Question 2.8: A mixture of CO and air is 10% fuel rich by volume and at 8....
A mixture of CO and air is 10% fuel rich by volume and at 8.28 bar and 555K. The mixture burns adiabatically and at constant volume. Calculate the products temperature neglecting dissociation.
The representative stoichiometric reaction equation gives the ideal reaction, which shows the required oxygen and nitrogen to thoroughly combust all the fuel:
CO + \frac{1}{2}O_{2} + \frac{1}{2} \times 3.76N_{2} \rightarrow CO_{2} + \frac{1}{2} \times 3.76N_{2}We know that the reaction is not stoichiometric, because we are told that the mixture is 10% rich – i.e. there is 10% more CO than can be burnt with the stoichiometric air in the above equation. Therefore, the true representative reaction equation will have 1.1CO, thus:
1.1CO + \frac{1}{2}O_{2} + \frac{1}{2} \times 3.76N_{2} \rightarrow CO_{2} + \frac{1}{2} \times 3.76N_{2} + 0.1COAlternatively, divide by 1.1 to get the reaction in terms of combustion of 1 kmol of fuel:
CO + 0.91(\frac{1}{2}O_{2} + \frac{1}{2} \times 3.76N_{2} ) \rightarrow 0.9CO_{2} + 1.71N_{2} + 0.0COIn order to determine the final temperature, we need to analyse the closed system with the method described above:
The internal energy of combustion can be obtained from the tabulated values of enthalpy of combustion and the equation to convert between the two:
\Delta H_{0} = \Delta U_{0} + \Delta (n_{p} – n_{R}) RT_{0}\Delta \tilde{h_{0}} for CO is –283 000 kJ/kmol for the reaction CO + \frac{1}{2}O_{2} \rightarrow CO_{2}
\Delta \tilde{u}_{0} = \Delta \tilde{h}_{0} – (n_{p} – n_{R}) \tilde{R}T\Delta \tilde{u}_{0} = -283 000 – (1 – 1.5)8.314 \times 298
Therefore the internal energy of combustion per kmol of CO is – 281 761 kJ/kmol.
To put this in specific terms, divide by the molecular mass of CO, 28 kg/kmol, to give –10 063 kJ/kg.
The first law can now be applied, for adiabatic (Q = 0) and constant volume (W = 0) to give:
U_{2} – U_{1} = 0
Where U_{2} is the total internal energy of the products at the end of the process and U_{1} is the total internal energy of the reactants before combustion. Knowing the temperature of the reactants, we can work out the internal energy of the reactants before combustion.
We don’t know the final temperature of the products – this is what is asked for. Using equation (2.44):
And reading values of molar internal energy from the tables (Rogers and Mayhew, 1995), to complete the following table of reactants:
| Species | n_{i} kmol | \tilde{u}_{i,555K} kJ/kmol | n_{i}\tilde{u}_{i,555K} kJ/kmol |
| CO | 1 | 2984 | 2984 |
| O_{2} | 0.455 | 3233 | 1471 |
| N_{2} | 1.71 | 2944 | 5034 |
| Σ | 9489 |
where the internal energy at 555K is found from the tables by interpolation. Values are stated at 400K and 600K for a number of gaseous species. The interpolation is done thus:
\tilde{u}_{i,555K} = \frac{555 – 400}{600 – 400}(\tilde{u}_{i,600K} – \tilde{u}_{i,400K}) + \tilde{u}_{i,400K}The reader can easily confirm that the values in the table are correct.
There is no direct way to obtain the product temperature since it is determined by balancing the equation. The method used is to guess a temperature and see how far out of balance the equation is. In this case, guess 3000K.
Making the same table as above for the products
| Species | n_{i} kmol | \tilde{u}_{i,3000K} kJ/kmol | n_{i}\tilde{u}_{i,3000K} kJ/kmol |
| O_{2} | 0.91 | 127 920 | 116 407 |
| CO | 0.09 | 68 598 | 6174 |
| N_{2} | 1.71 | 67 795 | 115 929 |
| Σ | 238510 |
Inserting the values in the equation:
U_{2} – U_{1} = 238 510 – 9489 + (–281 761) = –52 740 kJThis is negative, and shows that the products were not hot enough. Therefore try the next temperature up in the table, i.e. 3200K:
| Species | n_{i} kmol | \tilde{u}_{i,3200K} kJ/kmol | n_{i}\tilde{u}_{i,3200K} kJ/kmol |
| O_{2} | 0.91 | 138 720 | 126 235 |
| CO | 0.09 | 74 391 | 6695 |
| N_{2} | 1.71 | 73 555 | 125 779 |
| Σ | 258 709 |
And the equation now is
U_{2} – U_{1} = 258 709 – 9489 + (-281 761) = – 32 540 kJThis is still too cool so the process is repeated until the answer is acceptably close to zero.