Known Liquid octane and the theoretical amount of air enter an internal combustion engine operating at steady state in separate streams at 77°F, 1 atm. Combustion is complete and the products exit at 1140°F. The power developed by the engine and fuel mass flow rate are specified.
Find Determine the rate of heat transfer from the engine, in Btu/s.
Schematic and Given Data:
Engineering Model
1. The control volume identified by a dashed line on the accompanying figure operates at steady state.
2. Kinetic and potential energy effects can be ignored.
3. The combustion air and the products of combustion each form ideal gas mixtures.
4. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen. The nitrogen is inert and combustion is complete.
Analysis The balanced chemical equation for complete combustion with the theoretical amount of air is obtained from the solution to Example 13.1 as
C _{8} H _{18}+12.5 O _{2}+47 N _{2} \rightarrow 8 CO _{2}+9 H _{2} O ( g )+47 N _{2}
The energy rate balance reduces, with assumptions 1–3, to give
\frac{\dot{Q}_{ cv }}{\dot{n}_{ F }}=\frac{\dot{W}_{ cv }}{\dot{n}_{ F }}+\bar{h}_{ P }-\bar{h}_{ R }
1 =\frac{\dot{W}_{ cv }}{\dot{n}_{ F }}+\left\{8\left[\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right]_{ CO _{2}}+9\left[\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right]_{ H _{2} O ( g )}+47\left[\bar{h}_{ f }^{\circ\nearrow0}+\Delta \bar{h}\right]_{ N _{2}}\right\}
-\left\{\left[\bar{h}_{ f }^{\circ}+\Delta \bar{h}^{\nearrow0}\right]_{ C _{8} H _{18}( l )}+12.5\left[\bar{h}_{ f }^{\circ \nearrow0}+\Delta \bar{h}^{\nearrow0}\right]_{ O _{2}}+47\left[\bar{h}_{ f }^{\nearrow0}+\Delta \bar{h}^{\nearrow0}\right]_{ N _{2}}\right\}
where each coefficient is the same as the corresponding term of the balanced chemical equation and Eq. 13.9 has been used to evaluate enthalpy terms. The enthalpy of formation terms for oxygen and nitrogen are zero, and \Delta \bar{h}=0 for each of the reactants because the fuel and combustion air enter at 77°F.
\bar{h}(T, p)=\bar{h}_{ f }^{\circ}+\left[\bar{h}(T, p)-\bar{h}\left(T_{ ref }, p_{ ref }\right)\right]=\bar{h}_{ f }^{\circ}+\Delta \bar{h} (13.9)
With the enthalpy of formation for C _{8} H _{18}(l) from Table A-25E
\bar{h}_{ R }=\left(\bar{h}_{ f }^{\circ}\right)_{ C _{8} H _{18}(l)}=-107,530 Btu / lbmol (\text { fuel })
With enthalpy of formation values for CO _{2} \text { and } H _{2} O ( g ) from Table A-25E, and enthalpy values for N _{2}, H _{2} O , \text { and } CO _{2} from Table A-23E
\begin{aligned}\bar{h}_{ P }=& 8[-169,300+(15,829-4027.5)]+9[-104,040\\&+(13,494.4-4258)]+47[11,409.7-3729.5] \\=&-1,752,251 Btu / lbmol (\text { fuel })\end{aligned}
Using the molecular weight of the fuel from Table A-1E, the molar flow rate of the fuel is
\dot{n}_{ F }=\frac{0.004 lb ( fuel ) / s }{114.22 lb (\text { fuel }) / lbmol ( fuel )}=3.5 \times 10^{-5} lbmol ( fuel ) / s
Inserting values into the expression for the rate of heat transfer
\dot{Q}_{ cv }=\dot{W}_{ cv }+\dot{n}_{ F }\left(\bar{h}_{ P }-\bar{h}_{ R }\right)
=(50 hp )\left|\frac{2545 Btu / h }{1 hp }\right| \left|\frac{1 h }{3600 s }\right|
+\left[3.5 \times 10^{-5} \frac{ lbmol ( fuel )}{ s }\right][-1,752,251
-(-107,530)] \frac{\text { Btu }}{\text { lbmol(fuel) }}
= -22.22 Btu/s
1 These expressions correspond to Eqs. 13.12b and 13.15b, respectively.
\frac{\dot{Q}_{ cv }}{\dot{n}_{ F }}-\frac{\dot{W}_{ cv }}{\dot{n}_{ F }}=\bar{h}_{ P }-\bar{h}_{ R } (13.12b)
\frac{\dot{Q}_{ cv }}{\dot{n}_{ F }}-\frac{\dot{W}_{ cv }}{\dot{n}_{ F }}=\sum_{ P } n_{e}\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)_{e}-\sum_{ R } n_{i}\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)_{ i } (13.15b)
Skills Developed
Ability to…
• balance a chemical reaction equation for complete combustion of octane with theoretical air.
• apply the control volume energy balance to a reacting system.
• evaluate enthalpy values appropriately.
Quick Quiz
If the density of octane is 5.88 lb/gal, how many gallons of fuel would be used in 2 h of continuous operation of the engine? Ans. 4.9 gal.
