ELECTROTECHNICS N5 28 Jun 2017 12 54 06

SPOT ON ELECTROTECHNICS N5 QUESTIONS AND SOLUTIONS Downloaded from hlayiso.com, ag 00

= =. higher education =—= = & training — " a Department: ey Higher Education and Training pn REPUBLIC OF SOUTH AFRICA T500(E)(A7)T APRIL EXAMINATION NATIONAL CERTIFICATE ELECTROTECHNICS N5 (8080085) 7 April 2015 (Y-Paper) 13:00-16:00 Calculators may be used. This question paper consists of 5 pages and a formula sheet of 2 pages. Downloaded from hlayiso.com Copyright re, “Served Please turn over

808085) nm “BOUE (AT J DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ELECTROTECHNICS N5 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 the numbering system used in this question paper. 4. Write neatly and legibly. Downloaded from hlayiso.com

QUESTION 1 i 1.1 Brief “ tape, asthe why the terminal voltage of a DC shunt excited generator : Current supplied by the machine is increased. (2) 1.2 What is the functi Fee: | tion of th i i i a DC machine? commutating poles and compensating windings In 2 1.3 An eight pole DC motor has a wave connected armature with 800 conductors The brushes are displaced through five angular degrees from the geometrical axis. The armature current is 270 A, Calculate the following: 1.3.1 The demagnetising and cross-magnetising ampere turns per pole. (4) 1.3.2 The additional field current required neutralising this demagne- tisation if the field winding has 1600 turns/pole. (2) 1.4 A 535 V shunt motor draws an armature current of 40 A while running at 550 r/min. The armature circuit has a resistance of 0, 5 ©. If the magnetic flux is decreased by 30 % and the torque developed by the armature increases by 40 %. 4 . Calculate the following : 1.4.1 The armature current. (3) 1.4.2 The speed (7) : [20] QUESTION 2 2.1 Two circuits are connected in parallel to a 300 V, 50 Hz supply. The total current taken by the combination is 25 A, at unity power factor. Circuit A consists of a 8,80 resistor and a 200 uf capacitor connected in series. Circuit B consists of a resistor and an inductive reactance in series. Calculate for circuit B : 2.1.4 The current (4) 2.1.2 The power factor (2) 218 The impedance (2) 2.1.4 The reactance (1) 2.1.5 The resistance (1) Copyright reserved Downloaded from hlayiso.com Please turr. over - .y

A constan consisting t voltage at @ frequency of 2.5 MHz is applied across a circuit an inductor in series with a variable capacitor. When the ca capaciten iS set to 370 pf, the current is at its maximum value. When the nce is reduced to 340 pf, the current is 0,707 of its maximum value. Find the inductance and the resistance of the inductor. (10) [20] QUESTION 3 3.1 A three phase delta connected load, each phase of which has an inductive reactance of 75 Q and a resistance of 50 Q, is supplied from the secondary of a three phase star connected transformer, which has a phase voltage of 260 V. Calculate the following: L 3.1.1 The potential difference across each phase of the load. , (2) 3.4.2 The current in each phase of the load. \ (2) 3.1.3 The current in the transformer secondary windings. (1) 3.1.4 Total power drawn from the supply and its power factor (3) 3.2 A 15 kVA, 3500/700 volt single-phase transformer, operating at no-load, 4 has resistances and leakage reactance as follows: Primary winding: Resistance 7,5, reactance 150 Secondary winding: Resistance 0,5 Q, reactance 0,65 0 Determine the approximate value of the secondary voltage at full-load, with a power factor of 0,8 (lagging), when the primary supply voltage is 3 SOOV. ; (4) 3.3 A three-phase, 500 V, star-connected motor has an output of 100 kW, with an : efficiency of 80 % and a power factor of 0,8. J fa Calculate the following: ; / oy 4 } \ 3.3.1 The line current “(. ‘~ J (2) 3.3.2 If the motor windings were connected in delta, what would be the correct voltage suitable for a three-phase motor? (2) 3.4 \f the phase voltage of a three phase star connected alternator is 360 V, what would the line voltages be when: 3.4.1 The phase is correctly connected. (2) DOWARSSUCe Hoth alayisestom 20} Pp) ase turn ove! 3.4.2 Copyr'gnt “eserveg

(60 Ge, ~5U0(E A” 7 QUESTION 4 4 pereuiate the inductance and capacitance per phase of a 35km, three F phase overhead line having solid copper conductors of diameter 0,6 cm . that are spaced on the corners of a triangle having sides of length t 125cem, 165¢m and 235cm. (6) 3 4.2 An overhead, single phase transmission line delivers 2 200 kW at 31 kV. The F power factor is 0, 85 lagging. The total resistance of the line is 22 © and the total inductive reactance is 300. Calculate the following: e ie 4.2.1 The sending line voltage (4) ' 4.2.2 The per unit regulation (2) 4.2.3 The transmission efficiency (2) 4.3 Each branch of a three-phase star connected load consists of a coil of a resistance 5 Q and reactance 6 Q. The load is supplied at a line voltage of 400V, 50 Hz. The power supplied to the load measured by the two wattmeter method. Calculate their separate readings. (6) [20] QUESTION 5 5.1 A three-phase star-connected alternator supplies a 650 kW, 3,8 kV delta- connected induction motor, with an efficiency of 85 % and a full-load power factor of 0,8. Calculate the KVA output of the alternator and the value of the current in the alternator and motor windings. (7) 5:2 A three-phase, 50 Hz induction motor has 4 poles and runs at a speed of 4000 r/min when the total torque developed by the rotor is 170 Nm. Calculate the following: 5.2.1 The total input power to the rotor (4) 5.2.2 The rotor copper loss in watts (3) §.3 Determine the number of stator conductors per slot for a three-phase, 50 Hz, areata, if the winding is star-connected and has to supply a line voltage of iseueen the machine is on an open circuit. The flux per pole is 0, 4 wb. — seed ae coils and the stator to have three slots per pole per phase. is 375 r/ min and the distribution factor is 0,96. (6) [20] D TOTAL: 100 Ccopynight reserved ownloaded from hlayiso.com

Gos higher education | slee & training a Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA MARKING GUIDELINE | NATIONAL CERTIFICATE | APRIL EXAMINATION | ELECTROTECHNICS N5 _.. TAPRIL 2015 | This marking guideline consists of 9 pages. | = 5 r Downloaded from hlayi atom a we oo" fea

a ins | SR QUESTION 14 1.4 Due to the resistance loss in the armature circuit, ¥ the shunt current also decreases causing a further terminal voltage drop. “ (2) 1.2 Commutating poles improve commutation. ~ (1) Compensating windings minimize armature reaction. “ (1) 1.3 1.3.1 p=4c=2Z = 800 mech deg. = 5° and Elect deg =p x5 = 20° la = 270 A Field winding = 1600 t/p Demag At/p = 0.5x la/c x Z/2p x ( 40 / 360%) [soo auinv> =0,5 x 270/2 x 800/8 x( 4 x 20/360) “~ =\15 Cross-magAt/p =0,5x270/2x800/8x[1-(4x 20/360) ] v = 5250 Atv 42. (2) Sup 1.32 MMF = NIiv | = 1500 / 1600 = 094A v 2) 1.4 144 V = 535 Vla=40 A N12 550 rpm Ra =0,5061= 100% (1) 62 =70% (0,7) T1= 100% (1) T2 =140 % (1,4) (2) tit - Ie % 270X700 y+ T1/7T2 . = lai o1 / ta2z 2 v —— : , a = 14x40x1/1x0,7 v = 80 A“ (3) 1.4.2 =a =V—laRa = 535 — (40 x 0,5) # =515V v E2 =V-laz Ra = 535— (80x 0,5) ¥ = 405V v E1/E2 = N1¢61 / N2 $2 N2 = E2N1 01 / E1 a2 v = 495X550X1 / 515X007 Vv 7 = 755.2 rpm ¥ 20 Downloaded from hlayiso.com

QUESTION 2 2.2 yrignt reserved 2.1.1 ZA | x fo} Bj (12xfe) B—j (1/22 50 x 200 x 10%) 8} 15.92 nono tt £8 go ce 0 18,19 <-61,07 ¥ VIZA 300< 0/ 18,19 <- 61,07 16,49< 61,07 A mow oa IT- Toei - 16,49 < 64,07 25 +j0-(7,98 +)14,43), ¥ 47,02 -j14,43" 22.34 <-40,29 AiY 4 ily 2.1.2 Cos ¢ Cos 40:29 (2) iN} 201.3 ZB VB /IB: 300< 0 / 22,31 <- 40,29 ¥ @ 13.45 < 40,29V = (10,26 + 8,7) ) - il ie " ZB 2.14 XL = B TOY (from ZB = 10,26 +487) % ™ 2.1.5 R iT 40, rh BAM (from ZB = 10,26 +) 8,7) C1 =370x 10-!? for (Irnax) : Co = 340x101? for le = 0,707 Im * ; \ \_ xCr= 1/2nfo- oO = 1012 (20 2,5x108x 370 = 172,06 2¥ xe, ANObLLY XL) =XC1=172,06 resonance v Mia E12 i772 1.08 a XL: =2nfb rae i areae L =XL/2xnf = 172,06 /2 2x 10% 2,5 £ 40,95 yH EN XCz=1/ 2nfC2 = 107 | on x 10°x340x2,5 = 7,242 LH. © eo Increase in XC = 187,24 ~ 172,06 = 1518Q¥ V2 =V/Z Z1=R (resonance) v pyica ® a = \0 , a5 X10 © Downloadell fram hlayiso.com = **** turn over | tad f Yc + iS) P= 10, 1S Ug Not ge 7

a Oryole i L Alias Ade aor = Lt se US MFKInG GuireLine 9, - ©} -. q2 \ L ELECIRUTECHMICS Nz uy | vo l2 = 0,707 iy = O.707V/Z1 = Wr Z2=VR2+(15,18)2 v 0,707 V/2Z, = VIZ O7O7V/IR = VIVR? + (15,18) (0,707 V)? / R? = V2/ R2 + (15,18)? [R? + (15,18)2]/R2 = V2/(0,707V)2 [R? + (15,18)2]/R2 = 1 / 0,7072 R2 + (15,18)? =R?2/0,7072 v 0,7072 R2 + (0,7072)(15,18 2) = R? . R2 — 0,7072 R2 = (0,7077)(15,18 2. R2 (1~ 0,7072) = (0,707)(45,18 2) R2 = (0,7072)(15,182) / (1- 0,7072) R = 15,18 v. Downloaded from hlayiso.com

QUESTION 3 3.1 Transformer load & STAR | vp = 260 \ 0 VL=vp 4 N a= 50+ joe" ( Z= 90,14< 56.34 3.1.1 _ _ VL=ev3VP = V3 x 260 =450.33V ¥ = VP = 4 2 » 43 50,33 v (2) IP(load) =VP/ZP : = 480,33 < 0/ 90,185,56,31 v =5 <-56,34°% 3.1.3 ; F (2) aad IL =V3x5 SB gC AY (A) P = V3 ILVL'Cos 4 = V3 x 8,66 x 450,33 X 0,55 v = 3,72 kW (2) Cosh = = C0s56;31 of” & 0,55"140% () 3.2 single ph Trt. 15 kVA ware = 35001 700 = Ni NE Ra1=7,52 R2= 0,52 X1= 152 X2 = 0,652 vit = 3500V Cos 6 = 0,8 = 15000 / 3500 . =4,29A Re Ri +R2(N1/N2) = 7,57 0,5( 5) = 202 ge (Kl 2x2 (N1/N2)? = 15+ 0,65 (5) # 31,25 2 ne Je = J (Re) 2+ (Xe)’ aH = J (20)? + (31. 25)? = 37,1 Le roceryved D ownloaded fronvhlayisé.com niease tin ove"

ee Bee TN IS REG. =1;(Recos ¢ + Xesina)/Vi = 4.29 { (20)(0,8)+(31,25)(0,6)] 7 350 = 4.29[16 + 18,75 ]/3500 1 = 0,043 per unit v FULL V2 = NO LOAD V2 - CHANGE (700) —(N/L V2 x reg ) 700—( 700 x 0,043 ) 700 — 30,1 669,9 Vv noe wou 3.3 3.3.1 3 PH. Star con motor VL = 500V Pout = 100 kW = 08 COS 6 = 0,8 : Pin = 100/08 0 = 125 kW Te Ao Ve P =V3 ILVLCOS$ & wh 125000 = Y3IL(500)(9,8) v = Vow ; IL = 180,42A Oy i A 4 FOR DELTA CON. MOTOR WINDING : a IL = V3 IP = V3 18042 = 312.5 v e\ P= V3 INL COS 6 _ 26000 + V3x.312;5° x V x08 VL = 288,68 V v 4 VRN < 90 a a UN VYN < - 30 VBN< -150 Vp =360 3.4.1 iil Correctly connected vRY = Vr + Vy Gn = 360<0 + 360<60 _ = (0 +] 360) + (180 + | 311.77) v = 540 + | 311.77 | = 623.53<30 v | VBY=VytVy =360<0+360<120 \ =1804)311.77 +(-1804j311.77) v \ 20+ 623.5 , S =623.5<90V Digase turn Over Downloaded from hlayiso.com

CUROTECHINICS NZ SOU EWAT - QUESTION 4 {20} a d1 = 125x107 ; d2 = 165x102 dia =0,6cm r =0,3x10-2m ° d3 = 235x102 de = *°V1,25x 165x235 =1,69m v L = km [0,05 + 0,2 loge (de/r) ] mH = 35[ 0,05 + 0,2 loge ( 1,690,003) mH v = 46,09 mH v Cc = 35 [1/18 loge ( 1,69 / 0,003) | =0,307uf v sah (6) 4.2 4.2.1 R = 22 ohms X = 30 ohms P =2200 kW Vr $33.1 KV. Cos @ = 0,85 = 31,79 | = P/VCos¢ = 2200000 / 31000 x 0,85 = 83,49A v , =V<g tIR+ Ix < 90 Vs = 31000 < 31,87 + 83,49 X 22 + 83,49 X 56326,7 + j 16367,8 + 1836.78 +) =, & \ \ 28163,48 G1s8725! an (4) 33902,1< 33,83 kvY u Wou Vs - Wr / Ve , 34900 / 33902 — 31000: 0,094 v 31000 ¥ (2) Wom ow 4.2.2 Reg + Losses Po/ Po 3492 x 22 v Y ~ 1000 it Downloades¢ from hlayiso.com (2)

Tooyright aL DELINE 43 Zz = _ E-ECTRO- ECHI Te =3*i6 = 7.81< 50.19 igi di =400/\3 L = 400V 6 = 50,19 lag = 23094 v IP =VP/Zp = 2309<0/781< 5 _ Y 0,19 = 28.56 <-50,19 v IL =IP = 20.56<-50,19 v . oy nnd U3 Cfe-Fi P2+ P1 =V3 VLIL COS 6 p 2+P, =V3x400x 29,56 COS 50.19 | S13 412KW veccceee (1) Po —Pi 2 Tand x Pi TANé = =V3(P2—P1) /P2+P1 = P2-P1 = TAN 50,19x13,112/ V3. v ‘ = = 9,08 KW ie, 2P1 = 4.03 5 P1 =2,02 kW P2> = 13,112 -2,02 =11,092kW ov Ry QUESTION 5 54 Po = 650 kW V=3800V-, n= 85% (0,85) Cos 6 =08 Pin =P out / 1 = 650000/0,85 Vv 2 = 76470588W ; IL = Pin/V3 VL Cos $ = 764705,88/V3x3800x0.8 Vv = 145,23 4 1 (alt) = 145,23 A ¥ = IL /V3 , (motor) = 145,23 1V3 = 83,85A v = V3 VEIL S = V3 x 3800x 145.23 ¥ = 955873,81VA 7) = 955,87 KVA’ v < = Nr = 1000 rpm T = 470Nm 6.2 5.2.1 fe50Hz pa4/2=2 NW P pis wal Otis from hlayiso.com = 1500 rpm Pieage turn Over reserved

eaRkhIS CUIDCL NE g : T50CE) 477 ELE OTROTECHNICS Va Rotor input P= 2TINsT/ 60 =211x 1500 x 170 / 60 v = 26703.5 W = 26,7 kW v (< 5.2.2 Rotor Cu Loss = S x Rotor input S = Ns-Nr/ Ns = 1500~1000/1500 =v = 0,33 v Rotor Cu Loss = 0,33 x 26703,5 = 8812 2Wv (3) 53 t= 50 Hz ¢ = 0,4 wh kd=0,96 N=375 rpm. VL =V3Vp Vp =20/V3 2s = 11,557 kV - v Vp = 2 kf kd kp of2X Nn Ut Vp L.2kf kd of A547 7-2x4,11 x 0,96 x 1x 0,4 x50 v = 270.63 conductors / ph i “= Np'/ 60 Pp = 60:£/ N =.60 X50 / 375 8 pole pairs = 16 poles v Slots / phase Fee sgn 3X 16 slots / phase” * = 48 slots / phase Pe Cond / Slot I (cond. / ph) / (slots / ph) 270.63 / 48 (6) 5.625(say 6 v (say conductors) [20] L Nn I> i= a © ao Downloaded from hlayiso.com

—— ————— ——_— —— ——= eh ———$— ———— —————. —_—SS= SSS = higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA T480(E)(N18)N NOVEMBER EXAMINATION | NATIONAL CERTIFICATE ELECTROTECHNICS N5 (8080085) 18 November 2014 (Y-Paper) 13:00-16:00 This question paper consists of 5 pages, and a formula sheet of 2 pages. Downloaded from hlayiso.com

(8C8G085 = TALO(E sD 18)N DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE ELECTROTECHNICS N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. fn ating din this Number the answers correctly according to the numbering system use! 3. question paper. 4. Keep subsections of questions together. 5. Write neatly and legibly. “Ye Downloaded from hlayiso.com

QUESTION 1 41.1 State TWO characteristics of DC motors. (2) 1.2 4.2.1 Calculate the number of series turns per pole, required on a Compound generator, for it to maintain a constant voltage at 630 v, between no-load and full-load of 450 kW. With no series winding, it was found that the shunt current has to be 6 A on no-load and 7, 5 A on full-load, to maintain the voltage constant at 630 V. Number of turns per pole on the shunt winding is 2 400. (5) 4.2.2 lf the series coils were wound with 8 turns/pole and had a total resistance of 0, 06 Q: Calculate the value of the diverter resistance required to give level-compounding. (5) 13 A 30,5 kW, 440 V, eight -pole DC motor has a wave-wound armature with 7 200 conductors , and the commutator has 150 segments. The full-load efficiency is 85% and the shunt current is 1, 5 A. The brushes are shifted backwards through 1,4 segments from the geometric neutral . Calculate the demagnetising and cross-magnetising ampere-turns per pole. (8) [20] QUESTION 2 2.1 A coil dissipates 1 800 W and draws a current of 30 when it is connected to a 280 V, 50 Hz alternating current supply. When anot r coil is connected in parallel with it, the total supply current is 55 A, at a pow factor of 0,85. Calculate the current and power factor when the coils are connected in series across the same supply. (14) a2 A circuit, with a resistance of 8,5 ©, an inductance of 0,7 H and a variable capacitance in series, is connected across a 180 V, 50 Hz supply. Calculate: aa The capacitance to give resonance. (2) (4) [20] 222 The voltages across the inductance and the capacitance. Downloaded from hlayiso.com Copyright reserved Please turn over

QUESTION 3 3.1 A 310 kVA transf former h . secondary. The primar 138 880 turns on the primary and 220 turns on the and the leakage aie secondary resistances are 0,8 Q and 0,06 Q voltage is 2 700 V. e are 1,7 Q and 0,07 Q respectively. The supply Calculate : cent x) 3.1.4 The equivalent impedance referred to the primary circuit. C (3) 3.1.2 Voltage regulation for power factor of 0,8 lagging, and the secondary terminal voltage on full load. _ (-) (7) aa 3.1.3 Voltage regulation for power factor of 0,8 leading and secondary terminal voltage on full load. (4) 3.2 A three phase transformer has 510 turns on the primary and 30 turns on the secondary winding. The supply voltage is 2 600 V. . Calculate the secondary line voltage on no-load when the transformer is connected in: 3.2.1 star — delta (3) 3.2.2 delta — star (3) [20] QUESTION 4 4.1 Determine the total resistance, inductance and capacitance of a single- phase, 33 km, overhead line with solid conductors of 1,4 cm diameter and spaced 0, 7 m between centres. Take resistivity of conductor material as 1,7 pQem and ignore skin effect. (6) 4.2 Calculate the inductance and capacitance per phase, of 35 km of three phase overhead line, having solid copper conductors with a diameter of 41, 5m. When the overhead line is: 4.2.1 spaced 50 cm between adjacent centres in flat regular spacing. (8) 4.2.2 spaced on the corners of a triangle having sides of length 55 cm: 85cm: 110 cm. (6) 4.3 Calculate the inductance per phase of a 200 km, three phase transmission line, havingan equi ca con fro. spacing of 10 m and a conductor diameter of 30M MLOaacea TF i om hlayiso.com [20]

(c0@9U0ES, on T480(E (34 QUESTION 5 5.1 5.2 5.3 A three phase i nas a vator eatin gage core induction motor, with a star-connected rotor 28 O The EMF Kstwecn Ht pel Base and at standstill the reactance is 1380 r/min. e slip-rings is 260 V. Full-load speed is Calculate 5.1.1 th i i e fractional slip (3) 5.1.2 the EMF induced in each phase of the rotor (2) 5.1.3 the rotor reactance per phase (1) 5.1.4 the rotor current and power factor (if rings are short —circuited ) (3) 5.1.5 the rotor frequency (1) With reference to induction motors, what is meant by the term slip? Explain why slip is necessary for an induction motor to operate. (4) A three-phase, 50 Hz, eight pole induction motor has a slip of 0,05 per unit, when the output is 45 kW. The frictional loss is 375 WwW. Calculate: 2 5.3.1 _ Rotor speed 2) (4) [20] 5.3.2 Rotor copper loss TOTAL: 400 Downloaded from hlayiso.com

.} higher education Le & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA MARKING GUIDELINE | NATIONAL CERTIFICATE NOVEMBER EXAMINATION ELECTROTECHNICS N5 48 NOVEMBER 201 4 This memorandum consists of 9 pages. Downloaded from hlayiso.com a please tern over

QUESTION 1 14 e e 1.2.1 1.2.2 ELEcTa ote - Tn ECTROTECHwI7.g NGS F430 E),N18) 7 ao Poles in the field winding x eating Magnetic flux. , reat ma SS due to heavy wound armature. (Any 2 x 1) (2) Ampere tur! ‘ Y p MS required on no load = 6x2400 = 14400 Ampere turns required on fullload = A : load == 75x 2400 = 18000 v ‘atone turns required on series field on full load = 18000 - j = 36 v Full load line current IL = pw 00 = 450 X 10% 630 =71429A v Constant field current lf = 6A ° , la = IL+If =720,29A v Turns required on series field ~~..= 3600 720,29 =5 v (6) With series winding of 8 t/p and resistance of 0,06Q Ampere turns required i = ae : ith 6 turns , series field current = mi = 450A v Z Current through diverter Rx = 720,29 — 450 =270A Vv Voltage across diverter = voltage across series field “ a (Rx) = If (series) x R (series) (Ix) = 450 x 0.06 ¥ Oo ds me 264.29 CX = = Downloaded from hiayiso.com ):. — (4)

MARKING SIC ELINE ELEC TROT SOriNICL NS 1.3 Pi = 30500 = 35882.35Ww v r- i] = 35882.35 v 440 81,55 A 81,55-1,5 80,05A iw o nouow Total At/p = laZ/4pc 80.05 x 1200 4x4x2 3001,88 v ii} re) = 360°. __—sx:~s shift xp segments = 360/150x 14x 4 = 1344 ag Atjp = At/ p x 49 Demag p 360 = 3001,88 x4x 13.44 _“ 360 = 448,28 Vv Cross- mag At/p = At/p - Demag Atip 3001,88 — 448,28 2553,6 Downloaded from hlayiso.com 1430. S)(N1E 4

mk TR Ay QUESTION 2 TECHIES yg m4 OIE 2.1 Pp _ =1V cos b cos = Da 280 Cos 4 p = PR lagging v R = Py = 180 = Cos = R/Z Or 202 = 459 v Zi = Ricos¢ = 4.5 /0,324 - Zt s44< 71428 » 42 4 OR 21 = VA = 280<9 20 < -71,28 =14< 71,28 =4,49 +) 13,26 PARALLEL i = 20<-71,28 = 6,41 -j 18,94 IT = 55 <- 31,79 = 46,75-j29 v l2 SIT-h v = (46,75 —j 29) — (6,41 -j 18,94) = 46,75 —j 29-6.41 + j 18,94 v = 40,34 —j 10,06 = 4158<-14 A v Z2 = Vik~ = 280 _< 0 = 41,58< -14 v =673 < 14 v = 653+ J1,630 SERIES F ZT = 21+22 = 4,49 +j 13,26+ 6,53+ J 1,63 = 11,02 +j 14,89 = 1852<5352 | =wzT = 280<0 18,52 < 53,5 % = 15,12< -53,5 A ; v - lagging. 14) cos (Downloaded fréini filayiso.com Pleas? tun evel l= 4512A Coy. PYINE reserve

SLY STROOS Deni 2.2, 224 SNS T48. (E)IN3 2nflL=4/on¢fe )T G = 4 (2 x nx 50)2 x = 0,7 14,47 we v Y 22.2 | - yy) R (2) = 180/85 =21,18A v V = L 2x 0x50 x0,7 x 24,18 v = 4657.73 V Vv VC =VL= 4657,73V v (4) QUESTION 3 [20] 3.1.1 Re = 0,8 + 0,06 (660/220)2 = 1,34 0 v Xe = 1,7 + 0,07 (660/220)? = 2,330 v Ze = V(1,34)?+ (2,33) = 2692 Vv 3) 3.1.2 cos $2 = 0,8 singe = 0.6 v | 4 ( FULL LOAD) = VA = 310000 = 114,81A v V 2700 V reg.( Pf. 0,8 lagging) = h (Re cos #2 + Xe sin o2_) , M1 = 114,81(1,34 x 0.8 + 2.33 _X 0.6 ) 2700 v = 0,105 p.u. v = = g00V ¥ V2 (no load) = 2700 x 220/660 = 9 Decrease of V 2 ( bet. no load and full load) = 945V = 900 x 0,105 v Downloaded from hjayise.com- 505.5" vy V 2 ( full load) - ‘ please turn Over Copyright reserved

MARNING Vet eee’ E-SSTROTECHNICE 15 T48N(EVNIS)T 3.1.3 V reg.( Pf. 0,8 ieadi (PT. 0,8 leading) = } v4 ) ‘1 (Re cos $2 - Xe sin 42 ) = 114,81(1,34x 0,8 -2,33 x 0,6) 7 2700 Vv =0,0139pu. v Increase of V 2 ( bet. no load and full load) = 900 x 0,0139 = 12,51V v V 2 ( full load) “= 900 + 12,51 TeV (4) 32 321 Vph 1 = VLI 3 es SEN wrye00 11,732 = tortiv v SNE Ee ae Vph 2. = Vph 1 x N2/N1 iil | = 74501,11X 30/510 Nun i. = = 88,3 V v = secondary line voltage (3) 3.22 Vph 1 = 2600 Vph 2 = 2600 xN2/N1 = 2600 X 30/510 = 152,94V v vig =282vx73 * a = 2649 V v [20] QUESTION 4 41 Loop resistance = ol a = 1,7x66x10° 108% nx0,7? ¥ = 7,292 ¥ v Loop inductance = 66[0,05 + 0,2 log © dir] mH = 66 [ 0,05 + 0,2 log € ( 70/0,7)] mH = 64,01 mH 4 Total capacitance = 33 36 loge ( d=!) a r = 36 loge (70 0.7) 0,7 (6) = 0,2165 pF vi Copyright pe urn over Piease WU a “et Downloaded from hlayiso.com

4.2.2 de _ a = Ndxdx2q ima En = N50x50x100 om yy 63 x 10-2 m Y Km [ 0,05 + 0,2 loge (de/r) 35 [ 0,05 + 0,2 log e (0,63/0,75 x 19-2 )] 32,76 mH per phase v oO " Km [ 1/18 log e@ ( de/r ) 35 [ 1/18 log e(0,63/0,75x10-2)) vv = 044 ,5F v de =*Vd1xd2xq@3 = 3/55 x 85x 110 v = 80,12x 10-7m v = Km [0,05 + 0,2 log e ( de/r) ] = 35 [0,05 + 0,2 log e (80,12 x 10-7)] v ~ 0,75 x 10-? 34,45 mH per phase it] [ 4/18 log e (del )] 12x 10-)} ¥ 18 log e (80.12% 35{1! 0, Cc = Km 75 x 10-7 _ 9416p F per phase Downloaded from hlayiso.com aN please turn over (6) (6)

43 Km=200 3ph. d = 10m r= 15x 10-3 m L = Km[0,05 +02 log e (dir) |} mH = 200 [0,05 + 0,2 loge (10/15 x 10-3) ]mH 270,01 mH v QUESTION 5 5,1 5.1.1 Syn. speed =ffp = 50/2 = 251s Vv S) Syn. speed - Actual speed Syn. Speed 25 - 23 25 0,08 ¥ 450,11 V / = 0,08 X 150,11 =1201V an = N a i) it} N Oo) oO ~ rae w it SE2 W 0,96 lagging i St 0,08 x 50 4Hz Y cos + 5.1.5 Rotor frequency wu Downloaded from hlayiso.com 'g'y reserved (2) (20] Pleas. 2 tur. over

52 53 Copyright ieseved The difference between the synchronous spe c ed (Ns) v speed (N) of the rotor is known as slip. ve NS) and tne actual For an induction motor to develop torque, EMF and current must be induced in the rotor. v This will only occur when there is a relative speed difference between the rotating magnetic field and the rotor (i.e. The motor has to have a slip) “ Hence an induction motor must have slip to develop torque . 5.3.1 f = nip 69 50 = nix4 60 nit. = 750 r/min v n2 =ni-snt = 750 - 0,05 ( 750) = 712,5 r/min v 5.3.2 Gross P = 2nxn2xT 60 45375 = 27x 7125 xTv 60 T = 608,14 Nmv Rotor input = 2nxm xT 60 = 272x750 x 608,14 60 = 47763,2 W v Rotor cu. Loss s x rotor input 0,05 x 47763,2 238816 W v Downloaded from hlayiso.com TOTAL: (4) [20] 100

UMN higher education training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA T480(E)(J28)T AUGUST EXAMINATION NATIONAL CERTIFICATE ELECTROTECHNICS NS ~ (8080085) 28 July 2014 (Y-Paper) 43:00—16:00 Downloaded from hlayiso.com

oo Tw! 3U(), J28:- 14 1.2 1.3 1.4 1.5 y p n ho you would Cnar ge t e directio; of rotatio f DC e ota Otor. A DC motor can be self-regulating due to back EMF ° Briefly discuss this statement. | (4) An eight-pole generator has a lap-con The ratio of the pole arc per p ohne mat ; ure with 640 ole pitch is 08. conductors. Calculate the ampere-turns ; ; per pole of a com i indi i uniform air gap density pensating winding to give when the total armature current is 960 A, (4) An 88 kW, 540 V shunt generator has 1 800 turns on each pole of its field winding. On no-load a current of 9A in the field winding produces a terminal voltage of 542 V. but on full-load the shunt current has to be increased to 12 A for the same terminal voltage at the same speed. Calculate the number of series field turns per pole required for level 3 compounding. (8) . sant indings i hines. 2 Name the practical applications of lap and wave windings in DC machin ae QUESTION 2 2.1 e taken during a half cycle of a symmetrical ; inates wer Sacre dt R between The following Eb pane The current varies in a linear manner alternating cur 7 successive points: 120 150 180 30 60 90 Phase angle in degrees O 126 24,4 31 29 244 0 Current in amperes ' i e graph: Determine the following without plotting the g 2.1.1 The mean value 212 The RMS value Downloaded from hlayiso.com Piease turn over

nr ho A resonant circuit of 60 ohms in Of a coil of j Parallel with = ,... Ol! Of inductance ¢ : resistor of 8 900 ohms " The we capacitor. ig conn ui and a resistance frequency of 1.5 MHz : © supply ACross this sedi 4 ee with a ; IS V, with a "FS EMJ28,7 COMprisin Calculate the: 2.2.1 Value of the Capacitor at resonance (4) 2.2.2 Impedance of the parallel circuit 2.2.3 Current in each branch QUESTION 3 3.1 Name FOUR methods of reducing leakage flux in transformers. oh Two single-phase transformers are connected in parallel to a load of 900 A, at a power factor of 0,8 lagging. Test data: _ Open-circuit: 42 000/2 800 V for each transformer short-circuit with high voltage winding shortages: “TRH ‘ J 7 yn B 250 AS oF former A: secondary input 400 V, 450 A, 30 KW me a Transtor : put 250 V, 450 A, 35 kW in Transformer B: secondary Calculate the: (10) voltage secondary 3.2.1 ; f transformer A (3) t and power factor O 3.2.2 Outpu wer factor of transformer B (3) Output and po [20] 3.2.3 Downloaded from hlayiso.com

Ai QueSTION 4 The input power to a 3 350 V thre @-phase d i i i 180 kW. The power factor of the motor is 0, orate Hed mevetion motor i Calculate the: on Line and phase currents e 4.1.2 Input power readings on the TWO watt-meters 6) 41.3 kVA rating of the motor (2) 42 A three-phase transmission line supplies a 2,45 MW star-connected load, having a power factor of 0,8 lagging at a line voltage of 36 kV. The line has a resistance of 90 ohms per phase and an inductive reactance of 160 ohms per phase. Calculate the: 4.2.1 Voltage at the sending end (6) 4.2.2 Regulation - i (2) 4.2.3 Efficiency of the line rea QUESTION 5 ; 54 D n induction motor develop torque when running at synchronous a oes a speed? : as Discuss the reason for your answer . It tors are operating In parallel. 52 TWO similar three-phase. star-connected alternato! f 6,5 Q per phase and nchronous reactance OF 9, Each machine ee a is excited to generate an EMF of 2 675 V per negligible re machines have a phase displacement of 30 electrical degrees hase. Them relative to each other. Calculate the: (5) 5.2.1 Circulating current rs 6.224 Terminal voltage per phase oO 2: Power supplied from ONE machine to another. @ 5.2.3 Copyright .. ownloaded from hlayiso.com Please turn over

A Star-connecreg, has to generate a 1480(i:)(28)T 500 r/min, and . The stator has onductors per slot. full-pitched. kd = 0,96 Calculate the: 5.3.1 Number of poles (2) (4) 5312 Useful flux per pole 120} TOTAL: 100 Downloaded from hlayiso.com

NATIONAL CERTIFICATE AUGUST EXAMINATION This marking guideline consists of 9 pages. i | ) Downloaded from hlayiso.com — °° ""°%

SLECTROTEChinice a Ve QUESTION 1 ABE cde T 1.1 4.2 1.4 through the ma ic nt gnetic fi lux j Opposition to the aalisd wit : speed up until the ne bea will cause ah igh armature speed up the armature. v Ams rie mill sup change. v Theb Any change in th a ack i eb regulating, v EMF acts like a governor and therefore the motor is self is self- Pole =6 p=3 c = 2p = 2x4=8 v Z = 640 Pole arc = Pole pitct = 08 1 = 960A At/p = IZ x Pole argv 2c 2p Pole pitch 162,96 Av MW = 12Ax 1800t/p ¥ = 21600 Atip v Atip for full load Mt tip for no load = 9A x 1800 tp ¥ 16 200 At/p v At / p for series winding 24.600 - 16 200 v 54 No. of series UP = 33 Or Short hunt = 540041. + I) = 5400/174,96 = 30.86 “ Downloadedfrom hlayiso.com Please turn over

eee MS TEP NISS tys vanewound - for low Lap-woun for jow voltage and high current v’ oF farae macj ‘ jarge machines EES Wave-wound - for high voltage and low current. v or small machines eee tee (1) (4) [20] QuesTiON 2 4 2.1.1 ii 0+12.6/2 26+ 24,4/2 24,44 31/2 31+ 29/2. 29+ 244 /2 24,4+0/2 6,34 18.5 v 27,7 30v 12,2 nououou wu u rs? nyu nw Woo lave =ititistintis tis / nv =6,3+ 18,5+27,7+30726 =121,4 /6 ' ’ 5 = 20.23 AY (5) 2.122 lrms = Viti? Tita + Ppt Ps (NY \ 6,3?+ 18 52 + 27,77 + 302 + 26,717 + 12,27 /n v 2.2.1 XL = anit 49% = 6126,112% 2.2 49% 690 Xt pan VbC MW _ 47 ¥ 2010 x15 (ai2nt0 6 1,5) 7 Wow g50 x 10° 2 173x107 | 2173x107 FY LICR parallel 22 Downloaded ftir tifiyiso. com ; p = 626,20 please tun over (2)

SoS LE SANICS NE TB (EF) SL3)T 2.2.9 : = V/ZT = 80/ 9900 + 626204 = 80/636104 = 0,126 mAv pd across coil = 0,126 x 10-3 x 626204 = 789Vv lcoil == 78.9/¥ (60)? + (6126,11)2 = 12,88 mA v loap = = 22x1,5x10°x 17.3 x 10-#2x 78,9 = 129mA ¥ . es QUESTION 3 : n Se Making the transformer window long a Arranging the eamary and secondary windings. (4) Sandwiching the P uction. “# Using shell-tyPe constr = 280 3.2 322-1 s esc lL A i a6 2 = 0,889 < 80,39 ¥ = 400 <0 = VII 450< - 80,39 ZA 0,148 + j 0,88 Downloaded from hlayiso.com re

TRE.B SES ROTECHIRICS Hr - P=1VCOS¢ “ABOE). 28) ° 35 000 = 450 x 250 x COS @ COS @ = 0,31 lag o =-71,94 1B. = 450< -71.94 v ZB = Vil = 2 = 250< 0 = 250 < 0 = 0 450 <- 71.94 556 < 71,94 v _ = o17+j053 ZA+ZB = 0,148 +j 0,88+ 0,17 +} 0,53 | = 0,318+j14 = 144< 77,2 ¥ f | Zt = ZAXZB = 0,889 < 80,39 X 0.556 < 71,94 ZA + ZB 1,44 < 77,2 = 0,343 < 75.13% . —~ VD IL x ZT 900 < - 36.8 0,343 < 75,13 Sie 36,26°Y nou i 2 567.61 -j 191.18 9 564,74 < 427% 2 At 14x 0.558 < 71,94 = 7 sa = S&2E- p08 an 773 zs ie 46.4KVA > = 0,69 lagging % SB S75 4,44 < 77,2 _ 37,99 KvA ¥ 3) Downloaded fr _ on fre wauittayiso. com iu cos ¢

MARKING GUIDE: INE QUESTION 4 a 4.4.1 4.1.2 4.4.3 4.2 =! SCTROTECHNICS N5 PT = Y3 VL IL COS6 IL = 4180000 V3 x 3350 x 0,85 = 36,5A v 1P = [IL = 365/V3 = 2107AV v3 W1 = VL IL Cos (30+) = 3350x 36,5 cos (30+ 31,79) ¥ = 3350 x 36,5 cos ( 61,79) ‘ = 57,8 kWY W2 = VL IL COS (30-9) = 3350 x 36,5 COS (30 - 31,79) v = 122,22 kW ¥ 1 000 =245 MW = 2.45 MW_ per phase Load P = 2.45 ae uv = 36000 (cos $+jsind) ¥ 13 ih 36.000. ( 0,8 +) 9.) =e v3 = 46 627,69 + J12470,77 ¥ Downloaded from hlayiso.com Taba yi23y, (2) -6- ; . (8)

MA anitle GUIDELINE Te ELESTPO vech NICS NS T4280), jo9 4 = 40154, ¢ Inet = 245 ww i} = 442035 + | 7858 4v Vs Vr_+ Vd 16 627.69 + [7247077 + 4420.35 + | 7856.4 21048.04+ j 2032917 ie} dat " 2926252 <44 v Vs ling voltage = \8x2026252= 50.692 kV“, , 6) 4.2.2 Re = Vs-Vr Vr OR 50692 - 36 000 36 000 = 29266.97 - 36 QOON3® 36 000 /NV3 = 0408p.u 0,408 p .u. 816.67 kW Output = 2.45 MW £3 per phase Powerjoss = _/*R... = 217,105 kW x Input = output + losses = 616,67 + 217,105 4.2.3 Efficiency = outout Input 4033,78 Downloaded from hlayiso.com

ance © JILL e SLEUTROVE CHM Stat : ONC ASME oat QuesTION 5 a be 4 If the rotor were to run at synchronous rela we speed difference between NOUS Speed. then there would be hence no emi will be induced in Alans magnetic field and the 10 Y ced in the rotor v ; rotor rotor current and so no torque. v he rotor and therefore there will be no (4) av) 5.2.1 XS : 5,5O per phase 6 = 30 = j EA— (EB sin 6 +j EB cos 8) v = }2675-(2675sin 30 + j 2675 cos 30) | 2675-1 337, - | 2316.62 v iy = -1 337,5 +§j 358,38 484,68 ¥ = 1384,68 igo Circulating! = x65 ¥ 5.2.2 Terminal V/IP_ = ca * cet = 2583.85 V% FSI VL : {3 VP= V3 X IS xA106.5 * 5 = 4475.36 V ¥ 5.2.3 P 9583,8 < C wou 475,36 W, cs (3) _— 25,540 kW ¥ ‘ = 50 = 4500 r/min f VEL = B00 N 53 5.3.1 = Np/60 go = 1p00xE” 60 pa? v _ = 4 No of poles 22x2 Downloa ded from hlayiso.com Please turn over -_cerved

SENS TERY 2 No. of slots per phase = 2y4 8 (or 24 slots for 3 plases; No of conductors per phase ° NOES) 8x4 = 32Vv (or 96 conductors for 3 phase) ty EMF per phase = 800 Ep = 2,22kd kp o f Z/ph 461,88 = 2,22 x 0,96 x 1x x 50x 32 ¥ = 461. 88 ; ° 2,22 x 0,96 x 1x 50x 32 (.2= 32 per phase or 96/3) = v TOTAL: 100 Downloaded from hlayiso.com