A 3-phase 400 V 50 Hz 6 pole cage induction motor of rating 20 kW, with slip at

Question

A 3-phase 400 V 50 Hz 6 pole cage induction motor of rating 20 kW, with slip at rated load of 4% has a starting toque equal to 120% of the rated torque. It is coupled to a pump through a reduction gear box of 2: 1 ratio of efficiency 85%. The breakaway torque is 24 N.m at the motor shaft. Assume that the torque available for acceleration remains constant for the entire duration and the motor accelerates to rated speed in 6 seconds. What are the values of rated torque and starting torque of the motor? What is the torque available for acceleration? What is the rated speed of the motor? What is the total moment of inertia of the rotating parts of the system (seen at the motor shaft)?

Explanation / Answer

for a 3phase induction motor

torque equation=   3V2sr/(ws(r2+s2x2))=ksr/(r2+s2x2)wherek=4583.66  

Now Breakdown torque=   k/2x=24%u22172/.85=56.47==>x=40.58ohm  

Starting torque=   kr/(r2+x2)=1.2ksr/(r2+s2x2),wheres=0.04  

solving we get   r=xs%u221A=8.116ohm  

therefore

STARTING TORQUE=21.72 Nm

RATED TORQUE=18.1Nm which is the torque availabel for acceleration.........................ANS

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