Induction motor:Corresponding Slip

Corresponding Slip

Example 34.54. The maximum torque of a 3-phase induction motor occurs at a slip of 12%. The motor has an equivalent secondary resistance of 0.08 W/phase. Calculate the equivalent load resistance RL, the equivalent load voltage VL and the current at this slip if the gross power output is 9,000 watts.

Solution. RL = R2[ (1/s) – 1] = 0.08 [ (1/0.12) – 1] = 0.587 W /phase.

As shown in the equivalent circuit of the rotor in Fig. 34.53, V is a fictitious voltage drop equivalent to that consumed in the load connected to the secondary i.e. rotor. The

Example 34.55. A 3-phase, star-connected 400 V, 50-Hz, 4-pole induction motor has the following per phase parameters in ohms, referred to the stators.

R1 = 0.15, X1 = 0.45, R2¢ = 0.12, X2¢ = 0.45, Xm = 28.5

Compute the stator current and power factor when the motor is operated at rated voltage and frequency with s = 0.04. (Elect. Machines, A.M.I.E. Sec. B, 1990)

Solution. The equivalent circuit with all values referred to stator is shown in Fig. 34.54.

Example 34.58. A 400 V, 3-f, star-connected induction motor has a stator exciting impedance of (0.06 + j 0.2) W and an equivalent rotor impedance of (0.06 + j 0.22) W. Neglecting exciting current, find the maximum gross power and the slip at which it occurs.

(Elect. Engg.-II, Bombay Univ. 1987)

Example 34.60. The equivalent circuit of a 400 V, 3-phase induction motor with a star- connected winding has the following impedances per phase referred to the stator at standstill:

Stator : (0.4 + j 1) ohm; Rotor : (0.6 + j 1) ohm; Magnetising branch : (10 + j 50 ) ohm.

Find (i) maximum torque developed (ii) slip at maximum torque and (iii) p.f. at a slip of 5%. Use approximate equivalent circuit. (Elect. Machinery-III, Bangalore Univ. 1987)

Solution. (ii) Gap power transferred and hence the mechanical torque developed by rotor would be maximum when there is maximum transfer of power to the resistor R2¢/s shown in the approximate equivalent circuit of the motor in Fig. 34.59. It will happen when R2¢/s equals the impedance looking back into the supply source. Hence,

Tutorial Problem No. 34.4.

1. A 3-phase, 115-volt induction motor has the following constants : R2 = 0.07 W ; R2¢ = 0.08 W, X1 = 0.4 W and X2¢ = 0.2 W. All the values are for one phase only. At which slip the gross power output will be maximum and the value of the gross power output ?

[11.4% ; 8.6 kW]

2. A 3-phase, 400-V, Y-connected induction motor has an equivalent T-circuit consisting of R1 = 1 W, X1 = 2 W, equivalent rotor values are R2¢ = 1.2 W, X2¢ = 1.5 W. The exciting branch has an impedance of (4 + j 40) W. If slip is 5% find (i) current (ii) efficiency (iii) power factor (iv) output. Assume friction loss to be 250 W.

[ (i) 10.8 A (ii) 81% (iii) 0.82 (iv) 5 kW]

3. A 50 HP, 440 Volt, 3-phase, 50 Hz Induction motor with star-connected stator winding gave the following test results:

(i)No load test: Applied line voltage 440 V, line current 24 A, wattmeter reading 5150 and 3350 watts. (ii)Blocked rotor test: applied line voltage 33.6 volt, line current 65 A, wattmeter reading 2150 and 766 watts.

Calculate the parameters of the equivalent circuit.

[Rajiv Gandhi Technical University, Bhopal, 2000] [ (i) Shunt branch : Ro = 107.6 ohms, Xm = 10.60 ohms (ii) Series branch : r = 0.23 ohm, x = 0.19