Question

A solute A is being absorbed from a binary gas mixture of A and C into a pure solvent B in a packed tower with a cross sectional area of 0.2 m² at a temperature of 300 K and pressure of 1 atm. The inlet gas contains 3 mol% of A and its desired to remove 85% percent of A from the gas stream. The feed gas flowrate is 14 kmol/h and the solvent flowrate is 45 kmol/h. k_y’a_h = 0.04 kmol/m³s and k_x’a_h = 0.06 kmol/m³s. The equilibrium relationship can be assumed y = 1.2x.

What is the packing tower height?

Answer 1

Binary gas mixture ( A + C) in which A is a solute is to be absorbed by pure solvent B.

Cross sectional area of packed tower, A = 0.2 m2

Inlet gas having mole fraction of A, y1= 0.03

Mole ratio of A in inlet gas mixture, Y1 = y1/(1-y1) = 0.0309 molA/molC

We want removing 85 % of A which is present in inlet gas mixture then outlet gas left only 15 % of the inlet then mole ratio of A in the outlet, Y2 = 0.15*Y1= 0.00463 molA/molC

Mole fraction of A in outlet gas, y2 = Y2/(1+Y2) = 0.0046

Feed gas flow rate G = 14 kmol/h

Feed gas flow rate (A free basis) , Gs = G(1-y1) = 14*(1-0.03) = 13.58 kmol/h

Solvent inlet flow rate Ls = 45 kmol/h

For countercurrent absorber inlet mole ratio of A in liquid, X2 = 0 molA/molB

Outlet mole ratio in liquid = X1 molA/molB

Individual Mass transfer coefficient is given :

For gas phase, kya'= 0.04 kmol/m3.s

Liquid phae, kxa'= 0.06 kmol/m3.s

Overall mass transfer coefficient,

1/Kya'= 1/kya' + m/kxa'

1/Kya'= 1/0.04 + 1/0.06

1/Kya' = 41.666

Kya'= 0.024 kmol/m3.s

Equilibrium relationship, y = 1.2x

Then compare with y = mx, m = 1.2

#) Finding HTU:

HTU =( Gs/A)/Kya'

Gs = 13.58 kmol/h *(1h/3600s) = 3.7722*10^-3 kmol/s

HTU = (3.7722*10^-3kmol/s)/(0.2m2 * 0.024kmol/m3.s) = 0.785 m

#) Finding NTU:

absorbtion factor A = L/mG = 45/1.2*14 = 2.678

NTU = ln[{y1 - mx2) /(y2 - mx2) }(1-1/A) + 1/A]/ lnA

NTU = ln[{(0.03 - 0)/(0.0046 - 0)}(1-1/2.678) + 1/2.678]/ln(2.678)

NTU = 1.51

Packing tower height Z = HTU*NTU = 0.785*1.51 = 1.19 m

ans : 1.19 m