Cooling Tower estimation 3

[TMEldridge]

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?

 

[briand2]

In order to give about 5.5 degrees Celsius temperature drop, you need to evaporate about 1% of the cooling tower water circulation rate. You will need to make up that 1%, plus perhaps another 33% of that, i.e. about 1.33% in total. This all depends upon your cooling tower selection criteria, water conditions, etc.

Regards,

Brian

 

[imok2]

Cooling tower formulas

http://www.ctdoc.com/GlosCT.html

 

[25362]

For mechanical-draft cooling towers, the site brought by imok2 is OK, except probably by the estimation of the drift losses (for an unknown tower) which would be, say, 0.2% (not 0.002%) of the evaporation losses.

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.

 

[VEEKRISH]

If the range is about 8 deg F, then an amount of 1.5% of the total water in circulation is adequate to account for drift. evaporation and gland leakages in the system,
Veekrish

 

[25362]

For a natural draft tower add about 0.5% on the circulating rate o/a drift/windage losses over and above those of a mechanical draft tower.

 

[mbeychok]

.
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)
X_M = Concentration of chlorides in make-up water (M), in ppmw
X_C = Concentration of chlorides in circulating water (C), in ppmw
Cycles = Cycles of concentration = X_C / X_M
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 (X_M) = D (X_C) + W (X_C) = X_C (D + W)

and, therefore:

X_C / X_M = Cycles = M / (D + W) = M / (M – E) = 1 + {E / (D + W)}

From a simplified heat balance around the cooling tower:

(E) = (C) ([Δ]T) (c_p) / H_V

where:
H_V = latent heat of vaporization of water = ca. 1,000 Btu/pound
[Δ]T = temperature difference from tower top to tower bottom, in [°]F
c_p = 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)
.

 

[AEBuck]

The previous response definitely deserves a star for content, accuracy, completeness and clarity, ... Well done!!!

 

[lilliput1]

Also don't forget the cooling tower heat balance:

cooling load + compressor input = heat rejected by evapoaration of water

in consistent units , like Btu/hr.