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How to calculate cycles, blowdown, evaporation, makeup

Cooling towers

How to calculate cycles, blowdown, evaporation, makeup

by  mbeychok  Posted    (Edited  )
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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:

[img http://www.air-dispersion.com/CoolingTower.png]

In the customary USA units:

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 à usually chlorides)
X[sub]M[/sub] = Concentration of chlorides in make-up water (M), in ppmw
X[sub]C[/sub] = Concentration of chlorides in circulating water (C), in ppmw
Cycles = Cycles of concentration = X[sub]C[/sub] / X[sub]M[/sub]
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[sub]M[/sub]) = D (X[sub]C[/sub]) + W (X[sub]C[/sub]) = X[sub]C[/sub] (D + W)

and, therefore:

X[sub]C[/sub] / X[sub]M[/sub] = Cycles = M / (D + W) = M / (M û E) = 1 + {E / (D + W)}

From a simplified heat balance around the cooling tower:

(E) = (C) ([Δ]T) (c[sub]p[/sub]) / H[sub]V[/sub]

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

In SI metric units:

M = Make-up water in m[sup]3[/sup]/hr
C = Circulating water in m[sup]3[/sup]/hr
D = Draw-off water in m[sup]3[/sup]/hr
E = Evaporated water in m[sup]3[/sup]/hr
W = Windage loss of water in m[sup]3[/sup]/hr
X = Concentration in ppmw (of any completely soluble salts à usually chlorides)
X[sub]M[/sub] = Concentration of chlorides in make-up water (M), in ppmw
X[sub]C[/sub] = Concentration of chlorides in circulating water (C), in ppmw
Cycles = Cycles of concentration = X[sub]C[/sub] / X[sub]M[/sub]
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[sub]M[/sub]) = D (X[sub]C[/sub]) + W (X[sub]C[/sub]) = X[sub]C[/sub] (D + W)

and, therefore:

X[sub]C[/sub] / X[sub]M[/sub] = Cycles = M / (D + W) = M / (M û E) = 1 + {E / (D + W)}

From a simplified heat balance around the cooling tower:

(E) = (C) ([Δ]T) (c[sub]p[/sub]) / H[sub]V[/sub]

where:
H[sub]V[/sub] = latent heat of vaporization of water = ca. 2260 kJ / kg
[Δ]T = water temperature difference from tower top to tower bottom, in [°]C
c[sub]p[/sub] = specific heat of water = 4.184 kJ / kg / [°]C

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 or less 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.)

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