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Fuel consumption of multicylinder gasoline engine

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The rate of revolution of a multicylinder gasoline engine in an airplane, operating at 2500 rev/min, takes in energy 7.89*10^3 J and exhausts 4.58*10^3 J for each revolution of the crankshaft. How many liters of fuel does it consume in 1.00 h of operation if the heat of combustion is 4.03 * 10^7 J/L, and what is the torque exerted by the crankshaft on the load?

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Solution Summary

A detailed, step-by-step solution is provided. The hourly fuel consumption of the engine and the torque exerted by the crankshaft are computed. Units and unit conversions that are used in the solution are reviewed.

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The amount of energy consumed by the engine in 1.00 h of operation is the product of the energy intake per revolution of the crankshaft (7.89*10^3 J) and the number of revolutions of the crankshaft in 1.00 h.

There are 2500 rev/min, and there are 60 min/h, so the number of revolutions in 1.00 h is 2500*60 = 1.50*10^5.

Thus the total energy intake in 1.00 h is

(7.89*10^3 J)*(1.50*10^5) = 1.18*10^9 J

The total energy intake in 1.00 h is the product of the heat of combustion and the volume V of fuel which is consumed in 1.00 h. Thus

1.18*10^9 J = (4.03*10^7 J/L)*V

Solving this equation ...

Solution provided by:
  • AB, Hood College
  • PhD, The Catholic University of America
  • PhD, The University of Maryland at College Park
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