Page 112 - Entrophy Analysis in Thermal Engineering Systems

Irreversible engines—Open cycles 105

by-product of fuel oxidation should be calculated based on its concentration
in the combustion products stream.
_
A combination of Eqs. (8.1) and (8.2) to eliminate Q yields
0

f p p f _
a a T 0 Φ (8.3)
0
0 0 0 + T 0 S S S 0
0
_ W net ¼ H + H H
which reduces to the following expression if the system is fully reversible

f p p f
a a (8.4)
0 0 0 + T 0 S S S 0
0
0
_ W rev ¼ H + H H
If Eq. (8.3) is described per unit molar flowrate of fuel burned _n f , we have
w net ¼ w rev T 0 SEG (8.5)
_
where SEG ¼ Φ=_n f denotes the specific entropy generation, and
1 T 0
f p p a f
a
w rev ¼ + (8.6)
0
0 0 0 _ n f S S S 0
0
H + H H
_ n f
In the next section, we will mathematically show that the specific reversible
work, w rev , is approximately a fixed quantity under different operational
conditions. In order to reach this goal, the enthalpy and entropy terms in
Eq. (8.6) must be evaluated for a generic combustion reaction.

8.3 Proof that w rev is relatively constant

The proof to be presented is based on the following set of common
assumptions: (i) air consists of 21% oxygen and 79% nitrogen (molar basis),
(ii) fuel contains only carbon, hydrogen, and oxygen, to be represented by
C x H y O z , (iii) air and the combustion products behave as ideal gas mixtures,
and (iv) the combustion products consist of carbon dioxide, water, oxygen,
and nitrogen. As Eq. (8.6) requires calculation of the enthalpy and entropy of
the products at ambient conditions, water vapor condensation is also
accounted for in the products stream. The overall combustion reaction
can be represented as

C x H y O z + Λ O 2 +3:76N 2 Þ ! xCO 2 + n v H 2 O vðÞ + Λ x + z O 2
y
ð
4 2

+3:76ΛN 2 + y n v H 2 O lðÞ
2