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Unformatted text preview: tes and compositions of the exiting streams by the group (Kremser) method.
Analysis: Apply Kremser's method, with stripping factors, S = KV/L , computed for L =
entering feed = 1000 kmol/h, V = entering steam rate = 100 kmol/h, and Kvalues from Fig. 2.8 at
the average of the two entering temperatures, (250 + 300)/2 = 275oF and a pressure of 50 psia.
The feed gas composition and the resulting Kvalues and values of S are as follows:
Component
C1
C2
C3
nC 4
nC 5
nC10
Total: f, kmol/h
0.3
2.2
18.2
44.7
85.9
848.7
1000.0 Kvalue
67
26
12.2
6.0
2.85 S = KV/L
6.7
2.6
1.22
0.60
0.285 The fraction of a gas component not stripped, φS, is given by the Kremser Eq. (550), and by
material balance the amounts stripped and not stripped follow:
S −1
φ S = N +1
and l1 = φ S f and v3 = f − l1
S −1
The results are as follows: Component
C1
C2
C3
nC 4
nC 5
Total: φS
0.0028
0.036
0.181
0.46
0.72 l1, kmol/h
0.0
0.1
3.3
20.6
61.8
85.8 The exiting gas is 65.5 + 100 = 165.5 kmol/h.
The exiting liquid is 85.8 + 848.7 = 934.5 kmol/h. v3, kmol/h
0.3
2.1
14.9
24.1
24.1
65.5 Exercise 9.26
Subject: Liquidliquid extraction of a hydrocarbon mixture by diethylene glycol (DEG) by the
group (Kremser) method.
Given: 100 kmol/h of an equimolar mixture of benzene (B), toluene (T), nhexane (C6), and
nheptane (C7). Extraction at 150oC with 300 kmol/h of DEG. Extractor has 5 equilibrium
stages. Distribution coefficients given below.
Find: Flow rates and compositions of extract and raffinate.
Analysis: Treat the problem like a stripper. Apply Kremser's method, with extraction factors,
Eq. (939), E = KDV/L , computed for L = entering feed = 100 kmol/h, V = entering DEG solvent
rate = 300 kmol/h, and given values of KD = mole fraction in extract/mole fraction in raffinate.
The extract is like the vapor in stripping and the raffinate is like the liquid in stripping. The feed
composition and the Kvalues and values of S are as follows:
Component
Benzene
Toluene
nHexane
nHeptane
Total: f, kmol/h
25
25
25
25
100 KD
0.33
0.29
0.050
0.043 E = KDV/L
0.99
0.87
0.15
0.129 The fraction of a feed component not extracted, φE , is given by the Kremser Eq. (944), and by
material balance the amounts stripped and not stripped follow:
E −1
φ E = N +1
and l1 = φ E f and v3 = f − l1
E
−1
The results are as follows: Component
Benzene
Toluene
nHexane
nHeptane
Total: φE
0.177
0.230
0.850
0.871 l1, kmol/h
4.3
5.8
21.2
21.8
53.1 v3, kmol/h
20.7
19.2
3.8
3.2
46.9 Now compute the transfer of DEG to the raffinate using Eqs. (940) and (943), using KD = 30
U = L/KDV = 100/[30(300)] = 0.0111. φU = 0.9889. Therefore 0.9889(300) = 296.7 kmol/h of
DEG leaves in the extract and 300  296.7 = 3.3 kmol/h in the raffinate.
The exiting extract flow rate = 46.9 + 296.7 = 343.6 kmol/h
The exiting raffinate flow rate = 53.1 + 3.3 = 56.4 kmol/h. Exercise 9.27
Subject: Reboiled stripping of a normal paraffin hydrocarbon feed liquid by the group method.
Given: Feed liquid at 39oF and 300 psia and of composition giv...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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