1. EEK260 -Electrical MachinesBy
Dr. Mohamad Kamarol Mohd Jamil
Room: 2.32
Ext: 6064

2. Practical Transformers
What is practical transformer?
The winding of practical transformer have resistance (Wildi, pg 197)
The cores are not infinitely permeable (Wildi, pg 197)
The flux produced by the primary is not completely captured by the secondary. (Wildi, pg 197)
Produced leakage flux, eddy-current and hysteresis losses, which contribute to the temperature rise. (Wildi, pg 197)

3. Practical transformer
Over/Eddy- current loss, hysteresis loss and iron loss
Copper loss at coil resistance
Magnetic flux
Magnetic flux leakage
Fig. 1 Practical transformer

4. Exciting/magnetization current in practical transformer
When an AC power source is connected to a transformer, the current flows in the coils to establish a sinusoidal flux in the real ferromagnetic core. This current is called exciting current, iΦ.
No hysteresis
With hysteresis

5. No hysteresis Let consider a B-H characteristic with no hysteresis.
It can be rescaled to Φ-i curve for the core as shown below
The exciting current is purely lagging current
The exciting winding can be represented by a pure inductance
iΦ lag the voltage e by 90°
No power loss is involved

6. With hysteresis Magnetization of core Core loss
The exciting current is nonsinusoidal as well as nonsymmetrical with respect to the voltage waveform
The exciting current can be split into two components,
1.Core-loss current ic which is in phase with voltage e – modeled by a resistance Rc
2. Magnetization current im which is in phase with Φ and symmetrical with respect to e and lagging by 90°- modeled by reactance Lm or Xm

7. An imperfect core of practical transformer
Pm: iron losses (W)
Qm: reactive power needed to set up the mutual flux Φm (Var)

8. Example 1(10-1)
A large transformer operating at no-load draws an exciting current I0 of 5 A when the primary is connected to a 120 V, 60 Hz source (Fig. 2). From a wattmeter test it is known that the iron losses are equal to 180 W.
Calculate,
The reactive power absorbed by the core
The value of Rm and Xm
The value of If, Im and I0
Fig. 2

9. Solution for example 1(10-1)
a) The apparent power supplied to the core is
The iron losses are
The reactive power absorbed by the core is
b) The impedance corresponding to the iron losses is

10. Solution for example 1(10-1)
The magnetizing reactance is
c) The current needed to supply the iron losses is
If =1.5 A
120 V
The magnetizing current is
Im =4.8 A
I0=5 A
The exciting current is Φm

11. The equivalent circuit of a transformer
The losses that occur in real transformers have to be accounted for any accurate model of transformer behavior.
The major items to be considered in construction of such model are
1. Copper losses (I2R).
- Are the resistive heating losses in the primary and secondary windings of the transformer. Proportional to the square of the current in the windings.
2. Eddy current losses.
- Are the resistive heating losses in the core of the transformer. Proportional to the square of the voltage applied to transformer

12. The equivalent circuit of a transformer
3. Hysteresis losses.
- Are associated with the rearrangement of the magnetic domains in the core during each cycle.
4. Leakage flux .
- The fluxes which escape the core and pass through only one of the transformer windings. It produce a self-inductance in the primary and secondary coils, and the effects of this inductance must be accounted for.

13. The equivalent circuit of a practical transformer
Leakage flux
Leakage flux
Copper losses
Copper losses
Ideal transformer
Core excitation effects

14. Example 2 (10-2)
The secondary winding of a transformer possesses 180 turns. When the transformer is under load, the secondary current has an effective value of 18 A, 60 Hz. Furthermore, the mutual Φm has a peak value of 20mWb. The secondary leakage flux Φf2 has a peak value of 3mWb.
Calculate
The voltage induced in the secondary winding by its leakage flux
The value of the secondary leakage reactance
The value of E2 induced by the mutual flux Φm
= V
= 8 Ω
= 959 V

15. Simplifying the equivalent circuit
Fig.3 Complete equivalent circuit of a transformer at no load
Simplify to
I2=0, I1=0
Only I0 flows in R1 and Xf1. However these impedances are so small that the voltage drop across them is negligible

16. Simplifying the equivalent circuit
Fig.4 Complete equivalent circuit of a transformer at full-load
Simplify to
Ip >20I0
Only I0 can be neglected. However these impedances are so small that the voltage drop across them is negligible

17. Simplifying the equivalent circuit
Simplify the circuit by shifting everything to the primary side

18. Simplifying the equivalent circuit
Produces an internal voltage drop when the transformer is loaded. It affects the voltage regulation of the transformer
Transformer above 500 kVA possess a leakage reactance XP that is at least 5 times greater than RP
In such transformer RP can be neglected as far as voltage and currents are concerned

19. Determination of equivalent circuit parameters
R1, R2, Xf1, Xf2, Rm, Xm
These parameters can be directly and more easily determined by performing tests that involve little power consumption.
No-load test or open-circuit test
2. Short-circuit test

20. No-load test (or Open-circuit test)
This test is performed by applying a voltage to either the high-voltage side or low voltage side, whichever is convenient.
Possible to determine
the excitation branch
the magnitude & the angle of excitation impedance
the power factor of the input current

21. No-load test (or Open-circuit test)
The test result give the following information:
Pm = active power absorbed by core
EP I0 = Sm = apparent power absorbed by
core
Qm = reactive power absorbed by core
Resistance Rm corresponding to the core loss is
Magnetizing reactance Xm is

22. No-load test (or Open-circuit test)
The easiest way to calculate the values of Rm and Xm is to look first at the admittance of the excitation branch
Therefore the total excitation admittance is

23. No-load test (or Open-circuit test)
The magnitude of the excitation admittance can be found from the open circuit test voltage and current
The angle of the admittance can be found from the knowledge of the circuit power factor
Where the PF is always lagging for a real transformer

24. Short-circuit test
Secondary winding is short-circuited and the primary terminals are connected to a fairly low-voltage source. (lower than normal, less than 5% of rated voltage)
ISC < its nominal value
To prevent
Overheating
Rapid change in winding resistance RP XP
ESC
ISC

25. Short-circuit test
The input voltage is so low during the short-circuit test, thus the current flows (I0) through the excitation branch can be negligible RP XP
ESC
ISC
If the excitation current is ignored, then all the voltage drop in the transformer can be attributed to the series elements in the circuit
The magnitude of the series impedances referred to the primary side of the transformer

26. Short-circuit test
Total transformer impedance referred to the primary side is
The PF of the current given by
And is lagging. Therefore the total impedance is
Total transformer resistance and leakage reactance referred to the primary side

27. Transformer rating The losses appear in the form of heat yield to
1) Increase in temperature
2) Drop the efficiency
To keep the transformer level at an acceptable level, both of the applied voltage and current drawn by the load must be limited.
This two limits determine the nominal voltage Enp and nominal current Inp
Temperature rise of a transformer is directly related to the apparent power that flows through it

28. Transformer rating So what are the meanings of these rating?
The voltage ratings indicate that the transformer has two windings,
Rated for 1100 V
Rated for 110 V
These voltages are proportional to their respective number of turns, therefore the turns ratio also represent the voltage ratio.
For example, 10 kVA rating means that each winding is designed for 10 kVA.
Therefore, the current rating for HV and LV winding is

29. Actual transformer values
Rated power
Rated voltage
Rated current
Theodore Wildi, “Electrical machines, drives and power systems”, pg 202

30. Voltage regulation
Since a real transformer has series impedance within it, the output voltage of a transformer varies with the load even if the input voltage remains constant. To conveniently compare transformers in this respect, it is usual to define a quantity called voltage regulation
Voltage regulation is a quantity that compares the output voltage of the transformer at no load with the output voltage at full load
Secondary voltage at no-load
Secondary voltage at full-load
Usually it is a good practice to have as small a voltage regulation as possible.

31. Voltage regulation The voltage regulation can be expressed as
How can the voltage regulation of a transformer be determined?
To determine the VR of a transformer, it is necessary to understand the voltage drops within it. ES ES
Approximate transformer model referred to secondary side

32. Voltage regulation
VR of a transformer depends both on the magnitude of this series impedance and on phase angle of the current flowing through the transformer
The easiest way to determine the effect of the impedances and the current phase angles on the transformer VR is to examine a phasor diagram ES
VR > 0
A phasor diagram of a transformer operating at a lagging power factor.

33. Voltage Regulation
PF = unity
VR > 0
PF = leading
VR < 0

34. Simplified VR calculation

35. Transformer Efficiency
The efficiency of a transformer at a given load, can be calculated as follows
Where,
Since the output power is given as
The efficiency can be expressed as

36. Example 3 (10-4)
The nameplate of a distribution transformer indicates 250 kVA, 60 Hz, primary 4160 V, secondary 480 V.
Calculate the nominal primary and secondary currents
If we apply 2000 V to the 4160 V primary, can we still draw 250 kVA from the transformer

37. Example 4 (10-5)
A single-phase transformer rated at 3000 kVA, 69 kV/4.16 kV, 60 Hz has a total internal impedance Zp of 127 Ω, referred to the primary side.
Calculate
The rated primary and secondary currents
The voltage regulation from no-load to full-load for a 2000kW resistive load, knowing that the primary supply voltage is fixed at 69 kV
The primary and secondary currents if the secondary is accidentally short-circuited

38. Solution for example 4 a. Rated primary current
Rated secondary current
b. Leakage reactance, where resistance can be ignored due to the rated voltage exceeds 500kVA
Determine the load Z

39. Solution for example 4 Load impedance referred to primary side
c. If the secondary is accidentally short-circuited,
Voltage at full-load

40. Example 5(10-6)
During a short-circuit test on a transformer rated 500 kVA, 69 KV/4.16 kV, 60 Hz, the following voltage, current and power measurements were made. Terminal X1, X2 were in short-circuit as shown in figure below. The measurement results are as follows
ESC = 2600 V
ISC = 4 A
PSC = 2400 W
Calculate the value of the reactance and resistance of the transformer referred to the HV side

41. Solution for Example 5
Transformer impedance referred to the primary side is
Resistance referred to the primary is
Leakage reactance referred to the primary is

42. Example 6 (10-7)
An open-circuit test was conducted on the transformer given in example 3. The following results were obtained when the lo-voltage winding was excited. (in some cases, such as in a repair shop, a 69 kV voltage may not be available and the open-circuit test has to be done by exciting the LV winding at its rated voltage)
ES = 4160 V, I0 = 2 A, Pm = 5000 W
Using this information and the transformer characteristics found in example 3, calculate:
The values of Xm and Rm on the primary side
The efficiency of the transformer when it supplies a load of 250 kVA, whose p.f is 80% lagging.

43. Solution of example 6(a)
Referred to secondary side
The values of excitation impedance referred to the primary side will be a2 times greater

44. Solution of example 6(b)
S = 250 kVA
The load current is
The turns ratio is
The total losses are
The current on the primary side is
The active power delivered by the transformer to the load
The total copper loss (primary and secondary) is
The active power received by the transformer is
The efficiency

45. The per-unit system of measurement
Per-unit system is another approach to solving circuit containing transformer which eliminate the need of explicit voltage level conversion. The circuit containing many transformer can be solved easily with less chance of error.
Instead of dealing with ohm, amperes, volts and kilowatts, each electrical quantity is measured in a decimal fraction

46. Per-unit system
To define a given per unit system, the voltage and power (or apparent power) should be selected
Once the base values of S (or P) and V have been selected, all other base values can be computed easily

47. Example 7
A simple power system contains a 480 V generator connected to an ideal 1:10 step-up transformer, a transmission line, an ideal 20:1 step-down transformer, and a load. The impedance of the transmission line is 20+j60 Ω, and the impedance of the load is ° Ω. The base value for this system are chosen to be 480 V and 10 kVA at the generator.
Find the base voltage, current, impedance and apparent power at every point in the power system
Convert this system to its per-unit equivalent circuit
Find the power supplied to the load in this system
Find the power lost in the transmission line

48. Solution for example 7(a)
a. In the region 1,

49. Solution for example 7(a)
Turns ratio of first transformer (T1) is a=1/10
Therefore the voltage in the region 2 (transmission line region) is

50. Solution for example 7(a)
Turns ratio of first transformer (T2) is a=20/1
Therefore the voltage in the region 3 (load region) is

51. Solution for example 7(b)
To convert a power to a per-unit system, each component must be divided by its base value in its region of the system
The generator region (region 1)
The line impedance in transmission region is
The load impedance in load region

52. Solution for example 7(c)
Power of the load
Actual power to the load

53. Solution for example 7(d)
Per-unit power lost in transmission line is
Actual power lost in transmission line is

54. Exercise 1
The equivalent circuit impedances of a 20 kVA, 8000/240 V, 60 Hz transformer are to be determined. The open circuit test and short circuit test were performed on the primary side of the transformer and following data were taken
Open circuit test (on primary)
Short circuit test (on primary)
Voc= 8000 V
Vsc = 489 V
Ioc= A
Isc= 2.5 A
Poc = 400 W
Psc = 240 W
Find the impedance of the approximate equivalent circuit referred to the primary side and sketch the circuit
Convert the circuit determined in (a) to per-unit.

55. Transformer taps
To maintain the secondary voltage of ± 2% of it rated value
Why, ?
Due to voltage drop in transmission line, where the voltage may consistently be lower than normal.
Thus, a distribution transformer having ratio of 2400 V/120 V may be connected to a transmission line where the voltage is never higher than 2000V. Therefore under this situation the voltage across the secondary is considerably less than 120.
Distribution transformer with taps at 2400 V, 2292 V, 2184 V, and 2076 V.
To over come this problem taps are provided on the primary windings of distribution transformer.

56. Example 8
A 500 kVA 13,200/480 V distribution transformer has four 2.5 percent taps on its primary winding. What are the voltage ratios of this transformer at each tap setting
The five possible voltage ratings of this transformer are 5%
13,860/480 V
2.5%
13,530/480 V
13,200/480 V
-2.5%
12,870/480 V
-5%
12,540/480 V