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Repeat Prob. 14–79 for a total pressure of 88 kPa for air.
(a)
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The rate of heat transfer.
Answer to Problem 81P
The rate of heat transfer is 451.7 kJ/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=P1−Pν1 (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=0.622Pν1P1−Pν1 (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=P2−Pν2 (VIII)
Here, final pressure is P2.
Express final specific volume.
v2=RaT2Pa2 (IX)
Here, final temperature is T2.
Express final specific humidity.
ω2=0.622Pν2P2−Pν2 (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.
˙ν1=V1A1=V1[πD24] (XII)
Here, initial volume and area is V1 and A1 respectively, and diameter is D.
Express the mass flow rate of air at inlet.
˙ma1=˙ν1v1 (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.
˙ma1=˙ma2=˙ma
Here, mass flow rate of dry air at exit is ˙ma2 and mass flow rate of dry air is ˙ma.
Express water mass balance to the combined cooling to obtain the mass flow rate of water.
∑˙mw,i=∑˙mw,e˙ma1ω1=˙ma2ω2+˙mw˙mw=˙ma1(ω1−ω2) (XIV)
Here, mass flow rate of water at inlet and exit is ˙mw,i and ˙mw,e respectively, specific humidity at state 1 and 2 is ω1 and ω2 respectively and mass flow rate of water is ˙mw.
Express the cooling rate when the condensate leaves the system by applying an energy balance on the humidifying section.
˙Ein−˙Eout=Δ˙Esystem˙Ein−˙Eout=0˙Ein=˙Eout∑˙mihi=˙Qout+∑˙mehe
˙Qout=˙ma1h1−(˙ma2h2+˙mwhw)=˙ma1(h1−h2)−˙mwhw (XV)
Here, rate of heat rejected or cooling rate when the condensate leaves the system is ˙Qout, the rate of total energy entering the system is ˙Ein, the rate of total energy leaving the system is ˙Eout, the rate of change in the total energy of the system is Δ˙Esystem, initial and exit mass flow rate is ˙mi and ˙me respectively, enthalpy at inlet and exit is hi and he respectively, enthalpy at state 1 and 2 is h1 and h2 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=(x2−x1)(y3−y1)(x3−x1)+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°C and 35°C for x1, x2, and x3 respectively, 4.2469 kPa for y1 and 5.6291 kPa for y3 in Equation (XVI).
y2=(32°C−30°C)(5.6291 kPa−4.2469 kPa)(35°C−30°C)+4.2469 kPa=4.76 kPa=Psat@32°C
Thus, the saturation pressure at temperature of 32°C is,
Psat@32°C=4.76 kPa
Substitute 0.7 for ϕ1 and 4.76 kPa for Psat@32°C in Equation (I).
Pν1=(0.7)(4.76 kPa)=3.332 kPa
Substitute 3.332 kPa for Pν1 in Equation (II).
Tdp=Tsat@3.332 kPa (XVII)
Here, saturation temperature at pressure of 3.332 kPa is Tsat@3.332 kPa.
Refer Table A-5, “saturated water-pressure table”, and write saturation temperature at pressure of 3.332 kPa 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@3.332 kPa=25.8°C
Substitute 25.8°C for Tsat@3.332 kPa in Equation (XVII).
Tdp=25.8°C
Substitute 88 kPa for P1 and 3.332 kPa for Pν1 in Equation (III).
Pa1=88 kPa−3.332 kPa=84.67 kPa
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=0.287 kPa⋅m3/kgcp=1.005 kJ/kg⋅°C
Substitute 0.287 kPa⋅m3/kg for Ra, 32°C for T1 and 84.67 kPa for Pa1 in Equation (IV).
v1=(0.287 kPa⋅m3/kg)(32°C)84.67 kPa=(0.287 kPa⋅m3/kg)[(32+273) K]84.67 kPa=(0.287 kPa⋅m3/kg)(305 K)84.67 kPa=1.034 m3/kg dry air
Substitute 88 kPa for P1 and 3.332 kPa for Pν1 in Equation (V).
ω1=0.622(3.332 kPa)85 kPa−3.332 kPa=0.02448 kg H2O/kg dry air
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=2559.2 kJ/kg
Substitute 1.005 kJ/kg⋅°C for cp, 32°C for T1, 0.02448 for ω1 and 2559.2 kJ/kg for hg1 in Equation (VI).
h1=(1.005 kJ/kg⋅°C)(32°C)+(0.02448)(2559.2 kJ/kg)=94.80 kJ/kg dry air
Refer Table A-4, “saturated water-temperature table”, and write saturation pressure at temperature of 20°C.
Psat@20°C=2.339 kPa
Substitute 1 for ϕ2 and 2.339 kPa for Psat@20°C in Equation (VII).
Pν2=(1)(2.339 kPa)=2.339 kPa
Substitute 88 kPa for P2 and 2.339 kPa for Pν2 in Equation (VIII).
Pa2=88 kPa−2.339 kPa=85.661 kPa
Substitute 0.287 kPa⋅m3/kg for Ra, 20°C for T2 and 85.661 kPa for Pa2 in Equation (IX).
v2=(0.287 kPa⋅m3/kg)(20°C)85.661 kPa=(0.287 kPa⋅m3/kg)[(20+273) K]85.661 kPa=(0.287 kPa⋅m3/kg)(293 K)85.661 kPa=0.9817 m3/kg dry air
Substitute 88 kPa for P2 and 2.339 kPa for Pν1 in Equation (X).
ω2=0.622(2.339 kPa)85 kPa−2.339 kPa=0.01699 kg H2O/kg dry air
Refer Table A-4, “saturated water-temperature table”, and write the specific enthalpy saturated vapor at temperature of 20°C.
hg2@20°C=hg2=2537.4 kJ/kg
Substitute 1.005 kJ/kg⋅°C for cp, 20°C for T2, 0.01699 for ω2 and 2537.4 kJ/kg for hg2 in Equation (XI).
h2=(1.005 kJ/kg⋅°C)(20°C)+(0.01699)(2537.4 kJ/kg)=63.20 kJ/kg dry air
Refer Table A-4, “saturated water-temperature table”, and write the enthalpy of water at temperature of 20°C.
hw=hf@20°C=83.915 kJ/kg
Here, specific enthalpy saturation liquid at temperature of 20°C is hf@20°C.
Perform unit conversion of diameter from cm to m.
D=40 cm=40 cm[m100 cm]=0.4 m
Substitute 120 m/min for V1 and 0.4 m for D in Equation (XII).
˙ν1=(120 m/min)[π(0.4 m)24]=15.08 m3/min
Substitute 15.08 m3/min for ˙ν1 and 1.034 m3/kg dry air for ν1 in Equation (XIII).
˙ma1=15.08 m3/min1.034 m3/kg dry air=14.59 kg/min
Substitute 14.59 kg/min for ˙ma1, 0.02448 kg H2O/kg dry air for ω1 and 0.01699 kg H2O/kg dry air for ω2 in Equation (XIV).
˙mw=(14.59 kg/min)(0.02448−0.01699)=0.1093 kg/min
Substitute 14.59 kg/min for ˙ma1, 94.80 kJ/kg dry air for h1, 63.20 kJ/kg dry air for h2, 0.1093 kg/min for ˙mw and 83.915 kJ/kg for hw in Equation (XV).
˙Qout=[(16.87 kg/min)[(94.80−63.20) kJ/kg dry air]−(0.1093 kg/min)(83.915 kJ/kg)]=451.7 kJ/min
Hence, the rate of heat transfer is 451.7 kJ/min.
(b)
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The mass flow rate of the water.
Answer to Problem 81P
The mass flow rate of the water is 18.01 kg/min.
Explanation of Solution
Express the mass flow rate of the water.
˙mcooling water=˙QwcpΔT (XVIII)
Here, mass flow rate of the water is ˙Qw, 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=4.18 kJ/kg⋅°C
Substitute 451.7 kJ/min for ˙Qw, 4.18 kJ/kg⋅°C for cp and 6°C for ΔT in Equation (XVIII).
˙mcooling water=451.7 kJ/min(4.18 kJ/kg⋅°C)(6°C)=18.01 kg/min
Hence, the mass flow rate of the water is 18.01 kg/min.
c)
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The exit velocity of the airstream.
Answer to Problem 81P
The exit velocity of the airstream is 113.9 m/min.
Explanation of Solution
Express the exit velocity of the airstream.
V2=ν2ν1V1 (XIX)
Conclusion:
Substitute 1.034 m3/kg dry air for ν1, 0.9817 m3/kg dry air for ν2 and 120 m/min for V1 in Equation (XIX).
V2=0.9817 m3/kg dry air1.034 m3/kg dry air(120 m/min)=113.9 m/min
Hence, the exit velocity of the airstream is 113.9 m/min.
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