Water Mass that Will Condence out of Air Stream

rcl5011

I am currently attempting to determine the amount (mass) of water that is condensing during a temperature change inside of a heat exchanger. Inititial Temp is 300°F, Initial Pressure is 19.696psia (5psig) in the air stream, Initial %RH is 5%. Final Temperature is 129°F and becomes 100% RH at this point. I have steam tables and can determing saturation pressures at these temperatures. Psat Initial=66.98psia, Psat Final is 2.1719psia. I know the flow rate is 936FT^3/min.

I know I can determine the actual vapor pressure by multiplying the %relative humidity by the saturation pressure, and that I need to determine the actual vapor pressure initial.

input temp of 300F = 759.67R

Partial Vapor Pressure Input= 5%(66.98psia)=3.349psia
Partial Gas Pressure Input= 19.696-3.349=16.347psia

mol fraction of vapor=[R #mole/10.73159 psi ft^3]*3.349psia*936ft^3 / 759.67R= .3849 #mol

water vapor is 18.01528#/1#mol, so 18.01528*.3849#mol= 6.927#

So we can expect to have 6.927# of water condensing onto my heat exchanger coils every minute. What I will now do it use this mass flow rate to determine the heat transfer from condensation onto the coils by using Q=m*Δh, and I can determine the Air(gas) mass flow rate the same way as I just did the vapor, and use Q=m*CpΔT in order to find heat transfer from from gas contact with coils. total heat transfer would be the sum of the two Q's. My question is it reasonable to assume that if the exchanger coils are long enough, All 6.927# of water will condense every minute as long as the output temperature on the heat exchanger drops below the dew point? Again, thanks for all of your help.

-Ryan