Evaporation
Evaporation
(OP)
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
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





RE: Evaporation
RE: Evaporation
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?
RE: Evaporation
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.
RE: Evaporation
RE: Evaporation
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
RE: Evaporation
Have a look at the following link, to see whether it could be useful for your estimation:
http://www
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
RE: Evaporation