IVE(TY) Department of Engineering E&T2520 Electrical Machines 1 Miscellaneous Exercises

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1 TRANSFORMER Q1 IE(TY) Dpartmnt of Enginring E&T50 Elctrical Machins 1 Miscllanous Exrciss Q Q3 A singl phas, 5 ka, 0/440, 60 Hz transformr gav th following tst rsults. Opn circuit tst (440 sid opn): A 650 W Short-circuit tst (0 sid shortd): A 950 W (a) Driv th approximat quivalnt circuit in pr-unit valus. (b) Dtrmin th voltag rgulation at full load, 0.8 PF lagging and draw th phasor diagram. Q4. A 10kA, 00/400 50Hz, singl-phas transformr has th following quivalnt circuit paramtrs. Equivalnt Rsistanc for cor loss rfr to 00 sid R c1 =30Ω Magntizing Ractanc rfr to 00 sid X m1 =165Ω Primary winding (00 sid) rsistanc and ractanc R 1 =0.03Ω, X 1 =0.15Ω Scondary winding (400 sid) rsistanc and ractanc R = 0.1Ω, X =0.5Ω Calculat th quivalnt circuit paramtrs rfr to 400 sid. Find th prcntag voltag rgulation and fficincy whn supplying full load at ratd voltag 400 and 0.8 p.f. lading. P.1

2 DC MACHINES Q5 Q6 Q7 Q8B A dc machin is connctd across a 40-volt lin. It rotats at 100 rpm and is gnrating 30 volts. Th armatur currnt is 40A. (a) Is th machin functioning as a gnrator or as a motor? (b) Dtrmin th rsistanc of th armatur circuit. (c) Dtrmin powr loss in th armatur circuit rsistanc and th lctromagntic powr. (d) Dtrmin th lctromagntic torqu in nwton-mtrs. () If th load is thrown off, what will th gnratd voltag and th rpm of th machin b, assuming (i) No armatur raction. (ii) 10% rduction of flux du to armatur raction at 40 amps armatur currnt. Q8D. A dc shunt machin (3 kw, 30, 1500 rpm) has R a = 0.1 Ω. (i). Th dc machin is connctd to a 30 supply. It runs at 1500 rpm at no-load and 1480 rpm at full-load armatur currnt. (a) Dtrmin th gnratd voltag at full load. (b) Dtrmin th prcntag rduction of flux in th machin du to armatur raction at full-load condition. (ii). Th dc machin now oprats as a sparatly xcitd gnrator and th fild currnt is kpt th sam as in part 1. It dlivrs full load at ratd voltag. (a) Dtrmin th gnratd voltag at full load. (b) Dtrmin th spd at which th machin is drivn. (c) Dtrmin th trminal voltag if th load is thrown off. P.

3 INDUCTION MACHINES Q9 Q10 Q11. Th powr supplid to a thr-phas induction motor is 40 kw and th corrsponding stator losss ar 1.5 kw. Calculat (i) th total mchanical powr dvlopd and rotor I R loss whn th slip is 0.04 pr unit (ii) th output kw of th motor and (iii) th fficincy of th motor, if th friction and windag losss ar 0.8 kw. Nglct rotor iron and coppr losss. Q1 P.3

4 SYNCHRONOUS MACHINES Q13 Q14 Q15. (a) Whn a 50 ka, 3-phas,440-, 60 Hz, star connctd synchronous gnrator is drivn at its ratd spd, it is found that th opn-circuit trminal voltag is 440 lin-to-lin with a fild currnt of 7 A. Whn th stator trminals ar short circuitd, ratd currnt is producd by a fild currnt of 5.5 A. Dtrmin th synchronous ractanc pr phas. (b) If th synchronous in (a) is usd to supply an indpndnt load of 40 kw with 0.85 lagging powr factor at a potntial of 440, (i) (ii) Dtrmin th fild currnt rquird; and if th load is rducd to 0 kw at 0.75 lagging powr factor, to what valu will th fild currnt hav to b rducd to maintain ratd load potntial? Q16. Th 3 phas synchronous motor of 10 MA, 14k, star-connctd, R s =0.07ohm/phas and X s =16.5 ohm.phas is connctd to a 3 phas, 14 k, 60 Hz infinit bus and draws 5 MW at 0.85 lading powr factor. Dtrmin th valus of th stator currnt (I a ), th xcitation voltag (E f ), and th fild currnt (I f ). Draw th phasor diagram. (Ans. Q6 I a =4.6A, E f =10.73k, I f =65A) P.4

5 Q1 TRANSFORMERS (SOLUTIONS) Opn Circuit Tst L sid input voltag, = 00, L sid input currnt, I = 4A, L sid input powr, P oc = 15W P oc = Rc = 30 (L Sid) I m = I ( ) X m = = (L sid) Rc Rc I m Short Circuit Tst H sid input voltag = 60, H sid input currnt = 10A, H sid input powr = P oc =80W Psc = I R R =.8Ω (H sid) X = ( ) R X = 5. 3Ω (H sid) I R = R ( ) = 0.08Ω, X = X ( ) = Ω 1 1 R X R c X m Q4. Full load output powr = 0kA 0.8 = 16kW, th full load fficincy is Output powr 16k η = 100%, η = 100% = 97.53% Output powr + Iron loss + Coppr loss 16k X = M X M 1 = 660Ω, R = C = RC1 180Ω 1 1 X = + X X 1 = 0.85Ω R = R + R1 1 1 = 0.4Ω BASE1 =00 BASE = 400 I BASE1 = 50A I BASE =5A S BASE1 =10kA S BASE =10kA Z BASE1 =( BASE1 ) /S BASE1 =4Ω Z BASE =( BASE ) /S BASE =16Ω X (pu)= X /Z BASE =0.053 R (pu)= R /Z BASE =0.015 X M (pu)= X M /Z BASE = 41.5 R C (pu)= R C /Z BASE =80 R X E R c X m 1 NL = + I Z = 0 + ( R + jx ) xi (cos ( p. f.)) NL = 0.98 (pu) NL R = 100% = -% P =Cor Loss = / R C =15W, P cu =Coppr Loss=I R =150W P out =Output Powr= I cosφ=8000w, = Pout Efficincy = P + P + P % out cu P.5

6 Q. TRANSFORMERS (SOLUTIONS) P.6

7 Q3 TRANSFORMERS (SOLUTIONS) P.7

8 Q5 DC MACHINES (SOLUTIONS) Q6. E a = t Ia Ra = x 0.1 = 99.4 No load loss = Ea x Ia = 99.4 x 6 = 596.4W At full load, Ea (fl) = t IaRa = x 0.1 = 88 Ea ( fl) K aφ flω fl ω fl 88 N fl = = Thrfor, = N fl = rpm E ( ) K φ ω ω Q7 a T = a E I a a ω a = = 113.9Nm π 60 Ea ( fl) K aφ flω fl φ flω fl = = E ( ) K φ ω φ ω T = E I a a ω a Thrfor, = = Nm π 60 Ea= 5+110(1) = 115 Wm = 100 x π / 60 = rad/s. KaΦ = Ea/Wm = 115/15.71 = 0.915s/rad Ia = Ea / (Ra+R L ) = 115/ (0. +) = 5.7A T = KaΦIa = x 5.7 = 47.83Nm P = I L R L = 5.7 x = W N fl = N fl = rpm P.8

9 Q8B DC MACHINES (SOLUTIONS) P.9

10 Q8D P.10

11 Q9 INDUCTION MACHINES (SOLUTIONS) Q1 No Load Tst =380 phas voltag = p=19.4 I =Input currnt = 9A Powr Loss = P = 067/3=689W (pr phas) p p p P = Rc = Ω I m = I ( ) X m = = 7. 6Ω Rc Rc I m I o =/R c +j/(jx m )=3.14 j8.7a Lock rotor Tst Lin Input voltag = 188 =108.5(pr phas) Input currnt = 104A Thr phas powr loss= 340W P=1080W (pr phas) P I R R = 0.1, But R s =0.1/ = 0.05 sc = Rr = 0.05Ω I L R s X = ( ) R X = 1. 04Ω I X =X s +X r ' I o I 1 R r ' R c X m R r ' s R r '(1-s) s Spd = 1455 s =( )/1500=0.03 I 1 =/Z=19.4/(1.71+j1.04)=109.6 (-31.5 o )A Rotor Powr loss=p r =3x109.6 x0.05=1.8kw Z=R s +R r /s+jx=1.71+j1.04ω I L =I 1 +I o =96.8-j85.7=19.3 (-cos -1 (0.75))A Torqu=P r /s/(π1500/60)=38 Nm P.11

12 Q10 INDUCTION MACHINES (SOLUTIONS) Q11 Assum stator loss including stator coppr and cor loss Airgap powr = Input powr stator loss = = 38.5 kw=3 (I ) R / s 3 (I ) R = Coppr Loss = s 38.5 = 0.04 x 38.5 = 1.54 kw Gross Mchanical Powr Output = 3 (I ) R (1-s) / s = 1.54 (1-s) / s = 1.54 x 0.96 / 0.04 =36.96kW Actual Mchanical Powr Output = Gross Mchanical Output-Rotational Loss= =36.16kW Efficincy = Output / Input x 100% = / 40 x 100% = 90.4% SYNCHRONOUS MACHINE (SOLUTION) Q13 At no load, th gnratd mf 10 k E E f = = 5.77k Zs = j 15 = f 5770 I = = = 83.13A a 3 Z s k 1 Full load currnt, I a = cos 0. 8 = A 3 10k E f = t + I a Z s E f = = E f is proportional to I f, hnc % incras in I f = ( )/5770 x 100% = 10.8% P.1

13 Q14 SYNCHRONOUS MACHINES (SOLUTIONS) 1500= x50x60 / P, P = 4 T = / (1500xπ/60) = 63, Nm, Pin = 6600 x 3 1/ x90 x 1 = kW, Pout = 10MW. Efficincy = 10/ x 100 % = 95%, Rotational Loss = Pin Pout = 517 kw Powr loss in fild = 100 x 5.5 = 550W Q15 Q16 P.13