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Mass of Condensed Water Out of a Moist Air Stream 
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#1
Feb1213, 09:03 PM

P: 9

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 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 and final and subtract the two to get a delta P to use in the PV=nRT equation and solve for n and muliply that by the molar mass to finally come up with a mass. What I am unsure of is if I need to multiply the vapor presures by the ratio of my air stream pressure over the ambient air pressure to come up with an actual vapor pressure due to the fact that the air is at 5psig. I am also unsure as to whether the final vapor pressure should be equal to the Psat or is it PsatXPambient/19.696psia? In addition to that, What would I use for Volume and T in PV=nRT? I know the flow rate is 936FT^3/min. If someone could help me set this up correctly I would greatly appreciate it. Thanks for your time and willingness to share some insight. Ryan 


#2
Feb1213, 10:15 PM

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PF Gold
P: 5,245

The initial pressure (19.696) is the sum of the air partial pressure (dry basis) and the water vapor partial pressure. You know the water vapor partial pressure from the saturation vapor pressure and the relative humidity. From this information, you know the initial mole fractions of air (dry basis) and water vapor in the gas. How many lbmoles are there in 936 ft^3 of an ideal gas at 19.696 psi and 300F? This tells you the initial number of lbmoles/min of air (dry basis) and water vapor in the stream. The amount of air in the gas stream doesn't change, but the amount of water vapor in the gas stream does change. See if this gets you started.



#3
Feb1313, 09:28 AM

P: 9

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= .05(66.98)1.0(2.1719)= 31.3181PSIA = 4509.8064 lb/Ft^2 Delta T= 300129=171°F = 630.67°R n=PV/RT V=936ft^3 R=2760 ftlb/slug°R n= (4509.8046*936)/(2760*630.67) = 2.425 moles M=18.01528 grams/mol mass = 18.01528*2.425 moles = 43.687 grams = 0.0963133 pounds or water per minute. Please verify that this is correct process and units. Thank you for your time you have been very helpful 


#4
Feb1313, 01:33 PM

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PF Gold
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Mass of Condensed Water Out of a Moist Air Stream
This solution looks hinky to me. Please go back and carry out the steps that I suggested in my previous response. You have 936 ft^{3} of gas at 300 F and 19.696 psia. How many lbmoles is that? Use the gas constant 10.73159 psi ft^{3}/degR/lbmole. From the total pressure and the partial pressure of H20 vapor, what are the mole fractions of air(dry basis) and water. How many moles of water is that? How many pounds of water is that? This is how many pounds of water would condense out if all of it condensed out. Your final answer should be close to this number.



#5
Feb1313, 08:06 PM

P: 9

input temp of 300F = 759.67R
Partial Vapor Pressure Input= 5%(66.98psia)=3.349psia Partial Gas Pressure Input= 19.6963.349=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. 


#6
Feb1313, 08:45 PM

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Your equation calculated the number of moles of water vapor entering during 1 min in the inlet stream, not the mole fraction. But this number of moles was calculated correctly. You have enough information to calculate how much water vapor remains in the gas phase at the outlet of the exchanger. Just assume that the final total pressure is the same as the initial total pressure. You know the weight of dry air (and number of moles) entering per minute, and this is the same at the exit. You also know the partial pressure of water vapor in the exit stream, so you know the mole fractions of air and water vapor at the exit. (The mole fractions are the partial pressures divided by the total pressure). Since you know the number of moles of dry air in the exit stream and the mole fraction of water vapor, you know the number of moles of water vapor in the exit stream (per minute). Subtract this from the entering water vapor inlet flow, and this will give you the amount that condenses (per minute). So the trick to doing the exit stream is to assume that the total outlet pressure is the same as the inlet pressure.



#7
Feb1413, 07:07 AM

P: 9

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# _______________________________________________________________________ __ Output temp of 129F= 588.67R Psat=2.1719psia Partial Vapor Pressure Output= 100%(2.1719psia)=2.1719psia mols of vapor Out=[R #mole/10.73159 psi ft^3]*2.1719psia*936ft^3 / 588.67R= .3218 #mol water vapor is 18.01528#/1#mol, so 18.01528*.3218#mol= 5.797# So to determine condensed water, we need to know the difference (or loss) in air stream vapors= 6.928#5.797# = 1.131# of water condensing every minute. 


#8
Feb1413, 08:24 AM

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PF Gold
P: 5,245

This isn't quite right. The volume of the gas in the final condition has changed because of change in temperature and the condensation of some of the water. What is the same in the final state is the number of moles of air on a dry basis. This is 1.88 moles. In the final state, the mole fractions are 0.89 and 0.11. The number of moles of water vapor is down to 0.232, which is 4.18 lb.



#9
Feb1413, 09:30 AM

P: 9

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= 19.6963.349=16.347PSIA 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# mols of Gas in=[R #mole/10.73159 psi ft^3]*16.347psia*936ft^3 / 759.67R= 1.876 #mol _______________________________________________________________________ __ Output temp of 129F= 588.67R Psat=2.1719psia Partial Vapor Pressure Output= 100%(2.1719psia)=2.1719psia Partial Gas Pressure Output= 19.696=17.5241psia Ratio of partial pressure of vapor to gas is 2.1719/17.5241=.123937 So moles of Gas are constant and are 1.876 in and out; 1.876*.123937=.2325#moles of vapor .2325#mol *18.01528#/1#mol= 4.18868# total vapor out. 6.927#4.18868#=2.7383# of condensing water per minute 


#10
Feb1413, 09:52 AM

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PF Gold
P: 5,245

Yes. I think now you have it. A couple of things to remember:
1. The mole fraction of a species in a gas (ideal gas) is equal to the partial pressure divided by the total pressure. 2. You could have calculated that total moles of gas first using the ideal gas law, and then multiplied by the mole fractions to get the number of moles each species. Chet 


#11
Feb1413, 09:55 AM

P: 9

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



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