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Agasoline truck engine takes in 10,000 J of heat and delivers 2000 J of mechanical work per cycle. The heat is obtained by burning gasoline with heat of combustion

(a) What is the thermal efficiency of this engine?

(b) How much heat is discarded in each cycle?

(c) If the engine goes through 25 cycles per second, what is its power output in watts? In horsepower?

(d) How much gasoline is burned in each cycle?

(e) How much gasoline is burned per second? Per hour?

1 Answer

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Answer

given,

Q₁ = 10,000 J

W = 2000 J

heat of combustion, Lc = 5 x 10⁴ J/g

W = Q₁ - Q₂

Q₂ = Q₁ - W

Q₂ = 10,000 - 2000

Q₂ = 8000 J

Q₂ is the rejected to heat.

a) thermal efficiency of this engine


\eta = (W)/(Q_1)* 100


\eta = (2000)/(10000)* 100


\eta =20 \%

b) discarded heat is equal to 8000 J.

c) Power output

P = N W

N = 25 cycle/s

P = 25 x 2000 = 50000 W

1 W = 746 Horsepower

Power in horsepower


P = (50000)/(746)

P = 67 HP

d) gasoline burnt is equal to

Q₁ = m Lc


m = (Q_1)/(L_c)


m = (10000)/(5* 10^4)

m = 0.2 gram/cycle

e) gasoline burnt per hour


\dot{M} = N.m


\dot{M} = 25* 0.2


\dot{M} =5 gm/s

1 Hr = 3600 s


\dot{M} =5 * 3600


\dot{M} =18000\ g/hr

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