Geology Reference
In-Depth Information
The enthalpy of gaseous fuels can be calculated with Eq. (D.27):
P
i=1
x
i
h
i
(T) h
i
(T
0
)
d
1
H
f;i
= H
f;i
+
(D.27)
Where h
i
(T) is obtained with the method of Zelenik and Gordon (1961):
+ a
3
T
2
3
+ a
4
T
3
4
+ a
5
T
4
5
a
1
+ a
2
T
2
+
a
6
T
h
i
(T) = RT
(D.28)
The entropy of gaseous fuel can be calculated with Eq. (D.29):
P
i=1
x
i
s
i
(T) Rln
P
i
P
0
s =
(D.29)
d
1
Where s
i
(T) is obtained through Zelenik and Gordon (1961) with Eq. (D.30)
a
1
lnT + a
2
T + a
3
T
2
2
+ a
4
T
3
3
+ a
5
T
4
4
s
i
(T) = R
+ a
7
(D.30)
Coe
cients a
1
through a
7
for the calculation of h
(T) and s
i
(T) are provided
in Table D.6 for the different constituents composing the gaseous fuel.
From Eqs. D.4 through D.18, one can conclude that the exergy of fuels is a
function of ambient conditions. This means, that if the temperature, pressure or
the concentration of CO
2
for instance change, so will the exergy of a given fuel.
Table D.6 shows the coe
cients a
1
through to a
7
for ideal gases required for the
calculation of h
(T) and s
(T) in Eqs. D.28 and D.30, according to Zelenik and
Gordon (1961).
Additionally, Table D.7 is the resolution of Eq. (D.15) for the calculation of the
chemical potential of elements included in hydrocarbons.
Table D.6 Coe
cients a
1
through to a
7
(Zelenik and Gordon, 1961)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
CH
4
2.928
0.002569
7.844E-06
-4.91E-09
2.04E-13
-10054
4.634
C
2
H
6
1.463
0.01549
5.781E-06
-1.26E-08
4.59E-12
-11239
14.43
C
3
H
8
0.8969
0.02669
5.431E-06
-2.13E-08
9.24E-12
-13955
19.36
C
4
H
10
1.522
0.03429
8.101E-06
-2.92E-08
1.27E-11
-17126
18.35
C
5
H
12
1.878
0.04122
0.00001253
-3.70E-08
1.53E-11
-20038
18.77
N
2
3.704
-0.001422
2.867E-06
-1.20E-09
-1.40E-14
-1064
2.234
CO
2
2.401
0.008735
-6.607E-06
2.00E-09
6.33E-16
-48378
9.695
Search WWH ::
Custom Search