Unformatted text preview: 55 kmol/h (reflux ratio =
155/45), gives:
kmol/h:
Component
Feed Distillate Bottoms
C3
5
4.996
0.004
iC 4
15
14.987
0.013
nC 4
25
24.587
0.413
iC 5
20
0.354
19.646
nC 5
35
0.077
34.923
Total
100
45.001
54.999 Exercise 5.25
Subject: Degrees of freedom analyses for a partial reboiler and a total condenser. Given: A partial reboiler with exiting vapor and liquid in equilibrium. A total
condenser.
Find: Number of variables, number of independent equations, and degrees of freedom.
Analysis: Partial reboiler:
Have 3 streams with C + 3 variables each, plus the heat transfer rate.
Therefore, NV = 3(C + 3) + 1 = 3C + 10
The independent relationships are:
C
Component material balances
1
Enthalpy balance
1
Equality of temperatures of exiting vapor and liquid
1
Equality of pressures of exiting vapor and liquid
C
Component phase equilibrium between exiting vapor and liquid
3
Mole fraction summations for 3 streams
Therefore, NE = 2C + 6
From Eq. (567), ND = NV  NE = (3C + 10)  (2C + 6) = C + 4
Total condenser:
Have 2 streams with C + 3 variables each, plus the heat transfer rate.
Therefore, NV = 2(C + 3) + 1 = 2C + 7
The independent relationships are:
C
Component material balances
1
Enthalpy balance
2
Mole fraction summations for 2 streams Therefore, NE = C + 3
From Eq. (567), ND = NV  NE = (2C + 7)  (C + 3) = C + 4 Exercise 5.26
Subject:
Given: Degrees of freedom analyses for a stream mixer and a stream divider.
A steam mixer and a stream divider, each with heat transfer. Find: Number of variables, number of independent equations, and degrees of freedom.
Analysis: Stream mixer with two inlet streams:
Have 3 streams with C + 3 variables each, plus the heat transfer rate. Therefore, NV = 3(C + 3) + 1 = 3C + 10
The independent relationships are:
C
Component material balances
1
Enthalpy balance
3
Mole fraction summations for 3 streams
Therefore, NE =C + 4
From Eq. (567), ND = NV  NE = (3C + 10)  (C + 4) = 2C + 6
Stream divider with two exiting streams:
Have 3 streams with C + 3 variables each, plus the heat transfer rate.
Therefore, NV = 3(C + 3) + 1 = 3C + 10
The independent relationships are:
1
Total material balance
Component mole fraction equalities for exiting streams relative to the feed
2(C  1)
1
Enthalpy balance
1
Equality of temperatures of exiting streams
1
Equality of pressures of exiting streams
3
Mole fraction summations for 3 streams
Therefore, NE = 2C + 5
From Eq. (567), ND = NV  NE = (3C + 10)  (2C + 5) = C + 5 Exercise 5.27
Subject: Degrees of freedom analysis for the specifications of a distillation column. Given: Feed of 10 mol% benzoic acid and 90 mol% maleic anhydride. Column has a
total condenser and a partial reboiler. Operation is at 100 torr with a reflux ratio of 1.2
times minimum. Desired products are a distillate of 99.5 mol% maleic anhydride and a
bottoms of 0.5 mol% anhydride.
Find: Whether problem is completely specified.
Analysis: From operation (b) in Table 5.4,
N D = 2N + C + 9 = 2N + 11
The given specifications are equivalent to:
Pressure at each stage, including partial reboil...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details