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A 4-cylinder gasoline engine has an efficiency of 28% and delivers 180J of work

ID: 1296996 • Letter: A

Question

A 4-cylinder gasoline engine has an efficiency of 28% and delivers 180J of work per cycle per cylinder.
Part A: if the engine runs at 25 cycles per second (1500rpm), determine the work done per second.
Part B: what is the total heat input per second from the gasoline?
Part C: if the energy content of gasoline is 130 MJ per gallon, how long does one gallon last? A 4-cylinder gasoline engine has an efficiency of 28% and delivers 180J of work per cycle per cylinder.
Part A: if the engine runs at 25 cycles per second (1500rpm), determine the work done per second.
Part B: what is the total heat input per second from the gasoline?
Part C: if the energy content of gasoline is 130 MJ per gallon, how long does one gallon last?
Part A: if the engine runs at 25 cycles per second (1500rpm), determine the work done per second.
Part B: what is the total heat input per second from the gasoline?
Part C: if the energy content of gasoline is 130 MJ per gallon, how long does one gallon last?

Explanation / Answer

1) Engine fires at 25 cycles/s:

180 J / cycle/ cylinder * 4 cylinders * 25 cycles/second = 18000 J/s
Work = 18000 Joules every second

2) Total heat equals:

18000 J = 28% efficiency so "X" many Joules equals 100% effiicency?
28% / 100% = 18000 / X
solve for X :
X =64285.714 Joules of heat per second

3) 130MJ = 130,000,000 Joules

the engine uses 64285.714 Joules of heat every second so:

130,000,000 J / 64285.714 J = 2022.22 seconds of running time or about 33.7 minutes