Combustion
define some units
combustion of dodecane a parafin of type
C
n
⋅
H
2n
kN
:=
10
3
⋅
N
kPa
:=
10
3
⋅
Pa
(represents diesel fuel) in stochiometric proportions:
6
3
MPa
:=
10 Pa
kJ
:=
10
⋅
J
⋅
yH
2
⋅
O
⋅
+
heat
3
C
12
⋅
H
26
+
xO
2
=
⋅
+
zC
⋅
O
2
kmol
:=
overkill in this case but general method represented by solution of simultaneous equations from elements involved
initial values for given
x
:=
1
y
:=
1
z
:=
1
Given
construct
element:
C
z
=
1
2
O
x2
=
y
+
⋅
⋅
2 z
26
=
2 y
H
⋅
x
y
:=
Find x
y
,
z)
x
18.5
(
,
y
=
13
z
z
12
so combustion equation is
37
C
12
⋅
H
26
+
2
=
⋅
⋅
O
+
⋅
O
2
+
heat
→
C
12
⋅
H
26
+
2
⋅
O
2
=
⋅
⋅
O
+
12
C
⋅
O
2
⋅
2
⋅
13
H
2
⋅
+
heat
O2 comes with nitrogen: 21% by volume and 23.3% by weight in air (79% N2 by volume and 76.7% N2 by
weight  ~ 1% Ar lumped with N2) so .
.. need 79/21 atoms (volume) N2 for each O2
79
18.5
⋅
=
69.6
21
and combustion is .
..
this is on a stochiometric basis
(mole basis i.e. 1 mole of
C
12
⋅
H
26
+
18.5
O
2
+
⋅
=
⋅
⋅
O
+
⋅
⋅
O
2
+
⋅
C12H26 combines with 18.5
⋅
69.6 N
2
13 H
2
12 C
69.6 N
2
+
heat
moles of O2 etc.) or volume basis
to convert to weight use molecular
kg
kg
kg
weights
mw_O2
:=
32
mw_C12_H26
:=
(144
+
26)
⋅
mw_N2
:=
28
kmol
kmol
kmol
kg
mw_H2_O
:=
(2
+
16
)
⋅
kmol
mw_C_O2
:=
(12
+
32)
⋅
kg
kmol
1kmolC
12
⋅
H
26
+
⋅
⋅
O
2
+
⋅
⋅
N
2
=
⋅
⋅
H
2
⋅
O
+
⋅
⋅
O
2
+
⋅
18.5
kmol
69.6
kmol
13
kmol
12kmol
C
69.6kmol
N
2
+
LHV
⋅
1kmol
mw_C12_H26
⋅
⋅
...
=
13
mw_H2_O
⋅
kmol
⋅
H
2
⋅
O
...
+
LHV
this is divided by 170  the
170
C
12
H
26
170
molecular weight of C12H26 to
⋅
12kmol
mw_C_O2
18.5
kmol.mw_O2
⋅
express on a per 1 kg fuel basis
+
⋅
O
2
...
+
⋅
C
⋅
O
2
...
170
170
⋅
69.6kmol
mw_N2
+
69.6
kmol
⋅
mw_N2
⋅
N
2
+
⋅
⋅
N
2
)
mw_C12_H26
=
170kg
170
170
(1kmol
⋅
10/11/2004
1
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View Full Documentfor symbolic calculation
result is .
.. combustion of C12H26 by weight .
..
1kg
C
12
3.48
kg
11.46
kg
1.38
kg
⋅
⋅
⋅
⋅
⋅
⋅
⋅
=
⋅
⋅
11.46
kg
weight of air : weight of fuel = airfuel ratio
air_fuel_ratio
:=
3.48
+
11.46
air_fuel_ratio
⋅
⋅
H
26
O
2
N
2
H
2
⋅
O
3.11
k
⋅
C
⋅
O
2
N
2
+
heat
+
+
+
+
=
14.94
In order to insure complete combustion,air is usually supplied in excess, see example below. Products would
include air i.e. O2 and N2
to analyze combustion process use first law .
..
steady state, steady flow process .
..
m_dot
=
flow_rate
(5.46)
2
h
e
2
⋅
gz
e
V
i
V
e
d
d
+
d
dt
W
c_v
+
∑
∑
(5.47)
Q
c_v
m_dot
i
h
i
m_dot
e
+
+
+
+
2
i
n
n
⋅
⋅
=
⋅
2
t
n
n
work, KE and PE = 0 .
.
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 Fall '06
 DavidBurke
 Enthalpy, Combustion, LHV, kmol kg kmol, kg kmol kg

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