Page 277 - Advanced thermodynamics for engineers

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12.7 EXAMPLES OF SIGNIFICANCE OF K p

Evaluating the energy which is contained in the products at 2900 and 3000 K, allowing for the
variation in a as the temperature changes, gives.

T P ¼ 2900 K

Constituent CO 2 CO O 2 N 2
u 2900 131933.3 74388.9 78706.6 73597.5
n 0.871875 0.2281 0.0641 1.88

U P ðT P Þ¼ 275405 kJ

T P ¼ 3000 K

Constituent CO 2 CO O 2 N 2
137320 77277 81863 76468
u 3000
n 0.8281 0.2719 0.08594 1.88
U P ðT P Þ¼ 285521 kJ

These values are plotted in Fig. 12.3, and it can be seen that, if the variation of the energy terms was
linear with temperature, the temperature of the products after dissociation would be 2958 K. The
calculation will be repeated to show how well this result satisfies both the energy and dissociation
equations.
First, it is necessary to evaluate the degree of dissociation which will occur at this products’
temperature. At T P ¼ 2950 K, K p ¼ 3.62613 and this can be substituted into Eqn (12.78) to give
a ¼ 0.1494.
Hence, the chemical equation, taking account of dissociation, is
1
1:1CO þ ðO 2 þ 3:76N 2 Þ/0:8506CO 2 þ 0:2494CO þ 0:0747O 2 þ 1:88N 2
2
Applyingtheenergyequation,theenergyreleasedbythecombustionprocessgivesaproductsenergyof
U P ðT P Þ¼ð1 aÞ 279523 þ 43993 ¼ 281755 kJ
This energy is contained in the products as shown in the following table.

Constituent CO 2 CO O 2 N 2
135670 76079 80488 75299
u 2958
n 0.8506 0.2494 0.0747 1.88

U P ðT P Þ¼ 281947 kJ
The equations are balanced to within 0.068% and this is close enough for this example.