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Vlad1618 [11]
3 years ago
11

A 4.4 HP electric motor spins a shaft at 2329 rpm. Find: The torque load carried by the shaft is closest to: Select one: a)-27.0

6 ft*lb b. 19.24 ft*lb ? c)-31.17 ftlb d) 119.07 ftlb e)-9.92 ft*lb
Engineering
1 answer:
harina [27]3 years ago
3 0

Answer:

Load carried by shaft=9.92 ft-lb

Explanation:

Given:    Power P=4.4  HP

                    P=3281.08 W

<u><em>Power:  </em></u>Rate of change of work with respect to time is called power.

We know that P=Torque\times speed

     \omega=\frac{2\pi N}{60} rad/sec

So that P=\dfrac{2\pi NT}{60}

So   3281.08=\dfrac{2\pi \times 2329\times T}{60}

      T=13.45 N-m         (1 N-m=0.737 ft-lb)

 So T=9.92 ft-lb.

Load carried by shaft=9.92 ft-lb

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