Unformatted text preview: ent flow rate is given by Eq. (612). Assume that because the Kvalue of B2 is the lowest of SO2, B2, and B3, that B2 controls. Further assume that essentially
all of the B3 is stripped so that the 0.5 mol% of (B2 + B3) is entirely B2. The solutefree liquid
rate = flow rate of BS = L' = 100 lbmol/h and the fraction of B2 stripped = (2.0  0.503)/2.0 =
0.748
L'
100
'
(fraction stripped) =
(0.748) = 24.9 lbmol/h
Vmin =
KSO2
3.01
Now use this value to compute the values of fraction stripped of SO2 and B3.
KV ' 6.95(24.9)
=
> 1.0
100
L'
KV ' 4.53(24.9)
Fraction stripped of B3 = ' =
> 1.0
L
100
Fraction stripped of SO 2 = Therefore, the assumption that B2 controls is correct at the minimum stripping gas rate.
Assume an operational nitrogen flow rate of 1.5 times minimum. Therefore V’ =1.5(24.9) = 37.4
lbmol./h. The stripping factor, given by Eq. (616), is for B2,
S B2 = K B2V ' 3.01(37.4)
=
= 1.13
L'
100 Using Eq. (614), Fraction B2 stripped = 0.748 =
Solving, N = 2.4 equilibrium stages. S N +1 − S 113 N +1 − 113
.
.
=
N +1
N +1
S
−1
1.13 − 1 Analysis: (continued) Exercise 6.12 (continued) Fraction B3 stripped = S N +1 − S 1.7 2.4 +1 − 17
.
=
= 0.863
N +1
2 . 4 +1
−1
S −1
1.7 Fraction SO 2 stripped = S N +1 − S 2.612.4 +1 − 2.61
=
= 0.936
S N +1 − 1
2.612.4 +1 − 1 These fractions do not meet specifications. Therefore, the number of stages and/or the nitrogen
flow rate must be increased.
First try increasing the number of stages.
S = KV/L
Fraction stripped:
N=5
N=7
N = 10 SO2
2.61 B2
1.13 B3
1.7 0.995
0.9993
0.99996 0.880
0.922
0.954 0.970
0.990
0.998 For N = 5 stages, have in the exit liquid in lbmol/h, 0.05 SO2, 0.24 B2, and 0.24 B3. These
values correspond to a total of 100.53 lbmol/h of exit liquid. This results in 0.05 mol% SO2 and
0.48% (B2 + B3). Therefore, N = 5 stages is sufficient with a feed gas rate, V' , of 37.4 lbmol/h.
Other combinations of N and V' could be used.. Exercise 6.13
Subject:
Absorption of a light hydrocarbon gas mixture by ndecane as a function of
pressure and number of equilibrium stages at 90oF.
Given: Light hydrocarbon gas containing in lbmol/h, 1,660 C1, 168 C2, 96 C3, 52 nC4, and 24
nC5 for a total of 2,000 lbmol/h = V. Absorbent of L = 500 lbmol/h of nC10. Kvalue of nC10 =
0.0011.
Find: Component flow rates in the lean gas and rich oil for:
(a) N = 6 and P = 75 psia.
(b) N = 3 and P = 150 psia.
(c) N = 6 and P = 150 psia.
Analysis: Use the Kremser method with Eqs. (548) for fraction not absorbed, φA, (550 for
fraction not stripped, φA ,(538) for absorption factor, A, (551) for stripping factor, S , (545) and
(553) for component flow rates in the exit gas, υ1 , and (544) and (552) for component flow
rates in the exit liquid, lN, since no light hydrocarbons enter with the absorbent. Use Fig. 2.8 or
other source for Kvalues of C1 through nC5. For nC10 , assume the Kvalue is inversely
proportional to pressure.
(a) N = 6 and P = 75 psia: Component
C1
C2
C3
nC 4
nC 5
nC10
Total
(b) Kvalue
2.9
6.5
1.95
0.61
0.19
0.0011 A
0.0...
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 Spring '11
 Levicky
 The Land

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