Revised October 6, EEL , H. Zmuda. 3. Transformers 1

Transcription

1 Transformers Revised October 6, Transformers 1

2 The Ideal Transformer N N i rimary econdary Winding Winding Basic Transformer 3. Transformers

3 The Ideal Transformer - A primary current produces a flux in the core. A N i t Ni e t i t N N rimary Winding econdary Winding 3. Transformers 3

4 Recall lide et 1: i t Applied flux + e induced d N dt d t dt 0 _ Induced (opposing) flux 3. Transformers 4

5 Ideal Transformer The primary current gives rise to an induced voltage across the primary winding. + i t A 1 N i t N i t e Note: Open v t N N rimary Winding d N einduced t v t N dt e dt di t Circuit econdary Winding 3. Transformers 5

6 elf-inductance of rimary the output (secondary) is open-circuited similarly, di d N v t N dt e dt di t N L, L c N dt e d N di t v t N dt e dt t di t N L, L c N dt e 3. Transformers 6

7 Ideal Transformer the flux produced by the primary winding is coupled to the secondary winding and induces a voltage across the secondary. i t v + t A 1 N i t N i t e N N v t N N di t d p e induced t v t N dt e dt 3. Transformers 7

8 Mutual Inductance N N di t d p v t N dt e dt di N t N, p M M dt e N N L L M L L e e, 3. Transformers 8

9 Ideal Transformer N di t v t dt N p L v t e N N di t N L e dt p v t v t N N p Turns Ratio 3. Transformers 9

10 Ideal Transformer A load across the secondary winding will cause a current to flow that opposes the change in flux of the primary. i t v + t A N i t N Ni e N t i t v t N N di t d p e induced t v t N dt e dt 3. Transformers 10

11 Ideal Transformer The flux produced by the secondary current will coupled back to the primary winding. i t i t v t + N N v t A N i t Ni e t 3. Transformers 11

12 Ideal Transformer The flux produced by the secondary will be coupled back to the primary winding inducing a voltage on the primary. A 1 N i t N i t e i t i t v t N N v t + di d NN einduced t N dt e dt t 3. Transformers 1

13 The relationship between the primary and secondary currents can be determined from the total mmf: i t i t v t + N N v t mmf mmf Ni N i t t e e 3. Transformers 13

14 For the Ideal Transformer: ttl total mmf : Ni t Ni t e e 0 i t N i t N total power :,Re call : v t N v t N i t v t N N 1, p i t v t N N p 0 3. Transformers 14

15 v i t i t t N Ideal Transformer N v t v t N di t N di N t e dt e dt imilarly, v t N N di t N di t dt e dt e 3. Transformers 15

16 Ideal Transformer v i t i t t N Transformer N v t BLACK BOX 3. Transformers 16

17 Ideal Transformer i t i t v t N Transformer N v t 3. Transformers 17

18 Ideal Transformer i t i t v t N Transformer N v t 3. Transformers 18

19 v Ideal Transformer i t i t t N N v t Dot Convention: When the current on the primary/secondary enters the dotted terminal, the polarity of the voltage on the secondary/primary is positive. 3. Transformers 19

20 v Ideal Transformer i t i t t N N i t i t v t v t v t Circuit Representation 3. Transformers 0

21 hasor Representation V jli jmi V j MI j L I 1 N N N N L L M e e e I I 1,, V V V 3. Transformers 1

22 Impedance Transformation Z in V I I I V V V N I N,, V V N I N Z in I Z I L N N V ZLI V N N N I N N I I N N N Z L V Z LI Z L 3. Transformers

23 Review EEL 3111 as needed for circuit analysis techniques with transformers. (Chapter 13 in Alexander and adiku, Fundamentals of Electric Circuits) 3. Transformers 3

24 Real (Non-Ideal) Transformers 3. Transformers 4

25 Revisit Flux Linkage Note et 1, lide 64: Flux Linkage: d d e induced N dt dt Flux Linkage: N But each turn does not necessarily link the same amount of flux. 3. Transformers 5

26 Flux Linkage Taken from: J.D. Kraus & D.A. Fleisch, Electromagnetics with Applications, Fifth Edition, McGraw-Hill, 1999, page Transformers 6

27 Flux Linkage d d e induced t N dt dt e 1 N induced einduced t dt t dt Average flux per turn Hence we really mean: d e induced t N dt How does this flux effect the secondary? 3. Transformers 7

28 Flux Linkage When a voltage is applied across the primary of a transformer the average flux in the winding will be: 1 N v t dt How does this flux effect the secondary? 3. Transformers 8

29 Not all the primary flux is coupled to the secondary I I Leakage Flux (ideally small) Leakage Flux (ideally small) Mutual Flux Mutual M L Leakage rimary 3. Transformers 9

30 imilarly, Leakage Flux (ideally small) Mutual Flux I I Leakage Flux (ideally small) Mutual M L Leakage econdary 3. Transformers 30

31 Re-express Faraday s Law as: d v t N dt d M N N dt e t e t d v t N dt il l imilarly: L d dt L d M N N dt e t e t L d dt L 3. Transformers 31

32 The primary voltage due to the mutual flux is: d e t N dt The secondary voltage due to the mutual flux is: Hence: M d e t N dt e t e t e t v t M N dt N e t N v t d N M 3. Transformers 3

33 Magnetization Current in a Real Transformer i t N N rimary Winding econdary Winding Note again that when a primary source is connected, current flows in the primary even the secondary is open circuited. This is the current required to produce the flux in the (real) ferromagnetic core. 3. Transformers 33

34 Magnetization Current in a Real Transformer The applied current consists of two components: Magnetization current; the current required to produce the flux in the core, and The core loss current to account for eddy current and hysteresis loss. 3. Transformers 34

35 Eddy Current Loss A changing flux will induce a current in the magnetic core just as is does for a wire. Eddy currents flowing in the core. 3. Transformers 35

36 ince the eddy current loss is proportional to the path length, making the core using the laminated, layered structure shown minimizes these losses. 3. Transformers 36

37 Magnetization Current: Ignoring leakage, recall from lide 8 that the average flux in the core is: If the primary voltage is: 1 v t dt N then v t V cost V M N M sint Note that the flux (and hence the magnetization current) lags the applied voltage by 90 o. 3. Transformers 37

38 Example uppose the magnetization curve for a real core can be modeled as 7 Ni Ni( ) 3. Transformers 38

39 If the sinusoidal flux variation is small, the magnetization current is nearly sinusoidal as well. 0.1sin() t Ni ( ) t Ni( 0.1sin() t ) 0 Ni t Transformers 39

40 Once the peak flux reaches the saturation point in the core, the magnetization current is clearly on 0 1.0sin() t longer sinusiodal. Note also that even a small additional increase in the flux will cause a large corresponding increase in the 10 Ni( ) t 10 magnetization current. (ee next slide as well) Ni( 1.0sin() t ) 0 N Ni 10 t Transformers 40

41 .0sin() t Ni ( ) t Ni (.0sin ( t ) ) t Transformers 41

42 Hysteresis and Eddy Current Loss The other component of the no-load current is current required to supply power for the hysteresis and eddy current losses. These are core losses. As before, assume that the flux in the core is sinusoidal. By Faraday s law, the eddy currents in the core will be proportional to i eddy d The eddy currents will be largest when the core flux is passing through zero. ince core losses are resistive in nature, the eddy current will be in-phase with the applied primary voltage. dt 3. Transformers 4

43 i excitation i magnetization hysteresis eddy i 4 Ni ( 1.1sin ( t ) ) 1cos ( t ) Ni( 1.1sin() t ) 0 cos() t t 3. Transformers 43

44 Linear Circuit Model for a Real Transformer How do we model a linear real transformer? Copper losses Eddy current losses Hysteresis losses Leakage flux Not the magnetic saturation ti effects these are non linear. 3. Transformers 44

45 Linear Circuit Model for a Real Transformer How do we model a real transformer? Copper losses Add a resistance (R and R ) in the primary and secondary circuit, respectively. 3. Transformers 45

46 Linear Circuit Model for a Real Transformer Copper Losses Model R R V I N I N V IDEAL 3. Transformers 46

47 Linear Circuit Model for a Real Transformer How do we model a real transformer? Leakage flux 3. Transformers 47

48 Leakage Flux Recall from lide 31 that, d e t N L dt d e t N dt L L L ince much (most) of the leakage flux path is through air, and since air has a constant reluctance which his much greater than that of the core, the primary/secondary leakage flux is proportional to the primary/secondary ycoil current, respectively, or Reluctance of flux path Ni L c e Ni L c e Ni Ni ermeance of flux path 3. Transformers 48

49 Leakage Flux d di di e t N N c L dt dt dt d di di L el t N N c L dt dt dt L L 3. Transformers 49

50 Recall lide 31, d d v t el t e t N N dt dt L M d e t N dt M which suggests an inductor as a model. 3. Transformers 50

51 di e L t L dt R jx Flux Leakage Model di e L t L dt X L V I N I N V IDEAL R R d L d v t el t e t N N dt dt M jx 3. Transformers 51

52 How do we model these effects? Magnetization current Hysteresis (core) losses 3. Transformers 5

53 Recall (lide 37) that the magnetization current (in the unsaturated region) is proportional to the voltage applied to the core and lags the applied voltage by 90 o, v t V cost M V M N sint Ni e 1 d v t dt v t N N dt Hence it can be modeled by a reactance (an inductor) connected across the primary voltage source. 3. Transformers 53

54 Magnetization Current R jx R jx V I jx M N I N V IDEAL d d L v t el t e t N N dt dt L M 3. Transformers 54

55 The core loss current i hysteresis + eddy is a current proportional to the voltage applied to the core that is in-phase with the applied voltage, hence it can be modeled by a resistance R C connected across the primary voltage source. Remember, the core effects are really nonlinear, and the circuit model dlwe have produced dprovides only a linear approximation in this complicated situation. 3. Transformers 55

56 R jx Core Losses (Eddy + hysteresis) R jx V I jx M R C N I N V IDEAL 3. Transformers 56

57 R jx Why here... R jx jx M R C N N R jx IDEAL R jx jx M R C N N... and not here? IDEAL 3. Transformers 57

58 Because the voltage actually applied to the core is equal to the applied voltage less the internal voltage drops of the windings. R jx R jx N N jx M R C IDEAL 3. Transformers 58

59 implifications of the Model Recall from lide 8: o that Also recall from lide : v t v t N N p V N p V N a N p in L L N Z Z a Z Turns Ratio 3. Transformers 59

60 V implifications of the Model R I R jx jx M jx R C Original Model N N ar R I ja X jx V V V I I jx M R Model referred to its primary side C 1 I V av a 3. Transformers 60

61 V V implifications of the Model R jx Original Model I N jx M R C N I V R jx a V R a ai X j a R jx X M R j C I a V a Model referred to its secondary side 3. Transformers 61

62 Approximations to the Model eries Impedance R jx R jx V V I jx M R C N N I V V hunt (Excitation) Impedance For large transformers (several kilovoltamperes or more), the shunt impedance is much greater than the series impedance, and some approximations can be made. For example Transformers 6

63 Approximations to the Model R jx ar ja X V V I jx 1 M R C Model referred to its primary side I R jx a R jx R 1 M C Convince yourself of this a a I ja X I av av Approximation 3. Transformers 63

64 Further Approximations to the Model R a R X ja X jx M R C Neglect this entirely R a R X ja X 3. Transformers 64

65 Determination of Values of Components in the Model (EEL 401, Lab 6) Only two tests required: Open Circuit Test hort Circuit Test 3. Transformers 65

66 Open Circuit Test Measures average power vt V Ammeter Wattmeter I A i t OC OC OC + V v Transformer Under Test t Voltmeter 3. Transformers 66

67 Open Circuit Test ince the magnetizing impedance is generally much greater that the series primary impedance, the series impedance can be neglected. R jx This has no effect neglect this jx M R C R jx IDEAL 1 1 ZE R jx jx R jx R M C M C 3. Transformers 67

68 Open Circuit Test ZE, YE j 1 1 Z R X R jx Y E I C E C M M OC, F cos OC V V I OC OC OC Y E The power factor is IOC IOC cos 1 F always lagging for a V V OC OC real transformer 3. Transformers 68

69 Open Circuit Test Y E Y E j Z R X E C M 1 1 R C X M 1 1 X M 1 RC tan tan 1 X M R C 3. Transformers 69

70 R 1 1 C cos F tan X M 1 RC XM tan cos F Y E 1 1 R C X M XM tan cos F X M X 1 1 M tan cos 1 F 1 3. Transformers 70

71 1 1 YE X M tan cos F X 1 M 1 tan cos 1 F tan sec cos tan cos X M cos 1 F F F cos cos F 1 1 X M cos cos F sin cos F 3. Transformers 71

72 1 1 YE 1 X M sin cos F X M Z E sin cos 1 F 1 RC XM tan cos F ZE 1 ZE tan cos F 1 1 sin cos F cos cos F Z E F 3. Transformers 7

73 X M Z sin cos E, 1 F R C Z E F ZE jx R M C jx M R C F R C X C 0 (open) Z E 1 Z E (open) 3. Transformers 73

74 hort Circuit Test vt V I A ave i t C C C + V v Transformer Under Test t i t vt Begin with a small voltage for and increase it until is at its rated value. i t 3. Transformers 74

75 hort Circuit Test R jx R jx V I jx M R C N N I V R jx ar ja X V I jx M R C 3. Transformers 75

76 hort Circuit Test R jx ar ja X V I jx M R C Neglect this ince the shunt magnetizing impedance is generally large compared with the series impedance R + jx, the shunt impedance can be neglected.. 3. Transformers 76

77 hort Circuit Test R jx a R ja X V I R a R jx ja X Z R a R j X a X in 3. Transformers 77

78 Transformer Voltage Regulation and Efficiency ince any real transformer has an internal impedance (resistance), the output voltage varies with the load even if the input voltage remains constant. To quantify this effect, it is customary to define the voltage regulation (VR) of the transformer as: VR V V econdary, no load econdary, full load V econdary, full load 100% 3. Transformers 78

79 Transformer Voltage Regulation and Efficiency VR V Vecondary, no load V econdary, no load econdary, full load V econdary, full load 100% Vrimary a V rimary Vecondary, full load VR a 100% V V econdary, full load 3. Transformers 79

80 Transformer Voltage Regulation and Efficiency out in out 100% out loss 100% Losses: Copper (I R) ) Hysteresis (R C ) Eddy Current 3. Transformers 80

81 Transformer Voltage Regulation and Efficiency out out loss 100% VI cos 100% VI cos Copper Hysteresis Eddy core losses VI cos VI cos Copper Core 100% 3. Transformers 81

82 Transformer hasor Diagrams A useful visualization tool Recall our earlier model: V a R a ai X j a R jx X M R C I V a a j a Model referred to its secondary side 3. Transformers 8

83 Transformer hasor Diagrams A useful visualization tool Recall our earlier model: V a R a ai X j a R jx X M R C I V a a j a The excitation branch will have little effect on voltage regulation. 3. Transformers 83

84 Transformer hasor Diagrams A useful visualization tool Recall our earlier model: V a ai R eq R R a X jx eq jx j a X M R C I V a a j a V a Now this has no effect neglect it I R ji X V eq eq 3. Transformers 84

85 Transformer hasor Diagrams A useful visualization tool for this equation: V Case 1: Lagging ower Factor IReq jixeq V a V I 3. Transformers 85

86 Transformer hasor Diagrams A useful visualization tool for this equation: V Case 1: Lagging ower Factor IReq jixeq V a V I I R eq 3. Transformers 86

87 Transformer hasor Diagrams A useful visualization tool for this equation: V Case 1: Lagging ower Factor IReq jixeq V a I V I R eq ji X eq 3. Transformers 87

88 Transformer hasor Diagrams A useful visualization tool for this equation: V Case 1: Lagging ower Factor IReq jixeq V a V V I Req I V V gg g V V a VR 0 a V For lagging loads: a ji X 3. Transformers 88 eq

89 Transformer hasor Diagrams A useful visualization tool for this equation: V Case : Unity ower Factor IReq jixeq V a I V 3. Transformers 89

90 Transformer hasor Diagrams A useful visualization tool for this equation: V Case : Unity ower Factor For unity F: IReq jixeq V a V I IReq V V V V a VR a V a V ji X eq 0 Note that VR is smaller here than in the previous case. 3. Transformers 90

91 Transformer hasor Diagrams A useful visualization tool for this equation: Case 3: Leading ower Factor I V a ji X eq V I R eq For leading loads: V V V V a VR a V 0 Note that VR is could be negative. 3. Transformers 91

92 Example V : V A 50 kva, 400:40 V, 60 Hz distribution ib i transformer has a leakage impedance of j 0.9 Ohms in the high-voltage winding and j Ohms in the low-voltage winding. At the rated frequency, the impedance of the shunt exciting branch (the impedance of magnetization branch R C and jx M in parallel) is 63+j 6.3 j 43.7 Ohms when viewed from the low-voltage side. Draw the equivalent circuit referred to a) the high-voltage h side and b) the low-voltage side. 3. Transformers 9

93 R jx olution 10 :1 R jx 0.7 j0.9 jx M R C N N j j43.7 R jx olution art (a) ar ja X 0.7 j j jx M R C 63 j j Transformers 93

94 R jx olution 10 :1 R jx 0.7 j0.9 jx M R C N N j j 4370 olution art (b) R a X j a R jx j j jx M R C 6.3 j Transformers 94

95 Example Continued olution If 400 V rms is applied to the high voltage side of the transformer, calculate the current in the magnetizing impedance in parts (a) and (b), respectively. olution (a): ZT j4370 j j I M amps rms 3. Transformers 95

96 Example Continued olution If 400 V rms is applied to the high voltage side of the transformer, calculate the current in the magnetizing impedance in parts (a) and (b), respectively. olution (b): ZT 6.37 j I M amps rms 3. Transformers 96

97 Example Continued Consider the equivalent T circuit found with the impedances referred to the high-voltage side (below). (a)draw the cantilever equivalent circuit with the shunt branch at the high-voltage terminal. (b) With the low-voltage terminal an open-circuited and 400 V applied to the high-voltage h terminal, calculate l the voltage at the low-voltage terminal as predicted by each equivalent circuit. R jx a R ja X j j jx M R C 63 j Transformers 97

98 Example Continued olution (a)draw the cantilever equivalent circuit with the shunt branch at the high-voltage terminal. R a R jx ja X 1.4 j 1.8 jx M R C 63 j Transformers 98

99 Example Continued olution (b) With the low-voltage terminal an open-circuit and 400 V applied to the high-voltage terminal, calculate the voltage at the low-voltage terminal as predicted by each equivalent circuit. R jx ar ja X j j0.9 jx M R C 63 j4370 V V Open 3. Transformers 99

100 Example Continued (b) For the T : ZM V 400 Z Z M 63 j j j olution 63 j i 63.7 j j j arg 63 j j Transformers 100

101 Example Continued (b) For the T : olution Reflected back to the secondary side gives, V V V 3. Transformers 101

102 Example Continued (b) For the Cantilever : olution R a R jx ja X 400 jx M R C 63 j j1.8 Open V V V 400 V V error 100% 0.05% 40 Well within engineering accuracy 3. Transformers 10

103 Example Continued uppose that the 400 V feeder has an internal impedance of j1.60 Ohms. Find the voltage at the secondary terminals of the transformer when the load connected to the secondary draws rated current from the transformer and the power factor of the load is 0.80 lagging. Neglect the voltage drops in the transformer and the feeder caused dby the exciting i current. olution 0.30 j j Load jx M R C Neglect 63 j 4370 I L 3. Transformers 103 V V

104 Example Continued j j1.8 I olution L Load V This is the high-voltage end: 50-kVA I L 0.83 A cos lagging g 3. Transformers 104

105 Example Continued V 1.7 j3.4 olution I A 400 Load V ZI 1.7 j V L V 400 V j j ( j ) 0.8 e L V 3. Transformers 105

106 Example Continued olution At the load, V 39 V V L Transformers 106

107 Example Continued With instrumentation on the high-voltage side and the output shortcircuited, the short-circuit test readings for this example are 48 V, 0.8 A, and 617 W. An open circuit test with the low-voltage side energized gives instrument readings on that side of 40 V, 5.41 A, and 186 W. Determine the efficiency and the voltage regulation at full load, 0.80 power factor lagging. vt A ave V I i t C C C + V v Transformer Under Test t i t 3. Transformers 107

108 Example Continued olution R jx From lide 77 a R ja X Z R a R j X a X in Z in V I I Req Req Transformers 108

109 Example Continued olution in Z R a R X a X X a X Zin R a R eq in eq X Z R Transformers 109

110 Example Continued olution Operation at full load, 0.80 power factor, lagging, corresponds to a primary current of I VI 50,000-kVA V 400 and an output power of 0.8 A VI cos 50,000 VA ,000 W av 3. Transformers 110

111 Example Continued olution The loss in the windings is I R windings eq W The core loss is determined from the open-circuit test, The total loss is winding core 186 W given W core The power supplied is thus ave loss 40, ,803 W 3. Transformers 111

112 Example Continued olution The efficiency is out in out 100% out loss 100% 40, % 98.03% 40, Transformers 11

113 Example Continued Voltage Regulation: olution VR V V econdary, no load econdary, full load V econdary, full load 100% R a R jx ja X I L V full load 400 Load F lagging 3. Transformers 113

114 Example Continued olution Voltage Regulation: VI j 50, j36.9 I L e e V, j j Transformers 114

115 Example Continued olution The required input voltage is V Z I V in eq L full load 1.4 j j j13v R a R jx ja X I L I V in V full load 400 Load 3. Transformers 115

116 Example Continued olution The required input voltage is V Z I V in eq L full load 1.4 j j j13v V in Transformers 116

117 Example Continued olution Under loaded conditions, the load voltage is 400 volts, by design. If the load were removed the load voltage would jump to 446 volts. Thus VR V V econdary, no load econdary, full load V econdary, full load % % 1.9% 100% 3. Transformers 117

118 The er Unit ystem An Industry-tandard Normalization Computations relating to machines and transformers are often carried out in per-unit form. This is where all quantities expressed as fractions of appropriately chosen base values. All the usual computations are then carried out in these per unit values instead of the familiar volts, amperes, ohms, etc. There are a number of advantages to the system. One is that the parameter values of machines and transformers typically fall in a reasonably narrow numerical range when expressed in a per-unit system based upon their rating. The correctness of their values is thus subject to a rapid approximate check. 3. Transformers 118

119 The er Unit ystem A second advantage is that when transformer equivalent-circuit parameters are converted to their per-unit values, the ideal transformer turns ratio becomes es 1:1 and hence the ideal transformer can be eliminated. This greatly simplifies analyses. For complicated systems involving many transformers of different turns ratios, this simplification is a significant one in that a possible cause of serious mistakes is removed. 3. Transformers 119

120 The er Unit ystem Quantifies such as voltage V, current I, power,, reactive power Q, voltamperes VA,, resistance R,, reactance X, impedance Z, conductance G, susceptance B, and admittance Y can be translated to and from per-unit form as follows: er unit value actual value base value ofquantity Once V base and I base have been chosen, then, Q, VA V I Vbase Rbase, Xbase, Zbase I base base base base base Clearly only two independent base values can be chosen arbitrarily. base 3. Transformers 10

121 The er Unit ystem Typically VA base and V base are chosen, then the remaining quantities are uniquely established as per the previous slide. The value of VA base must be the same over the entire system of analysis. When a transformer is encountered, the values of V base differ on each side and should be chosen to have the same ratio as the turns ratio of the transformer. 3. Transformers 11

122 The er Unit ystem Usually the rated or nominal voltages of the respective sides are chosen as base values. If the base voltages of the primary and secondary are chosen to be in the ratio of the turns of the ideal transformer, the per-unit ideal transformer will have a unity turns ratio and hence can be eliminated. If these rules are followed, the procedure for performing system analyses in per-unit can be summarized as follows: 3. Transformers 1

123 The er Unit ystem We will illustrate via example, but the procedure is: 1. elect a VA base and a base voltage at some point in the system.. Convert all quantities to per unit on the chosen VA base and with a voltage base that transforms as the turns ratio of any transformer which is encountered as one moves through the system. 3. erform a standard electrical analysis with all quantities in per unit. 4. When the analysis is completed, all quantities can be converted back to real units (e.g., volts, amperes, watts, etc.) by multiplying their per-unit values by their corresponding base values. 5. The procedure is best illustrated with an example. 3. Transformers 13

124 The er Unit ystem Example The equivalent circuit for a 100-MVA, 7.97-kV:79.7-kV transformer is shown below. Convert this equivalent circuit to per unit using the transformer rating as base j j kV : 79.7-kV j Transformers 14

125 The er Unit ystem Example olution The base quantities for the transformer are: Low-voltage side: VA base = 100-MVA, V base =797-kV 7.97-kV R base Vbase X base VA base High-Voltage g side: VA base = 100-MVA, V base = 79.7-kV R base Vbase X 63.5 base VA base 3. Transformers 15

126 The er Unit ystem Example Compute the per unit values: 0.04 X X X M R R lti olution 3. Transformers 16

127 The er Unit ystem Example olution Compute the per unit turns ratio: 7.97-kV 79.7-kV Turns Ratiopu : 1:1 797kV 7.97-kV 797kV 79.7-kV There is no need to keep the transformer as far as analysis is concerned. 3. Transformers 17

128 The er Unit ystem Example in per unit values olution j : j j j j j Transformers 18

129 The er Unit ystem Example Consider again our earlier example of a 50-kVA, 400:40-V transformer. We found earlier that the current on the low voltage side is 5.43 amps (lide 96) and that its equivalent impedance referred to the high voltage side is j1.8 Ohms (lide 98). Using the transformer rating as the base, express in per unit on the low- and dhigh-voltage h sides (a) the exciting current and (b) the equivalent impedance. 3. Transformers 19

130 The er Unit ystem Example olution The base values are these give V I 400 V V 40 V base, base, 0.8 A I 08 A base, base, Z Vbase base,, 3. Transformers 130

131 The er Unit ystem Example olution Using the values found in lides 95 and 96, I I M, M, per unit per unit 08 As expected, the per-unit values are the same referred to either side corresponding to a unity turns ratio for the ideal transformer. This is a direct consequence of the choice of base voltages in the ratio of the transformer turns ratio and the choice of a constant volt-ampere base. 3. Transformers 131

132 The er Unit ystem Example olution From lide 100: j Zeq, j per unit j Zeq, j per unit 1.15 Again the values are the same, with the effect of the turns ratio accounted for by choice of base values. 3. Transformers 13

133 The er Unit ystem Example A 15-kVA 10:460-V transformer has an equivalent series impedance of j0.04 per unit. Calculate the equivalent series impedance in ohms (a) referred to the lowvoltage side and (b) referred to the high-voltage side. olution 10 Z base, L ,000/ Zbase, H ,000/ Transformers 133

134 The er Unit ystem Example olution Z Z, j0.04 eq L j , j0.04 eq H 0.54 j Transformers 134

135 Transformer Ratings Transformers have four major ratings: apparent power voltage current frequency 3. Transformers 135

136 Transformer Ratings The voltage rating serves two purposes; the first is to insure that breakdown does not occur between the transformer windings. The second is prevent the core from becoming saturated which, as we have seen (see results of our earlier example below), can result in very large magnetization currents Ni(.0sin() t ) 0.0 sin() t Ni( ) t t Transformers 136

137 Transformer Ratings If a steady-state voltage v(t) is applied to the transformer s primary winding, this gives the flux as vt V t max sin d t vt N t vtdt 1 dt N V N max t cos t 3. Transformers 137

138 Transformer Ratings From V max t cost N N we see that there are two ways to drive the core into saturation. One is to increase the voltage (our example). Another is to reduce the operating frequency. Accordingly, a (European) transformer rated at 50 Hz may be operated at a 0% high voltage at 60 Hz assuming that this does not cause insulation breakdown. 3. Transformers 138

139 Transformer Ratings The purpose of the apparent power rating it that, together with the voltage rating, it sets the current flow through the windings. The current flow is important because it controls the i R losses which in turn controls the heating of the coils. Heating significantly reduces insulation lifetime. 3. Transformers 139

140 The roblem of Current Inrush a transient effect uppose that the moment the transformer is connected to the power line the voltage across the primary winding is, vt V t max sin The maximum flux value reached during the first half-cycle of the applied voltage depends d on the value of. For = 90 o, and vt V sin t90 V cost max max V t sint max N max max or just the maximum value at steady state. V N 3. Transformers 140

141 The roblem of Current Inrush a transient effect Now suppose that = 0 o, then vt V t max sin and the maximum flux during the first half-cycle is T 1 f 1 t Vmax sintdt N t V max N max 0 cost V max N N 0 3. Transformers 141

142 The roblem of Current Inrush a transient effect Now the maximum flux is twice its steady-state value. Recall again our earlier example showing what happens if, near saturation, the flux is further increased by even a small amount. A doubling of flux could result (as a transient effect) is a huge inrush of current when the transformer is first plugged din. In fact, for this fist part of a cycle the primary really looks like a short. The primary winding must be designed to handle this effect. There is lots of engineering involved in designing magnetic circuits! This holds even truer for machines our next topic. 3. Transformers 14

143 Autotransformers (Text ection 3.9) and Three hase Transformers (Text ection 3.10) are left as a reading exercise. 3. Transformers 143