1. Electromechanical Systems
Unit 4
Power Supplies 1 1

2. Aim and Learning Outcomes
This unit explains
Generation of single-phase and three-phase AC supplies.
Production of rectified (uni-directional) supplies.
Power dissipation in star- and delta- connected three phase.
After completing this unit you should be able to
Calculate the current flow, voltage distribution and power dissipation in star-connected and delta-connected systems.
Calculate the average DC output voltage of controlled and uncontrolled half- and full- wave rectifiers.
Describe the principles of smoothing. 2

3. Content Single Phase Generators Three Phase Generator
Three Phase Connections
Star Connection
Delta Connection
Power Dissipation
Rectified Supplies 3

4. Mechanical World (Generators) Step Down Transformers
Power Supply
Mechanical World (Generators)
Transmission Line
Step Down Transformers 4

5. The Father of Generators
Michael Faraday is one of the giants of the history of science. A self-made, self-educated man, his public life was underpinned by his devout membership of a small Christian sect, whose rigid attitudes shadowed him at every turn, culminating in a crisis that tested his resolve as a scientist, his faith as a Christian and even the balance of his mind. Yet he became the greatest scientist of his day, and the central figure of an extraordinary scientific renaissance in London.
At the age of 21 Faraday secured a position as laboratory assistant to Sir Humphrey Davy at the Royal Institution. He rapidly overtook Davy as Britain’s most celebrated scientist, and his work at the Institution as a gifted experimenter and inspiring lecturer gave unprecedented impetus to public understanding of science over the course of nearly half a century. 5

6. Single Phase Generators6

7. Faraday’s Law:
“when a conductor is moved relative to a magnetic field, an emf will be induced in the conductor”
“when a conductor is moved relative to a magnetic field, an emf will be induced in the conductor”
When a coil is rotated in a magnetic field, sinusoidal emfs will be induced in each conductor arm. 7

8. Induced EMF Assume uniform magnetic field between the poles
Zero flux linkage with the coil when its plane is parallel to the field i.e. when  = 0,
Maximum flux linkages (m) when the coil is perpendicular to the field i.e. when  = 90 (/2 radians)
Flux linkages,  = m sin 
 = m sin t
Where the coil rotates at n rev/sec.
The resulting angular velocity,  = 2n rads/sec
 for t sec angular velocity is t 8

9. From Faraday’s Law
emf induced in the coil =
emf = m cos t
This is the principle of a basic single-phase alternating current (a.c.) generator. 9

10. Single-phase generator
Consider a permanent magnet:
Its North and South revolving at constant speed inside a stationary iron ring.
The magnet is driven by external mechanical source, such as turbine
Induced voltage across the inductors is given by the equation:
emf = m cos t 10

11. Power output of a single-phase generator11

12. Two-phase generator 12

13. Three-phase generator
Three-phase is similar to 2-phase, except that the stator has three identical windings instead of two.
The three windings A1, B2, and C3 are placed at 120 to each other
When the magnet is rotated at a constant speed, the voltages induced in the three windings have the same effective values but the peaks occur at different times as shown in Figure (b) 13

14. Power output of a 3-phase generator
Connecting the three windings of the generator to three identical resistors:
This arrangement requires 6 wires to deliver power to individual single-phase loads.
The resulting currents Ia, Ib, and Ic are respectively in phase with voltages Ea1, Eb2, and Ec3.
They are mutually out of phase by 120
The instantaneous power supplied to each resistor is again composed of a power wave that surges between zero and a maximum value Pm. 14

15. Three Phase Circuits
Electric power is generated, transmitted, and distributed in the form of 3-phase power.
Homes and small establishment are wired for single-phase power, which represents a tap-off from the basic 3-phase system.
Three-phase power is preferred over single-phase power for several important reasons:
Three phase motors, generators and transformers are simpler, cheaper, and more efficient.
Three-phase transmission lines deliver more power for a given weight and cost.
The voltage regulation of 3-phase transmission lines is inherently better.
In three-phase electrical system, the three phases are identical, but they deliver power at different times.
Three-phase system is basically composed of three single-phase systems that operate in sequence
Thus, one phase may be used to represent the behavior of all three 15

16. Types of Connections There are two types of connections:
Wye “Y” or “Star” connection
“” or “Delta” connection
Wye “Y” or Star
Delta “” or Mesh 16

17. Wye Connection “Y”
The ends of the return conductors 1, 2 and 3 are joined together to form the neutral point N
Load
Generator
(a)
(b) 17

18. (a)
(b)
(a)
(b) 18

19. The ends of the return conductors R’, Y’ and B’ are joined together to form the neutral point N.
The voltage between the line terminals R, Y and B, called the line voltage,
Phasor diagrams are used to Draw the Voltage
Taking careful note of voltage polarities, the line voltages are found using vector algebra.
Why? 19

20. Based on the fact that the length of the line-to-neutral phasors is Eln, we have the following:
The line-line voltage (line voltage) is times the line-to-neutral voltage:
Where,
EL = effective value of the line voltage [V]
ELN = effective value of the line-to-neutral voltage [V]
= a constant [approximate value = [1.73]. 20

21. Wye Connection “Y”- Adding a Load
When a balanced three phase load is connected to the generator, the currents will flow from the phases into the lines ILine= IPhase
The magnitude of the currents and their phase angle are calculated from the phase voltage and complex impedance, Z. 21

22. So for example if each phase comprises resistance and inductance in series, i.e. have a complex impedance, Z = R + jL 22

23. So for example if each phase comprises resistance and inductance in series, i.e. have a complex impedance, Z = R + jL
The phase angle  by which each current lags its respective phase voltage is given by
The figure shows the three current phasors each lagging their respective phase voltage. 23

24. At the load star point IR, IY and IB add to form IN.
Applying Kirchoff’s Current Law at N.
IN = IR + IY + IB
As the currents on the right hand side of this equation are not in phase, IN must be found by vector algebra. 24

25. Delta “” Connection
Here, the generator coils are connected as shown in Figure
This time ERY = ERR’ ; EYB = EYY’ ; EBR = EBB’ VL = VPh 25

26. This time ERY = ERR’ ; EYB = EYY’ ; EBR = EBB’ VL = VPh
Line voltage =phase voltage
When a delta connected load is added, the line currents divide at the load as shown in Figure, such that
IR = IRY – IBR
IY = IYB – IRY
IB = IBR - IYB 26

27. The line currents are then found by vector algebra (see previous page)
The magnitude and phase angle of the phase currents are calculated from the phase voltage and complex impedance of the load e.g.
The line currents are then found by vector algebra (see previous page)
i.e. line current = 3 phase current or IL= 3 IP
There is a 30 phase displacement between line current and phase current. 27

28. Delta “” Connection-Resistive Load
Let us determine the voltage and current relationships in such a delta connection, assuming a resistive load.
The resistors are connected across the line; consequently, resistor currents I1, I2, and I3 are in phase with respective line voltages Eab, Ebc, and Eca. Furthermore, according to KCL 28

29. Call lengths of phasors I1, I2, and I3 as Iz.
Call length of line current Ia, Ib, and Ic = IL
Using simple trigonometry, we can now write
The line-line Current is therefore times the current in each branch of delta-connected load:
Where,
IL = effective value of the line current [A]
IP = effective value of the current in one branch of a delta-connected load [A]
= a constant [approximate value = 1.73]. 29

30. Power Transmitted by a 3-Phase “Y” Line=
Generally,
P = VI cos 
(4.6.1)
Total power transfer = 3VpIp cos 
Where Vp = phase voltage
Ip = phase current
Start Connected: VL = 3Vp and IL = Ip
total power transfer =
Where, IL is the line Current
And EL is the line voltage 30

31. Power Transmitted by a 3-Phase “” Line
Similarly in delta-connection:
Apparent Power|Y= Apparent Power|
Where
S = total apparent power delivered by a 3-phase line [VA]
E = effective line voltage [V]
I = effective line current [A]
= a constant [approximate value = [1.73]. 31

32. Active, Reactive, and Apparent Power in 3-Phase Circuits
The relation between:
Active Power “P”,
Reactive Power “Q”, and
Apparent Power “S”
is the same for balanced 3-phase circuits as for single-phase circuits.
We therefore have
and
cos = P/S
where
 S = total 3-phase apparent power [VA]
P = total 3-phase active power [W]
Q = total 3-phase reactive power [var]
cos = power factor of the 3-phase load
 = phase angle between the line current and the line-to-neutral voltage [] 32

33. Rectified Supplies Rectifier:
A device used to convert alternating voltage (current) to direct voltage (current).
By using diodes, alternating currents (ac) supplies can be rectified to produce direct current (dc) supplies.
Half Wave Rectification
Full Wave Rectification
Bridge Rectification
Thyristors are used in a controlled rectification 33

34. Uncontrolled Half Wave Rectifier
Average DC output voltage, 34

35. Uncontrolled Full Wave Rectifier
Average DC output voltage = 35 =

36. Controlled Half Wave Rectification
Average DC output voltage, 36

37. Controlled Full Wave Rectification
Average DC output voltage = 37

38. Smoothing an Uncontrolled HWR
Filters: The direct output voltages from the rectifiers is pulsating so it is necessary to add a “smoothing” capacitor in the output circuit.
Capacitor charges to maximum when the diode conducts and discharges when it is not conducting, thus smoothes the output.
The variation in the output voltage is called as Ripples 38

39. Smoothing a Full Wave Rectifier39

40. Exercises
Ex1: A 3-phase 415V (line voltage) 50Hz supply feeds three star-connected 20F capacitors. Calculate the magnitude of the line currents.
Ex2: A 3-phase 415V 50Hz supply feeds three delta-connected 20F capacitors. Calculate the magnitude of the line currents.
Ex3: A 200kVA 3300V (line voltage) star-connected alternator operating at full load supplies a balanced 3-phase load at a power factor of 0.9 lagging. Calculate the magnitude of the line current.
kVA rating =
where = line current at full load.
Total power supplied by alternator = (kVA rating)  (power factor)
Ex4: A 500kVA 6600V delta connected alternator delivers full output to a balanced 3-phase load with a power factor of 0.8 lagging. Calculate the mangitude of the current in each phase of the alternator. 40

41. Problems
Q4.1 A balanced 3-phase 415V (line) 50Hz supply feeds three coils each with an inductance of 20mH and resistance 4 connected in star. Calculate the line current, phase angle and total power supplied.
32.2A; ; kW
Q4.2 The same supply feeds the same three coils reconnected in delta. Calculate the new phase current, power factor, line current and total power supplied.
55.7A; 0.54; 96.5A; 37.29kW
Q4.3 A 120Vrms 60Hz single phase supply feeds a diode bridge rectifier with a resistive load. Calculate the average DC output voltage.
108V
Q4.4 A 240Vrms 50Hz single phase supply feeds a controlled (thyristor) half wave rectifier with a resistive load. Calculate the maximum and minimum average DC output voltages and the corresponding delay angles, 
108V, 0V 41