Chapter 2 Power System Fundamentals: Balanced Three-Phase Circuits

Transcription

1 Chapter 2 Power System Fundamentals: Balanced Three-Phase Circuits This chapter reviews the fundamentals of balanced three-phase alternating current (ac) circuits. First, we define positive and negative balanced three-phase sequences. Second, we analyze balanced three-phase voltages and currents. Third, the different types of power are defined and measurements techniques for power are briefly reviewed. Fourth, we provide an overview of the analysis of balanced three-phase circuits using the per-unit system. This chapter provides an appropriate background of three-phase power for the unfamiliar reader, establishing the link between the physical reality and analytical techniques. It can be skipped by readers with knowledge of three-phase circuit analysis. 2.1 Introduction Power systems are generally based on three-phase alternating current (ac) circuits. This chapter describes the fundamentals of this type of circuits and is organized as follows. Section 2.2 defines balanced three-phase sequences. Section 2. describes balanced three-phase voltage and currents, as well as the two different symmetrical connections of system components and the equivalence among them. Section 2.4 defines instantaneous, active, reactive, and apparent powers and explains how to measure them. Section 2.5 clarifies why three-phase power is generally preferred over single phase-phase power. Section 2.6 defines the per-unit system, which is used in the remaining chapters of this book. Section 2.7 summarizes the chapter and suggests some references for further study. Finally, Sect. 2. proposes some exercises for further comprehending the concepts addressed in this chapter. Springer International Publishing AG 201 A.J. Conejo, L. Baringo, Power System Operations, Power Electronics and Power Systems, 17

2 1 2 Power System Fundamentals: Balanced Three-Phase Circuits 2.2 Balanced Three-Phase Sequences There are two ways of representing an ac source: 1. Using a sinusoidal representation: a.t/ D p 2Asin.!t C / (2.1) where: A is the root mean square (RMS) value of the source,! is its angular frequency (also known as angular speed) measured in radians per second, and is its initial phase angle. The RMS value of the source is computed as: A D s Z 1 T a T 2.t/dt (2.2) 0 where T is the period (measured in seconds). The angular frequency! is defined as the rate of change of the phase of the sinusoidal source and is computed as:! D 2 T D 2f (2.) where f is the ordinary frequency (measured in Hertz). 2. Using a phasorial representation: NA D A : (2.4) Figure 2.1 illustrates the relationship between a sinusoidal ac source (left plot) and a rotating vector or phasor (right plot). Observe that the projection of the rotating vector on the imaginary axis (right-hand-side of the figure) renders the sinusoidal form of the source: a.t/ D p 2Asin.!t C /, shown on the left-hand side of the figure. If three ac sinusoidal sources (or phasors) have equal magnitude and equal angle separation. 2 rad or120ı /, then they constitute a balanced three-phase sequence. For example, the following three ac sources constitute a balanced three-phase sequence:

3 2.2 Balanced Three-Phase Sequences 19 a(t) Im 2A ( w t y) y w t y Re Fig. 2.1 Relationship between a sinusoidal AC source (left) and a rotating vector (right) a A.t/ D p 2Asin.!t C / a B.t/ D p 2Asin!t C 2 ˆ: a C.t/ D p 2Asin (2.5)!t C C 2 : Since a A.t/, a B.t/, and a C.t/ constitute a balanced three-phase sequence, then we have: a A.t/ C a B.t/ C a C.t/ D 0: (2.6) The reader is encouraged to verify that (2.6) is correct. We may also represent the balanced sinusoidal sources in (2.5) using phasors, i.e.: NA A D A NA B D A 2 ˆ: (2.7) NA C D A C 2 : Figure 2.2 shows a balanced three-phase sequence using phasors, where the initial phase is 0. Note that the phasor denoted by NA A is leading 2= rad the phasor denoted by NA B and lagging 2= rad the phasor denoted by NA C. In this case, the balanced three-phase sequence is denominated positive sequence. If phases B and C are swapped, we obtain the so-called negative sequence that is shown in Fig. 2. and represented by the following phasors:

4 20 2 Power System Fundamentals: Balanced Three-Phase Circuits Fig. 2.2 Balanced three-phase positive sequence A C 2p rad 2p rad A A 2p rad A B Fig. 2. Balanced three-phase negative sequence A B 2p rad 2p rad A A 2p rad A C NA A D A 0 NA B D A C 2 ˆ: (2.) NA C D A 2 : In power systems, the reference phasor is generally indicated using the letter R, the phasor lagging 120 ı (or 2 rad) using the letter S, and the phasor leading 120ı rad) using the letter T. That is: (or 2

5 2. Balanced Three-Phase Voltages and Currents 21 NA R D A 0 ı NA S D A 120 ı ˆ: NA T D A C120 ı : Therefore, in power systems, a balanced three-phase positive sequence is generally represented as RST, while a negative one as RTS. 2. Balanced Three-Phase Voltages and Currents In this section, we analyze voltages and currents in balanced three-phase circuits with different connections Balanced Three-Phase Voltages In a balanced three-phase circuit, we identify two balanced voltage sequences, namely phase voltages and line voltages, as described below and illustrated in Fig Phase voltages are defined as the voltages between each phase and a reference point known as common star point, usually denoted as N (upper plot of Fig. 2.4). That is: Fig. 2.4 Phase (upper plot) and line (lower plot) voltages in a balanced three-phase network U R U S U T R S T N R S T U RS U ST U TR

6 22 2 Power System Fundamentals: Balanced Three-Phase Circuits NU R D NU RN D U F 0 ı NU S D NU SN D U F 120 ı ˆ: NU T D NU TN D U F C120 ı (2.9) where U F is the magnitude of the phase voltage. Since phase voltages constitute a balanced sequence, we have: NU R C NU S C NU T D 0: (2.10) On the other hand, each line voltage is defined as the difference of two phase voltages (lower plot of Fig. 2.4). That is: NU RS D NU R NU S D p U F 0 ı NU ST D NU S NU T D p U F 90 ı ˆ: NU TR D NU T NU R D p (2.11) U F 150 ı : Note that: NU RS C NU ST C NU TR D 0 (2.12) i.e., the line voltages also constitute a balanced sequence. Phasor diagrams for both phase voltages and line voltages are shown in Fig Fig. 2.5 Balanced phase and line voltages U TR U T U RS U R U S 0 U ST

7 2. Balanced Three-Phase Voltages and Currents Balanced Three-Phase Currents In a balanced three-phase network, the line currents constitute a balanced sequence, i.e.: NI R D I L. ' 0/ ı NI S D I L. ' 120/ ı (2.1) ˆ: NI T D I L. ' C 120/ ı where: I L is the magnitude of the line current and ' is the angle of a phase voltage with respect to the corresponding line current. Line currents are shown in Fig Note that: NI R C NI S C NI T D 0: (2.14) Illustrative Example 2.1 Currents in a balanced three-phase delta-connected load We consider the balanced three-phase delta-connected load (impedance NZ per phase) depicted in Fig We show below that if this load is supplied by a balanced three-phase line-current sequence (NI R, NI S, NI T ), then the delta currents, i.e., the currents inside the delta (NI RS, NI ST, NI TR ), constitute a balanced sequence as well. From Fig. 2.7, we obtain: Fig. 2.6 Balanced line currents Ī R Ī S Ī T Fig. 2.7 Balanced delta currents R I R I RS Z Z S T I S I T I ST Z I TR

8 24 2 Power System Fundamentals: Balanced Three-Phase Circuits NI R C NI TR D NI RS NI S C NI RS D NI ST ˆ: NI T C NI ST D NI TR : (2.15) Since: NI R C NI S C NI T D 0 and: NI S C NI T C NI R D 0 subtracting the above two expressions renders: Considering (2.15) and (2.16), we obtain: < NI RS C NI ST D NI S NI : ST C NI TR D NI T NI RS C NI ST C NI TR D 0 whichinmatrixformis: Solving for the delta currents yields: 2 4 NI RS NI ST NI TR NI RS C NI ST C NI TR D 0: (2.16) NI RS NI ST NI TR D NI S 5 D 4 NI T 5 : 0 NI S 4 NI T or: 2 4 NI RS NI ST NI TR D NI S NI T 5 :

9 2. Balanced Three-Phase Voltages and Currents 25 Thus, we can express delta currents as a function of line currents as: NI RS D 1 2 NI S NI T I ST D 1 I NS NI T ˆ: I TR D 1 I NS C 2NI T : Since NI R C NI S C NI T D 0, we get: NI RS D 1 I NR NI S NI ST D 1 I NS NI T ˆ: NI TR D 1 I NT NI R or: NI RS D 1 NI R C0 p ı NI ST D 1 NI S C0 p ı ˆ: NI TR D 1 NI T C0 p ı or finally: NI RS D p 1 I L. ' C 0/ ı NI ST D p 1 I L. ' 90/ ı ˆ: NI TR D p 1 I L. ' C 150/ ı : We conclude that if the line currents used to supplied a balanced delta-connected load constitute a balanced three-phase sequence, then the delta currents NI RS, NI ST, and NI TR have equal magnitude and equal angle separation, i.e., the delta currents constitute a balanced three-phase sequence as well. Figure 2. visualizes the relationship between line currents and delta currents in a balanced three-phase delta-connected load.

10 26 2 Power System Fundamentals: Balanced Three-Phase Circuits Fig. 2. Balanced line and delta currents in a balanced delta-connected load Ī T Ī TR 0 Ī RS 0 Ī R 0 Ī S Ī ST R Ī TR Ī R Ī RS Ī S Ī ST Fig. 2.9 Balanced three-phase delta-connected generator Ī T S T Illustrative Example 2.2 Currents in a balanced three-phase delta-connected generator We obtain below the relationships between the line and delta currents in a balanced delta-connected generator as the one in Fig Note that a simplified circuit diagram is used to show exclusively the reference directions for the currents. If instead of a delta-connected load we consider a delta-connected generator, the current balance equations become (Fig. 2.9): NI R C NI TR D NI RS NI S C NI RS D NI ST ˆ: NI T C NI ST D NI TR : Since NI RS C NI ST C NI TR D 0, we get (see (2.16)): NI RS D 1 I NS NI R NI ST D 1 I NT NI S ˆ: NI TR D 1 I NR NI T

11 2. Balanced Three-Phase Voltages and Currents 27 or: NI RS D 1 NI R C0 p ı NI ST D 1 NI S C0 p ı or similarly: and finally: ˆ: NI TR D 1 NI T C0 p ı NI RS D 1 NI R 150 p ı NI ST D 1 NI S 150 p ı ˆ: NI TR D 1 NI T 150 p ı NI RS D 1 NI S 0 p ı NI ST D 1 NI T 0 p ı ˆ: NI TR D 1 NI R 0 p ı : (2.17) Considering: NI R D I G 0 ı NI S D I G 120 ı ˆ: NI T D I G C120 ı (2.1) where I G is the magnitude of the line current, we have: NI RS D p 1 I G 150 ı NI ST D p 1 I G C90 ı ˆ: NI TR D p 1 I G 0 ı : (2.19) From Eqs. (2.1) and (2.19) above, we conclude that the line and delta currents in a balanced three-phase delta-connected generator are balanced sequences.

12 2 2 Power System Fundamentals: Balanced Three-Phase Circuits Fig Voltage sources: wye (upper plot) and delta (lower plot) connected Z y E R I R R N Z y E S I S S Z y E T I T T E C I TR E A Ī R R Z d Z d E B Z d I ST I RS I S I T S T 2.. Equivalence Wye-Delta To preserve balance, three balanced voltage sources can be either wye (y) ordelta (d) connected, as shown in the upper and lower plots of Fig. 2.10, respectively. In the wye connection, point N, known as the common start point, is considered as the reference for phase voltages. We consider the balanced voltage source sequence: NE R D E F 0 ı NE S D E F 120 ı (2.20) ˆ: NE T D E F C120 ı where E F is the magnitude of each voltage source, while the balanced voltage-source current sequence is: NI R D I L. ' 0/ ı NI S D I L. ' 120/ ı (2.21) ˆ: NI T D I L. ' C 120/ ı

13 2. Balanced Three-Phase Voltages and Currents 29 where I L is the magnitude of the line current. We derive below equivalence conditions for the wye and delta connections considering the circuits in Fig On the one hand, these circuits should be equivalent under no load conditions, i.e., if currents are null: NI R D NI S D NI T D 0: (2.22) Then: NE R NE S D NE A D p NE R 0 ı D p E F 0 ı NE S NE T D NE B D p NE S 0 ı D p E F 90 ı ˆ: NE T NE R D NE C D p NE T 0 ı D p E F 150 ı : (2.2) On the other hand, these circuits should be equivalent under load conditions as well. Thus: NI TR NI RS D NI R NI RS NI ST D NI S (2.24) ˆ: NI ST NI TR D NI T : Considering (2.17) and (2.1) renders: NI RS D p 1 NI S 0 ı NI ST D p 1 NI T 0 ı (2.25) ˆ: NI TR D p 1 NI R 0 ı or: NI RS D p 1 I G 150 ı NI ST D p 1 I G C90 ı ˆ: NI TR D p 1 I G 0 ı : (2.26) Under load conditions, line voltages in both connections should be equal. Thus, from Fig. 2.10, wehave: NE A C NI RS NZ d D NE S C NZ y NI S NZ y NI R C NE R (2.27)

14 0 2 Power System Fundamentals: Balanced Three-Phase Circuits or: or: Considering (2.2), we obtain: NE A C NI RS NZ d D NE R NE S C NI S NI R N Z y : (2.2) NE A C NI RS NZ d D NE A C p NI R 150 ı NZ y (2.29) 1 p NI R 150 ı NZ d D p NI R 150 ı NZ y (2.0) as well as: 1 p NZ d D p NZ y (2.1) and finally: NZ d D NZ y : (2.2) We conclude that the equivalence conditions for the wye and delta connections in the circuits in Fig are as follows: NE A D p NE R 0 ı NE B D p NE S 0 ı ˆ: NE C D p (2.) NE T 0 ı and: NZ d D NZ y (2.4) or alternatively: NE R D p 1 NE A 0 ı NE S D p 1 NE B 0 ı ˆ: NE T D p 1 NE C 0 ı (2.5)

15 2. Balanced Three-Phase Voltages and Currents 1 Fig Illustrative Example 2.: balanced Z Ē R iy three-phase circuit Ī 1 Ī 2 Ū R Ī R ĪRS Z cd Ī TR Z cd Z iy Ē S Ū S Ī S Ī ST Z cd Z iy Ē T Ū T Ī T Ī and: NZ y D 1 N Z d : (2.6) Illustrative Example 2. Balanced three-phase circuit We consider the circuit depicted in Fig in which voltage sources constitute a known balanced three-phase positive sequence and impedances NZ iy and NZ cd are also known. We compute below: 1. Currents NI R, NI S, and NI T. 2. Currents NI RS, NI ST, and NI TR.. Voltages NU R, NU S, and NU T. Since voltage sources constitute a balanced three-phase positive sequence, we have: NE R NE S DN 2 NE R ˆ: NE T DN NE R where N D ı. Additionally note that N 2 D N D ı The circuit in Fig is solved below using the mesh-current method [5]: 2 2 NZ iy C NZ cd NZ cd NZ iy 2 NI 1 2 NE R NE S 4 NZ cd NZ cd NZ cd 5 4 I N2 5 D NZ iy NZ cd 2 NZ iy C NZ cd NI NE S NE T

16 2 2 Power System Fundamentals: Balanced Three-Phase Circuits or: NZ iy C NZ cd NZ cd NZ iy NI 1 NE R 1 N 2 4 NZ cd NZ cd NZ cd 5 4 I N2 5 D : NZ iy NZ cd 2 NZ iy C NZ cd NI NE R N 2 N Solving for the currents: 2 NI NZ iy C NZ cd NZ cd NZ iy 4 NI 2 5 D 4 NZ cd NZ cd NZ cd 5 NI NZ iy NZ cd 2 NZ iy C NZ cd NE R 1 N NE R 1 2 and: or: NI 1 4 NI D 6 Z NZ NI iy C NZ 4 1 1C N iy 1 N cd NZ 5 E N R cd N 2 N NE R NI 1 D NZ iy C NZ cd NI 2 D.1 N / NE R NZ iy C NZ cd ˆ: NI D N NE R : NZ iy C NZ cd Thus, currents NI R, NI S, and NI T are: NE R NI R D NI 1 D NZ iy C 1 N Z cd NI S D NI NI 1 D. N 1/ NE R NZ iy C 1 Z N cd ˆ: NI T D NI D N NE R NZ iy C 1 N Z cd D N 2 NE R NZ iy C 1 N Z cd (2.7)

17 2. Balanced Three-Phase Voltages and Currents or: NI R D NI S D NE R NZ iy C 1 N Z cd NE S NZ iy C 1 N Z cd ˆ: NI T D NE T NZ iy C 1 N Z cd : On the other hand, currents NI RS, NI ST, and NI TR are computed as: NI RS D NI 1 NI 2 D E R. 1 CN / NZ iy C NZ cd E NI ST D NI NI 2 D N R. N 1 CN / NZ iy C NZ cd E ˆ: NI TR D NI 2 D N R. N 1/ NZ iy C NZ cd or: NI RS D N E R 1 N 2 NZ iy C NZ cd E NI ST D N R N 2 N D NZ iy C NZ cd E ˆ: NI TR D N R. N 1/ D NZ iy C NZ cd N 2 NE R 1 1 N E D N S NZ iy C NZ cd N NE R 1 1 N E D N T NZ iy C NZ cd 1 N 2 NZ iy C NZ cd 1 N 2 NZ iy C NZ cd : Note that the relationship between currents NI RS, NI ST, NI TR and NI R, NI S, NI T is as follows: NI RS NI R D N I ST NI S D N I TR NI T D 1 N 2 D p 0 ı D 1 p 0 ı : Finally, voltages NU R, NU S, and NU T are computed as: NU R D NE R NI R NZ iy NU S D NE S NI S NZ iy ˆ: NU T D NE T NI T NZ iy (2.)

18 4 2 Power System Fundamentals: Balanced Three-Phase Circuits or: NU R D NE R NU S D NE S NE R NZ iy NZ iy C 1 N Z cd NE S NZ iy NZ iy C 1 N Z cd ˆ: NU T D NE T NE T NZ iy NZ iy C 1 N Z cd and finally: ˆ: NU R D NE R NU S D NE S NU T D NE T NZ iy 1 A NZ iy C 1 Z N cd 1 NZ iy NZ iy C 1 Z N A cd 1 NZ iy NZ iy C 1 Z N A : cd Note that (NI R NI S NI T ), (NI RS NI ST NI TR ), and ( NU R NU S NU T ) constitute balanced threephase positive sequences. Currents and voltages have been obtained by analyzing the three-phase circuit depicted in Fig However, note that the resulting equations for line currents (2.7) and phase voltages (2.) are decoupled per phase. Thus, instead of using the three-phase circuit in Fig. 2.11, it is possible to use the three equivalent single-phase circuits, one per phase, depicted in Fig Illustrative Example 2.4 Equivalent single-phase circuits We consider the balanced three-phase circuit depicted in Fig Impedances NZ id and NZ cy, as well as voltage sources are known. Moreover, voltage sources constitute a balanced three-phase positive sequence and, thus: NE A NE B DN 2 NE A (2.9) ˆ: NE C DN NE A :

19 2. Balanced Three-Phase Voltages and Currents 5 Fig Illustrative Example 2.: equivalent single-phase circuits Z iy E R I R 1 Z cd I S Z iy E S 1 Z cd I T Z iy E T 1 Z cd Ē A Ī TR Ē B Ū R Ī R Z cy Z id Ī 1 Z id Ī 2 Ē C Z id I ST Ī Ī RS Ū S U T Ī S Ī T Z cy Z cy N Fig. 2.1 Illustrative Example 2.4: balanced three-phase circuit We compute below: 1. Currents NI R, NI S, and NI T. 2. Currents NI RS, NI ST, and NI TR.. Phase voltages NU R, NU S, and NU T.

20 6 2 Power System Fundamentals: Balanced Three-Phase Circuits We solve the circuit in Fig. 2.1 using the mesh-current method [5]: NZ id NZ id NZ id NI NZ id 2 NZ cy C NZ id NZ cy 5 4 I N2 5 D 4 N 2 NE A 5 : NZ id NZ cy 2 NZ cy C NZ id NI N NE A Solving for the currents: 2 2 NI 1 NZ id NZ id NZ id 4 NI 2 5 D 4 NZ id 2 NZ cy C NZ id NZ cy 5 NI NZ id NZ cy 2 NZ cy C NZ id 0 4 N 2 NE A 5 N NE A 1 2 and: and finally: 2 NI 1 4 NI 2 5 D NI C N Z cy NZ id 7 4 NZ id C NZ 4 cy NI 1 D NI 2 D ˆ: NI D 1 NZ id C NZ cy NE A 1 NZ id C NZ cy NE A 1 N 2 1 NZ id C NZ cy NE A.1 N / : 0 N 2 NE A N NE A 5 Then, we can compute currents NI R, NI S, and NI T as follows: NI R D NI 2 D N E A 1 N 2 NZ id C NZ cy NI S D NI NI 2 D N E A N 2 N NZ id C NZ cy E ˆ: NI T D NI D N A. N 1/ NZ id C NZ cy or:

21 2. Balanced Three-Phase Voltages and Currents 7 NI R D N E A 1 N 2 NZ id C NZ cy NI S D N E B 1 N 2 NZ id C NZ cy E ˆ: NI T D N C 1 N 2 : NZ id C NZ cy Next, mesh currents NI RS, NI ST, and NI TR are computed as: NI RS D NI 1 NI N 2 NE A 2 D NZ id C NZ cy N NE A NI ST D NI 1 NI D NZ id C NZ cy ˆ: NI TR D NI 1 D NE A NZ id C NZ cy or: NI RS D NI ST D NE B NZ id C NZ cy NE C NZ id C NZ cy ˆ: NI TR D NE A NZ id C NZ cy : Note that line currents (NI R, NI S, NI T ) and mesh currents (NI RS, NI ST, NI TR ) constitute balanced three-phase positive sequences. The relationship between line and mesh currents is as follows: NI RS I R D N I ST NI S D N I TR NI T D N 2 1 N 2 D 1 p 150 ı : Note that phases R, S, and T are decoupled. Thus, instead of analyzing the threephase circuit in Fig. 2.1, it is possible to consider the three equivalent single-phase circuits depicted in Fig Using these equivalent single-phase circuits, we obtain that phase voltages are equal to: NU R D NI R NZ cy NU S D NI S NZ cy ˆ: NU T D NI T NZ cy

22 2 Power System Fundamentals: Balanced Three-Phase Circuits Fig Illustrative Example 2.4: equivalent single-phase circuits 1 Z id 1 Ī R E A 1 a 2 ) Z cy 1 Z id 1 Ī S E B 1 a 2 ) Z cy 1 Z id 1 Ī T E C 1 a 2) Z cy or: NU R D NE NZ cy A 1 N 2 NZ id C NZ cy NU S DN 2 NE NZ cy A 1 N 2 NZ id C NZ cy ˆ: NU T DN NE A 1 N 2 NZ cy NZ id C NZ cy : As in the case of line and mesh currents, the phase voltages NU R, NU S, NU T constitute also a balanced three-phase positive sequence Common Star Connection We consider the balanced three-phase network depicted in Fig In this circuit, we have:

23 2. Balanced Three-Phase Voltages and Currents 9 Fig Common star connection: analysis of circuit with connection NN 0 Ē R Ē S Ē T Z Ī T N N Ī N Z Z Ī R Ī S NE R NZNI R D 0 NE S NZNI S D 0 ˆ: NE T NZNI T D 0 (2.40) and: E NI R D N R NZ E NI S D N S NZ E ˆ: NI T D N T NZ : (2.41) Adding these currents, we obtain: NI N D NI R C NI S C NI T D 1 NZ. NE R C NE S C NE T /: (2.42) Since voltage sources constitute a balanced sequence, we obtain: and, thus: NI N D 0 (2.4) NU NN 0 D 0: (2.44) That is, connection NN 0 is immaterial. Next, we analyze the same network, but in this case without connection NN 0, as depicted in Fig Solving this circuit by the mesh-current method [5], we obtain:

24 40 2 Power System Fundamentals: Balanced Three-Phase Circuits Fig Common star connection: analysis of circuit without connection NN 0 E R E T Z I R Z I 1 N N E S Z I S I 2 I T NZ 2 1 I N1 1 N I N D NE R : (2.45) 2 N 2 N Thus: and: and finally: I N1 I N2 D N E R NZ I N1 I N N 2 (2.46) 1 2 N 2 N D N E R NZ N 2 (2.47) 1 2 N 2 N E NI 1 D N R NZ ˆ: NI 2 D N NE (2.4) R Z D N E T NZ : Then, line currents can be computed as: NI R D NI 1 D 1 E NZ N R NI S D NI 2 NI 1 D 1 E NZ N R. N 1/ ˆ: NI T D NI 2 D 1 E NZ N T (2.49)

25 2. Balanced Three-Phase Voltages and Currents 41 or: NI R D 1 E NZ N R NI S D 1 E NZ N S (2.50) ˆ: NI T D 1 E NZ N T which is the same result previously obtained in Eq. (2.41) for the circuit with NN 0 connection. Note also that: NU N 0 N D NE R NI R NZ D NE R 1 NZ N E R NZ D 0 (2.51) as expected. We note once more that phase equations are decoupled: the equation of a given phase depends only on variables and constants of that phase. That is: NE R D NZNI R NE S D NZNI S (2.52) ˆ: NE T D NZNI T or, in matrix form: NE R NZ 0 0 NI R 4 E N S 5 D 4 0 NZ I NS 5 : (2.5) NE T 0 0 NZ NI T In other words, the impedance matrix is diagonal, which verifies phase decoupling. Thus, we can consider three independent single-phase networks as depicted in Fig Typically, the R equivalent single-phase circuit is used to represent the balanced three-phase circuit. Note that such single-phase circuit includes all required information to characterize the balanced three-phase circuit. The other two equivalent single-phase circuits replicate the R one circuit S lagging 120 ı, and circuit T leading 120 ı.

26 42 2 Power System Fundamentals: Balanced Three-Phase Circuits Fig Equivalent single-phase circuits Z Ē R Ī R Z Ē S Ī S Z Ē T Ī T 2.4 Instantaneous, Active, Reactive, and Apparent Power One of the most important magnitudes in three-phase circuits is the power, which is analyzed in this section Definitions The instantaneous power at any point of a three-phase circuit is defined as: p.t/ D 4 p R.t/ C p S.t/ C p T.t/ D u R.t/i R.t/ C u S.t/i S.t/ C u T.t/i T.t/: (2.54) Thus, considering a balanced three-phase circuit, p.t/ can be computed as: p.t/ D p 2U F sin.!t/ p 2I L sin.!t '/ C p 2U F sin!t 2 p2il sin!t 2 '

27 2.4 Instantaneous, Active, Reactive, and Apparent Power 4 C p 2U F sin!t C 2 p2il sin!t C 2 ' (2.55) where: U F is the RMS value of the phase voltage and I L is the RMS value of the line current. Rearranging terms: p.t/ D U F I L cos' cos.2œ!t 0 '/ C U F I L cos' cos 2!t 2 C U F I L cos' cos 2!t C 2 ' ' (2.56) and: p.t/ D U F I L cos' D U L p I L cos': (2.57) That is: p.t/ D p U L I L cos' (2.5) where U L is the RMS value of the line voltage. Thus, the instantaneous power at any point of a balanced three-phase circuit is time invariant. The three-phase active power, denoted by P, is equal to the instantaneous power and, thus: P 4 D p.t/ D p U L I L cos': (2.59) The fact that in three-phase power systems the three-phase active power is timeinvariant makes these systems preferable over single-phase systems, in which the active power has a nonzero average value, but is alternating. Alternating active power results in vibrations and long-term mechanical issues, while time-invariant active power does not. This is indeed a key reason for using three-phase systems instead of single-phase ones. The per-phase complex power is computed as: ˆ: NS R D NU R NI R D U L p I L ' NS S D NU S NI S D U L p I L 2 C 2 C ' D U L p I L ' NS T D NU T NI T D U L 2 p I L 2 C ' D U L p I L ': (2.60)

28 44 2 Power System Fundamentals: Balanced Three-Phase Circuits Then, the three-phase complex power, NS, is computed as: where: NS D NS R C NS S C NS T D p U L I L ' D p U L I L cos' C j p U L I L sin ' D P C jq (2.61) Q 4 D p UL I L sin' (2.62) is the three-phase reactive power. The magnitude of the three-phase complex power (S) is the so-called three-phase apparent power How to Measure Power? Note that the active power can be computed using either of the two expressions below: or: P D U R I R cos' R C U S I S cos' S C U T I T cos' T (2.6) P D p U L I L cos': (2.64) Equation (2.64) requires the system to be balanced, while Eq. (2.6) does not. Active power is generally measured using a watt-meter that multiplies three terms: the RMS value of current, the RMS value of the voltage, and the cosine of the angle between these two signals (see (2.64)). On the other hand, reactive power is measured as active power, but using varmeters that, instead of multiplying by the cosine, multiply by the sine. Finally, apparent power is measured using a volt-meter and an amp-meter. If we need to measure energy, then we should use an energy meter, which is a watt-meter that integrates over time. 2.5 Why Three-Phase Power? We illustrate below the economic advantage of three-phase power versus singlephase power through an example.

29 2.5 Why Three-Phase Power? 45 We consider the transfer of apparent power S [MVA] over a distance d [km] with a phase-to-neutral voltage U [kv]. Additionally, we consider that the available conductor admits a maximum current density ı [A=cm 2 ]. Using single-phase ac power, we have: S 1 D S U 1 D U (2.65) I 1 S U : Then, the conductor section should be: and the required material is: while losses are: A 1 D I 1 ı D S ıu (2.66) M 1 D 2A 1 d D 2 I 1 ı d D 2 Sd ıu (2.67) P L 1 2I2 1 d D 2 S2 A 1 U ıud D 2 S ıd (2.6) 2 S U where is the resistivity of the material used. Using three-phase ac power, we have: In this case, the conductor section should be: S D S U D p U (2.69) I S p D S U U : A D I ı D S ıu (2.70) and the required material is: M D A d D S ıu d D Sd ıu (2.71) while losses in this case are: P L I2 d D S2 A 9U ıud D S ıd: (2.72) 2 S U

30 46 2 Power System Fundamentals: Balanced Three-Phase Circuits Note that, on the one hand: M 1 M D 2 (2.7) i.e., the material required to transmit apparent power S [MVA] over a distance d [km] with a phase-to-neutral voltage U [kv] considering a single-phase ac line is about twice the material needed if a three-phase ac line is used. On the other hand, we have: P L 1 P L D 2 (2.74) i.e., the losses of transmitting apparent power S [MVA] over a distance d [km] with a phase-to-neutral voltage U [kv] considering a single-phase ac line are about twice the losses if a three-phase ac line is used. This simple back-of-the-envelope analysis illustrates the economic advantage of building/using a three-phase transmission line over a single-phase one. 2.6 Per-Unit System This section defines and describes the per-unit system, which is important in power systems spanning different voltage levels Motivation Power transformers interconnect power system areas with different voltage levels. This is a problem at the time of analyzing these systems since all magnitudes need to be transformed to a single voltage level. However, if a per-unit analysis is performed, this problem disappears and a unique voltage level is obtained. This greatly simplifies the subsequent analysis Per-Unit System Definition Any electrical variable or parameter (voltage, current, power, impedance) can be expressed as a function of its own units or with respect to a reference value known as base value, i.e.:

31 2.6 Per-Unit System 47 Fig. 2.1 Illustrative Example 2.5: single-phase circuit using real magnitudes I V = 220 V Z= 55 W m D M M B (2.75) where: m is the per-unit value, M is the value of the variable/parameter in its own units, and M B is the base value. Then, instead of analyzing a circuit using actual values, it is possible to analyze it using per-unit values. This generally simplifies the subsequent analysis. Illustrative Example 2.5 Illustration of per-unit analysis We consider the single-phase circuit depicted in Fig Taking into account that V D 220 V and Z D 55, we obtain the current I. To do so, we analyze the circuit using the per-unit system considering a base-voltage value of 220 V and a base-current value of 2 A. First, we obtain the equivalent per-unit circuit by transforming the voltage and impedance values to per-unit values. On the one hand, we compute the per-unit voltage v as follows: v D V V D 220 B 220 D 1 puv: In order to obtain the per-unit impedance, first we need to compute the baseimpedance value, which is obtained as the base-voltage value divided by the basecurrent value, i.e.: Z B D VB I D 220 D 110 : B 2 Then, we obtain the per-unit impedance as: z D Z Z D 55 D 0:5 pu: B 110 Finally, we derive the equivalent single-phase circuit using per-unit values and depict it in Fig

32 4 2 Power System Fundamentals: Balanced Three-Phase Circuits Fig Illustrative Example 2.5: single-phase circuit using per-unit magnitudes i v = 1 puv z = 0.5puW Next we can compute the per-unit current i as: i D v z D 1 0:5 D 2 pua: Finally, we can obtain the actual current I as: I D ii B D 2 2 D 4 A: Using per-unit analysis in Illustrative Example 2.5 is not convenient since it is possible to directly compute current I from the circuit in Fig. 2.1 as V Z D D 4 A. However, the use of a per-unit system to analyze power systems with multiple voltage levels is most convenient for two reasons (provided that the per-unit system is properly defined): 1. Power transformers disappear from the equivalent single-phase circuit. This is further analyzed and shown in Sect.. of Chap.. 2. Voltage values are close to 1 puv, which allows detecting errors. Besides these two important advantages, there is an additional advantage of using a per-unit analysis for three-phase power systems:. The per-unit impedances of machines generally take values within tight bounds, independently of their nominal values, which facilitates their characterization Definition of Base Values An appropriate definition of the base values is as follows: 1. We select a common single-phase base power, typically 1= of the rated power of the component of highest rated power, i.e.: S B D 1 S Nk (2.76)

33 2.6 Per-Unit System 49 where: S B is the single-phase base power and S Nk is the three-phase rated power of component k. 2. Recalling that power transformers separate the voltage zones of the network, we select the base voltage in one zone as the rated phase voltage of one component in that zone, i.e.: U Bi D 1 p U Nj (2.77) where: U Bi is the base-voltage value at zone i and U Nj is the rated three-phase voltage of component j in zone i.. The base-voltage values in other zones are determined strictly complying with the transformation ratios of the power transformers, which makes transformers disappear from equivalent single-phase circuits, i.e.: U Bi D U Bj U i U j (2.7) where: U Bi is the base-voltage value in zone i, U Bj is the base-voltage value in zone j, and U i =U j is the three-phase transformation ratio of the power transformer coupling zones i and j. 4. We define the base-current value and the base-impedance value per zone as and: I Bi D S B U Bi (2.79) respectively, where: I Bi is the base-current value at zone i and Z Bi is the base-impedance value at zone i. Z Bi D U2 Bi S B (2.0) 5. We specify phase shifts due to power transformers including delta and zigzag connections. We explain transformer phase shifts in Chap..

34 50 2 Power System Fundamentals: Balanced Three-Phase Circuits Fig Illustrative Example 2.6: power system G T 1 L T 2 M Illustrative Example 2.6 Per-unit analysis: base values We consider the three-phase power system depicted in Fig This system comprises a generator, two power transformers, a transmission line, and a motor. The rated powers and voltages of these components are provided below: Generator G: 60MVA,11kV. Transformer T 1 : 60 MVA, 12/12 kv. Line L: 70 MVA, 12 kv. Transformer T 2 : 60 MVA, 125/5 kv. Motor M: 0MVA,5kV. Using these data, we determine below the number of voltage zones, and the base values of each zone. Each power transformer divides the system into two voltage zones. Thus, we have three voltage zones, namely, generator zone (zone 1), line zone (zone 2), and motor zone (zone ). Next, we follow the procedure described above to determine the base values in each zone: 1. We select the single-phase base power as 1/ of the rated power of the motor, which has the highest rated power, i.e., S B D 0 D 26:667 MVA. Note that this base-power value is the same for all voltage zones. 2. We fix the base-voltage value of one of the zones as the rated phase voltage of one of the components of that zone. For example, we select the base-voltage value in the generator zone as the rated phase voltage of the generator, i.e., U B1 D p 11 D 6:51 kv.. We compute the base-voltage values in other zones by using the transformation ratios of transformers that separate each zone, i.e., U B2 D p D 69:59 kv 12 and U B D p D 2:794 kv We define the base-current value and the base-impedance value per zone using the corresponding base-power and base-voltage values, as well as Eqs. (2.79) and (2.0), respectively. For example, the base-current value in the generator 26: zone is I B1 D D 4:199 ka, while the base-impedance value in the 6: : line zone is Z B2 D D 1:01. 26: Table 2.1 summarizes the base values in each zone.

35 2.6 Per-Unit System 51 Table 2.1 Illustrative Example 2.6: base values Value Generator zone Line zone Motor zone S B [MVA] V B [kv] I B [ka] Z B [] Note that the base values defined in this section are single-phase base values. However, three-phase base values, equivalent to single-phase ones, are similarly defined. If we define the three-phase base-power value as: and the line base-voltage value in each zone as: S B 4 D S B (2.1) U Bi 4 D p UBi (2.2) then single-phase and three-phase base-current and single-phase and three-phase base-impedance values coincide. That is: and: I Bi D S B S B p D p p D S B D I Bi (2.) UBi UBi U Bi pubi 2 Z Bi D U2 B S B D S B D U2 Bi S B D Z Bi (2.4) respectively Per-Unit Analysis Procedure The procedure to analyze power systems using a per-unit system comprises the five steps below: 1. Define base values as explained in Sect Transform the three-phase power system into an equivalent single-phase circuit in which impedances are expressed in per unit.. Apply the operating conditions (in per unit). 4. Solve the circuit (in per unit). 5. Obtain actual values by multiplying per-unit values by the corresponding base values.

36 52 2 Power System Fundamentals: Balanced Three-Phase Circuits A number of examples to illustrate this procedure to analyze power systems are provided in Sect..6 of Chap Summary and Further Reading This chapter provides an overview of the fundamentals of power systems. First, we define balanced sequences, balanced voltages and currents, and powers. Second, we illustrate why three-phase power systems are used instead of single-phase ones. Finally, we provide an introduction to the analysis of power systems considering a per-unit system, which is used and further analyzed in the following chapters of this book. Basic references regarding power system analysis include Kothari and Nagrath [4] and Duncan Glover et al. [2]. Advanced references include Gómez-Expósito et al. [] and Bergen and Vittal [1]. 2. End-of-Chapter Exercises 2.1 Why three-phase power systems are used instead of single-phase ones? 2.2 List the advantages of analyzing power systems using a per-unit system. 2. Consider the three-phase circuit depicted in Fig Voltage sources constitute a balanced three-phase positive sequence: NE R D ı V NE S D ı V ˆ: NE T D ı V: Impedances NZ i are equal to j5, impedances NZ` are equal to j15, and impedances NZ RS D NZ ST D NZ TR are equal to j0. Using these data: 1. Compute currents NI R, NI S, and NI T, as well as currents NI RS, NI ST, and NI TR. 2. Compute voltages NU R1, NU S1, and NU T1, as well as voltages NU R2, NU S2, and NU T2.. Recompute the above voltages and currents if an impedance of j20 is connected between nodes N and N Compute the above voltages and currents if impedances NZ RS, NZ ST, and NZ TR are equal to 10, j10, and j10, respectively. Is it possible in this case to use the equivalent single-phase circuit? 2.4 Consider a balanced three-phase load that is supplied at a line voltage of 400 kv and absorbs a line current of ı A:

37 2. End-of-Chapter Exercises 5 Z i Ē R U R1 I R Z l U R2 I RS Z TR I TR Z RS N Z i Ē S U S1 I S Z l U S2 I Z ST ST N U T 2 Z i Ē T U T 1 I T Z l Fig Exercise 2.: three-phase circuit G 1 1 T 1 2 L 1 T 2 4 G 2 L 2 L 5 7 T T 4 6 C M Fig Exercise 2.5: power system 1. Compute the instantaneous power consumed in phases R, S, and T. 2. Compute the instantaneous three-phase power consumed by the load.. Compute the reactive and apparent power consumed by the load. 2.5 Consider the three-phase power system depicted in Fig The rated powers and voltages of the system components are provided below: Generators G 1 and G 2 : 500 MVA, 20 kv. Transformers T 1 and T 2 : 200 MVA, 500/1 kv. Transformers T and T 4 : 150 MVA, 500/20 kv. Motor M: 111 MW, cos D 0: (inductive), 20 kv.

38 54 2 Power System Fundamentals: Balanced Three-Phase Circuits Fig. 2.2 Exercise 2.6: power system G 1 T 1 L 1 G 2 T 2 L 2 Using these data, determine the number of voltage zones and the base values of each zone. 2.6 Consider the three-phase power system depicted in Fig The rated powers and voltages of the system components are provided below: Generator G 1 :50MVA,12kV. Generator G 2 : 100 MVA, 15 kv. Transformer T 1 : 50 MVA, 10/1 kv. Transformer T 2 : 100 MVA, 15/1 kv. Using as base values the rated parameters of generator G 2, determine the number of voltage zones and the base values of each zone. References 1. Bergen, A.R., Vittal, V.: Power Systems Analysis, 2nd edn. Prentice Hall, Upper Saddle River, NJ (1999) 2. Duncan Glover, J., Sarma, M.S., Overbye, T.: Power System Analysis and Design, 5th edn. Cengage Learning, Stamford, CT (200). Gómez-Expósito, A., Conejo, A.J., Cañizares, C.: Electric Energy Systems: Analysis and Operation. Taylor and Francis, Boca Raton, FL (200) 4. Kothari, D.P., Nagrath, I.J.: Modern Power System Analysis, 4th edn. Tata McGraw Hill Education Private Limited, New Delhi (2011) 5. Nilsson, J.W., Riedel, S.A.: Electric Circuits, 10th edn. Pearson, Upper Saddle River, NJ (2014)

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