1

AlternAtors, AC InduCtIon motors, synChronous motors, dC mAChInes, Power fACtor CorreCtIonInduction motors:

1. General lay-out of losses in induction motors:

Input power to stator windings

I2R loss instator

windings

Iron lossin stator

core

Synchronouspower

transferred torotor via

magnetic fieldin air gap.

[2πnsT Watts]

I2R loss inrotor

windings

Iron lossin rotor

core

Total mechanicalpower developed

by rotor[2πnrT Watts]

Useful mechanicalpower at output

shaft

Windageloss

Frictionloss at

bearingsand

sliprings

2

2. Tree-diagram of induction motor losses:

Stator

Rotor

Shaft

Copper losses(Pcu-s)

Iron (core) losses(Piron-s)

Copper losses(Pcu-r)

Iron (core) losses(Piron-r)

Windage losses Friction losses

Outputpower(Pout)

Inputpower(Pin)

ssphasescu RIP .).(3 2−− =

sironscuinouts PPPP −−− −−=

rrphasercu RIP .).(3 2−− =

φcos...3 linelinein IVP =

ηout

inPP =

outsrcu PsP −− = .

( )frictionwindagerironrcuoutsininout PPPPPPPP ++++−== −−−.η

3

3. p.u slip (s) = n

n ns

s r- where ns is the synchronous speed of

the motor and nr is the rotor speed of the motor.

4. Slip-speed = s ns$

5. n n slipspeedr s= =

6. f pn=

7. f s p nr $ $=

8. Rotor emf/phase at standstill: 2.22E k fp0 $ $ $ z= where kd is the distribution factor with:

. sin

sink

n

n

2

2d a

a

= and kp is the pitch factor with cosk2p}

=

n= number of slots/pole/phase

a= slots per pole

1800

}= (1 – coil pitch as a fraction of the pole pitch) x 180°

9. Z R Xrotor rotor2 2= + where Xrotor is the leakage reactance

per phase.

10. X s Xrotor 0$=

11. IZE

Zs E

rotorrotor

rotor

rotor

0$= = and tan

RX

Rs X

rotorrotor 0$

z = =

and cosZR

rotorrotor

z =

where: E0 = voltage per phase of stator X0 = reactance per phase at no-load

12. IR X

E0 2

02

0=+

13. Total I R2 losses in rotor = s Pcu out$ - = slipspeed T T2 rotor rotor 1 2$ $r ~ ~= -^ h

Where: n2 S1 $ $~ r= and n2 r2 $ $~ r=

14. Torque characteristics:

• p.u torque reduction to rotor:

TR s X

s E R

R s X

s R2

02

02

20

2

2

\$

$ $\

$

$ $z

+ +^ ^h h

• At maxumum torque:

R s X0$=

• Torque on rotor varies as (stator applied voltage)2

• Direct switching to supply:

TT

II

sfull load

starting

full load

short circuitfull load

2

$=- -

--c m

•• Star-delta starter:

TT

II

s31

full load

starting

full load

short circuitfull load

2

$=- -

--c m

• Motor speed at maximum torque = pf

s1 -^ h rev/s where

f is the rotational frequency, p is the pair of poles and s is the p.u. slip of the motor.

15. Phasor diagram:

Ø1

Ø2

Ф

I1I1'

I0

E1

V1I1X1

I1R1

I1Z1

E2

I2

I1 = stator currentI0 = Stator no-load currentI'1 = Stator demagnetizing currentI2 = Rotor currentU = flux developed by statorV1 = terminal voltageE2 = rotor EMFE1 = stator EMF

1Q = staor phase angle2Q = rotor phase angle

4

16. Circle diagram:

O

P

SK

B

F

D

E

JXM

LN

RC

GH

AI0

V

Torque line

Output line

Parallel to torque lineParallel to output line

Isc (projected)

Powerfactor

quartercircle

Phase angle at no-load

Phase angle at short-circuit

Phase angle at full-load

Rotor Copper losses

Stator Copper losses

Watt component of ironlosses + wind + friction

• Calculate power factor at no-load ( cos s 0z ) and short-circuit

( cos scz )

• cosV I

P

30

0

0

$ $z = & cos

V I

P

3sc

sc sc

sc

$ $z =

• Project value of Isc against normal voltage: IVV I

( )sc projectedsc

0$=

• Draw OV vertical and OE horizontal.

• Draw power factor quarter circle at convenient scale and calibrate on vertical axis OV. ( cos0 1< <z ) [at increments of say 0.01]

• Mark cos 0z on vertical axis and project horizontally to where it meets the power factor quarter circle.

• Construct scz by connecting origin O with intersection at the power factor quarter circle.

• Construct OB and connect AB. The line AB is now referred to as the output line.

• Construct a centre-perpendicular line on AB and extend it to meet AD at X. The point X forms the centre of the half circle ABJ. ( AJ OE' )

• Construct the half circle ABJ with X as centre point.

• Draw BD perpendicular to AJ and extend to E.

• Bisect BD to establish F. The line AF is now referred to as the torque line.

• BF represents the rotor copper losses, DF represents the stator copper losses, whilst DE represents the watt component of the iron losses, wind losses and friction losses.

• Mark cos full loadz - on vertical axis and project horizontally to where it meets the power factor quarter circle.

• Construct cos full loadz - and connect OP where P is the intersection with the half circle.

Typical values that can be deduced from the circle diagram:

• P V PC3output line$ $=

•PHPC

h = with GH being the watt component of the no-load

input power.

• sKMLM=

• P V SN3max line $=

• T BFstarting \

• Synchronous watts = V BF3 line$ $

5

Slip-ring induction motors:

Slip ring

n n-1 n-2 2 1

R0

Rn-1

Rn-2

R2

R1

R0 = rotor resistance per phaser = resistance per step

17. ..........sR

sR

sRn1

1

2

2 0= = = where sn is the slip at full load.

18. ..........sR

sR

sR

n

n

2

1

3

2 1= = = -

19. .......... ..........ss

ss

RR

RR

RRn1

2

2

3

1

2

2

3

1

0= = = = =-

20. ; ;RsR

RRR

R RRR

Rn n n

10

21

01 3

1

02= = =

- -c cm m

21. ; ; ; ..r R R r R R r R R r R R etcn n n n n1 1 2 2 2 3 1 1 0 2 2 1= - = - = - = -- - - - -

22. R R kn 1 0 $=- where kRRn

0

11= -

23. Rotor resistance per phase = P

s s E

s Is P3 1

3 1 r0

02

20$ $-

=-

^^

hh

Where:

sn

n n

synchronous

synchronous rotor=

- with n

pf60

synchronous

$=

E0 = rotor voltage at standstill = rotor volts3

P0 = power consumed at no-load

Ir = full load rotor current

6

24. f p n$=

25. RMS value of EMF per phase = . k k Z f2 22 d p$ $ $ $ $ z

26. sin

sink

n

n

2

2d

$

$

a

a

=

where:

n = the number of slots per pole per phasea = pole pitch

27. cosk2p}

= where } is the angle of short pitch coil.

28. degrees electrical = p x degrees mechanical

29. Parallel operation:

E1

E2

V

Z S1

Z S2

I1

I2

I total

30. V E I ZS1 1 1$= - or V E I ZS2 2 2$= -

31. IZ

E VS

11

1=-

and IZ

E VS

22

2=-

32. IZ Z Z Z Z

E Z E Ztotal

S S S S

S S

1 2 1 2

1 2 2 1=+ +

+^ h

33. Circulating current: IZ ZE E

CS S1 2

1 2=+-

DC Machines

34. Ec

p Z N60

2 $ $ $ $z= and T

cI Z pa

$

$ $ $

r

z=

35. P N TE I

602

a$ $ $

$r= =

36. Shunt generator:

Armature Rshunt

RA

IA

Ioutput Ishunt

Voutput

E V I RA A= +

I I IA output shunt= +

37. Shunt motor:

Armature Rshunt

RA

IA

Iinput Ishunt

Vinput

E V I RA A= -

I I Iinput A shunt= +

7

38. Series generator:

Armature

RA Rseries

Terminal voltage (V)

Induced EMF (E)

E V I R RA A series= + +^ h

39. Series motor:

Armature

RA Rseries

Terminal voltage (V)

Induced EMF (E)

E V I R RA A series= - +^ h

40. Compound short-shunt motor:

Armature

Rseries

Rshunt

RA

Vinput

E

Vseries

Vshunt

Iseries

Ishunt

IA

V V Vinput series shunt= +

I I I Iinput series A shunt= = +

E V I Rshunt A A= -

8

41. Compound short-shunt generator:

Armature

Rseries

Rshunt

RA

Voutput

E

Vseries

Vshunt

Iseries

Ishunt

IA

V V Voutput series shunt= +

I I I I IA shunt series shunt output= + = +

E V I Rshunt A A= -

42. Compound long-shunt motor:

Armature

Rseries

Rshunt

RA

Vinput

E

Vseries

Iseries

Ishunt

IA

IinputVA

V V Vinput series A= +

I I I I Iinput shunt series shunt A= + = +

E V I RA A A= -

43. Compound long-shunt generator:

Armature

Rseries

Rshunt

RA

Voutput

E

Vseries

Iseries

Ishunt

IA

IoutputVA

V V Voutput A series= +

I I I IA series shunt output= = +

E V I RA A A= -

9

44. Articulated scenarios:

Ec

p Z N60

2 $ $ $ $z= and T

cI Z pa

$

$ $ $

r

z=

keeping p, Z & c constant:

E NEE

NN

2

1

2 2

1 1&\ z

zz

= and T ITT

II

aa

a

2

1

2

1

2

1

&\ $$$

zzz

=

Keeping the flux constant as well:

E NEE

NN

2

1

2

1&\ =

45. Efficiencies of DC machines:

efficiencyinputoutput

= and input output losses- =

• Direct or rope-brake test:

W

r

S

PNT60

2out

r= and P V Iin $=

V IN w s r

602

$ $

$ $h

r=

-^ h

• Indirect or Swinburne method:

Exposition of losses

Variable losses

Armature losses: no-load (Pa0 ) I Ra a20 $

Armature losses: full load (Pa ) I Ra a2 $

Shunt losses (Pshunt ) I Vshunt $

Constant losses - friction, wind, iron etc.. (P0 ) I Va0 $

TOTAL LOSSES (Plosses total-Sum of above

PP P

motorin

in losses totalh =

- - where P V Iin $=

P PP

generatorout losses total

outh =

+ - where P V Iout $=

Maximum efficiency is where variable losses = constant losses

10

• Back-to-back or Hopkinson method:

V

Supp

ly

M G

SI2

I1

IaG

IsG

IaM

IsM

I1 + I2

RsM RsG

RaM RaG

I I I Ia S1 2M M= + - and I I Ia S1G G= +

I2 is the input current from the mains and is proportional to the combined total losses in the M-G set.

IVPsupply

out1 = where Pout is the rated output power capacity of the

shunt machine to be tested as given.

I II

combination1 2

1h =

+

exposition of losses:For the motor:

Variable losses

Armature losses (PaM ) I Ra a2

M M$

Shunt losses (PSM ) I VsupS plyM $

Constant losses - friction, wind, iron etc..(P0M ) See foot note

TOTAL LOSSES (Plosses totalM- ) Sum of above

exposition of losses:For the generator:

Variable losses

Armature losses (PaG ) I Ra a2G G$

Shunt losses (PSG ) I VsupS plyG $

Constant losses - friction, wind, iron etc..(P0G ) See foot note

TOTAL LOSSES (Plosses totalG- ) Sum of above

Foot note:

The constant losses are normally shared equally between the motor and the generator.

To determine the constant losses per machine, the total loss for the M-G set is calculated: P I Vsuptotal losses ply2M G $=- -

Then the sum total of variable losses for both motor and generator is obtained.

The total for constant losses for the M-G set is now calculated:

P P Sum total for variable losses

of MG combination

total losses0M G M G= --- -

P PP20 00

M G

M G= = -

11

Individual efficiencies:

Due to the motor being mechanically linked to the generator, the rated output of the generator = rated input to the motor = the rated power capacity of the M-G set as given.

Thus: P P rated power capacity of MGsetinput outputM G= =

PP P

motorinput

input losses total

M

M M

h =- -

P P

Pgenerator

input losses total

output

G G

G

h =+ -

Power factor correction:

46. Capacitors

Series arrangement: ....C C C C1 1 1 1total n1 2

= + + +

Parallel arrangement: .......C C C Ctotal n1 2= + + +

Q C V$=

Time constant: R C$x =

Transient period: 5x (average time to charge or discharge)

General instantaneous value:

v V V V efinal initial final

t= + + x

-^ h

i I I I efinal initial final

t= + + x

-^ h

Instantaneous value during charging (beginning at zero):

v V e1final

t= - x

-^ h

i I e1final

t= - x

-^ h

Instantaneous value during discharging (ending at zero):

v V einitial

t

$= x-

i I einitial

t

$= x-

47. Conditions for resonance: X XC L=

48. Xf C21

C $r= and X f L2L $r=

49. Resonant frequency: fL C2

1r

$r=

50. Harmonics in complex waves:

• Basics on phase displacement:

:

cos sin sin

cos sin

t t t

t t where f

2 2

22

&~ ~r

~r

~ ~r

~ r

= + =- -

- = - =

` `

`

j j

j

• Influence of odd-harmonics on reactance: ( f = fundamental frequency).

Fundamental: Xf C21

C1 $r= and X f L2L1 $r=

3rd Harmonic: Xf C2 3

1C3 $ $r= and X f L2 3L3 $ $r=

5th Harmonic: Xf C2 5

1C5 $ $r= and X f L2 5L5 $ $r=

7th Harmonic: Xf C2 7

1C7 $ $r= and X f L2 7L7 $ $r=

9th Harmonic: Xf C2 9

1C9 $ $r= and X f L2 9L9 $ $r=

etc......

• Influence of odd-harmonics on overall power: Refer to notation of instantaneous values:

...

sin sin

sin

e E t E t

E t etc

3

5

max max

max

3

5

1 3

5

~ ~ b

~ b

= + +

+ +

^

^

h

h

and:

..

sin sin

sin

i I t I t

I t etc

3

5

max max

max

1 3

5

1 2

5

~ a ~ a

~ a

= + + +

+ +

^ ^

^

h h

h

Now:

Power of fundamental:

cosP E I2

1max max1 11 1$ $ z= ^ h where cos 1 1 1z b a= -

Power of 3rd harmonic:

cosP E I2

1max max3 33 3$ $ z= ^ h where cos 3 3 3z b a= -

Power of 5th harmonic:

cosP E I2

1max max5 55 5$ $ z= ^ h where cos 5 5 5z b a= -

etc....

12

Total power supplied:

.....P P P P Ptotal n1 3 5= + + + +

.......V

E E E E2RMS

2 2 2 2max max max max5 n1 3=+ + + +

.......I

I I I I2RMS

2 2 2 2max max max max5 n1 3=+ + + +

Overall power factor:

cosV I

PRMS RMS

total

$z =

51. Resonant frequency of an electrical cable network:

f x fr $= with f being the fundamental frequency and x being a multiple of the fundamental frequency.

xMVAMVA

XX

C R

SC R

SC

C= =-

- where MVASC R- is the short circuit

duty at the point of connection; MVAC R- is the capacitor rating at the system voltage; XC is the capacitive reactance of the capacitors at fundamental frequency, and XSC is the short circuit reactance at the point of connection.

52. Active current: cosI z

53. Reactive current: sinI z

54. Active power in 3-phase circuits: cosP V I3active line line$ z=

55. Apparent power in 3-phase circuits: P V I3apparent line line$ $=

56. Power factor: PPapparent

active

57. Reactive component for a balanced inductive load (3-wire):

φ

φcos.3 linelineactive IVP =

line

line

apparent

IV

P

..3=

φsin

.3

lineline

reactiveI

VP

=

sinP V I3reactive line line$ $ $ z=

58. Combined R-L load with power factor correction instituted:

S1

S2

S3

S 4

S5

Effecti

ve co

rrecti

on effo

rt

Resultant apparent powerResultant KVAr

Active powerline

1φ

2φ4φ

Rφ

1rkVA

2rkVA 4r

kVA

5rkVA

1activeP2activeP

3activeP4activeP

Resultant active power

S1 = apparent power of inductive load no. 1 (e.g. induction motor), together with its reactive power kVAr1 , its active power Pactive1 and its lagging phase angle 1z .

S2 = apparent power of inductive load no. 2 (e.g. induction motor), together with its reactive power kVA2 , its active power Pactive2 and its lagging phase angle 2z .

S3 = apparent power of resistive load no. 3 (e.g. lighting system). A pure resistance load does not have any reactive component and the phase angle is zero. Therefore S Pactive3 3

= .

S4 = apparent power of a synchronous motor or –condenser. The reactive power component kVAr4 is opposite in direction to that of an induction motor, implying a leading power factor and therefore a leading phase angle 4z . The active component of the power is indicated as Pactive4 .

S5 = apparent power of a capacitor. The active power is zero. The reactive power is indicated by kVAr5 .

The resultant apparent power = vector sum of S1 + S2 + S3 + S4 + S5 at a lagging phase angle of Rz and a resultant active power P P P Pactive active active active1 2 3 4

+ + + .

The magnitude of the resultant reactive power =kVA kVA kVA kVAr r r r1 2 4 5

+ - +^ h

13

59. Combined inductive (motor, star connected) and resistive(lighting) load in a 3-phase 4-wire unbalanced distribution system operating at cosz .

E3

E1E2

I3

I1

I2

-X X

Y

-Y

120°

120°

120°

3φ

2φ1φ

I2

I 3

I n

33 sinφI

11 sinφI

22 sinφI

Resistive load (Lighting, being connected between line and neutral):

IV

P 3R

line

R1

1 $=

IV

P 3R

line

R2

2 $=

IV

P 3R

line

R3

3 $=

Inductive load (motor):

Line current: cos

IV

P

3line

line

input

$ $ z=

Active current: cosI IM lineactive $ z=

Reactive current: sinI IM linereactive $ z=

Combined load:

Line1: I I Iactive R Mactive1 1= +

I Ireactive Mreactive1=

I I Iactive reactive12 21 1

- + at tanIIactive

reactive1

1

1

1

z = -

Line 2: I I Iactive R Mactive2 2= +

I Ireactive Mreactive2=

I I Iactive reactive22 22 2

- + at tanIIactive

reactive2

1

2

2

z = -

Line 3: I I Iactive R Mactive3 3= +

I Ireactive Mreactive3= ; I I Iactive reactive3

2 23 3

- + at

tanIIactive

reactive3

1

3

3

z = -

14

Value of neutral current:

I I I Ineutral 1 2 3= + + (vector sum: via polar or rectangular notation)

or alternatively via basic calculations:

Sum of horizontal components via X-X:

line Calculation detail Value

1 cosI 301 1z+^ h

2 cosI 302 2z- +^ h

3 cosI 903 3z= +^ h

Sum total of horizontal components( HC/ ) above/

Sum of horizontal components via Y-Y:

line Calculation detail Value

1 sinI 301 1z- +^ h

2 sinI 302 2z- +^ h

3 sinI 903 3z= +^ h

Sum total of horizontal components( VC/ ) above/

HC j VCIneutral = + ^^ hh// converted to its polar equivalent

notation.

60. Average power factor = kWh kVA h

kWh

r2 2+^ ^h h