1. Utilisation of Electrical Energy
Introduction

2. Last time … 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.
13/01/2019
UEE

3. Introduction - objectives
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.
13/01/2019
UEE

4. Full-load calculations recap
These calculations ignore the magnetising branch current initially as Vs is the same.
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.
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UEE

5. Full-load induction motor modelVS
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UEE

6. 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

7. Zeq calculations Zeq = equivalent impedance.
Zeq = (Rs + R`r /s) + j(Xs + Xr).
Where s = slip.
Remember for inductive loads, current lags the voltage.
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UEE

8. Example 5 (from last week)
An induction motor has the following parameters: Rs = 0.6W, R`r = 0.4W, Xs = Xr = 0.6W, slip s = 0.04.
Zeq (full load) = ( /0.04) + j( )
= j1.2 W.
To be continued….
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9. Argand diagrams Plots the real (Re) and imaginary (Im) parts of Z.
Z = j1.2
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10. Calculations Zeq (full load) = 10.6 + j1.2 W.
This has both a real and an imaginary component plotted on the Argand diagram.
It can be expressed as the magnitude of Z, often seen as |Z| (r on the diagram) which is ( ) followed by the phase angle which is found by tan-1(1.2 / 10.6).
j1.2 W = ∟6.8°.
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UEE

11. Example 1
The magnitude of Z is known as the Modulus (Mod) and often seen as |Z|.
The angle q is known as the Argument (Arg).
Find the modulus and argument of the complex number j3.3.
|Z| = ( ) = 6.41W.
Arg(Z) = tan-1(3.3 / 5.5) = °.
Z =6.41∟ 31 ° W.
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UEE

12. Power factor
Now you have the argument, it is easy to calculate the power factor (PF) for an impedance.
PF = cos(Arg(z))
E.g. what is the power factor for Z =6.41∟ ° W?
PF = cos(30.96) = 0.86.
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UEE

13. Converting Mod Arg to complex impedance
Sometimes it is necessary to convert quantities expressed as Mod∟Arg to the complex form.
|Z| ∟q = |Z| cosq + j|Z| sinq W.
45.2 ∟75.1 ° = 45.2 cos(75.1) + j45.2 sin(75.1)
= j43.68 W.
Try worksheet questions 7 – 9 to practise this. Remember that sin and cos values can be negative.
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UEE

14. Worked example 01 (star wiring)
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15. First steps – branch current Io & PF
Ic = Vs / Rc = 240 / 120 = 2.0 A.
Im = Vs / Xm = 240 / 25 = 9.6 A.
From the diagram, Io = Ic – jIm = 2.0 – j9.6 A.
jIm is –ve because current lags voltage for inductors.
Io = 9.81∟-78.23° A.
Now we also have the PF = cos (-78.23) =
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16. Now F/L stator & rotor impedance
Zeq = (Rs + R`r /s) + j(Xs + X`r).
At full load (F/L) slip s = 0.04.
Zeq at full load = (Rs + R`r /0.04) + j(Xs + X`r).
Zeq = ( / 0.04) + j( )
= ( ) + j(1.3)
= j1.3 W
= 4.16 ∟18.2° W.
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UEE

17. Calculating I`r We have Vs and Zeq so I`r can be calculated.
I`r = Vs / Zeq = 240 / (4.16 ∟18.2° )
= 57.7 ∟-18.2° (angle becomes negative because 1 / j = -j).
57.7 ∟-18.2° needs to be expressed as real and imaginary parts again.
I`r = |Z| cosq + j|Z| sinq
= 57.7 cos (-18.2) + j57.7 sin (-18.2)
= j18.02 A.
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UEE

18. Calculating total current Is
Is is the sum of Io and I`r – already calculated.
Is = 2.0 – j9.6 A j18.02 A.
We can simply add the real and imaginary parts separately.
Is = – j27.62 A.
Convert back to the Mod ∟ Arg form.
Is = ∟-25.9 °.
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UEE

19. Power factor at rated Full-load
Is = ∟-25.9 °.
PF = cosq = cos -25.9° = 0.90.
Remember for inductive loads, current lags the voltage.
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UEE

20. Calculating power of induction motors
The load is represented by R`r / s.
This varies with the speed of the motor represented by s. Full load slip s = 0.04 (given earlier).
Power = I2R. Phase power = PS (stator to rotor).
Rotor power Ps = (I`r)2(R`r / s).
I`r = 57.7 ∟-18.2° but we only need to use the real part i.e A.
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UEE

21. Induction motor power contd.
Ps = x (0.15 / 0.04) = 3329 x 3.75 = 12,485 W.
There are 3 phases and windings so total rotor power = 3Ps = 3 x 12,485 = 37,455 W = kW.
Mechanical power = Pm.
Pm = (1 - s) Ps = (1 – 0.04) x 12,485 = 11,986 W.
Gross mechanical power = 3Pm = 3 x 11,986 = 35,958 W = 36.0 kW.
Shaft power = 3Pm – mechanical losses.
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UEE

22. Shaft power & efficiency
Shaft power = 3Pm – mechanical losses.
Assume mechanical losses of 2 kW.
Shaft power = = 34.0 kW.
Efficiency h = power out / power in x 100%.
Power in = 3VsIs cosq. (PF calculated previously).
Pin = 3 x 240 x x 0.90 = 40,934 W = 40.9 kW.
Efficiency h = 34.0 / 40.9 x 100 = 83.1%.
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UEE

23. Locked rotor test
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

24. Torque of induction motor
When calculating torque, the slip s must be taken into consideration.
At start, slip s = maximum = 1.
This changes the value of the load R`s to its minimum possible value.
At rated speed, slip is the lowest value so the value of the load R`s is the maximum value.
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UEE

25. 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.
s – slip.
Note: convert all speeds to rev.s-1. when calculating.
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26. Starting torque for ns = 15
Tstart = 3Vs2 /(2pns) x R`r / s[(Rs + R`r /s) + (Xs + X`r)] but s =1 in this case (start-up) so can be cancelled.
Tstart = 3Vs2 /(2pns) x R`r / [(Rs + R`r ) + (Xs + X`r)].
Using the given values for the stator and rotor….
Tstart = 3 x 2402 /(2p x 15) x 0.15 / [( ) + ( )]
= 1,833 x = Nm.
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UEE

27. Slip at max torque = sTmax
sTmax = R`r / (Rs2 + (Xs + Xr)2)
= 0.15 / ( )
= 0.15 / = 0.11
When maximum torque is delivered, the slip is 0.11 or 11%.
Rotor rpm at max torque = ns(1 - s)
= ns(1 – 0.11) = ns x 0.89 = 15 x 0.89 = rev.s-1 = 801 rpm.
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UEE

28. Next time Torque continued. Determining equivalent circuits.
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UEE

29. Summary 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.
13/01/2019
UEE