GAS MIXTURES AND PSYCHROMETRICS - Taibah U

G A S M I X T U R E S A N DP S Y C H R O M E T R I C S

Up to this point, we have limited our consideration to thermodynamicsystems that involve a single pure substance such as water, refrigerant-134a, or nitrogen. Many important thermodynamic applications, how-

ever, involve mixtures of several pure substances rather than a single puresubstance. Therefore, it is important to develop an understanding of mixturesand learn how to handle them.

In this chapter, we deal with nonreacting gas mixtures. A nonreacting gasmixture can be treated as a pure substance since it is usually a homogeneousmixture of different gases. The properties of a gas mixture obviously will de-pend on the properties of the individual gases (called components or con-stituents) as well as on the amount of each gas in the mixture. Therefore, it ispossible to prepare tables of properties for mixtures. This has been done forcommon mixtures such as air. It is not practical to prepare property tables forevery conceivable mixture composition, however, since the number of possi-ble compositions is endless. Therefore, we need to develop rules for deter-mining mixture properties from a knowledge of mixture composition and theproperties of the individual components. We do this first for ideal-gas mix-tures and then for real-gas mixtures. The basic principles involved are also ap-plicable to liquid or solid mixtures, called solutions.

At temperatures below the critical temperature, the gas phase of a substanceis frequently referred to as a vapor. The term vapor implies a gaseous state thatis close to the saturation region of the substance, raising the possibility of con-densation during a process.

When we deal with a gas–vapor mixture, the vapor may condense out of themixture during a process, forming a two-phase mixture. This may complicatethe analysis considerably. Therefore, a gas–vapor mixture needs to be treateddifferently from an ordinary gas mixture.

Several gas–vapor mixtures are encountered in engineering. In this chapter,we consider the air–water-vapor mixture, which is the most commonly en-countered gas–vapor mixture in practice. We also discuss air-conditioning,which is the primary application area of air–water-vapor mixtures.

1

ONLINE CHAPTER

24CONTENTS

1 Composition of a GasMixture: Mass and MoleFractions 2

2 P-v-T Behavior of GasMixtures: Ideal and RealGases 4

3 Properties of Gas Mixtures:Ideal and Real Gases 6

4 Dry and Atmospheric Air8

5 Specific and RelativeHumidity of Air 10

6 Dew-Point Temperature12

7 Adiabatic Saturation andWet-Bulb Temperatures13

8 The Psychrometric Chart17

9 Human Comfort andAir-Conditioning 18

10 Air-Conditioning Processes 20

Summary 31

References and SuggestedReadings 33

Problems 33

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1 COMPOSITION OF A GAS MIXTURE: MASS ANDMOLE FRACTIONS

To determine the properties of a mixture, we need to know the composition ofthe mixture as well as the properties of the individual components. There aretwo ways to describe the composition of a mixture: either by specifying thenumber of moles of each component, called molar analysis, or by specifyingthe mass of each component, called gravimetric analysis.

Consider a gas mixture composed of k components. The mass of the mix-ture mm is the sum of the masses of the individual components, and the molenumber of the mixture Nm is the sum of the mole numbers of the individualcomponents* (Figs. 1 and 2). That is,

(1a, b)

The ratio of the mass of a component to the mass of the mixture is called themass fraction mf, and the ratio of the mole number of a component to themole number of the mixture is called the mole fraction y:

(2a, b)

Dividing Eq. 1a by mm or Eq. 1b by Nm, we can easily show that the sum of themass fractions or mole fractions for a mixture is equal to 1 (Fig. 3):

The mass of a substance can be expressed in terms of the mole number Nand molar mass M of the substance as m � NM. Then the apparent (or aver-age) molar mass and the gas constant of a mixture can be expressed as

(3a, b)

The molar mass of a mixture can also be expressed as

(4)

Mass and mole fractions of a mixture are related by

(5)mfi �mi

mm�

Ni Mi

NmMm� yi

Mi

Mm

Mm �mm

Nm�

mm

�mi/Mi

�1

�mi/(mm Mi)�

1

�k

i�1

mfi

Mi

Mm �mm

Nm�

�mi

Nm�

�Ni Mi

Nm� �

k

i�1

yi Mi and Rm �Ru

Mm

�k

i�1

mfi � 1 and �k

i�1

yi � 1

mfi �mi

mm and yi �

Ni

Nm

mm � �k

i�1

mi and Nm � �k

i�1

Ni

■

2FUNDAMENTALS OF THERMAL-FLUID SCIENCES

EXAMPLE 1 Mass and Mole Fractions of a Gas Mixture

Consider a gas mixture that consists of 3 kg of O2, 5 kg of N2, and 12 kg ofCH4, as shown in Fig. 4. Determine (a) the mass fraction of each component,(b) the mole fraction of each component, and (c) the average molar mass andgas constant of the mixture.

*Throughout this chapter, the subscript m will denote the gas mixture and the subscript iwill denote any single component of the mixture.

FIGURE 1The mass of a mixture is equal to thesum of the masses of its components.

H2

6 kg

O2

32 kg

H2 + O2

38 kg+

FIGURE 2The number of moles of a nonreactingmixture is equal to the sum of thenumber of moles of its components.

H2

3 kmol

O2

1 kmol

H2 + O2

4 kmol+

FIGURE 3The sum of the mole fractions of amixture is equal to 1.

H2 + O2

yH2 =

yO2 =

0.750.251.00

FIGURE 4Schematic for Example 1.

3 kg O2

5 kg N2

12 kg CH4

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ONLINE CHAPTER 243

SOLUTION The schematic of the gas mixture is given in Fig. 4. We note thatthis is a gas mixture that consists of three gases of known masses.

Analysis (a) The total mass of the mixture is

Then the mass fraction of each component becomes

(b) To find the mole fractions, we need to determine the mole numbers of eachcomponent first:

NO2� � 0.094 kmol

NN2� � 0.179 kmol

NCH4� � 0.750 kmol

Thus,

and

(c) The average molar mass and gas constant of the mixture are determinedfrom their definitions,

Mm � � � 19.6 kg/kmol

or

Mm �

� (0.092)(32) � (0.175)(28) � (0.733)(16) � 19.6 kg/kmol

Also,

Discussion When mass fractions are available, the molar mass and mass frac-tions could also be determined directly from Eqs. 4 and 5.

Rm �Ru

Mm�

8.314 kJ/(kmol � K)19.6 kg/kmol

� 0.424 kJ/(kg � K)

�yi Mi � yO2MO2

� yN2MN2

� yCH4MCH4

20 kg1.023 kmol

mm

Nm

yCH4�

NCH4

Nm�

0.750 kmol1.023 kmol

� 0.733

yN2�

NN2

Nm�


� 0.175

yO2�

No2

Nm�


� 0.092

Nm � NO2� NN2

� NCH4� 0.094 � 0.179 � 0.750 � 1.023 kmol

mCH4

MCH4

�12 kg

16 kg/kmol

mN2

MN2

�5 kg

28 kg/kmol

mo2

Mo2

�3 kg

32 kg/kmol

mfCH4�

mCH4

mm�

12 kg20 kg

� 0.60

mfN2�

mN2

mm�

5 kg20 kg

� 0.25

mfO2�

mo2

mm�

3 kg20 kg

� 0.15

mm � mO2� mN2

� mCH4� 3 � 5 � 12 � 20 kg

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2 P-v-T BEHAVIOR OF GAS MIXTURES: IDEAL ANDREAL GASES

An ideal gas is defined as a gas whose molecules are spaced far apart so thatthe behavior of a molecule is not influenced by the presence of other mole-cules—a situation encountered at low densities. We also mentioned that realgases approximate this behavior closely when they are at a low pressure orhigh temperature relative to their critical-point values. The P-υ-T behavior ofan ideal gas is expressed by the simple relation Pυ � RT, which is called theideal-gas equation of state. The P-υ-T behavior of real gases is expressed bymore complex equations of state or by Pυ � ZRT, where Z is the compressi-bility factor.

When two or more ideal gases are mixed, the behavior of a molecule nor-mally is not influenced by the presence of other similar or dissimilar mole-cules, and therefore a nonreacting mixture of ideal gases also behaves as anideal gas. Air, for example, is conveniently treated as an ideal gas in the rangewhere nitrogen and oxygen behave as ideal gases. When a gas mixture con-sists of real (nonideal) gases, however, the prediction of the P-υ-T behavior ofthe mixture becomes rather involved.

The prediction of the P-υ-T behavior of gas mixtures is usually based ontwo models: Dalton’s law of additive pressures and Amagat’s law of additivevolumes. Both models are described and discussed below.

Dalton’s law of additive pressures: The pressure of a gas mixture is equalto the sum of the pressures each gas would exert if it existed alone at themixture temperature and volume (Fig. 5).

Amagat’s law of additive volumes: The volume of a gas mixture is equal tothe sum of the volumes each gas would occupy if it existed alone at the mix-ture temperature and pressure (Fig. 6).

Dalton’s and Amagat’s laws hold exactly for ideal-gas mixtures, but onlyapproximately for real-gas mixtures. This is due to intermolecular forces thatmay be significant for real gases at high densities. For ideal gases, these twolaws are identical and give identical results.

Dalton’s and Amagat’s laws can be expressed as follows:

Dalton’s law: (6)

Amagat’s law: (7)

In these relations, Pi is called the component pressure and Vi is called thecomponent volume (Fig. 7). Note that Vi is the volume a component wouldoccupy if it existed alone at Tm and Pm, not the actual volume occupied by thecomponent in the mixture. (In a vessel that holds a gas mixture, each compo-nent fills the entire volume of the vessel. Therefore, the volume of each com-ponent is equal to the volume of the vessel.) Also, the ratio Pi /Pm is called thepressure fraction and the ratio Vi /Vm is called the volume fraction of com-ponent i.

Vm � �k

i�1

Vi(Tm, Pm)

¶ exact for ideal gases,approximatefor real gases

Pm � �k

i�1

Pi(Tm,Vm)

■


FIGURE 5Dalton’s law of additive pressures fora mixture of two ideal gases.

+

GasmixtureA + BV , T

PA + PB

Gas BV, T

PB

Gas AV, T

PA

FIGURE 6Amagat’s law of additive volumes fora mixture of two ideal gases.

+

Gas mixtureA + BP , T

VA + VB

Gas A P, T

VA

Gas B P, T

VB

FIGURE 7The volume a component wouldoccupy if it existed alone at themixture T and P is called thecomponent volume (for ideal gases, itis equal to the partial volume yiVm).

O2 + N2

100 kPa400 K1 m3

O2

100 kPa400 K0.3 m3

N2

100 kPa400 K0.7 m3

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Ideal-Gas MixturesFor ideal gases, Pi and Vi can be related to yi by using the ideal-gas relation forboth the components and the gas mixture:

Therefore,

(8)

Equation 8 is strictly valid for ideal-gas mixtures since it is derived by assum-ing ideal-gas behavior for the gas mixture and each of its components. Thequantity yiPm is called the partial pressure (identical to the component pres-sure for ideal gases), and the quantity yiVm is called the partial volume (iden-tical to the component volume for ideal gases). Note that for an ideal-gasmixture, the mole fraction, the pressure fraction, and the volume fraction of acomponent are identical.

The composition of an ideal-gas mixture (such as the exhaust gases leavinga combustion chamber) is frequently determined by a volumetric analysis(called the Orsat Analysis) and Eq. 8. A sample gas at a known volume, pres-sure, and temperature is passed into a vessel containing reagents that absorbone of the gases. The volume of the remaining gas is then measured at theoriginal pressure and temperature. The ratio of the reduction in volume to theoriginal volume (volume fraction) represents the mole fraction of that partic-ular gas.

Real-Gas MixturesDalton’s law of additive pressures and Amagat’s law of additive volumes canalso be used for real gases, often with reasonable accuracy. This time, how-ever, the component pressures or component volumes should be evaluatedfrom relations that take into account the deviation of each component fromideal-gas behavior. One way of doing that is to use more exact equations ofstate (van der Waals, Beattie–Bridgeman, Benedict–Webb–Rubin, etc.) in-stead of the ideal-gas equation of state. Another way is to use the compress-ibility factor (Fig. 8) as

PV � ZNRuT (9)

The compressibility factor of the mixture Zm can be expressed in terms of thecompressibility factors of the individual gases Zi by applying Eq. 9 to bothsides of Dalton’s law or Amagat’s law expression and simplifying. We obtain

Zm � yiZi (10)

where Zi is determined either at Tm and Vm (Dalton’s law) or at Tm and Pm

(Amagat’s law) for each individual gas. It may seem that using either law willgive the same result, but it does not.

The compressibility-factor approach, in general, gives more accurate resultswhen the Zi’s in Eq. 10 are evaluated by using Amagat’s law instead of

�k

i�1

Pi

Pm�

Vi

Vm�

Ni

Nm� yi

Vi(Tm, Pm)Vm

�Ni RuTm�Pm

NmRuTm�Pm�

Ni

Nm� yi

Pi(Tm, Vm)Pm

�Ni RuTm�Vm

NmRuTm�Vm�

Ni

Nm� yi

ONLINE CHAPTER 245

FIGURE 8One way of predicting the P-υ-T

behavior of a real-gas mixture is to usecompressibility factors.

Pm Vm = Zm Nm Ru Tm

Zm = ∑ yi Zi

k

i = 1

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Dalton’s law. This is because Amagat’s law involves the use of mixture pres-sure Pm, which accounts for the influence of intermolecular forces between themolecules of different gases. Dalton’s law disregards the influence of dissim-ilar molecules in a mixture on each other. As a result, it tends to underpredictthe pressure of a gas mixture for a given Vm and Tm. Therefore, Dalton’s law ismore appropriate for gas mixtures at low pressures. Amagat’s law is more ap-propriate at high pressures.

Note that there is a significant difference between using the compressibil-ity factor for a single gas and for a mixture of gases. The compressibility fac-tor predicts the P-υ-T behavior of single gases rather accurately, as discussedin Chap. 3, but not for mixtures of gases. When we use compressibility fac-tors for the components of a gas mixture, we account for the influence of likemolecules on each other; the influence of dissimilar molecules remainslargely unaccounted for. Consequently, a property value predicted by this ap-proach may be considerably different from the experimentally determinedvalue.

3 PROPERTIES OF GAS MIXTURES:IDEAL AND REAL GASES

Consider a gas mixture that consists of 2 kg of N2 and 3 kg of CO2. The totalmass (an extensive property) of this mixture is 5 kg. How did we do it? Well,we simply added the mass of each component. This example suggests a sim-ple way of evaluating the extensive properties of a nonreacting ideal- or real-gas mixture: Just add the contributions of each component of the mixture(Fig. 9). Then the total internal energy, enthalpy, and entropy of a gas mixturecan be expressed, respectively, as

(11)

(12)

(13)

By following a similar logic, the changes in internal energy, enthalpy, andentropy of a gas mixture during a process can be expressed, respectively, as

(14)

(15)

(16)

Now reconsider the same mixture, and assume that both N2 and CO2 are at25°C. The temperature (an intensive property) of the mixture is, as you wouldexpect, also 25°C. Notice that we did not add the component temperatures to

�Sm � �k

i�1

�Si � �k

i�1

mi�si � �k

i�1

Ni� si (kJ/K)

�Hm � �k

i�1

�Hi � �k

i�1

mi�hi � �k

i�1

Ni�hi (kJ)

�Um � �k

i�1

�Ui � �k

i�1

mi�ui � �k

i�1

Ni�ui (kJ)

Sm � �k

i�1

Si � �k

i�1

misi � �k

i�1

Ni si (kJ/K)

Hm � �k

i�1

Hi � �k

i�1

mihi � �k

i�1

Ni hi (kJ)

Um � �k

i�1

Ui � �k

i�1

miui � �k

i�1

Ni ui (kJ)

■


FIGURE 9The extensive properties of a mixtureare determined by simply adding theproperties of the components.

2 kmol A6 kmol B

UA = 1000 kJ

UB = 1800 kJ

Um = 2800 kJ

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determine the mixture temperature. Instead, we used some kind of averagingscheme, a characteristic approach for determining the intensive properties ofa gas mixture. The internal energy, enthalpy, and entropy of a gas mixture perunit mass or per unit mole of the mixture can be determined by dividing theequations above by the mass or the mole number of the mixture (mm or Nm).We obtain (Fig. 10)

um � mfi ui and m � yi i (kJ/kg or kJ/kmol) (17)

hm � mfihi and m � yi i (kJ/kg or kJ/kmol) (18)

sm � mfisi and m � yi i (kJ/kg � K or kJ/kmol � K) (19)

Similarly, the specific heats of a gas mixture can be expressed as

Cυ, m � mfiCυ, i and υ, m � yi υ, i (kJ/kg � �C or kJ/kmol � �C)

(20)

Cp, m � mfiCp, i and p, m � yi p, i (kJ/kg � �C or kJ/kmol � �C) (21)

Notice that properties per unit mass involve mass fractions (mfi) and proper-ties per unit mole involve mole fractions (yi).

The relations just given are exact for ideal-gas mixtures, and approximate forreal-gas mixtures. (In fact, they are also applicable to nonreacting liquid andsolid solutions especially when they form an “ideal solution.”) The only majordifficulty associated with these relations is the determination of properties foreach individual gas in the mixture. The analysis can be simplified greatly, how-ever, by treating the individual gases as an ideal gas, if doing so does not intro-duce a significant error.

C�k

i � 1

C�k

i � 1

C�k

i � 1

C�k

i � 1

s�k

i � 1

s�k

i � 1

h�k

i � 1

h�k

i � 1

u�k

i � 1

u�k

i � 1

ONLINE CHAPTER 247

FIGURE 10The intensive properties of a mixture

are determined by weighted averaging.

2 kmol A3 kmol B

u-A = 500 kJ/kmol

u-B = 600 kJ/kmol

u-m = 560 kJ/kmol

EXAMPLE 2 Mixing Two Ideal Gases in a Tank

An insulated rigid tank is divided into two compartments by a partition. Onecompartment contains 7 kg of oxygen gas at 40°C and 100 kPa, and the othercompartment contains 4 kg of nitrogen gas at 20°C and 150 kPa. Now thepartition is removed, and the two gases are allowed to mix. Determine (a) themixture temperature and (b) the mixture pressure after equilibrium has beenestablished.

SOLUTION We take the entire contents of the tank (both compartments) as thesystem (Fig. 11). This is a closed system since no mass crosses the boundaryduring the process. We note that the volume of a rigid tank is constant and thusv2 � v1 and there is no boundary work done.Assumptions 1 We assume both gases to be ideal gases, and their mixture tobe an ideal-gas mixture. This assumption is reasonable since both the oxygenand nitrogen are well above their critical temperatures and well below their crit-ical pressures. 2 The tank is insulated and thus there is no heat transfer.3 There are no other forms of work involved. FIGURE 11

Schematic for Example 2.

O2

7 kg40°C

100 kPa

N2

4 kg20°C

150 kPa

Partition

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4 DRY AND ATMOSPHERIC AIRAir is a mixture of nitrogen, oxygen, and small amounts of some other gases.Air in the atmosphere normally contains some water vapor (or moisture) andis referred to as atmospheric air. By contrast, air that contains no water vaporis called dry air. It is often convenient to treat air as a mixture of water vaporand dry air since the composition of dry air remains relatively constant, but theamount of water vapor changes as a result of condensation and evaporationfrom oceans, lakes, rivers, showers, and even the human body. Although theamount of water vapor in the air is small, it plays a major role in hu-man comfort. Therefore, it is an important consideration in air-conditioningapplications.

■


Analysis (a) Noting that there is no energy transfer to or from the tank, the en-ergy balance for the system can be expressed as

Ein � Eout � �Esystem

0 � �U � �UN2� �UO2

[mCυ (Tm � T1)] N2� [mCυ (Tm � T1)] O2

� 0

By using Cv values at room temperature (from Table A–2a), the final tempera-ture of the mixture is determined to be

(4 kg)(0.743 kJ/kg � �C)(Tm � 20�C) � (7 kg)(0.658 kJ/kg � �C)(Tm � 40�C) � 0

Tm � 32.2�C

(b) The final pressure of the mixture is determined from the ideal-gas relation

PmVm � NmRuTm

where

NO2�

NN2�

Nm � NO2� NN2

� 0.219 � 0.143 � 0.362 kmol

and

VO2�

VN2�

Vm � VO2� VN2

� 5.70 � 2.32 � 8.02 m3

Thus,

Pm � � 114.5 kPa

Discussion We could also determine the mixture pressure by using PmVm

� mmRmTm, where Rm is the apparent gas constant of the mixture. This wouldrequire a knowledge of mixture composition in terms of mass or mole fractions.

NmRuTm

Vm�

(0.362 kmol)(8.314 kPa � m3�kmol � K)(305.2 K)

8.02 m3

aNRuT1

P1b

N2

�(0.143 kmol)(8.314 kPa � m3�kmol � K)(293 K)

150 kPa� 2.32 m3

aNRuT1

P1b

O2

�(0.219 kmol)(8.314 kPa � m3�kmol � K)(313 K)

100 kPa� 5.70 m3

mN2

MN2

�4 kg

28 kg�kmol� 0.143 kmol

mO2

MO2

�7 kg

32 kg�kmol� 0.219 kmol

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The temperature of air in air-conditioning applications ranges from about �10 to about 50°C. In this range, dry air can be treated as an ideal gas with a constant Cp value of 1.005 kJ/(kg · K) with negligible error (under 0.2 percent), as illustrated in Fig. 12. Taking 0°C as the reference temperature, the enthalpy and enthalpy change of dry air can be determinedfrom

hdry air � CpT � (1.005 kJ/kg · �C)T (kJ/kg) (22a)

and

�hdry air � Cp �T � (1.005 kJ/kg · �C) �T (kJ/kg) (22b)

where T is the air temperature in �C and �T is the change in temperature. Inair-conditioning processes, we are concerned with the changes in enthalpy �h,which is independent of the reference point selected.

It certainly would be very convenient to also treat the water vapor in the airas an ideal gas and you would probably be willing to sacrifice some accuracyfor such convenience. Well, it turns out that we can have the conveniencewithout much sacrifice. At 50�C, the saturation pressure of water is 12.3 kPa.At pressures below this value, water vapor can be treated as an ideal gas withnegligible error (under 0.2 percent), even when it is a saturated vapor. There-fore, water vapor in air behaves as if it existed alone and obeys the ideal-gasrelation Pυ � RT. Then the atmospheric air can be treated as an ideal-gas mix-ture whose pressure is the sum of the partial pressure of dry air* Pa and that ofwater vapor Pυ:

P � Pa � Pυ (kPa) (23)

The partial pressure of water vapor is usually referred to as the vapor pres-sure. It is the pressure water vapor would exert if it existed alone at the tem-perature and volume of atmospheric air.

Since water vapor is an ideal gas, the enthalpy of water vapor is a functionof temperature only, that is, h � h(T). This can also be observed from the T-sdiagram of water given in Fig. A–9 and Fig. 13 where the constant-enthalpylines coincide with constant-temperature lines at temperatures below 50°C.Therefore, the enthalpy of water vapor in air can be taken to be equal to theenthalpy of saturated vapor at the same temperature. That is,

hυ(T, low P) � hg(T) (24)

The enthalpy of water vapor at 0�C is 2501.3 kJ/kg. The average Cp valueof water vapor in the temperature range �10 to 50�C can be taken to be1.82 kJ/kg · �C. Then the enthalpy of water vapor can be determined approxi-mately from

hg(T) � 2501.3 � 1.82T (kJ/kg) T in �C (25)

or

hg(T) � 1061.5 � 0.435T (Btu/lbm) T in �F (26)

ONLINE CHAPTER 249

DRY AIR

T,°C Cp, kJ/kg · °C

–100

1020304050

1.00381.00411.00451.00491.00541.00591.0065

FIGURE 12The Cp of air can be assumed to beconstant at 1.005 kJ/kg · °C in the

temperature range �10 to 50°Cwith an error under 0.2 percent.

T, °C

s

h = const.

50

FIGURE 13At temperatures below 50�C, theh � constant lines coincide with

the T � constant lines in thesuperheated vapor region of water.

*Throughout this chapter, the subscript a will denote dry air and the subscript v will denotewater vapor.

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in the temperature range �10 to 50�C (or 15 to 120�F), with negligible error,as shown in Fig. 14.

5 SPECIFIC AND RELATIVE HUMIDITY OF AIRThe amount of water vapor in the air can be specified in various ways. Proba-bly the most logical way is to specify directly the mass of water vapor presentin a unit mass of dry air. This is called absolute or specific humidity (alsocalled humidity ratio) and is denoted by v:

v � (kg water vapor/kg dry air) (27)

The specific humidity can also be expressed as

v� � � � 0.622 (28)

or

v � (kg water vapor/kg dry air) (29)

where P is the total pressure.Consider 1 kg of dry air. By definition, dry air contains no water vapor, and

thus its specific humidity is zero. Now let us add some water vapor to this dryair. The specific humidity will increase. As more vapor or moisture is added,the specific humidity will keep increasing until the air can hold no more mois-ture. At this point, the air is said to be saturated with moisture, and it is calledsaturated air. Any moisture introduced into saturated air will condense. Theamount of water vapor in saturated air at a specified temperature and pressurecan be determined from Eq. 29 by replacing Pυ by Pg, the saturation pressureof water at that temperature (Fig. 15).

The amount of moisture in the air has a definite effect on how comfortablewe feel in an environment. However, the comfort level depends more on theamount of moisture the air holds (mυ) relative to the maximum amount ofmoisture the air can hold at the same temperature (mg). The ratio of these twoquantities is called the relative humidity f (Fig. 16)

f� � � (30)

where

Pg � Psat @ T (31)

Combining Eqs. 29 and 30, we can also express the relative humidity as

f� and v� (32a, b)

The relative humidity ranges from 0 for dry air to 1 for saturated air. Note thatthe amount of moisture air can hold depends on its temperature. Therefore, therelative humidity of air changes with temperature even when its specific hu-midity remains constant.

0.622fPg

P � fPg

vP(0.622 � v)Pg

Pυ

Pg

PυV/(RυT )

PgV/(RυT )

mυmg

0.622 Pυ

P � Pυ

Pυ

Pa

Pυ /Rυ

Pa /Ra

PυV/(RυT )

PaV/(RaT )

mυma

mυ

ma

■


WATER VAPOR

–100

1020304050

2482.92501.32519.82538.12556.32574.32592.1

2483.12501.32519.52537.72555.92574.12592.3

–0.20.00.30.40.40.2

–0.2

hg, kJ/kg

T, °C Table A-4 Eq. 9-25 Difference,

kJ/kg

FIGURE 14In the temperature range �10 to50�C, the hg of water can bedetermined from Eq. 25 withnegligible error.

AIR25°C, 100 kPa

(Psat, H2O @ 25°C = 3.169 kPa)

Pυ = 0Pυ < 3.169 kPaPυ = 3.169 kPa

dry airunsaturated airsaturated air

FIGURE 15For saturated air, the vapor pressureis equal to the saturation pressureof water.

AIR25°C, 1 atm

ma = 1 kgmυ =

mυ,max = 0.01 kg0.02 kg

Specific humidity: ω = 0.01

Relative humidity: φ = 50%

kg H2O

kg dry air

FIGURE 16Specific humidity is the actualamount of water vapor in 1 kg of dryair, whereas relative humidity is theratio of the actual amount of moisturein the air to the maximum amount ofmoisture air can hold at thattemperature.

cen54261_ch24.qxd 4/2/08 5:38 PM Page 10

Atmospheric air is a mixture of dry air and water vapor, and thus the en-thalpy of air is expressed in terms of the enthalpies of the dry air and the wa-ter vapor. In most practical applications, the amount of dry air in theair–water-vapor mixture remains constant, but the amount of water vaporchanges. Therefore, the enthalpy of atmospheric air is expressed per unit massof dry air instead of per unit mass of the air–water-vapor mixture.

The total enthalpy (an extensive property) of atmospheric air is the sum ofthe enthalpies of dry air and the water vapor:

H � Ha � Hυ � ma ha � mυhυ

Dividing by ma gives

h � � ha � hυ � ha � vhυ

or

h � ha � vhg (kJ/kg dry air) (33)

since hυ � hg (Fig. 17).Also note that the ordinary temperature of atmospheric air is frequently re-

ferred to as the dry-bulb temperature to differentiate it from other forms oftemperatures that shall be discussed.

mυ

ma

Hma

ONLINE CHAPTER 2411

moistureω kg

hg

Dry air1 kgha

(1 + ω) kg ofmoist air

h = ha + ωhg, kJ/kg dry air

FIGURE 17The enthalpy of moist (atmospheric)air is expressed per unit mass of dry

air, not per unit mass of moist air.

T = 25ºCP = 100 kPaφ = 75%

ROOM

5 m × 5 m × 3 m


EXAMPLE 3 The Amount of Water Vapor in Room Air

A 5-m 5-m 3-m room shown in Fig. 18 contains air at 25°C and 100 kPaat a relative humidity of 75 percent. Determine (a) the partial pressure of dryair, (b) the specific humidity, (c) the enthalpy per unit mass of the dry air, and(d ) the masses of the dry air and water vapor in the room.

SOLUTION A sketch of the room is given in Fig. 18. Both the air and the vaporfill the entire room, and thus the volume of each gas is equal to the volume ofthe room.Assumptions The dry air and the water vapor in the room are ideal gases.Analysis (a) The partial pressure of dry air can be determined from

Pa � P � Pυ+

where

Pυ � fPg � fPsat @ 25�C � (0.75)(3.169 kPa) � 2.38 kPa

Thus,

Pa � (100 � 2.38) kPa � 97.62 kPa

(b) The specific humidity of air is

v� � � 0.0152 kg H2O/kg dry air(0.622)(2.38 kPa)(100 � 2.38) kPa

0.622Pυ

P � Pυ

cen54261_ch24.qxd 4/2/08 5:38 PM Page 11

6 DEW-POINT TEMPERATUREIf you live in a humid area, you are probably used to waking up most summermornings and finding the grass wet. You know it did not rain the night before.So what happened? Well, the excess moisture in the air simply condensed onthe cool surfaces, forming what we call dew. In summer, a considerableamount of water vaporizes during the day. As the temperature falls during thenight, so does the “moisture capacity” of air, which is the maximum amountof moisture air can hold. (What happens to the relative humidity during thisprocess?) After a while, the moisture capacity of air equals its moisture con-tent. At this point, air is saturated, and its relative humidity is 100 percent. Anyfurther drop in temperature results in the condensation of some of the mois-ture, and this is the beginning of dew formation.

The dew-point temperature Tdp is defined as the temperature at which con-densation begins when the air is cooled at constant pressure. In other words,Tdp is the saturation temperature of water corresponding to the vapor pressure:

Tdp � Tsat @ Pυ (34)

This is also illustrated in Fig. 19. As the air cools at constant pressure, the va-por pressure Pυ remains constant. Therefore, the vapor in the air (state 1) un-dergoes a constant-pressure cooling process until it strikes the saturated vapor

■


T

s

T1

Tdp

P υ =

con

st.

2

1

FIGURE 19Constant-pressure cooling of moist airand the dew-point temperature on theT-s diagram of water.

(c) The enthalpy of air per unit mass of dry air is determined from Eq. 33,where hg is taken from Table A–4:

h � ha � vhυ � CpT � vhg

� (1.005 kJ/kg · �C)(25�C) � (0.0152)(2547.2 kJ/kg)

� 63.8 kJ/kg dry air

The enthalpy of water vapor (2547.2 kJ/kg) could also be determined from theapproximation given by Eq. 25:

hg @ 25�C � 2501.3 � 1.82(25) � 2546.8 kJ/kg

which is very close to the value obtained from Table A–4.

(d ) Both the dry air and the water vapor fill the entire room completely. There-fore, the volume of each gas is equal to the volume of the room:

Va � Vυ � Vroom � (5)(5)(3) � 75 m3

The masses of the dry air and the water vapor are determined from the ideal-gasrelation applied to each gas separately:

ma � � � 85.61 kg

mυ � � � 1.3 kg

The mass of the water vapor in the air could also be determined from Eq. 27:

mυ � vma � (0.0152)(85.61 kg) � 1.3 kg

(2.38 kPa)(75 m3)

(0.4615 kPa � m3/kg � K)(298 K)

PυVυ

RυT

(97.62 kPa)(75 m3)

(0.287 kPa � m3/kg � K)(298 K)

PaVa

RaT

cen54261_ch24.qxd 4/2/08 5:38 PM Page 12

line (state 2). The temperature at this point is Tdp, and if the temperature dropsany further, some vapor condenses out. As a result, the amount of vapor in theair decreases, which results in a decrease in Pυ. The air remains saturated dur-ing the condensation process and thus follows a path of 100 percent relativehumidity (the saturated vapor line). The ordinary temperature and the dew-point temperature of saturated air are identical.

You have probably noticed that when you buy a cold canned drink from avending machine on a hot and humid day, dew forms on the can. The forma-tion of dew on the can indicates that the temperature of the drink is below thedew-point temperature of the surrounding air (Fig. 20).

The dew-point temperature of room air can be determined easily by coolingsome water in a metal cup by adding small amounts of ice and stirring. Thetemperature of the outer surface of the cup when dew starts to form on the sur-face is the dew-point temperature of the air.

ONLINE CHAPTER 2413

MOISTAIR

Liquid waterdroplets(dew)

S O D AT < Tdp

FIGURE 20When the temperature of a cold drink isbelow the dew-point temperature of the

surrounding air, it “sweats.”

COLDOUTDOORS10°C

AIR20°C, 75%

Typical temperaturedistribution

16°C18°C

20°C 20°C 20°C18°C

16°C


EXAMPLE 4 Fogging of the Windows in a House

In cold weather, condensation frequently occurs on the inner surfaces of thewindows due to the lower air temperatures near the window surface. Consider ahouse, shown in Fig. 21, that contains air at 20�C and 75 percent relative hu-midity. At what window temperature will the moisture in the air start condens-ing on the inner surfaces of the windows?

SOLUTION The interior of a house is maintained at a specified temperature andhumidity. The window temperature at which fogging will start is to be determined.Analysis The temperature distribution in a house, in general, is not uniform.When the outdoor temperature drops in winter, so does the indoor temperaturenear the walls and the windows. Therefore, the air near the walls and the win-dows remains at a lower temperature than at the inner parts of a house eventhough the total pressure and the vapor pressure remain constant throughoutthe house. As a result, the air near the walls and the windows will undergo aPv � constant cooling process until the moisture in the air starts condensing.This will happen when the air reaches its dew-point temperature Tdp. The dewpoint is determined from Eq. 34 to be

Tdp � Tsat @

where

Pυ � fPg @ 20�C � (0.75)(2.339 kPa) � 1.754 kPa

Thus,

Tdp � Tsat @ 1.754 kPa � 15.4�C

Therefore, the inner surface of the window should be maintained above 15.4�Cif condensation on the window surfaces is to be avoided.

Pυ

7 ADIABATIC SATURATION ANDWET-BULB TEMPERATURES

Relative humidity and specific humidity are frequently used in engineeringand atmospheric sciences, and it is desirable to relate them to easily measur-able quantities such as temperature and pressure. One way of determining the

■

cen54261_ch24.qxd 4/2/08 5:38 PM Page 13

relative humidity is to determine the dew-point temperature of air, as dis-cussed in the last section. Knowing the dew-point temperature, we can deter-mine the vapor pressure Pυ and thus the relative humidity. This approach issimple, but not quite practical.

Another way of determining the absolute or relative humidity is related toan adiabatic saturation process, shown schematically and on a T-s diagram inFig. 22. The system consists of a long insulated channel that contains a poolof water. A steady stream of unsaturated air that has a specific humidity of v1

(unknown) and a temperature of T1 is passed through this channel. As the airflows over the water, some water will evaporate and mix with the airstream.The moisture content of air will increase during this process, and its tempera-ture will decrease, since part of the latent heat of vaporization of the water thatevaporates will come from the air. If the channel is long enough, the airstreamwill exit as saturated air (f � 100 percent) at temperature T2, which is calledthe adiabatic saturation temperature.

If makeup water is supplied to the channel at the rate of evaporation at tem-perature T2, the adiabatic saturation process described above can be analyzedas a steady-flow process. The process involves no heat or work interactions,and the kinetic and potential energy changes can be neglected. Then the con-servation of mass and conservation of energy relations for this two-inlet, one-exit steady-flow system reduces to the following:

Mass balance:

m· � m· � m· a (The mass flow rate of dry air remains constant)

m· � m· f � m· (The mass flow rate of vapor in the air increases by an amount equal

or to the rate of evaporation m· f)m· av1 � m· f � m· av2

w2w1

a2a1


FIGURE 22The adiabatic saturation process andits representation on a T-s diagramof water.

T

s

P υ1

2

1

Adiabaticsaturationtemperature

Dew-pointtemperature

Unsaturatedair

T1, 1ωφ1

Saturated airT2, 2ω

φ2 = 100%1 2

Liquid waterat T2

Liquid water

cen54261_ch24.qxd 4/2/08 5:38 PM Page 14

Thus,

m· f � m· a(v2 � v1)

Energy balance:

E·in � E

·out (since Q

·� 0 and W

·� 0)

m· a h1 � m· f h � m· a h2

or

m· a h1 � m· a (v2 � v1)h � m· a h2

Dividing by m· a gives

h1 � (v2 � v1)h � h2

or

(CpT1 � v1h ) � (v2 � v1)h � (CpT2 � v2h )

which yields

v1 � (35)

where, from Eq. 32b,

v2 � (36)

since f2 � 100 percent. Thus, we conclude that the specific humidity (andrelative humidity) of air can be determined from Eqs. 35 and 36 by measuringthe pressure and temperature of air at the inlet and the exit of an adiabatic sat-urator.

If the air entering the channel is already saturated, then the adiabatic satura-tion temperature T2 will be identical to the inlet temperature T1, in which caseEq. 35 yields v1 � v2. In general, the adiabatic saturation temperature will bebetween the inlet and dew-point temperatures.

The adiabatic saturation process discussed above provides a means of de-termining the absolute or relative humidity of air, but it requires a long chan-nel or a spray mechanism to achieve saturation conditions at the exit. A morepractical approach is to use a thermometer whose bulb is covered with acotton wick saturated with water and to blow air over the wick, as shown inFig. 23. The temperature measured in this matter is called the wet-bulb tem-perature Twb, and it is commonly used in air-conditioning applications.

The basic principle involved is similar to that in adiabatic saturation. Whenunsaturated air passes over the wet wick, some of the water in the wick evap-orates. As a result, the temperature of the water drops, creating a temperaturedifference (which is the driving force for heat transfer) between the air andthe water. After a while, the heat loss from the water by evaporation equals theheat gain from the air, and the water temperature stabilizes. The thermometerreading at this point is the wet-bulb temperature. The wet-bulb temperature

0.622Pg2

P2 � Pg2

Cp(T2 � T1) � v2hfg2

hg1� hf2

g2f2g1

f2

f2

f2

ONLINE CHAPTER 2415

Ordinarythermometer

Wet-bulbtemperature

Airflow

Liquidwater

Wick

FIGURE 23A simple arrangement to measure

the wet-bulb temperature.

cen54261_ch24.qxd 4/2/08 5:38 PM Page 15

can also be measured by placing the wet-wicked thermometer in a holder at-tached to a handle and rotating the holder rapidly, that is, by moving the ther-mometer instead of the air. A device that works on this principle is calleda sling psychrometer and is shown in Fig. 24. Usually a dry-bulb thermometeris also mounted on the frame of this device so that both the wet- and dry-bulbtemperatures can be read simultaneously.

Advances in electronics made it possible to measure humidity directly ina fast and reliable way. It appears that sling psychrometers and wet-wickedthermometers are about to become things of the past. Today, hand-held elec-tronic humidity measurement devices based on the capacitance change in a thinpolymer film as it absorbs water vapor are capable of sensing and digitally dis-playing the relative humidity within 1 percent accuracy in a matter of seconds.

In general, the adiabatic saturation temperature and the wet-bulb tempera-ture are not the same. However, for air–water-vapor mixtures at atmosphericpressure, the wet-bulb temperature happens to be approximately equal to theadiabatic saturation temperature. Therefore, the wet-bulb temperature Twb canbe used in Eq. 14 in place of T2 to determine the specific humidity of air.


EXAMPLE 5 The Specific and Relative Humidity of Air

The dry- and the wet-bulb temperatures of atmospheric air at 1 atm(101.325 kPa) pressure are measured with a sling psychrometer and deter-mined to be 25 and 15�C, respectively. Determine (a) the specific humidity,(b) the relative humidity, and (c) the enthalpy of the air.

SOLUTION Dry- and wet-bulb temperatures are given. The specific humidity,relative humidity, and enthalpy are to be determined.Analysis (a) The specific humidity v1 is determined from Eq. 35,

v1 �

where T2 is the wet-bulb temperature and v2 is

v2 � �

� 0.01065 kg H2O/kg dry air

Thus,

v1 �


(b) The relative humidity f1 is determined from Eq. 32a to be

f1 � � � 0.332 or 33.2%

(c) The enthalpy of air per unit mass of dry air is determined from Eq. 33:

h1 � h � v1h � CpT1 � v1h

h1 � [1.005 kJ/(kg · �C)](25�C) � (0.00653)(2547.2 kJ/kg)


g1υ1a1

(0.00653)(101.325 kPa)(0.622 � 0.00653)(3.169 kPa)

v1P2

(0.622 � v1)Pg1

(1.005 kJ/kg � °C)[(15 � 25)°C] � (0.01065)(2465.9 kJ/kg)(2547.2 � 62.99) kJ/kg

(0.622)(1.705 kPa)(101.325 � 1.705) kPa

0.622Pg2

P2 � Pg2


hg1� hf2

Dry-bulbthermometer

Wet-bulbthermometerwick

Wet-bulbthermometer

FIGURE 24Sling psychrometer.

cen54261_ch24.qxd 4/2/08 5:38 PM Page 16

8 THE PSYCHROMETRIC CHARTThe state of the atmospheric air at a specified pressure is completely speci-fied by two independent intensive properties. The rest of the properties canbe calculated easily from the relations already given. The sizing of a typicalair-conditioning system involves numerous such calculations, which mayeventually get on the nerves of even the most patient engineers. Therefore,there is clear motivation to computerize calculations or to do these calcula-tions once and to present the data in the form of easily readable charts. Suchcharts are called psychrometric charts, and they are used extensively in air-conditioning applications. A psychrometric chart for a pressure of 1 atm(101.325 kPa) is given in Fig. A–33 in SI units. Psychrometric charts at otherpressures (for use at considerably higher elevations than sea level) are alsoavailable.

The basic features of the psychrometric chart are illustrated in Fig. 25. Thedry-bulb temperatures are shown on the horizontal axis, and the specific hu-midity is shown on the vertical axis. (Some charts also show the vapor pres-sure on the vertical axis since at a fixed total pressure P there is a one-to-onecorrespondence between the specific humidity v and the vapor pressure Pυ, ascan be seen from Eq. 29.) On the left end of the chart, there is a curve (calledthe saturation line) instead of a straight line. All the saturated air states are lo-cated on this curve. Therefore, it is also the curve of 100 percent relative hu-midity. Other constant relative-humidity curves have the same general shape.

Lines of constant wet-bulb temperature have a downhill appearance to theright. Lines of constant specific volume (in m3/kg dry air) look similar, exceptthey are steeper. Lines of constant enthalpy (in kJ/kg dry air) lie very nearlyparallel to the lines of constant wet-bulb temperature. Therefore, the constant-wet-bulb-temperature lines are used as constant-enthalpy lines in some charts.

For saturated air, the dry-bulb, wet-bulb, and dew-point temperatures areidentical (Fig. 26). Therefore, the dew-point temperature of atmospheric air atany point on the chart can be determined by drawing a horizontal line (a lineof v � constant or Pυ � constant) from the point to the saturated curve. Thetemperature value at the intersection point is the dew-point temperature.

The psychrometric chart also serves as a valuable aid in visualizing the air-conditioning processes. An ordinary heating or cooling process, for example,will appear as a horizontal line on this chart if no humidification or dehumidi-fication is involved (that is, v � constant). Any deviation from a horizontalline indicates that moisture is added or removed from the air during theprocess.

■

ONLINE CHAPTER 2417

Dry-bulb temperature

Spec

ific

hum

idity

,ω

Satu

ratio

n lin

e,

φ

= 10

0%φ

= co

nst.

Twb = const.

h = const.

υ= const.

FIGURE 25Schematic for a psychrometric chart.

Saturation line

Tdp = 15°C

Tdb

= 1

5°C

Tw

b = 15°C

15°C

15°C

FIGURE 26For saturated air, the

dry-bulb, wet-bulb, and dew-point temperatures are identical.

EXAMPLE 6 The Use of the Psychrometric Chart

Consider a room that contains air at 1 atm, 35�C, and 40 percent relative hu-midity. Using the psychrometric chart, determine (a) the specific humidity,(b) the enthalpy, (c) the wet-bulb temperature, (d ) the dew-point temperature,and (e) the specific volume of the air.

SOLUTION At a given total pressure, the state of atmospheric air is completelyspecified by two independent properties such as the dry-bulb temperature andthe relative humidity. Other properties are determined by directly reading theirvalues at the specified state.

cen54261_ch24.qxd 4/2/08 5:38 PM Page 17

9 HUMAN COMFORT AND AIR-CONDITIONINGHuman beings have an inherent weakness—they want to feel comfortable.They want to live in an environment that is neither hot nor cold, neither humidnor dry. However, comfort does not come easily since the desires of thehuman body and the weather usually are not quite compatible. Achievingcomfort requires a constant struggle against the factors that cause discomfort,such as high or low temperatures and high or low humidity. As engineers, it isour duty to help people feel comfortable. (Besides, it keeps us employed.)

It did not take long for people to realize that they could not change theweather in an area. All they can do is change it in a confined space such asa house or a workplace (Fig. 28). In the past, this was partially accomplishedby fire and simple indoor heating systems. Today, modern air-conditioningsystems can heat, cool, humidify, dehumidify, clean, and even deodorize theair—in other words, condition the air to peoples’ desires. Air-conditioningsystems are designed to satisfy the needs of the human body; therefore, it isessential that we understand the thermodynamic aspects of the body.

The human body can be viewed as a heat engine whose energy input isfood. As with any other heat engine, the human body generates waste heat thatmust be rejected to the environment if the body is to continue operating. Therate of heat generation depends on the level of the activity. For an average

■


Analysis (a) The specific humidity is determined by drawing a horizontal linefrom the specified state to the right until it intersects with the v axis, as shownin Fig. 27. At the intersection point, we read

v� 0.0142 kg H2O/kg dry air

(b) The enthalpy of air per unit mass of dry air is determined by drawing a lineparallel to the h � constant lines from the specific state until it intersects theenthalpy scale. At the intersection point, we read

h � 71.5 kJ/kg dry air

(c) The wet-bulb temperature is determined by drawing a line parallel to theTwb � constant lines from the specified state until it intersects the saturationline. At the intersection point, we read

Twb � 24�C

(d ) The dew-point temperature is determined by drawing a horizontal line fromthe specified state to the left until it intersects the saturation line. At the inter-section point, we read

Tdp � 19.4�C

(e) The specific volume per unit mass of dry air is determined by noting the dis-tances between the specified state and the v � constant lines on both sides ofthe point. The specific volume is determined by visual interpolation to be

υ � 0.893 m3/kg dry air

22�C

�10�C

FIGURE 28We cannot change the weather, but wecan change the climate in a confinedspace by air-conditioning.

T = 35°C

Tdp

υ

Twbh

φ = 40

%

ω


cen54261_ch24.qxd 4/2/08 5:38 PM Page 18

adult male, it is about 87 W when sleeping, 115 W when resting or doingoffice work, 230 W when bowling, and 440 W when doing heavy physicalwork. The corresponding numbers for an adult female are about 15 percentless. (This difference is due to the body size, not the body temperature. Thedeep-body temperature of a healthy person is maintained constant at 37�C.)A body will feel comfortable in environments in which it can dissipate thiswaste heat comfortably (Fig. 29).

Heat transfer is proportional to the temperature difference. Therefore in coldenvironments, a body will lose more heat than it normally generates, whichresults in a feeling of discomfort. The body tries to minimize the energydeficit by cutting down the blood circulation near the skin (causing a palelook). This lowers the skin temperature, which is about 34�C for an averageperson, and thus the heat transfer rate. A low skin temperature causes discom-fort. The hands, for example, feel painfully cold when the skin temperaturereaches 10�C. We can also reduce the heat loss from the body either by puttingbarriers (additional clothes, blankets, etc.) in the path of heat or by increasingthe rate of heat generation within the body by exercising. For example,the comfort level of a resting person dressed in warm winter clothing in aroom at 10�C is roughly equal to the comfort level of an identical person doing moderate work in a room at about �23�C. Or we can just cuddle up and put our hands between our legs to reduce the surface area through whichheat flows.

In hot environments, we have the opposite problem—we do not seem to bedissipating enough heat from our bodies, and we feel as if we are going toburst. We dress lightly to make it easier for heat to get away from our bodies,and we reduce the level of activity to minimize the rate of waste heat genera-tion in the body. We also turn on the fan to continuously replace the warmerair layer that forms around our bodies as a result of body heat by the cooler airin other parts of the room. When doing light work or walking slowly, abouthalf of the rejected body heat is dissipated through perspiration as latent heatwhile the other half is dissipated through convection and radiation as sensibleheat. When resting or doing office work, most of the heat (about 70 percent)is dissipated in the form of sensible heat whereas when doing heavy physicalwork, most of the heat (about 60 percent) is dissipated in the form of latentheat. The body helps out by perspiring or sweating more. As this sweat evap-orates, it absorbs latent heat from the body and cools it. Perspiration is notmuch help, however, if the relative humidity of the environment is close to 100percent. Prolonged sweating without any fluid intake will cause dehydrationand reduced sweating, which may lead to a rise in body temperature and a heatstroke.

Another important factor that affects human comfort is heat transfer by ra-diation between the body and the surrounding surfaces such as walls and win-dows. The sun’s rays travel through space by radiation. You warm up in frontof a fire even if the air between you and the fire is quite cold. Likewise, in awarm room you will feel chilly if the ceiling or the wall surfaces are at a con-siderably lower temperature. This is due to direct heat transfer between yourbody and the surrounding surfaces by radiation. Radiant heaters are com-monly used for heating hard-to-heat places such as car repair shops.

The comfort of the human body depends primarily on three factors: the(dry-bulb) temperature, relative humidity, and air motion (Fig. 30). The tem-perature of the environment is the single most important index of comfort.

ONLINE CHAPTER 2419

23°C

f = 50%

Air motion

15 m/min

FIGURE 30A comfortable environment.

(Reprinted with special permission ofKing Features Syndicate.)

23°C

Waste heat

37°C

FIGURE 29A body feels comfortable when it

can freely dissipate its wasteheat, and no more.

cen54261_ch24.qxd 4/2/08 5:38 PM Page 19

Most people feel comfortable when the environment temperature is between22 and 27�C. The relative humidity also has a considerable effect on comfortsince it affects the amount of heat a body can dissipate through evaporation.Relative humidity is a measure of air’s ability to absorb more moisture. Highrelative humidity slows down heat rejection by evaporation, and low relativehumidity speeds it up. Most people prefer a relative humidity of 40 to 60 per-cent.

Air motion also plays an important role in human comfort. It removes thewarm, moist air that builds up around the body and replaces it with fresh air.Therefore, air motion improves heat rejection by both convection and evapo-ration. Air motion should be strong enough to remove heat and moisture fromthe vicinity of the body, but gentle enough to be unnoticed. Most people feelcomfortable at an airspeed of about 15 m/min. Very-high-speed air motioncauses discomfort instead of comfort. For example, an environment at 10�Cwith 48 km/h winds feels as cold as an environment at �7�C with 3 km/hwinds as a result of the body-chilling effect of the air motion (the windchillfactor). Other factors that affect comfort are air cleanliness, odor, noise, andradiation effect.

10 AIR-CONDITIONING PROCESSESMaintaining a living space or an industrial facility at the desired temperatureand humidity requires some processes called air-conditioning processes.These processes include simple heating (raising the temperature), simple cool-ing (lowering the temperature), humidifying (adding moisture), and dehumid-ifying (removing moisture). Sometimes two or more of these processes areneeded to bring the air to a desired temperature and humidity level.

Various air-conditioning processes are illustrated on the psychrometric chartin Fig. 31. Notice that simple heating and cooling processes appear as hori-zontal lines on this chart since the moisture content of the air remains constant(v � constant) during these processes. Air is commonly heated andhumidified in winter and cooled and dehumidified in summer. Notice howthese processes appear on the psychrometric chart.

Most air-conditioning processes can be modeled as steady-flow processes,and thus the mass balance relation m· in � m· out can be expressed for dry air andwater as

Mass balance for dry air: m· a, i � m· a, e (kg/s) (37)

Mass balance for water: m· w, i � m· w, e or m· a, ivi � m· a, eve

(38)

where the subscripts i and e denote the inlet and the exit states, respectively.Disregarding the kinetic and potential energy changes, the steady-flow energybalance relation E

·in � E

·out can be expressed in this case as

Q·

in � W·

in � m· i hi � Q·

out � W·

out � m· e he (39)

The work term usually consists of the fan work input, which is small relativeto the other terms in the energy balance relation. Next we examine some com-monly encountered processes in air-conditioning.

��

��

■


Cooling

Heating

Hum

idif

ying

Deh

umid

ifyi

ng

Coolin

g an

d

dehu

mid

ifyin

g

Heatin

g an

d

hum

idify

ing

FIGURE 31Various air-conditioning processes.

cen54261_ch24.qxd 4/2/08 5:38 PM Page 20

Simple Heating and Cooling (V � constant)Many residential heating systems consist of a stove, a heat pump, or an elec-tric resistance heater. The air in these systems is heated by circulating itthrough a duct that contains the tubing for the hot gases or the electric re-sistance wires, as shown in Fig. 32. The amount of moisture in the air remainsconstant during this process since no moisture is added to or removed from theair. That is, the specific humidity of the air remains constant (v � constant)during a heating (or cooling) process with no humidification or dehumidifica-tion. Such a heating process will proceed in the direction of increasing dry-bulb temperature following a line of constant specific humidity on thepsychrometric chart, which appears as a horizontal line.

Notice that the relative humidity of air decreases during a heating processeven if the specific humidity v remains constant. This is because the relativehumidity is the ratio of the moisture content to the moisture capacity of air atthe same temperature, and moisture capacity increases with temperature.Therefore, the relative humidity of heated air may be well below comfortablelevels, causing dry skin, respiratory difficulties, and an increase in staticelectricity.

A cooling process at constant specific humidity is similar to the heatingprocess discussed above, except the dry-bulb temperature decreases and therelative humidity increases during such a process, as shown in Fig. 33. Cool-ing can be accomplished by passing the air over some coils through which arefrigerant or chilled water flows.

The conservation of mass equations for a heating or cooling process that in-volves no humidification or dehumidification reduce to m· � m· � m· a fordry air and v1 � v2 for water. Neglecting any fan work that may be present,the conservation of energy equation in this case reduces to

Q·

� m· a(h2 � h1) or q � h2 � h1

where h1 and h2 are enthalpies per unit mass of dry air at the inlet and the exitof the heating or cooling section, respectively.

Heating with HumidificationProblems associated with the low relative humidity resulting from simpleheating can be eliminated by humidifying the heated air. This is accomplishedby passing the air first through a heating section (process 1-2) and thenthrough a humidifying section (process 2-3), as shown in Fig. 34.

The location of state 3 depends on how the humidification is accomplished.If steam is introduced in the humidification section, this will result in humid-ification with additional heating (T3 T2). If humidification is accomplishedby spraying water into the airstream instead, part of the latent heat of vapori-zation will come from the air, which will result in the cooling of the heatedairstream (T3 � T2). Air should be heated to a higher temperature in the heat-ing section in this case to make up for the cooling effect during the humidifi-cation process.

a2a1

ONLINE CHAPTER 2421

φ2 = 80% φ1 = 30%

12

12°C 30°C

ω = constantCooling

FIGURE 33During simple cooling, specifichumidity remains constant, but

relative humidity increases.

Air

Heating coils

ω2 = ω1

1 2 3

Heatingsection

Humidifyingsection

ω3 > ω2

Humidifier

FIGURE 34Heating with humidification.

EXAMPLE 7 Heating and Humidification of Air

An air-conditioning system is to take in outdoor air at 10�C and 30 percent rel-ative humidity at a steady rate of 45 m3/min and to condition it to 25�C and

ω2 = ω1

Heating coils

Heat

Air

φ2 < φ1

T2

T1, ω1, φ1

FIGURE 32During simple heating, specifichumidity remains constant, but

relative humidity decreases.

cen54261_ch24.qxd 4/2/08 5:38 PM Page 21


60 percent relative humidity. The outdoor air is first heated to 22�C in the heat-ing section and then humidified by the injection of hot steam in the humidify-ing section. Assuming the entire process takes place at a pressure of 100 kPa,determine (a) the rate of heat supply in the heating section and (b) the massflow rate of the steam required in the humidifying section.

SOLUTION We will take the system to be the heating or the humidifying sec-tion, as appropriate. The schematic of the system and the psychrometric chartof the process are shown in Fig. 35. We note that the amount of water vapor inthe air remains constant in the heating section (v 1 � v 2) but increases in thehumidifying section (v3 v2).Assumptions 1 This is a steady-flow process and thus the mass flow rate of dryair remains constant during the entire process. 2 Dry air and water vapor areideal gases. 3 The kinetic and potential energy changes are negligible.Analysis (a) Applying the mass and energy balances on the heating sectiongives

Dry air mass balance: m· � m· � m· a

Water mass balance: m· v1 � m· v2 → v1 � v2

Energy: Q·

in � m· a h1 � m· a h2 → Q·

in � m· a (h2 � h1)

The psychrometric chart offers great convenience in determining the propertiesof moist air. However, its use is limited to a specified pressure only, which is1 atm (101.325 kPa) for the one given in the appendix. At pressures other than1 atm, either other charts for that pressure or the relations developed earliershould be used. In our case, the choice is clear:

a2a1

a2a1

FIGURE 35Schematic and psychrometric chartfor Example 7.

AIR

Heating coils

φ 1 = 30%

1 2 3

V1 = 45 m3/min

10°C 22°C 25°C

1 2

3

φ 1 = 30

%

φ 3= 60

%

T1 = 10°C

T2 = 22°C

T3 = 25°Cφ3 = 60%

Humidifier

cen54261_ch24.qxd 4/2/08 5:38 PM Page 22

Cooling with DehumidificationThe specific humidity of air remains constant during a simple cooling pro-cess, but its relative humidity increases. If the relative humidity reaches un-desirably high levels, it may be necessary to remove some moisture from theair, that is, to dehumidify it. This requires cooling the air below its dew-pointtemperature.

The cooling process with dehumidifying is illustrated schematically and onthe psychrometric chart in Fig. 36 in conjunction with Example 8. Hot, moistair enters the cooling section at state 1. As it passes through the cooling coils,its temperature decreases and its relative humidity increases at constant spe-

ONLINE CHAPTER 2423

P � f1P � f1Psat @ 10�C � (0.3)(1.2276 kPa) � 0.368 kPa

P � P1 � P � (100 � 0.368) kPa � 99.632 kPa

υ1 � � � 0.815 m3/kg dry air

m· a � � � 55.2 kg/min

v1 � � � 0.0023 kg H2O/kg dry air

h1 � CpT1 � v1h � (1.005 kJ/kg · �C)(10�C) � (0.0023)(2519.8 kJ/kg)


h2 � CpT2 � v2h � (1.005 kJ/kg · �C)(22�C) � (0.0023)(2541.7 kJ/kg)


since v2 � v1. Then the rate of heat transfer to air in the heating sectionbecomes

Q·

in � m· a (h2 � h1) � (55.2 kg/min)[(28.0 � 15.8) kJ/kg]

� 673 kJ/min

(b) The mass balance for water in the humidifying section can be expressed as

m· v2 � m· w � m· v3

or

m· w � m· a(v3 � v2)

where

v3 � �


Thus,

m· w � (55.2 kg/min)(0.01206 � 0.0023)

� 0.539 kg/min

0.622(0.60)(3.169 kPa)[100 � (0.60)(3.169)] kPa

0.622f3Pg3

P3 � f3Pg3

a3a2

g2

g1

0.622(0.368 kPa)(100 � 0.368) kPa

0.622Pυ1

P1 � Pυ1

45 m3/min0.815 m3/kg

V·

1

υ1

(0.287 kPa � m3/kg � K)(283 K)99.632 kPa

RaT1

Pa1

υ1a1

g1υ1

cen54261_ch24.qxd 4/2/08 5:38 PM Page 23

cific humidity. If the cooling section is sufficiently long, air will reach its dewpoint (state 2, saturated air). Further cooling of air results in the condensationof part of the moisture in the air. Air remains saturated during the entire con-densation process, which follows a line of 100 percent relative humidity untilthe final state (state 3) is reached. The water vapor that condenses out of theair during this process is removed from the cooling section through a separatechannel. The condensate is usually assumed to leave the cooling section at T3.

The cool, saturated air at state 3 is usually routed directly to the room,where it mixes with the room air. In some cases, however, the air at state 3may be at the right specific humidity but at a very low temperature. In suchcases, air is passed through a heating section where its temperature is raised toa more comfortable level before it is routed to the room.



AIR

Cooling coils

φ2 = 100%

2 1

V1 = 10 m3/min

14°C 30°C

1

2

φ1 = 80%

φ2 = 100%

T2 = 14°C T1 = 30°Cφ1 = 80%

Condensateremoval14°C

Condensate

EXAMPLE 8 Cooling and Dehumidification of Air

Air enters a window air conditioner at 1 atm, 30�C, and 80 percent relative hu-midity at a rate of 10 m3/min, and it leaves as saturated air at 14�C. Part of themoisture in the air that condenses during the process is also removed at 14�C.Determine the rates of heat and moisture removal from the air.

SOLUTION We take the cooling section to be the system. The schematic of thesystem and the psychrometric chart of the process are shown in Fig. 36.

cen54261_ch24.qxd 4/2/08 5:38 PM Page 24

Evaporative CoolingConventional cooling systems operate on a refrigeration cycle, and they can beused in any part of the world. But they have a high initial and operating cost.In desert (hot and dry) climates, we can avoid the high cost of cooling by us-ing evaporative coolers, also known as swamp coolers.

Evaporative cooling is based on a simple principle: As water evaporates, thelatent heat of vaporization is absorbed from the water body and the surround-ing air. As a result, both the water and the air are cooled during the process.

ONLINE CHAPTER 2425

We note that the amount of water vapor in the air decreases during the process(v2 � v1) due to dehumidification.Assumptions 1 This is a steady-flow process and thus the mass flow rate of dryair remains constant during the entire process. 2 Dry air and the water vapor areideal gases. 3 The kinetic and potential energy changes are negligible.Analysis Applying the mass and energy balances on the cooling and dehumid-ification section gives


Water mass balance: m· v1 � m· v2 � m· w → m· w � m· a(v1 � v2)

Energy balance: m· i hi � Q·

out � m· e he →

Q·

out � m· a(h1 � h2) � m· w hw

The inlet and the exit states of the air are completely specified, and the totalpressure is 1 atm. Therefore, we can determine the properties of the air at bothstates from the psychrometric chart to be

h1 � 85.4 kJ/kg dry air

v1 � 0.0216 kg H2O/kg dry air

υ1 � 0.889 m3/kg dry air

and



Also,

hw � hf @ 14�C � 58.8 kJ/kg (Table A–4)

Then

m· a � � � 11.25 kg/min

m· w � (11.25 kg/min)(0.0216 � 0.0100) � 0.131 kg/min

Q·

out � (11.25 kg/min)[(85.4 �39.3) kJ/kg] � (0.131 kg/min)(58.8 kJ/kg)

� 511 kJ/min

Therefore, this air-conditioning unit removes moisture and heat from the air atrates of 0.131 kg/min and 511 kJ/min, respectively.

10 m3/min0.889 m3/kg dry air

V·

1

υ1

��a2a1

a2a1

cen54261_ch24.qxd 4/2/08 5:38 PM Page 25

This approach has been used for thousands of years to cool water. A porousjug or pitcher filled with water is left in an open, shaded area. A small amountof water leaks out through the porous holes, and the pitcher “sweats.” In a dryenvironment, this water evaporates and cools the remaining water in thepitcher (Fig. 37).

You have probably noticed that on a hot, dry day the air feels a lot coolerwhen the yard is watered. This is because water absorbs heat from the air as itevaporates. An evaporative cooler works on the same principle. The evaporativecooling process is shown schematically and on a psychrometric chart in Fig. 38.Hot, dry air at state 1 enters the evaporative cooler, where it is sprayed with liq-uid water. Part of the water evaporates during this process by absorbing heatfrom the airstream. As a result, the temperature of the airstream decreases andits humidity increases (state 2). In the limiting case, the air will leave the evap-orative cooler saturated at state 2�. This is the lowest temperature that can beachieved by this process.

The evaporative cooling process is essentially identical to the adiabaticsaturation process since the heat transfer between the airstream and the sur-roundings is usually negligible. Therefore, the evaporative cooling processfollows a line of constant wet-bulb temperature on the psychrometric chart.(Note that this will not exactly be the case if the liquid water is supplied ata temperature different from the exit temperature of the airstream.) Sincethe constant-wet-bulb-temperature lines almost coincide with the constant-enthalpy lines, the enthalpy of the airstream can also be assumed to remainconstant. That is,

Twb � constant (40)

and

h � constant (41)

during an evaporative cooling process. This is a reasonably accurate approxi-mation, and it is commonly used in air-conditioning calculations.


HOT, DRYAIR

2 1

1

2

2'

COOL, MOISTAIR

Liquidwater

Twb = const. h = const.~

FIGURE 38Evaporative cooling.

EXAMPLE 9 Evaporative Cooling of Air by a Swamp Cooler

Air enters an evaporative (or swamp) cooler at 1 atm/35°C, and 20 percent rel-ative humidity, and it exits at 80 percent relative humidity. Determine (a) theexit temperature of the air and (b) the lowest temperature to which the air canbe cooled by this evaporative cooler.

SOLUTION The schematic of the evaporative cooler and the psychrometricchart of the process are shown in Fig. 39.Analysis (a) If we assume the liquid water is supplied at a temperature notmuch different from the exit temperature of the airstream, the evaporative cool-ing process follows a line of constant wet-bulb temperature on the psychro-metric chart. That is,

Twb � constant

The wet-bulb temperature at 35°C and 20 percent relative humidity is deter-mined from the psychrometric chart to be 18.9°C. The intersection point of the

Hot, dryair Water that

leaks out

FIGURE 37Water in a porous jug left in an open,breezy area cools as a resultof evaporative cooling.

cen54261_ch24.qxd 4/2/08 5:38 PM Page 26

ONLINE CHAPTER 2427

Adiabatic Mixing of AirstreamsMany air-conditioning applications require the mixing of two airstreams. Thisis particularly true for large buildings, most production and process plants,and hospitals, which require that the conditioned air be mixed with a certainfraction of fresh outside air before it is routed into the living space. The mix-ing is accomplished by simply merging the two airstreams, as shown inFig. 40.

The heat transfer with the surroundings is usually small, and thus the mix-ing processes can be assumed to be adiabatic. Mixing processes normally in-volve no work interactions, and the changes in kinetic and potential energies,if any, are negligible. Then the mass and energy balances for the adiabaticmixing of two airstreams reduce to

Mass of dry air: m· � m· � m· (42)

Mass of water vapor: v1m· � v2m· � v3m· (43)

Energy: m· h1 � m· h2 � m· h3 (44)

Eliminating m· from these relations, we obtain

� � (45)

This equation has an instructive geometric interpretation on the psychro-metric chart. It shows that the ratio of v2 � v3 to v3 � v1 is equal to the ratio of m· to m· . The states that satisfy this condition are indicated by thedashed line AB. The ratio of h2 � h3 to h3 � h1 is also equal to the ratio of m· to m· , and the states that satisfy this condition are indicated by the dashedline CD. The only state that satisfies both conditions is the intersection pointof these two dashed lines, which is located on the straight line connectingstates 1 and 2. Thus we conclude that when two airstreams at two differentstates (states 1 and 2) are mixed adiabatically, the state of the mixture (state3) will lie on the straight line connecting states 1 and 2 on the psychrometricchart, and the ratio of the distances 2-3 and 3-1 is equal to the ratio of massflow rates m· and m· .

The concave nature of the saturation curve and the conclusion above lead toan interesting possibility. When states 1 and 2 are located close to the satura-tion curve, the straight line connecting the two states will cross the saturation

a2a1

a2a1

a2a1

h2 � h3

h3 � h1

v2 � v3v3 � v1

m� a1

m� a2

a3

a3a2a1

a3a2a1

a3a2a1

Twb � 18.9°C and the f � 80 percent lines is the exit state of the air. The tem-perature at this point is the exit temperature of the air, and it is determinedfrom the psychrometric chart to be

T2 � 21.3°C

(b) In the limiting case, air will leave the evaporative cooler saturated (f � 100percent), and the exit state of the air in this case will be the state where theTwb � 18.9°C line intersects the saturation line. For saturated air, the dry- andthe wet-bulb temperatures are identical. Therefore, the lowest temperature towhich air can be cooled is the wet-bulb temperature, which is

Tmin � T2� � 18.9°C

1

2

3A

h2 – h3

h3 – h1

C

h2

h3

h1

D

ω2 – ω3

ω3 – ω1

ω2

ω3

ω1

B

2

1

3Mixingsection

ω1 h1

ω2 h2

ω 3 h3

FIGURE 40When two airstreams at states 1 and 2

are mixed adiabatically, the state ofthe mixture lies on the straight line

connecting the two states.

2' 1

1

22'

AIR

φ 2 =

80%

φ 1 =

20%

Tmin T2 35°C

2

T1 = 35°C

φ = 20%

P = 1 atm

FIGURE 39Schematic and psychrometric chart

for Example 9.

cen54261_ch24.qxd 4/2/08 5:38 PM Page 27

curve, and state 3 may lie to the left of the saturation curve. In this case, somewater will inevitably condense during the mixing process.


EXAMPLE 10 Mixing of Conditioned Air with Outdoor Air

Saturated air leaving the cooling section of an air-conditioning system at 14�Cat a rate of 50 m3/min is mixed adiabatically with the outside air at 32�C and60 percent relative humidity at a rate of 20 m3/min. Assuming that the mixingprocess occurs at a pressure of 1 atm, determine the specific humidity, therelative humidity, the dry-bulb temperature, and the volume flow rate of themixture.

SOLUTION We take the mixing section of the streams as the system. Theschematic of the system and the psychrometric chart of the process are shownin Fig. 41. We note that this is a steady-flow mixing process.

Assumptions 1 Steady operating conditions exist. 2 Dry air and water vapor areideal gases. 3 The kinetic and potential energy changes are negligible. 4 Themixing section is adiabatic.

Analysis The properties of each inlet stream are determined from the psychro-metric chart to be




and




Then the mass flow rates of dry air in each stream are

m· � � � 60.5 kg/min

m· � � � 22.5 kg/min

From the mass balance of dry air,

m· � m· � m· � (60.5 � 22.5) kg/min � 83 kg/min

The specific humidity and the enthalpy of the mixture can be determinedfrom Eq. 45,

� �

� �

which yield



79.0 � h3

h3 � 39.40.0182 � v3

v3 � 0.01060.522.5

h2 � h3

h3 � h1

v2 � v3v3 � v1

m� a1

m� a2

a2a1a3


V·

2

υ2a2


V·

1

υ1a1

T2 = 32°Cφ2 = 60%V2 = 20 m3/min

Saturated airT1 = 14°CV1 = 50 m3/min

2

1

3MixingsectionP = 1 atm

V3ω3φ3T3

231

14°C 32°C

φ 2 =

60%

φ 1= 1

00%


cen54261_ch24.qxd 4/2/08 5:38 PM Page 28

Wet Cooling TowersPower plants, large air-conditioning systems, and some industries generatelarge quantities of waste heat that is often rejected to cooling water fromnearby lakes or rivers. In some cases, however, the water supply is limited orthermal pollution is a serious concern. In such cases, the waste heat must berejected to the atmosphere, with cooling water recirculating and serving as atransport medium for heat transfer between the source and the sink (the atmo-sphere). One way of achieving this is through the use of wet cooling towers.

A wet cooling tower is essentially a semienclosed evaporative cooler. Aninduced-draft counterflow wet cooling tower is shown schematically in Fig.42. Air is drawn into the tower from the bottom and leaves through the top.Warm water from the condenser is pumped to the top of the tower and issprayed into this airstream. The purpose of spraying is to expose a large sur-face area of water to the air. As the water droplets fall under the influence ofgravity, a small fraction of water (usually a few percent) evaporates and coolsthe remaining water. The temperature and the moisture content of the air in-crease during this process. The cooled water collects at the bottom of thetower and is pumped back to the condenser to pick up additional waste heat.Makeup water must be added to the cycle to replace the water lost by evapo-ration and air draft. To minimize water carried away by the air, drift elimina-tors are installed in the wet cooling towers above the spray section.

The air circulation in the cooling tower just described is provided by a fan,and therefore it is classified as a forced-draft cooling tower. Another populartype of cooling tower is the natural-draft cooling tower, which looks like alarge chimney and works like an ordinary chimney. The air in the tower has ahigh water-vapor content, and thus it is lighter than the outside air. Conse-quently, the light air in the tower rises, and the heavier outside air fills the va-cant space, creating an airflow from the bottom of the tower to the top. Theflow rate of air is controlled by the conditions of the atmospheric air. Natural-draft cooling towers do not require any external power to induce the air, butthey cost a lot more to build than forced-draft cooling towers. The natural-draft cooling towers are hyperbolic in profile, as shown in Fig. 43, and someare over 100 m high. The hyperbolic profile is for greater structural strength,not for any thermodynamic reason.

ONLINE CHAPTER 2429

These two properties fix the state of the mixture. Other properties of the mixtureare determined from the psychrometric chart:

T3 � 19.0�C

f3 � 89%


Finally, the volume flow rate of the mixture is determined from

V·

3 � m· υ3 � (83 kg/min)(0.844 m3/kg) � 70.1 m3/min

Discussion Notice that the volume flow rate of the mixture is approximatelyequal to the sum of the volume flow rates of the two incoming streams. This istypical in air-conditioning applications.

a3

COOL

WATER

AIR EXIT

WARM

WATER

AIRINLET

FAN

FIGURE 42An induced-draft counterflow

cooling tower.

WARM

WATER

COOL

WATER

AIR EXIT

AIRINLET

FIGURE 43A natural-draft cooling tower.

cen54261_ch24.qxd 4/2/08 5:38 PM Page 29

The idea of a cooling tower started with the spray pond, where the warmwater is sprayed into the air and is cooled by the air as it falls into the pond, asshown in Fig. 44. Some spray ponds are still in use today. However, they re-quire 25 to 50 times the area of a cooling tower, water loss due to air drift ishigh, and they are unprotected against dust and dirt.

We could also dump the waste heat into a still cooling pond, which isbasically a large artificial lake open to the atmosphere. Heat transfer fromthe pond surface to the atmosphere is very slow, however, and we wouldneed about 20 times the area of a spray pond in this case to achieve the samecooling.


FIGURE 44A spray pond (photo by YunusÇengel).

COOLWATER

30°Cφ2 =100%

WARMWATER

AIR

35°C100 kg/s

100 kg/s

Systemboundary

4

1

3

2

1 atm20°Cφ1 = 60%V1

Makeup water

22°C


EXAMPLE 11 Cooling of a Power Plant by a Cooling Tower

Cooling water leaves the condenser of a power plant and enters a wet coolingtower at 35�C at a rate of 100 kg/s. The water is cooled to 22�C in the coolingtower by air that enters the tower at 1 atm, 20�C, and 60 percent relativehumidity and leaves saturated at 30�C. Neglecting the power input to the fan,determine (a) the volume flow rate of air into the cooling tower and (b) the massflow rate of the required makeup water.

SOLUTION We take the entire cooling tower to be the system, which is shownschematically in Fig. 45. We note that the mass flow rate of liquid water de-creases by an amount equal to the amount of water that vaporizes in the towerduring the cooling process. The water lost through evaporation must be madeup later in the cycle to maintain steady operation.Assumptions 1 Steady operating conditions exist and thus the mass flow rateof dry air remains constant during the entire process. 2 Dry air and the water va-por are ideal gases. 3 The kinetic and potential energy changes are negligible.4 The cooling tower is adiabatic.Analysis Applying the mass and energy balances on the cooling tower gives


Water mass balance: m· 3 � m· v1 � m· 4 � m· v2

or m· 3 � m· 4 � m· a (v2 � v1) � m· makeup

Energy balance: m· i hi � m· e he → m· h1 � m· 3 h3 � m· h2 � m· 4 h4

or

m· 3 h3 � m· a (h2 � h1) � (m· 3 � m· makeup)h4

Solving for m· a gives

m· a �

From the psychrometric chart,




m· 3(h3 � h4)

(h2 � h1) � (v2 � v1) h4

a2a1��

a2a1

a2a1

cen54261_ch24.qxd 4/2/08 5:38 PM Page 30

ONLINE CHAPTER 2431

and



From Table A–4, h3 � hf @ 35�C � 146.68 kJ/kg H2O

h4 � hf @ 22�C � 92.33 kJ/kg H2O

Substituting,

m· a � � 96.9 kg/s

Then the volume flow rate of air into the cooling tower becomes

V·

1 � m· aυ1 � (96.9 kg/s)(0.842 m3/kg) � 81.6 m3/s

(b) The mass flow rate of the required makeup water is determined from

m· makeup � m· a (v2 � v1) � (96.9 kg/s)(0.0273 � 0.0087) � 1.80 kg/s

Discussion Note that over 98 percent of the cooling water is saved and recir-culated in this case.

(100 kg/s)[(146.68 � 92.33) kJ/kg][(100.0 � 42.2) kJ/kg] � [(0.0273 � 0.0087)(92.33) kJ/kg]

SUMMARY

A mixture of two or more gases of fixed chemical compositionis called a nonreacting gas mixture. The composition of a gasmixture is described by specifying either the mole fraction orthe mass fraction of each component, defined as

mfi � and yi �

where

mm � mi and Nm � Ni

The apparent (or average) molar mass and gas constant of amixture are expressed as

Mm � � yiMi and Rm �

Also,

mfi � yi and Mm �

Dalton’s law of additive pressures states that the pressure ofa gas mixture is equal to the sum of the pressures each gaswould exert if it existed alone at the mixture temperatureand volume. Amagat’s law of additive volumes states that the

volume of a gas mixture is equal to the sum of the volumeseach gas would occupy if it existed alone at the mixture tem-perature and pressure. Dalton’s and Amagat’s laws hold exactlyfor ideal-gas mixtures, but only approximately for real-gasmixtures. They can be expressed as

Dalton’s law: Pm � Pi(Tm, Vm)

Amagat’s law: Vm � Vi (Tm, Pm)

Here Pi is called the component pressure and Vi is called thecomponent volume. Also, the ratio Pi /Pm is called the pressurefraction and the ratio Vi /Vm is called the volume fraction ofcomponent i. For ideal gases, Pi and Vi can be related to yi by

The quantity yiPm is called the partial pressure and the quantityyiVm is called the partial volume. The P-υ-T behavior of real-gas mixtures can be predicted by using generalized compress-ibility charts. The compressibility factor of the mixture can beexpressed in terms of the compressibility factors of the indi-vidual gases as

Zm � yiZi�k

i � 1

Pi

Pm�

Vi

Vm�

Ni

Nm� yi

�k

i � 1

�k

i � 1

1

�k

i�1

mfi

Mi

Mi

Mm

Ru

Mm�

k

i � 1

mm

Nm

�k

i � 1�

k

i � 1

Ni

Nm

mi

mm

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where Zi is determined either at Tm and Vm (Dalton’s law) or atTm and Pm (Amagat’s law) for each individual gas.

The extensive properties of a gas mixture, in general, can bedetermined by summing the contributions of each componentof the mixture. The evaluation of intensive properties of a gasmixture, however, involves averaging in terms of mass or molefractions:

Um � Ui � miui � Ni i (kJ)

Hm � Hi � mihi � Ni i (kJ)

Sm � Si � misi � Ni i (kJ/K)

and

um � mfiui and m � yi i (kJ/kg orkJ/kmol)

hm � mfihi and m � yi i (kJ/kg orkJ/kmol)

sm � mfisi and m � yi i

[kJ/(kg � K) or kJ(kmol � K)]

Cυ, m � mfiCυ, i and υ, m � yi υ, i

Cp, m � mfiCp, i and p, m � yi p, i

These relations are exact for ideal-gas mixtures, and approxi-mate for real-gas mixtures. The properties or property changesof individual components can be determined by using ideal-gasor real-gas relations developed in earlier chapters.

The air in the atmosphere normally contains some water va-por, and it is referred to as atmospheric air. By contrast, air thatcontains no water vapor is called dry air. In the temperaturerange encountered in air-conditioning applications, both thedry air and the water vapor can be treated as ideal gases. Theenthalpy change of dry air during a process can be determinedfrom

�hdry air � Cp �T � [1.005 kJ/(kg · �C)] �T (kJ/kg)

The atmospheric air can be treated as an ideal-gas mixturewhose pressure is the sum of the partial pressure of dry air Pa

and that of the water vapor Pυ,

P � Pa � Pυ (kPa)

The enthalpy of water vapor in the air can be taken to be equalto the enthalpy of the saturated vapor at the same temperature:

hυ(T, low P) � hg(T) � 2501.3 � 1.82T (kJ/kg) T in �C

� 1061.5 � 0.435T (Btu/lbm) T in �F

in the temperature range �10 to 50�C (15 to 120�F).The mass of water vapor present in 1 unit mass of dry air is

called the specific or absolute humidity v,

v � � (kg H2O/kg dry air)

where P is the total pressure of air and Pυ is the vapor pressure.There is a limit on the amount of vapor the air can hold at agiven temperature. Air that is holding as much moisture as itcan at a given temperature is called saturated air. The ratio ofthe amount of moisture air holds (mυ) to the maximum amountof moisture air can hold at the same temperature (mg) is calledthe relative humidity f,

f � � �

where Pg � Psat @ T. The relative and specific humidities canalso be expressed as

f� and v �

Relative humidity ranges from 0 for dry air to 1 for satu-rated air.

The enthalpy of atmospheric air is expressed per unit massof dry air, instead of per unit mass of the air–water-vapormixture, as

h � ha � vhg (kJ/kg dry air)

The ordinary temperature of atmospheric air is referred to asthe dry-bulb temperature to differentiate it from other forms oftemperatures. The temperature at which condensation begins ifthe air is cooled at constant pressure is called the dew-pointtemperature Tdp:

Tdp � Tsat @

Relative humidity and specific humidity of air can be deter-mined by measuring the adiabatic saturation temperature ofair, which is the temperature air attains after flowing over waterin a long adiabatic channel until it is saturated,

v1 �

where

v2 �0.622Pg2

P2 � Pg2


hg1� hf2

Pυ

0.622fPg

P � fPg

vP(0.622 � v)Pg

Pυ

Pg

PυV/(RυT)

PgV/(RυT)

mυ

mg

0.622Pυ

P � Pυ

mυ

ma

C�k

i � 1

C�k

i � 1

C�k

i � 1

C�k

i � 1

s�k

i � 1

s�k

i � 1

h�k

i � 1

h�k

i � 1

u�k

i � 1

u�k

i � 1

s�k

i � 1�

k

i � 1�

k

i � 1

h�k

i � 1�

k

i � 1�

k

i � 1

u�k

i � 1�

k

i � 1�

k

i � 1

cen54261_ch24.qxd 4/2/08 5:38 PM Page 32

ONLINE CHAPTER 2433

and T2 is the adiabatic saturation temperature. A more practicalapproach in air-conditioning applications is to use a ther-mometer whose bulb is covered with a cotton wick saturatedwith water and to blow air over the wick. The temperature mea-sured in this manner is called the wet-bulb temperature Twb,and it is used in place of the adiabatic saturation temperature.The properties of atmospheric air at a specified total pressureare presented in the form of easily readable charts, called psy-chrometric charts. The lines of constant enthalpy and the linesof constant wet-bulb temperature are very nearly parallel onthese charts.

Most air-conditioning processes can be modeled as steady-flow processes, and therefore they can be analyzed by applying

the steady-flow mass (for both dry air and water) and energybalances,

Dry air mass: m· a, i � m· a, e

Water mass: m· w, i � m· w, e

or m· a, ivi � m· a, eve

Energy: Q·

in � W·

in � m· ihi � Q·

out � W·

out � m· ehe

where subscripts i and e denote inlet and exit states, respec-tively. The changes in kinetic and potential energies are as-sumed to be negligible.

��

REFERENCES AND SUGGESTED READINGS

1. ASHRAE. 1981 Handbook of Fundamentals. Atlanta,GA: American Society of Heating, Refrigerating, and Air-Conditioning Engineers, 1981.

2. W. Z. Black and J. G. Hartley. Thermodynamics. NewYork: Harper & Row, 1985.

3. S. M. Elonka. “Cooling Towers.” Power, March 1963.

4. D. C. Look, Jr. and H. J. Sauer, Jr. Engineering Thermo-dynamics. Boston: PWS Engineering, 1986.

5. W. F. Stoecker and J. W. Jones. Refrigeration and AirConditioning. 2nd ed. New York: McGraw-Hill, 1982.

6. K. Wark and D. E. Richards. Thermodynamics. 6th ed.New York: McGraw-Hill, 1999.

7. L. D. Winiarski and B. A. Tichenor. “Model of NaturalDraft Cooling Tower Performance.” Journal of theSanitary Engineering Division, Proceedings of theAmerican Society of Civil Engineers, August 1970.

PROBLEMS*

Composition of Gas Mixtures

1C What is the apparent gas constant for a gas mixture? Canit be larger than the largest gas constant in the mixture?

2C Consider a mixture of two gases. Can the apparent molarmass of this mixture be determined by simply taking the arith-metic average of the molar masses of the individual gases?When will this be the case?

3C What is the apparent molar mass for a gas mixture? Doesthe mass of every molecule in the mixture equal the apparentmolar mass?

4C Consider a mixture of several gases of identical masses.Will all the mass fractions be identical? How about the molefractions?

5C The sum of the mole fractions for an ideal-gas mixture isequal to 1. Is this also true for a real-gas mixture?

6C What are mass and mole fractions?

7C Using the definitions of mass and mole fractions, derive arelation between them.

8C Somebody claims that the mass and mole fractions for amixture of CO2 and N2O gases are identical. Is this true? Why?

9C Consider a mixture of two gases A and B. Show thatwhen the mass fractions mfA and mfB are known, the mole frac-tions can be determined from

yA � and yB � 1 � yA

where MA and MB are the molar masses of A and B.

10 The composition of moist air is given on a molar basis tobe 78 percent N2, 20 percent O2, and 2 percent water vapor. De-termine the mass fractions of the constituents of air.

11 A gas mixture has the following composition on a molebasis: 60 percent N2 and 40 percent CO2. Determine the gravi-metric analysis of the mixture, its molar mass, and gas con-stant.

MB

MA(1/mfA � 1) � MB

*Problems designated by a “C” are concept questions, andstudents are encouraged to answer them all. Problems with a CD-EES icon are solved using EES, and complete solutions togetherwith parametric studies are included on the enclosed CD. Problemswith a computer-EES icon are comprehensive in nature, and areintended to be solved with a computer, preferably using the EESsoftware that accompanies this text.

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12 Repeat Prob. 11 by replacing N2 by O2.

13 A gas mixture consists of 5 kg of O2, 8 kg of N2, and 10kg of CO2. Determine (a) the mass fraction of each component,(b) the mole fraction of each component, and (c) the averagemolar mass and gas constant of the mixture.

14 Determine the mole fractions of a gas mixture that con-sists of 60 percent CH4 and 40 percent CO2 by mass. Also, de-termine the gas constant of the mixture.

15 A gas mixture consists of 8 kmol of H2 and 2 kmol of N2.Determine the mass of each gas and the apparent gas constantof the mixture. Answers: 16 kg, 56 kg, 1.155 kJ/kg · K

P-v-T Behavior of Gas Mixtures

16C Is a mixture of ideal gases also an ideal gas? Give an ex-ample.

17C Express Dalton’s law of additive pressures. Does thislaw hold exactly for ideal-gas mixtures? How about nonideal-gas mixtures?

18C Express Amagat’s law of additive volumes. Does thislaw hold exactly for ideal-gas mixtures? How about nonideal-gas mixtures?

19C How is the P-υ-T behavior of a component in an ideal-gas mixture expressed? How is the P-υ-T behavior of a com-ponent in a real-gas mixture expressed?

20C What is the difference between the component pressureand the partial pressure? When are these two equivalent?

21C What is the difference between the component volumeand the partial volume? When are these two equivalent?

22C In a gas mixture, which component will have the higherpartial pressure—the one with the higher mole number or theone with the larger molar mass?

23C Consider a rigid tank that contains a mixture of twoideal gases. A valve is opened and some gas escapes. As a re-sult, the pressure in the tank drops. Will the partial pressure ofeach component change? How about the pressure fraction ofeach component?

24C Consider a rigid tank that contains a mixture of twoideal gases. The gas mixture is heated, and the pressure andtemperature in the tank rise. Will the partial pressure of eachcomponent change? How about the pressure fraction of eachcomponent?

25C Is this statement correct? The volume of an ideal-gasmixture is equal to the sum of the volumes of each individualgas in the mixture. If not, how would you correct it?

26C Is this statement correct? The temperature of an ideal-gasmixture is equal to the sum of the temperatures of each individ-ual gas in the mixture. If not, how would you correct it?

27C Is this statement correct? The pressure of an ideal-gasmixture is equal to the sum of the partial pressures of each in-dividual gas in the mixture. If not, how would you correct it?

28C Explain how a real-gas mixture can be treated as apseudopure substance using Kay’s rule.

29 A rigid tank contains 8 kmol of O2 and 10 kmol of CO2

gases at 290 K and 150 kPa. Estimate the volume of the tank.Answer: 289 m3

30 Repeat Prob. 29 for a temperature of 350 K.

31 A rigid tank contains 0.5 kmol of Ar and 2 kmol of N2 at250 kPa and 280 K. The mixture is now heated to 400 K. De-termine the volume of the tank and the final pressure of themixture.

32 A gas mixture at 300 K and 200 kPa consists of 1 kg ofCO2 and 3 kg of CH4. Determine the partial pressure of eachgas and the apparent molar mass of the gas mixture.

33 A 0.3-m3 rigid tank contains 0.6 kg of N2, and 0.4 kg ofO2 at 300 K. Determine the partial pressure of each gas and thetotal pressure of the mixture.Answers: 178.1 kPa, 103.9 kPa, 282.0 kPa

34 A gas mixture at 350 K and 300 kPa has the followingvolumetric analysis: 65 percent N2, 20 percent O2, and 15 per-cent CO2. Determine the mass fraction and partial pressure ofeach gas.

35 A rigid tank that contains 2 kg of N2 at 25°C and 200 kPais connected to another rigid tank that contains 3 kg of O2 at25°C and 500 kPa. The valve connecting the two tanks isopened, and the two gases are allowed to mix. If the final mix-ture temperature is 25°C, determine the volume of each tankand the final mixture pressure.Answers: 0.884 m3, 0.465 m3, 303.4 kPa

Properties of Gas Mixtures

36C Is the total internal energy of an ideal-gas mixture equalto the sum of the internal energies of each individual gas in themixture? Answer the same question for a real-gas mixture.

37C Is the specific internal energy of a gas mixture equal tothe sum of the specific internal energy of each individual gas inthe mixture?

38C Answer Probs. 36C and 37C for entropy.

39C Is the total internal energy change of an ideal-gas mix-ture equal to the sum of the internal energy changes of each in-dividual gas in the mixture? Answer the same question for areal-gas mixture.

O2

3 kg25°C

500 kPa

N2

2 kg25°C

200 kPa

FIGURE P35

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ONLINE CHAPTER 2435

40C When evaluating the entropy change of the componentsof an ideal-gas mixture, do we have to use the partial pressure ofeach component or the total pressure of the mixture?

41C Suppose we want to determine the enthalpy change of areal-gas mixture undergoing a process. The enthalpy change ofeach individual gas is determined by using the generalized enthalpy chart, and the enthalpy change of the mixture is determined by summing them. Is this an exact approach?Explain.

42 An insulated rigid tank is divided into two compartmentsby a partition. One compartment contains 0.5 kmol of CO2 at27°C and 200 kPa, and the other compartment contains 7.5kmol of H2 gas at 40°C and 400 kPa. Now the partition is re-moved, and the two gases are allowed to mix. Determine (a)the mixture temperature and (b) the mixture pressure afterequilibrium has been established. Assume constant specificheats at room temperature for both gases.

43 A 0.9-m3 rigid tank is divided into two equal compart-ments by a partition. One compartment contains Ne at 20°Cand 100 kPa, and the other compartment contains Ar at 50°Cand 200 kPa. Now the partition is removed, and the two gasesare allowed to mix. Heat is lost to the surrounding air duringthis process in the amount of 15 kJ. Determine (a) the finalmixture temperature and (b) the final mixture pressure.Answers: (a) 16.2˚C, (b) 138.9 kPa

44 Repeat Prob. 43 for a heat loss of 8 kJ.

45 An equimolar mixture of helium and argon gases is to beused as the working fluid in a closed-loop gas-turbine cycle.The mixture enters the turbine at 1.2 MPa and 1300 K and ex-pands isentropically to a pressure of 200 kPa. Determine thework output of the turbine per unit mass of the mixture.

46 A mixture of 80 percent N2 and 20 percent CO2

gases (on a mass basis) enters the nozzle of a turbo-jet engine at 600 kPa and 1000 K with a low velocity, and it ex-pands to a pressure of 85 kPa. If the isentropic efficiency of thenozzle is 92 percent, determine (a) the exit temperature and (b)the exit velocity of the mixture. Assume constant specific heatsat room temperature.

47 Reconsider Prob. 46. Using EES (or other) software,first solve the stated problem and then, for all other

conditions being the same, resolve the problem to deter-mine the composition of the nitrogen and carbon dioxide that is required to have an exit velocity of 800 m/s at the nozzleexit.

48 A piston-cylinder device contains a mixture of 0.2 kg ofH2 and 1.6 kg of N2 at 100 kPa and 300 K. Heat is now trans-ferred to the mixture at constant pressure until the volume isdoubled. Assuming constant specific heats at the average tem-perature, determine (a) the heat transfer and (b) the entropychange of the mixture.

49 An insulated tank that contains 1 kg of O2 at 15°C and300 kPa is connected to a 2-m3 uninsulated tank that containsN2 at 50°C and 500 kPa. The valve connecting the two tanks isopened, and the two gases form a homogeneous mixture at25°C. Determine (a) the final pressure in the tank, and (b) theheat transfer.Answers: (a) 444.6 kPa, (b) 187.2 kJ loss

50 Reconsider Prob. 49. Using EES (or other) software,compare the results obtained assuming ideal-gas be-

havior with constant specific heats at the average temperature,and using real-gas data obtained from EES by assuming vari-able specific heats over the temperature range.

Dry and Atmospheric Air: Specific and Relative Humidity

51C Is it possible to obtain saturated air from unsaturated airwithout adding any moisture? Explain.

52C Is the relative humidity of saturated air necessarily 100percent?

53C Moist air is passed through a cooling section where it iscooled and dehumidified. How do (a) the specific humidity and(b) the relative humidity of air change during this process?

54C What is the difference between dry air and atmosphericair?

55C Can the water vapor in air be treated as an ideal gas? Ex-plain.

56C What is vapor pressure?

57C How would you compare the enthalpy of water vapor at20�C and 2 kPa with the enthalpy of water vapor at 20�C and0.5 kPa?

H2

7.5 kmol40°C

400 kPa

CO2

0.5 kmol27°C

200 kPa

FIGURE P42

1.2 MPa1300 K

200 kPa

He-Arturbine

FIGURE P45

cen54261_ch24.qxd 4/2/08 5:38 PM Page 35


58C What is the difference between the specific humidityand the relative humidity?

59C How will (a) the specific humidity and (b) the relativehumidity of the air contained in a well-sealed room change asit is heated?

60C How will (a) the specific humidity and (b) the relativehumidity of the air contained in a well-sealed room change asit is cooled?

61C Consider a tank that contains moist air at 3 atm andwhose walls are permeable to water vapor. The surrounding airat 1 atm pressure also contains some moisture. Is it possible forthe water vapor to flow into the tank from surroundings?Explain.

62C Why are the chilled water lines always wrapped withvapor barrier jackets?

63C Explain how vapor pressure of the ambient air is deter-mined when the temperature, total pressure, and the relativehumidity of air are given.

64 A tank contains 21 kg of dry air and 0.3 kg of water vaporat 30�C and 100 kPa total pressure. Determine (a) the specifichumidity, (b) the relative humidity, and (c) the volume of thetank.

65 Repeat Prob. 64 for a temperature of 35�C.

66 A room contains air at 20�C and 98 kPa at a relative hu-midity of 85 percent. Determine (a) the partial pressure of dryair, (b) the specific humidity of the air, and (c) the enthalpy perunit mass of dry air.

67 Repeat Prob. 66 for a pressure of 85 kPa.

68 Determine the masses of dry air and the water vapor con-tained in a 240-m3 room at 98 kPa, 23�C, and 50 percent rela-tive humidity. Answers: 273 kg, 2.5 kg

Dew-Point, Adiabatic Saturation,and Wet-Bulb Temperatures69C What is the dew-point temperature?

70C Andy and Wendy both wear glasses. On a cold winterday, Andy comes from the cold outside and enters the warmhouse while Wendy leaves the house and goes outside. Whoseglasses are more likely to be fogged? Explain.

71C In summer, the outer surface of a glass filled with iced water frequently “sweats.” How can you explain thissweating?

72C In some climates, cleaning the ice off the windshield ofa car is a common chore on winter mornings. Explain how iceforms on the windshield during some nights even when there isno rain or snow.

73C When are the dry-bulb and dew-point temperaturesidentical?

74C When are the adiabatic saturation and wet-bulb temper-atures equivalent for atmospheric air?

75 A house contains air at 25�C and 65 percent relative humidity. Will any moisture condense on the inner surfaces ofthe windows when the temperature of the window drops to 10�C?

76 After a long walk in the 8�C outdoors, a person wearingglasses enters a room at 25�C and 40 percent relative humidity.Determine whether the glasses will become fogged.

77 Repeat Prob. 76 for a relative humidity of 70 percent.

78 A thirsty woman opens the refrigerator and picks up acool canned drink at 5°C. Do you think the can will “sweat” asshe enjoys the drink in a room at 25°C and 50 percent relativehumidity?

79 The dry- and wet-bulb temperatures of atmospheric air at95 kPa are 25 and 20�C, respectively. Determine (a) the spe-cific humidity, (b) the relative humidity, and (c) the enthalpy ofthe air, in kJ/kg dry air.

80 The air in a room has a dry-bulb temperature of 22°C anda wet-bulb temperature of 16�C. Assuming a pressure of 100kPa, determine (a) the specific humidity, (b) the relative hu-midity, and (c) the dew-point temperature.Answers: (a) 0.0091 kg H2O/kg dry air, (b) 54.0 percent, (c) 12.3�C

81 Reconsider Prob. 80. Determine the required proper-ties using EES (or other) software. What would the

property values be at a pressure of 300 kPa?

Psychrometric Chart

82C How do constant-enthalpy and constant-wet-bulb-temperature lines compare on the psychrometric chart?

83C At what states on the psychrometric chart are the dry-bulb, wet-bulb, and dew-point temperatures identical?

84C How is the dew-point temperature at a specified statedetermined on the psychrometric chart?

85C Can the enthalpy values determined from a psychromet-ric chart at sea level be used at higher elevations?

86 The air in a room is at 1 atm, 32°C, and 60 percentrelative humidity. Determine (a) the specific humid-

ity, (b) the enthalpy (in kJ/kg dry air), (c) the wet-bulb temper-ature, (d ) the dew-point temperature, and (e) the specificvolume of the air (in m3/kg dry air). Use the psychrometricchart or available software.

87 Reconsider Prob. 86. Determine the required proper-ties using EES (or other) software instead of the py-

chrometric chart. What would the property values be at alocation at 1500 m altitude?

88 A room contains air at 1 atm, 26°C, and 70 percent rela-tive humidity. Using the psychrometric chart, determine (a) the

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ONLINE CHAPTER 2437

specific humidity, (b) the enthalpy (in kJ/kg dry air), (c) thewet-bulb temperature, (d ) the dew-point temperature, and (e) the specific volume of the air (in m3/kg dry air).

89 Reconsider Prob. 88. Determine the requiredproperties using EES (or other) software instead of

the pychrometric chart. What would the property values be at a location at 2000 m altitude?

90 The air in a room has a pressure of 1 atm, a dry-bulbtemperature of 24�C, and a wet-bulb temperature of 17�C.Using the psychrometric chart, determine (a) the specifichumidity, (b) the enthalpy (in kJ/kg dry air), (c) the relativehumidity, (d ) the dew-point temperature, and (e) the specificvolume of the air (in m3/kg dry air).

91 Reconsider Prob. 90. Determine the requiredproperties using EES (or other) software instead of

the pychrometric chart. What would the property values be at alocation at 3000 m altitude?

Human Comfort and Air-Conditioning

92C What does a modern air-conditioning system do besidesheating or cooling the air?

93C How does the human body respond to (a) hot weather,(b) cold weather, and (c) hot and humid weather?

94C What is the radiation effect? How does it affect humancomfort?

95C How does the air motion in the vicinity of the humanbody affect human comfort?

96C Consider a tennis match in cold weather where bothplayers and spectators wear the same clothes. Which group ofpeople will feel colder? Why?

97C Why do you think little babies are more susceptible tocold?

98C How does humidity affect human comfort?

99C What are humidification and dehumidification?

100C What is metabolism? What is the range of metabolicrate for an average man? Why are we interested in themetabolic rate of the occupants of a building when we dealwith heating and air-conditioning?

101C Why is the metabolic rate of women, in general, lowerthan that of men? What is the effect of clothing on theenvironmental temperature that feels comfortable?

102C What is sensible heat? How is the sensible heat lossfrom a human body affected by the (a) skin temperature,(b) environment temperature, and (c) air motion?

103C What is latent heat? How is the latent heat loss from the human body affected by the (a) skin wettedness and (b) relative humidity of the environment? How is the

rate of evaporation from the body related to the rate of latentheat loss?

104 An average person produces 0.25 kg of moisture whiletaking a shower and 0.05 kg while bathing in a tub. Consider afamily of four who each shower once a day in a bathroom thatis not ventilated. Taking the heat of vaporization of water to be2450 kJ/kg, determine the contribution of showers to the latentheat load of the air conditioner per day in summer.

105 An average (1.82 kg) chicken has a basal metabolic rate of 5.47 W and an average metabolic rate of 10.2 W (3.78 W sensible and 6.42 W latent) during normal activity.If there are 100 chickens in a breeding room, determine the rate of total heat generation and the rate of moisture pro-duction in the room. Take the heat of vaporization of water to be 2430 kJ/kg.

106 A department store expects to have 120 customers and15 employees at peak times in summer. Determine the contri-bution of people to the total cooling load of the store.

107 In a movie theater in winter, 500 people, each generatingheat at a rate of 100 W, are watching a movie. The heat lossesthrough the walls, windows, and the roof are estimated to be140,000 kJ/h. Determine if the theater needs to be heated orcooled.

108 For an infiltration rate of 1.2 air changes per hour(ACH), determine sensible, latent, and total infiltration heatload of a building at sea level, in kW, that is 20 m long, 13 mwide, and 3 m high when the outdoor air is at 32�C and 50 per-cent relative humidity. The building is maintained at 24�C and50 percent relative humidity at all times.

109 Repeat Prob. 108 for an infiltration rate of 1.8 ACH.

Simple Heating and Cooling

110C How do relative and specific humidities change duringa simple heating process? Answer the same question for a sim-ple cooling process.

111C Why does a simple heating or cooling process appearas a horizontal line on the psychrometric chart?

112 Air enters a heating section at 95 kPa, 15�C, and 30 per-cent relative humidity at a rate of 4 m3/min, and it leaves at25�C. Determine (a) the rate of heat transfer in the heating sec-tion and (b) the relative humidity of the air at the exit.Answers: (a) 46.2 kJ/min, (b) 16.1 percent

113 A heating section consists of a 38-cm-diameter duct thathouses a 4-kW electric resistance heater. Air enters the heatingsection at 1 atm, 10°C, and 40 percent relative humidity at avelocity of 8 m/s. Determine (a) the exit temperature, (b) theexit relative humidity of the air, and (c) the exit velocity.Answers: (a) 13.5°C, (b) 31.7 percent, (c) 8.1 m/s

114 Air enters a 40-cm-diameter cooling section at 1 atm,32�C, and 30 percent relative humidity at 18 m/s. Heat is

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removed from the air at a rate of 1200 kJ/min. Determine(a) the exit temperature, (b) the exit relative humidity of the air,and (c) the exit velocity.Answers: (a) 24.4�C, (b) 46.6 percent, (c) 17.6 m/s

115 Repeat Prob. 114 for a heat removal rate of 800 kJ/min.

Heating with Humidification

116C Why is heated air sometimes humidified?

117 Air at 1 atm, 15�C, and 60 percent relative humidity isfirst heated to 20�C in a heating section and then humidified byintroducing water vapor. The air leaves the humidifying sectionat 25�C and 65 percent relative humidity. Determine (a) theamount of steam added to the air, and (b) the amount of heattransfer to the air in the heating section.Answers: (a) 0.0065 kg H2O/kg dry air, (b) 5.1 kJ/kg dry air

118 An air-conditioning system operates at a total pressure of1 atm and consists of a heating section and a humidifier thatsupplies wet steam (saturated water vapor) at 100�C. Air entersthe heating section at 10�C and 70 percent relative humidity ata rate of 35 m3/min, and it leaves the humidifying section at20�C and 60 percent relative humidity. Determine (a) the tem-perature and relative humidity of air when it leaves the heatingsection, (b) the rate of heat transfer in the heating section, and(c) the rate at which water is added to the air in the humidify-ing section.

119 Repeat Prob. 118 for a total pressure of 95 kPa for theairstream.Answers: (a) 19.5�C, 37.7 percent, (b) 391 kJ/min,(c) 0.147 kg/min

Cooling with Dehumidification

120C Why is cooled air sometimes reheated in summerbefore it is discharged to a room?

121 Air enters a window air conditioner at 1 atm, 32°C, and70 percent relative humidity at a rate of 3 m3/min, and it leavesas saturated air at 12�C. Part of the moisture in the air that con-denses during the process is also removed at 12�C. Determinethe rates of heat and moisture removal from the air.Answers: 170 kJ/min, 0.112 kg/min

122 An air-conditioning system is to take in air at 1 atm,34�C, and 70 percent relative humidity and deliver it at 22�Cand 50 percent relative humidity. The air flows first over thecooling coils, where it is cooled and dehumidified, and thenover the resistance heating wires, where it is heated to thedesired temperature. Assuming that the condensate is removedfrom the cooling section at 10�C, determine (a) the temperatureof air before it enters the heating section, (b) the amount of heatremoved in the cooling section, and (c) the amount of heattransferred in the heating section, both in kJ/kg dry air.

123 Air enters a 30-cm-diameter cooling section at1 atm, 35�C, and 60 percent relative humidity at 120

m/min. The air is cooled by passing it over a cooling coilthrough which cold water flows. The water experiences atemperature rise of 8�C. The air leaves the cooling sectionsaturated at 20�C. Determine (a) the rate of heat transfer, (b)the mass flow rate of the water, and (c) the exit velocity of theairstream.

124 Reconsider Prob. 123. Using EES (or other) soft-ware, develop a general solution of the problem in

which the input variables may be supplied and parametric stud-ies performed. For each set of input variables for which thepressure is atmospheric, show the process on the psychromet-ric chart.

125 Repeat Prob. 123 for a total pressure of 95 kPa for air.Answers: (a) 293.3 kJ/min, (b) 8.77 kg/min, (c) 113 m/min

1200 kJ/min

1 atm

AIR32°C, 30%

18 m/s

FIGURE P114

AIR

Heating coils Humidifier

10°C70%35 m3/min P = 1 atm

20°C60%

Sat. vapor100°C

FIGURE P118

Cooling coils

AIR

T + 8°CT

20°CSaturated

35°C60%120 m/min

Water

FIGURE P123

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ONLINE CHAPTER 2439

Evaporative Cooling

126C Does an evaporation process have to involve heat trans-fer? Describe a process that involves both heat and mass trans-fer.

127C During evaporation from a water body to air, underwhat conditions will the latent heat of vaporization be equal tothe heat transfer from the air?

128C What is evaporative cooling? Will it work in humid cli-mates?

129 Air enters an evaporative cooler at 1 atm, 36°C, and 20percent relative humidity at a rate of 4 m3/min, and it leaveswith a relative humidity of 90 percent. Determine (a) the exittemperature of the air and (b) the required rate of water supplyto the evaporative cooler.

130 Air enters an evaporative cooler at 95 kPa, 35°C, and 30percent relative humidity and exits saturated. Determine theexit temperature of air. Answer: 21.1�C

131 Air enters an evaporative cooler at 1 atm, 32°C, and 30percent relative humidity at a rate of 2 m3/min and leaves at22�C. Determine (a) the final relative humidity and (b) theamount of water added to air.

132 What is the lowest temperature that air can attain in anevaporative cooler if it enters at 1 atm, 29�C, and 40 percentrelative humidity? Answer: 19.3�C

133 Air at 1 atm, 15�C, and 60 percent relative humidity isfirst heated to 30�C in a heating section and then passedthrough an evaporative cooler where its temperature drops to25�C. Determine (a) the exit relative humidity and (b) theamount of water added to air, in kg H2O/kg dry air.

Adiabatic Mixing of Airstreams134C Two unsaturated airstreams are mixed adiabatically. Itis observed that some moisture condenses during the mixingprocess. Under what conditions will this be the case?

135C Consider the adiabatic mixing of two airstreams. Doesthe state of the mixture on the psychrometric chart have to beon the straight line connecting the two states?

136 Two airstreams are mixed steadily and adiabatically. Thefirst stream enters at 32�C and 40 percent relative humidity at arate of 20 m3/min, while the second stream enters at 12�C and90 percent relative humidity at a rate of 25 m3/min. Assumingthat the mixing process occurs at a pressure of 1 atm, deter-mine the specific humidity, the relative humidity, the dry-bulbtemperature, and the volume flow rate of the mixture.Answers: 0.0096 kg H2O/kg dry air, 63.4 percent, 20.6�C,45.0 m3/min

137 Repeat Prob. 136 for a total mixing-chamber pressure of95 kPa.

138 During an air-conditioning process, 25 m3/min of condi-tioned air at 18°C and 30 percent relative humidity is mixedadiabatically with 8 m3/min of outside air at 27°C and 90 per-cent relative humidity at a pressure of 1 atm. Determine (a) thetemperature, (b) the specific humidity, and (c) the relative hu-midity of the mixture.Answers: (a) 20.1°C, (b) 0.0077 kg H2O/kg dry air, (c) 52.3 percent

139 Reconsider Prob. 138. Using EES (or other) soft-ware, develop a general solution of the problem in

which the input variables may be supplied and parametric stud-ies performed. For each set of input variables for which thepressure is atmospheric, show the process on the psychromet-ric chart.

140 A stream of warm air with a dry-bulb temperature of40�C and a wet-bulb temperature of 32�C is mixed adiabati-cally with a stream of saturated cool air at 18�C. The dry airmass flow rates of the warm and cool airstreams are 8 and6 kg/s, respectively. Assuming a total pressure of 1 atm, deter-mine (a) the temperature, (b) the specific humidity, and (c) therelative humidity of the mixture.

141 Reconsider Prob. 140. Using EES (or other) soft-ware, determine the effect of the mass flow rate of

saturated cool airstream on the mixture temperature, specific

AIR

Humidifier

1 atm36°Cφ1 = 20%

Watermω

φ2 = 90%

FIGURE P129

2

3P = 1 atm

AIRφ3T3

ω3

132°C40%

12°C90%

FIGURE P136

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humidity, and relative humidity. Vary the mass flow rate of sat-urated cool air from 0 to 16 kg/s while maintaining the massflow rate of warm air constant at 8 kg/s. Plot the mixture tem-perature, specific humidity, and relative humidity as functionsof the mass flow rate of cool air, and discuss the results.

Wet Cooling Towers

142C How does a natural-draft wet cooling tower work?

143C What is a spray pond? How does its performance com-pare to the performance of a wet cooling tower?

144 The cooling water from the condenser of a power plantenters a wet cooling tower at 40�C at a rate of 90 kg/s. The wa-ter is cooled to 25�C in the cooling tower by air that enters thetower at 1 atm, 23�C, and 60 percent relative humidity andleaves saturated at 32�C. Neglecting the power input to the fan,determine (a) the volume flow rate of air into the cooling towerand (b) the mass flow rate of the required makeup water.

145 A wet cooling tower is to cool 60 kg/s of water from 40to 26�C. Atmospheric air enters the tower at 1 atm with dry-and wet-bulb temperatures of 22 and 16�C, respectively, andleaves at 34�C with a relative humidity of 90 percent. Using thepsychrometric chart, determine (a) the volume flow rate of airinto the cooling tower and (b) the mass flow rate of the re-quired makeup water. Answers: (a) 44.9 m3/s, (b) 1.16 kg/s

146 A wet cooling tower is to cool 50 kg/s of cooling waterfrom 40 to 25�C at a location where the atmospheric pressure is96 kPa. Atmospheric air enters the tower at 20�C and 70 per-cent relative humidity and leaves saturated at 35�C. Neglectingthe power input to the fan, determine (a) the volume flow rateof air into the cooling tower and (b) the mass flow rate of therequired makeup water.Answers: (a) 33.2 m3/s, (b) 1.04 kg/s

Review Problems

147 Air has the following composition on a mole basis: 21percent O2, 78 percent N2, and 1 percent Ar. Determine thegravimetric analysis of air and its molar mass. Answers: 23.2percent O2, 75.4 percent N2, 1.4 percent Ar; 28.96 kg/kmol

148 Using EES (or other) software, write a program todetermine the mole fractions of the components of

a mixture of three gases with known molar masses when themass fractions are given, and to determine the mass fractions ofthe components when the mole fractions are given. Run theprogram for a sample case, and give the results.

149 Using EES (or other) software, write a program todetermine the apparent gas constant, constant-vol-

ume specific heat, and internal energy of a mixture of threeideal gases when the mass fractions and other properties of theconstituent gases are given. Run the program for a sample case,and give the results.

150 The condensation of the water vapor in compressed-airlines is a major concern in industrial facilities, and thecompressed air is often dehumidified to avoid the problemsassociated with condensation. Consider a compressor thatcompresses ambient air from the local atmospheric pressure of92 kPa to a pressure of 800 kPa (absolute). The compressed airis then cooled to the ambient temperature as it flows throughthe compressed-air lines. Disregarding any pressure losses, de-termine if there will be any condensation in the compressed-airlines on a day when the ambient air is at 25�C and 40 percentrelative humidity.

151 The relative humidity of air at 27°C and 1 atm is in-creased from 25 percent to 75 percent during a humidificationprocess at constant temperature and pressure. Determine thepercent error involved in assuming the density of air to have re-mained constant.

152 Dry air whose molar analysis is 78.1 percent N2,20.9 percent O2, and 1 percent Ar flows over a water body untilit is saturated. If the pressure and temperature of air remainconstant at 1 atm and 25�C during the process, determine(a) the molar analysis of the saturated air and (b) the density ofair before and after the process. What do you conclude fromyour results?

153 Determine the mole fraction of the water vapor at thesurface of a lake whose surface temperature is 15°C andcompare it to the mole fraction of water in the lake, which isvery nearly 1.0. The air at the lake surface is saturated, and theatmospheric pressure at lake level can be taken to be 95 kPa.

154 Determine the mole fraction of dry air at the surface of alake whose temperature is 18°C. The air at the lake surface issaturated, and the atmospheric pressure at lake level can betaken to be 100 kPa.

COOLWATER

34°C90%

WARMWATER

AIRINLET

60 kg/s40°C

1 atmTdb = 22°CTwb = 16°C

Makeup water

26°C

AIREXIT

FIGURE P145

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ONLINE CHAPTER 2441

155 Consider a room that is cooled adequately by an air con-ditioner whose cooling capacity is 7500 kJ/h. If the room is tobe cooled by an evaporative cooler that removes heat at thesame rate by evaporation, determine how much water needs tobe supplied to the cooler per hour at design conditions.

156 The capacity of evaporative coolers is usually expressedin terms of the flow rate of air in ft3/min (or cfm), and a practi-cal way of determining the required size of an evaporativecooler for an 8-ft-high house is to multiply the floor area of thehouse by 4 (by 3 in dry climates and by 5 in humid climates).For example, the capacity of an evaporative cooler for a 30-ft-long, 40-ft-wide house is 1200 4 � 4800 cfm. Develop anequivalent rule of thumb for the selection of an evaporativecooler in SI units for 2.4-m-high houses whose floor areas aregiven in m2.

157 A cooling tower with a cooling capacity of 100 tons (440kW) is claimed to evaporate 15,800 kg of water per day. Is thisa reasonable claim?

158 The U.S. Department of Energy estimates that 190,000barrels of oil would be saved per day if every household in theUnited States raised the thermostat setting in summer by 6�F(3.3�C). Assuming the average cooling season to be 120 daysand the cost of oil to be $20/barrel, determine how muchmoney would be saved per year.

159 The thermostat setting of a house can be lowered by 1°Cby wearing a light long-sleeved sweater, or by 2°C by wearinga heavy long-sleeved sweater for the same level of comfort. Ifeach °C reduction in thermostat setting reduces the heating costof a house by 7 percent at a particular location, determine howmuch the heating costs of a house can be reduced by wearingheavy sweaters if the annual heating cost of the house is $600.

160 The air-conditioning costs of a house can be reduced byup to 10 percent by installing the outdoor unit (the condenser)of the air conditioner at a location shaded by trees and shrubs.If the air-conditioning costs of a house are $500 a year, deter-mine how much the trees will save the home owner in the 20-year life of the system.

161 A 5-m3 tank contains saturated air at 25�C and 97 kPa.Determine (a) the mass of the dry air, (b) the specific humidity,and (c) the enthalpy of the air per unit mass of the dry air.Answers: (a) 5.49 kg, (b) 0.0210 kg H2O/kg dry air,(c) 78.62 kJ/kg dry air

162 Reconsider Prob. 161. Using EES (or other) soft-ware, determine the properties of the air at the ini-

tial state. Study the effects of heating the air at constant volumeuntil the pressure is 110 kPa. Plot the required heat transfer, inkJ, as a function of pressure.

163 Air at 105 kPa, 16°C, and 50 percent relative humidityflows in an 20-cm-diameter duct at a velocity of 15 m/s. Deter-

mine (a) the dew-point temperature, (b) the volume flow rate ofair, and (c) the mass flow rate of dry air.

164 Air enters a cooling section at 97 kPa, 35°C, and 20 per-cent relative humidity at a rate of 9 m3/min, where it is cooleduntil the moisture in the air starts condensing. Determine (a)the temperature of the air at the exit and (b) the rate of heattransfer in the cooling section.

165 Outdoor air enters an air-conditioning system at 10°Cand 40 percent relative humidity at a steady rate of 22 m3/min,and it leaves at 25°C and 55 percent relative humidity. The out-door air is first heated to 22°C in the heating section and thenhumidified by the injection of hot steam in the humidifyingsection. Assuming the entire process takes place at a pressureof 1 atm, determine (a) the rate of heat supply in the heatingsection and (b) the mass flow rate of steam required in the hu-midifying section.

166 Air enters an air-conditioning system that usesrefrigerant-134a at 30�C and 70 percent relative humidity at arate of 4 m3/min. The refrigerant enters the cooling section at700 kPa with a quality of 20 percent and leaves as saturatedvapor. The air is cooled to 20�C at a pressure of 1 atm. Deter-mine (a) the rate of dehumidification, (b) the rate of heat trans-fer, and (c) the mass flow rate of the refrigerant.

167 Repeat Prob. 166 for a total pressure of 95 kPa for air.

168 An air-conditioning system operates at a total pressure of1 atm and consists of a heating section and an evaporativecooler. Air enters the heating section at 10�C and 70 percentrelative humidity at a rate of 30 m3/min, and it leaves the evap-orative cooler at 20�C and 60 percent relative humidity. Deter-mine (a) the temperature and relative humidity of the air whenit leaves the heating section, (b) the rate of heat transfer in theheating section, and (c) the rate of water added to air in theevaporative cooler.Answers: (a) 28.3�C, 23.0 percent, (b) 696 kJ/min, (c) 0.13 kg/min

169 Reconsider Prob. 168. Using EES (or other) soft-ware, study the effect of total pressure in the range

94 to 104 kPa on the results required in the problem. Plot theresults as functions of total pressure.

170 Repeat Prob. 168 for a total pressure of 96 kPa.

171 Conditioned air at 13°C and 90 percent relative humidityis to be mixed with outside air at 34°C and 40 percent relativehumidity at 1 atm. If it is desired that the mixture have a rela-tive humidity of 60 percent, determine (a) the ratio of the dryair mass flow rates of the conditioned air to the outside air and(b) the temperature of the mixture.

172 Reconsider Prob. 171. Determine the desired quan-tities using EES (or other) software instead of the

psychrometric chart. What would the answers be at a locationat an atmospheric pressure of 80 kPa?

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173 A natural-draft cooling tower is to remove 50 MWof waste heat from the cooling water that enters the

tower at 42�C and leaves at 27�C. Atmospheric air enters thetower at 1 atm with dry- and wet-bulb temperatures of 23 and18�C, respectively, and leaves saturated at 37�C. Determine (a)the mass flow rate of the cooling water, (b) the volume flowrate of air into the cooling tower, and (c) the mass flow rate ofthe required makeup water.

174 Reconsider Prob. 173. Using EES (or other) soft-ware, investigate the effect of air inlet wet-bulb

temperature on the required air volume flow rate and themakeup water flow rate when the other input data are the statedvalues. Plot the results as functions of wet-bulb temperature.

Design and Essay Problems

175 Prolonged exposure to mercury even at relatively low buttoxic concentrations in the air is known to cause permanentmental disorders, insomnia, and pain and numbness in thehands and the feet, among other things. Therefore, the maxi-mum allowable concentration of mercury vapor in the air atworkplaces is regulated by federal agencies. These regulationsrequire that the average level of mercury concentration in theair does not exceed 0.1 mg/m3.

Consider a mercury spill that occurs in an airtight storageroom at 20°C in San Francisco during an earthquake. Calculatethe highest level of mercury concentration in the air that canoccur in the storage room, in mg/m3, and determine if it iswithin the safe level. The vapor pressure of mercury at 20°C is0.173 Pa. Propose some guidelines to safeguard against the for-mation of toxic concentrations of mercury vapor in air in stor-age rooms and laboratories.

176 The condensation and even freezing of moisture in build-ing walls without effective vapor retarders are of real concernin cold climates as they undermine the effectiveness of the in-sulation. Investigate how the builders in your area are copingwith this problem, whether they are using vapor retarders or va-por barriers in the walls, and where they are located in thewalls. Prepare a report on your findings, and explain the rea-soning for the current practice.

177 The air-conditioning needs of a large building can be metby a single central system or by several individual windowunits. Considering that both approaches are commonly used inpractice, the right choice depends on the situation on hand.Identify the important factors that need to be considered indecision making, and discuss the conditions under which anair-conditioning system that consists of several window units ispreferable over a large single central system, and vice versa.

178 Identify the major sources of heat gain in your house insummer, and propose ways of minimizing them and thusreducing the cooling load.

179 Write an essay on different humidity measurement de-vices, including electronic ones, and discuss the advantagesand disadvantages of each device.

180 Design an inexpensive evaporative cooling system suit-able for use in your house. Show how you would obtain a wa-ter spray, how you would provide airflow, and how you wouldprevent water droplets from drifting into the living space.

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