1

ELECTRICAL MACHINES - II

Emf induced in the armature conductors:

The voltage induced in the conductors is an alternating one and is converted into DC using

commutator segments as shown in the figure.

The brush 1 always makes contacts with a conductor moving under north pole and the brush 2

makes contact with the conductor moving under south pole and hence the emf collected across

brushes 1 and 2 is always of unipolar nature as given below:

When there are four conductors (making two turns ) we have four commutator segments as shown

below :

2

As the number of conductors and the commutator segments increases, the DC output emf also

increases and becomes more and more smoother. The emf induced in the armature is always an

AC emf. Using commutator segments, the DC emf is collected and AC emf is collected using slip

rings.

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UNIT – I : SYNCHRONOUS GENERATORS

Each brush makes contact with the ring which is permanently connected to a particular conductor.

As the conductor moves through north and south poles the voltage collected through the brushes

becomes AC emf.

For every pair of poles, one cycle of AC emf is induced.

For the above case two cycles of voltage is induced. Hence the frequency of emf induced is

120/PNf c/s

where N is speed in rpm.

1 Deg. Mechanical = P/2 Deg. Electrical.

4

Constructional details of synchronous machines:

The armature power handling capacity is always high in synchronous machines compared to dc

machines.Moreover field windings require less DC power.Hence in synchronous machines

armature winding is placed in stator and the field winding is placed in the rotor getting its supply

through slip rings. If the armature winding is in the rotor the following problems arise :

Higher electro dynamic forces

Higher mechanical losses

Large power at nearly high current and voltage has to be taken through slip rings

Sparking and higher contact losses

Hence the armature is placed in stator and the field is made rotating in synchronous machines.

Rotor winding in synchronous machines:

DC machine

Stator- Field poles

Rotor -Armature windings

Syn. machines

Stator-Armature windings

Rotor -Field poles

Rotor field

Salient poles -projecting pole

rotor

wound rotor -non salient pole

rotor

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Emf equation of alternator:

Let φ The total flux per pole in weber varying sinusoidally as φmsinωt

P No. of Poles

T No. of turns/ phase

f The frequency

Emf induced/turn = e = - mdt

dd

dtsin ωt

Emf/turn = ω Φm cosωt volts

Emf/phase = ω T Φm cosωt volts

Erms/phase = 2πfT Φm /√2 = 4.44 π f T volts

Erms/phase =4.44 π f T volts

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The above equation is complete provided all the conductors are in the same slot and there is no

short pitch in the coil.

But, the windings are distributed over the different slots and are not concentrated. Moreover the

coil pitch is not always π degrees but it is short pitched . Hence we have to take distribution

factor(kd) and the pitch factor (kp) into account:

Distribution factor:

Assume,

g - no. of slots/pole/phase

p- no. of poles

α- slot angle ( slot pitch )

Arithmetic sum of the voltage in g slots

=g*e=(R sin2

) *2*g =2Rg sin 2

Vector sum of voltage in g slots =2[R sin (g 2

)]

Distribution factor= Voltage induced in the distributed winding /Voltage induced in the

concentrated winding

Kd=

2sin2

)2

sin(2

Rg

gR

=

2sin

)2

sin(

g

g

For nth order harmonic emf, Kdn=

2sin

)2

sin(

ng

gn

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Pitch factor : Emf in short pitched coil /Emf of full pitched coil

The pitch of the winding=(π-ρ)

Short chorded angle= ρ

Pitch factor, k p =e

e

2

2cos2

=cos2

For nth harmonic emf, the pitch factor kpn

kpn= cos2

n

In addition to distribution factor and pitch factor one other factor (i.e.)skew factor also should be

taken into account, when the winding is skewed.

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Hence the complete emf equation in alternate is ,

E1=4.44φ1fTkd1kp1ks1 volts

.

.

.

En=4.44φn (nf)Tkdnkpnksn volts

For flux waveform given below,

Φ=φ1sinωt+ φ3sin3ωt+ φ5sin5ωt+……. Φnsinnωt

The emf equation is as follows:

e= E1msinωt+ E3msin3ωt+ E5msin5ωt+……. Enmsinnωt

Erms= 2 2 2 2

1 3 5 nE rms + E rms + E rms +.......+E rms .Though the induced emf in the individual

phases contain harmonics , we can get almost pure sinusoidal emf as output in the following

ways:

By connecting the winding in star, the third harmonic and multiples of third harmonics

are eliminated.

By short chording the windings by 360 , the fifth harmonic and multiples of fifth are

eliminated

i.e., Kp5=cos (5*36/2)=cos 900=0

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By properly distributing the winding, 7th harmonic can be eliminated

Kd7=2

7sin

27

sin

g

g

=0

(i.e.) 2

7 g=1800

By properly skewing , 11th harmonic can be eliminated.

Next order of harmonics is only from 13th

The 13th order flux will be almost = φ1/(13)2 = φ1/169

which is less than 1 percent in magnitude and is insignificant hence the emf becomes pure

sinusoidal even though the flux is not exactly sinusoidal.

Alternate method to derive the emf equation:

The emf equation for alternator can also be derived from the average voltage induced in armature

as in the case of DC.

The speed of alternator - N rpm

Flux/pole - φ Wb.

No. of poles - p

No. of turns / phase - T

Average emf per conductor =

pN60

=

60

pN

Average emf per turn = 60

2 pN =

60*2

4 pN volts

The rms value of emf per phase = 120

**11.1*4 TpN

(where 1.11 is the form factor for sinusoidal wave)

E1 = 120

**44.4 TpN = 4.44 φfT volts

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En = 4.44φn(nf)T volts

Taking into account the winding factors kpn,kdn,ksn.,

En = 4.44φn(nf)T kdnkpn,ksn volts

=4.44φn(nf)T kwn volts

where,

Kwn is the winding factor ; kwn= kdnkpnksn

Regulation of Alternators:

Regulation of alternators is defined as the change in the terminal voltage from no load to full

load at a given power factor the field current kept constant,expressed in terms of rated terminal

voltage.

i.e. %regulation=[(E-V)/V]*100% at a given load

Methods for predetermination of Regulation:

(1) Synchronous impedance method / emf method ---(pessimistic method)

(2) Ampere –turns method / mmf method ---(optimistic method)

(3) Zero power factor method / ZPF method

(4) ASA method / American Standards Association method

Synchronous impedance: is defined as the ratio of rated emf to that of the short circuit current at

a given field current (corresponding to rated terminal voltage on no-load).

Zs=Erated/Isc at constant If

Synchronous impedance method: In this method the effect of armature reaction and leakage

reactance is accounted as voltage drops.

Xs=(Zs2 - Rac

2)

The effective armature resistance R ac is taken as 1.5 times the DC armature resistance.

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Regulation for lagging p.f load :

0 * sE V I Z

0 sE V IZ = cos( ) sin( )s sV IZ jIZ

sin( )tan

cos( )

s

s

IZ

V IZ

Using approximate formula (neglecting the quadrature component since it is very less)

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E = ( cos sinV IR IX ) volts

Hence percentage regulation,

% Reg = ( E – V / V ) * 100 % = cos sin

*100%IR IX

V

For leading pf the requlation is,

E = ( cos sinV IR IX ) volts

% Reg = IRcos±IXsin

V 100%

+ ------- for lagging pf

- ------- for leading pf

The exact value of E for leading pf is,

0 * sE V I Z = cos( ) sin( )s sV IZ jIZ

Max regulation occurs at, θ = β

cos sin[% Reg]= [ ]

d d IR IX

d d V

= IRsin±𝐼𝑋𝑐𝑜𝑠

Equating to 0 ,we get , cos sinIR IX

tan tanIX

IR : i.e.,

This method results in a very high regulation and it is called pessimistic method.

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Regulation by Ampere-turns method: (MMF)

For lagging power factor:

If1 ---- Field current corresponds to emf E’

If2 ---- Field current to circulate start circuit full load current full load current Isc =IFL

2 2

1 2 1 22 * cos(90 )f f f fI I I I

For leading power factor:

FI = 2 2

1 2 1 22 * cos(90 )f f f fI I I I

The regulation obtained by MMF method is less than that obtained by syn. Imp method and hence

is called optimistic method.

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Regulation by Zero p.f method or Potier triangle method:

- Conduct OCC and plot the characteristics (E vs If)

- Conduct ZPF load test : Vary the inductive load and plot the variation of terminal voltage

against field current, keeping the load current constant at full load.

- Draw OCC and SC characteristics (Field corresponding to full load SC current)

- Draw a horizontal line from the rated phase voltage (Ep) to meet the ZPF curve at point Q

- Make PQ =OA [The short circuit field current]

- From point P draw a line parallel to OC the slope of OCC at start. Let it meet the OCC at point

S. From S draw a perpendicular line to PQ to meet at R.

Then, QR = Field representing the effect of armature reaction.

SR = The IX drop (Reactance drop alone)

The regulation by ZPF method is given by ,

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% Reg = 100E V

V

%

Regulation by ASA method:

fscI ----- Field current for full load short circuit characteristics

1fI ------ Field Corresponds to rated voltage V without saturation.

2fI ------Field corresponds to saturation for voltage of E

Load Characteristics of Alternator:

1. Terminal voltage (vs) loads current at constant Field current IF.

2. Excitation (vs) load current at constant V.

3. Terminal voltage (vs) excitation at constant IL.

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V (vs) IL at constant IF

V =E-IX for lagging pf

V = E+IX for leading pf , assuming R X

E (vs) IL at constant V

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E = V+IX (lag)

E = V-IX (lead)

V (vs) IF at constant IL

Parallel operation of Alternators:

Reason for going for parallel operation:

When the load goes on increasing, connect more alternators in parallel.

If any one is faulty, the remaining alternators can share the load.

Servicing and overhauling can be done as individual machines, with the others supplying

the load.

Conditions to be satisfied for parallel operation:

1. Incoming alternator should have the same voltage of the existing one

2. Frequency of incoming machine and existing supply should be same.

3. Polarities of both should be same.

4. Phase sequence should be same

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Infinite bus:

Means the bus bar maintaining a constant terminal voltage irrespective of infinite loads

connected.

Synchronising of Alternators:

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We can use synchroscope for synchronising or through three dark lamp method or two bright and

one dark lamp method.

Procedure to synchronise : three dark lamp method

Run the alternator using prime mover to the rated speed

Check the voltage of the existing bus by throwing the switch to position 1

Measure the voltage of the incoming alternator by throwing the switch to switch2.Adjust

the field so that the alternator terminal voltage is same as bus bar voltage

If the polarity and phase sequence are same then the three lamps will be dark over a

considerable period.

When all the three lamps are simultaneously dark close the TPST by shorting all the

lamps and the alternator is synchronized with the bus.

In the two bright and one dark lamp method, one lamp is connected between R1 & R2 , and the

other two lamps are interchanged as shown below:

When one lamp is dark, the other two will always be bright. This method is better than 3 dark

lamps method, as we can check for both dark and bright simultaneously.

Synchronising power & torque:

Referring to the fig. in page 18 , the circulating current between E1 & E2 is given by,

The circulating current Is= (E1-E2)/2Xs

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The circulating current will be in phase with E1 and against E2 , and the power is

P=E1Is

And the synchronizing torque=(E1Is)/2πns N-m

ns –synchronous speed in rps.

This syncronising power will result in reduction of speed of prime mover 1 and increase with

speed of prime mover 2. Hence, both the machines will restore back to synchronism.

Load sharing of alternators in parallel:

E1-I1Z1= V I1Z1=I2Z2

E2-I2Z2=V I1Z1=(I-I1)Z2

I1Z1= I2Z2 I1(Z1+Z2)=IZ2

I1/I2=Z2/Z1 I1=IZ2/(Z1+Z2)

I2=IZ21/(Z1+Z2)

Parallel operation theorem:

E1-I1Zs1= E2-I2Zs2= E3-I3Zs3=………. En-InZsn=V

I1+ I2+ I3+……… +In= IL

(E1-V)/Zs1)+ (E2-V)/Zs2)+………. [(En-V)/Zsn)=V/Zl

(E1/Zs1)+ (E2/Zs2)+……….+ (En/Zsn)=(V/Zl)+ (V/Zs1)+ (V/Zs2)+ ………..+(V/Zsn)

=V(1/Zl)+ (1/Zs1)+ (1/Zs2)+…….. (1/Zsn) Is1+ Is2+……+

Isn=V(1/Z)

V=Isc(Z)

where , (1/Z)= (1/Zl)+ (1/Zs1)+ (1/Zs2) ….+(1/Zsn)

Isc=Isc1+ Isc2+……+ Iscn

Problem:

Two parallel winding synchronous generators has emf of 1000 0 and 1000 10 V per phase.

Phase synchronous Impedance are,

1 (0.1 2.0)SZ j

2 (0.2 3.2)SZ j

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They supply a load of impedance Zl = (2+j1)Ω/phase. Find terminal voltage, load currents and

power output of each machine.

1 1 1 1

(2 1) (0.1 2.0) (0.2 3.2)z j j j

1 1 1

(0.365 0.830) 0.906 66.3z j

1.103 66.3

(1000 0) 985 173.64

(0.1 2.0) 0.2 3.2SC

j jI

j j

(25 498) ( 35 310)

10 808

j j

j

808 90.7 A

667 303 730 24.4j V

11

1

1000 667 303

0.1 2.0s

E V jI

Z j

100 159j

224 44.9 A

22

2

985 174 667 303

0.2 3.2s

E V j jI

Z j

46 96j

2 106 64.3I

Power factor of machine 1 = cos (44.9 – 24.4 ) =0.937

Power factor of machine 2= cos(64.3-24.4) =0.767

Power supplied by generator 1 = 1 1 1cosV I

P1 = 153kW & P2=64kW

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Armature reaction in alternators:

The effect of armature reaction in alternators under UPF load is similar to that of DC generators.

The armature flux acts in the inter polar zone and the flux under LPT is weakened and under the

TPT it is strengthened.

For UPF (cross magnetising)

For ZPF lag load (demagnetising)

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For ZPF leading load (magnetizing)

Armature reaction in 3φ alternators and rotating magnetic field:

24

At instant 1,

Resultant flux ΦR= NIm +2[(NIm/2)cos 600]

= NIm +2[(NIm/2)*0.5]=1.5NIm

At instant 2,

ΦR=1.5NIm

When the poles rotate by 1200 , the resultant flux ΦR also rotates by 1200 and its magnitude

remains constant at 1.5NIm. This is called as rotating magnetic flux.

The rotating magnetic field has the same no. of poles as that of the main field poles.

25

The direction of current is shown corresponding to instant 1 , in fig as page 25

The no. of poles of the RMF is the same as the no. of poles for which the armature winding is

wound.

When a 3Φ balanced voltage is applied to the 3Φ balanced winding (i.e. winding displaced

by 1200 electrically in space) a Rotating Magnetic Field is produced.

Regulation of salient pole Alternators & Blandul’s two reaction theory:

In salient pole alternators, the synchronous reactance Xd under pole axis is different from the

synchronous reactance Xq under interpolar (quadrature axis), whereas both the reactances are

same in cylindrical rotors and Xd= Xq= Xs ; since the air gap is same along the entire periphery.

Armature reaction field due to partly lagging loads

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The armature reaction flux ΦA is resolved into two components (i.e. two sine waves)with

ΦA cosθ magnitude acting in the interpolar zone and ΦA sinθ magnitude acting in the polar axis

demagnetizing the main flux.

The load current I should be resolved into two components Id & Iq responsible to cause voltage

drops in the direct and quadrature axis.

IdXd- voltage drop in the direct axis

IqXq- voltage drop in the quadrature axis

The values of Xd and Xq are determined by conducting slip test on the salient pole alternator.

Slip test: The alternator rotor is run by prime mover at a speed slightly less or more than the

synchronous speed.

*The field of the salient pole is not excited

* the armature is supplied with low ac voltage through auto transformer.

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Slip test :

Apply different levels of ac voltage and at each level note down the maximum and minimum

values of voltage oscillations in the voltmeter and similarly the maximum and minimum

oscillations in the armature current.

Compute the value of Vmax/Imin=Xd

& Vmin/Imax=Xq

for each application.Take the average values of Xd & Xq for finding out the regulation.

When the rotating magnetic field and DC field poles align , we get maximum permeance and the

reactance is Xd and when they dealign we get minimum permeance and the reactance is Xq.

cos cos '

cos cos( ) sin( )

a a d d

a a a d

E V I R I X

V I R I X

cos sintan

cos sin

a q a a

a a a q

I X I R

V I R I X

The regulation is % Reg = 100E V

V

%

Power developed (Pδ) and power output(P0):

Assuming cylindrical rotor,

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Power output, P0=VIcosθ

Power developed, Pδ=EIcos(θ+δ)

Power developed – Armature cu. Loss = P0

P0=VIcosθ =I[Vcosθ]

Pδ=EIcosθ= I[Vcosθ+IR] = VIcosθ+I2R = Po+ I2R

0

s s s

E V E VI

Z Z Z

I has two components , (i.e) (Icosθ - jIsinθ)

cos [ cos( ) cos ] sin [ sin( ) sin ]s s s s

E V E VE E

Z Z Z Z

[cos( ) sin( )] [cos sin ]s s

E Vj j

Z Z

= I cos + j I sin

cos cos( ) cos

sin sin( ) sin

s s

s s

E VI

Z Z

E VI

Z Z

Power is developed when the voltage and current are in the same phase

cos( )

( cos cos ) ( sin sin )

P EI

E I E I

Negative power

cos cos( ) cos sin sin( ) sins s s s

E V E VE E

Z Z Z Z

=𝐸2

𝑍𝑠cos 𝛿 ∗ cos(𝛽 − 𝛿) −

𝑉𝐸

𝑍𝑠cos 𝛽 ∗ cos 𝛿 −

𝐸2

𝑍𝑠sin 𝛿 ∗ sin(𝛽 − 𝛿) +

𝑉𝐸

𝑍𝑠sin 𝛽 ∗ sin 𝛿

=𝐸2

𝑍𝑠

[cos 𝛿 ∗ cos(𝛽 − 𝛿) − sin 𝛿 ∗ sin(𝛽 − 𝛿)] +𝑉𝐸

𝑍𝑠[sin 𝛽 ∗ sin 𝛿 − cos 𝛽 ∗ cos 𝛿]

=𝐸2

𝑍𝑠[cos(𝛿 + 𝛽 − 𝛿)] −

𝑉𝐸

𝑍𝑠cos(𝛽 + 𝛿)

𝑃𝛿 =𝐸2

𝑍𝑠cos(𝛽) −

𝑉𝐸

𝑍𝑠cos(𝛽 + 𝛿)

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Power output P0 =VIcosθ

=V[Icosθ]

cos( ) coss s

E VV

Z Z

𝑃0 =𝑉𝐸

𝑍𝑠cos(𝛽 − 𝛿) −

𝑉2

𝑍𝑠cos(𝛽)

When Ra is neglected, as Ra<<Xs β=90°

𝑃𝛿 =𝑉𝐸

𝑍𝑠sin 𝛿 = 𝑃0 = 𝑉𝐼 cos 𝜃

Where V = E

2

sins

VP

X

ive for generator

ive for motor

Power developed in salient pole alternator:

𝑃𝛿 =𝑉𝐸

𝑍𝑠sin 𝛿 = 𝑃0 = 𝑉𝐼 cos 𝜃 i.e. 𝑃𝛿 = 𝑉[𝐼𝑞 cos 𝛿 + 𝐼𝑑 sin 𝛿]

sinq

q

VI

X

-----------------------------(1)

cosd dE I X V

cosd dI X E V

cosd

d

E VI

X

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

Substituting for Id and Iq,

sin coscos sin

q d

V E VP V

X X

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2 sin cos sin cossinE

q d d

VP V

X X X

2 sin cossinE

d q

d q d

VVX X

X X X

𝑃𝛿 =𝑉𝐸

𝑋𝑑sin 𝛿 +

𝑉2

2𝑋𝑑𝑋𝑞(𝑋𝑑 − 𝑋𝑞) sin 2𝛿

The Pδ equation shows that, there is an additional power developed due to saliency of the rotor, at

twice the frequency.

Even when the field is not excited (i.e E =0) little power is developed due to the saliency (i.e)

2

( )sin 22

d q

d q

VX X

X X

On no load the RMF and the field poles align and the load angle is almost zero.

As the load current increases E advances with reference to terminal voltage V and δ, increases.

Sub-transient, transient and steady state reactance’s of synchronous machines:

When sudden short circuit takes place the armature current shoots up to very high values. As per

constant flux linkage theory, the flux cannot suddenly increase, and hence to maintain constant flux

the inductance and hence the reactance of the system decreases.

The flux is associated with damper winding, field winding and the armature winding and armature

reaction.

(i) Sub-transient reactance, x’’, is due to -- Armature winding inductance and combination of

damper winding , field winding and armature reaction in parallel

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(ii) Transient reactance, x’ ,is due to -- Armature winding inductance and combination of

armature reaction and field winding in parallel.

(iii) Steady state synchronous reactance, xs , is due to --Armature winding inductance and

armature reaction inductance.

Symmetrical short circuit current:

*****