Question

Hot air containing 1 mole% water vapor flows into a textile dryer at 350°F and 6.2 psig at a rate of 30,000 ft3/hr, and emerges at 220°F and 1 atm absolute pressure (i.e., it is vented outside the building) containing 10 mol% H2O. Calculate the following:a. The mass flow rate of the entering hot air stream in lbm/hrb. The rate of evaporation of the water from the textiles in lbm/hrc. The volumetric flow rate of the emerging air stream in ft3/hr

Answer 1

Given data

Temperature T = 350°F = 449.8 K

Pressure P = 6.2 psi x 1atm/14.7psi = 0.4217 atm

Volumetric flow rate V = (30000 ft3/hr) x (28.317L/ft3)

= 849510 L/hr

Moles of hot wet air n = PV/RT

= 0.4217*849510/0.0821*449.8

= 9700.84 mol/hr

Moles of water vapor = 0.01*9700.84 = 97 mol/hr

Moles of hot dry air = 9700.84 - 97 = 9603.84 mol/hr

Mol fraction of water vapor X1 = 0.01

Mol fraction of hot dry air X2 = 0.99

Average molecular weight

M = X1M1 + X2M2

= 0.01*18 + 0.99*29 = 28.89 g/mol

Part a

Mass flow rate of entering hot air stream

= Moles of hot wet air x average molecular weight

= 9700.84 mol/hr x 28.89 g/mol

= 280257.2676 g/hr x 1lb/453.592g

= 617.86 lbm/hr

Part b

Emerging stream

Temperature T = 220°F = 377.6 K

Pressure P = 1 atm

Moles of hot dry air n = 9603.84 mol/hr

Total Moles of emerging air = 9603.84*(1.1) = 10564.224 mol/hr

Moles of water vapor = 1056.42 mol/hr

Rate of evaporation = 1056.42 - 97 = 959.42 mol/hr

Part C

Volumetric flow rate

V = nRT /P

= 10564.224*0.0821*377.6/1

= 327501.085 L/hr x (1ft3/28.317L)

= 11565.53 ft3/hr