CHE425: Problem set #10
1)1 Acetone in air is being absorbed by water in a packed tower having a cross-sectional area
of 0.250 m2 at 293 K and 1 atm. At these conditions, the equilibrium relation is given by ye =
1.2xe. The inlet air contains 3.0 mole % acetone and the outlet 0.5 %. The air flow is 14.6
kmol/hr and the water inlet flow is 40.5 kmol/hr. The water inlet contains 0.1 mole %
acetone. Film coefficients for the given flow in the tower are kya = 3.8×10-2 kmol/s∙m3 and
kxa = 6.2×10-2 kmol/s∙m3. Determine the tower height.
Ans: 2.03 m
2) An absorber containing seven theoretical plates is used to process a natural gas stream
having the composition listed below:
Component
Mol %
C1
76.524
C2
13.110
C3
4.938
nC4
2.126
nC5
2.09
nC6
1.208
The gas is available for processing at a pressure of 450 psia. If 65% of the propane in the gas
stream is to be absorbed, what will be the composition of the gas stream leaving the absorber.
The average absorber temperature may be assumed to be 560oR.
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(1) ln K = A/T2 + B  C ln(P) + D/P2
(2) ln K = A/T2 + B  C ln(P) + D/P, where P is in psia, T is in oR
Compound
A
B
C
D
Form
=============================================================
Methane
292860
8.2445
.8951
59.8465
(1)
Ethane
687248.2
7.90694
.866
49.02654
(1)
Propane
970688.6
7.15059
.76984
6.90224
(2)
n-Butane
1280557
7.94986
.96455
0
(1)
n-Pentane
1524891
7.33129
.89143
0
(1)
n-Hexane
1778901
6.96783
.84634
0
(1)
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Ans:
The composition of the gas stream leaving the absorber is given in the following table
Component
Moles in feed
Moles exit stream
Mole% in exit stream
1
C1
76.524
72.2543
85.9395
C2
13.110
10.0919
12.0033
C3
4.938
1.7282
2.0555
nC4
2.126
0.0014
0.0016
nC5
2.09
0
0
nC6
1.208
0
0
C. J. Geankoplis, Transport Processes and Separation Process Principle, Prentice Hall, 2003, pg. 672
3. 2Solute A is to be removed from an inert gas B in a multi-stage countercurrent absorption
tower. The gas enters the tower at a rate of 200 kmol/h and contains 25 mol% A. The solvent
enters the tower at a rate of 800 kmol/h and is initially free of solute. Determine (a) the
concentration of the exiting gas stream and (b) the number of stages, if the exiting liquid
stream contains 5.0 mol% A. The equilibrium relationship is yA = 4.0xA. Assume that the
carrier gas is insoluble in the solvent and the solvent is nonvolatile.
Ans:
Yt = 0.053, yt = 0.050
Number of equilibrium stages = 3.62657
4. 3Solute A is to be stripped from a liquid stream by contacting the liquid with a pure gas.
The liquid enters the stripping tower at an A-free rate of 150 kmol/h and contains 30 mol%
A. The gas enters the column at a rate of 500 kmol/h. Determine the number of stages
required to reduce the concentration of A in the exiting liquid stream to 1.0 mol%. The
distribution of A in the gas and liquid phase is expressed as yA = 0.4xA.
Yt = 0.126, yt = 0.112
Number of equilibrium stages = 11.6379
2
3
Hines, A. L. and Maddox R. N., Mass Transfer: Fundamentals and Applications, Prentice Hall, 1985, pg. 280
Hines, A. L. and Maddox R. N., Mass Transfer: Fundamentals and Applications, Prentice Hall, 1985, pg. 280
5. Using the enthalpy-composition diagram for the methanol-water system (Figure 1),
calculate the number of stages required and the feed plate location to produce 98 wt%
methanol liquid product and 5 wt% methanol bottoms product from a feed of 45 wt%
methanol entering the column at 50oC. The column has a total condenser and a partial
reboiler, and operates at 1.5 times minimum reflux. What is the reboiler duty in Btu/hr if the
feed rate is 5,000 lb/hr? What is the condenser duty?
Ans:
QD = 2.37×106 Btu/hr
QR = 2.77×106 Btu/hr
From the chart, need 8 eq. trays + 1 reboiler.
6. Solve the following set of linear equations1 using the Thomas algorithm.
0
2  3 0
1 2  1 0 


0 4  1 1 


2  1
0 0
 y1    4 
y   2 
 2 =  
 y3   9 
   
 y4   2 
List the coefficient matrix after the elimination step. The matrix should have the following
form
1
0

0

0
p1
0
1
0
0
p2
1
0
0   y1   q1 
0   y2   q2 

p3   y3   q3 
   
1   y4   q4 
Ans:
1
0

0

0
0   y1   2 
1 3.5
0   y2  1.14 

0 1 0.143  y3   31 
  

0 0
1   y4   4 
2
0
y4 = 4
1
=4
y3 = (3  c3 y4)/3
= [31  1(4)]/0.143
=3
y2 = (2  c2 y3)/2
= [1.14  (1)(3)]/3.5
=2
y1 = (1  c1 y2)/1
= [2  (3)(2)]/2
=1
Riggs, J. B. An Introduction to Numerical Methods for Chemical Engineers, Texas Tech University Press,
1994, p. 39