Question 13.16: Evaluating Exergetic Efficiency of an Internal Combustion En...

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, and burn completely. The combustion products exit at 1140°F. The power developed by the engine is 50 horsepower, and the fuel mass flow rate is 0.004 lb/s.

Find Devise and evaluate an exergetic efficiency for the engine using the fuel standard chemical exergy value from Table A-26  (Model II).

Schematic and Given Data: See Fig. E13.4.

Engineering Model 

1. See the assumptions listed in the solution to Example 13.4.

2. The environment corresponds to Model II of Table A-26.

3. Air entering the engine at 25°C, 1 atm with the composition 21% O_{2}, 79% N_{2} has negligible exergy.

1 Analysis An exergy balance can be used in formulating an exergetic efficiency for the engine: At steady state, the rate at which exergy enters the engine equals the rate at which exergy exits plus the rate at which exergy is destroyed within the engine. With assumption 3, exergy enters the engine only with the fuel. Exergy exits the engine accompanying heat and work and with the products of combustion.

If the power developed is taken to be the product of the engine, and the heat transfer and exiting product gas are regarded as losses, an exergetic efficiency expression that gauges the extent to which the exergy entering the engine with the fuel is converted to the product is

2 \varepsilon=\frac{\dot{W}_{ cv }}{\dot{ E }_{ F }}

where \dot{E} _{ F } denotes the rate at which exergy enters with the fuel.

Since the fuel enters the engine at 77°F and 1 atm, which correspond to the values of T_{0} \text { and } p_{0} of the environment, and kinetic and potential energy effects are negligible, the exergy of the fuel is just the chemical exergy. There is no thermomechanical contribution. Thus, with data from Table A-1 and Table A-26 (Model II)

The exergetic efficiency is then

3 \varepsilon=\left(\frac{50 hp }{81.5 Btu / s }\right)\left|\frac{2545 Btu / h }{1 hp }\right|\left|\frac{1 h }{3600 s }\right|

= 0.434 (43.4%)

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 55 Btu 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} (13.41)

2 The exergy of exhaust gas and engine coolant of internal combustion engines may be utilizable for various purposes—for instance, additional power might be produced using bottoming cycles as considered in Problem 9.10D. In most cases, such additional power would be included in the numerator of the exergetic efficiency expression. Since a greater portion of the entering fuel exergy is utilized in such arrangements, the value of ε would be greater than that evaluated in the solution.

3 Approximating the chemical exergy by the higher heating value of liquid octane from Table A-25E, 20,610 Btu/lb, we get \dot{ E }_{ F }=82.4 Btu / s \text { and } \varepsilon=0.429(42.9 \%).

Skills Developed

Ability to…

• devise and evaluate an exergetic efficiency for an internal combustion engine.

Quick Quiz

Using a rationale paralleling that for the internal combustion engine, devise and evaluate an exergetic efficiency for the gas turbine power plant of Example 13.5. A ns. 0.332 (33.2%).