1. ELECTRICAL SYSTEM TECHNOLOGY
Chapter 1
Transformer

2. ASSESMENTS (a) Coursework: 50% (i) 30% from Practical
# 20%: Lab Reports
# 10%: Lab Test
(ii) 20% from Written
# 15%: Test 1 & Test 2
# 5%: Tutorials, Attendance & Quizzes
(b) Final Exam: 50%
(c) TOTAL: 100%

3. Sensor and Transducers
COURSE OUTLINES
Electrical Machines
Transformer
DC Machines
AC Machines
Instrumentations
DC and AC Meters
DC and AC Bridges
Sensor and Transducers

4. LIST OF EXPERIMENTS Lab 0 – Lab Introduction
Lab 1 – Single Phase Transformer; Voltage and Current Ratio
Lab 2 – DC Series Motor
Lab 3 – Three Phase AC Induction Motor
Lab 4 – d’Arsonval Galvanometer
Lab 5 – The Basic Voltmeter Design
Lab 6 – The Wheatstone Bridge

6. 1. INTRODUCTION
A transformer is a device that changes ac electric power at one voltage level to another voltage level through the action of a magnetic field.
Power Distribution System
DC System
Thomas A. Edison
For incandescent bulb
AC System
(Transformer)
Low Voltage (120-V)
High Voltage
(12 to 25-kV)
Step Up Voltage
(for Transmission over Long Distance)
110 to 1000-kV
Step Down Voltage
(for Final Use)
12 to 34.5-kV
High Current
Huge Power Loss in Transmission Line
Decrease Current
Very Low Power Loss in Transmission Line

7. INTRODUCTION

8. 2. TRANSFORMER CONSTRUCTION
Basic design of a transformer consists of
one core and two coils.
The core can be air, soft iron or steel.
The core types of a transformer:
Core form
Shell form
Core form is a construction of
two-legged laminated core with
the coils wound on two legs.
Shell form is a construction of
three-legged laminated core with
the coils wound on center legs.
Windings of shell form are arranged
concentrically to minimize the flux
leakages.
(a)
(b)
Fig.1-1 The transformer construction of (a) Core-form and (b) shell-form

9. 2. TRANSFORMER CONSTRUCTION
(b)
Figure 1-2 Core construction of (a) hollow type and (b) shell-type.

10. 3. SYMBOLS & SIMPLE CIRCUIT
Air core
Iron core
Figure 1-3 Schematic symbols of transformer ~ RL
Primary windings
Secondary windings
Figure 1-4 Simple circuit of transformer

11. 4. FUNCTION OF TRANSFORMER'S PARTS
Table 1-1: Principle Parts of a Transformer
Part
Function
Core
Provides a path for the magnetic lines of flux
Primary winding
Receives energy from the ac source
Secondary winding
Receives the energy from primary winding and delivers it to the load
Enclosure
Protects the above components from dirt, moisture, and mechanical damage
Core
Primary Windings
Secondary Windings ~
AC Source RL ►

12. Fig.1-5 Paths of mutual and leakage flux
5. MAGNETIC FLUX
The flux produced by a transformer is divided into two components:
Mutual flux:
the portion of flux that goes from
the primary winding to secondary
winding through the core.
It remains in the core and links
both windings.
Leakage flux:
the portion of flux that goes
through one of the windings but
not the other one. It returns
through the air. ►
ΦLP
ΦLS ~ N1 N2 RL ΦM
Fig.1-5 Paths of mutual and leakage flux

13. 5. MAGNETIC FLUX The total average primary flux: (1-1)
where ФM = flux component linking both primary and secondary coils
ФLP = primary leakage flux
The total average secondary flux:
(1-2)
where ФLP = secondary leakage flux

14. 6. TRANSFORMER LOSSES
Total power losses in a transformer are a combination of four types of losses:
Leakage flux: fluxes which escape the core and pass through only one of the transformer windings.
Copper losses: resistive heating losses in the primary and secondary windings of the transformer. They are proportional to the square of the current in the windings.
Eddy current losses: resistive heating loss in the core of transformer. They are proportional the square of the voltage applied to the transformer.
Hysteresis losses: losses caused by the rearrangement of the magnetic domains in the core.

15. Ideal Transformer at No-Load
By definition, an ideal transformer is a device that has no core loss and no leakage flux due to its core is infinitely permeable. Any flux produced by the primary is completely linked by the secondary, and vice versa.
Ideal Transformer at No-Load
No current is(t) flowing out of the secondary side.
Relationship between voltage applied to the primary side of the transformer, VP(t) and the voltage produced on the secondary side, VS(t) is equal to the ratio of the numbers of turns on the primary and on the secondary.
Fig.1-6 Schematic symbol of an ideal transformer at no-load. + - ►
iP(t) NP
iS(t) = 0 NS VP
vS(t) ~
vP(t) VS
(1-3)
where a is defined to be the turns ratio of the transformer.

16. Ideal Transformer under Load
There is current iS(t) flowing out of the secondary side. The relationship between the current iP(t) flowing into the primary side of the transformer and the current iS(t) flowing out of the secondary side is ►
(1-4)
iP(t) NP NS
iS(t) + + ► ►
where, iP(t) = primary current (A)
iS(t) = secondary current (A) ~ VP VS RL
In terms of phasor quantities, the equation 1-2 and 1-3 can be written as follow: - - ►
Fig.1-7 Schematic symbol of an ideal transformer under load.
(1-5)
Phase angle of VP is the same as the angle of VS and the phase angle of IP is the same as the phase angle of IS.

17. 7. IDEAL TRANSFORMER ~ - - R
The dots appearing at one end of each winding in
Fig. 1-3 tell/”state” the polarity of the voltage and
current on the secondary side of the transformer.
This utilize the dot convention.
Dot convention:
If the primary voltage is positive at the dotted end of the winding with respect to the undotted end, then the secondary voltage will be positive at the dotted end also. Voltage polarities are same with respect to the dots on each side of the core.
If the primary current of the transformer flows into the dotted end of the primary winding, the secondary current will flow out of the dotted end of the secondary winding. ►
iP(t) NP NS
iS(t) + + ► ► ~ VP VS R RL - - ►
Fig.1-7 Schematic symbol of an ideal transformer under load.

18. Power in an Ideal Transformer►
iP(t)
The power supplied by the ac generator to the transformer is given by the equation: NP NS
iS(t) + + ► ► ~
(1-6) VP VS R RL
where θp is the angle between the primary voltage and the primary current.
The power supplied by the secondary transformer to its load is given by the equation: - - ►
Fig.1-7 Schematic symbol of an ideal transformer under load.
(1-7)
where θs is the angle between the secondary voltage and the secondary current.
Cos θp = power factor of the primary windings and cos θs = power factor of the primary windings.

19. 7. IDEAL TRANSFORMER ~ Pout = Pin - - R
For an ideal transformer, θP = θS. Therefore, applying the turns-ratio equations gives VS = VP/a and IS = aIP, so the output power of an ideal transformer is equal to its input power: ►
iP(t) NP NS
iS(t) + + ► ► ~ VP VS R RL
Pout = Pin
(1-8) - - ►
Fig.1-7 Schematic symbol of an ideal transformer under load.

20. 7. IDEAL TRANSFORMER Impedance Ratio • + – + – VS VP VP IP IS
VAC • ZL IS VS VP
Impedance Ratio
The impedance of a device or an element is defined as the ratio of the voltage across it to the current flowing through it.
Thus, the impedance of the load connected to the secondary winding of a transformer can be expressed as: IP
VAC VP + –
(1-9)
The impedance of the primary circuit of the transformer is given by:
(1-10)
Fig.1-8 (a) Impedance scaling through a transformer. (b) Definition of impedance.

21. 8. SHIFTING IMPEDANCES FROM SECONDARY TO PRIMARY+ – I1
VAC • Z1 V2 I2 V4 V3 Z2 Z3 Z4 I4 I3
aV2
I2/a
a2Z2
(a)
(b)
Objectives:
To solve for all the voltages across each element.
To calculate the current flowing each element.
Fig.1-9
The actual circuit showing the actual voltages and currents on the secondary side.
b. Impedance Z2 is shifted to the primary side. Note the corresponding change in V2 and I2.

22. 8. SHIFTING IMPEDANCES FROM SECONDARY TO PRIMARY
aV3 + – I1
VAC Z1
aV2
I2/a V4 Z4 I4
a2Z3
I3/a •
a2Z2
aV4
I4/a
a2Z4
I = 0
Fig.1-9
Impedance Z3 is shifted to the primary side. Note the corresponding change in V3 and I3.
Impedance Z4 is shifted to the primary side. Note the corresponding change in V4 and I4. The currents in T are now zero.
(c)
(d)

23. 8. SHIFTING IMPEDANCES FROM SECONDARY TO PRIMARY+ – I1
VAC Z1
aV2
I2/a
aV4
I4/a
aV3
a2Z3
I3/a
a2Z2
a2Z4
Fi.1-9
All the impedances are now transferred to the primary side and the transformer is no longer needed.
With shifting impedances from secondary to primary, we will find that:
The impedance values is multiplied by a2.
The real voltage across the transferred impedance increases by a factor a.
The real current flowing the transferred impedance decreases by a factor a.

24. 9. SHIFTING IMPEDANCES FROM PRIMARY TO SECONDARY+ – I1
VAC • Z1 V2 I2 V4 V3 Z2 Z3 Z4 I4 I3
aI1
Z1/a2
V1/a
Fig.1-10
The actual circuit showing the actual voltages and currents on the primary side.
b. Impedance Z1 is transferred to the secondary side. Note the corresponding change in V1 and I1.
(a)
(b) V1

25. 9. SHIFTING IMPEDANCES FROM PRIMARY TO SECONDARY
VAC
aI1
Z1/a2 V2 I2 V4 V3 Z2 Z3 Z4 I4 I3
V1/a + – •
Fig.1-10
The source is transferred to the secondary side. Note the corresponding change in VAC. Note also that the currents in T are zero.
All the impedances and even the source are now on the secondary side. The transformer is no longer needed because the currents are zero.
(c)
V1/a V3
Z1/a2 Z3
aI1 I3 + – Z2 Z4 I2 V2 V4 I4
VAC/a
(d)

26. 9. SHIFTING IMPEDANCES FROM PRIMARY TO SECONDARY
With shifting impedances from primary to secondary, we will find that:
The impedance values is divided by a2.
The voltage across the transferred impedance is lower than the real voltage.
The real current flowing the transferred impedance is higher than the real current.

27. Fig.1-11 AC power system (a) without and (b) with transformer+ –
Iline jX R
Zload
Iload T1 T2 • IG
VAC
Zline 1 2 3
Vload
Fig AC power system (a) without and (b) with transformer
Based on the use in AC power system, a transformer is differed in a various names:
Unit transformer:
a transformer connected to the output of a generator and used to step up the voltage to transmission line (110+ kV).
2. Substation transformer:
a transformer at the end of the transmission line, which steps down the voltage from transmission levels to distribution levels (from 2.3 to 34.5 kV).
Distribution transformer:
a transformer that takes the distribution voltage and steps down it to the final voltage at which the power is actually used (110, 208, 220 V, etc.).
All these transformers are essentially same—the only difference among them is their intended
use.

28. 11. ANALYSIS OF CIRCUITS CONTAINING IDAL TRANSFORMERS
For your first exercise, open the text book of “Electrical Machinery Fundamentals-Fourth Edition by Stephen J. Chapman, Pg (Example 2-1). You try to understand the example first and after that you just change the values of V, Zload and Zline and then you answer all questions like in the example. OK…
Remember….I’ll test you 

29. 11. ANALYSIS OF CIRCUITS CONTAINING IDAL TRANSFORMERS+ –
Iline jX R
Zload
Iload
Vload T1 T2 • IG
VAC
Zline
(a)
(b)
Fig The power system (a) without and (b) with transformer

30. 11. ANALYSIS OF CIRCUITS CONTAINING IDAL TRANSFORMERS
VAC + –
Iline jX R T1 • IG
Zline
Equivalent circuit • •
VAC jX R + – IG
Equivalent circuit • •
Fig.1-13 (a) System referred to the transmission system voltage level. (b) System and transmission line referred to the generator’s voltage level

31. 12. REAL TRANSFORMER MODELο VS I0 VP IP
VAC + –
IS = 0 RP RS • ► NP NS
jXP
jXS
(a) VS I0 VP IP
VAC + –
IS ≠ 0 RP RS • ► NP NS
jXP
jXS
(b)
Fig Complete circuits of a real transformer for (a) open and (b) short circuits

32. 12. REAL TRANSFORMER MODEL
To construct a complete circuit of a real transformer:
each imperfection (core losses) and permeability are taken in to
account;
their effects are included in the transformer model.
Imperfection and permeability can be represented by the values of resistance and reactance, respectively, by doing two tests:
open-circuit test
short-circuit test

33. 12. REAL TRANSFORMER MODEL
Open-Circuit Test I0 VP
IP = 0 ο IS
VAC + –
Excitation branch RC
jXM Im If VS • ► RP
jXP RS
jXS
Fig An imperfect core represented by a reactance Xm and a resistance Rm
The resistance, RC shows the core losses.
Magnetizing reactance, XM represents a measure of the permeability of
the transformer core.
The current, Im represents the magnetizing current needed to create the
magnetic flux Φm.

34. 12. REAL TRANSFORMER MODEL
Open-Circuit Test
IOC VP
IP = 0 ο IS
VAC + –
Excitation branch RC
jXM Im If VS • ► RP
jXP RS
jXS
Fig An imperfect core represented by a resistance RC and a reactance XM
Because of RP and XP are too small in comparison to RC and XM to cause a significant voltage drop, so all the input voltage is dropped across the excitation branch.
Therefore, RC and XM can be obtained with measuring the magnitude of input voltage (VOC), input current (IOC) and input power (POC).

35. 13. MEASURING THE VALUES OF COMPONENTS IN A TRANSFORMER
Wattmeter
ip (t)
v (t) V A + -
vp (t)
is (t) ~
(a)
Wattmeter
ip (t)
v (t) A + -
vp (t)
is (t) V ~
(b)
Fig (a) open-circuit test and (b) short-circuit test

36. 13. MEASURING THE VALUES OF COMPONENTS IN A TRANSFORMER
Open-Circuit Test
Using the input voltage, input current and input power of the open-circuit test , power factor (PF) can be determined as follows:
(1-11)
The power factor angle (θ) is given by
(1-12)
The power factor is always lagging for a real transformer, so the angle of the current always lags the angle of the voltage by θ degrees. Therefore, the excitation admittance YE is
(1-13)

37. 13. MEASURING THE VALUES OF COMPONENTS IN A TRANSFORMER
Open-Circuit Test
Beside that, the excitation admittance can be also obtained using the formula:
(1-14)
where, GC = conductance of the core-loss resistor
BM = susceptance of the magnetizing inductor
RC = resistance in the transformer
XM = reactance in the transformer
Thus, we find
(1-15)
(1-16)

38. 13. MEASURING THE VALUES OF COMPONENTS IN A TRANSFORMER
From the values of resistance (RC) and reactance (XM), we can determine the core losses and reactive power following equations:
(1-17)
(1-18)
where, RM = resistance representing the core losses (Ω)
Xm = magnetizing reactance of the primary winding (Ω)
Vp = primary voltage (V)
PM = core losses (W)
QM = reactive power needed to set up the mutual flux Φm (W)

39. 13. MEASURING THE VALUES OF COMPONENTS IN A TRANSFORMER
Short-Circuit Test VS I0 VP IP
VAC + –
IS ≠ 0 RP RS • ► NP NS
jXP
jXS
Fig Complete circuit of a real transformer for short-circuit
Apart from the magnitude of input voltage (VSC), input current (ISC) and input power (PSC), resistance and inductance values can be obtained with constructing the equivalent circuit.

40. 13. MEASURING THE VALUES OF COMPONENTS IN A TRANSFORMER
Short-Circuit Test I0 RP
jXP
a2RS
ja2XS
IS/a
VAC + –
Fig Equivalent circuit of a real transformer for short-circuit
Equivalent resistance and reactance are
(1-19)
(1-20)

41. 13. MEASURING THE VALUES OF COMPONENTS IN A TRANSFORMER
Short-Circuit Test
The series impedance ZSE is equal to
(1-21)
The power factor of the current is given by:
(1-22)
and is lagging. The current angle is thus negative, and the overall impedance angle θ is positive:
(1-23)

42. 13. MEASURING THE VALUES OF COMPONENTS IN A TRANSFORMER
Short-Circuit Test
Series impedance can be calculated as follows:
(1-24)
Therefore, we get
(1-25)
(1-26)

43. 14. PER-UNIT SYSTEM OF MEASUREMENTS
Per-unit (pu) system of measurements is an another method to solve the circuits containing transformers. In this method, the impedance transformations can be avoid. Thus, circuits containing many transformers can be solved easily with less chance of error.
In pu system, the voltages, currents, powers, impedances, and other electrical quantities are not measured in SI units (volts, amperes, watts, ohms, etc.) but measured in decimal fraction.
Any quantity can be expressed on a per-unit basis by the equation:
(1-27)
where actual value is a value in volts, amperes, ohms, etc.

44. 14. PER-UNIT SYSTEM OF MEASUREMENTS
To define a pu system, two quantities need to select. The ones usually selected are voltage and real or apparent power.
In a single-phase system, these relationships are
(1-28)
(1-29)
(1-30)

45. 15. TRANSFORMER VOLTAGE REGULATION
Voltage regulation is a quantity that compares the secondary voltage of a transformer at no load with the secondary voltage at full load. It is defined by the equation:
(1-31)
where, VS,nl = secondary voltage at no-load (V)
VS,fl = secondary voltage at full-load (V)
Since at no load, VS = VP/a, the voltage regulation can also be expressed as
(1-32)

46. 15. TRANSFORMER VOLTAGE REGULATION
If the transformer equivalent circuit is in the per-unit system, then voltage regulation can be expressed as
(1-33)
For an ideal transformer, VR = 0.

47. 16. TRANSFORMER EFFICIENCY
The efficiency of a devices (motors, generators as well as transformers) is defined by the equation:
(1-34)
(1-35)
At full-load, a transformer has the total loss (Ploss) in Watts:
(1-36)

48. 15. VOLTAGE REGULATION where,
Pcore = input power in Watts on the open-circuit test = core loss
PCu = input power in Watts on the short-circuit test with full-load currents
= I2R loss on full load
and the output power (Pout) in Watts is given by:
(1-37)
So, the efficiency of the transformer can be written as
(1-38)