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The power output of a car engine running at 2500 rpm is 500 kW . You may want to review (Pages 574 - 577) . Part A How much work is done per cycle if the engine's thermal efficiency is 20.0 %

Respuesta :

Answer:

Explanation:

2500 rpm = 2500 / 60 rps ( revolution per second )

= 41.67 rps .

power output = 500 kW

thermal efficiency = 20 %

power input = (100 /20) x 500 kW

= 2500 kW .

work done per second = 2500 kJ

work done per cycle = 2500 / 41.67

= 60 kJ .

okpalawalter8

We have that the  workdone per cycle is mathematically given as

Wc=12.00KJ

 Workdone per cycle of engine

Question Parameters:

The power output of a car engine running at 2500 rpm is 500 kW

Generally the work done per cycle  is mathematically given as

Wc=500*1/41.666

Where

=2500rev/60sec

=41.666

Meaning one cycle takes 1/41.666

Hence

Wc=500*1/41.666

Wc=12.00KJ

For more information on work

https://brainly.com/question/756198

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