Page 274 - Advanced thermodynamics for engineers

262 CHAPTER 12 CHEMICAL EQUILIBRIUM AND DISSOCIATION

In this reaction, the amount of substance in the products is equal to the amount of substance in the
reactants and there is no effect of pressure in the dissociation equation. Comparing the results for the
carbon monoxide (Eqn (12.1)) and the water gas reactions (Eqn (12.73)), it can be seen that if a mixture
of products of the reaction in Eqn (12.1) was subjected to a change in pressure the chemical
composition of the mixture would change, whereas the chemical composition of the products of the
water–gas reaction would be the same at any pressure.
These points are returned to later in this section.

12.7.3 EXAMPLE 3: A ONE DEGREE OF DISSOCIATION EXAMPLE
A spark-ignition engine operates on a 10% rich mixture of carbon monoxide and air. The conditions
at the end of compression are 8.5 bar and 600 K, and it can be assumed that the combustion is
adiabatic at constant volume. Calculate the maximum pressure and temperature achieved if
dissociation occurs.
(There are two different approaches for solving this type of problem; both of these will be outlined
below. The first approach, which develops the chemical equations from the degrees of dissociation, is
often the easier method for hand calculations because it is usually possible to estimate the degree of
dissociation with reasonable accuracy, and it can also be assumed that the degree of dissociation of the
water vapour is less than that of the carbon dioxide. The second approach is more appropriate for
computer programs because it enables a set of simultaneous (usually nonlinear) equations to be
defined.)

General considerations
The products of combustion with dissociation have to obey all of the laws which define the conditions
of the products of combustion without dissociation, described in Chapter 10, plus the ratios of con-
. This means that the problem becomes one with two
stituents defined by the equilibrium constant, K p r
iterative loops: it is necessary to evaluate the degree of dissociation from the chemical equation and the
equilibrium constant, and to then ensure that this obeys the energy equation (i.e. the First Law of
Thermodynamics). The following examples show how this can be achieved.

Solution:
Note that there is only one reaction involved in this problem, and that the carbon monoxide
converts to carbon dioxide in a single step. Most combustion processes have more than one chemical
reaction.
Stoichiometric combustion equation
1
CO þ ðO 2 þ 3:76N 2 Þ / CO 2 þ 1:88N 2 (12.75)
2
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n R ¼3:38 n P ¼2:88
Rich combustion equation
1
1:1CO þ ðO 2 þ 3:76N 2 Þ / CO 2 þ 0:1CO þ 1:88N 2 (12.76)
2
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n R ¼3:48 n P ¼2:98