Chapter 12: Three-Phase Circuits

Transcription

1 Chater 1: Three-Phase Circuits 1.1 ntroduction 1. Balanced Three-Phase oltages 1.3 Balanced Wye-Wye connection 1.4 Balanced Wye-Delta Connection 1.7 Power in a Balanced System 1.1 NTRODUCTON A single-hase ac ower system consists of a generator connected through a air of wires (i.e., a transmission line) to a load as shown in figure below: Circuits or systems in which the ac sources oerate at the same frequency but different hases are known as olyhase. The following figures below shows a two-hase three-wire system and a three-hase four-wire system. As distinct from a single-hase system, a two-hase system is roduced by a generator consisting of two coils laced erendicular to each other so that the voltage generated lags the other by 90º. Similarly, a three-hase system is roduced by a generator consisting of three sources having a same amlitude and frequency but out of hase with each other by 10º. Since the three-hase system is by far the most revalent and most economical olyhase system, hence all the discussion in this chater will be mainly on three-hase systems. EEEB13 Chater 1: Three-Phase Circuits Dr. C.S.Tan

2 Three-hase systems are imortant for several reasons: Nearly all electric ower is generated and distributed in three-hase at the oerating frequency of 60Hz or ω=377rad/s (i.e., US) or 50Hz or ω=314rad/s (i.e., East Jaan). When single-hase or two-hase inuts are required, they can be taken from the threehase system rather than generated indeendently. f more hases like 48 hases are required for industry usage, three-hase sulied can be maniulated to rovide 48 hases The instantaneous ower in a three-hase system can be constant (i.e., not ulsating).this results in uniform ower transmission and less vibration of three-hase machines. For the same amount of ower, the three-hase system is more economical than the single-hase system. The amount of wire required for a three-hase system is less than that required for an equivalent single-hase system. 1. BAANCED THREE_PHASE OTAGES This section will begin with a discussion of balanced three-hase voltages. Three-hase voltages are often roduced with a three-hase ac generator (or alternator) whose cross-sectional view is shown below. The generator basically consists of a rotating magnet called the rotor surrounded by a stationary winding called the stator. Three searate windings or coils with terminal a-a, b-b and c-c are hysically laced 10º aart around the stator. As the rotor rotates, its magnetic field cuts the flux from the three coils and induces voltages in the coils. Since the coils are laced 10º aart, the induced voltages in the coils are equal in magnitude but out of hase by 10º shown below. EEEB13 Chater 1: Three-Phase Circuits Dr. C.S.Tan

3 Since each coil can be regarded as a single-hase generator by itself, the three-hase generator can suly ower to both single-hase and three-hase loads. A 3-hase voltage system is equivalent to three single-hase systems. The voltage sources (generator connection) can be connected in wye-connected shown in (a) or delta-connected as in (b) The voltages an, bn and cn are between lines a, b and c and the neutral line, n resectively. an, bn and cm are called hase voltages. t is called balanced if they are same in magnitude and frequency and are out of hase from each other by 10º such that an + bn + cn = 0.. (1.1) an = bn = cn.. (1.) There are two ossible combinations; ositive sequence and negative sequence Positive/ abc Sequence (Balanced) Negative / acb Sequence (Balanced) et an = 0 vector diagram for hase abc et an = 0 vector diagram for hase acb hase sequence gives hase sequence gives Assuming an as the reference hence = 0 and = θ an Where =amlitude of hase voltage in rms Thus; = 0 = θ an bn cn = an 10 = ( θ 10 ) = ( θ + 40 ).(1.3) = an 40 = ( θ 40 ) = ( θ + 10 ) This sequence is roduced when the rotor rotates in the counterclockwise direction. Assuming an as the reference hence = 0 and = θ an Where =amlitude of hase voltage in rms Thus; = 0 = θ an bn = an 40 = ( θ 40 ) = ( θ + 10 ). (1.4) cn = an 10 = ( θ 10 ) = ( θ + 40 ) This sequence is roduced when the rotor rotates in the clockwise direction. EEEB13 Chater 1: Three-Phase Circuits Dr. C.S.Tan

4 Mathematically, both hase sequences in Eq. (1.3) and Eq. (1.4) satisfy Eq. (1.1). For examle, from Eq. (1.4) + + = θ + ( θ 40 ) + ( θ 10 ) an bn cn ( ) = θ + + ( ) = θ 1 + ( j0.866) + ( 0.5 j0.866) = 0 ike the generator connections, loads can also be connected in Y or. Balanced load is where the hase imedances are equal in magnitude and are in hase Y-connected oad (Balanced) Three-hase load configuration: -connected oad (Balanced) Three-hase load configuration: For balanced Y-connected load: Where Z1 = Z = Z3 = ZY.. (1.5) Z is the Total load imedance er hase Y For balanced Y-connected load: Z a Zb Zc ZΔ = = =.. (1.6) Where Z Δ is the Total load imedance er hase Recall from earlier chater 9, Eq. (9.69) that 1 Z Δ = 3 Z Y or Z Y = Z Δ..(1.7) 3 Hence, Y-connected load can be transformed into a -connected load or vice versa using Eq. (1.7) Since both sources and loads can be connected in either Y or, there are 4 ossible connections: Y-Y connection (i.e., Y-connected source with Y-connected load) Y- connection (i.e., Y-connected source with -connected load) - connection (i.e., -connected source with -connected load) -Y connection (i.e., -connected source with Y-connected load) EEEB13 Chater 1: Three-Phase Circuits Dr. C.S.Tan

5 1.3 BAANCED WYE-WYE CONNECTON The balanced four-wire Y-Y system shown below has a Y-connected source on the left and Y- connected load on the right. Assume a balanced load (i.e., imedances are equal). Z Y = Total load imedance er hase, Ω/hase Z s = Source imedance er hase, Ω/hase Z l = ine imedance er hase, Ω/hase Z = oad imedance er hase, Ω/hase Z n = imedance of the neutral line er hase, Ω/hase Thus in general Z Y = Z s + Zl + Z.. (1.8) Z s and Z l are often very small comared with Z therefore, one can assume that Z Y = Z if no source or line imedance is given. By luming the imedances together, the Y-Y system shown above can be simlified as below oltages: Assuming the source rotates in ositive sequence, the hase voltages (i.e., line-to-neutral voltages) are (from Eq. (1.3)) where = θ = 0 = θ, an an = 10 or = ( θ 10 ) = ( θ + 40 ) an bn = + 10 = ( θ 40 ) = ( θ + 10 ) an cn.. (1.9) EEEB13 Chater 1: Three-Phase Circuits Dr. C.S.Tan

6 From the hasor diagram, the line voltages (i.e., line to line voltages) ab, bc and ca are related to the hase voltages. For examle, = + = = θ θ 10 ab an nb an bn ( ) ( θ )( ) = θ ( ) = 1 ( 0.5 j0.866) ( θ )( 1.5 j0.866) ( θ )( 3 30 ) = + = ( θ ) = ( )( an ) = 3 30 ( roved) Similarly, we can get bc and as follows: ca ab = an = θ = θ + ( ) ( )( ) ( ) ( θ )( ) ( ) ( θ )( ) ( 30 ) = 3 30 = ( 10 ) 3 30 = 3 ( θ 90 ) bc bn = 3 30 = ( + 10 ) 3 30 = 3 ( θ ) ca cn Note: The magnitude of the line voltages is.. (1.10) 3 times the magnitude of the hase voltages and the line voltages lead their corresonding hase voltage by 30 (See hasor diagram below) = 3 Where = an = bn = cn and = = = ab bc ca From Phasor diagram (Positive Sequence) ab = an + nb = + bc bn nc = + Phase oltages: an = θ ca cn na bn = an 10 = an + 40 = ( θ 10 ) = ( θ + 40 ) cn = an 40 = an + 10 = ( θ 40 ) = ( θ + 10 ).. (1.11).. (1.1) EEEB13 Chater 1: Three-Phase Circuits Dr. C.S.Tan

7 Currents: Balanced Y-Y connection: Alying K to each hase, we obtain the line currents as: an a = ZY bn an 10 b = = = a 10.. (1.1) ZY ZY cn an + 10 c = = = a + 10 ZY ZY And (See hasor diagram on the left) ine Current, a = Phase Current, AB. ine Current = Current in each line Phase Current= Current in each hase of the source or load a + b + c = 0 ( ) n = a + b + c = 0 or nn = Z n n = 0.. (1.13) This imlies that the voltage across the neutral wire is equal to zero. Therefore the neutral line can be removed without affecting the system (balanced Y-Y). Since it is a balanced Y-Y system, the system can be reresented or analyzed using a single hase equivalent circuit with or without neutral line (see below). The single-hase analysis yields the line current, a as a an =.. (1.14) Z Y EEEB13 Chater 1: Three-Phase Circuits Dr. C.S.Tan

8 1.4 BAANCED WYE-DETA CONNECTON Balanced Y- connection is a 3-hase system with balanced Y-connected sources sulying - connected load. From the figure above, there is no neutral connection from source to load in this tye of connection. Assuming ositive sequence, the hase voltages Balanced Y- connection: (source) are (i.e., Eq. (1.9)) an = θ, = ( θ 10 ) = ( θ + 40 ) bn = ( θ 40 ) = ( θ + 10 ) cn Phasor Diagram: Relationshi between hase and line voltages As shown and derive in Section 1.3, the line voltages across the source are (i.e., Eq. (1.10)) ab = 3 ( θ + 30 ) ab = 3 ( θ + 30 ) bc = 3 ( θ 90 ) or bc = ab 10 ca = 3 ( θ ) ca = ab + 10 ooking at the balanced Y- connection, it shows that the line voltages (at the source) are equal to the voltage across the load imedances. Assuming ositive sequence, ab = AB = 3 an 30 bc = BC = 3 bn 30 = 3 an 90.. (1.15) ca = CA = 3 cn 30 = 3 an From these voltages (at the load), we can obtain the hase currents as AB BC CA AB =, BC =, CA =.. (1.16) Z Z Z Note: Using K, the hase currents can be determined. For examle, alying K around loo aabbna to find hase currents AB Thus, taking AB as the reference AB 0 AB = Z BC AB 10 BC = = = AB 10.. (1.17) Z Z CA AB + 10 CA = = = AB + 10 Z Z EEEB13 Chater 1: Three-Phase Circuits Dr. C.S.Tan

9 Balanced connection at the load To obtain the line currents, a, b, and c. Alying KC at nodes A, B, and C we obtain At node A: = a AB CA At node B : = b BC AB At nodec : = c CA BC.. (1.18) Note: = line current, a, b, c = hase current,,, AB BC CA Phasor Diagram: Relationshi between hase and line currents Thus a = AB CA = AB( ) = AB( j0.866) Similarly, b, Generally, = 3 AB 30 and c, gives b c = 3BC 30 = 3CA 30 = 3.. (1.19) = a = b = c and = = = AB BC CA Note: The magnitude of the line current is 3 times the magnitude of the hase current and the line currents lag their corresonding hase currents by 30 (See hasor diagram beside) Another way to analyze the Y- circuit is by transforming the -connected load to an equivalent Y-connected load. Using the transformation formula in Eq. (1.7), we will have a Y-Y system. Hence, the three-hase Y- system can be relaced by a single-hase equivalent circuit as shown below. where Z Y = Z 3.. (1.0) This will allows us to calculate the line currents. The hase currents are then obtained using Eq. (1.19) with the fact that each of the hase currents leads the corresonding line current by 30º. EEEB13 Chater 1: Three-Phase Circuits Dr. C.S.Tan

10 1.7 POWER N A BAANCED SYSTEM The total instantaneous ower in the load is the sum of instantaneous owers in the three hases: that is = a + b+ c = v i + v i + v i AN a BN b CN c ( cosωt)( cos( ωt θ) ) ( cos( ωt 10 ))( cos( ωt θ 10 )) + ( cos( ωt+ 10 ))( cos( ωt θ + 10 )) = + Alying trigonometric identity, gives (f you are interested to derive lease referred to.50) = 3cosθ.. (1.1) where = hase voltage = hase current θ = angle of the load imedance = angle between the hase voltage and the hase current From Eq. (1.1), one can notice that the total instantaneous ower in a balanced three-hase system is constant. n other words, it does not change with time as one can see in the instantaneous ower of each hase. This result is true for both Y and -connected loads. Since the total instantaneous ower is indeendent of time, the average ower er hase P either for -connected load or Y-connected load is /3, yields P = /3= cosθ.. (1.) And the average reactive ower er hase is Q = sinθ.. (1.3) The average aarent ower er hase is S =.. (1.4) Total real ower in a balanced 3-hase system is: P = Pa + Pb + Pc = 3cosθ.. (1.5) Total reactive ower in a balanced 3-hase system is: Q = Qa + Qb + Qc = 3sinθ.. (1.6) Total aarent/comlex ower in a balanced 3-hase system is: 3 S = Sa + Sb + Sc = 3 = 3 Z = Z.. (1.7) where Z = Z φ = load imedance er hase Alternatively, we may write Eq. (1.7) as S = P+ jq EEEB13 Chater 1: Three-Phase Circuits Dr. C.S.Tan

11 Power Balance Y-Y Connection Power Balanced for Y- Connection Assuming abc sequence = 3 30 = 3 30 ower absorbed by the load is = = Z = Z S3φ Z... (1.8) = 3 = 3 Or the above 3-hase can be reresented as er hase diagram since they are balanced Assuming abc sequence = 3 30 = 3 30 ower absorbed by the load is = = Z = Z 1 Z = Z = 3 3 S3φ Z... (1.9) = 3 = Or the above 3-hase can be reresented as er hase diagram since they are balanced Where = an From the single-hase one-line diagram above, ower absorbed by the load is = Z Where = a S = 3S 3φ 1φ 3φ = 3 = 3 a S Z Z Where = an From the single-hase one-line diagram above, ower absorbed by the load is Z = 3 S3φ = 3 Where = a S3φ = Z = a Z EEEB13 Chater 1: Three-Phase Circuits Dr. C.S.Tan