Soluções de Problemas de Física Elétrica - Capítulo 1

1 PROBLEM SOLUTIONS Chapter 1 Problem 11 Part a Rc lc µAc lc µrµ0Ac 0 AWb Rg g µ0Ac 1017 106 AWb part b Φ NI Rc Rg 1224 104 Wb part c λ NΦ 1016 102 Wb part d L λ I 6775 mH Problem 12 part a Rc lc µAc lc µrµ0Ac 1591 105 AWb Rg g µ0Ac 1017 106 AWb part b Φ NI Rc Rg 1059 104 Wb part c λ NΦ 8787 103 Wb part d L λ I 5858 mH WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 2 Problem 13 part a N 9 iiot urns HoAc part b B T core 166 A LoNg Problem 14 part a L c c r ve gtleuom Lg letourto 121 turns boAc HoAc part b B 182 A HoNg letoH Problem 15 part a 25 15 95 500 1000 1500 2000 2500 H Avr part b 3499 lp 1 EEE 730 V1 0047228 rp iit 658 A poN WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 3 part c 25 ea 0 Problem 16 part a NI A x He Ti Be 38 Be 1 3H part b Equations 2gHHele NI Bg Ag BeAc and Be pole B pH can be combined to give 2 NI NI eo 0 Ak 0 a me eem Be arn Problem 17 part a gt 2 lc Ip IB 215 A poN part b 1199 u no 1 eepaege 1012 g 2 lc Ip I B 302 A poN WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 4 part c 2 a 18 aw 16 a 14 Pw E12 B41 208 06 04 02 i 1 2 FS 4 i 6 Current A Problem 18 N72 A g er 4 I 0353 mm L Le Problem 19 part a I 2nRo Ri g 357 cm Ac Ro Rih 12 cm part b R L 133x 10 AWb R0 AWb HoAc part c N2 L 0319 H R R m part d p BalRet RelAe 331 4 N part e NBAc 105 mWb Problem 110 part a Same as Problem 19 part b 9 7 te 5 Re loc 133 x 10 AWb Re nAe 316 x 10 AWb 5 part c L N 2 Rg Rg 0311 mH part d I BgRc RgAc N 338 A part e Same as Problem 19 Problem 111 Minimum µr 340 Problem 112 L µ0N 2Ac g lcµr Problem 113 L µ0N 2Ac g lcµr 305 mH Problem 114 part a Vrms ωNAcBpeak 2 192 V rms part b Irms Vrms ωL 167 A rms Wpeak 05L 2 Irms2 850 mJ WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 6 Problem 115 part a R3 R R23 427 cm part b 2 p Hoa Los nt gt 2 Ie part c For w 2760 radsec and Apeak NAgBpeak 0452 Wh i Viems WApeak 171 V rms ii lIims me 181 Arms iii Wpoeak 05LV2Ims 0817 J part d For w 2750 radsec and Apeak N Ag Bpeak 0452 Wh i Viems WApeak 142 V rms ii lIims me 181 Arms iii Wpoeak 05LV2Ims 0817 J Problem 116 part a Voltag 5 Feo 1 E d E WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 7 part b Emax 4f NAcBpeak 345 V Problem 117 part a LI NI I N Ba turns g FB 036 mm part b From Eq321 Weap 2e9Bist 0207 J Weore AeleBsat 9045 J gap 2110 core Qu Thus Wot Weap Weore 0252 J From Eq 147 12LI 0252 J QED Problem 118 part a Minimum inductance 4 mH for which g 00627 mm N 20 turns and Vims 678 V part b Maximum inductance 144 mH for which g 499 mm N 1078 turns and Vims 224 V Problem 119 part a 2 n72 p ote N 560 mH 2nr part b Core volume Veore 27rra 400 m3 Thus B W Voore 487 J 20 part c For T 30 sec di 2rrBuoN a T 292x 10 Asec di L163 V eT Problem 120 part a Acu fwabd Voleu 2abw h 2a part b WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 8 JouAcu rove g part c NI Jeu 7 Acu part d Paiss Voleu pJeu part e a 2 rs Wma Vol a wh gap g D0 part f L 5 LP Wag 2Wrmag powhA2 R 4RP 4 Pass Paiss pg Vole Problem 121 Using the equations of Problem 120 Paiss 115 W IT324A N 687 turns R1080 T 618 msec Wire size 23 AWG Problem 122 part a NL NI i B aft B e ify 91 g2 A A ii Aq Ni Ai By Ao Bo pio N i 91 g2 wa Ay iii Ag No A2Bo ptoN1 No we qi WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 9 part b i By0 B wee g2 A2 ii At N A2 Bo LoN1 No 2 Ip eee 2 Ag iii AQ No Ao Bo LoN5 Tp part c i B HoNitli B HoNi ti 4 Ho N22 gn g2 g2 A A A ii At NiAi By Ao Bo Lio N qi LoN1 No Ip 1 g2 g2 ve A A iii AQ No A2 Bo LoN1 No qi jo N35 Tp 92 92 part d A A A A LniN Loo poN3 Lie poNi No 91 92 92 g2 Problem 123 E C R Ry Pe Ri 0O fe Tg ta fh 2 9 Ra TA Ris A Ro TA Re A part a Li N N yA RitRo2Rygt Ra2 ly lo1y2 9 Lpo WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 10 N NpAc lath tlog ubMo La Lapa qe wee Ra t RalRi Re Ry In la 2 12 g uHo part b Lap Lea NRi Ro Rg N Ac litle 49 uMo RaRa 2Ri Re Ry la la 2 tla 9 uj0 NN NNpAc ADIDAS BES IIB Rat 2Ri RaRy lat 2 l 9 up0 part c arin Lorin Lar fia in U dt AltA B1B HAL dt tA 2B QED Problem 124 part a NN Ly2 CC Dw z g part b v daz I dLy2 Ni NopoD da dt dt 2g dt Se coswt 2g 2 Problem 125 part a H Nyy Nyi4 InRo Ri2 aRo Ri part b d dB U2 dt NotnAB N2tnA part c Vo G v dt GNotnAB 11 Problem 126 Rg g µ0Ag 442 105 AWb Rc lc µAg 333 µ AWb Want Rg 005Rc µ 12 104µ0 By inspection of Fig 110 this will be true for B 166 T approximate since the curve isnt that detailed Problem 127 part a N1 Vpeak ωtRo RiBpeak 57 turns part b i Bpeak Vopeak GN2tRo Ri 0833 T ii V1 N1tRo RiωBpeak 625 V peak Problem 128 part a From the M5 magnetization curve for B 12 T Hm 14 Am Similarly Hg Bµ0 954 105 Am Thus with I1 I2 I I HmlA lC g Hgg N1 382 A part b Wgap gAgapB2 2µ0 321 Joules part c λ 2N1AAB 0168 Wb L λ I 439 mH Problem 129 part a WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 12 part b Area 191 Joules part c Core loss 150 Wkg Problem 130 Bins 11 T and f 60 Hz Vims WNABims 467 V Core volume A 105 x 107 m3 Mass density 765 x 103 kgm Thus the core mass 105 x 10765 x 10 803 kg At B 11 T rms 156 T peak core loss density 13 Wkg and rms VA density is 20 VAkg Thus the core loss 13 x 803 104 W The total exciting VA for the core is 20 x 803 160 VA Thus its reactive component is given by V160 1042 122 VAR The rms energy storage in the air gap is AB Weap gecms 361 Joules Ho corresponding to an rms reactive power of VARgap WWeap 1361 Joules Thus the total rms exciting VA for the magnetic circuit is VArms sqrt104 1361 122 1373 VA and the rms current is Tims VArmsVims 294 A Problem 131 parta Area increases by a factor of 4 Thus the voltage increases by a factor of 4 to e 1096cos377t part b J doubles therefore so does the current Thus I 026 A part c Volume increases by a factor of 8 and voltage increases by a factor of 4 There Ig ms doubles to 020 A part d Volume increases by a factor of 8 as does the core loss Thus P 128 W Problem 132 From Fig 119 the maximum energy product for samariumcobalt occurs at approximately B 047 T and H 360 kAm Thus the maximum energy product is 169 x 10 Jm Thus 08 Am ss 2m 340 cm and WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 13 08 lm 02 035 om x am om Thus the volume is 340 x 035 120 cm which is a reduction by a factor of 509121 49 Problem 133 From Fig 119 the maximum energy product for neodymiumironboron occurs at approximately B 063 T and H 470kAm Thus the maximum energy product is 290 x 10 Jm Thus Am 98 9 om 254m 063 and 08 lm 02 027 a oa x am Thus the volume is 254 x 025 0688 cm which is a reduction by a factor of 5090688 74 Problem 134 From Fig 119 the maximum energy product for samariumcobalt occurs at approximately B 047 T and H 360 kAm Thus the maximum energy product is 169 x 10 Jm Thus we want B 12 T Bn 047 T and Ay 360 kAm Al B hm g g 2 a ar 65 mm B B Am Ag 3 2rRh 3 260 cm Am Rm 14 287 cm T Problem 135 From Fig 119 the maximum energy product for neodymiumironboron oc curs at approximately By 063 T and Hy 470 kAm The magnetization curve for neodymiumironboron can be represented as By PRA B where B 126 T and wr 106719 The magnetic circuit must satisfy WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 14 Hmd Hegg Ni BmAm Bg Ag part a For i 0 and B 05 T the minimum magnet volume will occur when the magnet is operating at the maximum energy point B Am 3 Ag 476 cm ef d FE a 109 mm part b dA Bd Be Se al N For B 075 i 179 A For B 025 i 60 A Because the neodymiumironboron magnet is essentially linear over the op erating range of this problem the system is linear and hence a sinusoidal flux variation will correspond to a sinusoidal current variation WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 15 PROBLEM SOLUTIONS Chapter 2 Problem 21 At 60 Hz w 1207 primary Vimsmax NiwABrmsmax 2755 Vrms secondary Vimsmax NowAcBrmsmax 172 Vrms At 50 Hz w 1007 Primary voltage is 2295 V rms and secondary voltage is 143 V rms Problem 22 V2Vims N 1 t DABoak 67 turns Problem 23 n 23 turns 8 Problem 24 Resistance seen at primary is Ry Ni N2R2 6250 Thus Yi i16 A 1 Ry and No 4 V2 Vi 0 V Problem 25 The maximum power will be supplied to the load resistor when its im pedance as reflected to the primary of the ideal transformer equals that of the source 2 kQ Thus the transformer turns ratio N to give maximum power must be Rs N 632 Rioad Under these conditions the source voltage will see a total resistance of Rtot 4 kQ and the current will thus equal J VRtot 2 mA Thus the power delivered to the load will equal Proad IN Rioad 8 mW WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 16 Here is the desired MATLAB plot 8 7 7 6 5 Ea 2 1 5 2 FS 4 3 6 a 8 9g 10 Number of turns Problem 26 The maximum power will be supplied to the load resistor when its im pedance as reflected to the primary of the ideal transformer equals that of the source 2 kQ Thus the transformer turns ratio N to give maximum power must be R N 632 Rioad Under these conditions the source voltage will see a total impedance of Zor 2 72 kQ whose magnitude is 22 kQ The current will thus equal J VsZtot 22 mA Thus the power delivered to the load will equal Proad I N Rioaa 16 mW Here is the desired MATLAB plot 16 14 12 10 Es 2 6 4 2 5 2 FS 4 3 6 a 8 9g 10 Number of turns WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 17 Problem 27 Xm x Problem 28 part a Referred to the secondary Lin Lma 3 150 mH partb Referred to the secondary Xm wlm2 5679 X 848 mQ and Xj 693 mQ Thus Xm i VYiN 7960 V i W 2 and V2 V2 ii l 1730 A X sc Xi XmX1 Problem 29 part a Vi Xm i 347 A V2 NV 2398 V 1X Xm gtx part b Let X XN and Xsc X1 XmXm X For Lratea 50 kVA120 V 417 A Yi LratedX sc 231 V 1 X I Dra ed 1 A oN x ted 157 Problem 210 Proaa I 555 A L Vi and thus L Iq w 106 A VaNVy 4 jXuly 2381296 V The power factor is cos 96 0986 lagging 18 Problem 211 part a part b ˆIload 30 kW 230 V ejφ 938 ejφ A where φ is the powerfactor angle Referred to the high voltage side ˆIH 938 ejφA ˆVH ZH ˆIH Thus i for a power factor of 085 lagging VH 2413 V and ii for a power factor of 085 leading VH 2199 V part c WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 19 Problem 212 part a part b Following methodology of Problem 211 i for a power factor of 085 lagging VH 4956 V and ii for a power factor of 085 leading VH 4000 V part c Problem 213 part a Iload 160 kW2340 V 684 A at cos1 089 271 ˆVtH N ˆVL ZtIL which gives VH 337 kV part b ˆVsend N ˆVL Zt ZfIL WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 20 which gives Vgenq 334 kV part c Ssend Psend JQsend Vsenalxng 164 kW j645 kVAR Thus Psend 164 kW and Qsend 645 kKVAR Problem 214 Following the methodology of Example 26 efficiency 984 percent and regulation 125 percent Problem 215 part a Vise Zequ 2 1078 mOQ Isc Pye Regt SS 478 mQ Tee XegqL 4Zeqt Rea 1077 mQ and thus ZogL 48 7108 mQ part b Req N Regt 0455 2 Xequ NXeq1 1024 2 ZeqH 103 7046 mQ part c From the opencircuit test the coreloss resistance and the magne tizing reactance as referred to the lowvoltage side can be found Rew 2 3110 oh PoeL 7 SocL VocLlocL 497 kVA Qoon Veen Poo 452 kVAR and thus 21 XmL V 2 ocL QocL 141 Ω The equivalentT circuit for the transformer from the lowvoltage side is thus part d We will solve this problem with the load connected to the high voltage side but referred to the lowvoltage side The rated lowvoltage current is IL 50 MVA8 kV 625 kA Assume the load is at rated voltage Thus the lowvoltage terminal voltage is VL Vload ZeqLIL 8058 kV and thus the regulation is given by 805388 00072 072 percent The total loss is approximately equal to the sum of the opencircuit loss and the shortcircuit loss 393 kW Thus the efficiency is given by η Pload Pin 500 5039 0992 992 percent part e We will again solve this problem with the load connected to the highvoltage side but referred to the lowvoltage side Now ˆIL 625 258 kA Assume the load is at rated voltage Thus the lowvoltage terminal voltage is VL Vload ZeqLˆIL 7758 kV and thus the regulation is given by 775888 00302 302 percent The efficiency is the same as that found in part d η 992 percent Problem 216 The core length of the second transformer is is 2 times that of the first its core area of the second transformer is twice that of the first and its volume is 2 2 times that of the first Since the voltage applied to the second transformer is twice that of the first the flux densitities will be the same Hence the core loss will be proportional to the volume and Coreloss 2 23420 967 kW WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 22 The magnetizing inductance is proportional to the area and inversely pro portional to the core length and hence is 2 times larger Thus the noload magnetizing current will be 2 times larger in the second transformer or Tnoload V2 493 697 A Problem 217 part a Rated current at the highvoltage side is 20 kVA24 kV 833 A Thus the total loss will be Poss 122 257 379 W The load power is equal to 08 x 20 16 kW Thus the efficiency is 16 0977 977 percent 1 T6379 Pewee part b First calculate the series impedance Zequ ReqH jXeqH of the transformer from the shortcircuit test data PscH Reg p 369 D scH SscH Voc HIscH 613 x 833 511 kVA Thus QscH 4 Se 34 Poy 442 VAR and hence XeqH eo 6350 Teen The regulation will be greatest when the primary and secondary voltages of the transformer are in phase as shown in the following phasor diagram ON 1 oY yi v 4 L Thus the voltage drop across the transformer will be equal to AV NoaaZeqH 612 V and the regulation will equal 612 V24 kV 0026 26 percent Problem 218 For a power factor of 087 leading the efficiency is 984 percent and the regulation will equal 348 percent Problem 219 part a The voltage rating is 2400 V2640 V part b The rated current of the high voltage terminal is equal to that of the 240V winding ratea 30 x 10240 125 A Hence the kVA rating of the transformer is 2640 x 125 330 kVA 23 Problem 220 part a part b The rated current of the high voltage terminal is equal to that of the 120V winding Irated 104120 833 A Hence the kVA rating of the transformer is 600 833 50 kVA part c The full load loss is equal to that of the transformer in the con ventional connection Ploss 1 0979 10 kW 210 W Hence as an auto transformer operating with a load at 085 power factor Pload 08550 kW 425 kW the efficiency will be η 425 kW 4271 kW 0995 995 percent Problem 221 part a The voltage rating is 78 kV86 kV The rated current of the high voltage terminal is equal to that of the 8kV winding Irated 50 1068000 625 kA Hence the kVA rating of the transformer is 86 kV 625 kA 5375 MVA part b The loss at rated voltage and current is equal to 393 kW and hence the efficiency will be η 5375 MW 5381 MW 09993 9993 percent Problem 222 No numerical result required for this problem Problem 223 part a 797 kV23 kV 191 A651 A 1500 kVA part b 138 kV133 kV 109 A1130 A 1500 kVA part c 797 kV133 kV 191 A1130 A 1500 kVA part d 138 kV23 kV 109 A651 A 1500 kVA Problem 224 part a i 239 kV115 kV 300 MVA ii Zeq 00045 j019 Ω iii Zeq 0104 j430 Ω WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 24 part b i 239 kV664 kV 300 MVA ii Zeq 00045 7019 2 iii Zeq 00347 71472 Problem 225 Following the methodology of Example 28 Vioaq 236 V linetoline Problem 226 The total series impedance is Zo Zp Z 7117 011 7220 011 7139Q The transformer turns ratio is N 9375 The load current as referred to the transformer highvoltage side will be 25 MVA Toad N a ec 781 kA V3 24 kV where cos093 216 The linetoneutral load voltage is Vioaa 243 kV part a At the transformer highvoltage terminal V V3NVioad LoaaZt 2317 kV linetoline part b At the sending end V V3NVioad MoadZtot 2333 kV linetoline Problem 227 240 235 230 Z ns 220 215 i oo 40 30 20 10 oO 10 20 30 40 50 pf angle degrees Problem 228 First calculate the series impedance ZeqH Regu jXeqH Of the trans former from the shortcircuit test data ZogH 048 7118 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 25 The total imedance between the load and the sending end of the feeder is Zo4 Z5 ZogH 0544 7920580 The transformer turns ration is N 2400120V3 116 part a The referred load voltage Vioaq and current Ijoaq will be in phase and can be assumed to be the phase reference Thus we can write the phasor equation for the sendingend voltage as Ve Vioad Droaa Ztot We know that Vz 2400sqgrt3 1386 V and that Iicagd 100 kVAV324 kV Taking the magnitude of both sides of the above equation gives a quadradic equation in Vioaa View 2ReotLioad Vioad Z0tTioaa V2 which can be solved for Vioaa Vioad Rtotlioad V V2 XtotLioaa 1338 kV Referred to the lowvoltage side this corresponds to a load voltage of 1338 kVN 116 V linetoneutral or 201 V linetoline part b 24 Feeder current ml 651 A V3Zt0t 1 HV winding current Got 376 A V3 LV winding current 651N 752 kA Problem 229 part a The transformer turns ratio is N 7970120 664 The sec ondary voltage will thus be A Vi jXm V2 1197420101 oN acon part b Defining Rj N2Ry N71 kQ 441 MOQ and Zeq jXmRb Ry 7X 1343 77581 kO the primary current will equal 7970 1032 7987 mA Ri 9X1 Zeq WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 26 The secondary current will be equal to 7 7 jXm IhNI 119720054 A aaa and thus V2 Rplz 119720054 V part c Following the methodology of part b V2 119620139 V Problem 230 This problem can be solved iteratively using MATLAB The minimum reac tance is 291 Q Problem 231 part a Solid line Rb dotted line Xb 02 03 weg 04 for Fos ie 206 2 07 H 039 j 1 al 9 500 1000 1500 2000 2500 3000 Burden Ohms part b Solid line Rb dotted line Xb 05 04 i o3 e 02 SS 2 O41 UT ost avrsccsnrnncccteesccceteessis 5 0 z 01 02 03 ao 45 500 1000 1500 2000 2500 3000 Burden Ohms WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 27 Problem 232 part a The transformer turns ratio N 2005 40 For I 200 A I jXm I 498720024 ON x 5Xm 7D part b Defining Ry N250uQ 049 bh4 jm 498720210 2 ON REF REF Gm t XA Problem 233 part a Solid line Rb dotted line Xb OT O3p 035 oo 5 04 E oss oo 08 a aes 06 06D 200 300 400 500 600 700 800 900 1000 Burden uOhms part b Solid line Rb dotted line Xb 022 02 018 016 B04 B012 508 006 004 00795 200 300 400 600600700 00 900 1000 Burden uOhms Problem 234 VoaseL ZbaseL ae 1800 p he Prase ViraseH ZbaseH ae 2452 p a Prase 28 Thus R1 00095ZbaseL 171 mΩ X1 0063ZbaseL 113 mΩ Xm 148ZbaseL 266 Ω R2 00095ZbaseH 233 Ω X2 0063ZbaseH 154 Ω Problem 235 part a i ZbaseL 797 1032 75 103 0940 Ω XL 012ZbaseL 0113 Ω ii ZbaseH 79702 75 103 847 Ω XH 012ZbaseH 102 Ω part b i 797 V138 kV 225 kVA ii Xpu 012 iii XH 102 Ω iv XL 0339 Ω part c i 460 V138 kV 225 kVA ii Xpu 012 iii XH 102 Ω iv XL 0113 Ω Problem 236 part a In each case Ipu 1012 833 pu i IbaseL Pbase 3 VbaseL 225 kVA 3 797 V 163 A IL IpuIbaseL 1359 A ii IbaseH Pbase 3 VbaseH 225 kVA 3 138 kV 94 A IH IpuIbaseH 784 A part b In each case Ipu 1012 833 pu i IbaseL Pbase 3 VbaseL 225 kVA 3 460 V 282 A IL IpuIbaseL 2353 A ii IbaseH Pbase 3 VbaseH 225 kVA 3 138 kV 94 A IH IpuIbaseH 784 A WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 29 Problem 237 part a On the transformer base Phase t 800 MVA Xeen 157 157 127 8 F aK pe part b On the transformer base the power supplied to the system is Pout 700850 0824 pu and the total power is Sout Pourpf 0825095 0868 pu Thus the per unit current is J 08682 where cos 095 182 i The generator terminal voltage is thus Ve 104 1Z 1032394 pu 2682394 kV and the generator internal voltage is Veen 10 1Z Zeon 2072443 pu 5372443 kV ii The total output of the generator is given by Sgen VI 08262 03361 Thus the generator output power is Psen 08262 x 850 7022 MW The correspoinding power factor is PgenSgen 0926 lagging 30 PROBLEM SOLUTIONS Chapter 3 Problem 31 By analogy to Example 31 T 2B0Rl I1 sin α I2 cos α 663 102 I1 sin α I2 cos α Nm Thus part a T 0530 cosα Nm part b T 0530 sinα Nm part c T 0530 I1 sin α I2 cos α Nm Problem 32 T 05304 Nm Problem 33 Can calculate the inductance as L Nφ I 1000 013 10 13 H Thus Wfld 1 2 LI2 650 Joules Problem 34 part a For x 09 mm L 295 mH and thus for I 6 A Wfld 0531 Joules part bFor x 09 mm L 196 mH and thus for I 6 A Wfld 0352 Joules Hence Wfld 0179 Joules Problem 35 For a coil voltage of 04 V the coil current will equal I 04011 37 A Under the assumption that all electrical transients have died out the solution will be the same as that for Problem 34 with a current of 37 A instead of 60 A part a Wfld 0202 Joules part b Wfld 0068 Joules Problem 36 For x x0 L L0 30 mH The rms current is equal to Irms I0 2 and thus part a Wfld 1 2LI2 rms 0227 Joules WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 31 part b Paiss 7R 163 W Problem 37 part a Be poNi 29 B B Waa 5 x Airgap volume x 2gAe 210 240 pio N Ao 1 40 2 72 7 Ag T part b 2 2 L 2a HoN Ao a 2g 7 Here is the MATLAB plot 2 19 18 V7 16 5 15 13 12 tg 20 10 oO 10 20 30 theta degrees Problem 38 part a vct Voe7 r RC part b Waa q2C Cv22 Thus CVe Waa0 Waaoo 0 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 32 part c vat V2e72t7 int 9 Passlt OR oo CVS Waiss Paiss t dt 0 2 Problem 39 part a VY L iit Re T p part b VoL Waa0 2R2 Waaoo 0 part c V2e72t7 Pyisst i7tR R V2L iss Paisst dt 0 Wa diss t 2 R Problem 310 Given L 2 T R 48sec R13 MW Thus 1 1 R T 5 whi 5E R i R 624 MJoules Problem 311 part a Four poles part b OWig ad IG py Taa dom dm E Lo L2 cos 26m Ip Lz sin 20m WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 33 Problem 312 part a Ni B Molva g 9 R2h where gj is the length of the fixed gap is its length and R is the radius of the solenoid Here is the MATLAB plot 16 14 12 Eos Bos s 04 02 5 05 1 15 2 25 Variablegap length cm part b B2 Waa 7Rg 240 Here is the MATLAB plot 45 40 36 30 34 20 15 10 5 5 056 1 15 2 25 YVariablegap length cm WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 34 part c L 2Waai Here is the MATLAB plot 18 16 14 1 6 4 2 5 05 1 15 2 25 Yariablegap length cm Problem 313 If the plunger is moved very slowly ie idLdt Ldidt the current will be essentially constant and all of the change in stored energy will come from the mechanical work applied to the plunger Thus part a Work Waag 02 cm Waag 225 cm 467 pJoules part b The battery will supply only the energy dissipated in the coil Problem 314 The coil inductance is equal to L jo NA2g and hence the lifting force is equal to dL pio N2 A 2 faa 5 i 2 dg 4g where the minus sign simply indicates that the force acts in the direction to reduce the gap and hence lift the mass The required force is equal to 931 N the mass of the slab times the acceleration due to gravity 98 msec Hence setting g gmim and solving for i gives 29min faa min A i 70 J 385 m and Umin lminR 108 V Problem 315 part a a 913071 x 107 ag 0124209 a3 281089 a4 105582 b 968319 x 107 by 137037 x 1077 b3 632831 x 107 bg 171793 x 1073 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 35 part b i Here is the MATLAB plot 16 14 LS wl Sy zg L 4 Lf 25 05 1 15 2 25 3 Winding current A ii Waa 130 Joules and W4 137 Joules Assuming no core relctance Wag 118 Joules and Wj 130 Joules part c i Here is the MATLAB plot 16 14 ow 210 4 25 4 6 8 10 12 14 16 18 20 22 Winding current A ii Waa 142 Joules and Wi 148 Joules Assuming no core relctance Wag 139 Joules and Wi 147 Joules Problem 316 N72 A i dL L g 2 dg 29 The timeaveraged force can be found by setting i Itms where Irms VimswL Thus L V2 rms rms 115 N Sa 2gwL 2w UgN2Ae WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 36 Because the inductor is being driven by a voltage source the gap flux density remains constant independent of the airgap length and hence the force also remains constant Problem 317 part a p 8 part b Hoxl os Bsrl 8 part c Note that as the coil moves upward in the slot the energy associated with the leakage flux associated withing the coil itself remains constant while the energy in the leakage flux above the coil changes Hence to use the energy method to calculate the force on the coil it is necessary only to consider the energy in the leakage flux above the slot B li Waa Ss dV Mor a 210 2s Because this expression is explicity in terms of the coil current 7 and becasue the magnetic energy is stored in air which is magnetically linear we know that Wha Waa We can therefore find the force from f dWig gli id dx 2s This force is positive acting to increase x and hence force the coil further into the slot part d faq 181 Nm Problem 318 H2 2N2 Waa x coil volume 3 je Thus fe dW fla Lomo N P dro h 0 and hence the pressure is f boN 2 P I Qrroh 2h 0 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 37 The pressure is positive and hence acts in such a direction as to increase the coil radius ro Problem 319 part a q Waalae f vaeda 0 part b Ow faa 9 ra v q part c Wha vg dWaa dWhg adv faadr Thus Ow Waa qvxdu faa an Problem 320 part a q 2 2 Waa dq q xq fd oq 2dg oa DeoA Cv eg Av Waa qv xdv 5 On part b hha OWig Cv Av id Ox 2 Qe and thus eo AV faaVo 5 262 Problem 321 part a V2 dC Rd Ta dc 2 0 5 fo Go part b In equilibrium Taa Tspring 0 and thus WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 38 Rd 2 0 A rr Vie Here is the plot 40 35 30 wt Boo 2 15 10 5 55 500 1000 1500 Vde V Problem 322 part a N2A N2A Ly 1 Loy 2 290 290 part b Lio HoNiN2A 12 2 Jo part c 1 1 A Waa Luiz Lois Lygiiig Mo Miia Noiz 2 2 4g part d Ow oA 2 faa Nyt Nota Ogo i1i2 Age Problem 323 part a 1 ol 2 2 2 Wia hut 9 baat Lygiqig I Ly Log 2042 sin wt Taa 42x 1032 siné sin wt Nm 1142 39 part b Tfld 21 103I2 sin θ Nm part c Tfld 021 Nm part d part e The curve of spring force versus angle is plotted as a straight line on the plot of part d The intersection with each curve of magnetic force versus angle gives the equilibrium angle for that value of current For greater accuracy MATLAB can be used to search for the equilibrium points The results of a MATLAB analysis give I θ 5 525 707 353 10 213 part f WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 40 Problem 324 part a dL Tha irig 0 28ii2sin Nm dé L 2 05i2 i 112i1 cos Lo Therefore for 7 10 sinwt Taa 314i7 sin cos6 314sin wt sin 6 cos 0 7851cos2wt sin20 Nm part b Tha 785 Nm part c It will not rotate It will come to rest at angular positions where dT Tha0 and 29 dé ie at 90 or 6 270 Problem 325 part a Winding 1 produces a radial magnetic which under the assumption that g 19 N Bri hom ay g The zdirected Lorentz force acting on coil 2 will be equal to the current in coil 2 multiplied by the radial field B and the length of coil 2 2 NN fi 2nr9pN2By 112 mokoaes 4412 g part b The self inductance of winding 1 can be easily written based upon the winding1 flux density found in part a QarolpgN lyi TTotlolVy g The radial magnetic flux produced by winding 2 can be found using Amperes law and is a function of z 0 Ozu2 B HoNataea eszcath Hollate rthzl WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM Al Based upon this flux distribution one can show that the self inductance of coil 2 is 2 N32 2h Lo2 TOKO 2 12 g 3 part c Based upon the flux distribution found in part b the mutual inductance can be shown to be 2 NLN h Ly42 TOMO we g 2 part d d 1 1 rropoNs 2QrropoN No faa Lisi Looid Lygiyi2 ono 3 oHom ivig dx 2 2 g g Note that this force expression includes the Lorentz force of part a as well as a reluctance force due to the fact that the self inductance of coil 2 varies with position x Substituting the given expressions for the coil currents gives N32 2 NN faa TOKO i cos wt mono 2 Ty Ip cos wt g g Problem 326 The solution follows that of Example 38 with the exception of the magnet properties of samariumcobalt replaced by those of neodymiumboroniron for which pr 106u0 Hi 940 kAm and B 125 T The result is faa 203N atxOcm fd 151N atx 05 cm Problem 327 part a Because there is a winding we dont need to employ a fictitious winding Solving Hyd Hggo Ni BnwD BhD in combination with the constitutive laws Bm rRHm He Bg pbollg gives me du wg be hn Note that the flux in the magnetic circuit will be zero when the winding current is equal to I HdN Hence the coenergy can be found from WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 42 integrating the flux linkage of the winding from an initial state where it is zero ie with i Jp to a final state where the current is equal to i The flux linkages are given by AX NwDB and hence a UgwDN Ni Hed Waali 2 Ai x di w Hi To fet hea oN The force is then dW powDNgo Ni Ad faa fd Ho a dx Hoda wgo 2 2N i for i 0 f dWig How Dgo Hed fid 2 dx a Hoa da wgo where the minus sign indicates that the force is acting upwards to support the mass against gravity ii The maximum force occurs when h DHd fimax Ho Mynaxa 2 where a is the acceleration due to gravity Thus Monex pow DHd 2a part b Want Max LgwDHd Iminxh PUnmingen 4 5 4 Substitution into the force expression of part a gives Imin 2 V2Hed 059Hd Problem 328 part a Combining Hyd Heg 0 mr Bm 2nrolBy Bs Lo He By Lr Am Hyme gives WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 43 Hd By Poa a i part b The flux linkages of the voice coil can be calculate in two steps First calculate the differential flux linkages of a differential section of the voice coil of dN2 turns at height z above the bottom of the voice coil which is at zZ2 l Hdo2 l dz dNo B2nrodz Heduo2nrol 2 any z g 12 27 R TO Recognizing that dN2z N2hdz we can now integrate over the coil to find the total flux linkages ath N2Hd 2 la2h2 dv 2h HHo ro a h2 a tn rn part c Note from part a that the magnet in this case can be replaced by a winding of Ni21 Hd ampereturns along with a region of length d and permeability wR Making this replacement from part a the self inductance of the winding can be found 2rrohN7d Au N14 2rrohNi Bg m0 1 CHO a a in Ce and thus 2rrohN7duo fu bb 2ld a ie rn Similar the mutual inductance with the voice coil can be found from part b as Li A2 Ni A2 N2NoMUo02779 I xX h2 ta 15 28 We can now find the coenergy ignoring the term L22i32 1 2 aa Waa 9 Lyi Li212 LoHdnrgh 4 Lio No Hd 27rodl py ee ie LR ro LR ro WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 44 part d dW Ho N2Hed 2rd faa dx 4 Ho 2la g UR ro Problem 329 part a Hintm Hyx Hog 0 7R3 R3Bm TRIBy 27RihBy Bs Hog By poly Bn HRHm HH where pop 1059 and HL 712 kAm Solving gives Ri HA etm B Ho Rs cin 0562 T Qha gh jun R2R2 and 2h B By 0535 T m part b We can replace the magnet by an equivalent winding of Ni Htm The flux linkages of this equivalent winding can then be found to be 27 pohR N A N2rRih Bg mu orn t it The force can then be found as i dL 27 9 hR1Ni faa 2 dx 2p9 R2 ht 2 2 htm 2he gRi pute na 2mp0hR1Hetm 7 mHohR yy 00158 N 2h R 20 Rehtm er gi wntre 18 part c WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 45 Problem 330 part a If the plunger is stationary at x 09a the inductance will be constant at LD 01L Thus it 8 e tt where tT LR The force will thus be 2 dL Lo Vo 2 7 aa 2 dx 2a 3 part b Xo 09a4 7 090 7 8 oe Kyo 2aKy R Problem 331 part a Since the current is fixed at i Ip 4 A the force will be constant at f ILo2a 145 N Thus Xp 09ea 7 156 em Ko part b dx dx M 2 7 f Ko09a x 02 de 548 3502 N dL Lo dx dx vIRIo IpR a dt v 6 0182 part c The equations can be linearized by letting Xo 2t and vVovt The result is d x de 17502 and dx 0182 dt part d For in meters xt e coswt m where 1750 418 radsec and WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 46 ut 76le sinwt V Problem 332 part a For a de voltage of Vo 6 V the corresponding dc current will be Io VoR 4 A the same as Problem 331 Hence the equilibrium position will be the same Xo 156 cm part b For a fixed voltage the dynamic equations become d x di Lo dx j Li iRL 1 Yo iR Li iR Lo a dt ir or di dx 1514x107914 0182 i 6 5i 4 x 107 Ox 0 82i and dea i L M 2 f Ko09a Ko09a 2 or x 2 02 de 009092 693 350a part c The equations can be linearized by letting Xo 2t and i Ip 7t The result is Xo di Lo dx L 1 I Oath o i dt or di dx 015i15x 107 0728 PO at dt and dx IoLo M dt a Kow or dx 2 02 de 07271 3502 Problem 333 part a Following the derivation of Example 31 for a rotor current of 8 A the torque will be give by T To sina where Ty 00048 Nm The stable equilibrium position will be at a 0 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM A7 part b d2 J iP To sina part c The incremental equation of motion is la J dt2 Toa and the natural frequency is Ti w i 062 radsec J corresponding to a frequency of 0099 Hz Problem 334 As long as the plunger remains within the core the inductance is equal to L jioda N 5 ag 2 where z is the distance from the center of the solenoid to the center of the core Hence the force is equal to i dL podnN7i7x fad 9 ae x ag Analogous to Example 310 the equations of motor are dx dx podnNix M B Ka19 I dt dt 2 to ag The voltage equation for the electric system is vy ip Hoden S 2 di 2podnNa dx ag 2 dt ag dt These equations are valid only as long as the motion of the plunger is limited so that the plunger does not extend out of the core ie ring say between the limits a2 x a2 48 PROBLEM SOLUTIONS Chapter 4 Problem 41 part a ωm 1200 π30 40π radsec part b 60 Hz 120π radsec part c 1200 56 1000 rmin Problem 42 The voltages in the remaining two phases can be expressed as V0 cos ωt 2π3 and V0 cos ωt 2π3 Problem 43 part a It is an induction motor parts b and c It sounds like an 8pole motor supplied by 60 Hz Problem 44 part a part b WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 49 part c part d Problem 45 Under this condition the mmf wave is equivalent to that of a singlephase motor and hence the positive and negativetraveling mmf waves will be of equal magnitude Problem 46 The mmf and flux waves will reverse direction Reversing two phases is the procedure for reversing the direction of a threephase induction motor Problem 47 F1 Fmax cos θae cos ωet Fmax 2 cos θae ωt cos θae ωt F2 Fmax sin θae sin ωet Fmax 2 cos θae ωt cos θae ωt and thus Ftotal F1 F2 Fmax cos θae ωt WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 50 Problem 48 For n odd 2 ae cos n6dé nb 12 sin Joey2 cos nd0 2 For 3 576 nO 097 n1 sin 9 0 n3 026 n5 Problem 49 part a Rated speed 1200 rmin part b 1g Bagipeak poles I 113 A ApiokN part c 2 Op 3 RBagipeak 0937 Wh Problem 410 From the solution to Problem 49 p 0937 Wb Vim N 804 kV V2 Problem 411 From the solution to Problem 49 p 0937 Wb wkyN Vims 104 kV V2 Problem 412 The required rms linetoline voltage is Vims 1303 751 kV Thus 2 Nz ve 39 turns Problem 413 part a The flux per pole is 2IRBagipeak 00159 Wh The electrical frequency of the generated voltage will be 50 Hz The peak voltage will be 51 Vpeak ωNΦ 388 V Because the spacefundamental winding flux linkage is at is peak at time t 0 and because the voltage is equal to the time derivative of the flux linkage we can write vt Vpeak sin ωt where the sign of the voltage depends upon the polarities defined for the flux and the stator coil and ω 120π radsec part b In this case Φ will be of the form Φt Φ0 cos2 ωt where Φ0 00159 Wb as found in part a The stator coil flux linkages will thus be λt NΦt NΦ0 cos2 ωt 1 2 NΦ01 cos 2ωt and the generated voltage will be vt ωΦ0 sin 2ωt This scheme will not work since the dccomponent of the coil flux will produce no voltage Problem 414 Fa iaA1 cos θa A3 cos 3θa A5 cos 5θa Ia cos ωtA1 cos θa A3 cos 3θa A5 cos 5θa Similarly we can write Fb ibA1 cos θa 120 A3 cos 3θa 120 A5 cos 5θa 120 Ia cos ωt 120A1 cos θa 120 A3 cos 3θa A5 cos 5θa 120 and Fc icA1 cos θa 120 A3 cos 3θa 120 A5 cos 5θa 120 Ia cos ωt 120A1 cos θa 120 A3 cos 3θa A5 cos 5θa 120 The total mmf will be WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 52 Fiot FatFoFe alt cos 6 wt As cos 56 wt 3 wt ofalAt cos 8 wt As cos 5 G 5 We see that the combined mmf contains only a fundamental spaceharmonic component that rotates in the forward direction at angular velocity w and a 5th spaceharmonic that rotates in the negative direction at angular velocity w5 Problem 415 The turns must be modified by a factor of 18 1200 9 ey Gry 14 064 Problem 416 30E P Npolesn 625 mW Problem 417 part a 2 2Boeaklr 2 x 2 x 125 x 021 x 09522 125 mWb Pp poles peak 4 oe Vims 1 230V3 x 4 Nph poles v x 43 turns V2tfmekwPp W227 x 60 x 0925 x 00125 part b From Eq B27 2 p Holr ie 212 mH Tg poles Problem 418 part a Vers 108 mWb VJ2 TNph Bycak op 0523 T WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 53 part b TB eakY p P 065 A Quok Ny part c Na eak V2 Vimsw Da P H f Ik I 069 Problem 419 No numerical solution required Problem 420 2D1 eak B ea peak ros Peals 4k N I max F peak X poles 7 oles Tpeak 2 2 peak lt peak 439 x 10 Nm Ppoeak TpeakWm 828 MW Problem 421 2D1 Preak Byea peak ros Peak 4k N I max F eak peak x poles 1 oles Tpeak 9 5 PyeakFrpeak 161 Nm Poeak Tpeakwm 606 kW Problem 422 part a dMa AMys T ta lalt do tiple do Mig ip cos Oo ig sin 0 54 This expression applies under all operating conditions part b T 2MI2 0cos θ0 sin θ0 2 2 MI2 0 sin θ0 π4 Provided there are any losses at all the rotor will come to rest at θ0 π4 for which T 0 and dtdθ0 0 part c T 2 MIaIfsin ωt cos θ0 cos ωt sin θ0 2 MIaIf sin ωt θ0 2 MIaIf sin δ part d va Raia d dt Laaia Mafif 2 IaRa cos ωt ωLaa sin ωt ωMIf sin ωt δ vb Raib d dt Laaib Mbfif 2 IaRa sin ωt ωLaa cos ωt ωMIf cos ωt δ Problem 423 T MIfib cos θ0 ia sin θ0 2 MIf Ia I2 sin δ I2 sin 2ωt δ The timeaveraged torque is thus T 2 MIfIa I2 sin δ Problem 424 part a T i2 a 2 dLaa dθ0 i2 b 2 dLbb dθ0 iaib dLab dθ0 iaif dMaf dθ0 ibif dMbf dθ0 2 IaIfM sin δ 2I2 aL2 sin 2δ part b Motor if T 0 δ 0 Generator if T 0 δ 0 part c For If 0 there will still be a reluctance torque T 2I2 aL2 sin 2δ and the machine can still operate WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 55 Problem 425 part a v 25 msec part b The synchronous rotor velocity is 25 msec part c For a slip of 0045 the rotor velocity will be 1 0045 x 25 239 msec Problem 426 Ins Pe 3 3 1 N V2 Lo 3 4 kw Noh 3 145 93 x 10 2 2x7 o18 A V2 Lo 3 4 091 x 280 Problem 427 part a Defining 27wavelength nB 2wB w Bpeax cos Bada 3 148 mWb 0 part b Since the rotor is 5 wavelengths long the armature winding will link 10 poles of flux with 10 turns per pole Thus Apeak 100 0148 Wb part c w Gv and thus Vine AP 346 Virms V2 56 PROBLEM SOLUTIONS Chapter 5 Problem 51 Basic equations are T ΦRFf sin δRF Since the field current is constant Ff is constant Note also that the resultant flux is proptoortional to the terminal voltage and inversely to the frequency ΦR Vtf Thus we can write T Vt sin δRF f P ωfT Vt sin δRF part a Reduced to 311 part b Unchanged part c Unchanged part d Increased to 396 Problem 52 part a The windings are orthogonal and hence the mutual inductance is zero part b Since the two windings are orthogonal the phases are uncoupled and hence the flux linkage under balanced twophase operation is unchanged by currents in the other phase Thus the equivalent inductance is simply equal to the phase selfinductance Problem 53 Lab 1 2 Laa Lal 225 mH Ls 3 2 Laa Lal Lal 708 mH Problem 54 part a Laf 2 Vllrms 3ωIf 794 mH part b Voltage 5060 154 kV 128 kV Problem 55 part a The magnitude of the phase current is equal to Ia 40 103 085 3 460 591 A and its phase angle is cos1 085 318 Thus WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 57 ˆIa 591ej318 Then ˆEaf Va jXs ˆIa 460 3 j415 591ej318 136 568 V The field current can be calculated from the magnitude of the generator voltage If 2Eaf ωLaf 113 A part b ˆEaf 266 381 V If 153 A part c ˆEaf 395 278 V If 202 A Problem 56 The solution is similar to that of Problem 55 with the exception that the sychronous impedance jXs is replaced by the impedance Zf jXs part a ˆEaf 106 666 V If 122 A part b ˆEaf 261 437 V If 163 A part c ˆEaf 416 312 V If 220 A Problem 57 part a Laf 2 Vllrms 3ωIf 498 mH part b ˆIa 600 103 3 2300 151 A ˆEaf Va jXs ˆIa 177 413 V If 2Eaf ωLaf 160 A WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 58 part c See plot below Minimum current will when the motor is operating at unity power factor From the plot this occurs at a field current of 160 A Problem 58 part a Zbase V 2 base Pbase 26 1032 750 106 0901 Ω Ls XspuZbase ω 488 mH part b Lal XalpuZbase ω 043 mH part c Laa 2 3Ls Lal Lal 340 mH Problem 59 part a SCR AFNL AFSC 0520 part b Zbase 26 1032800 106 0845 Ω Xs 1 SCR 219 pu 185 Ω part c Xsu AFSC AFNL ag 192 pu 162 Ω WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 59 Problem 510 part a AFNL 114 SCR AFSC part b Zpase 416075000 x 10 346 O xX 111 386 s 1 U 9 SCR P part c AFSC Xs u yman vV Q AFNL ag 088 pu 305 Problem 511 No numerical solution required Problem 512 part a The total power is equal to S Ppf 4200 kW087 4828 kVA The armature current is thus 4828 x 103 i 2 cos 087 6702295 A V3 4160 Defining Z Ra 7X5 0038 7481 2 41 Ear Va Zslal A260 Z1 4349 V line to neutral V3 Thus I AFNL 5 306 A 4160V3 part b If the machine speed remains constant and the field current is not reduced the terminal voltage will increase to the value corresponding to 306 A of field current on the opencircuit saturation characteristic Interpolating the given data shows that this corresponds to a value of around 4850 V linetoline WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 60 Problem 513 260 250 240 230 220 e 210 200 190 180 170 160 600 1000 1500 2000 2500 3000 3500 4000 4500 5000 Power kVV Problem 514 At rated power unity power factor the armature current will be J 5000 kW3 4160 V 694 A The power dissipated in the armature winding will then equal Parm 3 x 694 x 0011 159 kW The field current can be found from Al Eat Va Zla A160 Z1 3194 V linetoneutral v3 and thus 194 Ie AFNL ss 319A 4160V3 At 125C the fieldwinding resistance will be 2345 125 Re 0279 0324 0 2345 75 and hence the fieldwinding power dissipation will be Preig I Rs 211 kW The total loss will then be Prot Poore Parm Prrictionwindage Piela 120 kW Hence the output power will equal 4880 kW and the efficiency will equal 48805000 0976 976 61 Problem 515 part a part b AFNL 736 A AFSC 710 A part c i SCR 104 ii Xs 0964 per unit and iii Xsu 117 per unit Problem 516 For Va 10 per unit Eafmax 24 per unit and Xs 16 per unit Qmax Eafmax Va Xs 0875 per unit Problem 517 part a Zbase V 2 base Pbase 529 Ω Xs 1 SCR 0595 perunit 315 Ω WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 62 part b Using generator convention for current part c Eaf 150 420 0357 perunit For Va 10 perunit ˆIa Eaf Va jXs 108 90 perunit 136 90 kA using Ibase 1255 A part d It looks like an inductor part e Eaf 700 420 167 perunit For Va 10 perunit ˆIa Eaf Va jXs 112 90 perunit 141 90 kA In this case it looks like a capacitor WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 63 Problem 518 a 125 12 Zn i 1 105 1 400 200 300 400 500 600 700 Ifld A Problem 519 part a It was underexcited absorbing reactive power part b It increased part c The answers are the same Problem 520 part a 22 Xs 0268 perunit part b P 0875 and S P09 0972 both in per unit The power factor angle is cos 09 258 and thus J 08752 258 Eat Va Xslq 1152116 perunit The field current is Jf AFNLE 958 A The rotor angle is 116 and the reactive power is Q VS P 424 MVA part c Now Eas 10 per unit PX 6 sin7 E 136 sin V and thus Er 102136 Ba Va i Sto 08812679 JXs Q ImagVf 0104 perunit 104 MVAR WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 64 Problem 521 EarVa Va Eat I a Ix X sin 6 j cos 6 The first term is a constant and is the center of the circle The second term is a circle of radius Eaf Xs A Ei A My IXST Jo Ww bof Xs i Xs TWN y qe OM 7 oo a Locus of qa Problem 522 part a i A far a A Aiba 4 We Cs Nols Vo ii Ve Voo 10 per unit P 375650 0577 per unit Thus 6 sin 126 65 and ˆIa Vtejδt V jX 0578 393 perunit Ibase Pbase 3 Vbase 1564 kA and thus Ia 904 kA iii The generator terminal current lags the terminal voltage by δt2 and thus the power factor is pf cos1 δt2 0998 lagging iv ˆEaf V jX XsˆIa 150 perunit 360 kVlinetoline part b i Same phasor diagram ii ˆIa 0928 632 perunit Ia 145 kA iii pf 0994 lagging iv Eaf 206 per unit 494 kV linetoline Problem 523 part a WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 66 part b part c Problem 524 part a From the solution to Problem 515 Xs 0964 per unit Thus with V Eaf 10 per unit WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 67 Veo Ea Ponax 104 perunit noindent Hence full load can be achieved This will occur at X 6sin 746 Eas Vintty part b The generator base impedance is 131 2 Thus X 014131 0107 per unit Now P VoFat 1 04 it 0934 it 135 MW max 1 erunlt VU erunlit Xn X peru peru Problem 525 Follwing the calculation steps of Example 59 Fa 135 per unit Problem 526 Now Xq 964 per unit and Xg 0723 per unit Thus part a VooE at v2 1 1 P sin d x sin 26 1037 sind 0173 sin 26 An iterative solution with MATLAB shows that maximum power can be achieved at 6 536 part b Letting Xp Xa Xoo and XQ Xqt Xoo VooEat v2 1 1 P sind x x sin 26 0934sin 6 0136 sin 26 An iterative solution with MATLAB shows that maximum power that can be achieved is 141 Mw which occurs at a power angle of 75 Problem 527 03 a8 oe eo as na c a XY SN 505 VEG Nw 5 i Ss g 04 2 part a XQ MA a ff part b QO 03 i pan c QO fy part d 02 iy an iy a OF f Y oF 20 40 60 80 100 120 140 160 180 Power angle degrees WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 68 Problem 528 La 4 V 1 RoTe J Xo Ta More Morarion y negate at deaue k bane bap Problem 529 ba qeanis a J KT le N L RTa Geneerne Norenow 9 negahve as deawn Problem 530 For Fat 0 V2 1 1 Pax 021 21 5 ag 2 This maximum power occurs for 6 45 Ty V2289 9786 perunit Xa VYsind Ig xX 109 perunit WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 69 and thus I Ij 2 134 per unit S VI 134 perunit Hence Q VS P 132 perunit Problem 530 VooEat v2 1 1 P fo 2 X sind 5 sm 6 The generator will remain synchronized as long as Pnax P An iterative search with MATLAB can easily be used to find the minimum excitation that satisfies this condition for any particular loading part a For P 05 must have E 0327 per unit part b For P 10 must have Fas 0827 per unit Problem 532 part a vt af i Gf Ke A LO J Xu De Z V a a la part b We know that P 095 per unit and that VooVi P sin Ot and that I Vi Vo JX It is necessary to solve these two equations simultaneously for VY V 264 so that both the required power is achieved as well as the specified power factor 70 angle with respect to the generator terminal voltage This is most easily done iteratively with MATLAB Once this is done it is straightforward to calculate Vt 102 perunit Eaf 205 perunit δ 466 Problem 533 part a Define XD Xd Xbus and XQ Xq Xbus i Eafmin Vbus XD 004 perunit Eafmax Vbus XD 196 perunit ii part b WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 71 part c 1 y og 08 07 3 06 E 06 04 o3 02 01 5 02 04 06 08 1 12 14 16 18 2 Perunit excitation Problem 534 nxpoles 3000x6 P T99 qa9 10 He Problem 535 part a Because the load is resistive we know that 12 2 L354 3V 3192 part b We know that E 2083 120 V Solving Eur VV2 Xsla for X gives JE V2 X Via Va 3410 La part c The easiest way to solve this is to use MATLAB to iterate to find the required load resistance If this is done the solution is V 108 V linetoneutral 187 V linetoline Problem 536 i Ea wks Rat Rpt jwl RaRytjwLa Thus Z wks Ka VOB FROS eLaye 7 4 Bact Clearly will remain constant with speed as long as the speed is sufficient to insure that w Ra RpLa 72 PROBLEM SOLUTIONS Chapter 6 Problem 61 part a Synchronous speed is 1800 rmin Therefore s 1800 1755 1800 0025 25 part b Rotor currents are at slip frequency fr s60 15 Hz part c The stator flux wave rotates at synchronous speed with respect to the stator 1800 rmin It rotates at slip speed ahead of the rotor s1800 45 rmin part d The rotor flux wave is synchronous with that of the stator Thus it rotatesat synchronous speed with respect to the stator 1800 rmin It rotates at slip speed ahead of the rotor s1800 45 rmin Problem 62 part a The slip is equal to s 08950 00178 The synchronous speed for a 6pole 50Hz motor is 1000 rmin Thus the rotor speed is n 1 s1000 982 rmin part b The slip of a 4pole 60Hz motor operating at 1740 rmin is s 1800 1740 1800 00333 333 The rotor currents will therefore be at slip frequency fr 60 00333 2 Hz Problem 63 part a The synchronous speed is clearly 1200 rmin Therefore the motor has 6 poles part b The fullload slip is s 1200 1112 1200 00733 733 part c The rotor currents will be at slip frequency fr 60 00733 44 Hz part d The rotor field rotates at synchronous speed Thus it rotates at 1200 rmin with respect to the stator and 12001112 88 rmin with respect to the rotor Problem 64 part a The wavelenth of the fundamental flux wave is equal to the span of two poles or λ 4512 0375 m The period of the applied excitation is T 175 1333 msec Thus the synchronous speed is WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 73 r Us 281 msec 1013 kmhr part b Because this is an induction machine the car in this case will never reach synchronous speed part c 1013 95 00622 622 s 013 006 622 The induced track currents will be a slip frequency f s75 466 Hz part d For a slip of 622 and a car velocity of 75 kmhr the synchronous velocity must be 75 Us Ts 800 kmhr Thus the electrical frequency must be 79 80 592 H f sins and the track currents will be at a frequency of sf 368 Hz Problem 65 part a For operation at constant slip frequency f the applied electrical frequency f is related to the motor speed in rmin n as poles frn om h and thus since the slip frequency jf remains constant we see that the applied electrical frequency will vary linearly with the desired speed Neglecting the voltage drop across the armature leakage inductance and winding resistance the magnitude of the armature voltage is proportional to the airgap flux density and the frequency Hence the magnitude of the armature voltage must vary linearly with electrical frequency and hence the desired speed part b The electrical frequency of the rotor currents is equal to the slip frequency and hence will remain constant Since the rotor will be operating in a constant flux which varies at a constant frequency the magnitude of the rotor currents will be unchanged part c Because the rotor airgap flux density and the rotor currents are unchanged the torque will remain constant Problem 66 part a Since the torque is proportional to the square of the voltage the torquespeed characteristic will simply be reduced by a factor of 4 part b Neglecting the effects of stator resistance and leakage reactance having both the voltage and frequency maintains constant airgap flux Hence 74 the torquespeed characteristic looks the same as the original but the synchro nous speed is halved Problem 67 part a Four poles part b Counterclockwise part c 1800 rmin part d Four poles part e No There will be dc flux linking the inductionmotor rotor wind ings but there will be no resultant voltage at the slip rings Problem 68 part a 1500 rmin part b The induction motor rotor is rotating at 1500 rmin in the clock wise direction Its stator flux wave is rotating at 3000 2poles 1000 rmin in the counterclockwise direction Thus the rotor sees a flux wave rotating at 2500 rmin Noting that a flux wave rotating at 1000 rmin would produce 50Hz voltages at the slip rings we see that in this case the rotor frequency will be fr 50 25001000 125 Hz part c Now the stator flux wave will rotate at 1000 rmin in the clockwise direction and the rotor will see a flux wave rotating at 500 rmin The induced voltage will therefore be at a frequency of 25 Hz Problem 69 part a R1 will decrease by a factor of 104 to 0212 Ω part b Xm will increase by a factor of 185 to 538 Ω part c R2 will decrease by a factor of 3558 to 0125 Ω part d All values will decrease by a factor of 3 Problem 610 This problem can be solved by direct substitution into the equations in chapter 6 which can inturn be easily implemented in MATLAB The following table of results was obtained from a MATLAB script which implemented the equivalentcircuit equations assuming the coreloss resistance Rc is in parallel with the magnetizing reactance Rc was calculated as Rc 4602 220 962 Ω slip 10 20 30 speed rmin 1782 1764 1746 Tout Nm 85 165 234 Pout kW 85 165 234 Pin kW 458 896 128 power factor 081 087 085 efficiency 933 944 938 Problem 611 part a 1741 rmin WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 75 part b 1799 rmin part c part d Problem 612 For a speed of 1725 rmin Pout 426 kW Pin 457 kW pf 0751 lagging and η 933 Problem 613 It is necessary to find find the value of R2 This can be easily done by writing a MATLAB script to iteratively find that value of R2 which fullload internal torque at a slip of 35 The result is R2 00953 Ω Once this is done the same MATLAB script can be used to sustitute the machine parameters into the equations of chapter 6 to find Tmax 177 Nm at a slip of 182 and Tstart 716 Nm Problem 614 This problem is readily solved once the value of R2 has been found as dis cussed in the solution to Problem 613 The impedance of the feeder must be added in series with the armature resistance R1 and leakage reactance X1 A MATLAB script can then be written to find the desired operating point The WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 76 result is that the motor achieves rated output at a slip of 367 and a terminal voltage of 2216 V linetoline Problem 615 part a For R 0 Riq 0 and thus from Eq 634 R ae Xieq X2 SmaxT From Eq 636 Trax 05rpnViPeq wWsX1 eq Xp and from Eq633 with s 1 NphVieg he Tstart 3 TT oor wsR3 X1eq X2 Noting that Tmax 220 163 Tetart 135 we can take the ratio of the above equations R 2 Tmax 163 R3 Xieq X2 x 1 Totart Ro X1eq X2 oe From Eq634 with Regi 0 Smaxt RoX1q X2 Hence 2 1 0 5SiraxT 163 SmaxT which can be solved to give Smaxr 0343 343 part b From Eq 633 with Regi 0 and with s Sratea T MphVieqR2Srated eee WsR2Sratea Xt1eq X2 and thus Tmax 91 05R2Sratea Xeq1 X2 Trated R2SratedXeq1 Xp 051 SmaxT Srated SmaxTSrated This can be solved to give WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 77 Srated 02405maxr 00824 824 part c I Vive Vive start DO Ro 9Xeq1 X2 Xeq1 X2Smaxt J Ty 4 Vieq Vieq rate R2Srated jXeq1 Xp Xeq1 X2SmaxTSrated j Thus I star max rate 41 ostart SmaxrSrated J 416 FIT 4 95 gos I2ratea SmaxT JI 343 J Problem 616 Given Tinax 2379 SmaxT 055 and sq 0087 start by taking the ratio of Eqs 636 and 633 Tmax 05Rijeq Rosa Xijeq X2 Th Rieg Re oq Xi0q X2 Rasn Substituting Eq 634 gives 2 2 2 Rieq 1 1 rang 09 8 2 4 Gs TT fuga yl Th Rs SmaxT Substituting given values and solving gives Req 1 1315 Ro From Eq 633 we can write 2 2 Req 1 XeqitX2 Tstart 5 Ro Re ert sg A tr Ta Req 1 7 XeqatX2 4eoG e From Eq 634 Xiea Xa 1 Bica Roz 7 SmaxT Re and thus we can solve for 78 Tstart 126Tfl Problem 617 part a Using MATLAB to solve the equivalentcircuit equations from the equivalentcircuit at a slip of 335 the power applied to the shaft can be calculated to be 5032 kW Thus the rotational loss is 32 kW Similarly the input power to the equivalent circuit is 5280 kW Based upon an efficiency of 94 the actual motor input power is 500 kW094 5319 kW Thus the core losses are equal to 5319 5280 39 kW part b The equivalent circuit is solved in the normal fashion For ease of calculation the core loss can be accounted for by a resistor connected at the equivalentcircuit terminals based upon the results of part a this corresponds to a resistance of 147 kΩ The shaft input power is equal to the negative of the shaft power calculated from the equivalent circuit plus the rotational loss power The electrical output power is equal to the negative of the input power to the equivalent circuit The result is using MATLAB i Generator output power 512 kW ii efficiency 916 iii power factor 089 part c Basically the same calculation as part b The impedance of the feeder must be added to armature impedance of the induction motor The result is using MATLAB i Power supplied to the bus 498 kW ii Generator output power 508 kW Problem 618 Problem 619 part a Given I2 2maxTR2 90I2 2flR2 Thus I2maxT 30I2fl Ignoring R1 R1eq 0 and we can write WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 79 Voq L Ros JXeq X2 and thus Ton jXeq X2 RoSmaxT To max jXeq X2 Rosa Substitution from Eq 634 R X1eq X2 SmaxT gives Ton jt To max J SmaxT SA and thus Ioq Ij 1 v2 12 maxT j SmaxT Sai 1 Smaxt Sa Finally we can solve for SmaxT Smaxt 412sq 00948 948 part b Taking the ratio of Eqs 636 and 633 with Riq 0 and substi tution of Eq 634 gives Tmax 05Ro sa X1eq X2 051 Smaxtf 218 Ts X1eq X2R2sa SmaxT SA In other words Tinax 218 per unit part c In a similar fashion one can show that Tstart 1 SmaxtSa ostart ss 041 Ta A 1 So axT 0 In other words Titart 041 per unit Problem 620 part a T x I3R2s Thus Tstart In start 132 Ta A Io 4 3 and thus Tytart 132 per unit part b As in the solution to Problem 615 neglecting the effects of Ry WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 80 Z start SmaxtT SA j 12 rated SmaxT j This can be solved for smaxt 1 Ip start Iq maxT 4 a 0224 224 r A sq lostart Ia 1 0 7 Again from the solution to Problem 615 2 Tmax 051 SmaxtSa 319 Trated SmaxT SA and thus Tmax 312 per unit Problem 621 part a Solving the equations of chapter 6 with s 1 for starting with MATLAB yields I start 233 A T start 791 Nm part b i When the motor is connected in Y the equivalentcircuit para meters will be three times those of the normal A connection Thus Ry 01385 Q Ry 0162 Q X 087 9 X2 0840 Xm 288 0 ii Tstart 776 A T start 263 Nm Problem 622 part a Prot Py 312 R1 2672 W part b The parameters are calculated following exactly the procedure found in Example 65 The results are Ry 1110 X 390 2 Ro 1340 X2 390 2 Xm 168 2 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 81 part c Solving the equations of chapter 6 using the equivalentcircuit parameters of part b I 291 A P 106 kW power factor 091 lagging Pout 100 kW efficiency 945 Problem 623 Because this is a blockedrotor test one can ignore the magnetizing reactance Xm As a result the motor input impedance can be approximated as Zin y Ry Ry iX Xp Ry can be calculated from the blockedrotor power and current Bpi Ro R 3B which gives Motor 1 Rp 0174 2 Motor 2 Ro 0626 2 The motor starting torque is proportional to JR2 and thus the torque ratio is given by Tmotor2 13 motor2 R2motor2 nai 13 motor2 Tmotor1 13 motor1 R2motort R2motor1 13 motor1 Thus for the same currents the torque will be simply proportional to the resistance ratio and hence Tmotor2 0278 Tmotor1 From the given data we see that for the same voltage the current ratio will be 12 motor2L2motor1 994747 139 and hence Tmotor2 0492 Tmotor1 Problem 624 Rotational loss 3120 W RR 0318Q Re 0605 2 Xy2280 X23420Q0 Xp 6340 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 82 Problem 625 Neglecting R and hence Reg1 gives from Eq 635 Ro SmaxT X1 eq Xo and from Eq 636 T 05npnVPeq 051ph VpeqgSmaxT max Ws X1eq Xp WsRe If the frequency is reduced from 60 to 50 Hz X1q X2 will drop by a factor of 56 and hence smaxr will increase by a factor of 65 to Smaxr 18 corresponding to a speed of 10001 018 820 rmin Tmax Will increase as Tinax 190230 Tinaxs0 19023065 5 ggg Tmax 60 56 or Tmax 50 283 Problem 626 Smaxt Rg Therefore 11 Ry WHiomAM 207 2 SmaxT Rexi11SmarT Rex0 1 Problem 627 part a From the solution to Problem 615 Tmax 051 Smaxtsa Ta SmaxT SA Given that TraxTa 225 and Smaxr 016 this can be solved for sg 00375 375 part b The rotor rotor power dissipation at rated load is given by SA Protor Prated 29kW 1 sq part c From the solution to Problem 619 Tstart 1 SmaxtSa 070 Ta A 1 SosaxT Rated torque is equal to 75 kWwm4 where wmq 6071sg 1814 radsec Thus Trateq 413 Nm and Tetart 070 per unit 290 Nm part d If the rotor resistance is doubled the motor impedance will be the same if the slip is also doubled Thus the slip will be equal to s 2sg 750 part e The torque will equal to fullload torque Thus T 413 Nm WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 83 Problem 628 250 150 E 100 50 95 200 400 600 800 1000 1200 1400 1600 1 sho Speed rpm Problem 629 part a Protor Prated A 363 kW 1 sq part b From the solution to Problem 615 Tmax 051 SmaxtSa Ta SmaxT SA Given that TinaxTq 310 and sg 1200 11581200 0035 this can be solved for Smaxt 0211 211 This corresponds to a speed of 12001 0211 947 rmin part c Sufficient resistance must be inserted to increase 5yax7 from 0211 to 10 Thus Ro tot 017211 0806 2 and hence the added resistance must be Rext 0806 0211 0635 2 part d The applied voltage must be reduced by a factor of 56 to 383 V linetoline part e From Eq 635 Smaxt ReX1q X2 If the frequency de creases by a factor of 56 the reactances will also decrease by a factor of 56 and hence Sm axT Will increase by a factor of 65 to 0042 Hence the corresponding speed will be 10001 0042 958 rmin Problem 630 If the impedance of the motor at starting is made equal to that of the motor at a slip of 56 the starting current will be equal to 200 of its rated value This can be done by increasing the rotor resistance for 902 45 mQ to 0045 Rotor 804 mQ ator 0056 and hence the requierd added resistance will be Rex 804 45 759 mf The starting torque under this condition will be 190 of the fullload torque WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 84 Problem 631 The synchronous speed of this motor is 36008 450 rmin 4712 radsec Thus its fullload speed is 4501 0041 4316 rmin The corresponding torque will be 250 x 1047121 0041 553 x 10 Nm At a speed of 400 rmin the torque will be 553 x 104004316 475 x 10 Nm With no external resistance the slope of the torquespeed characteristic is thus 553 x 1034316 1281 The slope of the desired torquespeed charac teristic is 475 x 10400 1188 Since the initial slope of the torquespeed characteristic is inversely proportional to the rotor resistance the total rotor resistance must be 1281 Root a 5 264m Therefore the required added resistance is 264 245 19 mQ 85 PROBLEM SOLUTIONS Chapter 7 Problem 71 part a ωm Va part b ωm 1 If part c ωm will be constant Problem 72 part a For constant terminal voltage the product nIf where n is the motor speed is constant Hence since If 1Rf Rf 1180 Rf 5 1250 and hence Rf 842 Ω part b 1380 rmin Problem 73 Check this part a ωm halved Ia constant part b ωm halved Ia doubled part c ωm halved Ia halved part d ωm constant Ia doubled part e ωm halved Ia reduced by a factor of 4 Problem 74 part a Rated armature current 25 kW250V 100 A part b At 1200 rmin Ea can be determined directly from the magneti zation curve of Fig 727 The armature voltage can be calculated as Va Ea IaRa and the power output as Pout VaIa With Ia 100 A If A Ea V Va V Pout kW 10 150 164 164 20 240 254 254 25 270 284 284 part c The solution proceeds as in part b but with the generated voltage equal to 9001200 075 times that of part b If A Ea V Va V Pout kW 10 112 126 126 20 180 194 194 25 202 216 216 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 86 Problem 75 part a part b i ii WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 87 Problem 76 part a Va Es Pshatt Es 2 and thus V2 E Va V2 4 Psnatt Ra 2 The motor speed n can then be found from FE 12 i n 1200 a x mn rmin Here is the desired plot produced by MATLAB 1200 1190 1180 1170 1160 1150 1140 1130 1205 3 10 is 20 25 Power kVV part b The solution for EF proceeds as in part a With the speed constant at 1200 rmin and hence constant w solve for If as Ea I Ke where K 150D VA Here is the desired plot produced by MATLAB 17 Pp 16 oO 3 10 is 20 25 Power kVV WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 88 Problem 77 The solution is similar to that of Problem 76 with the exception that the assumed straightline magnetization characteristic is replaced by the nonlinear characteristic of Fig 727 MATLAB with the spline function used to represent the nonlinear characteristic of Fig 727 then produces the following plots part a 1210 1200 1190 1180 1170 5 1160 1150 1140 1130 120 5 10 15 20 5 Power kVV part b 225 22 mas 21 205 2 1 o 5 10 15 20 5 Power kVV Problem 78 part a From the load data the generated voltage is equal to 254 627 x 0175 265 A From the magnetizing curve using the spline function of MATLAB the corresponding field current is 154 A Hence the demagnetizing effect of this armature current is equal to 195 154500 204 Aturnspole part b At the desired operating point the generator output power will be 250 V x 615 A 154 kW Therefore the motor speed will be 154 n 1195 55 1139 rmin Because the machine terminal voltage at no load must equal 230 V from the magnetizing curve we see that the shunt field under this operating condition WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 89 must equal 105 A and hence the shunt field resistance must be 219 2 Hence under this loading condition with a terminal voltage of 250 V the armature voltage will be 250 615 x 0065 2508 V the shunt field current will equal 2508219 115 A and thus the armature current will equal 615 115 627 A The generated voltage can now be calculated to be 2508 6270175 286 V The corresponding voltage on the 1195 rmin mag curve will be E 28611951139 285 V and hence the required net field ampereturns is using the MATLAB spline function 1042 Aturns The shuntfield ampereturns is 115 x 500 575 Aturns the demagnetizing armature ampturns are 204 A turns and hence the required series turns are 1042 204 Ns 1042 575 204 106 11 turns 615 Problem 79 From the given data the generated voltage at J 90A and n90 975 rmin is E90 V IRa Rs 230 90011 008 2129 V Similarly the generated voltage at I 30 A is E30 230 30011 008 2243 V Since Ey x n F30 n30 30 E90 n90 90 Making use of the fact that 3090 048 we can solve for n30 E30 90 30 n90 2140 n00 90 Fa ea tmin Problem 710 1500 1450 1400 1350 Series field adds 1300 o 1250 ety Series field opposes 1 1005 10 20 30 40 50 60 70 80 90 100 Power kVV 90 Problem 711 part a For constant field current and hence constant field flux constant torque corresponds to constant armature current Thus for speeds up to 1200 rmin the armature current will remain constant For speeds above 1200 rmin ignoring the voltage drop across the armature resistance the motor speed will be inversely proportional to the field current and hence the field flux Thus the armature current will increase linearly with speed from its value at 1200 rmin Note that as a practical matter the armature current should be limited to its rated value but that limitation is not considered in the plot below part b In this case the torque will remain constant as the speed is in creased to 1200 rmin However as the field flux drops to increase the speed above 1200 rmin it is not possible to increase the armature current as the field flux is reduced to increase the speed above 1200 rmin and hence the torque track the field flux and will decrease in inverse proportion to the change in speed above 1200 rmin Problem 712 part a With constant terminal voltage and speed variation obtained by field current control the field current and hence the field flux will be inversely proportional to the speed Constant power operation motor A will then require WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 91 constant armature current Constant torque motor B will require that the armature current variation be proportional to the motor speed Thus motor A J 125 A motor B I 1255001800 347 A part b motor A J 125 A motor B I 1251800125 450 A part c Under armature voltage control and with constant field current the speed will be proportional to the armature voltage The generated voltage will be proportional to the speed Constantpower operation motor A will re quire aramture current that increases inversely with speed while constant torque operation motor B will require constant armature current For the conditions of part a motor A I 1251800125 450 A motor B J 125 A For the conditions of part b motor A I 1255001800 347 A motor B J 125 A Problem 713 Fa VaTaRa wm K0a Kaba 1 KPa Thus 1 v y TR Kaq Kaa The desired result can be obtained by taking the derivative of wy with g dim 1 2T Ra V dq K4 KPa 1 TR 21a a Va R82 21Ra Va 1 sa VV 2E K From this we see that for FE 05V dwada 0 and for E 05Va dwda 0 QED WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 92 Problem 714 part a Synchronous machine 1 3 Tajac 30 x 10 x 10 377 A V3 460 Eat Vaac i Xslaac 460V3 7513 x 377 3284 Vjln DC machine P Eglaac 30 kW Es Vade TaacRa Thus Vade 4 V2 40 4PRa E 1 226 V 2 part b Increase the dcmotor field excitation until Ey Vaac 230 V in which case the dc motor input current will equal zero and it will produce no shaft power The ac machine will operate at a power angle of zero and hence its terminal current will be Fg Va ac Taac 122 A Xz part c If one further increases the dcmachine field excitation the dc machine will act as a generator In this case defining the dc generator current as positive out of the machine P Eglaac 30 kW Es Vade TaacRa Thus Vade Vi2a 4PRa E 1 226 V 2 and Ey Va Cc Tnde 128A WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 93 The ac machine will now be operating as a motor The armature current will be the negative of that of part a and hence the power factor will be unity Its magnitude will be 377 A Problem 715 First find the demagnetizing mmf At rated load E Va IaRtot 600 250 x 0125 5688 V Using the MATLAB spline function the corresponding field current on the 400 rmin magnetizing curve is Tp 232 A Thus the demagnetizing mmf at a current of 250 A is equal to 250 232 18 A and in general the effective seriesfield current will be equal to I Ig og I 18 ref 8 For a starting current of 460 A the effective series field current will thus equal 399 A Using the MATLAB spline function this corresponds to a generated voltage of 474 V from the 400 rmin magnetization curve The corresponding torque which will be the same as the starting torque due to the same flux and armature current can then be calculated as Fal 474 x 560 T 5200N wm 400730 Problem 716 At no load Fant 230 635 x 011 2293 V At full load Fag 230 115 011 2174 V But E n thus Ey fl On 2174 1 n 21 21 ma Mal 0 5 a 68 rmin Problem 717 The motor power is given by P F where Ea Kagum and where from Eq 73 polesC 4 x 666 Ka 212 27m 27 x 2 Thus for 6g 001 FE Kaquy 212w yp Vi Es Ra WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 94 Using MATLAB and its spline function to represent the fan character istics an iterative routine can be written to solve for the operating point the intersection of the motor and fan characteristics The result is that the motor will operate at a speed of 999 rmin and an output power of 839 kW Problem 718 part a Assuming negligible voltage drop across the armature resistance at no load the field current can be found from the 1300 rmin magnetization curve by setting E 230 V This can be most easily done using the MATLAB spline function The result is J 167 A This corresponds to Nel 2500 aturns of mmf part b At rated load FE Va 1IR 230465 x 017 2221 V From the noload 1300 rmin magnetization curve the corresponding field current is 150 A again obtained using the MATLAB spline function Thus the effective armature reaction is Armature reaction 16715 A x 1500 turnspole 251 Aturnspole part c With the series field winding Riot Ra Rs 0208 2 Thus under this condition E VIR 2203 This corresponds to a 1300 rmin generated voltage of 2367 V and a corresponding field current determined from the magnetization curve using the MATLAB spline function of 184 A corresponding to a total of 2755 Aturns Thus the required series field turns will be 2755 N 755 2500 251 108 465 or rounding to the nearest integer N 11 turnspole part d Now the effective field current will be 2500 251 20 x 465 Log Jo 212A 1500 From the 1300 rmin magnetization curve E 2461 V while the actual E Va Rtotla 2203 V Hence the new speed is 2203 n 1300 5 1164 rmin Problem 719 part a At full load 1185 rmin with a field current of 0554 A Ea Va Ia Rtot 2214 V where Riot 021 0035 0245 9 95 An 1825 rmin magnetization curve can be obtained by multiplying 230 V by the ratio of 1185 rmin divided by the given speed for each of the points in the data table A MATLAB spline fit can then be used to determine that this generated voltage corresponds to a fieldcurrent of 0527 A Thus the armature reaction is 0554 05272000 534 Aturnspole part b The fullload torque is T EaIa ωm 2214 342 1185π30 628 N m part c The maximum field current is 230310 0742 Ω The effective field current under this condition will therefore be Ieff 2000 0742 160 2000 0662 A From the 1185 rmin magnetization curve found in part b this corresponds to a generated voltage of 245 V Thus the corresponding torque will be T EaIa ωm 245 65 1185π30 128 N m part d With the addition of 005 Ω the total resistance in the armature circuit will now be Rtot 0295 Ω The required generated voltage will thus be Ea Va IaRtot 2196 V This corresponds to 219611851050 2478 V on the 1185 rmin magnetiza tion curve and a corresponding effective field current of 0701 A As can be seen from the data table a noload speed of 1200 rmin corre sponds to a field current of 0554 A Thus the seriesfield Aturns must make up for the difference between that required and the actual field current as reduced by armature reaction Ns NfIfeff If Armature reaction Ia 20000701 0554 534 352 98 turns Problem 720 part a From the demagnetization curve we see that the shunt field current is 055 A since the noload generated voltage must equal 230 V The fullload generated voltage is Ea Va IaRa 2194 V and the corresponding field current from the demagnetization curve obtained using the MATLAB spline function is 0487 A Thus the demagnetization is equal to 2000055 0487 127 Aturns WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 96 part b The total effective armature resistance is now Rio 0150028 0178 Q Thus the fullload generated voltage will be Ea Va Ia Rtot 2174 V The net effective field current is now equal to 048770842000 0628 A The corresponding voltage at 1750 rmin found from the magnetization curve using the MATLAB spline function is 2427 V and hence the fullload speed is 2174 n 1750 Gad 1567 rmin part c The effective field current under this condition will be Teg 055 12542000 2302000 0685 A From the 1750 rmin magnetization curve using the MATLAB spline func tion this corresponds to a generated voltage of 249 V Thus the corresponding torque will be Fala 249 x 125 T 170N wm 1750730 Problem 721 part a For a constant torque load changing the armature resistance will not change the armature current and hence J 60 A partb FE Va Rala Thus without the added 100 resistor EF 216 V and with it E 156 V Thus 1 Speed ratio me 072 Problem 722 parts a and b 1000 rn 800 600 met 400 au 2 Re03 0 x 6 oan Pa swe Fenos 5 400 x 600 hoo 1000 0 1000 2000 3000 4000 Torque Nm WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 97 Operation in the fourth quadrant means that the motor torque is positive while the speed is negative In this case the motor is acting as a generator and absorbing energy from the lowering load which would otherwise accelerate due to the effects of gravity part c 473 rmin Problem 723 part a At rated load E 230 122 times0064 222 V Thus rated load speed is 222 n 1150 1133 rmin part b The maximum value of the starting resistance will be required at starting 230 2 x 122 244 Rat Rmax and thus Rmax 0878 2 part c For each value of Riot Ra Rext the armature current will reach its rated value when the motor reaches a speed such that E 230 122 Rtoto1a At this point Ryo will be reduced such that the armature current again reaches 122 A Based upon this alogrithm the external resistance can be controlled as shown in the following table Step number RextQ Eamin V Mmin rmin V Eamax V max rmin 1 0878 0 0 115 587 2 0407 115 587 173 882 3 0170 173 882 202 1030 4 0051 202 1030 216 1101 5 0 216 1101 Problem 724 part a At no load Zant Kmwmn1 Va Thus Va 85 Wmnl K 021 405 radsec Hence the fullload speed is Wm1307 3865 rmin part b At zero speed the current will be J VaRa 447 A and the corresponding torque will be T KyJ 94 Nm part c KinVa Ea KinVa Kmnwm T kK 1 Ra Ra 98 Here is the desired plot obtained using MATLAB Problem 725 part a At no load ωmnl 11 210π30 1174 radsec and Eanl Va IanlRa 494 V Thus Km Eanl ωmnl 421 103 Vradsec part b The no load rotational losses are Protnl EanlIanl 62 mW part c At zero speed the current will be Ia VaRa 109 A and the corresponding torque will be T KmIa 46 mNm part d The output power versus speed characteristic is parabolic as shown below An iterative MATLAB scripts can easily find the two desired operating points 2761 rmin for which the efficiency is 243 and 8473 rmin for which the efficiency is 728 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 99 Problem 726 No numerical solution required for this problem Problem 727 Based upon the calculations of Problem 725 at 8750 rmin the rotational losses will be 294 mW Thus the total required electromechanical power will be P 779 mW The generated voltage will be Ea Kmωm 386 V and the armature current will thus be Ia PEa 0202 A Thus the desired armature voltage will be Va Ea RaIa 479 V WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 100 PROBLEM SOLUTIONS Chapter 8 Problem 81 part a In this case β 45 π4 rad and Lmax N 2µ0βRD 2g 96 mH and there is a 15 overlap region of constant inductance part b Tmax1 LmaxI2 1 2β 61 102I2 1 N m Tmax1 LmaxI2 2 2β 61 102I2 2 N m WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 101 part c i1 i2 5 A i ω 0 Tnet 0 ii ω 45 Tnet 0 iii ω 75 Tnet 153 Nm Problem 82 When a single phase is excited magnetic circuit analysis can show that all the mmf drop occurs across the two air gaps associated with that phase Thus there is no additional mmf available to drive flux through the second phase Problem 83 Same argument as in the solution of Problem 83 Problem 84 part a and b Lmax DRαµ0N 2 2g 215 mH WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 102 Le 2 prow 4 i pose p hue Max y y f e 7 N L oY XN os 126 2 ve 79 60 Jo o 30 60 972 Ie 1 Bo eo part c 2gB Ip 696 A oN part d I0 Lmax Tmax 1 N 9 2 99 m T Tmax 12 150 bo 30 90 feo 120 30 60 160 BO 5 103 part e phase 1 ON 120 θ 90 30 θ 0 60 θ 90 150 θ 180 phase 2 ON 150 θ 120 60 θ 30 30 θ 60 120 θ 150 phase 3 ON 180 θ 150 90 θ 60 0 θ 30 90 θ 120 part f The rotor will rotate 90 in 30 msec n 14 r 35 msec 714 rsec 429 rmin The rotor will rotate in the clockwise direction if the phase sequence is 1 2 3 1 Problem 85 When the rotor is aligned with any given pole pair it is clearly medway between the other two pole pairs Thus rotation in one direction will increase the inductance of one set of poles and decrease the inductance of the remaining set Thus depending on which of the remaining poles is excited it is possible to get torque in either direction Problem 86 The rotor will rotate 15 as each consecutive phase is excited Thus the rotor will rotate 1 revolution in 24 sequences of phase excitation or 8 complete sets of phase excitation Thus the rotor will rotate 1 revolution in 8 15 120 msec Thus it will rotate at 1012 833 rsec 500 rmin Problem 87 part a If phase 1 is shut off and phase 2 is turned on the rotor will move to the left by 2β3 429 Similarly turning off phase 2 and turning on phase 3 will cause the rotor to move yet another 429 Thus starting with phase 1 on to move 214429 5 steps the sequence will be 1 ON 1 OFF 2 ON 2 OFF 3 ON 3 OFF 1 ON 1 OFF 2 ON 2 OFF 3 ON part b Clockwise is equivalent to rotor rotation to the right The required phase sequence will be 1 3 2 1 3 2 The rotor will rotate 429step and hence the rotor speed will be WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 104 125 rmin 360 r 1 step 429 105 104 stepsmin Thus the required step time is time step 1 min 8400 step 60 sec min 572 msecstep Problem 88 part a When phase 1 is energized the rotor will be aligned as shown in the following figure From the figure we see that if phase 1 is turned off and phase 2 is energized the rotor will rotate 461 degrees to the right clockwise to align with the phase2 pole Similarly if phase 3 is excited after phase 1 is turned off the rotor will rotate 461 degrees to the left counterclockwise part b 80 rmin 360 r 1 step 461 625 103 stepsmin time step 1 min 625 103 step 60 sec min 96 msecstep The required phase sequence will thus be 1 3 2 1 3 2 1 Problem 89 part a For time in which the current is building up i1t 100t 0005 575t A WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 105 This expression is valid until time t 222 msec at which point i1t 167 A part b i1t 022 200t 222 103 005 575444 103 7 This expression is valid until time t 332 msec part c Here are the desired plots The integral under the torque curve is 238104 Nmsec while the positive torque integral is 329 Nmsec Thus there is a 257 reduction in torque due to negative torque production during the currentdecay period Problem 810 part a For time in which the current is building up i1t 100t 0005 575t A This expression is valid until time t 222 msec at which point i1t 167 A part b i1t 022 250t 222 103 005 575444 103 7 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 106 This expression is valid until time t 310 msec part c Here are the desired plots The integral under the torque curve is 259104 Nmsec while the positive torque integral is 320 Nmsec Thus there is a 190 reduction in torque due to negative torque production during the currentdecay period Problem 811 part a The phase inductance looks like the plot of Problem 84 part a with the addition of the Lleak 45 mH leakage inductance Now Lmax 215 45 260 mH WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 107 Le Lmex Lose 20 I8 Il Ge co t0 0 0 60 fo Me 50 Wo part b For 30 6 0 dL 215 mH Wo a3tad 205 mHrad dO 1750r 2z7rad Imin dt min x r 60sec 83 radsec dL dL dt Wm do 3762 The governing equation is di dL vtRL i i Noting that dL dt R we can approximate this equation as dLi oe at and thus t dt i L0 Lt Substituting vt 75 V and Lt 45 x 10 376 then gives t it ee 45 x 1073 376 108 which is valid over the range 0 t 571 msec at which time it 165 A Here is the desired plot part c During this time starting at time t 571 msec vt 75 V and Lt 260 103 376t 571 103 Thus it 165 75 t 571 103 260 103 376t 571 103 which reaches zero at t 884 msec Here is the plot of the total current transient part d The torque is given by T i2 2 dL dθ Here is the plot WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 109 Problem 812 part a The plot of Lθ is shown below Here from Examples 81 and 83 Lleak 5 mH and Lmax 133 mH part b The solution for 675 θ 75 0 t 25 msec is exactly the same as part a of Example 83 it 100t 0005 512t A For 75 θ 75 25 msec t 313 msec dLdt 0 and thus v iR Ldi dt 100 15i 0133 di dt This equation has an exponential solution with time constant τ LR 887 msec i 667 686et0002500887 At t 313 msec it 139 A Following time t 313 msec the solution proceeds as in Example 83 Thus it 1468 100 313 103 0005 512t 563 103 The current reaches zero at t 425 msec Here is the corresponding plot produced by MATLAB WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 110 part c T i2 2 dL dθ Problem 813 part a WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 111 part b Inverter volt ampere rating Net output power 155 part c Pmech 2968 W Problem 814 part a Based upon the discussion in the text associated with Fig 818 the following table can be produced θm bit pattern i1 i2 0 1000 I0 0 45 1010 I0 I0 90 0010 0 I0 135 0110 I0 I0 180 0100 I0 0 225 0101 I0 I0 270 0001 0 I0 315 1001 I0 I0 part b There will be 8 pattern changes per revolution At 1200 rmin 20 rsec there must be 160 pattern changes per second corresponding to a time of 625 msec between pattern changes Problem 815 part a The rotor will rotate 2 counter clockwise part b The phase excitation will look like with T 417 msec WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 112 part c 8r 2880 48 2 min min sec 417 msec The frequency will be 1 i H f TP 6 Hz Problem 816 part a The displacement will be 3603 x 14 8571 part b There will be one revolution of the motor for every 14 cycles of the phase currents Hence 1 mi 14 fH 900 min cycles 210 Hz min 60 sec r 113 PROBLEM SOLUTIONS Chapter 9 Problem 91 part a ˆImain ˆV Zmain 138 564 A ˆIaux ˆV Zaux 99 492 A part b We want the angle of the auxiliarywinding current to lead that of the main winding by 90c π2 rad Thus defining Z aux Zaux jXC XC 1ωC we want Z aux tan1 ImZaux XC ReZaux Zmain π 2 Thus XC 145 Ω and C 183 µF part c ˆImain ˆV Zmain 138 564 A ˆIaux ˆV Zaux 126 336 A Problem 92 The solution is basically the same as for Problem 91 but now with Zmain 482 j604 Ω and Zaux 795 j768 Ω and ω 100π part a ˆImain 155 514 A ˆIaux 109 440 A part b C 227 µF part c ˆImain 155 514 A ˆIaux 118 386 A Problem 93 No numerical solution required WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 114 Problem 94 Imain 348 A power factor 059 Pout 132 W speed 1719 rmin Torque 0732 Nm efficiency 585 Problem 95 The solution follows that of Example 92 Imain 438 A power factor 065 Pout 204 W speed 1724 rmin Torque 113 Nm efficiency 630 Problem 96 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 115 Problem 97 part a From Eq 46 the peak amplitude in space and time of the mmf waves are given by 4 ky Non F eak L ea peaks Peaks Thus 4 42 Fnainpeak T 207 v2 391 A turns T and 4 Fauxpeak 111 v2 340 A turns T 4 part b The auxiliary winding current must be phase shifted by 90 from that of the main winding and the mmf amplitudes must be equal Hence Iaux should be increased to N Laux I i J 128 A main Neux 8 Problem 98 The internal torque is proportional to Rms Rp and thus is equal to zero when R Rp From Example 92 X main 1 Re a ss X22 8Q2 main 1sQ2main and X main 1 Rp a X22 2 8Q2main 12 Q2main We see that Rp Rp if 2 sQa main 18Q2 main OF 1 s11 Q2main and thus 1 yH 1 1 n ns1 s n Q2main where ns is the synchronous speed in rmin WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 116 Problem 99 The solution follows that of Example 93 part a I 3452 323 A Ig 712300 A part b Puech 6700 W Problem 910 part a Following the calculations of Example 93 with s 1 Tinech 148 Nm part b Setting Va 230 V Vg 230e79 V gives Tinech 164 Nm part c Letting Va Va and Vg jV gives Vo a Vir IVs Y Va jVe 2 2 Let Z Ry 5X1 Z Thus 1M VatiVa Ve VaJVp ZW OF Pyap Psapp ReI TP Re T gap gapb f by VaV Ws Ws Z B Clearly the same torque would be achived if the phase voltages were each equal in magnitude to VaVg Problem 911 The impedance Z must be added to the impedances of the motor of Problem 99 The solution then proceeds as in Example 93 The terminal voltage can be found as Vico Vo IZ Vie Ve 1pZ For Va 23520 Vg 212278 a MATLAB analysis gives View 2052 80 Vig 194273 which is clearly more balanced than the applied voltage WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 117 Problem 912 part a slip s 0035 Le Vi 5 Va jVa 21484 125 V Lea 5 YW gVa 9Vg 4762775 V Ze and Zp can be calculated from Eqs 94 and 95 with s 0035 Vi Ig 299 Z 640 Rit iki tZ Ys lh 448207 Ri GX Z Pyapt 217 Rp 784 W Paap 217 Ry 659 W Teap Per Pep 381Nm Ws part b Repeating the analysis of part a with s 1 gives Tytart 120nm part c Now we have a twophase machine operating under balanced two phase conditions We can now apply the analysis of Section 65 jXmRi jX1 Vieqg Vi 208 V ves a jX1 Xm and Riq jX16q j7Xm in parallel with R 7X1 0698 7502 2 Thus R SmaxT 348 Ri cg X1cq X and 05npnV Tmax i a Mh eq 203Nm s Req Ri eg X1cq X part d Now we have a singlephase machine operating with V 220 V and s 004 a Va Le 5782 528 A Ri jX105Z Zp 118 Pgap 05I2 αRf Rb 691 W Pmech Pgap1 s 664 W and finally Pout Pmech Prot 599 W part e ˆIf ˆIb 05ˆIα 282 523 ˆVf R1 jX1 ZfˆIf ˆVb R1 jX1 ZbˆIb and thus ˆVβ jVf Vb 05j ˆIαZf Zb 167 813 V In other words the opencircuit voltage across the β winding will be 167 V Problem 913 This problem can be solved using a MATLAB script similar to that written for Example 95 part a Tstart 028 Nm part b Imain 193 A Iaux 32 A part c I 213 A and the power factor is 099 lagging part d Pout 2205 W part e Pin 2551 W and η 864 Problem 914 This problem can be solved using a MATLAB script similar to that written for Example 95 An iterative search gives C 704 µF and an efficiency of 871 Problem 915 This problem can be solved using a MATLAB script similar to that written for Example 95 An iterative search shows that the minimum capacitance is 809 µF WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 119 Problem 916 part a parts b and c WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 120 PROBLEM SOLUTIONS Chapter 10 Problem 101 part a Up t Y Y T sy ie hh 4 part b 17 2 T4 AVot Vo rms t dt dt vi Vz f v2 1 Vz f z 867 V part c paiss V2R 9 mW Problem 102 part a The diode does not turn on until the source voltage reaches 06 V which occurs at time t 064VoT T60 oY mM 6 T Th 38y 2 v SK VY ho Whe part b 2 74 4x 84t Vine z dt 343 V T J r60 L part c paiss V2R 78 mW Problem 103 v2 AVot bee 4 te m VTS UT 3 e r Problem 104 35 30 25 3m Bis 10 a 95 65 6 45 4 35 3 25 2 InC Problem 105 part a Peak Vp 310 V part b Ripple voltage 257 V part c Timeaveraged power dissipated in the load resistor 177 W part d Timeaveraged power dissipated in the diode bridge 041 W Problem 106 If ust 0 diode D1 is ON diode D2 is off and the inductor current is governed by the following differential equation di Lo Ri v5t of ust 0 diode D1 is off and diode D2 is on and the inductor current is governed by the differential equaion di L di Ri 0 A simple integration implemented in MATLAB produces the following plot WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 122 6 ey 3 1 time msec Problem 107 7 yi 5 OF 20 40 60 80 100 120 time msec Problem 108 part a Letting T 27w Vac 7 Vo sin wt M q3y T Jo T part b Vae Tac 29A WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 123 part c 3 30 25 20 le 10 ey oO 5 10 ae 2 4 6 8 10 12 14 16 18 20 time g msec part d a ney 2 15 1 o5 OF 2 4 6 8 10 12 14 16 18 20 time g msec Problem 109 part a Letting T 27w 1 7 Vi Vac z Vo sinwt 2 1 coswtg T Jig 21 part b Vac Vo Iae 1 d R oR 1 coswtg WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 124 part c a ney S18 o5 OF 1 2 3 4 5 6 7 8 9 10 delay time msec Problem 1010 part a Letting T 27w 9 Tle Vac z Vosinwt Yo 1 coswtq T Jeg T part b Vac Vo Igg 1 d R GR 1 coswta part c 6 ey 1 PN delay time msec WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 125 part d 5 4 3 2 1 0 FT z Fa 89 2 4 6 8 10 12 14 16 18 2 time msec Problem 1011 part a i 1 08 06 04 2 02 3 04 06 085 1 2 3 4 5 6 7 alpha ii 1 54 Vac Vo sind do cosas T m4 Tv rV2 iii Volac Proa V ol c load dcd rV2 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 126 part b 3 08 06 04 02 8 oO 2 o2 04 06 08 19 1 2 3 4 5 6 7 alpha ii 1 ops Vi Vi Vac Vo sin dd cosag WT I3n4 T V2 iii Volac Proa v cl ct TE load deta Vd The power is negative hence energy is being extracted from the load Problem 1012 part a From Eq 1011 2Vo Ig 183 A de FR Quy and from Eq 108 Ci Ls t cos 1 mets 312 msec Ww Vo part b For L 0 2Vo Igc 236 A d 7wR Problem 1013 part a At 1650 rmin the generated voltage of the dc motor is equal to 1650 a 1 E 85 813 V WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 127 The motor input power will then be Py 1E IRa 665 W part b 2 2V2 x 115 Vac 7 COS Qq 2 cosaq 1035 cosag V T T Thus for Vac Fa InRa Ri 905 V ag 2919 Problem 1014 The rated current of this motor is Pratea 1000 Lrated 118A eo Vaated 85 The controller must limit Ig to twice pateq or 236 A Under this condition Vae InRa Ry 285 V From part b of the solution to Problem 1013 Vac 1035 cosag V and thus the controller must set wg 740 Problem 1015 The required dc voltage is Vp I Rp 277 V From Eq 1019 Vi Viciems 204 V rms 3V2 Problem 1016 The required dc voltage is Vp Rp 231 V From Eq 1020 1 1 VE Qq cos 390 Q a vi Problem 1017 part a The magnet resistance is sufficiently small that its voltage drop can be ignored while the magnet is being charged The desired charge rate is di ho 80 A25 sec 32 Asec Thus the required dc voltage will be di L 15 Va Ti 57 V WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 128 Thus Qa cos Vac cos SS 17 393 32 Viirms 3V2 x 15 part b Constant current simply requires a dc voltage of Vae Rlae 36 x 107 x 80 029 V Thus c 15 Qa cos Vac cos of 892 3V2 Vitems 3V2 x 0288 Problem 1018 part a 2 7 Qnt 8 perl Qnt Y z ut cos az f Vo cos dt Yo sin7D 515 V dn part b Harmonic number Peak amplitude V 1 515 2 0 3 66 4 0 5 127 6 0 7 28 8 0 9 57 10 0 Problem 1019 part a vy 2 wtcos 8 at 2 sin nD 3 v S G sin 3m Az 0 for D 13 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 129 part b Harmonic number Peak amplitude V 1 551 2 0 3 0007 4 0 5 110 6 0 7 79 8 0 9 0007 10 0 Problem 1020 part a Time period Sl 52 83 4 0wtaq DT2 ON OFF ON OFF ag DT2wt maqDT2 ON ON OFF OFF taqDT2wtataqDT2 OFF ON OFF ON tag DT2wt 2r7aqDT2 OFF OFF ON ON 27 aq DT2 wt 27 ON OFF ON OFF part b By analogy to the solution of Problem 1018 part a fo sina D An and by inspection aq part c Vat Vat pt ituzt cos y COS Qg Problem 1021 From Eq 1034 2D 1Vo it ave R 1765 A From Eq 1029 T1D 2L iL min 3 ase te 1745A R Ge and from Eq 1020 DT 2 ts 1 2 7 e it max 11784 A it 2 Ge Finally Ripple iL max iL min 039 A WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 130 PROBLEM SOLUTIONS Chapter 11 Problem 111 part a From the noload data Ean 120 Kp a 0953 Wmantent 1718730 x 07 Combining Ela T3 Wm and Va Ey InRa gives E 05 Ya v2 dun T Ra 05 120 1202 41800730 x 152 x 0145 1164V Thus E I Y 0648 A Wm Ke and defining Ip max 120104 114 A It D 0567 I max part b Ig 0782 A and D 0684 part c 07 068 066 064 062 06 058 o SF00 1550 1600 1650 1700 1750 1800 rpm WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 131 Problem 112 part a Ig 0645 A and D 0565 part b Jp 0784 A and D 0686 part c 07 068 066 064 062 06 058 a Soo 1550 1600 1650 1700 1750 1800 rpm Problem 113 part a From part a of Problem 111 kp 0953 For E 120 V and n 1300 rmin wm 13007 30 and thus E Ip 0925 A i and It D 0809 I max where Imax 114 A as found in Problem 111 part b I2 DI max 0686 A Ea Wm TK 1837 rmin and thus n 30wy7 part c igt Ipo Ip Ip2e7 0686 0239e77 where 7 LsRe 352 msec part d din Va Keitwm J Keit EB Keigt it rigt rigt R where it is as given in part c 132 Problem 114 part a Eanl Va RaInl 2397 V The rotational loss is given by Prot EanlIanl 374 W Based upon If VaRf 181 A ωmnl Eanl IfKf 3125 rmin and thus nnl 30ωmnlπ 2984 rmin part b part c WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 133 part d 500 450 400 350 See s 250 200 150 100 ao 5 10 15 Pout kVV Problem 115 part a First convert Ky to SI units V 1rmin Ky 218 x 10 2 8 x 10 rmin 730 radsec 08 mV radsec Teta 0094 ozcdotin 664 x 104 N m At stall Tastal TstanKm 0319 A Thus Va 3 Ra 941 0 I 0319 part b Wm Mn 35 1299 radsec Thus Fant Wmnthma 270 V Va Es n Tan an 315 mA and thus the noload rotational loss is Prot Fanitani 853 mA part c 11000 10000 9000 8000 7000 6000 000 4000 Suny 12 14 16 18 2 22 24 26 28 3 Va V WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 134 Problem 116 part a Wnt ni 3 3749 radsec Fant Va LaniRa 240 V Es nh Km 2 639 mVradsec Wmnl part b Prot Fanitam 113 W part c D I A rmin Proaa W 080 1335 3393 293 075 1270 3179 261 070 1205 2964 231 065 1140 2749 203 060 1070 2535 176 055 1005 2320 151 050 930 2107 127 Problem 117 The rotor acceleration is governed by the differential equation em 7 Kay dt Converted to kgm the moment of inertia is 452 x 10 kgm Thus to get to the final speed wm 12 x 104 730 1257 radsec Jim 45 x 1079 x 1257 Kale 208 103 x01 7 23 msec Problem 118 part a Lajrated Frated Prot 8 812 A Wmrated Km 30007 30 x 0465 Trated Kmtarated 378 Nm part b Thoaa KinIa Trot Here Trot 873000730 027 Nm Thus Tioagd 261 Nm and Poad Tioadm 729 W part c The differential equation governing the motor speed is dw mJ dn I dt Timech Trot Thoad WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 135 Here n is the motor speed in rmin Tinech KmJa 0465 x 70 326 Nm and from part b Tioad 729 02730 N load 2670 v nm m Thus the differential equation is dn 4 Ts 4094 n 1252 x 10 0 and the solution is n 3051 381e7 rmin where T 0255 sec Problem 119 From the solution to Problem 118 Ia rateq 812 A Neglecting rotational losses the motor speed can be calculated from the differential equation dw Timec Kyla rate J it h wrated and thus 4 nist Wm Wmo t J and thus the motor will reach zero speed at time Jwm0 186 x 1073 x 3000730 Kinlasated 0465 x 812 0155 sec Problem 1110 part a Rated speed 120fpoles 1800 rmin part b Prated Lrated 1380A V3 Vrated part c In per unit V 10 and P 10001100 0909 Thus J 0909 and Eat Va 9 Xslq 1552 497 per unit Thus Jp 155 x 85 131 A part d The inverter frequency will equal f 6013001800 433 Hz and the motor power input will be P 100013001800 443 W If one scales the base voltage and base power with frequency then the base impedance WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 136 will scale with frequency Hence under this new operating condition the per unit terminal voltage generated voltage and synchronous reactance will be un changed while the perunit power will be P 4431100 x 43360 0558 per unit Thus PX 6 sin 279 and Bap 15527 Va Pag I 06882268 7x 0688 2268 Thus the power factor is cos1268 089 leading part e Continuing with the base quantities of part d I 0558 per unit and Fag Va 7 Xela 1202 470 per unit and thus J 120 x 85 102 A Problem 1111 part a No numerical calculation required part b 1500 rmin V 383 kV11 Puax 833 kW Ig 131 A 2000 rmin V 460 kV 11 Pax 1000 kW Ig 126 A Problem 1112 Ls 523 mH Lap 631 mH Trated 531 Nm Problem 1113 part a Lat V2 Vise 634 iH V3 Whase AFNL L can be calculated from the perunit value of X WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 137 2 Zpase Vase 2120 Prase and Lpase ZbaseWbase 56 mH Thus Xs Xs 1970 Zpase Xs Ls a 523 mH Lpase 23 m part b Wmbase Wbase2poles 607 and Thase PoaseWmbase 531 Nm Thus T 05 Thase 265 Nm 2 2 T lIg 100A Q 00 2Q I 708 A rms V2 part c aseLia Ls Egg Pesealt 935 v V2 Because ip 0 I and Eng both lie along the quadrature axis Thus the terminal voltage magnitude will be given by Va Far 7 Xs1 274 V ln 474 V 11 Problem 1114 part a The various machine parameters were calculated in the solution to Problem 1113 T 075Thase 398 Nm and wy 14757 30 1545 radsec Thus P wT 615 kW part b 2 2 T lig 1451A Q 7 2Q I 1026 A rms V2 part c fe 6014751800 492 Hz WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 138 part d aseLia Ls Exp WbaseHaff 999 V V2 Because ip 0 I and Eng both lie along the quadrature axis Thus the terminal voltage magnitude will be given by Va For 7 Xsl 260 V ln 450 V 11 Problem 1115 part a The various machine parameters were calculated in the solution to Problem 1113 With Ty increased to 08Tbase 424 Nm 2 2 Trot 1603A Q 603 2Q I 1133 A rms V2 aseLia Ls Bag Pesesatt 9353V V2 Because ip 0 I and Eng both lie along the quadrature axis Thus the terminal voltage magnitude will be given by Va For 7 XsL 3245 V Ln 5621 V 1 122 per unit part b The required calculations follow those of Example 119 i The terminal voltage will be set equal to 460 V 11 10 per unit ii The linetoneutral terminal voltage is Vajn 460sqrt3 2656 V Thus mt re I wee 1004 A rms iii Wm Ls 6tan 267 an Vajln Thus inrmg V2 I sind 1138 A and inmp V2 I sind 849 A WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 139 iv 2 2 Trot p 197A 3 sam case Problem 1116 T 4431 Nm Ie 1217 A ip 784A iq 955 A I 874 A rms V 1993 V rms 11 Problem 1117 part a V2V 2 230V3 Apo NA Marated bn v2 230Vv3 0512 Wb Wmrated 35007 30 part b The frequency will be 60 Hz and hence X wells 324 2 Eag 36003500 2303 1366 V Ta rated 2000sqrt3 230 502 A The armature current is equal to Le Va rated ln Bat JXs Although the magnitude of Eng Eng Zo is known its angle 6 required to give I Ia rated is not A MATLAB script can be used to easily iterate to find that 6 673 o The motor power is then given by 3Eat V p Se sind 196 kW Xs Then pat 522Nm Wm and 2 2 T to5 Faiz am iD 2I ated 1 205 A WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 140 Problem 1118 As calculated in the solution to Problem 1117 Apy 0512 Wb At 4000 rmin fe 4000360060 667 Hz and thus we 2a f 4189 radsec The rms linetoneutral armature flux linkages under this operating condition will be V2V ha NA arated In 0448 Wb We and maximum torque will correspond to operating the motor at its rated arma ture current I Ia ratead 502 A Solving 2 2 AD AQ 2 Ls Apm Lsiq 2 2LsIa 2L sip Apm Apyy 7 2 for ip gives 202 2LsIa Abn 4D OL Apu 38 Thus ig 1212 if 562 A Thus the maximimum torque will be given by 1 Tmax 2 PS Apwig 432 Nem 2 2 and for a speed of 4000 rmin wm 4189 radsec Pax WmT max 1810 W Problem 1119 The rated current of this motor is Pra er arated 377 A v3 Va rated 11 v2 Va rated in V2230sqrt3 Apy 0235 Wb We 76207 30 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 141 part a The torque will be maximized when ig is maximized In this case this will occur when ip 0 and thus 1Qmax V2 Ta rated 532 A and thus 1 Tinax 2 PS Apatiqanax 188 Nm 2 2 part b The peak linetoneutral flux linkages are Aa Aby rN 1 Ady Lsiqumax 0257 Wb Thus to avoid exceeding rated terminal voltage the electrical frequency of the motor must be limited to v2 Varated Ln Wemax 7319 radsec and the corresponding motor speed will be nN Wemax 6989 rmin T part c At 10000 rmin we 100007 30 1047 radsec In order to maintain rated terminal voltage the peak linetoneutral armature flux linkages must now be limited to v2 Varated Ln Xamax 0179 Wb We Thus solving the peak linetoneutral armature flux linkages Namax AB AQ Ls Apm Lsiq 2LsIaratea 2LsipApmM Mane 2 for ip gives r2 2LsIn rated Abu p aac eee ee 373 A p 2LsApmM and thus WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 142 iqg 222 rated it 380A The motor torque is then given by 3 1 TP2 dpmig 134Nm 2 2 Since this is a twopole machine and wy We the corresponding power will be PmT 141 kW The motor power factor will be P power factor 09877 v3 Va rated 11 Tarated Problem 1120 15 10 a OG 1000 2000 3000 4000 5000 6000 7000 8000 3000 10000 Motor speed rpm 19 18 17 16 15 14 135 1000 2000 3000 4000 5000 6000 7000 8000 3000 10000 Motor speed rpm Problem 1121 part a Following the analysis of Chapter 6 jXmRi jX1 Zieq Rieg X1eq 0099 71082 Beg Ned Sh RY Xi Xm J WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 143 05npnV 2 Tmax oieh ea neq 234Nm wWsX1 eq Xp R Snax 0102 102 Ri eg Xiq X2 part b wy 2rf2poles 607 1885 radsec At s 0029 wn 1 sws 1830 radsec The torque is given by 1 Nph V2R2s T ph ical 28 196Nm ws Wien os Xq FX and the power is PwyT 231 kW part c With the frequency reduced from 60 Hz to 30 Hz ws the terminal voltage and each reactance must be scaled by the factor 3560 The torque expression can be solved for the slip This can most easily be done iteratively with a MATLAB script The result is s 0051 51 the speed is 997 rmin and the output power is 131 kW Problem 1122 part a 250 200 150 E ico Elle Soa oO 200 400 600 800 1000 1200 1400 1600 1800 rpm part b The same MATLAB script can be used to iteratively find the drive frequency for which smaxr 10 The result is a drive frequency of 544 Hz and a torque of 151 Nm 144 Problem 1123 Problem 1124 The motor torque is a function of the ratio R2s The slip with R2ext 0 is s0 1200 1157 1200 00358 and that with R2ext 087 Ω is s1 1200 1072 1200 01067 Thus solving R2 s0 R2 087 s1 for R2 gives R2 044 Ω Problem 1125 The motor torque is a function of the ratio R2s The slip with R2ext 0 is s0 1200 1157 1200 00358 The desired operating speed corresponds to a slip of s1 1200 850 1200 02917 Thus substituting the value of R2 found in the solution to Problem 1124 and solving for R2ext R2 s0 R2 R2ext s1 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 145 for Raext gives Raext 314 2 Problem 1126 part a If Ri is assumed negligible the torque expression becomes T i ph Vieq R2s ws Res X16q X2 Substituting the corresponding expressions for Tiax Pe i 05rpnVireq mass Ws X16q Xo 8 fa maxT Xieqg X2 gives 2 T Tax SmaxT s Defining the ratio of fullload torque to maximum torque as Tq 1 k 0472 Tmax 212 O47 the fullload slip can then be found as k 00414 414 4 Smax 0 4 4 V1 part b The full load rotor power dissipation is given by SA Prose Pa 3240 1 sq part c At rated load wm rated 1 59ws 1807 radsec The rated torque is Trated Prated Wmrated 415 Nm Setting s 1 gives 2 Tstart Tmax 681 283 Nm SmaxT SmaxT part d If the stator current is at its full load value this means that R2s is equal to its full load value and hence the torque will be equal to the fullload torque 415 Nm WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 146 part e The slip will be twice the original full load slip or 826 Problem 1127 part a 1200 1169 002 Sq 1300 00258 and thus the fullload rotor power dissipation is equal to SA Protor Pa 928 W 1 sq part b Defining Tmax k 24 Tq and using the derivation found in the solution to Problem 1126 gives SmaxT f V k2 1 01211 Thus the motor speed at maximim torque is Nmax 12001Smaxr 1055 rmin part c We want Smaxr to increase by a factor of 101211 826 Thus the rotor resistance must increase by this factor In other words Ro Ravext 826R2 which gives Ro ext 167 2 part d The 50Hz voltage will be 56 that of 60Hz Thus the applied voltage will be 367 V linetoline part e If the frequency and voltage are scaled from their rated value by a factor kp the torque expression becomes Te 1 Mph keVieq R2s kewso J Ros ktX1eq X2 where wo is the ratedfrequency synchronous speed of the motor Clearly the torque expression will remain constant if the slip scales invesely with ke Thus 60 450 5 460 0031 The synchronous speed at 50 Hz is 1000 rmin and thus nas0 LO001 sa50 969 rmin WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 147 Problem 1128 part a From the data given in Problem 1123 the motor inductances are 1 313 mH L2350mH Ly 128 mH and thus Lg 1 Ly 1315 mH and DR Le Lm 1319 mH R R 108 mQ and Rag Ro 296 mQ Finally the rated motor torque The peak flux linkages corresponding to rated voltage linetoneutral voltage are given by 2 Viase 2 2400 Nestea v2 Voase V2 2400 gg wp V3 Wbase V3 12077 The required torque can be determined from the given power and speed as Pech 400 x 10 Treen O Tyan 730 827 Nom Setting Apr Aratea gives 2 2 IR Tmech 1461A 0 5 a ADR and r ip an 405 A part b i2 i2 I 22 1072 A 2 part c 1 Wme Wm ES 3606 radsec We Wine 52 2 3687 radsec LR ip and thus WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 148 fo 2 587 Hz 20 part d 2 Vv Ratio weLsg ia Raia weLgip 2 1421 V ln 2461 V 11 Problem 1129 part a From the given data the motor inductances are 0915 mH Ly0936 mH L 4934 mH and thus Lg 1 Lm 5025 mH and DR lg Ly 5027 mH R Ry 322 mO and Rag Ro 703 mQ Finally the rated motor torque The peak flux linkages corresponding to rated voltage linetoneutral voltage are given by 2 Voase 2 230 Arated V2 Voase V2230 9 s98 wr V3 Wbase V3 12077 The motor torque is Tynech 8513001800 614 Nm Setting ApR Arated We can solve for tq and ip 2 2 LR Tmech n 419A 9 5 as 3 ADR and r ip an 101A The motor mechanical velocity in electrical radsec is 1 Wme Wm Ee 2723 radsec and thus WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 149 We Wme 2 2781 radsec IR 4D and fo 2 443 Hz 27 part b i2 i I 22 304A 2 L2 Vv Ratp weLs Tig Raia weLgip 2 1018 V ln1763 V l part c Sin V3 Vala 930 kVA part d 5 120 115 ia 110 105 100 95 390 85 80 85 90 95 100 105 110 115 120 125 LambdaDR percent of rated Problem 1130 The motor parameters are calculated in the solution to Problem 1129 part a The motor torque is Tmech 8514501800 685 Nm Setting ADR Arated We can solve for ig and ip 2 2 IR Tmech 467A a 5 saa ADR r ip an 101A WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 150 and if 22 I 29 338 A 2 part b The motor mechanical velocity in electrical radsec is Wme Wm Ee 3037 radsec and thus We We 2 3101 radsec IR 4D and fo 2 494 Hz 27 part c ig is now increased to 514 A and hence with Ipr and hence Apr unchanged 3 poles Lm Tmec a 7 3 N h 5 2 2 roxie 753 Mm Thus the motor speed is Timec n 1800 2S 1595 rmin 85 and wy nm 30 1670 radsec Pech Tmechwm 114 kW part d The terminal voltage is V Ratio weLsg 44 Jig Raia weLgip 2 1256 V ln 2175 V Ll The drive frequency can be found from Wme Wm Ee 3341 radsec We Wme 2 3412 radsec IR 4D WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 151 and fo 7 543 Hz 27 part e Sin V3 Vala 139 kVA part f Iteration with a MATLAB script gives App 953 of Arated Problem 1131 part a From the given data the motor inductances are 496 mH L2z602 mH L 1183 mH and thus Lg 1 Lm 1233 mH and DR Lo Ly 1243 mH R Ry 212 mQ and Rag Rp 348 m Finally the rated motor torque The peak flux linkages corresponding to rated voltage linetoneutral voltage are given by 2 Vase 241 Nestea W2 Mowe V24160 9 1 wry V3 Whase V3 1207 At a power output of 1050 kW and a speed of 828 rmin wm 843 radsec Tmech 125 x 104 Nm Setting ApR Arated gives 2 2 T tgs 2421 A 3 poles Ap ip 225 vated 1 762 A if 2 I 422 1795 A 2 The terminal voltage is 12 Vv Ratp weLs Tig Raia weLgip 2 2415 V ln 4183 V 11 WWWCOMUNIDADEELETRICIDADEBLOGSPOTCOM 152 The drive frequency can be found from 1 Wme Wm Ee 3372 radsec We Wme 2 3461 radsec LR ip and fo 2 551 Hz 2a part b The equivalentcircuit of Chapter 6 can be analyzed readily using MATLAB as follows All the reactances must be scaled from their 60Hz values to 551 Hz The rms input voltage must be set equal to 2415 V linetoneutral The slip must be calculated based upon a synchronous speed of ns 60f2poles 826 rmin If this is done the equivalent circuit will give exactly the same results as those of part a