Question

A four cylinder gasoline engine has an efficiency of 0.28 and delivers 200 J of work per cycle per cylinder. The engine fires at 25 cycles per second.

(a) Determine the work done per second.

_____J/s (answer here is not 5000, i've tried it)

(b) What is the total heat input per second from the gasoline?

_____J/s

(c) If the energy content of gasoline is 130 MJ per gallon, how long does one gallon last?

_____min

Answer 1

Part A.

We know that work done per second is also known as power, So

P = Work-done/time = number of cylinders*work per cycle per cylinder*fire cycle per second

P = 4*200*25

P = 20000 J/s

Part B.

we know that efficiency is given by:

n = P_out/P_in

P_in = heat input per second = P_out/n

P_in = 20000/0.28

P_in = 71428.57 J/s = 7.14*10^4 J/s

Part C.

Given that energy content of gasoline per gallon is 130 MJ, So

P_in = Q_in/t

t = Q_in/P_in

t = 130*10^6/71428.57

t = 1820 sec = 1820/60

t = 30.33 min

Let me know if you've any query.