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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