Liquid ctane is burned steadily. The adiabatic flame temperature is to be determined for different cases.
Assumptions 1 This is a steady-flow combustion process. 2 The combustion chamber is adiabatic. 3 There are no work interactions. 4 Air and the combustion gases are ideal gases. 5 Changes in kinetic and potential energies are negligible.
Analysis (a) The balanced equation for the combustion process with the theoretical amount of air is
\mathrm{C}_{8} \mathrm{H}_{18}(\ell)+12.5\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2}\right) \rightarrow 8 \mathrm{CO}_{2}+9 \mathrm{H}_{2} \mathrm{O}+47 \mathrm{~N}_{2}
The adiabatic flame temperature relation H_{\text {prod }}=H_{\text {react }} in this case reduces to
\sum N_{p}\left(\bar{h}_{f}^{\circ}+\bar{h}-\bar{h}^{\circ}\right)_{p}=\sum N_{r} \bar{h}_{f, r}^{\circ}=\left(N \bar{h}_{f}^{\circ}\right)_{\mathrm{C}_{8} \mathrm{H}_{\mathrm{18}}}
since all the reactants are at the standard reference state and \bar{h}_{f}=0 for \mathrm{O}_{2} and \mathrm{N}_{2}. The \bar{h}_{f}^{\circ} and h values of various components at 298 K are
| Substance |
\bar{h}_{f}^{\circ} kJ/kmol |
\bar{h}_{298 \mathrm{K}} kJ/kmol |
| \mathrm{C}_{8} \mathrm{H}_{18}(\ell) |
-249,950 |
— |
| \mathrm{O}_{2} |
0 |
8682 |
| \mathrm{N}_{2} |
0 |
8669 |
| \mathrm{H}_{2} \mathrm{O}(g) |
-241,820 |
9904.0 |
| \mathrm{C} \mathrm{O}_{2}(\ell) |
-393,520 |
9364 |
Substituting, we have
\left(8 \mathrm{kmol} \mathrm{CO}_{2}\right)\left[\left(-393,520+\bar{h}_{\mathrm{CO}_{2}}-9364\right) \mathrm{kJ} / \mathrm{kmol} \mathrm{CO}_{2}\right]
+\left(9 \mathrm{kmol} \mathrm{H}_{2} \mathrm{O}\right)\left[\left(-241,820+\bar{h}_{\mathrm{H}_{2} \mathrm{O}}-9904\right) \mathrm{kJ} / \mathrm{kmol} \mathrm{H}_{2} \mathrm{O}\right]
+\left(47 \mathrm{kmol} \mathrm{N}_{2}\right)\left[\left(0+\bar{h}_{\mathrm{N}_{2}}-8669\right) \mathrm{kJ} / \mathrm{kmol} \mathrm{N}_{2}\right]
=\left(1 \mathrm{kmol} \mathrm{C}_{8} \mathrm{H}_{18}\right)\left(-249,950 \mathrm{~kJ} / \mathrm{kmol} \mathrm{C}_{8} \mathrm{H}_{18}\right)
which yields
8 \bar{h}_{\mathrm{CO}_{2}}+9 \bar{h}_{\mathrm{H}_{2} \mathrm{O}}+47 \bar{h}_{\mathrm{N}_{2}}=5,646,081 \mathrm{~kJ}
It appears that we have one equation with three unknowns. Actually we have only one unknown-the temperature of the products T_{\text {prod }}- since h=h(T) for ideal gases. Therefore, we have to use an equation solver such as EES or a trial-and-error approach to determine the temperature of the products.
A first guess is obtained by dividing the right-hand side of the equation by the total number of moles, which yields 5,646,081/(8 + 9 + 47) = 88,220 kJ/kmol. This enthalpy value corresponds to about 2650 K for \mathrm{N}_2, 2100 K for \mathrm{H}_2 \mathrm{O}, and 1800 K for \mathrm{CO}_2 . Noting that the majority of the moles are \mathrm{N}_2 , we see that T_\text{prod} should be close to 2650 K, but somewhat under it. Therefore, a good first guess is 2400 K. At this temperature,
8 \bar{h}_{\mathrm{CO}_{2}}+9 \bar{h}_{\mathrm{H}_{2} \mathrm{O}}+47 \bar{h}_{\mathrm{N}_{2}} =8 \times 125,152+9 \times 103,508+47 \times 79,320
=5,660,828 \mathrm{~kJ}
This value is higher than 5,646,081 kJ. Therefore, the actual temperature is slightly under 2400 K. Next we choose 2350 K. It yields
8 × 122,091 + 9 × 100,846 + 47 × 77,496 = 5,526,654
which is lower than 5,646,081 kJ. Therefore, the actual temperature of the products is between 2350 and 2400 K. By interpolation, it is found to be T_\text{prod} = 2395 K
(b) The balanced equation for the complete combustion process with 400 percent theoretical air is
\mathrm{C}_{8} \mathrm{H}_{18}(\ell)+50\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2}\right) \rightarrow 8 \mathrm{CO}_{2}+9 \mathrm{H}_{2} \mathrm{O}+37.5 \mathrm{O}_{2}+188 \mathrm{~N}_{2}
By following the procedure used in (a), the adiabatic flame temperature in this case is determined to be T_\text{prod} = 962 K.
Notice that the temperature of the products decreases significantly as a result of using excess air.
(c) The balanced equation for the incomplete combustion process with 90 percent theoretical air is
\mathrm{C}_{8} \mathrm{H}_{18}(\ell)+11.25\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2}\right) \rightarrow 5.5 \mathrm{CO}_{2}+2.5 \mathrm{CO}+9 \mathrm{H}_{2} \mathrm{O}+42.3 \mathrm{~N}_{2}
Following the procedure used in (a), we find the adiabatic flame temperature in this case to be T_\text{prod} = 2236 K.
Discussion Notice that the adiabatic flame temperature decreases as a result of incomplete combustion or using excess air. Also, the maximum adiabatic flame temperature is achieved when complete combustion occurs with the theoretical amount of air.
