Symmetrical Components

Transcription

1 Prof. Dr.-Ing. Ralph Kennel Technische Universität München Lehrstuhl für Elektrische Antriebssysteme und Leistungselektronik München, 06. März

2 symmetrical 3phase system U 1 U 2 = U 1 e -j2/3π U 3 = U 1 e -j4/3π versors U 3 U 2 U 2 = U 1 a² U 3 = U 1 a 2

3 the basic idea is that an asymmetrical set of N phasors can be expressed as a linear combination of N symmetrical sets of phasors by means of a complex linear transformation Dr. rer. nat. Erika Mustermann (TUM) kann beliebig erweitert werden Infos mit Strich trennen 3

4 usual strategies in electrical engineering calculation(s) might be possible as well Real World (time domain) initial situation transformation usually it is easier this way! Dream World (xxx-domain) initial situation calculation(s) Real World (time domain) result but more complex back transformation Dream World (xxx-domain) result

5 all components are rotating in the same direction (within the complex plane)!!! 5

6 any 3phase system can be split into the 6

7 what is the purpose? 7

8 what is the purpose? 2nd step : calculate the response for each of the 1st step : split the system into the 3rd step : recombine the into real system 8

9 what is the purpose? 2nd step : calculate the response for each of the 1st step : split the system into the 3rd step : recombine the into real system 9

10 what is the purpose? 2nd step : calculate the response for each of the 1st step : split the system into the 3rd step : recombine the into real system 10

11 what is the purpose? 2nd step : calculate the response for each of the 1st step : split the system into the 3rd step : recombine the into real system 11

12 what is the purpose? 2nd step : calculate the response for each of the 1st step : split the system into the 3rd step : recombine the into real system 12

13 13

14 what is the purpose? electrical AC machines show different impedances in 14

15 Graphic Decomposition in Symmetrical Components 15

16 Graphic Decomposition in Symmetrical Components 16

17 Graphic Decomposition in Symmetrical Components 17

18 all in one Graphic Decomposition in Symmetrical Components 18

19 Multiphase Systems in General the number of phases is not really fixed to 3 19

20 for 3 phase systems 20

21 for 4 phase systems 21

22 for 5 phase systems 22

23 for 6 phase systems 23

24 What about 2phase Systems??? the so-called 2phase system is not really a 2phase system because the phase shift is not 180 based on the phase shift of 90 between the phases it is a 4phase system with only 2 phases used for 4phase systems 24

25 Example: Transformer U 1 = Z 1 I 1 U 2 = Z 2 I 2 U 0 = Z 0 I 0 in rotating electrical machines Z 1, Z 2 and Z 0 are different in stationary electrical machines at least Z 1 and Z 2 are equal in a transformer Z 1 = Z 2 = Z K2 equivalent circuit 25

26 Example: Transformer in a transformer the zero impedance Z 0 depends on the design of the transformer in case of a star connection on primary and secondary side there cannot be any zero sequence current on the primary side (due to Krichhoff s law) U 0 = Z 0 I 0 the equivalent circuit contains the secondary side only the quantity of the mutual inductance depends on the design 3-leg-core mutual inductance is low (magnetic flux has to go through the air) 26

27 Example: Transformer in a transformer the zero impedance Z 0 depends on the design of the transformer in case of a star connection on primary and secondary side there cannot be any zero sequence current on the primary side (due to Krichhoff s law) U 0 = Z 0 I 0 the equivalent circuit contains the secondary side only the quantity of the mutual inductance depends on the design 5-leg-core mutual inductance is higher (magnetic flux goes through the outer legs) 27

28 Example: Transformer in a transformer the zero impedance Z 0 depends on the design of the transformer in case of a delta connection on primary and a star connection secondary side a zero sequence current on the primary side is possble U 0 = Z 0 I 0 the equivalent circuit contains primary and secondary side The quantity of the mutual inductance is negligible in comparison to the leakage inductances 28

29 Example: single phase load of a transformer U u = U z = I z Z I u = - I z I u = I 1 + I 2 + I 0 = - I z I v = I w = 0 U u = U 1 + U 2 + U 0 = U z U 1 = U 20u + Z 1 I 1 U 2 = Z 2 I 2 U 1 = Z 0 I 0 we assume, that the grid source voltage contains a U 20u component only 29

30 Example: single phase load of a transformer U 1 = U z = U 20u + ZU 1 u = I 1 U+ z Z= 2 I z I 2 Z+ Z 0 I 0 U I u = - I 1 = U z = U 20u + 1/3 (Z 1 z + Z 2 + Z 0 ) I u I 1 I 2 + I 0 = - I U I u = - I z 1 = U z = U 20u - 1/3 (Z 1 z + Z 2 + Z 0 ) I z I v = I w = 0 U 1 = Z 1 I 1 U 2 = Z 2 I 2 U 0 = Z 0 I 0 I 1 = 1/3 (I u + I v a + I v a²) U u = U 1 + U 2 + U 0 = U z U 1 = U 20u + Z 1 I 1 U 2 = Z 2 I 2 U 1 = Z 0 I 0 I 1 = I 2 = I 0 = 1/3 I u = - 1/3 I z we assume, that the grid source voltage contains a U 20u component only 30

31 Example: single phase load of a transformer U 1 = U z = U 20u + 1/3 (Z 1 + Z 2 + Z 0 ) I u U 1 = U z = U 20u - 1/3 (Z 1 + Z 2 + Z 0 ) I z in case of a delta connection : Z 1 = Z 2 Z 0 in case of a double star connection : Z 1 = Z 2 Z 0 and consequently : Z 1 = Z 2 1/3 (Z 1 + Z 2 + Z 0 ) with respect to high 1/3 (Z 1 + Z 2 + Z 0 ) U 1 = U z = U 20u - 1/3 (Z 1 + Z 2 + Z 0 ) I z will be very low 31

32 Conclusion simplify the investigation in unsymmetrical loads 32

33 Thank you!