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CONTENTS
CONTENTS

+ 0 ) 2 6 - 4

Learning Objectives
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Synchronous Motor-General
Principle of Operation
Method of Starting
Motor on Load with Constant
Excitation
Power
Flow
within
a
Synchronous Motor
Equivalent Circuit of a
Synchronous Motor
Power Developed by a
Synchronous Motor
Synchronous Motor with
Different Excitations
Effect of increased Load with
Constant Excitation
Effect of Changing Excitation
of Constant Load
Different Torques of a
Synchronous Motor
Power Developed by a
Synchronous Motor
Alternative Expression for
Power Developed
Various Conditions of Maxima
Salient Pole Synchronous Motor
Power Developed by a Salient
Pole Synchronous Motor
Effects of Excitation on
Armature Current and Power
Factor
Constant-Power Lines
Construction of V-curves
Hunting or Surging or Phase
Swinging
Methods of Starting
Procedure for Starting a
Synchronous Motor
Comparison between
Synchronous and Induction
Motors
Synchronous Motor
Applications

CONTENTS
CONTENTS

!&

SYNCHRONOUS
MOTOR

Ç

Rotary synchronous motor for lift
applications

1490

Electrical Technology

38
...
Synchronous Motor—General
A synchronous motor (Fig
...
1) is electrically identical with an alternator or a
...
generator
...
c
...
Most
synchronous motors are rated between
150 kW and 15 MW and run at speeds
ranging from 150 to 1800 r
...
m
...
It runs either at synchronous speed
or not at all i
...
while running it maintains a constant speed
...

2
...
It has
to be run upto synchronous (or near
synchronous) speed by some means,
before it can be synchronized to the
supply
...
It is capable of being operated under
a wide range of power factors, both lagging and leading
...


38
...
Principle of Operation
As shown in Art
...
7, when a 3-φ winding is fed by a 3-φ supply, then a magnetic flux of
constant magnitude but rotating at synchronous speed, is produced
...
38
...
With the rotor
rings
position as shown,
Exciter
suppose the stator poles
are at that instant situated
at points A and B
...

Fig
...
1

Synchronous Motor

1491

But half a period later, stator poles, having rotated around, interchange their positions i
...
N S is at
point B and S S at point A
...
Hence, rotor tends to
rotate clockwise (which is just the reverse of the first direction)
...
e
...
Owing to its large inertia, the rotor cannot instantaneously respond to such
quickly-reversing torque, with the result that it remains stationary
...
38
...
38
...
38
...
38
...
The stator and rotor poles are attracting each
other
...
38
...
Here, again
the stator and rotor poles attract each
other
...
e
...
38
...


38
...
Method of Starting
The rotor (which is as yet unexcited) is speeded up to synchronous
/ near synchronous speed by some arrangement and then excited by the
d
...
source
...

excited, it is magnetically locked into
position with the stator i
...
, the rotor
poles are engaged with the stator poles and both run synchronously in the same direction
...
The synchronous speed is given by the usual relation N S = 120 f / P
...
As the load on the motor is increased, the rotor progressively tends to fall
back in phase (but not in speed as in d
...
motors) by some angle (Fig
...
4) but it still continues to
run synchronously
...
In other words, the torque developed by the motor depends on
this angle, say, α
...
38
...
38
...
In Fig
...
5 are shown two pulleys P and Q transmitting power from the driver to the
load
...
When Q is loaded, it slightly falls behind owing
to the twist in the shaft (twist angle corresponds to α in motor), the angle of twist, in fact, being a
measure of the torque transmitted
...


38
...
Motor on Load with Constant Excitation
Before considering as to what goes on inside a synchronous motor, it is worthwhile to refer
briefly to the d
...
motors
...
29
...
c
...
m
...
Eb is set up in its armature conductors
...
The value of Eb depends, among other factors, on the speed of
the rotating armature
...


Fig
...
6

Fig
...
7

Fig
...
8

Similarly, in a synchronous machine, a back e
...
f
...
This back e
...
f
...
c
...
The net voltage in armature (stator) is the vector difference (not
arithmetical, as in d
...
motors) of V and Eb
...
c
...


Synchronous Motor

1493

Fig
...
6 shows the condition when the
motor (properly synchronized to the supply)
is running on no-load and has no losses
...

It is seen that vector difference of Eb and V is
zero and so is the armature current
...
In other words, the motor just
floats
...
38
...
**
If, now, the motor is loaded, then its rotor will further fall back in phase by a greater value of
angle α − called the load angle or coupling angle (corresponding to the twist in the shaft of the
pulleys)
...
38
...


38
...
Power Flow within a Synchronous Motor
Let
then

R a = armature resistance / phase ; XS = synchronous reactance / phase
ZS = Ra + j X S ;

Ia =

ER V − Eb
=
; Obviously, V = Eb + Ia ZS
ZS
ZS

The angle θ (known as internal angle) by which Ia lags behind ER is given by tan θ = X S / R a
...

Motor input = V Ia cos φ
—per phase
Here, V is applied voltage / phase
...
IL cos φ
...
m
...
× armature current × cosine of the angle between the two i
...
,
angle between Ia and Eb reversed
...
Fig
...
8
Out of this power developed, some would go to meet iron and friction and excitation losses
...

2
Out of the input power / phase V Ia cos φ, and amount Ia R a is wasted in armature***, the rest
2
(V
...
If power input / phase of the motor is P, then
P = Pm + Ia2 R a
or

*

mechanical power in rotor
Pm = P − Ia R a
2
For three phases
Pm =
3 V L IL cos φ − 3 Ia R a
The per phase power development in a synchronous machine is as under :
2

—per phase

This figure is exactly like Fig
...
74 for alternator except that it has been shown horizontally rather than
vertically
...
Its magnitude will change
only when rotor dc excitation is changed i
...
, when magnetic strength of rotor poles is changed
...


1494

Electrical Technology
Power input/phase in stator
P = V Ia cos φ
Armature (i
...
, stator) Cu loss
2
= Ia R a

Mechanical power in armature
Pm = Eb Ia cos (α − φ)

Iron, excitation & friction losses

Output power Pout

Different power stages in a synchronous motor are as under :

A C Electrical
Power Input to
Stator (Armature)

Stator
Cu
Loss

Pin

Gross Mechanical
Power Developed
in Armature
Pm

Net Mechanical
Power Output at
Rotor Shaft,

Iron Friction
& Excitation
Loss

Pout

38
...
Equivalent Circuit of a Synchronous Motor
Fig
...
9 (a) shows the equivalent circuit model for one armature phase of a cylindrical rotor
synchronous motor
...
38
...
m
...
i
...
, −Eb and the impedance drop Ia ZS
...
The angle α* between
the phasor for V and Eb is called the load angle or power angle of the synchronous motor
...
C
...
38
...
7
...
Hence, the equivalent circuit for the motor becomes as
shown in Fig
...
10 (a)
...
38
...


1495

Synchronous Motor
∴

EbV
sin α
XS
EV
= 3 b sin α
XS

Pin =


...
for three phases

Since stator Cu losses have been
neglected, Pin also represents the gross
mechanical power {Pm} developed by the
motor
...
55 Pm / N s N–m
...


+Ia Xs

A
a
f

+
V

E

Xs
Ia

Eb

Ia

B

(b)

(a)
Fig
...
10

Example 38
...
A 75-kW, 3-φ, Y-connected, 50-Hz, 440-V cylindrical rotor synchronous motor
operates at rated condition with 0
...
f
...
The motor efficiency excluding field and stator
losses, is 95% and X S = 2
...
Calculate (i) mechanical power developed (ii) armature current
(iii) back e
...
f
...

Solution
...
95 = 78,950 W
(ii) Since power input is known
∴

3 × 440 × Ia × 0
...
Let V = 254
∠0º as shown in Fig
...
11
...
9º × 2
...
9º = 254 − 322 (cos
126
...
9º) = 254 − 322 (− 0
...
8) = 516 ∠−
∠−30º
(iv) ∴
α = −30º
(v) pull-out torque occurs when α = 90º

Ia
f
O

o

V=254ÐO
d
516

Ð -30

Ia Xs
o

Eb
Fig
...
11

256 × 516
EbV
sin δ = 3
= sin 90º = 157,275 W
2
...
55 × 157, 275/1500 = 1,000 N-m
maximum Pm = 3

38
...
Synchronous Motor with Different Excitations
A synchronous motor is said to have normal excitation when its Eb = V
...
In both these conditions, it has a lagging power
factor as shown in Fig
...
12
...
c
...
38
...
There will be some value of excitation for
which armature current will be in phase with V , so that power factor will become unity, as shown in
Fig
...
13 (b)
...
m
...
Eb can be found with the help of vector diagrams for various
power factors, shown in Fig
...
14
...
38
...
38
...
f
...
38
...
f
...
14 (b)]
Eb = V + Ia Z S cos [180º − (θ + φ)] + j Ia Z S sin [180º − (θ + φ)]

I a Z S sin [180º − (θ + φ)]
V + I a Z S cos[180º − (θ + φ)]
(iii) Unity p
...
[Fig
...
14 (c)]
Here,
OB = Ia Ra and BC = Ia XS

−1  I X
∴
Eb = (V − Ia Ra) + j Ia XS ; α = tan  a S 
V
I
R
−

a a 
α = tan−1

C

C
C

a
A

B

b

a

q
f

a
V

O
(b)

a
q

A

Ia
(a)

E

Zs

Ia

ER =I

s

Z
R =I
a

E

f

a

s

O

B

Z
=I a
ER

q
a

Eb

Eb

O

B

Ia

V

A

(c)

Fig
...
14

38
...
Effect of Increased Load with Constant Excitation
We will study the effect of increased load on a synchronous motor under conditions of normal,
under and over-excitation (ignoring the effects of armature reaction)
...
Whatever the value of excitation, it
would be kept constant during our discussion
...
e
...

(i) Normal Excitation
Fig
...
15
...

Now, suppose that load on the motor
a 2
a 1
V
V
is increased as shown in Fig
...
15
O
f 2
O
f 1
Ia1
(b)
...

(b)
(a)
Unlike a d
...
motor, a synchronous
Fig
...
15
motor cannot increase its Ia by
decreasing its speed and hence Eb because both are constant in its case
...
rotor falls back in phase i
...
,
load angle increases to α2 as shown in
Fig
...
15 (b),
2
...
as a result, Ia1 increases to Ia2,
thereby increasing the torque developed by the motor,
4
...

Since increase in Ia is much greater
than the slight decrease in power factor, Geared motor added to synchronous servo motor line offers a
wide range of transmission ratios, and drive torques
...
It will be
seen that essentially it is by increasing its Ia that the motor is able to carry the extra load put on it
...
38
...
38
...


1498

Electrical Technology

(ii) Under-excitation
As shown in Fig
...
16 (b), with a small load and hence, small torque angle α1, Ia1 lags behind V
by a large phase angle φ1 which means poor power factor
...
That is why
Ia1 of Fig
...
16 (b) is larger than Ia1 of Fig
...
15 (a)
...
f
...
f
...
Due to increase both in Ia and p
...
, power
generated by the armature increases to meet the increased load
...

(iii) Over-excitation
ER2
When running on light load, α1 is small but
Ia1
Eb
Ia1 is comparatively larger and leads V by a larger
E
R1
angle φ1
...
The armature current also
a 1
V
increases thereby producing the necessary
O
increased armature power to meet the increased
Over Exicitation
applied load (Fig
...
17)
...
f
...
38
...

Summary
The main points regarding the above three cases can be summarized as under :
1
...

2
...
f
...

3
...
f
...

4
...
f
...


ER

O

f

E=
b 40
0V

1

Ia

0V

Solution
...
5º (mech) Displacement [Fig 38
...
) =
(mech)
2
∴ α (elect)
20 × 0
...
2
...
It has armature ressistance per phase of zero and synchronous
reactance of 10 Ω
...
5º (mechanical) from its synchronous position, compute
...
Machines, AMIE Sec
...
38
...
5 + j 35 = 35 ∠ 87
...
5º/10 ∠ 90º = 3
...
5º A/phase
Obviously, Ia lags behind V p by 2
...
5 × cos 2
...
Hence 4197 W also represent
power developed by armature
...
38
...
4 = 338
...
9º
(iii) I a = 338
...
9º/10 ∠ 90º = 33
...
1º A/phase
(iv) motor power/phase = V p Ia cos φ = 400 × 33
...
1º = 12,244 W
Total power = 3 × 12,244 = 36,732 W = 36
...
rotor displacement increases from 5º (elect) to 50º (elect) i
...
Eb falls back in phase
considerably
...
ER increases from 35 V to 338 V/phase
3
...
5 A to 33
...
angle φ increases from 2
...
1º so that p
...
decreases from 0
...
906 (lag)
5
...

Obviously, increase in Ia is much more than decrease in power factor
...

Special Illustrative Example 38
...
f
...
Motoring - mode : 400 V/Ph, 32 A/Ph, Unity p
...
, XS = 10 ohms
...


Fig
...
19 (a) Generator-mode

1500

Electrical Technology

Solution
...
38
...
66º
400
Total power in terms of parameters measurable at terminals (i
...
, V , I, and φ)
= 3 V ph Iph cos φ = 3 × 400 × 32 = 38
...
25, δ = tan

400 × 512
...
66º ) × 10−3 = 38
...

For motoring mode :
V = O A = 400, − IXS = A B = 320
E = OB = 512
...
38
...
66°, as before
...
It is +ve for
generator and –ve for motor
...
f
...
f
...

As before, power can be calculated in two ways and it will be electrical power input to motor and
also the mechanical output of the motor
...
4 kW
= 3×

Fig
...
19 (b) Motoring mode

38
...
Effect of Changing Excitation on Constant Load
As shown in Fig
...
20 (a), suppose a synchronous motor is operating with normal excitation
(Eb = V ) at unity p
...
with a given load
...
f
...
The armature is drawing a power of V
...
Now, let us discuss the effect of decreasing or
increasing the field excitation when the load applied to the motor remains constant
...
38
...
m
...
is reduced to Eb1
at the same load angle α1
...

Even though Ia1 is larger than Ia in magnitude it is incapable of producing necessary power V Ia for
carrying the constant load because Ia1 cos φ1 component is less than Ia so that V Ia1 cos φ1 < V Ia
...
It increases back e
...
f
...
Consequently, armature
current increases to Ia2 whose in-phase component produces enough power (V Ia2 cos φ2) to meet the
constant load on the motor
...
38
...
The resultant
voltage ER1 causes a leading current Ia1 whose
in-phase component is larger than Ia
...
Accordingly, load angle decreases
from α1 to α2 which decreases resultant voltage
from ER1 to ER2
...
In that case, armature
develops power sufficient to carry the constant
load on the motor
...


38
...
Different Torques of a
Synchronous Motor
Fig
...
20

Various torques associated with a synchronous motor are as follows:
1
...
running torque
3
...
pull-out torque
(a) Starting Torque
It is the torque (or turning effort) developed
by the motor when full voltage is applied to its
stator (armature) winding
...
Its value may be as low
as 10% as in the case of centrifugal pumps and as
high as 200 to 250% of full-load torque as in the
case of loaded reciprocating two-cylinder compressors
...
It is
determined by the horse-power and speed of the driven machine
...
The motor must have a breakdown or a maximum running torque greater than this value in order to avoid stalling
...
Afterwards, excitation is switched on and the rotor pulls into step with the synchronouslyrotating stator field
...


1502

Electrical Technology

(d) Pull-out Torque
The maximum torque which the motor can develop without pulling out of step or synchronism is
called the pull-out torque
...
Motor develops maximum torque when its rotor is retarded by an
angle of 90º (or in other words, it has shifted backward by a distance equal to half the distance
between adjacent poles)
...


38
...
Power Developed by a Synchronous Motor
In Fig
...
21, O A represents supply voltage/phase and Ia = I is the armature current, AB is back
e
...
f
...
OB gives the resultant voltage ER = IZS (or I XS if R a is negligible)
...
Line
a
CD is drawn at an angle of θ to A B
...

f Eb I aX s
+
Mechanical power per phase developed in the rotor
V
is
E
Ia
B
Pm = Eb I cos ψ

...

38
...
(ii)
ZS
ZS
 ZS
 ZS
This is the expression for the mechanical power developed in terms of the load angle (α) and the
internal angle (θ) of the motor for a constant voltage V and Eb (or excitation because Eb depends on
excitation only)
...
N = 9
...

EV
d Pm
∴
= − b sin (θ − α) = 0 or sin (θ − α) = 0
∴ θ=α
ZS
dα
*

Since R a is generally negligible, Z S = X S so that θ ≅ 90º
...


1503

Synchronous Motor
2

2

EbV Eb
EV E
cos α or (Pm )max = b − b cos θ
...
e
...
Maximum value of
θ (and hence α) is 90º
...
Equation (ii) is
plotted in Fig
...
22
...
(iv) (Pm )max = b
...
This corresponds to the ‘pull-out’ torque
...
38
...
13
...
38
...
e
...
m
...
per phase i
...
, Eb at an angle α with O A
...

Mechanical power developed is,
Pm = Eb
...
38
...
(i)

Now, ER and functions of angles θ and γ will be eliminated as follows :
From ∆ OAB ; V /sin γ = ER / sin α
∴ sin γ = V sin α / ER
From ∆ OBC ; ER cos γ + V cos α = Eb
∴ cos γ = (Eb − V cos α)/ER
Also
cos θ = R a / Z S and sin θ = X S / Z S
Substituting these values in Eq
...
E  R E − V cos α X S V sin α 

...
b
ZS  ZS
ER
ZS
ER 
2
EV
E R
= b (Ra cos α + X S sin α) − b a

...

Note
...
(ii), we get
Pm =

2
E2
E Z cos θ
EV
EbV
(Z S cos θ cos α + Z S sin θ sin α) − b S
= b cos (θ − α) − b cos θ
ZS 2
ZS
ZS
ZS2

It is the same expression as found in Art
...
10
...
37
...


1504

Electrical Technology

38
...
Various Conditions of Maxima
The following two cases may be considered :
(i) Fixed Eb, V, Ra and XS
...
Differentiating Eq
...
38
...
(ii), we get
2
EV
EV 
R
X  E R
E2 R
(Pm )max = b2 (R a cos θ + X S sin θ) − b 2 a = b 2  Ra
...
S  − b 2 a
ZS
ZS 
Zs
Zs 
Zs
Zs
2
2
2
2
EbV  Ra + X s  Eb Ra EbV Eb Ra
=
−

 −
=

...

Solving for Eb from Eq
...
(Pm )max 
2R 
a

The two values of Eb so obtained represent the excitation limits for any load
...
In this case, Pm varies with excitation or Eb
...
m
...
Eb which is necessary for maximum power possible
...
(i) above may be differentiated with respect to Eb and equated to zero
...
(ii)
∴
2R a
ZS
ZS2
d Eb
Putting this value of Eb in Eq
...
15
...
It is due to the fact that cylindrical-rotor motors have a uniform air-gap, whereas in salientpole motors, air-gap is much greater between the poles than along the poles
...

Hence, salient-pole theory is required only when very high degree of accuracy is needed or when
problems concerning transients or power system stability are to be handled
...
38
...
m
...
to give maximum power, but it is not the maximum possible value of the
generated voltage, at which the motor will operate
...
The motor has d-axis reactance X d and q-axis
reactance X q
...
The complete
phasor diagram of a salient-pole synchronous motor, for a lagging power factor is shown in
Fig
...
24 (a)
...
38
...
e
...
m
...
Eb is given by
Eb = V cos α − Iq R a − Id Xd
Similarly, as proved earlier for a synchronous generator, it can also be proved from Fig
...
24 (b)
for a synchronous motor with R a = 0 that
I a X q cos φ
tan α =
V − I a X q sin φ
In case R a is not negligible, it can be proved that
I a X q cos φ − I a R a sin φ
tan α =
V − I a X q sin φ − I a Ra cos α

38
...
Power Developed by a Salient Pole Synchronous Motor
The expression for the power developed by a salient-pole synchronous generator derived in
Chapter 35 also applies to a salient-pole synchronous motor
...
per phase

E V

V 2 (X d − Xq)
b

3
sin
sin 2 α 
...
NS in rps
...
55 Pm / N S
As explained earlier, the power consists of two components, the first component is called excitation power or magnet power and the second is called reluctance power (because when excitation is
removed, the motor runs as a reluctance motor)
...
4
...
Neglecting losses, calculate the torque developed by the
motor if field excitation is so adjusted as to make the back e
...
f
...

V = 2300 / 3 = 1328 V ; Eb = 2 × 1328 = 2656 V
EV
2656 × 1328
Excitation power / phase = b sin α =
sin 16º = 30,382 W
Xd
32

Solution
...
55 × 117,425/1000 = 1120 N-m

1506

Electrical Technology

Example 38
...
A 3300-V, 1
...
Neglecting all losses, calculate the excitation e
...
f
...
f
...
(Similar Example, Swami Ramanand Teertha Marathwada Univ
...


V = 3300 / 3 = 1905 V; cos φ = 1; sin φ = 0 ; φ = 0º
6
I a =1
...
4125 ; ψ = − 22
...
4º) = 22
...
4º) = − 100 A; Iq = 262 cos (− 22
...
4º) − (− 100 × 4) = 2160 V
= 1029 sin α + 151 sin 2 α

Pm =

2
V (X d − X q)
EbV
sin α +
sin 2α
2 Xd Xq
Xd


...
kW/phase

...
285 ; α = 73
...
4º + 151 sin 2 × 73
...
6
...

The effective resistance and synchronous reactance per phase are respectively 1 ohm and 30 ohm
...
8 leading
...
Engg
...
(i) Motor power input = 3 × 11000 × 60 × 0
...
8 kW
Pm = P2 − rotor Cu loss = 915 − 10
...
2 kW

0V

180

−1
Vp = 11000/ 3 = 6350 V ; φ = cos 0
...
9º ;
−1
θ = tan (30/1) = 88
...
38
...
1º + 36
...
572

∴

Eb = 7528 V ; line value of Eb = 7528 ×

3 = 13042

E=
b 7
528
V
o

88
...
9
6350V
Fig
...
25

A

Synchronous Motor

1507

Special Example 38
...
Case of Salient - Pole Machines
A synchronous machine is operated as below :
As a Generator : 3 -Phase, Vph = 400, Iph = 32, unity p
...

As a Motor : 3 - Phase, Vph = 400, Iph = 32, unity p
...

Machine parameters :
Xd = 10 Ω, Xq = 6
...


Fig
...
26 (a) Generator-action

Solution
...
5 = 208 V
OB =

4002 + 2082 = 451 V,

−1 AB
−1
= tan 208 = 27
...
5 = 51
...
8 V
Currents : I = OC = 32, Iq = I cos δ = OD = 28
...
, Id = DC = I sin δ = 14
...
E leads V
in case of generator, as shown in Fig
...
26 (a)
Power (by one formula) = 3 × 400 × 32 × 10−3 = 38
...
8

Power (by another formula) = 3 × 
sin 27
...
5 × sin 55º 
10
2
65



= 38
...
38
...
5º as before but now E lags behind V
...
8 V in the direction shown
...
8 V as before
Currents : OC = 32 amp
...
4 amp
...
8 amp
...
4 amp
...
8 amp
Power (by one formula) = 38
...
44 kW
OB =

Note
...
f
...


Example 38
...
A 500-V, 1-phase synchronous motor gives a net output mechanical power of
7
...
9 p
...
lagging
...
8 Ω
...
Calculate the
commercial efficiency
...
1988)
Solution
...
46 kW = 7,460 W
= Pout + iron and friction losses + excitation losses
= 7460 + 500 + 800 = 8760 W

...
38
...
9 ± (500 × 0
...
8 × 3760
2 × 0
...
7 32
...
2 A
1
...
6
1
...
2 × 0
...
8206

or 82
...


Example 38
...
A 2,300-V, 3-phase, star-connected synchronous motor has a resistance of
0
...
2 ohm per phase
...
5 power factor leading with a line current of 200 A
...
m
...
per
phase
...
Engg
...
1993)
φ
θ
(θ + φ)
cos 144
...
Here,
∴

=
=
=
=

cos− 1 (0
...
2/0
...
8º
84
...
8º
− cos 35
...
209 = 442 V
The vector diagram is shown in Fig
...
27
...
2 + 2
...
209 Ω
2

Eb
q

442

3 =1328 volt

Fig
...
27

2
V + E R2 − 2 V
...
2º = 1708 Volt / Phase

Example 38
...
A 3-phase, 6,600-volts, 50-Hz, star-connected synchronous motor takes 50 A
current
...

Find the power supplied to the motor and induced emf for a power factor of (i) 0
...
8 leading
...
Engg
...
1988)
Solution
...
f
...
8 lag (Fig
...
28 (a))
...
8 = 457,248 W
Supply voltage / phase = 6600 / 3 = 3810 V
−1
−1
φ = cos (0
...
8′
ZS =
∴

202 + 12 = 20 Ω (approx
...
m
...
= 3263 × 3 = 5651 V
(ii) Power input would remain the same
...
38
...
38
...
38
...
m
...
= 3 × 4447
= 7,700 V
Note
...


1510

Electrical Technology

Example 38
...
A synchronous motor having 40% reactance and a negligible resistance is to
be operated at rated load at (i) u
...
f
...
8 p
...
lag (iii) 0
...
f
...
What are the values of induced
e
...
f
...

(Electrical Machines-II, Indore Univ
...
Let
(i) At unity p
...


V = 100 V, then reactance drop = Ia X S = 40 V
θ = 90º,

Here,

Eb =

1002 + 402 = 108 V


...
38
...
f
...
8 (lag
...
5 V, as in Fig
...
29 (b)
B

B

V

100 V

Ia

f

A

Eb
a

M

A

100A

M

q

V

O

O

q

40

q

B

Eb

40

Eb

40 V

a
f
100 V

Ia
(b)

(a)

Ia
A

(c)

Fig
...
29

Alternatively,

AM 2 + MB 2 = 762 + 322 = 82
...
f
...
8 (lead
...
9º = 126
...
9º = 128 V
2
2
2
2
2
Again from Fig
...
29 (c), Eb = (OM + O A) + M B = 124 + 32 ; Eb = 128 V
...
12
...
5 Ω and 40 Ω respectively
...
m
...

and angular retardation of the rotor when fully loaded at (a) unity p
...
(b) 0
...
f
...
8 p
...
leading
...
Engineering-II, Bangalore Univ
...
38
...


Full-load armature current = 1,000 × 1,000 /

3 × 11,000 = 52
...
8 ∴ φ = 36º53′
Armature resistance drop / phase = Ia Ra = 3
...
5 = 184 V
reactance drop / phase = Ia XS = 40 × 52
...
)
2

2

−1

tan θ = X S / Ra ∴ θ = tan (40 / 3
...
f
...
38
...
9961
sin α

sin α = 2,100 × 0
...
3212
∴ α = 18º44′′
(b) At p
...
0
...
38
...
36
...
190
sin α
sin 48º 7′ 0
...
7443/5190 = 0
...
f
...
8 leading [Fig
...
30 (c)]
∠ BOA = θ + φ = 85º + 36º53′ = 121º53′
2
2
2
∴
Eb = 6,351 + 2,100 − 2 × 6,351 × 2,100 × cos 121º53′
∴
Eb = 7,670 volt per phase
...
m
...
8493
sin α
sin α = 2,100 × 0
...
2325 ∴ α = 13º27′′

Special Example 38
...
Both the modes of operation : Phase - angle = 20º Lag
Part (a) : A three phase star-connected synchronous generator supplies a current of 10 A
having a phase angle of 20º lagging at 400 volts/phase
...
5 ohms
...
Calculate voltage
regulation
...
The phasor diagram is drawn in Fig 38
...
38
...
8 V
AF = I Xq = 10 × 6
...
8 V
OF =
DC
OD
FE
E

2
2
376 + 201
...
7 V, δ = 8
...
22º = 4
...
22º = 8
...
73 × 3
...
56 V
...
7 + 16
...
3 V
443 − 400
× 100% = 10
...


Fig
...
31 (b) Motoring-mode

AF = − Ia X q = − 65 V, AB = 136
...
8 V
OB = 400 cos 20º = 376 V
OF =
20º − δ
FE
E
Currents :
Ia
Iq

3762 + 71
...
8 V
−1

−1

= tan BF/OB = tan 71
...
8º, δ = 9
...
874 × (3
...
56 volts, as shown in Fig
...
31 (b)
= OE = OF + FE = 382
...
56 = 376
...
8º = 9
...
8º = 1
...
Id is in downward direction
...
e
...

Thus, Excitation emf = 376
...
2°
(Note
...
)

Power (by one formula) = 11276 watts, as before
Power (by another formula)
2
= 3 [(V E/X d) sin δ + (V /2) {(1/X q) sin 2δ}]
= 3 [(400 × 376
...
2º + (400 × 400/2) (3
...
4º]
= 3 × [2406 + 1360] = 11298 watts
...
]

1513

Synchronous Motor

Example 38
...
A 1-φ alternator has armature impedance of (0
...
866)
...
The iron and friction losses
amount to 500 W
...
m
...

(Elec
...
1990)
Solution
...
Cu loss/phase = Ia R a = 50 × 0
...
f
...
775 ∴ φ = 39º lag or lead
...
866 / 0
...
38
...
38
...
52 + 0
...


In Fig
...
32 (b), ∠ BOA = 60º + 39º = 99º
∴

AB = Eb =

(2002 + 50 2 ) − 2 × 200 × 50 cos 99º ; Eb = 214 V
...
15
...
4 + j 6) ohm/phase
...
When the motor is loaded, the rotor is retarded by 3º mechanical
...
What is the maximum power the motor can supply without falling out of step?
(Power Apparatus-II, Delhi Univ
...
Per phase Eb = V = 2200/ 3 = 1270 V
α = 3º (mech) = 3º × (8/2) = 12º (elect)
...
33 (a)
...
42 + 62 = 6
...
013 = 44
...
From ∆ OAB,
Fig
...
33 (a)
266
1270
we get,
=
sin
12º
sin (θ − φ)
∴
sin (θ − φ) = 1270 × 0
...
9926
∴ (θ − φ) = 83º
−1
−1
Now,
θ = tan (X S / R a) = tan (6 / 0
...
18º
φ = 86
...
18º
∴ p
...
= cos 3
...
998(lag)
Total motor power input = 3 V Ia cos φ = 3 × 1270 × 44
...
998 = 168 kW

1514

Electrical Technology
Total Cu loss = 3 Ia R a = 3 × 44
...
4 = 2
...
34 = 165
...
4
−
=
−
= 250 kW
2
ZS
6
...
013

Example 38
...
A 1−φ, synchronous motor has a back e
...
f
...
The synchronous reactance of the armature is 2
...
Find the power factor at which the motor is operating and state whether the
current drawn by the motor is leading or lagging
...
As induced e
...
f
...

In the vector diagram of Fig
...
33 (b), OA represents
applied voltage, A B is back e
...
f
...
OB
represents resultant of voltage V and Eb i
...
ER
In ∆ OBA,
ER =
=
Now,
∴
Now
∴

ER
sin 30º
sin (θ + φ)
tan θ
p
...
of motor

B
25

0V

C

Ia

ER
q

D

o

f

30

A

200 V

O

Fig
...
33 (b)

(V + Eb − 2 V Eb cos 30º )
2

2

2
2
(220 + 250 − 2 × 200 × 250 × 0
...
5
= 125/126 (approx
...
5
∴ θ = 68º12′
∴ φ = 90º − 68º12′ = 21º48′
= cos 21º48′ = 0
...
17
...
For a certain load, the input is 900 kW and the induced line e
...
f
...
(line value)
...
Neglect resistance
...
Machines, Nagpur Univ
...
Applied voltage / phase = 6,600 / 3 = 3,810 V
Back e
...
f
...
I cos φ = 900,000

∴
I cos φ = 9 × 105 / 3 × 6,600 = 78
...
38
...
X S = 10 I
BC = OB cos φ = 10 I cos φ
B
= 10 × 78
...
4 V
2
2
2
∴
5,140 = 787
...
4 = 1
...
2º, cos φ = 0
...
74 ; I = 78
...
527 = 149
...
38
...
18
...
Its synchronous reactance is 20 ohms per phase and armature resistance
negligible when the input power is 1000 kW, the power factor is 0
...
Find the power angle
and the power factor when the input is increased to 1500 kW
...
Machines, AMIE Sec
...
When Power Input is 1000 kW (Fig
...
35 (a))

3 × 6600 × Ia1 × 0
...
3 A
−1
ZS = X S = 20 Ω ; Ia1 Z S = 109
...
8 = 36
...
9º)
2
2
= 3810 + 2186 − 2 × 3810 × 2186 × − cos 53
...
e
...

When Power Input is 1500 kW :
2

2

2

B

B

2624 V

E
E
3 × 6600 × I a2 cos φ 2 =
b =5
b =5
41
Ia
41
I
0V
0
2
a1
1500,000; Ia2 cos φ2 = 131
...
38
...
2 = 2624 V
2
2
In ∆ ABC, we have, A B = AC
Fig
...
35
2
2
2
2
+ BC or 5410 = A C + 2624
∴
AC = 4730 V; OC = 4730 − 3810 = 920 V
tan φ2 = 920 / 2624 ; φ2 = 19
...
f
...
4º = 0
...
19
...
Its synchronous reactance
is 10 Ω per phase and resistance is negligible
...

(Elect
...
B, 1990)

q
a
f

0

3 × 400 × Ia cos φ = 5472 ; Ia cos φ = 7
...
38
...


AC =

231 − 79
2

2

A

231V

Fig
...
36

= 217 V; OC = 231 − 217=14 V

tan φ = 14 / 79; φ = 10º; cos φ = 0
...
9; Ia = 7
...
985 = 8 A; tan α = BC / A C = 79 / 217; α = 20º
Example 38
...
A 2,000-V, 3-phase, star-connected synchronous motor has an effective resistance and synchronous reactance of 0
...
2 Ω respecB
tively
...
m
...
is 2,500 V
...

43
V
ER
(Elect
...
A
...
I
...
T
...
, June 1992)
Solution
...
m
...
is greater than the applied voltage, the motor must be running with a leading p
...
If
the motor current is I, then its in-phase or power component is I
cos φ and reactive component is I sin φ
...
38
...
2 / 0
...
8º
BC is ⊥ AO produced
...
2 + j2
...
2 + j2
...
2 − 2
...
2 + 0
...
2 − 2
...
2 + 0
...
2 + 0
...
2 − 2
...
m
...
/ phase
In Fig
...
37
OA
AB
ER

I =

2
2
2
2
I1 + I 2 = 231 + 71 = 242 A

p
...
= I1/I = 231/242 = 0
...
21
...
46 kW
from the three phase mains
...
5 ohm
...
f
...
75 lag
...
The excitation loss is
650 watts
...

Calculate
...
A 3 -phase synchronous motor receives power from two sources :
(a) 3-phase a
...
source feeding power to the armature
...
C
...

Thus, power received from the d
...
source is utilized only to meet the copper-losses of the field
winding
...
c
...

In case of the given problem

3 × Ia × 440 × 0
...
052 amp
2
Total copper-loss in armature winding = 3 × 13
...
50 = 255 watts
Power supplied to the motor = 7460 + 650 = 8110 watts
Output
efficiency of the motor =
Input
Output from shaft = (Armature Input) − (Copper losses in armature winding)
− (friction and iron losses)
= 7460 − 255 − 500 = 6705 watts
Efficiency of the motor = 6705 × 100% = 82
...


B

Ia

497

3V

V
58
20

Example 38
...
Consider a 3300 V delta connected
synchronous motor having a synchronous reactance per
phase of 18 ohm
...
707 when
drawing 800 kW from mains
...

(Elect
...
1993)

90o

45o

O

A

3300 V
Fig
...
38

3 × 3300 × Ia × 0
...
3 A;
ZS = 18 Ω ; Ia ZS = 114
...
707; φ = 45º; θ = 90º;
cos (θ + φ) = cos 135º = − cos 45º = − 0
...
38
...
707
∴
Eb = 4973 V
From ∆ OAB, we get 2058/sin α = 4973/sin 135º
...
23
...
04 ohm and 0
...
Compute for
full-load 0
...
f
...
m
...
per phase and mechanical power developed
...
5%
...
Machines AMIE Sec
...
Motor input = 75,000 / 0
...
8 =146
...
04 2 + 0
...
402 Ω

B

Ia ZS = 146
...
402 = 58
...
4 / 0
...
3º; φ = cos 0
...
9º; (θ + φ) = 121
...
36
...
8 − 2 × 231 × 58
...
2; Eb / phase = 266 V
2
Stator Cu loss for 3 phases = 3 × 146
...
04 = 2570 W;
Ns = 120 × 50/40 = 1500 r
...
m
...
55 × 78510/1500 = 500 N-m
...
3
36
...
38
...
24
...
5 Ω and 4 Ω per phase respectively
...
If the load torque is increased until the line
current is increased to 60 A, the field current remaining unchanged, calculate the gross torque
developed and the new power factor
...
Machines, AMIE Sec
...
The conditions corresponding to the first case are shown in Fig
...
40
...
5 = 3
...
968/0
...
936) = 81
...
m
...
is increased
...
7)
...
38
...

q
q
O
A
a
f
Let φ be the new phase angle
...

38
...

38
...

Since the field current remains constant, the value of Eb remains the same
...
4325 or 81º 48′ − φ = 64º24′
∴
φ = 81º48′ − 64º24′ = 17º24′
...
f
...
954 (lag)
Motor input = 3 × 400 × 60 × 0
...
5 = 5,400 W
Electrical power converted into mechanical power = 39,660 − 5,400 − 34,260 W
NS = 120 × 50/6 = 1000 r
...
m
...
55 × 34,260/1000 = 327 N-m
Example 38
...
A 400-V, 10 h
...
(7
...
Determine the minimum current and the
corresponding induced e
...
f
...
Assume an efficiency of 85%
...
C
...
1987)
Solution
...
e
...
The vector diagram is as shown in Fig
...
42
...
85 = 8,775 W

Eb =

263

126
...
67 A
Impedance drop = Ia XS = 10 × 12
...
7 V
Voltage / phase = 400 / 3 = 231 V


...
38
...
72 = 263
...
26
...
5 kW, star-connected synchronous motor has a
full-load efficiency of 88%
...
2 + j 1
...

If the excitation of the motor is adjusted to give a leading power factor of 0
...
m
...

(ii) the total mechanical power developed
(Elect
...
M
...
E
...
B, 1989)
Solution
...
5/0
...
61 kW; Ia = 42,610 / 3 × 400 × 0
...
3 A
V = 400 / 3 = 231 V; Z S = 0
...
6 = 1
...
87º
ER = Ia ZS = 68
...
612 = 110 V;
−1
φ = cos (0
...
84º
Now, (φ + θ) = 25
...
87º = 108
...
71º = − 0
...
71º or Eb = 286 V
Line value of excitation voltage = 3 × 285 = 495 V
(b) From ∆ OAB, (Fig
...
43) ER / sin α = Eb / sin (φ + θ), α = 21
...
4º = 14,954 W
1
...
8

8
q =
O

f = 25
...
38
...
27
...
Its synchronous reactance is 20 ohm per phase and armature resistance
negligible
...
8 leading
...

(Elect
...
M
...
E
...
B, 1991)
Solution
...
8 = 109
...
38
...
Since R a is
negligible, θ = 90º
B
E
b =5
Ia
410
ER = Ia XS = 109
...
8, φ = 36
...
87º) ;
f = 36
...
38
...
3521
Eb
cos (α + φ)
cos φ
cos φ
φ = 19
...
39º = 0
...
28
...
5 ohm/phase
...
866 p
...
leading
...
Excitation, friction, windage and iron losses total 2 kW
...
1990)

2

(Pm)max

EV E R
394 × 231 394 2 × 0
...
03
ZS
ZS 2
4
...

–Art
...
12
Maximum power developed in armature for 3 phases
= 3 × 17,804 = 52,412 W
Net output = 52,412 − 2,000 = 50,412 W = 50
...
V = 400/ 3 = 231 V/phase; Z S = 0
...
03 ∠ 82
...
9º
Ia Z S = 60 × 4
...
866
φ = 30º (lead)
B
As seen from Fig
...
45,
E
2
2
2
b=
Eb = 231 + 242 − 2 × 231 × 242 cos 112
...
9

o

f = 30
O

Ia
o

231 V

C

Fig
...
45

Example 38
...
A 6-pole synchronous motor has an armature impedance of 10 Ω and a resistance
of 0
...
When running on 2,000 volts, 25-Hz supply mains, its field excitation is such that the e
...
f
...
Calculate the maximum total torque in N-m developed before the
machine drops out of synchronism
...
Assuming a three-phase motor,
V = 2000 V, Eb = 1600 V ; R a = 0
...
5/10 = 1/20
Using equation (iii) of Art
...
power for 3 phases is
2
EbV Eb
2000 × 1600 16002 × 1
cos θ =
−
−
(Pm)max =
= 307,200 watt
10
10 × 20
ZS
ZS
Now,
NS = 120 f /P = 120 × 25/6 = 500 r
...
m
...
max be the maximum gross torque, then
307200
T g max = 9
...
38
...
A 2,000-V, 3-phase, 4-pole, Y-connected synchronous motor runs at 1500
r
...
m
...
The
resistance is negligible as compared with synchronous reactance of 3 Ω per phase
...

(Elet
...
-I, Nagpur Univ
...


B
11

0V

50

60

Voltage/phase = 2000/ 3 = 1150 V
Induced e
...
f
...
38
...
Since R a is negligible, θ = 90º
...
2605; φ = 16
...
f
...
2º = 0
...
38
...
965 = 668
...
p
...

∴ T g = 9
...


Power input =
NS

Example 38
...
A 3-φ, 3300-V, Y-connected synchronous motor has an effective resistance and
synchronous reactance of 2
...
0 Ω per phase respectively
...
m
...
is 3800 V between lines, calculate (i) the maximum total mechanical power that the motor can
develop and (ii) the current and p
...
at the maximum mechanical power
...
Gujarat Univ
...
θ = tan−1 (18/2) = 83
...
38-10)
2
2
1/2
ER = (1905 + 2195 − 2 × 1905 × 2195 × cos 83
...
11 Ω
I a = 2744/18
...
11
ZS
Z 2
18
...
855 (lead)
...
32
...
m
...
is 520 V
...
5 + j4
...
If the friction and iron
losses are constant at 1000 W, calculate the power output, line current, power factor and efficiency
for maximum power output
...
Machines-I, Madras Univ
...
As seen from Art
...

−1
−1
Now,
θ = tan (4/0
...
90º = α
ER =
Now,

4152 + 5202 − 2 × 415 × 520 × cos 82
...
52 = 4
...
03 = 155 A

3 × 155 = 268
...
5
=
= 45,230 W
− b 2a =
−
ZS
4
...
25
ZS

Line current =
(Pm)max

Max
...
69 kW
2
Total Cu loss = 3 × 155 × 0
...
5 × cos φ = 171,770 ; cos φ = 0
...
7845 or 78
...
1
1
...
5 A at a power factor of 0
...
Calculate the
power supplied and the induced e
...
f
...
25 + j3
...

[29
...
The input to a 11-kV, 3φ, Y-connected synchronous motor is 60 A
...
Find (a) power supplied to the
motor and (b) the induced e
...
f
...
f
...
8 leading
...
I
...
T
...
Dec
...
A 2,200-V, 3-phase, star-connected synchronous motor has a resistance of 0
...
Find the generated e
...
f
...
8 leading
...
21 kV; 14
...
62 kV; 12
...
A 3-phase, 220-V, 50-Hz, 1500 r
...
m
...
It receives an input line current of 30 A at a leading power factor of 0
...

Find the line value of the induced e
...
f
...

If the mechanical load is thrown off without change of excitation, determine the magnitude of the
current under the new conditions
...

[268 V; 6º, 20
...
A 400-V, 3-phase, Y-connected synchronous motor takes 3
...
Calculate the current and p
...
if the induced e
...
f
...

[6
...
86 lead] (Electrical Engineering, Madras Univ
...
The input to 6600-V, 3-phase, star-connected synchronous motor is 900 kW
...
If the generated voltage is 8,900 V

1522

Electrical Technology
(line), calculate the motor current and its power factor
...
See solved Ex
...
17 ] (Electrotechnics, M
...
Univ
...
A 3-phase synchronous motor connected to 6,600-V mains has a star-connected armature with an
impedance of (2
...
The excitation of machine gives 7000 V
...
Find the maximum output of the motor
...
68 kW]
8
...
Operating with an excitation corresponding to an e
...
f
...
To what open-circuit e
...
f
...

[4 kV]
9
...
6 kV, star-connected, 3-phase, synchronous motor works at constant voltage and constant excitation
...
0 + j 20) per phase
...
8 leading
...

[0
...
B Advanced Elect
...
A 2200-V, 373 kW, 3-phase, star-connected synchronous motor has a resistance of 0
...
0 Ω per phase respectively
...
m
...
per phase if the
motor works on full-load with an efficiency of 94 per cent and a p
...
of 0
...

[1510 V] (Electrical Machinery, Mysore Univ
...
The synchronous reactance per phase of a 3-phase star-connected 6600 V synchronous motor is 20
Ω
...
m
...
is 8,942 V
...
f
...

[97 A; 0
...
A synchronous motor has an equivalent armature reactance of 3
...
The exciting current is adjusted
to such a value that the generated e
...
f
...
Find the power factor at which the motor would
operate when taking 80 kW from a 800-V supply mains
...
965 leading] (City & Guilds, London)
13
...
The effective resistance
and synchronous reactance per phase are respectively 1 ohm and 30 ohms
...
8 leading
...
5 kW, 13 kV] (Elect
...
M
...
E
...
B, 1990)
14
...
5 ohm per phase and 4-ohm per phase respectively
...
If the load torque is increased until the
line current is 60 A, the field current remaining unchanged, find the gross torque developed, and the
new power factor
...
93] (Elect
...
AMIETE Dec
...
The input to a 11,000-V, 3-phase, star-connected synchronous motor is 60 amperes
...
Find the power
supplied to the motor and the induced e
...
f
...
8 (a) leading and (b) lagging
...
36 kV] (Elect
...
1981)
16
...

What is the output corresponding to a maximum input to a 3-φ delta-connected 250-V, 14
...
m
...
is 320 V ? The effective resistance and synchronous
reactance per phase are 0
...
5 Ω respectively
...
Give values for (i) output (ii) line current
(iii) p
...

[(i) 47
...
804] (Elect
...
Feb
...
A synchronous motor takes 25 kW from 400 V supply mains
...
Calculate the power factor at which the motor would operate when the field excitation is so adjusted that the generated EMF is 500 volts
...
666 Leading] (Rajiv Gandhi Technical University, Bhopal, 2000)

Synchronous Motor

1523

38
...
Effect of Excitation on Armature Current and Power Factor
The value of excitation for which back e
...
f
...
We will now discuss what happens when motor is either over-excited or
under-exicted although we have already touched this point in Art
...

Consider a synchronous motor in which the mechanical load is constant (and hence output is also
constant if losses are neglected)
...
38
...
38
...
e
...
The armature current I lags
behind V by a small angle φ
...
e
...

In Fig
...
47 (b)* excitation is less than 100% i
...
, Eb < V
...
We note that the magnitude of I is
increased but its power factor is decreased (φ has increased)
...
e
...
Hence, as excitation is decreased, I will increase but p
...
will decrease so that
power component of I i
...
, I cos φ = OA will remain constant
...

Incidentally, it may be noted that when field current is reduced, the motor pull-out torque is also
reduced in proportion
...
38
...
e
...
Here, the resultant
voltage vector ER is pulled anticlockwise and so is I
...
It may also happen for some value of excitation, that I may be in phase with V i
...
, p
...
is unity
[Fig
...
47 (d)]
...

Two important points stand out clearly from the above discussion :
(i) The magnitude of armature current varies with excitation
...
In between, it has minimum value corresponding to a certain excitation
...
38
...

(ii) For the same input, armature current varies over a wide range and so causes the power factor
also to vary accordingly
...
f
...
f
...
In between, the p
...
is unity
...
f
...
38
...


1524

Electrical Technology

Fig
...
48

are shown in Fig
...
48 (b)
...
f
...
It would be noted that minimum armature current corresponds to unity
power factor
...
38
...
This property of the motor renders it
extremely useful for phase advancing (and so power
factor correcting) purposes in the case of industrial
loads driven by induction motors (Fig
...
49) and
lighting and heating loads supplied through
transformers
...

Especially on light loads, the power drawn by them
has a large reactive component and the power factor
has a very low value
...
By using
synchronous
motors
in
conjunction with
induction motors
and transformers,
Synch
Induction
Motor
Motors
the
lagging
reactive power
Fig
...
49
required by the
latter is supplied locally by the leading reactive
component taken by the former, thereby relieving
the line and generators of much of the reactive
component
...
When used in this
way, a synchronous motor is called a synchronous
capacitor, because it draws, like a capacitor,
leading current from the line
...


Example 38
...
Describe briefly the effect of varying
excitation upon the armature current and p
...

Inductor motor
of a synchronous motor when input power to the
motor is maintained constant
...
3 kW, star-connected synchronous motor has a full-load efficiency of
88%
...
2 + j 1
...
If the excitation of the
motor is adjusted to give a leading p
...
of 0
...
m
...
(b) the
total mechanical power developed
...
Voltage / phase = 400 / 3 = 231 V;

Synchronous Motor

1525

(1
...
22 ) = 1
...
88 × 0
...
5V
F
A
∴
I ZS = 1
...
5 V
0
231V
With reference to Fig
...
50
tan θ = 1
...
2 = 8, θ = 82º54′
Fig
...
50
cos φ = 0
...

∴
(θ + φ) = 82º54′ + 25º50′ = 108º44′
Now cos 108º44′ = − 0
...
5 − 2 × 231 × 109
...
3212) = 285
...
6 V
ZS =

Line value of Eb = 3 × 285
...
88 = 42,380 W
2
Total Cu losses = 3 × I R a = 3 × 682 × 0
...
3 kW
...
34
...
8 lagging
...
m
...
is increased by 30%, the power taken remaining the
same
...
f
...

Solution
...
000/ 3 × 693 × 0
...
8, sin φ = 0
...
38
...
In ∆ OAB,
2
2
2
Fig
...
51
Eb = 400 + 100 − 2 × 400 × 100 × cos (90º − φ)
2
2
2
= 400 + 100 − 2 × 400 × 100 × 0
...

The vector diagram for increased e
...
f
...
38
...
Now, Eb = 1
...

It can be safely assumed that in the second case, current is leading V by some angle φ′
...
As power input
remains the same and V is also constant, I cos φ should be the same far the same input
...
8 = 40 = I′ cos φ′
2
2
2
In
∆ ABC, A B = AC + BC
I¢= 40A
B
Now
BC = I′ X S cos φ′ (∵ OB = I′ XS )
Eb =
= 40 × 2 = 80 V
f¢
454
V
2
2
2
90°
∴
454 = A C + 80 or A C = 447 V
f¢
A
C
∴
OC = 447 − 400 = 47 V
O
V=400 V
∴
tan φ′ = 47/80, φ′ = 30º26′
Fig
...
52
∴ New
p
...
= cos 30º26′
= 0
...
8623 = 46
...


1526

Electrical Technology

Example 38
...
A synchronous motor absorbing 60 kW is connected in parallel with a factory
load of 240 kW having a lagging p
...
of 0
...
If the combined load has a p
...
of 0
...
f
...
1990)

3-Phase
300 kW 333
...
f
...
9(lag)

Solution
...
38
...

Total load = 240 + 60 = 300 kW; combined p
...
= 0
...
8º, tan φ = 0
...
4834 = 145(lag)
Factory Load
cos φL = 0
...
9º, tan φL = 0
...
75 = 180 (lag)
or
load kVA = 240/0
...
6 = 180
∴ leading kVAR supplied by synchronous motor = 180 − 145 = 35
...
5kVA
35kVAR
f

240W

m

60W

60kW
p
...

Lead

f

145 kVAR

180 kVAR

LOAD
240kW
0
...
f
...
3kVA
300kVA

Fig
...
53

For Synchronous Motor
kW = 60, leading kVAR = 35, tan φm = 35/60 ; φm = 30
...
3º = 0
...
f
...
863 (lead)
...
5
...
18
...
38
...
m
...
/ phase,
Eb
...
The armature
current is OI lagging behind ER by an angle θ =
−1
tan X S / R a
...
Since Z S is constant,
ER or vector OB represents (to some suitable scale)
the main current I
...
B L is drawn perpendicular
to OX which is at right angles to O Y
...

Hence,
OB cos φ = I cos φ = B L
The power input / phase of the motor
= V I cos φ = V × B L
...
38
...
If motor is working with a constant intake, then
locus of B is a straight line || to OX and ⊥ to O Y i
...
line EF for which BL is constant
...
Similarly, a series
of such parallel lines can be drawn each representing a definite power intake of the motor
...
for equal increase in intake, the power lines are parallel and equally-spaced
2
...
the perpendicular distance from B to O X (or zero power line) represents the motor intake
4
...
e
...
In other words, locus of B is a circle with radius = AB and centre at A
...
Any further
increase in load on the motor will bring point B down to a lower line
...
The area to the right of AY 1 represents unstable conditions
...


38
...
Construction of V-curves

No
Lo
ad

Armature Current

The V -curves of a synchornous motor show how armature current varies with its field current
when motor input is kept constant
...
c
...
c
...

p
...
(lag)
38
...
There is a family of such curves, each
corresponding to a definite power intake
...
f
...
Power
2L
1/
input to motor is kept constant at a definite
value
...
P
...
Field
Current
noted
...
Similar curves
0
Field Current
can be drawn by keeping motor input constant
at different values
...
38
...
38
...

Detailed procedure for graphic construction of V -curves is given below :
1
...
38
...

2
...
are drawn
where A B, A B1, A B2, etc
...
m
...
The
intersections of these circles with lines of constant power give positions of the working points
for specific loads and excitations (hence back e
...
fs)
...
,
represent different values of ER (and hence currents) for different excitations
...
m
...

vectors A B, A B1 etc
...
38
...


1528

Electrical Technology

3
...
m
...
, are projected on the magnetisation
and corresponding values of the field (or exciting) amperes are read from it
...
The field amperes are plotted against the corresponding armature currents, giving us ‘V ’
curves
...
38
...
20
...
Hunting may also be caused if supply frequency is pulsating (as in the case of generators driven by reciprocating internal combustion engines)
...
), its
Salient
Field Poles
rotor falls back in phase by the coupling
angle α
...
If now, there is sudden
decrease in the motor load, the motor is
Shaft
immediately pulled up or advanced to a
new value of α corresponding to the new
load
...

Winding
Rings
In this way, the rotor starts oscillating (like
Fig
...
57
a pendulum) about its new position of

Synchronous Motor

1529

equilibrium corresponding to the new load
...
37
...
The
amplitude of these oscillations is built up
to a large value and may eventually become
so great as to throw the machine out of
synchronism
...
These dampers consist of shortcircuited Cu bars embedded in the faces of
the field poles of the motor (Fig
...
57)
...

But it should be clearly understood that
Salient - poled squirrel eage motor
dampers do not completely prevent hunting because their operation depends upon the presence of some oscillatory motion
...


38
...
Methods of Starting
As said above, almost all synchronous motors are equipped with dampers or squirrel cage windings consisting of Cu bars embedded in the pole-shoes and short-circuited at both ends
...
The procedure is as follows :
The line voltage is applied to the armature (stator) terminals and the field circuit is left unexcited
...
c
...
At that moment the stator and rotor poles get engaged or interlocked with
each other and hence pull the motor into synchronism
...
At the beginning, when voltage is applied, the rotor is stationary
...
m
...
in the rotor during the starting period, though the
value of this e
...
f
...

Normally, the field windings
are meant for 110-V (or 250
V for large machines) but
during starting period there
are many thousands of volts
induced in them
...

Fig
...
58

1530

Electrical Technology

2
...
In some cases, this may not
be objectionable but where it is, the applied voltage at starting, is reduced by using autotransformers (Fig
...
58)
...
Usually, a value of 50% to 80% of the full-line voltage is satisfactory
...
38
...
For reducing the supply voltage, the
switches S 1 are closed and S 2 are kept open
...


38
...
Procedure for Starting a Synchronous Motor
While starting a modern synchronous motor provided with damper windings, following procedure
is adopted
...
First, main field winding is short-circuited
...
Reduced voltage with the help of auto-transformers is applied across stator terminals
...

3
...
c
...
If excitation is sufficient, then the
machine will be pulled into synchronism
...
Full supply voltage is applied across stator terminals by cutting out the auto-transformers
...
The motor may be operated at any desired power factor by changing the d
...
excitation
...
23
...
For a given frequency, the synchronous motor runs at a constant average speed whatever the
load, while the speed of an induction motor falls somewhat with increase in load
...
The synchronous motor can be operated over a wide range of power factors, both lagging
and leading, but induction motor always runs with a lagging p
...
which may become very
low at light loads
...
A synchronous motor is inherently not self-starting
...
The changes in applied voltage do not affect synchronous motor torque as much as they
affect the induction motor torque
...

5
...
c
...

6
...
p
...
) because their power
factor can always be adjusted to 1
...
However, induction motors
are excellent for speeds above 600 r
...
m
...
Synchronous motors can be run at ultra-low speeds by using high power electronic converters
which generate very low frequencies
...


Synchronous Motor

1531

38
...
Synchronous Motor Applications
Synchronous motors find extensive application for the following classes of service :
1
...
Constant-speed, constant-load drives
3
...
38
...
f
...

(b) Constant-speed applications
Because of their high efficiency and high-speed, synchronous motors (above 600 r
...
m
...

Low-speed synchronous motors (below 600 r
...
m
...

(c) Voltage regulation
The voltage at the end of a long transmission line varies greatly especially when large inductive
loads are present
...
By installing a synchronous motor with a
field regulator (for varying its excitation), this voltage rise can be controlled
...
f
...
If, on the other hand, line voltage rises due to line
capacitive effect, motor excitation is decreased, thereby making its p
...
lagging which helps to maintain
the line voltage at its normal value
...
1
...

Q
...

Ans
...
3
...

Q
...

Ans
...
5
...

Q
...

Ans
...

The power factor ?
Yes
...
However, when
underexcited, it has lagging power factor
...

Which has more efficiency; synchronous or induction motor ?
Synchronous motor
...
constant speed load service
2
...
power factor correction
4
...


1532

Electrical Technology

Q
...
What is a synchronous capacitor ?
Ans
...

Q
...
What are the causes of faulty starting of a synchronous motor ?
Ans
...
voltage may be too low – at least half voltage is required for starting
2
...
static friction may be large – either due to high belt tension or too tight bearings
4
...
field excitation may be too strong
...
9
...
It is usually due to the following reasons :
1
...
some faulty connection in auxiliary apparatus
3
...
open-circuit in one phase or short-circuit
5
...

Q
...
A synchronous motor starts as usual but fails to develop its full torque
...
1
...
field spool may be reversed 3
...

Q
...
Will the motor start with the field excited ?
Ans
...

Q
...
Under which conditions a synchronous motor will fail to pull into step ?
Ans
...
no field excitation
2
...
excessive load inertia

OBJECTIVE TESTS – 38
1
...
In a synchronous motor, the magnitude of stator
back e
...
f
...
c
...
An electric motor in which both the rotor and
stator fields rotates with the same speed is called
a/an
...

(a) d
...

(b) chrage

(c) synchronous
(d) universal
4
...
m
...
in rotor field winding by
stator flux
(d) compulsion due to Lenz’s law
5
...
When running under no-load condition and with
normal excitation, armature current Ia drawn
by a synchronous motor
(a) leads the back e
...
f
...

7
...
angle
...
If load angle of a 4-pole synchronous motor is
8º (elect), its value in mechanical degrees is

...
5
(d) 0
...
The maximum value of torque angle a in a synchronous motor is
...

(a) 45
(b) 90
(c) between 45 and 90
(d) below 60
10
...
m
...

(d) armature current
...
When load on a synchronous motor running
with normal excitation is increased, armature
current drawn by it increases because
(a) back e
...
f
...
When load on a normally-excited synchronous
motor is increased, its power factor tends to

1533

(a) approach unity
(b) become increasingly lagging
(c) become increasingly leading
(d) remain unchanged
...
The effect of increasing load on a synchronous
motor running with normal excitation is to
(a) increase both its Ia and p
...

(b) decrease Ia but increase p
...

(c) increase Ia but decrease p
...

(d) decrease both Ia and p
...

14
...

15
...
Ignoring the effects of armature reaction, if
excitation of a synchronous motor running with
constant load is decreased from its normal
value, it leads to
(a) increase in but decrease in Eb
(b) increase in Eb but decrease in Ia
(c) increase in both I a and p
...
which is
lagging
(d) increase in both Ia and φ
17
...
f
...
f
...
f
...
f
...
f
...
-II, Delhi Univ
...
The V -curves of a synchronous motor show
relationship between
(a) excitation current and back e
...
f
...
f
...
c
...
c
...


1534

Electrical Technology

19
...
If main field current of a salient-pole
synchronous motor fed from an infinite bus and
running at no-load is reduced to zero, it would
(a) come to a stop
(b) continue running at synchronous speed
(c) run at sub-synchronous speed
(d) run at super-synchronous speed
21
...
c
...
-II, Delhi Univ
...
1987)

22
...
c
...
A synchronous machine is called a doublyexcited machine because
(a) it can be overexcited
(b) it has two sets of rotor poles
(c) both its rotor and stator are excited
(d) it needs twice the normal exciting current
...
Synchronous capacitor is
(a) an ordinary static capacitor bank
(b) an over-excited synchronous motor driving mechanical load
(c) an over-excited synchronous motor running without mechanical load
(d) none of the above 623
(Elect
...
M
...
E
...
B, 1993)

ANSWERS
1
...
b
23
...
d
13
...
c

3
...
a

4
...
a

5
...
d

6
...
b

7
...
c

8
...
d

9
...
b

10
...
d

11
...
d

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