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

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12.8.2 Conditioning Moist Air at Constant Composition

Building air-conditioning systems frequently heat or cool a moist air stream with no

change in the amount of water vapor present. In such cases the humidity ratio v remains

constant, while relative humidity and other moist air parameters vary. Example 12.10

gives an elementary illustration using the methodology of Sec. 12.8.1.

Heating Moist Air in a Duct

c c c c EXAMPLE 12.10 c

Moist air enters a duct at 108C, 80% relative humidity, and a volumetric flow rate of 150 m

/min. The mixture

is heated as it flows through the duct and exits at 308C. No moisture is added or removed, and the mixture pres-

sure remains approximately constant at 1 bar. For steady-state operation, determine (a) the rate of heat transfer,

in kJ/min, and (b) the relative humidity at the exit. Changes in kinetic and potential energy can be ignored.

SOLUTION

Known:

Moist air that enters a duct at 108C and f 5 80% with a volumetric flow rate of 150 m

/min is heated

at constant pressure and exits at 308C. No moisture is added or removed.

Find: Determine the rate of heat transfer, in kJ/min, and the relative humidity at the exit.

Schematic and Given Data:

Engineering Model:

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

5 0.

3. The entering and exiting

moist air streams are regarded

as ideal gas mixtures adhering

to the Dalton model.

Fig. E12.10a

)

Boundary

= 30°C

= 150

= 10°C

= 80%

(AV)

___

min

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

5 m





dry air

5 m





water vapor

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

and m

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

5 v

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

0 5 Q

2 W

1 1m

1 m

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 Q

5 m

2 h

1 m

2 h

12.8 Analyzing Air-Conditioning Processes 743

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744 Chapter 12

Ideal Gas Mixture and Psychrometric Applications

Noting that m

5 vm

, where v is the humidity ratio, the expression for

can be written in the form

➊

5 m

31h

2 h

21 v1h

2 h

24 (a)

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

, respectively: h

5 283.1 kJ/kg, h

5 303.2 kJ/kg. The specific enthalpies of the water vapor are found using

< h

and data from Table A-2 at T

and T

, respectively: h

5 2519.8 kJ/kg, h

5 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, y

is the specific volume of the dry air evaluated at T

and the partial pressure of the dry air

. Using the ideal gas equation of state

M2T

The partial pressure p

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

vapor p

: p

5 p 2 p

. To find p

, use the given inlet relative humidity and the saturation pressure at 108C

from Table A-2

5 f

0.8

0.01228 bar

5 0.0098 bar

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

5 0.9902 bar. The specific volume of the dry air is then

8314 N ? m

28.97 kg ? K

1283 K2

0.9902 3 10

5 0.82 m

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

150 m

min

0.82 m

5 182.9 kg

min

The humidity ratio v can be found from

v 5 0.622

p 2 p

5 0.622

0.0098

1 2 0.0098

5 0.00616

vapor

dry air

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

5 182.9

303.2 2 283.1

0.00616

2556.3 2 2519.8

717 k

min

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

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

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

5 p

5 0.0098 bar. The relative

humidity at the exit is then

➋

0098

0.04246

5 0.231123.1%2

where p

is from Table A-2 at 308C.

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 f

5 80% and a dry-bulb temperature of 108C. 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

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12.8 Analyzing Air-Conditioning Processes 745

Mixture enthalpy

10°C30°C

Dry-bulb temperature

= 100%

= 80%

Fig. E12.10b

Using the psychrometric chart, what is the dew point tempera-

ture, in 8C, for the moist air entering? At the exit? Ans.

78C, same.

5 v

and a dry-bulb temperature of 308C. 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 rearranging Eq. (a) of part (a) to read

5 m

31h

1 vh

2 1h

1 vh

4 (b)

To evaluate Q

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

1 vh

)

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

1 vh

)

25.7 kJ/kg(dry air), (h

1 vh

)

5 45.9 kJ/kg(dry air).

Using the specific volume value y

at the inlet state read from the chart together with the given volumetric

flow rate at the inlet, the mass flow rate of the dry air is found as

150 m

min

0.81 m

dry air

5 185

kg1dry air

min

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

5 185

dry air

min

145.9 2 25.72

dry air

5 3737

min

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

➊

The first underlined term in this equation for Q

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 c

for dry air varies only slightly

over the interval from 10 to 308C (Table A-20), the specific enthalpy change of

the dry air could be evaluated alternatively with c

5 1.005 kJ/kg ? K.

➋

No water is added or removed as the moist air passes through the duct at

constant pressure; accordingly, the humidity ratio v and the partial pressures

and p

remain constant. However, because the saturation pressure

increases as the temperature increases from inlet to exit, the relative humid-

ity decreases: f

, f

➌

The mixture pressure, 1 bar, differs slightly from the pressure used to construct

the psychrometric chart, 1 atm. This difference is ignored.

Ability to…

❑

apply psychrometric termi-

nology and principles.

❑

apply mass and energy bal-

ances for heating at con-

stant composition in a con-

trol volume at steady state.

❑

retrieve necessary property

data.

✓

Skills Developed

➌

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746 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

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 will

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 circulates. 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 condenses 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 at state 2 might be uncomfortable for direct use in occu-

pied spaces. However, by passing the stream through a following heating section, it can

be brought to a condition—state 3—most occupants would regard as comfortable.

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

refrigerant circulates. This requires the use of mass and energy rate balances for the

dehumidifier section. They are developed next.

Mass Balance

The mass flow rate of the condensate m

can be related to the mass flow rate of the

dry air m

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

ing through the dehumidifier section. At steady state

5 m

(dry air)

5 m

1 m

(water)

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

. Solving for the mass flow

rate of the condensate

5 m

2 m

Introducing m

5 v

and m

5 v

, the amount of water condensed per unit

mass of dry air passing through the device is

5 v

2 v

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

Fig. 12.11 Dehumidification. (a) Equipment schematic. (b) Psychrometric chart representation.

TAKE NOTE...

A dashed line on a property

diagram signals only that

a process has occurred

between initial and final

equilibrium states, and

does not define a path for

the process.

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This expression requires the humidity ratios v

and v

. Because no moisture is added

or removed in the heating section, it can be concluded from conservation of mass

that v

5 v

, so v

can be used in the above equation in place of v

Energy Balance

The mass flow rate of the refrigerant through the cooling coil m

can be related to

the mass flow rate of the dry air m

by means of an energy rate balance applied to the

dehumidifier section. With W

5 0, negligible heat transfer with the surroundings,

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

reduces at steady state to

0 5 m

2 h

1 m

2 m

1 m

where h

and h

denote the specific enthalpy values of the refrigerant entering and

exiting the dehumidifier section, respectively. Introducing m

5 v

, m

5 v

and m

2 v

0 5 m

2 h

1 m

2 h

1 v

2 v

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

saturated vapor values corresponding to T

and T

, respectively. Since the condensate

is assumed to exit as a saturated liquid at T

, h

5 h

. Solving for the refrigerant

mass flow rate per unit mass of dry air flowing through the device

2 h

1 v

2 v

2 h

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 v

, to state 2, where

the mixture is saturated (f

5 100%), the temperature is T

, T

, and the humidity

ratio is v

, v

. During the subsequent heating process, the humidity ratio remains

constant, v

5 v

, and the temperature increases to T

. Since all states visited are not

equilibrium states, these processes are indicated on the psychrometric chart by dashed

lines.

The example that follows provides an illustration involving dehumidification where

one of the objectives is the refrigerating capacity of the cooling coil.

Assessing Dehumidifier Performance

c c c c EXAMPLE 12.11 c

Moist air at 308C and 50% relative humidity enters a dehumidifier operating at steady state with a volumet-

ric flow rate of 280 m

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

exits the dehumidifier saturated at 108C. Saturated moist air exits in a separate stream at the same tem-

perature. 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 308C and 50% relative humidity with a volumetric flow rate of 280 m

/min.

Condensate and moist air exit in separate streams at 108C.

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.

12.8 Analyzing Air-Conditioning Processes 747

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748 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Schematic and Given Data:

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

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

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

1AV2

21T

The partial pressure of the dry air p

can be determined from p

5 p

2 p

. Using the relative humidity at the

inlet f

and the saturation pressure at 308C from Table A-2

5 f

5 10.5210.0424625 0.02123 bar

Thus, p

5 1.013 2 0.02123 5 0.99177 bar. Inserting values into the expression for m

gives

1280 m

min210.99177 3 10

18314

28.97 N ? m

kg ? K21303 K2

5 319.35 kg

min

(b) Conservation of mass for the water requires m

5 m

1 m

. With m

5 v

and m

5 v

, the rate at

which water is condensed per unit mass of dry air is

5 v

2 v

The humidity ratios v

and v

can be evaluated using Eq. 12.43. Thus, v

5 0.662 a

2 p

b5 0.622a

0.02123

0.99177

b5 0.0133

kg1vapor

kg1dry air2

Since the moist air is saturated at 108C, p

equals the saturation pressure at 108C: p

5 0.01228 bar from Table

A-2. Equation 12.43 then gives v

5 0.0076 kg(vapor)/kg(dry air). With these values for v

and v

5 0.0133 2 0.0076 5 0.0057

kg1conde

nsate2

kg1dry air2

Heating coilCooling coil

Condensate,

saturated at

= 10°C

Control volume

13 2

Saturated

mixture

10°

(AV)

= 280 m

/min

= 30°C

= 50%

Fig. E12.11a

Engineering Model:

The control volume shown in the accompanying fig-

ure operates at steady state. Changes in kinetic and

potential energy can be neglected, and W

5 0.

2. There is no significant heat transfer to the sur-

roundings.

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

temperature T

5. The moist air streams are regarded as ideal gas

mixtures adhering to the Dalton model.

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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

0 5 Q

1 1m

1 m

22 m

2 1m

1 m

2 (a)

With m

5 v

, m

5 v

, and m

5 1v

2 v

, this becomes

5 m

31h

2 h

22 v

1 v

1 1v

2 v

4 (b)

which agrees with Eq. 12.55. In Eq. (b), the specific enthalpies of the water vapor at 1 and 2 are evaluated at

the saturated vapor values corresponding to T

and T

, 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, Eq. (b)

reads

5 1319.35231283.1 2 303.222 0.013312556.321 0.007612519.821 0.0057142.0124

5211,084 kJ

min

Since 1 ton of refrigeration equals a heat transfer rate of 211 kJ/min (Sec. 10.2.1), the required refrigerating

capacity is 52.5 tons.

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

As shown on the sketch of the psychrometric chart, Fig. E12.11b, the state of the moist air at the inlet 1 is defined

by f 5 50% and a dry-bulb temperature of 308C. At 2, the moist air is saturated at 108C. Rearranging Eq. (a),

we get

5 m

1 vh

1 1v

2 v

4 (c)

The underlined terms and humidity ratios, v

and v

, can be read directly from the chart. The mass flow rate of

the dry air can be determined using the volumetric flow rate at the inlet and y

read from the chart. The specific

enthalpy h

is obtained (as above) from Table A-2: h

at T

. The details are left as an exercise.

Using the psychrometric chart, determine the wet-bulb tem-

perature of the moist air entering the dehumidifier, in °C. Ans.

22°C.

Specific enthalpy

of moist air, in

kJ/kg(dry air)

f = 50%

f = 100%

10°C

-bulb temperature

30°C

+ w h

)

+ w h

)

Fig. E12.11b

12.8 Analyzing Air-Conditioning Processes 749

Ability to…

❑

apply psychrometric termi-

nology and principles.

❑

apply mass and energy

balances for a dehumidifica-

tion process in a control

volume at steady state.

❑

retrieve property data for

dry air and water.

❑

apply the psychrometric

chart.

✓

Skills Developed

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750 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Analyzing a Steam-Spray Humidifier

c c c c EXAMPLE 12.12 c

Moist air with a temperature of 228C and a wet-bulb temperature of 98C enters a steam-spray humidifier. The

mass flow rate of the dry air is 90 kg/min. Saturated water vapor at 1108C 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 8C.

SOLUTION

Known:

Moist air enters a humidifier at a temperature of 228C and a wet-bulb temperature of 98C. The mass flow

rate of the dry air is 90 kg/min. Saturated water vapor at 1108C 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 8C.

Schematic and Given Data:

Moist air

= ?

90 kg/min

= 22°C

= 9°C

Saturated water vapor

at 110°C, m

= 52 kg/h

Boundary

Fig. E12.12a

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

remains valid at this pressure.

4. The moist air streams are regarded as ideal gas mixtures adher-

ing to the Dalton model.

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 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 are increased. This is illustrated by the

accompanying psychrometric chart of Fig. 12.12b. If liquid water is 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.

> T

< T

ωω

Dry-bulb temperatureDry-bulb temperature

Moist air

ωω

Water injected

(vapor or liquid)

(a)(b)(c

)

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

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Analysis:

(a)

The humidity ratio at the exit v

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

5 m



1dry air2

1 m

5 m



1water2

With m

5 v

, and m

5 v

, where m

is the mass flow rate of the air, the second of these becomes

5 v

Using the inlet dry-bulb temperature, 228C, and the inlet wet-bulb temperature, 98C, the value of the humidity ratio

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

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

This value should be verified as an exercise. Inserting values into the expression for v

5 0.002 1

152 kg

1 h

60 min

90 kg

min

5 0.0116

g1vapor2

kg1dry air2

(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

0 5 h

2 h

1 v

1 1v

2 v

(a)

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

values, and h

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

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

➊

1 vh

5 1h

1 vh

1 1v

2 v

(b)

As shown on the sketch of the psychrometric chart, Fig. E12.12b, the first term on the right of Eq. (b) can be obtained

from Fig. A-9 at the inlet state, defined by the intersection of the inlet dry-bulb temperature, 228C, and the inlet wet-bulb

temperature, 98C; the value is 27.2 kJ/kg(dry air). The second term on the right can be evaluated using the known humid-

ity ratios v

and v

and h

from Table A-2: 2691.5 kJ/kg(vapor). The value of the second term of Eq. (b) is 25.8 kJ/kg

(dry air). The state at the exit is then fixed by v

and (h

1 vh

)

5 53 kJ/kg(dry air), calculated from the two values

just determined. Finally, the temperature at the exit can be read directly from the chart. The result is T

< 23.58C.

Specific enthalpy of moist air,

in kJ/kg(dry air)

= 23.5°C

+ w h

)

= 53 kJ/kg(dry air)

= 9°C

= 22°C

+ w h

)

= 27.2 kJ/kg(dry air)

Fig. E12.12b

Alternative IT Solution:

➋

The following program allows T

to be determined using IT, where m

is denoted as mdota, m

is denoted as

mdotst, w1 and w2 denote v

and v

, respectively, and so on.

12.8 Analyzing Air-Conditioning Processes 751

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752 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

// Given data

T1 5 22 // 8C

Twb1 5 9 // 8C

mdota 5 90 // kg/min

p 5 1 // bar

Tst 5 110 // 8C

mdotst 5 (52 / 60) // converting kg/h to kg/min

// Evaluate humidity ratios

w1 5 w_TTwb (T1,Twb1,p)

w2 5 w1 1 (mdotst / mdota)

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

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

0 5 h1 2 h2 1 (w2 2 w1)*hst

// Evaluate enthalpies

h1 5 ha_Tw(T1,w1)

h2 5 ha_Tw(T2,w2)

hst 5 hsat_Px(“Water/Steam”,psat,1)

psat 5 Psat_T(“Water/Steam ”,Tst)

Using the Solve button, the result is T

5 23.48C, which agrees closely with the

values obtained above, as expected.

➊

A solution of Eq. (b) using data from Tables A-2 and A-22 requires an

iterative (trial) procedure. The result is T

5 248C, as can be verified.

➋

Note the use 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.

63%.

TAKE NOTE...

Evaporative cooling takes

place at a nearly constant

wet-bulb temperature.

12.8.5 Evaporative Cooling

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

This involves 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 humid-

ity of the moist air entering at state 1, part of the injected water evaporates. The

energy for evaporation is provided 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 additional moisture carried by the exiting moist air

stream is normally beneficial.

For negligible heat transfer with the surroundings, no work W

, 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 this special case

of Eq. 12.55:

1 v

2 v

1 1h

1 v

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

Ability to…

❑

apply psychrometric termi-

nology and principles.

❑

apply mass and energy

balances for a spray humidi-

fication process in a control

volume at steady state.

❑

retrieve necessary property

data using the psychromet-

ric chart.

❑

apply IT for psychrometric

analysis.

✓

Skills Developed

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