Chapter 11 Balanced Three-Phase Circuits

Transcription

1 Chapter 11 Balanced Three-Phase Circuits Three-Phase Systems 11.3 Analysis of the Y-Y Circuit 11.4 Analysis of the Y- Circuit 11.5 Power Calculations in Balanced Three-Phase Circuits 11.6 Measuring Average Power in Three- Phase Circuits 1

2 Overview An electric power distribution system looks like: where the power transmission uses balanced three-phase configuration. 2

3 Why three-phase? Three-phase generators can be driven by constant force or torque (to be discussed). ndustrial applications, such as high-power motors, welding equipments, have constant power output if they are three-phase systems (to be discussed). 3

4 Key points What is a three-phase circuit (source, line, load)? Why a balanced three-phase circuit can be analyzed by an equivalent one-phase circuit? How to get all the unknowns (e.g. line voltage of the load) by the result of one-phase circuit analysis? Why the total instantaneous power of a balanced three-phase circuit is a constant? 4

5 Section 11.1, 11.2 Three-Phase Systems 1. Three-phase sources 2. Three-phase systems 5

6 One-phase voltage sources One-phase ac generator: static magnets, one rotating coil, single output voltage v(t)= m cost. ( 6

7 Three-phase voltage sources Three static coils, rotating magnets, three output voltages v a (t), v b (t), v c (t). 7

8 deal Y- and -connected voltage sources Neutral 8

9 Real Y- and -connected voltage sources nternal impedance of a generator is usually inductive (due to the use of coils). 9

10 Balanced three-phase voltages Three sinusoidal voltages of the same amplitude, frequency, but differing by 120 phase difference with one another. There are two possible sequences: 1. abc (positive) sequence: v b (t) lags v a (t) by acb (negative) sequence: v b (t) leads v a (t) by

11 abc sequence v b (t) lags v a (t) by 120 or T/3. a m 0, b m 120, c m

12 Three-phase systems (Y or ) (Y or ) Source-load can be connected in four configurations: Y-Y, Y-, -Y, - t s sufficient to analyze Y-Y, while the others can be treated by -Y and Y- transformations. 12

13 Section 11.3 Analysis of the Y-Y Circuit 1. Equivalent one-phase circuit for balanced Y-Y circuit 2. Line currents, phase and line voltages 13

14 General Y-Y circuit model Ref. The only essential node. 14

15 Unknowns to be solved Line (line-to-line) voltage: voltage across any pair of lines. Phase (line-toneutral) voltage: voltage across a single phase. Line voltage Line current Phase current Phase voltage For Y-connected load, line current equals phase current. 15

16 Solution to general three-phase circuit No matter it s balanced or imbalanced three- phase circuit, KCL leads to one equation: 0 N 0 aa ga bb a n 1a cc N, A b n gb 1b N B c n gc 1c N C (1), mpedance of neutral line. Total impedance along line aa. Total impedance along line bb. Total impedance along line cc. which is sufficient to solve N (thus the entire circuit). 16

17 Solution to balanced three-phase circuit For balanced three-phase circuits, 1. { a'n, b'n, c'n } have equal magnitude and 120 relative phases; 2. { ga = gb = gc }, { 1a = 1b = 1c }, { A = B = C }; total impedance along any line is the same ga + 1a + A = =. Eq. (1) becomes: N an N 0 bn an N cn 0, bn N N 0. cn N, 17

18 Meaning of the solution N = 0 means no voltage difference between nodes n and N in the presence of 0. Neutral line is both short (v = 0) and open (i = 0). The three-phase circuit can be separated into 3 one-phase circuits (open), while each of them has a short between nodes n and N. 18

19 Equivalent one-phase circuit Phase voltage of source Line current Phase voltage of load nn =0 aa Directly giving the line current & phase voltages: a n N aa, A, aa an aa1 a A. ga 1a A Unknowns of phases b, c can be determined by the fixed (abc or acb) sequence relation. 19

20 The 3 line and phase currents in abc sequence Given aa a n, the other 2 line currents are: bn 120 cn bb aa, 120 cc aa, which still follow the abc sequence relation. cc aa bb 20

21 The phase & line voltages of the load in abc seq. an A, BN bn B 120, CN 120. AB BN 120 (abc sequence) Line voltage 3 30, BC CA , 150. Phase voltage 21

22 The phase & line voltages of the load in acb seq. AB BC CA BN 120,, 150. (acb sequence) Phase voltage Line voltage Line voltages are 3 times bigger, leading (abc) or lagging (acb) the phase voltages by

23 Example 11.1 (1) Q: What are the line currents, phase and line voltages of the load and source, respectively? ga 1a (abc sequence) Phase voltages A = ga + 1a + A =40 + j30. 23

24 Example 11.1 (2) The 3 line currents (of both load & source) are: aa bb cc ga aa aa an 1a A j A. The 3 phase voltages of the load are: 2.4 A, A, BN aa A j ,. CN

25 Example 11.1 (3) The 3 line voltages of the load are: AB BC CA , AB AB , 25

26 Example 11.1 (4) The 3 phase voltages of the source are: j0 5 an a n aa ga 120. bn cn an an, , The three line voltages of the source are: ab bc ca 330 an , , ab ab

27 Section 11.4 Analysis of the Y- Circuit 27

28 Load in configuration Line current Phase current Line voltage = Phase voltage 28

29 29 -Y transformation for balanced 3-phase load The impedance of each leg in Y-configuration ( Y ) is one-third of that in -configuration ( ):.,, c b a b a c b a a c c b a c b. 3 3 Y

30 Equivalent one-phase circuit The 1-phase equivalent circuit in Y-Y config. continues to work if A is replaced by /3: Line current Line-to-neutral voltage Phase voltage Line voltage a n directly giving the line current: aa, and line-to-neutral voltage:. aa ga A 1a A 30

31 The 3 phase currents of the load in abc seq. Can be solved by 3 node equations once the 3 line currents aa, bb, cc are known: aa AB CA, bb BC AB, cc CA BC. Line current (abc sequence) Phase current Phase current Line current 31

32 Section 11.5 Power Calculations in Balanced Three-Phase Circuits 1. Complex powers of one-phase and the entire Y-Load 2. The total instantaneous power 32

33 Average power of balanced Y-Load The average power delivered to A is: P A aa cos, L, L 3, (rms value) A. The total power delivered to the Y-Load is: P tot 3PA 3 cos 3 L L cos. 33

34 34 Complex power of a balanced Y-Load The reactive powers of one phase and the entire Y-Load are:. sin 3 sin 3, sin L L tot Q Q The complex powers of one phase and the entire Y-Load are: ; * j L L j tot j e e S S e jq P S

35 One-phase instantaneous powers The instantaneous power of load A is: p A ( t) v ( t) iaa( t) m m cost cos( t ). (abc sequence) The instantaneous powers of A, C are: p ( t) v ( t) i ( t) p B C ( t) BN m m bb cos t cos t , m m cos t 120 cos t

36 Total instantaneous power The instantaneous power of the entire Y-Load is a constant independent of time! p tot ( t) p A 1.5 ( t) p ( t) ( t) 1.5 cos 2 2 cos 3 cos. B The torque developed at the shaft of a 3-phase motor is constant, less vibration in machinery powered by 3-phase motors. The torque required to empower a 3-phase generator is constant, need steady input. p C m m 36

37 Example 11.5 (1) Q: What are the complex powers provided by the source and dissipated by the line of a-phase? The equivalent one-phase circuit in Y-Y configuration is: 1a S (rms value) 37

38 Example 11.5 (2) The line current of a-phase can be calculated by the complex power is: S * aa, j A. * aa, The a-phase voltage of the source is: an aa 1 a j

39 Example 11.5 (3) The complex power provided by the source of a- phase is: San an The complex power dissipated by the line of a- phase is: S * aa aa aa a ka j0.025 ka. 39

40 Key points What is a three-phase circuit (source, line, load)? Why a balanced three-phase circuit can be analyzed by an equivalent one-phase circuit? How to get all the unknowns (e.g. line voltage of the load) by the result of one-phase circuit analysis? Why the total instantaneous power of a balanced three-phase circuit is a constant? 40