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Assume that the conversion of energy into mechanical work (at the wheel) in an internal combustion engine is 20%. Calculate gallons of gasoline required to deliver 30 horsepower at the wheel, for one hour. 1 HP = 746 Watts
1 HP for 1 hour is 0.746 kWh
1 Kwh = 3 412 BTU

Respuesta :

Answer:

amount require of fuel = 10.917 L or 2.884 gallon

Explanation:

given data

power output = 20%

power = 30 horsepower

1 HP = 746 Watts

1 HP for 1 hour = 0.746 kWh

1 Kwh = 3 412 BTU

solution

we know here power is 30  horsepower

so P = 22.38 kW

we get here heat supply that is express as

heat supply = [tex]\frac{P}{0.20}[/tex]

heat supply = [tex]\frac{22.38}{0.20}[/tex]  

heat supply = 111.9 kW

so here total energy will be as = 111.9 kW × 1 × 3600

total energy = 402.84 mg

we know that calorific value is roughly 36.9 MJ/L

so amount require of fuel is here as

amount require of fuel = [tex]\frac{402.84}{36.9}[/tex]  

amount require of fuel = 10.917 L or 2.884 gallon

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