Question

Air containing 2.1 mol% sulfur dioxide (SO2) is scrubbed with pure water in an absorption tower of 1.5 m2 cross-sectional area and 3.6 m height packed with structured packing materials at a pressure of 1.2 atm and a temperature of 20˚C. The surface area per unit volume of the packing, a, is 200 m2/m3.The total gas flow rate is 60 mol/s, the liquid flow rate is 1800 mol/s and the mole fraction of SO2 in the outlet gas stream is 0.3%. At the column temperature, the equilibrium relationship is given by:
y* = 25.0x
(a) Calculate NOG for the desired purification.
(b) Calculate mass transfer coefficient of liquid film (kx) if mass transfer coefficient of gas film, kp = 7.5 mol/m2-atm-s.
(c) If the packing height of the column is increased to 4.0 m (column diameter and packing materials unchanged), and the same inlet gas stream is purified by the same liquid solvent, how much is the mole fraction of SO2 in the leaving liquid and gas streams?

Answer 1

Air containing SO2 mole fraction, y1 = 0.021

Mole ratio of so2 inlet air, Y1 = y1/(1-y1) = 0.0214

Mole fraction of SO2 in outlet air, y2 = 0.003

Mole ratio of SO2 in outlet air, Y2 = y2/(1-y2) = 0.003

Inlet lqiuid mole ratio of so2, X2 = 0

Total pressure of system P = 1.2 atm and temperature T = 20 C.

Total gas flow rate G = 60 mol/s

Gas flow rate so2 free basis Gs = G(1-y1) = 60*(1-0.021) = 58.74 mol/s

Pure Liquid flow rate Ls = L = 1800 mol/s

Cross sectional area A = 1.5 m2

Height of packed tower z = 3.6 m

Surface area per unit volume packing factor a = 200 m2/m3

Equilibrium relation y* = 25x.

Applying material balance over packed bed,

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

X1 = 58.74(0.0214 - 0.003)/1800 = 6*10^-4

Mole ratio of outlet so2 in liquid X2 = 6*10^-4

a) Calculate NOG

Absorption factor A = Ls/mGs = 1800/25*58.74 = 1.225

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

NOG = ln[{(0.021 -0)/(0.003 -0)*(1-1/1.225) + 1/1.225]/ln(1.225)

NOG = 3.66

b) calculate mass transfer coefficient in liquid film kx:

Height of packed bed z = 3.6m

HOG * NOG = 3.6

HOG*3.6 = 3.6

HOG = 1 m

Gs/(A*Kya) = 1 m

(58.74mol/s)/1.5m2 *Kya = 1 m

Kya = 39.16 mol/s.m3

Ky *200 m2/m3 = 39.16 mol/s.m3

Ky = 0.1958 mol/s.m2

Given indvidual mass transfer coefficient for gas phase

kp = 7.5 mol/m2.satm

kya = kp * P = 7.5 mol/m2.s.atm * 1.2 atm = 9 mol/s.m2

We know formula for overall mass transfer coefficient

1/Ky = 1/ky + m/kx

1/0.1958 = 1/9 + 25/kx

kx = 5 mol/s.m2

Mass transfer coefficient in liquid film, kx =5 mol/m2.s

c) if packing height will be changed z =4 m

Column diameter and packing material unchanged.

Liquid and gas flow rates are same.

Ls = 1800 mol/s

Gs = 58.74 mol/s

HOG = Gs/A*Kya

HOG =1 fixed because Gs, Ky, a, A remain same.

NOG = z/HOG = 4/1 = 4

A = 1.225 ( remain same) only y2 and x1 changes.

ln[{(0.0214 -0)/(y2 - 0)}( 1- 1/1.225) + 1/1.225] = 4

0.00385/y2 + 0.816 = exp(4)

y2 = 7.15*10^-5

Outlet gas having so2 mole fraction y2 = 7.3*10^-5

Mole ratio Y2 = 7.15*10^-5

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

58.74(0.0214 - 7.15*10^-5 ) = 1800(X1 - 0)

X1 = 6.96*10^-4

Mole fraction of SO2 in the leaving liquid x1 = 6.96*10^-4

Mole fraction of SO2 in leaving gas y2 = 7.15*10^-5

Here it is very dilute so that mole fraction = mole ratio

X1 = x 1and Y2 = y2.