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

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596 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Assumptions:

1. The control volume shown in the accompanying figure operates at steady state.

2. The changes in kinetic and potential energy between inlet and exit can be ignored and

3. The entering and exiting moist air streams can be regarded as ideal gas mixtures.

Analysis:

(a) The heat transfer rate can be determined from the mass and energy rate balances. At steady state, the amounts of dry

air and water vapor contained within the control volume cannot vary. Thus, for each component individually it is necessary

for the incoming and outgoing mass flow rates to be equal. That is

For simplicity, the constant mass flow rates of the dry air and water vapor are denoted, respectively, by and . From these

considerations, it can be concluded that the humidity ratio is the same at the inlet and exit: 

 

The steady-state form of the energy rate balance reduces with assumption 2 to

In writing this equation, the incoming and outgoing moist air streams are regarded as ideal gas mixtures of dry air and water

vapor.

Solving for

Noting that , where  is the humidity ratio, the expression for can be written in the form

(1)

To evaluate from this expression requires the specific enthalpies of the dry air and water vapor at the inlet and exit, the

mass flow rate of the dry air, and the humidity ratio.

The specific enthalpies of the dry air are obtained from Table A-22 at the inlet and exit temperatures T

and T

, respec-

tively: h

 283.1 kJ/kg, h

 303.2 kJ/kg. The specific enthalpies of the water vapor are found using h

 h

and data from

Table A-2 at T

and T

, respectively: h

 2519.8 kJ/kg, h

 2556.3 kJ/kg.

The mass flow rate of the dry air can be determined from the volumetric flow rate at the inlet (AV)

In this equation, v

is the specific volume of the dry air evaluated at T

and the partial pressure of the dry air p

. Using the

ideal gas equation of state

The partial pressure p

can be determined from the mixture pressure p and the partial pressure of the water vapor p

: p



p  p

. To find p

, use the given inlet relative humidity and the saturation pressure at 10C from Table A-2

Since the mixture pressure is 1 bar, it follows that p

 0.9902 bar. The specific volume of the dry air is then



8314

28.97

b1283 K2

10 .9902  10

N/m

 0.82 m

/kg

 f

 10.82 10.01228 bar2 0.0098 bar





M2 T



1AV2

 m

31h

 h

2 v

 h

 vm

 m

 h

2 m

 h

0  Q

 W

 1m

 m

2 1m

 m

 m

1water2

 m

1dry air2

 0.

❶

12.8 Analyzing Air-Conditioning Processes 597

Using this value, the mass flow rate of the dry air is

The humidity ratio  can be found from

Finally, substituting values into Eq. (1) we get

(b) The states of the water vapor at the duct inlet and exit are located on the accompanying T–v diagram. Both the

composition of the moist air and the mixture pressure remain constant, so the partial pressure of the water vapor at the

exit equals the partial pressure of the water vapor at the inlet: p

 p

 0.0098 bar. The relative humidity at the exit

is then

where p

is from Table A-2 at 30C.

Alternative Psychrometric Chart Solution: Let us consider an alternative solution using the psychrometric chart. As shown

on the sketch of the psychrometric chart, Fig. E12.10b, the state of the moist air at the inlet is defined by 

 80% and a

dry-bulb temperature of 10C. From the solution to part (a), we know that the humidity ratio has the same value at the exit

as at the inlet. Accordingly, the state of the moist air at the exit is fixed by 

 

and a dry-bulb temperature of 30C. By

inspection of Fig. A-9, the relative humidity at the duct exit is about 23%, and thus in agreement with the result of part (b).

The rate of heat transfer can be evaluated from the psychrometric chart using the following expression obtained by rear-

ranging Eq. (1) of part (a) to read

(2)

To evaluate from this expression requires values for the mixture enthalpy per unit mass of dry air (h

 h

) at the in-

let and exit. These can be determined by inspection of the psychrometric chart, Fig. A-9, as (h

 h

)

 25.7 kJ/kg (dry

air), (h

 h

)

 45.9 kJ/kg (dry air).

Using the specific volume value v

from the chart at the inlet state together with the given volumetric flow rate at the in-

let, the mass flow rate of the dry air is found as

Substituting values into the energy rate balance, Eq. (2), we get

which agrees closely with the result obtained in part (a), as expected.

 3737

min

 185

kg1dry air2

min

145.9  25.72

kg1dry air2



150 m

/min

0.81 m

/kg1dry air2

 185

kg1dry air2

min

 m

31h

 vh

 1h

 vh



0.0098

0.04246

 0.231

123.1%2

 3717 kJ/min

 182.9 31303.2  283.12 10.006162 12556.3  2519.824

 0.00616

kg 1vapor2

kg 1dry air2

v  0.622

p  p

b 0.622 a

0.0098

1  0.0098



150 m

/min

0 .82 m

/kg

 182.9 kg/min

❷

❸

598 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Mixture enthalpy

10°C30°C

Dry-bulb temperature

= 100%

= 80%

The first underlined term in this equation for is evaluated with specific enthalpies from the ideal gas table for air,

Table A-22. Steam table data are used to evaluate the second underlined term. Note that the different datums for enthalpy

underlying these tables cancel because each of the two terms involves enthalpy differences only. Since the specific heat

for dry air varies only slightly over the interval from 10 to 30C (Table A-20), the specific enthalpy change of the dry

air could be evaluated alternatively with c

 1.005 kJ/kg K.

No water is added or removed as the moist air passes through the duct at constant pressure; accordingly, the humidity ra-

tio  and the partial pressures p

and p

remain constant. However, because the saturation pressure increases as the tem-

perature increases from inlet to exit, the relative humidity decreases: 

 

The mixture pressure, 1 bar, differs slightly from the pressure used to construct the psychrometric chart, 1 atm. This dif-

ference is ignored.

 Figure E12.10b

❶

❷

❸

 12.8.3 Dehumidification

When a moist air stream is cooled at constant mixture pressure to a temperature below its

dew point temperature, some condensation of the water vapor initially present would occur.

Figure 12.11 shows the schematic of a dehumidifier using this principle. Moist air enters at

state 1 and flows across a cooling coil through which a refrigerant or chilled water circu-

lates. Some of the water vapor initially present in the moist air condenses, and a saturated

moist air mixture exits the dehumidifier section at state 2. Although water would condense

at various temperatures, the condensed water is assumed to be cooled to T

before it exits

the dehumidifier. Since the moist air leaving the humidifier is saturated at a temperature lower

than the temperature of the moist air entering, the moist air stream might be unsuitable for

direct use in occupied spaces. However, by passing the stream through a following heating

section, it can be brought to a condition most occupants would regard as comfortable. Let

us sketch the procedure for evaluating the rates at which condensate exits and refrigerant

circulates.

MASS BALANCE. The mass flow rate of the condensate can be related to the mass flow

rate of the dry air by applying conservation of mass separately for the dry air and water

passing through the dehumidifier section. At steady state

 m

 m

1water2

 m

1dry air2

12.8 Analyzing Air-Conditioning Processes 599

The common mass flow rate of the dry air is denoted as . Solving for the mass flow rate

of the condensate

Introducing and , the amount of water condensed per unit mass of

dry air passing through the device is

This expression requires the humidity ratios 

and 

. Because no moisture is added or re-

moved in the heating section, it can be concluded from conservation of mass that 

 

so 

can be used in the above equation in place of 

ENERGY BALANCE. The mass flow rate of the refrigerant through the cooling coil can

be related to the mass flow rate of the dry air by means of an energy rate balance applied

to the dehumidifier section. With negligible heat transfer with the surroundings,

and no significant kinetic and potential energy changes, the energy rate balance reduces at

steady state to

where h

and h

denote the specific enthalpy values of the refrigerant entering and exiting the

dehumidifier section, respectively. Introducing and  (

 

)

where the specific enthalpies of the water vapor at 1 and 2 are evaluated at the saturated va-

por values corresponding to T

and T

, respectively. Since the condensate is assumed to exit

as a saturated liquid at T

, h

 h

. Solving for the refrigerant mass flow rate per unit mass

of dry air flowing through the device



 h

2 v

 v

 1v

 v

 h

0  m

 h

2 m

31h

 h

2 v

 v

 1v

 v

 v

, m

 v

0  m

 h

2 1m

 m

2 m

 1m

 m

 0,

 v

 v

 v

 m

 m

 Figure 12.11 Dehumidification. (a) Equipment schematic. (b) Psychrometric chart

representation.

Initial dew

point

= 100%

Dry-bulb temperature

Cooling

coil

ie 32

Heating

coil

= 100%

< T

> T

Condensate –

saturated at T

(Dehumidifier section) (Heating section)

(a)(b)

Moist air

p = 1 atm

600 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

The accompanying psychrometric chart, Fig. 12.11b, illustrates important features of the

processes involved. As indicated by the chart, the moist air first cools from state 1, where

the temperature is T

and the humidity ratio is 

, to state 2, where the mixture is saturated

(

 100%), the temperature is T

 T

, and the humidity ratio is 

 

. During the

subsequent heating process, the humidity ratio would remain constant, 

 

, and the

temperature would increase to T

. Since the states visited would not be equilibrium states,

these processes are indicated on the psychrometric chart by dashed lines. The example to

follow provides a specific illustration.

EXAMPLE 12.11 Dehumidifier

Moist air at 30C and 50% relative humidity enters a dehumidifier operating at steady state with a volumetric flow rate of

280 m

/min. The moist air passes over a cooling coil and water vapor condenses. Condensate exits the dehumidifier saturated

at 10C. Saturated moist air exits in a separate stream at the same temperature. There is no significant loss of energy by heat

transfer to the surroundings and pressure remains constant at 1.013 bar. Determine

(a) the mass flow rate of the dry air, in

kg/min,

(b) the rate at which water is condensed, in kg per kg of dry air flowing through the control volume, and (c) the

required refrigerating capacity, in tons.

SOLUTION

Known: Moist air enters a dehumidifier at 30C and 50% relative humidity with a volumetric flow rate of 280 m

/min. Con-

densate and moist air exit in separate streams at 10C.

Determine: Find the mass flow rate of the dry air, in kg/min, the rate at which water is condensed, in kg per kg of dry air,

and the required refrigerating capacity, in tons.

Schematic and Given Data:

Heating coilCooling coil

Condensate,

saturated at

= 10°C

Control volume

13 2

Saturated

mixture

10°C

(AV)

= 280 m

/min

= 30°C

= 50%

Analysis:

(a) At steady state, the mass flow rates of the dry air entering and exiting are equal. The common mass flow rate of the dry

air can be determined from the volumetric flow rate at the inlet



1AV2

 Figure E12.11

Assumptions:

1. The control volume shown in the accompanying

figure operates at steady state. Changes in kinetic and

potential energy can be neglected, and

2. There is no significant heat transfer to the

surroundings.

3. The pressure remains constant throughout at

1.013 bar.

4. At location 2, the moist air is saturated. The con-

densate exits at location 3 as a saturated liquid at tem-

perature T

 0.

12.8 Analyzing Air-Conditioning Processes 601

The specific volume of the dry air at inlet 1, v

, can be evaluated using the ideal gas equation of state, so

The partial pressure of the dry air p

can be determined from p

 p

 p

. Using the relative humidity at the inlet 

and

the saturation pressure at 30C from Table A-2

Thus, p

 1.013  0.02123  0.99177 bar. Inserting values into the expression for gives

(b) Conservation of mass for the water requires With and the rate at which water

is condensed per unit mass of dry air is

The humidity ratios 

and 

can be evaluated using Eq. 12.43. Thus, 

Since the moist air is saturated at 10C, p

equals the saturation pressure at 10C: p

 0.01228 bar from Table A-2. Equa-

tion 12.43 then gives 

 0.0076 kg(vapor)kg(dry air). With these values for 

and 

(c) The rate of heat transfer between the moist air stream and the refrigerant coil can be determined using an energy rate

balance. With assumptions 1 and 2, the steady-state form of the energy rate balance reduces to

With , and , this becomes

where the specific enthalpies of the water vapor at 1 and 2 are evaluated at the saturated vapor values corresponding to T

and

, respectively, and the specific enthalpy of the exiting condensate is evaluated as h

at T

. Selecting enthalpies from Tables A-2

and A-22, as appropriate

Since 1 ton of refrigeration equals a heat transfer rate of 211 kJ/min (Sec. 10.2), the required refrigerating capacity is 52.5 tons.

If a psychrometric chart were used to obtain data, this expression for would be rearranged to read

The underlined terms and humidity ratios 

and 

would be read directly from the chart; the specific enthalpy h

would

be obtained from Table A-2 as h

at T

 m

31h

 vh

 1h

 vh

 1v

 v

11,084 kJ/min

 0.0076

12519.82 0.0057 142.0124

 1319.352 31283.1  303.22 0.0133 12556.32

 m

31h

 h

2 v

 v

 1v

 v

 1v

 v

 v

, m

 v

0  Q

 1m

 m

2 m

 1m

 m

 0.0133  0.0076  0.0057

kg1condensate2

kg1dry air2

 0.622 a

 p

b 0.622 a

0.02123

0.99177

b 0.0133

kg1vapor2

kg1dry air2

 v

 v

 v

 m

 m



1280 m

/min2 10.99177  10

N/m

18314



28.97 N

m/kg

K2 1303 K2

 319.35 kg/min

 f

 10.52 10.042462 0.02123 bar



1AV2



❶

 12.8.4 Humidification

It is often necessary to increase the moisture content of the air circulated through occupied

spaces. One way to accomplish this is to inject steam. Alternatively, liquid water can be

sprayed into the air. Both cases are shown schematically in Fig. 12.12a. The temperature of

602 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

> T

< T

ωω

Dry-bulb temperatureDry-bulb temperature

Moist air

ωω

Water injected

(vapor or liquid)

(a)(b)(c)

 Figure 12.12 Humidification. (a) Control volume. (b) Steam injected. (c) Liquid injected.

the moist air as it exits the humidifier depends on the condition of the water introduced.

When relatively high-temperature steam is injected, both the humidity ratio and the dry-bulb

temperature would be increased. This is illustrated by the accompanying psychrometric chart

of Fig. 12.12b. If liquid water was injected instead of steam, the moist air may exit the

humidifier with a lower temperature than at the inlet. This is illustrated in Fig. 12.12c. The

example to follow illustrates the case of steam injection. The case of liquid water injection

is considered further in the next section.

EXAMPLE 12.12 Steam-Spray Humidifier

Moist air with a temperature of 22C and a wet-bulb temperature of 9C enters a steam-spray humidifier. The mass flow rate

of the dry air is 90 kg/min. Saturated water vapor at 110C is injected into the mixture at a rate of 52 kg/h. There is no heat

transfer with the surroundings, and the pressure is constant throughout at 1 bar. Using the psychrometric chart, determine at

the exit

(a) the humidity ratio and (b) the temperature, in C.

SOLUTION

Known: Moist air enters a humidifier at a temperature of 22C and a wet-bulb temperature of 9C. The mass flow rate of

the dry air is 90 kg/min. Saturated water vapor at 110C is injected into the mixture at a rate of 52 kg/h.

Find: Using the psychrometric chart, determine at the exit the humidity ratio and the temperature, in C.

Schematic and Given Data:

Moist air

= ?

90 kg/min

= 22°C

= 9°C

Saturated water vapor

at 110°C, m

= 52 kg/h

Boundary

 Figure E12.12

Assumptions:

1. The control volume shown in the accompanying figure operates

at steady state. Changes in kinetic and potential energy can be neg-

lected and

2. There is no heat transfer with the surroundings.

3. The pressure remains constant throughout at 1 bar. Figure A-9

remains valid at this pressure.

 0.

Analysis: (a) The humidity ratio at the exit 

can be found from mass rate balances on the dry air and water individually. Thus

 m

 m

1water2

 m

1dry air2

12.8 Analyzing Air-Conditioning Processes 603

With and where is the mass flow rate of the air, the second of these becomes

Using the inlet dry-bulb temperature, 22C, and the inlet wet-bulb temperature, 9C, the value of the humidity ratio 

can be

found by inspection of the psychrometric chart, Fig. A-9. The result is 

 0.002 kg(vapor)kg(dry air). This value should

be verified as an exercise. Inserting values into the expression for 

(b) The temperature at the exit can be determined using an energy rate balance. With assumptions 1 and 2, the steady-state

form of the energy rate balance reduces to a special case of Eq. 12.55. Namely

(1)

In writing this, the specific enthalpies of the water vapor at 1 and 2 are evaluated as the respective saturated vapor values, and

denotes the enthalpy of the saturated vapor injected into the moist air.

Equation (1) can be rearranged in the following form suitable for use with the psychrometric chart.

(2)

The first term on the right can be obtained from Fig. A-9 at the state defined by the intersection of the inlet dry-bulb tem-

perature, 22C, and the inlet wet-bulb temperature, 9C: 27.2 kJ/kg(dry air). The second term on the right can be evaluated

with the known humidity ratios 

and 

and the value of h

from Table A-2: 2691.5 kJ/kg(vapor). The state at the exit is

fixed by 

and (h

 h

)

 53 kJ/kg(dry air). The temperature at the exit can then be read directly from the chart. The

result is

Alternative Solution:

The following program allows T

to be determined using IT, where is denoted as mdota, is denoted as mdotst, w1

and w2 denote 

and 

, respectively, and so on.

// Given data

T1 = 22 // C

Twb1 = 9 // C

mdota = 90 // kg/min

p = 1 // bar

Tst = 110 // C

mdotst = (52 / 60) // converting kg/h to kg/min

// Evaluate humidity ratios

w1 = w_TTwb(T1,Twb1,p)

w2 = w1 + (mdotst / mdota)

// Denoting the enthalpy of moist air at state 1 by

// h1, etc., the energy balance, Eq. (1), becomes

0 = h1 – h2 + (w2 – w1)*hst

// Evaluate enthalpies

h1 = ha_Tw(T1,w1)

h2 = ha_Tw(T2,w2)

hst = hsat_Px(“Water/Steam”,psat,1)

psat = Psat_T(“Water/Steam”,Tst)

Using the Solve button, the result is T

 23.4C, which agrees closely with the values obtained above, as expected.

A solution of Eq. (2) using data from Tables A-2 and A-22 requires an iterative (trial) procedure. The result is T

 24C,

as can be verified.

Note the use of special Moist Air functions listed in the Properties menu of IT.

 23.5°C.

 vh

 1h

 vh

 1v

 v

0  h

 h

 v

 1v

 v

 0.002 

152 kg/h201 h



60 min 0

90 kg/min

 0.0116

kg1vapor2

kg1dry air2

 v



 v

❶

❷

❶

❷

604 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

 12.8.5 Evaporative Cooling

Cooling in hot, relatively dry climates can be accomplished by evaporative cooling. This in-

volves either spraying liquid water into air or forcing air through a soaked pad that is kept

replenished with water, as shown in Fig. 12.13. Owing to the low humidity of the moist air

entering at state 1, part of the injected water evaporates. The energy for evaporation is pro-

vided by the air stream, which is reduced in temperature and exits at state 2 with a lower

temperature than the entering stream. Because the incoming air is relatively dry, the addi-

tional moisture carried by the exiting moist air stream is normally beneficial.

For negligible heat transfer with the surroundings, no work , and no significant changes

in kinetic and potential energy, the steady-state forms of the mass and energy rate balances

reduce for the control volume of Fig. 12.13a to

where h

denotes the specific enthalpy of the liquid stream entering the control volume. All

the injected water is assumed to evaporate into the moist air stream. The underlined term

accounts for the energy carried in with the injected liquid water. This term is normally much

smaller in magnitude than either of the two moist air enthalpy terms. Accordingly, the enthalpy

of the moist air varies only slightly, as illustrated on the psychrometric chart of Fig. 12.13b.

Recalling that lines of constant mixture enthalpy are closely lines of constant wet-bulb tem-

perature (Sec. 12.7), it follows that evaporative cooling takes place at a nearly constant wet-

bulb temperature.

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

 v

2 1v

 v

 1h

 v

Moist air

, T

< T

ωω

Water at T

Soaked

pad

(a) (b)

Dry-bulb temperature

Mixture enthalpy

per unit mass

of dry air

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

representation.

EXAMPLE 12.13 Evaporative Cooler

Air at 38C and 10% relative humidity enters an evaporative cooler with a volumetric flow rate of 140 m

/min. Moist air exits

the cooler at 21C. Water is added to the soaked pad of the cooler as a liquid at 21C and evaporates 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 38C and   10% enters an evaporative cooler with a volumetric flow rate of 140 m

/min. Moist air exits

the cooler at 21C. Water is added to the soaked pad of the cooler at 21C.

12.8 Analyzing Air-Conditioning Processes 605

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:

Analysis:

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

where is the mass flow rate of the water to the soaked pad. To find requires 

, , and 

. These will now be evalu-

ated in turn.

The humidity ratio 

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 

and p

at T

from Table A-2, we have p

 

 0.0066 bar. With

this, 

 0.00408 kg(vapor)kg(dry air).

The mass flow rate of the dry air can be found as in previous examples using the volumetric flow rate and specific vol-

ume of the dry air. Thus

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

 0.887 m

/kg(dry air).

Inserting values, the mass flow rate of the dry air is

To find the humidity ratio 

, reduce the steady-state forms of the mass and energy rate balances using assumption 1 to

obtain

With the same reasoning as in previous examples, this can be expressed as

where h

denotes the specific enthalpy of the water entering the control volume at 21C. Solving for 

where c

 1.005 kJ/kg K. With h

, h

, and h

from Table A-2

 .011

kg1vapor2

kg1dry air2



11.0052

138  212 0.00408 12570.7  88.142

2451.8



 h

 v

 h



 T

2 v

 h

0  1h

 vh

 1v

 v

 1h

 vh

0  1m

 m

2 m

 1m

 m



140 m

/min

0.887 m

/kg1dry air2

 157.8

kg1dry air2

min



1AV2

 m

 v

Water at 21C

= 21C

Boundary

Soaked pad

= 38C

= 10%

___

min

(AV)

= 140

 Figure E12.13

Assumptions:

1. The control volume shown in the accompanying figure oper-

ates at steady state. Changes in kinetic and potential energy can

be neglected and .

2. There is no heat transfer with the surroundings.

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

orates fully into the moist air.

4. The pressure remains constant throughout at 1 atm.

 0

❶

❷