Chapter 12: Three Phase Circuits

Transcription

1 Chapter 12: Three Phase Circuits 12.1 What Is a Three Phase Circuit? 12.2 Balance Three Phase Voltages 12.3 Balance Three Phase Y to Y Connection 12.4 Other Balance Three Phase Connections 12.5 Power in a Balanced System 12.6 Unbalanced Three Phase Systems 12.7 Design & Applications Three Phase Power Measurement & Residential Wiring 12.8 Summary Nikola Tesla 10/07/1856 7/01/1943 was a Serbian American inventor, electrical engineer, mechanical engineer, physicist, and futurist best known for his contributions to the design of the modern ac electricity supply system. A three phase system with rotating magnetic fields 1

2 12.1 What is a Three Phase Circuit? (1) Single Phase Systems Two wire type Three wire type Two Phase Three Wire System Normal household system is a singlephase three wire system: The same magnitude and the same phase of the terminal voltages The connection of both 120 V and 240 V appliances. 2

3 12.1 What is a Three Phase Circuit? (2) Three Phase System: produced by a generator consisting of three sources having the same amplitude and frequency but out of phase with each other by 120. Three sources with 120 out of phase Four wired system Advantages: 1. Most of the electric power is generated and distributed in three phase. 2. The instantaneous power in a three phase system can be constant. 3. The amount of power, the three phase system is more economical that the single phase. 4. In fact, the amount of wire required for a three phase system is less than that required for an equivalent single phase system. 3

4 12.2 Balance Three Phase Voltages (1) A three phase generator consists of a rotating magnet (rotor) surrounded by a stationary winding (stator). Three Phase Generator Generated Voltages Three phase voltage sources: (l) Y connected ; (r) Δ connected 4

5 12.2 Balance Three Phase Voltages (2) Balanced phase voltages are equal in magnitude and are out of phase with each other by 120. The phase sequence is the time order in which the voltages pass through their respective maximum values. A balanced load is one in which the phase impedances are equal in magnitude and in phase. Balanced Y connected load, Z 1 = Z 2 = Z 3 = Z Y Balanced Δ connected load, Z a = Z b = Z c = Z Δ Z Δ = 3Z Y 5

6 12.3 Balance Three Phase Y to Y Connection (1) Four possible connections 1. Y Y connection (Y connected source with a Y connected load) 2. Y Δ connection (Y connected source with a Δ connected load) 3. Δ Δ connection 4. Δ Y connection Abalanced Y Y system is a three phase system with a balanced y connected source and a balanced y connected load. Phase (line to neutral) voltages: Line (line to line) voltages: 6

7 12.3 Balance Three Phase Y to Y Connection (2) ZY Z s +Z l +ZL Line currents: (Phase currents: same as) Hint: The neutral line can be removed w/o affecting the system. 7

8 12.3 Balance Three Phase Y to Y Connection (3) Example: Calculate the line currents in the three wire Y Y system. 8

9 12.4 Other Balance Three Phase Connections (1) A balanced Y Δ system is a (most practice) three phase system with a balanced y connected source and a balanced Δ connected load. Phase currents: Line currents: lag by 30 o I L I I 3 I, where p I I I L a b c I I I p AB BC CA 9

10 12.4 Other Balance Three Phase Connections (2) Abalanced Δ Δ system is a three phase system with a balanced Δ connected source and a balanced Δ connected load. Phase voltages: (Line voltages: same as) Phase currents: Line currents: Each line current lags the corresponding phase current by 30 o, themagnitudeofthelinecurrentis 3 times the magnitude I p of the phase current, I 3I L p 10

11 12.4 Other Balance Three Phase Connections (3) Abalanced Δ Y system is a three phase system with a balanced y connected source and a balanced y connected load. Phase voltages: (Line voltages: same as) Line currents: (Phase currents: same as) 11

12 12

13 12.5 Power in a Balanced System (1) For a Y connected load, the phase voltages and currents are Note: ZY Z Total real power in the load: The total instantaneous power in a balanced threephase system is constant. Total complex power: 13

14 12.5 Power in a Balanced System (2) Example (Line losses): Figure shows a balanced three wire Y to Y circuit. Determine average power delivered by the three phase source, delivered to the three phase load, and absorbed by the three phase line. (a) A balanced three wire Y to Y circuit and (b) the per phase equivalent circuit. The line current is calculated: I aa 100 ( ) A 50 j(377)(0.045)

15 The phase voltage at the load is The power delivered by the source is calculated as So I aa V AN ( ) (40 j(377)(0.04)) I ( ) 81 2 V ( ) A and V ( ) V (100)(1.894) Pa cos(18.7 ) 89.7W 2 The power delivered to the load is calculated as I The power lost in the line is calculated as I ( ) A and R 40, so P W 2 aa A A ( ) A and R 10, so P W 2 aa aa aa The three phase load receives 3P A = W, and 3P aa =53.7Wis lost in the line. A total of 80 percent of the power supplied by the source is delivered to the load. The other 20 percent is lost in the line. The three phase source delivers 3P a = W. aa an

16 12.5 Power in a Balanced System (4) Example (Reducing line losses): 80% of the power supplied by the source is delivered to the load, and the other 20% is lost in the line in the last example. The loss in the line can be reduced by reducing the current in the line. Reducing the current in the load would reduce the power delivered to the load. Transformers provide a way of reducing the line current without reducing the load current. In this example, two three phase transformers are added to the three phase circuit. A transformer at the source steps up the voltage and steps down the current. Conversely, a transformer at the load steps down the voltage and steps up the current. Because the turns ratios of these transformers are reciprocals of each other, the voltage and current at the load are unchanged. The current in the line will be reduced to reduce the power lost in line. The line voltage will increase. The higher line voltage will require increased insulation and increased attention to safety.

17 Figure shows the per phase equivalent circuit of the balanced threewire Y to Y circuit that includes the two transformers. Determine the average power delivered by the three phase source, delivered to the three phase load, and absorbed by the three phase line. (a) A per phase equivalent circuit for a balanced Y to Y circuit with stepup and step down transformers and (b) the corresponding frequency domain circuit used to calculate the line current. To analyze the per phase equivalent circuit in (a), notice that: The secondary voltage of the left hand transformer is 10 times the primary voltage, that is, 1000 cos (377t). The impedance connected to the secondary of the right hand transformer can be reflected to the primary of this transformer by multiplying by 100. The result is a 4000 Ω resistor in series with a 4 H inductor.

18 These observations lead to the one mesh circuit shown in (b). The mesh current in this circuit is the line current of the three phase circuit. This line current is calculated as The current into the dotted end of the secondary of the left hand transformer in (a) is I aa, so the current into the dotted end of the primary of this transformer is The current into the dotted end of the primary of the right hand transformer is I aa (ω), so the current into the dotted end of the secondary is The phase voltage at the load is I aa 1000 ( ) A 4010 j(377)(4.005) V I a AN ( ) 10( I ( )) A I A The power delivered by the source is calculated as aa ( ) ( 10 I ( )) A aa ( ) (40 j(377)(0.04)) I ( ) V Ia ( ) A and (100)(2.334) Van ( ) A so Pa cos(20.6 ) 109.2W 2 A

19 The power delivered to the load is calculated as I ( ) A and R 40, so P W 2 A A A The power lost in the line is calculated as I ( ) A and R 10, so P W 2 aa aa A Now 98% of the power supplied by the source is delivered to the load. Only 2% lost in the line.

20 12.5 Power in a Balanced System (8) Comparing the power loss: P P I R R V 2 2 L loss 2 L 2 2 L P P I R R R L L loss 3( ) L VL VL P Ploss 2R 2r material for single-phase 2( r l) 2r P loss R r material for three-phase 3( r l) 3r 3 If same power loss is tolerated in both system, three phase system use only 75% of materials of a single phase system. 20

21 Example: 12.5 Power in a Balanced System (9) 21

22 12.6 Unbalanced Three Phase Systems (1) An unbalanced system is due to unbalanced voltage sources or an unbalanced load. VAN VBN VCN I a, I b, I c, Z Z Z A B C I (I I I ) n a b c To calculate power in an unbalanced three-phase system requires that we find the power in each phase. The total power is not simply three times the power in one phase but the sum of the powers in the three phases. 22

23 12.6 Unbalanced Three Phase Systems (2) Example: For the unbalanced circuit, find: (a) the line currents, (b) the total complex power absorbed by the load, and (c) the total complex power absorbed by the source. 23

24 12.7 Design & Applications (1) Design (Power factor correction): Figure shows a three phase circuit. The capacitors are added to improve the power factor of the load. We need to determine the value of the capacitance, C, requiredto obtain a power factor of 0.9 lagging. The circuit is excited by sinusoidal sources all having the same frequency, 60 Hz or 377 rad/s. The circuit is at steady state. The circuit is a linear circuit. Phasors can be used to analyze this circuit. The circuit is a balanced three phase circuit. A per phase equivalent circuit can be used to analyze this circuit. The load consists of two parts. The part comprising resistors and inductors is connected as a Y. The part comprising capacitors is connected as a Δ. A Δ to Y transformation can be used to simplify the load.

25 The per phase equivalent circuit is shown as State the Goal: Determine the value of C required to correct the power factor to 0.9 lagging. Generate a Plan: A formula was provided for calculating the reactance, X 1,neededtocorrectthe power factor of a load 2 2 R X X1 1 R tan(cos pfc) X where R and X are the real and imaginary parts of the load impedance before the powerfactoriscorrectedandpfcisthecorrectedpowerfactor.afterthisequation is used to calculate X 1, the capacitance, C, canbecalculatedfromx 1.Noticethat X 1 will be the reactance of the equivalent Y connected capacitors. We will need to calculate the Δ connected capacitor equivalent of the Y connected capacitor.

26 Act on the Plan: We note that: Z = R +jx =20+j75.4Ω. Therefore, the reactance, X 1, needed to correct the power factor is 2 2 X tan(cos 0.9) The Y connected capacitor equivalent to the Δ connected capacitor can be calculated from Z Y = Z Δ /3. Therefore, the capacitance of the equivalent Y connected capacitor is 3C. Finally, because X 1 =1/(3Cω), we have 1 1 C F 3 X 377 3( 92.6) 1 Verify the Proposed Solution: When C = μf, the impedance of one phase of the equivalent Y connected load will be 1 Z Y (20 j75.4) j377 3 C j (20 j75.4) j377 3 C The value of the power factor is pf cos tan so the specifications have been satisfied.

27 12.7 Design & Applications (4) 27

28 12.7 Design & Applications (5) 28

29 12.8 Summary (1) The generation & transmission of electrical power are more efficient in three phase systems employing 3 voltages of the same magnitude & frequency and differing in phase by 120 from each other. The three phase source consists of either three Y connected sinusoidal voltage sources or three Δ connected sinusoidal voltage sources. Similarly, the circuit elements that comprise the load are connected to form either a Y or a Δ. The transmission line connects the source to the load and consists of either three or four wires. Analysis of three phase circuits using phasors and impedances will determine the steady state response of the three phase circuit. We are particularly interested in the power the three phase source delivers to the three phase load.

30 12.8 Summary (2) The current in the neutral wire of a balanced Y to Y connection is zero; thus, the wire may be removed if desired. The key to the analysis of the Y to Y circuit is the calculation of the line currents. When the circuit is not balanced, the first step in the analysis of this circuit is to calculate V nn, the voltage at the neutral node of the three phase load with respect to the voltage at the neutral node of the three phase source. For a Δ load, we converted the Δ load to a Y connected load by using the relation Δ to Y transformation. Then we proceeded with the Y to Y analysis. The line current for a balanced Δ load is 3 times the phase current and is displaced 30 in phase. The line to line voltage of a Δ load is equal to the phase voltage.