1

Unit-4: Synchronous Machines

2

• Important Parts of the Synchronous Machines are-Stator (Armature windings), Rotor (Field windings) andDamper windings (To prevent hunting in Generator and to provide starting torque in Motors)

• The most commonly used machine for the generation of electric power is the Synchronous generator called an alternator, as it generates ac.

• The armature winding is placed on the stator and the field poles on the rotor.

• When the rotor is rotated by a prime-mover, it works as an alternator (Synchronous generator).

• To run it as a synchronous motor, -Three-phase supply is connected to the stator winding

(Armature), and -A dc supply (for the field) to the rotor winding to produce

magnetic poles.

Unit-4: Synchronous Machines

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Construction of Stator and Rotor

4

Advantages of having armature winding on the stator and field winding on the Rotor

• It is easier to provide insulation to armature winding for high voltages, as the stationary winding is not subjected to mechanical stress due to centrifugal forces and also more space is available.

• The external three-phase circuit can directly be connected with fixed terminals on the stator, without the need of slip-rings.

• For dc supply to the rotor field winding, only two slip-rings, each capable of handling much smaller current and requiring insulation for much lower voltages, are needed.

• The revolving field system is light in weight, and therefore canrun with high speed.

5

(a) Non-salient or Cylindrical Type (b) Salient or Projected Poles Type

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Main rotating flux:

The rotating flux generates the induced voltage in the stator at a frequency , . the induced emf per phase in the stator due to the field is and lags by

w. r. to

ns =f

P 2+fffffffff Es t

` a

= + N st Φrot ω sin ωt` a

= N st Φrot ω cos ωt@90ob c

ω = 2 Aπ ns Est =N sta Φrot ω

2pwwwwwwwwwwwwwwwwwwwwwwwwwwwffffffffffffffffffffffffff

Φlink t` a

=Φrot cos ωt` a

Est = 2pwwwwwwwwwwwwwwwwwwwwwwwwwwwAπ f N sΦ f AK w , Φ f = Φrot

Es t` a

= N stdΦlink t

` a

dtffffffffffffffffffff

f = P2ffffns

E f

Aπ2ffffor 90o Φ f =Φrot

Φ f

Induced EMF per phase in the Stator (Armature Winding) of the alternator:

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Pitch-Factor Kp of a short-pitched coil• The factor by which the emf per coil is reduced because of the pitch

being less than full-pitch is known as pitch factor (or coil span factor) kp. Thus,

phasor sum of the coil-side emfs 2 cos ( / 2) cos ( / 2)arithmetic sum of the coil-side emfs 2p

EkEβ β= = =

6 cos (6 / 2) 0 6 / 2 90 or 30pk β β β≡ = ⇒ = ° = °

Suppose that we wish to remove 6th harmonic altogether.

(a) Full-pitch coil in a 4-pole machine.(b) Phasor sum of coil-side emfs. (c) Short-pitch coil in a 4-pole machine.

(d) Phasor sum of coil-side emfs.For the nth harmonic, the pitch factor is given by

cos ( / 2)pnk nβ=

10

Distribution Factor (kd) of a distributed windings• The distribution factor (or breadth factor) kd for q slots per pole

per phase is defined as •

or

If q (the number of slots per pole per phase) is very large, the angle α becomes very small, then

• The total angle qα(expressed in electrical radians) is called the phase spread (σ).

phasor sum of component emfs 2 2 sin ( / 2)arithmetic sum of component emfs 2 2 sin ( / 2)d

AD AF R qkq BC q BG q R

αα

×= = = =

× × × ×sin ( / 2)

sin ( / 2)dqk

qαα

=

sin ( / 2)/ 2d

qkqα

α=

Calculation of distribution factor, for q = 3.

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The miff distribution along the air gap is a square or rectangular wave wave due to slotting, which is main cause of harmonics in the induced voltage.

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15

Synchronous Generator loading and armature reaction:

• The generator is loaded , flows in the armature winding.

• The load current produces a rotating flux, due to mmf .

• is called armature reaction, which rotates at synchronous speed and in the direction of the rotor.

•Thus the resultant mmf will be Phasor sum Fr=Ff + Fa

•This rotating flux Фa induces a ac three phase voltage in the stator winding.

•This voltage is-- subtracted from the induced voltage.- represented by a voltage drop on the synchronous reactance.

•Thus the equivalent circuit of a synchronous generator is a voltage source and a reactance connected in series.

I a

Φa F a

F a

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F r lags F f by an angleδ and Er lags E f by the same angle δF r = F f + F a

Φr =Φ f + Φa

but Ea = j I a X a [ Er = E f @ j I a X a

If is the leakage reactance, then total reactance,is called synchronous reactance then,

In generating action : the field poles are driven ahead of the resultant field by an angle δ by the prime mover.In case of motor : the field poles lags behind the resultant field by an angle δdue to the load.“δ’ is also called power angle or torque angle.

Torque developed due to interaction of two fields:

X l X s = X a + X L

V t = Er@ I a Ra + jX l

b c

= E f @ j I a X a@ I a Ra + jX l

b c

V t = E f @ I a Zs , where Zs = Ra + jX s = synchronous impedance A

F rF f

F f F r

T = π2ffffP

2fffff g

2

Φr F f sinδ and T∝ F f AF r A sinδ

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18

Nature of armature reaction: Assuming that the armature resistance and leakage reactance are negligible so that

In motoring action cases will exactly reversed in all the three cases as the direction of is reversed.

V t = Er due to Φr

I a

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OCC, SCC and MEASUREMENT OF SYNCHRONOUS IMPEDANCE

• The synchronous impedance Zs of an alternator can be determined by plotting its open-circuit and short-circuit characteristics.

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21

22

Zpu=ZΩ/ZB

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Alternator phasor diagram for constant E f = 1.0 Pu

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Alternator characteristics

(a)Terminal voltage Vs field current- the field current equivalent to OF’ is required to compensate Fa and Pal i.e. armature reaction miff and leakage flux equivalent miff .

OF = Fa + Fal

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(b) Variation in Terminal Voltage Vs Load current(Effect of armature reaction on terminal voltage)

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(c) Compounding curves: variation of alternator excitation if Vs load current at constant speed & terminal voltage.

Compounding curves: V- curves of synchronous m/c constant load (real power) and variable excitation (reactive power).

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Voltage Regulation of Alternators

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Air gap line

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We have used,φaα Fa and Ea α φa, but it will be correct only before saturation .After saturation , the used value of Xs is more than the actual , because used Xs is the ratio of voltage of w.r.t air gap line and current w.r.t scc , instead of occ and scc. Thus more value of Ef is used which gives % regulation more than the actual one (pessimistic result).

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Phasor diagram of synchronous machines (alternator):Taking armature current as reference:

Case 1 Lagging pf load:

Case 2 Leading pf load:

E f =V t + I a Ra + j I a X s

E f = V t2 +V z

2 + 2V t V z cosαqwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

whereα =Ψ@ θαis the angle between the vectorV t andV z the drop voltage

b c

in this case , α =Ψ + θthus

E f = V t2 +V z

2 + 2V t V z cosαqwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

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Phasor diagram of the alternator (at lagging pf):Phasor diagram taking as reference and is taken into account:It is clear from the phasor diagram,

V tRa

E f

L

L

L

M

M

M> V t

L

L

L

M

M

M

for the lagging Pf

magnitude of excitation emf :

E f = V t + Ia Ra cosθ + Ia Xs sinθb c2

+ Ia Xs cosθ@ Ia Ra sinθb c2s

wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

E f = V t + Ia Xs sinθb c2

+ Ia Xs cosθb c2s

wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwif Ra is negligible

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Alternator or generator under load (Leading pf):

isolated loads with a leading pf current leads the terminal voltage, EThe voltage drop across the synchronous reactance, leads the

current by 90The induced voltage, generated by the flux φ is equal to the phasor

sum of E and

note that always leads E by the angle δFor lagging loads is greater than EFor leading loads E is greater than

Ex

EφEx

EoEo

Eo

E f = V t + I a Ra cosθ@ I a X s sinθb c2

+ I a X s cosθ + I a Ra sinθb c2s

wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

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If2 may not be in the direction of Ia.

Thus , used If2 is of lesser magnitude compare to actual value and Ef is less which gives lower value of regulation than the actual one(optimistic result).

35

If1

Ff1

900Vt

φ

If2

Ia

Fa

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• ZPFC is drawn by loading the machine as a generator with pure inductive load or synchronizing the machine with the mains and regulating its excitation to yield ZPF operation.•ZPFC is a plot between armature terminal voltage Vt and field current for constant value of armature current and speed.•In the diagram, OA is the field current required to circulate the short circuit current equal to the rated current of the armature .•B point is obtained by noting the field current for the rated voltage Vt at ZPF load.•Only two points A and b are sufficient to draw ZPFC.

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aat

lat

RIVXIV

++

=−

ϕϕψ

cossintan

1

E=Er

jIa Xl

IaRaVt

Ia

ψφ

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i.e. BH=OA

r t t

i.e. Ife

=Ef

FG=BE=BH-EH

αtanDEEH =

αtanDEEH = α=initial slope of OCC

=Slope of air gap line

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40

A 220V, 50Hz, 6 pole star connected alternator with ohmic resistance of 0.06Ω/phase gave the following data for open circuit, short circuit and full load zero-power factor characteristics-

If mA 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.8 2.2 2.6 3.0 3.4

Ef in V (measured between phase to N)

16.73 33.5 50.2 67.0 84.3 99.3 112 134 151 164 173.2 179.0

Isc in A 6.6 13.2 20.0 26.5 32.4 40.0 46.3 59.0 ‐ ‐ ‐ ‐

Zpf terminal voltage (phase to neutral)

‐ ‐ ‐ ‐ ‐ 0 16.73 50.8 80.8 102 120 132.7

Find the percentage voltage regulation at full load current of 40A at 0.8 power factor lagging by(a)EMF method(b) MMF method(c) ZPF (potier) method

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Solution: Rated per phase voltage ,(a) EMF method: minimum value of corresponding to maximum value of SC current should be used.

Vt =220

3pwwwwwwwwwwwwwwwwwfffffffffff=127V

ZS

ZS =13459fffffffffff=2.271

X S = 2.271` a2

@ 0.06` a2q

wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww=2.27Ω phase*

takingcurrent Ia asreference,

E f = V t cosφ + Ia Ra

b c2+ V t sinφ + Ia X s

b c2swwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

= 127B0.8 + 40B0.06` a2+ 127B0.6 + 40B2.27

` a2qwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

=196.7V

%regulation =196.7@127127

ffffffffffffffffffffffffffffffffffffff=54.9%

Ef

jIa XS

δ

φ

Vt

42

(b) MMF methodField current to produce rated voltage is approximately 1.69A from OCC.

= 1.69AField current required to circulate the rated armature current is 1.2A from SCC

= 1.2 A

I f 1Vt=127V

If1

I f 2I f = I f 1

b c2+ I f 2

b c2+ 2 I f 1

I f 2cos 90@φb cr

wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

φ = cos@ 1 0.8` a

= 36.8

I f = 1.69` a2 + 1.2

` a2 + 2B1.69B1.2Bcos 53.2` aq

wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww= 2.59A

corresponding to I f = 2.59 E f = 163.5V from OCC A

% regulation = 163.5@127127

fffffffffffffffffffffffffffB100% = 28.75%

I f

I f 2

φ900 90@φ

43

(C) ZPF method: potier triangle BDF is drawn. = rated terminal voltage BH=OA=1.2A is drawn HD parallel to air gap line. DE perpendicular to BH line.Voltage drop (from actual graph)Armature leakage reactance,

V t

I a X L = DE = 30V

X l =30I a

ffffff= 3040ffffff= 0.75Ω potier reactance

b c

E = Vcosφ + I a Ra

b c2+ Vsinφ + I a X l

b c2rwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

= 148.6V

andψ = tan@ 1 Vsinφ + I a X l

Vcosφ + I a Ra

ffffffffffffffffffffffffffffffF G

= 45.60

now, Fa = BE = BH@EH = 1.2@ DEtanfffffffffα = 1.2@ 30

tanffffffffα

f g

= 0.84

I f = 2.134` a2 + 0.84

` a2 + 2B2.134B0.84Bcos 90@ψb c

rwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

I f = 2.79A, from OCC, E f = 169V

% regulation = 169@127127

fffffffffffffffffffffffB100 = 33%

44

Que. A 3-ø Y- connected synchronous generator rated at 10 kVA and 230.0 V has a synchronous reactanace of 1.2Ω/phase and an armature resistance of 0.5Ω/phase. Calculate the following-(i)The % voltage regulation at full load with 0.8 pf(lagging) (ii) the pf of the load such that the voltage regulation is zero on full load.

full load current I a =10B103

3pwwwwwwwwwwwwwwwwB230

ffffffffffffffffffffff= 25.1 A

i` a

Rated terminal voltage V t =230

3pwwwwwwwwwwwwwwwwfffffffff= 132.8V

E f = V t cosφb c

+ I a Rarwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww2

+ V t sinφ + I a X S

b c2

E f = 11.8` a2 + 109.8

` a2qwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

= 161.8 V

% regulation = 161.8@132.8132.8

fffffffffffffffffffffffffffffffB100 %

= 21.8 %

45

ii` a

taking voltage as reference,from phasor diagram,

E f = V t + I a Ra cosθ + I a X S sinθb c

+ j I a X S cosθ@ I a Ra sinθb c

for voltage regulation to be zero AE f @V t = 0either I a Ra cosθ + I a X S sinθ = 0…A 1

` a

or I a X S cosθ@ I a Ra sinθ = 0…AA 2` a

from 1` a

tanθ =@ Ra

X S

fffffff[ cosθ = 0.92 leading

b c

from 2` a

tanθ = X S

Ra

fffffffwhich gives lagging pf which is not possible for zero regulation A

46

Ques. An star connected 3-ø alternator delivers a 3-ø star connected load at a pf of 0.8 (lagging).A wire connects the load and the alternator. The terminal voltage at no load is 2500 and at full load & 1460 kW is 2200. determine the terminal voltage, when it delivers a 3-ø star connected load having a resistance at 6Ω and reactance 8Ω per phase respectively. Assume constant field excitation.Solution-

I L =1460B 10 3

3pwwwwwwwwwwwwwwwwB 2200B 0.8

fffffffffffffffffffffffffffffffffffff= 478.9 A

I L =2500 ∠ 5 @ 2200 ∠ 0` a

3pwwwwwwwwwwwwwwwwwwwwwwwww*

Zfffffffffffffffffffffffffffffffffffffffffffffffffff

Z =2500 cos 5 + j 2500 sin5 @ 2200 3p

wwwwwwwwwwwwwwwwwwwwwwwww*

478.9 0.8 0.6+b c

ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff2500/√3

Z IL

47

=290.5 + j 217.9 3p

wwwwwwwwwwwwwwwwwwwwwwwww*

478.9 0.8 j 0.6*d e

ffffffffffffffffffffffffffffffffffffff

=167.7 + j 125.8383.12 j 287.34*fffffffffffffffffffffffffffffffff

0.1225 + j 0.42b c

Ω

V t = 2500B6 + j 8b c

6 + j 8b c

+ 0.1225 + j 0.42b c

fffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

= 2401@ j 35.5= 2401.26 ∠@0.84

IZ= 1.08Ω

2500/√3 Vt

R Dubey (9810381076) 48

R Dubey (9810381076) 49

R Dubey (9810381076) 50

51

Solution: rated per phase voltage

(a) EMF method: minimum value of corresponding to maximum value of SC current should be used.

V t =220

3pwwwwwwwwwwwwwwwwwfffffffffff= 127 V

ZS

ZS =13459fffffffffff= 2.271

X S = 2.271` a2

@ 0.06` a2q

wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww= 2.27 Ω

phasefffffffffffffffffff

taking current I a as reference,

E f = V t cosφ + I a Ra

b c2+ V t sinφ + I a X s

b c2swwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

= 127B0.8 + 40B0.06` a2 + 127B0.6 + 40B2.27

` a2qwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

= 196.7 V

%regulation =196.7@ 127

127ffffffffffffffffffffffffffffffffffffff

B100% = 54.9%

Ef

j I a X S

V tδφ

I a Ra

I a

52

V tI f 1

R Dubey (9810381076) 53

54

55

R Dubey (9810381076) 56

Power Transferred by the Alternator

R Dubey (9810381076) 57

58

IES 2005Ques. A synchronous generator supplies rated power at 0.8 pf (lagging). Its resistance and synchronous reactance are 0.1 and 1.0 pu. Calculate the terminal voltage if the open circuit voltage is 2.1 pu.

Solution- rated power means rated current = 1 pu

E f = V t cosφ + I a Ra

b c2d e

+ V t sinφ + I a X S

b c2swwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

2.1 = V tB0.8 + 1B0.1b c2

+ V tB0.6 + 1B1b c2r

wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

2.1` a2 = V t

2 + 1.36 V t + 1.01on solving V t = 1.05 pu

δ

Ef

J Ia Xs

Ia

Ia Ra

Vt

R Dubey (9810381076) 59

60

Operation of alternator with constant voltage and with varying load and varying excitation:Constant load means

is constantEf sinδ= constant as Pe is constant because Vt and Xs are constants.

Then Ef is varied by varying field current If . The power angle δ varies such that Ef sin δ is constant.

on the phasor diagram the tip of Ef moves on a line parallel to Vt and at a distancesin δ =

since Ia cosθ will be constant, the projection of current phasor on Vt remains constant. i.e. the tip of the current phasor traces a line perpendicular to Vt & at a distance Ia cosθ =Pe/Vt

t

se

VXP

s

tfe

XVEP δsin

=

61

The excitation corresponding to UPF is normal excitation while excitation larger than this is over excitation (Lagging Pf) and less than this is under excitation (Leading Pf) in case of alternator. Reverse is the case for motor.

Normal excitation

t

sef

VXPE =δsin

t

sef

VXPE =δsin

jIa1 Xs

jIa2Xs

62

•Effect of field current as an alternator connected to infinity –bus.

1.0 puIa&Pf

Pf

Pf=1

IfLeading Lagging

R Dubey (9810381076) 63

64

65

Operation with a Large System- infinite bus

The power system is modelled as an infinite bus which maintains constant frequency and constant voltage. Here, if we increase the mechanical drive, we do not increase the frequency as in stand-alone system; rather, we contribute larger real power to the grid. Likewise, if we increase the dc field current, we do not increase the output voltage as in stand-alone system; rather, we change the reactive power contributed to the system.

66

If we increase the mechanical drive to the alternator, the rotor power angle δR increases and as a result the real power, delivered to the infinite bus increases. The reactive power delivered by the alternator can be controlled by controlling the dc exciting current If.

If we increase If, the magnitude of excitation voltage E increases. If we keep the mechanical drive constant (i.e., the angle δR unaltered), three conditions by just varying the dc excitation current If:

R Dubey (9810381076) 67

68

69

IES 2001Q1. An star connected alternator is synchronized at no load with an infinite bus of 11kV, its steam input is then increased till its output power is 15MW. Now when its excitation emf is increased to 130 % the machine starts operating at a pf of 0.8 lagging. Compute the synchronous reactance of the machine. Neglect armature resistance.Determine the pf, load angle and armature current of the machine before the excitation emf is increased.

Solution . Alternator is synchronized at no load, thus

When excitation is increased to 130%Ef2=1.3Ef1 & θ2=cos-1(0.8)=370

123.0sin

5sin3

11 315sin.

1

1

1111

=

=

==

s

s

s

tff

X

X

XVE

VE

δ

δ

δEf2

jIa2 Xs

Ia2

δ2

θ2

70

From the phasor diagram,

putting the values of

Armature current, before the excitation is increased,

E f 2= V t cosθ2

b c2+ V t sinθ2 + I a2

X S

b c2rwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

but 3pwwwwwwwwwwwwwwwwwwwwwwwwwwwA11 A cosθ2 A I a2

= 15B103

I a2= 984 A

E f 2= 1.3B 11

3pwwwwwwwwwwwwwwwwwwwwwwwwwwwfffffffffand cosθ2 = 0.8 sinθ2 = 0.6V t =

113pwwwwwwwwwwwwwwwwwwwwwwwwwwwfffffffffand I a2

= 984 A

we get X S = 2.73Ω

then sinδ1 = 0.123 X S [ δ = 19.78 [ θ1 =19.78

2fffffffffffff

Pf = cosθ1 = 0.985 leadingb c

3pwwwwwwwwwwwwwwwwwwwwwwwwwwwA11 A cosθ1 =

15B103

I a1

fffffffffffffffffff[ I a1

= 15B103

3pwwwwwwwwwwwwwwwwwwwwwwwwwwwA11 A cosθ1

fffffffffffffffffffffffffffffffff=799.3A

R Dubey (9810381076) 71

72

Two alternators connected in parallel:Frequency i.e. speed power characteristics of the two alternator connected in parallel is shown in fig..Active and reactive power supplied to the common load by each generator, are controlled respectively by their prime movers throttles and field currents.

Condition 1: total load Condition 2 : in order to increase load on and reduce the load on , the frequency power ch. & must be raised and the frequency power ch. & must be lowered to keep the frequency constant , by the adjustment of the governor setting.Now, total load ,

G2

G1G2

G1

PL = P1. + P2.

PL = P1+ P2

73

IES 1997 Q2. Two synchronous generators are supplying a common load. Generator 1 has a no-load frequency of 510.5 Hz and regulation of 1 MW/Hz. Generator 2 has a no load frequency of 51.0 Hz and regulation of 1MW/Hz. The total load 2.5 MW at 0.8 Pf lagging are required to be shared.(a)At what frequency, are the generators supplying this load and how much power is supplied by each generator?(b) A additional load of 1MW is attached to this system. What will be the new frequency and power generator of each generator?(c)How much is governor set- point of generator 2 to be adjusted to bring the system frequency at 50 Hz for 3.5 MW system load?

74

f 1=51.5@1BP1 and F2 =51.0@1BP2

but the machine are constrained to operate at common frequency A i Ae Af 1= f 2 = f say

` a

51.5@P1=51@P2 [ P1@P2 =0.5a` a

P1+P2 =2.5[ P1=1.5 MW & P2 =1.0 MWthen f =51.5@1.5 =50 Hzb` a

P1+P2 =3.5[ P1=2.0 MW & P2 =1.5 MWnewfrequency f . =51.5@2 =49.5 Hzc` a

in this case the systemfrequency is 50 Hziε A51.5@P1=50 [ P1=1.5 MW thus P2 =2.0 MWfor 2nd generator, on load speed is to be adjusted@f 02@P2 =50[ f 02 =52 Hz

Solution : frequency power ch. Equation of the two alternators-

R Dubey (9810381076) 75

Salient pole generatorThe air gap is not uniform along d-axis and q-axis, thus a constant can not be taken

like non-salient (cylindrical type ) generators.Phasor diagram can be drawn using two reaction theory (by Blondel) in which armature reaction mmf Fa can be resolved into two sinusoidal components Fad along d-axis and Faq along q- axis.

X S

I a = I q + I d

F a = F q + F d

Φa =Φad + Φaq

φaq

76

Phasor diagram - is taken as reference,Where (internal Pf angle)If load is leading then

θ = Pf angle δ = power angle

Where Xd=Xad+Xal and Xq=Xaq+Xal and

i.e. reactance due to armature reaction and leakage reactance in d-axis and q-axis respectively.

E f

ψψ

cossin

aq

ad

IIII

==

|| δθψ +=

qaddaaft

qqddaatf

XjIXjIRIEVXjIXjIRIVE

−−−=+++=

ψcos22 q

qdIIIIa =+=

Vt

|| δθψ −=

77

Calculation of voltage regulation:

AC must be a reactance drop as it is perpendicular to that means . Let it is, then ,

The term is due to saliency and reduces to zero for non- salient pole machine as

I a Ra I a

I a X

AD = AC cosΨ = I a X cosΨ = X I q = X q I q let X = X q

b c

E f . =V t + I a Ra + j I a X and CE = j I d X d@ j I a X sinΨ

= j I d X d@ X q

b c

E f = E f . + j I d X d@ X q

b c

[ E f = E f . + I d X d@ X q

b c

E f =V t + I a Ra + j I a X + j I d X d@ X q

b c

jI d X d@ X q

b c

X d = X q

Voltage regulation =E f @V t

V t

fffffffffffffffffffB100

78

Ex.- A salient pole synchronous generator has the following per unit parameterscalculate the excitation voltage on per unit basis and

regulation in % , when the generator is delivering rated KVA at rated voltage and at a pf of (a) 0.8 lagging (b) 0.8 leading .

Solution .

X d = 1.2 , X q = 0.8 , Ra = 0.025 E f

takingV t as a reference,V t = 1.0 + j 0.0a` a

I a = 0.8@ j 0.6 = 1 ∠@36.9o[ θ = 36.9o

I a Ra = 0.02@ j 0.015

j I a X q = j 0.8@ j 0.6b c

0.80` a

= 0.48 + j 0.64

E f . =V t + j I a X q + I a Ra = 1.5 + j 0.625 = 1.625 ∠ 22.2o

δ = 22.2o thus Ψ = δ + θ = 59.52o

I d = 1.0 sinΨ = 0.862I q = 1.0 cosΨ = 0.507

E f = 1.625 + I d X d@ X q

b c

= 1.9698 ∠ 22.2o

% regulation = 1.9698@ 1.01.0

ffffffffffffffffffffffffffffffB100% = 96.98 %

79

b` a

V t = 1.0 + j 0.0 and I a = 0.8 + j 0.6 = 1 ∠ 36.9o

I a Ra = 0.02 + j 0.015j I a X q =@ 0.48 + j 0.64 =@ 0.048 + j 0.64E f . =V t + I a Ra + j I a X q = 0.54 + j 0.655 = 0.849 ∠ 50.0o

Ψ = θ@ δ = 50.5o@36.9o =13.6o

I d = 1.0 sin13.6o = 0.235I q = 1.0 cos13.6o

then E f = E f . + I d X d@ X q

b c

= 0.849 + 0.235 0.4` a

E f = 0.943 pu

% regulation = 0.943@ 11

fffffffffffffffffffffffB100 % =@ 5.7 % negative

b c

80

Power angle characteristics of salient pole machine

After neglecting , the phasor diagram can be drawn as-For a lagging power factor angle θ

The per phase power is given as-

Ra

81

i.e. fundamental component and second harmonics with respect to the load angle The second component exists even it is because of different reluctance along the d-axis and q-axis, called the reluctance power.

δ

P = I d V t sinδb c

+ I q V t cosδb c

V t sinδ = ab = dc = I q X q & I q =V t sinδ

X q

fffffffffffffffff

similarly , V t cosδ = E f @ I d X d & I d =E f @V t cosδ

X d

fffffffffffffffffffffffffffffff

putting in the equation of power,

P =E f V t sinδ

X d

fffffffffffffffffffffffff+ V t2

2fffffff 1

X q

ffffffff@

1X d

ffffffffF G

sin2δ

E f = 0

82

83

84

Synchronous MotorHow it is different from an Induction Motor ?

1. A synchronous motor always runs at synchronous speed, whereas an induction motor runs at a speed slightly less than the synchronous speed. Thus, it is constant speed motor in true sense.

2. In synchronous motor, there are magnetic poles on the rotor. Induction motor has no magnetic poles on the rotor.

3. Induction motors are self starting but synchronous motors are not self starting.

85

The starting torque of synchronous motor is zero:When a 3Ø balanced voltage is applied to a 3Ø balanced winding of synchronous motor, a rotating magnetic field with synchronous speed is produced.If rotor is stationary (at starting), then stator field is equal to a stator field rotating at a synchronous speed w. r. to the rotor field. Thus the angle between the two fields λ= ωstis a function of time.The total field energy if-

the average value of this torque is Zero, as the average value of sinωst is zero.

W fld =12fffμ0 A π rl

gffffffffffffffffffF r

2

W fld =12fffμ 0 π rl

gfffffffffffffffffF s

2 + F R2 + 2F sF cosω st

d e

torque, T e =@∂W fld

∂ffffffffffffffλ =

μ0 π rlg

fffffffffffffffffAF s F R sinλ

where g = length of air gapr = radius of the rotorl = axial length of rotor

T e = K AF s AF R A sinω s t letμ 0 πrl

gfffffffffffffff= K

stationary

rotating

86

This is because, the rotor poles are stationary with N & S-poles on it. Stator poles with N & S poles on it are rotating with synchronous speed. When the N-pole of stator comes near to S-pole of the rotor, rotor experiences the maximum torque and when the S-pole of the stator comes to S-pole of the rotor, experiences the maximum torque in the opposite torque due to repulsion. Thus , the average value of the torque is zero.

87

88

Starting of Synchronous Motors(a) Starting by Using Damper Winding

(b) Starting by Using a Separate Induction Motor.(c) Starting By Using small DC Motor (Pony Motors)

89

90

91

92

93

94

95

96

Phasor Diagram of a Synchronous Motor

(a)Floating. (b) Motor on (c) Motor at on bus-bar no-load. Mechanical Load.

97

Effect of Change of ExcitationPhasor Diagrams for a Synchronous Motor

driving a constant load for (a) Normal excitation, (b) Over excitation, (c) Under excitation.

98

99

100

101

102

103

104

105

Effect of Excitation on Power Factor and Armature Current

106

An overexcited synchronous motor is used as a phase modifier or compensator.

107

108

109

110