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Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooling and dehumidifying coil with an enthalpy of 85 kJ/kg of dry air and a humidity ratio of 19 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio of 8 grams/kg of dry air. If the condensate water leaves the coil with an enthalpy of 67 kJ/kg, the required cooling capacity of the coil in kW is
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- 75.0
- 123.8
- 128.2
- 159.0
Correct Option: B
Mass flow rate of condensate,
ṁw = ṁa (w1 - w2) 3 × (.019 - .008) = .033kg/s
By applying energy balance equation we get
ṁa h1 = ṁa h2 + ṁa hfg + Q
3 × 85 = 3 × 43 + 0.033 × 67 + Q
Q = 123.8 kW