406 Balanced Three-Phase Circuits e) Calculate the phase voltages at the terminals of the generator, V an , V bn , and V cn . f) Calculate the line voltages V ab , V bc , and V ca at the terminals of the generator. g) Repeat (a)-(f) for a negative phase sequence. Solution a) Figure 11.10 shows the single-phase equivalent circuit. b) The a-phase line current is 120 /0° L A = <= dA (0.2 + 0.8 + 39) + /(0.5 + 1.5 + 28) 120 /0° 40 + /30 = 2.4/-36.87° A. a' 0.2(2 /0-5 O a o.8(2 /1-50 A f 'VW ^VYY>_ '.i A © 120/0° V V an 39 n 1/28(1 Figure 11.10 A The single-phase equivalent circuit for Example 11.1. For a positive phase sequence, I bB = 2.4 /-156.87° A, I cC = 2.4 /83.13° A. c) The phase voltage at the A terminal of the load is VAN = (39 + /28)(2.4/-36.87°) = 115.22/-1.19° V. For a positive phase sequence, V BN = 115.22/-121.19° V, V CN = 115.22/118.81° V. d) For a positive phase sequence, the line voltages lead the phase voltages by 30°; thus VAB = (V3/30°)V A N = 199.58 /28.81° V, V BC = 199.58/-91.19° V, V C A = 199.58 /148.81° V. e) The phase voltage at the a terminal of the source is V an = 120 - (0.2 + /0.5)(2.4/-36.87°) = 120 - 1.29/31.33° = 118.90 -/0.67 = 118.90/-0.32° V. For a positive phase sequence, V bn = 118.90/-120.32° V, V cn = 118.90/119.68° V. f) The line voltages at the source terminals are Vab = (V5/30")V a n - 205.94 /29.68° V, Vbc = 205.94 /-90.32° V, V ca = 205.94/149.68° V. g) Changing the phase sequence has no effect on the single-phase equivalent circuit. The three line currents are IaA = 2.4 /-36.87° A, I,, B = 2.4 /83.13° A, I cC = 2.4/-156.87° A. The phase voltages at the load are V AN = 115.22/-1.19° V, V BN = 115.22/118.81° V, V CN = 115.22/-121.19° V. For a negative phase sequence, the line voltages lag the phase voltages by 30°: V AB = (V3/-30°)V AN - 199.58 /-31.19° V, V BC = 199.58 /88.81° V, VCA = 199.58/-151.19° V. The phase voltages at the terminals of the gener- ator are V an - 118.90/-0.32° V, V bll = 118.90/119.68° V, V cn = 118.90/-120.32° V. The line voltages at the terminals of the genera- tor are V ab = (V3/-30°)V an = 205.94/-30.32° V, V bc = 205.94 /89.68° V, V,,, = 205.94/-150.32° V. 11.4 Analysis of the Wye-Delta Circuit 407 ^ASSESSMENT PROBLEMS Objective 1—Know how to analyze a balanced, three-phase wye-wye circuit 11.1 The voltage from A to N in a balanced three- phase circuit is 240 /—30° V. If the phase sequence is positive, what is the value of V BC ? Answer: 415.69/-120 11.2 V. The c-phase voltage of a balanced three-phase Y-connected system is 450 /-25° V. If the phase sequence is negative, what is the value ofV AB ? Answer: 779.42 /65° V. 11.3 The phase voltage at the terminals of a bal- anced three-phase Y-connected load is 2400 V. The load has an impedance of 16 + /12 il/cp and is fed from a line having an impedance of 0.10 + /0.80 Cl/cj). The Y-connected source at the sending end of the line has a phase NOTE: Also try Chapter Problems 11.7, 11.9, and 11.11. sequence of acb and an internal impedance of 0.02 -I- /0.16 n/<£. Use the a-phase voltage at the load as the reference and calculate (a) the line currents I aA , I bB , and I cC ; (b) the line volt- ages at the source, V ab , V bc , and V ca ; and (c) the internal phase-to-neutral voltages at the source, V a <n, V b - n , and V c - n . Answer: (a) I aA = 120 /-36.87° A, I bB = 120 /83.13° A, and I cC = 120/-156.87° A; (b) V ab = 4275.02 /-28.38° V, V bc = 4275.02 /91.62° V, and V ca = 4275.02 /-148.38° V; (c) V a . n = 2482.05 /1.93° V, V b - n = 2482.05 /121.93° V, and V c ' n = 2482.05/-118.07° V. 11.4 Analysis of the Wye-Delta Circuit If the load in a three-phase circuit is connected in a delta, it can be trans- formed into a wye by using the delta-to-wye transformation discussed in Section 9.6. When the load is balanced, the impedance of each leg of the wye is one third the impedance of each leg of the delta, or Relationship between three-phase delta-connected and wye-connected -4 impedance z - ~± 3 (11.21) which follows directly from Eqs. 9.51-9.53. After the A load has been replaced by its Y equivalent, the a-phase can be modeled by the single- phase equivalent circuit shown in Fig. 11.11. We use this circuit to calculate the line currents, and we then use the line currents to find the currents in each leg of the original A load. The relationship between the line currents and the currents in each leg of the delta can be derived using the circuit shown in Fig. 11.12. When a load (or source) is connected in a delta, the current in each leg of the delta is the phase current, and the voltage across each leg is the phase voltage. Figure 11.12 shows that, in the A configuration, the phase voltage is identical to the line voltage. To demonstrate the relationship between the phase currents and line currents, we assume a positive phase sequence and let A/> represent the magnitude of the phase current. Then AB cr Ipc =/ A /-HO". l rA = At/120". V a - n Zga a Zu c A z A N Figure 11.11 A A single-phase equivalent circuit. (11.22) (11.23) Figure 11.12 • A circuit used to establish the relationship between line currents and phase currents in (11-24) a balanced A load. 408 Balanced Three-Phase Circuits Figure 11.13 • Phasor diagrams showing the relationship between line currents and phase currents in a A-connected load, (a) The positive sequence, (b) The negative sequence. In writing these equations, we arbitrarily selected I AB as the reference phasor. We can write the line currents in terms of the phase currents by direct application of Kirchhoff s current law: IaA = IAB ~ ICA = I A /Q° -U/120* = V3I 4 /-30°, IbB = IBC - IAB = U /-120° - U. = ^/^,/-150% (11.25) (11.26) IcC _ IcA ~~ IBC = /,,/120° -/^/-120° = y/3I A /90°. (11.27) Comparing Eqs. 11.25-11.27 with Eqs. 11.22-11.24 reveals that the magni- tude of the line currents is V3 times the magnitude of the phase currents and that the set of line currents lags the set of phase currents by 30°. We leave to you to verify that, for a negative phase sequence, the line currents are V3 times larger than the phase currents and lead the phase currents by 30°. Thus, we have a shortcut for calculating line currents from phase currents (or vice versa) for a balanced three-phase A-connected load. Figure 11.13 summarizes this shortcut graphically. Example 11.2 illustrates the calculations involved in analyzing a balanced three-phase circuit having a Y-connected source and a A-connected load. Example 11.2 Analyzing a Wye-Delta Circuit The Y-connected source in Example 11.1 feeds a A-connected load through a distribution line hav- ing an impedance of 0.3 + /0.9 Q,/4>. The load impedance is 118.5 + /85.8 Cl/4>. Use the a-phase internal voltage of the generator as the reference. a) Construct a single-phase equivalent circuit of the three-phase system. b) Calculate the line currents I aA , I bB , and I cC . c) Calculate the phase voltages at the load terminals. d) Calculate the phase currents of the load. e) Calculate the line voltages at the source terminals. Solution a) Figure 11.14 shows the single-phase equivalent circuit. The load impedance of the Y equivalent is 118.5 + /85.8 = 39.5 + /28.6 ft/0. a' 0.212 /0.511 a 0.3 ft /0.9 0 A C_J 120/0° 39.5 ft /28.6 ft N Figure 11.14 • The single-phase equivalent circuit for Example 11.2. b) The a-phase line current is 120 /C IaA = (0.2 + 0.3 + 39.5) + /(0.5 + 0.9 + 28.6) 120/0° rr ~- = 2.4 /-36.87° A. 40 + /30 z 11.4 Analysis of the Wye-Delta Circuit 409 Hence I bB = 2.4/-156.87° A, I cC = 2.4 /83.13° A. c) Because the load is A connected, the phase volt- ages are the same as the line voltages. To calcu- late the line voltages, we first calculate V AN : VAN = (39.5 + /28.6)(2.4/-36.87°) = 117.04/-0.96° V. Because the phase sequence is positive, the line voltage VAB * S VAB = (V5/30')V A N = 202.72 /29.04° V. Therefore V BC = 202.72 /-90.96° V, V CA = 202.72 /149.04° V. d) The phase currents of the load may be calculated directly from the line currents: l AB 1 V3 /30' aA = 1.39/-6.87° A. Once we know I AB , we also know the other load phase currents: I BC = 1.39/-126.87° A, I CA = 1.39/113.13° A. Note that we can check the calculation of I AB by using the previously calculated V A B ar >d the impedance of the A-connected load; that is, V AB _ 202.72/29.04° IAB ~" ~Z^ ' 118.5 +/85.8 = 1.39/-6.87° A. e) To calculate the line voltage at the terminals of the source, we first calculate V an . Figure 11.14 shows that V an is the voltage drop across the line impedance plus the load impedance, so V an = (39.8 + /29.5)(2.4/-36.87°) = 118.90/-0.32° V. The line voltage V ab is V ab = (V3~/30°)V an , or V ab = 205.94 /29.68° V. Therefore V bc = 205.94/-90.32° V, V ca = 205.94/149.68° V. ^ASSESSMENT PROBLEMS Objective 2—Know how to analyze a balanced, three-phase wye-delta connected circuit 11.4 The current I CA in a balanced three-phase A-connected load is 8 /—15° A. If the phase sequence is positive, what is the value of I cC ? Answer: 13.86/-45° A. 11.5 A balanced three-phase A-connected load is fed from a balanced three-phase circuit. The reference for the b-phase line current is toward the load. The value of the current in the b-phase is 12 /65° A. If the phase sequence is negative, what is the value of I A B? Answer: 6.93 /-85° A. 11.6 The line voltage V A B at the terminals of a bal- anced three-phase A-connected load is 4160 /0° V. The line current I aA is 69.28/-10° A. a) Calculate the per-phase impedance of the load if the phase sequence is positive. b) Repeat (a) for a negative phase sequence. Answer: (a) 104 /-20° H; (b) 104/+40° a. 11.7 The line voltage at the terminals of a balanced A-connected load is 110 V. Each phase of the load consists of a 3.667 II resistor in parallel with a 2.75 Cl inductive impedance. What is the mag- nitude of the current in the line feeding the load? Answer: 86.60 A. NOTE: Also try Chapter Problems 11.12,11.13, and 11.16. 410 Balanced Three-Phase Circuits IaA %B *• Ice' B + v BN ^B z A — C i Ts » i\ Z c Figure 11.15 • A balanced Y load used to introduce average power calculations in three-phase circuits. 11.5 Power Calculations in Balanced Three-Phase Circuits So far, we have limited our analysis of balanced three-phase circuits to determining currents and voltages. We now discuss three-phase power calculations. We begin by considering the average power delivered to a balanced Y-connected load. Average Power in a Balanced Wye Load Figure 11.15 shows a Y-connected load, along with its pertinent currents and voltages. We calculate the average power associated with any one phase by using the techniques introduced in Chapter 10. With Eq. 10.21 as a starting point, we express the average power associated with the a-phase as ^A = |V AN ||I aA |cos(t9 vA -M, (11.28) where 8„ A and 0, A denote the phase angles of V AN and I uA , respectively. Using the notation introduced in Eq. 11.28, we can find the power associ- ated with the b- and c-phases: Pa = |V B Nl|IbBl«>s(0vB-0/B); Pc = !V CN ||I cC |cos(f\ c -0; C ). (11.29) (11.30) In Eqs. 11.28-11.30, all phasor currents and voltages are written in terms of the rms value of the sinusoidal function they represent. In a balanced three-phase system, the magnitude of each line-to-neutral voltage is the same, as is the magnitude of each phase current. The argu- ment of the cosine functions is also the same for all three phases. We emphasize these observations by introducing the following notation: and V d , = |V AN | = |VBNI = IVCNI, (11.31) (11.32) (11.33) Moreover, for a balanced system, the power delivered to each phase of the load is the same, so PA = PB = PC = P* = V^ cos ^, (11.34) where P^ represents the average power per phase. The total average power delivered to the balanced Y-connected load is simply three times the power per phase, or P T = 3P 4) = 3V <h I (!) cosO (11.35) Expressing the total power in terms of the rms magnitudes of the line volt- age and current is also desirable. If we let V L and I L represent the rms magnitudes of the line voltage and current, respectively, we can modify Eq. 11.35 as follows: Total real power in a balanced three-phase load • 3 (^) /LC ° ! 4> V3V L / L cos^. (11.36) 11.5 Power Calculations in Balanced Three-Phase Circuits 411 In deriving Eq. 11.36, we recognized that, for a balanced Y-connected load, the magnitude of the phase voltage is the magnitude of the line volt- age divided by V5, and that the magnitude of the line current is equal to the magnitude of the phase current. When using Eq. 11.36 to calculate the total power delivered to the load, remember that 0$ is the phase angle between the phase voltage and current. Complex Power in a Balanced Wye Load We can also calculate the reactive power and complex power associated with any one phase of a Y-connected load by using the techniques intro- duced in Chapter 10. For a balanced load, the expressions for the reactive power are 2r = 3Q^ = V3V L / L sin^. (11.37) (11.38) Equation 10.29 is the basis for expressing the complex power associated with any phase. For a balanced load, S<t> - ^AN^aA _ ^BNlbB ~ V CN IcC v; (11.39) where V^ and 1^, represent a phase voltage and current taken from the same phase. Thus, in general, (11.40) (11.41) < Total reactive power in a balanced three-phase load -4 Total complex power in a balanced three- phase load Power Calculations in a Balanced Delta Load If the load is A-connected, the calculation of power—reactive or complex- is basically the same as that for a Y-connected load. Figure 11.16 shows a A-connected load, along with its pertinent currents and voltages. The power associated with each phase is ^A = |V A BI|IAB|COS(0 VAB - 0 /AB ), PB = |VBCI|IBC|C°S(0 V BC " 0 '-BC)» ^c = |V C AI|ICA|COS(0 VCA -0 (CA ). For a balanced load, |V AB | = IVBCI = IVCAI = V* HABI = IIBCI = IICAI = /* 0<AB = #t>BC ~ ^/BC = ^«CA e vAB rCA = 0., and P A = PB = Pc = Pd> = K^cos^. Note that Eq. 11.48 is the same as Eq. 11.34. Thus, in a balanced load, regardless of whether it is Y- or A-connected, the average power per phase is equal to the product of the rms magnitude of the phase voltage, the rms magnitude of the phase current, and the cosine of the angle between the phase voltage and current. B L Z A V,- CA (11.42) (11.43) (11.44) VM (11.45) Figure 11.16 • A A-connected load used to discuss (11.46) power calculations. (11.47) (11.48) The total power delivered to a balanced A-connected load is P r = 3^ = 31^ cos 0* = V3V L / L cos0*. (11-49) Note that Eq. 11.49 is the same as Eq. 11.36. The expressions for reactive power and complex power also have the same form as those developed for the Y load: Q^ = V 4> I lf> sme 4> ; (11.50) Qr = 3Q* = 3VV 0 sin $+; (11.51) S<i> = P<i> + jQ^> = V^lJ; (11.52) S T = 3S lb = V3V L I L /$£. (11.53) Instantaneous Power in Three-Phase Circuits Although we are primarily interested in average, reactive, and complex power calculations, the computation of the total instantaneous power is also important. In a balanced three-phase circuit, this power has an inter- esting property: It is invariant with time! Thus the torque developed at the shaft of a three-phase motor is constant, which in turn means less vibra- tion in machinery powered by three-phase motors. Let the instantaneous line-to-neutral voltage v AN be the reference, and, as before, 0^ is the phase angle 0„ A — 0 /A . Then, for a positive phase sequence, the instantaneous power in each phase is PA = v AN i aA = V m I m cos art cos (a>t - 0 4 ), PB = V BNA>B = V,„I m cos (art - 120°) cos (a>t -0,/,- 120°), Pc = UCN'CC = V in I m cos(a)t + 120°) cos (art -0^,+ 120°), where V m and I m represent the maximum amplitude of the phase voltage and line current, respectively. The total instantaneous power is the sum of the instantaneous phase powers, which reduces to 1.5V m I m cos 9^, that is, PT = PA + PB + Pc = 1.5V lfI / m cos00. Note this result is consistent with Eq. 11.35 since V m = V2~V'$ and I m = V2/^ (see Problem 11.26). Examples 11.3-11.5 illustrate power calculations in balanced three- phase circuits. V3 cos0 rf 11.5 Power Calculations in Balanced Three-Phase Circuits 413 Example 11.3 Calculating Power in a Three-Phase Wye-Wye Circuit a) Calculate the average power per phase delivered to the Y-connected load of Example 11.1. b) Calculate the total average power delivered to the load. c) Calculate the total average power lost in the line. d) Calculate the total average power lost in the generator. e) Calculate the total number of magnetizing vars absorbed by the load. f) Calculate the total complex power delivered by the source. Solution a) From Example 11.1, V^ = 115.22 V, I 4 = 2.4 A, and 0^ = -1.19 - (-36.87) = 35.68°. Therefore ?$ = (115.22)(2.4) cos 35.68' = 224.64 W. The power per phase may also be calculated from l\R^ or P+ = (2.4) 2 (39) = 224.64 W. b) The total average power delivered to the load is p T = 3/^ = 673.92 W. We calculated the line voltage in Example 11.1, so we may also use Eq. 11.36: P T = V3(199.58)(2.4) cos 35.68° = 673.92 W. c) The total power lost in the line is P Vme = 3(2.4) 2 (0.8) = 13.824 W. d) The total internal power lost in the generator is gen 3(2.4)2(0.2) = 3.456 W. c) The total number of magnetizing vars absorbed by the load is Q T = V5( 199.58)(2.4) sin 35.68° = 483.84 VAR. f) The total complex power associated with the source is S T = 3St = -3(120)(2.4) /36.87° = -691.20 -/518.40 VA. The minus sign indicates that the internal power and magnetizing reactive power are being deliv- ered to the circuit. We check this result by calcu- lating the total and reactive power absorbed by the circuit: P = 673.92 + 13.824 + 3.456 = 691.20 W (check), Q = 483.84 + 3(2.4) 2 (1.5) + 3(2.4) 2 (0.5) = 483.84 + 25.92 + 8.64 = 518.40 VAR(check). Example 11.4 Calculating Power in a Three-Phase Wye-Delta Circuit a) Calculate the total complex power delivered to the A-connected load of Example 11.2. b) What percentage of the average power at the sending end of the line is delivered to the load? Solution a) Using the a-phase values from the solution of Example 11.2, we obtain V^ = V AB = 202.72 /29.04° V, Ia = I AB = 139/-6.87 a A. Using Eqs. 11.52 and 11.53, we have S T = 3(202.72 /29.04° )(1.39 /6.87°) = 682.56 + /494.21 VA. b) The total power at the sending end of the distri- bution line equals the total power delivered to the load plus the total power lost in the line; therefore /» input = 682.56 + 3(2.4)2(0.3) = 687.74 W. The percentage of the average power reaching the load is 682.56/687.74, or 99.25%. Nearly 100% of the average power at the input is delivered to the load because the impedance of the line is quite small compared to the load impedance. 414 Balanced Three-Phase Circuits Example 11.5 Calculating Three-Phase Power with an Unspecified Load A balanced three-phase load requires 480 kW at a lagging power factor of 0.8. The load is fed from a line having an impedance of 0.005 + /0.025 0/<£. The line voltage at the terminals of the load is 600 V. a) Construct a single-phase equivalent circuit of the system. b) Calculate the magnitude of the line current. c) Calculate the magnitude of the line voltage at the sending end of the line. d) Calculate the power factor at the sending end of the line. Solution a) Figure 11.17 shows the single-phase equivalent circuit. We arbitrarily selected the line-to-neutral voltage at the load as the reference. 0.005 n /0.025 n a • 'VW— —<'Y^'Y-^ 9 A + fa* •— "C 600 + —« »N 160 kW at 0.8 lag Figure 11.17 • The single-phase equivalent circuit for Example 11.5. b) The line current I* A is given by 600 V3 IaA = (160 + /120)10 3 , or IaA = 577.35 /36.87° A. Therefore, I aA = 577.35 /-36.87° A. The mag- nitude of the line current is the magnitude of I aA : I L = 577.35 A. We obtain an alternative solution for I L from the expression p r h. = V3V L I L cosB p = V3(600)/ L (0.8) = 480,000 W; 480,000 V3(600)(0.8) 1000 V3 = 577.35 A. c) To calculate the magnitude of the line voltage at the sending end, we first calculate V an . From Fig. 11.17, V n AN 600 + Z,l (HA -z= + (0.005 + /0.025)(577.35/-36.87°) = 357.51 /1.57° V. Thus V L = V3|V : J = 619.23 V. d) The power factor at the sending end of the line is the cosine of the phase angle between V an and I aA : pf = cos [1.57° - (-36.87°)] = cos 38.44° = 0.783 lagging. An alternative method for calculating the power factor is to first calculate the complex power at the sending end of the line: S (b = (160 + /12())10 3 + (577.35) 2 (0.005 + /0.025) = 161.67 +/128.33 kVA = 206.41 /38.44° kVA. The power factor is pf = cos 38.44° = 0.783 lagging. Finally, if we calculate the total complex power at the sending end, after first calculating the magnitude of the line current, we may use this value to calculate V L . That is, V3VJ L = 3(206.41) x 10 3 , 3(206.41) X 10 3 Vi = V3(577.35) ' = 619.23 V. 11.6 Measuring Average Power in Three-Phase Circuits 415 ^ASSESSMENT PROBLEMS Objective 3—Be able to calculate power (average, reactive, and complex) in any three-phase circuit 11.8 The three-phase average power rating of the central processing unit (CPU) on a mainframe digital computer is 22,659 W. The three-phase line supplying the computer has a line voltage rating of 208 V (rms). The line current is 73.8 A (rms).The computer absorbs magnetizing VARs. a) Calculate the total magnetizing reactive power absorbed by the CPU. b) Calculate the power factor. Answer: (a) 13,909.50 VAR; (b) 0.852 lagging. NOTE: Also try Chapter Problems 11.22 and 11.24. 11.9 The complex power associated with each phase of a balanced load is 144 + /192 kVA. The line voltage at the terminals of the load is 2450 V. a) What is the magnitude of the line current feeding the load? b) The load is delta connected, and the imped- ance of each phase consists of a resistance in parallel with a reactance. Calculate R and X. c) The load is wye connected, and the imped- ance of each phase consists of a resistance in series with a reactance. Calculate R and X. Answer: (a) 169.67 A; (b)R = 41.68 ft, X = 31.26 ft; (c) R = 5 ft, X = 6.67 ft. 11.6 Measuring Average Power in Three-Phase Circuits The basic instrument used to measure power in three-phase circuits is the electrodynamometer wattmeter. It contains two coils. One coil, called the current coil, is stationary and is designed to carry a current proportional to the load current. The second coil, called the potential coil, is movable and carries a current proportional to the load voltage. The important features of the wattmeter are shown in Fig. 11.18. The average deflection of the pointer attached to the movable coil is proportional to the product of the effective value of the current in the cur- rent coil, the effective value of the voltage impressed on the potential coil, and the cosine of the phase angle between the voltage and current. The direction in which the pointer deflects depends on the instantaneous polar- ity of the current-coil current and the potential-coil voltage.Therefore each coil has one terminal with a polarity mark —usually a plus sign—but some- times the double polarity mark ± is used. The wattmeter deflects upscale when (1) the polarity-marked terminal of the current coil is toward the source, and (2) the polarity-marked terminal of the potential coil is con- nected to the same line in which the current coil has been inserted. The Two-Wattmeter Method Consider a general network inside a box to which power is supplied by n conducting lines. Such a system is shown in Fig. 11.19. If we wish to measure the total power at the terminals of the box, we need to know n — 1 currents and voltages. This follows because if we choose one terminal as a reference, there are only n - 1 independent voltages. Likewise, only n — 1 independent currents can exist in the n con- ductors entering the box. Thus the total power is the sum of n — 1 product terms; that is, p = v x i\ + v 2 t2 + • • • + t> fl _ r L_j. Watt scale Current-coil terminals Potential-coil terminals Pointer Figure 11.18 • The key features of the electrodynamometer wattmeter.