Single phase Transformer

Electrical Machines

1. Vivekananda College of Engineering for Women Department of Electrical & Electronics Engineering Electrical Machines – I [ Sub.Code - U15EE413 – Regulation 2015 ] UNIT – IV - TRANSFORMERS Present by Mr. A. Johny Renoald M.E, Ph.D

2. UNIT – IV Transformers - Principle of operation Types Basic construction Equivalent circuit Regulation and efficiency Auto transformer

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8. Transformer Equation Faraday’s Law states that, If the flux passes through a coil of wire , a voltage will be induced in the turns of wire. This voltage is directly proportional to the rate of change in the flux with respect of time. If we have N turns of wire, Lenz’s Law 8 Dr.B.GOPINATH, Prof / EEE

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28. Transformer Equation For an ac sources , Let V(t) = V m sin t i (t) = i m sin t Since the flux is a sinusoidal function; Then: Therefore: Thus: 28 Dr.B.GOPINATH, Prof / EEE

29. Transformer Equation For an ideal transformer In the equilibrium condition, both the input power will be equaled to the output power, and this condition is said to ideal condition of a transformer. From the ideal transformer circuit, note that, Hence, substitute in ( i ) ………………… (i) 29 Dr.B.GOPINATH, Prof / EEE

30. Dr.B.GOPINATH, Prof / EEE Transformer Equation Where, ‘ a ’ is the Voltage Transformation Ratio ; which will determine whether the transformer is going to be step-up or step-down E 1 > E 2 For a >1 For a <1 E 1 < E 2 30

31. Dr.B.GOPINATH, Prof / EEE Transformer Rating Transformer rating is normally written in terms of Apparent Power . Apparent power is actually the product of its rated current and rated voltage . Where, I 1 and I 2 = rated current on primary and secondary winding. V 1 and V 2 = rated voltage on primary and secondary winding. Rated currents are actually the full load currents in transformer 31

32. Conservator Tank of a Transformer This is a cylindrical tank mounted on supporting structure on the roof of the transformer's main tank. When transformer is loaded, the temperature of oil increases and consequently the volume of oil in the transformer gets increased. Again; when ambient temperature is increased, the volume of oil is also increased. The conservator tank of a transformer provides adequate space for expansion of oil. Conservator tank of transformer also acts as a reservoir of oil. 32 Dr.B.GOPINATH, Prof / EEE

33. Silica Gel Breather of Transformer Whenever electrical power transformer is loaded, the temperature of the transformer insulating oil increases, consequently the volume of the oil is increased. The color of silica gel crystal is dark blue but, when it absorbs moisture; it becomes pink. 33 Dr.B.GOPINATH, Prof / EEE

34. Explosion Vent of Transformer The purpose of the explosion vent in a transformer is to prevent damage of the transformer tank be releasing any excessive pressure generated inside the transformer. 34 Dr.B.GOPINATH, Prof / EEE

35. Construction of Buchholz Relay Buchholz relay in transformer is an oil container housed the connecting pipe from main tank to conservator tank . It has mainly two elements. The upper element consists of a float. The float is attached to a hinge in such a way that it can move up and down depending upon the oil level in the Buchholz relay Container. One mercury switch is fixed on the float. The alignment of mercury switch hence depends upon the position of the float. The lower element consists of a baffle plate and mercury switch. This plate is fitted on a hinge just in front of the inlet (main tank side) of Buchholz relay in transformer in such a way that when oil enters in the relay from that inlet in high pressure the alignment of the baffle plate along with the mercury switch attached to it, will change. 35 Dr.B.GOPINATH, Prof / EEE

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38. Practical Transformer (Equivalent Circuit ) V 1 = primary supply voltage V 2 = 2 nd terminal (load) voltage E 1 = primary winding voltage E 2 = 2 nd winding voltage I 1 = primary supply current I 2 = 2 nd winding current I 1 ’ = primary winding current I o = no load current I c = core current I m = magnetism current R 1 = primary winding resistance R 2 = 2 nd winding resistance X 1 = primary winding leakage reactance X 2 = 2 nd winding leakage reactance R c = core resistance X m = magnetism reactance V 1 I 1 R 1 X 1 R C I c X m I m I o E 1 E 2 V 2 I 1 ’ N 1 : N 2 R 2 X 2 Load I 2 38 Dr.B.GOPINATH, Prof / EEE

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40. R 2 ’ = R 2 / K 2 X 2 ’ = X 2 / K 2 V 2 ’ = V 2 / K Z L = Z L / K 2 I 2 ’ = I 2 K 40 Dr.B.GOPINATH, Prof / EEE

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42. Single Phase Transformer (Referred to Primary) Actual Method V 1 I 1 R 1 X 1 R C I c X m I m I o E 1 E 2 V 2 I 2 ’ N 1 : N 2 R 2 ’ X 2 ’ Load I 2 42 Dr.B.GOPINATH, Prof / EEE

43. Single Phase Transformer (Referred to Primary ) Approximate Method V 1 I 1 R 1 X 1 R C I c X m I m I o E 1 E 2 V 2 I 2 ’ N 1 : N 2 R 2 ’ X 2 ’ Load I 2 43 Dr.B.GOPINATH, Prof / EEE

44. Single Phase Transformer (Referred to Primary ) Approximate Method V 1 I 1 R 01 X 01 aV 2 In some application, the excitation branch has a small current compared to load current, thus it may be neglected without causing serious error. 44 Dr.B.GOPINATH, Prof / EEE

45. Single Phase Transformer (Referred to Secondary ) Actual Method I 1 ’ R 1 ’ X 1 ’ R C ’ I c X m ’ I m I o I 2 R 2 X 2 V 2 45 Dr.B.GOPINATH, Prof / EEE

46. Single Phase Transformer (Referred to Secondary ) Approximate Method I 1 ’ R 02 X 02 V 2 Neglect the excitation branch 46 Dr.B.GOPINATH, Prof / EEE

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49. Transformer Losses Generally, there are two types of losses; Iron losses :- occur in core parameters Copper losses :- occur in winding resistance Iron Losses Copper Losses P oc and P sc will be discusses later in transformer test 49 Dr.B.GOPINATH, Prof / EEE

50. Transformer Efficiency To check the performance of the device, by comparing the output with respect to the input. The higher the efficiency, the better the system. Where, if ½ load, hence n = ½ , ¼ load , hence n = ¼ , 90% of full load, n =0.9 Where P cu = P sc P c = P oc 50 Dr.B.GOPINATH, Prof / EEE

51. Power Factor Power factor = angle between Current and voltage , cos  V I   = -ve V I   = +ve V I  = 1 Lagging Leading unity Power factor always lagging for real transformer. 51 Dr.B.GOPINATH, Prof / EEE

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54. Dr.B.GOPINATH, Prof / EEE Voltage Regulation The purpose of voltage regulation is basically to determine the percentage of voltage drop between no load and full load. Voltage Regulation can be determine based on 3 methods: Basic Definition Short – circuit Test Equivalent Circuit 54

55. Dr.B.GOPINATH, Prof / EEE Voltage Regulation (Basic Definition) In this method, all parameter are being referred to primary or secondary side. Can be represented in either Down – voltage Regulation Up – Voltage Regulation 55

56. Voltage Regulation (Short – circuit Test) In this method, direct formula can be used. If referred to primary side If referred to secondary side Note that: ‘–’ is for Lagging power factor ‘+’ is for Leading power factor I sc must equal to I FL 56 Dr.B.GOPINATH, Prof / EEE

57. Voltage Regulation (Equivalent Circuit ) In this method, the parameters must be referred to primary or secondary If referred to primary side If referred to secondary side Note that: ‘+’ is for Lagging power factor ‘–’ is for Leading power factor j terms ~0 57 Dr.B.GOPINATH, Prof / EEE

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61. S.no I OC V OC W OC 1. 0.7 230 40 61 Dr.B.GOPINATH, Prof / EEE

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67. S.NO I SC V SC W SC 1. 4.5 15 90 P = VI ; 1KVA, 220/110V ; Ip = 1000/220 = 4.5 Amps Is = 1000 / 110 = 0.09 Amps 67 Dr.B.GOPINATH, Prof / EEE

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75. Example 1 : A 5 KVA, 500/250 V, 50 Hz, single phase transformer gave the following readings, O.C. Test : 500 V, 1 A, 50 W (L.V. side open) S.C. Test : 25 V, 10 A, 60 W (L.V. side shorted) Determine : i ) The efficiency on full load, 0.8 lagging p.f . ii) The voltage regulation on full load, 0.8 leading p.f . iii) The efficiency on 60% of full load, 0.8 leading p.f . iv) Draw the equivalent circuit referred to primary and insert all the values in it. Solution : In both the tests, meters are on H.V. side which is primary of the transformer. Hence the parameters obtained from test results will be referred to primary. From O.C. test,      V o = 500 V,    I o = 1 A,     W o = 50 W . . .                         Cos Φ o = W oc /V oc I oc = 50/(500x1) = 0.1 . . .                          I w = I oc cos = 1 x 0.1 = 0.1 A and                       I m = I o sin Φ o = 1 x 0.9949 = 0.9949 A . . .                           R o =V o / I c = 500/0.1 = 5000 Ω and                        X o = V o / I m = 500/0.9949 = 502.52 Ω and    W o = P i = iron losses = 50 W 75 Dr.B.GOPINATH , Prof / EEE

76. From S.C. test,      V sc = 25 V, I sc = 10 A, W sc = 60 W . . .                            R sc = W sc /I sc 2 = 60/(10) 2 = 0.6 Ω                                Z sc = V sc / I sc = 25/10 = 2.5 Ω . . .                           X sc = √(2.5 2 - 0.6 2 ) = 2.4269 Ω                            ( I 1 ) F.L. = VA rating/V 1                                          = (5 x 10 3 )/500 = 10 A and                           I sc = (I 1 ) F.L. . . .                             W sc = ( P cu ) F.L. = 60 W i ) η  on full load, cos = 0.8 lagging 76 Dr.B.GOPINATH, Prof / EEE

77. I ii) For 60% of full load, n = 0.6 and cos Φ 2 = 0.8 leading] . . .                   P cu = copper loss on new load = n 2 x ( P cu ) F.L.                              = (0.6) 2 x 60 = 21.6 W                              = 97.103 % iv) The equivalent circuit referred to primary is shown in the Fig. 77 Dr.B.GOPINATH, Prof / EEE ii) Regulation on full load, cos Φ 2 = 0.8 leading = - 1.95 %

78. Example 2 : The open circuit and short circuit tests on a 10 KVA, 125/250 V, 50 Hz, single phase transformer gave the following results :  O.C. test : 125 V, 0.6 A, 50 W (on L.V. side) S.C. test : 15 V, 30 A. 100 W (on H.V. side) Calculate : i ) copper loss on full load ii) full load efficiency at 0.8 leading p.f . iii) half load efficiency at 0.8 leading p.f . iv) regulation at full load, 0.9 leading p.f . Solution : From O.C. test we can weite ,                       W o = P i = 50 W = Iron loss        From S.C. test we can find the parameters of equivalent circuit. Now S.C. test is conducted on H.V. side i.e. meters are on H.V. side which is transformer secondary. Hence parameters from S.C. test results will be referred to secondary. 78 Dr.B.GOPINATH, Prof / EEE

79.   V sc = 15 V, I sc = 30 A, W sc = 100 W . . .             R sc = W sc /( I sc ) 2 =10/(30) 2 = 0.111 Ω                Z SC = V sc / I sc = 15/30 = 0.5 Ω . . .            X SC = √(Z sc 2 - R SC 2 ) = 0.4875 Ω i ) Copper loss on full load                  (I 2 ) F.L. = VA rating/V 2 = (10 x 10 3 )/250 = 40 A        In short circuit test, I sc = 30 A and not equal to full load value 40 A.        Hence W sc does not give copper loss on full load . . .              W sc = P cu   at 30 A = 100 W Now          P cu α I 2                  ( P cu at 30 A)/( P cu at 40 A) = (30/40) 2                  100/( P cu at 40 A) = 900/1600                   P cu at 40 A = 177.78 W . . .                ( P cu ) F.L. = 177.78 W 79 Dr.B.GOPINATH, Prof / EEE

80. ii) Full load η , cos Φ 2 = 0.8 iii) Half load η , cos Φ 2 = 0.8                   n = 0.5 as half load,            (I 2 ) H.L. = 0.5 x 40 = 20 = 97.69% 80 Dr.B.GOPINATH, Prof / EEE

81. iv) Regulation at full load, cos Φ = 0.9 leading = -1.8015% 81 Dr.B.GOPINATH, Prof / EEE

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