Three Phase Circuits. Three Phase Circuits

Transcription

1 Three Phase Circuits 1 Three Phase Circuits Chater Objectives: Be familiar with different three-hase configurations and how to analyze them. Know the difference between balanced and unbalanced circuits earn about ower in a balanced three-hase system Know how to analyze unbalanced three-hase systems Be able to use PSice to analyze three-hase circuits Aly what is learnt to three-hase measurement and residential wiring 2

2 Three hase Circuits An AC generator designed to develo a single sinusoidal voltage for each rotation of the shaft (rotor) is referred to as a single-hase AC generator. If the number of coils on the rotor is increased in a secified manner, the result is a Polyhase AC generator, which develos more than one AC hase voltage er rotation of the rotor In general, three-hase systems are referred over single-hase systems for the transmission of ower for many reasons. 1. Thinner conductors can be used to transmit the same ka at the same voltage, which reduces the amount of coer required (tyically about 25% less). 2. The lighter lines are easier to install, and the suorting structures can be less massive and farther aart. 3. Three-hase equiment and motors have referred running and starting characteristics comared to single-hase systems because of a more even flow of ower to the transducer than can be delivered with a single-hase suly. 4. In general, most larger motors are three hase because they are essentially selfstarting and do not require a secial design or additional starting circuitry. 3 Single Phase, Three hase Circuits a) Single hase systems two-wire tye b) Single hase systems three-wire tye. Allows connection to both 120 and 240. Two-hase three-wire system. The AC sources oerate at different hases. 4

3 Three-hase Generator The three-hase generator has three induction coils laced 120 aart on the stator. The three coils have an equal number of turns, the voltage induced across each coil will have the same eak value, shae and frequency. 5 Balanced Three-hase oltages Three-hase four-wire system Neutral Wire A Three-hase Generator oltages having 120 hase difference 6

4 Balanced Three hase oltages Neutral Wire a) Wye Connected Source b) Delta Connected Source an bn cn = 0 = 120 = 240 an bn cn = 0 = = a) abc or ositive sequence b) acb or negative sequence 7 Balanced Three hase oads A Balanced load has equal imedances on all the hases a) Wye-connected load b) Delta-connected load Balanced Imedance Conversion: Conversion of Delta circuit to Wye or Wye to Delta. Z = Z = Z = Z Y Z = Z = Z = Z a b c Z = 3Z Z = Y Y 1 Z 3 8

5 Three hase Connections Both the three hase source and the three hase load can be connected either Wye or DETA. We have 4 ossible connection tyes. Y-Y connection Y- connection - connection -Y connection Balanced connected load is more common. Y connected sources are more common. 9 Balanced Wye-wye Connection A balanced Y-Y system, showing the source, line and load imedances. Source Imedance ine Imedance oad Imedance 10

6 Balanced Wye-wye Connection ine current I n add u to zero. Neutral current is zero: I n = -(I a + I b + I c )= 0 Phase voltages are: an, bn and cn. The three conductors connected from a to A, b to B and c to C are called INES. The voltage from one line to another is called a INE voltage ine voltages are: ab, bc and ca Magnitude of line voltages is 3 times the magnitude of hase voltages. = 3 11 Balanced Wye-wye Connection ine current I n add u to zero. Neutral current is zero: I n = -(I a + I b + I c )= 0 Magnitude of line voltages is 3 times the magnitude of hase voltages. = 3 = 0, = 120, = an bn cn = + = = ab an nb an bn = = bc bn cn = = + = ca cn an an bn

7 Balanced Wye-wye Connection Phasor diagram of hase and line voltages = = = ab bc ca = 3 = 3 = 3 = 3 an bn cn = = = an bn cn 13 Single Phase Equivalent of Balanced Y-Y Connection Balanced three hase circuits can be analyzed on er hase basis.. We look at one hase, say hase a and analyze the single hase equivalent circuit. Because the circuıit is balanced, we can easily obtain other hase values using their hase relationshis. I a = Z an Y 14

8 15 Balanced Wye-delta Connection Three hase sources are usually Wye connected and three hase loads are Delta connected. There is no neutral connection for the Y- system. ine currents are obtained from the hase currents I AB, I BC and I CA I = I I = I a AB CA I = I I = I I b BC AB = I I = ICA 3 30 I = 3I c CA BC AB BC I I I AB BC CA I = I = I = I a b c I = I = I = I = Z = Z = Z AB BC CA AB BC CA 16

9 Balanced Wye-delta Connection Phasor diagram of hase and line currents I = I = I = I a b c I = I = I = I AB BC CA I = 3I Single hase equivalent circuit of the balanced Wye-delta connection Z 3 17 Balanced Delta-delta Connection Both the source and load are Delta connected and balanced. AB BC I AB =, I BC =, ICA = Z Z Z I = I I, I = I I, I = I I a AB CA b BC AB c CA BC CA 18

10 Balanced Delta-wye Connection Transforming a Delta connected source to an equivalent Wye connection Single hase equivalent of Delta Wye connection Power in a Balanced System The total instantaneous ower in a balanced three hase system is constant. v = 2 cos( ωt) v = 2 cos( ωt 120 ) v = 2 cos( ωt ) AN BN CN i = 2I cos( ωt θ ) i = 2I cos( ωt θ 120 ) i = 2I cos( ωt θ ) a b b = + + = v i + v i + v i a b c AN a BN b CN c = 2 I cos( ωt) cos( ωt θ ) + cos( ωt 120 ) cos( ωt θ 120 ) + cos( ωt+ 120 ) cos( ωt θ ) [ ] 1 cos Acos B = [cos( A + B) + cos( A B)] Using the identity and simlifying 2 = 3 I cos θ The instantenous ower is not function of time. The total ower behaves similar to DC ower. This result is true whether the load is Y or connected. The average ower er hase P =. 3 P = = I 3 cos θ 20

11 Power in a Balanced System The comlex ower er hase is S. The total comlex ower for all hases is S. = 3 I cosθ 1 1 P = = I cos θ Q = = I sin θ S = I 3 3 S = P + jq = I Comlex ower for each hase P = P + P + P = 3P = 3 I cosθ = 3 I cosθ Q = 3Q = 3 I sinθ = 3 I sinθ S=3S a b c 3 = 3 I = 3I Z = Z 2 S = P + jq = 3 I θ 2 Total comlex ower, I, and I are all rms values, θ is the load imedance angle 21 Power in a Balanced System S=3S = 3 I = 3 I Z = 2 S = P + jq = 3 I θ 3 Z 2 Toal comlex ower, I, and I are all rms values, θ is the load imedance angle Notice the values of,, I, I for different load connections. = 3 I = I = I = 3 I I I I I I I Y connected load. connected load. 22

12 Power in a Balanced System 23 Single versus Three hase systems Three hase systems uses lesser amount of wire than single hase systems for the same line voltage and same ower delivered. a) Single hase system b) Three hase system 2 2 Wire Material for Single hase 2( π r l) 2r 2 = = = '2 '2 (2) = 1.33 Wire Material for Three hase 3( π r l) 3r 3 If same ower loss is tolerated in both system, three-hase system use only 75% of materials of a single-hase system 24

13 25 =840 (Rms) I Caacitors for f Correction 26

14 I S = = = 50.68A Without Pf Correction Unbalanced Three Phase Systems An unbalanced system is due to unbalanced voltage sources or unbalanced load. In a unbalanced system the neutral current is NOT zero. 27 Unbalanced three hase Y connected load. ine currents DO NOT add u to zero. I n = -(I a + I b + I c ) 0 28

15 29 Three Phase Power Measurement Two-meter method for measuring three-hase ower 30

16 Residential Wiring Single hase three-wire residential wiring 31 Problem Determine the current in the neutral line. UNBAANCED OAD NEUTRA CURRENT IS NOT ZERO 32

17 Problem Solve for the line currents in the Y- circuit. Take Z = Ω Ω. SINGE PHASE EQUIAENT CIRCUIT 33 Problem Find the line currents I a, I b, and I c in the three-hase network below. Take Z = 12 - j15ω, Z Y = 4 + j6 Ω, and Z l = 2 Ω. ONE DETA AND ONE Y CONNECTED OAD IS CONNECTED TWO oads are arallel if they are converted to same tye. Delta connected load is converted to Y connection. 34

18 Problem For the balanced circuit below, ab = Find the line currents I aa, I bb, and I cc. BAANCED Y CONNECTED OAD. Source voltage given is line to line, obtain the line to neutral voltage. 35 Problem The following three arallel-connected three-hase loads are fed by a balanced threehase source. oad 1: 250 ka, 0.8 f lagging oad 2: 300 ka, 0.95 f leading oad 3: 450 ka, unity f If the line voltage is 13.8 k, calculate the line current and the ower factor of the source. Assume that the line imedance is zero. 36

19 Problem A rofessional center is sulied by a balanced three-hase source. The center has four lants, each a balanced three-hase load as follows: oad 1: 150 ka at 0.8 f leading oad 2: 100 kw at unity f oad 3: 200 ka at 0.6 f lagging oad 4: 80 kw and 95 kar (inductive) If the line imedance is j0.05 Ω er hase and the line voltage at the loads is 480, find the magnitude of the line voltage at the source. 37 Problem The Figure dislays a three-hase delta-connected motor load which is connected to a line voltage of 440 and draws 4 ka at a ower factor of 72 ercent lagging. In addition, a single 1.8 kar caacitor is connected between lines a and b, while a 800-W lighting load is connected between line c and neutral. Assuming the abc sequence and taking an = 0, find the magnitude and hase angle of currents I a, I b, I c, and I n. Total load is UNBAANCED. INE CURRENTS I a, I b, I c ARE NOT ERQUA Single hase, 800 W lighting load connected to hase C only. Pf for lighting loads is unity. I = I = 1 S 3 38

20 440 S I1 = = ( θ + 30 ) 3 39