Electro N6 study pack hlayiso.com

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higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA alculators may be used. tion paper consists of 5 pages and a formula sheet of 5 pages. Copyright reserved Downloaded from hlayiso.comPecs: turn over |

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(8080096) -2- T490(E)(J30)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ELECTROTECHNICS N6 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to paper. Copyright reserved Downloaded from hlayiso.com Please turn over

(8080096) -3- T490(E)(J30)T QUESTION 1 1.1 Explain how the speed of a DC shunt motor may be changed to the following: 1.41 Above normal speed 1.4.2 Below normal speed (4) 1.2 A DC shunt motor delivers 8 kW at 750 r/min from a 480) armature circuit resistance is 1,2 ohms and the shunt i 800 ohms. The efficiency at this load is 83%. Calculate the following: 4.2.1 The no-load armature current. (12) 1.2.2 The armature current wh is 60 Nim. Assume the flux remains consta (5) Hint: [21] QUESTION 2 A 380 v, unbale hase, delta-connected load, takes the following phase s phasor reference and assume a phase rotation of R-Y-B Za Calculate the following. 2.1 The impedance of each load .(6) 2.2 The current in each line (6) [12] Copyright reserved Downloaded from hlayiso.compeesse turn over

(8080096) -4- T490(E)(J30)T QUESTION 3 3.4 Explain how the constant losses in a transformer are kept to an economical minimum. (4) 3.2 The percentage impedance of a 125 kVA, single-phase, 3 000/250 V transformer is (0,5 + j5) per cent. At a power factor of 0,8 lagging, calculate: 3.2.1 The turns ratio (2) 3.2.2 The percentage regulation (2) 3.2.3 (1) 3.2.4 The full-load copper loss (3) 3.2.5 The power factor at whigi i (3) 3.2.6 The voltage to current on sho (2) 117] QUESTION 4 41 Explain, with th (5) 4.2 The synchronous reactance (4) The percentage voltage regulation at a power factor of 0,8 lagging I WH Copyright resevee Downloaded from hlayiso.com ease tum over

(8080096) ~6- T490(E)(J30)T QUESTION 5 5.1 What is a synchronous capacitor? 5.2 A three-phase synchronous motor is connected in parallel with a load of 500 kW at a power factor of 0,75 lagging. The excitation of the motor is adjusted until the total power factor is 0,9 lagging. The input of the motor is 100 kW. 5.2.1 Calculate the kVA input to the motor by also ni king sketch. & §2.2 Calculate the power factor of the mctor. QUESTION 6 6.1 Name TWO primary parts of a thre: 6.2 A three-phase, 500 V inducti supply. The slip is 5% windage/friction losses ai Calculate the following by using the T-method: 7A 7.2 7.3 The sending current The sending voltage The power factor at the sending end Important: Draw the T-method circuit diagram. TOTAL: (2) (7) (1) {10} (2) (5) (2) (2) [11] (11) (3) (1) [15] 100 Copyrghtreseved =~ Downloaded from hlayiso.com

(8080096) -1- T490(E)(J30)T FORMULA SHEET GS-MASJIENE DC MACHINES E=V-IaRa AMO Ey Ny, ®, _ A TF, 1, %, SPOEDBEHEER pay -ta[ R Rse + Rse E=V —-Ia Ra- TOETSING TESTING DIREKTE METODE DIRECT METHOD SWINBURNE- SWINBURNE METODE METHOD HOPKINSON. HOPKINSON RENDEMEN. EFFICIENCIES THE SAME IRON LOSS Vv Nu, +13) Rat (tly — 14)? Ra + (5 + 14) y} AB n LY generator 1 4 (+)? Ra + Ig ¥ + € (+ h)¥ - {a 4+Ip—lyy Rat IgV + ‘I motor — GQ +h)V Copyright reserved Downloaded from hlayiso.com Please turn over

(8080098) WS-BELASTING STERSTELSELS GEBALANSEERDE KRING DELTASTELSELS DRIEDRAAD- STELSELS * °2- T490(E)(J30)T = Vor AC LOADS IRa Td STAR SYSTEMS — ¥J-120° ~ ¥en = VERWYSING ly = ——— REFERENCE Zyy o = _ ¥|120° ip= Zan $3 Ty =Int+Ip+Ty THREE-WIRE SYSTEMS Vox = Vos + Vn Von =Vos + V sw Ven = Pes + Voy Copyright reserved Downloaded from hlayiso.com Please turn over

(8080096) 3. T490(E)(J30)T KOMPLEKSE GOLFVORMS e, =£,, Sin of COMPLEX WAVE FORMS e, = Ky E,, Sin2 of e; = Ky Ey Sin3 at e=E,, (Sin af + ky Sin 2 of + ky Sin3 af) 2 2 2 2 pe Eq lt Eq2+ Eq3 +t EY N 2R p= (P14 224+ 1234..+BN)R je Pi+P2+.402N 2 ke |x + B22 4..+ BLN 2 TRANSFORMATORS TRANSFORMERS Po+ Psc Enige waarde van belast vollas Any value of load at k of full-load Po MAXIMUM EFFICIENCY _ kS Cos ¢ k S Cos $+ Po+k? Pse Copyright reserved Downloaded from hlayiso.com Please turn over

(8080096) 4 T490(E)(J30)T FORMULES “sR iRe FORMULAE 0 = Vv I Xe %X= V %Z,=WR, +f WX, Vo =1Z, 2 Pro =I? Ry Psc LV se Cos $, = AC MACHINES WS-MASHENE ALTERNATORS ALTERNATORS _ ana Sin — 2 Kd = n Sin 2 2 Ry = Cos (p = Cos E=2Kf Kd Kp f OZ E= yl Cos $+ IRY + Sin p+ IXY E=V + IR Cos $= IX Sing E=E|g¢+IR|o + x|90 Reg = 7 SINCHRONE MOTOR FeE=Er Er=2 SYNCHRONOUS MOTOR E=V|— 6+ JR|180° + 1¥ |- 90° Copyright reseved Downloaded from hlayiso.com Pease tum over

(8080096) -5- INDUKSIEMOTOR yy cA s Xy =SXo Zy = 4 RZ + (SX0P Zo =R? + Xo? SEo RZ + (SX0) MAKSIMUM RENDEMENT d= Ry =SXo Rotorkoperverlies = S rot Rotor copper loss = S rétor Copyright reserved 7490(E\(J30)T INDUCTION MOTOR MUM EFFICIENCY Downloaded from hlayiso.com

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Electrotechnics N6 Memorandum August 2014 QUESTION 1 1.1 1.1.1 Above normal speed : decrease the field current by inserting a resistance in the fieid circuit. vv 1.1.2 Below normal speed : decrease the voltage across the armature by inserting a resistance in the armature circuit, VV (4) 1.2 Pout=8kW N= 750 r/minV = 480 V R,=120 Rgy= 8000 n=83% iron,friction & wind losses 0) Pinput 121 1 = — L=—_ Vv Vv = ROWE 199 — 9638,554 Pinput 480 Pinput = NPY" y 100 1, =2008a ([) = 222" «100 ln = “a3 * : sh Reh Pinput = 9638,554 w (1) = 180 B00 Total losses = P input — P output =0,6A @ = 9638,554 — 8000 () I, = In Ish = 1638554 W @) -20,08-0,6 =19,48A (j) Iron, friction & wind = Total losses - I,” Ry - Ign-V = 1638,554- 19,482, 1,2 -0,6.480 = 895,19 W (1) 895,19 lag = =" = L86EA (1) 1.2.2 Ey =V—la Ra = 480-19,48 1,2 © = 456,624 V (7) 2uNT EL, = a | 60 2n750 60 L= —— a 60. 456,624 =10,32A Oo) 0) Downloaded from hlayiso.com (12) (5)

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QUESTION 2 Vi =380V Aconmnected V = Vpn i, = v3 Inn Iny = 204j0A =20|0°A lygp = 25-~j10A = 26,926 |-21,8°A Ipp = 304+j15A = 33,541 [26,577 A = VR = va = VBR 21 Zpy= ie vp = Zer = TeR _ 380 [0° () _ 380 |=120¢ 0) ____ 380 |120° (D 20 {0° © 26,926 | - 21,8°-120° ~ 33,541 [2657°+120" =19|{0°o ‘0 = 14,113 [21,8° 0 oO) = 11,329 |- 26,57° 9 0) 22 Ip = Ipy— Ibe ly = Tyg — Ipy =20+j0-30-j15 = 25-j10—20-jo =-10-j15 (1) =5-j10 o = 18,028 -123,69° A (i) = 11,18 |- 63,43°A 0) Is = Ibn Wye =30+j15-25+J10 =5+j25 0 : re = 25,495 |78,69°A © bil | 23] Downloaded from hlayiso.com

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QUESTION 3 3.1 Core losses — which can be reduced: (a) by using high quality core material with low hysteresis loss @) (b) by laminating the core. Oo) Dielectric losses — which can be reduced by using insulating material that is non) Be] hydroscopic and has a high dielectric strength. 0) 3.2 $=125kVA Vz = 3000V V2, = 250V %2=(0,5+j5)% cos pz = 0.8 lagging %Z=5,025 |84,29° % Ni Va 3.2.1 MV Ni _ 3000 Ny 250 O I =22 NiiN,=2211 O 14Ze1 C08 (©,~@, } aA ‘ 3.2.2 %Reg = eee IiZer = eV = 150,75 eos (84,29"~3687") y 4 4g (0) =2825 ang ~ 3000 ““s00 * [2] =3,4% 0) = 150,75 V 3.2.3 Vz= Ez, — Reg2 Reg, = Reg, X a Reg, = 0,034x3000 a = E,— Reg, = 102K = = 102 V = 250-85 =85V . fi 22415 V @ 1,.R, 324 Pao = Ly. Reg L== % R = 2“ x 100 V4 Vy %R 3 ——XV. = 44,6677 .0,36 = tixie = tot 3000 iy 6 5 x 3000 = 625,01 Watt (i) = 41,6674 (/ = 400____ aa @ 3! Downloaded from hlayiso.com

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’ 3.2.5 @2 = O_ at max regulation om Cosy = cos 36877 ~o;6tagging (1) 3.2.6 Vee= ly. Zea 0) = 150,75 V (i) QUESTION 4 A To exciter Short-circuit the machine terminals through ammeters and taking readings of short) : circuit current and field current up to twice full-load current. Cd) [Ss] 4.2 S=30KVA Vz = 380 V e=3,5A E,=220Vat3,5A Rpn = 0,50 2 2 Eph ©n open circuit Ss 4.2.1 Xg = -_ Zp = I= 5 J ds Roh Ss Ipn With the same If LT Vv. = [2,7872 — 0,52 @ - ral = 30 x 103 45,58 ¥3 . 380 =2,7410 Q) =2,7870 6 -45,580 (1) [4] 4.2.2 cos =0,8 lagging Zs = 0,5 +4 2,741 Vpn = 219,4 | 0° % Reg = =— x ioo =2,787 |79,66°0 (]) Downloaded from hlayiso.com

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Eon = Von + LZ % Reg = =—* x ioo 324,313 -219,4 + ty = xiool = 219,4 [0° + 45,58 |- 36,87°. 2,787 |79,66° rer = 219,4 [0° + 127,031 |42,79° % Reg = 47,82 % D = 219,4+j0+93,222 +] 86,294 = 312,622 +] 86,294 @ = 324,313 [15,43°V oO) 5] QUESTION 5 | J 5.1 If the motor is running on no-load and the excitation is increased, the current taken by the motor will lead the supply voltage by 90°(very nearly}. This means that the - motor can then act like a capacitor, and in such a case it is called a synchronous capacitor. A synchronous motor {over excited} may therefore be used to improve the power factor ofaload. {(] —— Qa 5.2 Py =S500KW P;=600kW cos, = 0,75 lagging cos; = 0,9 lagging > NB P3 ' * O , OW - So = S3— Sy X Sem 83 Sh = 500 = £00 0,75 0,9 = 666,667 |- 41,41° kVA = 666,667 |- 25,84" kVA = 500 - j 440,962 kVA @ = 600 — j 290,593 kVA @) Sq = 600 —j 290,593 - 500 + j 440,962 kVA =100+j150,369kvA_ (2) = 180,585 |56,37° kVA (D Downloaded from hlayiso.com

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5.2.2 cos dz = cos 56,37° = 0,554 leadin, a — QUESTION 6 Li OL] 6.1. The stator and the rotor. j2 J 6.2 V,=500V 1,=80A 5=0,05 cos @ = 0,8 lagging Stator losses = 2kW_ Wind & friction losses = 750 W 6.2.1 Rotor cu losses = S. Rotorinput Rotorinput = Pinpur — Statorlosses Oo V3ViIL cos (D = 1000 —Statorlosses / _ ¥3. 500. 80. 0,8 2 () ~ 1000 Rotor cu losses = 0,05 . 53,426 [5 | =2,671kw (I) 6.2.2 Poutpur = Rotor output — friction & wind Poutput = 50,755 ~ 0,75 Rotor output = (1 —$) Rotor input =50 kW = (1~0,05 ) 53,426 = 50,755 kW @) [2] 6.2.3 n= eet x 100 input 50 = Ema *100 () sas () Li] Downloaded from hlayiso.com

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os, QUESTION 7 R L R L D Is 2 3 A 2 3 IL E Vs c V, oO 7 ° P,=60MW cos@, = 0,8 lagging V,; = 110kV R=0,3 x 200 = 600 L= 0,195 x 200 = 390 mH C = 0,0093 x 200 = 1,86 pF < = 300 4 = 195 mH 2 71 Zona = Zax = 30+) 2.0.50.195x 107% = 30+] 61,261 = 68,212 |63,91° 0 o Zap =O) y= ee 2.7. 50. 1,86 x10 V3 =0~-j1711,3430 = 76210,236 |0°V (O = 1711,343 |-90°9 (WD lag = Lp = a Var = Tag: Zaz ¥3.Vr. cosdr 20 x10 Ve. 432%108 08 = 109,347 |- 36,87°. 68,212 |63,9152' = 109,347 |-36,87°A (0) = 7458,778 |27,04 V = 87,477 - {65,608 A t = 6643,454 +j 3390,853 V (1) Van = Ve + Vas Tap = Tan 92923,048 |2,34° = 76210,236 +j0 + 6643,454 + j 3390,853 = 2202s,048 |7,54° 1711,343 |- 90° = 82853,690 + j 3390,853 = 48,455 [92,34°A ~ = 82923,048 |2,34° V () =-1,978+]48,415A () Downloaded from hlayiso.com

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Ts = Ipa = Ing t+ Le =- 1,978 + j48,415_ + 87,477 ~-j 65,608 WD = 85,499 -j 17,193 A Is = 87,21 |-14,37° A (1) 7.2 Voa = Ig-Zpa = 87,21 {- 11,37" . 68,212 |63,91° = $948,811 |52,54° V = 3618,133 + j 4722,02 V Vs = Voa + Vag = 3618,133 + j 4722,02 + 82853,690 + j 3390,853 = 86471,823 + j 8112,872 = 86851568 |5,36°V _¥3 .Vs SL“ "1000 _¥3 X 86851,568 1000 (D Vey, =:150,434 kV | 7.3. Cos = Cos (5,36° + 11,37°) = 0,96 lagging (D [31 [iy [is] Tolle | Downloaded from hlayiso.com

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Te “ZO uy & Northlink College THE EDUCATION CONNECTION CORRESPONDENCE STUDIES ELECTROTECHNICS N6 INSTRUCTIONS: 1. Answer ALL the questions, 2. Read ALL the questions carefully. 3. Number the answers correctly. 4. Start each question on a NEW page. 5. Keep subsections of questions together. 6. Write neatly and legibly please. Stu0'| QUESTIONS GUIDE QUESTION 1 1.1. State which losses occur in a DC machine and distinguish between constant 12 13 and variable losses. (4) Explain how the speed of a DC shunt motor may be varied both above-and below normal speed. ; (4) A 500 V, DC series motor takes 100 A from the supply when running at 500 r/min. The armature resistance is 0,1 ohms and the series field resistance is 0,05 ohms. Calculate the speed at which the motor will run if the armature current is 0,7 of the full-load current and a resistance of 0,1 ohms is connected in parallel with the field winding. Assume the flux is proportional to the field current. (10) [18] Downloaded from hlayiso.com

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QUESTION 2 A resistance of 950 ohms is connected in parallel with a capacitor of 3,185 pF. A voltage of 150 Sin 314t + 70 Sin ( 942t + 60° ) V is applied across the circuit, ; Calculate the following: 2.1. Anexpression for the instantaneous value of the current. - (0) 2.2 The RMS value of the current and voltage (2) 2.3. The power dissipated by the circuit (2) 24 The energy dissipated in the circuit during 2 milliseconds qt) 2.5 The power factor of the cireuit Q) [17] TOTAL: [35] QUESTION 3 3.1 Explain, with the aid of a neat diagram, how a short-circuit test is carried out on a single-phase transformer, (5) 3.2 A500 kVA, 11000/500 V, single phase transformer has its maximum efficiency at 75 % of full-load. The maximum efficiency is 94,7 per cent at a power factor . of 0,85 lagging, Calculate : 3.2.1 The iron losses (6)) 3.2.2 The full-load copper losses (2) 3.2.3 The full-load efficiency at 0,85 power factor lagging (3) 3.2.4 The percentage voltage regulation at full-load and at unity power factor ; (3) [18] Downloaded from hlayiso.com

QUESTION 4 4.1 Explain armature reaction in alternators. (3) 42 Explain the term pitch factor (2) 43 A [500kVA, 11 kV, 50 Hz, three-phase, star connected alternator has a resistance of 1 ohm per phase and a synchronous reactance of 20 ohms per phase, Calculate the percentage regulation when it delivers full-load at a power-factor of 0,8 lagging and at normal rated voltage. (8) [13] QUESTION 5 5.1 Explain, with the aid of phasor diagrams, what happens when the load torque of a synchronous motor is increased. (5) 5.2 The line current taken by a three-phase, star-connected, synchronous motor is 190 A. The applied line voltage is 11 kV. The excitation of the motor is such that it produces an EMF of 13,5 kV, The impedance of the motor is (5+]j 20) ohms. The load angle is 30° electrical. Calculate the power factor at which the motor is operating. (9) [14] QUESTION 6 The following are the results of a test on a 30,31 kW, 500 V, 50 Hz, three-phase, six pole, star-connected induction motor : No-load test : 500 V 1oA 1730 W Locked rotor test : 150 °V 42A 3273 W Draw the circle diagram using a scale of 7 A= 1 cm. If the stator resistance per phase is 0,295 ohms, determine : 61. the stator copper loss at standstill ; (10) 6.2 the rotor copper loss at standstill (i) 6.3 the maximum torque in N.m (i) 6.4 — the slip at maximum torque () 6.5 the power factor at maximum torque qd) [14] Downloaded from hlayiso.com

QUESTION 7 Two 6,6 kV, star-connected alternators operate in parallel and supply the following loads : 400 kW at a power factor of unity 400 kW at a power factor of 0,85 lagging 300 kW at a power factor of 0,8 lagging 800 kW at a power factor of 0,7 lagging. . The armature current of machine one is 100 A at power factor of 0,9 lagging. Calculate the power output for machine twa. : (6) [6] TOTAL: [65] TOTAL MARKS = 100 Downloaded from hlayiso.com

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ELECTROTECHNICS N6 Memo “V2 -2ar i QUESTION... ee 1.1. Constant losses I. Friction and wind losses v Jl, Iron losses v IH. Shunt field copper loss % Variable losses J. Brush contact loss * I. Armature copper loss v 4) 1,2 Above normal speed : decrease the field current by inserting a resistance in the field cirouit. vw Below normal speed : decrease the voltage across the armature by inserting a resistance in the armature cirovit, VV 4) 13 V=500V tysla=lq=100A N,=S00rmin = -R, = 0,10 Rp = 0,05 0 Ine = 0,7 Jaa R, = 0,10 Oren f= BE BV ba GREE Re) Y No = “om 1 eee Y= 500-70 (C6 Sr +01) E, =V—las (Rat RD =46667A VY =490667V WW “= 500 — 100 (0,1 + 0,05) ¥ =485 V v Na = 490,667, 100. 500 46,667. 485 = 1083.94 r/min V (10) {18] Downloaded from hlayiso.com

QUESTION Z 150 Sin 314t + 70 Sin ( 942t + 60°) V R=9500 C=3,185 pF . ni. . _ . 1 Fa 950450 Zp = 0 ~j soe 3,185 X10 = 950 |0° 2 =0—j 999,91 — = 999,91 |-90° 0 = Ym, = Jaébs AT Tmt = ty Za t Zox 150 {0° y 980 |o*, 999,91 | -90° 950+) 04+ 0-j 999,91 ~ 688,72 |—43,53° __949914,5 |-90" C - S499285 |=908 . = 0,218 |43,53° A V = Sprazs [-a8a = 688,72 |-43,53° 0 Vmg Jing = 2g Za. Bos 70 [60° > 2, y 3 Lat Za 314,508 | -70,67° 950 | 0°. 333,303 | —90° = 0 t = 200 FO See 0e | 0.223 | 130,67" 4 * 950+) 0 + 0-j 333,303 _ 316637,85 |-90° 1006,773 |-19,33° = 314,508 |- 70,6772 1 i= 0,218 Sin (314t + 43,53° ) +0223 Sin (942t + 130,677) AV (10) 22 ve [Ya Vins” I= fi” bua” , - 2 2 _ {1502-4707 _ | 0,21.8?-+ 0,223? J a J 2 V =0221A V (2) = 117,05 V Downloaded from hlayiso.com

72 230 Pe & R _ 147,08? . 350. ‘ =14422W Vv (2) 24 Energy =P.t = 14,422.2. 1079 = 0,0288 joules ¥ 3) C 2.5 -cosp= rar 14,422 ( = 71 417,05 0,224 =0,56 leading 9 (2) [17] TOTAL [35] PARTA © Downloaded from hlayiso.com

ELECTROTECHNICS N6 Memo QUESTION 3 . toe cee ee eee eee ee i Fee 341 C) {oN mor | oor x The secondary winding is short-cirenited through a suitable ammeter, A low voltage is applied to the primary side. This voltage is varied until full-load current flows through the primary and secondary, WW (5) 3.241 S=SO0kKVA V,=11000V V, =500V Nnax = 94,7 % k=0,75 cos = 0,85 lagging _, _k.S.cosd, 100 y= k, S. cos@ + 2(Pq) 0,75. 500, 103. 0,85. 100 ¥ 0,75. 500.103. 0,85 + 2(P9) 94,7 = 0,947 [0,75. 500. 10%. 0,85 + 2(P9)] = 0,75. 500. 10%. 0,85 V 301856,25 + 1,894 Py = 318750 ¥ 318750 -301856,25 | 2 1,894 = 8919,615 W - ~$92kW oY (5) k?. Peo Ke 3.2.2 Pye = _ 8,92 0,75? = 18,858kW V (2) Downloaded from hlayiso.com

_ S cos @ .100 fl 3.23 = S$ .cos d+ Pot Pe x 100 ‘ _ 500, 0,85. 100 ' 500.0,85 + 8,92 + 15,858 oy -oooe 7 = 94,49 % ¥ V, - 3.2.4 Reg = Ee) ¥ 100 Ik at 4 Vy V. 403 0 = ‘SC1. COS be =n? =: 300. 10 “Reg vy & 100 (pz = 0°) 11000 P P = LY 100 iL = Vye1. cos de) = 45,455 A — 15.858. 10°. 100 m 45,455. 11000 =3172% ¥ QUESTION 4 4.1 Itis the effect of the stator ampere turns on the value and the distribution of the magnetic flux in the air gaps between the poles and the stator core. When the alternator is loaded, the magnetic effect of the armature current has two components: the component which distorts the main flux is called armature reaction; the other component is known as leakage reactance, 4,2 The pitch factor is the ratio of the actual e.m-f to the em.f, that would be obtained ifall the coils were full pitched, 43. S=I1500KVA Ver ALkV Rpp=tQ Xs= 200 cos = 0,8 lagging ‘f= 50 Hz E-V s % Reg = —— .10 = % Reg Vv 0 i Eu Vv h= 23090 ' _ 1500. 108 _ . Zg = 14520 =F a0 = 6350;853 V ¥ 20,025 |a7d4° QV =A VY Downloaded from hlayiso.com Q) (13) @) Q)

Eps = V +12 = 6350,853 + j 0+ 78,73 | -36,87° . 20,025 | 87,14° = 6350,853 +j 0+ 1576,568 |50,27°. ¥. . = 6350,853 + j 0 + 1007,696 + j 1212,483 = 7358,549 + j 1212,483 =7487,772 |9,362V B, =i12917¥ ¥ wReg- MBE yoo ¥ at = 17.43% V ®) {13] QUESTION 5 5.4 1 Ey »Vv E Er Li. Vv v E Er | v ] v When the motor Is synchronised, E is equal and opposite in phase to V. Vv When the driving torque of the motor is removed, E is retarded by an angle but remains equal V. Er now lets a current | flow in the circuit which exerts a driving torque on the motor, v The load torque is increased, E retards further, Er, land $ increases, but the motor continues to run at synchronous speed, v {5) 52 4,=190A VW =11kV BL=135kV 2,65 45200 a = 30° E; po Aer Downloaded from hlayiso.com

See graph (5) Z, = 20,616 [15,962 AV Eph = “z Von = % = 7794229 lise v oY = 6350,853 [oy ¥ - =- 6750 + {3897115 ¥ ¥ = 6350,853+j)0V R,= V+E = 6350,853 + j 0 - 6750 + j 3897,115 = - 399,147 +] 3897,115 Vv = 3917,502 [95,85° V ¥ = 95,85° — 75,96° = 19,89° y cos b = Cos 19,89° = 0,94 leading v¥ 9) [14] QUESTION 6 Poutput = 30,31 kW V, = 500V f= 50 Hz Ip=10A Vee = 150 V Ine 42A Py =1730W Pi, = 3273 W Ron = 0,295 Q Scale: lom=7A cosy = eh COS Pge = Bon = 1730 = 3273 ¥3. 500. 10 ¥3. 150. 42 =O2Jagging V =O3 lagging V I. at normal voltage = a Lee: DF = a =140A V = 2,86.om v Downloaded from hlayiso.com

6.1 Stator cu losses at standstill = /3.¥,.DF.scale = V3 .500.2,86.7 = 17337,829 W =17338kw oY 6.2 Rotor culosses at standstill = /3.V,.BF. scale = 3 500 .2,6.7 = 15761,662 W = 15,762 kW. ¥ n __ Rotorinputat Tipax _ Rotor cu losses at Ty 63° Tmax = Qe. hy 64 Som = “Roterinput at Tn ty = : = = 16,667 r/min = © 100 _ V¥3..V,.LN. scale — 42 ‘Tmax 2,7. ny ~ gaa +100 ¥3 . 500. 8,28. 7 _ = woe = 14,569 2.7, 16,667 156%, \ = 479,316 Nm v co 6.5 cosd=0,72 lagging [14] Downloaded from hlayiso.com

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10 QUESTION? Load : $1 = 400 kVA = 400 | 0° PA PB Pi=Si . =400+{0 kVA | | | | | s2=22 = 470,588 |-31,79° : = 400 - 5 247,909 kVA ¥ SA SB $2 83 S4 ~ s3 = = 375 |-36,87° . = 300 - {225 kVA SA=93.¥,-h S4= 52 = 1142,857 |-45,57° = ¥3 . 6600. 100 , = 800,043 — {816,121 kVA_ ¥ = 1143,154 kVA SA = 1028,855 - j 498,255 kVA ¥ SB= (S1+82+83+84)-SA = (400 + j 0+ 400 ~j 247,909 + 300 —j 225 + 800,043 —j 816,121) - 1028,855 + j 498,255 = 1900,043 —j 1289,03 — 1028,855 + j 498,255 = 871,188 —j 790,775 v SB =1176,56 |-42,23° kVA oy PB=SB.coso = 1176,56 . 0,74 = 8712 kW v (6) TOTAL PART B (65) TOTAL (A+B) [100] Downloaded from hlayiso.com

INSTRUCTIONS: (6 Northlink College oe THE EDUCATIGN CONNECTION CORRESPONDENCE STUDIES ELECTROTECHNICS N6 Answer ALL the questions. Read ALL the questions carefully. Number the answers correctly. Start each question on a NEW page. Keep subsections of questions together, Write neatly and legibly please. AwWswNe QUES TIONS Stu) Gunoe. To- ZO QUESTION 1 Lt 12 Give a brief explanation of why the current supplied to a d.c. motor increases when the motor is mechanically loaded. (65) A 45 kW, 450 V d.c, shunt motor takes a current of 4 A when running light at a speed of 620 r/min. The resistance of the armature circuit ( brushes included ) is 0,25 ohms and that of the field circuit is 265 ohms. If the motor is fully loaded, calculate: 1.2.1 The input current : (6) — 1.2.2 The speed (4) 1.2.3 ‘The armature current when the efficiency is a maximum (3) . [18] Downloaded from hlayiso.com

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QUESTION 2 2.1 A three-phase, four-wize, star-comected load consists of the following: R-N : 450 A al unity power factor Y-N : 100 kW at 0.8 power factor lagging B-N : 115 kVA at 0,9 power factor leading Take a phase sequence R-Y-B and a line voltage of 433 V. Calculate the current in the nentral wire and its phase angle with respect to Vay. (8) 2.2 The current flowing in a circuit is represented by: i= 5 Sin 314t+ 0,5 Sin 942t(A) and the voltage by v= 180 Sin (314t +5) + 20 Sin ( 942t + 2) (¥). Caleulate the following: 2.2.1 The total power supplied (3) 2,2.2 The RMS value of the voltage and current (4) 2.2.3 The power factor (2) (a7) TOTAL: [35] QUESTION 3 3.1 Explain how the constant losses in a transformer are Kept to an economical roinimum. (3) 3.2 Name auy other losses which occur in a transformer and state how this losses varies with the load (2) 3.3 A 150 kVA, 2000/400 V, single phase transformer has a primary resistance of 0,17 ohms a secondary resistance of 0,0084 ohms. The iron losses is 1,5 kW. Calculate the following: 3.3.1 The equivalent resistance referred to the secondary (2) 3.3.2 The full-load efficiency at 0,8 power factor lagging 4) Downloaded from hlayiso.com

power factor @) [16] QUESTION 4 4.1 Define the regulation of an alternator. (2) 4.2 Define distribution factor of a winding (2) 4,3 The armature of a 12 pole star connected three phase alternator having a flux per pole of 0,05 weber, has 144 slots. There are four conductors in each slot and the coil pitch is 0,75 of the pole pitch, If the alternator runs at a speed of 500 y/min and the form factor is 1,17, calculate the open circuit line voltage. (8) [12] QUESTION 5 , 5.1 Show with the aid of phasor diagrams that the power factor of a synchronous motor working on a constant mechanical load depends on its excitation. (5) 5.2 A.200-kVA six pole, 2,3 kV, star-connected synchronous motor has a synchronous impedance of ( 8 + j 56 )%. This motor is fully loaded at a power factor of 0,8 leading. Calculate the following: 5.2.1 The ohmic values of the phase resistance and reactance. (4) 5.2.2 The EMF to which the machine is excited. 4) 5.2.3 The load angle in electrical and mechanical degrees. @) [16] QUESTION 6 3.3.3 The maximum efficiency at 0,85 power factor lagging Q) 3.3.4 The percentage voltage regulation at full-load and at unity A 480 V, six pole, three-phase, star-connected induction motor has a rotor impedance of 0,16 +j 1,1 ohms per phase at standstill. The standstill EMF between slip rings is 290 V. Calculate the following: 6.1 The torque developed at a full-load slip of 5 per cent. (6) 6.2 The full-load power output if the friction and windage losses are 880 watts. QB) 6.3. The speed at maximum torque (3) . [12] Downloaded from hlayiso.com

QUESTION 7 A three-phase induction motor with an input of 500 kW at 0,75 power factor lagging is connected in parallel with a three-phase synchronous motor with an input of 300 kVA at 0,95 power factor leading. ‘The supply voltage is 3 200 V. Calculate the following: 7.1 The line current of the total load. (8) 7.2 The power factor of the total load. (1) 19] TOTAL: [65] TOTAL MARKS = 100 Downloaded from hlayiso.com

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(s) W OL] See SEE mY esto f+ psy'06e == 87897 f+ STORE — GOP'9GI ~ 8086S —-0 f+ OSt = NOy op My of ME = Ny VSCSET T+ STORE ‘V 60961 1 B03 6S - = VPSTTY] 9b VolSDET-T 00S = oS STF OZE] OOP = oS 98 — OEl-| 00S = Ost a S0'08s g0T STE 20%" 0OT. Nog, e500" SAA my = Vorros= ty v ol ogr = Ny sme = Sy Aost= Lay Actee A tt TNOLSaIO sx] ©) VEST = szo yrosy} = Az:4 oral *4 SOSSO} O[QULIEA = sesso] 1uersvoo Aousionra XBU IY CTT ®) ASLECGPR = WE SSS STO" 68°801 - OS = every ig cgejet wes ecrzey = N RET As OF {1 Ne APP GRD = STO CET— Sh = Vauet= ey As Z e i S671 b= Rey = ON trom _ ta — 0% a pe 1 — Of Ory mou Ry CEE 9) VtiiH Tt 869'T +68°301 = T VRB Ht — SSS’S6c — Ost ez moayprae 0 = G78'SE09) +81 OSE ~ PI STO BISTO+ OSPXY + OLX Sa Ze +L OS VY LE] + O] A+ MOg = (ust + OE) A, ‘SPSSO] D[QUUZA + SOSSO] JUBPSUOD + Meg = Nduig Ystt Cle CET G89 US USTO mea HUE 029 ="N Ven 0 ANSP A MASH smog TT @ “manms un aseazout ue Spusabssuds pur ‘PL Ur Ssecusep © OL yNSOL yoIyAA Saseorosp posds stp “poproy St soJOUL sUp LOA w Camessuoo st & sumssz) Nw» F Fel a (q) powsouad sur yoeq —(q) () x) sourisisxumono samme —_(e) suodn spusdp soyour oy Aq uavexp quasi YET TNOLESING OMSK INSOINDOTLONIO GTS Downloaded from hlayiso.com

22.1 22.2 i= 5 Sin 314t+0,5 Sin 942t (A) v=180Sin (B1dt+Z) +20Sin{9420+5) M4. in nn a 200. S$ cosd 20.08 cosp . 2 + 2 =433+25 Py =A3S,5 watts, @} Vena ?+ Vina” ve ax tas _, [200% #207 “ 2 may) 23552 @ P cos a 4385 1A2,227 3,553 = 0,362 lagging {2} TOTAL 3] PARTA ELECTROTECENICS N6 Memo QUESTION 3 31 32 33 Constant losses: A. Core losses ~ which can be reduced: {By using high quality core material with low hysteresis loss T. By Jaminating the core. : B. Dielectric losses — which can be reduced by using insulating materi that is non-hydroscopic and has a high dielectric strength. @) Variable losses: i. Copper losses ( load * I. Eddy current iosses & ( load )* Q) S=ISOKVA = 2000V 1 =400¥ 2, 50,172 R= 0,008¢0 2 331 Rey = Ret Ry 2) 2 0,084 + 9,17 (22) = 0,0152. 0 2) $.cosd 400 332 15 3.6050 POF Pag b= = 250. 20. 0,8. 100 = 250-208 150, 105, 0,8 445004 (375%, 0,0152) 400 = 97.06 % =SI5A ®@ aan _- _KS.c0s@. 100 P35 TS cose uP) mz 0838 450 10% 085 409 0,838, 150. 105. 0,85 +2( 3500) = 97.27 % = 0838 8) Downloaded from hlayiso.com

@ ABET ILE = 39 A OLSTIT TTS = ESET IVTST Carts = oor “gh aoEe 9595 ~ oor . versie I Zia sz |, AoASTRT 6959S =Z% AwWs8 cus ®) OTUs = Ep 40z'08 ‘00% 008 “9S Ligot XH” FRwOcts = OSTT = oosz “gh EN ‘0x08 “oot 50% 02 o0E “8 = x 3Ash Toot AS a1 nas > Yo res Sulpes| 80 =o sop. w9s!+3)=Z% METSU fed YANOOTH=5 vse (9) “ATpSTOLYNS PaseasOUl St uOLMAOND It “pes] 03 udm WwEIIASS) 3} SosMeD UE J SaseaJONT UOHEIIOXS uy aseasouy Surpeat y Bucy] y I A A—_— a ae “a az Wh vs FNOLISHO 9 Ia] ® Revel = (pre 08 °80 “a56'0 “Fz6'0 “att “z) EM =a Reo FEO = as + z eed — ae 5rS0O™ =, . — = Py : 2509 = ty was & t= Fa tt £°2E “AOS = zt oat -_” og Os = saseyd ssajod jra=s sojed “gat =? — sr Cie Open z) gp a AUT = PI WIL GOS = N Pah osT's70 =a youd sod x ¢z‘9 = qoud fog FH MOIS/PUCD — FHT = SOs WSV=o asd ged Cy @ ROIS 9U0 UE PerequsqUoS Tuam SOPES [100 98 eZ poutAgo 9q pom jem Furs oy O Surpuum @ Ur peompur Yura ou uosmag OMe: 21 St 10:98 UORRGLASTD Ou Ty &) “Wisuoo surewas tusamo pray pur psods au opm ‘SUORIPTOS peo]-|[ny pue PRO]-OU UDeMteq SSeTOA feUTUIO, Urssuayp EL Ly FNOUSHTG (ou @ r= 00% Cot (o* zsto’o) sz Az rE erT aE BUN YEE $ Downloaded from hlayiso.com

E =-361,196 +} 658,645 ~ 1327.906 +7 0 = + 1689102 +j 658,645 = 1812,975 115872 V E = 3140.17 Vv. 523 Getege = 180°- 158.7" meer = 5 = 215° electrion, =38 = 7,12 mech @ (6) QUESTION 6 We 480V p=3 2 =O16+j11 Ey = 290V $=0,05 61 T= Rotorinput Rak ay ‘ Rotor cu losses Rotor input nr 83. Roe 5 $.E9 yj, ate ee 0.08 25? 10% p.t64j 0.05. 24 = 8372 10° 9169 148,97 * Ip 74988 1-18.97 24 3, 4948". O16 ‘Rotor input OnE = 23503.4 Watt n= win m2 3 = 16,667 r/min pwn 228034 22%. 16,867 T - 224441 Nm ® 6.2 — Friction & wind losses » 880 watt P output = Rotor output — Friction & wind Rotor output * Rotor input (I -S ) = 93503,4 (1~ 0,05) = 22398.23 watt 22328,23-860 1000 P output = 21,45 LW @) 63 At Tox Stm- Xp 7 Ra Ry Sra Pouput = O26 at a Sra = 255 Nem 27 CL Spud 60 = 16,667 { 1+ 0.145 } 60 = 854,545 r/min ®) 2} Downloaded from hlayiso.com

too) so] ‘6 @ @® ‘TWLOL TLavd TWLOL SUDSUT SIO O = 98'ET SOS= p50 ZL VESTS = ooze sh s0T "6res3 ~ tnegn 1 *s Vato o8Se7] CEESS =S VAN 8CLeE I~ SEL we $L9'C6 [+ 98% + TIS OPE S-OOS = a's = tg VAN SLYSG T+ S8Z ms VAT. OTST] OE = *5 VAT O60 T- WE = VAACIT TFT PO = a0 oes s02 ay a's A0KE= TA Bupeat cg'g=G soo WAX QOE™ % Suse ec'g=Psoo MA COS= ETL TNOLMSANO Downloaded from hlayiso.com

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