A 220 v four pole 10 hp 60 hz y connected three phase...

A 220 V four-pole 10-hp 60-Hz Y-connected three-phase induction motor develops its full-load mechanical torque at 3.6% slip. The per phase circuit model parameters for the motor are: R1 = 0.34 Ω Xm = 15.2 Ω X1 = 0.46 Ω X2 = 0.46 Ω Neglect rotational losses. Determine: a. the value of the rotor resistance R2 b. the maximum torque and the slip value at which it occurs. 1 hp = 746 W

Transcript

00:01 See over here we are given a circuit and for that circuit we need to find the value of rotor resistor.

00:09 So, let's see what is the circuit over here given to us.

00:21 So, here we are having a combination of resistors and inductors in the given series.

00:29 One inductor is kept in parallel to these two combinations.

00:35 So, over here we have provided a voltage.

00:45 Here on the left we have r1 and x1, on the right we have r2, x2 whereas this connected over here is xm.

00:55 So, we are provided the voltage of 220 volts and 10 horse power.

01:04 We will see that 1 horse power equals to 746 watt, then how much will be the 10 horse power? it will be equals to 7460 watt that is the power contained by our circuit.

01:21 Then we are having a 60 hertz of frequency given to this particular circuit.

01:28 Now, the value of r1, x1, r2, x2 and xm are defined over here.

01:40 R1 is 0 .34 ohms, x1 is 0 .46 ohm, r2 is the rotor resistance whose value we are not aware and we need to find out and x2 is again 0 .46 ohm and whereas xm is 15 .2 ohms.

02:03 So, we have got a set of equation like this to find our rotor resistor that is the impedance equals to 15 .2 like this particular resistor which is in inductor which is in parallel with r2 in series with 0 .46 inductor that is our x1 or x2.

02:30 Then this thing we will take into the another equation that is zp equals to j15 .2 into r2 upon 5 plus j0 .46 divided by r2 upon 5 plus j0 .46 plus j15 .6...

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