Page 239 - Advanced thermodynamics for engineers

226 CHAPTER 10 THERMODYNAMICS OF COMBUSTION

and

p 1 V 1 ¼ n 1 <T 1 ¼ n R <T 1
giving

n P <T 3 V 1
p 3 ¼ p 1
n R <T 1 V 3
3097
¼ 8 1 ¼ 82:59 bar:
300

Example 3: nonadiabatic combustion in an engine
In the previous calculation, example 2, the combustion was assumed to be adiabatic. If 10% of the
energy liberated by the fuel was lost by heat transfer, calculate the final temperature and pressure after
combustion.
Solution
The effect of heat transfer is to reduce the amount of energy available to raise the temperature of the
products. This can be introduced in Eqn (10.30) to give Eqn (10.31).

U P T P ¼ ðQ v Þ Q HT þ U R T R U R T s þ U P T s (10.31)
s
In this case Q HT ¼ 0.1(Q v ) s and hence

U P T P ¼ 0:9ðQ v Þ þ U R T R U R T s þ U P T s (10.32)
s
Substituting values gives
4
U P ðT P Þ¼ 0:9 50144 16 þ 9:8696 10 þ 66802:2
¼ 887571:8kJ:
Using a similar technique to before, the first estimate of U P (T P ) is

887571:8
U P ðT P Þ¼ ¼ 84369:9kJ kmol:
10:52
From the previous calculation the value of T P lies between 2500 and 3000 K, and an estimate is

84369:9 76848:7
T P ¼ 2500 þ 500 ¼ 2712 K:
94623:3 76848:7
Try T P ¼ 2700 K to check for energy balance.

Constituent CO 2 H 2 O N 2
121,807.5 95,927.5 67,877.4
u 2700
n 1 2 7.52
nu 121,807.5 191,854.9 510,438.1

U P ðT P Þ¼ 823993:7kJ: