Answer to Question #126099 in Thermal Power Engineering for Sahil

Question #126099
Air at 1 atm absolute pressure, 38oC and 80% relative humidity is to be cooled to 18oC
and fed into a plant area at a rate of 510 m3
/min (STP). Calculate the rate (kg/min) at which
water condenses
1
Expert's answer
2020-07-15T06:44:55-0400

To Calculate the rate (kg/min) at which water condenses, we need to apply the concepts that are related to ideal gas laws, and mass flow rate.


PV=mRT:TheGasIdealLawPV=mRT : The Gas Ideal Law


Where,Where,

P=PressureP = Pressure

V=VolumeV = Volume

m=Massm=Mass

R=GasConstantR = Gas Constant

T=TemperatureT = Temperature


Data are given by;


T1=38oCT_1 = 38^oC

T2=18oCT_2 = 18^oC

η=0.8\eta=0.8

Flow rate (air water vapor mixture)= 510m3/min(STP)510 m^3/min (STP)


Note that the 1 atm absolute pressure is 1.01325bar1.01325 bar

Now, to calculate the mass flow;

PV=mRTPV=mRT


m=PVRT=(1.01325105)510)(291311)=570.99645kg/minm=\frac{PV}{RT} =\frac{(1.01325*10^5)*510)}{(291*311)}=570.99645kg/min


To the mass flow rate (kg/min)(kg/min) at which water condenses

η=μm\eta=\frac{\mu}{m}


    μ=ηm    0.8570.99645kg/min=456.79716kg/min\implies \mu= \eta*m \implies 0.8*570.99645kg/min = 456.79716kg/min


Answer=456.79716kg/minAnswer = 456.79716kg/min



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