Question

A four-cylinder gasoline engine has an efficiency of 0.22 and delivers 200 J of work per cycle per cylinder. The engine fires at 27 cycles per second.

A) Determine the work done per second.

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

C) If the energy content of gasoline is 150 MJ per gallon, how long does one gallon last?

Answer 1

(a) {Work Done Per Second} =

= {Work_Done/cycle/cyl}*{4 cyls}*{27 cycles/sec}

= {200 Joules/cycle/cyl}*{4 cyls}*{27 cycles/sec}

= 21600 Joules/sec

(b) {Fuel Heat Input Per Second} =

= {Work Done Per Second}/{Efficiency}

= {21600 Joules/sec}/{0.220}

= 98181.8 Joules/sec

(c) {Fuel Consumption(liters/sec)} =

= {Fuel HeatInput Per Second}/{Gas Energy Content PerLiter}

= {98181.8 Joules/sec}/{150*10^6 Joules/liter}

= 0.0006545 liters/sec

{Time To Consume 1 Liter} =

= {1 Liter}/{0.002511 liters/sec}

= 1527.8 sec