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Part 11 1001

dc generator

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2K views40 pages

Part 11 1001

dc generator

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lhan
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Solutions to Test 14 399 PNZ® _ PNZ@ = NZo 60a 60(mP) 6Om pe. OO 14288) 76.707 Z® —728(25x 10") N= 377rpm =—— =0.005A Me «~~ 2500 E=V+l, R, =10+0.005(1000) 60Ea___—«6O(15)(2)— © PZ® ~ 2(1200)(40,000) x 10* N= 1562.5rpm he ee 0 = SoA Mw 110 Vieeder = Rteoaer = 50(0.04)=2V Ven = Ve + Veeder =110+2=112V Veo tte ly =o = = 224A a Ro SO tly, = 50+ 2.24 2.244 E=V,, +1,R, =112+52.24(0.15) E=119.84=120V = Pt. 50,000 _ 297.274 M220 Nan = Meg 220 423A Ric be A, = +14, = 227.27+4.23 1, =231.5A E=V, +1,R, = 220 + 231.5(0.026) = 226.02 400 1001 Solved Problems in Electrical Engineering by R_ Rojas Jr _ Py _ 180,000 co Pues anos =k +l, =720+6= 726A E=\, +1R, = 250+ 726(0.02) = 264.52 V Py = El, = 264.52(726) = 192.04 kW PL__180 =f. 100% "=p, ~ 792.04 * n= 93.73% P.__ 200,000 — Pr _ 200,000 _ g6q 56.0 i My 230 : la apeeo — 30,000 = 4090.91 Ww . Solutions to Test 14 405 Preyer = 9-15 Posse, = 0.15 (4090.91) = 613.636 W ie = 0.85 Pros, = 0.85 (4090 91) = 3477.273 W Note: For a series generator, at maximum efficien: : ; cy, the copper loss is equivalent to the stray power loss. ‘em = 613.636 W PL = x"Pan, 613.636 =0.42 Po 3477.273 Praten = XPoute, = 0-42(30,000) = 12600 W Pas a 12600 Postm +2Pax, 12600 + 2(613.636) Nm = 91.12% sa | P, =kyN P, tae Ne: Tm = =0.9112 2 2 N: Naat N,? P., -°. (Re) = 109f, } 0.0001N2" => @ 1000, Substitute Eq.1 and Eq.2 inEq.3: 0.25N, + 0.0001N,* = 0.5(250 + 100) N,? +25001N, — 1750000 = 0 Using the quadratic formula : 2500 + ,/(2500)? — 4(1) 21) 750000) _ - 2500+ 3640 re 2 106 __ 1007 Solved Problems in Electrical Bae °: = ke a se | Pe, LNB 2 NBrg Ree Pal NiBiny P,, - 1350. 2 (1200(2,500,000) )* = 600) 4000(2,000,000) } Pron = Pa +Pr Pre = keN? +k,N At1000 rpm 410 =k, (1000)? +k, (1000) k, =0.01-1000k, >@ At 750mm 6 =k, (750)? +k,,(750) k, = 0.008 750k, =@ Equate Eq.1toEq.2: 0.01-1000k, = 0.008 - 750k, k, =8x10% k, =0.01-1000/8 x 10% 2x10* At 500 rpm: Prone = KgN? +k,N = 8 x 10*(500)? + 2x 10°°(500) Pose = KW 1, = Pe _ 20,000 we 400 Van = Vi +hRy = 400+ 50(0. 1) Van = 405 V =50A Mv, 405 ee hin =F a5 74-7644 I, =|. +len = 50+ 4.764 = 54.764 4 Solutions to Test 14 407 B= Vin tl(Re +e) + Vy = 405 + 54.764(0.14 0.06) +2 E=415.76V 1 Seay =a E=\Y+ (Re +R) =O Substitute Eq.1 in Eq.2: 10000 | V, =v, +|—— ++ Jo.1+0. 264 +f ca 10 1+0.05) (204, «4922.-ocor26 Jv, i 1.00125, 7 +1500-264V, =0 V2 -263.67V, +1498.127 =0 Using the quadratic formula : Mi ae 263.67 + (263.67) = 4(1)(1498.127) f 2¢4) a 263.67 + 252.05 2 V.'* = 257.86 volts Ma Eg io... ~ additionainumber of ampere- turns required tomake the generator flat compounded. } © as = Nunlan, = (1000)(0.2) = 200 AT Note: This additional flux shall be supplied by the series field windings when the Generator is connected as a compound machine. ® veces = Noelse 1, = Peers _ 200 _ soa eg nee 408 100i Solved Problems in Electrical Engineering by R. Rojas Jr By CDT: Vs eo RR tye Ralla — tee) = beaF ve Ia f p= 80.028) _ 9.0277 | 95 — 50 = 250(8.33) = 2082.5 W = (508.33)?(0.025) = 6459.98 W 2. = (508.33)? (0.01) + 2583.99 W P oe =, Pry = + Pa + Pan + Pee + Pury + Para ton = 125,000 + 6459.98 + 2082.5 + 2583.99 + 5000 + 0.0(125,000) P,, = 142,376.47 W 4hp P, = 142376.47 W oe “746 W = 190.85 hp Note: The input of the generator is equivalent to the brake horsepower rating) the motor. EI ban = a - 230-254 1, =I +1, =100+25=102 5A Re Py = Vil, = 250(100) = 25 kW P, =1,7R, =(102.5)?(0.1) Py =1.05kW Pp =la7Rip = (102.5)7(0.02) Pp =0.21kW aa, Solutions to Test 14 409 Pan = Vilen = 250(2.5) = 0.625 KW Pye ha Raw © (102.5)? (0.025) ~ 0.263 kW P WPL + Pat Pe + Pa, + Pus + Pavey 25 x 100% "25+ 1.05 +0.21+0.626+0.263+18 1 = 86.36% Note: Interpole windings (Rup) are always connected In series with the armature windings. Asa shunt motor at no - load Be we, Pwray = (135)(2.5) — (2.5)? (0.2) = 336.25 W As.along shunt compound motor ha ze = 1 =25A Weak thn = 4442.5 = 46.58 Py, = Vila, = 125(2.5) = 312.5 W PY =1,?R, = (46.5)7(0.2) = 432.45 W Pu R yp = (48.5)? (0.25) = 540.56 W Pe Pe 5500 x 100% 5500 + 432.45 + 312.5 + 336.25 = 83.57% Gd, - 2% - 20000 504 : Vo = Vi +h Ry +Ry) = 400 + 50(0.06 + 0.1) Ven = 408 V Man _ 408 me Greet 8 410 _ 1001 Solved Problems in Electrical Engineering by R. Rojas Jr I =h +h =50+4.8=54.8A E= Vy, +1R, + Vy = 408 + 54.8(0.1) +2 E-415.48V ete 20 100A M200 Van = Vi +h (Reo + Rr) = 200 + 100(0.03 + 0.1) Vex = 213 volts =a —265A 80 1 + lan = 100 + 2.66 02.664 E = Vy, +1,R, = 213 + 102.66(0.04) = 217.106 volts Po= El, = 217.106(102.66) P, = 22.29kW Ea E=V, +L, RathRe 510 = 500 + 0.021, + 0.031, 1, = 500-15, >@ tah +h Me +h Roo Ray 500 +0.031, ah + Ls I, = 6.25 + 1.0003751, >@ Equate Eq.1to Eq.2: 500-1.51, = 6.25+1.0003751, I, =197.47A fa Rr, - Resa _ (0.015)(0.03) 9 Rw +R, 0.015 +0.03 Ry, = 0.019 I =k thn =150425 I, =152.5A “i __ Solutions to Test 14411 Ee Ve the th Reg = 290+ (182.5)(0.032) + (150)(0.01) E= 236.38 V ee Pe El _ (2300150) (236.38)(152.5) 195.7% x 100% Note: A diverter resistor is always connected actoss or in parallel with the series field winding, the purpose of which is to control the currént flow in the series field winding. At110 V operating voltage : 100,000 110 150,000 ——— = 1363.64A 110 Bao P4. | sl, +12 =909.1+ 1363.64 = 2272.74A a =909.1A Using their characteristic triangles : For generator G2: Al, = 1363.64.av 20 = 68.182 AV Al +Al, = Al 8091.4 + 68.182 av = 2272.74-2000 Weatriay ® Since the load decreases, the new operating voltage will increase: 412 1001 Solved Problems in Electrical Engineering by R_ Rojas Jr Myer = Mosg + AV = 11041.7 ew = 111.7 volts Massa = View + Vrs = 111-7410 VpowedG: = 121.7 volts, + Verop = 111.7420 Vino onde = Vnew Veoweaa2 = 131.7-volts Generator A Generator B Note: As given the current in the field windings are negligible. By KCL at node a: hay hea =h Ey = eM Eee Re Ra 450-V, , 460-V, _ 320,000 0.02 0.025 Ve 22,500- SOV, +18,400-4ay, ~ 320,000 A 40/900 V, ~-90V,7 = 320,000 V7 — 454.44V, + 3555.55 -0 ‘Using the quadratic formula : in) _ 454.44 + 438: uy = SES ane.475 M =46.5V _ 454.444 438.51 2 annie Note: Referto the circuit diagram of problem #517 6 py KCL at node a tae Th thane + tana Solutions to Test 14 413 i) —V, , 235-V, 290-M , So we 001 0.01 50" 50 200 V, + 46,500 = 2500 + 0.04 Vv, Vv = 219.95 220V Mv Note: The concept used in problems # 516 & 517 are still applicable when there are three or more generators in parallel By KCL at node a = Igy + hag + 4a = he + lens + bana + Sena Eo-M Ee-Y Ea-Y, Ea, WM ay : Res Re Ral * Rw, Bae Ben 240-V, 242-V, | 245-Vi _ MOM Me Sen’ one 002 a0 40 40 ~150 V, + 36350 = 2000+ 0.075 V, \, = 228.88.V Eg Ve MM _ 240-226.88 _ a Pesan Ny hey la = Ra Re 0.02 _ 242 - 228.88 _ 228.0 exon 0.02 40 — 228.88 - 228.88 _ 500A os Note: Refer to the circuit diagram of problem #516. BYKCL at node a: Nat thee =k + lynn + lane Ses vi 3 al Spr Seat ay othe Rea ee #4 1001 Solved Problems in Electrical Engineering by R. Rojas Jr 450-V, | 450-V, MA .. ————_._ = 4000 + E+ a oot ” 007 60 * 60 — 200 V, + 91000 = 4000 + 0.033 V, M = 434.93 hy het hen = Vi Me. 450-434.93 _ 434.93 Re Ran 0.01 60 i, =1499.75 = 1500A Ee-M MY _ 460— 434.93 434.93 Ra Raz 0.01 60 i, = 2499.75 = 25004 Wy = tao — tena = Pa = Vil, =(434.93)(1500) = 652.4 kW Py = Vil = (434.93)(2500) = 1087.3 kW ee uno 420-0.011, -V, =0 pene By ratio and proportion : = a4 ON 425 c |, ~ 480=M 0.16 h =h +l, S09 = 220= ML aeO =, 0.01 0.16 MV. = 417.65 volts j= SeNe Wisc Bl eo ma Solutions to Test 14 415 h, _ 235 Share = '» _ 235 % Share I, Bop ™ 190% % Share = 47% FL Veron = %ReEg(V,, ) =(0.10)(220) = 22v E, = 220+20=240V E, =(115)(2) = 230 v fe =(115)(0.01) = 1.150 P.__ 50,000 = = = 227.278 'o Va 220 a 0.0968 2 By KVL: E,-l(r, +%)-E, =0 j- Fas __ 240-230 fo+t 0.0968+1.15 1=8.02A Pet _ 600,000 _ 5608.69 ea 230 P, 800,000 Noa = — O88. = SEES = 3200 A a ev 250 tru “lag = AV _ VR) Al Al, - Vn.) _ 2608.69av %VR(Ve.) ~ (0.05)(230) Al, = 226.84 av ‘Assume gener: off from the line. The change in the ately tripped machine. 16 1001 Solved Problems in Electrical Engineering by R. Rojas Jr 4h = Alas 226.84 AV = 1600 AV = 7.05 volts Vines = Vigig — AV = 250 - 7.05 = 242.95 Vines = 243V Note: Since generator #1 will increase its load, its terminal voltage will drop. =e = 28200 Al, leoace 2000 Al, = 4004V>@ BV _ %VR2(Vimea) _ 0.04(250) jae 1600 Al, = 160AV =>=@ Al, + Al, = AI=@® Substitute Eq. 1 and Eq.2inEq.3: 400AV + 160AV = 3000-1500 AV = 2.678 View = Viets + AV = 250+ 2.678 = 252.678 Vinee 2 253 El Note: Since the busload decreases, terminal voltage increases. Solutic ay AVEC, uutions to Test 14 417 ar ) o. i, tes = 2.03(250) Al, = 266.67 AV > @ AV _ ®YR2(Vreteag ) Al Vretect Al, = 240 AV >@ ~ 0.04(250) Al, + Al, = Al=> @ Substitute Eq.1andEq.2inEq.3: 266.67AV + 240 AV = 600 AN =1.184 hat, = 2008 - 12008 Ningy = 1200 + Al, = 1200 + 266.6701 184) Nw = 1515.73 A Tagen = 1200+ Alp = 1200 + 240(1.184) 418 100} Solved Problems in Electrical Engineerint by R. Rojas Jr _ DC Motors SPEED CHARACTERISTICS OF ADC Mi oa aS where: N= speed (rpm) a = number of armature current paths P = number of poles Z = number of conductors @ = flux per pole (weber) Es = back emf or counter emf (volt) k = proportionality constant TORQUE DEVELOPED IN THE ARMATURE rose) To Sass Pa Ele Tako where: T = torque developed (newton-meter) ls = armature current (ampere) 'N = speed of armature rotation (rpm) P.= power developed in the armature (watt) a = number of armature current paths P = number of poles Z = number of conductors ® = flux per pole (weber) k = proportionality constant ____ DC Motors 419 CAL POWER OUTPUT 2aNT Hp. Hi 2xNT 33,000 P ~ ya760 here: Hp = mechanical power o1 ut (horsepower N = speed of armature Seapets 2 T = torque developed jote; If the given torque ts in pound-foot use the constant 33,000 and if it is in newton-meter, use the constant 44760. he armature coils and the shunt field coils are connected in parallel. Note: In a shunt motor, unless otherwise specified, the flux is assumed tu be constant. The armature coils and the shunt field coils are connected in series. 420 1001 Solved Problems in Electrical Enginecrine by R. Rojas Jr i na sertes mator, Aux is directh Lone SHUNT COMPOUND MOTOR The senes field cols are connected in series with the armature coils while the shun fald cols are connected across the series combination. tee = te a &, =V.-1(Ry +R) The series field coils are connected in series with the supply voltage while the shunt freid the armature coils are connected in parallel. t i tea = tm ten He ten Co mmtadananince, _—— _DCMotors 421 r “i At starting, the armature draws a high current due to the back emf of the motor at the instant of starting is equal to zero (Ey = O). g Armature starting current without a starting rheostat lee = Mv, R, 0 Armature starting current with a starting rheostat inserted in the armature aRY where: les = armature starting current (ampere) R,'= equivalent resistance of the armature circuit (ohm) V. = supply voltage (volt) R=resistance of the starting rheostat (ohm) © Speed relation is the percentage rise in speed when the mechanical load of the motor is removed. ae ~ Main «100% = no-load speed of the motor Ne. = full-oad speed of the motor 3 Nets For power losses, refer'to the formulas in DG generators: 422 1001 Solved Problems in Electrical neering by R. Rojas Jr Run the machine as a motor at no load with the same operating speed Fu ~ Vina ton ma, Re P, = rotational or constant losses (watt) Vaqu. = Supply voltage during the test (volt) \ armature current during the test (ampere) armature equivalent resistance (ohm) PONY BRAKE TEST mMeTeR This test is used to determine the output horsepower of the machine. 2aNT « Fa(scalomading-deadweight)xiengthotam . "P= where: T = torque exerted by the motor during the test N = speed of the motor shaft during the test (rpm) Hp = output horsepower of the motor k = constant for conversion 33,000 if torque (T) is in pound-foot (Ib-ft) = 44760 if torque (T) is in Newton-meter (N-m) The direction of rotation of a self-excited DC motor can be reversed using any of the ff. techniques: Q by interchanging the armature terminals Q by interchanging the field terminais Note: If both of them were interchanged, the direction of rotation will not change. ‘SPEED CONTROL OF A SELF-EXCITED DC MOTOR The speed of a self-excited DC motor can be varied using ff. techniques Q by inserting a rheostat in the armature circuit one we Q by inserting a rheostat in the field circuit Q_byusing a potentiometer to vary the supply voltage applied to the moto! _ pot Test 15 423. S25: A simplex lap wound armature has 580 conductors and carries a current of 125 amperes per armature current path. if the flux per pole is 20 mWb. Calculate the tlectromagnetic torque developed by the armature? 210.64 N-m 252.72 N-m 230.77 N-m 207.63 N-m oom, Problem 526: A€pole lap wound shunt motor takes 300 A when the speed is 500 rpm. The flux per pole is 0.06 Wb and the armature has 870 tums. Neglecting the shunt field current, calculate the brake horsepower of the motor. Assume a constant loss of 4%. A 175 B. 168 e.-172 D. 165 Problem A 220 V shunt motor driving its normal load draws an armature current of 50 A from a 220 V de source. The armature resistance of this motor including brushes is 0.25 ohm. How much armature current (minimum) will this motor draw from a 200 V fe source when driving the same load with the field adjusted to maintain the same Teblem $26: EE Board October 199° DC shunt motor has a full load rating of 15 hp, 230 volts, 57.1 amperes, 1400 | Rig he armature circuit resistance atid hm and the field circuit resistance is ‘Sunent™® Neglecting the effect of reaction, determine the no-load line 1001 Solved Problems ir al Engineering by R_ Rojas Jr io D 864A Problem $29: rated load, a shunt motor draws an armature current of 50 A from a 230 V ae, pain wens Tunning at 1000 rpm. At no load the armature current drops to 5 A. If the effect of armature reaction has weaken the flux by 2% from no-load to full ioag, determine speed of the motor at no-load. Assume the armature resistance to be 0.15 ohm. A. 1202 rm B. 1221 rpm ©. 1122 rpm D. 1010 mpm Problem 530: EE Board October 1990 The nameplate rating of a shunt motor is 150 hp, 600 volts, 205 A, 1700 rpm. The resistance of the shunt field circuit is 240 ohms and the total armature circuit resistance is 0.15 ohm. Calculate the speed regulation of the motor. ba 4.07% B. 489% Cc. 525% D. 5.18% Problem 531: A DC shunt motor runs at 600 rpm on a 240-volt supply while drawing a line current of 30 A. Its armature and field resistances are 0.5 Q and 120 Q respectively. What resistance should be placed in series with the armature circuit in order to reduce the speed to 400 rpm. Assume no changes in the armature or field current. A 26992 B. 24192 Cc. 2079 D. 2.632 Problem $32: A 120-¥ shunt motor has an armature equivalent of 0.5 9 and a field resista’ of 60 ©. At full loac, the motor takes 10 A and the speed is 1000 rpm. At what SI must this motor maybe driven as a generator to detiver 10 A to an external load § 120? A. 1158 rpm B. 1086 rpm Cc. 1262 rpm D. 1045 rpm Test 15 425. $33: EE Board April 1990 A belt-driven 150 KW shunt wound DC generator is running at 450 rpm and is full load to a bus bar at 240 V_ At what speed will it run if the belt breaks and the machine continues to run taking 6.5 kW from the bus bar? The armature and field resistances are 0.05 chm and 85 ohms respectively. The brush contact drop is 4.5 volt per brush. Neglect armature reaction A 305 rpm B88 333 A-shunt motor draws a current of 40 A from a 120 V source and runs at 1200 pm at rated load. The armature and field circuit resistances are 0.1 © and 60 ©. respectively. Determine the speed of the motor at half load. Assume a brush drop of 2V at rated load and 1 V at half load A 1180 spm B. ©. 1230pm D. Problem 535: EE Board October 1990 __ ADC shunt motor has a full load rating of 15 hp, 230 volts, 57.1 amperes, 1400 fpm. The armature circuit resistance is 0.13 ohm and the field circuit resistance is ‘M5 chms. Neglecting the effect of armature reaction, determine the no-load speed. 3 srmaure and field rosstances are 0.2 chm and 110 ohms, respectively Sipe Semaing resistance such thet tive: etaning: armel. tclmant Soe aad ‘times the full load current 426 ___1001 Solved Problems in Electrical Engineering by R. Rojas Jr__ Problem 537: ‘A 120-V DC motor rated at 5-hp has a full load efficiency of 86%. The field armature resistances are 60 ohms and 0.75 ohm, respectively. What st fesistance will be required to limit the armature current at starting to 200% of Fated armature current? A. 1.02 ohms B. 1.12 ohms Cc. 1.00 ohm D. 1.22 ohms Problem 538: A 10-hp 230 V DC motor of 85% full load efficiency is located 450 ft from supply mains. If the motor's starting current is 1.75 times the full load current, what the smallest cross-sectional area of copper wire required when the allowable drop in the feeder at starting is limited to 24 volts? A 30MCM B. 28MCM Cc. 26MCM D. 27MCM Problem $39: EE Board October 1990 The namepiate rating of a shunt motor is 150 hp, 600 volts, 205 A, 1700 The resistance of the shunt field circuit is 240 ohms and the total armature resistance is 0.15 ohm. if the motor is to be deliver full-load torque at 1,200 what value of resistance must be added to the armature circuit? A. 0.833 ohm B. 0.802 ohm Cc. 0.827 ohm D. 0.862 ohm Problem 540: A 120 V shunt motor draws a current of 77 A at rated load. The ai shunt field resistances are 0.2 2 and 60 ohms respectively. A multitap theostat is inserted in the armature circuit to limit the starting current. If the has a resistance of 1.732 ohms, determine the percentage tap used such starting current will be limited to 150% of its full load value? A. 60% 8B. 50% Cc. 55% D. 45% 2 Test 15427 A 10-hp power 220 Vv DC pril 1985 unt motor has an armature and field resistance of 0.25 ohm and 100 ohms res; tae Noe necivan The full load efficiency is 83%. Determine the Earcentol fulciond value, ler that the starting current will not exceed 200 A 2.889 B. 2599 Cc. 2.159 Dp. 2459 450 HP, 550 V shunt wound motor draws a line cui rrent of 4.5 A at no load. The shunt field resistance is 275 ohms and the armature resistance exclusive of brushes, is 0.3 ohm, the brush drop at full load is 2 V. At full load, the motor draws a line current of 84 A. Calculate the efficiency at full load. A. 92.0% B. 88.5% C. 91.2% D. 89.9% _ Problem 543: ‘A 100-volt shunt motor is developing 6 hp while operating at an overall efficiency of 86%. The armature and shunt field resistances are 0.06 and 50 ohms respectively. Determine stray power losses. Board October 1998 of a shunt motor is 7.8 hp. It draws 50 A from 120 V. The field 1.2.A. What is the efficiency of the motor? ering by R. Rojas Jr etrical Engineer 428 1001 Solved Prot 8B 260hms Cc 240hms D. 280nhms Problem 546: EE Board March 1998 A cettain shunt motor has an armature resistance of 0.05 ohm. It draws 50 a terminal voltage of 120 V. Assume other miscellaneous losses at 1%. Dete; ‘the output horsepower of the motor A BShp B 71hp Cc 68hp D. 7.8hp Problem 547: The armature and field resistances of a shunt machine are 0.2 ohm and ohms respectively. While running as a generator, the generated emf is 250 V at ff rpm. If the machine is run as a shunt motor, it takes 4 A at 220 V. At a certain lo the motor takes 30 A at 220 V. However on load, the armature reaction weakens field by 2 %. Find the motor speed at this load. 1120 mm 1050 rpm 1042 rpnp 1025 cpm ooo, Problem $48: EE Board October 1990 A shunt motor, which has a field resistance of 220 ohms and an ai resistance of 0.8 ohm takes 26 A from a 260 V supply when running at 500 full load. in order to control the speed of the motor a 1.2-ohm resistor is conne: series with the armature. Calculate the s i eat peed at which the motor will run A 438 rpin B 445:pm C. 4241pm D 472 rpm Problem $49: EE Board October 1992 ‘The input and output powers of a 220 V. shunt respectively. The field and armature circ stan aoe pare 20 taking the same current Series with supply voltage? Assume the motor is. A 599 rpm 6B. 601 rpm Cc. 572mm D. 583 rpm Problem $51: The equivalent armature resistance of a series motor is 0.1 ohm. When connected across a 110-V mains, the armature takes 20 A and its speed is 1000 tpm. Determine its speed when the armature takes 50 A from the sarrie mains, with the field increased by 10% A 809 rpm 8, 954 rpm C. 856rpm D. 884 rpm Problem $52: A 400 V series motor has a field resistance of 0.2 ohm and an armature resistance of 0.1 ohm. The motor takes 30 A of current al 1000 rpm while developing full load torque. What is the motor speed when this motor develops 60% of full load torque. A 1302.4 rpm B 1256.2 rpm ©. 1297.6 rpm D. 1135.5 mpm Problem A 10 hp, 230-V, 1200 rpm series motor having rated load efficiency of 85.5%. armature resistance including brushes is 0.3 ohm. The fieid resistance is 0.25 | Assuming the flux varies directly as the armature current, what value of Tesistance should be placed in series with this motor when starting, in order that the Current maybe limited to a value that will exert a starting torque equal to Of its rated torque? 430 __ 1001 Solved Problems in El Engineering by R. Rojas Jr Problem 55.4: The field and armature resistances of a 220-V series motor are 0.2 © and @) respectively. The motor takes 30 A of current while running at 700 rpm. If the: iron and friction losses are 350 W, determine the motor efficiency A 90.6% B. 915% Cc. 89.4% D. 92.2% Problem sss: A 100-V series motor is used to drive a’ load through a pulley. This maching, an armature resistance of 0.2 ohm and a series field resistance of 0.25 ohm. torque of 25 N-m is applied to the pulley, the speed is 600 rpm. If stray power at this load is 300 W, calculate the armature current. A 20.62A 6B. 21.89A C. 2272A D 23.41A Problem 556: @ 400 V series motor, having an equivalent armature circuit resistance of | ohm, takes 44 A of current while running at 650 rpm. What is the motor speed line current of 36 A? A. 803 rpm B. 822 rpm C. 812 rpm D. 806 rpm Problem 557: The resistance of each of the two coils of a series motor is 0.04 ohm, a resistance, 0.04 ohm. The motor takes 50 A at 100 V while running at 800 rpm! the coils are in series. What will it speed if the coils are re-connected in parallel the load torque is doubled? A. 800 rpm B. 400 rpm C. 200 rpm D. 650 rpm Problem $58: On full load @ 500 V series motor takes 100 A and runs at 820 mpm: armature and seri¢s field resistances are 0.1 and 0.04 ohm respectively. be its speed when developing half-tull load torque and with a 0.08-ohm connected across the series field winding. Assume flux is proportional to courrent. ———— a oat? __ fest 15431 A. 1563 rpm B. 1634 rpm ¢. 1359 pm D. 1429 rpm Problem 559: A 400 V series motor working with unsaturat led field is taking 60 A and runnin, at 840 rpm. The total resistance of the motor is 0.1 ohm. At what speed will the motor run when developing halt-full load torque. A 1193 1m 6B. 1202 rpm C. 1167 rpm D. 1352 rpm Problem 560: A long shunt compound motor takes a current of 42 A from a 230 V source. The generator parameters are: Ra = 0.1 Q, Rse = 0.2 9 and Rsh = 50 Q. If the friction and windage losses amount to 400 W, determine the overall efficiency of the machine. A 82.24% B. 80.56% C. 81.22% D. 79.53% Problem 561: A 230 V, long shunt machine has the following parameters: Ra = 2.0 2, Rsh = 460 and Rse = 0.25 9. When the machine is run at no-load at its normal speed and rated voltage, the armature draws 0.6 A. Determine the armature current drawn ifthe machine delivers an output of 5 bhp. 24.358 21220 20.364 22.834 Problem A25 long nt compound motor takes 5 A when running light and funs at ee a Tee eae, 0.15 9, Shunt field resistance, 200 Q and Series field resistance 0.05 2. At rated load the mojor takes 86.5 A and runs at 720 TPM. Determine the ratio of torque developed from no-léad to full load. gom> 432 1001 Solved Problems in Elecirical Engineering by R. Rojas Jr Problem 563: field resistances of 0.15 © a has armature and shunt : eeu aa incl Sanne light, it takes 6 A and runs at 1200 rpm. A seri fans one is added to make it long shunt cumulatively compounded. Thy china baled ® fam per pole by 25% when the motor is taking its full 1g curre bs compound motor. A. 887 mpm B. 890 mpm C. 868 pm D. 865 rpm Problem 564: f 0.4 ohm, a shunt field DC motor has an armature resistance o} ‘ ees ohms and a series field resistance of 4 ohms. This machine is connected as a short shunt compound motor to 220 V mains. Calculate the power developed by the armature if the armature current is 30 A. Neglect the brush drop. A. 2254.6 W B. 25728W Cc. 26526Ww D. 2150.5 Ww Problem $65: The input current to a 220 V, short shunt ‘compound motor at no load is 6 A. shunt field circuit resistance is 220 ohms: the armature resistance is 0.10 ohm ai the series field resistance is 0.08 ohm. What is the stray power loss? A. 1153.31 W B. 1066.22 W C. 1232.54W D. 1073.25Ww ——— $67: EE Board Octo! ——$$$__ _Test 15 433 T1986 in a brake test of an elevator doo; 19k tobe 1.09 kg, What is the output hp of the doc Meee weight of the arm is found a 940 np g. 10.38 hp c. 8.26 hp 1p. 9.58 hp Problem 568: A shunt motor with an armature and field resistance of 0.055 and 32 ohms, respectively, is to be tested for its mechanical efficiency by means of a rope brake. When the motor is running at 1400 rpm, the longitudinal pull on the 6-inch diameter pulley is 57 pounds. The readings of the line ammeter and voltmeter are 35 and 105, respectively. Calculate the efficiency. A. 77.10% B. 75.32% C. 78.28% D. 79.12% Problem 569: A shunt motor was tested by means of a pony brake having a length arm of 3.6 feet and a tare weight of 5.7 Ibs. The current drawn by the machine from a 240 V line was 50.9 A when the scale reading was 24 lbs and the speed of the motor was 1215 ‘pm. Calculate the rotational losses of the motor. The armature and shunt field fesistances of the machine are 0.25 © and 120 Q, respectively. 7 a s age sete i Cat 570: EE Board October 1993 3 are > — @hp, force ill be exerted on the scale in a pony when | oa 400 Soe teasing et full load, The length of the brake arm is 3 ft. “id the tare weight of the brake is 3.75 ID. bee | ian > 3633 Bs os Be, 434 1001 1 Engineering by R. Rojas Jr wed Problems in Elect ANSWER KEY C 535. C 545. C 555. A 565. D B 536. A 546. D 556. A 566 B A 537. C 547. B 557. A 567. C A C 548 A 558. D 568. A 4yL-4b Topnotcher D C 549. C 559. A 569. A D B 550. A 560. B 570. B 32-40 - Passer A B 551.0 561.B B D 552. C 562.8 23-31 - Conditional Reare 6 cee acces see = Failed Michael Faraday, an English chemist and physicist made one of the most significant Giscoveries in the history of electricity, the “ELECTROMAGNETIC INDUCTION". a Faraday, one of 10 children of a blacksmith was born near London. He was first Sppremmced 10 a bookbinder, but at the age of 22 realized his boyhood dream by becoming i f E i 3 e ener rere dscowage Faraday. Once he said * FAILURES ARE JUST AS |MPORTANT SUCCESSES, FAILURES ALSO TEACHES: Newas hailed as one of Europe's top scientist ——— ———— - ___Solutions to Test est IS 435 ‘Sofutions te Test a) o Note: The meaning of *N* in the formula below is speed in rpm” _PNZOIl, _ PNZOI, eet Be eg “eomP~ 60m =O 2nNT Pa=—G9 =O Equate Eq. 1to Eq.2: (#2): _ 2xNT 60m 60 Te Zol, _ (580)(20 x 10°)(125) © 2am 2n(1) T =230.77N-m BI e, = PNZ@ _ PNZ@ _NZ@ _ 500(870)(0.06) _ 455, “60a 60mP 60m 60(1) E,l, 435(300) Pye = 20) eh 745 (746 “i Pirate = (0.96)(175) Prato = 168 hp Note: Pirate = output power of the motor Ea Note: For the same load and speed, the power developed in the armature for both conditions are the same. Eat, Eyles + lage ay = (Ve — la Fe Maz ; (220 ~50(0.25))50 = (200 -1,, (0.25) lag 9-254,” ~ 2001, = 10375 la’ ~8001,, +.41500 = 0 Using the quadratic formula : i 800+ ee 4(1)(41500) _ soos 688-47 B: TOO} Solved Problems in Electrical Engineering by R. Rojas Jr. ty he lay = 57.1-2 = 551A P, =12R, = (65.1)? (0.13) = 394.68 W . Nan = (230)(2) = 460 W MV, = 230(57.1) = 13,133 W Pa, = 15(746) = 11190 W Py, = Poe +Pa + Pan + Prwray 43,133 = 11190 + 394.68 + 460 + Paray Pyrey = 1088.32 W Porm = Ving len —'ea F 1088.32 = 230 lang —loyy (9-13) 2 Ing 7 — 1769-231, + 8371.692 = 0 Using the quadratic formula : _ 1769.23 + Y(1769.23)" — 4(1)(6371.692) = 2(1) _ 1769.23 + 1759.74 A 2 lag? = 4-745-A sea. Yampa = Naga tet = 4-745 42 Vong, = 6-745. A Ee = Vs “laa = 230-50(0.15) = 222.5 E., = Ws -leg Re = 230 -5(0.15) = 229.25 V = N wees Ny Ex% N, Ey, ®y E,, ®. me=n{ 2 =) =1000 229.25 (0.980 SOO "Ee, ®2 222.60, >) = 1008.73 Nz = 1010 rpm i To __ Solutions to Test 35 087 a. Re 240-294 Nef eh. =205-25-202.54 P, =1,? - = (202.5)?(0.15) = 6150.937 W Pay = Valen = (600)(2.5) = 1500 W Py, = Im. = 205(600) = 123,000 w Poy = 150(746) = 111900 Ww Pr = Pas + Pa + Pan + Parry 123,000 = 111900 + 6150.937 + 1500+ Pra, Pyray = 3449 W Paray = Vor lon — layg 7 3449 = 6001,,,, —14,, (0.15) 2 ln ~ 40001,,, +22993.33 = 0 ‘Using the quadratic formula : 241) 2 =5.757A Eo = Ve -legy Ra = 600 - 5.757(0.15) = 599.136 V Eee, = Ve “lag Re = 600 - 202.5(0.15) = 569.625 V ; 36 Nw = Np EE J 1700 ($5 S)) 1788.07 pm 41788,07-1700 , 19% a 4700 Note: Since not specified, flux is assumed constant for both conditions. 438 1001 Sotved Problems in Electrical Engineering by R. Rojas Jr. ln = 30-2 = 28A Ee, = Va -laiFte = 240-28(0.5) E,, = 226V N=k,E Na _ Foz Ny Eo, Note: Since the field current as given is constant, the Aux in the air gap will a become constant. 6 2(R, +R) 40 — 28(0.5 +R) In =10-2=8A V.—1LR, =120-8(0.5) E, = 116 volts As agenerator : \ hk +hen =104+2=124 E,=V. +R. =120+12(0.5) E, =126 volts E=kN NE, 4 N, = —Z=2 = 1000(1 agin One) = 1086.2 N, =1086 rpm = (400 mi le 400 | = 150.657 V En -Fo() 220( 555)

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