Answer to Question #155102 in Chemical Engineering for munashe

Question

Air with a humidity, YM, of 0.012 kg H2O / kg dry air enters a recirculating drier, where it removes some water from a solid material (S) and leaves with a humidity, YP, of 0.030 kg H2O/kg dry air. A fraction of the wet air leaving the drier is removed from the process and the rest is recycled. Fresh make-up air with a humidity, YF, of 0.004 kg H2O/kg dry air is mixed with the recycled air. The water (W) removed from the solid material S leaves in the system in stream P? Calculate (a) The mass of water removed per kg of feed air supplied (4 marks) (b) The percentage of wet air removed after the drier. (4 marks)

Answer 1

(a) Mass of water = 0.030 - 0.012 - 0.004 = 0.014 kg H₂O / kg feed air

(b) % Wet air = 0.014 / 0.034 x 100 = 41%