Lecture (11) Three Phase Transformers

Transcription

1 Lecture (11) Three Phase Transformers 11-1

2 High Low I = ( ) = / 4.16 Kv The line current in the low voltage 1000 kv 0.8 p.f. lagging 11-2

3 For different connections, I2 V 1 = = a I VN Connection a Y-Y - 69/4.16 = Y- -Y 1 2 Mohammed, ll rights reserved 10-

4 PRIMRY = HIGH VOLTGE PHSE CONNECTION Y / Y SECONDRY = LOW VOLTGE Where the Load is / Y / / Y

5 I ph I L = I ph B B C C I ph = = This is the current in the phase for the side 11-5

6 When asked to find the phase value, the angle will change (shifted). The angle of the line value lags the phase value by Ref Phase CCW Line

7 Source 69 kv : 4.16 kv B B C C I L = I ph = The phase current in the delta is 0 degrees [HED OF] the line current 11-7

8 WYE CONNECTED LOD DIRECTION OF ROTTION OF PHSORS Counter Clock Wise V B I V N LOOKING DIRECTION -V BN =V NB ϕ -V CN =V NC ϕ = POWER FCTOR NGLE I C V CN ϕ 0 O 0 O ϕ V BN V BC -B-C SEQUENCE V C LINE VLUE -V N =V N I B LINE VLUE OF VOLTGE LEDS THE PHSE VLUE BY 0 O Mohammed, ll rights reserved 11-8

9 DELT CONNECTED LOD DIRECTION OF ROTTION OF PHSORS V B I B I LOOKING DIRECTION -I BC =I CB I C ϕ ϕ 0 O 0 O ϕ -I C =I C I BC I C V BC -B-C SEQUENCE LINE VLUE OF CURRENT LGS THE PHSE VLUE BY 0 O V C -I B =I B I B LINE VLUE ϕ = POWER FCTOR NGLE Mohammed, ll rights reserved 11-9

10 For our example, I 2 69 kv = I I 2 = I I = = I 2 I 1 = in each _ I in the line = (4.8) leg =

11 Example Source Z lin = j2 Line Y / 200 / 20 V (10 kv) Load Load Made of two Components 1. 2 kw Heating load / phase 2. IM--> 20 kv, 0.8 p.f., 20 V. x10 kv transformers 200/20 Z eq = Ω For the transformer 11-11

12 1. What s the voltage at the source 2. What is the voltage regulation of the transmission Line. Z T.L. = j2 Ω = R 2 + X 2 = = Z TL Ω Per Phase 11-12

13 Load Combo: 6 kw heating 6 kw + j0 = S Heat 20 (0.8 + j0.6) = S Motor 16 + j12 kv = S Motor 1 hp = 746 w hp = = This is the load hp value 11-1

14 S Motor + S Heat = S Total 6 kw + j j12 kv = S Total 22 + j 12 kv = S Total S = kv Sin 28.6 = cos 28.6 =

15 The line current for the total load I = * = a = V V For the transformer Z eq = j0.01 Ω Z eq = a 2 ( j0.01) Ω =0.166+j 0. Ω 1 2 = = 5.77 (ref. to the low side) (ref. to the high side) 11-15

16 The Whole System V S? V L =20V T. Line Transformer ~ j j

17 dding the TL Impedance Z eq = Z TLine + Z Transformer Z eq = j2. = Ω I 2 V S = av V S a ( ) V S = = j V Z eq = ( 5.77)( 20) 0 + ( ) Phase Value 11-17

18 Neglecting core loss, calculate everything... Cost of power loss Voltage regulation Current on both sides of the transformer Voltage on both sides of the transformer, etc dd this to HW.. But Submit Next Class before in Class Solution 11-18

19 OPEN DELT CONNECTION Three Phase Connection / (25 KV) S = * V ph * I ph 11-19

20 - Open Delta Connection B C B C S = V L I L 11-20

21 / V ph =V L S = V L I L S = V L I L Open S S = = V L V I L L I L 11-21

22 Open / = V 1 L I L = V I L L Open rating = 1 / Rating = 57.8% of 75 kv 11-22