396 Sinusoidal Steady-State Power Calculations Figure P10.61 255/0! rl V (rms) 40 ft ;200 f) 720 ft j 1500 ft N 2 rVi Ideal 6800 ft 10.65 The ideal transformer connected to the 10 ft load in Problem 10.64 is replaced with an ideal trans- former that has a turns ratio of a:\. a) What value of a results in maximum average power being delivered to the 10 ft resistor? b) What is the maximum average power? 10.62 The variable load resistor R L in the circuit shown in PSPICE Fig. PI0.62 is adjusted for maximum average power MULTISIM transfer to i? L . a) Find the maximum average power. b) What percentage of the average power developed by the ideal voltage source is delivered to R L when R L is absorbing maximum average power? c) Test your solution by showing that the power developed by the ideal voltage source equals the power dissipated in the circuit. Figure P10.62 80 ft 20 ft 40 ft 1 1:2 i 500/0° f + V (rms) 360 ft 10.63 Repeat Problem 10.62 for the circuit shown in PSPICE Fig. P10.63. MULTISIM Figure P10.63 40/0! f + V(vms)\- 10.64 Find the average power delivered to the 10 ft resis- tor in the circuit of Fig. P10.64. Figure P10.64 2.5 ft a l :201 - 7|30:11 ( i i \ ) \ Ideal I • [ideal] 10ft Sections 10.1-10.6 10.66 The hair dryer in the Practical Perspective uses a piwrecnvi 60 Hz smusoidal voltage of 120 V (rms). The heater element must dissipate 250 W at the LOW setting, 500 W at the MEDIUM setting, and 1000 W at the HIGH setting. a) Find the value for resistor R2 using the specifica- tion for the MEDIUM setting, using Fig. 10.31. b) Find the value for resistor Ri using the specifica- tion for the LOW setting, using the results from part (a) and Fig. 10.30. c) Is the specification for the HIGH setting satisfied? 10.67 As seen in Problem 10.66, only two independent PRACTICAL power specifications can be made when two resis- PERSPECTIVE * r PSPICE tors make up the healing element of the hair dryer. a) Show that the expression for the HIGH power rating (P H ) is PH = Pi PA pr where P M = the MEDIUM power rating and P L = the LOW power rating. b) If P L = 250 W and P M = 750 W, what must the HIGH power rating be? 10.68 Specify the values of R\ and R 2 in the hair dryer cir- pRAcncAt cu it in Fig. 10.29 if the low power rating is 240 W PERSPECTIVE b r 0 PSPICE and the high power rating is 1000 W Assume the MULTISIM supply voltage is 120 V (rms). (Hint: Work Problem 10.67 first.) 10.69 If a third resistor is added to the hair dryer circuit in PRACTICAL Fig. 10.29, it is possible to design to three independ- PSPICE ent power specifications. If the resistor R$ is added MULTISIM j n ser i es w ith the Thermal fuse, then the correspon- ding LOW, MEDIUM, and HIGH power circuit dia- grams are as shown in Fig. P10.69. If the three Problems 397 power settings are 600 W, 900 W, and 1200 W, respectively, when connected to a 120 V (rms) sup- ply, what resistor values should be used? Figure P10.69 LOW MEDIUM HIGH 10.70 You have been given the job of redesigning the hair PRACTICAL dryer described in Problem 10.66 for use in 'ERSPECTIVE J DESIGN England. The standard supply voltage in England is PROBLEM ± M. -mf i~J f PSPICE 220 V (rms). What resistor values will you use in MULTISIM your design to meet the same power specifications? 10.71 Repeat Problem 10.68 using single resistor values from Appendix H. Calculate the resulting low, medium, and high power ratings. 10.72 Repeat Problem 10.70 using single resistor values from Appendix H. Calculate the resulting low, medium, and high power ratings. ^.J. I H i^m CHAPTER CONTENTS 11.1 Balanced Three-Phase Voltages p. 400 11.2 Three-Phase Voltage Sources p. 401 11.3 Analysis of the Wye-Wye Circuit p. 402 11.4 Analysis of the Wye-Delta Circuit p. 407 11.5 Power Calculations in Balanced Three-Phase Circuits p. 410 11.6 Measuring Average Power in Three-Phase Circuits p. 415 /CHAPTER OBJECTIVES 1 Know how to analyze a balanced, three-phase wye-wye connected circuit. 2 Know how to analyze a balanced, three-phase wye-delta connected circuit. 3 Be able to calculate power (average, reactive, and complex) in any three-phase circuit. 398 Balanced Three-Phase Circuits Generating, transmitting, distributing, and using large blocks of electric power is accomplished with three-phase circuits. The comprehensive analysis of such systems is a field of study in its own right; we cannot hope to cover it in a single chapter. Fortunately, an understanding of only the steady-state sinusoidal behavior of balanced three-phase circuits is sufficient for engi- neers who do not specialize in power systems. We define what we mean by a balanced circuit later in the discussion. The same cir- cuit analysis techniques discussed in earlier chapters can be applied to either unbalanced or balanced three-phase circuits. Here we use these familiar techniques to develop several short- cuts to the analysis of balanced three-phase circuits. For economic reasons, three-phase systems are usually designed to operate in the balanced state. Thus, in this introduc- tory treatment, we can justify considering only balanced circuits. The analysis of unbalanced three-phase circuits, which you will encounter if you study electric power in later courses, relies heav- ily on an understanding of balanced circuits. The basic structure of a three-phase system consists of volt- age sources connected to loads by means of transformers and transmission lines. To analyze such a circuit, we can reduce it to a voltage source connected to a load via a line. The omission of the transformer simplifies the discussion without jeopardizing a basic understanding of the calculations involved. Figure 11.1 on page 400 shows a basic circuit. A defining characteristic of a bal- anced three-phase circuit is that it contains a set of balanced three-phase voltages at its source. We begin by considering these voltages, and then we move to the voltage and current relation- ships for the Y-Y and Y-A circuits. After considering voltage and current in such circuits, we conclude with sections on power and power measurement. Practical Perspective Transmission and Distribution of Electric Power In this chapter we introduce circuits that are designed to handle large blocks of electric power. These are the circuits that are used to transport electric power from the generating plants to both industrial and residential customers. We intro- duced the typical residential customer circuit as used in the United States as the design perspective in Chapter 9. Now we introduce the type of circuit used to deliver electric power to an entire residential subdivision. One of the constraints imposed on the design and oper- ation of an electric utility is the requirement that the utility maintain the rms voltage level at the customer's premises. Whether lightly loaded, as at 3:00 am, or heavily loaded, as at midaftemoon on a hot, humid day, the utility is obligated to supply the same rms voltage. Recall from Chapter 10 that a capacitor can be thought of as a source of magnetizing vars. Therefore, one technique for maintaining voltage levels on a utility system is to place capacitors at strategic loca- tions in the distribution network. The idea behind this tech- nique is to use the capacitors to supply magnetizing vars close to the loads requiring them, as opposed to sending them over the lines from the generator. We shall illustrate this concept after we have introduced the analysis of bal- anced three-phase circuits. 399 400 Balanced Three-Phase Circuits Three-phase Three-phase voltage • source line / \ \ Three-phase load Figure 11.1 • A basic three-phase circuit. 11.1 Balanced Three-Phase Voltages A set of balanced three-phase voltages consists of three sinusoidal volt- ages that have identical amplitudes and frequencies but are out of phase with each other by exactly 120°. Standard practice is to refer to the three phases as a, b, and c, and to use the a-phase as the reference phase. The three voltages are referred to as the a-phase voltage, the b-phase voltage, and the c-phase voltage. Only two possible phase relationships can exist between the a-phase voltage and the b- and c-phase voltages. One possibility is for the b-phase voltage to lag the a-phase voltage by 120°, in which case the c-phase volt- age must lead the a-phase voltage by 120°. This phase relationship is known as the abc (or positive) phase sequence. The only other possibility is for the b-phase voltage to lead the a-phase voltage by 120°, in which case the c-phase voltage must lag the a-phase voltage by 120°. This phase relationship is known as the acb (or negative) phase sequence. In phasor notation, the two possible sets of balanced phase voltages are V h = V IM /-120' v c = y m /+120- (11.1) and v m /o\ V h = 1/,,,/ + 120°, V c = ^,,,/-120°. (11.2) Equations 11.1 are for the abc, or positive, sequence. Equations 11.2 are for the acb, or negative, sequence. Figure 11.2 shows the phasor dia- grams of the voltage sets in Eqs. 11.1 and 11.2. The phase sequence is the clockwise order of the subscripts around the diagram from V a . The fact that a three-phase circuit can have one of two phase sequences must be taken into account whenever two such circuits operate in par- allel. The circuits can operate in parallel only if they have the same phase sequence. Another important characteristic of a set of balanced three-phase voltages is that the sum of the voltages is zero. Thus, from either Eqs. 11.1 or Eqs. 11.2, Figure 11.2 A Phasor diagrams of a balanced set of three-phase voltages, (a) The abc (positive) sequence, (b) The acb (negative) sequence. V fl + V„ + V r = 0. (11.3) Because the sum of the phasor voltages is zero, the sum of the instanta- neous voltages also is zero; that is, v & + v h + v c = 0. (11.4) Now that we know the nature of a balanced set of three-phase volt- ages, we can state the first of the analytical shortcuts alluded to in the introduction to this chapter: If we know the phase sequence and one voltage in the set, we know the entire set. Thus for a balanced three- phase system, we can focus on determining the voltage (or current) in one phase, because once we know one phase quantity, we know the others. NOTE: Assess your understanding of three-phase voltages by trying Chapter Problems 11.2 and 11.3. 11.2 Three-Phase Voltage Sources 401 11.2 Three-Phase Voltage Sources A three-phase voltage source is a generator with three separate wind- ings distributed around the periphery of the stator. Each winding com- prises one phase of the generator. The rotor of the generator is an electromagnet driven at synchronous speed by a prime mover, such as a steam or gas turbine. Rotation of the electromagnet induces a sinusoidal voltage in each winding. The phase windings are designed so that the sinusoidal voltages induced in them are equal in amplitude and out of phase with each other by 120°. The phase windings are stationary with respect to the rotating electromagnet, so the frequency of the voltage induced in each winding is the same. Figure 11.3 shows a sketch of a two- pole three-phase source. There are two ways of interconnecting the separate phase windings to form a three-phase source: in either a wye (Y) or a delta (A) configura- tion. Figure 11.4 shows both, with ideal voltage sources used to model the phase windings of the three-phase generator. The common terminal in the Y-connected source, labeled n in Fig. 11.4(a), is called the neutral terminal of the source. The neutral terminal may or may not be available for exter- nal connections. Sometimes, the impedance of each phase winding is so small (com- pared with other impedances in the circuit) that we need not account for it in modeling the generator; the model consists solely of ideal voltage sources, as in Fig. 11.4. However, if the impedance of each phase winding is not negligible, we place the winding impedance in series with an ideal sinusoidal voltage source. All windings on the machine are of the same construction, so we assume the winding impedances to be identical. The winding impedance of a three-phase generator is inductive. Figure 11.5 shows a model of such a machine, in which is the winding resistance, and X w is the inductive reactance of the winding. Because three-phase sources and loads can be either Y-connected or A-connected, the basic circuit in Fig. 11.1 represents four different configurations: Source Y Y A A Load Y A Y A We begin by analyzing the Y-Y circuit. The remaining three arrangements can be reduced to a Y-Y equivalent circuit, so analysis of the Y-Y circuit is the key to solving all balanced three-phase arrangements. We then illus- trate the reduction of the Y-A arrangement and leave the analysis of the A-Y and A-A arrangements to you in the Problems. Axis of a-phase winding Axis of c-phasc windine \ Axis of b-phase winding Stator Figure 11.3 A A sketch of a three-phase voltage source. Figure 11.4 A The two basic connections of an ideal three-phase source, (a) A Y-connected source, (b) A A-connected source. 402 Balanced Three-Phase Circuits R, .Ifiw R,r jx w iv b fX* R,r ^vw- JX W JXu (a) (b) Figure 11.5 • A model of a three-phase source with winding impedance: (a) a Y-connected source; and (b) a A-connected source. 11.3 Analysis of the Wye-Wye Circuit Figure 11.6 illustrates a general Y-Y circuit, in which we included a fourth conductor that connects the source neutral to the load neutral. A fourth conductor is possible only in the Y-Y arrangement. (More about this later.) For convenience, we transformed the Y connections into "tipped- over tees." In Fig. 11.6, Z ga , Z gb , and Z gc represent the internal impedance associated with each phase winding of the voltage generator; Z la , Z lb , and Z lc represent the impedance of the lines connecting a phase of the source to a phase of the load; Z 0 is the impedance of the neutral conductor con- necting the source neutral to the load neutral; and ZA, Z B , and ZQ repre- sent the impedance of each phase of the load. We can describe this circuit with a single node-voltage equation. Using the source neutral as the reference node and letting V N denote the node voltage between the nodes N and n, we find that the node-voltage equation is Z 0 + V N -V a ZA + Z ia + -¾ b'n Zn + ZII, + Z lb + 'gb VN ~ Ve'n Zc + Z lc + Z fiC = 0. (11.5) v • 'an V • * c n v 1 t C T 7 ^6- ) V b - n Zgb a I n b c Zia z„ Zib Zlc A laA N i t ;c Z B Z A Z c Figure 11.6 • A three-phase Y-Y system. 11.3 Analysis of the Wye-Wye Circuit 403 This is the general equation for any circuit of the Y-Y configuration depicted in Fig. 11.6. But we can simplify Eq. 11.5 significantly if we now consider the formal definition of a balanced three-phase circuit. Such a circuit satisfies the following criteria: 1. The voltage sources form a set of balanced three-phase voltages. In Fig. 11,6, this means that V a - n , V b < n , and V c < n are a set of bal- anced three-phase voltages. 2. The impedance of each phase of the voltage source is the same. In Fig. 11.6, this means that Z ga = Z gb = Z gc . 3. The impedance of each line (or phase) conductor is the same. In Fig. 11.6, this means that Zi a = Z^ = Z\ c . 4. The impedance of each phase of the load is the same. In Fig. 11.6, this means that Z A = Z B = ZQ. There is no restriction on the impedance of a neutral conductor; its value has no effect on whether the system is balanced. If the circuit in Fig. 11.6 is balanced, we may rewrite Eq. 11.5 as < Conditions for a balanced three-phase circuit 1^ + A) • a r T\ l * IVn I * r (11.6) where Z& - Z A + Z la + Zga - Z B + Z ]b + Zgb — Z c + Z lc + Z gc- The right-hand side of Eq. 11.6 is zero, because by hypothesis the numera- tor is a set of balanced three-phase voltages and Z^ is not zero. The only value of V N that satisfies Eq. 11.6 is zero. Therefore, for a balanced three- phase circuit, VM = 0. (11.7) Equation 11.7 is extremely important. If V N is zero, there is no differ- ence in potential between the source neutral, n, and the load neutral, N; consequently, the current in the neutral conductor is zero. Hence we may either remove the neutral conductor from a balanced Y-Y configuration (It, = 0) or replace it with a perfect short circuit between the nodes n and N (V N = 0). Both equivalents are convenient to use when modeling bal- anced three-phase circuits. We now turn to the effect that balanced conditions have on the three line currents. With reference to Fig. 11.6, when the system is balanced, the three line currents are f laA h\i I Va'n" Z A + Z la v„„ - ZB + Z\b v c - n - v N + z ga V N + z gb V N »cC Z<i> \w b'n V c < n %C + Z\c + Zee Z„ (11.8) (11.9) (11.10) We see that the three line currents form a balanced set of three-phase cur- rents; that is, the current in each line is equal in amplitude and frequency and is 120° out of phase with the other two line currents. Thus, if we calcu- late the current I aA and we know the phase sequence, we have a shortcut Balanced Three-Phase Circuits a' T X :) I z •^ga a Z ia c A Z A —o v 0 .„i Figure 11.7 A A single-phase equivalent circuit. N for finding I bB and I cC . This procedure parallels the shortcut used to find the b- and c-phase source voltages from the a-phase source voltage. We can use Eq. 11.8 to construct an equivalent circuit for the a-phase of the balanced Y-Y circuit. From this equation, the current in the a-phase conductor line is simply the voltage generated in the a-phase winding of the generator divided by the total impedance in the a-phase of the circuit. Thus Eq. 11.8 describes the simple circuit shown in Fig. 11.7, in which the neutral conductor has been replaced by a perfect short circuit. The circuit in Fig. 11.7 is referred to as the single-phase equivalent circuit of a bal- anced three-phase circuit. Because of the established relationships between phases, once we solve this circuit, we can easily write down the voltages and currents in the other two phases. Thus, drawing a single- phase equivalent circuit is an important first step in analyzing a three- phase circuit. A word of caution here. The current in the neutral conductor in Fig. 11.7 is I aA , which is not the same as the current in the neutral conduc- tor of the balanced three-phase circuit, which is I, LA + IhR + I, la A bB IcC- (11.11) +-f- VAB ' ? + T + V BC + V AN VCN Z B ZA - — iN Zc Figure 11.8 A Line-to-line and line-to-neutral voltages. Thus the circuit shown in Fig. 11.7 gives the correct value of the line cur- rent but only the a-phase component of the neutral current. Whenever this single-phase equivalent circuit is applicable, the line currents form a balanced three-phase set, and the right-hand side of Eq. 11.11 sums to zero. Once we know the line current in Fig. 11.7, calculating any voltages of interest is relatively simple. Of particular interest is the relationship between the line-to-line voltages and the line-to-neutral voltages. We establish this relationship at the load terminals, but our observations also apply at the source terminals. The line-to-line voltages at the load termi- nals can be seen in Fig. 11.8. They are V AB , V BC , and VCA> where the dou- ble subscript notation indicates a voltage drop from the first-named node to the second. (Because we are interested in the balanced state, we have omitted the neutral conductor from Fig. 11.8.) The line-to-neutral voltages are V AN , V BN , and V CN . We can now describe the line-to-line voltages in terms of the line-to-neutral voltages, using Kirchhoff s voltage law: AB AN VBN, (11.12) BC VRN V CN , (11.13) CA 'CN AN- (11.14) To show the relationship between the line-to-line voltages and the line-to-neutral voltages, we assume a positive, or abc, sequence. Using the line-to-neutral voltage of the a-phase as the reference, VAN = V*/01, BN VA/-120-, VCN = v,/+i2o; (11.15) (11.16) (11.17) 11.3 Analysis of the Wye-Wye Circuit 405 where V^ represents the magnitude of the line-to-neutral voltage. Substituting Eqs. 11.15-11.17 into Eqs. 11.12-11.14, respectively, yields AB BC Vs /0° - ^/-120° = V3V, h /30°, (11.18) V* /-120° - V* /120° = s/Wj, /-90°, (11.19) Y CA = J^/12GT - ^/0£ = V3> <6 /150°. (11.20) Equations 11.18-11.20 reveal that 1. The magnitude of the line-to-line voltage is V3 times the magni- tude of the line-to-neutral voltage. 2. The line-to-line voltages form a balanced three-phase set of voltages. 3. The set of line-to-line voltages leads the set of line-to-neutral volt- ages by 30°. We leave to you the demonstration that for a negative sequence, the only change is that the set of line-to-line voltages lags the set of line-to-neutral voltages by 30°. The phasor diagrams shown in Fig. 11.9 summarize these observations. Here, again, is a shortcut in the analysis of a balanced sys- tem: If you know the line-to-neutral voltage at some point in the circuit, you can easily determine the line-to-line voltage at the same point and vice versa. We now pause to elaborate on terminology. Line voltage refers to the voltage across any pair of lines; phase voltage refers to the voltage across a single phase. Line current refers to the current in a single line; phase current refers to current in a single phase. Observe that in a A connec- tion, line voltage and phase voltage are identical, and in a Y connection, line current and phase current are identical. Because three-phase systems are designed to handle large blocks of electric power, all voltage and current specifications are given as rms val- ues. When voltage ratings are given, they refer specifically to the rating of the line voltage. Thus when a three-phase transmission line is rated at 345 kV, the nominal value of the rms line-to-line voltage is 345,000 V. In this chapter we express all voltages and currents as rms values. Finally, the Greek letter phi (<f>) is widely used in the literature to denote a per-phase quantity. Thus V^„ I, /( , Z^, P^ and Q^ are interpreted as voltage/phase, current/phase, impedance/phase, power/phase, and reactive power/phase, respectively. Example 11.1 shows how to use the observations made so far to solve a balanced three-phase Y-Y circuit. Figure 11.9 A Phasor diagrams showing the relation- ship between line-to-tine and line-to-neutral voltages in a balanced system, (a) The abc sequence, (b) The acb sequence. Example 11.1 Analyzing a Wye-Wye Circuit A balanced three-phase Y-connected generator with positive sequence has an impedance of 0.2 + y'0.5 il/4 and an internal voltage of 120 V/<f>. The generator feeds a balanced three-phase Y-connected load having an impedance of 39 + /28 fl/4>. The impedance of the line connect- ing the generator to the load is 0.8 + /1,5 (l/4>. The a-phase internal voltage of the generator is speci- fied as the reference phasor. a) Construct the a-phase equivalent circuit of the system. b) Calculate the three line currents I aA , I bB , and I cC . c) Calculate the three phase voltages at the load. VAN, VBN< and V CN . d) Calculate the line voltages V AB , V 1}C > and \ C A at the terminals of the load. . considering only balanced circuits. The analysis of unbalanced three-phase circuits, which you will encounter if you study electric power in later courses, relies heav- ily on an understanding of. considering voltage and current in such circuits, we conclude with sections on power and power measurement. Practical Perspective Transmission and Distribution of Electric Power In this chapter. introduce circuits that are designed to handle large blocks of electric power. These are the circuits that are used to transport electric power from the generating plants to both industrial