First-Law Analysis of Steady-Flow Combustion Liquid propane (C3H8) enters a combustion chamber at 25°C at a rate of 0.05 kg/min where it is mixed and burned with 50 percent excess air that enters the combustion chamber at 7°C, as shown in Fig. 15-23. An analysis of the combustion gases reveals

Question

First-Law Analysis of Steady-Flow Combustion
Liquid propane ([latex]\mathrm{C}_{3} \mathrm{H}_{8}[/latex]) enters a combustion chamber at 25°C at a rate of 0.05 kg/min where it is mixed and burned with 50 percent excess air that enters the combustion chamber at 7°C, as shown in Fig. 15-23. An analysis of the combustion gases reveals that all the hydrogen in the fuel burns to [latex]\mathrm{H}_{2} \mathrm{O}[/latex] but only 90 percent of the carbon burns to CO[latex]_2[/latex], with the remaining 10 percent forming CO. If the exit temperature of the combustion gases is 1500 K, determine (a) the mass flow rate of air and (b) the rate of heat transfer from the combustion chamber.

Accepted Answer

Liquid propane is burned steadily with excess air. The mass flow rate of air and the rate of heat transfer are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air and the combustion gases are ideal gases. 3 Kinetic and potential energies are negligible.

Analysis We note that all the hydrogen in the fuel burns to [latex]\mathrm{H}_{2} \mathrm{O}[/latex] but 10 percent of the carbon burns incompletely and forms CO. Also, the fuel is burned with excess air and thus there is some free [latex]\mathrm{O}_{2}[/latex] in the product gases.
The theoretical amount of air is determined from the stoichiometric reaction to be

[latex]\mathrm{C}_{3} \mathrm{H}_{8}(\ell)+a_{\mathrm{th}}\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight) ightarrow 3 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O}+3.76 a_{\mathrm{th}} \mathrm{N}_{2}[/latex]

[latex]a_{\mathrm{th}}=3+2=5[/latex]

Then the balanced equation for the actual combustion process with 50 percent excess air and some CO in the products becomes

[latex]\mathrm{C}_{3} \mathrm{H}_{8}(\ell)+7.5\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2} ight) ightarrow 2.7 \mathrm{CO}_{2}+0.3 \mathrm{CO}+4 \mathrm{H}_{2} \mathrm{O}+2.65 \mathrm{O}_{2}+28.2 \mathrm{~N}_{2}[/latex]

(a) The air-fuel ratio for this combustion process is

[latex]\mathrm{AF}=\frac{m_{ ext {air }}}{m_{ ext {fuel }}} =\frac{(7.5 imes 4.76 \mathrm{kmol})(29 \mathrm{~kg} / \mathrm{kmol})}{(3 \mathrm{kmol})(12 \mathrm{~kg} / \mathrm{kmol})+(4 \mathrm{kmol})(2 \mathrm{~kg} / \mathrm{kmol})} [/latex]

[latex]=25.53 \mathrm{~kg} ext { air } / \mathrm{kg} ext { fuel }[/latex]

Thus,

[latex]\dot{m}_{ ext {air }} =(A F)\left(\dot{m}_{ ext {fuel }} ight) [/latex]

[latex]=(23.53 \mathrm{~kg} ext { air } / \mathrm{kg} ext { fuel })(0.05 \mathrm{~kg} ext { fuel } / \mathrm{min}) [/latex]

[latex]=1.18 \mathrm{~kg} ext { air } / \mathrm{min}[/latex]

(b) The heat transfer for this steady-flow combustion process is determined from the steady-flow energy balance [latex]E_{ ext {out }}=E_{ ext {in }}[/latex] applied on the combustion chamber per unit mole of the fuel,

[latex]Q_{ ext {out }}+\sum N_{p}\left(\bar{h}_{f}^{\circ}+\bar{h}-\bar{h}^{\circ} ight)_{p}=\sum N_{r}\left(\bar{h}_{f}^{\circ}+\bar{h}-\bar{h}^{\circ} ight)_{r}[/latex]

or

[latex]Q_{ ext {out }}=\sum N_{r}\left(\bar{h}_{f}^{\circ}+\bar{h}-\bar{h}^{\circ} ight)_{r}-\sum N_{p}\left(\bar{h}_{f}^{\circ}+\bar{h}-\bar{h}^{\circ} ight)_{p}[/latex]

Assuming the air and the combustion products to be ideal gases, we have h = h(T), and we form the following minitable using data from the property tables:

Substance	[latex]\bar{h}_{f}^{\circ}[/latex] kJ/kmol	[latex]\bar{h}_{280k}[/latex] kJ/kmol	[latex]\bar{h}_{290k}[/latex] kJ/kmol	[latex]\bar{h}_{1500k}[/latex] kJ/kmol
[latex]\mathrm{C}_{3} \mathrm{H}_{8}(\ell)[/latex]	-118,910	—	—	—
[latex]\mathrm{O}_{2}[/latex]	0	8150	8682	49,292
[latex]\mathrm{N}_{2}[/latex]	0	8141	8669	47,073
[latex]\mathrm{H}_{2} \mathrm{O}(g)[/latex]	-241,820	—	9904	57,999
[latex]\mathrm{CO}_{2}[/latex]	-393,520	—	9364	71,078
CO	-110,530	—	8669	47,517

The [latex]\bar{h}_{f}^{\circ}[/latex] of liquid propane is obtained by subtracting the [latex]\bar{h}_{f g}[/latex] of propane at [latex]25^{\circ} \mathrm{C}[/latex] from the [latex]\bar{h}_{f}^{\circ}[/latex] of gas propane. Substituting gives

[latex]Q_{ ext {out }}=\left(1 \mathrm{kmol} \mathrm{C}_{3} \mathrm{H}_{8} ight)\left[\left(-118,910+h_{298}-h_{298} ight) \mathrm{kJ} / \mathrm{kmol} \quad\mathrm{C}_{3} \mathrm{H}_{8} ight] [/latex]

[latex]+\left(7.5 \mathrm{kmol} \mathrm{O}_{2} ight)\left[(0+8150-8682) \mathrm{kJ} / \mathrm{kmol} \quad\mathrm{O}_{2} ight] [/latex]

[latex]+\left(28.2 \mathrm{kmol} \mathrm{N}_{2} ight)\left[(0+8141-8669) \mathrm{kJ} / \mathrm{kmol} \quad\mathrm{N}_{2} ight] [/latex]

[latex]-\left(2.7 \mathrm{kmol} \mathrm{CO}_{2} ight)\left[(-393,520+71,078-9364) \mathrm{kJ} / \mathrm{kmol} \quad\mathrm{CO}_{2} ight] [/latex]

[latex]-(0.3 \mathrm{kmol} \mathrm{CO})[(-110,530+47,517-8669) \mathrm{kJ} / \mathrm{kmol} \quad\mathrm{CO}] [/latex]

[latex]-\left(4 \mathrm{kmol} \mathrm{H}_{2} \mathrm{O} ight)\left[(-241,820+57,999-9904) \mathrm{kJ} / \mathrm{kmol} \quad\mathrm{H}_{2} \mathrm{O} ight] [/latex]

[latex]-\left(2.65 \mathrm{kmol} \mathrm{O}_{2} ight)\left[(0+49,292-8682) \mathrm{kJ} / \mathrm{kmol} \quad\mathrm{O}_{2} ight] [/latex]

[latex]-\left(28.2 \mathrm{kmol} \mathrm{N}_{2} ight)\left[(0+47,073-8669) \mathrm{kJ} / \mathrm{kmol} \quad\mathrm{N}_{2} ight] [/latex]

[latex]= 363,880 \mathrm{~kJ} / \mathrm{kmol} \mathrm{of} \quad\mathrm{C}_{3} \mathrm{H}_{8}[/latex]

Thus 363,880 kJ of heat is transferred from the combustion chamber for each kmol (44 kg) of propane. This corresponds to 363,880/44=8270 kJ of heat loss per kilogram of propane. Then the rate of heat transfer for a mass flow rate of 0.05 kg/min for the propane becomes

[latex]\dot{Q}_{ ext {out }}=\dot{m} q_{ ext {out }}=(0.05 \mathrm{~kg} / \min )(8270 \mathrm{~kJ} / \mathrm{kg})=413.5 \mathrm{~kJ} / \mathrm{min}=6.89 \mathrm{~kW}[/latex]

Question 15.6: First-Law Analysis of Steady-Flow Combustion Liquid propane ...

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