Electrical Machines : 1 induction phenomena . -

.. Dynamically induced e.m.f .. E = BL volts (1) Where :B := Flux densitywebers/m2 L := Conductor Length m := Conductor Speed m/s ( 1 ) E = BLsin volts (2) .. - .. Statically induced e.m.f. .. E = -Tc (d/dt)volts (3) Where Tc: = number of turns in the electric circuit: = magnetic flux webers ( 1 ( ) 3 . ) 2 Interaction Phenomena e )( i ( ) B ( /2 ) F ( ) F = BLiNewtons (4) F F .ntuBu sin =nFig (1) ( 2 - ) ( 2 - ) ( . 2 - ) . . .. . ,. . . .. . ( 3 . ) : ( 4 ) .. ( 2 ) ( ( 4 ) .. ) .. . . Gramme-ring generator . ( 5 ) 120 ( 5 - . ) Stationary armature : ( 4 ) 50 / Frequency 50 / t t 2 f = pn cycle/sec (5) Where: f: = Frequency ,cps p: = no of pole pairs = P/2 n: = speed rps, =60N : 3000 / :

120) 3000 )( 10 (120 = = PNf 250 cp:s : 50 /: 120fN =P = 600 rpm.(120)(50)=10 Synchronous Genr. Synchronous speed N Synch. Speed ,rpm No. of poles 25cps50cps60cps1500300036002 750150018004 500100012006 3757509008 30060072010 25050060012 37516 30020 20030 15040 " Dynamo " 1 Direct current generator 2 Direct current Motor Essential Construction of a.d.c machine 1 - : The Field Poles , , , . . Pole Core , , Yoke . 2 - The Armature armature Core Commutator . , ( . 1 - ) ( 1 - ( ) 1 - ( . ) 1 - ) micanite . ( 2 ) air gap , , ( 16 / 1 4 / 1 ) . . . : .. .. , . , . .. , .. . . : , . . Gramme ring armature ( 3 - . ) , , .. . .. . ( 4 - ) , . . - .. .. .. , . .. : 1 - . 2 - , .3 - . 4 - . . Drum winding , Lapwinding WaveWinding coil . coil side. inductor . .. front end connection . back end connection . : ( Y) Pitch . , Pole pitch . (fY ) front pitch . (bY ) back pitch Commutator pitch Yc 1 singly reentrant winding The simplex lap ( 6 ( )A ) ( . B ) ( A ) ( B ) ( A ) ( A )( B). ( C ( ) B ( ) B )( C ) . ( A ). ( A ) 1 7 ( B ) 2 8 ( C ) 3 9 CoilSlot Occupied CoilSlot Occupied A1 and 7F6 and 12 B2 and 8G7 and 13 C3 and 9H8 and 14 D4 and 10I9 and 15 E5 and 11J10 and 16 etc, 7 ( A ( ) G )8 ( B ( ) H ) 9 ( C ( ) I ) . ( B ( ) A ( ) C ( ) B ) progressive winding ( . B ( ) A ( ) C ( ) B ) retrogressive winding ( 6 ) ( B ) ( A ( ) C ( ) B .. ) simplex ( B ( ) A ( ) C ( ) B )Fig (6) single reentrant 2 ( A ) ( B ( ) A ) ( A ( ) B ) ( . C ) ( B ( ) B ) .. The duplex lap doubly reentrant winding ( Tow windings) ( 7 ) 3 : Duplex lap singly reentrant winding ( B ) ( A ) ( A ) ( A ) ( A . ) ( A ) ( 8 ). 4 The triplex lap winding ( B ) ( A . ) Singly reentrant Triply reentrant( 9 ) . 1 The simplex wave singly reentrant winding ( 10 ( ) A ) ( B ) ( C ) ( C ( ) A ( ) D ( ) C ) ( B . ) ( A ) . . . 2 : . : ( 11 ) . . chorded winding . 80 % . .. Equalizer Connections ( 12 - ) .. 1 - . 2 - . 3 - . .. circulatingcurrent . ( 12 ) .. . .. . winding calculations - : Yb Yf Yf = yb 2min terms of coil sides (1-a) and for only 2 coil sides per slot, it becomes Yf = yb min terms of slots(1-b) Where: Yf := front pitchin coil sides or slots, Yb:= back pitch in coil sides or slots m :=multiplicity of winding ( 1 for simplex , 2 for duplex, 3for triplex ,etc.) ( ) . ( - ) Yf Yb . Yf Yb. 1 - simplex . 2 - duplex 3 - triplex .... . . Example: A simplex lap winding is to be installed in a four-pole dynamo that requires 21 inductors per path. If there are 3 turns per coil and 2 coil sides per slot, determine (a) number of coils,(b) number of slots ; (c) number of commutator segments ; (d) back pitch;(e)frontpitch,(f)commutatorpitchanddrawadevelopedviewofthe proposed winding. Solution: a) No of parallel paths = no of poles = 4 Total no of inductors = 4 paths x 21 inductors/path = 84 inductors Each coil has 3x2 = 6 inductors no of coils =84 = 14 coils 6 b) No of coil sides = 14 x 2 = 28 coil sides, each slot has 2 coil sides: no of slots =28 = 14 slots 2 c) Simplex lap winding requires only one commutatar segment per coil, then the no commuter segments is 14. d) back pitch Yb ~a pole pitch = 28 coil sides = 7 coil sides 4poles e) Front pitch Yf = Yb 2m, m = 1= 9 coil sides for retrogressive winding = 5 coil sides for progressive windingf) The commutator pitch is the same as the multiplicity of the winding: Yc = 1commutator segment. : , .. . , . - : : Y= Yb + Yf (2) 2 , PY = ( P , ) PY = 2 ,Or Y= 2 (3) P : , Y= 2m In coil sides (4) P P = 2p NC= 2 : NC = PY m(5) YC Y . YC= NC m In commutator segments(6) P Example: A simplex wave winding is to be installed in a four-pole dynamo whose armature has 14 slots. Determine (a) Yb , (b) Yf , (c) Yc and draw the developed winding Solution : (As in the lap winding Yb~ one pole span) Since there are 14 slots , assume that 14 coils will be needed and 14 commutator segments are required : YC = NC m = 141 =7 1 Or6 1 Segments P222 Thecommutatorpitchisnotawholenumber;therefour14coilscannotbeused. Reduce the no. of coils by 1 and repeatYC= 13 1 =7 segments for progressive winding2 In slot(4) NC mandY =2NC2m Y =P 2p = 6segments for retrogressive windingThe no. of coils required = 13 cols Yb 13*2 =6 1 Can be taken as 7 coil side42 = 7 or 6 26 2= 2 Y =4 PYf = 2Y-Yb = 14-7 = 7 for progressive winding = 12-7 = 5 for retrogressive windingYb = 7,Yf =7, Yc =7 ( progressive ) Winding Table Dead or dummy coils , . : , .. 600 . . No ofslot=14 , no of commutater segment=13 200 , . e.m.f. induced in the armature ( B) /2 e = BL Volt L, / , ( 13 ) . ) ( .. , ( .. 13 ) e , N S . .. .. ( 13 )( B ) ab .. : Eav = BLVolt Where: B : = average flux density per pole pitch , weber / m L : =active length of the conductor , in m, : = velocity of conductor, in m/sec= ab t t : = time required for the conductor to travel the distance ab , equal to the pole pitch . Volts| = Bl(ab)Eav = t t Where : | BL(ab) : = total flux per pole , andt= 1 Sec.np the average e.m.f. per conductor isEav = | |np volts1/np and the total induced e.m.f. between brushes isE = |np Z Volts ( 7)a WhereZ: = total no. of conductorson the armature a : = number of parallel paths , and Z/a : = no. of conductor in seriesIf the speed in rpm , eq (7) becomesE =|NPZ Volts ( 7`)60a Example: A 900-rpm 6-pole generator has a simplex lap winding. There are 300 conductors on the armature . the flux per pole is 0.05 weber . Determine the e.m.f. induced between brunches . Solution: | = 0.05 weber , n= 900 = 15 rps 60 no. of paths = no. of poles = 6 E =0.05 * 15 *6 * 300 225 volts 6 : | , | =a E ( 8 ) n p Z ( 7 . ) | . ( 2 , ) . ( 14 ) | , , ( mmf ) abcdef( 14 ). Ampere turns Length,meter Ampere turns per meter Flux density weber/m2 Area square meter Flux , weber MaterialPart H1.L1L1H1K1(|/A1)A1| Dynamo steel sheet Core ab H2.L2L2H20.5 |/A2A20.5| Cast steel Yoke bc H1.L1L1H1K1(|/A1)A1| Dynamo steel sheet Core cd H3.L3L3H3|/A3A3 | AirGap de H4.L4L4H4) 0.5| K2(A40.5| Dyn. Steel sheet Armature ef A4 H3.L3L3H3|/A3A3 | AirGap fQ IN Total ampere-turns for 2 ples =IN/2 Total ampere- turns per pole Where: : = leakage coefficient =Core Flux 1.1 1.2 Armature Flux K : = Correction factor to allow for the thickness of the oxide on the surface of the lamination and the air-duct speed = 1.1-1.25 D.C. Generators : Types of Field Excitation : , . ( 15 - ) Separately Excited d. c. Generators ( 15 - ,, ) Self Excited d. c. Generators , b r If : 1 Shunt generator : If . If r ( 15 ) 2 Series Generator( : 15 ) . 3 Compound Generator : Fsh Fse . ( 15 ) . : Characteristic of Generators curvesCharacteristic . . , 1 - : internal characteristic curve , open circuit characteristic , or no-load magnetization curve , : E = K|N(9) Where : : = constant 1. P .Z K = 60 a| :=Flux per pole , in weber N:=speed of armature , in rpm . ( 16 - ) . E , If- .. , ( 1 ) ( 16 - ) residual magnetism . 2 3 .. 3 4 . ( 4 , )retentively . 4 5 Hystresis loop , . At N1rpm the induced voltage is E1 = K|N1and At N2rpm the induced voltage is E2 = K|N2 E1 = K|N1 and E1 = N1 E2K|N2E2N2 K , | ( 16 - ) ( 16 - ) , .. . .. Field resistance line ( 17 - , ) ( 17 - ) - . , . . : The build up process ( 18 - ,) oa. , ab ob` .. cb` . , , od` .. ed` .... F , g 0g` ,Kg` gg` f . , . : Critical Field Resistance oa . a` oa . ob b ob ( 18 - ) : 1 - . 2 - . 3 - .. , 4 - . 5 - . : Armature Reaction , . , : ( 19 - ) . F . ( 19 - ) , , FA . , , . , Trailing Pole Tips , Leading PoleTips ( 19 - ) FR .. . ( 20 -) FA: FD : = demagnetizing component of armature reaction FC : = Cross-magnetizing component of armature reaction ( . 20 - ) : 2 , . .( 21 ) FR , FC,FD FA , F FR FA. Example: A4-polegeneratorhas288surfaceconductors.Thearmatureislapwound,and thearmaturecurrentis120amp.Thebrunchesareadvanced15spacedegrees. Determine demagnetizing and cross-magnetizing armature ampere-turns. Solution: = 15

2=30

No. of brunches = no. of poles ( lap winding ) = 4total number of degrees covered by the demagnetizing conductors = 120

, this gives that one-third the conductors on the armature , or 96 conductors , are demagnetizing conductors.

no. paths = 4, current per path = 120/4 = 30 amp Demagnetizing ampere conductors.= 30 * 96 = 2880 Demagnetizing ampere turns= 2880 =1440 2 no. of cross magnetizing conductors =2of the no. of conductors on the armature 3 cross magnetizing amperes-turns =192*30 =2880 2 : ( 22 - ) , N` , S , N , , ( 22 - ) - ( .. 23 - , ) qr st I j k h ( 23 - ) b d f . ( 23 . ) ( 24 ) . : - . - FD. - ( N` ( 23 ) (r`s` instability. : FA . : 1 - : ( 19 - ( ) 23 ) : : - ( 25 ) . : ( 25 ) . . - : ( 25 ) . 2 , . : - : compensating windings ( 26 ). : Interpoles .. . ( 27 - ) . : Commutation .. .. . ideal commutation : 1 - . . 2 - . ( 28 ) straightline commutation 1 10 . 20 40 . ) ( 10 . 1 3 20 4 10 20 7 10 . Fig(28) current coil undergoing commutation condition ( 28 - ( ) +20 ( ) - 20 ) t , ( 28 - . ) ( 29 30 ) Fig (29) commutation with braches Fig (30) commutation with too far back of neutral planebraches too far ahead 45 5 50 40 20 ( 29 30 ) ( 29 ) 25 ( +20 ) ( -20) t t1( . 30 ) ... ( 31 ). . : .. . .. commutating Zone( 32 ) . ( 28 29 30 ) ( 32 ) |c 2|c ... ec = -N d|c Volts dt = -N 2|c volts (10) t Where: N : = no. of turns in the coil t :=time to across the commutating zone , sec. .. ... . .. .. . .. . .. .. .. .. Commutating poles Interpoles : .. .. .. . ( 33 ) B1 B2. .. 2 - : External characteristics - Shunt generator characteristics hb ( 34 ) e breakdownpoint f ... h` h fgh` hysteresis : 1 - : Ra Ia , : , V=E-IaRavolts(11) Where : V: = terminal voltage , E: = generated voltage . 2 - . .. 3 - : . .. ( 34 - ) : , .. , .. . ( 35 ) 1 230 , 900 / 1200 / . ( ) ( 34 ) 20 IL V If . : Generator Regulation . "" regulation : *100(12)VNL-VFLPercent voltage RegulatorVFLWhere: VNL: = no load terminal voltage - volts VFL : = Full load terminal voltage volts Example : A 10-kw 230-volt shunt generator delivers rated current at rated voltage. When loadiscompletelyremoved,theterminalvoltagerisesto250volts.Determinethe voltage regulation. Solution: VFL = 230 volts , VNL = 250 volts*100 = 8.7 250-230Regulation%230 . : 1 . 2 . Total characteristic ... E Ia: Ia = IL +IF(13)Where : IL : = load current ,If : = shunt field current E : = V + IaRa(14) : qr ( 36 ) oa ob ) ( . oY oa . oe ef oe` . e`f` e` of` Ia f`g` . fg IaRa e`f ( e`f=oe ) g . . : : Pg = E.Iawatt(15)Where:E: = induced e.m.f. , Ia : = total or armature current Example: A 20 kw 220-volt shunt generator has an armature resistance of 0.07O and shunt field resistance of 200 ohms . Determine power developed in armature when it delivers its rated output. Solution: = 90.9amp 20*103 Rated current I = 220 = 1.1amp 220Field current If = 200Armature current Ia = 90.9 + 1.1 = 92.0 ampInduced voltage E = 220+ (92.0*0.07)= 226.4 volts Power developed in armature Pg = 226.4 * 92.0= 20.83 kw The same result may be obtained by adding power losses as follows := 242 watt (220)2 Field lossPf = 200Armature loss Pa=(92.0 )2 * 0.07 = 592 watt Power developed in armature Pg = 20000 + 242 + 592= 20834= 20.83 kw Compound generator characteristics . ( 37 ) . . V = E IaRa IsRs(16) Where : Is: = series-field current Rs : = series field resistance . cumulative compound generator differential- compound generator constant current generator . ( 38 . ) degree of compounding : 1 Over compounding2 Flat Compounding 3 under compounding ( 38 . ) . Diverter . : 1 If1 2 If2. Nf ( If2-If1) Nf(If2-If1)=NsIs (17)Where :Nf: = no. of turns in shunt field ,If1 : = shunt-field current at rated current and shunt characteristic,If2 : = shunt-field current at rate load and the desired degree of compoundingNs : = no. of turns in series field ,Is: = series field current at rated load (17`) (If2-If1)Nfand Ns =IsIn case of short compounding Is = IL , and in case of long compoundingIs = IL+If2=Ia at rated load and the desired degree of compounding Example: A 250-kw 250 volt four-pole shunt generator requires a field excitation of 2.7 amp when delivering a rated load of 150 amp . A field current of 5.0 amp is necessary to raise the terminal voltage to the desired value at rate load. If the shunt field has 500 turns per pole and the series field 10 turns per pole,findtheresistancevalueofthediverterwhenthegeneratoristooperateasacumulative compound generator. The resistance of the series field is 0.005 O . Solution: Nf ( If2-If1) = NsIs =115amp 500(5.0-2.7) series field current required Is =10To field the resistance of the diverter : IdRd =IsRs (150-115)Rd = 115* 0.005 O = = 0164 . 035005 . 0 * 115dR : - , . , . .. , . . . .. ( . 40 ) , . oa d " " . bc , Booster ( 41 ) Feeders . : , , : : - , , : . " Floating on the bus bars .. " . , , , . , , . ( 42 - ( ) 1 ) ) ( ( a ) , ( 1 ( ) 2 )( 2 ( ) b ) . . , - . : - ( 43 - ) ) ( ( 43 - . ) ( 1 ) .. ( a ) , ( 2 ) ... ( 1 ( ) 2 , )( 1 ) . , ( 43 - ( ) 1 )( 2 ) ( 1 ) ( 2 )( 2 ) ( 1 ) , . D.C. Motors , . . Principle of the Motor ( 44 - )( 44 - . )F=BLI newtons ( 18) Where: B : = flux density of main field , in weber/m2 L : = Length of the conductor , in meter I : = current , in amperesOrF= BL I Dynes (18`) 10 Where:B in lines /cm2 or gauss, L in centimeters and I in amperes F ( 44 , ) , Fleming'sleft-handrule , . , . , ( 45 ) , . ( 18 ) : B, . . . Fig(45) torque produced by coil T = FxrN-m (19) Where: T : = torque , in N-m F : = Force , newtonR:=distancemeasuredperpendicularlyfrom direction of F to center ofrotation , m ( 45 , )( 46 - ( ) 1 ) ( 2 ( , ) 3 ) r ( , 3 . )( 4 ) ( 3 ( . ) 4 )( 46 - ) . ( 46 - ) Commutator , , ( 46 - ). Torque developed by a motor ( 46 ) . : From eq.s ( 18, 19) the torque developed by an armature can be obtained as follows : T =(d ) (ZBavLI) N-m( i ) 2 Where : D : = diameter of the armature , in meter . The flux entering the armature of the one pole . | =BaL ( d ) ( ii ) P P : = no. of polesSubstituting in (i) the value of Bav from (ii) T =(d )( |P )ZLI 2Ld = P . 1 .|ZI 2 But I = Ia/a, a : = no. of the parallel pathsThen : T=PZ Ia|N-m 2a T = KtIa| N.m (20) Where Kt : = constant= PZ/2a ( 20 ) . electromagnetictorqueorinternaltorque . Example: Whenamotorarmatureistaking50ampfromtheline,itdevelops60IB-ft torque.Thefieldstrengthisreducedto75percentofitsoriginalvalueandthe current increases to 50 amp. Determine new value of torqueSolution: For the two conditions, we have60 = K`t|.50 (a) Tnew = K`t 0.75|*80 (b) Dividing eq (b) by eq (a)Tnew = K`t 0.75|*80 60K`t|*50 Tnew = 60 * 0.75 * (80/50) = 72 lb.ft [10 HP 110-volt motor, Ra=0.05O , Ia=110/0.05 =2200 amp but Ia=90A only. .. ] Counter Electromotive Force Ra ... . ( 47 . ) : Ia =Vt E Amp.( 21 ) Ra Where: Ia : = armature current Vt : = motor terminal voltage Ra : = armature-circuit resistanceE : = Counter emfRewriting eq ( 21) we haveVt = E+ IaRa (22) ( 11 )E = V+ IaRa (11) .. IaRa.. IaRa. .. .. ( 7 : ) E =|NPZ volts( 7` ) 60a As Z,P and a are constant for any given motor , the counter emf isE = K|N Solving for speedN = K`E (23) | Where K` = 1/K Substituting for E in eq (23)its value given in (22) , the speed becomesN = K`V- IaRa (24) | ( 23 .. )( . 24 ). Example: In a motor the armature resistance is 0.1 ohm. When connected across 110-volt mains the armature takes 20 amp, and its speed is1200rpm. Determine its speed when the armature takes 50 amp prom this same mains, with the field increased 10 per cent. Solution: Applying eq.(24)N2 = K` 110-50*0.1 = 105 . |1|2 N1K`110-20*0.1108|2 |1 But N1 = 1200 rpm and |2 = 1.10|1 Therefore N2 = 1200 * 105 ( |1 ) = 1061 rpm 1081.1|1 ( 23 ) | ( 20 ) . .. : Ia = Vt E Ra . .Ia: T = ktIa| . . : . VtIa Ia2Ra , : Pm = VtIa - Ia2Ra= ( Vt - IaRa )Iawattsbut the counter emfE = Vt -IaRa Pm = EIa (25) . . . Pm . : ( 48 - ) |R . ( 48 - ) |R . . Vt E . : , , . , , . : , |f . ( 20 )( , , 49 - : ) T = Kt|Ia In a shunt motor | = |f = constantThenT = K`t Ia

T Ia : : .. E = K|fN: , .. Ia = Vt E Ra . ( .. ) , . ( 49 - ) . N =K` Vt - IaRa

|f Ia , IaRa 2 6 % ( , 49 -- ,) . ( 49 - ) . : ( 50 - ) , .T=K|tIa Substituting Ia for |, thenT=K"tIa2andT Ia2 ( 49 ) . ( 24 ) S = K`Vt- Ia (Ra+ Rs) (26) | Where:Rs : = series field resistance Ia | : .. Ia ( Ra+Ia ) 3 - 8 % Vt ) ( | Ia ( 50 ) . . ( 50 - ) . . Comulative CompoundMotor differential CompoundMotor : ( 51 : )1 2 3 Kinetic energy . ( 51 . ) . . : : Conveyers, Cranes Starting Direct current Motors .. .. .. Starter ( 52 ) Three-Point Starter ( 52 ) Four Point Starter. . . . releaseOver Load ( 53 - ) ( 53 . ) ) ( . ( 52 53 ) . ( 54 ) ) ( Controller. : 150 - 200 % .. . ( ... . 55 ) . .. : E =Vt Ia-FL Rstep x( i ) mIa-FL .. E =Vt mIa-FL Rstep x+1 ( i i) equating eqs ( i ) and ( i i)mIa-FL Rstep x+1 = Ia-FL Rstep x Rstep x+1 = Ia-FL *Rx(27) mIa-FL ( 27 ) . Example: Determine the resistance of each step of a starter for the following motor : 10 HP,240volts,armature-circuitresistance0.5ohm,Full-loadcurrent45amp. Starting current to be 150% percent of full-load current Solution: R1 = Vt = 240 = 3.56 O Imax67.5 R2 = 100 *3.56= 2.37 O 150 R3 = 100 *2.37= 1.58 O 150 R4 = 100 *1.58= 1.052 O 150 R5 = 100 *1.052= 0.704 O 150 R6 = 100 *0.704= 0.468 O 150 ( R6 ) : Speed control : N Vt - IaRa

| : Vt | . ( I2R ) , . ( 56 ( , ) 57 :: ) , Smoth acceleration . , - ( 58 . ) R R R . Motor Testing , . - 50 , . Fig (57) ward-leonard system Prony brake ( 59 ) F r . : Work done in one revolution = F ( 2 r ) kg-m Where:F : = net force , kg and r : = brake arm , m Then ,Work done in N rpm =F ( 2 r ) kg-m The hours powerHp= 2 (Fr)N (28) 60*75 (1Hp = 75 kg-m/sec) ( or= 550 lb ft / sec ) but torque T = Fr , thenHp= 2 TN (29) 4500 . , : : Efficiency = Output (30) Input , : Efficiency = Output (31) Output + losses andEfficiency = Input- losses (32) Input . ( , 31 ) ( 32 ). : - R2I : , , : 1 - : ( , 60 . ) V1 . V1 V1 V2 . Pa = I2a Ra (33-a) . 2 - : IF Vf , Vf Pf =VFIF(33-b) . , 3 - : , IS : , PS= IS2RS RS. . - : Stray Power Loss : , 1 - : Core Loss Eddy Current Loss Hysteresis Loss - .. , | N- . . - : , "Steinmetz Formula"PH NB1.6 2 - Friction and windage losses . . : Stray load loss 200 1 % . - : For a generator = VI (34) VI+Ia2Ra2+If2Rf+Is2Rs+S.P.+PSL for a motor = VI- (Ia2Ra2+If2Rf+Is2Rs+S.P.+PSL)

(35)VI Where:S.P. : = Stray power lossPS.L : = Stray load loss Rt R 75 R = ( 234.5+75 ) Rt 234.5 +t ( 34 ( ) 35 ) S.P. ( : ) : - | N | . : E = K |N( (36) E)1and then | =N K N. . E | ( 61 () ) IaRa :Va ~E ( E/N )VaI=Va(Ia+If) = VaIa+VaIf S.P. VaIa+VaIf=VaIf+Ia2Ra+S.P S.P. = VaIa- Ia2Ra(37) : S.P. = f`(|, N) | If ( 36 ) K : ,N) (38) E S.P.=f`(N S.P. ( 62 : )Fig 62 Va E E/N E, E/N S.P. Swinburn`s TestorStraypowermethod : 1 - . 2 - . : . : ( 34 : )1 - Ia ~I .2 - Is = Ia ~I .3 - ( ) PSL4 - ( 34 : ) = VI VI+I2(Ra+ Rs) +I2fRf +S.P. = V V+I(Ra+Rs)+ 1 (I2pRf+S.P.) I d [V + I(Ra+Rs) + 1 (I2fRf+S.P.)]=0 dII From which , (Ra+Rs) - 1 (I2fRf+S.P.) =0 I2 I2(Ra+Rs) = I2fRf +S.P. (39) . Fig 62 : Opposition Test ( Kapp Method , Hopkinson Test ) S.P. . VI (63 : ) Motor-armature loss = I21R1 Generator-armature loss = I22R2 Motor stray Power= (S.P)1 Generator stray Power = (S.P.)2 R1R2. .. .. . | . : .(S.P.)1 = E1 (40) (S.P.)2E2 Where E1 = V-I1R1(41-a) andE2 = V+I2R2 (41-b) : (S.P)1 +(S.P.)2 = VI I21R1 I22R2(42) : 1 - Psl . 2 - . 3 - . ( 40 ( ) S.P. ). 1-Foreachtypeofthedifferentd.c.generators,discussthepossibilityofits operation as : (a) a practical voltage source(b) a practical current source. 2-givethesalientconditionwhichmustbemaintainedwhilemeasuringeachthe following characteristics of the d.c. generator : (a) the internal characteristics,(b) the load characteristics , and(c) the regulation characteristics . 3Inacommulativecompoundgenerator,explainhowcouldthedegreeof compounding be varied. Specify thethreedistinguishes degrees of compoundingof this generator . 4 State the condition which must be hold when connecting a d.c. generator to the commonbus-bars.Whatismeantby"afloatingd.c.machineonthecommonbus-bars?Doessuchamachineneedaprime-mover?Howcanafloatingmachinebe forced to operate as : ( i ) motor , ( ii ) generator . 5 After the parallelizing of a d.c. generator to an another , explain how to share the load between them. Discuss the steady. State stability of the parallel operation , if the two generators are : (a) shunt generators. (b) commulative compound generators give the solution which may be used to prevent any instability. 6 Describe and explain the function of each of the following special d.c. machines : (a) Rototrol(b) Regulex7Whatismeantbycross-Fieldmachines?Nameageneratorwhichbelongsto these machines and explain its function as a rotating amplifier.8Buildeachofthefollowingmachines(i)Rototrol,(ii)Regulex,and(iii) Metadyne ; in a control system which may be used to regulate : (a) the voltage of d.c. generator (b) the speed of d.c. motor , and (c) the voltage of an alternatorgive the schematic diagram in each case, and explain how the voltage or speed could be maintained constant9Ad.c.generatorsuppliesthesameratedfull-loadcurrentatthesameterminal voltage when the machine is connected either self-excited shunt, or separately exited compare between the steady-state short-circuit current in both case if the machine is short-circuited by increasing the load. 10 Explain how to get the shunt characteristics of a d. c. generator from its open-circuit characteristics;(i) neglecting the armature reaction(ii) taking the armature reaction into consideration : 1 - . ( . ) . 2 - ) ( ( . ) . starter3 - . . . braking4 - system outputs inputs : 1 - open loop control. . 2 - closed loop control. . . . . . 1 - 2 - 3 - 1 - starter ...V=E+IaRa aaRE VI=1 ... E=0 2 aaRVI = 3s aaR RVI+= ( 1 ) 5 90 % 200 E 0.2 : 48 ... V R I V EA IIVIHPinputoutputARVIa aaaaaas86 . 195 2 . 0 * 72 . 20 20072 . 20 100 *90 * 200746 * 5100 ** 200746 * 590% 100 *746 *% 100 * %10002 . 0200= = == === == = =q : starter . ) ( R . 4 3 2 1R R R R RR RVIaas+ + + =+= R Ias 1.5 . ( 5 ( ) 6 .) ( 1 ) off a( 6 )( 5 ) brass ( 4 .) ON ( 2 ) . . . Diodes . . . Thyristor . 6o cosoV V = Vo cos(0) =1 .7 o ocos V( 1 & 2 ) Ias V 1.5 ... . V . 4 5Fig( 2 )( 1 ) Thyristor . V 1.5 . o Vo=200 V=200ooa asV VVolt VR I V88031 . 0200216 . 6coscos * 200 216 . 6cos216 . 62 . 0 * 72 . 20 * 5 . 1== ===== =oooo 88o. .2 - : : 1 - Dynamic Braking 2 - Counter Current Braking3 - Regenerative Braking1 - R . ... . . I E/R . 2 - plugging rolling mills . : ) ( aaaR RE VI++=8 . . 3 - overhauling )( electricaltrain cageofahoist E V .)( . ( ) )( )( )( : ( 4 ) . . 180 inverter . . 3 - . ) ( V= E+IaRa ... | Kn E = | KR I Vna a=9 - V - Ra. - | - . : WardLeonard .) ( ( . 7 . ) - M1-G M1 . R . M G . M G . M1 . R G M . . : chopper . 1 - : . ) ( . ( . 8 ) regenerative . tachogenerator tachogenerator 10 1000 ./ Fs )/(/ 1450 / tacho 14.5 Vs Vref Es Es Amp1 A1 VA1 s As ref sse A VV V enF* 1V1s= == Ia VI DCCT Ia 20 100 FI / VI VA1 er Amp2 A2101112 Amp2 V . IarI A rI a Ie A VV V eF I V* 21= == ( 3 ) 20 300 1500 / 0.4 ( 7 ) A1=100 Amp1 10 ( ) Fr=1/5(V/A)&Fs=1/100(V/rpm)&A2=2000 50 . V ( ) Ia er & es - ) = ( - 500 / - 100 / 20 1500 / 20 % V . : Vref = 15 1500 rpm n=0&Eb=0 50AV=IaRa=50*0.4 V=20V 20V Amp2 V e A VV V V eV nF VVAVeS AS ref SS SI1500 15 * 100 * 115 0 150 01 . 0 * 001 . 020002021= = == = == = == = = Amp1 10V VA1 10V 1500V . eI & VAI 14VI=10-0.01=9.99V 13 AFVIIIa95 . 495199 . 9= = VI=20V, Ia=49.95A, er=0.01, es=15V 500 rpm E=V-Ia*RaE=300-20*0.4=292V n Eo 131415V EnnE EnnEE33 . 971500500* 2922121 21212== == V Ia 50A V e A VV V V eV nF VVAVeV R I E VS AS ref SS SIa a1000 10 * 100 * 110 5 1551001* 5000587 . 0200033 . 117233 . 117 4 . 0 * 50 33 . 971= = == = == = == = == + = + = VA1 10 V . VS = 15-0.1=14.9V 1490 rpm 1490 rpm VA1 10V 10V 1490rpm .V e e I V VFVIV e V VS r aIIaI AI I10 & 0587 . 0 & 707 . 49 & 33 . 117707 . 49519413 . 99413 . 9 0587 . 0 10= = = == = == = = 00 rpm 10 500 rpm E2=149.7V V=214.7V VS=10V ES=5V EI=0.0173 VI=9.8927V Ia=49.46A =20AaI 300V V VV F I VVAVeAII a II15 . 4 15 . 0 4451* 2015 . 020003002= + == = == = = % 277 . 0 % 100 *15001496 1500% 100149610019585 . 149585 . 14 0415 . 0 150415 . 010015 . 4=== = == ==speederrorrpmFVnV VV eSSSS )( 20 %f aI I To If IaV VA IIIaa551* 252520 *1002520= ==+ = Eb % 343 . 0 % 100 *15009 . 1494 1500%9 . 149451949 . 14949 . 14 05151 . 0 1505151 . 0100151 . 5151 . 5 151 . 0 5551* 25151 . 020003022302 4 . 0 * 25 292=== = == == == + == = == = == + = + =speederrorrpmFVnV VV eV VV F I VVAVeV R I E VSSSSAII a IIa a 2 - Chopper Thy T1 T2 . V Vav 2 11t ttV Vav+= 16 - ratedspeed )( . )( rated current I2a*Rad . . - : ( 9 ). : . : )( Rsf N )(. Rsa N . ( 4 ) 300 0.5 300 1000 / 3 30 )( 3 ) ( 20 %A IA IA Iaaof29 1 302 1 31300300= == == = Iao Ia. rpm nnnnEEI Irpm nEEnnV Erpm nnEEnnV EV Ebbof fbobbbobbbo8 . 829* 8 . 0 ** 10005 . 198299* 8 . 0 *** 8 . 0* 8 . 09 . 6632995 . 198* 10005 . 198 5 . 3 * 29 3008 . 9542995 . 28510005 . 285 5 . 0 * 29 300299 5 . 0 * 2 3001111 11 21 22122212====== === ===== == =||||| | - - ( 5 ) 200 0.08 0.12 30 1500 / 3000 / 9 30 O = =O == == === = O =+ = + =09 . 9 2 . 0 29 . 929 . 94 . 116 * 9 2004 . 11630 * 15009 * 3000* 194&194 2 . 0 * 30 2002 . 008 . 0 12 . 0222 21 1211adff bbffbbf bf bbs aRRR EV EI nI nEEnI EI n EV ErR R Ro|o | o ( 6 ) 240 40 800 / 0.15 50 % 0.2 0.15 A IIIITTIII II IIITTII TI I I Iaaaaaaa fa faafaf f a f24 . 693 * ) 40 (25 . 12 / 1& 2 / 122 2222212122212 21 12 21 1212 1 2 2==== = == =|||oo| 50 % T1 T2=1.5*T1