Environmental Engineering Reference
In-Depth Information
Solution
We will construct the Gibbs function for the mixture and obtain the equilibrium com-
position by minimization of the function subject to the elemental mass balance
constraints.
At 1 bar and 1000 K, the assumption of ideal gas is justified, and we can write
Equation (5.36) for each component involved in the process:
4
61
R
u
T
+ln
:
n
CH
4
+
λ
C
R
u
T
+
4
λ
H
R
u
T
=0
X
i
n
i
CH
4
:
40
604
R
u
T
:
+ln
n
C
2
H
2
+
2
R
u
T
+
2
λ
C
λ
H
R
u
T
=0
X
i
n
i
C
2
H
2
:
28
249
R
u
T
:
+ln
n
C
2
H
4
+
2
R
u
T
+
4
λ
C
λ
H
R
u
T
=0
X
i
n
i
C
2
H
4
:
26
R
u
T
+ln
n
C
2
H
6
:
13
+
2
R
u
T
+
6
λ
C
λ
H
R
u
T
=0
X
i
n
i
C
2
H
6
:
CO
2
:
−
61
R
u
T
94
:
n
CO
2
+
λ
C
R
u
T
+
2
λ
O
R
u
T
=0
X
i
n
i
+ln
:
λ
C
R
u
T
+
λ
O
R
u
T
=0
:
−
47
942
R
u
T
n
CO
X
i
n
i
CO
+ln
+
n
O
2
+
2
λ
O
R
u
T
=0
X
i
n
i
O
2
:
ln
n
H
2
+
2
λ
H
R
u
T
=0
X
i
n
i
H
2
:
ln
!
+
2
:
R
u
T
+
λ
O
λ
H
:
−
03
R
u
T
46
n
H
2
O
X
i
n
i
H
2
O
+ln
R
u
T
=0
The total number of each type of atommust be the same as the number entering the
reactor. This conservation of elements forms equality constraints on the equilib-
rium composition. The conservation equations are:
C
:
n
CH
4
+2n
C
2
H
4
+2n
C
2
H
2
+n
CO
2
+n
CO
+2n
C
2
H
6
=2
H
:
4n
CH
4
+4n
C
2
H
4
+2n
C
2
H
2
+2n
H
2
+2n
H
2
O
+2n
C
2
H
6
=14
O
:
2n
CO
2
+n
CO
+2n
O
2
+n
H
2
O
=4
X
n
i
=n
CH
4
+n
C
2
H
4
+n
C
2
H
2
+n
CO
2
+n
CO
+n
O
2
+n
H
2
+n
H
2
O
+n
C
2
H
6
i
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