1. Utilisation of Electrical Energy
Introduction

2. Last time … Full-load power factor and branch current continued.
Equivalent impedance, Zeq.
Argand diagrams.
Full-load power and efficiency.
Locked rotor tests.
Torque of induction motors.
Assignment 4 (induction motors) working time.
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UEE

3. Introduction - objectives
Torque of induction motors.
Magnetization branch calculations.
Locked rotor calculations.
Changing the circuit diagram from referred to rotor to referred to stator.
Power & torque calculations.
Assignment 4 (induction motors) working time.
30/12/2018
UEE

4. Gross output torque Formula for gross output torque
= Pm / (2pnr) = Ps / (2pns)
Where Ps = stator power.
ns = synchronous speed.
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UEE

5. Determining circuit values
No-load test allows calculation of the magnetising branch current.
Consider just one phase.
Calculation of Vs (here shown as Vo to differentiate it) is made from the line voltage.
Vo = line voltage / 3.
Po = No-load power / 3.
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UEE

6. Equivalent circuit
For no-load problems, the induction motor appears as in the diagram, all current flowing in the magnetisation branch.
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UEE

7. Phase angle at no load
Phase angle can be calculated with knowledge of the power and the measured current and voltage.
Cosqo = Po / (VoIo).
Magnetising branch current can be calculated now as all current is assumed to be travelling through the magnetising branch only.
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UEE

8. Magnetising branch currents
These can be calculated as the real and virtual parts of the total current measured.
Ic = Io cos qo
Im = Io sin qo
Now the values of Rc and Xm can be calculated.
Rc = Vo / Ic Xm = Vo / Im
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UEE

9. Worked example 1 – mag. branch
A no-load test on a 30.0 kW, 1000 rpm, V, 3-ph, 50 Hz, star-connected cage induction motor yielded the following results.
No-load line voltage 500 V.
No-load current A.
No-load power W.
Determine the values of Rc and Xm.
30/12/2018
UEE

10. Worked example 1 Vo = line voltage / 3 = 500 / 3 = 288.6 V.
Po = No-load power / 3 = 1980 / 3 = 660 W.
Cosqo = Po / (VoIo) = 660 / (288.6 x 18.5)
= so qo = 82.9°.
Ic = Io cos qo = 18.5 x = A.
Im = Io sin qo = 18.5 x = 18.4 A.
30/12/2018
UEE

11. Worked example 1 Ic = 2.287 A. Im = 18.4 A.
Rc = Vo / Ic = / = W.
Xm = Vo / Im = / 18.4 = W.
This has now provided the values for Xc and Xm of the magnetising branch.
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UEE

12. Example for you Please attempt question 1 on worksheet 08.
Always draw a circuit diagram when answering electrical circuit problems.
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UEE

13. Locked rotor tests
The locked rotor test is carried out on an induction motor.
Also known as short circuit test, locked rotor test or stalled torque test.
From the locked rotor test, short circuit current at normal voltage, power factor on short circuit, total leakage reactance, and starting torque of a motor can be calculated.
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UEE

14. Locked rotor equivalent circuit
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UEE

15. Locked rotor calculations
Vsc is the locked rotor phase voltage.
Psc is the locked rotor phase power.
Vsc = locked rotor line voltage / 3.
Psc = locked rotor power / 3.
If the current Isc is known, now the phase angle at locked rotor can be calculated.
30/12/2018
UEE

16. Calculating qsc – locked rotor phase angle
Cosqsc = Psc / (VscIsc).
With knowledge of the locked rotor phase angle, the real and imaginary components of the impedance can be calculated.
Since slip, s = 1, the component R`r / s = R`r.
Total real impedance (resistance) = Rs + R`r.
This resistance can be calculated from the values of Vsc, Isc and cosqsc.
30/12/2018
UEE

17. Total series resistance
Rs + R`r = Vsc cosqsc / Isc W.
The motor is connected in star configuration.
The resistance measured between two rotor terminals in star configuration measures the stator resistance of two stator resistances i.e 2Rs.
If this resistance is known, the value of R`r can be calculated as the total resistance is already known.
R`r = total series resistance – Rs.
30/12/2018
UEE

18. Total series impedance
With knowledge of the locked rotor voltage and current and the phase angle, the imaginary component of the current that flows through the motor can be calculated.
Xs + X`r = Vsc sinqsc / Isc W.
For induction motors, it is generally assumed that Xs = X`r so the individual values can be calculated.
Now these component values have been calculated and can be entered onto the equivalent circuit.
30/12/2018
UEE

19. Worked example 2 – locked rotor
A no-load test on A 30.0 kW, 1000 rpm, V, 3-ph, 50 Hz, star-connected cage induction motor yielded the following results.
Locked-rotor voltage 188 V.
Locked rotor current = Isc 54.5 A.
Locked-rotor power W.
The resistance measured between two stator terminals = 0.8 W.
Determine the values of Rs, R`r, Xs and X`r.
Assume that Xs = X`r .
30/12/2018
UEE

20. Worked example 2 Vsc = locked rotor line voltage / 3 = 188 / 3
Psc = locked rotor power / 3 = 6300 / 3
= 2100 W.
Cosqsc = Psc / (VscIsc) = 2100 / (108.5 x 54.5)
= so qsc = 71.3°.
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UEE

21. Worked example 2 Total resistance = Rs + R`r.
Rs + R`r = Vsc cosqsc / Isc
= x / 54.5 = W.
The resistance between two stator terminals = 0.8 W = 2Rs so Rs = 0.4 W.
R`r = = 0.24 W.
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UEE

22. Worked example 2 Xs + X`r = Vsc sinqsc / Isc
=108.5 x / 54.5 = 1.89 W.
For induction motors, it is generally assumed that Xs = X`r so the individual values can be calculated.
Xs = X`r = 1.89 / 2 = W.
Now these component values can be entered onto the equivalent circuit.
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UEE

23. Example for you Please attempt question 2 on worksheet 08.
Always draw a circuit diagram when answering electrical circuit problems.
30/12/2018
UEE

24. Converting the circuit diagram
What is the circuit, referred to the stator?
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UEE

25. Relationship between R`s and Rs
Circuit diagrams can show the component values referred to either the stator or rotor.
The previous circuit shows values referred to the rotor.
kr represents the turns ratio of the motor.
R`s = Rs / kr so Rs = R`s x kr
X`s = Xs / kr so Xs = X`s x kr
R`r = kr x Rr
X`r = kr x Xr
30/12/2018
UEE

26. Calculating values referred to stator
kr = 5.
Rs = R`s x kr = 0.03 x 5 = 0.15 W.
Xs = X`s x kr = 0.18 x 5 = 0.90 W.
R`r = kr x Rr = 0.15 x 5 = 0.75 W.
X`r = kr x Xr = 0.9 x 5 = 4.5 W.
Remember that the load is represented by the term Rr (or R`r) and needs to be divided by slip s when necessary to provide its value at a particular speed.
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UEE

27. Circuit referred to stator
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28. Simplifying the circuit, s = 0.05
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29. Adding similar terms
This leaves the simplified circuit which can be solved for Is if Vs is known.
PF angle = tan-1(5.4 / 15.15) = 19.62°.
Total resistance
= Vs cosqs / Is
Is= Vs cosqs / RT
= 1000 x / = 62.2A
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UEE

30. Power calculations Power delivered to rotor = VsIs cosq
= 1000 x 62.2 x = 58,568 W = Ps.
Mechanical power = Pm.
Pm = (1 - s) Ps = (1 – 0.05) x 58,568
= 55,640 W.
There are three phases so output power = 3 x Pm = 3 x 55,640 = 167 kW.
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UEE

31. Torque calculations Formula for gross output torque
= Pm / (2pnr) = Ps / (2pns) .
ns = synchronous speed
= supply frequency/ number of pole pairs
50 / 3 = revs-1.
Ps for one phase = 58,568 W.
Total stator power = 3 x 58,568 = 175,700 W
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UEE

32. Gross output torque Gross output torque = Ps / (2pns)
= 175,700 / (2 x p x 16.67)
= 175,700 / (104.7)
= 1678 Nm.
30/12/2018
UEE

33. Summary Torque of induction motors was examined.
Magnetization branch calculations were carried out.
Locked rotor calculations were made.
Changing the circuit diagram from referred to rotor to referred to stator was tackled.
Power & torque calculations were reviewed.
Assignment 4 (induction motors) working time was given.
30/12/2018
UEE