Cooling Tower estimation
Cooling Tower estimation
(OP)
I need to estimate required make-up flow to an existing and very undocumented cooling tower. I've got tonnage of heat rejection, chilled water flow, etc. Are there any rule-of-thumb estimates for drift and evaporation losses so I have some basis for starting?





RE: Cooling Tower estimation
Regards,
Brian
RE: Cooling Tower estimation
http://www.ctdoc.com/GlosCT.html
RE: Cooling Tower estimation
See also
http://www.cheresources.com/ctowerszz.shtml
BTW, "cycles" refer to cycles of concentration, which generally mean the ratio of the chlorides content in the circulating water to that of the make-up water. If such data is unknown, a "safer" assumption would be to take, for example, 3 (not 5) cycles to reach a quick ballpark estimate of the make-up rate needed.
RE: Cooling Tower estimation
Veekrish
RE: Cooling Tower estimation
RE: Cooling Tower estimation
I realize that this posting is some months after the original posting in this thread. However, it may be useful to others who may need the same type of information in the future.
Here are the governing relationships for the makeup flow rate, the evaporation and windage losses, the draw-off rate, and the concentration cycles in an evaporative cooling tower system:
M = Make-up water in gal/min
C = Circulating water in gal/min
D = Draw-off water in gal/min
E = Evaporated water in gal/min
W = Windage loss of water in gal/min
X = Concentration in ppmw (of any completely soluble salts … usually chlorides)
XM = Concentration of chlorides in make-up water (M), in ppmw
XC = Concentration of chlorides in circulating water (C), in ppmw
Cycles = Cycles of concentration = XC / XM
ppmw = parts per million by weight
A water balance around the entire system is:
M = E + D + W
Since the evaporated water (E) has no salts, a chloride balance around the system is:
M (XM) = D (XC) + W (XC) = XC (D + W)
and, therefore:
XC / XM = Cycles = M / (D + W) = M / (M – E) = 1 + {E / (D + W)}
From a simplified heat balance around the cooling tower:
(E) = (C) (ΔT) (cp) / HV
where:
HV = latent heat of vaporization of water = ca. 1,000 Btu/pound
ΔT = temperature difference from tower top to tower bottom, in °F
cp = specific heat of water = 1 Btu/pound/°F
Windage losses (W), in the absence of manufacturer's data, may be assumed to be:
W = 0.3 to 1.0 percent of C for a natural draft cooling tower
W = 0.1 to 0.3 percent of C for an induced draft cooling tower
W = about 0.01 percent of C if the cooling tower has windage drift eliminators
Concentration cycles in petroleum refinery cooling towers usually range from 3 to 7. In some large power plants, the cooling tower concentration cycles may be much higher.
(Note: Draw-off and blowdown are synonymous. Windage and drift are also synonymous.)
Milton Beychok
(Contact me at www.air-dispersion.com)
.
RE: Cooling Tower estimation
RE: Cooling Tower estimation
cooling load + compressor input = heat rejected by evapoaration of water
in consistent units , like Btu/hr.