Given:
Moles of methane:
nf ≔ 1 mol
Stoichiometric air moles:
ast ≔ 2 mol
Equivalence ratio:
phi ≔ 1.1
Actual air moles:
ast
= 1.818 mol
a ≔ ――
phi
Rich combustion reaction of methane is given as:
CH4 + a (O2 + 3.76 N2) ----> b CO2 + c H2O + d N2 + f CO + g H2
Carbon balance:
1＝b + f
Hydrogen balance:
4＝2 ⋅ c + 2 ⋅ g
or,
2＝c + g
Oxygen balance:
2 ⋅ a＝2 ⋅ b + c + f
Nitrogen balance:
a ⋅ 3.76 ⋅ 2 ＝ 2 ⋅ d
or,
d ＝ a ⋅ 3.76
Looking at the carbon, hydrogen, and oxygen balances, there are 3 equations and 4
unknowns. Therefore, we need a fourth equation to solve this problem.
For this, revisiting ME 636 notes from almost a decade ago, one could use the Water-Gas Shift
reaction to set up a fourth equation.
The Water-Gas Shift reaction is given as:
CO2 + H2 <----> CO + H2O
f⋅c
Equilibrium constant:
Kp ＝ ――
b⋅g
or,
NOTE: The original equation from the text
is mapped to the coefficients currently in
use.
f ⋅ c - Kp ⋅ b ⋅ g ＝ 0
Now, we can throw this in the mix and solve using fsolve (in Matlab). But wait, where's Kp.
1.761 1.611 0.2803
+ ―――
ln ⎛⎝Kp⎞⎠ ＝ 2.743 - ――- ――
t
t2
t3
where,
T
t ＝ ――
1000
1.761 1.611 0.2803
+ ―――
ln ⎛⎝Kp⎞⎠ ＝ 2.743 - ――- ――
t
t2
t3
b ≔ 0 mol
c ≔ 0 mol
Guess Values
f ≔ 0 mol
g ≔ 0 mol
T ≔ 600 K
T
t ≔ ―――
1000 K
⎛
1.761 1.611 0.2803 ⎞
+ ―――
Kp ≔ exp ⎜2.743 - ――- ――
⎟
t
t2
t3 ⎠
⎝
Constraints
b + f ＝ 1 mol
2 ⋅ c + 2 ⋅ g ＝ 4 mol
2 ⋅ b + c + f - 2 ⋅ a＝0
Solver
f ⋅ c - Kp ⋅ b ⋅ g ＝ 0
⎡b⎤
⎢c⎥
⎢ ⎥ ≔ find ((b , c , f , g))
⎢f⎥
⎣g⎦
Final Output:
Carbon dioxide moles:
b = 0.993 mol
Water moles:
c = 1.644 mol
Carbon monoxide moles:
f = 0.007 mol
Hydrogen moles:
g = 0.356 mol
T
t ＝ ――
1000