Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 14.7, Problem 128RP

Air enters a cooling section at 97 kPa, 35°C, and 30 percent relative humidity at a rate of 6 m3/min, where it is cooled until the moisture in the air starts condensing. Determine (a) the temperature of the air at the exit and (b) the rate of heat transfer in the cooling section.

(a)

Expert Solution
Check Mark
To determine

The temperature of the air at the exit.

Answer to Problem 128RP

The temperature of the air at the exit is 14.8°C.

Explanation of Solution

As the process is a steady flow and thus the mass flow rate of dry air remains constant during the entire process.

m˙a1=m˙a2=m˙a

Here, the mass flow rate of air at inlet is m˙a1, mass flow rate of dry air at exit is m˙a2 and mass flow rate of dry air is m˙a.

The amount of moisture in the air remains constant as it flows through the heating section as process involves no dehumidification or humidification.

ω1=ω2 (I)

Here, specific humidity at state 1 and 2 is ω1andω2 respectively.

Express initial partial pressure.

Pν1=ϕ1Pg1=ϕ1Psat@35°C (II)

Here, relative humidity at state 1 is ϕ1, initial vapor pressure is Pg1 and saturation pressure at temperature of 35°C is Psat@35°C.

Express initial humidity ratio.

ω1=0.622Pν1P1Pν1 (III)

Here, pressure at state 1 is P1.

Express initial enthalpy.

h1=cpT1+ω1hg1@35°C (IV)

Here, specific heat at constant pressure is cp and initial specific enthalpy saturated vapor at temperature of 35°C is hg1@35°C.

Express specific volume at state 1.

v1=RaT1Pa1=RaT1P1Pν1 (V)

Here, gas constant of air is Ra, partial pressure of air at state 1 is Pa1 and temperature at state 1 is T1.

As the air at the final state is saturated and the vapor pressure during the process will remain constant, thus the dew point temperature is the exit temperature.

T2=Tdp=Tsat@Pv (VI)

Here, exit temperature is T2, dew point temperature is Tdp and saturation temperature at vapor pressure is Tsat@Pv.

Conclusion:

Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure and initial specific enthalpy saturated vapor at temperature of 35°C.

Psat@35°C=5.629kPahg1@35°C=2564.6kJ/kg

Substitute 0.3 for ϕ1 and 5.629kPa for Psat@35°C in Equation (II).

Pν1=(0.3)(5.629kPa)=1.69kPa

Substitute 1.69kPa for Pν1 and 97kPa for P1 in Equation (III).

ω1=0.622(1.69kPa)97kPa1.69kPa=0.0110kgH2O/kgdryair

Substitute 0.0110kgH2O/kgdryair for ω1 in Equation (I).

ω2=0.0110kgH2O/kgdryair

Refer Table A-2, “ideal-gas specific heats of various common gases”, and write the properties of air.

cp=1.005kJ/kg°CRa=0.287kPam3/kgK

Substitute 1.005kJ/kg°C for cp, 35°C for T1, 0.0110kgH2O/kgdryair for ω1 and 2564.6kJ/kg for hg1@35°C in Equation (IV).

h1=(1.005kJ/kg°C)(35°C)+(0.0110)(2564.6kJ/kg)=63.44kJ/kgdryair

Substitute 0.287kPam3/kgK for Ra, 35°C for T1, 1.69kPa for Pν1 and 97kPa for P1 in Equation (V).

v1=(0.287kPam3/kgK)(35°C)97kPa1.69kPa=(0.287kPam3/kgK)(35+273)K95.31kPa=(0.287kPam3/kgK)(308K)95.31kPa=0.927m3/kgdryair

Substitute 1.69kPa for Pv in Equation (VI).

T2=Tsat@1.69kPa (VII)

Refer Table A-5, “saturated water-pressure table”, and write the saturation temperature or exit temperature at pressure of 1.69kPa using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2x1)(y3y1)(x3x1)+y1 (VIII)

Here, the variables denote by x and y is pressure and exit or saturation temperature respectively.

Show the specific enthalpy at state 2 corresponding to specific entropy as in Table (1).

Pressure

P(kPa)

Saturation or exit temperature

Tsat(°C)

1.5 (x1)13.02 (y1)
1.69 (x2)(y2=?)
2 (x3)17.50 (y3)

Substitute 1.5kPa,1.69kPaand2kPa for x1,x2andx3 respectively, 13.02°C for y1 and 17.50°C for y3 in Equation (VIII).

y2=(1.69kPa1.5kPa)(17.50°C13.02°C)(2kPa1.5kPa)+13.02°C=14.8°C=Tsat@1.69kPa

Substitute 14.8°C for Tsat@1.69kPa in Equation (VII).

T2=14.8°C

Hence, the temperature of the air at the exit is 14.8°C.

(b)

Expert Solution
Check Mark
To determine

The rate of heat transfer in the cooling section.

Answer to Problem 128RP

The rate of heat transfer in the cooling section is 134kJ/min.

Explanation of Solution

Express the enthalpy of air at exit.

h2=cpT2+ω2hg2@14.8°C (IX)

Here, temperature at exit is 14.8°C and initial specific enthalpy saturated vapor at temperature of 14.8°C is hg2@14.8°C.

Express mass flow rate of air.

m˙a=ν˙1v1 (X)

Here, volume flow rate at inlet is ν˙1.

Express the rate of heat transfer in the cooling section.

Q˙out=m˙a(h1h2) (XI)

Conclusion:

Refer Table A-4, “saturated water-temperature table”, and write the final specific enthalpy saturated vapor at temperature of 14.8°C using an interpolation method.

hg2@14.8°C (XII)

Show the final specific enthalpy saturated vapor corresponding to exit temperature as in Table (2).

Exit temperature

T2(°C)

Final specific enthalpy

saturated vapor

hg2(kJ/kg)

10 (x1)2519.2 (y1)
14.8 (x2)(y2=?)
15 (x3)2528.3 (y3)

Use excels and tabulates the values from Table (2) in Equation (VIII) to get,

hg2@14.8°C=2528.1kJ/kg

Substitute 2528.1kJ/kg for hg2@14.8°C in Equation (XII).

hg2@14.8°C=2528.1kJ/kg

Substitute 1.005kJ/kg°C for cp, 14.8°C for T1, 0.0110kgH2O/kgdryair for ω2 and 2528.1kJ/kg for hg2@14.8°C in Equation (IX).

h2=(1.005kJ/kg°C)(14.8°C)+(0.0110)(2528.1kJ/kg)=42.78kJ/kgdryair

Substitute 6m3/min for ν˙1 and 0.927m3/kgdryair for v1 in Equation (X).

m˙a=6m3/min0.927m3/kgdryair=6.47kg/min

Substitute 6.47kg/min for m˙a, 63.44kJ/kgdryair for h1 and 42.78kJ/kgdryair for h2 in Equation (XI).

Q˙out=(6.47kg/min)(63.44kJ/kg42.78kJ/kg)=134kJ/min

Hence, the rate of heat transfer in the cooling section is 134kJ/min.

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Chapter 14 Solutions

Thermodynamics: An Engineering Approach

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