Determine the air-fuel ratio of C3H8 with 150 percent theoretical air supplied
Answers
Answer:
The chemical reaction in the combustion of propane is as follows,
C
3
H
8
+
1.5
a
(
O
2
+
3.76
N
2
)
→
c
C
O
2
+
d
H
2
O
+
5.64
a
N
2
+
0.5
a
O
2
Where, 'a' is the air coefficient, 'c' is the coefficient of the carbon and 'd' is the coefficient of hydrogen.
By the balancing of equations, we find out that,
c
=
3
,
d
=
4
and for a, balancing the coefficient of oxygen atoms as,
3
a
=
2
c
+
d
+
a
2
a
=
2
c
+
d
Substitute the values of c, d in the above equation therefore,
a
=
5
The equation can be rewritten as,
C
3
H
8
+
7.5
(
O
2
+
3.76
N
2
)
→
3
C
O
2
+
d
H
2
O
+
28.2
N
2
+
2.5
O
2
First calculating the molar mass of the propane,
C
3
H
8
=
12
×
3
+
1
×
8
=
44
And the molar mass of air is considered as, 29.
Therefore calculating the air fuel ratio,
A
F
=
m
a
m
f
=
n
a
×
M
a
n
p
r
o
p
a
n
e
×
M
p
r
o
p
a
n
e
=
(
7.5
×
4.76
)
×
29
1
×
44
=
23.529
k
g
o
f
a
i
r
k
g
o
f
f
u
e
l
Thus,
23.529
k
g
o
f
a
i
r
k
g
o
f
f
u
e
l
is the amount of mixture required for complete combustion of propane.
The chemical reaction in the combustion of propane is as follows,
C3H8+1.5a(O2+3.76N2)→cCO2+dH2O+5.64aN2+0.5aO2
Where, 'a' is the air coefficient, 'c' is the coefficient of the carbon and 'd' is the coefficient of hydrogen.
By the balancing of equations, we find out that,
c=3,
d=4
and for a, balancing the coefficient of oxygen atoms as,
3a=2c+d+a
2a=2c+d
Substitute the values of c, d in the above equation therefore,
a=5
The equation can be rewritten as,
C3H8+7.5(O2+3.76N2)→3CO2+dH2O+28.2N2+2.5O2
First calculating the molar mass of the propane,
C3H8=12×3+1+8
And the molar mass of air is considered as, 29.
Therefore calculating the air fuel ratio,
AF=
=×
/
×
=(7.5×4.76)×29/1×44=23.529kg of fuel/kg of fuel
Thus,
23.529 kg of air/kg of fuel is the amount of mixture required for complete combustion of propane.