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

Explanation:

Humidity Ratio at inlet (ω₁) = 19 g/kg of dry air = 19 × 10^-3 kg/kg of dry air

Humidity Ratio at outlet (ω₂) = 8 × 10^-3 kg/kg of dry air

Mass of water vapour at inlet m_vi = ω₁ × m_a

 = 19 × 10^-3 × 3 kg/s

= 57 × 10^-3 kg/s

Mass of water vapour at outer m_vo = ω₂ × m_a

= 8 × 10^-3 × 3

= 24 × 10^-3 kg/s

Mass of water that condensed = m_vi - m_vo

= (57 - 24) × 10^-3

= 33 × 10^-3 kg/s

= 0.033 kg/s

Cooling capacity of coil = h_i - h₀ - h_c

Where h_i = Inlet enthalpy, h₀ = outlet enthalpy, h_c = enthalpy of condensate;

A = [(85 - 43) kJ/kg × 3 kg/s] - [67 kg/kg × 0.033]