Practical Transformer on Load

Transcription

1 Practical Transformer on Load We now consider the deviations from the last two ideality conditions : 1. The resistance of its windings is zero. 2. There is no leakage flux. The effects of these deviations become more prominent when a practical transformer is put on load. Monday, August 01, 2011 Transformers 1

2 (1) Effect of Winding Resistance Current flow through the windings causes a power loss called I 2 R loss or copper loss. This effect is accounted for by including a resistance R 1 in the primary and resistance R 2 in the secondary Monday, August 01, 2011 Transformers 2

3 (2) Effect of Flux Leakage The difference between the total flux linking with the primary and the useful mutual flux Φ u linking with both the windings is called the primary leakage flux, Φ L1. Similarly, Φ L2 represents the secondary leakage flux. Flux leakage results in energy being alternately stored in and discharged from the magnetic fields with each cycle of the power supply. It is not directly a power loss, but causes the secondary voltage to fail to be directly proportional to the primary voltage, particularly under heavy loads. Monday, August 01, 2011 Transformers 3

4 Leakage flux in a transformer (a) Its definition. (b) Its effect accounted for. The useful mutual flux Φ u is responsible for the transformer action. The leakage flux Φ L1 induces an emf E L1 in the primary winding. Monday, August 01, 2011 Transformers 4

5 Similarly, flux Φ L2 induces an emf E L2 in the secondary. Hence, we include reactances X 1 and X 2 in the primary and secondary windings, in the equivalent circuit. The paths of leakage fluxes Φ L1 and Φ L2 are almost entirely due to the long air paths and are therefore practically constant. The reluctance of the paths being very high, X 1 and X 2 are relatively small even on full load. However, the useful flux Φ u remains almost independent of the load. Monday, August 01, 2011 Transformers 5

6 Monday, August 01, 2011 Transformers 6

7 Equivalent Circuit of a Transformer It is merely a representation of the following KVL equations : V I R ji X E I ( R jx ) E E I R ji X V I ( R jx ) V Monday, August 01, 2011 Transformers 7

8 Points to Draw Phasor Diagram 1. Resistive voltage drop in phase with current phasor. 2. Inductive voltage drop in quadrature with current. 3. To get V 1, add I 1 Z 1 to E Add V 2 and I 2 Z 2, to get E Current I 1 is vector sum of I 0 and I Angle between V 1 and I 1 gives the power factor angle of the transformer. Monday, August 01, 2011 Transformers 8

9 Phasor Diagram for Practical Transformer on Resistive Load V 1 I 1 Z 1 I 1 X 1 I 1 R 1 I ' 1 -E 1 I 1 V E E I ( R jx ) V I ( R jx ) O I 0 m I 2 V 2 E 1 I 2 R 2 I 2 Z 2 I 2 X 2 E 2 Monday, August 01, 2011 Transformers 9

10 Practical Transformer on Inductive Load Monday, August 01, 2011 Transformers 10

11 Practical Transformer on Capacitive Load Monday, August 01, 2011 Transformers 11

12 Simplified Equivalent Circuit The no-load current I 0 is only about 3-5 % percent of the full-load current. The exciting circuit R 0 -X 0 in is shifted to the left of impedance R 1 -X 1. Transforming the impedances from the secondary to the primary side. Monday, August 01, 2011 Transformers 12

13 Equivalent resistance and reactance referred to the primary side R R ( R / K ) and X X ( X / K ) 2 2 e1 1 2 e1 1 2 Monday, August 01, 2011 Transformers 13

14 Approximate Equivalent Circuit Monday, August 01, 2011 Transformers 14

15 Example 5 A single-phase, 50-kVA, 4400-V/220-V, 50-Hz transformer has R 1 = 3.45 Ω, R 2 = Ω, X 1 = 5.2 Ω and X 2 = Ω. Calculate (a) the R e as referred to the primary, (b) the R e as referred to the secondary, (c) the X e as referred to the primary, (d) the X e as referred to the secondary, (e) the Z e as referred to the primary, (f) the Z e as referred to the secondary, and (g) the total copper loss. Monday, August 01, 2011 Transformers 15

16 Solution : Full-load primary current, Full-load secondary current, K V V 2 1 I kva A V (a) 2 2 (b) (c) e1 1 2 I 2 kva A V 220 R R ( R / K ) 3.45 [0.009/(0.05) ] 7.05 Ω R K R R Ω 2 2 e2 1 2 (0.05) X X X K 11.2 Ω 2 2 e1 1 ( 2 / ) 5.2 [0.015/(0.05) ] 2 (d) 2 2 Xe2 K X1 X 2 (0.05) Ω Monday, August 01, 2011 Transformers 16

17 Ze1 Re1 X e1 (7.05) (11.2) Ω (e) (f) Ze2 Re2 X e2 (0.0176) (0.028) (g) Total copper loss Ω I R 909 W I2 R2 (11.36) 3.45 (227) Alternatively, by considering equivalent resistances, total copper loss IR W e1 (11.36) 7.05 IR 909 W e2 (227.27) Monday, August 01, 2011 Transformers 17

18 Voltage Regulation The voltage regulation of a transformer is defined as the change in its secondary terminal voltage from no load to full load, the primary voltage being assumed constant. and V 2(0) = secondary terminal voltage at no load, V 2 = secondary terminal voltage at full load. The voltage drop V 2(0) - V 2 is called the inherent regulation. V ( i) Per unit regulation down V 2(0) 2 V 2(0) 2(0) 2 % regulation down 100 V V V 2(0) Monday, August 01, 2011 Transformers 18

19 V2(0) V2 ( ii) Per unit regulation up V2 V2(0) V2 % regulation up 100 V 2 Normally, when nothing is specified, regulation means regulation down. Monday, August 01, 2011 Transformers 19

20 Approximate Voltage Drop The secondary terminal voltage at no load, V2(0) E2 KE1 KV1 Exact voltage drop = V2(0) V2 OC OA OG OA AG AF+FG Monday, August 01, 2011 Transformers 20

21 Approximate voltage drop, AF AE EF AE BD I R In case of leading power factor, cos I X 2 e2 2 e2 Approximate voltage drop, AF AE EF AE BD In general, I R cos I X 2 e2 2 e2 2 e2 2 e2 sin sin Approximate voltage drop I R cos I X sin Monday, August 01, 2011 Transformers 21

22 I R 2 e2 2 e2 % Regulation 100 V r cos I X V cos V 2(0) x sin sin Use + sign for lagging power factor and sign for leading power factor. Condition for Zero Regulation : I R Possible only if the load has leading power factor. Re 2 cos I X sin 0 tan X 2 e2 2 e2 e2 Monday, August 01, 2011 Transformers 22

23 Condition for Maximum Regulation Maximum regulation can occur only for inductive load. The voltage drop is maximum when d ( I 2 R e2 cos I 2 X e2 sin ) 0 d ( I2R 2 sin e I2 Xe 2 cos ) 0 X e2 tan R e2 Monday, August 01, 2011 Transformers 23

24 Example 6 Solution : Monday, August 01, 2011 Transformers 24

25 Example 7 Solution : the load voltage, V V 2 Monday, August 01, 2011 Transformers 25

26 Example 8 A single-phase, 40-kVA, 6600-V/250-V, transformer has primary and secondary resistances R 1 = 10 Ω and R 2 = 0.02 Ω, respectively. The equivalent leakage reactance as referred to the primary is 35 Ω. Find the full-load regulation for the load power factor of (a) unity, (b) 0.8 lagging, and (c) 0.8 leading. Monday, August 01, 2011 Transformers 26

27 Solution : Given : R 1 = 10 Ω; R 2 = 0.02 Ω; X e1 = 35 Ω 250 the turns-ratio, K the full-load current, I2 160 A 250 R K R R 2 2 e2 1 2 (0.0379) and X K X (0.0379) e2 2 2 e1 (a) For power factor, cos = 1; sin = 0. Hence, I R cos I X sin V 2 e2 2 e2 % Regulation 100 Monday, August 01, 2011 Transformers 27 2(0) % 250

28 (b) For power factor, cos = 0.8 (lagging, positive); 2 sin 1 cos 0.6 I R cos I X sin V 2 e2 2 e2 % Regulation 100 2(0) % 250 (c) For power factor, cos = 0.8 (leading, negative); I R cos I X V 2 e2 2 e2 % Regulation 100 2(0) sin sin % 250 Monday, August 01, 2011 Transformers 28