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Unformatted text preview: 72 mol/L, x1 = 0.016 mol/L
% extracted = y1S/ xFQ x 100% = (0.0672)(500)/(0.05)(1,000) x 100% = 67.2% (3) (b) For three crosscurrent stages, can use Eq. (521) in the following form,
x3
1
=
xF
1+ E / N 3 (4) where from Eq. (514), E = KS/Q = (4.2)(500)/1,000 = 2.1 and N = 3. Frm Eq. (4),
Percent extraction = 1 − x3
1
× 100% = 1 −
× 100% = 79.6%
3
xF
(1 + 2.1/ 3) (c) ) For threestage countercurrent extraction, Eq. (528) applies in the form,
x3
1
1
=
=
= 0.058
2
3
xF
1+ E + E + E
1 + 2.1 + 2.12 + 2.13
% extraction = (1  0.058) x 100% = 94.2% (5) Exercise 5.13
Subject:
Given: Extraction of pdioxane (B) from water (A) with solvent (C).
Extraction factor, E = 0.9 compared to 2.4 in Example 5.2. Assumptions: Water and solvent are mutually insoluble. Equilibrium stages.
Find: The effect on % extraction of the number and configuration of stages.
Analysis: For a single stage and the cocurrent arrangement with any number of stages,
N, Eq. (513) applies, where the B subscript has been dropped:
X (N)
1
1
=
=
= 0.526
(F)
X
1 + E 1 + 0.9
Therefore, the % extraction = (1  0.526) x 100% = 47.4% for any number of cocurrent
stages.
For a crosscurrent cascade, Eq. (521) applies: X (N)
1
=
(F)
X
1+ E / N 1 = N 1 + 0.9 / N N (1) For a countercurrent cascade, Eq. (529) applies:
X (N)
=
X (F) 1
N
n=0 E =
n 1
N 0.9 (2)
n n=0 Results from Eqs. (1) and (2) are used to calculate % extraction from: X (N)
× 100%
X (F)
Using Eqs. (1) to (3), the following results are obtained:
% extraction = 1 − N
2
3
4
5 % extraction,
crosscurrent
52.4
54.5
55.6
56.3 % extraction,
countercurrent
63.1
70.9
75.6
78.7 (3) Exercise 5.13 (continued)
We see that for E < 1, the % extraction increases only slowly with increasing number of
stages. It is desirable to have the situation where E > 1. Exercise 5.14
Subject: Multistage absorption of a hydrocarbon gas with a highmolecularweight oil.
Given: Hydrocarbon feed gas and absorbent oil of Example 5.3 in Fig. 5.11. Absorber
pressure of 400 psia and average temperature of 97.5oF, with Kvalues at these
conditions.
Assumptions: Applicability of Kremser's method.
Find: % absorption of components and total gas and % stripping of oil for N = 1, 3, 10,
and 30 equilibrium stages. Effect of N on each component.
Analysis: The same values of the absorption factors, A, and stripping factors, S, apply.
Also Eqs. (548) and (550) are used to compute the fractions not absorbed, φΑ, and not
stripped, φB, respectively. Then Eq. (554) is used to compute the component flow rates
in the exit vapor. Then an overall material balance is used to compute the component
flow rates in the exit liquid:
l N = l0 + υ N +1 − υ 1 Let the % absorption of a component in the entering gas = l N / l0 + υ N +1 × 100%
The % stripping of the oil in the absorbent = υ 1 oil / l0 oil × 100% Using a spreadsheet, the following results are obtained for the % absorption of C1, C2, C3,
nC4, and nC5, the total % absorption of these light hydrocarbons, and % stripping of oil. N
1
3
6
10
30 C1
3.01
3.10
3.10
3.10
3.10 C2...
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 Spring '11
 Levicky
 The Land

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