A 60 Hz, three-phase. 6 – pole, Y-connected induction motor produces 100 hp at 1150 rpm. The
friction and windage losses of the motor are 1.2 kW, the stator copper and core losses are 2.1.kW
Compute the input power and the efficiency of the motor
Input powerPin=Tω63025Pin=2πfNω63025Pin=2π∗60∗100∗115063025Pin=687.885Input \space power\\ P_{in}= \frac{T \omega}{63025}\\ P_{in}= \frac{2 \pi fN \omega}{63025}\\ P_{in}= \frac{2 \pi * 60*100 *1150}{63025}\\ P_{in}= 687.885\\Input powerPin=63025TωPin=630252πfNωPin=630252π∗60∗100∗1150Pin=687.885
Efficiencyη=687.885−1.2−2.1687.885∗100η=99.5%Efficiency \\ η= \frac{687.885-1.2-2.1}{687.885}*100\\ η= 99.5 \%Efficiencyη=687.885687.885−1.2−2.1∗100η=99.5%
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