Electrical Machines :

induction phenomena

1

: . :

Dynamically induced e.m.f

- ..

n

.. : E = BL )1( volts : Where B := Flux density 2webers/m L := Conductor Length m := Conductor Speed m/s ) 1 ( : E = BLsin )2( volts ..

.Statically induced e.m.f

- ..

t n = sin .. )1( Fig )E = -Tc (d/dt )3( volts Where Tc: = number of turns in the electric circuit : = magnetic flux webers )1( )3( .2 Interaction Phenomena

) e( ) i( )B/ 2( ) F( (F = BLi 4( Newtons

. F F

)2-( ) 2-( . )2- ( . .

2 Fig

.. . , . .

..

. . )3( .

: )4( .. )2(

2

) ) 4 ( .. ( .. . . Gramme-ring generator . ) 5 ( 021 ) 5- ( .

Stationary armature

:

)4( : 05 / Frequency 05 / : f = pn :Where f: = Frequency ,cps 2/p: = no of pole pairs = P n: = speed 06 = ,rpsN

)5( cycle/sec

: :

0003 / PN ) 0003 () 01( = f = 021 021

250 cp: s

05 / =N 120f P )021() 05( 01

:

:

=

= 006 .rpm

.Synchronous Genr Synchronous speed N : Synch. Speed ,rpm 60cps 50cps 25cps 0063 0003 0051 0081 0051 057 0021 0001 005 009 057 573 027 006 003 006 005 052 573

No. of poles 2 4 6 8 01 21 61

20 30 40

300 200 150

" " Dynamo : 1 Direct current generator 2 Direct current Motor

Essential Construction of a.d.c machine :

1 - : The Field Poles , , , . . , Pole Core , Yoke .

The Armature

2-

armature Core Commutator . ,

. )1- ( ) 1- ( )1- ( . ) 1- ( micanite . )2( , air gap , ) 1/61 1/4 ( . .

.

: .. , .. .. , .. , .. . . , . . Gramme ring armature ) 3 - , ( .

:

.. , . .. . )4 - , ( : . . - .. .. , .. .. .

: 1- . 2- , . 3- . 4- . .

Drum winding

, Lap winding Wave Winding

coil . coil side . inductor : .. . front end connection . back end connection . : )Y ) Pitch , . Pole pitch . )Yf) front pitch . )Yb) back pitch Commutator pitch Yc

1 The simplex lap singly reentrant winding )6( ) ( A . ) ( B ) ( A ) B ( ) ( A ) ( A ) ( B . ) ( C ) ( B ) ( B ) ( C . ) ( A . ) ( A 1 7 ) ( B 2 8 ) ( C 3 9 : Coil A B C D E Slot Occupied 7 1 and 8 2 and 9 3 and 01 4 and 11 5 and Coil F G H I J Slot Occupied 21 6 and 31 7 and 41 8 and 51 9 and ,10 and 16 etc

)6( Fig

: 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 single reentrant

2

) ( A ) ( B ) ( A ) ( A ) ( B . ) ( C ) ( B ) ( B .. The duplex lap doubly reentrant winding ) ( Tow windings ) 7(

Duplex lap singly reentrant winding

3 :

) ( 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 : )01( ) ( A ) ( B ) ( C ) ( C ) ( A ) ( D ) ( C ) . ( B ) ( A .

. .

2 : .

:

)11( . .

chorded winding . 08% .. .

Equalizer Connections )21 - (

.. : 1- . 2- . . 3- .. circulating current . )21 ( .. . .. .

winding calculations :

- : Yb Yf : Yf = yb 2m )in terms of coil sides (1-a and for only 2 coil sides per slot, it becomes Yf = yb m in terms of slots )(1-b

Where: Yf := front pitch in 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.) ) ( . . Yb Yf ) - ( .Yb Yf : simplex -1 . duplex -2 triplex -3 .... . . 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) front pitch,(f) commutator pitch and draw a developed view of the 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 84 no of coils = 14 coils 6 = b) No of coil sides = 14 x 2 = 28 coil sides, each slot has 2 coil sides: 28 = 14 slots no of slots = 2 c) Simplex lap winding requires only one commutatar segment per coil, then the no commuter segments is 14. 28 coil sides = 7 coil sides

4 poles = d) back pitch Yb a pole pitch e) Front pitch Yf = Yb 2m 1=,m = 9 coil sides for retrogressive winding = 5 coil sides for progressive winding :f) The commutator pitch is the same as the multiplicity of the winding .Yc = 1 commutator segment

: .. , , . .

- :: Yb + Yf = Y )2( 2 , PY = ) P , ( , 2PY = Or 2 )3( P , : = Y

Y =

2m P 2 NC =

In coil sides (4) P = 2p

=Y

2NC2m 2p

= and Y

NC m P

In slot

( ) 4

: NC = PY m (5) . Y YC YC = 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 : NC m 141 1 1 YC = = 7 Or 6 Segments P 2 2 2 = The commutator pitch is not a whole number; therefour 14 coils can not be used. Reduce the no. of coils by 1 and repeat 13 1 YC = = 7 segments for progressive winding 2 =6 segments for retrogressive winding The no. of coils required = 13 cols 13*2 1 Yb = 6 Can be taken as 7 coil side 4 2 2 P 2 26 4 NC m P In commutator segments (6)

=Y

=

= 7 or 6

Yf = 2Y-Yb 7 = Yb Winding Table

7-41 = = 7 for progressive winding = 12-7 = 5 for retrogressive winding 7= , Yf ) , Yc =7 ( progressive

31=No of slot=14 , no of commutater segment

Dead or dummy coils , .

: .. , 006 . . 002 , . e.m.f. induced in the armature

)( B / 2 e = BL Volt L , / , )31( ) ( . .. , .. )31( , e N .S .. .. ) 31 ( ) ( B ab .. : Eav = BL Volt :Where B : = average flux density per pole pitch L : = active length of the conductor , weber / m ,, in m ab : = velocity of conductor = , in m/sec t . t : = time required for the conductor to travel the distance ab , equal to the pole pitch )Bl(ab = Volts = Eav t t : Where BL(ab) : = total flux per pole , and 1 = t .Sec np the average e.m.f. per conductor is

np volts 1/np and the total induced e.m.f. between brushes is Z Volts E = np a Where Z: = total no. of conductors on the armature a : = number of parallel paths , and Z/a : = no. of conductor in series If the speed in rpm , eq (7) becomes NP E= Volts Z 60a Eav = Example:

( 7)

( 7`)

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 60 = 15 rps

no. of paths = no. of poles = 6 E = 0.05 * 15 *6 * 300 6 225 volts

: , aE (8) = npZ . ( 7 ) .

) 2 ( , . ) 41 ( , , ) (mmf abcdef )41( .Part Core ab Yoke bc Core cd Gap de Armature ef Gap fQ Material Dynamo steel sheet Cast steel Dynamo steel sheet Air .Dyn Steel sheet Air Area , Flux square weber meter Flux density 2weber/m Ampere turns ,Length per meter meter Ampere turns

0.5 0.5

1A 2A 1A 3A 4A 3A

)1K1( /A 20.5 /A )1K1( /A 3 /A 5.0 K2( 4A 3 /A

1H 2H2 L 1H 3H ) 4H 3H

1L

1H1.L 2H2.L

1L 3L 4L 3L

1H1.L 3H3.L 4H4.L 3H3.L IN 2/IN

= Total ampere-turns for 2 ples Total ampere- turns per pole

:Where Core Flux 2.1 1.1 Armature Flux K : = Correction factor to allow for the thickness of the oxide on the surface of the 52.1-1.1 = lamination and the air-duct speed = : = leakage coefficient

D.C. GeneratorsTypes of Field Excitation :

: , . ) 51- ( Separately Excited d. c. Generators ) 51-,, ( Self Excited d. c. Generators

, b r If

: 1 : Shunt generator If . If r ) 51 ( 2 : Series Generator ) 51 ( . 3 : Compound Generator Fsh . Fse ) 51 ( .

: Characteristic of Generators curves Characteristic . , .

1 - :internal characteristic curve , open circuit characteristic , or no-load magnetization curve , : E = K N )9( : Where Z 1 = K ..P : = constant a 06 : = Flux per pole , in weber N: = speed of armature , in rpm . ) 61- ( .

, E - If , .. )1( )61- ( residual magnetism . 2 3 .. 3 4 . )4( , retentively . 4 5 Hystresis loop , . 1At N1 rpm the induced voltage is E1 = K N and 2At N2 rpm the induced voltage is E2 = K N 1E 1E 1N K N 2E =1

K N2

and

2E

=

2N

K , )61-( )61- ( .. , .. .

Field resistance line

) 71- ( , )71- ( -. , . .

: The build up process

)81- (, oa . , ab `ob .. . `cb , , `od .. .... `ed F , g , `0g `Kg `gg f .

, .

: Critical Field Resistance oa . `a oa . ob b ob )81- ( : 1 . .2 3 , .. 4 . .5

Armature Reaction

:

, , .

: ) 91- ( . F . ) 91- ( ,

, FA . , , . , , Trailing Pole Tips Leading Pole Tips )91- ( FR .. . )02- ( FA : FD : = demagnetizing component of armature reaction FC : = Cross-magnetizing component of armature reaction

. )02- ( : ,2 . . )12( FR , FC, FD FA , F FR FA . :Example A 4-pole generator has 288 surface conductors. The armature is lap wound, and .the armature current is 120 amp. The brunches are advanced 15 space degrees .Determine demagnetizing and cross-magnetizing armature ampere-turns :Solution

51 = 03 = 2 4 = ) No. of brunches = no. of poles ( lap winding total 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 0882 = 69 * 03 = .Demagnetizing ampere conductors 0882 Demagnetizing ampere turns = 0441= 2 = no. of cross magnetizing conductors = cross magnetizing amperes-turns 2 3 03*291 2 of the no. of conductors on the armature 0882=

: )22- ( , N` , S , N , , ) 22- ( - .. ) 32- ( , qr st I j k h ) 32- ( . b d f ) 32 ( . )42( .

:

. FD . ) `N ) 32 ( ) `r`s instability . : FA . :

1 - : ) 91- ( ) 32 ( :

: - ) 52 ( .

: ) 52 ( .

.

- : ) 52 ( .

2 , . :

- : compensating windings )62( .

: Interpoles .. .

)72-( .

: Commutation .. .. . ideal commutation : 1- . . . 2- )82( straight line commutation 1 01 . 02 04 . ) ( 01. 1 3 02 4 01 02 7 01.

Fig(28) current coil undergoing commutation condition )82-( ) +02( )- 02( t , ) 82- ( . ) 92 03(

Fig (29) commutation with braches too far back of neutral plane

Fig (30) commutation with braches too far ahead 54 5 05 04 02 ) 92 03 ( ) 92 ( 52 ) +02 ( ) -02 ( t 1 t . ) 03 ( .. . ) 13 ( .

. : .. .. . commutating Zone ) 23 ( . ) 82 92 03 ( ) 23 ( c 2 c .. .

ec = -N = -N

d c dt 2 tc

Volts volts )01(

:Where N : = no. of turns in the coil .t : = time to across the commutating zone , sec

.. . .. . .. .. .. .. .. .. .. .. Commutating poles Inter poles : .. .. .. . ) 33 ( 1 B 2 B .. .

2- : External characteristics

- Shunt generator characteristics hb ) 43 ( e break down point f .. . `h h `fgh hysteresis

: : 1- Ra Ia , , : V=E-IaRa )11( volts Where : V: = terminal voltage . , E: = generated voltage .. . 2- : 3- .. . )43- ( : , .. .. , .

) 53 ( 1 032 , 009 / 0021 / . ) ( ) 43 ( 02 IL V If .

: Generator Regulation . "" regulation : Percent voltage Regulator :Where 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 load is completely removed, the terminal voltage rises to 250 volts. Determine the .voltage regulation :Solution , VNL = 250 volts 032-052 % Regulation 001* 7.8 = 032 . : 1 . 2 . VFL = 230 volts VNL-VFL VFL )21( *001

Total characteristic .. . E Ia : (Ia = IL +IF 31( : Where IL : = load current , If : = shunt field current E : = V + IaRa )41( : qr ) 63 ( ) oa ob ( . oY oa . oe ef `oe . `e`f `e `of Ia `f`g . fg IaRa ( e`f ( e`f =oe g . .

:(Pg = E.Ia : watt 51(

:Where .E: = induced e.m.f , Ia : = total or armature current :Example A 20 kw 220-volt shunt generator has an armature resistance of 0.07 and shunt field resistance of 200 ohms . Determine power developed in armature when it .delivers its rated output :Solution 02*301 Rated current =I 9.09 = amp 022 Field current = If 022 002 1.1 = amp

Armature current Ia = 90.9 + 1.1 = 92.0 amp Induced 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 2 )022( Field loss = Pf 242 = watt 002 Armature loss Pa = (92.0 )2 * 0.07 = 592 watt Power developed in armature Pg = 20000 + 242 + 592 = 20834 = 20.83 kw

Compound generator characteristics . ) 73 ( . .

V = E IaRa IsRs )61( : Where Is : = series-field current Rs : = series field resistance

. cumulative compound generator differential- compound generator constant current generator . ) 83 ( . degree of compounding : 1 2 Over compounding Flat Compounding 3 under compounding ) 83 ( . . Diverter .

: 1 1 If 2 2. If

( 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 compounding Ns : = no. of turns in series field , Is : = series field current at rated load If2-If1)Nf) and Ns = (17`) Is In case of short compounding Is = IL , and in case of long compounding Is = 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 ,find the resistance value of the diverter when the generator is to operate as a cumulative . compound generator. The resistance of the series field is 0.005

Solution: Nf ( If2-If1) = NsIs (5.0-2.7)500 series field current required Is = 10 To field the resistance of the diverter : IdRd =IsRs (150-115)Rd = 115* 0.005Rd =

= 115 amp

115 * 0.005 = 0.0164 35

: - , . , .. , ..

.. .. . )_04 ( , . oa " d " . , bc Booster )14( Feeders .

: , , :

: -

, , : . " " Floating on the bus bars .. , . , , . , , . )24- ( )1( ) ( ) (a , )1( )2( )2( ) (b . , . - .

: -

)34- ( ) ( )34- ( . )1( .. ) (a , )2( ... )1( )2( , )1( . , ) 34- ( )1( )2( )1( )2( )2( )1( , .

D.C. Motors , . .

Principle of the Motor )44- ( )44- ( . F=BLI newtons )81 ( :Where 2B : = flux density of main field , in weber/m L : = Length of the conductor , in meter I : = current , in amperes Or I F= BL Dynes )`81( 01 :Where B in lines /cm2 or gauss, L in centimeters and I in amperes F )44 , ( , Fleming's left-hand rule , . , . , ) 54( , . )81( : , B . . .

Fig(45) torque produced by coil

T = Fxr N-m )91( :Where T : = torque , in N-m F : = Force , newton R : = distance measured perpendicularly from direction of F to center of rotation , m )54( , )64- ( )1( ) 2( , )3( r , )3( . )4( )3( . )4( )64- ( . )64- ( Commutator , , ) 64- ( .

Torque developed by a motor ) 64 ( . : : From eq.s ( 18, 19) the torque developed by an armature can be obtained as follows d (= T )) (ZBavLI ) N-m ( i 2 : Where . D : = diameter of the armature , in meter . The flux entering the armature of the one pole d ) ) ( ii ( =BaL P

P : = no. of poles Substituting in (i) the value of Bav from (ii) d P T =( )( )ZLI 2 Ld P 1 . . ZI 2 , a : = no. of the parallel paths = PZ I 2a a

But I = Ia/a Then :

T=

N-m

T = KtIa N.m (20) Where Kt : = constant = PZ/2a ) 02 ( . electromagnetic torque or internal torque . Example: When a motor armature is taking 50 amp from the line , it develops 60 IB-ft torque . The field strength is reduced to 75 per cent of its original value and the current increases to 50 amp. Determine new value of torque Solution: For the two conditions, we have 60 = K`t .50 Tnew = K`t 0.75 *80 Tnew = 60 (a) (b)(Dividing eq (b) by eq (a

K`t 0.75 *80 K`t *50 Tnew = 60 * 0.75 * (80/50) = 72 lb.ft[10 HP 110-volt motor, Ra=0.05 , Ia=110/0.05 =2200 amp but Ia=90A only ] . ..

Counter Electromotive Force

Ra . ... ) 74( . : = Ia Vt E Ra .Amp ) 12 ( :Where Ia : = armature current Vt : = motor terminal voltage Ra : = armature-circuit resistance E : = Counter emf Rewriting eq ( 21) we have Vt = 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 is E = K N Solving for speed E `N = K )32( Where K` = 1/K Substituting for E in eq (23) its value given in (22) , the speed becomes V- IaRa `N = K )42( )32( .. . )42( . :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 is 1200rpm. Determine its speed when the .armature takes 50 amp prom this same mains, with the field increased 10 per cent

:Solution )42(.Applying eq 1.0*05-011 2N `K 2 = = 1N 1.0*02-011 `K 1 1 But N1 = 1200 rpm and 2 = 1.10 Therefore 1 * 0021 = 2N 1601 = ) 1.1 ( rpm 1 )32( )02( .. . : Vt E = Ia Ra 501 801 . . Ia : T = ktIa . 501 . 801 1

2

. . : VtIa ,Ia2Ra : 2 Pm = VtIa - Ia Ra = ) Vt - IaRa )Ia watts but the counter emf E = Vt -IaRa Pm = EIa )52( . . . Pm .

:

) 84- ( R . ) 84-( R . . Vt E .

: , , , . , . :

, f . )02( , , )94- ( : T = Kt Ia In a shunt motor = f = constant Then T = K`t Ia T Ia : .. : E = K fN .. , : Vt E = Ia Ra . ) .. , ( . )94- ( . Vt - IaRa `N =K f Ia , IaRa 2 6 % , )94-- (, . )94- ( .

: )05- ( ,

. T=K tIa Substituting Ia for , then 2T=K"tIa2 and T Ia ) 94 ( . ) 42( `S = K )Vt- Ia (Ra+ Rs )62(

:Where Rs : = series field resistance Ia : .. Ia ( ( Ra+Ia 3-8% Vt ) ( Ia ) 05 (.

. )05- ( .

. Comulative Compound Motor differential Compound Motor : ) 15 ( : 1 2 3 Kinetic energy . ) 15( . . .

: Conveyers, Cranes

:

Starting Direct current Motors .. .. .. Starter ) 25 ( Three-Point Starter ) 25 ( . Four Point Starter

. . . release Over Load ) 35- ( )35 ( .

) ( .

) 25 35 ( . ) 45 ( ) ( Controller .

: 051-002% .. .. ... ) 55 ( .

.. : E =Vt Ia-FL Rstep x )(i -m Ia FL .. 1+E =Vt mIa-FL Rstep x )( i i )equating eqs ( i ) and ( i i mIa-FL Rstep x+1 = Ia-FL Rstep x Ia-FL = 1+Rstep x *Rx )72( mIa-FL

) 72 ( . :Example 01 : Determine the resistance of each step of a starter for the following motor . HP , 240 volts , armature-circuit resistance 0.5 ohm , Full-load current 45 amp Starting current to be 150% percent of full-load current :Solution

Vt 042 = = 3.56 Imax 5.76 001 = 2R * 3.56 = 2.37 051 001 = 3R * 2.37 = 1.58 051 001 = 4R * 1.58 = 1.052 051 001 = 5R * 1.052 = 0.704 051 001 = 6R * 0.704 = 0.468 051 ) 6 R ( = 1R

: Speed control : Vt - IaRa : Vt . ) ( I2R , N

. )65( , ) 75 ( :: , Smoth acceleration .

, - ) 85 ( . R R R Fig (57) ward-leonard system .

Motor Testing

, . - 05 , .

Prony brake ) 95 ( F r . :

) Work done in one revolution = F ( 2 r :Where F : = net force , kg , Then Work done in N rpm The hours power = Hp and r : = brake arm , m ) = F ( 2 r 2 (Fr)N 57*06 kg-m )82( )(1Hp = 75 kg-m/sec ) ( or = 550 lb ft / sec but torque T = Fr , then = Hp 2 TN 0054 )92(

kg-m

, .

: : Output = Efficiency )03( Input , : Output = Efficiency )13( Output + losses and Input- losses = Efficiency )23( Input

. , ) 13 ( ) 23 ( .

: - : I2R , , : : 1- , ) 06 ( . 1 V . 1 V 1 V 2 V . 2 Pa = I a R a )(33-a . : 2- IF , Vf Vf Pf =VFIF )(33-b , . : 3- , IS , : 2 PS = IS RS RS . .

Stray Power Loss

- :

, : : Core Loss 1-

Eddy Current Loss Hysteresis Loss -

.. , -N . . : -

, "Steinmetz " Formula 6.1 PH NB Friction and windage losses 2- . .

:Stray load loss 002 1% . - : For a generator VI = )43( 2 2 2 VI+Ia Ra +If Rf+Is2Rs+S.P.+PSL for a motor )VI- (Ia2Ra2+If2Rf+Is2Rs+S.P.+PSL 3( = )5 VI :Where S.P. : = Stray power loss PS.L : = Stray load loss Rt R 57 57+5.432 (= R ) Rt 234.5 +t

)43( )53( ) .S.P ( :

: - N . : E=K N 1 E ( ( )63( = and then K N N . . E )16( ) ( IaRa : Va E ) (E/N 26 Fig VaI=Va(Ia+If) = VaIa+VaIf .S.P VaIa+VaIf=VaIf+Ia2Ra+S.P S.P. = VaIa- Ia2Ra )73(

: )S.P. = f`( , N If )63( K : )`S.P.=f E ),N )83( N .S.P ) 26( :

26 Fig

Va E E/N E , E/N .S.P Swinburn`s Test or Stray power method : . 1- . 2- : . : )43( : . Ia I 1- .Is = Ia I 2- ) ( 3- PSL )43( : 4- VI = 2 .VI+I (Ra+ Rs) +I2fRf +S.P = V 2 1 +)V+I(Ra+Rs ).(I pRf+S.P I

d 2 1 + )[V + I(Ra+Rs 0= ]).(I fRf+S.P dI I , From which )(Ra+Rs1 ).(I2fRf+S.P 2I 0= )93(

. I2(Ra+Rs) = I2fRf +S.P

.

: ( Opposition Test ( Kapp Method , Hopkinson Test .S.P . ) VI 36 ( : Motor-armature loss 1= I21R Generator-armature loss 2= I22R Motor stray Power 1)= (S.P Generator stray Power 2).= (S.P 2R1 R . . .. .. .. : 1).(S.P 1E = )04( 2).(S.P 2E Where 1E1 = V-I1R )(41-a and 2E2 = V+I2R )(41-b : 2(S.P)1 +(S.P.)2 = VI I21R1 I22R )24(

: Psl 1- . . 2- . 3- )04( ) (.S.P .

1 - For each type of the different d.c. generators , discuss the possibility of its operation as : (a) a practical voltage source (b) a practical current source. 2 - give the salient condition which must be maintained while measuring each the following characteristics of the d.c. generator : (a) the internal characteristics, (b) the load characteristics , and (c) the regulation characteristics . 3 In a commulative compound generator , explain how could the degree of compounding be varied. Specify the three distinguishes degrees of compounding of this generator . 4 State the condition which must be hold when connecting a d.c. generator to the common bus-bars. What is meant by "a floating d.c. machine on the common busbars ? Does such a machine need a prime-mover ? How can a floating machine be 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) Regulex 7 What is meant by cross-Field machines? Name a generator which belongs to these machines and explain its function as a rotating amplifier. 8 Build each of the following machines (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 alternator give the schematic diagram in each case, and explain how the voltage or speed could be maintained constant 9 A d.c. generator supplies the same rated full-load current at the same terminal 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 opencircuit characteristics; (i) neglecting the armature reaction (ii) taking the armature reaction into consideration

: 1- . ) ( . . 2- ) ( . ) ( . starter 3- . . . braking 4-

system outputs inputs : open loop control 1- . .

closed loop control 2- . . . .

.

. 1- 2- 3-

1- starter ... V=E+IaRa= Ia V E Ra

1 2 3 ... 0= E

V = Ia Ra V = Ia Ra + R s

)1( 5 09% 002 E 2.0 := I as V 002 = = 1000 A Ra 2.0 output 647 * HP = % 001 * % 001 * input VI a

= %= 09 = Ia

647 * 5 001 * 200 * I a

647 * 5 * 100 = 20 .72 A 09 * 002 E = V I a Ra = 200 20 .72 * 0.2 = 195 .86V

84 ...

: starter ) ( . R . = I as V Ra + R

4 5

4R = R1 + R2 + R3 + R

R Ias 5.1 . )5( )6(. )1( off a )6( )5( 4( .(brass ON )2( . . .

Diodes . . . Thyristor . V = Vo cos 6 Vo 1= )0(.cos V cos 7 )1 & 2( Ias V 5.1 . ... V .

Fig

2.0 * 27. 02 * 5.1= V = I as Ra V = 6.216 Volt V = Vo cos 6.216 = 200 * cos 612.6 = cos 130.0 = 002 = 88 o

)2( )1( .Thyristor V 5.1 . 002= Vo 002=V

.88o .

2- : : Dynamic Braking 1- Counter Current Braking 2- Regenerative Braking 3- 1- R . ... . . I E/R . 2- plugging rolling mills . ) ( := Ia V + Ea R + Ra

8

. . 3- overhauling )( electrical train cage of a hoist E V )(. . )1( )( )( )( : )4( . . 081 inverter

. .

. ) ( : V= E+IaRa ... E = Kn =n V I a Ra K

3-

9

: - V - Ra . -

-

. : Ward Leonard ) (. . )7( . - M1-G 1 M . R . M .G M G . 1 M . R G M .

.

: chopper . 1- : ) ( . . . )8(

regenerative . tachogenerator tachogenerator 01 0001/. Fs / )/( 0541 / tacho 5.41 Vs Vref

Vs = nF s

Es Es 1 Amp 1 A 1VA :01 11 21

e s = Vref V s V A1 = A1 * e s

VI = I a FI

Ia VI DCCT Ia 02 001 FI / VI 1VA er 2Amp 2 A 2 Amp V . Ia:er = V A1 VI

V = A2 * er

31 41 51

)3( 02 003 0051/ 4.0 )7( 001=1 A 1 Amp 01 ) ( & )Fr=1/5(V/A) & Fs=1/100 (V/rpm 0002=2 A 05 . )V ( Ia er & es : ) = ( 1- - 005/ - 001/ 02 0051/ 02% V . : 51 = Vref 0051 rpm 0= n=0&Eb 50A 4.0*05=V=IaRa V=20V 20V 2 Amp :

V 02 = = 0.01V 0002 2A VS = nFS = 0 * 0.01 = 0V = eI eS = Vref VS = 15 0 = 15V V A1 = A1 * eS = 100 * 15 = 1500V

1 Amp 10V 1 VA 10V . 1500V eI & VAI 41 VI=10-0.01=9.99V 31 := Ia 99.9 V I = 49 .95 A 1 FI 5

: VI=20V, Ia=49.95A, er=0.01, es=15V 005 rpm : E=V-Ia*Ra E=300-20*0.4=292V En 2E 2 n = 1E1 n 1E 2 = E 2n 005 * 292 = 1n 0051

E 2 = 97.33V

V = E + I a Ra = 97 .33 + 50 * 0.4 = 117 .33V V 33. 711 = = 0.0587 V 2A 0002 1 * 005 = V S = nF S = 5V 001 e S = Vref V S = 15 5 = 10V = eI V A1 = A1 * e S = 100 * 10 = 1000 V

V Ia 50A :

1 VA 01 V . : VS = 15-0.1=14.9V 0941 rpm 0941 rpm 1VA 10V 10V . 1490rpmV I = V AI e I = 10 0.0587 = 9.9413V = Ia VI 3149.9 = 707.94 = 1 FI 5 V = 117 .33V & I a = 49 .707 & er = 0.0587 & e S = 10V

0001 rpm

005 : rpm E2=149.7V V=214.7V VS=10V ES=5V 3710.0=EI VI=9.8927V Ia=49.46A Ia=20A : 300VV 003 = = 0.15V 0002 2A 1 V I = I a FI = 20 * = 4V 5 V AI = 4 + 0.15 = 4.15V = eI

51.4 0.0415V 001 VS = 15 0.0415 = 14.9585V = eS =n 5859.41 VS = = 1496rpm 1 FS 001 6941 0051 = %001speederror %772.0 = %001* 0051

)( 02% :T a I f I

If Ia+ 02 = I a I a = 25 A * 52 = V I 1 = 5V 5 52 02 * 001

V = E + I a Ra = 292 + 25 * 0.4 = 302V V 203 = = 0.151V 0002 2A 1 V I = I a FI = 25 * = 5V 5 V AI = 5 + 0.151 = 5.151V = eI 151.5 = 0.05151V 001 VS = 15 0.05151 = 14.949V = eS =n

Eb :

949.41 VS = = 1494 .9rpm 1 FS 5 9. 4941 0051 = % speederror % 343.0 = %001 * 0051

2- Chopper Thy 1 T 2T . V Vav :Vav = V 1t 2 t1 + t

61

- rated speed )( )( .

rated current *I2a Rad . .

- : )9( . :

. :

)( Rsf N .)( Rsa N .

)4( 003 5.0 003 0001 / 3 03 )( 3 ) ( 02% = If I ao 003 = 1A 003 = 3 1 = 2 A

I a = 30 1 = 29 A

E bo = 300 2 * 0.5 = 299 V E 2n = b 1n E bo 2n 5. 582 = 0001 992 n 2 = 954 .8rpm

Iao Ia .

E b = 300 29 * 0.5 = 285 .5V

E b = 300 29 * 3.5 = 198 .5V E 2n = b 1n E bo * 0001 = 2 n 2If 5. 891 = 663 .9rpm 992 1 = 0.8 * I f

-

12 = 0.8 * E bo 1n1 * = Eb 1n * 0.8 * 11000 * 992 = 1198 .5 n * 0.8 * n = 829 .8rpm

-

80.0 + 21.0 = R = Ra + Rs r = 0.2

)5( 002 80.0 21.0 03 0051/ 0003 / 9 03

E b1 = 200 30 * 0.2 = 194 V E bn & I f E bnIf

1 n1 I f 1E b = 2 Eb 2 n2 I f * 491 = 2 E b 2 Eb 9 * 0003 = 116 .4V 03 * 0051 4. 611 = = 200 9 * R f

R f = 9.29 Rad = 9.29 0.2 = 9.09

2 I f 2 = 1 / 2I a 2 & I f 1 = I f T a I

)6( 042 04 008 / 51.0 05% 2.0 51.0

I f2 1I f 1 I a I 1I a 1T = 1= 1 a = 2 2 T2 2 I a 2 I f 2 I a 2 1 / 2I a

2 2I 1T 1= 2a 21.5T 2I a2 3 * 2 ) 04( = 2 I a

05% 1 T 1T2=1.5*T

I a 2 = 69 .24 A