CENGEL'S 9TH EDITION OF THERMODYNAMICS:
CENGEL'S 9TH EDITION OF THERMODYNAMICS:
9th Edition
ISBN: 9781260917055
Author: CENGEL
Publisher: MCG
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Chapter 14.7, Problem 81P

Repeat Prob. 14–79 for a total pressure of 88 kPa for air.

(a)

Expert Solution
Check Mark
To determine

The rate of heat transfer.

Answer to Problem 81P

The rate of heat transfer is 451.7kJ/min.

Explanation of Solution

Express initial partial pressure.

Pν1=ϕ1Pg1=ϕ1Psat@32°C (I)

Here, relative humidity at state 1 is ϕ1, initial vapor pressure is Pg1 and saturation pressure at temperature of 32°C is Psat@32°C.

Express the dew point temperature of the incoming air.

Tdp=Tsat@Pν1 (II)

Here, initial specific humidity is ϕ1.

Express initial partial pressure.

Pa1=P1Pν1 (III)

Here, initial pressure is P1.

Express initial specific volume.

v1=RaT1Pa1 (IV)

Here, universal gas constant of air is Ra and initial temperature is T1.

Express initial specific humidity.

ω1=0.622Pν1P1Pν1 (V)

Express initial enthalpy.

h1=cpT1+ω1hg1 (VI)

Here, initial specific enthalpy at saturated vapor is hg1, specific heat at constant pressure of air is cp and temperature at state 1 is T1.

Express final partial pressure.

Pν2=ϕ2Pg2=ϕ2Psat@20°C (VII)

Here, relative humidity at state 2 is ϕ2, final vapor pressure is Pg2 and saturation pressure at temperature of 20°C is Psat@20°C.

Express final partial pressure.

Pa2=P2Pν2 (VIII)

Here, final pressure is P2.

Express final specific volume.

v2=RaT2Pa2 (IX)

Here, final temperature is T2.

Express final specific humidity.

ω2=0.622Pν2P2Pν2 (X)

Express final enthalpy.

h2=cpT2+ω2hg2 (XI)

Here, final specific enthalpy at saturated vapor is hg2, specific heat at constant pressure of air is cp and temperature at state 2 is T2.

Express initial volume rate of air.

ν˙1=V1A1=V1[πD24] (XII)

Here, initial volume and area is V1andA1 respectively, and diameter is D.

Express the mass flow rate of air at inlet.

m˙a1=ν˙1v1 (XIII)

Here, initial specific volume is v1.

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

Express water mass balance to the combined cooling to obtain the mass flow rate of water.

m˙w,i=m˙w,em˙a1ω1=m˙a2ω2+m˙wm˙w=m˙a1(ω1ω2) (XIV)

Here, mass flow rate of water at inlet and exit is m˙w,iandm˙w,e respectively, specific humidity at state 1 and 2 is ω1andω2 respectively and mass flow rate of water is m˙w.

Express the cooling rate when the condensate leaves the system by applying an energy balance on the humidifying section.

E˙inE˙out=ΔE˙systemE˙inE˙out=0E˙in=E˙outm˙ihi=Q˙out+m˙ehe

Q˙out=m˙a1h1(m˙a2h2+m˙whw)=m˙a1(h1h2)m˙whw (XV)

Here, rate of heat rejected or cooling rate when the condensate leaves the system is Q˙out, the rate of total energy entering the system is E˙in, the rate of total energy leaving the system is E˙out, the rate of change in the total energy of the system is ΔE˙system, initial and exit mass flow rate is m˙iandm˙e respectively, enthalpy at inlet and exit is hiandhe respectively, enthalpy at state 1 and 2 is h1andh2 respectively, and enthalpy of water is hw.

Conclusion:

Refer Table A-4, “saturated water-temperature table”, and write saturation pressure at temperature of 32°C using an interpolation method.

Write the formula of interpolation method of two variables.

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

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

Show the saturation pressure corresponding to temperature as in Table (1).

Temperature

T(°C)

Saturation pressure

Psat(kPa)

30 (x1)4.2469 (y1)
32 (x2)(y2=?)
35 (x3)5.6291 (y3)

Substitute 30°C,32°Cand35°C for x1,x2,andx3 respectively, 4.2469kPa for y1 and 5.6291kPa for y3 in Equation (XVI).

y2=(32°C30°C)(5.6291kPa4.2469kPa)(35°C30°C)+4.2469kPa=4.76kPa=Psat@32°C

Thus, the saturation pressure at temperature of 32°C is,

Psat@32°C=4.76kPa

Substitute 0.7 for ϕ1 and 4.76kPa for Psat@32°C in Equation (I).

Pν1=(0.7)(4.76kPa)=3.332kPa

Substitute 3.332kPa for Pν1 in Equation (II).

Tdp=Tsat@3.332kPa (XVII)

Here, saturation temperature at pressure of 3.332kPa is Tsat@3.332kPa.

Refer Table A-5, “saturated water-pressure table”, and write saturation temperature at pressure of 3.332kPa using an interpolation method.

Show the saturation temperature corresponding to pressure as in Table (2).

Pressure

P(kPa)

Saturation temperature

Tsat(°C)

3 (x1)24.08 (y1)
3.332 (x2)(y2=?)
4 (x3)28.96 (y3)

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

Tsat@3.332kPa=25.8°C

Substitute 25.8°C for Tsat@3.332kPa in Equation (XVII).

Tdp=25.8°C

Substitute 88kPa for P1 and 3.332kPa for Pν1 in Equation (III).

Pa1=88kPa3.332kPa=84.67kPa

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

Ra=0.287kPam3/kgcp=1.005kJ/kg°C

Substitute 0.287kPam3/kg for Ra, 32°C for T1 and 84.67kPa for Pa1 in Equation (IV).

v1=(0.287kPam3/kg)(32°C)84.67kPa=(0.287kPam3/kg)[(32+273)K]84.67kPa=(0.287kPam3/kg)(305K)84.67kPa=1.034m3/kgdryair

Substitute 88kPa for P1 and 3.332kPa for Pν1 in Equation (V).

ω1=0.622(3.332kPa)85kPa3.332kPa=0.02448kgH2O/kgdryair

Refer Table A-4, “saturated water-temperature table”, and write the specific enthalpy saturated vapor at temperature of 32°C using an interpolation method.

Show the specific enthalpy saturated vapor corresponding to temperature as in Table (3).

Temperature

T(°C)

specific enthalpy

saturated vapor

hg1(kJ/kg)

30 (x1)2555.6 (y1)
32 (x2)(y2=?)
35 (x3)2564.6 (y3)

Use excels and tabulates the values from Table (3) in Equation (XVI) to get,

hg1@32°C=hg1=2559.2kJ/kg

Substitute 1.005kJ/kg°C for cp, 32°C for T1, 0.02448 for ω1 and 2559.2kJ/kg for hg1 in Equation (VI).

h1=(1.005kJ/kg°C)(32°C)+(0.02448)(2559.2kJ/kg)=94.80kJ/kgdryair

Refer Table A-4, “saturated water-temperature table”, and write saturation pressure at temperature of 20°C.

Psat@20°C=2.339kPa

Substitute 1 for ϕ2 and 2.339kPa for Psat@20°C in Equation (VII).

Pν2=(1)(2.339kPa)=2.339kPa

Substitute 88kPa for P2 and 2.339kPa for Pν2 in Equation (VIII).

Pa2=88kPa2.339kPa=85.661kPa

Substitute 0.287kPam3/kg for Ra, 20°C for T2 and 85.661kPa for Pa2 in Equation (IX).

v2=(0.287kPam3/kg)(20°C)85.661kPa=(0.287kPam3/kg)[(20+273)K]85.661kPa=(0.287kPam3/kg)(293K)85.661kPa=0.9817m3/kgdryair

Substitute 88kPa for P2 and 2.339kPa for Pν1 in Equation (X).

ω2=0.622(2.339kPa)85kPa2.339kPa=0.01699kgH2O/kgdryair

Refer Table A-4, “saturated water-temperature table”, and write the specific enthalpy saturated vapor at temperature of 20°C.

hg2@20°C=hg2=2537.4kJ/kg

Substitute 1.005kJ/kg°C for cp, 20°C for T2, 0.01699 for ω2 and 2537.4kJ/kg for hg2 in Equation (XI).

h2=(1.005kJ/kg°C)(20°C)+(0.01699)(2537.4kJ/kg)=63.20kJ/kgdryair

Refer Table A-4, “saturated water-temperature table”, and write the enthalpy of water at temperature of 20°C.

hw=hf@20°C=83.915kJ/kg

Here, specific enthalpy saturation liquid at temperature of 20°C is hf@20°C.

Perform unit conversion of diameter from cmtom.

D=40cm=40cm[m100cm]=0.4m

Substitute 120m/min for V1 and 0.4m for D in Equation (XII).

ν˙1=(120m/min)[π(0.4m)24]=15.08m3/min

Substitute 15.08m3/min for ν˙1 and 1.034m3/kgdryair for ν1 in Equation (XIII).

m˙a1=15.08m3/min1.034m3/kgdryair=14.59kg/min

Substitute 14.59kg/min for m˙a1, 0.02448kgH2O/kgdryair for ω1 and 0.01699kgH2O/kgdryair for ω2 in Equation (XIV).

m˙w=(14.59kg/min)(0.024480.01699)=0.1093kg/min

Substitute 14.59kg/min for m˙a1, 94.80kJ/kgdryair for h1, 63.20kJ/kgdryair for h2, 0.1093kg/min for m˙w and 83.915kJ/kg for hw in Equation (XV).

Q˙out=[(16.87kg/min)[(94.8063.20)kJ/kgdryair](0.1093kg/min)(83.915kJ/kg)]=451.7kJ/min

Hence, the rate of heat transfer is 451.7kJ/min.

(b)

Expert Solution
Check Mark
To determine

The mass flow rate of the water.

Answer to Problem 81P

The mass flow rate of the water is 18.01kg/min.

Explanation of Solution

Express the mass flow rate of the water.

m˙coolingwater=Q˙wcpΔT (XVIII)

Here, mass flow rate of the water is Q˙w, specific heat at constant pressure of water is cp and rise in temperature is ΔT.

Conclusion:

Refer Table A-2, “ideal-gas specific heats of various common gases”, and write specific heat at constant pressure of water.

cp=4.18kJ/kg°C

Substitute 451.7kJ/min for Q˙w, 4.18kJ/kg°C for cp and 6°C for ΔT in Equation (XVIII).

m˙coolingwater=451.7kJ/min(4.18kJ/kg°C)(6°C)=18.01kg/min

Hence, the mass flow rate of the water is 18.01kg/min.

c)

Expert Solution
Check Mark
To determine

The exit velocity of the airstream.

Answer to Problem 81P

The exit velocity of the airstream is 113.9m/min.

Explanation of Solution

Express the exit velocity of the airstream.

V2=ν2ν1V1 (XIX)

Conclusion:

Substitute 1.034m3/kgdryair for ν1, 0.9817m3/kgdryair for ν2 and 120m/min for V1 in Equation (XIX).

V2=0.9817m3/kgdryair1.034m3/kgdryair(120m/min)=113.9m/min

Hence, the exit velocity of the airstream is 113.9m/min.

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Chapter 14 Solutions

CENGEL'S 9TH EDITION OF THERMODYNAMICS:

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