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 : The Gas Ideal Law"


"Where,"

"P = Pressure"

"V = Volume"

"m=Mass"

"R = Gas Constant"

"T = Temperature"


Data are given by;


"T_1 = 38^oC"

"T_2 = 18^oC"

"\\eta=0.8"

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


Note that the 1 atm absolute pressure is "1.01325 bar"

Now, to calculate the mass flow;

"PV=mRT"


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


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

"\\eta=\\frac{\\mu}{m}"


"\\implies \\mu= \\eta*m \\implies 0.8*570.99645kg\/min = 456.79716kg\/min"


"Answer = 456.79716kg\/min"



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