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A three-phase decapolar motor consumes a power of 250 KW when working at full load at...

A three-phase decapolar motor consumes a power of 250 KW when working at full load
at 550 rpm It has an efficiency of 82.5% and is connected to a network of 380 V at 50 (Hz). The
company where the kilowatt hour is invoiced at $ 183.95. When replaced by a
engine of similar characteristics but with 94% efficiency, the power [kW saved
it will be

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Answer #1

The question seems fairly straight forward.

For the first motor with 82.5 % efficiency, input power, Pin1 = 250 kW

Therefore, output power, Pout = efficiency * Pin1 = 0.825 * 250 kW = 206.25 kW

Pout is the only useful power.

In the second case the engine has 94 % efficiency. But the output power to the load is constant.

Therefore, Pout = 206.25 kW

Therefore, power consumed by the second motor = Pout / Efficiency of the second motor = 206.25 / 0.94

=  219.415 kW

Therefore, power saved by using second machine = Power consumed by 1st machine - Power consumed by 2nd machine = 250kW - 219.415 kW = 30.585 kW ==> ans

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