Unformatted text preview: rom air at 25oC by 5N aqueous triethanolamine Given: Feed gas containing 10 mol% CO2 and 90 mol% air. Absorbent of 5N aqueous
triethanolamine containing 0.04 moles of CO2 per mole of amine solution. Column with 6
equilibrium stages. Exit liquid to contain 78.4% of the CO2 in the feed gas. Therefore, exit gas
contains 21.6% of the CO2 in the entering gas. Equilibrium data for CO2 at 25oC in terms of
mole ratios.
Assumptions: Negligible absorption of air and stripping of amine and water.
Find: (a) Moles of amine solution required per mole of feed gas.
(b) Exit gas composition.
Analysis: Use the nomenclature and type of plot shown in Fig. 6.11(a). Therefore, for CO2,
X0 = 0.04 mol CO2/mol amine solution
YN+1 = Y7 = 10/90 = 0.1111 mol CO2/mol air
Y1 = 0.216(10)/90 = 0.024 mol CO2/mol air
(b) Therefore, the exit gas composition is 0.024 mol CO2/mol air or 0.024/(1 + 0.024) x 100% =
2.34 mol% CO2 and 97.66 mol% air.
(a) A plot of the equilibrium data as Y vs. X is given below. The operating point (X0, Y1) at the
top of the column is included. A straight operating line through this point is found by trial and
error to give 6 equilibrium stages, when using Y7 = 0.1111. The resulting XN = X6 = 0.085.
From Eq. (63), the slope of the operating line = L'/V' = (0.1111  0.024)/(0.085  0.04) = 1.936
mol triethanolamine solution/mol air. The feed gas contains 9 mol air/10 mol feed gas.
Therefore, mols of amine solution/mol feed gas = 1.936(0.9) = 1.74.
See plot on next page. Exercise 6.7 (continued) Exercise 6.8
Subject:
Absorption of acetone from air by water at 20oC and 101 kPa (760 torr) in a
valvetray column.
Given: 100 kmol/h of feed gas containing 85 mol% air and 15 mol% acetone. Pure water is
the absorbent. Overall tray efficiency is 50%. Absorb 95% of the acetone. Equilibrium px data
for acetone are given as listed below.
Assumptions: Negligible absorption of air and stripping of water.
Find: (a) Minimum ratio, L'/V' of moles of water/mole of air.
(b) Number of equilibrium stages for L'/V' = 1.25 times minimum.
(c) Concentration of acetone in the exit water.
Analysis: Use the nomenclature and type of plot shown in Fig. 6.11(a). Then, the operating line
will be straight. For acetone,
X0 = 0.0 mol acetone/mol entering water
YN+1 = 0.15/0.85 = 0.1765 mol acetone/mol air in entering gas
Flow rate of acetone in exit gas = (1  0.95)(15) = 0.75 kmol/h. With 85 kmol/h of air,
Y1 = 0.75/85 = 0.00882 mol acetone/mol air
Convert the px equilibrium data to mole ratio, YX data, using y = p/P, Y = y/(1 y), X = x/(1x)
p, torr
x
y
X
Y
30.0 0.033 0.0395 0.0341 0.0411
62.8 0.072 0.0826 0.0776 0.0901
85.4 0.117 0.1124 0.1325 0.1266
103.0 0.171 0.1355 0.2063 0.1568
(a) With the type of curvature in the YX equilibrium curve, shown below, the minimum
absorbent rate is determined by a straight operating line that passes through the point (Y1 , X0 )
and is drawn tangent to the equilibrium curve, as shown. From Eq. (63), the slope of the
operating line = L'/V' = 1.06 mol water/mol of air on an acetonefree basis = minimum ratio.
(b) For 1.25 times minimum, L'/V' = 1.25(1.06) = 1.325. Now a stra...
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 Spring '11
 Levicky
 The Land

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