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Question 15.5: Evaluation of the Enthalpy of Combustion Determine the entha...

Evaluation of the Enthalpy of Combustion

Determine the enthalpy of combustion of liquid octane (\mathrm{C}_{8} H_{18}) at 25°C and 1 atm, using enthalpy-of-formation data from Table A-26. Assume the water in the products is in the liquid form.

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The enthalpy of combustion of a fuel is to be determined using enthalpy of formation data.

Properties The enthalpy of formation at 25°C and 1 atm is -393,520 kJ/kmol for \mathrm{CO}_{2},-285,830 \mathrm{~kJ} / \mathrm{kmol} for \mathrm{H}_{2} \mathrm{O}(\ell), and -249,950 \mathrm{~kJ} / \mathrm{kmol} for \mathrm{C}_{8} \mathrm{H}_{18}(\ell) (Table A-26).

Analysis The combustion of \mathrm{C}_{8} \mathrm{H}_{18} is illustrated in Fig .15-20. The stoichiometric equation for this reaction is

\mathrm{C}_{8} \mathrm{H}_{18}+a_{\text {th }}\left(\mathrm{O}_{2}+3.76 \mathrm{~N}_{2}\right) \rightarrow 8 \mathrm{CO}_{2}+9 \mathrm{H}_{2} \mathrm{O}(\ell)+3.76 a_{\mathrm{th}} \mathrm{N}_{2}

 

Both the reactants and the products are at the standard reference state of 25°C and 1 atm. Also, \mathrm{N}_{2} and \mathrm{O}_{2} are stable elements, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of \mathrm{C}_{8} \mathrm{H}_{18} becomes (Eq. 15-6)

 

\bar{h}_{C} =H_{\text {prod }}-H_{\text {react }}

 

=\sum N_{p} \bar{h}_{f, p}^{\circ}-\sum N_{r} \bar{h}_{f, r}^{\circ}=\left(N \bar{h}_{f}^{\circ}\right)_{\mathrm{CO}_{2}}+\left(N h_{f}^{\circ}\right)_{\mathrm{H}_{2} \mathrm{O}}-\left(N \bar{h}_{f}^{\circ}\right)_{\mathrm{C}_{8} \mathrm{H}_{18}}

 

Substituting,

\bar{h}_{C}=(8 \mathrm{kmol})(-393,520 \mathrm{~kJ} / \mathrm{kmol})+(9 \mathrm{kmol})(-285,830 \mathrm{~kJ} / \mathrm{kmol})

 

-(1 \mathrm{kmol})(-249,950 \mathrm{~kJ} / \mathrm{kmol})

 

=-5,471,000 \mathrm{~kJ} / \mathrm{kmol} \mathrm{C}_{8} \mathrm{H}_{18}=-47,891 \mathrm{~kJ} / \mathrm{kg} \mathrm{C}_{8} \mathrm{H}_{18}

 

which is practially identical to the listed value of 47,890 kJ/kg in Table A-27. Since the water in the products is assumed to be in the liquid phase, this h_{C} value corresponds to the HHV of liquid \mathrm{C}_{8} \mathrm{H}_{18}.

 

 

Discussion it can be shown that the result for gaseous octane is -5,512,200 kJ/kmol or -48,255 kJ/kg.

15.20

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