Evaporation

[zaphod1]

I realize that this may be a simple question but...
In a open (Not actually open but not pressurized or blanketed) tank of water heated to 150 deg. F, air is around 120 degrees F in the spaces above the top of the tank, There is a exhaust fan drawing 5200 scfm from the around the top of the tank, the tank is 75 ft^2- The make-up air is approx. 80 degrees F and the RH is 65-70%. Can anyone help me calculate the evaporation rate or direct me to a reference?
Thanks in advance

 

[epoisses]

Use either 150 or 120 deg F whichever makes you err on the conservative side, don't spend more time on it. I assume 80 deg F is only the temperature of the air supplied but not the temp inside the tank; same for the RH mentioned. Find the corresponding water vapor pressure, divide by 1 bar. Times 5200 scfm.

 

[siretb]

at 150°F the partial preesure of water is about 25.6 kPa. the "air" contains about 26% water, by volume.
5200 cfm is a lot. By the way you mention a fan of 5200 scfm, but the fan is operating at a higher than normal temperature.
Saturation would tranlate to almost 1.8 ton/hr evaporation. This looks too much. 120°F would about half this figure.
Looks still high.
What is the tank cross section? is it agitated? is it heated?

 

[25362]

Without an external heat supply, the evaporation would be accompanied by a cooling of the water mass and a reduction of vapor pressure and evaporation rate.

 

[zaphod1]

Sorry about the delay- I have measure the cfm output of the stack, and although the fan is rates at 5200 scfm, my measurements indicate 1617 cfm- I am supposing the difference is due to the lower density of the air. The tank is 75 ft^2. The application is in a spray washer, which is spraying the water at approx. 140gpm over a moving chain belt that carries parts. The tank is heated with an electric immersion heater. If you would, please let me know the equations that you are using so that I can determine losses at different temps. Thanks

 

[siretb]

I did another calculation. This one is based on evaporation rates
As by data posted by KINETICO
E=exp(-5.95-0.0266T°F) E in gallons per ft2 per hour
at 150°F water temp this yields 10.6 gallons/hour

As per Heat losse from open tank (cheresources.com)
heat losses about 2300 W/m2 translating into 25 kilos/hour. They quote that this figure could be dooubled if there is "wind velocity" That could be the case with your high air rate from fan.

The initial calculattion I did was asssuming that the mixture above the tank was saturated with air. The equations were the ones of vapour pressure versus temperature. With the revised air rate you give this would yield
1617 cfm --> 2747 m3/hr at 120°F the air contains 11.5% water (wet basis), so this is 2747*0.115=315 m3/hr ; under normal (0°C temp) that is 268 Nm3/hr. So we would loose 268/22.4=12 kilomoles/hour water 215 kilo/hour

I think that the former calculations based on evaporation rates are better, and therefore that your air rate is high so that the air is not saturated with water. I'd go for about 40-50 kilos/hour. Tell me if you think this makes sense

 

[25362]

Have a look at the following link, to see whether it could be useful for your estimation:

http://www.engineeringtoolbox.com/evaporation-water-surface-d_690.html

Using the formula therein and the data you supplied, I got evaporation rates as follows:

no air velocity, 34.3 kg/h
air at 1 m/s, 60.5 kg/h
air at 2 m/s, 86.7 kg/h

 

[zaphod1]

Thanks all for your valuable input. We performed a test last night to determine losses due to evaporation- The measurements indicate approximately 60 gallons per hour are being evaporated. At this point I’m still trying to reconcile this mathematically.