Question #172734

The evaporative condenser of an ammonia refrigeration plant has a water flow rate of 230 kg/s and enters the natural draft cooling tower at 40 C. The water is cooled to 30 C by air entering at 36 C db and 38% RH. The air leaves the tower as saturated at 45 C db. Calculate:

(a) Amount of makeup water in kg/hr.

(b) The cooling tower efficiency (%).

Expert's answer

*a) Water make-up (M ) = Total water losses = Drift Losses ( D) + Evaporation Losses (E ) + Blow down Losses (B)*

D = **0.1 to 0.3** percent of Circulating water (C ) for an induced draft cooling tower

C = Circulating water in m^{3}/hr

"\\lambda" = Latent heat of vaporization of water = 540 kcal/kg or 2260 kJ / kg

T_{i} – T_{0} = water temperature difference from tower top to tower bottom in °C ( cooling tower inlet hot water and outlet cold water temperature difference)

C_{p} = specific heat of water = 1 kcal/kg / °C (or) 4.184 kJ / kg / °C

C = 230 kg/s = (230 kg / 998 kg/m^{3}) / (1/3600 h) = 830 m^{3}/h

E = (830 m^{3}/h x (40 - 30°C) x 4.184 kJ / kg / °C ) / 2260 kJ / kg = 15.4 m^{3}/h

M = 830 x 0.1 + 15.4 = 98.4 m^{3}/h or 98 203 kg/h

*b)* Cooling tower efficiency can be expressed as

*μ = (t*_{i}* - t*_{o}*) 100 / (t*_{i}* - t*_{wb}*) *

*where*

*μ = cooling tower efficiency (%) - the common range is between 70 - 75%*

*t*_{i}* = inlet temperature of water to the tower (*^{o}*C)*

*t*_{o}* = outlet temperature of water from the tower (*^{o}*C)*

*t*_{wb}* = wet bulb temperature of air (*^{o}*C, *^{o}*F)*

*μ = (40 - 30) / (40 - 26) * 100 = 71.4%*

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