Moran M.J., Shapiro H.N. Fundamentals of Engineering Thermodynamics

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Moist air

, T

< T

ωω

Water at T

Soaked

pad

(a)

(b)

Dry-bulb temperature

Mixture enthalpy

per unit mass

of dry air

Fig. 12.13 Evaporative cooling. (a) Equipment schematic. (b) Psychrometric chart

representation.

mixture enthalpy are closely lines of constant wet-bulb temperature (Sec. 12.7), it

follows that evaporative cooling takes place at a nearly constant wet-bulb tem-

perature.

In the next example, we consider the analysis of an evaporative cooler.

12.8 Analyzing Air-Conditioning Processes 753

c c c c EXAMPLE 12.13 c

Water at 70°F

= 70°F

Boundar

Soaked pad

(AV)

= 100°F

= 10%

= 5000

___

min

Fig. E12.13

Engineering Model:

The control volume shown in the accompanying figure oper-

ates at steady state. Changes in kinetic and potential energy

can be neglected and W

5 0.

2. There is no heat transfer with the surroundings.

3. The water added to the soaked pad enters as a liquid and

evaporates fully into the moist air.

4. The pressure remains constant throughout at 1 atm.

5. The moist air streams are regarded as ideal gas mixtures

adhering to the Dalton model.

Considering an Evaporative Cooler

Air at 1008F and 10% relative humidity enters an evaporative cooler with a volumetric flow rate of 5000 ft

/min.

Moist air exits the cooler at 708F. Water is added to the soaked pad of the cooler as a liquid at 708F and evap-

orates fully into the moist air. There is no heat transfer with the surroundings and the pressure is constant

throughout at 1 atm. Determine (a) the mass flow rate of the water to the soaked pad, in lb/h, and (b) the

relative humidity of the moist air at the exit to the evaporative cooler.

SOLUTION

Known:

Air at 1008F and f 5 10% enters an evaporative cooler with a volumetric flow rate of 5000 ft

/min.

Moist air exits the cooler at 708F. Water is added to the soaked pad of the cooler at 708F.

Find: Determine the mass flow rate of the water to the soaked pad, in lb/h, and the relative humidity of the

moist air at the exit of the cooler.

Schematic and Given Data:

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754 Chapter 12

Ideal Gas Mixture and Psychrometric Applications

Analysis:

(a)

Applying conservation of mass to the dry air and water individually as in previous examples gives

5 m

2 v

where m

is the mass flow rate of the water to the soaked pad. To find m

requires v

, m

, and v

. These will

now be evaluated in turn.

The humidity ratio v

can be found from Eq. 12.43, which requires p

, the partial pressure of the moist

air entering the control volume. Using the given relative humidity f

and p

at T

from Table A-2E, we have

5 f

5 0.095 lbf/in.

With this, v

5 0.00405 lb(vapor)ylb(dry air).

The mass flow rate of the dry air m

can be found as in previous examples using the volumetric flow rate and

specific volume of the dry air. Thus

1AV2

The specific volume of the dry air can be evaluated from the ideal gas equation of state. The result is y

5 14.2 ft

/lb

(dry air). Inserting values, the mass flow rate of the dry air is

5000 ft

min

14.2 ft

lb1dry air2

5 352.1

lb1dry air

min

To find the humidity ratio v

, reduce the steady-state forms of the mass and energy rate balances using

assumption 1 to obtain

0 5

1 m

With the same reasoning as in previous examples, this can be expressed as the following special form of Eq. 12.55:

0 5 1h

1 vh

1 1v

2 v

2 1h

1 vh

(a)

where h

denotes the specific enthalpy of the water entering the control volume at 708F. Solving for v

2 h

1 v

2 h

2 T

1 v

2 h

where c

5 0.24 Btu/lb ? 8R. With h

, h

, and h

from Table A-2E

0.24

100 2 70

1 0.00405

1105 2 38.1

1092 2 38.1

5 0.0109

vapor

dry air

Substituting values for m

, v

, and v

into the expression for m

5 c352.1

lb1dry a

ir2

min

60 m

1 h

`d 10.0109 2 0.004052

lb1water

dry air

5 144.7

water

(b) The relative humidity of the moist air at the exit can be determined using Eq. 12.44. The partial pressure of

the water vapor required by this expression can be found by solving Eq. 12.43 to obtain

1 0.622

Inserting values

10.01092114.696 lbf

in.

0.0109 1 0.622

5 0.253 lbf

in.

➊

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At 708F, the saturation pressure is 0.3632 lbf/in.

Thus, the relative humidity at the exit is

0.253

363

5 0.697169.7%2

Alternative Psychrometric Chart Solution: Since the underlined term in Eq. (a)

is much smaller than either of the moist air enthalpies, the enthalpy of the moist

air remains nearly constant, and thus evaporative cooling takes place at a nearly

constant wet-bulb temperature. See Fig. 12.13b and the accompanying discus-

sion. Using this approach with the psychrometric chart, Fig. A-9E, determine

humidity ratio and relative humidity at the exit, and compare with the previ-

ously determined values. The details are left as an exercise.

➊

A constant value of the specific heat c

has been used here to evaluate the

term (h

2 h

). As shown in previous examples, this term can be evaluated

alternatively using the ideal gas table for air.

Using steam table data, what is the dew point temperature at

the exit, in 8F? Ans. 59.68F.

Ability to…

❑

apply psychrometric termi-

nology and principles.

❑ apply mass and energy

balances for an evaporative

cooling process in a control

volume at steady state.

❑

retrieve property data for

dry air and water.

✓

Skills Developed

12.8.6 Adiabatic Mixing of Two Moist Air Streams

A common process in air-conditioning systems is the mixing of moist air streams, as

shown in Fig. 12.14. The objective of the thermodynamic analysis of such a process

is normally to fix the flow rate and state of the exiting stream for specified flow rates

and states of each of the two inlet streams. The case of adiabatic mixing is governed

by Eqs. 12.56 to follow.

The mass rate balances for the dry air and water vapor at steady state are, respec-

tively,

1 m

5 m





dry air

1 m

5 m





water vapor

(12.56a)

12.8 Analyzing Air-Conditioning Processes 755

(b)

Mixture enthalpy

per unit mass

of dry air

+ w h

)

+ w h

)

+ w h

)

Insulation

1 m

, T

2 m

, T

(a)

Fig. 12.14 Adiabatic mixing of two moist air streams. (a) Equipment representation.

(b) Psychrometric chart representation.

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756 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

With m

5 vm

, the water vapor mass balance becomes

1 v

5 v





water vapor

(12.56b)

Assuming

5 W

5 0 and ignoring the effects of kinetic and potential energy,

the energy rate balance reduces at steady state to

1 v

1 m

1 v

5 m

1 v

(12.56c)

where the enthalpies of the entering and exiting water vapor are evaluated as the

saturated vapor values at the respective dry-bulb temperatures.

If the inlet flow rates and states are known, Eqs. 12.56 are three equations in three

unknowns: m

, v

, and

1 v

. The solution of these equations is illustrated by

Example 12.14.

Let us also consider how Eqs. 12.56 can be solved geometrically with the psychro-

metric chart: Using Eq. 12.56a to eliminate m

, the mass flow rate of dry air at 3,

from Eqs. 12.56b and 12.56c, we get

2 v

1 v

22 1h

1 v

(12.57)

From the relations of Eqs. 12.57, we conclude that on a psychrometric chart state 3 of

the mixture lies on a straight line connecting states 1 and 2 of the two streams before

mixing (see end-of-chapter Prob. 12.93). This is shown in Fig. 12.14b.

Analyzing Adiabatic Mixing of Two Moist Air Streams

c c c c EXAMPLE 12.14 c

A stream consisting of 142 m

/min of moist air at a temperature of 58C and a humidity ratio of 0.002 kg(vapor)/

kg(dry air) is mixed adiabatically with a second stream consisting of 425 m

/min of moist air at 248C and 50%

relative humidity. The pressure is constant throughout at 1 bar. Determine (a) the humidity ratio and (b) the

temperature of the exiting mixed stream, in 8C.

SOLUTION

Known:

A moist air stream at 58C, v 5 0.002 kg(vapor)/kg(dry air), and a volumetric flow rate of 142 m

/min is

mixed adiabatically with a stream consisting of 425 m

/min of moist air at 248C and f 5 50%.

Find: Determine the humidity ratio and the temperature, in 8C, of the mixed stream exiting the control volume.

Schematic and Given Data:

kg (vapor)

__________

kg (dry air)

(AV)

= 142 m

/min

= 5°C

= 0.002

(AV)

= 425 m

/min

= 24°C

= 50%

= ?

Insulation

Fig. E12.14

Engineering Model:

The control volume shown in the accompanying

figure operates at steady state. Changes in kinetic

and potential energy can be neglected and W

5 0.

2. There is no heat transfer with the surroundings.

3. The pressure remains constant throughout at 1 bar.

4. The moist air streams are regarded as ideal gas

mixtures adhering to the Dalton model.

Analysis:

(a)

The humidity ratio v

can be found by means of mass rate balances for the dry air and water vapor,

respectively

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1 m

5 m





dry air

1 m

5 m





water vapor

With m

5 v

, m

5 v

, and m

5 v

, the second of these balances becomes (Eq. 12.56b)

1 v

5 v

Solving

1 v

Since m

5 m

1 m

, this can be expressed as

1 v

1 m

To determine

requires values for

, m

, and m

. The mass flow rates of the dry air, m

and m

, can be

found as in previous examples using the given volumetric flow rates



The values of y

, y

, and

are readily found from the psychrometric chart, Fig. A-9. Thus, at v

5 0.002 and

5 58C, y

5 0.79 m

dry air

. At f

5 50% and T

5 248C, y

5 0.855 m

air

and v

0094

. The

mass flow rates of the dry air are then m

5 180 k

air

min and m

5 497 k

air

min. Inserting values

into the expression for

0.002

180

0.0094

497

180 1 497

5 0.0074

vapor

dry air

(b) The temperature

of the exiting mixed stream can be found from an energy rate balance. Reduction of the

energy rate balance using assumptions 1 and 2 gives (Eq. 12.56c)

1 vh

1 m

1 vh

5 m

1 vh

(a)

Solving

1 vh

1 m

1 vh

1 m

(b)

With

1 vh

5 10 kJ

dry air

and

1 vh

5 47.8 kJ

dry air

from Fig. A-9 and other known values

1 vh

180

1 497

47.8

180 1 497

5 37.7

dry air

➊

This value for the enthalpy of the moist air at the exit, together with the previously determined value for

fixes the state of the exiting moist air. From inspection of Fig. A-9, T

5 198C.

Alternative Solutions:

The use of the psychrometric chart facilitates the solution for T

. Without the chart, an iterative solution of Eq. (b)

using data from Tables A-2 and A-22 could be used. Alternatively, T

can be determined using the following IT

program, where

is denoted as phi2, the volumetric flow rates at 1 and 2 are denoted as AV1

and AV2, respec-

tively, and so on.

// Given data

T1 5 5 // 8C

w1 5 0.002 // kg(vapor) / kg(dry air)

AV1 5 142 // m

/min

T2 5 24 // 8C

phi2 5 0.5

AV2 5 425 // m

/min

p 5 1 // bar

12.8 Analyzing Air-Conditioning Processes 757

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758 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

// Mass balances for water vapor and dry air:

w1 * mdota1 1 w2 * mdota2 5 w3 * mdota3

mdota1 1 mdota2 5 mdota3

// Evaluate mass flow rates of dry air

mdota1 5 AV1 / va1

➋

va1 5 va_Tw(T1, w1, p)

mdota2 5 AV2 / va2

va2 5 va_Tphi(T2, phi2, p)

// Determine w2

w2 5 w_Tphi(T2, phi2, p)

// The energy balance, Eq. (a), reads

mdota1 * h1 1 mdota2 * h2 5 mdota3 * h3

h1 5 ha_Tw(T1, w1)

h2 5 ha_Tphi(T2, phi2, p)

h3 5 ha_Tw(T3, w3)

Using the Solve button, the result is T

5 19.018C and v

5 0.00745 k

(vapor)/

kg (dry air), which agree with the psychrometric chart solution.

➊

A solution using the geometric approach based on Eqs. 12.57 is left as an

exercise.

➋

Note the use here of special Moist Air functions listed in the Properties

menu of IT.

Using the psychrometric chart, what is the relative humidity

at the exit? Ans. < 53%.

Ability to…

❑

apply psychrometric termi-

nology and principles.

❑

apply mass and energy

balances for an adiabatic

mixing process of two moist

air streams in a control

volume at steady state.

❑

retrieve property data

for moist air using the

psychrometric chart.

❑

apply IT for psychrometric

analysis.

✓

Skills Developed

12.9 Cooling Towers

Power plants invariably discharge considerable energy to their surroundings by heat

transfer (Chap. 8). Although water drawn from a nearby river or lake can be

employed to carry away this energy, cooling towers provide an alternative in loca-

tions where sufficient cooling water cannot be obtained from natural sources or

where concerns for the environment place a limit on the temperature at which

cooling water can be returned to the surroundings. Cooling towers also are fre-

quently employed to provide chilled water for applications other than those involv-

ing power plants.

Cooling towers can operate by natural or forced convection. Also they may be

counterflow, cross-flow, or a combination of these. A schematic diagram of a forced-

convection, counterflow cooling tower is shown in Fig. 12.15. The warm water to be

cooled enters at 1 and is sprayed from the top of the tower. The falling water usually

passes through a series of baffles intended to keep it broken up into fine drops to

promote evaporation. Atmospheric air drawn in at 3 by the fan flows upward, coun-

ter to the direction of the falling water droplets. As the two streams interact, a frac-

tion of the entering liquid water stream evaporates into the moist air, which exits at

4 with a greater humidity ratio than the incoming moist air at 3, while liquid water

exits at 2 with a lower temperature than the water entering at 1. Since some of the

incoming water is evaporated into the moist air stream, an equivalent amount of

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makeup water is added at 5 so that the return mass flow rate of the cool water equals

the mass flow rate of the warm water entering at 1.

For operation at steady state, mass balances for the dry air and water and an

energy balance on the overall cooling tower provide information about cooling tower

performance. In applying the energy balance, heat transfer with the surroundings is

usually neglected. The power input to the fan of forced-convection towers also may

be negligible relative to other energy rates involved. The example to follow illustrates

the analysis of a cooling tower using conservation of mass and energy together with

property data for the dry air and water.

Fan

Warm water inlet

, m

Return water

< T

5 Makeup

water

Liquid

Discharged moist air

, T

ωω

Atmospheric ai

, T

Fig. 12.15 Schematic of a cooling

tower.

c c c c EXAMPLE 12.15 c

Determining Mass Flow Rates for a Power Plant Cooling Tower

Water exiting the condenser of a power plant at 388C enters a cooling tower with a mass flow rate of 4.5 3 10

kg/h.

A stream of cooled water is returned to the condenser from a cooling tower with a temperature of 308C and

the same flow rate. Makeup water is added in a separate stream at 208C. Atmospheric air enters the cooling

tower at 258C and 35% relative humidity. Moist air exits the tower at 358C and 90% relative humidity. Determine

the mass flow rates of the dry air and the makeup water, in kg/h. The cooling tower operates at steady state.

Heat transfer with the surroundings and the fan power can each be neglected, as can changes in kinetic and

potential energy. The pressure remains constant throughout at 1 atm.

SOLUTION

Known:

A liquid water stream enters a cooling tower from a condenser at 388C with a known mass flow rate. A

stream of cooled water is returned to the condenser at 308C and the same flow rate. Makeup water is added at

208C. Atmospheric air enters the tower at 258C and f 5 35%. Moist air exits the tower at 358C and f 5 90%.

Find: Determine the mass flow rates of the dry air and the makeup water, in kg/h.

12.9 Cooling Towers 759

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760 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Schematic and Given Data:

Engineering Model:

The control volume shown in the accompanying figure

operates at steady state. Heat transfer with the surround-

ings can be neglected, as can changes in kinetic and

potential energy; also W

5 0.

2. To evaluate specific enthalpies, each liquid stream is

regarded as a saturated liquid at the corresponding speci-

fied temperature.

3. The moist air streams are regarded as ideal gas mixtures

adhering to the Dalton model.

4. The pressure is constant throughout at 1 atm.

Fig. E12.15

Makeup water

= 20°C

Atmospheric air

= 25°C,

= 35%

Liquid water, T

= 38°C

= 4.5 × 10

kg/h

Liquid water, T

= 30°C

= 4.5 × 10

kg/h

Moist air

= 35°C

= 90%

Analysis: The required mass flow rates can be found from mass and energy rate balances. Mass balances for the

dry air and water individually reduce at steady state to

5 m



dry air

1 m

5 m

1 m





water

The common mass flow rate of the dry air is denoted as m

. Since m

5 m

, the second of these equations

becomes

5 m

2 m

With m

5 v

and m

5 v

5 m

2 v

Accordingly, the two required mass flow rates, m

and m

, are related by this equation. Another equation relating

the flow rates is provided by the energy rate balance.

Reducing the energy rate balance with assumption 1 results in

0 5 m

1 m

2 m

1 m

Evaluating the enthalpies of the water vapor as the saturated vapor values at the respective temperatures and

the enthalpy of each liquid stream as the saturated liquid enthalpy at each respective temperature, the energy

rate equation becomes

0 5 m

1 m

2 m

1 m

Introducing m

5 m

, m

5 m

2 v

, m

5 v

, and m

5 v

and solving for m

2 h

1 v

2 v

2 1v

2 v

(a)

The humidity ratios

and

required by this expression can be determined from Eq. 12.43, using the partial

pressure of the water vapor obtained with the respective relative humidity. Thus, v

5 0.00688 kg(vapor)/kg(dry

air) and v

5 0.0327 kg(vapor)/kg(dry air).

With enthalpies from Tables A-2 and A-22, as appropriate, and the known values for v

, and m

, the expres-

sion for m

becomes

14.5 3 10

21159.21 2 125.792

308.2 2 298.2

0.0327

2565.3

0.00688

2547.2

0.0258

83.96

5 2.03 3 10

Finally, inserting known values into the expression for m

results in

5 12.03 3 10

210.0327 2 0.0068825 5.24 3 10

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Alternative Psychrometric Chart Solution: Equation (a) can be rearranged to

read

2 h

1 v

22 1h

1 v

22 1v

2 v

The specific enthalpy terms h

, h

, and h

are obtained from Table A-2, as

above. The underlined terms and v

and v

can be obtained by inspection of a

psychrometric chart from the engineering literature providing data at states

3 and 4. Figure A-9 does not suffice in this application at state 4. The details

are left as an exercise.

Using steam table data, determine the partial pressure of the

water vapor in the entering moist air stream, p

, in bar. Ans. 0.0111 bar.

Ability to…

❑

apply psychrometric termi-

nology and principles.

❑

apply mass and energy

balances for a cooling tower

process in a control volume

at steady state.

❑

retrieve property data for

dry air and water.

✓Skills Developed

In this chapter we have applied the principles of thermodynamics

to systems involving ideal gas mixtures, including the special

case of psychrometric applications involving air–water vapor mix-

tures, possibly in the presence of liquid water. Both closed sys-

tem and control volume applications are presented.

The first part of the chapter deals with general ideal gas mix-

ture considerations and begins by describing mixture composi-

tion in terms of the mass fractions or mole fractions. The Dalton

model, which brings in the partial pressure concept, is then intro-

duced for the p–y–T relation of ideal gas mixtures.

Means are also introduced for evaluating the enthalpy, inter-

nal energy, and entropy of a mixture by adding the contribution

of each component at its condition in the mixture. Applications

are considered where ideal gas mixtures undergo processes at

constant composition as well as where ideal gas mixtures are

formed from their component gases.

In the second part of the chapter, we study psychrometrics.

Special terms commonly used in psychrometrics are intro-

duced, including moist air, humidity ratio, relative humidity,

mixture enthalpy, and the dew point, dry-bulb, and wet-bulb

temperatures. The psychrometric chart, which gives a graphi-

cal representation of important moist air properties, is intro-

duced. The principles of conservation of mass and energy are

formulated in terms of psychrometric quantities, and typical

air-conditioning applications are considered, including dehu-

midification and humidification, evaporative cooling, and mix-

ing of moist air streams. A discussion of cooling towers is also

provided.

The following list provides a study guide for this chapter.

When your study of the text and end-of-chapter exercises has

been completed, you should be able to

write out the meanings of the terms listed in the margin

throughout the chapter and understand each of the related

concepts. The subset of key concepts listed below is particu-

larly important.

describe mixture composition in terms of mass fractions or

mole fractions.

relate pressure, volume, and temperature of ideal gas mix-

tures using the Dalton model, and evaluate U, H, c

, c

, and

S of ideal gas mixtures in terms of the mixture composition

and the respective contribution of each component.

apply the conservation of mass and energy principles and the

second law of thermodynamics to systems involving ideal gas

mixtures.

For psychrometric applications, you should be able to

evaluate the humidity ratio, relative humidity, mixture enthalpy,

and dew point temperature.

use the psychrometric chart.

apply the conservation of mass and energy principles and the

second law of thermodynamics to analyze air-conditioning

processes and cooling towers.

c CHAPTER SUMMARY AND STUDY GUIDE

c KEY ENGINEERING CONCEPTS

mass fraction, p. 706

gravimetric analysis, p. 706

mole fraction, p. 706

molar (volumetric) analysis, p. 707

apparent molecular weight, p. 707

Dalton model, p. 710

partial pressure, p. 710

psychrometrics, p. 727

moist air, p. 727

humidity ratio, p. 728

relative humidity, p. 729

mixture enthalpy, p. 729

dew point temperature, p. 731

dry-bulb temperature, p. 738

wet-bulb temperature, p. 738

psychrometric chart, p. 740

Key Engineering Concepts 761

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762 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

c KEY EQUATIONS

Ideal Gas Mixtures: General Considerations

5 m

m (12.3) p. 706

Analysis in terms of mass fractions

1 5

(12.4) p. 706

5 n

n (12.6) p. 706

1 5

(12.7) p. 707

Analysis in terms of mole fractions

M 5

(12.9) p. 707

Apparent molecular weight

(12.12) p. 710

p 5

(12.13) p. 710

u 5

(12.21) p. 712

h 5

(12.22) p. 712

s 5

(12.27) p. 713

y,i

(12.23) p. 712

Mixture specific heats on a molar basis

p,i

(12.24) p. 712

Psychrometric Applications

v 5

5 0.622

(12.42, 12.43) p. 729 Humidity ratio

f 5

(12.44) p. 729 Relative humidity

5 h

1 vh

(12.46) p. 729 Mixture enthalpy per unit mass of dry air

Partial pressure of component i and relation to mixture

pressure p

Internal energy, enthalpy, and entropy per mole of mixture.

—

and

—

evaluated at mixture temperature T.

—

evaluated at T and

partial pressure p

c EXERCISES: THINGS ENGINEERS THINK ABOUT

1. In an equimolar mixture of O

and N

, are the mass fractions

equal? Explain.

2. If two different ideal gases mix spontaneously, is the process

irreversible? Explain.

3. Which component of the fuel–air mixture in a cylinder of an

automobile engine would have the greater mass fraction?

4. A rigid, insulated container is divided into two compartments

by a partition, and each compartment contains air at the

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