Answer to Question #126099 in Thermal Power Engineering for Sahil

Question

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

Answer 1

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

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

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

$Answer = 456.79716kg/min$