Nizwa College of Technology Electrical Power TechnologyChapter -2

CHAPTER 2TRANSFORMERS2.1 What is a Transformer?A device that transfer energy from one system to another. It can accept energy at one voltage and deliver it at another voltage. It permits electrical energy to be generated at relatively low voltages and transmitted at high voltages and low currents, thus reducing line losses, and to be used at safe voltages. Transformer is a device which uses the phenomenon of Mutual Induction to change the values of alternating voltages and currents. Main advantages of AC transmission and distribution ease alternating voltage can be increased or decreased by transformer. Losses are low & efficiency is high.Being static they have a long life and very stable . The range in size from the miniature units used in electronics applications & to large power size in power station. Principle of operation is the same for each. 2.2 TRANSFORMER CONSTRUCTION(i) CoreTransformer is consists of two electrical circuits linked by a common ferromagnetic core as shown in figure 2.1 Broadly classifying, the core construction can be separated into core type and shell type. In a core type construction the winding surrounds the core. In a shell type on the other hand the iron surrounds the winding.

(a) Transformer coil and core (b) Circuit symbolFigure 2.1 Transformer construction

Figure 2.2 Transformer Core(ii) Windings In a two winding transformer two windings would be present. The one which is connected to a voltage source and creates the flux is called as a primary winding. The second winding where the voltage is induced by induction is called a secondary winding. If the secondary voltage is less than that of the primary the transformer is called a step down transformer. If the secondary voltage is more then it is a step up transformer. (iii) InsulationThe insulation used in the case of electrical conductors in a transformer is varnish or enamel in dry type of transformers. In larger transformers to improve the heat transfercharacteristics the conductors are insulated using un-impregnated paper or cloth and the whole core-winding assembly is immersed in a tank containing transformer oil. The transformer oil thus has dual role. It is an insulator and also a coolant. Cooling is achieved by air in small transformers and oil in large transformers.

2.3 Principle of OperationWhen the secondary is open circuited and an alternating voltage V1 is applied to the primary winding, a small current called no load current Io, flows, sets up a magnetic flux in the core.This alternating flux links with both primary and secondary coils and induces EMFs in them of E1 and E2 respectively by mutual induction. Induced EMF E in a coil of N turns is given by voltswhere (d/dt) - rate of change of flux. In an ideal transformer , the rate of change of flux is the same for both primary and secondary and thus

(ie), induced EMF per turn is constant . Assuming no losses, E1= V1 & E2=V2 Hence

Where, V1/ V2 - Voltage ratio N1/N2 - Turns ratio or transformation ratioif N2 is less than N1 then V2 is less than V1 and then the device is termed as step down transformer. If N2 is greater than N1 then V2 is greater than V1 and then the device is termed as step up transformer.when a load is connected across the secondary winding I2 flows. In an ideal transformer losses are neglected and transformer is considered to be a 100 percent efficient. Hence Input power = Output power or V1I1= V2I2 In an ideal transformer, primary and secondary ampere turns are equal . Thus

Combining equations,

Rating of a transformer is stated in terms of the volt- amperes that it can transform without overheating. The transformer rating is either V1I1 or V2I2, where I2 is the full load secondary current.

2.4 EMF EQUATION OF A TRANSFORMERThe magnetic flux set up in the core of a transformer when an alternating voltage is applied to its primary winding is also alternating and is sinusoidal. Let m be the maximum value of the flux and f be the frequency of the supply. The time for 1 cycle of the alternating flux is the periodic time T, where T =(1/f ) seconds. The flux rises sinusoidally from zero to its maximum value in (1/4) cycle, and the time for (1/4) cycle is (1/4f ) seconds. Hence,the average rate of change of flux = = 4 f m Wb/s, and Since 1 Wb/s=1 volt, The average e.m.f. induced in each turn = 4 fm volts. As the flux varies sinusoidally, then a sinusoidal e.m.f. will be induced in each turn ofboth primary and secondary windings. For a sine wave, r.m.s. valueForm factor = ___________________ = 1.11 average value

Hence r.m.s. value = form factor X average value = 1.11 X average value Thus r.m.s. e.m.f. induced in each turn = 1.11 X 4 f m volts = 4.44 f m voltsTherefore, r.m.s. value of e.m.f. induced in primary,E1 = 4.44 f m N1 volts -------- (1)and r.m.s. value of e.m.f. induced in secondary,E2 = 4.44 f m N2 volts -------- (2)Dividing the equations (1) and (2) gives:

2.5 Transformer No-load Phasor diagramCore flux is common to both primary and secondary windings in a transformer and is thus taken as the reference phasor diagram. On no load , primary winding takes a small current called no load current (Io). By neglecting the losses, then primary winding are pure inductor, this current lags the applied voltage V1 by 90. For no loss transformer, shown in figure 2.3(a) Io produces the flux and is in phase with the flux.

Figure 2.3 Transformer on no-load phasor diagramPrimary induced EMF in opposition to V1 (By Lenzs Law) and is shown 180 out of phase with V1 and equal in magnitude. Secondary EMF is shown for 2:1 turns ratio transformer. No load phasor diagram for a practical transformer is shown in fig(b). When losses are considered then no-load current (Io) is the phasor sum of two components:(i) Im- magnetizing component ( in phase with flux)(ii) Ic - core loss component (supplying hysteresis & eddy current loss ) 90 out of phase with flux.No-load current, I0 = IM2+IC2

whereIM =I0 sin 0 and IC =I0 cos 0.Power factor on no-load= cos 0 = (IC/I0).The total core losses (i.e. iron losses)= V1I0 cos 0

2.6 TRANSFORMER ON-LOAD PHASOR DIAGRAMIf the voltage drop in the windings of a transformer is assumed negligible, then the terminal voltage V2 is the same as the induced e.m.f. E2 in the secondary. Similarly, V1 =E1. Assuming an equal number of turns on primary and secondary windings, then E1 =E2, andlet the load have a lagging phase angle 2. In the phasor diagram of Fig. 2.4 , current I2 lags V2 by angle 2. When a load is connected across the secondary winding, a current I2 flows in the secondary winding. The resulting secondary e.m.f. acts so as to tend to reduce the core flux. However this does not happen since reduction of the core flux reduces E1, hence a reflected increase in primary current I1 occurs which provides a restoring m.m.f. Hence at all loads, primary and secondary m.m.f.s are equal, but in opposition, and the core flux remains constant. I1 is sometimes called the balancing current and is equal, but in the opposite direction, to current I2 as shown in Fig. 2.4. I0, shown at a phase angle 0 to V1, is the no-load current of the transformer. The phasor sum of I1 and I0 gives the supply current I1 and the phase angle between V1 and I1 is shown as 1.

Figure 2.4 Transformer on load phasor diagram

2.7 REGULATION OF A TRANSFORMERWhen the secondary of a transformer is loaded, the secondary terminal voltage, V2, falls. As the power factor decreases, this voltage drop increases. This is called the regulation of the transformer and it is usually expressed as a percentage of the secondary no-load voltage, E2. For full-load conditions:

Regulation = X 100%The fall in voltage, (E2 V2), is caused by the resistance and reactance of the windings. Typical values of voltage regulation are about 3% in small transformers and about 1% in large transformers.

2.8 TRANSFORMER LOSSES AND EFFICIENCY

There are broadly two sources of losses in transformers on load, these being copper losses and iron losses. (a) Copper losses are variable and result in a heating of the conductors, due to the fact that they possess resistance. If R1 and R2 are the primary and secondary winding resistances then theTotal copper loss = I12R1 +I22R2(b) Iron losses are constant for a given value of frequency and flux density and are of two types hysteresis loss and eddy current loss.(i) Hysteresis loss is the heating of the core as a result of the internal molecular structure reversals which occur as the magnetic flux alternates. The loss is proportional to the area of the hysteresis loop and thus low loss nickel iron alloys are used for the core since their hysteresis loops have small areas.(ii) Eddy current loss is the heating of the core due to e.m.f.s being induced not only in the transformer windings but also in the core. These induced e.m.f.s set up circulating currents, called eddy currents. Owing to the low resistance of the core, eddy currents can be quite considerable and can cause a large power loss and excessive heating of the core. Eddy current losses can be reduced by increasing the resistivity of the core material or, more usually, by laminating the core (i.e. splitting it into layers or leaves) when very thin layers of insulating material can be inserted between each pair of laminations. This increases the resistance of the eddy currentpath, and reduces the value of the eddy current.

Transformer efficiency,

Transformer efficiency is usually expressed as a percentage. It is not uncommon for power transformers to have efficiencies of between 95% and 98%Output power= V2I0 cos2Total losses=copper loss + iron losses, andinput power=output power + lossesMaximum efficiencyIt may be shown that the efficiency of a transformer is a maximum when the variable copper loss (i.e. I12R1 +I22R2) is equal to the constant iron losses.

2.9 TRANSFORMER TEST(a) OPEN CIRCUIT TESTThe purpose of this test is to determine no-load losses or core losses and no load current I0. It is helpful in finding X0 and R0. To carry out open circuit test it is the LV side of the transformer where rated voltage at rated frequency is applied and HV side is left opened as shown in the circuit diagram The readings on voltmeter, ammeter and the wattmeter which are connected to the primary of the transformer are taken. As there is no load in secondary, a small current called no-load current I0 flows in primary. Since I0 is very small copper loss I2R is negligible. Hence wattmeter gives only core loss under no load condition.

Figure 2.5 Open circuit testNo load power factor cos 0 = W / VI Magnetizing current Im = I0 sin 0 Core loss component Ic = I0 cos 0 Reactance X0 = V1/ ImResistance R0 = V1/ Ic

(b) Short circuit testEquivalent impedance (Z01 0r Z02), leakage reactance (X01 or X02)and total reactance (R01 or R02)of the transformer can be determined using the short circuit test This test is used to find the copper loss at full load. This loss is used to calculate the efficiency of transformer. Regulation of the transformer can also be determined using this test.

Figure 2.5 Short circuit testPrimary winding is connected to supply. Secondary winding is solidly short circuited. Voltage in the primary winding is slowly increased till the full load currents are flowing in primary and secondary winding. As very small voltage is applied core losses are very small. The wattmeter gives the full load copper loss I2R for the whole transformer ie both primary Cu loss and secondary Cu loss.If Vsc is the voltage required to circulate the rated load current, thenZ01 = Vsc / I1Ro1 = W / I12X01 = (Z012 R012)2.10 THREE PHASE TRANSFORMERThree-phase double-wound transformers are mainly used in power transmission and are usually of the core type. They basically consist of three pairs of single phase windings mounted on one core, as shown in Fig. 21.17, which gives a considerable saving in the amount of iron used. The primary and secondary windings in Fig. 21.17 are wound on top of each other in the form of concentric cylinders, similar to that shown in Fig. 21.6(a). The windings may be with the primary delta-connected and the secondary star-connected, or star-delta, star-star or delta-delta, depending on its use.

Figure 2.6 Three phase transformer

Figure 2.7 Three phase transformer connection types

2.11 TYPES OF TRANSFORMER1. Auto TransformerAn auto transformer is a transformer which has part of its winding common to the primary and secondary circuits. Fig. 21.14(a) shows the circuit for a double wound transformer and Fig. 21.14(b) that for an auto transformer. The latter shows that the secondary is actually part of the primary, the current in the secondary being (I2 I1). Since the current is less in this section, the cross-sectional area of the winding can be reduced, which reduces the amount of material necessary. Figure 21.15 shows the circuit diagram symbol for an auto transformer.

Figure 2.8 Auto transformerAdvantages of auto transformersThe advantages of auto transformers over double wound transformers include:1. saving in cost since less copper is needed 2. less volume, hence less weight3. higher efficiency, resulting from lower I2R losses4. continuously variable output voltage is achievable if a sliding contact is used5. smaller percentage voltage regulation.Disadvantages of auto transformersThe primary and secondary windings are not electrically separate, hence if an open-circuit occurs in the secondary winding the full primary voltage appears across the secondary.Uses of auto transformersAuto transformers are used for reducing the voltage when starting induction motors and for interconnecting systems that are operating at approximately the same voltage.

2. ISOLATING TRANSFORMERSTransformers not only enable current or voltage to be transformed to some different magnitude but provide a means of isolating electrically one part of a circuit from another when there is no electrical connection between primary and secondary windings. An isolating transformer is a 1:1 ratio transformer with several important applications, including bathroom shaver-sockets, portable electric tools, model railways, and so on.

PROBLEMS ON TRANSFORMER RATIO1. A transformer has 500 primary turns and 3000 secondary turns. If the primary voltage is 240V, the secondary voltage, assuming an ideal transformer ?PROBLEMS ON TRANSFOMER ON LOAD2. A transformer takes a current of 0.8A, when its primary is connected to a 240V, 50Hz supply, the secondary being open circuit. If the power absorbed is 72 watts, determine (a) the iron loss current, (b) the power factor on no-load, and (c) the magnetizing current.PROBLEMS ON EMF EQUATION3. A 100 KVA, 4000V / 200V, 50 Hz single phase transformer has 100 secondary turns. Determine (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum value of the flux.4. A single-phase, 50 Hz transformer has 25 primary turns and 300 secondary turns. The cross-sectional area of the core is 300 cm2. When the primary winding is connected to a 250V supply, determine (a) the maximum value of the flux density in the core, and (b) the voltage induced in the secondary winding.5. A 60 kVA, 1600V/100V, 50 Hz, single-phase transformer has 50 secondary windings. Calculate (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum value of the flux. [(a) 37.5A, 600A (b) 800 (c) 9.0 mWb]6. A single-phase, 50 Hz transformer has 40 primary turns and 520 secondary turns. The cross-sectional area of the core is 270 cm2. When the primary winding is connected to a 300 volt supply, determine (a) the maximum value of flux density in the core, and (b) the voltage induced in the secondary winding. [(a) 1.25T (b) 3.90 kV]7. A single-phase 800V/100V, 50 Hz transformer has a maximum core flux density of 1.294T and an effective cross-sectional area of 60 cm2. Calculate the number of turns on the primary and secondary windings. [464, 58]8. A 3.3 kV/110V, 50 Hz, single-phase transformer is to have an approximate e.m.f. per turn of 22V and operate with a maximum flux of 1.25T. Calculate (a) the number of primary and secondary turns, and (b) the cross-sectional area of the core. [(a) 150, 5 (b) 792.8 cm2]PROBLEMS ON REGULATION9. The open circuit voltage of a transformer is 240V. A tap changing device is set to operate when the percentage regulation drops below 2.5%. Determine the load voltage at which the mechanism operates.

PROBLEMS ON EFFICIENCY

10. In a 25KVA, 2000 / 200 V power transformer the iron and full load copper losses are 350W and 400W respectively. Calculate the efficiency at unity power factor at (i) full load and (ii) half load.11. A 220 / 400V, 10KVA, 50Hz single phase transformer has at full load, a copper loss of 120W. If it has an efficiency of 98% at full load, unity power factor, determine the iron losses. What would be the efficiency of the transformer at half full load at 0.8 power factor lagging?12. The efficiency of a 400 KVA, single phase transformer is 98.77% when delivering full load at 0.8 power factor and 99.13% at half load and unity power factor. Calculate (i) the iron loss (ii) the full load copper loss.

Short answer questions on Transformer1. What is a transformer?2. Explain briefly how a voltage is induced in the secondary winding of a transformer3. Draw the circuit diagram symbol for a transformer4. State the relationship between turns and voltage ratios for a transformer5. How is a transformer rated?6. Briefly describe the principle of operation of a transformer7. Draw a phasor diagram for an ideal transformer on no-load8. State the e.m.f. equation for a transformer9. Draw an on-load phasor diagram for an ideal transformer with an inductive load10. Name two types of transformer construction11. What core material is normally used for power transformers12. Name three core materials used in r.f. transformers13. State a typical application for (a) a.f. transformers (b) r.f. transformers14. How is cooling achieved in transformers?15. State the expressions for equivalent resistance and reactance of a transformer, referred to the primary 16. Define regulation of a transformer17. Name two sources of loss in a transformer18. What is hysteresis loss? How is it minimized in a transformer?19. What are eddy currents? How may they be reduced in transformers?20. How is efficiency of a transformer calculated?21. What is the condition for maximum efficiency of a transformer?22. What does resistance matching mean?23. State a practical application where matching would be used24. Derive a formula for the equivalent resistance of a transformer having a turns ratioof N1:N2 and load resistance RL25. What is an auto transformer?26. State three advantages and one disadvantage of an auto transformer compared witha double-wound transformer27. In what applications are auto transformers used?28. What is an isolating transformer? Give two applications29. Describe briefly the construction of a threephase transformer30. For what reason are current transformers used?31. Describe how a current transformer operates32. For what reason are voltage transformers used?33. Describe how a voltage transformer operates

Multi-choice questions on transformers1. The e.m.f. equation of a transformer of secondary turns N2, magnetic flux density Bm, magnetic area of core a, and operating at frequency f is given by:(a) E2 =4.44N2Bm a f volts (b) E2 =4.44(N2Bmf)/a volts(c) E2 =(N2Bmf) / a volts(d) E2 =1.11N2Bm a f volts2. A step-up transformer has a turns ratio of 10. If the output current is 5A, the inputcurrent is:(a) 50A (b) 5A(c) 2.5A (d) 0.5A3. A440V/110Vtransformer has 1000 turns on the primary winding. The number of turns on the secondary is:(a) 550(b) 250(c) 4000 (d) 254. An advantage of an auto-transformer is that:(a) it gives a high step-up ratio(b) iron losses are reduced(c) copper loss is reduced(d) it reduces capacitance between turns5. A 1 kV/250V transformer has 500 turns on the secondary winding. The number of turnson the primary is:(a) 2000 (b) 125(c) 1000 (d) 2506. The core of a transformer is laminated to:(a) limit hysteresis loss(b) reduce the inductance of the windings(c) reduce the effects of eddy current loss(d) prevent eddy currents from occurring7. The power input to a mains transformer is 200W. If the primary current is 2.5A, the secondary voltage is 2V and assuming no losses in the transformer, the turns ratio is:(a) 40:1 step down (b) 40:1 step up(c) 80:1 step down (d) 80:1 step up8. A transformer has 800 primary turns and 100 secondary turns. To obtain 40V from the secondary winding the voltage applied to the primary winding must be:(a) 5V (b) 320V(c) 2.5V (d) 20V9. A 100 kVA, 250V/10 kV, single-phase transformer has a full-load copper loss of 800Wand an iron loss of 500W. The primary winding contains 120 turns. For the statements in questions 10 to 16, select the correct answer from the following list:

(a) 81.3kW (b) 800W (c) 97.32% (d) 80kW (e) 3 (f) 4800(g) 1.3kW (h) 98.40% (i) 100kW(j) 98.28% (k) 200W (l) 101.3kW(m) 96.38% (n) 400W10. The total full-load losses11. The full-load output power at 0.8 power factor12. The full-load input power at 0.8 power factor13. The full-load efficiency at 0.8 power factor14. The half full-load copper loss15. The transformer efficiency at half full-load, 0.8 power factor16. The number of secondary winding turns17. Which of the following statements is false?(a) In an ideal transformer, the volts per turn are constant for a given value of primaryvoltage(b) In a single-phase transformer, the hysteresis loss is proportional to frequency(c) A transformer whose secondary current is greater than the primary current is astep-up transformer(d) In transformers, eddy current loss is reduced by laminating the core18. An ideal transformer has a turns ratio of 1:5 and is supplied at 200V when the primarycurrent is 3 A. Which of the following statements is false?(a) The turns ratio indicates a step-up transformer(b) The secondary voltage is 40V(c) The secondary current is 15A(d) The transformer rating is 0.6 kVA(e) The secondary voltage is 1 kV(f) The secondary current is 0.6A19. Iron losses in a transformer are due to:(a) eddy currents only(b) flux leakage(c) both eddy current and hysteresis losses(d) the resistance of the primary and Secondary windings20. A load is to be matched to an amplifier having an effective internal resistance of 10ohm via a coupling transformer having a turns ratio of 1:10. The value of the load resistance for maximum power transfer is:(a) 100 ohm (b) 1 kohm(c) 100m ohm (d) 1m ohm

Further problems on the transformer principle of operation1. A transformer has 600 primary turns connectedto a 1.5 kV supply. Determine the number of secondary turns for a 240V output voltage, assuming no losses. [96]2. An ideal transformer with a turns ratio of 2:9 is fed from a 220V supply. Determine its output voltage. [990V]3. A transformer has 800 primary turns and 2000 secondary turns. If the primary voltageis 160V, determine the secondary voltage assuming an ideal transformer. [400V]4. An ideal transformer with a turns ratio of 3:8 has an output voltage of 640V. Determine its input voltage. [240V]5. An ideal transformer has a turns ratio of 12:1 and is supplied at 192V. Calculate thesecondary voltage. [16V]6. A transformer primary winding connected across a 415V supply has 750 turns. Determine how many turns must be wound on the secondary side if an output of 1.66 kV is required. [3000 turns]7. An ideal transformer has a turns ratio of 15:1 and is supplied at 180V when the primary current is 4A. Calculate the secondary voltage and current. [12V, 60A]8. A step-down transformer having a turns ratio of 20:1 has a primary voltage of 4 kV and a load of 10kW. Neglecting losses, calculate the value of the secondary current. [50A]9. A transformer has a primary to secondary turns ratio of 1:15. Calculate the primaryvoltage necessary to supply a 240V load. If the load current is 3A determine the primary current. Neglect any losses. [16V, 45A]10. A 10 kVA, single-phase transformer has a turns ratio of 12:1 and is supplied from a2.4 kV supply. Neglecting losses, determine (a) the full load secondary current, (b) theminimum value of load resistance which can be connected across the secondary winding without the kVA rating being exceeded, and (c) the primary current.[(a) 50A (b) 4 (c) 4.17A]11. A 20 resistance is connected across the secondary winding of a single-phase power transformer whose secondary voltage is 150V. Calculate the primary voltage andthe turns ratio if the supply current is 5A, neglecting losses. [225V, 3:2]

Further problems on the no-load phasor diagram1. A 500V/100V, single-phase transformer takes a full load primary current of 4A. Neglecting losses, determine (a) the full load secondary current, and (b) the rating of the transformer. [(a) 20A (b) 2 kVA]2. A 3300V/440V, single-phase transformer takes a no-load current of 0.8A and the iron loss is 500W. Draw the no-load phasor diagram and determine the values of the magnetizing and core loss components of the no-load current. [0.786A, 0.152A]3. A transformer takes a current of 1A when its primary is connected to a 300V, 50 Hz supply, the secondary being on open-circuit. If the power absorbed is 120 watts, calculate (a) the iron loss current, (b) the power factor on no-load, and (c) the magnetising current.[(a) 0.40A (b) 0.40 (c) 0.917A]

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