Question

A gas stream of flow rate 10.0 lbmole/ft2 h contains
6.0% sulfur dioxide (SO2) by volume in air. It is desired to
reduce the SO2 level in the treated gas to no greater than 0.5% in a countercurrent packed tower operating at 30°C and 1.0 atm total system pressure, using water containing no dissolved SO2 as the absorption solvent. The desired solvent flow rate is 2.0 times the minimum solvent flow rate. At these flow conditions, the film mass-transfer coefficients and kxa =250 lbmole/ft3 h, and kya= 15 lbmole/ft3 h, for the liquid and gas films, respectively. Equilibrium distribution data for SO2 in water at 30°C is provided in the table below.

Determine the height of packing required to accomplish the separation or operation at 2.0 times the minimum solvent flow rate.

Answer 1

Gas stream molar flow rate G = 10 lbmol/ft2.h

Total pressure of the gas P = 1 atm = 760 mmHg

As we know that for ideal gas, % mol = % volume

If we assume gas treated as ideal gas then

Sulfur dioxide SO2 initially present in gas, y1 = 0.06(mol basis)

Mole ratio of SO2 inlet gas, Y1 = y1/(1-y1) = 0.0638

Gas molar flow rate (SO2 free basis), Gs = G(1-y1) = 10*(1-0.06) = 9.4 lbmol/ft2.h

Final outlet air having SO2 mole fraction, y2 = 0.005

Mole ratio of SO2 outlet gas, Y2 = y2/(1-y2) = 0.005025

Pure liquid flow rate inlet having no SO2, X2 = 0

Firstly we have to find out the equilibrium relation by drawing equilibrium data

Xa = mole ratio of so2 in liquid

Ya - mole ratio of so2 in gas = Pa/Pb = Pa/(P - Pb)

Ya = mXa

By graph we can see that equilibrium relation,

Ya = 0.108Xa

m = 0.108

Applying material balance for finding minimum liquid flow rate,

Gs(Y1 - Y2) = Lmin(X1 - X2)

As we know that for countercurrent operation X1 is equilibrium with Y1,

X1 = Y1/0.108 = 0.0638/0.108 = 0.5907

Lmin = 9.4(0.0638 - 0.005025)/(0.5907 - 0)

Lmin = 0.935 lbmol/ft2.h

We want packed tower height when Ls = 2Lmin =2*0.935 = 1.87 lbmol/h.ft2

Individual Mass transfer coefficient is given

kxa = 250 lbmol/h.ft3

kya = 15 lbmol/h.ft3

Overall mass transfer coefficient,

1/Kya = 1/kya + m/kxa

1/Kya = 1/15 + 0.108/250 = 0.067

Kya = 14.9 lbmol/h.ft3

Finding HTU:

HTU = Gs/Kya =( 9.4lbmol/h.ft2)/(14.9lbmol/h.ft3) = 0.6307 ft

HTU = 0.1922 m

Finding NTU:

m = 0.108

Absorption factor, A = Ls/mGs = 1.87/(0.108*9.4) = 1.842

NTU = ln[{(Y1 - mX2)/(Y2 -mX2)}(1-1/A) + 1/A]/lnA

NTU = ln[{(0.0638 - 0)/(0.005025 - 0)(1- 1/1.842) + 1/1.842]/ln(1.842)

NTU = 3.025

Packed tower height,

Z = NTU*HTU = 3.025 * 0.6307 = 1.9 ft

packed tower height Z = 1.9 ft

or

Z = 0.58 m