Q2 (a) For an induction motor, define; (i) Synchronous...

Q2 (a) For an induction motor, define; (i) Synchronous speed. (ii) Rotor speed. (iii) Slip speed. (b) A per phase IEEE recommended equivalent circuit of a 6-pole squirrel-cage induction motor that operates from a 415 V line voltage at 50 Hz, slip of 2% is given below in Figure Q2(b). Parameters of the motors are as follows: R1 = 0.15 ?, X1 = 0.45 ?, X2' = 0.2 ?, R2' = 0.2 ? and Xm = 15 ?. Calculate; (i) The stator current, (ii) The input power, (iii) The output power, (iv) Torque and (v) Efficiency. The fixed winding and friction losses is 800 W. Neglect the core loss. (c) Calculate the starting current and compare it to calculated stator current in Q2(b). Explain the difference and state a method to avoid this behaviour.

Transcript

00:01 High.

00:02 So we are given that number of poles p is equals to 6, voltage applied 415 of volts.

00:11 So in a part for synchronous motor, we have to find the synchronous speed.

00:18 So this can be given by ns is equals to 120 f by p.

00:22 Plugging the values we get 120 into 50 by 6 will comes out equals to 1000 of revolutions per minute.

00:34 Now in the same for the second part, we have to find the rotor speed which can be given by nr is equals to synchronous speed times of 1 minus of s whereas indicating the slip which we have been given 2 percent that means 0 .02.

00:50 So this will become 1000 into the bracket 1 minus 0 .02.

00:56 So on simplifying the rotor speed will comes out 980 revolutions per minute.

01:04 Now in the same first third part, we have to find the slip speed.

01:11 So basically slip speed can be given as ns minus of nr.

01:16 So this should be equals to 1000 minus of 980 will comes out equals to 20 revolutions per minute.

01:29 Now finally in the p part, let us draw the circuit first this.

01:34 So let us calculate first the equivalent impedance at the final.

01:39 So this can be equals to jota of xm.

01:43 See this is parallel to this r2 dash by s plus of jota of this x2 dash means there is combination of this two.

01:52 So let us plug in the values we get 6 .7968 plus jota of 4 .6690 after plugging the values.

02:04 In similar way as you can see zin this should be equals to r1 plus jota of x1 plus the z bar of f.

02:15 So this should be equals to 0 .15 plus jota of 0 .45 plus 6 .7968 plus jota of 4 .6690.

02:26 So input impedance zin will comes out 6 .9468 plus jota of 5 .119 ohms.

02:36 So in the b's first part, we have to calculate the stator current.

02:42 So stator current can be given as i1 is equals to v1 by the input impedance that is zin.

02:49 So this should be equals to v1 we have been given 415 by inductor as converting in line voltage divided by 6 .9468 plus jota of 5 .119.

03:06 Now convert this rectangular into polar part and solve it.

03:11 You will get the stator current i1 is equals to 27 .7664 angle negative of 36 .39 degree of amperes.

03:24 Now in the same b's second part, we have to find the input power.

03:30 So this can be given by the formula p in is equals to under root 3 times of 415 into 27 point means under root 3 times of vi.

03:41 So this should be equals to under root 3 into vl 415 into 27 .7664 times of cosine of 36 .39...

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