Electric Circuits, 9th Edition P45 doc

416 Balanced Three-Phase Circuits Applying this general observation, we can see that for a three- conductor circuit, whether balanced or not, we need only two wattmeters to measure the total power. For a four-conductor circuit, we need three wattmeters if the three-phase circuit is unbalanced, but only two wattmeters if it is balanced, because in the latter case there is no current in the neutral line. Thus, only two wattmeters are needed to measure the total average power in any balanced three-phase system. The two-wattmeter method reduces to determining the magnitude and algebraic sign of the average power indicated by each wattmeter. We can describe the basic problem in terms of the circuit shown in Fig. 11.20, where the two wattmeters are indicated by the shaded boxes and labeled W\ and W 2 . The coil notations cc and pc stand for current coil and poten- tial coil, respectively. We have elected to insert the current coils of the wattmeters in lines aA and cC. Thus, line bB is the reference line for the two potential coils. The load is connected as a wye, and the per-phase load impedance is designated as Z^ = \Z\ /jh This is a general representation, as any A-connected load can be represented by its Y equivalent; further- more, for the balanced case, the impedance angle 0 is unaffected by the A-to-Y transformation. We now develop general equations for the readings of the two wattmeters. We assume that the current drawn by the potential coil of the wattmeter is negligible compared with the line current measured by the cur- rent coil. We further assume that the loads can be modeled by passive circuit elements so that the phase angle of the load impedance (0 in Fig. 11.20) lies between -90° (pure capacitance) and +90" (pure inductance). Finally, we assume a positive phase sequence. From our introductory discussion of the average deflection of the wattmeter, we can see that wattmeter 1 will respond to the product of |V A JJ|, |IaA^ an d the cosine of the angle between V A B an d IaA- M we denote this wattmeter reading as W h we can write Wi -IVABIPUCOS*, = VIA cos 0 X . (11.54) It follows that W2 = |VCBI|ICC|COS02 = VJ L COS0 2 . (11.55) In Eq. 11.54,0] is the phase angle between V AB and I aA , and in Eq. 11.55, 0 2 is the phase angle between V CB and I c( > To calculate Wj and W 2 , we express 0] and 0 2 in terms of the imped- ance angle 0, which is also the same as the phase angle between the phase voltage and current. For a positive phase sequence, 0 } = 0 + 30° = 0^ + 30°, (11.56) 0 2 = 0 - 30° = 0$ - 30°. (11.57) The derivation of Eqs. 11.56 and 11.57 is left as an exercise (see Problem 11.34). When we substitute Eqs. 11.56 and 11.57 into Eqs. 11.54 and 11.55, respectively, we get Wi = K L / L cos(^ + 30°), (11.58) W 2 = K L / L COS(0A - 30°). (11.59) h t + v i h + v 2 /3 + v 3 m 1 • 1 • Z * General network Figure 11.19 • A general circuit whose power is supplied by n conductors. IaA cc A Wi pc4 Z^ = |Z|Z0 V AN b«- 'CN c+ -•— -¾ c( N Figure 11.20 • A circuit used to analyze the two-wattmeter method of measuring average power delivered to a balanced load. 11.6 Measuring Average Power in Three-Phase Circuits 417 (11.60) To find the total power, we add W ] and W 2 ; thus p T = W ] + W 2 = 2VL/LCOS^COS30" = V3VL/ L COS^, which is the expression for the total power in a three-phase circuit. Therefore we have confirmed that the sum of the two wattmeter readings yields the total average power. A closer look at Eqs. 11.58 and 11.59 reveals the following about the readings of the two wattmeters: 1. If the power factor is greater than 0.5, both wattmeters read positive. 2. If the power factor equals 0.5, one wattmeter reads zero. 3. If the power factor is less than 0.5, one wattmeter reads negative. 4. Reversing the phase sequence will interchange the readings on the two wattmeters. These observations are illustrated in the following example and in Problems 11.41-11.51. | Computing Wattmeter Readi Calculate the reading of each wattmeter in circuit in Fig. 11.20 if the phase voltage at load is 120 V and (a) Z+ = 8• +; (b) Z 0 = 8 - ;6 ft; (c) Z^ = 5 + ;5 V3 12; (d) Z s = 10 /-75° ft. (e) Verify for (a)-(d) the sum of the wattmeter readings equals the power delivered to the load. Solution a) Z (b = 10 /36.87° ft, V L = 120 V3V, and / L = 120/10 = 12 A. W ] = (120V3~)(12) cos (36.87° + 30°) = 979.75 W, W 2 = (120 V3)(12) cos (36.87° - 30°) = 2476.25 W. b) Z & = 10 /-36.87° ft, V L = 120V^ V, and / L = 120/10 = 12 A. Wi = (120V3)(12) cos (-36.87° + 30°) = 2476.25 W, W 2 = (120V3)(12) cos (-36.87° - 30°) = 979.75 W. ngs in the the 6 ft; and that total Three-Phase Circuits c) Z^ = 5(1 + jV3) = 10 /60° O, V L = 120V3 V, and/ L = 12 A. Wi = (120V5)(12) cos (60° + 30°) = 0, W 2 = (120 V3)(12) cos (60° - 30°) = 2160 W. d) Z 4> = 10 /-75° ft, V L = 12QV3" V, and / L = 12 A. Wi = (120V3)(12)cos(-75° + 30°) = 1763.63 W, W 2 = (120V5)(12) cos (-75° - 30°) = -645.53 W. e) P r (a) = 3(12) 2 (8) - 3456 W, W } +W 2 = 979.75 + 2476.25 = 3456 W, P r (b) = P r (a) = 3456 W, Wi + W 2 = 2476.25 + 979.75 = 3456 W, /V(c) = 3(12) 2 (5) = 2160 W, W t + W 2 = 0 + 2160 = 2160W, P r (d) = 3(12) 2 (2.5882) = 1118.10 W, Wi + W 2 = 1763.63 - 645.53 = 1118.10W. NOTE: Assess your understanding of the two-wattmeter method by trying Chapter Problems 11.43 and 11.44. 418 Balanced Three-Phase Circuits 3-0 line -) Generator Plant 4 *) / Sub Station Figure 11.21 A A substation connected to a power plant via a three-phase line. 0.6 O -AW- /4.8 n ^YYYV A v.„ 13,800 /r\0 y ^3 *~ 1.2 MW 1.2 MVAR n N Figure 11.22 • A single phase equivalent circuit for the system in Fig. 11.21. Practical Perspective Transmission and Distribution of Electric Power At the start of this chapter we pointed out the obligation utilities have to maintain the rms voltage level at their customer's premises. Although the acceptable deviation from a nominal level may vary among different utilities we will assume for purposes of discussion that an allowable tolerance is ± 5.8%. Thus a nominal rms voltage of 120 V could range from 113 to 127 V. We also pointed out that capacitors strategically located on the sys- tem could be used to support voltage levels. The circuit shown in Fig. 11.21 represents a substation on a Midwestern municipal system. We will assume the system is balanced, the line-to-line voltage at the substation is 13.8 kV, the phase impedance of the distribu- tion line is 0.6 + /4.811, and the load at the substation at 3 PM on a hot, humid day in July is 3.6 MW and 3.6 magnetizing MVAR. Using the line-to-neutral voltage at the substation as a reference, the single-phase equivalent circuit for the system in Fig. 11.21 is shown in Fig. 11.22. The line current can be calculated from the expression for the complex power at the substation. Thus, 13,800 V3 I* A = (1.2 + /1.2)10* It follows that or I* A = 150.61 + /150.61 A I aA = 150.61 - /150.61 A. The line-to-neutral voltage at the generating plant is 13,800 V3 '0° + (0.6 + /4.8)(150.61 - /150.61) = 8780.74 + /632.58 = 8803.50/4.12° V. Therefore the magnitude of the line voltage at the generating plant is |V ab | = V3(8803.50) = 15,248.11V. We are assuming the utility is required to keep the voltage level within ± 5.8% of the nominal value. This means the magnitude of the line-to-line voltage at the power plant should not exceed 14.6 kV nor be less than 13 kV. Therefore, the magnitude of the line voltage at the generating plant could cause problems for customers. When the magnetizing vars are supplied by a capacitor bank connected to the substation bus, the line current I aA becomes I aA = 150.61 + /0 A. Summary 419 Therefore the voltage at the generating plant necessary to maintain a line- to-line voltage of 13,800 V at the substation is 13,800 Van = -^- /O: + (0-6 + /4.8)(150.61 + /0) = 8057.80 + /722.94 = 8090.17/5.13° V. Hence |V ab | = V3(8090.17) = 14,012.58 V. This voltage level falls within the allowable range of 13 kV to 14.6 kV. NOTE: Assess your understanding of this Practical Perspective by trying Chapter Problems 11.52(a)-(b) and 11.53, 11.56, and 11.57. Summary • When analyzing balanced three-phase circuits, the first step is to transform any A connections into Y connections, so that the overall circuit is of the Y-Y configuration. (See page 402.) • A single-phase equivalent circuit is used to calculate the line current and the phase voltage in one phase of the Y-Y structure. The a-phase is normally chosen for this purpose. (See page 404.) • Once we know the line current and phase voltage in the a-phase equivalent circuit, we can take analytical short- cuts to find any current or voltage in a balanced three- phase circuit, based on the following facts: • The b- and c-phase currents and voltages are identi- cal to the a-phase current and voltage except for a 120° shift in phase. In a positive-sequence circuit, the b-phase quantity lags the a-phase quantity by 120°, and the c-phase quantity leads the a-phase quantity by 120°. For a negative sequence circuit, phases b and c are interchanged with respect to phase a. • The set of line voltages is out of phase with the set of phase voltages by ±30°. The plus or minus sign corre- sponds to positive and negative sequence, respectively. • In a Y-Y circuit the magnitude of a line voltage is V3 times the magnitude of a phase voltage. • The set of line currents is out of phase with the set of phase currents in A-connected sources and loads by T30°. The minus or plus sign corresponds to positive and negative sequence, respectively. • The magnitude of a line current is V3 times the mag- nitude of a phase current in a A-connected source or load, (See pages 404-405 and 407-408.) • The techniques for calculating per-phase average power, reactive power, and complex power are identical to those introduced in Chapter 10. (See page 410.) • The total real, reactive, and complex power can be deter- mined either by multiplying the corresponding per phase quantity by 3 or by using the expressions based on line current and line voltage, as given by Eqs. 11.36,11.38, and 11.41. (See pages 410 and 411.) • The total instantaneous power in a balanced three-phase circuit is constant and equals 1.5 times the average power per phase. (See page 412.) • A wattmeter measures the average power delivered to a load by using a current coil connected in series with the load and a potential coil connected in parallel with the load. (See page 415.) • The total average power in a balanced three-phase cir- cuit can be measured by summing the readings of two wattmeters connected in two different phases of the circuit. (See page 415.) 420 Balanced Three-Phase Circuits Problems AH phasor voltages in the following Problems are stated in terms of the rms value. Section 11.1 11.1 Verify that Eq. 11.3 is true for either Eq. 11.1 or Eq.11.2. 11.2 What is the phase sequence of each of the following sets of voltages? a) y a = 208 cos (OJ? + 27°) V, ^ = 208 cos (cot + 147°) V, v c = 208 cos (cot - 93°) V. b) v. a = 4160 cos (cot - 18°) V, v b = 4160 cos (cot - 138°) V, ?; c = 4160 cos (cot + 102°) V. 11.3 For each set of voltages, state whether or not the volt- PSPICE a ges form a balanced three-phase set. If the set is bal- anced, state whether the phase sequence is positive or negative. If the set is not balanced, explain why. a) v a = 139 cos 377? V, v b = 139 cos (377? + 120°) V, v c = 139 cos (377? - 120°) V. b) v. d = 381 cos 377? V, v b = 381 cos (377? + 240°) V, v c = 381 cos (377? + 120°) V. c) v a = 2771 sin (377? - 30°) V, v b = 2771 cos 377? V, v c = 2771 sin (377? + 210°) V. d) v A = 170 sin (cot + 30°) V, v b = -170 cos cot V, v c = 170 cos (cot + 60°) V. e) v a = 339 cos cot V, % = 339 cos (cot + 120) V, v c = 393 cos (cot - 120°) V. f) v. A = 3983 sin (cot + 50°) V, v b = 3983 cos (cot - 160°) V, v c = 3983 cos (cot + 70°) V. Section 11.2 11.4 Refer to the circuit in Fig. 11.5(b). Assume that there are no external connections to the terminals a,b,c. Assume further that the three windings are from a balanced three-phase generator. How much current will circulate in the A-connected generator? Section 11.3 11.5 A balanced three-phase circuit has the following characteristics: • Y-Y connected; • The line voltage at the source, V ab , is 240 V3~/90° V; • The phase sequence is negative; • The line impedance is 4 -I- /5 £l/<f>', • The load impedance is 76 + /55 ft/cf). a) Draw the single phase equivalent circuit for the a-phase. b) Calculated the line current in the a-phase. c) Calculated the line voltage at the load in the a-phase. 11.6 Find the rms value of I„ in the unbalanced three- phase circuit seen in Fig. PI 1.6. Figure PI 1.6 0.1 n /0.8ft a 0.4 ft /3.2 ft A 59.5 ft /76 ft ->WV »****"¥"> 0 /yy*^ (VYY> 0 'WW 0.1ft /0.8 ft b 0.4 ft /3.2 ft B 39.5 ft /26 ft -• "VS/V rrvY> 9 VW— M^ 240/- 240/-120° V 0.1ft /0.8 ft c 0.4 ft /3.2 ft C 19.5 ft /lift -• /vW «^w> » VW rvw>_ N Problems 421 11.7 The time-domain expressions for three line-to-neutral voltages at the terminals of a Y-connected load are v AN = 7620 cos (W + 30°) V, v m = 7620 cos (a)t + 150°) V, '» CN = 7620 cos (cot - 90") V. What are the time-domain expressions for the three line-to-line voltages v Aii , I>BO and VQA? 11.8 a) Is the circuit in Fig. PI 1.8 a balanced or unbal- PSPICE anced three-phase system? Explain. MULTISIM b) Find I„. Figure PI 1.8 ^2()0 /0!V 11.9 The magnitude of the line voltage at the terminals of a balanced Y-connected load is 6600 V. The load impedance is 240 - /70 0/<£. The load is fed from a line that has an impedance of 0.5 + /4 0/(£. a) What is the magnitude of the line current? b) What is the magnitude of the line voltage at the source? 11.10 a) Find I„ in the circuit in Fig. PI 1.10. PSPICE b) FindV AN . c) FindV AB . d) Is the circuit a balanced or unbalanced three- phase system? 11.11 The magnitude of the phase voltage of an ideal balanced three-phase Y-connected source is 125 V. The source is connected to a balanced Y-connected load by a distribution line that has an impedance of 0.1 + /0.8 Cl/<j). The load impedance is 19.9 + /14.2 Cl/4>. The phase sequence of the source is acb. Use the a-phase voltage of the source as the reference. Specify the magnitude and phase angle of the following quantities: (a) the three line currents, (b) the three line voltages at the source, (c) the three phase voltages at the load, and (d) the three line voltages at the load. Section 11.4 11.12 A balanced, three-phase circuit is characterized as follows: • Y-A connected; • Source voltage in the c-phase is 20/-90° V; • Source phase sequence is abc; • Line impedance is 1 + /3 CL/<f>; • Load impedance is 117 — /99 12/(/). a) Draw the single phase equivalent for the a-phase. b) Calculate the a-phase line current. c) Calculate the a-phase line voltage for the three- phase load. 11.13 An acb sequence balanced three-phase Y-connected source supplies power to a balanced, three-phase A- connected load with an impedance of 12 + /9 fl/4>. The source voltage in the b-phase is 240/-50° V. The line impedance is 1 + /1 fl/4>. Draw the single phase equivalent circuit for the a-phase and use it to find the current in the a-phase of the load. 11.14 A balanced A-connected load has an impedance of 864 — /252 Q,/(f). The load is fed through a line hav- ing an impedance of 0.5 4- /4 Q/<f). The phase volt- age at the terminals of the load is 69 kV. The phase sequence is positive. Use V AB as the reference. a) Calculate the three phase currents of the load. b) Calculate the three line currents. c) Calculate the three line voltages at the sending end of the line. Figure P11.10 0.4 ft /212 a 1.6 n -vw- /4 ft 7811 -AlVir /54 a O Till (FV /1.6 b 2.6 0 /2.4 ft B 77 fl /56 O —• WV ooor> 1 w^ 0.2 ft /1.2 ft c 0.8 ft /3.8 ft C 79 ft /55 ft —• VA/ orvv> 0 VvV rwYV N 422 Balanced Three-Phase Circuits 11.15 A balanced Y-connected load having an imped- ance of 72 + /21 Cl/(f) is connected in parallel with a balanced A-connected load having an imped- ance of 150/0° Q,/<f>. The paralleled loads are fed from a line having an impedance of /1 0/4>. The magnitude of the line-to-neutral voltage of the Y- load is 7650 V. a) Calculate the magnitude of the current in the line feeding the loads. b) Calculate the magnitude of the phase current in the A-connected load. c) Calculate the magnitude of the phase current in the Y-connected load. d) Calculate the magnitude of the line voltage at the sending end of the line. 11.16 In a balanced three-phase system, the source is a bal- anced Y with an abc phase sequence and a line voltage V ab = 208/50° V. The load is a balanced Y in paral- lel with a balanced A.The phase impedance of the Y is 4 + /3 il/cf) and the phase impedance of the A is 3 - /9 0./4). The line impedance is 1.4 -I- /0.8 Oj4>. Draw the single phase equivalent circuit and use it to calculate the line voltage at the load in the a-phase. 11.17 The impedance Z in the balanced three-phase cir- cuit in Fig. PI 1.17 is 100 - /75 O. Find a) IAB> IBO and I CA , b) IaA* \^ and I cC , c) I ba , I cb , and I ac . Figure P11.17 13.2/-120° kV 11.18 For the circuit shown in Fig. PI 1.18, find PSPKE a) the phase currents I AB , I RC , and I CA MULTISIM / r ^° OK - Kl ^ b) the line currents I aA , I bB , and F cC when Z, = 2.4 - /0.7 ft, Z 2 = 8 + /6 Q, and Z 3 = 20 + /0 ft. Figure P11.18 a A 480/-120° V 11.19 A three-phase A-connected generator has an inter- nal impedance of 9 -(- /90 mft/<jf>. When the load is removed from the generator, the magnitude of the terminal voltage is 13,800 V. The generator feeds a A-connected load through a transmission line with an impedance of 20 + /180 mft/$. The per-phase impedance of the load is 7.056 + /3.417 ft. a) Construct a single-phase equivalent circuit. b) Calculate the magnitude of the line current. c) Calculate the magnitude of the line voltage at the terminals of the load. d) Calculate the magnitude of the line voltage at the terminals of the source. e) Calculate the magnitude of the phase current in the load. f) Calculate the magnitude of the phase current in the source. 11.20 A balanced three-phase A-connected source is shown in Fig. PI 1.20. a) Find the Y-connected equivalent circuit. b) Show that the Y-connected equivalent circuit delivers the same open-circuit voltage as the original A-connected source. c) Apply an external short circuit to the terminals A, B, and C. Use the A-connected source to find the three line currents I aA , I bB , and I cC . d) Repeat (c) but use the Y-equivalent source to find the three line currents. Figure PI 1.20 • A 2.7 O /13.511 4156/-120° V • C 11.21 The A-connected source of Problem 11.20 is con- nected to a Y-connected load by means of a bal- anced three-phase distribution line. The load Problems 423 impedance is 1910-/636 fl/<f). and the line imped- ance is 9.1 + /71.5(2/0. a) Construct a single-phase equivalent circuit of the system. b) Determine the magnitude of the line voltage at the terminals of the load. c) Determine the magnitude of the phase current in the A-source. d) Determine the magnitude of the line voltage at the terminals of the source. 11.27 The three pieces of computer equipment described below are installed as part of a computation center. Each piece of equipment is a balanced three-phase load rated at 208 V. Calculate (a) the magnitude of the line current supplying these three devices and (b) the power factor of the combined load. • Hard Drive: 4.864 kW at 0.79 pf lag • CD/DVD drive: 17.636 kVA at 0.96 pf lag . CPU: line current 73.8 A, 13.853 kVAR 11.28 Calculate the complex power in each phase of the unbalanced load in Problem 11.18. Section 11.5 11.22 In a balanced three-phase system, the source has an abc sequence, is Y-connected, and V an = 120/20° V. The source feeds two loads, both of which are Y-connected. The impedance of load 1 is 8 + /6 il/cf). The complex power for the a-phase of load 2 is 600/36° VA. Find the total complex power supplied by the source. 11.23 A balanced three-phase source is supplying 60 kVA at 0.6 lagging to two balanced Y-connected parallel loads. The distribution line connecting the source to the load has negligible impedance. Load 1 is purely resistive and absorbs 30 kW. Find the per-phase impedance of Load 2 if the line voltage is 120 V5 V and the impedance components are in series. 11.24 A three-phase positive sequence Y-connected source supplies 14 kVA with a power factor of 0.75 lagging to a parallel combination of a Y-connected load and a A-connected load. The Y-connected load uses 9 kVA at a power factor of 0.6 lagging and has an a-phase current of 10/-30° A. a) Find the complex power per phase of the A-connected load. b) Find the magnitude of the line voltage. 11.29 A balanced three-phase distribution line has an impedance of 1 + /8 fl/</>. This line is used to sup- ply three balanced three-phase loads that are con- nected in parallel. The three loads are L! = 120 kVA at 0.96 pf lead, L 2 = 180 kVA at 0.80 pf lag, and L 3 = 100.8 kW and 15.6 kVAR (magnetizing). The magnitude of the line voltage at the terminals of the loads is 2400 V3 V. a) What is the magnitude of the line voltage at the sending end of the line? b) What is the percent efficiency of the distribution line with respect to average power? 11.30 The line-to-neutral voltage at the terminals of the balanced three-phase load in the circuit shown in Fig. PI 1.30 is 1200 V. At this voltage, the load is absorbing 500 kVA at 0.96 pf lag. a) Use V AN as the reference and express I na in polar form. b) Calculate the complex power associated with the ideal three-phase source. c) Check that the total average power delivered equals the total average power absorbed. d) Check that the total magnetizing reactive power delivered equals the total magnetizing reactive power absorbed. 11.25 The total apparent power supplied in a balanced, three-phase Y-A system is 4800 VA. The line volt- age is 240 V. If the line impedance is negligible and the power factor angle of the load is -50°, deter- mine the impedance of the load. 11.26 Show that the total instantaneous power in a bal- anced three-phase circuit is constant and equal to 1,5V m I m cos 0$, where V m and I m represent the maximum amplitudes of the phase voltage and phase current, respectively. Figure PI 1.30 0.18 fi J iA4n # Vs/V ^VY-r>_ :-/180 a 0.18 0 /1.44« —/yyV e-v-Y-v>_ 0.18 a /1-440 * WV i^r>nr\_ A 500 kVA 0.% pf lag -•C 424 Balanced Three-Phase Circuits PSPICE MULTISIM 11.31 a) Find the rms magnitude and the phase angle of I CA in the circuit shown in Fig. PI 1.31. b) What percent of the average power delivered by the three-phase source is dissipated in the three- phase load? Figure PI 1.31 /2 ft a 1.5 ft -•—'vw- f^J 1365/0! 6 565/0° V /2 0 b i. 5 r> 1365/-120° V 1365/120° V /2 ft :85.5 0 /114 a B 85.5 XI !yii4n :85.5 ft ;/ii4 n c 1.5 ft 11.32 At full load, a commercially available 100 hp, three- phase induction motor operates at an efficiency of 97% and a power factor of 0.88 lag. The motor is sup- plied from a three-phase outlet with a line-voltage rating of 208 V. a) What is the magnitude of the line current drawn from the 208 V outlet? (1 hp = 746 W.) b) Calculate the reactive power supplied to the motor. 11.33 Three balanced three-phase loads are connected in parallel. Load 1 is Y-connected with an impedance of 400 + /300 ft/0; load 2 is A-connected with an impedance of 2400 - /1800 ft/<£; and load 3 is 172.8 + /2203.2 kVA. The loads are fed from a dis- tribution line with an impedance of 2 + /16 fl/cj). The magnitude of the line-to-neutral voltage at the load end of the line is 24 V3 kV. a) Calculate the total complex power at the send- ing end of the line. b) What percentage of the average power at the sending end of the line is delivered to the loads? 11.34 A three-phase line has an impedance of 0.1 + /0.8 Cl/(f). The line feeds two balanced three-phase loads connected in parallel. The first load is absorbing a total of 630 kW and absorbing 840 kVAR magnetizing vars. The second load is Y-connected and has an impedance of 15.36 - /4.48 0,/4>. The line-to-neutral voltage at the load end of the line is 4000 V. What is the magnitude of the line voltage at the source end of the line? 11.35 A balanced three-phase load absorbs 96 kVA at a lagging power factor of 0.8 when the line voltage at the terminals of the load is 480 V. Find four equiva- lent circuits that can be used to model this load. 11.36 The total power delivered to a balanced three- phase load when operating at a line voltage of 2400 V3 V is 720 kW at a lagging power factor of 0.8. The impedance of the distribution line sup- plying the load is 0.8 + /6.4 Q,/<fi. Under these operating conditions, the drop in the magnitude of the line voltage between the sending end and the load end of the line is excessive. To compen- sate, a bank of A-connected capacitors is placed in parallel with the load. The capacitor bank is designed to furnish 576 kVAR of magnetizing reactive power when operated at a line voltage of 2400 V3 V. a) What is the magnitude of the voltage at the sending end of the line when the load is operat- ing at a line voltage of 2400 V3 V and the capac- itor bank is disconnected? b) Repeat (a) with the capacitor bank connected. c) What is the average power efficiency of the line in (a)? d) What is the average power efficiency in (b)? e) If the system is operating at a frequency of 60 Hz, what is the size of each capacitor in microfarads? 11.37 A balanced bank of delta-connected capacitors is connected in parallel with the load described in Assessment Problem 11.9. The effect is to place a capacitor in parallel with the load in each phase. The line voltage at the terminals of the load thus remains at 2450 V. The circuit is operating at a fre- quency of 60 Hz.The capacitors are adjusted so that the magnitude of the line current feeding the paral- lel combination of the load and capacitor bank is at its minimum. a) What is the size of each capacitor in microfarads? b) Repeat (a) for wye-connected capacitors. c) What is the magnitude of the line current? 11.38 A balanced three-phase source is supplying 150 kVA at 0.8 pf lead to two balanced Y-connected parallel loads. The distribution line connecting the source to the load has negligible impedance. The power associ- ated with load 1 is 30 + /30 kVA. a) Determine the types of components and their impedances in each phase of load 2 if the line voltage is 2500 V3 V and the impedance compo- nents are in series. b) Repeat (a) with the impedance components in parallel. Problems 425 11.39 The output of the balanced positive-sequence three-phase source in Fig. PI 1.39 is 41.6 kVA at a lagging power factor of 0.707. The line voltage at the source is 240 V. a) Find the magnitude of the line voltage at the load. b) Find the total complex power at the terminals of the load. Figure P11.39 Balanced three-phase' source 0.04 n /°- 03 n WV /-Y-YV^_ 0.04 a /°- 03 n —Vs/V /"YVY-N— 0.04 0 /0-°3 U —VvV or-or^ Balanced "•three-phase load 11.43 The two wattmeters in Fig. 11.20 can be used to compute the total reactive power of the load. a) Prove this statement by showing that V3(W 2 - Wj) = V3V L / L siiiV b) Compute the total reactive power from the wattmeter readings for each of the loads in Example 11.6. Check your computations by cal- culating the total reactive power directly from the given voltage and impedance. 11.44 The two-wattmeter method is used to measure the power at the load end of the line in Example 11.1. Calculate the reading of each wattmeter. 11.45 The wattmeters in the circuit in Fig. 11.20 read as follows: W x = 40,823.09 W, and W 2 = 103,176.91 W. The magnitude of the line voltage is 2400 V5 V. The phase sequence is positive. Find Z^. Section 11.6 11.40 Derive Eqs. 11.56 and 11.57. 11.41 In the balanced three-phase circuit shown in Fig. PI 1.41, the current coil of the wattmeter is con- nected in line aA, and the potential coil of the wattmeter is connected across lines b and c. Show that the wattmeter reading multiplied by V3 equals the total reactive power associated with the load. The phase sequence is positive. Figure PI 1.41 11.46 a) Calculate the reading of each wattmeter in the circuit shown in Fig. PI 1.46. The value of Z^ is 40/-30° a. b) Verify that the sum of the wattmeter readings equals the total average power delivered to the A-connected load. Figure PI 1.46 24Q/W V | a • K m D • r- C ' ± cc [ [ W 1 1 pcj A B C Z<b z* z* N bl± Wi -1-) »_| /-YVV^ 240/120 o ~V~ 4^ 240/-120° V -• W, +™~ l-\ c 11.42 The line-to-neutral voltage in the circuit in Fig. PI 1.41 is 680 V, the phase sequence is positive, and the load impedance is 16 - j 12 0/$. a) Calculate the wattmeter reading. b) Calculate the total reactive power associated with the load. 11.47 The two-wattmeter method is used to measure the power delivered to the unbalanced load in Problem 11.18. The current coil of wattmeter 1 is placed in line aA and that of wattmeter 2 is placed in line bB. a) Calculate the reading of wattmeter 1. b) Calculate the reading of wattmeter 2. c) Show that the sum of the two wattmeter read- ings equals the total power delivered to the unbalanced load. . circuit for the system in Fig. 11.21. Practical Perspective Transmission and Distribution of Electric Power At the start of this chapter we pointed out the obligation utilities have to maintain. Problems 11.52(a)-(b) and 11.53, 11.56, and 11.57. Summary • When analyzing balanced three-phase circuits, the first step is to transform any A connections into Y connections, so that the overall