Question

An experiment on the growth rate of certain organisms requires an environment of humid air enriched in oxygen. Three input streams are fed into an evaporation chamber to produce an output stream with the desired composition. A: Liquid water, fed at a rate of 80 cm3/min B: Air (21 mole% O2, the balance N2) C: Pure oxygen, with a molar flow rate 0.49 times the molar flow rate of stream B The output gas is analyzed and is found to contain 1.2 mole% water. Use a block flow diagram to help determine the mole fraction of oxygen in the outlet stream.

Answer 1

Flow diagram

Writing the overall balance

F= A+B+C ( where A is water, B is Air and C is Pure oxygen) (1)

Given C=0.49B   Eq.1 becomes F= A+B+0.49B= A+1.49B (2)

Assuming air density to be 1 g/cc , Water flow rate =80*1= 80 g/min and molar flow rate of water= 80/ 18 gmoles/min =4.444 gmoles/min

Total water in the outlet. Let outlet be M. M*1.2/100= 4.44 M= 4.44*100/1.2 =370 moles/min

Oxygen + ai r in the outlet= 370-4.44 =365.5 moles/min

B+C= 365.6 but C=0.49B there fore (0.49+1)B= 365.5, 1.49B= 365.5, B =365.5/1.49=245.3 Moles/min

C= 0.49*245.3 =120.2 moles/min

B contains 21 mole % oxygen Oxygen from B= 0.21*245.3=51.5 moles/min and pure oxygen= 120.2 moles/min

Total oxygen in the outlet= 51.5+120.2 =171.7 moles/min

Mole fraction of oxygen at the outlet= 100*(171.7/370)=46.4%