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topic 14924

Hot Tank Evaporation Rates



A discussion started in 2002 & continuing through 2017 -- add your Q to bring it back to the Hot Topics page.

(2002)

Q. I am working on a study to minimize wastewater in an aircraft plant metal finishing line. I am considering fog rinses above hot process tanks to minimize dragout losses. Are their standard calculations for the evaporation losses from hot process tanks (150 to 190 °.) that can reliably predict the evaporation that could be offset by the fog spray. Are there problems with this approach other than the obvious problems of too much water used and the physical arrangement of the nozzles?

Douglas Mull
engineers - Salina, Kansas


(2002)

A. Hi Douglas,

I was intrigued with your mention of 'fog' sprays ... and, obviously, as you pointed out, water consumption.

People tend to euphemistically promote products. FOG spray. ATOMIZATION of droplets. But if you research manufacturers' tech. data, you'll notice that with 'jets', the smaller the jet size (for a given gpm), the smaller the droplets will be.

And from a viewpoint of mass transfer and lower water consumption you need small droplets. The best are the so-called air atomizing nozzles but, ah, the best they can do is around 60 microns (going by memory) but that's a big, big step compared to the 300 micron sized droplets of, say, most smaller jets.

My own preference (considering potential jet blockage,spray angle and droplet size) was the solid cone wide angle injection moulded (plastic) jets, l/8" sizing.

Food for thought, eh?

freeman newton portrait
Freeman Newton
White Rock, British Columbia, Canada

(It is our sad duty to
advise that Freeman passed away
April 21, 2012. R.I.P. old friend).



(2002)

A. For still tanks, the following equation works fairly well.

E = exp -(7.2-.03236T) where E = the evaporation rate in gallons/ft2/hour, and T = temperature in degrees Fahrenheit.

For air agitated tanks, the evaporation is much higher and the equation is

E = exp -(5.95-.0265T)

Lyle Kirman
consultant - Cleveland, Ohio


May 11, 2011

Questions for Lyle Kirkman's equation: E = exp -(7.2-.03236T)
1. What does 'exp' stands for? Want to figure out how to calculate after substituting temperature...
2. Don't you consider the volume of the tank in the equation?

Mar Ventura
- San Diego


July 30, 2014

A. 'exp' stands for the exponential function, so everything in the parentheses is the exponent and the base is "e", which is roughly 2.718.

Also, when dealing with evaporation the volume of the tank is a non-issue. To get the evaporation rate for a specific tank you would multiply your result by the area of water-air contact (the water exposed to air).

Erik Anderson
- Minneapolis, Minnesota



July 11, 2017

Q. Hi,

If the hot water tank in question is a covered tank (12 ft diameter) with a 12 inch exhaust duct - natural draft

Would you presume that the only exit for the water vapor is the area of the 12" exhaust duct (0.785ft2) instead of the 113 ft2 top surface area?

Kent Thomas
- Chicago Heights, Illinois USA



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