Evaluating Exergetic Efficiency of a Reactor Fueled by Liquid Octane
For the reactor of Examples 13.8 and 13.9, determine the exergy destruction, in kJ per kmol of fuel, and devise and evaluate an exergetic efficiency. Consider two cases: complete combustion with the theoretical amount of air, and complete combustion with 400% theoretical air. For the fuel, use the standard chemical exergy value from Table A-26 (Model II).
Known Liquid octane and air, each at 25°C and 1 atm, burn completely in a well-insulated reactor operating at steady state. The products of combustion exit at 1 atm pressure.
Find Determine the exergy destruction, in kJ per kmol of fuel, and evaluate an exergetic efficiency for complete combustion with the theoretical amount of air and 400% theoretical air.
Schematic and Given Data: See Fig. E13.9.
Engineering Model
1. See assumptions listed in Examples 13.8 and 13.9.
2. The environment corresponds to Model II of Table A-26.
3. Air entering the reactor at 25°C, 1 atm with the composition 21 \% O _{2}, 79 \% N _{2} has negligible exergy.
1 Analysis An exergy rate balance can be used in formulating an exergetic efficiency: At steady state, the rate at which exergy enters the reactor equals the rate at which exergy exits plus the rate at which exergy is destroyed within the reactor. With assumption 3 exergy enters the reactor only with the fuel. The reactor is well insulated, so there is no exergy transfer accompanying heat transfer. There is also no work \dot{W}_{ cv }. Accordingly, exergy exits only with the combustion products, which is the valuable output in this case. The exergy rate balance then reads
\dot{ E }_{ F }=\dot{ E }_{\text {products }}+\dot{ E }_{ d } (a)
where \dot{E}_{ F } is the rate at which exergy enters with the fuel, \dot{ E }_{\text {products }} is the rate at which exergy exits with the combustion products, and \dot{E}_{ d } is the rate of exergy destruction within the reactor.
An exergetic efficiency then takes the form
\varepsilon=\frac{\dot{E}_{\text {products }}}{\dot{E}_{ F }} (b)
The rate at which exergy exits with the products can be evaluated with the approach used in the solution to Example 13.15. But in the present case effort is saved with the following approach: Using the exergy balance for the reactor, Eq. (a), the exergetic efficiency expression, Eq. (b), can be written alternatively as
\varepsilon=\frac{\dot{E}_{F}-\dot{E}_{d}}{\dot{E}_{F}}=1-\frac{\dot{E}_{d}}{\dot{E}_{F}} (c)
The exergy destruction term appearing in Eq. (b) can be found from the relation
\frac{\dot{ E }_{ d }}{\dot{n}_{ F }}=T_{0} \frac{\dot{\sigma}_{ cv }}{\dot{n}_{ F }}
where T_{0} is the temperature of the environment and \dot{\sigma}_{ cv } is the rate of entropy production. The rate of entropy production is evaluated in the solution to Example 13.9 for each of the two cases. For the case of complete combustion with the theoretical amount of air
\frac{\dot{ E }_{ d }}{\dot{n}_{ F }}=(298 K )\left(5404 \frac{ kJ }{ kmol \cdot K }\right)=1,610,392 \frac{ kJ }{ kmol }
Similarly, for the case of complete combustion with 400% of the theoretical amount of air
\frac{\dot{ E }_{ d }}{\dot{n}_{ F }}=(298)(9754)=2,906,692 \frac{ kJ }{ kmol }
Since the fuel enters the reactor at 25°C, 1 atm, which correspond to the values of T_{0} \text { and } p_{0} of the environment, and kinetic and potential effects are negligible, the exergy of the fuel is just the
standard chemical exergy from Table A-26 (Model II): 5,413,100 kJ/kmol. There is no thermomechanical contribution. Thus, for the case of complete combustion with the theoretical amount of air, Eq. (c) gives
\varepsilon=1-\frac{1,610,392}{5,413,100}=0.703(70.3 \%)
Similarly, for the case of complete combustion with 400% of the theoretical amount of air, we get
2 \varepsilon=1-\frac{2,906,692}{5,413,100}=0.463(46.3 \%)
1 The entering air has chemical exergy that can be calculated
from Eq. 13.41b using the known oxygen and nitrogen mole fractions together with their chemical exergies from Table A 26. The result is 129 kJ per kmol of air. Compared to the chemical exergy of the fuel, such a value is negligible.
\overline{ e }^{ ch }=\sum_{i=1}^{ j } y_{i}\overline{ e} _{i}^{ ch }+\bar{R} T_{0} \sum_{i=1}^{ j } y_{i} \ln y_{i} (13.41b)
2 The calculated efficiency values show that a substantial portion of the fuel exergy is destroyed in the combustion process. In the case of combustion with the theoretical amount of air, about 30% of the fuel exergy is destroyed. In the excess air case, over 50% of the fuel exergy is destroyed. Further exergy destructions would take place as the hot gases are utilized. It might be evident, therefore, that the overall conversion from fuel input to end use would have a relatively low exergetic efficiency. The vapor power plant exergy analysis of Sec. 8.6 illustrates this point.
Skills Developed
Ability to…
• determine exergy destruction for a reactor.
• devise and evaluate an appropriate exergetic efficiency.
Quick Quiz
For complete combustion with 300% of theoretical air, would the exergetic efficiency be greater than, or less than, the exergetic efficiency determined for the case of 400% of theoretical air? A ns. Greater than.
