How much water is expected to be consumed by a cooling tower per ton of refrigeration?
From the wikipedia article:
At first I misread the article as saying 15,000 Btu/hour would evaporate 3 gallons per minute. It is actually saying that 15,000 Btu/hour equals delta T of 10F at 3 gallons per minute.A ton of air-conditioning is the rejection of 12,000 Btu/hour (12,661 kJ/hour). The equivalent ton on the cooling tower side actually rejects about 15,000 Btu/hour (15,826 kJ/hour) due to the heat-equivalent of the energy needed to drive the chiller's compressor. This equivalent ton is defined as the heat rejection in cooling 3 U.S. gallons/minute (1,500 pound/hour) of water 10°F, which amounts to 15,000 Btu/hour, or a chiller coefficient-of-performance (COP) of 4.0 (or EER of 13.65).
To estimate evaporation rate needed to reject heat for per ton of refrigeration, I would do:
(15000 btu per hour) / (2260 kJ/litre) = 0.030 US gallons per minute
or 0.1167 litres per minute
or 0.168 m^3 per day
since 15000 Btu/hour is the heat dumped out by a chiller of COP:4.
and since 2260 kJ/litre is the heat of vaporization of water
and since I assume that all the heat rejection is due to evaporation... this is a big assumption, since cooling tower exhaust air is at higher temperature than inlet too.