Adiabatic Flame Temperature in Steady Combustion Liquid octane (C8H18) enters the combustion chamber of a gas turbine steadily at 1 atm and 25°C, and it is burned with air that enters the com-bustion chamber at the same state, as shown in Fig. 15-27. Determine the adiabatic flame temperature for

Question

Adiabatic Flame Temperature in Steady Combustion

Liquid octane ([latex]\mathrm{C}_{8} \mathrm{H}_{18}[/latex]) enters the combustion chamber of a gas turbine steadily at 1 atm and 25°C, and it is burned with air that enters the com-bustion chamber at the same state, as shown in Fig. 15-27. Determine the adiabatic flame temperature for (a) complete combustion with 100 percent theoretical air, (b) complete combustion with 400 percent theoretical air, and (c) incomplete combustion (some CO in the products) with 90 percent theoretical air.

Accepted Answer

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

[latex]\mathrm{C}_{8} \mathrm{H}_{18}(\ell)+12.5\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight) ightarrow 8 \mathrm{CO}_{2}+9 \mathrm{H}_{2} \mathrm{O}+47 \mathrm{~N}_{2}[/latex]

The adiabatic flame temperature relation [latex]H_{ ext {prod }}=H_{ ext {react }}[/latex] in this case reduces to

[latex]\sum N_{p}\left(\bar{h}_{f}^{\circ}+\bar{h}-\bar{h}^{\circ} ight)_{p}=\sum N_{r} \bar{h}_{f, r}^{\circ}=\left(N \bar{h}_{f}^{\circ} ight)_{\mathrm{C}_{8} \mathrm{H}_{\mathrm{18}}}[/latex]

since all the reactants are at the standard reference state and [latex]\bar{h}_{f}=0[/latex] for [latex]\mathrm{O}_{2}[/latex] and [latex]\mathrm{N}_{2}[/latex]. The [latex]\bar{h}_{f}^{\circ}[/latex] and h values of various components at 298 K are

Substance	[latex]\bar{h}_{f}^{\circ}[/latex] kJ/kmol	[latex]\bar{h}_{298 \mathrm{K}}[/latex] kJ/kmol
[latex]\mathrm{C}_{8} \mathrm{H}_{18}(\ell)[/latex]	-249,950	—
[latex]\mathrm{O}_{2}[/latex]	0	8682
[latex]\mathrm{N}_{2}[/latex]	0	8669
[latex]\mathrm{H}_{2} \mathrm{O}(g)[/latex]	-241,820	9904.0
[latex]\mathrm{C} \mathrm{O}_{2}(\ell)[/latex]	-393,520	9364

Substituting, we have

[latex]\left(8 \mathrm{kmol} \mathrm{CO}_{2} ight)\left[\left(-393,520+\bar{h}_{\mathrm{CO}_{2}}-9364 ight) \mathrm{kJ} / \mathrm{kmol} \mathrm{CO}_{2} ight] [/latex]

[latex]+\left(9 \mathrm{kmol} \mathrm{H}_{2} \mathrm{O} ight)\left[\left(-241,820+\bar{h}_{\mathrm{H}_{2} \mathrm{O}}-9904 ight) \mathrm{kJ} / \mathrm{kmol} \mathrm{H}_{2} \mathrm{O} ight] [/latex]

[latex]+\left(47 \mathrm{kmol} \mathrm{N}_{2} ight)\left[\left(0+\bar{h}_{\mathrm{N}_{2}}-8669 ight) \mathrm{kJ} / \mathrm{kmol} \mathrm{N}_{2} ight] [/latex]

[latex]=\left(1 \mathrm{kmol} \mathrm{C}_{8} \mathrm{H}_{18} ight)\left(-249,950 \mathrm{~kJ} / \mathrm{kmol} \mathrm{C}_{8} \mathrm{H}_{18} ight)[/latex]

which yields

[latex]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}[/latex]

It appears that we have one equation with three unknowns. Actually we have only one unknown-the temperature of the products [latex]T_{ ext {prod }}-[/latex] since [latex]h=h(T)[/latex] 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 [latex]\mathrm{N}_2[/latex], 2100 K for [latex]\mathrm{H}_2 \mathrm{O}[/latex], and 1800 K for [latex]\mathrm{CO}_2 [/latex]. Noting that the majority of the moles are [latex]\mathrm{N}_2 [/latex], we see that [latex]T_ ext{prod}[/latex] should be close to 2650 K, but somewhat under it. Therefore, a good first guess is 2400 K. At this temperature,

[latex]8 \bar{h}_{\mathrm{CO}_{2}}+9 \bar{h}_{\mathrm{H}_{2} \mathrm{O}}+47 \bar{h}_{\mathrm{N}_{2}} =8 imes 125,152+9 imes 103,508+47 imes 79,320 [/latex]

[latex]=5,660,828 \mathrm{~kJ}[/latex]

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 [latex]T_ ext{prod}[/latex] = 2395 K

(b) The balanced equation for the complete combustion process with 400 percent theoretical air is

[latex]\mathrm{C}_{8} \mathrm{H}_{18}(\ell)+50\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight) ightarrow 8 \mathrm{CO}_{2}+9 \mathrm{H}_{2} \mathrm{O}+37.5 \mathrm{O}_{2}+188 \mathrm{~N}_{2}[/latex]

By following the procedure used in (a), the adiabatic flame temperature in this case is determined to be [latex]T_ ext{prod}[/latex] = 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

[latex]\mathrm{C}_{8} \mathrm{H}_{18}(\ell)+11.25\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight) ightarrow 5.5 \mathrm{CO}_{2}+2.5 \mathrm{CO}+9 \mathrm{H}_{2} \mathrm{O}+42.3 \mathrm{~N}_{2}[/latex]

Following the procedure used in (a), we find the adiabatic flame temperature in this case to be [latex]T_ ext{prod}[/latex] = 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.

Question 15.8: Adiabatic Flame Temperature in Steady Combustion Liquid octa...

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