**Q:** Do you know of any papers or charts that give water evaporation rates? We have an opportunity to reduce our cleaner tank temperature from 180 to 80°F and want to know how much water we can save. R. T.

**A: **R.T., that’s an interesting question, because I am an environmental engineer, not a mechanical/process engineer. But I do have two sources.

In an old U.S. EPA publication “Summary Report: Control and Treatment Technology for the Metal Finishing Industry, In-Plant Changes” (EPA 625/8-82-008), published in 1982, Figure 5 shows “surface evaporation rate from plating baths with no aeration” assuming “ambient conditions at 75°F, 75% relative humidity, plating solution is 95% mole fraction H2O.” At 80°F, the water evaporation rate is estimated at 0.017 gal/hr/ft2, while at 180°F the water evaporation rate is estimated at 0.3 gal/hr/ft2, an increase of almost 18 times.

If my memory serves me correctly, it takes about 8,500 BTU to evaporate a gallon of water; considering inefficiencies of energy transfers, one can easily input 9,500 BTU of energy in order to obtain 8,500 BTU for actual evaporation. In terms of dollars, your energy savings will be much larger than water savings; by reducing your bath temperature from 180 to 80°F your energy savings due to evaporation alone (not considering losses through tank walls and parts) is 2700 BTU/hr/ ft^{2 }((0.3 - 0.17) x9500). For a small 10-ft2 tank, that is 27,000 BTU/hr and 648,000 BTU/day. Assuming you are heating with natural gas at a cost of $1/therm (100,000 BTU/therm), your savings for a 10-ft2 tank could be about $6.50 per day; that’s about $1,700 per year with five days per week production.

I also found an old article by M.D. Syed, Ph.D. and D.G. Strang, P.E., of D.W. Thomson Consultants Ltd. in an unknown publication that gave the following two equations that we have used over the years to estimate emissions of various volatile chemicals from open-top tanks:

W_{1 }= ((95 + 0.425V) ? (P_{w }- P_{a }) ? A)/L

and

W_{2 }= ((201 + 0.88V) ? (P_{w }- P_{a }) ? A)/L

where

W_{1 }= Rate of water evaporation with parallel air flow (lb/hr),

W_{2 }= Rate of water evaporation with transverse air flow (lb/hr),

A = Open tank surface area (square feet),

V = Air velocity over the water surface (ft/min),

L = Latent heat of evaporation (BTU/lb),

P_{w }= Vapor pressure taken at the temperature of the water, that is saturation pressure (inches of mercury),

P_{a }= Vapor pressure of moisture in the air at room temperature and relative humidity, that is partial pressure (inches of mercury).

I am sure our readers have other sources they can share as well.