CHAPTER 2. Transformers. Dr Gamal Sowilam

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1 CHAPTER Transformers Dr Gamal Sowilam

2 Introduction A transformer is a static machine. It is not an energy conversion device, it is indispensable in many energy conversion systems. A transformer essentially consists of two or more windings coupled by a mutual magnetic field. Ferromagnetic cores are used to provide tight magnetic coupling and high flux densities, known as iron core transformer. They are used in high-power applications. Air core transformers have poor magnetic coupling and are sometimes used in low-power electronic circuits.

3 Two types of core construction are normally used. 1. In the core type, The windings are wound around two legs of a magnetic core of rectangular shape.

4 In the shell type, the windings are wound around the center leg of three-legged magnetic core. To reduce core losses, the magnetic core is formed of a stack of thin laminations. L-shaped laminations are used for core-type construction and E- shaped laminations are used for shell-type construction.

5 Power Transformer

6 Distribution transformer

7 A schematic representation of a two-winding transformer is shown in the figure. Two vertical bars are used to signify tight magnetic coupling between the windings. One winding is connected to an ac supply, referred to as the primary winding. The other winding is connected to a load, referred to as the secondary winding.

8 Transformer Use Transformer primary function is to change voltage level. Transformers are used to step up and step down voltage at various stages of power transmission. Electric power is generated in a power house at about 30 kv and it is used in domestic house at 110 or 0 V. Electric power is transmitted at 00 kv to 500 kv.

9 Ideal Transformer Consider a two-windings transformer as shown in Fig. Ideal transformer has the following properties: Winding resistances are negligible. No leakage fluxes and core losses are negligible. Permeability of the core is infinite, the exciting current to establish flux in the core is negligible.

10 When the primary winding is connected to a voltage v 1 a voltage e 1 emf will be induced, v1 N1 The core flux also links the secondary winding and induce a voltage e, From the two equations, e e 1 v N d dt d dt v v 1 N N 1 a

11 Because the net mmf required to establish a flux in the ideal core is zero, N i i i N i N N N N 1 i i 1 a net mmf 0 The currents in the windings are inversely proportional to the number of turns. The instantaneous power is: The input VA at primary winding = The output VA at secondary winding v i v i 1 1

12 Example 0 There are 400 turns of wire in an iron-core coil. If this coil is to be used as the primary of a transformer, how many turns must be wound on the coil to form the secondary winding of the transformer to have a secondary voltage of 1 V if the primary voltage is 5 volts?

13 If the supply voltage v1 is sinusoidal, then the previous equation can be written in the terms of rms values:

14 Impedance transfer ازى تشوف المعاوقة من جھة الاولى او الثانیة ھام جدا so

15 An impedance from the primary side can also be transferred to the secondary side, and in that case its value has to be divided by the square of the turns ratio. An impedance Z connected in the secondary will appear as an impedance Z looking from the primary. An impedance from the primary side can also be transferred to the secondary side, and in that case its value has to be divided by the square of the turns ratio:

16 Example 1 A speaker of 9 resistive impedance is connected to a supply of 10 V with internal resistive impedance of 1, as shown in Figure. (a) Determine the power absorbed by the speaker. (b) To maximize the power transfer to the speaker, a transformer of 1: 3 turns ratio is used between source and speaker as shown in Figure. Determine the power taken by the speaker. [ (a) 9W (b) 5W

17 Polarity determination Consider the two windings shown. Terminals 1 & 3 are identical, because Currents entering these terminals produce fluxes in the same direction in the Core forms the common magnetic path.

18 Polarities of windings must be known if transformers are connected in parallel. To share a common load. Fig. shows the parallel connection of two single-phase Transformers.

19 Practical Transformer In a practical transformer, the windings have resistances, leakage flux exists, core losses occur, and the permeability of the core material is not infinite.

20 In the effects of winding resistance and leakage flux, the transformer windings are tightly coupled by a mutual flux. The effects of leakage flux can be accounted for by an inductance, called Leakage inductance.

21 In practical transformer, a magnetizing current is required to establish a flux in the core. This effect can be represented by a magnetizing inductance L m The core loss can be represented by a resistance R c. A practical transformer is therefore equivalent to an ideal transformer plus external impedances that represent imperfections of an actual transformer.

22 Equivalent Circuit The ideal transformer can be moved to the right or left by referring all quantities to the primary or secondary side. The equivalent circuit with the ideal transformer moved to the right is shown in the Figure.

23 Referred Equivalent Circuit For convenience, the ideal transformer is usually not shown and the equivalent circuit is drawn as shown in the Figure. All quantities referred to one side. The referred quantities are indicated with primes.

24 Approximate Equivalent Circuit The voltage drops I 1 R 1 and I 1 X l1 are normally small and E 1 V 1 (see Figure). Approximate Equivalent circuit simplifies computation and the winding resistances and leakage reactances can be lumped together.

25 Approximate Equivalent Circuit In a transformer, the exciting current I is a small percentage of the rated current. A further approximation can be made by removing excitation branch. Referred to side 1

26 Referred to side

27 Equivalent Circuit Parameters Referred equivalent circuit model can be used to predict the behavior of the transformer. The parameters R 1, X l1, R c1, X m1, R, X l and a (N 1 /N ) must be known so that the equivalent circuit can be used. These parameters can be easily determined by performing tests that involve little power consumption. A no-load test (open-circuit test) and short-circuit test will provide information for determining the parameters of the transformer equivalent circuit.

28 No-Load Test It is performed by applying a voltage to either primary or secondary winding side, whichever is convenient. If a 1100/110 V transformer is to be tested, the voltage would be applied to secondary winding, because a power supply of 110 V is readily available than 1100 V. Parameters R c and X m can be determined from the voltmeter, ammeter and wattmeter readings.

30 Short-Circuit Test It is performed by short-circuiting one winding and applying rated current to the other winding. If the secondary terminals are shorted, the high impedance of the shunt branch can be neglected. It is convenient to perform this test by applying a voltage to the high-voltage winding. Parameters R eq and X eq can be determined from the voltmeter, ammeter and wattmeter readings.

31 Example Tests are performed on a 1, 10 kva, 00/0 V, 60 Hz transformer. a) Derive the parameters for the approximate equivalent circuit referred to primary and secondary winding. b) Determine the power factor for the no-load and shortcircuit test.

32 a)

35 b)

36 Voltage Regulation Consider Fig..14a, where the transformer is represented by a series impedance Zeq. If a load is not applied to the transformer (i.e., open-circuit or no-load condition) the load terminal voltage is If the load switch is now closed and the load is connected to the transformer secondary, the load terminal voltage is

37 To reduce the magnitude of the voltage change, the transformer should be designed for a low value of the internal impedance Zeq. Voltage regulation is defined as the change in magnitude of the secondary voltage as the load current changes from the no-load to the loaded condition. Voltage regulation V NL V L V L For the equivalent circuit referred to the primary, Voltage regulation The load is normally taken as the rated voltage, therefore, V ' NL ' V L V ' L V ' L V ' rated

38 From the approximate equivalent circuit referred to the primary, V V I R ' ' ' 1 eq1 eq1 If the load is thrown off, V 1 will appear as V, hence, V NL ' V1 ji X So, Voltage regulation ( in percent) V 1 V V ' rated ' rated 100%

The voltage regulation depends on the power factor of the load.

The phasor diagram is drawn in Figure. The locus of V1 is a circle of radius.

That is, Therefore the maximum voltage regulation occurs: if the power factor

40 The voltage regulation depends on the power factor of the load. This can be appreciated from the phasor diagram of the voltages. The phasor diagram is drawn in Figure. The locus of V1 is a circle of radius. The magnitude of V1 will be maximum if the phasor is in phase with. That is, Therefore the maximum voltage regulation occurs: if the power factor angle of the load is the same as the transformer equivalent impedance angle and the load power factor is lagging.

41 Example 3 Consider the transformer in example. Determine the voltage regulation in percent for the following load conditions: a) 75% full load, 0.6 power factor lagging. [4.86%] b) 75% full load, 0.6 power factor leading. [-.8%]

43 The meaning of 4.86% voltage regulation is that if the load is thrown off, the load terminal voltage will rise from 0 to volts. In other words, when the 75% full load at 0.6 lagging power factor is connected to the load terminals of the transformer, the voltage drops from to 0 volts.

44 Note that the voltage regulation for this leading power factor load is negative. This means that if the load is thrown off, the load terminal voltage will decrease from 0 to volts. To put it differently, if the leading power factor load is connected to the load terminals, the voltage will increase from to 0 volts.

45 Efficiency Losses in transformers are small. Because the transformer is a static device, there are no rotational losses such as windage and friction losses in a rotating machine. In a well-designed transformer, the efficiency can be as high as 99%. The efficiency is out in The losses in the transformer are the core loss (P c ) and copper loss (P cu ). Therefore, P P P out P P P out out c Pout losses P cu P in losses P out

46 The copper losses can be determined as: The core loss (P c ) is almost constant and can be obtained fro the no-load test of a transformer, eq eq cu R I R I R I R I P cos cos cos eq c out R I P I V I V I V P

Maximum Efficiency For constant values of the terminal voltage V and load power factor angle, the maximum efficiency occurs when The condition for maximum efficiency

47 Maximum Efficiency For constant values of the terminal voltage V and load power factor angle, the maximum efficiency occurs when The condition for maximum efficiency is obtained when P I c eq That is core loss = copper loss. For full load condition, P I Maximum efficiency for any power factor: R cu, FL, FL eq R V VI cos I cos P c

48 For constant values of the terminal voltage V and load current I, the maximum efficiency occurs when If this condition is applied to efficiency equation, the condition for maximum efficiency is that is, load power factor = 1

49 Therefore, maximum efficiency in a transformer occurs when the load power factor is unity (i.e., resistive load) and load current is such that copper loss equals core loss.

50 ALL-DAY (OR ENERGY) EFFICIENCY Such transformers are called power transformers, and they are usually designed for maximum efficiency occurring near the rated output. A transformer connected to the utility that supplies power to your house and the locality is called a distribution transformer. Such transformers are connected to the power system for 4 hours a day and operate well below the rated power output for most of the time. It is therefore desirable to design a distribution transformer for maximum efficiency occurring at the average output power. A figure of merit that will be more appropriate to represent the efficiency performance of a distribution transformer is the "allday" or "energy" efficiency of the transformer.

51 This is defined as follows: If the load cycle of the transformer is known, the all-day efficiency can be determined.

52 Example 4 For the transformer in example, determine: a) Efficiency at 75% rated output and 0.6 PF. [95.3%] b) The value of maximum efficiency. At what percent of full load does this maximum efficiency occur? [97.15%], [68.% full load]

53 a)

54 b)

55 Other Method From Example 4 From Equation

56 Example 5 A 50 kva, V transformer has a core loss Pc = 00 W at rated voltage and a copper loss Pcu = 500 W at full load. It has the following load cycle. Determine the all-day efficiency of the transformer.

57 AUTOTRANSFORMER This is a special connection of the transformer from which a variable ac voltage can be obtained at the secondary. A common winding as shown is mounted on a core and the secondary is taken from a tap on the winding. In contrast to the two-winding transformer discussed earlier, the primary and secondary of an autotransformer are physically connected. Since all the turns link the same flux in the transformer core

58 If the secondary tapping is replaced by a slider, the output voltage can be varied over the range 0 < V < V1. The ampere-turns provided by the upper half (i.e., by turns between points a and b) are: The ampere-turns provided by the lower half (i.e., by turns between points b and c) are: For ampere-turn balance, from Equations the voltages and currents are related by the same turns ratio as in a two-winding transformer.

Chapter 2-1 Transformers

Principles of Electric Machines and Power Electronics Chapter 2-1 Transformers Third Edition P. C. Sen Transformer application 1: power transmission Ideal Transformer Assumptions: 1. Negligible winding