the t can be half to produce half the sensible cooling. Consider: if we have airflow = 2000, a summer peak supply air temperature of t db = 12°C, and a room condition of t db = 24°C, then Full sensible cooling:

q s full = 1210 × 0.94 m3/s × (24°C – 12°C) = 13 650 W

Half sensible cooling:

q s half = 1210 × 0.94 m3/s × (24°C – 12°C) × 0.5 = 6825 W

Simply by having the air handler supply t db = 18°C, we can provide the cooling necessary. See Figure 8-8 for the psychrometric analysis of this concept. Also note that the sensible heat ratio will change as the heat gain decreases in the off-peak load times of the year. A few words of caution are in order before you apply this strategy: •

• •

•

Data centers, IT rooms, and telecom rooms may have only slight decreases in the sensible load throughout the year and, therefore, cannot use supply air temperature reset. Process or industrial applications may never change in sensible load and so cannot use supply air temperature reset. VAV systems are very economical to run because the airflow varies as the sensible load goes down. But if you reset the supply air temperature upward too much, you will eat into or eliminate the fan horsepower savings derived from this system. Granted, you can probably reset the supply air t db = 12°C to 13°C or 14°C, but not up to 19°C to 21°C. This is a great case for energy modeling of the VAV system to see how high in reset tem perature you can go at the expense of fan energy usage. Remember, the higher the supply air temperature, the more refrigeration or cooling energy you can save. Be very careful about how much supply air temperature reset you do in very humid areas of the world. Remember, the supply air temperature from the cooling coil sets the required dew point to maintain the room relative

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humidity via the sensible heat ratio calculation. Many times in the off-peak cooling season you could reset the supply air temperature, but the outdoor humidity conditions force the cooling coil to always be in dehumidification mode. And be careful of rainy days in the off-peak cooling season, as the outdoor moisture content may take precedent over supply air temperature reset. All modern HVAC control systems should do some indoor relative humidity sensing as a standard benefit, so the building manager/operator has the ability to make the right decision in this event. Many other energy-conserving measures can be used on a building that do not involve the psychrometric process in the HVAC system and, therefore, are not discussed in this course.

References ASHRAE. 2013. Figure B1-1, Normative Appendix B, Building envelope climate criteria. In ANSI/ASHRAE/IES Standard 90.1-2013, Energy standard for buildings except low-rise residential buildings. Atlanta: ASHRAE. ASHRAE. 2012. Chapter 26, Air-to-air energy recovery equipment. In ASHRAE Handbook—HVAC Systems and Equipment .


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8-1

A heat wheel with a desiccant coating is a: a) Sensible heat recovery device b) Total enthalpy heat recovery device c) Sensible-to-total heat recovery device d) Total-to-sensible heat recovery device

8-2

When is preheating of the outdoor airstream necessary on a heat recovery device? a) When the outdoor air temperature is below –16°C. b) When the outdoor air dew point is below 0°C. c) When the exhaust airstream has a dew point above 0°C and the leaving air temperature is below 0°C. d) All of the above.

8-3

Heat recovery effectiveness is the actual amount of heat transferred versus the maximum amount that could be transferred. a) True b) False

8-4

Energy recovery involves the transfer of sensible heat from one airstream to the other airstream. a) True b) False

8-5

An air-side economizer should be considered on any/all air systems that have 100% outdoor air capability and high operation hours with an ambient air tem perature below 16°C and a demand for cooling. a) True b) False

8-6

Water-side economizers can be used on a chilled-water system with all terminal fan-coils and an air-cooled water chiller. a) True b) False

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There is a sensible heat recovery system between equal outdoor air and exhaust airstreams in Phoenix, Arizona, and the summer design outside is t db = 48°C and  = 10% rh. If the effectiveness is 75% of the heat recovery device and the exhaust airstream is t db = 24°C and  = 40% rh, what are the dry-bulb temperature and relative humidity of the outdoor airstream leaving the recovery device? a) t db = 32°C and  = 25% rh b) t db = 35°C and  = 20% rh c) t db = 38°C and  = 18% rh d) t db = 30°C and  = 28% rh

8-8

From Exercise 8-7, what are the leaving air conditions of the exhaust airstream with everything else being the same? a) t db = 30°C and  = 30% rh b) t db = 35°C and  = 25% rh c) t db = 40°C and  = 20% rh d) t db = 42°C and  = 16% rh

8-9

If the entering air conditions to a cooling tower are t db = 48°C and t wb = 19°C and the cooling tower has a full-load approach temperature of 4°C, what is the leaving water from cooling tower (at full load)? a) 40°C b) 35°C c) 23°C d) 29°C

8-10

Supply air temperature reset can be used on all air-conditioning systems, any time of the year in all parts of the world, regardless of the ambient air conditions. a) True b) False

Special Applications and Psychrometric Considerations Study Objectives After completing this chapter, you should be able to  understand the five special cases of psychrometric applications in the

HVAC industry and the psychrometric analysis of each,  select equipment for each of these systems, and  understand the effect of indirect and direct evaporative cooling in series.


Introduction This chapter discusses five special cases of psychrometric applications in the HVAC industry: cooling towers, cleanrooms, indoor swimming pools, direct evaporative cooling, and indirect evaporative cooling.

Cooling Towers Starting with cooling towers may seem strange because the function of a cooling tower is to cool water. However, it cools the water by rejecting the heat, through an evaporative/sensible process cooling, to the ambient or outdoor air. The cooling tower approach temperature is the difference between the leaving water temperature and the ambient air wet-bulb temperature. An example shows what happens to the ambient air and the entering water as they pass through the cooling tower. Consider a 1055 kW cooling tower that can cool 0.056 m3/s of water from 35°C to 30°C. The heat being rejected by the water is q in kW = 4710 kW/m3/s °C × water flow in m3/s t in °C = 4710 × 0.056 m3/s × 5°C = 1319 kW

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The entering ambient air to the cooling tower is t db = 40°C and t wb = 25°C. This tower moves 28.45 m3/s of air, which leaves the tower almost at the saturation line on the psychrometric chart. Therefore, the air must pick up 1319 kW, as shown by the total heat required equation discussed in Chapter 4: qt = 1.2 × airflow × (h1 – h 2) where h 1 is the enthalpy at t db = 40°C and t wb = 25°C, or h = 76 kJ/kgda. Therefore, 1319 kW = 1.2 × 28.45 m3/s × (h1 – h 2) 1319 kW = 1.2 × 28.45 m3/s × (76 – h 2) therefore h 2 = 114 kJ/kgda This matches the Figure 9-1 psychrometric chart. So, the leaving air temperature is t db = 33°C and t wb = 32.8°C, or almost saturated air. Note that the cooling tower approach is the difference between the leaving water temperature (28°C) and the 25°C entering wet-bulb ambient temperature, or 3°C. See Figure 9-1 for the details of the air condition as it flows through the cooling tower. Note that part of the process is sensible cooling, but the majority is latent heat being added to the ambient air as the tower water is cooled. Also note that the entering ambient air can be anywhere on the t wb = 25°C wet-bulb line and we will get the same results. The only difference is the

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amount of latent heat and the amount of sensible heating or cooling that takes place as the air moves through the cooling tower.

Indoor Swimming Pools From a design prospective, the indoor air dry-bulb temperature for indoor swimming pools or natatoriums used for recreational purposes should be the same temperature as the pool water temperature. That way, the amount of pool water lost to evaporation into the pool enclosure is reduced. However, if this is not possible, do not allow t db to be greater than +2°C above the water temperature. The range for recreational pool water temperature is 24°C to 19°C, with a recommended  = 50% rh to 60% rh. As an example in this section, we will design around t db = 26°C and  = 55% rh. A specially designed unit called a pool dehumidifier is used to provide dehumidification, reheat, and the proper amount of outdoor ventilation air as shown in Figure 9-2. The unit also has the capability to provide auxiliary heat (of wintertime outdoor air), pool water heat (energy saver), an external refrigerant condenser (reject heat outdoors), and energy or heat recovery devices as explained in Chapter 8. It is also acceptable to return the water condensed by the dehumidifying coil back to the swimming pool. The amount of outdoor ventilation air required is 2.4 L/s·m2 of total area, which comes from ANSI/ASHRAE Standard 62.1 (ASHRAE 2013). Total area is defined as the pool surface plus the deck area around the pool if it gets wet during normal operation. If this indoor pool area is connected to or is part of a larger building, then it should be at a slightly negative pressure to the rest of the building (–12 Pa of water). This will ensure that the chlorine odor and the highmoisture-content air do not get into the rest of the building.

Figure 9-2

Single-blower pool dehumidifier.

90


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80%

60%

100 30

40%

30 120

90 25 80

y l p a h 60 t E n

110 25

70

e u r t r a p e m 50 T e 20 Cooling n Coil i o Line t r a t u15 a S

40 30

H

100 u

m i d i t 20% 90 y 15 R a t i o

Indoor Pool Air

80

10

Pool SHR Line

10 20

5

0

20

Reheat Coil

10

12

20

70 5

26

30

40

60

50


Figure 9-3

Indoor pool dehumidification and reheat process.

Take care to ensure that the building envelope is designed to handle the high-dew-point indoor air. Exterior windows are discouraged in cold winter design areas, as they will sweat excessively and cause damage. For our example, the indoor pool is at t db = 26°C and  = 55% rh. Note that the indoor dew point is t dp = 17°C. The pool dehumidification unit cools and dehumidifies the air first and then reheats the air to meet the psychrometric needs. From a load calculation, the room sensible heat ratio (SHR) for this example is 0.5. So the air is cooled from t db = 26°C down to t db = 12°C and room moisture is removed in the process. Then the air is reheated from t db = 12°C up to t db = 19°C to intersect the SHR line on the psychrometric chart and balance the sensible and latent cooling processes. See Figure 9-3 for the actual pool dehumidification and reheat process. Also note that to provide for the full heating load at winter design, both the refrigeration reheat coil and the auxiliary heat will be used to warm the air to the design supply air temperature. Conversely, for the summer design, dehumidification will be needed. So the supply air temperature of t db = 19°C will cover the cooling design load without any reheat. This then requires a second refrigerant condenser to reject the heat to the ambient air.

Cleanrooms The need for cleanrooms has expanded greatly over time. They are used in manufacturing facilities for microprocessors, pharmaceuticals, medical products, and various electronic devices. The common requirement of these facili-


Figure 9-4

91

Makeup air handler.

ties is a clean area using high-efficiency particle arrestor air filtration with precise dry-bulb temperature and relative humidity control. A cleanroom HVAC system is divided into two subsystems with different functions. First is the makeup air system, which provides preconditioned outdoor air to the cleanroom area because a large amount of exhaust air is typically removed in the manufacturing process. The second subsystem is the recirculating room air handlers that slightly cool and filter the room air, at the same time maintaining an airflow rate in the room. As an example, a cleanroom is designed at summer conditions of t db = 21°C and  = 45% rh, which is means a t dp of 9°C. The design outdoor air is t db = 40°C and t wb = 25°C. Therefore, the makeup air handler must cool this hot/ humid summer design air to something less than t db = 9°C, because the recirculation air handlers perform only sensible cooling. Or, said another way, the makeup air has to remove all the outdoor air latent load plus any room latent load prior to the air being mixed into the cleanroom. Because most cleanrooms have very few people working in them at any given time and the manufacturing tool load is mostly a sensible load, the latent load from the cleanroom is typically small. For this example, by cooling the air down to t db = 8°C, we can handle the cleanroom latent load from our latent load calculations. The makeup air handler must also be able to add humidity to the air when the outdoor air is dry, as well as heat the air to near room condition in the winter. These are typically very large and long air handlers, because they perform many functions on the outdoor airstream. Note in Figure 9-4 all the components necessary to provide preconditioned outdoor air. Let’s look at the psychrometrics of this make-up air handler at the summer design conditions. The psychrometric chart for this unit is shown in Figure 9-5.

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80%

60%

40%

100 30

30 120

90 25 80

y l p a h 60 t E n

Glycol 40 Coil

110 25

70

r e t u a r p e e m20 50 T n i o t r a Chilled t u15 Water a S

20

Outdoor Air Design

Precool Coil

m i d i t 20% 90 y 15 R a t i o 80

Coil

30

H

100 u

10

10

70

20

5 5

Reheat

0

8 10 11

19 20

30

34

40

60

50


Figure 9-5

Makeup air handler psychrometrics at summer design conditions.

All three cooling coils and the reheat coil are used to precondition the air before mixing it with the cleanroom recirculation air. A summary of the four coils follows: 1. Precool coil takes the 100% outdoor air from t db = 40°C down to t db = 34°C with a process cooling loop water supply at 30°C. 2. Chilled-water coil then takes the air from t db = 34°C down to t db = 11°C with the chilled-water loop at 5°C. 3. Glycol (antifreeze) cooling coil then takes the air from t db = 11°C down to t db = 8°C with a water/glycol solution at 2°C. 4. Reheat coil then heats the air from t db = 8°C up to t db = 19°C with the return water from the process cooling loop that is at about 35°C. This is a huge energy-saving feature to use the return water as a heating source and thus cool the return water in the process. This preconditioned outdoor air is then mixed with the cleanroom return air that is at t db = 21°C and  = 45% rh at the inlet of the recirculation air-handling units. If the relative humidity gets to below 45%, then the glycol coil raises its supply air temperature a degree or two to bring the cleanroom back up to  = 45% rh. At any outdoor condition, this makeup air handler has the components to bring the outdoor air to the desired mixed condition before it is introduced into the cleanroom. The cleanroom recirculation units are very simple: they consist of a fan, a small chilled-water coil, and a set of air prefilters. The actual cleanroom high-


60%

100 30

40%

93

30 120

90 25 80

y l p a h 60 t E n

70

r e t u a r p e m 50 T e 20 n i o t r a t u15 a S

40 30

110 25 20

H

100 u

m i d i t 20% 90 y 15 R a t i o

Clean Room Condition

80

10

10 70

20

5

Coil Leaving Temperature from Recirculating AHU

0

10

20

30

40

5

60

50


Figure 9-6

Recirculation cleanroom air handler psychrometrics.

efficiency particle arrestor filters are in the ceiling of the cleanroom and provide airflow at a high velocity through the room. Because they are sensiblecooling-only units and have a very small temperature drop, t db = 21°C to t db = 19°C, or only 2°C of cooling, there is no temperature deviation in the cleanroom. The psychrometrics of the recirculation cleanroom air handler are shown in Figure 9-6. The mixing of the preconditioned makeup air and the return air from the cleanroom is interesting in that the ratio of the room/makeup air is typically 20/1 to 50/1, depending on the process in the cleanroom. Therefore, the mixed condition is only reduced a few tenths of a degree in the dry bulb temperature. The room temperature sensor controls the chilled-water valve in the recirculation air handler to change the leaving air temperature slightly if needed.

Direct Evaporative Cooling Direct evaporative cooling can be used very effectively in the hot and dry climates of the world to provide for human comfort. In direct evaporative cooling, the airstream is 100% outdoor air and in contact with water. As some water evaporates, it lowers the dry-bulb temperature of the airstream, cooling the air. The process of direct evaporative cooling is a constant-wet-bulb-temperature process, as shown in Figure 9-7. For example, consider outdoor air conditions of t db = 38°C and  = 5% rh, in which the process goes up and to the left on the 15°C wet-bulb line. If the direct evaporative cooling has a 95% efficiency, then the leaving air tempera-

94


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80%

60%

40%

100 30

30 120

90 25 80

y l p a h t 60 E n

110 25

70

e u r t r a p e m 50 T e 20 n SHR i o Line t a u r15 t S a

40 30

20

m i d i t 20% 90 y 15 R a t i o

Room Condition

80

10

10

70

20

Leaving Evaporation Section

5

0

H

100 u

10

17

Outdoor Air

20

30

38

40

5

60

50


Figure 9-7

Psychrometrics of direct evaporative cooling.

ture will be t db = 16°C and  = 90% rh. We can also calculate the condition as follows: Evaporative effect = (EATdb – EATwb) × Efficiency 22°C = (38°C – 15°C) × 0.95 LAT = EATdb – Evaporative effect = 38°C – 22°C = 16°C where EAT =

entering air temperature

LAT

leaving air temperature

=

So, if the room has a high sensible load and a very low latent load, we can keep the room conditions at t db = 24°C and  = 60% rh with a fairly flat SHR line.

Indirect Evaporative Cooling Indirect evaporative cooling is simply cooling the air with a cooling coil and then using the evaporative process to cool the water that goes through the cooling coil. By definition, then, indirect evaporative cooling is not as efficient as direct evaporative cooling because two heat transfers take place in the process. Take the direct evaporative cooling example: we can make 16°C water in this process and we waste the cool air back to ambient. We take this 16°C water to a


60%

100 30

40%

95

30 120

90 25 80

y l p a h 60 t E n

70

r e t u a r p e m 50 T e 20 n i o t a r t u15 S a

40 30

110 25 20

m i d i t 20% 90 y 15 R a t i o

Room Condition

80

10

10 20

70

Leaving Air

5

5

0

H

100 u

10

20

24

30

40

60

50


Figure 9-8

Psychrometrics of indirect evaporative cooling.

cooling coil and we can make t db = 20°C air with t db = 24°C air entering the coil. Again, if our cooling load is mostly/all sensible and our airflow is high enough, we can maintain the room at t db = 24°C and  = 50% rh. See Figure 98 for the psychrometrics of the indirect evaporative cooling process. However, when used together with 100% outdoor air, the leaving air tem perature can be lowered by 5°C. We use the same outdoor conditions of t db = 38°C and  = 5% rh, but our efficiency is only 45% at best. We can use the same formula as before so our indirect section can deliver Evaporative effect = (EATdb – EATwb) × Efficiency 10°C = (38°C – 16°C) × 0.45 LAT = EATdb – Evaporative effect = 38°C – 10°C = 28°C Indirect evaporative cooling can be used in series with direct evaporative cooling. Air from the indirect section can now enter the direct evaporative section at t db = 28°C and  = 10% rh and move up the wet-bulb line of 11.5°C with a leaving air condition of t db = 12°C and  = 90% rh. This is now a much better leaving air condition, as we can easily maintain room conditions of t db = 24°C and  = 50% rh. Evaporative effect = (EATdb – EATwb) × Efficiency 16°C = (28°C – 11.5°C) × 0.95

96

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Special Applications and Psychrometric Considerations 80%

60%

40%

100 30

30 120

90 25 80 70

y l p a h 60 t E n

110


40 30

10

25 20

20% 15

Room Condition

H

100 u

m i d i t 90 y R a t i o 80

10

S H R

70

20

5

Leaving Evaporation Section

0

10 12

Outdoor Air

Indirect

20

24

28 30

38 40

5

60

50


Figure 9-9

Psychrometrics of indirect and direct evaporative cooling in series.

Figure 9-10 Air handler with indirect and direct evaporative cooling secti ons.

LAT = EATdb – Evaporative effect = 28°C – 16°C = 12°C So, as you can see, the combination of both indirect and direct evaporative cooling in series can deliver air that can provide for a comfortable room without mechanical refrigeration. See Figure 9-9 for a plot of indirect and direct evaporative cooling in series. Figure 9-10 shows the component arrangement for an air handler with both indirect and direct evaporative cooling. The waste air is the air that provides cooling to the one side of the indirect heat exchanger.


97

Reference ASHRAE. 2013. ANSI/ASHRAE Standard 62.1-2013, Ventilation for acceptable indoor air quality. Atlanta: ASHRAE.

98

Chapter 9



9-1

A cooling tower needs to reject heat from 0.075 cu m/s of water entering at 35°C and leaving at 30°C. What is the total heat required to be rejected? a) 1766 kW b) 184 kW c) 1610 kW d) 1495 kW

9-2

From Exercise 9-1, if the cooling tower has an airflow of 47.2 m3/s and ambient air conditions of t db = 30°C and t wb = 24°C, what are the leaving air conditions of the tower? a) t db = 30°C, t wb = 28.5°C b) t db = 32°C, t wb = 31°C c) t db = 31°C, t wb = 30.5°C d) t db = 30°C, t wb = 29.7°C

9-3

What is the cooling tower approach temperature for the cooling tower in Exercise 9-2? a) 3°C b) 7°C c) 4°C d) 6°C

9-4

In the design of an indoor swimming pool, it is best to keep the swimming pool water temperature and the room temperature as far apart as comfortably possi ble. a) True b) False

9-5

In a cleanroom with design conditions of t db = 20°C and  = 40% rh, the makeup air must be cooled to what dry-bulb temperature or the relative humidity will not be met? a) t db = 12°C b) t db = 20°C c) t db = 5°C d) t db = 10°C


99

If we cool the air via direct evaporative cooling from t db = 43°C and  = 2% rh, what is the lowest leaving air temperature we can achieve? a) t db = 17°C b) t db = 20°C c) t db = 18°C d) t db = 12°C

9-7

In Exercise 9-6, if our evaporative efficiency is 80%, what are the leaving air conditions? a) t db = 20°C and  = 70% rh b) t db = 22°C and  = 70% rh c) t db = 21°C and  = 58% rh d) t db = 25°C and  = 50% rh

9-8

In Exercises 9-6 and 9-7, if the room sensible heat ratio is 0.9, what is the expected room relative humidity if the room is at t db = 24°C? a)  = 53% rh b)  = 60% rh c)  = 50% rh d) Cannot maintain room at t db = 24°C with this leaving condition

9-9

If we use the same outdoor conditions of t db = 43°C and  = 2% rh from Exercise 9-6 and an indirect evaporative cooling section of 40% efficiency, what are the leaving air conditions from this section? a) t db = 33°C and  = 3% rh b) t db = 31°C and  = 20% rh c) t db = 35°C and  = 5% rh d) t db = 19°C and  = 5% rh

9-10

If we add a direct evaporative cooling section in series downstream of the indirect section in Exercise 9-9 and the direct section has an efficiency of 70%, what are the leaving air conditions? a) t db = 15°C and  = 95% rh b) t db = 15°C and  = 65% rh c) t db = 19°C and  = 60% rh d) t db = 19°C and  = 52% rh

Appendix A— Thermodynamic Properties of Moist Air Table A-1 Temp., °C t

Thermodynamic Properties of Moist Air at Standard Atmospheric Pressure, 101.325 kPa

Humidity Ratio W s , kg w /kgda

Specific Volume, m3 /kgda

Specific Enthalpy, kJ/kgda

vda

vas

vs

hda

has

hs

Specific Entropy, kJ/(kgda ·K) Temp., °C t s s da

s

–60 –59 –58 –57 –56 –55 –54 –53 –52 –51 –50 –49 –48 –47 –46 –45 –44 –43 –42 –41 –40 –39 –38 –37 –36 –35 –34 –33 –32 –31 –30 –29 –28 –27 –26 –25 –24 –23 –22 –21

0.0000067 0.0000076 0.0000087 0.0000100 0.0000114 0.0000129 0.0000147 0.0000167 0.0000190 0.0000215 0.0000243 0.0000275 0.0000311 0.0000350 0.0000395 0.0000445 0.0000500 0.0000562 0.0000631 0.0000708 0.0000793 0.0000887 0.0000992 0.0001108 0.0001237 0.0001379 0.0001536 0.0001710 0.0001902 0.0002113 0.0002345 0.0002602 0.0002883 0.0003193 0.0003532 0.0003905 0.0004314 0.0004761 0.0005251 0.0005787

0.6027 0.6055 0.6084 0.6112 0.6141 0.6169 0.6198 0.6226 0.6255 0.6283 0.6312 0.6340 0.6369 0.6397 0.6425 0.6454 0.6482 0.6511 0.6539 0.6568 0.6596 0.6625 0.6653 0.6682 0.6710 0.6738 0.6767 0.6795 0.6824 0.6852 0.6881 0.6909 0.6938 0.6966 0.6994 0.7023 0.7051 0.7080 0.7108 0.7137

0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0002 0.0002 0.0002 0.0002 0.0003 0.0003 0.0003 0.0004 0.0004 0.0004 0.0005 0.0005 0.0006 0.0007

0.6027 0.6055 0.6084 0.6112 0.6141 0.6169 0.6198 0.6226 0.6255 0.6283 0.6312 0.6340 0.6369 0.6397 0.6426 0.6454 0.6483 0.6511 0.6540 0.6568 0.6597 0.6626 0.6654 0.6683 0.6711 0.6740 0.6769 0.6797 0.6826 0.6855 0.6883 0.6912 0.6941 0.6970 0.6998 0.7027 0.7056 0.7085 0.7114 0.7143

–60.341 –59.335 –58.329 –57.323 –56.317 –55.311 –54.305 –53.299 –52.293 –51.287 –50.281 –49.275 –48.269 –47.263 –46.257 –45.252 –44.246 –43.240 –42.234 –41.229 –40.223 –39.217 –38.212 –37.206 –36.200 –35.195 –34.189 –33.183 –32.178 –31.172 –30.167 –29.161 –28.156 –27.150 –26.144 –25.139 –24.133 –23.128 –22.122 –21.117

0.016 0.018 0.021 0.024 0.027 0.031 0.035 0.040 0.046 0.052 0.059 0.066 0.075 0.085 0.095 0.107 0.121 0.136 0.153 0.172 0.192 0.215 0.241 0.269 0.301 0.336 0.374 0.417 0.464 0.516 0.573 0.636 0.706 0.782 0.866 0.958 1.059 1.170 1.291 1.424

–60.325 –59.317 –58.308 –57.299 –56.289 –55.280 –54.269 –53.258 –52.247 –51.235 –50.222 –49.209 –48.194 –47.179 –46.162 –45.144 –44.125 –43.104 –42.081 –41.057 –40.031 –39.002 –37.970 –36.936 –35.899 –34.859 –33.815 –32.766 –31.714 –30.656 –29.593 –28.525 –27.450 –26.368 –25.278 –24.181 –23.074 –21.958 –20.831 –19.693

–0.2494 –0.2447 –0.2400 –0.2354 –0.2307 –0.2261 –0.2215 –0.2169 –0.2124 –0.2078 –0.2033 –0.1988 –0.1943 –0.1899 –0.1854 –0.1810 –0.1766 –0.1722 –0.1679 –0.1635 –0.1592 –0.1549 –0.1506 –0.1464 –0.1421 –0.1379 –0.1337 –0.1295 –0.1253 –0.1211 –0.1170 –0.1129 –0.1088 –0.1047 –0.1006 –0.0965 –0.0925 –0.0884 –0.0844 –0.0804

–0.2494 –0.2446 –0.2399 –0.2353 –0.2306 –0.2260 –0.2213 –0.2167 –0.2121 –0.2076 –0.2030 –0.1985 –0.1940 –0.1895 –0.1850 –0.1805 –0.1761 –0.1716 –0.1672 –0.1628 –0.1583 –0.1539 –0.1495 –0.1451 –0.1408 –0.1364 –0.1320 –0.1276 –0.1232 –0.1189 –0.1145 –0.1101 –0.1057 –0.1013 –0.0969 –0.0924 –0.0880 –0.0835 –0.0790 –0.0745

–60 –59 –58 –57 –56 –55 –54 –53 –52 –51 –50 –49 –48 –47 –46 –45 –44 –43 –42 –41 –40 –39 –38 –37 –36 –35 –34 –33 –32 –31 –30 –29 –28 –27 –26 –25 –24 –23 –22 –21

–20 –19 –18 –17 –16 –15 –14 –13 –12 –11

0.0006373 0.0007013 0.0007711 0.0008473 0.0009303 0.0010207 0.0011191 0.0012261 0.0013425 0.0014689

0.7165 0.7193 0.7222 0.7250 0.7279 0.7307 0.7336 0.7364 0.7392 0.7421

0.0007 0.0008 0.0009 0.0010 0.0011 0.0012 0.0013 0.0014 0.0016 0.0017

0.7172 0.7201 0.7231 0.7260 0.7290 0.7319 0.7349 0.7378 0.7408 0.7438

–20.111 –19.106 –18.100 –17.095 –16.089 –15.084 –14.078 –13.073 –12.067 –11.062

1.570 1.728 1.902 2.091 2.298 2.523 2.769 3.036 3.326 3.642

–18.542 –17.377 –16.198 –15.003 –13.791 –12.560 –11.310 –10.037 –8.741 –7.419

–0.0765 –0.0725 –0.0685 –0.0646 –0.0607 –0.0568 –0.0529 –0.0490 –0.0452 –0.0413

–0.0699 –0.0653 –0.0607 –0.0560 –0.0513 –0.0465 –0.0416 –0.0367 –0.0317 –0.0267

–20 –19 –18 –17 –16 –15 –14 –13 –12 –11

–10

0.0016062

0.7449

0.0019

0.7468

–10.056

3.986

–6.070

–0.0375

–0.0215

–10

102

Appendix A Thermodynamic Properties of Moist Air

Table A-1 Temp., °C t

Thermodynamic Properties of Moist Air at Standard Atmospheric Pressure, 101.325 kPa (Continued)




vda

vas

vs

hda

has

hs


s

–9 –8 –7 –6 –5 –4 –3 –2 –1

0.0017551 0.0019166 0.0020916 0.0022812 0.0024863 0.0027083 0.0029482 0.0032076 0.0034877

0.7478 0.7506 0.7534 0.7563 0.7591 0.7620 0.7648 0.7677 0.7705

0.0021 0.0023 0.0025 0.0028 0.0030 0.0033 0.0036 0.0039 0.0043

0.7499 0.7529 0.7560 0.7591 0.7622 0.7653 0.7684 0.7716 0.7748

–9.050 –8.045 –7.039 –6.034 –5.028 –4.023 –3.017 –2.011 –1.006

4.358 4.763 5.202 5.677 6.193 6.750 7.354 8.007 8.712

–4.692 –3.282 –1.838 –0.356 1.164 2.728 4.337 5.995 7.707

–0.0337 –0.0299 –0.0261 –0.0223 –0.0186 –0.0148 –0.0111 –0.0074 –0.0037

–0.0163 –0.0110 –0.0055 0.0000 0.0057 0.0115 0.0175 0.0236 0.0299

–9 –8 –7 –6 –5 –4 –3 –2 –1

0 1 2 3 4 5 6 7 8 9

0.0037900 0.004076 0.004382 0.004708 0.005055 0.005425 0.005819 0.006238 0.006684 0.007158

0.7733 0.7762 0.7790 0.7819 0.7847 0.7875 0.7904 0.7932 0.7961 0.7989

0.0047 0.0051 0.0055 0.0059 0.0064 0.0068 0.0074 0.0079 0.0085 0.0092

0.7780 0.7813 0.7845 0.7878 0.7911 0.7944 0.7978 0.8012 0.8046 0.8081

0.000 1.006 2.011 3.017 4.023 5.029 6.034 7.040 8.046 9.052

9.475 10.198 10.970 11.794 12.673 13.611 14.610 15.674 16.807 18.013

9.475 11.203 12.981 14.811 16.696 18.639 20.644 22.714 24.853 27.065

0.0000 0.0037 0.0073 0.0110 0.0146 0.0182 0.0219 0.0254 0.0290 0.0326

0.0364 0.0427 0.0492 0.0559 0.0627 0.0697 0.0769 0.0843 0.0919 0.0997

0 1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62

0.007663 0.008199 0.008768 0.009372 0.010013 0.010694 0.011415 0.012181 0.012991 0.013851 0.014761 0.015724 0.016744 0.017823 0.018965 0.020173 0.021451 0.022802 0.024229 0.025738 0.027333 0.029018 0.030797 0.032677 0.034663 0.036760 0.038975 0.041313 0.043783 0.046391 0.049145 0.052053 0.055124 0.058368 0.061795 0.065416 0.069242 0.073286 0.077561 0.082081 0.086863 0.091922 0.097278 0.102949 0.108958 0.115326 0.122080 0.129248 0.136858 0.144945 0.153545 0.162697 0.172446

0.8017 0.8046 0.8074 0.8103 0.8131 0.8159 0.8188 0.8216 0.8245 0.8273 0.8301 0.8330 0.8358 0.8387 0.8415 0.8443 0.8472 0.8500 0.8529 0.8557 0.8585 0.8614 0.8642 0.8671 0.8699 0.8727 0.8756 0.8784 0.8813 0.8841 0.8869 0.8898 0.8926 0.8955 0.8983 0.9011 0.9040 0.9068 0.9096 0.9125 0.9153 0.9182 0.9210 0.9238 0.9267 0.9295 0.9324 0.9352 0.9380 0.9409 0.9437 0.9465 0.9494

0.0098 0.0106 0.0113 0.0122 0.0131 0.0140 0.0150 0.0160 0.0172 0.0184 0.0196 0.0210 0.0224 0.0240 0.0256 0.0273 0.0291 0.0311 0.0331 0.0353 0.0376 0.0400 0.0426 0.0454 0.0483 0.0514 0.0547 0.0581 0.0618 0.0657 0.0698 0.0741 0.0788 0.0837 0.0888 0.0943 0.1002 0.1063 0.1129 0.1198 0.1272 0.1350 0.1433 0.1521 0.1614 0.1714 0.1819 0.1932 0.2051 0.2179 0.2315 0.2460 0.2615

0.8116 0.8152 0.8188 0.8224 0.8262 0.8299 0.8338 0.8377 0.8416 0.8457 0.8498 0.8540 0.8583 0.8626 0.8671 0.8716 0.8763 0.8811 0.8860 0.8910 0.8961 0.9014 0.9069 0.9124 0.9182 0.9241 0.9302 0.9365 0.9430 0.9498 0.9567 0.9639 0.9714 0.9791 0.9871 0.9955 1.0041 1.0131 1.0225 1.0323 1.0425 1.0531 1.0643 1.0759 1.0881 1.1009 1.1143 1.1284 1.1432 1.1587 1.1752 1.1925 1.2108

10.058 11.063 12.069 13.075 14.081 15.087 16.093 17.099 18.105 19.111 20.117 21.124 22.130 23.136 24.142 25.148 26.155 27.161 28.167 29.174 30.180 31.187 32.193 33.200 34.207 35.213 36.220 37.227 38.233 39.240 40.247 41.254 42.261 43.268 44.275 45.282 46.289 47.297 48.304 49.311 50.319 51.326 52.334 53.341 54.349 55.356 56.364 57.372 58.380 59.388 60.396 61.404 62.412

19.297 20.661 22.111 23.653 25.290 27.028 28.873 30.830 32.906 35.107 37.441 39.914 42.533 45.308 48.245 51.355 54.646 58.128 61.812 65.708 69.829 74.185 78.791 83.660 88.806 94.245 99.993 106.068 112.487 119.270 126.438 134.014 142.021 150.483 159.429 168.887 178.889 189.466 200.656 212.497 225.030 238.300 252.357 267.251 283.041 299.788 317.560 336.431 356.482 377.800 400.484 424.641 450.388

29.354 31.724 34.181 36.728 39.371 42.115 44.966 47.929 51.011 54.219 57.558 61.037 64.663 68.444 72.388 76.503 80.801 85.289 89.979 94.882 100.009 105.372 110.985 116.860 123.013 129.458 136.213 143.294 150.720 158.510 166.685 175.268 184.282 193.751 203.704 214.169 225.178 236.763 248.960 261.808 275.349 289.627 304.690 320.592 337.389 355.144 373.924 393.803 414.862 437.188 460.880 486.044 512.799

0.0362 0.0397 0.0432 0.0468 0.0503 0.0538 0.0573 0.0607 0.0642 0.0676 0.0711 0.0745 0.0779 0.0813 0.0847 0.0881 0.0915 0.0948 0.0982 0.1015 0.1048 0.1081 0.1115 0.1147 0.1180 0.1213 0.1246 0.1278 0.1311 0.1343 0.1375 0.1407 0.1439 0.1471 0.1503 0.1535 0.1566 0.1598 0.1629 0.1660 0.1692 0.1723 0.1754 0.1785 0.1816 0.1846 0.1877 0.1908 0.1938 0.1968 0.1999 0.2029 0.2059

0.1078 0.1162 0.1248 0.1337 0.1430 0.1525 0.1624 0.1726 0.1832 0.1942 0.2057 0.2175 0.2298 0.2426 0.2560 0.2698 0.2842 0.2992 0.3148 0.3311 0.3481 0.3658 0.3843 0.4035 0.4236 0.4447 0.4666 0.4895 0.5135 0.5386 0.5650 0.5925 0.6213 0.6514 0.6830 0.7162 0.7509 0.7874 0.8256 0.8658 0.9081 0.9525 0.9993 1.0485 1.1003 1.1549 1.2126 1.2734 1.3377 1.4056 1.4775 1.5537 1.6345

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62

Fundamentals of Psychrometrics (SI), Second Edition Table A-1 Temp., °C t

63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

Thermodynamic Properties of Moist Air at Standard Atmospheric Pressure, 101.325 kPa (Continued)


0.182842 0.193937 0.205794 0.218478 0.232067 0.246645 0.262309 0.279167 0.297343 0.316979 0.338237 0.361304 0.386399 0.413774 0.443727 0.476610 0.512842 0.552926 0.597470 0.647218 0.703089 0.766233 0.838105 0.920580 1.016105 1.127952 1.260579 1.420235

103



vda

vas

vs

hda

0.9522 0.9551 0.9579 0.9607 0.9636 0.9664 0.9692 0.9721 0.9749 0.9778 0.9806 0.9834 0.9863 0.9891 0.9919 0.9948 0.9976 1.0005 1.0033 1.0061 1.0090 1.0118 1.0146 1.0175 1.0203 1.0232 1.0260 1.0288

0.2780 0.2957 0.3147 0.3350 0.3568 0.3803 0.4056 0.4328 0.4622 0.4941 0.5287 0.5663 0.6072 0.6520 0.7010 0.7550 0.8145 0.8805 0.9539 1.0360 1.1283 1.2328 1.3519 1.4887 1.6473 1.8332 2.0539 2.3198

1.2302 1.2508 1.2726 1.2957 1.3204 1.3467 1.3748 1.4049 1.4372 1.4719 1.5093 1.5497 1.5935 1.6411 1.6930 1.7497 1.8121 1.8809 1.9572 2.0421 2.1373 2.2446 2.3665 2.5062 2.6676 2.8564 3.0799 3.3487

63.420 64.428 65.436 66.445 67.453 68.462 69.470 70.479 71.488 72.496 73.505 74.514 75.523 76.532 77.542 78.551 79.560 80.569 81.579 82.589 83.598 84.608 85.618 86.628 87.638 88.648 89.658 90.668

has

477.856 507.192 538.557 572.131 608.118 646.746 688.271 732.985 781.220 833.353 889.821 951.124 1017.843 1090.659 1170.366 1257.907 1354.402 1461.196 1579.917 1712.556 1861.573 2030.041 2221.858 2442.035 2697.127 2995.880 3350.228 3776.888

hs

541.276 571.620 603.993 638.576 675.572 715.208 757.741 803.464 852.707 905.850 963.326 1025.638 1093.367 1167.191 1247.907 1336.458 1433.962 1541.765 1661.496 1795.145 1945.171 2114.649 2307.476 2528.662 2784.764 3084.528 3439.885 3867.556


0.2089 0.2119 0.2149 0.2179 0.2208 0.2238 0.2268 0.2297 0.2326 0.2356 0.2385 0.2414 0.2443 0.2472 0.2501 0.2529 0.2558 0.2587 0.2615 0.2644 0.2672 0.2701 0.2729 0.2757 0.2785 0.2813 0.2841 0.2869

s

1.7203 1.8114 1.9084 2.0117 2.1220 2.2398 2.3659 2.5011 2.6464 2.8028 2.9715 3.1539 3.3517 3.5668 3.8014 4.0581 4.3401 4.6511 4.9956 5.3794 5.8091 6.2933 6.8430 7.4721 8.1987 9.0472 10.0508 11.2558

63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

Appendix B— Dimensions, Units, and Unit Conversion Factors Table B-1

Dimensions and Units Used in Air-Conditioning Applications

Dimension

SI Unit

I-P Unit

Acceleration

m/s2

ft/s2

Area

m2

ft2

Density

kg/m 3

lbm/ft3

Energy

N·m, joule (J)

Btu, ft·lb

Force

(kg·m)/s2, newton (N)

pound (lb f )

Length

metre (m)

foot (ft)

Mass

kilogram (kg)

pound mass (lb m)

Power

J/s, watt (W)

Btu/h

Pressure

N/m2, pascal (Pa)

pounds per square inch (psi)

Specific heat

J/(kg·°C), J/(kg·K)

Btu/lb m·°F

Time

second (s)

second (s)

Temperature (absolute)

kelvin (K)

degree Rankine (°R)

Temperature

degree Celsius (°C)

degree Fahrenheit (°F)

Thermal conductivity

W/(m·°C), W/(m·K)

Btu/h·ft·°F

Thermal flux density

W/m 2

Btu/h·ft2

Velocity

m/s

ft/s, ft/min, fpm

Volume

m3

ft3

Volume flow rate

m 3/s

ft3/s, ft3/min, cfm

106

Appendix B Dimensions, Units, and Unit Conversion Factors Table B-2

Unit Conversion Factors

Dimension

SI Unit

I-P Unit

Length

1 m = 3.281 ft

1 ft = 0.305 m

Area

1 m2 = 10.76 ft 2

1 ft 2 = 0.0929 m 2

Volume

1 m3 = 35.32 ft 3 1 m 3 = 1000 L

1 ft 3 = 0.0284 m 3 1 ft 3 = 7.481 gal

Mass

1 kg = 2.205 lb m

1 lb m = 0.454 kg

Force

1 N = 0.2248 lb f

1 lb f = 4.448 N

Energy

1 kJ = 0.9478 Btu 1 J = 0.7376 ft·lb f 1 kWh = 3.412 × 10 3 Btu

1 Btu = 778.2 ft·lb f = 1.055 kJ 1 ft·lb f = 1.356 J 1 Btu = 2.930 × 10 –4 kWh

1 kJ/kg = 0.4298 Btu/lb m

1 Btu/lbm = 2.326 kJ/kg

Power

1 W = 3.412 Btu/h 1 kW = 1.341 hp 1 kW = 0.2844 ton refrigeration

1 Btu/h = 0.293 W 1 hp = 2545 Btu/h = 0.746 kW 1 ton = 12,000 Btu/h = 3.517 kW

Pressure

1 Pa = 1.450 × 10 –4 psi 1 atm = 101 kPa

1 psi = 6.897 × 10 3 Pa 1 atm = 14.7 psi = 29.92 in. Hg

Temperature

1°C T = 9/5°F T y°C = [(9/5)y + 32]°F K = °C + 273.15

1°F T = 5/9°C T y°F = (y – 32)(5/9)°C °R = °F + 459.67

Velocity

1 m/s = 1.969 × 10 2 ft/min

1 ft/min = 5.079 × 10 –3 m/s

Mass density

1 kg/m 3 = 6.243 × 10 –2 lbm/ft3

1 lb m/ft3 = 16.02 kg/m 3

Mass flow rate

1 kg/s = 2.205 lb m/s 1 kg/s = 7.937 × 10 3 lbm/h

1 lb m/s = 0.4535 kg/s 1 lb m/h = 1.260 × 10 –4 kg/s

Volume flow rate

1 m 3/s = 2.119 × 10 3 cfm 1 m 3/s = 1.585 × 10 4 gal/min

1 cfm = 4.719 × 10 –4 m3/s 1 gal/min = 6.309 × 10 –5 m3/s

Thermal conductivity

1 W/(m·°C) = 0.5778 Btu/h·ft·°F

1 Btu/h·ft·°F = 1.731 W/(m·°C)

Heat transfer coefficient

1 W/(m2·°C) = 0.1761 Btu/h·ft 2·°F

1 Btu/h·ft 2·°F = 5.679 W/(m2·°C)

Specific heat

1 J/(kg·°C) = 2.389 × 10 –4 Btu/lbm·°F

1 Btu/lbm·°F = 4.186 × 10 3 J/(kg·°C)

Specific energy, Specific enthalpy

Appendix C— Climatic Design Information The climatic design information in this appendix is from Chapter 14 of the 2013 ASHRAE Handbook—Fundamentals.

Appendix D— Thermodynamic Properties of Water at Saturation Table D-1 Temp., °C t

–60 –59 –58 –57 –56 –55 –54 –53 –52 –51 –50 –49 –48 –47 –46 –45 –44 –43 –42 –41 –40 –39 –38 –37 –36 –35 –34 –33 –32 –31 –30 –29 –28 –27 –26 –25 –24 –23 –22 –21 –20 –19 –18 –17 –16 –15 –14 –13

Absolute Pressure pws, kPa

0.00108 0.00124 0.00141 0.00161 0.00184 0.00209 0.00238 0.00271 0.00307 0.00348 0.00394 0.00445 0.00503 0.00568 0.00640 0.00720 0.00810 0.00910 0.01022 0.01146 0.01284 0.01437 0.01607 0.01795 0.02004 0.02234 0.02489 0.02771 0.03081 0.03423 0.03801 0.04215 0.04672 0.05173 0.05724 0.06327 0.06989 0.07714 0.08508 0.09376 0.10324 0.11360 0.12490 0.13722 0.15065 0.16527 0.18119 0.19849

Thermodynamic Properties of Water at Saturation

Specific Volume, m3 /kgw Sat. Solid vi /v f

0.001081 0.001082 0.001082 0.001082 0.001082 0.001082 0.001082 0.001082 0.001083 0.001083 0.001083 0.001083 0.001083 0.001083 0.001083 0.001084 0.001084 0.001084 0.001084 0.001084 0.001084 0.001085 0.001085 0.001085 0.001085 0.001085 0.001085 0.001085 0.001086 0.001086 0.001086 0.001086 0.001086 0.001086 0.001087 0.001087 0.001087 0.001087 0.001087 0.001087 0.001087 0.001088 0.001088 0.001088 0.001088 0.001088 0.001088 0.001089

Evap. vig /v fg

90971.58 79885.31 70235.77 61826.23 54488.28 48077.54 42470.11 37559.49 33254.07 29474.87 26153.80 23232.03 20658.70 18389.75 16387.03 14617.39 13052.07 11666.02 10437.46 9347.38 8379.20 7518.44 6752.43 6070.08 5461.68 4918.69 4433.64 3999.95 3611.82 3264.15 2952.46 2672.77 2421.58 2195.80 1992.68 1809.79 1644.99 1496.36 1362.21 1241.03 1131.49 1032.38 942.64 861.34 787.61 720.70 659.94 604.72

Specific Enthalpy, kJ/kgw

Sat. Vapor v g

Sat. Solid hi /h f

90971.58 79885.31 70235.78 61826.24 54488.28 48077.54 42470.11 37559.50 33254.07 29474.87 26153.80 23232.04 20658.70 18389.75 16387.03 14617.39 13052.07 11666.02 10437.46 9347.38 8379.20 7518.44 6752.43 6070.08 5461.68 4918.69 4433.64 3999.95 3611.82 3264.16 2952.46 2672.77 2421.58 2195.80 1992.68 1809.79 1644.99 1496.36 1362.21 1241.03 1131.49 1032.38 942.65 861.34 787.61 720.70 659.94 604.73

–446.12 –444.46 –442.79 –441.11 –439.42 –437.73 –436.03 –434.32 –432.61 –430.88 –429.16 –427.42 –425.68 –423.93 –422.17 –420.40 –418.63 –416.85 –415.06 –413.27 –411.47 –409.66 –407.85 –406.02 –404.19 –402.36 –400.51 –398.66 –396.80 –394.94 –393.06 –391.18 –389.29 –387.40 –385.50 –383.59 –381.67 –379.75 –377.81 –375.88 –373.93 –371.98 –370.01 –368.05 –366.07 –364.09 –362.10 –360.10

Evap. hig /h fg

2836.27 2836.45 2836.63 2836.81 2836.97 2837.13 2837.28 2837.42 2837.56 2837.69 2837.81 2837.93 2838.04 2838.14 2838.23 2838.32 2838.39 2838.47 2838.53 2838.59 2838.64 2838.68 2838.72 2838.74 2838.76 2838.78 2838.78 2838.78 2838.77 2838.75 2838.73 2838.70 2838.66 2838.61 2838.56 2838.49 2838.42 2838.35 2838.26 2838.17 2838.07 2837.96 2837.84 2837.72 2837.59 2837.45 2837.30 2837.14

Specific Entropy, kJ/(kgw ·K)

Sat. Vapor h g

Sat. Solid si / s f

2390.14 2391.99 2393.85 2395.70 2397.55 2399.40 2401.25 2403.10 2404.95 2406.81 2408.66 2410.51 2412.36 2414.21 2416.06 2417.91 2419.76 2421.62 2423.47 2425.32 2427.17 2429.02 2430.87 2432.72 2434.57 2436.42 2438.27 2440.12 2441.97 2443.82 2445.67 2447.51 2449.36 2451.21 2453.06 2454.91 2456.75 2458.60 2460.45 2462.29 2464.14 2465.98 2467.83 2469.67 2471.51 2473.36 2475.20 2477.04

–1.6842 –1.6764 –1.6687 –1.6609 –1.6531 –1.6453 –1.6375 –1.6298 –1.6220 –1.6142 –1.6065 –1.5987 –1.5909 –1.5832 –1.5754 –1.5677 –1.5599 –1.5522 –1.5444 –1.5367 –1.5289 –1.5212 –1.5135 –1.5057 –1.4980 –1.4903 –1.4825 –1.4748 –1.4671 –1.4594 –1.4516 –1.4439 –1.4362 –1.4285 –1.4208 –1.4131 –1.4054 –1.3977 –1.3899 –1.3822 –1.3745 –1.3668 –1.3591 –1.3514 –1.3437 –1.3360 –1.3284 –1.3207

Evap. sig / s fg

13.3064 13.2452 13.1845 13.1243 13.0646 13.0054 12.9468 12.8886 12.8310 12.7738 12.7171 12.6609 12.6051 12.5498 12.4950 12.4406 12.3867 12.3331 12.2801 12.2274 12.1752 12.1234 12.0720 12.0210 11.9704 11.9202 11.8703 11.8209 11.7718 11.7231 11.6748 11.6269 11.5793 11.5321 11.4852 11.4386 11.3925 11.3466 11.3011 11.2559 11.2110 11.1665 11.1223 11.0784 11.0348 10.9915 10.9485 10.9058

Sat. Vapor s g

Temp., °C t

11.6222 11.5687 11.5158 11.4634 11.4115 11.3601 11.3092 11.2589 11.2090 11.1596 11.1106 11.0622 11.0142 10.9666 10.9196 10.8729 10.8267 10.7810 10.7356 10.6907 10.6462 10.6022 10.5585 10.5152 10.4724 10.4299 10.3878 10.3461 10.3047 10.2638 10.2232 10.1830 10.1431 10.1036 10.0644 10.0256 9.9871 9.9489 9.9111 9.8736 9.8365 9.7996 9.7631 9.7269 9.6910 9.6554 9.6201 9.5851

–60 –59 –58 –57 –56 –55 –54 –53 –52 –51 –50 –49 –48 –47 –46 –45 –44 –43 –42 –41 –40 –39 –38 –37 –36 –35 –34 –33 –32 –31 –30 –29 –28 –27 –26 –25 –24 –23 –22 –21 –20 –19 –18 –17 –16 –15 –14 –13

138

Appendix D Thermodynamic Properties of Water at Saturation Table D-1

Temp., °C t

–12 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0


0.21729 0.23771 0.25987 0.28391 0.30995 0.33817 0.36871 0.40174 0.43745 0.47604 0.51770 0.56266 0.61115

Thermodynamic Properties of Water at Saturation (Continued)


0.001089 0.001089 0.001089 0.001089 0.001089 0.001090 0.001090 0.001090 0.001090 0.001090 0.001091 0.001091 0.001091

Evap. vig /v fg

Sat. Vapor v g

554.51 508.81 467.19 429.25 394.66 363.09 334.26 307.92 283.82 261.78 241.60 223.10 206.15

554.51 508.81 467.19 429.26 394.66 363.09 334.26 307.92 283.83 261.78 241.60 223.11 206.15

Specific Enthalpy, kJ/kgw Sat. Solid hi /h f

Evap. hig /h fg


Sat. Vapor h g

Sat. Solid si / s f

Evap. sig / s fg

Sat. Vapor s g

Temp., °C t

–358.10 –356.08 –354.06 –352.04 –350.00 –347.96 –345.91 –343.86 –341.79 –339.72 –337.64 –335.56 –333.47

2836.98 2836.80 2836.62 2836.44 2836.24 2836.03 2835.82 2835.60 2835.37 2835.13 2834.88 2834.63 2834.36

2478.88 2480.72 2482.56 2484.40 2486.23 2488.07 2489.91 2491.74 2493.57 2495.41 2497.24 2499.07 2500.90

–1.3130 –1.3053 –1.2976 –1.2899 –1.2822 –1.2745 –1.2668 –1.2592 –1.2515 –1.2438 –1.2361 –1.2284 –1.2208

10.8634 10.8213 10.7795 10.7380 10.6967 10.6558 10.6151 10.5747 10.5345 10.4946 10.4550 10.4157 10.3766

9.5504 9.5160 9.4819 9.4481 9.4145 9.3812 9.3482 9.3155 9.2830 9.2508 9.2189 9.1872 9.1558

–12 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0

Transition from saturated solid to saturated liquid

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

0.6112 0.6571 0.7060 0.7581 0.8135 0.8726 0.9354 1.0021 1.0730 1.1483 1.2282 1.3129 1.4028 1.4981 1.5989 1.7057 1.8188 1.9383 2.0647 2.1982 2.3392 2.4881 2.6452 2.8109 2.9856 3.1697 3.3637 3.5679 3.7828 4.0089

0.001000 0.001000 0.001000 0.001000 0.001000 0.001000 0.001000 0.001000 0.001000 0.001000 0.001000 0.001000 0.001001 0.001001 0.001001 0.001001 0.001001 0.001001 0.001001 0.001002 0.001002 0.001002 0.001002 0.001003 0.001003 0.001003 0.001003 0.001004 0.001004 0.001004

206.139 192.444 179.763 168.013 157.120 147.016 137.637 128.927 120.833 113.308 106.308 99.792 93.723 88.069 82.797 77.880 73.290 69.005 65.002 61.260 57.760 54.486 51.421 48.551 45.862 43.340 40.976 38.757 36.674 34.718

206.140 192.445 179.764 168.014 157.121 147.017 137.638 128.928 120.834 113.309 106.309 99.793 93.724 88.070 82.798 77.881 73.291 69.006 65.003 61.261 57.761 54.487 51.422 48.552 45.863 43.341 40.977 38.758 36.675 34.719

–0.04 4.18 8.39 12.60 16.81 21.02 25.22 29.43 33.63 37.82 42.02 46.22 50.41 54.60 58.79 62.98 67.17 71.36 75.55 79.73 83.92 88.10 92.29 96.47 100.66 104.84 109.02 113.20 117.38 121.56

2500.93 2498.55 2496.17 2493.80 2491.42 2489.05 2486.68 2484.31 2481.94 2479.58 2477.21 2474.84 2472.48 2470.11 2467.75 2465.38 2463.01 2460.65 2458.28 2455.92 2453.55 2451.18 2448.81 2446.45 2444.08 2441.71 2439.33 2436.96 2434.59 2432.21

2500.89 2502.73 2504.57 2506.40 2508.24 2510.07 2511.91 2513.74 2515.57 2517.40 2519.23 2521.06 2522.89 2524.71 2526.54 2528.36 2530.19 2532.01 2533.83 2535.65 2537.47 2539.29 2541.10 2542.92 2544.73 2546.54 2548.35 2550.16 2551.97 2553.78

–0.0002 0.0153 0.0306 0.0459 0.0611 0.0763 0.0913 0.1064 0.1213 0.1362 0.1511 0.1659 0.1806 0.1953 0.2099 0.2245 0.2390 0.2534 0.2678 0.2822 0.2965 0.3108 0.3250 0.3391 0.3532 0.3673 0.3813 0.3952 0.4091 0.4230

9.1559 9.1138 9.0721 9.0306 8.9895 8.9486 8.9081 8.8678 8.8278 8.7882 8.7488 8.7096 8.6708 8.6322 8.5939 8.5559 8.5181 8.4806 8.4434 8.4064 8.3696 8.3331 8.2969 8.2609 8.2251 8.1895 8.1542 8.1192 8.0843 8.0497

9.1558 9.1291 9.1027 9.0765 9.0506 9.0249 8.9994 8.9742 8.9492 8.9244 8.8998 8.8755 8.8514 8.8275 8.8038 8.7804 8.7571 8.7341 8.7112 8.6886 8.6661 8.6439 8.6218 8.6000 8.5783 8.5568 8.5355 8.5144 8.4934 8.4727

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

30 31 32 33 34 35 36 37 38 39

4.2467 4.4966 4.7592 5.0351 5.3247 5.6286 5.9475 6.2818 6.6324 6.9997

0.001004 0.001005 0.001005 0.001005 0.001006 0.001006 0.001006 0.001007 0.001007 0.001007

32.881 31.153 29.528 28.000 26.561 25.207 23.931 22.728 21.594 20.525

32.882 31.154 29.529 28.001 26.562 25.208 23.932 22.729 21.595 20.526

125.75 129.93 134.11 138.29 142.47 146.64 150.82 155.00 159.18 163.36

2429.84 2427.46 2425.08 2422.70 2420.32 2417.94 2415.56 2413.17 2410.78 2408.39

2555.58 2557.39 2559.19 2560.99 2562.79 2564.58 2566.38 2568.17 2569.96 2571.75

0.4368 0.4506 0.4643 0.4780 0.4916 0.5052 0.5187 0.5322 0.5457 0.5591

8.0153 7.9812 7.9472 7.9135 7.8800 7.8467 7.8136 7.7807 7.7480 7.7155

8.4521 8.4317 8.4115 8.3914 8.3715 8.3518 8.3323 8.3129 8.2936 8.2746

30 31 32 33 34 35 36 37 38 39

40 41 42 43 44 45 46 47 48

7.3844 7.7873 8.2090 8.6503 9.1118 9.5944 10.0988 10.6259 11.1764

0.001008 0.001008 0.001009 0.001009 0.001009 0.001010 0.001010 0.001011 0.001011

19.516 18.564 17.664 16.815 16.012 15.252 14.534 13.855 13.212

19.517 18.565 17.665 16.816 16.013 15.253 14.535 13.856 13.213

167.54 171.72 175.90 180.08 184.26 188.44 192.62 196.80 200.98

2406.00 2403.61 2401.21 2398.82 2396.42 2394.02 2391.61 2389.21 2386.80

2573.54 2575.33 2577.11 2578.89 2580.67 2582.45 2584.23 2586.00 2587.77

0.5724 0.5858 0.5990 0.6123 0.6255 0.6386 0.6517 0.6648 0.6778

7.6832 7.6512 7.6193 7.5876 7.5561 7.5248 7.4937 7.4628 7.4320

8.2557 8.2369 8.2183 8.1999 8.1816 8.1634 8.1454 8.1276 8.1099

40 41 42 43 44 45 46 47 48

Fundamentals of Psychrometrics (SI), Second Edition Table D-1

139


Specific Volume, m3 /kgw



Temp., °C t


49

11.7512

0.001012

12.603

12.604

205.16

2384.39

2589.54

0.6908

7.4015

8.0923

49

50 51 52 53 54 55 56 57 58 59

12.3513 12.9774 13.6305 14.3116 15.0215 15.7614 16.5322 17.3350 18.1708 19.0407

0.001012 0.001013 0.001013 0.001014 0.001014 0.001015 0.001015 0.001016 0.001016 0.001017

12.027 11.481 10.963 10.472 10.006 9.5639 9.1444 8.7461 8.3678 8.0083

12.028 11.482 10.964 10.473 10.007 9.5649 9.1454 8.7471 8.3688 8.0093

209.34 213.52 217.70 221.88 226.06 230.24 234.42 238.61 242.79 246.97

2381.97 2379.56 2377.14 2374.72 2372.30 2369.87 2367.44 2365.01 2362.57 2360.13

2591.31 2593.08 2594.84 2596.60 2598.35 2600.11 2601.86 2603.61 2605.36 2607.10

0.7038 0.7167 0.7296 0.7424 0.7552 0.7680 0.7807 0.7934 0.8060 0.8186

7.3711 7.3409 7.3109 7.2811 7.2514 7.2219 7.1926 7.1634 7.1344 7.1056

8.0749 8.0576 8.0405 8.0235 8.0066 7.9899 7.9733 7.9568 7.9405 7.9243

50 51 52 53 54 55 56 57 58 59

60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79

19.9458 20.8873 21.8664 22.8842 23.9421 25.0411 26.1827 27.3680 28.5986 29.8756 31.2006 32.5750 34.0001 35.4775 37.0088 38.5954 40.2389 41.9409 43.7031 45.5271

0.001017 0.001018 0.001018 0.001019 0.001019 0.001020 0.001020 0.001021 0.001022 0.001022 0.001023 0.001023 0.001024 0.001025 0.001025 0.001026 0.001026 0.001027 0.001028 0.001028

7.6666 7.3418 7.0328 6.7389 6.4591 6.1928 5.9392 5.6976 5.4674 5.2479 5.0387 4.8392 4.6488 4.4671 4.2937 4.1281 3.9699 3.8188 3.6743 3.5363

7.6677 7.3428 7.0338 6.7399 6.4601 6.1938 5.9402 5.6986 5.4684 5.2490 5.0397 4.8402 4.6498 4.4681 4.2947 4.1291 3.9709 3.8198 3.6754 3.5373

251.15 255.34 259.52 263.71 267.89 272.08 276.27 280.45 284.64 288.83 293.02 297.21 301.40 305.59 309.78 313.97 318.17 322.36 326.56 330.75

2357.69 2355.25 2352.80 2350.35 2347.89 2345.43 2342.97 2340.50 2338.03 2335.56 2333.08 2330.60 2328.11 2325.62 2323.13 2320.63 2318.13 2315.62 2313.11 2310.59

2608.85 2610.58 2612.32 2614.05 2615.78 2617.51 2619.23 2620.96 2622.67 2624.39 2626.10 2627.81 2629.51 2631.21 2632.91 2634.60 2636.29 2637.98 2639.66 2641.34

0.8312 0.8438 0.8563 0.8687 0.8811 0.8935 0.9059 0.9182 0.9305 0.9428 0.9550 0.9672 0.9793 0.9915 1.0035 1.0156 1.0276 1.0396 1.0516 1.0635

7.0770 7.0485 7.0201 6.9919 6.9639 6.9361 6.9083 6.8808 6.8534 6.8261 6.7990 6.7720 6.7452 6.7185 6.6920 6.6656 6.6393 6.6132 6.5872 6.5613

7.9082 7.8922 7.8764 7.8607 7.8451 7.8296 7.8142 7.7990 7.7839 7.7689 7.7540 7.7392 7.7245 7.7100 7.6955 7.6812 7.6669 7.6528 7.6388 7.6248

60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79

80 81 82 83 84 85 86 87 88 89

47.4147 49.3676 51.3875 53.4762 55.6355 57.8675 60.1738 62.5565 65.0174 67.5587

0.001029 0.001030 0.001030 0.001031 0.001032 0.001032 0.001033 0.001034 0.001035 0.001035

3.4042 3.2780 3.1572 3.0415 2.9309 2.8249 2.7234 2.6262 2.5330 2.4437

3.4053 3.2790 3.1582 3.0426 2.9319 2.8259 2.7244 2.6272 2.5341 2.4448

334.95 339.15 343.34 347.54 351.74 355.95 360.15 364.35 368.56 372.76

2308.07 2305.54 2303.01 2300.47 2297.93 2295.38 2292.83 2290.27 2287.70 2285.14

2643.01 2644.68 2646.35 2648.01 2649.67 2651.33 2652.98 2654.62 2656.26 2657.90

1.0754 1.0873 1.0991 1.1109 1.1227 1.1344 1.1461 1.1578 1.1694 1.1811

6.5356 6.5100 6.4846 6.4592 6.4340 6.4090 6.3840 6.3592 6.3345 6.3099

7.6110 7.5973 7.5837 7.5701 7.5567 7.5434 7.5301 7.5170 7.5039 7.4909

80 81 82 83 84 85 86 87 88 89

90 91 92 93 94 95 96 97 98 99

70.1824 72.8904 75.6849 78.5681 81.5420 84.6089 87.7711 91.0308 94.3902 97.8518

0.001036 0.001037 0.001037 0.001038 0.001039 0.001040 0.001040 0.001041 0.001042 0.001043

2.3581 2.2760 2.1973 2.1217 2.0492 1.9796 1.9128 1.8486 1.7870 1.7277

2.3591 2.2771 2.1983 2.1228 2.0502 1.9806 1.9138 1.8497 1.7880 1.7288

376.97 381.18 385.38 389.59 393.81 398.02 402.23 406.45 410.66 414.88

2282.56 2279.98 2277.39 2274.80 2272.20 2269.60 2266.98 2264.37 2261.74 2259.11

2659.53 2661.16 2662.78 2664.39 2666.01 2667.61 2669.22 2670.81 2672.40 2673.99

1.1927 1.2042 1.2158 1.2273 1.2387 1.2502 1.2616 1.2730 1.2844 1.2957

6.2854 6.2611 6.2368 6.2127 6.1887 6.1648 6.1411 6.1174 6.0938 6.0704

7.4781 7.4653 7.4526 7.4400 7.4275 7.4150 7.4027 7.3904 7.3782 7.3661

90 91 92 93 94 95 96 97 98 99

100 101 102 103 104 105 106 107 108 109

101.4180 105.0910 108.8735 112.7678 116.7765 120.9021 125.1472 129.5145 134.0065 138.6261

0.001043 0.001044 0.001045 0.001046 0.001047 0.001047 0.001048 0.001049 0.001050 0.001051

1.6708 1.6161 1.5635 1.5129 1.4642 1.4174 1.3724 1.3290 1.2873 1.2471

1.6719 1.6171 1.5645 1.5140 1.4653 1.4185 1.3734 1.3301 1.2883 1.2481

419.10 423.32 427.54 431.76 435.99 440.21 444.44 448.67 452.90 457.13

2256.47 2253.83 2251.18 2248.52 2245.85 2243.18 2240.50 2237.81 2235.12 2232.41

2675.57 2677.15 2678.72 2680.28 2681.84 2683.39 2684.94 2686.48 2688.02 2689.55

1.3070 1.3183 1.3296 1.3408 1.3520 1.3632 1.3743 1.3854 1.3965 1.4076

6.0471 6.0238 6.0007 5.9777 5.9548 5.9320 5.9092 5.8866 5.8641 5.8417

7.3541 7.3421 7.3303 7.3185 7.3068 7.2951 7.2836 7.2721 7.2607 7.2493

100 101 102 103 104 105 106 107 108 109

110 111 112 113 114 115

143.3760 148.2588 153.2775 158.4348 163.7337 169.1770

0.001052 0.001052 0.001053 0.001054 0.001055 0.001056

1.2083 1.1710 1.1351 1.1005 1.0671 1.0349

1.2094 1.1721 1.1362 1.1015 1.0681 1.0359

461.36 465.60 469.83 474.07 478.31 482.55

2229.70 2226.99 2224.26 2221.53 2218.78 2216.03

2691.07 2692.58 2694.09 2695.60 2697.09 2698.58

1.4187 1.4297 1.4407 1.4517 1.4626 1.4735

5.8194 5.7972 5.7750 5.7530 5.7310 5.7092

7.2380 7.2268 7.2157 7.2047 7.1937 7.1827

110 111 112 113 114 115

Sat. Solid vi /v f

Evap. vig /v fg

Sat. Vapor v g

Sat. Solid hi /h f

Evap. hig /h fg

Sat. Vapor h g

Sat. Solid si / s f

Evap. sig / s fg

Sat. Vapor s g

Temp., °C t

140

Appendix D Thermodynamic Properties of Water at Saturation Table D-1

Temp., °C t




Evap. vig /v fg


Sat. Vapor v g

Sat. Solid hi /h f

Evap. hig /h fg

Sat. Vapor h g

Specific Entropy, kJ/(kgw ·K) Sat. Solid si / s f

Evap. sig / s fg

Sat. Vapor s g

Temp., °C t

116 117 118 119

174.7678 180.5090 186.4036 192.4547

0.001057 0.001058 0.001059 0.001059

1.0038 0.9739 0.9450 0.9171

1.0049 0.9750 0.9461 0.9182

486.80 491.04 495.29 499.53

2213.27 2210.51 2207.73 2204.94

2700.07 2701.55 2703.02 2704.48

1.4844 1.4953 1.5062 1.5170

5.6874 5.6658 5.6442 5.6227

7.1719 7.1611 7.1504 7.1397

116 117 118 119

120 122 124 126 128 130 132 134 136 138

198.6654 211.5782 225.1676 239.4597 254.4813 270.2596 286.8226 304.1989 322.4175 341.5081

0.001060 0.001062 0.001064 0.001066 0.001068 0.001070 0.001072 0.001074 0.001076 0.001078

0.8902 0.8392 0.7916 0.7472 0.7058 0.6670 0.6308 0.5969 0.5651 0.5353

0.8913 0.8403 0.7927 0.7483 0.7068 0.6681 0.6318 0.5979 0.5662 0.5364

503.78 512.29 520.80 529.32 537.85 546.39 554.93 563.49 572.05 580.62

2202.15 2196.53 2190.88 2185.19 2179.47 2173.70 2167.89 2162.04 2156.15 2150.22

2705.93 2708.82 2711.69 2714.52 2717.32 2720.09 2722.83 2725.53 2728.20 2730.84

1.5278 1.5494 1.5708 1.5922 1.6134 1.6346 1.6557 1.6767 1.6977 1.7185

5.6013 5.5587 5.5165 5.4746 5.4330 5.3918 5.3508 5.3102 5.2698 5.2298

7.1291 7.1081 7.0873 7.0668 7.0465 7.0264 7.0066 6.9869 6.9675 6.9483

120 122 124 126 128 130 132 134 136 138

140 142 144 146 148 150 152 154 156 158

361.5010 382.4271 404.3178 427.2053 451.1220 476.1014 502.1771 529.3834 557.7555 587.3287

0.001080 0.001082 0.001084 0.001086 0.001088 0.001091 0.001093 0.001095 0.001097 0.001100

0.5074 0.4813 0.4567 0.4336 0.4118 0.3914 0.3722 0.3541 0.3370 0.3209

0.5085 0.4823 0.4577 0.4346 0.4129 0.3925 0.3733 0.3552 0.3381 0.3220

589.20 597.79 606.39 615.00 623.62 632.25 640.89 649.55 658.21 666.89

2144.24 2138.22 2132.15 2126.04 2119.88 2113.67 2107.41 2101.10 2094.74 2088.32

2733.44 2736.01 2738.54 2741.04 2743.50 2745.92 2748.30 2750.64 2752.95 2755.21

1.7393 1.7600 1.7806 1.8011 1.8216 1.8420 1.8623 1.8825 1.9027 1.9228

5.1900 5.1505 5.1112 5.0723 5.0335 4.9951 4.9569 4.9189 4.8811 4.8436

6.9293 6.9105 6.8918 6.8734 6.8551 6.8370 6.8191 6.8014 6.7838 6.7664

140 142 144 146 148 150 152 154 156 158

160

618.1392

0.001102

0.3057

0.3068

675.57

2081.86

2757.43

1.9428

4.8063

6.7491

160

Skill Development Exercises To receive full continuing education credit, all questions must be answered and submitted at www.ashrae.org/sdlonline . Please log in using your student ID number and the SDL number. Your student ID number is composed of the last five digits of your Social Security Number or another unique five-digit number you create when first registering online. The SDL number for this course can be located near the top of the copyright page of this book.


s e s i c r e x E t n e m p o l e v e D l l i k S 1 r e t p a h C

Skill Development Exercises for Chapter 1 Total number of questions: 4 1-1

How many basic processes of air conditioning can be performed on moist air? a) Two b) Three c) Four

1-2

Which combination process will increase both the temperature and the moisture content? a) Cooling and dehumidification b) Heating and dehumidification c) Heating and humidification

1-3

Enthalpy is the total heat content of the air. a) True b) False

1-4

Change in elevation has no effect on the air density. a) True b) False



Dry-bulb temperature is measured with a wet sock around the sensing bulb. a) True b) False

2-2

Saturation temperature of air is the point at which the dry-bulb, wet-bulb, and dew-point temperatures are equal. a) True b) False

2-3

Relative humidity does not change as the dry-bulb temperature changes. a) True b) False

2-4

The dry-bulb temperature can be above the dew-point temperature. a) True b) False

2-5

According to Appendix A, what is the specific enthalpy h s of saturated air at 5°C? a) 15.231863 b) 18.63 c) 5.02 d) None of the above

2-6

According to Appendix A, under the same condition cited in Exercise 2-5, what is the specific volume v ? a) 0.811 b) 0.794 c) 0.006 d) None of the above

2-7

According to Appendix A, what is the specific enthalpy of dry air h da at 50°C? a) 52.33 b) 225.03 c) 50.31 d) None of the above

C h a p t e r 2

S k i l l D e v e l o p m e n t E x e r c i s e s



According to Appendix A, under the same condition cited in Exercise 2-7, what is the specific volume v ? a) 0.915 b) 1.24 c) 0.012 d) None of the above



On a psychrometric chart, the y-axis is humidity ratio and the x-axis is: a) Relative humidity b) Dew-point temperature c) Dry-bulb temperature d) Wet-bulb temperature

3-2

Using the psychrometric chart in Figure 3-4, determine the relative humidity of an air parcel with W = 6.4 and t db = 15°C. a) 60% rh b) 70% rh c) 80% rh d) 90% rh

3-3

Using the psychrometric chart in Figure 3-4, determine the dew-point temperature of an air parcel with t db = 21°C and  = 50% rh. a) 10°C b) 12°C c) 15°C d) 19°C

3-4

Using the psychrometric chart in Figure 3-4, determine the humidity ratio W of an air parcel with a saturation temperature of t db = 10°C. a) 6.5 b) 7.6 c) 30% d) 10°C

3-5

Using the psychrometric chart in Figure 3-4, determine the specific volume v of an air parcel with t db = 21°C and W = 10. a) 0.82 b) 0.846 c) 0.86 d) none of the above

3-6

According to the psychrometric chart in Figure 3-4, what is the enthalpy of t db = 25°C dry air? a) 22

C h a p t e r 3


Fundamentals of Psychrometrics (SI), Second Edition b) 35 s e s i c r e x E t n e m p o l e v e D l l i k S

c) 76 d) 25

3-7

According to the psychrometric chart in Figure 3-4, what is the wet-bulb tem perature of a moist air parcel with t db = 21°C and  = 50% rh air? a) 21°C b) 14°C c) 10°C d) 13°C

3-8

3 r e t p a h C

According to the psychrometric chart in Figure 3-4, what is the dew point of t db = 10°C saturated air? a) 10°C b) 4°C c) 0°C d) –5°C

3-9

According to the psychrometric chart in Figure 3-4, what is the wet-bulb tem perature of t db = 21°C dry air? a) 0°C b) –4°C c) 4°C d) 6.5°C

3-10

Using the psychrometric chart in Figure 3-4, plot the points t db = 21°C, h = 24, and t db = 21°C, t wb = 14°C, then connect the points with a line. Upon investigation of the line, which of the following is the best description? a) The line is almost vertical. b) The line has a slope of about 45° (angle). c) The line almost horizontal.



Moist air that is heated without humidification has the following change in relative humidity: a) Increase b) Decrease c) Stays the same d) Depends on the type of humidifier

4-2

What is the equation that converts enthalpy changes into capacity (kW)? a) 1210× airflow × (t 1 – t 2) b) 1.2 × airflow × (h1 – h 2) c) 3300 × airflow × (W 1 – W 2) d) None of the above

4-3

Which of the following is true concerning humidification by steam versus by (cold water) atomization? a) Atomization always maintains a constant relative humidity. b) Steam humidification adds no net energy to the airstream. c) Heat to make steam in the steam humidifier comes from the air entering the humidifier. d) Heat to evaporate water in the atomizer comes from the air entering the humidifier.

4-4

A heating coil can provide for both heating and humidification. a) True b) False

4-5

A cooling coil can provide for both cooling and dehumidification. a) True b) False

4-6

What is the change in enthalpy when dry air is heated from 10°C to 23°C? a) 10 b) 13 c) 16 d) 18

C h a p t e r 4




What is the enthalpy change when saturated air at 10°C is conditioned to be saturated air at 23°C? a) 39 b) 35 c) 13 d) 45

4-8

One day in Phoenix, Arizona, the temperature reaches 40.5°C with 20% rh. Water is sprayed into the air to cool it. What will the temperature of the air be when the relative humidity increases to 50% rh? a) 30°C b) 35°C c) 40°C d) 22°C

4-9

If the air entering a heating coil is dry and 21°C db and the leaving air is 43°C, how many watts of cooling are supplied by the coil at 2.35 m3/s if the fan is located at the coil inlet? a) 58 000 W b) 65 000 W c) 61 000 W d) 62 550 W

4-10

Air enters a cooling coil at 38°C and 40% rh and leaves saturated at a temperature of 7°C. What is the total watts of cooling required if a 2.35 m3/s fan is located at the inlet of the cooling coil? a) 190 200 W b) 176 840 W c) 160 000 W d) 158 960 W



The definition of sensible heat ratio (SHR) is the: a) Ratio of sensible to latent load b) Ratio of latent to sensible load c) Ratio of total load to sensible load d) Ratio of sensible load to total load

5-2

If the sensible load on a building is equal to the latent load, the value of SHR is: a) 2 b) 1 c) 0.5 d) –2

5-3

The psychrometric condition for supply air that will satisfy the requirements of a room depends on: a) The amount of outdoor air needed b) The desired room condition c) Room SHR d) All of the above e) Answers b and c only

5-4

Why is it possible to satisfy a room with a variety of “assumptions” about the temperature change across a coil (heating or cooling)? a) Because there is a corresponding airflow with every t . b) Because the heat/cool load calculation is never accurate. c) Because the comfort zone is large. d) Because there is a wide variety of methods for heating and cooling.

5-5

Which condition below is not possible to show on a psychrometric chart? a)

t db =

24°C, h = 54 kJ/kg

b)

t db =

32°C, t wb = 25°C

c)

t wb =

25°C, h = 84

d)

t db =

24°C,  = 50%

C h a p t e r 5




In a system, 1 m3/s of air at 15³C and 30% rh is mixed with 4 m3/s air at 27°C and 80% rh. Find the mixed-air temperature using the mixing equation. a) 18°C b) 17.5°C c) 16°C d) 25°C

5-7

In Exercise 5-6, what is the mixed-air relative humidity? a) 51% rh b) 40% rh c) 60% rh d) None of these

5-8

In a system, 1 m3/s of air at 4°C and 90% rh is adiabatically mixed with moist air at 26°C but unknown relative humidity. The final mixture is at 22°C and 50% rh. What is the relative humidity and airflow rate of the second airstream? a) 42% rh, 5 m3/s b) 42% rh, 3 m3/s c) 60% rh, 5 m3/s d) 35% rh, 6 m3/s

5-9

If the sensible load is 600 000 W and the latent load is 300 000 W, what is the SHR? a) 2.0 b) 1.0 c) 0.66 d) 0.76

5-10

If the room design is t db = 24°C and  = 50% rh and we mix in 25% outdoor air at t db = 48°C and  = 10% rh, what is the mixed-air dry-bulb temperature? a) 45°C b) 42°C c) 30°C d) Not possible

5-11

From Exercise 5-10, what is the mixed-air relative humidity? a) 33% rh b) 15% rh c) 21% rh d) 28% rh



Which type of humidification requires the change to not exceed the temperature rise capacity of a heating coil? a) Water spray b) Steam c) Both the same d) Neither has an impact

6-2

From the discussion of the psychrometrics of cooling coils, which “rule of thumb” will best select the cooling coil conditions? a) Temperature drop across a cooling coil should be about 10°C. b) Relative humidity off the coil should be 90%. c) Volume of air across a cooling coil should be kept to a minimum. d) Coil temperatures should be selected to be as low as possible.

6-3

Which of the following statements best describe why cooling coils cannot accommodate large latent loads with small sensible loads? a) Cooling coils rust if too much condensate forms. b) Cooling coils will freeze up if the coil temperature gets too low. c) Cooling coils tend to dehumidify first, then drop the air temperature. d) Condensation requires a drop in air temperature to the dew point.

6-4

Consider a room heating load with a 200 000 W sensible loss and 40 000 W latent loss, with room design conditions of t db = 22°C and approximately  = 40% rh. The air handler has an adiabatic humidifier downstream from a heating coil without any outdoor air. If the leaving air temperature is t db = 38°C after the humidifier, what is the airflow required to satisfy the load? a) 12 b) 10.3 c) 8 d) None of these

C h a p t e r 6




What is the leaving air temperature t db from the heating coil for the conditions listed in Exercise 6-4? a) 37°C b) 40°C c) 38°C d) None of these

6-6

What is the leaving relative humidity  from the heating coil for the conditions listed in Exercise 6-4? a) 15% rh b) 12% rh c) 20% rh d) 24% rh

6-7

What is the leaving relative humidity  from the adiabatic humidifier for the conditions listed in Exercise 6-4? a) 15% rh b) 25% rh c) 19% rh d) 28% rh

6-8

Using the air handler in Exercise 6-4 and 10.3 m3/s, adding a cooling coil to satisfy a room sensible heat gain of 146 kW and a room latent heat gain of 15 000 W, and room conditions of t db = 24°C and  = 40% rh and without outdoor air, what is the required leaving air temperature t db and  from the cooling coil? a) 12°C t db ,  = 90% rh b) 13°C t db ,  = 80% rh c) 12°C t db ,  = 75% rh

6-9

What is the room sensible heat ratio for the conditions listed in Exercise 6-8? a) 0.89 b) 0.95 c) 0.91 d) 1.0

6-10

Would you attempt to add humidity to the leaving airstream for the conditions listed in Exercise 6-8 in the cooling mode with an adiabatic humidifier? a) Yes b) No c) Not sure


Skill Development Exercises for Chapter 7 Total number of questions: 11 For all of the Skill Development Exercises for Chapter 7, consider three zones in a small office building that we are going to heat and cool. The cooling and heating loads are as follows: Zone

Sensible Cooling, W

LatentCooling, W

HeatingSensible, W

1

12 000

1700

6000

2

16 000

2000

8000

3

20 000

3300

10 000

Assume room design conditions of the following: Coolingt db = 24°C and  = 50% rh Heating t db = 21°C and  = 40% rh Use a sea-level psychrometric chart. 7-1

What is the sensible heat ratio for all three zones in order 1, 2, 3? (Round to two decimal places.) a) 0.87, 0.89, 0.86 b) 0.88, 0.9, 0.91 c) 0.87, 0.89, 0.88

7-2

If we provide 25% outdoor air for code-required ventilation to all three zones, what is the mixed air condition in the summer if the outdoor air is t db = 38°C and  = 25% rh? a) t db = 34.5°C and  = 30% rh b) t db = 29°C and  = 36% rh c) t db = 27.5°C and  = 42% rh

7-3

For Zone 1 only, if we use individual fan-coils for each zone, what is the required supply airflow? a) airflow = 0.9 m3/s b) airflow = 0.99 m3/s c) airflow = 1.03 m3/s

C h a p t e r 7




For Zone 1 only, what are the leaving air conditions from the cooling coil assuming we use 25% outdoor air from Exercise 7-2 and the correct supply airflow? a) t db = 12°C and  = 90% rh b) t db = 14°C and  = 88% rh c) t db = 16°C and  = 80% rh

7-5

For Zone 1 only, what is the total cooling capacity, q t , of the cooling coil with the correct airflow and leaving air conditions? a) 18 000 W b) 22 000 W c) 19 008 W

7-6

If all three zones were put on a central air handler with a constant-volume terminal reheat system, what would the airflow of all three zones be, in order 1, 2, 3? (Same outdoor design and percent outdoor air.) a) 0.99, 1.3, 1.6 b) 0.9, 1.4, 1.7 c) 0.99, 1.32, 1.65

7-7

If all three zones were put on a central air handler with a variable-air-volume reheat VAV box and 25% outdoor air, what are the required leaving air conditions from this air handler? a) t db = 13.5°C and  = 91% rh b) t db = 14.5°C and  = 88% rh c) t db = 16°C and  = 82% rh

7-8

With the system in Exercise 7-7, what are the new required airflows by zone in order 1, 2, 3 with the new leaving conditions? a) 0.94, 1.26, 1.57 m3/s b) 0.99, 1.26, 1.6 m3/s c) 1.03, 1.3, 1.55 m3/s

7-9

What is the reheat required by zone in order 1, 2, 3 to meet the total reheat load plus the winter heat loss load? (Use t db = 21°C for room condition and 25% outdoor air.) a) 14 000, 18 000, 25 000 W b) 14 530, 19 930, 24 240 W c) 16 000, 21 000, 26 000 W


From Exercise 7-7, with the correct leaving conditions and airflow, what is the total cooling capacity of the central air-handler cooling coil? a) 73 280 W b) 70 120 W c) 79 910 W

7-11

If the system in Exercise 7-7 were a constant-volume, dual-duct system, what would be the heat capacity of the hot-deck coil used in the central air handler? (Room at t db = 24°C.) a) 75 000 W b) 65 000 W c) 55 100 W d) 60 500 W

C h a p t e r 7



s e s i c r e x E t n e m p o l e v e D l l i k S


A heat wheel with a desiccant coating is a: a) Sensible heat recovery device b) Total enthalpy heat recovery device c) Sensible-to-total heat recovery device d) Total-to-sensible heat recovery device

8-2

8 r e t p a h C

When is preheating of the outdoor airstream necessary on a heat recovery device? a) When the outdoor air temperature is below –16°C. b) When the outdoor air dew point is below 0°C. c) When the exhaust airstream has a dew point above 0°C and the leaving air temperature is below 0°C. d) All of the above.

8-3

Heat recovery effectiveness is the actual amount of heat transferred versus the maximum amount that could be transferred. a) True b) False

8-4

Energy recovery involves the transfer of sensible heat from one airstream to the other airstream. a) True b) False

8-5

An air-side economizer should be considered on any/all air systems that have 100% outdoor air capability and high operation hours with an ambient air tem perature below 16°C and a demand for cooling. a) True b) False

8-6

Water-side economizers can be used on a chilled-water system with all terminal fan-coils and an air-cooled water chiller. a) True b) False


8-8

8-9

There is a sensible heat recovery system between equal outdoor air and exhaust airstreams in Phoenix, Arizona, and the summer design outside is t db = 48°C and  = 10% rh. If the effectiveness is 75% of the heat recovery device and the exhaust airstream is t db = 24°C and  = 40% rh, what are the dry-bulb temperature and relative humidity of the outdoor airstream leaving the recovery device? a)

t db =

32°C and  = 25% rh

b)

t db =

35°C and  = 20% rh

c)

t db =

38°C and  = 18% rh

d)

t db =

30°C and  = 28% rh

From Exercise 8-7, what are the leaving air conditions of the exhaust airstream with everything else being the same? a)

t db =

30°C and  = 30% rh

b)

t db =

35°C and  = 25% rh

c)

t db =

40°C and  = 20% rh

d)

t db =

42°C and  = 16% rh

If the entering air conditions to a cooling tower are t db = 48°C and t wb = 19°C and the cooling tower has a full-load approach temperature of 4°C, what is the leaving water from cooling tower (at full load)? a) 40°C b) 35°C c) 23°C d) 29°C

8-10

Supply air temperature reset can be used on all air-conditioning systems, any time of the year in all parts of the world, regardless of the ambient air conditions. a) True b) False

C h a p t e r 8





A cooling tower needs to reject heat from 0.075 cu m/s of water entering at 35°C and leaving at 30°C. What is the total heat required to be rejected? a) 1766 kW b) 184 kW c) 1610 kW d) 1495 kW

9-2

9-3

From Exercise 9-1, if the cooling tower has an airflow of 47.2 m3/s and ambient air conditions of t db = 30°C and t wb = 24°C, what are the leaving air conditions of the tower? a)

t db =

30°C, t wb = 28.5°C

b)

t db =

32°C, t wb = 31°C

c)

t db =

31°C, t wb = 30.5°C

d)

t db =

30°C, t wb = 29.7°C

What is the cooling tower approach temperature for the cooling tower in Exercise 9-2? a) 3°C b) 7°C c) 4°C d) 6°C

9-4

In the design of an indoor swimming pool, it is best to keep the swimming pool water temperature and the room temperature as far apart as comfortably possi ble. a) True b) False

9-5

In a cleanroom with design conditions of t db = 20°C and  = 40% rh, the makeup air must be cooled to what dry-bulb temperature or the relative humidity will not be met? a)

t db =

12°C

b)

t db =

20°C

c)

t db =

5°C

d)

t db =

10°C


9-7

9-8

If we cool the air via direct evaporative cooling from t db = 43°C and  = 2% rh, what is the lowest leaving air temperature we can achieve? a)

t db =

17°C

b)

t db =

20°C

c)

t db =

18°C

d)

t db =

12°C

In Exercise 9-6, if our evaporative efficiency is 80%, what are the leaving air conditions? a)

t db =

20°C and  = 70% rh

b)

t db =

22°C and  = 70% rh

c)

t db =

21°C and  = 58% rh

d)

t db =

25°C and  = 50% rh

In Exercises 9-6 and 9-7, if the room sensible heat ratio is 0.9, what is the expected room relative humidity if the room is at t db = 24°C? a)  = 53% rh b)  = 60% rh c)  = 50% rh d) Cannot maintain room at t db = 24°C with this leaving condition

9-9

9-10

If we use the same outdoor conditions of t db = 43°C and  = 2% rh from Exercise 9-6 and an indirect evaporative cooling section of 40% efficiency, what are the leaving air conditions from this section? a)

t db =

33°C and  = 3% rh

b)

t db =

31°C and  = 20% rh

c)

t db =

35°C and  = 5% rh

d)

t db =

19°C and  = 5% rh

If we add a direct evaporative cooling section in series downstream of the indirect section in Exercise 9-9 and the direct section has an efficiency of 70%, what are the leaving air conditions? a)

t db =

15°C and  = 95% rh

b)

t db =

15°C and  = 65% rh

c)

t db =

19°C and  = 60% rh

d)

t db =

19°C and  = 52% rh

C h a p t e r 9


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