Exercise 4.1
Subject:
Degrees of freedom analysis for a threephase equilibrium stage.
Given:
Equilibrium stage of Figure 4.35, with two feeds (one vapor, one liquid), vapor entering
from stage below, liquid entering from stage above, three exiting streams (one vapor, two liquid),
and heat transfer.
Assumptions:
Equilibrium stage
Find:
(a) List and count of variables.
(b) List and count of equations.
(c) Number of degrees of freedom.
(d) List of reasonable set of design variables.
Analysis:
(a) With 4 streams in and 3 streams out, and heat transfer,
Number of variables
=
N
V
= 7(
C
+3) + 1
= 7
C
+ 22
The variables are 7 total flow rates, 7 temperatures, 7 pressures, 1 heat transfer rate and
C
mole
fractions for each of the 7 streams.
(b) The equations are:
C
Component material balances
1
Energy balance
2
Pressure identity equations for the 3 exiting streams
2
Temperature identity equations for the 3 exiting streams
7
Mole fraction sums (one for each stream)
2
C
Phase equilibrium equations:
K
y
x
K
y
x
i
i
i
i
i
i
I
I
II
II
and
=
=
Total number of equations
=
N
E
=
3
C
+ 12
(c)
Degrees of freedom
=
N
D
= N
V
 N
E
= (7
C
+ 22)  (3
C
+ 12)
= 4
C
+10
(d)
A possible set of specifications is:
For each entering stream: Total flow rate, temperature, pressure, and
C
1 mole fractions,
which totals 4(
C
+ 2) = 4
C
+ 8
For the remaining 2 variables, choose any combination of
Q
, temperature of one of the
three exiting streams, and/or pressure of one of the three exiting streams.
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Exercise 4.2
Subject:
Determination of uniqueness of three different operations.
Given:
(a) An adiabatic equilibrium stage with known vapor and liquid feed streams, and
known stage temperature and pressure.
(b) Same as (a), except that stage is not adiabatic.
(c) Partial condenser using cooling water, with known vapor feed (except for flow
rate), outlet pressure, and inlet cooling water flow rate.
Assumptions:
Exiting streams in equilibrium.
Find:
(a) Whether composition and amounts of exiting vapor and liquid can be computed.
(b) Same as part (a).
(c) Whether cooling water rate can be computed.
Analysis:
(a) With two steams in and two out, number of variables =
N
V
=
4(
C
+ 3) = 4
C
+ 12
Equations are:
C
Component material balances
1
Energy balance
1
Pressure identity for 2 exiting streams
1
Temperature identity for 2 exiting streams
4
Mole fraction sums for 4 streams
C
Phase equilibrium equations
Therefore, number of equations =
N
E
=2
C
+ 7
Degrees of freedom =
N
D
= N
V
 N
E
= (4
C
+ 12)  (2
C
+ 7) = 2
C
+5
Given specifications are: 2
C
+ 4 variables for the two feed streams. Only one
specification left. Therefore
can not specify both
T
and
P
for exiting streams
.
(b) If stage in part (a) is not adiabatic, add
Q
as a variable to give
N
V
=
4
C
+ 13.
The number of equations stays the same, i. e.
N
E
=2
C
+ 7.
Thus, have one additional degree of freedom, giving,
N
D
= N
V
 N
E
= 2
C
+6.
Can now specify both
T
and
P
for exiting streams
.
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 Spring '11
 suh
 Equilibrium, pH

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