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
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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