Recent content by rcl5011

Thank you very much for your help on this. It is greatly appreciated.

Ok, so the ratio of partial pressures is directly proportional to the ratio of moles input temp of 300F = 759.67R Psat=66.98psia Partial Vapor Pressure Input= 5%(66.98psia)=3.349psia Partial Gas Pressure Input= 19.696-3.349=16.347PSIA mols of vapor in=[R #mole/10.73159 psi...

input temp of 300F = 759.67R Psat=66.98psia Partial Vapor Pressure Input= 5%(66.98psia)=3.349psia mols of vapor in=[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#...

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 the water vapor partial pressure is not dependant on the air pressure at all, and is constant based only on RH and saturation pressure? Also, is my final delta P = (5%RH * Saturation pressure at 300°) - (100%RH * saturation pressure at 129°)? What i would have then is: DeltaP=...

Vectors are added head to tail. You can draw them out starting with A being horizontal to the right. B is added to the arrow end of A up 30° from horizontal, and C is added to the end of B at 80° from the line of vector B or an effective 110° from horizontal. You need to sum the X and Y...

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...