1. Utilisation of Electrical Energy
Introduction

2. Last time … Assignment – Transformers contd.
Open & short circuit test recap.
Efficiency calculations.
Measured efficiency.
Maximum efficiency.
Assignment working time.
10/01/2019
UEE

3. Introduction - objectives
Induction motor introduction.
Poles, slip & rpm.
Wiring options.
Equivalent circuit of induction motor.
No-load power factor and branch current.
Equivalent impedance, Zeq.
Assignment 4 (induction motors) working time.
10/01/2019
UEE

4. Diagram of induction motor
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UEE

5. Induction motors Three-phase squirrel-cage induction motors.
Rugged, reliable, no electrical connection to rotor.
Windings on stator induce a rotating magnetic field in the rotor.
Causes the rotor to spin.
With no load, the rotor spins at the stator’s magnetic field rotation frequency.
10/01/2019
UEE

6. Stator, rotor and poles
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UEE

7. Key to abbreviations Supply frequency – f, Hz.
p - number of pole pairs (2 poles = 1 pole pair).
ns - synchronous speed, rev.s-1 or rpm.
nr - rotor speed, rev.s-1 or rpm.
s – slip.
Note: convert all speeds to rev.s-1. when calculating.
10/01/2019
UEE

8. Example 1
A 6-pole, 3-phase induction motor is connected to a 300 Hz supply. Determine the speed of rotation of the magnetic field produced by the stator.
Speed of rotation, = ns = supply frequency/ number of pole pairs.
= 300 / (6/2) = 100 rev.s-1 or 6000 rpm.
10/01/2019
UEE

9. Adding a load to the motor
In reality, the rotor spins at a lower speed than the stator frequency nr.
The difference between stator magnetic field speed and rotor speed is known as slip, s.
Slip, s = (ns – nr) / ns.
Often expressed as a percentage.
Slip %, s = (ns – nr) / ns x 100%
10/01/2019
UEE

10. Example 2
A 6-pole, 3-phase induction motor runs at 970 rpm at a certain load. If the stator is connected to a 50 Hz supply, find the percentage slip at this load.
ns = f / p = 50 / 3 = rev.s-1.
nr = 970 / 60 = rev.s-1.
s = (ns – nr) / ns = ( ) / = 0.03 = 3% slip.
10/01/2019
UEE

11. Motor wiring configurations
Can be star or delta wired.
Delta is the usual configuration for normal running.
Star is suitable for starting.
In delta, full supply voltage appears across the 3 windings, L1, L2 and L3.
In star, supply voltage / 3 appears across the windings.
10/01/2019
UEE

12. 10/01/2019
UEE

13. Example 3
A 3 phase induction motor is connected to a 750 V supply. What is the voltage across the windings L1 and L2 when:
A) It is star connected.
B) It is delta connected.
For star connections, voltage across windings = supply voltage / 3 = 750 / 1.73 = 433V.
For delta connection, voltage across windings = supply voltage = 750 V.
10/01/2019
UEE

14. Induction motor equivalent circuit
To all intents and purposes, the induction motor can be thought of as a rotating transformer.
The equivalent circuit for a transformer can be used to analyse an induction motor with a little alteration to represent the load.
10/01/2019
UEE

15. Key to symbols Vs = voltage across stator. Rc = core losses.
Xm = magnetising losses.
Rs = resistance of stator.
Rr = resistance of rotor.
Xs = impedance of stator.
Xr = impedance of rotor.
10/01/2019
UEE

16. Simplified induction motor circuit
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UEE

17. No-load current calculations
When there is no load, all current flows through the Rc and Xm components only.
Supply voltage is across Rc and Xm .
Ic = Vs / Rc
Im = Vs / Xm
These are the real and reactive parts of the total branch impedance.
Reactive part is inductive therefore lagging.
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UEE

18. Power factor calculations
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19. Branch current, Io
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20. Example 4
A star-connected three phase induction motor is supplied with 500 V. Rc is 75W and Xm is 25W.
Calculate the no-load power factor and Io.
In star connections, Vs = supply voltage / 3
= 500 / 3 = V.
Io = Ic + Im
10/01/2019
UEE

21. Example 4 continued Ic = Vs / Rc = 288.7 / 75 = 3.85 A.
Im = Vs / Xm = / 25 = A.
Power factor q = tan-1 Im / Ic = tan-1 (11.55 /3.85)
= tan-1 (3) = 71.6 degrees lagging.
Power factor = cos 71.6 degrees = lag.
Io = (Ic2 + Im2) = ( ) = 12.2 A.
10/01/2019
UEE

22. Full-load calculations
These calculations ignore the magnetising branch current initially.
This simplifies the model of the motor and leaves only series resistances and reactances which form the total impedance, Zeq.
It can be shown that the real part of the load Rr varies with the rotor speed but the reactance Xr remains constant.
A variable resistor represents the rotor speed.
10/01/2019
UEE

23. Full-load induction motor model
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UEE

24. Complex notation for Zeq
When adding the resistances and reactances, it is sometimes easier to use complex notation.
Zeq is the resultant of series impedances.
Ohmic part is the real part of the impedance.
Reactive part is the imaginary part of the impedance.
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UEE

25. Zeq calculations Zeq = equivalent impedance.
Zeq = (Rs + R`r/s) + j(Xs + Xr).
Where s = slip.
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UEE

26. Example 5
An induction motor has the following parameters: Rs = 0.6W, R`r = 0.4W, Xs = Xr = 0.6W, slip s = 0.04.
Zeq = ( /0.04) + j( )
= j1.2 W.
To be continued….
10/01/2019
UEE

27. Summary Induction motor introduction. Poles, slip & rpm.
Wiring options.
Equivalent circuit of induction motor.
No-load power factor and branch current.
Equivalent impedance, Zeq.
Assignment (induction motors) working time.
10/01/2019
UEE