Question

An air stream is saturated with a vapor (B) at 130F and 1 atm pressure. It is cooled down to 80F where some of the B is condensed and separated from the vapor stream. The remaining vapor stream is then reheated back to 130F.

a. What is the mole fraction of B in the air coming out of the process?

b. For an inlet flow rate of 2000 ft3/min of air saturated with B, what would be the flow rate of the liquid B stream (lb/min)?

c. If the process could cool the stream down to 50F instead of 80F, what would be the outlet mole fraction of B in this case?

Note:

Vapor pressure of B at 80F = 0.5067 psia

Vapor pressure of B at 130F = 2.221 psia MW of B = 20

Heat capacity of liquid B = 0.87 (btu)/(lb)(F)

Answer 1

1 atm =14.7 psia

At 130 deg.F, air stream is saturated. Partial pressure = vapor pressure = 2.221 Psia

Moles of water vapor/ moles of dry air = partial pressure of water vapor/ partial pressure of dry air = 2.221/(14.7-2.221)= 0.1779

Moles of water vapor = 0.1779* moles of dry air

Flow rate = 2000ft3/min Temperature= 130 deg.F = 130+460=590R, P= 14.7 Psia

R= 10.731 psift3

Total number of moles, n= PV/RT= 14.7*2000/(10.731*590) = 4.643 lb moles/min

Moles of water vapor + moles of dry air = 4.643

Molar flow rate of B vapor + moles of B vapor/0.1779= 4.643

Molar flow rate of B vapour (1+1/0.1779)= 4.643

Molar flow rate of B vapour = 4.643/6.621= .7012 lb moles/min = 0.7012*20 lb/min=14.024 lb/min

At exit the air is assumed to be saturated. Hence partial pressure = 0.5067 Psia

Moles of B vapor/ mole of dry air = 0.5067/(14.7-0.5067)= 0.0357

Moles of B vapor = 0.0357* moles of dry air

Moles of B condensed = moles of water vapor at 130deg.F- moles of water vapor at 80 deg.F= Moles of dry air*(0.1779-0.0357)= 0.1422* mole of dry air

Moles of B vapor at 80 deg.F= 0.0357* moles of dry air

Moles of B vapor/ moles of dry air =0.0357

moles of dry air/ mole of B vapor = 1/0.0357 = 28.01

1+ moles of dry air/mole of B vapor =29.01

Moles of wet air/ moles of B vapor = 29.01

Moles of B vapor/ moles of wet air = 1/29.01= 0.0344

Mole fraction of B vapor at 80 deg.F ( outlet)=0.0344

At 50 deg.F, vapor pressure of B need to be calculated. For that deltah/R need to be calculated. This is done by using Classis- Clayperon equation.

Ln(P2/P1)= (deltaH/R)*(1/T1-1/T2)

Ln (2.221/0.5067)= (deltaH/R)*(1/80+460 – 1/130+460)

deltaH/R = 9416.5

at 50 deg.F (T1)= 50+460= 510R

ln(2.221/P1)= 9416.5*(1/510-1/590)

P1= 0.1816 psia

So at 50deg.F, moles of B vapor = 0.1816/(14.7-0.1816)* moles of dry air

Moles of B vapor = 0.01251* moles of dry air

Mole of dry air/ mole of B vapor = 1/0.01251= 79.93

1+ moles of dry air/moles of B vapor = 80.93

Mole of wet air/ moles of B = 80.93

Mole of B/ total moles = 1/80.93= 0.0123

Mole fraction of B at 50 deg.F =0.0123