Download 1 Introduction to 3 phase induction motors

Introduction to 3 phase induction motors
Definition:
Induction motor is an a.c. motor in which currents in the
stator winding (which is connected to the supply ) set up a flux which
causes currents to be induced in the rotor winding ; these currents
interact with the flux to produce rotation . Also called asynchronous
motor. The rotor receives electrical power in exactly the same way as
the secondary of a two winding transformer receiving its power from
primary. That is why an induction motor can be called as a rotating
transformer i.e., in which primary winding is stationary but the
secondary is free to rotate.
Types of induction motors:
Depending on rotor there are two types of induction motors :1- Squirrel cage induction motor :
Rotors is very simple and consist of bars of aluminum (or copper)
with shorting rings at the ends.
2- wound rotor induction motor :
Three phase windings (star connected) with terminals brought out to
slip rings for external connections.
The first type is more common used compared to second one due to:
1- Robust : No brushes . No contacts on rotor shaft.
2- Easy to manufacture.
3- Almost maintenance-free , except for bearing and other
mechanical parts.
4- Since the rotor has very low resistance, the copper loss is low and
efficiency is high .
Construction:
There are two main types of components which are used in induction
motor manufacturing as follows:
1- Active components : which are classified into two categories:a- Magnetic materials (0.5 mm electrical steel)
b- Electrical materials ( copper wires ,insulations ,bars , end
rings ,slip rings ,brushes , and lead wires)
1
2- Constructional components: like frame , end shields , shaft ,
bearings , and fan . These components are shown in Figures 1 & 2 .
Figure- 1 parts of squirrel cage induction motor
Figure- 2 Axial view of squirrel cage induction motor
2
Stator construction :
The stator is made up of several thin laminations (0.5 mm )of
electrical steel (silicon steel) , they are punched and clamped
together to form a hollow cylinder ( stator core ) with slots , as
shown in Figure- 3. Coils of insulated wires are inserted into
these slots . Each grouping of coils , together with the core it
surrounds , forms an electromagnet (a pair of poles).The number
of poles of an induction motor depends on the internal connection
of the stator windings .
Rotor construction:
The squirrel cage rotor is made up of several thin electrical steel
lamination (0.5mm) with evenly spaced bars , which are made up
of aluminum or copper , along the periphery .In the most popular
type of rotor (squirrel cage rotor) , these bars are connected at ends
mechanically and electrically by the use of end rings as shown in
Fig.4. Almost 90 % of induction motors have squirrel cage rotors .
This is because the squirrel cage rotors has a simple and rugged
construction .The rotor slots are not exactly parallel to the shaft.
Instead , they are given a skew for two main reasons :The first
reason is to make the motor run quietly by reducing magnetic hum
and to decrease slot harmonics. The second reason is to help reduce
the locking tendency of the rotor (the rotor teeth tend to remain
locked under the stator teeth due to direct magnetic attraction
between the two). The rotor is mounted on the shaft using bearings
on each end ; one end of the shaft is normally kept longer than the
other for driving the load . Between the stator and the rotor, there
exist an air gap , through which due to induction , the energy is
transferred from the stator to the rotor.
3
The wound rotor has a set of windings on the rotor slots
which are not short circuited , but are terminated to a set of
slip rings . These are helpful in adding external resistors and
contactors , as shown in Figure 5.
4
Typical name plate of induction motor
A typical name plate of induction motor is shown in Figure 6, and
Table 1 .
Motor standards
NEMA : National Electrical Manufactures Association
IEC : International Electrotechnical Commission
5
Motor insulation class
Insulations have been standardized and graded by their resistance to
thermal aging and failure. Four insulation classes are in common use ,
they have been designated by the letters A , B , F, and H .
The temperature capabilities of these classes are separated from each other
by 25 °C increments. The temperature capabilities of each insulation class
is defined as being the maximum temperature at which the insulation can
be operated to yield an average life of 20,000 hours , as in Table below.
Insulation Class
Temperature Rating
A
105° C
B
130° C
F
155° C
H
180° C
Motor degree of protection
I P : International Protection , I P * #
*
0
1
2
3
4
5
6
Protection against ingress
of bodies
Non protected
#
Protected against ingress of
foreign solid bodies of 50
mm or greater.
Protected against ingress of
foreign solid bodies of 12
mm or greater.
1
Protected against ingress of
dripping water .
2
Protected against ingress of
foreign solid bodies of 2.5
mm or greater.
Protected against ingress of
foreign solid bodies of 1 mm
or greater.
Partially protected against
ingress of dust .
Totally protected against
ingress of dust.
3
Protection against ingress of
dripping water at maximum
angle of 150 degrees from
the vertical.
Protection against water
falling like rain.
0
Protection against ingress
of water
Non protected
4
Protection against splashing
water.
5
Protection against water jets
.
Protection against special
conditions on ship's board.
6
7
8
6
Protection against
immersion in water .
Protection against
prolonged immersion in
water .
Chapter (8)
Three Phase Induction Motors
Introduction
The three-phase induction motors are the most widely used electric motors in
industry. They run at essentially constant speed from no-load to full-load.
However, the speed is frequency dependent and consequently these motors are
not easily adapted to speed control. We usually prefer d.c. motors when large
speed variations are required. Nevertheless, the 3-phase induction motors are
simple, rugged, low-priced, easy to maintain and can be manufactured with
characteristics to suit most industrial requirements. In this chapter, we shall
focus our attention on the general principles of 3-phase induction motors.
8.1 Three-Phase Induction Motor
Like any electric motor, a 3-phase induction motor has a stator and a rotor. The
stator carries a 3-phase winding (called stator winding) while the rotor carries a
short-circuited winding (called rotor winding). Only the stator winding is fed
from 3-phase supply. The rotor winding derives its voltage and power from the
externally energized stator winding through electromagnetic induction and
hence the name. The induction motor may be considered to be a transformer
with a rotating secondary and it can, therefore, be described as a “transformertype” a.c. machine in which electrical energy is converted into mechanical
energy.
Advantages
(i)
(ii)
(iii)
(iv)
(v)
It has simple and rugged construction.
It is relatively cheap.
It requires little maintenance.
It has high efficiency and reasonably good power factor.
It has self starting torque.
Disadvantages
(i)
It is essentially a constant speed motor and its speed cannot be changed
easily.
(ii) Its starting torque is inferior to d.c. shunt motor.
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8.2 Construction
A 3-phase induction motor has two main parts (i) stator and (ii) rotor. The rotor
is separated from the stator by a small air-gap which ranges from 0.4 mm to 4
mm, depending on the power of the motor.
1.
Stator
It consists of a steel frame which encloses a
hollow, cylindrical core made up of thin
laminations of silicon steel to reduce
hysteresis and eddy current losses. A number
of evenly spaced slots are provided on the
inner periphery of the laminations [See Fig.
(8.1)]. The insulated connected to form a
Fig.(8.1)
balanced 3-phase star or delta connected
circuit. The 3-phase stator winding is wound for a definite number of poles as
per requirement of speed. Greater the number of poles, lesser is the speed of the
motor and vice-versa. When 3-phase supply is given to the stator winding, a
rotating magnetic field (See Sec. 8.3) of constant magnitude is produced. This
rotating field induces currents in the rotor by electromagnetic induction.
2.
Rotor
The rotor, mounted on a shaft, is a hollow laminated core having slots on its
outer periphery. The winding placed in these slots (called rotor winding) may be
one of the following two types:
(i) Squirrel cage type
(ii) Wound type
(i)
Squirrel cage rotor. It consists of a laminated cylindrical core having
parallel slots on its outer periphery. One copper or aluminum bar is placed
in each slot. All these bars are joined at each end by metal rings called end
rings [See Fig. (8.2)]. This forms a permanently short-circuited winding
which is indestructible. The entire construction (bars and end rings)
resembles a squirrel cage and hence the name. The rotor is not connected
electrically to the supply but has current induced in it by transformer action
from the stator.
Those induction motors which employ squirrel cage rotor are called
squirrel cage induction motors. Most of 3-phase induction motors use
squirrel cage rotor as it has a remarkably simple and robust construction
enabling it to operate in the most adverse circumstances. However, it
suffers from the disadvantage of a low starting torque. It is because the
rotor bars are permanently short-circuited and it is not possible to add any
external resistance to the rotor circuit to have a large starting torque.
182
Fig.(8.2)
Fig.(8.3)
(ii) Wound rotor. It consists of a laminated cylindrical core and carries a 3phase winding, similar to the one on the stator [See Fig. (8.3)]. The rotor
winding is uniformly distributed in the slots and is usually star-connected.
The open ends of the rotor winding are brought out and joined to three
insulated slip rings mounted on the rotor shaft with one brush resting on
each slip ring. The three brushes are connected to a 3-phase star-connected
rheostat as shown in Fig. (8.4). At starting, the external resistances are
included in the rotor circuit to give a large starting torque. These
resistances are gradually reduced to zero as the motor runs up to speed.
Fig.(8.4)
The external resistances are used during starting period only. When the motor
attains normal speed, the three brushes are short-circuited so that the wound
rotor runs like a squirrel cage rotor.
8.3 Rotating Magnetic Field Due to 3-Phase Currents
When a 3-phase winding is energized from a 3-phase supply, a rotating
magnetic field is produced. This field is such that its poles do no remain in a
fixed position on the stator but go on shifting their positions around the stator.
For this reason, it is called a rotating Held. It can be shown that magnitude of
this rotating field is constant and is equal to 1.5 φm where φm is the maximum
flux due to any phase.
183
To see how rotating field is produced, consider a 2-pole,
3i-phase winding as shown in Fig. (8.6 (i)). The three
phases X, Y and Z are energized from a 3-phase source
and currents in these phases are indicated as Ix, Iy and Iz
[See Fig. (8.6 (ii))]. Referring to Fig. (8.6 (ii)), the fluxes
produced by these currents are given by:
φ x = φ m sin ωt
Fig.(8.5)
φ y = φ m sin (ωt − 120°)
φ z = φ m sin (ωt − 240°)
Here φm is the maximum flux due to any phase. Fig. (8.5) shows the phasor
diagram of the three fluxes. We shall now prove that this 3-phase supply
produces a rotating field of constant magnitude equal to 1.5 φm.
Fig.(8.6)
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(i)
At instant 1 [See Fig. (8.6 (ii)) and Fig. (8.6 (iii))], the current in phase X is
zero and currents in phases Y and Z are equal
and opposite. The currents are flowing outward
in the top conductors and inward in the bottom
conductors. This establishes a resultant flux
towards right. The magnitude of the resultant
flux is constant and is equal to 1.5 φm as
proved under:
At instant 1, ωt = 0°. Therefore, the three
fluxes are given by;
φ x = 0;
φ y = φ m sin (− 120°) = −
φz = φ m sin (− 240°) =
Fig.(8.7)
3
φ ;
2 m
3
φ
2 m
The phasor sum of − φy and φz is the resultant flux φr [See Fig. (8.7)]. It is
clear that:
Resultant flux, φr = 2 ×
3
60°
3
3
φ m cos
= 2×
φm ×
= 1.5 φ m
2
2
2
2
(ii) At instant 2, the current is maximum
(negative) in φy phase Y and 0.5 maximum
(positive) in phases X and Y. The
magnitude of resultant flux is 1.5 φm as
proved under:
At instant 2, ωt = 30°. Therefore, the three
fluxes are given by;
φm
2
φ y = φ m sin (−90°) = −φ m
Fig.(8.8)
φ x = φ m sin 30° =
φ z = φ m sin (−210°) =
φm
2
The phasor sum of φx, − φy and φz is the resultant flux φr
φ
120° φ m
Phasor sum of φx and φz, φ'r = 2 × m cos
=
2
2
2
φ
Phasor sum of φ'r and − φy, φ r = m + φ m = 1.5 φ m
2
Note that resultant flux is displaced 30° clockwise from position 1.
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(iii) At instant 3, current in phase Z is zero and
the currents in phases X and Y are equal and
opposite (currents in phases X and Y arc
0.866 × max. value). The magnitude of
resultant flux is 1.5 φm as proved under:
At instant 3, ωt = 60°. Therefore, the three
fluxes are given by;
3
φ x = φ m sin 60° =
φ ;
2 m
3
φ y = φ m sin (− 60°) = −
φ ;
2 m
φz = φ m sin (− 180°) = 0
Fig.(8.9)
The resultant flux φr is the phasor sum of φx and − φy (Q φ z = 0 ) .
φr = 2 ×
3
60°
φ m cos
= 1.5 φ m
2
2
Note that resultant flux is displaced 60° clockwise from position 1.
(iv) At instant 4, the current in phase X is
maximum (positive) and the currents in phases
V and Z are equal and negative (currents in
phases V and Z are 0.5 × max. value). This
establishes a resultant flux downward as
shown under:
At instant 4, ωt = 90°. Therefore, the three
fluxes are given by;
Fig.(7.10)
φ x = φ m sin 90° = φ m
φm
2
φ
φz = φ m sin (−150°) = − m
2
φ y = φ m sin (−30°) = −
The phasor sum of φx, − φy and − φz is the resultant flux φr
φ
120° φ m
Phasor sum of − φz and − φy, φ'r = 2 × m cos
=
2
2
2
φ
Phasor sum of φ'r and φx, φ r = m + φ m = 1.5 φ m
2
Note that the resultant flux is downward i.e., it is displaced 90° clockwise
from position 1.
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It follows from the above discussion that a 3-phase supply produces a rotating
field of constant value (= 1.5 φm, where φm is the maximum flux due to any
phase).
Speed of rotating magnetic field
The speed at which the rotating magnetic field revolves is called the
synchronous speed (Ns). Referring to Fig. (8.6 (ii)), the time instant 4 represents
the completion of one-quarter cycle of alternating current Ix from the time
instant 1. During this one quarter cycle, the field has rotated through 90°. At a
time instant represented by 13 or one complete cycle of current Ix from the
origin, the field has completed one revolution. Therefore, for a 2-pole stator
winding, the field makes one revolution in one cycle of current. In a 4-pole
stator winding, it can be shown that the rotating field makes one revolution in
two cycles of current. In general, fur P poles, the rotating field makes one
revolution in P/2 cycles of current.
∴
or
Cycles of current =
Cycles of current per second =
P
× revolutions of field
2
P
× revolutions of field per second
2
Since revolutions per second is equal to the revolutions per minute (Ns) divided
by 60 and the number of cycles per second is the frequency f,
∴
or
f=
P Ns Ns P
×
=
2 60 120
Ns =
120 f
P
The speed of the rotating magnetic field is the same as the speed of the
alternator that is supplying power to the motor if the two have the same number
of poles. Hence the magnetic flux is said to rotate at synchronous speed.
Direction of rotating magnetic field
The phase sequence of the three-phase voltage applied to the stator winding in
Fig. (8.6 (ii)) is X-Y-Z. If this sequence is changed to X-Z-Y, it is observed that
direction of rotation of the field is reversed i.e., the field rotates
counterclockwise rather than clockwise. However, the number of poles and the
speed at which the magnetic field rotates remain unchanged. Thus it is necessary
only to change the phase sequence in order to change the direction of rotation of
the magnetic field. For a three-phase supply, this can be done by interchanging
any two of the three lines. As we shall see, the rotor in a 3-phase induction
motor runs in the same direction as the rotating magnetic field. Therefore, the
187
direction of rotation of a 3-phase induction motor can be reversed by
interchanging any two of the three motor supply lines.
8.4 Alternate Mathematical Analysis for Rotating
Magnetic Field
We shall now use another useful method to find
the magnitude and speed of the resultant flux due
to three-phase currents. The three-phase sinusoidal
currents produce fluxes φ1, φ2 and φ3 which vary
sinusoidally. The resultant flux at any instant will
be the vector sum of all the three at that instant.
The fluxes are represented by three variable
Fig.(8.11)
magnitude vectors [See Fig. (8.11)]. In Fig. (8.11),
the individual flux directions arc fixed but their magnitudes vary sinusoidally as
does the current that produces them. To find the magnitude of the resultant flux,
resolve each flux into horizontal and vertical components and then find their
vector sum.
φh = φ m cos t ωt − φ m cos(ωt − 120°) cos 60° − φ m cos(ωt − 240°) cos 60°
3
= φm cos ωt
2
3
φ v = 0 − φm cos(ωt − 120°) cos 60° + φ m cos(ωt − 240°) cos 60° = φ m sin ωt
2
The resultant flux is given by;
[
]
1/ 2
3
3
φr = φ 2h + φ 2v = φ m cos 2 ωt + (− sin ωt ) 2
= φ m = 1.5 φ m = Constant
2
2
Thus the resultant flux has constant magnitude (= 1.5 φm) and does not change
with time. The angular displacement of φr relative to
the OX axis is
∴
3
φ v 2 φ m sin ωt
= tan ωt
=
tan θ =
φh 3
φ cos ωt
2 m
θ = ωt
Fig.(8.12)
Thus the resultant magnetic field rotates at constant angular velocity ω( = 2 πf)
rad/sec. For a P-pole machine, the rotation speed (ωm) is
ωm =
2
ω rad / sec
P
188
2 π Ns 2
= × 2πf
60
P
or
∴
Ns =
... N s is in r.p.m.
120 f
P
Thus the resultant flux due to three-phase currents is of constant value (= 1.5 φm
where φm is the maximum flux in any phase) and this flux rotates around the
stator winding at a synchronous speed of 120 f/P r.p.m.
For example, for a 6-pole, 50 Hz, 3-phase induction motor, N, = 120 × 50/6 =
1000 r.p.m. It means that flux rotates around the stator at a speed of 1000 r.p.m.
8.5 Principle of Operation
Consider a portion of 3-phase induction motor as shown in Fig. (8.13). The
operation of the motor can be explained as under:
(i) When
3-phase
stator
winding is energized from
a 3-phase supply, a
rotating magnetic field is
set up which rotates round
the stator at synchronous
speed Ns (= 120 f/P).
(ii) The rotating field passes
through the air gap and
cuts the rotor conductors,
Fig.(1-)
which
as
yet,
are
stationary. Due to the relative speed between the rotating flux and the
stationary rotor, e.m.f.s are induced in the rotor conductors. Since the
rotor circuit is short-circuited, currents start flowing in the rotor
conductors.
(iii) The current-carrying rotor conductors are placed in the magnetic field
produced by the stator. Consequently, mechanical force acts on the rotor
conductors. The sum of the mechanical forces on all the rotor conductors
produces a torque which tends to move the rotor in the same direction as
the rotating field.
(iv) The fact that rotor is urged to follow the stator field (i.e., rotor moves in
the direction of stator field) can be explained by Lenz’s law. According
to this law, the direction of rotor currents will be such that they tend to
oppose the cause producing them. Now, the cause producing the rotor
currents is the relative speed between the rotating field and the stationary
rotor conductors. Hence to reduce this relative speed, the rotor starts
running in the same direction as that of stator field and tries to catch it.
189
8.6 Slip
We have seen above that rotor rapidly accelerates in the direction of rotating
field. In practice, the rotor can never reach the speed of stator flux. If it did,
there would be no relative speed between the stator field and rotor conductors,
no induced rotor currents and, therefore, no torque to drive the rotor. The
friction and windage would immediately cause the rotor to slow down. Hence,
the rotor speed (N) is always less than the suitor field speed (Ns). This difference
in speed depends upon load on the motor.
The difference between the synchronous speed Ns of the rotating stator field and
the actual rotor speed N is called slip. It is usually expressed as a percentage of
synchronous speed i.e.,
% age slip, s =
Ns − N
× 100
Ns
(i) The quantity Ns − N is sometimes called slip speed.
(ii) When the rotor is stationary (i.e., N = 0), slip, s = 1 or 100 %.
(iii) In an induction motor, the change in slip from no-load to full-load is
hardly 0.1% to 3% so that it is essentially a constant-speed motor.
8.7 Rotor Current Frequency
The frequency of a voltage or current induced due to the relative speed between
a vending and a magnetic field is given by the general formula;
Frequency =
NP
120
where N = Relative speed between magnetic field and the winding
P = Number of poles
For a rotor speed N, the relative speed between the rotating flux and the rotor is
Ns − N. Consequently, the rotor current frequency f' is given by;
( N s − N)P
120
s Ns P
=
120
f '=
N −N

Q s = s

Ns 

N P

Q f = s 
120 

= sf
i.e., Rotor current frequency = Fractional slip x Supply frequency
(i) When the rotor is at standstill or stationary (i.e., s = 1), the frequency of
rotor current is the same as that of supply frequency (f' = sf = 1× f = f).
190
(ii) As the rotor picks up speed, the relative speed between the rotating flux
and the rotor decreases. Consequently, the slip s and hence rotor current
frequency decreases.
Note. The relative speed between the rotating field and stator winding is Ns − 0
= Ns. Therefore, the frequency of induced current or voltage in the stator
winding is f = Ns P/120—the supply frequency.
8.8 Effect of Slip on The Rotor Circuit
When the rotor is stationary, s = 1. Under these conditions, the per phase rotor
e.m.f. E2 has a frequency equal to that of supply frequency f. At any slip s, the
relative speed between stator field and the rotor is decreased. Consequently, the
rotor e.m.f. and frequency are reduced proportionally to sEs and sf respectively.
At the same time, per phase rotor reactance X2, being frequency dependent, is
reduced to sX2.
Consider a 6-pole, 3-phase, 50 Hz induction motor. It has synchronous speed Ns
= 120 f/P = 120 × 50/6 = 1000 r.p.m. At standsill, the relative speed between
stator flux and rotor is 1000 r.p.m. and rotor e.m.f./phase = E2(say). If the fullload speed of the motor is 960 r.p.m., then,
s=
(i)
1000 − 960
= 0.04
1000
The relative speed between stator flux and the rotor is now only 40 r.p.m.
Consequently, rotor e.m.f./phase is reduced to:
E2 ×
40
= 0.04 E 2
1000
or
sE 2
(ii) The frequency is also reduced in the same ratio to:
50 ×
40
= 50 × 0.04
1000
or
sf
(iii) The per phase rotor reactance X2 is likewise reduced to:
X2 ×
40
= 0.04X 2
1000
or
sX 2
Thus at any slip s,
Rotor e.m.f./phase = sE2
Rotor reactance/phase = sX2
Rotor frequency = sf
where E2,X2 and f are the corresponding values at standstill.
191
8.9 Rotor Current
Fig. (8.14) shows the circuit of a 3-phase induction motor at any slip s. The rotor
is assumed to be of wound type and star connected. Note that rotor e.m.f./phase
and rotor reactance/phase are s E2 and sX2 respectively. The rotor
resistance/phase is R2 and is independent of frequency and, therefore, does not
depend upon slip. Likewise, stator winding values R1 and X1 do not depend
upon slip.
Fig.(8.14)
Since the motor represents a balanced 3-phase load, we need consider one phase
only; the conditions in the other two phases being similar.
At standstill. Fig. (8.15 (i)) shows one phase of the rotor circuit at standstill.
Rotor current/phase, I 2 =
Rotor p.f., cos φ 2 =
E2
E2
=
Z2
R 22 + X 22
R2
R2
=
Z2
R 22 + X 22
Fig.(8.15)
When running at slip s. Fig. (8.15 (ii)) shows one phase of the rotor circuit
when the motor is running at slip s.
Rotor current, I'2 =
sE 2
sE 2
=
Z' 2
R 22 + (sX 2 )2
192
Rotor p.f., cos φ'2 =
R2
R2
=
Z'2
R 22 + (sX 2 )2
8.10 Rotor Torque
The torque T developed by the rotor is directly proportional to:
(i) rotor current
(ii) rotor e.m.f.
(iii) power factor of the rotor circuit
∴
T ∝ E 2 I 2 cos φ 2
T = K E 2 I 2 cos φ 2
or
where
I2 = rotor current at standstill
E2 = rotor e.m.f. at standstill
cos φ2 = rotor p.f. at standstill
Note. The values of rotor e.m.f., rotor current and rotor power factor are taken
for the given conditions.
8.11 Starting Torque (Ts)
Let
E2 = rotor e.m.f. per phase at standstill
X2 = rotor reactance per phase at standstill
R2 = rotor resistance per phase
Rotor impedance/phase, Z 2 = R 22 + X 22
Rotor current/phase, I 2 =
Rotor p.f., cos φ 2 =
∴
...at standstill
E2
E2
=
Z2
R 22 + X 22
...at standstill
R2
R2
=
Z2
R 22 + X 22
...at standstill
Starting torque, Ts = K E 2 I 2 cos φ 2
= K E2 ×
=
K E 22 R 2
R 22 + X 22
193
E2
R 22 + X 22
×
R2
R 22 + X 22
Generally, the stator supply voltage V is constant so that flux per pole φ set up
by the stator is also fixed. This in turn means that e.m.f. E2 induced in the rotor
will be constant.
∴
Ts =
K1 R 2
R 22 + X 22
=
K1 R 2
Z 22
where K1 is another constant.
It is clear that the magnitude of starting torque would depend upon the relative
values of R2 and X2 i.e., rotor resistance/phase and standstill rotor
reactance/phase.
It can be shown that K = 3/2 π Ns.
∴
E 22 R 2
3
Ts =
⋅
2π N s R 22 + X 22
Note that here Ns is in r.p.s.
8.12 Condition for Maximum Starting Torque
It can be proved that starting torque will be maximum when rotor
resistance/phase is equal to standstill rotor reactance/phase.
Now
Ts =
K1 R 2
(i)
R 22 + X 22
Differentiating eq. (i) w.r.t. R2 and equating the result to zero, we get,

dTs
R 2 (2 R 2 ) 
1
=0
= K1  2
−
dR 2
 R 2 + X 22 R 2 + X 2 2 

2
2 
(
or
R 22 + X 22 = 2R 22
or
R 2 = X2
)
Hence starting torque will be maximum when:
Rotor resistance/phase = Standstill rotor reactance/phase
Under the condition of maximum starting torque, φ2 = 45° and rotor power
factor is 0.707 lagging [See Fig. (8.16 (ii))].
Fig. (8.16 (i)) shows the variation of starting torque with rotor resistance. As the
rotor resistance is increased from a relatively low value, the starting torque
increases until it becomes maximum when R2 = X2. If the rotor resistance is
increased beyond this optimum value, the starting torque will decrease.
194
Fig.(8.16)
8.13 Effect of Change of Supply Voltage
Ts =
K E 22 R 2
R 22 + X 22
Since E2 ∝ Supply voltage V
∴
Ts =
K2 V2 R 2
R 22 + X 22
where K2 is another constant.
∴
Ts ∝ V 2
Therefore, the starting torque is very sensitive to changes in the value of supply
voltage. For example, a drop of 10% in supply voltage will decrease the starting
torque by about 20%. This could mean the motor failing to start if it cannot
produce a torque greater than the load torque plus friction torque.
8.14 Starting Torque of 3-Phase Induction Motors
The rotor circuit of an induction motor has low resistance and high inductance.
At starting, the rotor frequency is equal to the stator frequency (i.e., 50 Hz) so
that rotor reactance is large compared with rotor resistance. Therefore, rotor
current lags the rotor e.m.f. by a large angle, the power factor is low and
consequently the starting torque is small. When resistance is added to the rotor
circuit, the rotor power factor is improved which results in improved starting
torque. This, of course, increases the rotor impedance and, therefore, decreases
the value of rotor current but the effect of improved power factor predominates
and the starting torque is increased.
(i)
Squirrel-cage motors. Since the rotor bars are permanently shortcircuited, it is not possible to add any external resistance in the rotor circuit
at starting. Consequently, the stalling torque of such motors is low. Squirrel
195
cage motors have starting torque of 1.5 to 2 times the full-load value with
starting current of 5 to 9 times the full-load current.
(ii) Wound rotor motors. The resistance of the rotor circuit of such motors
can be increased through the addition of external resistance. By inserting
the proper value of external resistance (so that R2 = X2), maximum starting
torque can be obtained. As the motor accelerates, the external resistance is
gradually cut out until the rotor circuit is short-circuited on itself for
running conditions.
8.15 Motor Under Load
Let us now discuss the behaviour of 3-phase induction motor on load.
(i) When we apply mechanical load to the shaft of the motor, it will begin to
slow down and the rotating flux will cut the rotor conductors at a higher
and higher rate. The induced voltage and resulting current in rotor
conductors will increase progressively, producing greater and greater
torque.
(ii) The motor and mechanical load will soon reach a state of equilibrium
when the motor torque is exactly equal to the load torque. When this
state is reached, the speed will cease to drop any more and the motor will
run at the new speed at a constant rate.
(iii) The drop in speed of the induction motor on increased load is small. It is
because the rotor impedance is low and a small decrease in speed
produces a large rotor current. The increased rotor current produces a
higher torque to meet the increased load on the motor. This is why
induction motors are considered to be constant-speed machines.
However, because they never actually turn at synchronous speed, they
are sometimes called asynchronous machines.
Note that change in load on the induction motor is met through the
adjustment of slip. When load on the motor increases, the slip increases
slightly (i.e., motor speed decreases slightly). This results in greater
relative speed between the rotating flux and rotor conductors.
Consequently, rotor current is increased, producing a higher torque to
meet the increased load. Reverse happens should the load on the motor
decrease.
(iv) With increasing load, the increased load currents I'2 are in such a
direction so as to decrease the stator flux (Lenz’s law), thereby
decreasing the counter e.m.f. in the stator windings. The decreased
counter e.m.f. allows motor stator current (I1) to increase, thereby
increasing the power input to the motor. It may be noted that action of
the induction motor in adjusting its stator or primary current with
196
changes of current in the rotor or secondary is very much similar to the
changes occurring in transformer with changes in load.
Fig.(8.17)
8.16 Torque Under Running Conditions
Let the rotor at standstill have per phase induced e.m.f. E2, reactance X2 and
resistance R2. Then under running conditions at slip s,
Rotor e.m.f./phase, E'2 = sE2
Rotor reactance/phase, X'2 = sX2
Rotor impedance/phase, Z'2 = R 22 + (sX 2 )2
Rotor current/phase, I'2 =
Rotor p.f., cos φ'm =
E '2
sE 2
=
Z' 2
R 22 + (sX 2 )2
R2
R 22 + (sX 2 )2
Fig.(8.18)
Running Torque, Tr ∝ E'2 I'2 cos φ'2
∝ φ I'2 cos φ'2
197
(Q E '2 ∝ φ)
∝ φ×
s E2
R 22 + (s X 2 )2
×
R2
R 22 + (s X 2 )2
φ s E2 R 2
∝
R 22 + (s X 2 )2
K φ s E2 R 2
=
R 22 + (s X 2 )2
K1 s E 22 R 2
=
(Q E 2 ∝ φ)
R 22 + (s X 2 )2
If the stator supply voltage V is constant, then stator flux and hence E2 will be
constant.
∴
Tr =
K2 s R2
R 22 + (s X 2 )2
where K2 is another constant.
It may be seen that running torque is:
(i) directly proportional to slip i.e., if slip increases (i.e., motor speed
decreases), the torque will increase and vice-versa.
(ii) directly proportional to square of supply voltage (Q E 2 ∝ V ) .
It can be shown that value of K1 = 3/2 π Ns where Ns is in r.p.s.
∴
s E 22 R 2
s E 22 R 2
3
3
Tr =
⋅
=
⋅
2π N s R 2 + (s X )2 2π N s (Z' )2
2
2
2
At starting, s = 1 so that starting torque is
E 22 R 2
3
Ts =
⋅
2π N s R 22 + X 22
8.17 Maximum Torque under Running Conditions
Tr =
K2 s R 2
(i)
R 22 + s 2 X 22
In order to find the value of rotor resistance that gives maximum torque under
running conditions, differentiate exp. (i) w.r.t. s and equate the result to zero i.e.,
[ (
)
]
dTr K 2 R 2 R 22 + s 2 X 22 − 2s X 22 (s R 2 )
=
=0
ds
2
2
2 2
R2 + s X2
(
198
)
or
(R
or
R 22 = s 2 X 22
or
R 2 = s X2
2
2
2+s
)
X 22 − 2s X 22 = 0
Thus for maximum torque (Tm) under running conditions :
Rotor resistance/phase = Fractional slip × Standstill rotor reactance/phase
Now
Tr ∝
s R2
… from exp. (i) above
R 22 + s 2 X 22
For maximum torque, R2 = s X2. Putting R2 = s X2 in the above expression, the
maximum torque Tm is given by;
Tm ∝
1
2 X2
Slip corresponding to maximum torque, s = R2/X2.
It can be shown that:
E 22
3
Tm =
⋅
N-m
2π N s 2 X 2
It is evident from the above equations that:
(i) The value of rotor resistance does not alter the value of the maximum
torque but only the value of the slip at which it occurs.
(ii) The maximum torque varies inversely as the standstill reactance.
Therefore, it should be kept as small as possible.
(iii) The maximum torque varies directly with the square of the applied
voltage.
(iv) To obtain maximum torque at starting (s = 1), the rotor resistance must
be made equal to rotor reactance at standstill.
8.18 Torque-Slip Characteristics
As shown in Sec. 8.16, the motor torque under running conditions is given by;
T=
K2 s R 2
R 22 + s 2 X 22
If a curve is drawn between the torque and slip for a particular value of rotor
resistance R2, the graph thus obtained is called torque-slip characteristic. Fig.
(8.19) shows a family of torque-slip characteristics for a slip-range from s = 0 to
s = 1 for various values of rotor resistance.
199
Fig.(8.19)
The following points may be noted carefully:
(i) At s = 0, T = 0 so that torque-slip curve starts from the origin.
(ii) At normal speed, slip is small so that s X2 is negligible as compared to
R2.
∴
T ∝ s/R2
∝s
... as R2 is constant
Hence torque slip curve is a straight line from zero slip to a slip that
corresponds to full-load.
(iii) As slip increases beyond full-load slip, the torque increases and becomes
maximum at s = R2/X2. This maximum torque in an induction motor is
called pull-out torque or break-down torque. Its value is at least twice the
full-load value when the motor is operated at rated voltage and
frequency.
(iv)
to maximum torque, the term
2 2
2
s X 2 increases very rapidly so that R 2 may be neglected as compared
to s 2 X 22 .
∴
T ∝ s / s 2 X 22
∝ 1/ s
... as X2 is constant
Thus the torque is now inversely proportional to slip. Hence torque-slip
curve is a rectangular hyperbola.
(v) The maximum torque remains the same and is independent of the value
of rotor resistance. Therefore, the addition of resistance to the rotor
circuit does not change the value of maximum torque but it only changes
the value of slip at which maximum torque occurs.
8.19 Full-Load, Starting and Maximum Torques
200
Tf ∝
Ts ∝
Tm ∝
s R2
R 22
+ (s X 2 )2
R2
R 22 + X 22
1
2 X2
Note that s corresponds to full-load slip.
(i)
∴
Tm R 22 + (s X 2 )2
=
Tf
2s R 2 X 2
Dividing the numerator and denominator on R.H.S. by X 22 , we get,
Tm (R 2 / X 2 )2 + s 2 a 2 + s 2
=
=
Tf
2s(R 2 / X 2 )
2a s
R2
Rotor resistance/phase
=
X 2 Standstill rotor reactance/phase
where
a=
(ii)
Tm R 22 + X 22
=
Ts 2 R 2 X 2
Dividing the numerator and denominator on R.H.S. by X 22 , we get,
Tm (R 2 / X 2 )2 + 1 a 2 + 1
=
=
Tf
2(R 2 / X 2 )
2a
where
a=
R2
Rotor resistance/phase
=
X 2 Standstill rotor reactance/phase
8.20 Induction Motor and Transformer Compared
An induction motor may be considered to be a transformer with a rotating shortcircuited secondary. The stator winding corresponds to transformer primary and
rotor winding to transformer secondary. However, the following differences
between the two are worth noting:
(i) Unlike a transformer, the magnetic circuit of a 3-phase induction motor has
an air gap. Therefore, the magnetizing current in a 3-phase induction motor
is much larger than that of the transformer. For example, in an induction
motor, it may be as high as 30-50 % of rated current whereas it is only 15% of rated current in a transformer.
(ii) In an induction motor, there is an air gap and the stator and rotor windings
are distributed along the periphery of the air gap rather than concentrated
201
on a core as in a transformer. Therefore, the leakage reactances of stator
and rotor windings are quite large compared to that of a transformer.
(iii) In an induction motor, the inputs to the stator and rotor are electrical but the
output from the rotor is mechanical. However, in a transformer, input as
well as output is electrical.
(iv) The main difference between the induction motor and transformer lies in
the fact that the rotor voltage and its frequency are both proportional to slip
s. If f is the stator frequency, E2 is the per phase rotor e.m.f. at standstill
and X2 is the standstill rotor reactance/phase, then at any slip s, these values
are:
Rotor e.m.f./phase, E'2 = s E2
Rotor reactance/phase, X'2 = sX2
Rotor frequency, f' = sf
8.21 Speed Regulation of Induction Motors
Like any other electrical motor, the speed regulation of an induction motor is
given by:
% age speed regulation =
where
N 0 − N F.L.
× 100
N F.L.
N0 = no-load speed of the motor
NF.L. = full-load speed of the motor
If the no-load speed of the motor is 800 r.p.m. and its fall-load speed in 780
r.p.m., then change in speed is 800 − 780 = 20 r.p.m. and percentage speed
regulation = 20 × 100/780 = 2.56%.
At no load, only a small torque is required to overcome the small mechanical
losses and hence motor slip is small i.e., about 1%. When the motor is fully
loaded, the slip increases slightly i.e., motor speed decreases slightly. It is
because rotor impedance is low and a small decrease in speed produces a large
rotor current. The increased rotor current produces a high torque to meet the full
load on the motor. For this reason, the change in speed of the motor from noload to full-load is small i.e., the speed regulation of an induction motor is low.
The speed regulation of an induction motor is 3% to 5%. Although the motor
speed does decrease slightly with increased load, the speed regulation is low
enough that the induction motor is classed as a constant-speed motor.
8.22 Speed Control of 3-Phase Induction Motors
202
N = (1 − s) N s = (1 − s)
120 f
P
(i)
An inspection of eq. (i) reveals that the speed N of an induction motor can be
varied by changing (i) supply frequency f (ii) number of poles P on the stator
and (iii) slip s. The change of frequency is generally not possible because the
commercial supplies have constant frequency. Therefore, the practical methods
of speed control are either to change the number of stator poles or the motor slip.
1.
Squirrel cage motors
The speed of a squirrel cage motor is changed by changing the number of stator
poles. Only two or four speeds are possible by this method. Two-speed motor
has one stator winding that may be switched through suitable control equipment
to provide two speeds, one of which is half of the other. For instance, the
winding may be connected for either 4 or 8 poles, giving synchronous speeds of
1500 and 750 r.p.m. Four-speed motors are equipped with two separate stator
windings each of which provides two speeds. The disadvantages of this method
are:
(i) It is not possible to obtain gradual continuous speed control.
(ii) Because of the complications in the design and switching of the
interconnections of the stator winding, this method can provide a
maximum of four different synchronous speeds for any one motor.
2.
Wound rotor motors
The speed of wound rotor motors is changed by changing the motor slip. This
can be achieved by;
(i) varying the stator line voltage
(ii) varying the resistance of the rotor circuit
(iii) inserting and varying a foreign voltage in the rotor circuit
8.23 Power Factor of Induction Motor
Like any other a.c. machine, the power factor of an induction motor is given by;
Power factor, cos φ =
Active component of current (I cos φ)
Total current (I)
The presence of air-gap between the stator and rotor of an induction motor
greatly increases the reluctance of the magnetic circuit. Consequently, an
induction motor draws a large magnetizing current (Im) to produce the required
flux in the air-gap.
(i) At no load, an induction motor draws a large magnetizing current and a
small active component to meet the no-load losses. Therefore, the
induction motor takes a high no-load current lagging the applied voltage
203
by a large angle. Hence the power factor of an induction motor on no
load is low i.e., about 0.1 lagging.
(ii) When an induction motor is loaded, the active component of current
increases while the magnetizing component remains about the same.
Consequently, the power factor of the motor is increased. However,
because of the large value of magnetizing current, which is present
regardless of load, the power factor of an induction motor even at fullload seldom exceeds 0.9 lagging.
8.24 Power Stages in an Induction Motor
The input electric power fed to the stator of the motor is converted into
mechanical power at the shaft of the motor. The various losses during the energy
conversion are:
1.
Fixed losses
(i) Stator iron loss
(ii) Friction and windage loss
The rotor iron loss is negligible because the frequency of rotor currents under
normal running condition is small.
2.
Variable losses
(i) Stator copper loss
(ii) Rotor copper loss
Fig. (8.20) shows how electric power fed to the stator of an induction motor
suffers losses and finally converted into mechanical power.
The following points may be noted from the above diagram:
(i) Stator input, Pi = Stator output + Stator losses
= Stator output + Stator Iron loss + Stator Cu loss
(ii) Rotor input, Pr = Stator output
It is because stator output is entirely transferred to the rotor through airgap by electromagnetic induction.
(iii) Mechanical power available, Pm = Pr − Rotor Cu loss
This mechanical power available is the gross rotor output and will
produce a gross torque Tg.
(iv) Mechanical power at shaft, Pout = Pm − Friction and windage loss
Mechanical power available at the shaft produces a shaft torque Tsh.
Clearly, Pm − Pout = Friction and windage loss
204
Fig.(8.20)
8.25 Induction Motor Torque
The mechanical power P available from any electric motor can be expressed as:
P=
where
2π N T
60
watts
N = speed of the motor in r.p.m.
T = torque developed in N-m
∴
T=
60 P
P
= 9.55
N-m
2π N
N
If the gross output of the rotor of an induction motor is Pm and its speed is N
r.p.m., then gross torque T developed is given by:
Tg = 9.55
Similarly,
Pm
N-m
N
Tsh = 9.55
Pout
N-m
N
Note. Since windage and friction loss is small, Tg = Tsh,. This assumption hardly
leads to any significant error.
8.26 Rotor Output
If Tg newton-metre is the gross torque developed and N r.p.m. is the speed of the
rotor, then,
Gross rotor output =
2π N Tg
60
watts
If there were no copper losses in the rotor, the output would equal rotor input
and the rotor would run at synchronous speed Ns.
205
2π N s Tg
60
∴
Rotor input =
∴
Rotor Cu loss = Rotor input − Rotor output
=
(i)
∴
watts
2π Tg
( N s − N)
60
Rotor Cu loss N s − N
=s
=
Rotor input
Ns
Rotor Cu loss = s × Rotor input
(ii) Gross rotor output, Pm = Rotor input − Rotor Cu loss
= Rotor input − s × Rotor input
∴
Pm = Rotor input (1 − s)
Gross rotor output
N
=1 − s =
Rotor input
Ns
Rotor Cu loss
s
=
Gross rotor output 1 − s
(iii)
(iv)
It is clear that if the input power to rotor is Pr then s Pr is lost as rotor Cu loss
and the remaining (1 − s)Pr is converted into mechanical power. Consequently,
induction motor operating at high slip has poor efficiency.
Note.
Gross rotor output
=1 − s
Rotor input
If the stator losses as well as friction and windage losses arc neglected, then,
Gross rotor output = Useful output
Rotor input = Stator input
∴
Useful output
= 1 − s = Efficiency
Stator output
Hence the approximate efficiency of an induction motor is 1 − s. Thus if the slip
of an induction motor is 0.125, then its approximate efficiency is = 1 − 0.125 =
0.875 or 87.5%.
8.27 Induction Motor Torque Equation
The gross torque Tg developed by an induction motor is given by;
206
Tg =
=
Rotor input
2π N s
... N s is r.p.s.
60 × Rotor input
2π N s
(See Sec. 8.26)
... N s is r.p.s.
2
Rotor Cu loss 3(I'2 ) R 2
=
Rotor input =
s
s
Now
(i)
As shown in Sec. 8.16, under running conditions,
I'2 =
where
s E2
R 22 + (s X 2 ) 2
=
K = Transformation ratio =
∴
Rotor input = 3 ×
s K E1
R 22 + (s X 2 ) 2
Rotor turn s/phase
Stator tur ns/phase
s 2 E 22 R 2
3 s E 22 R 2
1
×
=
R 22 + (s X 2 ) 2 s R 22 + (s X 2 ) 2
(Putting me value of I'2 in eq.(i))
s 2 K 2 E12 R 2
2 2
1 3 s K E1 R 2
Rotor input = 3 × 2
× =
R 2 + (s X 2 ) 2 s R 22 + (s X 2 ) 2
Also
(Putting me value of I'2 in eq.(i))
∴
s E 22 R 2
Rotor input
3
Tg =
=
×
2π N s
2π N s R 22 + (s X 2 ) 2
s K 2 E12 R 2
3
=
×
2π N s R 22 + (s X 2 ) 2
...in terms of E2
...in terms of E1
Note that in the above expressions of Tg, the values E1, E2, R2 and X2 represent
the phase values.
8.28 Performance Curves of Squirrel-Cage Motor
The performance curves of a 3-phase induction motor indicate the variations of
speed, power factor, efficiency, stator current and torque for different values of
load. However, before giving the performance curves in one graph, it is
desirable to discuss the variation of torque, and stator current with slip.
(i) Variation of torque and stator current with slip
Fig. (8.21) shows the variation of torque and stator current with slip for a
standard squirrel-cage motor. Generally, the rotor resistance is low so that full207
load current occurs at low slip. Then even at full-load f' (= sf) and. therefore, X'2
(= 2π f' L2) are low. Between zero and full-load, rotor power factor (= cos φ'2)
and rotor impedance (= Z'2) remain practically constant. Therefore, rotor current
I'2(E'2/Z'2) and, therefore, torque (Tr) increase directly with the slip. Now stator
current I1 increases in proportion to I'2. This is shown in Fig. (8.21) where Tr and
I1 are indicated as straight lines from no-load to full-load. As load and slip are
increased beyond full-load, the increase in rotor reactance becomes appreciable.
The increasing value of rotor impedance not only decreases the rotor power
factor cos φ'2 (= R2/Z'2) but also lowers the rate of increase of rotor current. As a
result, the torque Tr and stator current I1 do not increase directly with slip as
indicated in Fig. (8.21). With the decreasing power factor and the lowered rate
of increase in rotor current, the stator current I1 and torque Tr increase at a lower
rate. Finally, torque Tr reaches the maximum value at about 25% slip in the
standard squirrel cage motor. This maximum value of torque is called the
pullout torque or breakdown torque. If the load is increased beyond the
breakdown point, the decrease in rotor power factor is greater than the increase
in rotor current, resulting in a decreasing torque. The result is that motor slows
down quickly and comes to a stop.
Fig.(8.21)
In Fig. (8.21), the value of torque at starting (i.e., s = 100%) is 1.5 times the fullload torque. The starting current is about five times the full-load current. The
motor is essentially a constant-speed machine having speed characteristics about
the same as a d.c. shunt motor.
(ii) Performance curves
Fig. (8.22) shows the performance curves of 3-phase squirrel cage induction
motor.
208
Fig.(8.22)
The following points may be noted:
(a) At no-load, the rotor lags behind the stator flux by only a small amount,
since the only torque required is that needed to overcome the no-load
losses. As mechanical load is added, the rotor speed decreases. A decrease
in rotor speed allows the constant-speed rotating field to sweep across the
rotor conductors at a faster rate, thereby inducing large rotor currents. This
results in a larger torque output at a slightly reduced speed. This explains
for speed-load curve in Fig. (8.22).
(b) At no-load, the current drawn by an induction motor is largely a
magnetizing current; the no-load current lagging the applied voltage by a
large angle. Thus the power factor of a lightly loaded induction motor is
very low. Because of the air gap, the reluctance of the magnetic circuit is
high, resulting in a large value of no-load current as compared with a
transformer. As load is added, the active or power component of current
increases, resulting in a higher power factor. However, because of the large
value of magnetizing current, which is present regardless of load, the power
factor of an induction motor even at full-load seldom exceeds 90%. Fig.
(8.22) shows the variation of power factor with load of a typical squirrelcage induction motor.
Output
(c) Efficiency =
Output + Losses
The losses occurring in a 3-phase induction motor are Cu losses in stator
and rotor windings, iron losses in stator and rotor core and friction and
windage losses. The iron losses and friction and windage losses are almost
independent of load. Had I2R been constant, the efficiency of the motor
would have increased with load. But I2R loss depends upon load.
209
Therefore, the efficiency of the motor increases with load but the curve is
dropping at high loads.
(d) At no-load, the only torque required is that needed to overcome no-load
losses. Therefore, stator draws a small current from the supply. As
mechanical load is added, the rotor speed decreases. A decrease in rotor
speed allows the constant-speed rotating field to sweep across the rotor
conductors at a faster rate, thereby inducing larger rotor currents. With
increasing loads, the increased rotor currents are in such a direction so as to
decrease the staler flux, thereby temporarily decreasing the counter e.m.f.
in the stator winding. The decreased counter e.m.f. allows more stator
current to flow.
(e) Output = Torque × Speed
Since the speed of the motor does not change appreciably with load, the
torque increases with increase in load.
8.29 Equivalent Circuit of 3-Phase Induction Motor at
Any Slip
In a 3-phase induction motor, the stator winding is connected to 3-phase supply
and the rotor winding is short-circuited. The energy is transferred magnetically
from the stator winding to the short-circuited, rotor winding. Therefore, an
induction motor may be considered to be a transformer with a rotating secondary
(short-circuited). The stator winding corresponds to transformer primary and the
rotor finding corresponds to transformer secondary. In view of the similarity of
the flux and voltage conditions to those in a transformer, one can expect that the
equivalent circuit of an induction motor will be similar to that of a transformer.
Fig. (8.23) shows the equivalent circuit (though not the only one) per phase for
an induction motor. Let us discuss the stator and rotor circuits separately.
Fig.(8.23)
Stator circuit. In the stator, the events are very similar to those in the
transformer primal y. The applied voltage per phase to the stator is V1 and R1
and X1 are the stator resistance and leakage reactance per phase respectively.
The applied voltage V1 produces a magnetic flux which links the stator winding
(i.e., primary) as well as the rotor winding (i.e., secondary). As a result, self-
210
induced e.m.f. E1 is induced in the stator winding and mutually induced e.m.f.
E'2(= s E2 = s K E1 where K is transformation ratio) is induced in the rotor
winding. The flow of stator current I1 causes voltage drops in R1 and X1.
∴
V1 = − E1 + I1 (R 1 + j X1 )
...phasor sum
When the motor is at no-load, the stator winding draws a current I0. It has two
components viz., (i) which supplies the no-load motor losses and (ii)
magnetizing component Im which sets up magnetic flux in the core and the airgap. The parallel combination of Rc and Xm, therefore, represents the no-load
motor losses and the production of magnetic flux respectively.
I0 = Iw + Im
Rotor circuit. Here R2 and X2 represent the rotor resistance and standstill rotor
reactance per phase respectively. At any slip s, the rotor reactance will be s X2
The induced voltage/phase in the rotor is E'2 = s E2 = s K E1. Since the rotor
winding is short-circuited, the whole of e.m.f. E'2 is used up in circulating the
rotor current I'2.
∴
E ' 2 = I' 2 ( R 2 + j s X 2 )
The rotor current I'2 is reflected as I"2 (= K I'2) in the stator. The phasor sum of
I"2 and I0 gives the stator current I1.
It is important to note that input to the primary and output from the secondary of
a transformer are electrical. However, in an induction motor, the inputs to the
stator and rotor are electrical but the output from the rotor is mechanical. To
facilitate calculations, it is desirable and
necessary to replace the mechanical load by an
equivalent electrical load. We then have the
transformer equivalent circuit of the induction
motor.
It may be noted that even though the frequencies
of stator and rotor currents are different, yet the
magnetic fields due to them rotate at
synchronous speed Ns. The stator currents
produce a magnetic flux which rotates at a speed
Ns. At slip s, the speed of rotation of the rotor
field relative to the rotor surface in the direction
of rotation of the rotor is
=
120 f ' 120 s f
=
= s Ns
P
P
But the rotor is revolving at a speed of N relative
to the stator core. Therefore, the speed of rotor
211
Fig.(8.24)
field relative to stator core
= sN s + N = ( N s − N ) + N = N s
Thus no matter what the value of slip s, the stator and rotor magnetic fields are
synchronous with each other when seen by an observer stationed in space.
Consequently, the 3-phase induction motor can be regarded as being equivalent
to a transformer having an air-gap separating the iron portions of the magnetic
circuit carrying the primary and secondary windings.
Fig. (8.24) shows the phasor diagram of induction motor.
8.30 Equivalent Circuit of the Rotor
We shall now see how mechanical load of the motor is replaced by the
equivalent electrical load. Fig. (8.25 (i)) shows the equivalent circuit per phase
of the rotor at slip s. The rotor phase current is given by;
I'2 =
s E2
R 22 + (s X 2 ) 2
Mathematically, this value is unaltered by writing it as:
I' 2 =
E2
( R 2 / s) 2 + ( X 2 ) 2
As shown in Fig. (8.25 (ii)), we now have a rotor circuit that has a fixed
reactance X2 connected in series with a variable resistance R2/s and supplied
with constant voltage E2. Note that Fig. (8.25 (ii)) transfers the variable to the
resistance without altering power or power factor conditions.
Fig.(8.25)
The quantity R2/s is greater than R2 since s is a fraction. Therefore, R2/s can be
divided into a fixed part R2 and a variable part (R2/s − R2) i.e.,
R2
1
= R 2 + R 2  − 1
s
s 
212
(i)
The first part R2 is the rotor resistance/phase, and represents the rotor Cu
loss.
1
(ii) The second part R 2  −1 is a variable-resistance load. The power
s 
delivered to this load represents the total mechanical power developed in
the rotor. Thus mechanical load on the induction motor can be replaced by
1
a variable-resistance load of value R 2  −1 . This is
s 
∴
1
R L = R 2  − 1
s 
Fig. (8.25 (iii)) shows the equivalent rotor circuit along with load resistance RL.
8.31 Transformer Equivalent Circuit of Induction Motor
Fig. (8.26) shows the equivalent circuit per phase of a 3-phase induction motor.
Note that mechanical load on the motor has been replaced by an equivalent
electrical resistance RL given by;
1
R L = R 2  − 1
s 
(i)
Fig.(8.26)
Note that circuit shown in Fig. (8.26) is similar to the equivalent circuit of a
transformer with secondary load equal to R2 given by eq. (i). The rotor e.m.f. in
the equivalent circuit now depends only on the transformation ratio K (= E2/E1).
Therefore; induction motor can be represented as an equivalent transformer
connected to a variable-resistance load RL given by eq. (i). The power delivered
to RL represents the total mechanical power developed in the rotor. Since the
equivalent circuit of Fig. (8.26) is that of a transformer, the secondary (i.e.,
rotor) values can be transferred to primary (i.e., stator) through the appropriate
use of transformation ratio K. Recall that when shifting resistance/reactance
from secondary to primary, it should be divided by K2 whereas current should be
multiplied by K. The equivalent circuit of an induction motor referred to
primary is shown in Fig. (8.27).
213
Fig.(8.27)
Note that the element (i.e., R'L) enclosed in the dotted box is the equivalent
electrical resistance related to the mechanical load on the motor. The following
points may be noted from the equivalent circuit of the induction motor:
(i) At no-load, the slip is practically zero and the load R'L is infinite. This
condition resembles that in a transformer whose secondary winding is
open-circuited.
(ii) At standstill, the slip is unity and the load R'L is zero. This condition
resembles that in a transformer whose secondary winding is short-circuited.
(iii) When the motor is running under load, the value of R'L will depend upon
the value of the slip s. This condition resembles that in a transformer whose
secondary is supplying variable and purely resistive load.
(iv) The equivalent electrical resistance R'L related to mechanical load is slip or
speed dependent. If the slip s increases, the load R'L decreases and the rotor
current increases and motor will develop more mechanical power. This is
expected because the slip of the motor increases with the increase of load
on the motor shaft.
8.32 Power Relations
The transformer equivalent circuit of an induction motor is quite helpful in
analyzing the various power relations in the motor. Fig. (8.28) shows the
equivalent circuit per phase of an induction motor where all values have been
referred to primary (i.e., stator).
Fig.(8.28)
214
(i)
R'
1
Total electrical load = R '2  − 1 + R '2 = 2
s
s 
Power input to stator = 3V1 I1 cos φ1
There will be stator core loss and stator Cu loss. The remaining power will
be the power transferred across the air-gap i.e., input to the rotor.
3(I"2 )2 R '2
(ii) Rotor input =
s
2
Rotor Cu loss = 3(I"2 ) R '2
Total mechanical power developed by the rotor is
Pm = Rotor input − Rotor Cu loss
3(I"2 )2 R '2
1
=
3(I"2 )2 R '2 = 3(I"2 )2 R '2  − 1
s
s 
This is quite apparent from the equivalent circuit shown in Fig. (8.28).
(iii) If Tg is the gross torque developed by the rotor, then,
Pm =
2π N Tg
60
or
2π N Tg
1
3(I"2 )2 R '2  − 1 =
60
s 
or
1 − s  2π N Tg
3(I"2 )2 R '2 
 = 60
 s 
or
1 − s  2π N s (1 − s) Tg
3(I"2 )2 R '2 
=
60
 s 
∴
or
3(I"2 )2 R '2 s
Tg =
2π N s 60
[Q N = N s (1 − s)]
N-m
3(I"2 )2 R '2 s
Tg = 9.55
Ns
N-m
Note that shaft torque Tsh will be less than Tg by the torque required to meet
windage and frictional losses.
215
8.33 Approximate Equivalent Circuit of Induction Motor
As in case of a transformer, the approximate equivalent circuit of an induction
motor is obtained by shifting the shunt branch (Rc − Xm) to the input terminals
as shown in Fig. (8.29). This step has been taken on the assumption that voltage
drop in R1 and X1 is small and the terminal voltage V1 does not appreciably
differ from the induced voltage E1. Fig. (8.29) shows the approximate equivalent
circuit per phase of an induction motor where all values have been referred to
primary (i.e., stator).
Fig.(8.29)
The above approximate circuit of induction motor is not so readily justified as
with the transformer. This is due to the following reasons:
(i) Unlike that of a power transformer, the magnetic circuit of the induction
motor has an air-gap. Therefore, the exciting current of induction motor (30
to 40% of full-load current) is much higher than that of the power
transformer. Consequently, the exact equivalent circuit must be used for
accurate results.
(ii) The relative values of X1 and X2 in an induction motor are larger than the
corresponding ones to be found in the transformer. This fact does not
justify the use of approximate equivalent circuit
(iii) In a transformer, the windings are concentrated whereas in an induction
motor, the windings are distributed. This affects the transformation ratio.
In spite of the above drawbacks of approximate equivalent circuit, it yields
results that are satisfactory for large motors. However, approximate equivalent
circuit is not justified for small motors.
8.34 Starting of 3-Phase Induction Motors
The induction motor is fundamentally a transformer in which the stator is the
primary and the rotor is short-circuited secondary. At starting, the voltage
induced in the induction motor rotor is maximum (Q s = 1). Since the rotor
impedance is low, the rotor current is excessively large. This large rotor current
is reflected in the stator because of transformer action. This results in high
starting current (4 to 10 times the full-load current) in the stator at low power
216
factor and consequently the value of starting torque is low. Because of the short
duration, this value of large current does not harm the motor if the motor
accelerates normally. However, this large starting current will produce large
line-voltage drop. This will adversely affect the operation of other electrical
equipment connected to the same lines. Therefore, it is desirable and necessary
to reduce the magnitude of stator current at starting and several methods are
available for this purpose.
8.35 Methods of Starting 3-Phase Induction Motors
The method to be employed in starting a given induction motor depends upon
the size of the motor and the type of the motor. The common methods used to
start induction motors are:
(i) Direct-on-line starting
(ii) Stator resistance starting
(iii) Autotransformer starting
(iv) Star-delta starting
(v) Rotor resistance starting
Methods (i) to (iv) are applicable to both squirrel-cage and slip ring motors.
However, method (v) is applicable only to slip ring motors. In practice, any one
of the first four methods is used for starting squirrel cage motors, depending
upon ,the size of the motor. But slip ring motors are invariably started by rotor
resistance starting.
8.36 Methods of Starting Squirrel-Cage Motors
Except direct-on-line starting, all other methods of starting squirrel-cage motors
employ reduced voltage across motor terminals at starting.
(i) Direct-on-line starting
This method of starting in just what the name implies—the motor is started by
connecting it directly to 3-phase supply. The impedance of the motor at
standstill is relatively low and when it is directly connected to the supply
system, the starting current will be high (4 to 10 times the full-load current) and
at a low power factor. Consequently, this method of starting is suitable for
relatively small (up to 7.5 kW) machines.
Relation between starling and F.L. torques. We know that:
Rotor input = 2π Ns T = kT
Rotor Cu loss = s × Rotor input
But
∴
or
3(I'2 )2 R 2 = s × kT
T ∝ (I'2 )2 s
217
(Q I'2 ∝ I1 )
T ∝ I12 s
or
If Ist is the starting current, then starting torque (Tst) is
T ∝ I st2
(Q at starting s = 1)
If If is the full-load current and sf is the full-load slip, then,
Tf ∝ I f2 s f
2
∴
Tst  I st 
=   × sf
Tf  I f 
When the motor is started direct-on-line, the starting current is the short-circuit
(blocked-rotor) current Isc.
2
∴
Tst  I sc 
=   × sf
Tf  I f 
Let us illustrate the above relation with a numerical example. Suppose Isc = 5 If
and full-load slip sf =0.04. Then,
2
∴
2
Tst  I sc 
5 I 
=   × s f =  f  × 0.04 = (5) 2 × 0.04 = 1
Tf  I f 
 If 
Tst = Tf
Note that starting current is as large as five times the full-load current but
starting torque is just equal to the full-load torque. Therefore, starting current is
very high and the starting torque is comparatively low. If this large starting
current flows for a long time, it may overheat the motor and damage the
insulation.
(ii) Stator resistance starting
In this method, external resistances are connected in series with each phase of
stator winding during starting. This causes voltage drop across the resistances so
that voltage available across motor terminals is reduced and hence the starting
current. The starting resistances are gradually cut out in steps (two or more
steps) from the stator circuit as the motor picks up speed. When the motor
attains rated speed, the resistances are completely cut out and full line voltage is
applied to the rotor.
This method suffers from two drawbacks. First, the reduced voltage applied to
the motor during the starting period lowers the starting torque and hence
increases the accelerating time. Secondly, a lot of power is wasted in the starting
resistances.
218
Fig.(8.30)
Relation between starting and F.L. torques. Let V be the rated voltage/phase.
If the voltage is reduced by a fraction x by the insertion of resistors in the line,
then voltage applied to the motor per phase will be xV.
I st = x I sc
2
Now
Tst  Ist 
=   × sf
Tf  I f 
or
Tst
I 
= x 2  sc  × s f
Tf
 If 
2
Thus while the starting current reduces by a fraction x of the rated-voltage
starting current (Isc), the starting torque is reduced by a fraction x2 of that
obtained by direct switching. The reduced voltage applied to the motor during
the starting period lowers the starting current but at the same time increases the
accelerating time because of the reduced value of the starting torque. Therefore,
this method is used for starting small motors only.
(iii) Autotransformer starting
This method also aims at connecting the induction motor to a reduced supply at
starting and then connecting it to the full voltage as the motor picks up sufficient
speed. Fig. (8.31) shows the circuit arrangement for autotransformer starting.
The tapping on the autotransformer is so set that when it is in the circuit, 65% to
80% of line voltage is applied to the motor.
At the instant of starting, the change-over switch is thrown to “start” position.
This puts the autotransformer in the circuit and thus reduced voltage is applied
to the circuit. Consequently, starting current is limited to safe value. When the
motor attains about 80% of normal speed, the changeover switch is thrown to
219
“run” position. This takes out the autotransformer from the circuit and puts the
motor to full line voltage. Autotransformer starting has several advantages viz
low power loss, low starting current and less radiated heat. For large machines
(over 25 H.P.), this method of starting is often used. This method can be used
for both star and delta connected motors.
Fig.(8.31)
Relation between starting And F.L. torques. Consider a star-connected
squirrel-cage induction motor. If V is the line voltage, then voltage across motor
phase on direct switching is V 3 and starting current is Ist = Isc. In case of
autotransformer, if a tapping of transformation ratio K (a fraction) is used, then
phase voltage across motor is KV 3 and Ist = K Isc,
2
2
2
Tst  Ist 
KI 
I 
=   × s f =  sc  × s f = K 2  sc  × s f
Tf  I f 
 If 
 If 
Now
2
∴
Tst
I 
= K 2  sc  × s f
Tf
 If 
Fig.(8.32)
220
The current taken from the supply or by autotransformer is I1 = KI2 = K2Isc. Note
that motor current is K times, the supply line current is K2 times and the starting
torque is K2 times the value it would have been on direct-on-line starting.
(iv) Star-delta starting
The stator winding of the motor is designed for delta operation and is connected
in star during the starting period. When the machine is up to speed, the
connections are changed to delta. The circuit arrangement for star-delta starting
is shown in Fig. (8.33).
The six leads of the stator windings are connected to the changeover switch as
shown. At the instant of starting, the changeover switch is thrown to “Start”
position which connects the stator windings in star. Therefore, each stator phase
gets V 3 volts where V is the line voltage. This reduces the starting current.
When the motor picks up speed, the changeover switch is thrown to “Run”
position which connects the stator windings in delta. Now each stator phase gets
full line voltage V. The disadvantages of this method are:
(a) With star-connection during starting, stator phase voltage is 1 3 times the
(
)2
line voltage. Consequently, starting torque is 1 3 or 1/3 times the value
it would have with ∆-connection. This is rather a large reduction in starting
torque.
(b) The reduction in voltage is fixed.
This method of starting is used for medium-size machines (upto about 25 H.P.).
Relation between starting and F.L. torques. In direct delta starting,
Starting current/phase, Isc = V/Zsc where V = line voltage
Starting line current =
In star starting, we have,
3 Isc
Starting current/phase, I st =
V 3 1
=
I sc
Z sc
3
Now
2
 I sc
Tst  Ist 
=   × s f = 
Tf  I f 
 3 × If
or
Tst 1  Isc 
= 
 × sf
Tf 3  I f 
2

 × sf


2
where
Isc = starting phase current (delta)
If = F.L. phase current (delta)
221
Fig.(8.33)
Note that in star-delta starting, the starting line current is reduced to one-third as
compared to starting with the winding delta connected. Further, starting torque
is reduced to one-third of that obtainable by direct delta starting. This method is
cheap but limited to applications where high starting torque is not necessary e.g.,
machine tools, pumps etc.
8.37 Starting of Slip-Ring Motors
Slip-ring motors are invariably started by rotor resistance starting. In this
method, a variable star-connected rheostat is connected in the rotor circuit
through slip rings and full voltage is applied to the stator winding as shown in
Fig. (8.34).
Fig.(8.34)
(i)
At starting, the handle of rheostat is set in the OFF position so that
maximum resistance is placed in each phase of the rotor circuit. This
reduces the starting current and at the same time starting torque is
increased.
(ii) As the motor picks up speed, the handle of rheostat is gradually moved in
clockwise direction and cuts out the external resistance in each phase of the
rotor circuit. When the motor attains normal speed, the change-over switch
is in the ON position and the whole external resistance is cut out from the
rotor circuit.
222
8.38 Slip-Ring Motors Versus Squirrel Cage Motors
The slip-ring induction motors have the following advantages over the squirrel
cage motors:
(i) High starting torque with low starting current.
(ii) Smooth acceleration under heavy loads.
(iii) No abnormal heating during starting.
(iv) Good running characteristics after external rotor resistances are cut out.
(v) Adjustable speed.
The disadvantages of slip-ring motors are:
(i) The initial and maintenance costs are greater than those of squirrel cage
motors.
(ii) The speed regulation is poor when run with resistance in the rotor circuit
8.39 Induction Motor Rating
The nameplate of a 3-phase induction motor provides the following information:
(i) Horsepower
(ii) Line voltage
(iii) Line current
(iv) Speed
(v) Frequency
(vi) Temperature rise
The horsepower rating is the mechanical output of the motor when it is operated
at rated line voltage, rated frequency and rated speed. Under these conditions,
the line current is that specified on the nameplate and the temperature rise does
not exceed that specified.
The speed given on the nameplate is the actual speed of the motor at rated fullload; it is not the synchronous speed. Thus, the nameplate speed of the induction
motor might be 1710 r.p.m. It is the rated full-load speed.
8.40 Double Squirrel-Cage Motors
One of the advantages of the slip-ring motor is that resistance may be inserted in
the rotor circuit to obtain high starting torque (at low starting current) and then
cut out to obtain optimum running conditions. However, such a procedure
cannot be adopted for a squirrel cage motor because its cage is permanently
short-circuited. In order to provide high starting torque at low starting current,
double-cage construction is used.
Construction
As the name suggests, the rotor of this motor has two squirrel-cage windings
located one above the other as shown in Fig. (8.35 (i)).
(i) The outer winding consists of bars of smaller cross-section short-circuited
by end rings. Therefore, the resistance of this winding is high. Since the
223
outer winding has relatively open slots and a poorer flux path around its
bars [See Fig. (8.35 (ii))], it has a low inductance. Thus the resistance of
the outer squirrel-cage winding is high and its inductance is low.
(ii) The inner winding consists of bars of greater cross-section short-circuited
by end rings. Therefore, the resistance of this winding is low. Since the
bars of the inner winding are thoroughly buried in iron, it has a high
inductance [See Fig. (8.35 (ii))]. Thus the resistance of the inner squirrelcage winding is low and its inductance is high.
Fig.(8.35)
Working
When a rotating magnetic field sweeps across the two windings, equal e.m.f.s
are induced in each.
(i) At starting, the rotor frequency is the same as that of the line (i.e., 50 Hz),
making the reactance of the lower winding much higher than that of the
upper winding. Because of the high reactance of the lower winding, nearly
all the rotor current flows in the high-resistance outer cage winding. This
provides the good starting characteristics of a high-resistance cage winding.
Thus the outer winding gives high starting torque at low starting current.
(ii) As the motor accelerates, the rotor frequency decreases, thereby lowering
the reactance of the inner winding, allowing it to carry a larger proportion
of the total rotor current At the normal operating speed of the motor, the
rotor frequency is so low (2 to 3 Hz) that nearly all the rotor current flows
in the low-resistance inner cage winding. This results in good operating
efficiency and speed regulation.
Fig. (8.36) shows the operating characteristics of double squirrel-cage motor.
The starting torque of this motor ranges from 200 to 250 percent of full-load
torque with a starting current of 4 to 6 times the full-load value. It is classed as a
high-torque, low starting current motor.
224
Fig.(8.36)
8.41 Equivalent Circuit of Double Squirrel-Cage Motor
Fig. (8.37) shows a section of the double squirrel cage motor.
Here Ro and Ri are the per phase resistances of the outer cage
winding and inner cage winding whereas Xo and Xi are the
corresponding per phase standstill reactances. For the outer
cage, the resistance is made intentionally high, giving a high
Fig.(8.37)
starting torque. For the inner cage winding, the resistance is low
and the leakage reactance is high, giving a low starting torque
but high efficiency on load. Note that in a double squirrel cage motor, the outer
winding produces the high starting and accelerating torque while the inner
winding provides the running torque at good efficiency.
Fig. (8.38 (i)) shows the equivalent circuit for one phase of double cage motor
referred to stator. The two cage impedances are effectively in parallel. The
resistances and reactances of the outer and inner rotors are referred to the stator.
The exciting circuit is accounted for as in a single cage motor. If the
magnetizing current (I0) is neglected, then the circuit is simplified to that shown
in Fig. (8.38 (ii)).
Fig.(8.38)
225
From the equivalent circuit, the performance of the motor can be predicted.
Total impedance as referred to stator is
Z o1 = R 1 + j X1 +
Z' Z'
1
= R 1 + j X1 + i o
1 Z'i + 1 Z'o
Z'i + Z'o
226
Tests for finding parameters of 3Φ induction motor
To accurately model the three phase induction motor , the
equivalent circuit parameters (R1 , X1 , R2 , X2 , Rc , Xm)
should be known. A single-phase equivalent circuit with lumped
parameters is the most traditional model of an AC induction
motor .The steady-state operating characteristics of a three-phase
induction motor are often investigated using a per-phase
equivalent circuit . In this circuit R1, and X1 represent stator
resistance and leakage reactance, respectively; R2 and X2 denote
the rotor resistance and leakage reactance referred to the stator,
respectively; Rc resistance stands for core losses; Xm represents
magnetizing reactance. The equivalent circuit is used to facilitate
the computation of various operating quantities, such as stator
current, input power, losses, induced torque, and efficiency.
When power aspects of the operation need to be emphasized, the
shunt resistance is usually neglected; the core losses can be
included in efficiency calculations along with the friction,
windage, and stray losses.
The parameters of the equivalent circuit can be obtained from the
dc, no-load, and blocked-rotor tests .The DC Test is performed to
compute the stator winding resistance . A dc voltage is applied to
the stator windings of an induction motor. The resulting current
flowing through the stator windings is a dc current; thus, no
voltage is induced in the rotor circuit, and the motor reactance is
zero. The stator resistance is the only circuit parameter limiting
current flow. In No-Load Test , a rated, balanced ac voltage at a
rated frequency is applied to the stator while it is running at no
load, and input power, voltage, and phase currents are measured at
the no-load condition. In Blocked-Rotor Test, the rotor of the
induction motor is blocked, and a reduced voltage is applied to the
stator terminals so that the rated current flows through the stator
windings. The input power, voltage, and current are measured.
1
Notes
1– Experience has shown certain Design Types have certain values
of X1 and X2 as function of Xbr as follows:
•
•
•
•
•
Wound rotor :
Design A
:
Design B
:
Design C
:
Design D
:
X1 = X2 = 0.5Xbr
X1 = X2 = 0.5Xbr
X1 = 0.4Xbr and X2 = 0.6Xbr
X1 = 0.3Xbr and X2 = 0.7Xbr
X1 = X2 = 0.5Xbr
2- For some design-class induction motors, a blocked-rotor test is
conducted under a test frequency, usually less than the normal
operating frequency so as to evaluate the rotor resistance
appropriately. If the test frequency is different from the rated
frequency, one can compute the total equivalent reactance at the
normal operating frequency (Xbr) as follows since the reactance is
directly proportional to the frequency:
X br 
Where
f rated
'
 X br
f test
X´br is blocked-rotor reactance at the test frequency
6
Example : 208 V, 60 Hz, six-pole Y-connected 25-hp design class B
induction motor is tested in the laboratory, with the following results:
DC test
: 13.5 V , 64 A
No load test
: 208 V , 22.0 A , 1200 W , 60 Hz
Blocked-rotor test : 24.6 V , 64.5 A , 2200 W , 15 Hz
Find R1 , R2 , X1 , X2 , and Xm of this motor ?
Solution:
From DC test :
R dc = V dc / I dc = 13.5 V/ 64 A = 0.2109 Ω
For Y-connected induction motor R1 = R dc / 2 = 0.2109 / 2 = 0.1054 Ω
From No Load test :
I1nl  22 A , V1nl 
Q1nl 
V1nl  I1nl 
2
208
 120 V , Pnl  1200 W
3
P 
  nl 
 3 
2
7
2
 1200 
 120  22  
  2609.5
 3 
2
Q1nl
2
X nl 
V1nl
Q1nl
X nl 
120 2
 5.5182   X 1  X m
2609.5
From blocked-rotor test :
I1br  64.5 A , V1br 
Rbr 
P1br
I 1br
2

24.6
3
 14.2 V , P1br 
2200
 733.333 W
3
733.333
 0.1762 
64.5 2
R2  Rbr  R1  0.1762  0.1054  0.0708 
Q1br 
V1br  I1br 2  P1br 2
Q1br 
14.2  64.52  (733.333) 2
 548.721
X br 
Q1br 548.721

 0.1318 
2
64.52
I1br
X br 
f rated
60 Hz
'
 X br 
 0.1318  0.5272 
f test
15 Hz
'
at 15 Hz
For design B motor
X1 = 0.4 Xbr = 0.4 * 0.5272 = 0.2108 Ω
X2 = 0.6 Xbr = 0.6 * 0.5272 = 0.3163 Ω
So ,
Xm = Xnl – X1 = 5.5182 – 0.2108 = 5.3074 Ω
8
HOME WORK
Q1: Test carried out on 3phase, Y-connected, 5HP ,208V ,4Pole
SCIM , with the following results:
DC test
: 20 V , 25 A
No load test
: 208 V , 4 A , 250 W , 60 Hz
Blocked-rotor test : 35 V , 12 A , 450 W , 15 Hz
Find
R1 , R2 , X1 , X2 , and Xm of this motor ?
Q2: Test carried out on 3phase, Δ-connected, 30kW ,415V,
50 Hz SCIM , with the following results:
No load test
: 415 V , 21 A , 1250 W
Blocked-rotor test : 100 V , 45 A , 2730 W
Find
X1 , X2 , and Xm of this motor ?
9
Loading , braking & application of a 3phase induction motors
Introduction ;
• When selecting a 3-phase induction motor for an application several
motor types can fill the need , manufactures often specify the motor class
best suited to drive the load .
• 3-phase induction motors under 500 hp are standardized ,i.e. the
frames have standard dimensions. So, A 25 hp, 1725 rpm, 60 Hz motor
from one manufacture can be replaced by that of any other manufacturer
without having to change the mounting holes, shaft height, or the type of
coupling
• a motor must satisfy minimum requirements for starting torque, lockedrotor current, overload capacity, and temperature rise.
Motor characteristics under load
• High-inertia load stains a motor by prolonging the starting period.
– the starting currents in both the rotor and stator are high during starting.
– overheating from I²R losses becomes a problems.
– prolonged starting of very large motors will overload the utility
transmission network.
• Induction motors are often started on reduced voltage to:– limits the current drawn by the motor
– reduces the heating rate of the rotor
• Heat dissipated in the rotor during starting, from zero speed to rated
speed, is equal to the final kinetic energy stored in all the revolving
components .
LOAD CONSIDERATIONS
There are three basic load types, and these types are classified by the
relationship of horsepower and speed.
1- CONSTANT TORQUE LOAD.
2- CONSTANT HORSE POWER LOAD.
3- VARIABLE TORQUE LOAD.
1
Constant torque applications are those that have the same torque at all
operating speeds, and horsepower varies directly with the speed. About
90% of all applications, other than pumps, are constant torque loads.
Examples of this type include conveyors, hoisting loads, surface winding
machines, positive displacement pumps and piston and screw
compressors.
Constant horsepower applications have higher values of torque at lower
speeds, and lower values of torque at higher speeds. Examples include
lathes, milling machines, and drill presses .
The drills in the diagram below are an example of a constant horsepower
application. When a larger hole is being drilled, the drill is operating at
low speed, but it requires a very high torque to turn the large drill in the
material. When a small hole is being drilled, the drill is operated at a high
2
speed, but it requires a very low torque to run the small drill in the
material.
The last basic load type is variable torque. The torque required varies as
the square of its speed, and horsepower requirements increase as the cube
of the speed. Examples include centrifugal pumps, turbine pumps,
centrifugal blowers, fans and centrifugal compressors.
Motor Classifications according to speed
– For a given output power, a high-speed motor compared with a low-speed motor
• costs less
• is smaller sized
• has higher efficiency and power factor
–Low speed applications are often best served by using a high speed- motor
and a gear-box which is often required to modify operating speed especially
for very high-speed applications (> 3600 rpm).
3
Braking of a 3 phase induction motors:
1- Plugging :
• In some applications, the motor and its load must come to a quick stop
– such braking action can be accomplished by interchanging two
stator leads ,the lead switching causes the revolving field to turn in
the opposite direction. Kinetic energy is absorbed from the
mechanical load causing the speed to fall.The absorbed energy is
dissipated as heat in the rotor circuit.
• plugging produces I²R losses that exceed the locked rotor losses.
• High rotor temperatures will result that may cause the rotor bars to
overheat or even melt. The heat dissipated in the rotor during plugging,
from initial speed to zero, is three times the original kinetic energy of all
revolving parts.
2- DC Braking:
• An induction motor with high-inertial load can also come to a quick
stop by circulating dc current in the stator winding.
– any two stator terminals can be connected to a dc source.
– the dc current produces stationary N,S poles in the stator.
• the number of stationary poles are the same at the number of rotating
poles normally produced with ac currents.
– as the rotor bars sweeps past the dc field, an ac rotor voltage is induced
• the I²R losses produced in the rotor circuit come at the expense of the
kinetic energy stored in the revolving components.
– the motor comes to a rest by dissipating as heat all the kinetic energy.
• The benefit of dc braking is the far less heat that is produced.
– dissipated rotor losses is equal to the kinetic energy of the revolving
parts.
– energy dissipation is independent of the dc current magnitude
– the braking torque is proportional to the square of the dc braking
current.
Enclosure considerations
The enclosure of the motor must protect the windings, bearings, and other
mechanical parts from moisture, chemicals,mechanical damage and
abrasion from grit. NEMA standards define more than 20 types of
enclosures under the categories of open machines, totally enclosed
machines, and machines with encapsulated or sealed windings.
The most commonly used motor enclosures are open dripproof, totally
enclosed fan cooled and explosionproof.
4
Open Dripproof(ODP):
The open dripproof motor has a free exchange of air with the ambient.
Drops of liquid or solid particles do not interfere with the operation at any
angle from 0 to 15 degrees downward from the vertical. The openings are
intake and exhaust ports to accommodate interchange of air. The open
dripproof motor is designed for indoor use where the air is fairly clean
and where there is little danger of splashing liquid.
ODP MOTOR
Totally Enclosed Fan Cooled (TEFC):
This type of enclosure prevents the free exchange of air between the
inside and outside of the frame, but does not make the frame completely
airtight. A fan is attached to the shaft and pushes air over the frame
during its operation to help in the cooling process. The ribbed frame is
designed to increase the surface area for cooling purposes. There is
also a totally enclosed non-ventilated (TENV) design which does not use
a fan, but is used in situations where air is being blown over the motor
shell for cooling, such as in a propeller fan application.
The TEFC style enclosure is the most versatile of all. It is used on pumps,
fans, compressors, general industrial belt drive and direct connected
equipment. Special protection can be added to the TEFC motor to help it
withstand hostile environments such as chemical and pulp and paper
applications.
TEFC MOTOR
5
Explosion proof
The explosion proof motor is a totally enclosed machine and is designed
to withstand an explosion of specified gas or vapor inside the motor
casing and prevent the ignition outside the motor by sparks, flashing
or explosion. These motors are designed for specific hazardous purposes,
such as atmospheres containing gases or hazardous dusts. For safe
operation, the maximum motor operating temperature must be below the
ignition temperature of surrounding gases or vapors. Explosion proof
motors are designed, manufactured and tested under the rigid
requirements of the Underwriters Laboratories.
ENVIRONMENTAL CONSIDERATIONS :
AC motors that are properly selected and used should give many years of
satisfactory service. Motor life is prolonged by keeping the motor cool,
dry, clean and lubricated. Choosing the best motor for the application and
the environment will insure that long life. Because so many conditions
contribute toward short service life, it is impractical to set forth all
possibilities, but some discussion will help point out the need for
application analysis where trouble is experienced.
Overheating:
Heat is one of the most destructive stresses causing premature motor
failure. Overheating occurs because of motor overloading, low or
unbalanced voltage at the motor terminals, excessive ambient
temperatures, or poor cooling caused by dirt or lack of ventilation. If the
heat is not dissipated, insulation failure and possibly lubrication and
bearing failure can damage a motor.
Ambient Temperature:
For the purposes of standardization, a value of 40°C has been selected as
the standard industrial ambient. This value covers a large range of
practical ambient temperatures encountered throughout the world.
Temperature rise:
This aspect is related to the maximum allowable temperature increase
above the defined ambient that the winding cannot exceed when the
motor is operating at the name plate or rated power output.
6
Rise by Resistance : One method of determining winding temperature
rise (Δt)during motor testing is to measure the electrical resistance
between the leads of each phase of the winding. Each phase consists of a
set of windings (coils) made from a conductor (typically copper) that has
a resistance that is measured in ohms. This value is measured and
recorded prior to initiating a load test to determine temperature rise,
along with the actual ambient temperature, The test is performed by
applying a load equal to the designed power capability of the motor by
dynamometer or other means . Under test conditions, the winding
temperature of a continuous duty motor must achieve steady-state
stabilization while operating under the defined power output or loading
value. The heating developed under loading results in a resistance
change of the winding conductors.
 t  235 

t  Rhot  Rcold  *  1
 Rcold 
where t1  test room temperatur e
Service Factor :NEMA defines maximum allowable temperature rise for
power output conditions greater than the nominal rated conditions. The
commonly defined service factor is 1.15, but other multipliers can be
applied provided the maximum temperature rise is limited to that
specified for the service factor condition. A motor with a 1.15 service
factor is expected to deliver 15 percent more power than the normal full
load condition.
Moisture: Moisture should be kept from entering a motor. Water from
splashing or condensation seriously degrades an insulation system. The
water alone is conducting. The proper type of motor should be chosen for
use in a damp environment.
Contamination: No conducting contaminants such as factory dust and
sand gradually promote over-temperature by restricting cooling air
circulation. In addition, these may erode the insulation and the varnish,
gradually reducing their effectiveness.
Altitude: Standard motor ratings are based on operation at any altitude up
to 3300 feet (1000 meters). High altitude derating is required above 3300
feet because of lower air density.Less dense air reduces the ability for
heat to be dissipated into the air from the surfaces of equipment With
reduced thermal transfer ability, a motor cannot effectively dissipate heat
losses, which require adjustments in the temperature rating, and
corresponding power capability of a given motor as defined in Table
below.
7
Induction motor operating as a generator :
When run faster than its synchronous speed , an induction motor runs as a
generator called induction generator . It convertsthe mechanical energy it
receives into electrical energy and this energy is released by by the stator .
As soon as motor speed exceeds its synchronous speed , it starts delivering
active power ( P) to the three phase line . However , for creating its own
magnetic field ,it absorbs reactive power (Q)from the line to which it is
connected.The reactive power can also be supplied by a group of capacitors
connected across its terminals.If capacitance is insufficent , the generator
voltage will not build up .Induction generator is widely used today in wind
turbines ,which convert wind energy into electrical energy.
Complete torque -speed curve of a three phase induction machine
8
Introduction to Synchronous Machines
Definition:
A synchronous machine is an ac rotating machine whose speed under
steady state condition is proportional to the frequency of the current in its
armature. The magnetic field created by the stator currents rotates at the
synchronous speed ,and that created by the field current on the rotor is
rotating at the synchronous speed also, and a steady torque results. So,
these machines are called synchronous machines because they
operate at constant speeds and constant frequencies under steadystate conditions.
Synchronous machines are commonly used as
generators especially for large power systems, such as turbine generators
and hydroelectric generators in the grid power supply. Because the rotor
speed is equal to the synchronous speed of stator magnetic field,
synchronous motors can be used in situations where constant speed drive is
required. Since the reactive power generated by a synchronous machine can
be adjusted by controlling the magnitude of the rotor field current,
unloaded synchronous machines are also often installed in power systems
for power factor correction or for control of reactive kVA flow. Such
machines, known as synchronous condensers, and may be more
economical in the large sizes than static capacitors.
The bulk of electric power for everyday use is produced by polyphase
synchronous generators( alternators), which are the largest single-unit
electric machines in production. For instance, synchronous generators
with power ratings of several hundred megavolt-amperes (MVA) are
fairly common, and it is expected that machines of several thousand
megavolt- amperes will be in use in the near future. Like most rotating
machines, synchronous machines are capable of operating both as a
motor and as a generator. They are used as motors in constant-speed
drives, and where a variable-speed drive is required ,a synchronous
motor is used with an appropriate frequency changer such as an
inverter. As generators, several synchronous machines often operate
in parallel, as in a power station. While operating in parallel, the
generators share the load with each other; at a given time one of the
generators may not carry any load. In such a case, instead of shutting
down the generator, it is allowed to "float" on the line as a
synchronous motor on no-load.
1
Construction of synchronous machines :
The synchronous machine has 3 phase winding on the stator and a d.c. field winding
on the rotor.
1. Stator :
It is the stationary part of the machine and is built up of sheet-steel laminations
having slots on its inner periphery. A 3-phase winding is placed in these slots. The
armature winding is always connected in star and the neutral is connected to ground.
2. Rotor :
The rotor carries a field winding which is supplied with direct current through two slip
rings by a separate d.c. source. Rotor construction is of two types, namely;
(i) Salient (or projecting) pole type .
(ii) Non-salient (or cylindrical) pole type .
(i) Salient pole type:
In this type, salient or projecting poles are mounted on a large circular steel frame
which is fixed to the shaft of the alternator as shown in Fig. (1). The individual field
pole windings are connected in series in such a way that when the field winding is
energized by the d.c. exciter, adjacent poles have opposite polarities.
Low-speed alternators (120 - 400 r.p.m.) such as those driven by water turbines have
salient pole type rotors due to the following reasons:
(a) The salient field poles would cause .an excessive windage loss if driven at high
speed and would tend to produce noise.
(b) Salient-pole construction cannot be made strong enough to withstand the
mechanical stresses to which they may be subjected at higher speeds.
Since a frequency of 50 Hz is required, we must use a large number of poles on the
rotor of slow-speed alternators. Low-speed rotors always possess a large diameter
to provide the necessary space for the poles. Consequently, salient-pole type rotors
have large diameters and short axial lengths.
Fig. 1 salient pole rotor
2
(ii) Non-salient pole(cylindrical) type :
In this type, the rotor is made of smooth solid forged-steel radial cylinder having a
number of slots along the outer periphery. The field windings are embedded in
these slots and are connected in series to the slip rings through which they are
energized by the d.c. exciter. The regions forming the poles are usually left
unslotted as shown in Fig. (2). It is clear that the poles formed are non-salient i.e.,
they do not project out from the rotor surface.
Fig. 2 cylindrical rotor
High-speed alternators (1500 or 3000 r.p.m.) are driven by steam turbines and
use non-salient type rotors due to the following reasons:
(a) This type of construction has mechanical robustness and gives noiseless
operation at high speeds.
(b) The flux distribution around the periphery is nearly a sine wave and hence
a better e.m.f. waveform is obtained than in the case of salient-pole type.
Since steam turbines run at high speed and a frequency of 50 Hz is required, we
need a small number of poles on the rotor of high-speed alternators (also called
turboalternators). We can use not less than 2 poles and this fixes the highest
possible speed. For a frequency of 50 Hz, it is 3000 r.p.m. The next lower speed is
1500 r.p.m. for a 4-pole machine. Consequently, turboalternators possess 2 or 4
poles and have small diameters and very long axial lengths.
3
Classification of synchronous machines according to the form of excitation:
1. Brushes excitation systems:
The field structure is usually the rotating member of a synchronous machine
and is supplied with a dc-excited winding to produce the magnetic flux. This dc
excitation may be provided by a self-excited dc generator mounted on the same
shaft as the rotor of the synchronous machine. This dc generator is known as
exciter. The direct current generated inside exciter is fed to the synchronous
machine field winding. In slow-speed machines with large ratings, such as
hydroelectric generators, the exciter may not be self-excited. Instead, a pilot
exciter, which may be self-excited or may have a permanent magnet, activates the
main exciter.
Fig 1 Block Diagram of Excitation system by pilot &main
Fig. 1 Brushes Excitation system for a synchronous machine
2.Brushless systems:
This type of excitation has a shaft-mounted bridge rectifier, that rotate with the
rotor, thus avoiding the need for brushes and slip rings.
Fig.2 Brushless Excitation system for a synchronous machine
4
3.static systems:
This type of exctation is widely used in small size alternators, in which portion of
the AC from each phase of synchronous generator is fed back to the field winding
as a DC excitation through a system of transformer ,rectifiers and reactors.
External source of a DC is necessary for initial excitation of the field windings.
Advantages of stationary armature ;
The field winding of an alternator is placed on the rotor and is connected to d.c.
supply through two slip rings. The 3-phase armature winding is placed on the stator.
This arrangement has the following advantages:
(i) It is easier to insulate stationary winding for high voltages , because they are not
subjected to centrifugal forces .
(ii) The stationary 3-phase armature can be directly connected to load without going
through large, unreliable slip rings and brushes.
(iii) Only two slip rings are required for d.c. supply to the field winding on the rotor.
Since the exciting current is small, the slip rings and brush gear required are of light
construction.
(iv) Due to simple and robust construction of the rotor, higher speed of rotating d.c.
field is possible. This increases the output obtainable from a machine of given
dimensions.
Cooling :
Because synchronous machines are often built in extremely large sizes, they
are designed to carry very large currents. A typical armature current density
may be of the order of 10 A/mm 2 in a well-designed machine. Also, the
magnetic loading of the core is such that it reaches saturation in many regions.
The severe electric and magnetic loadings in a synchronous machine produce
heat that must be appropriately dissipated. Thus the manner in which the
active parts of a machine are cooled determines its overall physical structures.
In addition to air, some of the coolants used in synchronous machines include
water, hydrogen, and helium.
5
Damper Bars :
So far we have mentioned only two electrical windings of a synchronous machine:
the three-phase armature winding and the field winding. We also pointed out that,
under steady state, the machine runs at a constant speed, that is, at synchronous
speed. However, like other electric machines, a synchronous machine undergoes
transients during starting and abnormal conditions. During transients, the rotor may
undergo mechanical oscillations and its speed deviates from the synchronous speed,
which is an undesirable phenomenon. To overcome this, an additional set of
windings, resembling the cage of an induction motor, is mounted on the rotor. When
the rotor speed is different from the synchronous speed, currents are induced in
the damper windings .the damper winding acts like the cage rotor of an induction
motor, producing a torque to restore the synchronous speed. Also, the damper bars
provide a means of starting to the synchronous motors, which is otherwise not selfstarting. Fig6 shows the damper bars on a salient rotors
Fig .6 salient pole rotor showing the field winding and damper bars
6
Differences between three phase induction machine and synchronous machine
Three phase induction machine
Stator phases either star or delta
connected
Synchronous machine
Stator phase are star connected only
Rotor windings are not fed by electricity,
currents flow through rotor due to induction
process.
Rotor windings are fed by dc source.
Run below synchronous speed, as a motor
Run above synchronous speed, as a generator
Run at synchronous speed for both motor and
generator
Self starting , as a motor
Need damper bars to start , as a motor
Operate with lagging power factor only
Operate with lagging, leading, and unity power
factor
Simple in construction , rugged, low
maintenance, cheap , especially in squirrel cage
type.
Complex in construction , expensive.
Not active in low speed operation, as a motor
Active in both low and high speed operation
, as a motor
Not active in generating mode
Used widely in generation of electricity
Difficult in speed regulation, as a motor
Precise speed regulation, as a motor
7
THREE PHASE STATOR WINDINGS
The stator windings for alternating-current motors and generators are alike. It should be
noted that direct-current and alternating-current windings differ essentially by the former
being of the closed-circuit type (through commutator) ,while alternating-current windings
are of the open-circuit type.
Types of A-C Windings:
With reference to the arrangements of coils used in three phase stator, windings may be
divided into two general classes as follows:
I. Distributed Windings:
1. Spiral or chain.
2. Lap.
3. Wave.
II. Concentrated Windings:
1. Lap.
2. Wave.
Distributed Windings:
An armature winding which has its conductors of any one phase under a single pole placed
in several slots, is said to be distributed. When these conductors are bunched together in
one slot per pole per phase, the winding is called concentrated. It is usual in a distributed
winding to distribute the series conductors in any phase of the winding among two or more
slots under each pole. A distributed winding has two principal advantages, first, a distributed
winding generates a voltage wave that is nearly a sine curve, secondly, copper is evenly
distributed on the armature surface, therefore, heating is more uniform and this type of
winding is more easily cooled.
Concentrated Windings:
The concentrated winding gives the largest possible emf from a given number of conductors
in the winding. That is for a definite fixed speed and field strength in an alternator, the
concentrated winding requires a less number of conductors than a distributed winding, but
increases the number of turns per coil.
Lap and Wave Windings:
Both distributed and concentrated windings make use of lap and wave connections. These
arrangements are in principle the same as used in direct-current windings. The diagrams of
Figs. 1 and 2 show distributed lap and wave windings .
8
Fig. 1 Single-phase lap winding
Fig 2. Three-phase wave winding , using one slot per pole per phase, star-connected.
Chain Winding :
In this winding as shown in Fig. 3 there is only one coil side in a slot. An odd or even
number of conductors per slot may be used but several shapes of coils are required since the
coils enclose each other. The number of coils required in this winding is also small
compared with other windings. This type of winding is mainly used in alternating current
generators.
Fig. 3 Three-phase chain winding
9
Single layer and double layer winding :
In double layer winding two coil sides are located in each armature slot [Fig. 4 ].
If there is only one coil side located in each armature slot, the winding is called
single layer.
Fig. 4 A double layer winding. One side of the coil lies in the top of the slot and the other side in the
bottom of the of the slot.
Full pitch and fractional pitch windings:
For a three-phase winding the total number of coil sides or the total number of slots should
be just divisible by three (the number of phases) and sometimes by the number of poles.
This will result in a full pitch winding, that is, a winding in which a coil spans exactly the
distance between the centers of adjacent poles. If the coil spans less than this distance, so
that its two sides are not exactly under the centers of adjacent poles at the same time, it is
said to have a fractional-pitch. When a fractional pitch is used, the total number of slots
per phase must be a whole number. The fractional-pitch coils are frequently used in a.c.
machines for two main reasons. First, less copper is required per coil and secondly the
waveform of the generated voltage is improved .The number of stator slots, divided by the
number of poles gives a value of the pole pitch expressed in terms of the slots. Coil pitch
is expressed as a fraction of the pole pitch, in slots, or in electrical degrees. In the case of a
6 pole machine having 72 stator slots, and a double-layer winding, the pole pitch would be
12 slots(72 / 6). If the coil pitch were given as 2/3, this would be 120° ( 2/3 * 180°) or 8
slots (2/3 * 12). A full coil pitch for this winding would be 180 degrees, or 12 slots. A full
pitch winding is one in which the coil pitch is equal to the pole pitch, and a fractional pitch
winding is one in which the coil pitch is not equal to the pole pitch. For the coils used in
small machines round insulated wire is most employed. These coils are either wound in
the slots by hand or assembled by use of specially formed coils wound in forms and
insulated before being placed in the slots. Such formed coils are usually used except in
cases where the slots are closed or nearly closed. For large machines where the amperes to
be carried in each armature circuit is a large value, copper straps are frequently employed
for making up the armature coils. In very large machines a copper bar is used instead of
the copper straps. In such a case one bar serves as a conductor of a coil having one turn per
slot. The copper bars are connected to the end connections of the coils by brazing, welding
or bolting. In all cases, whatever the construction of the coil used, the slots must be
properly insulated with mica, polyester film or other suitable insulating material according
10
to the adequate class of insulation. The coil throw refers to the start of coil from one stator
slot to the finish of this coil at another stator slot . Figures 5 (a) , (b) , (c), and (d) show
four different coils, in each of them the pole pitch is of 12 slots while the coil pitch in (a) is
11 slots , in (b) is 13 slots, in (c) is 9 slots, and in (d) is 15 slots .
Fig. 5. Possible pitch for one coil per slot windings(pole pitch=24 slots/2poles=12) (a) coil throw 1 to 12,
coil pitch=12-1=11 (b) coil throw 1 to 14, coil pitch=14-1=13 (c) coil throw 1 to 10, coil pitch=10-1=9 ,
(d) coil throw 1 to 16, coil pitch=16-1=15 .
Phase belt and phase spread :
A group of adjacent slots belonging to one phase under one pole pair is known
as phase belt. The angle subtended by a phase belt is known as phase spread.
The 3-phase windings are always designed for 60° phase spread . Fig. (6)
shows a 2-pole, 3-phase double-layer, full pitch, distributed winding for the
stator of an alternator. There are 12 slots and each slot contains two coil sides.
The coil sides that are placed in adjacent slots belong to the same phase such
as( a1, a3 ) or (a2, a4 )constitute a phase belt. Since the winding has doublelayer arrangement, one side of a coil, such as (a1), is placed at the bottom of a
slot and the other side (- a1 ) is placed at the top of another slot spaced one pole
pitch apart. Note that each coil has a span of a full pole pitch or 180 electrical
degrees. Therefore. the winding is a full-pitch winding. Note that there are 12
total coils and each phase has four coils. The four coils in each phase are
connected in series so that their voltages aid. The three phases then may be
connected to form Y or -connection. Fig. (7) shows how the coils are
11
connected to form a Y-connection. Any star diagram can be readily changed
into a corresponding delta diagram by opening the star points and connecting
the inner end of phase A to the outer end of phase B, the inner end of phase B
to the outer end of phase C , and the inner end of phase C to the outer end of
phase A .
Fig. 6
Fig. 7
Simple rule for checking proper phase relationship in a three-phase winding:
A fundamental consideration when checking the instantaneous flow of current
in a three phase circuit, is to imagine that when the current flows in the same
direction in two legs of the circuit, it flows in the opposite direction in the third
leg. This principle can be applied to both motors and generators. In Figure 8, it
must be supposed that current flows in all three leads of the star connection
toward the point of the star connection. And that in the case of a delta
connection the current flows around the three sides of the delta in the same
direction. Then in either case for a three-phase winding, the polarity of each of
the pole-phase groups will alternate regularly around the winding and can be
indicated by arrows as in Fig. 8. By the use of this scheme there is no chance
12
for a reversal of a phase to be passed by not noticed when checking the
winding.
.
Fig. 8 Simple scheme of alternately reversing arrows of pole-phase groups to check correct phase
polarity of a 3-phase winding. It is supposed in this case that current flows in the three leads
toward the star points of the winding which are indicated thus ( *) .
Why the alternators have their windings connected in star?
Because the star connection has the following advantages over delta connection:
(i) To obtain a desired voltage between outside terminals of a generator, the voltage
built up in any one phase need be only 58 percent(1 / √3) of the terminal value.
Hence only 58 percent of the turns required for a Δ-connected armature are
necessary with a consequent lowering of insulation cost.
(ii) A star connected winding offers the advantage of a fourth or neutral lead
making possible the advantages of a four -wire system, with or without grounded
neutral.
(iii)The wave shape of a star-connected winding is improved, owing to the
elimination of third harmonics and multiples of third harmonics from the
terminal voltage. Fig.9 shows a star connected armature( stator), The terminal
voltages E12 , E23 and,E 31 are 120 electrical degrees apart. In the third harmonics
the e.m.f.'s of three phases (e01 ,e02 ,e03) are being in phase( 3 X 120º =
360º= 0º ),to obtain terminal voltages, the resultant is zero. Hence no third
harmonics appears between terminals.The circulating currents from the third
harmonics cause unnecessary losses and dangerous heating in Δ-connected
alternator, where it is not in the Y-connected alternator case. In addition the use of
13
a(5/6)th pitch winding in three-phase Y-connected generators reduces the fifth and
seventh harmonics, if pesent, to almost nil, so the lowest harmonics that can be
present is eleventh.
Fig. 9
14
Windings factors:
The stator winding of synchronous machine is distributed over the entire stator.
The distributed winding produces nearly a sine waveform and the heating is
more uniform. Likewise, the coils of armature winding are not full-pitched i.e.,
the two sides of a coil are not at corresponding points under adjacent poles. The
fractional pitched armature winding requires less copper per coil and at the
same time waveform of output voltage is improved. The distribution and
pitching of the coils affect the voltages induced in the coils. We shall discuss
two winding factors:
(i) Distribution factor (Kd).
(ii) Pitch factor (Kp).
(i) Distribution factor (Kd) :
A winding with only one slot per pole per phase is called a concentrated
winding. In this type of winding, the e.m.f. generated/phase is equal to the
arithmetic sum of the individual coil e.m.f.s in that phase. However, if the
coils/phase are distributed over several slots in space (distributed winding), the
e.m.f.s in the coils are not in phase (i.e., phase difference is not zero) but are
displaced from each by the slot angle (The angular displacement in electrical
agrees between the adjacent slots is called slot angle). The e.m.f./phase will be
the phasor sum of coil e.m.f.s. The distribution factor Kd is defined as:
The distribution factor can be determined by constructing a phasor diagram for
the coil e.m.f.s. Let m = 3. The three coil e.m.f.s are shown as phasors AB, BC
and CD [See Fig. 10 (i)] each of which is a chord of circle with centre at O and
subtends an angle at O. The phasor sum of the coil e.m.f.s subtends an angle
15
m(Here m = 3) at O. Draw perpendicular bisectors of each chord such as Ox,
Oy etc [See Fig. 10 (ii)].
Note that ( m is the phase spread.
Fig. 10
(ii) Pitch factor (Kp) :
A coil whose sides are separated by one pole pitch (i.e., coil span is 180electrical) is
called a full-pitch coil. With a full-pitch coil, the e.m.f.s induced in the two coil sides a in
phase with each other and the resultant e.m.f. is the arithmetic sum of individual e.m.fs.
However the waveform of the resultant e.m.f. can be improved by making the coil pitch
less than a pole pitch. Such a coil is called short-pitch coil. This practice is only possible
with double-layer type of winding The e.m.f. induced in a short-pitch coil is less than that
of a fullpitch coil. The factor by which e.m.f. per coil is reduced is called pitch factor K p. It
is defined as:
Consider a coil AB which is short-pitch by an angle Өelectrical degrees as shown in Fig.
(11). The e.m.f.s generated in the coil sides A and B differ in phase by an angle Өand can
16
be represented by phasors EA and EB respectively as shown in Fig. (12). The diagonal of the
parallelogram represents the resultant e.m.f. ER of the coil.
Fig. 11
Fig. 12
For a full-pitch winding, Kp = 1. However, for a short-pitch winding, Kp < 1.
Note that Өis always an integer multiple of the slot angle .
17
SYNCHRONOUS
GENERATOR
The machine which produces 3-phase electrical power from mechanical power is called
an alternator or synchronous generator. Alternators are the primary source of all the
electrical energy we consume. These machines are the largest energy converters found
in the world.
Alternator Operation :
Like the dc generator, a synchronous generator functions on the basis of Faraday's
law, which state if the flux linking the coil changes in time, a voltage is induced
in a coil. Stated in another form, a voltage is induced in a conductor if it cuts
magnetic flux lines. The rotor winding is energized from the d.c. exciter and alternate N
and S poles are developed on the rotor. When the rotor is rotated by a prime mover, the
stator or armature conductors are cut by the magnetic flux of rotor poles. Consequently,
e.m.f. is induced in the armature conductors due to electromagnetic induction. The
induced e.m.f. is alternating since N and S poles of rotor alternately pass the armature
conductors. The frequency of induced e.m.f. is given by;
Where
N = speed of rotor in r.p.m. ,
P = number of rotor poles.
The magnitude of the voltage induced in each phase depends upon the rotor flux, the
number and position of the conductors in the phase and the speed of the rotor.
[Fig. 1 (i)] shows star-connected armature winding and d.c. field winding. When the rotor
is rotated, a 3-phase voltage is induced in the armature winding.. The magnitude of e.m.f.
in each phase of the armature winding is the same. However, they differ in phase by 120°
electrical as shown in the phasor diagram [ Fig. 1 (ii)].
Fig. 1
18
Considering the round rotor synchronous generator in Fig.2,which shows a simple
consentrated stator winding ( 1 slot / pole / phase).
Fig .2 (a) A round rotor synchronous generator
(b) Air gap flux density distribution produced by the rotor excitation
and assuming that the flux density in the air gap is uniform implies that sinu soidally varying voltages will be induced in the three coils aa' ,bb' and cc' if the rotor,
rotates at a synchronous speed, Ns. Also if Φр is the flux per pole, ω is the angular
frequency, and T is the number of turns in phase a (coil aa'), then the voltage
induced in phase a is given as:
ea = ω T Φр sin ωt
= Em sin ωt
Where
E m = ω T Φр , and
ω =2 π f .
Because phases b and c are displaced from phase a by +_ 120º , the corresponding
voltages may be written as
eb = Em sin (ωt – 120º)
ec = Em sin (ωt + 120º)
These voltages are sketched in Fig. 3 and correspond to the voltages from a threephase generator.
19
Fig. 3 A three-phase voltage produced by a three-phase synchronous generator
Generated E.M.F. Equations :
The magnetic flux cut by one armature conductor, when the rotor of an alternator
is made to revolve through one revolution, is Φр * P, where Φр is the magnetic flux
per pole and P is the number of poles. If N is the speed in revolution per minute,
then the flux cut per second becomes
Φр * P *( N /60)
Since 1 volt is generated when 1 weber of flux is cut per second, the average
voltage generated in this conductor becomes
Eav = Φр * P *( N /60) volts
If the total number of conductors on the armature is Z and they are connected into
A parallel paths, the average voltage between terminals becomes
Eav per conductor * ( Z / A )
In case of alternating current generators the e.m.f. depends not only upon the total
flux cut per second but also upon the way in which the flux and the conductors
are distributed. A change in distribution of flux changes the relative values of the
maximum and effective (r.m.s.) e.m f.'s.
In addition, the e.m.f. built up in any one conductor, when considered
vectorially, cannot always be added directly to that of another as there may be a
phase displacement between them. The instantaneous values, though can be
added algebraically, but in adding the effective values it is necessary to consider
the phase differences between the different e.m.f.'s to be added. In order to take
these factors in account, the flux distribution and winding types must be known.
Fig.4(a) shows the space distribution of the airgap flux of an alternator. In this
case, the flux density B is assumed to be sinusoidal in space when measured
around the inside periphery of the stator. This flux density B can be expressed as
B=Bmax sinθ.
where θ is measured from the position midway between the poles ; B is the flux density
measured in webers per length of the field pole (L ) as shown in Fig.4(b) per length of
pole pitch arc ; and Bmax is the maximum flux density produced by a pole.
20
The total flux per pole is
Φp = ο ∫π LB max sin θ dθ = -| LB max cos θ |ο π =2 LB max Weber
...(i)
As this flux wave is moved around the air gap the conductors a and b of the stator coil
will have voltages induced in them.
Fig 4 Shapes of flux density waves
The voltage induced in conductor a will be
e a =BL v
...(ii)
where v is the velocity of the flux wave in radians per second, or
v = 2 π f , f being the frequency of the flux wave or e.m.f. induced.
The voltage in conductor a is
ea =B max L 2π f sin θ
But θ = ω t where ω is the angular velocity of the rotor in radians per second and t is the
time of rotation from the position shown in Fig.4(a). Therefore,
e a = B max L 2π f sin ω t
...(iii)
Substituting equation (i) in equation (iii)
e a = (Φp /2 ) *2π f sin ω t
...(iv)
Likewise, the voltage induced in conductor b
e b = (Φp /2 ) *2π f sin ω t
...(v)
21
and the voltage at the terminals of the turn made up of the conductors a and b is
e turn = Φp 2π f sin ω t
...(vi)
If the single turn on the armature (stator) were replaced by a coil of T turns, the voltage
per coil would be
e eoil = Φp 2π f T sin ω t
The effective value of generated e.m.f.
...(vii)
Er.m.s = Emax / √2 = (2π / √2) f Φp T
volts (r.m.s)
Er.m.s =4.44 f Φp T
volts (r.m.s)
...(viii)
...(ix)
As the r.m.s. value is related to the average value so,
Form factor (k f ) = (r.m.s. value) / (average value)
* For sinusoidal wave of e.m.f., k f = 1.11
E r . m. s . = 4 k f T f Φp
volts
Also, since the induced e.m.f.'s in the conductors of an alternator are not all in phase,
the above relation for Er.m.s. must be multiplied by k p and k d , the pitch factor and the
distribution (breadth) factor.
Therefore,
Er.m.s. =4 k f k p k d f T Φp
volts per phase
...(x)
* For full pitched and concentrated windings kd =1 and kP =1.
If the alternator is star connected, and neglecting the effect of armature reaction, the
alternator line voltage is,
EL = √ 3 (Er.m.s. per phase)
, so
EL = √ 3( 4 k f k p k d f T Φp)
volts.
22
Synchronous Generator Characteristics :
( i ) Magnetisation curve:
A plot of the exciting current versus terminal voltage of alternator is known as
the magnetisation curve. This magnetisation curve is obtained by passing different
values of currents in exciting windings, thereby giving, correspondingly
different values of terminal voltage. The no load magnetisation curve is
shown in Fig.5-I] ,which has the same general shape as B-H curve of armature
steel.
The full load magnetisation characteristics with unity power factor and with
0.8 lagging power factor have been shown at ( II ) and ( III ) in Fig.5.
Fig. 5 Magnetization Curve
Fig. 5 Magnetization curves for alternator at different loading conditions
( ii ) Load characteristics :
If the speed and exciting current remain constant, the terminal voltage of the
alternator changes with the load currents. The plot between the terminal voltage
and the load (or armature) current of an alternator is known as load
characteristics. An increase in the armature (or load) currents make the terminal
voltage drops. This has been shown in Fig. 6. The drop in terminal voltage is attributed
to many reasons but primarily because of the following:
( a ) Resistance and reactance of the armature (or stator) winding.
( b ) Armature reaction.
The resistance and reactance of the armature winding causes the drop in
generated e.m.f. (voltage) ,whereas the armature reaction weakens the magnetic
field and thereby decreases the generated e.m.f. (voltage).
23
The magnitude of the effect of armature reaction depends upon, the power factor
of load i.e. angle of lag or lead of the stator (armature) current. In case of a
unity power factor of load, each phase of alternator when connected to the
load takes a current which is in phase with its generated voltage. But the
magnetic field is strengthened, instead of weakening, if the load (or armature)
current, is leading. In the above case, when the power factor is leading, the drop
in voltage due to resistance and reactance of stator winding may be less than the
increase in voltage due to armature reaction. Thus the terminal voltage on
load may be more than that at no-load, if the angle of lead of the load (or
armature) current is sufficient.
Fig. 6. Load Characteristic of alternator
( iii ) Effect of variation of power factor on terminal voltage :
The load characteristics at different power factors with leading and lagging
armature currents are shown in Fig. 7. If the load, current and excitation are kept
constant, the terminal voltage falls on changing the power factor from leading
to lagging one.This effect is because of the armature flux helping the main flux,
in case p.f. is leading, hence generating more e.m,f. and the armature flux,
opposing the main flux, in case the p.f. is lagging, hence generating less e.m.f.,
Therefore, the terminal voltage at lagging power factor decreases from that on
leading p.f. because of decrease in generated e.m.f.
24
Fig. 7
Armature Leakage & Synchronous Reactances :
When the current pass through the stator conductors the flux is set up, and a
portion of this flux does not cross the air gap but completes the path inside the
stator as shown in Fig. 8. This flux is known as leakage flux,which sets up an
e.m.f. in the stator winding .This e.m.f. leads the load current by 90° and
proportional to the magnitude of load current. This e.m.f. is due to the
leakage inductance of the armature winding. The magnitude of the leakage
inductance in practical units,henrys, is given by the general equation :
L = (Flux in webers per amperes) * (No. turns) henrys
And leakage reactance per phase,
Xl = ω L = 2 π f L ohms/ph
Xl causes a voltage drop in alternator terminal voltage and this drop is equal
to an e.m.f. set up by the leakage flux. Also, there is another source causes
voltage drop , that is due to armature reaction which can be represented by
a fictitious reactance X a .
Xl + X a = Xs
Where Xs is the per phase synchronous reactance of armature winding.
Fig 8
25
Performance Of A Round-Rotor Synchronous Generator :
At the outset we wish to point out that we will study the machine on a per phase basis,
implying a balanced operation. Thus let us consider a round-rotor machine operating as
a generator on no-load. Variation of the terminal voltage with the exciting current
(field current) is shown in Fig. 5-I , and is known as the open-circuit characteristic
of a synchronous generator. Let the open-circuit phase voltage be Eo for a certain
field current If. Here Eo is the internal voltage of the generator. We assume that If is
such that the machine is operating under unsaturated condition. Next we short-circuit
the armature at the terminals, keeping the field current unchanged (at If), and
measure the armature phase current Ia In this case, the entire internal voltage Eo is
dropped across the internal impedance of the machine. In mathematical terms,
Eo = IaZs
and Zs is known as the synchronous impedance , which is equal to
Zs= R a + j XS
If the generator operates at a terminal voltage Vt, while supplying a load corresponding
to an armature current Ia, then
Eo = Vt + Ia (R a + j XS)
In an actual synchronous machine, except in very small ones, we almost always
have XS >> R a , in which case Zs ≈ j X S. Among the steady-state characteristics of a
synchronous generator, its voltage regulation and power-angle characteristics are the
most important ones.
We define the voltage regulation of a synchronous generator at a given load as
percent voltage regulation = (Eo - Vt) / Vt * 100 %
where Vt is the terminal voltage on load and Eo is the no-load terminal voltage.
The voltage regulation is dependent on the power factor of the load. the voltage
regulation for a synchronous generator may even become negative. The angle
between Eo and Vt is defined as the power angle, δ. Notice that the power angle, δ, is
not the same as the power factor angle, φ. To justify this definition, we consider
Fig. 9, from which we obtain
I a XS cos φ = Eo sin δ
26
…(i)
Fig. 9 Phasor diagram of round rotor generator
Now, from the approximate equivalent circuit (assuming that XS >> R a )as shown
in Fig 10-a,
the power delivered by the generator = power developed, Pd = Vt Ia cos φ ,
which follows from Fig. 9 also. Hence, in conjunction with the (i) equation, we get
Pd = (Eo Vt / XS ) sin δ
per phase
…(ii)
Pd = 3(Eo Vt / XS ) sin δ
for three phases
Which shows that the internal power of the machine is porportional to sin δ,
Equation (ii) is often said to represent the power-angle characteristic of a round
rotor synchronous machine.
Fig. 10-b shows that for a negative δ, the machine will operate as a motor.
Fig 10 : (a) an approximate equivalent circuit of synchronous machine
(b) power-angle characteristics of a round-rotor synchronous machine
27
Performance Of A Salient-Pole Synchronous Generator :
Because of saliency, the reactance measured at the terminals of a salient-rotor machine
will vary as a function of the rotor position. This is not so in a round-rotor machine.
Thus a simple definition of the synchronous reactance for a salient-rotor machine is
not immediately forthcoming. To overcome this difficulty, we use the two-reaction
theory proposed by "Andre Blondel". The theory proposes to resolve the given
armature mmf's into two mutually perpendicular components, with one located along
the axis of the rotor salient pole, known as the direct (or d )axis and with the other in
quadrature and known as the quadrature (or q )axis. Correspondingly, we may
define the d-axis and q-axis synchronous reactances, Xd and Xq for a salient-pole
synchronous machine. Thus, for generator operation, we draw the phasor diagram of
Fig. 11. Notice that Ia has been resolved into its d- and q-axis (fictitious) components,
Id and Iq With the help of this phasor diagram, we obtain
Id = Ia sin ( δ + φ )
Iq = Ia cos ( δ + φ )
Vt sin δ = Iq Xq = Ia Xq cos ( δ + φ )
From these we get
Vt sin δ = Ia Xq cos δ cos φ - Ia Xq sin δ sin φ
Or
(Vt + Ia Xq sin φ) sin δ = Ia Xq cos δ cos φ
Dividing both sides by cos δ and solving for tan δ yields
tan δ = (Ia Xq cos φ) / (Vt + Ia Xq sin φ)
…(iii)
With δ known (in term of φ), the voltage regulation may be computed from
Eo = Vt cos δ + Id Xd
Percent regulation = (Eo - Vt) / Vt *100%
Fig. 9 Phasor Diagram of a Salient-Pole Generator
Fig 11 Phasor diagram of salient-pole generator
28
In fact, the phasor diagram depicts the complete performance characteristics of the
machine.
Let us now use Fig. 11 to drive the power-angle characteristics of a salient-pole
generator. If armature resistance is neglected, Pd = Vt Ia cos φ, Now, from Fig. 11, the
projection of Ia on Vt is
Pd / Vt = Ia cos φ = Iq cos δ + Id sin δ
Solving
Iq Xq = Vt sin δ
and
Id Xd = Eo - Vt cos δ
For Iq and Id , and substituting in (i), gives
Pd = (Eo Vt / Xd ) sin δ + ( Vt2 / 2 ) [ 1/ Xq – 1/ Xd ] sin 2δ
Pd = 3 (Eo Vt / Xd ) sin δ + 3 ( Vt2 / 2 ) [ 1/ Xq – 1/ Xd ] sin 2δ
per phase ...(iv)
for three phases
The equation can also be established for a salient-pole motor (δ<0), the graph of above
equation is given in Fig. 12. Observe that for Xd = Xq = XS, reduces to the round-rotor
equation.
Fig. 12 Power-angle characteristic of salient-pole machine
29
Parallel Operation of Synchronous Generators
An electric power station often has several synchronous generators operating in
parallel with each other. Some of the advantages of parallel operation are :
1. In the absence of the several machines, for maintenance or some other reason, the
power station can function with the remaining units.
2. Depending on the load, generators may be brought on line, or taken off, and thus
result in the most efficient and economical operation of the station.
3. For future expansion, units may be added on and operate in parallel.
In order that a synchronous generator may be connected in parallel with a system
(or bus), the following conditions must be fulfilled:
1.
The frequency of the incoming generator must be the same as the frequency
of the power system to which the generator is to be connected.
2.
The magnitude of the voltage of the incoming generator must be the same
as the system terminal voltage.
3.
With respect to an external circuit, the voltage of the incoming generator
must be in the same phase as system voltage at the terminals.
4.
In a three-phase system, the generator must have the same phase sequence
as that of the bus.
The process of properly connecting a synchronous generator in parallel with a
system is known as synchronizing. Tow generators can be synchronized either by
using a synchroscope or lamps. Figure 1. shows a circuit diagram showing lamps as
well as synchroscope. The potential transformers (PTs) are used to reduce the
voltage for instrumentation. Let the generator G1 be already in operation with its
switch Sg1 closed. Other switches Sg2, S1, and S2 are all open.
After the generator G2 is started and brought up to approximately synchronous
speed, S2 is closed. Subsequently, the lamps La , Lb ,and Lc begin to flicker at a
frequency equal to the difference of the frequencies of G1 and G2. The equality
of the voltages of the two generators is ascertained by the voltmeter V, connected by
the double-pole double-throw switch S. Now, if the voltages and frequencies of the
two generators are the same, but there is a phase difference between the two
30
voltages, the lamps will glow steadily. The speed of G2 is then slowly adjusted
until the lamps remain permanently dark (because they are connected such that
two voltages through them are in opposition). Next, Sg2 is closed and S2 may be
opened.
Fig 1. Synchronizing Two Generators
In the discussion above, it has been assumed that G1 and G2 both have the same
phase rotation. On the other hand, let the phase sequence of G1 be abc
counterclockwise and that of G2 be a'b'c' clockwise. At the synchronous speed of
G1, a and a' may be coincident. This will be indicated by a dark La, but Lb and
Lc will have equal brightness,the phase rotation of G2 must be reversed. When
G2 runs at a speed slightly less than the synchronous speed, with reverse phase
sequence with respect to G1 ,the lamps will be dark and bright in the cyclical
order La,Lb and Lc, the phase rotation of G2 must be reversed with increasing of its
speed to synchronous speed. This process of testing the phase sequence is known as
phasing out.
31
A synchroscope is often used to synchronize two generators which have Previously
been phased out. A synchroscope is an instrument having a rotating pointer,
which indicates whether the incoming machine is slow or fast. One type of
synchroscope is shown schematically in Fig. 2. It consists of a field coil, F,
connected to the main busbars through a large resistance Rf to ensure that the
field current is almost in phase with the busbar voltage, V. The rotor consists of
two windings R and X, in space quadrature, connected in parallel to each other and
across the incoming generator.
Fig 2 A Synchroscope
The windings R and X are so designed that their respective currents are
approximately in phase and 90° behind the terminal voltage, Vi, of the incoming
generator. The rotor will align itself so that the axes of R and F are inclined at an
angle equal to the phase displacement between V and Vi. If there is a difference
between the frequencies of V and Vi, the pointer will rotate at a speed proportional
to this difference. The direction of rotation of the pointer will determine if the
incoming generator is running below or above synchronism. At synchronism, the
pointer will remain stationary at the index. In present-day power stations,
automatic synchronizers are used.
32
Circulating Current and Load Sharing
At the time of synchronizing (that is, when S2 of Fig.1 is closed), if G2 is
running at a speed slightly less than that of G1 the phase relationships of their
terminal voltages with respect to the local circuit are as shown in Fig.3(a). The
resultant voltage Vc acts in the local circuit to set up a circulating current Ic
lagging Vc by a phase angle φc For simplification, if we assume the generators to be
identical, then
tan φc = Ra / Xs
Ic = Vc / 2Zs
Where Ra + J Xs = synchronous impedance , Ra = armature resistance , Xs =
synchronous reactance.
Fig 3 Circulating Currents between Tow Generators
33
Notice from Fig. 3(a) that Ic has a component in phase with V1, and thus acts as a
load on G1 and tends to slow it down. The component of Ic in phase opposition to V2
aids G2 to operate as a motor and thereby G2 picks up speed. On the other hand, if
G2 was running faster than G1 at the instant of synchronization, the phase
relationships of the voltages and the circulating current become as shown in Fig.
3(b). Consequently, G2 will function as a generator and will tend to slow down;
and while acting as a motor, G1 will pick up speed. Thus there is an inherent
synchronizing action which aids the machines to stay in synchronism.
We now recall the power developed by a synchronous machine that Vt is the
terminal voltage, which is the same as the system busbar voltage. The voltage E
is the internal voltage of the generator and is determined by the field excitation.
As we have discussed earlier, a change in the field excitation merely controls the
power factor and the circulation current at which the synchronous machine operates.
The power developed by the machine depends on the power angle δ. For G2 to share
the load, for a given Vt and E the power angle must be increased by increasing the
prime-mover power. The load sharing between two synchronous generators is
illustrated by the following examples.
Example 1:
Two identical three-phase wye-connected synchronous generators share equally a
load of 10 MW at 33 kV and 0.8 lagging power factor. The synchronous
reactance of each machine is 6 Ω per phase and the armature resistance is
negligible.If one of the machines has its field excitation adjusted to carry 125 A
of lagging current, what is the current supplied by the second machine ? The
prime mover inputs to both machines are equal.
SOLUTION
The phasor diagram of current division is shown in Fig.4, where in I 1 = 125 A.
Because the machines are identical and the prime-mover inputs to both machines are
equal, each machine supplies the same true power:
Fig 4
34
I1 cos φ1= I2 cos φ2 = 0.5 I cos φ
I = 10 * 106 / (√3 * 33*103 * 0.8) = 218.7 A
I1 cos φ1= I2 cos φ2 = 0.5 * 218.7 * 0.8 = 87.5 A
The reactive current of the first machine is therefore
I1 | sin φ1 | = √ (1252 – 87.52) = 89.3 A
And since the total reactive current is
I | sin φ | = 218.7 * 0.6 = 131.2 A
The reactive current of the second machine is
I2 | sin φ2 | = 131.2 – 89.3 = 41.9 A
Hence
I2 = √ (87.52 + 41.92) = 97 A
Example 2:
Consider the tow machines of example 1. if the power factor of the first machine is
0.9 lagging and the load is shared equally by the tow machines, what are the power
factor and current of the second machine?
SOLUTION
Load:
Power = 10,000 KW, Apparent power =12,500 KVA, Reactive power = 7500 KVar
First machine:
Power = 5000 KW
φ1 = cos-1 0.9 = –25.8º
Reactive power = 5000 tan φ1 = –2422 KVar
Second machine:
Power = 5000 KW
Reactive power = –7500 – (–2422) = – 5078 KVar
tan φ2 = – 5078 / 5000 = – 1.02
cos φ2 = 0.7
I2 = 5000 / (√3 * 33 * 0.7) = 124.7 A
35
Determination of Synchronous Reactance from Open
Circuit and Short Circut Tests
i- Open Circuit Test:
The machine is run on no load and the induced
e.m.f. per phase is measured corresponding to various values of field
current and a curve between induced e.m.f. per phase, Eo and field current, If is
drawn which is known as open circuit characteristic (O.C.C.) and has been
illustrated in Fig 5.
Fig 5 Open Circuit Characteristic
ii- Short Circuit Test: The armature winding
is short-circuited through a
low resistance ammeter. The speed is kept constant during this test and
short-circuit current is measured corresponding to various values of field
current. The field current or excitation is increased to give short-circuit
current about twice the full load current. The short circuit characteristic
(S.C.C.) is drawn by plotting a curve between short circuit current Isc as
ordinate and field current, If as abscissa, as shown in Fig. 5.
Consider OC the normal field current, then BC gives short circuit current,
Isc corresponding to this value of field current on the S.C.C. and AB gives the
induced e.m.f. per phase on the O.C.C. for the same excitation. Since on
short circuit for field current OC,the whole of the induced e.m.f. AC is
utilized to create a short circuit current, Isc given by BC.
Hence synchronous Impedance,
Zs = AC (in volts) / BC (in amperes)
And synchronous reactance,
Xs = √ ( Zs2 – Ra2)
36
Where Ra is the effective armature resistance per phase which can be measured
directly by volt-meter and ammeter method or by using the wheat-stone bridge.
For normal working conditions the armature resistance measured so is increased by
60% or so. This is being done to allow for skin effect and thus effective armature
resistance Ra is obtained.
Measurement of Xd and Xq
The d-axis synchronous reactance is determined from O.C. and S.C. tests, the q-axis
synchronous reactance can be measured by many methods, one of these methods is
the slip test method.
Slip test method:
From this test, the value of Xd and Xq can be determined. The synchronous machine
is driven by a separate prime-mover (or motor) at a speed slightly different from
synchronous speed. The field windings are left open and positive sequence balanced
voltages of reduced magnitude (around 25% of rated value) and of rated frequency
are impressed across the armature terminals. Under these conditions, the
relative velocity between the field poles and the rotating armature m.m f.
waves is equal to the difference between synchronous speed and the rotor
speed,i.e. the slip speed. A sma ll A.C.voltage across the open field
winding indicates that the field poles and rotating m.m.f. wave, are revolving
in the same direction,and this is what is required in slip test. if field poles
revolve in a direction opposite to the rotating m.m.f. wave, negative sequence
reactance would be measured.
At one instant, when the peak of armature m.m.f. wave is in line with the field
pole or direct axis,the reluctance offered by the small air gap is minimum as shown
in fig. 6 (a). At this instant the impressed terminal voltage per phase divided by
the corresponding armature current per phase, gives d -axis synchronous
reactance Xd.
After one-quarter of slip cycle, the peak of armature m.m.f. wave acts on the
interpolar or q-axis of the magnetic circuit, Fig. 6 (b). and the reluctance
offered by long air-gap is maximum. At this instant, the ratio of armature
terminal voltage per phase to the corresponding armature current per
phase, gives q-axis synchronous react ance Xq. Oscillograms of armature
current, terminal voltage and the e.m.f. induced in the open-circuited field
winding are shown in Fig.7. A much larger slip than would be used in practice,
has been shown in Fig.7, merely for the sake of clarity.
37
Fig. 6 pertaining to the physical concepts of (a) Xd , (b) Xq
Fig. 7 Typical oscillograms in slip test
When the armature m.m.f. wave is along the direct axis, the armature flux
passing through open field winding is maximum, therefore, the induced field
e.m.f., i.e. dфa/dt is zero. The d-axis can, therefore, be located on the
oscillogram of Fig.7 where the induced field e.m.f. is zero. When armature
m.m.f. wave is along q-axis, the armature flux linking the field winding is zero,
therefore, the induced field e.m.f. dфa/dt is maximum. Thus the q-axis can also be
located on the oscillogram. If oscillograms can't be taken, then an ammeter
and a voltmeter are used as shown in the connection diagram of Fig.8. The primemover (or d.c. motor) speed is adjusted till ammeter and voltmeter pointers
38
swing slowly between maximum and minimum positions. Under this
condition, maximum and minimum readings of both ammeter and voltmeter are
recorded in order to determined Xd and Xq.
Fig 8 Slip-test connection diagram for obtaining Xd and Xq
Since the applied voltage is constant, the air-gap flux would be constant. When
crest of the rotating m.m.f. wave is in line with the field-pole axis, Fig. 6 (a),
minimum air-gap offers minimum reluctance, consequently the armature
current, required for the establishment of constant air-gap flux, must be
minimum. Constant applied voltage minus the minimum impedance voltage drop
(armature current being minimum) in the leads and 3-phase variac gives
maximum armature-terminal voltage. Thus the d-axis, synchronous reactance is
given by
Xd = (Max. armature terminal voltage per phase) / (Min. armature current per phase)
By a similar thought process
Xq = (Min. armature terminal voltage per phase) / (Max. armature current per phase)
During slip test, it would be observed that swing of the ammeter pointer is
very wide, whereas the voltmeter has only small swing because of the low
impedance voltage drop in the leads and 3-phase variac. Since low armatureterminal voltages are used, values of reactances obtained are unsaturated values.
When performing this test, the slip should be made as small as possible,
otherwise the currents induced in the amortisseur circuits would cause large
errors in the measurement of Xd and Xq (lower value of reactances for larger
slips). It is however quite difficult to maintain very small slips, as the
reluctance torque due to saliency tends to bring the rotor into synchronism with
the rotating armature m.m.f. wave. It is because of this reason that the slip test
must be conducted at low values of armature terminal voltage so that reluctance
torque due to saliency is low.
39
The advantages of oscillographic method over voltmeter-ammeter methods are
i-
elimination of the inertia effects of voltmeter and ammeter
ii-
the possibility of large slip-speed, which in turn allows higher armatureterminal voltages to be applied.
In practice, there may be error in reading the oscillograms. At the same time.
voltmeter ammeter readings are not very reliable because of their inertia effect. In
view of these shortcomings, slip test is conducted only to determine the ratio of
Xd / Xq.
Now, using the value of Xd computed from o.c. and s.c. tests, Xq can be determined as follow :
Xq = Xd / Xq (from slip test) * Xd (from O.C. and S.C. tests)
40
Voltage Regulation of Alternator
From the previous discussion, it has been known that, the terminal voltage
of an alternator changes from no load to full load.
The voltage regulation of an alternator is the difference between the no-load
voltage and the full-load voltage exprsed in per cent of full load voltage.
i.e. percentage regulation =( V no l oad – V f ul l l oad ) / V f ul l l oad * 100%
= (Eo – Vt) / Vt *100%
Where Eo is no load terminal voltage, Vt is the full load rated terminal
voltage
In case of leading power factor, Eo is less than Vt hence the regulation will be
negative.
The regulation of an alternator is usually much higher than that of power
transformers. This large regulation results from:
 the large amount of leakage reactance present in the alternator.
 the armature resistance.
 effect of armature reaction (this is the most predominant factor).
To make the generator voltage constant, automatic control on the field
current (excitation) is necessary. A change in lo ad will cause readjustment
of excitation in sufficient magnitude so as to provide the constant-potential
characteristic. The device used to control the field current (excitation) is
called a voltage regulator. The regulator changes the alternator from a variablevoltage machine to a constant- voltage machine.
1. Determination of Voltage Regulation by Synchronous
impedance method or E.M.F. method:
The magnitude of voltage regulation depends upon the load current and power
factor of the load. The vector diagram for any load at any power factor can
be drawn. This has been illustrated in Fig. 1.
Let I be the load current lagging behind the terminal voltage Vt by a phase angle φ.
 No load terminal voltage, Eo = OF
 Full load terminal voltage, Vt = OA
 Armature resistance drop, I R a= AC, this drop is always in phase with the
load current I.
 Synchronous impedance drop, I XS = CF, this drop is always perpendicular to
the load current and also the resistance drop.
41
Fig. 1 Vector Diagram of Alternator
Now
 OF2 = OD2 + DF2 = ( OB + BD )2 + ( DC + CF )2
Or
 Eo2 = ( Vt cos φ + I R a)2 + ( Vt sin φ + I XS)2
Since
 AC = BD = I R a
And AB = CD = Vt sin φ
Then
 Eo = √ [( Vt cos φ + I R a)2 + ( Vt sin φ + I XS)2]
 percentage regulation = (Eo – Vt) / Vt *100%
For leading power factor of load, the phase angle φ will be taken as
negative and the value Eo will be even less than Vt at large magnitude of
leading power factor and the percentage regulation will be negative. For
unity power factor taking phase angle φ as zero, the load current will be in
phase, with the terminal voltage, Vt .These cases have been illustrated in Fig. 2(a)
and 2 (b).
(a) Unity power factor of load
(b) Leading power factor of load
Fig 2
The regulation obtained by this method is always higher than the actual
values and is therefore a pessimistic method. However, the results are more
on the safe side and the method is theoretically accurate for non-salient pole
42
machines with distributed field windings when saturation is not considered.
The value of synchronous reactance varies with the saturation. At low
saturation its value is larger because of the effect of armature reaction is
greater than that higher saturation. Under short circuit conditions, saturation
is very low and the value of synchronous impedance (or reactance) measured is
higher than that in actual working conditions.
2. Zero power factor method:
This is also called the general method, Potier reactance (or triangle) method
of obtaining the voltage regulation.
In the e.m.f. method. the phasor diagram involving voltages is used, For the z.p.f
method, the e.m.fs. are handled as voltages and the m.m.fs. as field ampere-turns
or field amperes.
Zero-power-factor characteristic (z.p.f c.), in conjunction with O.C.C., is useful in
obtaining the armature leakage reactance Xl and armature reaction m.m.f. Fa For an
alternator, z.p.f c. is obtained as follows
 The synchronous machine is run at rated synchronous speed by the primemover.
 A purer inductive load is connected across the armature terminals and field
current is increased till full load armature current is flowing.
 The load is varied in steps and the field current at each step is adjusted to
maintain full-load armature current. The plot of armature terminal voltage and
field current recorded at each step, gives the zero -power- factor characteristic
at full-load armature current.
From fig 3(a), it can be seen that the terminal voltage Vt and the air-gap
voltage Er are very nearly in phase and are, therefore, related by the simple
algebraic equation
Vt = E r – I X l
The resultant m.m.f. Fr and the field m.m.f. Ff are also very nearly in phase and
are related by the simple algebraic equation
Ff = F r + F a
Assume that O.C.C. gives the exact relation between air-gap voltage E r
and the resultant m.m.f. Fr under load. Also, assume armature leakage
reactance to be constant.
The O.C.C. and z.p.f c. are shown in Fig.3(b). For field excitation Ff or
field current I f, equal to OP the open-circuit voltage is PK. With the field
excitation and speed remaining unchanged, the armature terminals are
connected to a purely inductive load such that full load armature current
flows. An examination of Fig. 3(a) and (b) reveals that under zero power.
factor load, the net excitation Fr is OF which is less than OP (=Ff)by Fa
According to the resultant m.m.f. OF the air-gap voltage Er is FC and if
43
CB=IXl is subtracted from Er= FC, the terminal voltage FB = PA = Vt is
obtained. Since z.p.f c. is a plot between the terminal voltage and field
current If or Ff , which has not changed from its no-load value of OP, the
point A lies on the z.p.f.c. The triangle ABC so obtained is called the Potier
triangle, where CB=IXl and BA = Fa. Thus, from the Potier triangle, the
armature leakage reactance Xl and armature reaction m.m.f. Fa can be
determined.
Fig 3
(a) General phasor diagram of Round-Rotor Alternator at zero power
factor
(b) O.C.C., z.p.f c. And potier triangle
If the armature resistance is assumed zero and the armature current is kept
constant, then the size of Potier triangle ABC remains constant and can be
shifted parallel to itself with its corner C, remaining on the O.C.C. and its
corner A, tracing the z.p.f.c. Thus the z.p.f.c. has the same shape as the
O.C.C. and is shifted vertically downward by an amount equal to IXl (i.e.,
ieakage reactance voltage drop) and horizontally to the right by an amount
equal to the armature reaction m.m.f. Fa.or the field current equivalent to
armature reaction m.m.f.
For determining IXl and Fa experimentally, it is not necessary to plot the
entire z.p.f.c. Only two points A and F' shown in Fig 3(b) are sufficimt.
The point A(PA=rated voltage) is obtained by actually loading the
alternator so that the rated armature current flows in the alternator. The other
point F' on the z.p.f.c. corresponds to the zero terminal voltage and can,
therefore, be obtained by performing short-circuit test. So here OF' is the field
current required to circulate short-circuit current equal to the armature current
(generally rated current) at which the point A is determined in the zero(near zero)
power-factor test.
44
Now draw a horizontal line AD, parallel and equal to F'O. Through point D,
draw a straight line parallel to the air-gap line, intersecting the O.C.C. at C. Draw
CB perpendicular to AD. Then ABC is the Potier triangle from which
BC = IXl
AB = Fa
Since the armature current I at which the point A is obtained, is Known, Xl can be
calculated.
Then determine the air-gap voltage Er by the relation,
Er = Vt + I ( R a + j X l )
According to the magnitude of Er obtain Fr from O.C.C. and draw it leading
Er by 90º, The armature reaction m.m.f. Fa and armature leakage reactance X l ,
can be determined from the Potier triangle, as explained before. Now Fa is
drawn in phase with I as shown in Fig. 3(a) Then
Ff = Fr – Fa
is obtained and corresponding to Ff, excitation voltage Ef (or Eo) is recorded
from O.C.C. and the voltage regulation obtained.
Z.p.f. method requires O.C.C. and z.p.f.c., and gives quite accurate results.
Example 1:
A three phase 6000 V Alternator gave the following open circuit characteristic at
normal speed:
Field current in amper
14
Terminal Voltage in volt 4000
18
23
30
43
5000
6000
7000
8000
With the alternator short circuited and full load current folwing, the field current is
16 A. Neglecting the armature resistance and using synchronous impedance method,
Determine the voltage regulation of the alternator supplying the full load of 200
KVA, at 0.8 power factor lagging.
Solution
The open circuit characteristics have been drawn in fig 4, and corresponding to an
excitation (field) current , If of 16 ampers, the open circuit voltage is 4700 V.
The phase O.C. voltage = 4700 / √3 = 2714 V
Full load current = (200 * 1000) / (√3 * 6000 * 0.8) = 24 A
45
Fig 4
Short curcuit current corresponding to field current of 16 ampers is 24 ampers
Synchronuos impedance ,
Zs = O.C. voltage / S.C. current
= 2714 / 24 = 113 Ω
Now rated voltage of alternator per phase,
V= 6000 / √3 = 3464 volts
And
Cos φ = 0.8
Sin φ = 0.6
Load current,
I = 24 ampers
Eo = √ [( Vt cos φ + I R a)2 + ( Vt sin φ + I XS)2]
= √[(3464 * 0.8 + 0)2 + (3464 * 0.6 +24 * 113)2]
= 5533 V
Therefore,
Percentage regulation at full load = (Eo – Vt) / Vt *100%
= (5533 – 3464) / 3464 *100
= 59.7%
Example 2:
A 220 V, 50 Hz, 6-pole star-connected alternator with, ohmic resistance of
0.06 Ω/ph, gave the following data for open-circuit, short-circuit and fullload zero-power-factor characteristics :
Find the percentage voltage regulation at full-load current of 40 amps at powerfactor of 0.8 lag by (a) e.m f. method (b) z. p. f .
46
Field current, A
0.2 0.4
0.8
1.0
1.2
1.4
1.8 2.2
O.C. voltage, V
29 58.0 87.0 116
146
172
194
232 261.5 284 300
S.C. current, A
6.6 13.2 20.0 26.5 32.4 40.0 46.3 59
z.p.f. terminl voltage, v -
-
0.6
-
-
-
0.0
29
-
88 140
2.6 3.0
-
-
177 208
Solution
Rated per phase voltage = 220 / √3 = 127 V
Per phase values for O.C.C. and z.p.f.c. are tabulated below and O.C.C., S.C.C. and
z.p.f.c. are plotted in fig 5.
Field current, A
0.2
1.4
1.8
2.2
2.6 3.0
O.C. voltage, V
16.7 33.5 50.2 67 84.3 99.3 112
134
151
164 173
z.p.f. terminl voltage,v -
0.4
-
0.6
-
0.8 1.0
-
-
1.2
0.0
16.73 50.8 80.8 102 120
A- E.M.F. Method:
Zs = O.C. voltage / S.C. current
From above tables, Zs ≈ 134 / 59 = 2.27 Ω
Here Xs ≈ Zs = 2.27 Ω, since Ra is quite small.
Eo = √ [( Vt cos φ + I R a)2 + ( Vt sin φ + I XS)2]
= √ [(127 * 0.8 + 40 * 0.06) 2 + (127 * 0.6 + 40 * 2.27)2]
= 196 volts
Percentage regulation = (Eo – Vt) / Vt *100%
= (196 – 127) / 127 *100 = 54%
B- Zero power factor Method:
First of all, the Potier triangle ABC is drawn as described before, Point A
corresponds to the rated voltage of 127 V on the Z. p.f.c. The line AD is drawn
parallel and equal to F'O = 1.2 A.Then DC is drawn parallel to the air-gap line,
meeting the O.C.C. at point C. Perpendicular CB on AD, lives IXl drop equal to
30 volts.
Armature leakage reactance Xl = 30 / 40 = 0.75 Ω
The air-gap voltage Er,
Er = Vt + I ( R a + j X l ) = 127(0.8+j0.6) + 40*(0.06+j0.75)= 148.6∟45.6º
Or = √ [( Vt cos φ + I R a)2 + ( Vt sin φ + I X l )2]
=√ [(127 * 0.8 + 40 * 0.06) 2 + (127 * 0.6 + 40 * 0.75)2]
= 148.6 volts
47
Fig 5
Corresponding to Er = 148.6 V, the field current Fr from O.C.C. is 2.134 A, the
armature m.m.f. Fa, from Potier triangle is AB = 0.84 A.
Fr = 2.134 ∟(45.6º+ 90º) = 2.134 ∟135.6º
Fa = 0.84 A
Ff = Fr – Fa = 2.134 ∟135.6º – 0.84
= 2.797 ∟147.7º A
For Ff = 2.797 A, the excitation voltage from O.C.C. is Eo=169.0 volts
Percentage regulation = (Eo – Vt) / Vt *100%
= (169 – 127) / 127 *100 = 33.1%
As already stated, z.p.f. method gives quite accurate results and here the
voltage regulation with this method is 33.1%.
The voltage regulation by e.m.f. method is 54% , This value is much higher
than the accurate value of 331 % and in view of this, this method may be
called pessimistic method.
48
The Automatic Voltage Regulator
The method of operation in the power station is to maintain the terminal
voltage of the alternator at the rated value and to adjust the excitation
with change in load current accordingly. As the variations in load may be
very violent both as regards magnitude and rapidity, it is clear that hand
regulation of the excitation is impossible and that automatic means must
be adopted. Now the flux per pole of a large turbo-alternator may amount
to several webers, and therefore the self-induction of the field winding will be
very high.
Consider, for example, a 15,000 kVA, 4 pole turbo-alternator having a flux
per pole of 1.15 webers, a rotor current of 600 amperes, and a rotor
winding of 66 turns per pole.
L(per pole)=(Flux per ampere) * (No. of turns)
= (1.15 / 600 )* 66 = 0.127 henry
And Inductance, L = 4 * 0.127 = 0.508 henry for the whole winding
Winding resistance, R=0.2 Ω
The time constant = L / R = 0.508 / 0.2 = 2.54
The time constant L / R of the field circuit gives the value in seconds, and is
the time taken for the flux to reach 0.6321 of its final value. With modern
large two-pole, turbo-alternators the time constant is considerably
greater. What is required in the power station that there shall be an almost
instantaneous increase in excitation to the desired value. An increase in load
calls for an increase in excitation. This increase is made much greater than
is ultimately required, and as a result the flux builds up very rapidly. Before
the flux can build up too far, the excitation is reduced again. This is
known as the "overshooting-the-mark" principle. There are two types of
quick acting regulator which work on this principle :
(a) the vibrator type, in which a fixed resistance is rapidly cut in and
out.
(b) the rheostat type in which the resistance is variable.
Transistor-Cruntrolled or Transistor Type-Automatic
Voltage Regulators
The automatic voltage regulation is affected by matching a quantity proportional to
the alternator voltage against a 'reference'. The difference between these two, called
'error' has to be rectified before it can be fed to the excitor, and this is affected by
means of transistor amplifiers. This is a 'brushless' method of voltage control
for an alternator. Here no slip rings, commutators and brush gear is required.
These types of regulators are also called electronic voltage regulators
49
The reference circuit is supplied by voltage feed back from the alternator, the
general scheme of control has been illustrated in block diagram of fig 6.
Fig 6 Block diagram of the control system of a transistor-controlled alternator
50
Three Phase Synchronous Motor
We know that the stator of a three-phase synchronous machine, carrying a
three-phase winding currents produces a rotating magnetic field in the air
gap of the machine. Referring to Fig. 1(a), we will have a rotating magnetic
field in the air gap of the salient pole machine when its stat or windings are
fed from a three-phase source. Let the rotor (or field) winding be unexcited.
The rotor will have a tendency to align with the rotating field at all times in
order to present the path of least reluctance. Thus if the field is rotating, the
rotor will tend to rotate with the field. From Fig 1(b) we see that a round
rotor will not tend to follow the rotating magnetic field because the uniform air
gap presents the same reluctance all around the air gap and the rotor does not
have any preferred direction of alignment with the magnetic field. This torque,
which we have in the machine of Fig 1(a), but not in the machine of Fig 1(b), is
called the reluctance torque. It is present by virtue of the variation of the
reluctance around the periphery of the machine.
(a)
(b)
Fig. 1 (a) A Salient-rotor machine
(b) A Round-rotor machine
Next, let the field winding [Fig. 1(a) or (b)] be fed by a dc source that produces
the rotor magnetic field of definite polarities, and the rotor will tend to align with
the stator field and will tend to rotate with the rotating magnetic field. We
observe that for an excited rotor, a round rotor, or a salient rotor, both will tend
to rotate with the rotating magnetic field, although the salient rotor will have an
51
additional reluctance torque because of the saliency. In the past lecture we derive
expressions for the electromagnetic torque in a synchronous machine attributable to
field excitation and to saliency.
So far we have indicated the mechanism of torque production in a round -rotor
and in a salient-rotor machine. To recapitulate, we might say that the stator rotating
magnetic field has a tendency to "drag" the rotor along, as if a north pole on the
stator "locks in" with a south pole of the rotor. However, if the rotor is at a
standstill, the stator poles will tend to make the rotor rotate in one direction and then
in the other as they rapidly rotate and sweep across the rotor poles. Therefore, a
synchronous motor is not self-starting. In practice, as mentioned earlier, the
rotor carries damper bars that act like the cage of an induction motor and thereby
provide a starting torque. The mechanism of torque production by the damper bars
is similar to the torque production in an induction motor. Once the rotor starts
running and almost reaches the synchronous speed, it locks into position with the
stator poles. The rotor pulls into step with the rotating magnetic field and runs at
the synchronous speed; the damper bars go out of action. Any departure from the
synchronous speed results in induced currents in the damper bars, which tend to
restore the synchronous speed.
Performance of a Round-rotor Synchronous Motor
Except for some precise calculations, we may neglect the armature resistance
as compared to the synchronous reactance. Therefore, the steady-state per
phase equivalent circuit of a synchronous machine simplifies to the one shown
in fig 2(a). in which we show the terminal voltage Vt, the internal voltage Eo, and
the armature current Ia, going "into" the machine or "out of" it, depending
on the mode of operation "into" for motor and "out of" for generator.
With the help of this circuit we will study some of the steady-state operating
Fig 2 : (a) an approximate equivalent circuit
(b) power-angle characteristics of a round-rotor synchronous machine
52
characteristics of a synchronous motor. In Fig. 2(b) we show the power-angle
characteristics as given by the power developed equation. Here positive
power and positive δ imply the generator operation, while a negative δ
corresponds to a motor operation. Because δ is the angle between Eo and Vt,
Eo is ahead of Vt in a generator, whereas in a motor, Vt is ahead of Eo. The
voltage-balance equation for a motor is, from Fig. 2(a),
Vt = Eo + j Ia XS
The per phase power developed is
Pd = (Eo Vt / XS ) sin δ
If the motor operates at a constant power,then from above equations required that
Ia XS cos φ = Eo sin δ
…(i)
We recall that Eo depends on the field current, I f . Consider tow cases:
(1) when I f is adjusted so that Eo< Vt, and the machine is underexcited.
(2) when I f is increased to a point that Eo > Vt , and the machine becomes
overexcited.
The voltage-current relationships for the two cases are shown in Fig. 3(a).
For Eo > Vt at constant power, δ is greater than the δ for Eo < Vt .Notice that
an underexcited motor operates at a lagging power factor (Ia lagging Vt),
whereas an overexcited motor operates at a leading power factor.
In both cases the terminal voltage and the load on the motor are the same.
Thus we observe that the operating power factor of the motor is
controlled by varying the field excitation, hence altering Eo. This is a very
important property of synchronous motors. The locus of the armature current
at a constant load, as given by (i), for varying field current is also shown in Fig
3(a). From this we can obtain the variations of the armature current Ia with the
field current, I f (corresponding to Eo ), and this can be done for different
loads, as shown in figures 3(b)and(c).
53
(c)
Fig 3 (a) phasor diagram for motor operation (Eo', Ia', φ', δ')for underexcited
operation,( Eo'', Ia'', φ'', δ'')for overexcited operation
(b), (c) V-Curves of a synchronous motor
These curves are known as the V- Curves of the synchronous motor. One of the
applications of a synchronous motor is in power factor correction, as demonstrated
by the following examples. In addition to the V curves, we have also shown the
curves for constant power factors. These curves are known as compounding
curves. In the preceding paragraph, we have discussed the effect of change in
the field current on the synchronous machine power factor. However, the load
supplied by a synchronous machine cannot be varied by changing the power factor.
Rather, the load on the machine is varied by instantaneously changing the speed (in
case of a generator, by supplying additional power by the prime mover), and thus
changing the power angle corresponding to the new load. In a synchronous
motor, a load change results in a change in the power angle.
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EXAMPLE:
A three-phase wye-connected load takes 50 A of current at 0.707 lagging power
factor at 220 V between the lines. A three-phase wye-connected, round-rotor
synchronous motor, having a synchronous reactance of 1.27 Ω per phase. is
connected in parallel with the load. The power developed by the motor is 33
kW at a power angle, δ, of 30°. Neglecting the armature resistance, calculate (a)
the reactive kilovolt-amperes (kVAR) of the motor and (b) the overall power
factor of the motor and the load.
Solution
The circuit and the phasor diagram on a per phase basis are shown in figures
4(a)and(b). From power developed equation, we have
Pd = 1/3 * 33,000 = 220/√3 * Eo/1.27 * sin 30º
Which yield Eo = 220
And from phasor diagram, Ia XS = 127, or Ia = 127/ 1.27 = 100 A, and φa = 30º
The reactive kilovolt-amperes of the motor = √3 * Vt Ia sin φa
=√3* 220/1000 * 100* sin 30 = 19 VAR
Notice that φa is the power-factor angle of the motor, φ L is the power angle of the
load, φ is the overall power-factor angle, are shown in fig 4(b). The power angle δ,
is also shown in this phasor diagram, from which
I = IL + Ia
Fig 4 (a) Circuit diagram (b) Phasor diagram
Or algebraically adding the real and reactive components of the currents, we obtain
Ireal = Ia cos φa + IL cos φL
Ireactive = Ia sin φa – IL sin φL
The overall power factor angle, φ, is thue given by
tan φ = [Ia sin φa – IL sin φL ] / [Ia cos φa + IL cos φL] = 0.122
Or φ = 7º and cos φ = 0.992 leading.
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TORQUE – SPEED curve of a synchronous motor:
In synchronous motor speed remains constant at Ns for all loads as in figure 5.
At any other speed (≠ Ns) ,there is no motor –action , and therefore no torque will
produced.
Fig - 5
Starting of Three Phase Synchronous Motor
As is clear from the construction of the motor, the stator of the synchronous motor
is wound and the winding is connected to a.c. mains while the rotor of the
motor is mostly of salient pole field construction except in special high speed
two pole motors where it is non-salient pole field construction. The rotor is supplied
from a d.c. source.
Consider Fig. 5, the stator for convenience of explanation is shown as having salient
poles N' and S'. When the rotor is excited from d.c. source, there develop N and S
poles on the rotor and this polarity is retained by the rotor throughout but the
polarity of stator poles changes because it is connected to a.c. mains and the
polarity alternates with the frequency of the a.c. supply.
First consider the rotor is stationary and in a position as shown in Fig. 6 (a). At
this instant the similar poles of rotor and stator repel each other and the rotor tends to
rotate in clockwise direction. But half a cycle later (i.e. 1/2f second) the polarityof
the stator poles is reversed [as in Fig 6 (b)] but the polarity of rotor poles remains the
same.
(a)
Fig 6
(b)
At this very instant unlike poles of stator and rotor attract each other and, therefore,
the rotor tends to rotate back in the anti-clockwise direction. This shows that the
torque acting on the rotor of the synchronous motor is not unidirectional but
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pulsating one. It is because of inertia of the rotor, it will not move in any
direction. That is why the synchronous motor is not self-starting one.
Now let the rotor (which is yet unexcited) is speeded up to synchronous or
near synchronous speed by some external arrangement and then it is excited
through the d.c. source. The moment the rotor running at nearly
synchronous speed is excited, it is magnetically locked into position with
the stator poles which runs synchronously and that both in the same
direction. Because of this interlocking between the rotor and stator poles, the
synchronous motor has either to run synchronously or not at all.
Methods of Starting Synchronous Motors
Since the three phase synchronous motor has no starting torque,so artificial
means must be provided for starting it as below:
1. External source. If the d.c. field of a synchronous motor is supplied by a
direct-connected exciter, this exciter can be used as a starting motor. This
method is now rarely used.
2. Induction-motor start. If a squirrel cage winding is constructed in the pole
faces of the synchronous motor (damper bars), it can be used to develop a
starting torque (as well as provide damping) similar to that of the ordinary
induction motor. As the motor reaches about 95 per cent of synchronous speed
(it is operating as an induction motor with 5 percent slip) its field is excited and the
motor pulls into step.
For such a set up, the problem of limiting the starting current without to low a value
of starting torque is met in several ways:
.
(a) Auto-transformers are used which are similar in design to those employed on
induction motors. The stator is connected to the reduced voltage supply of the
auto-transformer until synchronous or near synchronous speed is reached, and is
then connected to the full voltage.
(b) The voltage supplied to the stator can be reduced by using a series reactors
in the supply lines. These reactors give a large voltage drop and low P.F. at
starting.
(c) Various types of rotor windings are used like double squirrel
cage, or using special bar cross-sections (T bar, L bar, or simply deep,
narrow bars) to give high skin effect to the squirrel cage and thus limit
the starting current.
(d) Multiple winding. This involves a special arrangement of stator coils
so that two or more complete windings are paralleled for normal
operation. The paralleled windings have normal values of' reactance.
If one winding is left open during starting process the resistance will
double and the leakage reactance will be greater than the normal value and
will result in a reduced current without the utilization of an autostarter. Switching arrangement is shown in Fig. 7.
57
(a) For starting, the short-circuited switch is open and only winding 1 is utilized
(b) The short circuiting switch closes the neutral of the second winding for parallel
operation at normal load
Fig 7 Arrangement of the double-winding synchronous motor
Synchronous Condenser
When a three phase synchronous motor is used for power factor correction with no
mechanical output (no load), it is known as a synchronous condenser. The term
'condenser' applied to device that it draws leading current as does a static
condenser. There is a considerable increase in leading kilovolt- amperes
available when the horsepower load of a synchronous motor is less than its rated
load.
The synchronous condenser is especially designed so that practically all its rated
kVA are available for p.f. correction. Because of the absence of shaft load,
the mechanical design is modified from the ordinary motor standards. Because of
the p.f. adjustment possible, the synchronous condenser is particularly applicable to
transmission line control.
The ability to control the power factor of the synchronous motor by simply
varying the excitation is of very great practical importance. The importance lies in
the fact that the motor can be operated at a leading power factor when desired,
and consequently if the remaining installation has a low power factor, the
overall. power factor for the installation can be brought close to unity.
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Hunting and Damping of Synchronous Motor
When a synchronous motor is operating under steady-load conditions and
an additional load is suddenly applied, the developed torque is less than
that required by the load and the motor starts to slow down. A slight
reduction in speed increases the phase angle of the generated voltage
and permits more current to flow through the armature Some of the
kinetic energy of the rotating parts is given out to the load during speed
reduction . When the motor is slowing down, it cannot cease deceleration
at exactly the correct torque angle corresponding to the increased load. It
passes beyond this point, develops more than required torque, and
increases in speed. This is followed by a reduction in speed and a
repetition of the entire cycle. Such a periodic change in speed is called
hunting.
The mass of the rotating part and the 'spring effect' of the flux lines are
the necessary elements which give the rotating field structure a natural
oscillating period. If the load varies periodically with this same
frequency or some multiple of it, the tendency is for the amplitude of
these oscillations to increase cumulatively until the motor is thrown
out of step, causing corresponding current and power pulsations. The
mechanical stresses are likely to be severe, and the armature current
greatly increases when the motor leaves synchronism.
A solid pole gives a damping action but has the disadvantage of
excessive pole-face losses unless closed slots are used on the armature.
The most satisfactory arrangement from the standpoint of simplicity
seems to be laminated poles with the squirrel-cage 'amortisseur' or
damping winding. The effectiveness of damper depends upon its
resistance and to a lesser extent upon the length of the airgap. A low
resistance damping winding produces the stronger effect, but if the
synchronous motor is to be started by induction-motor action, using
this squirrel cage, the winding resistance should be fairly high to
produce good starting torque. Because of these opposing tendencies, a
compromise usually must be reached in damping winding design.
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SHEET 5
SOLVED EXAMPLES IN SYNCHRONOUS MACHINES
Example 1 :
Calculate the percent voltage regulation for a three-phase wye-connected 2500 kVA
6600-V turboalternator operating at full-load and 0.8 power factor lagging. The per
phase synchronous reactance and the armature resistance are 10.4 Ω and 0.071 Ω,
respectively?
Solution:
Clearly, we have XS >> R a, The phasor diagram for the lagging power factor
neglecting the effect of R a is shown in Fig. (a). The numerical values are as follows:
Vt = 6600 / √3 = 3810 V
Ia = ( 2500 * 1000 ) / ( √3 * 6600) = 218.7 A
Eo = 3810 + 218.7( 0.8 – j 0.6) j10.4 = 5485∟19.3º
Percentage regulation = ( 5485 – 3810 )/ 3810 *100 = 44%
Example 2 :
Repeat Ex.1 calculations with 0.8 power factor leading as shown in Fig. (b) ?
Solution:
Eo = 3810 + 218.7( 0.8 + j 0.6) j10.4 = 3048∟36.6º
Percentage regulation = ( 3048 – 3810 )/ 3810 *100 = -20%
Fff
H.W. : Repeat Ex.1 calculations with unity power factor , and draw the phasor
diagram for this case ?
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Example 3 :
A 20-KVA, 220 V, 60 Hz, way-connected three phase salient-pole synchronous
generated supplies rated load at 0.707 lagging power factor.The phase constants of
the machine are Ra = 0.5 Ω and Xd =2 Xq = 4 Ω. Calculate the voltage regulation at
the specified load. ?
Solution:
Vt = 220 / √3 = 127 V
Ia = 20000 / (√3 * 220) = 52.5 A
φ = cos-1 0.707 = 45º
tan δ = (Ia Xq cos φ) / (Vt + Ia Xq sin φ)
= 52.5 * 2 * 0.707 / ( 127 * 52.5 * 2 * 0.707)
= 0.37
δ = 20.6º
Id = 52.5 sin ( 20.6 + 45 ) = 47.5 A
Id Xd = 47.5 * 4 = 190 V
Eo = Vt cos δ + Id Xd
= 127 cos 20.6 + 190 = 308 V
Percent regulation = (Eo – Vt ) / Vt *100%
= ( 308 – 127 ) / 127 *100%
= 142 %
Example 4:
The stator core of a 4-pole, 3-phase a.c. machine has 36 slots. It carries a short pitch
3-phase winding with coil span equal to 8 slots. Determine the distribution and coil
pitch factors?
Solution:
Number of slots per pole = 36 / 4 = 9
Angular displacement between slots = β = 180º / 9 = 20º
Coil span = 20º * 8 =160º
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Therefore, pitch factor = k p = cos (θ/2)
k p = cos 10º = 0.985
,
θ = 180º – 160º = 20º
Now, number of slots per pole per phase,
m = 9 /3 = 3
Distribution factor,
kd = ( Length of long chord ) / (Sum of lengths of short chords )
= [sin (mβ/2)] / [m sin (β/2)]
= sin 30º / (3 sin 10º) = 0.96
Example 5:
3- phase, 50 Hz generator has 120 turns per phase. The flux per pole is 0.07
weber, assume sinusoidally distributed. Find (a) the e.m f. generated per phase. (b)
e.m.f. between the line terminals with star connection. Assume full pitch winding
and distribution factor equal to 0.96 ?
Solution:
(a) E.m.f. generated per phase = 4 k f k p k d f T Φp
volts
= 4 * 1.11 * 0.96 * 1 * 0.07 * 50 * 120
=1792 volts
(b) E.m.f. between line terminals= √ 3 E.m.f. generated per phase =√ 3 * 1792
= 3100 volts
Example 6:
A three phase, 16-pole, star-connected alternator, has 192 stator slots with eight
conductors per slot and the conductors of each phase are connected in series. The
coil span is 150 electrical degrees. Determine the phase and line voltages if the
machine runs at 375 r.p m. and the flux per pole is 64 mWb distributed sinusoidally
over the pole?
Solution:
The flux is sinusoidally distributed over the pole, hence from factor, k f = 1.11
Pitch factor , k p = cos [(180º - 150º) / 2 ] = cos 15º = 0.966
Distribution factor, kd = [sin (mβ/2)] / [m sin (β/2)]
m = Number of slots per pole per phase = 192 / (16 * 3) = 4
β = 180º / (No. of slots/pole) = 180º / (192/16) = 15º
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kd = [ sin (4 * 15º / 2)] / [4 * sin (15º/2)] = sin 30º / (4 * sin 7.5º) = 0.958
Now total number of conductors per phase
=Total number of slots per phase * number of conductors per slot.
= ( 192 / 3 ) * 8 = 512
Number of turns per phase = T = 512 / 2 = 256
Frequency of generated e.m.f. = f = P N / 120
=[ Number of poles * Synchronous Speed] / 120
=[ 16 * 375] / 120 = 50 cycle per second
Flux per pole = Φp=0.064 wb
Therefore, e.m.f. per phase = 4 k f k p k d f T Φp
volts
= 4 * 1.11 * 0.966 * 0.958 * 0.064 * 50 * 256
= 3367 volts
Hence for star connected alternator, Line voltage = √3 * 3367 = 5830 volts
Example 7:
A 2200 volt, star-connected, 50 cycle alternator has 12 poles. The stator has 108
slots each with 5 conductors per slot. Calculate the necessary flux per pole to give
2200 volts on no-load. The winding is concentric and the value of kf can be taken as
1.11 ?
Solution:
Slots per phase = 108 / 3 = 36
Therefore, conductors per phase = 36 * 5 = 180
and number of turns per phase = 180 / 2 = 90
Slots per pole per phase = 36 / 12 =12
Frequency,f =50 cycle per second
kf = 1.11
k p = 1.0 for concentric winding
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kd = [sin (mβ/2)] / [m sin (β/2)]
Where β = 180º / (No. Of slots/pole) = 180º / (108/12) = 20º
m = Number of voltage vectors= Number of slots per pole per phase
= 108 / (12*3) = 3
kd = [sin (3*20º/2)] / [ 3 sin (20º/2)] = [sin 30º] / [3 sin 10º] = 0.96
Also e.m.f. per phase = Eph = 2200 /√3 volts
The e.m.f. equation for Alternator is
Eph = 4 k f k p k d f T Φp
volts
Φp = Flux per pole = Eph / [4 k f k p k d f T ] =( 2200 /√3) / [ 4*1.11*1*0.96*50*90]
= 0.066 wb
Example 8:
Find the number of armature conductors in series per phase required for the
armature of a 3 phase, 50 Hz, 10 pole alternator with 90 slots. The winding is to
be star connected to give a line voltage of 11,000 volts. The flux per pole is 0.16
webers?
Solution:
Number of slots per pole = 90 / 10 = 9
Therefore,
β = 180º / 9 = 20º
Slots per pole per phase, m = 9 / 3 = 3
Distribution factor,
kd = [sin (mβ/2)] / [m sin (β/2)]
kd = [sin (3*20º/2)] / [ 3 sin (20º/2)] = [sin 30º] / [3 sin 10º] = 0.96
Since there is no mention about the type of winding, the full pitched winding is assumed.
Hence , k p = 1.0
Also , Eph = 11000 / √3 = 6352 volts.
Now , Eph = 4 k f k p k d f T Φp
volts
T = [Eph] / [4 k f k p k d f Φp]
= [6352] / [4 * 1.11 * 0.96 * 1 * 0.16 * 50]
= 1863
Number of Armature conductors in serues per phase = 1863 * 2 = 3726
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Example 9 :
A six-pole, 3-phase, 60 cycle alternator has 12 slots per pole and 4 conductors
per slot. The winding is 5/6 th pitched. There is 0.025 wb flux entering the
armature from each north pole and the flux is sinusoidally distributed along
the air gap. The armature coils are all connected in series. The winding is starconnected. Determine the open circuit e.m.f of the alternatorper phase?
Solution:
Here, number of conductors connected in series per phase
=12 * 6 * 4 / 3 = 96
Therefore, number of turns per phase , T = 96 / 2 = 48
Flux per pole , Φp = 0.025 wb
Frequency,
f = 5 0 c yc l e p e r s e c o n d .
k f = 1.11
k p = cos [180º(1 – 5/6) / 2] = cos 15º = 0.966.
Number of slots per pole per phase, m = 12 / 3 = 4
β = 180º / 12 = 15º
Distribution factor ,
kd = [sin (mβ/2)] / [m sin (β/2)]
kd = [sin (4*15º/2)] / [ 4 sin (15º/2)] = [sin 30º] / [4 sin 7.5º] = 0.958
Hence induced e.m.f.,
Eph = 4 k f k p k d f T Φp
volts
= 4 * 1.11 * 0.966 * 0.958 * 0.025 * 50 * 48
= 295 volts.
65