Question

Stripping of Acetone We will strip acetone from water into an air stream. Operation is isothermal at 20°C and at a total pressure of 278 mmHg. The entering liquid is 16 mole% acetone. The entering liquid flow rate is 10 kg moles/hr. We desire the exiting liquid to be 1 mole % acetone. The entering air is pure. Assume that the water is non-volatile and the air is insoluble. Equilibrium data at 20°C: a. Find the minimum flow rate of air b. If the air flow rate is 4 kgmole/hr, find the number of equilibrium stages

X,mole fraction acetone	partial pressure acetone,mmHg
0	0
0.0333	30.0
0.0720	62.8
0.1170	85.4
0.1710	103.0

Answer 1

In this problem acetone is stripped with air as stripping agent

T = 20°C

P = 278 mmHg

Given data

L = 10 kmol/h

xo = 0.16

Xo = 0.16/(1-0.16) = 0.1904

xn = 0.01

Xn = 0.0101

Equilibrium data

x( mole fraction)	X( mole ratio)	pA	y= pA/P	Y
0	0	0	0	0
0.0333	0.0344	30	0.1079	0.120
0.0720	0.0775	62.8	0.225	0.290
0.1170	0.1325	85.4	0.307	0.443
0.1710	0.206	103	0.370	0.587

Plotting the graph between X and Y

To find minimum gas flowrate, the outlet gas concentration will be in equilibrium with inlet liquid concentration.

The inlet mole ratiobof liquid is 0.1904

From graph we can find Ymax

From graph Y1max = 0.0510

Yn+1=0, since air is pure

Doing material balance we get

Ls = L/(1+Xo) = 10/(1+0.1904) = 8.40 kmol/h

Substituting all values we get

Gs min= G = 29.69 kmol/h

If G = 4 kmol/h= Gs

Then substituting all values in material balance equation =

Y1 = 0.3786

The operating line coordinates are

(0, 0.0101) and (0.1904, 0.3786)

From graph number of stages ~ 6

The graph is shown below

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