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# Class Note for ECE 3336 at UH

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Date Created: 02/06/15

G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Chapter 7 Instructor Notes Chapter 7 surveys all important aspects of electric power Coverage of Chapter 7 can take place immediately following Chapter 4 or as part of a later course on energy systems or electric machines The material in this chapter will be of particular importance to Aerospace Civil Industrial and Mechanical engineers who are concerned with the utilization of electric power The chapter permits very flexible coverage with sections 71 and 72 describing basic singlephase AC power ideas The ll examples contained in these two sections and the boxes Focus on Methodology Complex Power Calculations for a Single Load p 338 Focus on Methodology Complex Power CalculationforPower Factor Correction p 345 Focus on Measurements The Wattmeter pp 315353 and Focus on Measurements Power Factor p 354 will help the students master this basic material A survey course might only use this introductory material The next two sections discuss transformers and threephase power Sixexamples illustrate these ideas in detail Two descriptive sections 75 and 76 are also provided to introduce the ideas of residential wiring grounding and safety and the generation and distribution of AC power These sections can be covered independent of the transform er and threephase material The homework problems present a few simple applications in addition to the usual exercises meant to reinforce the understanding of the fundamentals Problems 719 and 72l24 present a variety of power factor correction problems Problem 720 illustrates the billing penalties incurred when electric loads have insufficient power factors this problem is based on actual data supplied by Detroit Edison Two advanced problems 740 741 discuss transformer test methods and problems 742 and 758 describe additional advanced applications these problems may be suitable in a second course in energy systems Those instructors who plan to integrate the threephase material into a course on power systems and electric machines will find that some of the threephase circuit problems 745759 can be assigned in conjunction with the material covered in Chapter 17 as part of a more indepth look at threephase machines Learning Objectives 1 Understand the meaning of instantaneous and average power master AC power notation and compute average power for AC circuits Compute the power factor of a complex load 2 Learn complex power notation compute apparent real and reactive power for complex loads Draw the power triangle and compute the capacitor size required to perform power factor correction on a load 3 Analyze ideal transform er39 compute primary and secondary currents and voltages and turns ratios Calculate re ected sources and impedances across ideal transformers Understand maximum power transfer 4 Learn threephase AC power notation39 compute load currents and voltages for balanced wye and delta loads 5 Understand the basic principles of residential electrical wiring and of electrical safety 7l G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Section 71 Power in AC Circuits Problem 71 Solution Known quantities Resistance value R 30 Q and the voltage across the soldering iron V 1 17 V Find The power dissipated in the soldering iron Analysis The power dissipated in the soldering iron is N2 2 V l 17 P 4563W R 30 Problem 72 Solution Known quantities Rated power P 1000 W and the voltage across the heating element l7 240V Find The resistance of the heating element Analysis The power dissipated in the electric heater is N2 2 RV240 576Q P 1000 Problem 73 Solution Known quantities Resistance value R 50 Q of the resistor Thredpower dissipated in the resistor if the current source connected to the resistor is a it 560slt50tgt A b it 5cos 50t 45 A c it 5 cos50t 260350t 45 A d it 560350t 2 A Analysis The average power can be expressed as 1 2 PavEIR 72 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 5250 a P 625 W av 2 b P 5 50 av 625W c By using phasor techniques I 540 2Z 45 5 14142jl4l42 35858jl4l42 3854642152 A Then the instantaneous current can be expressed as it 3 8546 coslOOt 2152 A Therefore the average power is 2 P 38546 50 av 3714W d The instantaneous voltage can be expressed as vt Rit 250 cos50t 100 V Then the instantaneous power can be written as pt vt it 25060350t 100 5 cos50t 2 1250 603250t 1000 cos50t 200 625 625 cos100t lOOOcoslt50tgt 200 w Therefore the average power is 13v 625 200 825W Problem 74 Solution Known quantities The current values Find The rrns value of each of the following currents a cos 450t 200s 450t b cos 5tsin 5t c cos 450t 2 d cos 5t cos5t n3 e cos 200t cos 400t Analysis The rrns current can be expressed as 12 Irms J5 N 1 2 a I 21213A J2 J2 b Using phasor analysis I cosStcoslt5t 90 140 14 90 1 j J22 45 A N J2 I 1A J2 73 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 c ii22707A J5 d Using phasor analysis I 140 1460 1 05 j0866 1732430 A i 1225A J5 N 1 1 e I 1414A J56 Problem 75 Solution Known quantities The current rms value 4 A the voltage source rms value 1 10 V the lag between the current and the voltage 60 Find The power dissipated by the circuit and the power factor Analysis The average power drawn by the circuit is V g110J524J P 700s cos60 220W The power factor is pf cos60 05 Problem 76 Solution Known quantities The voltage source rms value 120 V the source frequency 60 Hz the power consumption 12 kW and the power factor 08 Find a The rms current b The phase angle c The impedance d The resistance Analysis a The power is expressed as P if cos Thus the rms current is N P 1200 I 12 5 A T76030 12008 b The power factor is pf cos Thus the phase angle 6is 74 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 0 cos 10 8 3687 c The impedance Z is Z L 96 Q I 125 d The resistanceR is R Zcos0 7689 Problem 77 Solution Known quantities The rms values of the supply voltage and current 110 V and 14 A the power requirement 1 kW the machine efficiency 90 and the power factor 08 Find The AC machine efficiency Analysis The efficiency is MechanicalPower 1kW09 1111W quotmotor ElectricalPower coslt gt 1232W Problem 78 Solution Known quantities The waveform of a voltage source Find a The steady DC voltage that would cause the same heating effect across a resistance b The average current supplied to a 109 resistor connected across the voltage source c The average power supplied to a 19 resistor connected across the voltage source Analysis a VDCI7 j ig224v b 1mwi41m 210 I72 2 C PHVF 75 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Section 72 Complex Power Focus on Methodology Complex power calculation for a single load 1 Compute load voltage and current in ms phasor form using the AC circuit analysis methods presented in Chapter 4 and converting peak amplitude to rms values V V4 9V i 12 61 Compute the complex power S V1 and set ReS Pm ImS Draw the power triangle as shown in Figure 71 1 If Q is negative the load is capacitive39 if positive the load is reactive Compute the apparent power in units of VA V HeP N Focus on Methodology Complex power calculation for power factor correction 1 Compute the load voltage and current in ms phasor form using the AC circuit analysis methods presented in Chapter 4 and converting peak amplitude to rms values Compute the complex power S VI and set ReS Pm ImS Draw the power triangle for example as shown inFigure 719 Compute the power factor of the load pf cos6 If the reactive power of the original load is positive inductive load then the power factor can be brought to unity by connecting a parallel capacitor across the load such that l QC Q where Q is the reactance of the inductive load mC V HeP N Problem 79 Solution Known quantities The current and the voltage values Find The average power the reactive power and the complex power Analysis a P cos0cos20 21140W Q V73179 450 50 sin20 7696 VAR WE S V 11250420 VA b P V7cos0140585cos 30 7093w Q V7 710 140 5 85 sin 30 4095 VAR 76 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 S Vi 8194 30 VA c P cos050192cos 458 6688w Q fume 50192Sin 458 6887VAR S VF 9604 458 VA d P V7cos0 740108cos 859 45 60408W Q 17f Sin0 740108sin 859 45 52327VAR S VF 79924 409 VA Problem 710 Solution Known quantities The current and the voltage values or the impedance Find The power factor and state if it is leading or lagging Analysis a pf cos9i 19V cos212 0932 Leading b pf cos9i 9v cos 40 6 0759 Lagging c z39Lt 487 Sinltmt 274 487 Sinltoat 2 74 90 pf cos9i 19V cos67 0391 Leading d 0 tan 1 337 2 pf cose 9v COS 337 0832 Lagging Problem 711 Solution Known quantities The power factor or the values of the current and the voltage Find The kind of the load capacitive or inductive Analysis a Capacitive b Capacitive c Since iL I 18 603031 90 Inductive d Since the phase difference is zero Resistive 77 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Problem 712 Solution Known quantities The values of the resistance R 49 the capacitance C 118 HF the inductance L 2 H and the voltage source Find The real and reactive power supplied by the following sources a vst10cos3tV b vst100039tV Analysis N 10 a w3Z 396 39644 39OQI 177A T J J J 54 Pf2R125wQf2XOVAR N 10 b w9Z j18 j244j16 2 042A T 5165 Pf2R07w Qf2X282VAR Problem 713 Solution Known quantities The values of the resistances R1 8 Q R2 6 Q the reactances XC 12 Q XL 6 Q and the voltage sources v31 364 n3 v V 2440644v Find a The active and reactive current for each source b The total real power Analysis a From Figure P7l3 V31 R1 jXL I1 12 8 139611 139612 Vs2 XL 11 I2 R212 Xclz 1396116 f612 Substituting the values for the voltages sources gives 18 1312 8 j6Il 1612 i 192 1144 j6116 j6IZ Solving for 11 and Iz yields 11 0398 j338A 12 1091 j09llA Therefore the active and reactive currents for each source are IA10398A d IR1338A an IA21091A IR2091A 78 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 b PR21R1112614212834032105W Problem 714 Solution Known quantities The values of the resistors RL 259 R 19 the capacitor C 01mF the voltage source VS 230V and the frequency f 60 HZ Find a The source power factor b The current Is c The apparent power delivered to the load d The apparent power supplied by the source e The power factor of the load Analysis R R R 26 a pfSOW 1 10 07 Leading Z Z R1m2 Rlaad Xc J676 703396 N l7 b S s 2 Z 3714 Therefore IS 624 45 A Sload V131ad Qlad TSZVRZ Xc239 1401kVA d T3173 1426kVA R 25 load szm Zload 0 3645 39 F Problem 715 Solution Known quantities The values of the resistors RL 25 Q R 19 the inductor L 01H the voltage source VS 230V and the frequency f 60 HZ Find a The apparent power supplied by the source b The apparent power delivered to the load c The power factor of the load Analysis N2 N2 2 a39 SV V 230 1155kVA Z Rim Rm 2 X i 262 3772 2 2 b SW 4322 lem X 252 3772 1141kVA 79 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 R d 25 c 055 mm 21m 452 Problem 716 Solution Known quantities The values of the resistors RL 25 Q R 19 the capacitor C 01mF the inductor L 7035mH the voltage source 73 230V and the frequency f 60 HZ Find a The apparent power delivered to the load a The real power supplied by the source b The power factor of the load Analysis N2 NZ 2 a gV V 230 203kVA Z Rim 1197 2 XL XS 2 J262 265 2652 NZ 2 b XCXL P01203kw c XS XL 2 pf 1 Problem 717 Solution Known quantities The values of the resistor R 209 the capacitor C 01mF the voltage source 73 50V Find The apparent power the real power and the reactive power draw the power triangle Analysis NZ 2 s V 50 2500753VA R2 y 2 J202 2652 332 0C P Scos0 5 753 4536W Z 332 2 2 Q VS P 60VAR P4536W R For the power triangle I9 603quot1 53 Therefore the power triangle can be drawn as shown below s753 VA Q3960 VAR 710 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Problem 718 Solution Known quantities The values of the resistor R 209 the capacitor C 01mF the voltage source 73 50V Find The apparent power the real power and the reactive power in the cases of f 50 and 0 Hz Analysis For the frequency of 0 Hz SOHZ 0 POHZ 0 QOHZ 0 For the frequency of 50 Hz N2 S V 6715A SOHZ 2 1 R cocZ 20 P Scos6S 6715 36W Z 372 SOHZ QM S2 P2 567VA Problem 719 Solution Known quantities A singlephased motor connected across a 220V source at 50 Hz power factor pf 10 I 20 A and 11 25 A Find The capacitance required to give a unity power factor When connected in parallel with the load Analysis The magnitude of the current T2 is f2 1le f2 625 400 15A The voltage source can be expressed as 17 13 XC Therefore the required capacitor is 3 2 L 217 F V 60 2203 l4 7ll G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Problem 720 Solution Known quantities The currents and voltages required by an airconditioner a freezer a refrigerator and their power factors Find The power to be supplied by an emergency generator to run all the appliances Analysis In this problem we will use the following equations P T17003t9 Q INT7 31716 pf cost9 The real and reactive power used by the air conditioner are P1 9612009 10368 W Q1 96120 Sincos 109 50215 VAR The real and reactive power used by the freezer are P2 42120087 43848 W Q2 42120Sincos 1087 2485 VAR The real and reactive power used by the refrigerator are P3 3512008 336 w Q3 35 120 smcos 108 252 VAR The total real and reactive power P are P P1 P2 P3 181128 W Q Q1 Q2 Q3 100265 VAR Therefore the following power must be supplied S P jQ 181128 j100265 VA 2070342897 7 VA Problem 721 Solution Known quantities The schematics of the power supply module consisting of two 25kV singlephase power stations the power consumption by the train the DC power supply at a low speed operation the average power factor in AC operation the overhead line equivalent specific resistance and negligible rail resistance Find a The equivalent circuit b The locomotive current in the condition of a 10 voltage drop c The reactive power d The supplied real power overhead line losses and the maximum distance between two power station supplied in the condition of a 10 voltage drop when the train is located at the half distance between the stations e Overhead line losses in the condition of a 10 voltage drop when the train is located at the half distance between the stations assuming pf l The French TGV is designed with the state of art power compensation system f The maximum distance between the two power station supplied in the condition of a 10 percent voltage drop When the train is located at the half distance between the stations assuming the DC 15 kV operation at a quarter power Analysis a The equivalent circuit is 11 Line Line 12 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 b The locomotive current for the 10 voltage drop is 11MW N N N N N P I 1 1 21 21 2 L0 611A LOC 1 2 1 2 VS 10lcos0 225kV08 c The reactive power is 2 P2 17 10 212 P2 Q LOC LOC S 0 LOC LOC 1375 MVA2 11MW2 825 MVAR d The supplied real power is P1 S2 Q2 17 f2 Q2 1285Mw The overhead line power loss is PLine PLOC 1732 Q2 11MW 1527 MVA2 825 MVAR2 1 85 MW The maximum distance between the two power stations is Rm HRLW 5L2 59 2 Distancemax LOC e The overhead line power loss is PLOC 10I7S 1LOC cos0 2500 v 489 A 122 MW 025P f LOCAM2037A 90VSDC 1350V PM 10VS 1LOC 305kW 2RLine 100km 029km P RLine l l RLine M 0 0735 Q 2 LOC The maximum distance between the two power stations is 2R D1stancemax 15km 02 Qkm Problem 722 Solution Known quantities One hundred 40W lamps supplied by a 120V and 60Hz source the power factor of 065 the penalty at billing and the average prices of the power supply and the capacitors Find Number of days of operation for which the penalty billing covers the price of the power factor correction capacitor Analysis The capacitor value for pf 085 is 713 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 VCa wVC wVC l 2 2 I 2 2 513 333 392 333 19 420HF 337120 377120 Therefore the number of days of operation for which the penalty billing covers the price of the power factor correction capacitor is 420 uFSOi Number of Days mF 88 Days 4kw1hr001124 4 day Problem 723 Solution Known quantities Reference to the problem 722 and the network current decreasing with the power factor correction Find a The capacitor value for the unity power factor b The maximum number of lamps that can be installed supplementary without changing the cable network if a local compensation capacitor is used Analysis a lIC IL N l P T Pl 1 f i W CNL V 065 V 513 333 862HF Va Va 377120 b Initial cable network is f 513A Vcos0 120065 One lamp current for pf l is PLwn 40 I NP 0333A LW V 120 I The total number of lamps N 154 Lamp Therefore the number of supplementary lamps 154 100 54 Problem 724 Solution Known quantities The voltage and the current supplied by a source VS 74500 V TS 134 20 A G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Find a The power supplied by the source which is dissipated as heat or work in the load b The power stored in reactive components in the load c Determine if the circuit is an inductive or a capacitive load Analysis a S gm 7250 v132 20 A 455270 VA 1556 j4276VA PM 15 56 w b Q 4276 VAR c 9 91 9V 70 3 pf 6039 0342 Lagging The load is inductive Problem 725 Solution Known quantities The voltage supplied by a power plant VS 1 450 cosatV a 377 rads and the impedances of the plant Z 7 j 9 and ofthe power plant ZG 3 jO l lmQ Find Determine C so that the plant power factor is corrected to 1 Analysis Note ZG in uences only the phase difference between V5 and V0 and not the one between V0 and Z For this reason the result does not depend from Zg Z 22c RJXX1Xc XXCJRXC RJ39XXc 1 ZZC RX XC RjX XCR jX XC XXCR RXC X XC jR2XC XXCX XC R2X XC2 Req leq If 13 and V0 are in phase they have the same phase angle For this reason it must be ZeqReq XeqReq j0 2 Xeq0 2 RZXC XXCX XC0 2 2 Xc 509 1 2 C 1 1 513pF 60C wXC 377rads509 Problem 726 Solution Known quantities The voltage supplied by a power plant VS 1 450 cosatV a 377 rads and the impedances of the plant Z 74100 Q 715 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Find Determine C so that the plant power factor is corrected to 1 Analysis Note ZG in uences only the phase difference between V5 and V0 and not the one between V0 and Z For this reason the result does not depend from Zg Z 7410 Q689jl21 2 Z ZZC RJXX JXC XXCJRXC R J39X XC eq ZZc RJX1Xc RJXXcRJXXc XXcR RXCX XC JR2XC XXCX Xc R2X XC2 Req jX eq If 13 and V0 are in phase they have the same phase angle For this reason it must be 2 Zeq Req Xeq Req 0 2 Xeq 0 2 R XC XXCX XC 0 R2 X2 1 1 1 4031Q 2 C 658 F wC ch 377rads4031r2 H XC Problem 727 Solution Known quantities The voltage across a plant 70 450400 V f 60 HZ and the current through it without the capacitance in parallel with the plant is 174 10 A the value of the capacitance in parallel with the plant C 1740 HF Find The reduction of current which resulted from connecting the capacitor into the circuit Assumptions The impedance of the power plant is very small ZG m 0 Analysis Without capacitor is Tg 174 10 A v0 45040 v With capacitor being Z G E 0 the voltage across the plant does not change as well as the current v0 45040 v 2 ig 174 10 v KCL TSTgTC 0 is igic ig 174 10 M ZC ljoaC 1674 j2952j37717410761674j0A167440 A 716 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Problem 728 Solution Known quantities The voltage across a plant V0 1 170 603601 V f 60 HZ the current through it Without the capacitance in parallel with the plant is I 130 603601 l 1 A and the value of the capacitance in parallel with the plant C 387 AF Find The reduction of current which resulted from connecting the capacitor into the circuit Assumptions The impedance of the power plant is very small ZG m 0 Analysis Without capacitor is ig 1304 11 A v0 17040 v With capacitor being Z G E 0 the voltage across the plant does not change as well as the current v017040 v 2 i 1304 11 v g KCL TSTgTC0 N N N N 7 o ISIgICIg 01304 11 17M0 ZC ljoaC 12761 1392481j3771741076 12761390A1276140 A Problem 729 Solution Known quantities The values of the voltages and all the impedances Find The total average power the real power dissipated and the reactive power stored in each of the impedances Analysis SlV i1 will 20643430 kVA1788j1032kVAPav1jQ1 zl 074 30 Q 2 Wall 9633470 kVA 956 1391 17kVA Pav2 jQ2 22 1544 V31 7322 17040 170490 v2 Z 03 1049 24042445 v2 57845313 kVA3468 394624kVAP 054 5313 Q 1 ms JQs 7l7 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Problem 730 Solution Known quantities The voltage and the current supplied by a source VS 1704 9 V is 134160 A Find a The power supplied by the source which is dissipated as heat or work in the load b The power stored in reactive components in the load c Determine if the circuit is an inductive or a capacitive load Analysis a S vsi 1704 9 v13416 A 11054 25 VA 1001 1467 VA PW 1001w b Q 467VAR c 9 91 9V 25 3 pf 0039 0906 Leading The load is capacitive 718 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Section 73 Transformers Problem 731 Solution Known quanti es Each secondary connected to 5kW resistive load the primary connected to 120V rms Find a Primary power b Primary current Analysis a39 Pprim Psecl Pse02 N P b 1 M833A P V 120 im 83340 A Problem 732 Solution Known quanti es N N N 1 N The ratio between the secondary and the primary N m n and ml 2 mz E m prim Find a Vsec and V3901 if Vimm 220 V rms and n 11 b n if Vimm 110 V rms and Vsecg 5 V rms Analysis V a Vm z zEzzov n 11 Vsecl Keg 2 V b 211 ziVch 5 Problem 733 Solution Known quanti es The circuit and Vg 120 V rms Find a The total resistance seen by the voltage source b The primary current c The primary power Analysis 719 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 From the circuit shown on the right hand side vg11R1v1 12112R2 v2 nv1 i1 m392 V R 39 R a Rm g 1 1V1R1 2 2 11 11 n N V 120 b 11 g 60A Rtot c 1111127ng Problem 734 Solution Known quanti es The circuit and Vg 120 V rms Find a The secondary current b The installation efflciency Plum Pmme c The value of the load resistance which can absorb the maximum power from the given source Analysis From the circuit shown on the right hand side 11 60 a 12 15 A 4 4 b PM 122R2 2251636kvv05 39 PSOW vg 11 120 60 72kW 39 c For the maXimum power transfer Rprz39m R39sec R1 zizRW gt Rm 121 n2 219462169 71 Problem 735 Solution Known quanti es The voltage and the power that a transformer is rated to deliver to a customer V1 2 380 V Pm 460 kW Find a The current that the transformer supply to the customer b The maXimum power that the customer can receive if the load is purely resistive c The maXimum power that the customer can receive if the power factor is 08 lagging d The maXimum power that the customer can receive if the power factor is 07 lagging e The minimum power factor to operate if the customer requires 300 kW 720 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Analysis a From Sm 2 I722 2 Soul 11 zizwzizlkA V1 380 b For an ideal transformer P0 PM 6059 T7171 6059 For COS9 l P0 17171 460 kW c For COS9 08 the maximum power is PM 17171 cos9 368 kW d For COS9 07 the maximum power is PM 17171 cos9 322 kW e For Pout 300 kW the minimum power factor is cos9 h M 065 Pm 460 kW Problem 736 Solution Known quantities The voltage VS I 294 COS 3 77f V the resistances in a circuit containing a transformer and the ratio V0 I 1 n VS 5 Find a Primary current b V0 c Secondary power d The installation efficiency Pload Psource Analysis a The primary circuit is described in the figure lefthand side 100 Q The primary current is 171 294 1 Vs 77 9 11 2 082A 2 5 R1n 122 J2 10062525 11 1 144 cos377tA b The output voltage is 17 173 43171 2 208 100082 0 n n 25 v0 t 7lcos377tV 504V 721 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 c For the secondary power if pfl P2272170 71I70n082504251033W d The installation efficiency is PM P2 1033W 72 13mm 17371 208V082A Problem 737 Solution Known quantities The resistances RS 218009 RL 89 Find The turn s ratio that will provide the maximum power transfer to the load Analysis From Equation 741 for the re ected source impedance circuit we have RS eq N 2R3 Therefore the power is maximized if RsequL gt Nzln RLRS 20067 n 15 Problem 738 Solution Known quantities The voltage source and the resistances in the circuit Find a Maximum power dissipated by the load b Maximum power absorbing from the source c The installation eff1ciency Analysis All the impedances are resistances and therefore it is possible to consider the modules of voltages and currents a To maximize the power delivered to the 89 resistance 11 must be selected to maximize the load current 2 gt 0 Note that V1 21V2 and 1121112 11 KVL Mesh l vg 311V2 411 123n12 V2 4nI2 12 KVL Mesh 2 V2 812 412 11 812 412 n12 Rearranging the two mesh equations 722 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 1 7quot 4012 ZV2 Vg 2 7n 42 l12 4n12 vg 12 4n12 V2 quot 7t 1 2 7n2 8n12vg 112 7n2 8n12 n14n 8Vg 7n212 Vg dquot 7n2 8n122 7n2 8n122 5112 For the max1mum value of the load current 12 0 dn 7n2120 3 n1 131 The maximum load current is 71 I2 V 0122V 1347A 7n2 8n 12 g g The maximum power dissipated by the load is 2 Flood Rload12 b The maximum power absorbing from the source is 2 2 Pszmrcezpload3g39l 4QIl2 2 2 Pszmrce Pload3gn2 4Qn212 Psource 2540W934W70W 354kW c The installation efficiency is PM 245kW 0 7 397 2 13mm 354kW 39 Problem 739 Solution Known quantities The current and the voltage delivered by the transformer and the circuit of the transformer Find The efficiency of the installation Analysis 17W Wm 217m 220Vrms fW zljlrec zlfsec 25Arms 7t 2 VWE11A1ms 13920 20 is J 11 273Arms PS 19112 Pm 745W55kW 6245kW 723 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Therefore the efficiency of the installation is 7 Flood Psec Psource Psource 6245 kw Problem 740 Solution Known quantities The model for the circuit of a transformer and the results of two tests performed at n 377 rads 1 Opencircuit test I700 241v foo 095A PM 32w 2 Shortcucuittestzlic 5V in 525A PM 26W Find The value of the impedances in the equivalent circuit Analysis The power factor during the open circuit test is P pf 003 N of 01398 Laggmg C 0 The excitation admittance is given by Y 1 4cos 1pfmz 8L96 s00005511 j0003903s c as RC 18kQ RC zl39gkg X 3 Xc2562 2 LC 5 068H a The power factor during the short circuit test is P pf 00309 N S 09905 Leadmg SC SC The series impedance is given by ZW 7 4cos 1pjgciz7914 g 09476 1013119 1 525 RW0I9476Q Rw09476Q 3 XW013119 Lw 0348mH a Problem 741 Solution Known quantities The model for the circuit of a 460 kVA transformer and the results of two tests performed at f 60 Hz 1 Opencircuit test I700 4600V foo 07A PM 200w 724 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 2 Shortcircuit test T7 52 V Psc 50W Find The value of the impedances in the equivalent circuit Analysis The power factor during the open circuit test is P pfoc 004600 0062 Laggrng 00 OC The excitation admittance is given by I 1 07 6 3 YZCOS 2 4 86450829410 015210 S c Vac pfoc 4600 J R0210638k9 RC 106 3 8 kg 3 X X02658k 2 ch czl746H a The power factor during the short circuit test P pfSC COS6SC i z 0 SSC The series impedance has therefore imaginary part 0 1 Ideal transformer V2 1 rw 1 Z zizcos pf 20549 w W w Ps0 SC 0549 0 0 szosm RWZO Sm ICE LcI l gt X 10638 46 X 0 Q L w 0 17 w w 0 kg H I I Therefore the equivalent circuit is shown besides L J Problem 742 Solution Known quantities The circuit of the singlephase transformer with the high voltage regulation from five different slots in the primary winding the secondary voltage regulation in the range of 10 and the number of turns in the secondary coil Find The number of turns for each slot Analysis The secondary voltages are 725 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 17211723 012V108V1ms I722 I723 006V114Vrms I723120V1ms I724I7Z3006V126V1ms I725 l723012V132Vrms Therefore the number of turns for each slot is 7111 2037 X 2 E 408turns N n121929gtlt25386turns 3 n131833gtlt2z367turns m 7114 1746gtlt 2 E 349 turns n15 1666gtlt 2 E 333turns Problem 743 Solution Known quantities The pipe s resistance 00002 Q the secondary resistance 000005 Q the primary current 288 A and pf 091 Find a The plot number b The secondary reactance c The installation efficiency Analysis a The secondary current is Pprim Bras 3 Vprim prim 6030 Rsec Isec l7 IN cos 1m pm pm 220 288 091 4800A Rm 000025 Therefore the plot number is IS 4800 N 1666 3 Plot NumberS I 288 prim b The secondary resistance is pfprim pfsec 1 f 1 X R 1000025 lll4pQ sec sec pf2 c The installation efficiency is 2 7 Pload Rload 39Isec Rload 200 Q P R 250 19 2 Prim Rsec 39 Isec SEC 08 726 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Problem 744 Solution Known quantities A singlephase transformer converting 6 kV to 230 V with 095 efficiency the pf of 08 and the primary apparent power of 30 KVA Find a The secondary current b The transformer s ratio Analysis a The secondary current is Pm PMquot 47 Smquot cos0 17 3008095 228kW fm 124 A Vsec 39 6030 b The primary current is N Smquot 30000 25A 6000 rim N P V Therefore the transformer s ratio is prim Nlei248 prim 727 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Section 74 ThreePhase Power Problem 745 Solution Known quantities The magnitude of the phase voltage of a threephase wye system 220 V rms Find The expression of each phase in both polar and rectangular coordinates Analysis The phase voltages in polar form are van 20400 v vbquot 2204 120 v vquot 2204120 v The rectangular forms are van 220v vbquot 110 j19052V vquot 110j1905V The line voltages in polar form are vab vanz 150 3804 150 v vb 380490 v Vw 3804 30 V The line voltages in rectangular form are m 329 190V N7bc j380V v 329 190V Problem 746 Solution Known quantities The phase currents Tan 540 Tbquot 641500 Tm 44165quot Find The current in the neutral Wire Analysis The neutral current is in inquot ibn im 540 64150 441650 406 1404 573413514 A Problem 747 Solution Known quantities The voltage sources 7 12040 V VW 1204120 V 78 1204240 V Find a The voltages VRW 7 7 b The voltages VRW VWB VBR using ny VxxgZ 30quot c Compare the results obtained in a and b WB BR 39 728 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Analysis a VRW VR VW 12040 1204120 120 60 110392 20784 30 V VWB W VB 1204120 1204240 60 110392 60 110392 2078490 V VBR VB VR 1204240 12040 60 110392 120 20784 150 V b VRW VRJ34 30 120J34 30 20784 30 V VWB W J34 30 1204120 J 4 30 2078490 V VBR V854 30 1204240 J34 30 20784210 V 207 84 150 V c The two calculations are identical Problem 748 Solution Known quantities The voltage sources VR 11040 V VW 1104120 V VB 1104240 V and the three loads ZR 50g2 ZW j20 2 ZB j4SQ Find a The current in the neutral wire b The real power Analysis a NR amp11040 2240 A ZR 50 NB w 2444150 A 28 13945 iW VW 1104120 554210 A ZW 13920 TN i L L 22 554210 2444150 4924 1619 A b PRig50222242w Problem 749 Solution Known quantities The voltage sources VR 220400 V VW 22041200 V VB 22042400 V and the impedances RW RB RR 109 Find a The current in the neutral wire b The real power 729 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Analysis a TR 5 22040 22400 A RR NW V W 2204120 2241200 A RW 10 T8 E 2204240 2242400 A R 10 B Therefore the current in the neutral wire is TN TRTW TB 0A b The real power is v2 2202 PTgRiVZVRT R3RT 3f3 1452kW Problem 750 Solution Known quantities A threephase electric oven with a phase resistance of 10 Q connected at 3 X 380 V AC Find a The current owing through the resistors in Y and A connections b The power of the oven in Y and A connections Analysis a In Yconnection I 22 A W R 10 10 In Aconnection N V 380 1 38 A RD R 10 b In Yconnection PJ i J I7 TRYJ 380221451ltW me me me In Aconnection PJ IZWZWJ I7 TRDJ 380384331ltW me Problem 751 Solution Known quantities Apparent power of 50 kVA and supplied voltage of 380 V for a synchronous generator Find The phase currents the active powers and the reactive powers if a The power factor is 085 b The power factor is l 730 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Analysis a For the power factor of 085 SJ V7 3 fiN 76A EV J3 380 P Scos0 50000085 425kW Q J82 P2 J50 4252 263kVAR b For the power factor of 100 S P 2 T 76A P S cos0 50000 100 500kW Q SZ P20 Problem 752 Solution Known quantities The voltage sources V510 170 COSltCOIgtV VS2 I 170 603601 120 V v53t 170 cosat 120 v and the impedances Z1 05420 Q Z 03540 Q 23 17Z 90 Q the frequency f 60 HZ Find The current through Zl using a Loopmesh analysis b Node analysis c Superposition Analysis a Applying KVL in the upper mesh V52 V51 iZ1 i1 T2Zz 0 3 T1Z1 Zzi2 Applying KVL in the lower mesh 0 v i 11 1223 0 3 i1 22 w z v52 v53 3 2 For each mesh equation v v 17040 1704120 170 85 1147 2944 30 v 1 v 03 1704120 1704 120 85 j147 85 1147 294490 V 21 22 047 1390 171035 08384118 Q 22 23 035 j17174z 784 Q Therefore the current through 21 is NN T lt S751 i752 ZZ 294 30 03540 N v52 v53 22 23 294490 1744 784 1 21 22 ZZ 08384118 03540 ZZ 22 23 03540 1744 784 5124 1084 103490 41594 1129 2934418 A l464 666 012340 14164 712o b Choose the ground at the center of the three voltage source and let a be the center of the three loads The voltage between the node a and the ground is unknown 731 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Applying KCL at the node 11 Va Vsi Va Vsz Va V53 0 Z1 Z2 Z3 Rearranging the equation V51 Jr Jr 17040 17041200 1704 120 V221 Zz 230520quot 035410o 174 90o 1 1 1 1 1 1 21 Z 23 05420 035410 174 90o 3404 20 4864120D 1004330 30345730 24 20 28640 059490 474Z l2 Applying KVL the current through 21 is N N N N V 47 17040 6394585 Va Vsl IIZ1 0 3 I1 1 054200 c Superposition is not the method of choice for its great complexity 6394585 V 2934 418 A Problem 753 Solution Known quantities The voltage sources V1t 170 cosatV V20 17OCoswt120 V v30 l7OCosat 120 V and the impedances R 1009 C 047 11F L 100mH the frequency f 400 HZ Find The current throughR Analysis For each impedance Z1 100 2 10040 Q Z2 j j 1 j8466 284664 90 Q DC 2717 C Z3 jam j2nfL 125139 2513490 39 Applying KVL in the upper mesh V2 V1 1121 I1 12Z2 0 2 1121 ZZ IZ ZZV1 V2 Applying KVL in the lower mesh V3 V2 I2 11Z2 1223 0 2 11 ZZ 1222 ZV2 V3 For each mesh equation 71 72 170400 1704120D 170 85 jl47 2944 30 V V V3 1704120 1704 120 85 j147 85 1147 2944900 V 21 22 100 18466 85254 83312 2 23 j8466j2513 59534 9011 Therefore the current throughR is 732 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Yl Yz Z2 2944 30 84664 90 TV2 V3 22 294490 59534 90 1 ZlZz Z2 85254 8330 84664 90 7 Z2 22 84664 90 7 59534 900 17501034 120 248910340 22144 432 50751034 1733 71671034 180 22084 156quot 10034 276 A Problem754 Solution Known quantities The voltage sources Vslt 170 cosatV VS2I170 603601 120 V vs3t 170 cosat 120 V and the impedances Z1 340 Q 22 7490 Q Z3 0 jl IQ the frequency f 60 HZ Find The currents T1 T2 T3 Analysis Applying KVL in the upper mesh 22 Nsl 121 0 N V V 17040 1704120 2944 30 11 S1 S2 9814 30 A Z1 340 340 Applying KVL in the rightside mesh V3 V1 T222 0 N V V 17040 1704 120 294430 12 S1 S3 4214 60 A 22 7490 7490 Applying KVL in the lower mesh V3 V2 T323 0 N V2 V3 1704120 1704 120 294490 I3 2684 180 A Z3 114 90quot 114 90quot Problem 755 Solution Known quantities The voltage sources VRW 4164 30 V VWB 4164210 V VBR 416490 V and the impedances R1 R2 R3 409 L1 L2 L3 5 mH The frequency of each of the sources f 60 HZ 733 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Find The currents TW TB TR TN Analysis The line voltages are VRW 4164 30 v vWB 4164210 v vBR 416490v The phase voltages are vR 24040 v W 2404120 v v3 2404 120 v The currents are iRV R VR VR 240400 6z 27 A Zl R1Jwai R1127IL1 40 1377510 3 TWVWL203641173 A 22 40j377510 iB V B 2404 120 6z 1227 A Z3 40 j377510 3 TNTRTWTB0 Problem 756 Solution Known quantities The voltage sources VRW 4164 30 v vWB 41642100 v VBR 4164900 V and the impedances R1 R2 R3 409 L1 L2 L3 5 mH The frequency of each of the sources f 60 HZ Find a The power delivered to the motor b The motor39s power factor c The reason for which it is common in industrial practice not to connect the ground lead to motors of this type Analysis a The power delivered to the motor is P 3I7RTR cos0 32406cos 27 4315 w b The motor39s power factor is pf cos 2 7 09988 Lagging c The circuit is balanced and no neutral current flows thus the connection is unnecessary Problem 757 Solution Known quantities A threephase induction motor designed not only for Y connection operation in general but also for A connection at the nominal Y voltage for a short time operation Find The ratio between the powers 734 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Analysis The power for Y connection operation is N 1 N N N PY 3 VPhaseMomR Iliney 3 e Ilinel J3Vlme Ilmel The power for A connection operation is N N N Vline PA 3 Vline IphaseMomR PhaseMomR 339Vvline 39 39wz 339Vvline Iliney 339PY PhaseMomR Therefore the ratio between the powers is P Ratio A 3 P Y Problem 758 Solution Known quantities The voltage sources VR 12040 V VW 1204120 V VB 1204240 V and the impedances R1 R2 R3 SQ XLl XL2 XL3 69 ofthe motor Find a The total power supplied to the motor b The power converted to mechanical energy if the motor is 80 efficient c The power factor d The risk for the company to face a power factor penalty if all the motors in the factory are similar to this one Analysis a By virtue of the symmetry of the circuit we can solve the problem by considering just one phase The current TR is i VR 120400 R R1jX1 5j6 The total power supplied to the motor is P 3I7RTR cos0 31201536cos5019 35413w b The mechanical power is Pm 08P 283223 W c The power factor is pf cos9 064 d The company will face a 25 penalty 1536z 5019 A 735 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 7 Problem 759 Solution Known quantities The voltage source at 220 V ms of a residential fourWire system supplying power to the singlephase appliances ten 75W bulbs on the lst phase one 750W vacuum cleaner with pf 087 on the 2nd phase ten 40W lamps with pf 064 on the 3quoti phase Find a The current in the neutral Wire b The real reactive and apparent power for each phase Analysis a The current in the neutral Wire is N P 3 P 3 10 75 IA A J A 3440A Vline 3 80 N PB 75133 75750 VIM vmecos0 380087 N P 313 IC N C N Cm 284 290 A09539263A VIM Vmecos0 380064 TN iATB fC 095 j068A 2 TN 116A b The real reactive and apparent powers for the lst phase are SAPA750W QA0 The real reactive and apparent powers for the 2quotd phase are PB 750w SB P B 862VA QB SB smaB 431VAR 00393 087 The real reactive and apparent powers for the 3quoti phase are P PC 400 w SC C 63 625VA QC SC sinltr9c 478VAR 0030C 3924 150 A 34 j195A 736

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