A 12 pole, 3-phase alternator drives at a speed of 500 r.p.m supplies power to an 8 pole, three-phase induction motor. If the slip of the motor at full load speed is 4 %, calculate the full load of the motor.
"N_s=\\dfrac{120f}{p}\\\\\n500=\\dfrac{120f}{12}\\\\\nf=50Hz\\\\\nSpeed\\ of\\ induction\\ motor\\\\\nN_s=\\dfrac{120\\times {50}}{8}=750rpm\\\\\nN_r=N_s(1-s)\\\\\nN_r=750(1-0.04)=720rpm\\\\"
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