Answer to Question #310562 in Mechanical Engineering for Jhonny

Question

13. The evaporative condenser of an ammonia refrigeration plant has a water flow rate of 100 kg/s and enters a natural draft cooling tower at 43oC. The water is cooled to 30oC by air entering at 27oC db 24oC wet bulb. The air leaves the tower as 90% RH and 32oC db. Calculate the percent of make-up water needed the tower. At 27oC db and 24oC wb: h = 72.5 KJ/kg, w = 0.0178 At 90% RH and 32oC db: h = 102 KJ/kg, w = 0.0275.

Answer 1

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 m³/hr

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

T_i – T₀ = 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 = (100 kg / 998 kg/m³) / (1/3600 h) = 360.72 m³/h

E = (830 m³/h x (40 - 30°C) x 4.184 kJ / kg / °C ) / 2260 kJ / kg = 15.4 m³/h

M = 360.72 x 0.1 + 15.4 = 51.47 m³/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 (^oC)

t_o = outlet temperature of water from the tower (^oC)

t_wb = wet bulb temperature of air (^oC, ^oF)

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