ovounruly9346 ovounruly9346
  • 13-07-2019
  • Engineering
contestada

Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 seconds.

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wagonbelleville wagonbelleville
  • 22-07-2019

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

[tex]torque =\frac{power}{2\pi N}[/tex]

[tex]power=2\pi N\times T[/tex]

P=[tex]2\times \pi \times2500 \times 4.5[/tex]

P=70,685W

P=70.685 KW

power=[tex]\frac{work done}{time}[/tex]

work done = power * time

                  = 70.685*30=2120.55J

                  = 2.12 kJ

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