Download Induction Motors

Induction Motors
Introduction
• Three-phase induction motors are the most common
and frequently encountered machines in industry
– simple design, rugged, low-price, easy maintenance
– wide range of power ratings: fractional horsepower to 10
MW
– run essentially as constant speed from no-load to full load
– Its speed depends on the frequency of the power source
• not easy to have variable speed control
• requires a variable-frequency power-electronic drive for optimal
speed control
Construction
• An induction motor has two main parts
– a stationary stator
• consisting of a steel frame that supports a hollow,
cylindrical core
• core, constructed from stacked laminations,
• having a number of evenly spaced slots, providing the
space for the stator winding
Stator of IM
Construction
– a revolving rotor
• composed of punched laminations, stacked to create a series of rotor
slots, providing space for the rotor winding
• one of two types of rotor windings
• conventional 3-phase windings made of insulated wire (wound-rotor) »
similar to the winding on the stator
• aluminum bus bars shorted together at the ends by two aluminum rings,
forming a squirrel-cage shaped circuit (squirrel-cage)
• Two basic design types depending on the rotor design
– squirrel-cage: conducting bars laid into slots and shorted at both
ends by shorting rings.
– wound-rotor: complete set of three-phase windings exactly as the
stator. Usually Y-connected, the ends of the three rotor wires are
connected to 3 slip rings on the rotor shaft. In this way, the rotor
circuit is accessible.
Construction
Squirrel cage rotor
Wound rotor
Notice the
slip rings
Construction
Slip rings
Cutaway in a
typical woundrotor IM.
Notice the
brushes and the
slip rings
Brushes
Rotating Magnetic Field
• Balanced three phase windings, i.e.
mechanically displaced 120 degrees
form each other, fed by balanced three
phase source
• A rotating magnetic field with constant
magnitude is produced, rotating with a
speed
nsync
120 f e

P
rpm
Where fe is the supply frequency and
P is the no. of poles and nsync is called the
synchronous speed in rpm (revolutions
per minute)
Synchronous speed
P
50 Hz
60 Hz
2
3000
3600
4
1500
1800
6
1000
1200
8
750
900
10
600
720
12
500
600
Rotating Magnetic Field
Rotating Magnetic Field
Bnet (t )  Ba (t )  Bb (t )  Bc (t )
 BM sin(t )0  BM sin(t  120)120  BM sin(t  240)240
 BM sin(t )xˆ
3
BM sin(t  120)]yˆ
2
3
[0.5BM sin(t  240)]xˆ  [
BM sin(t  240)]yˆ
2
[0.5BM sin(t  120)]xˆ  [
Rotating Magnetic Field
Principle of operation
• This rotating magnetic field cuts the rotor windings and
produces an induced voltage in the rotor windings
• Due to the fact that the rotor windings are short circuited, for
both squirrel cage and wound-rotor, and induced current
flows in the rotor windings
• The rotor current produces another magnetic field
• A torque is produced as a result of the interaction of those
two magnetic fields
 ind  kBR  Bs
Where ind is the induced torque and BR and BS are the magnetic
flux densities of the rotor and the stator respectively
Induction motor speed
• At what speed will the IM run?
– Can the IM run at the synchronous speed, why?
– If rotor runs at the synchronous speed, which is the
same speed of the rotating magnetic field, then the
rotor will appear stationary to the rotating magnetic
field and the rotating magnetic field will not cut the
rotor. So, no induced current will flow in the rotor and
no rotor magnetic flux will be produced so no torque
is generated and the rotor speed will fall below the
synchronous speed
– When the speed falls, the rotating magnetic field will
cut the rotor windings and a torque is produced
Induction motor speed
• So, the IM will always run at a speed lower than
the synchronous speed
• The difference between the motor speed and the
synchronous speed is called the Slip
nslip  nsync  nm
Where nslip= slip speed
nsync= speed of the magnetic field
nm = mechanical shaft speed of the motor
The Slip
s
nsync  nm
nsync
Where s is the slip
Notice that : if the rotor runs at synchronous speed
s=0
if the rotor is stationary
s=1
Slip may be expressed as a percentage by multiplying the above
eq. by 100, notice that the slip is a ratio and doesn’t have units
Induction Motors and Transformers
• Both IM and transformer works on the principle of
induced voltage
– Transformer: voltage applied to the primary windings
produce an induced voltage in the secondary windings
– Induction motor: voltage applied to the stator windings
produce an induced voltage in the rotor windings
– The difference is that, in the case of the induction motor,
the secondary windings can move
– Due to the rotation of the rotor (the secondary winding of
the IM), the induced voltage in it does not have the same
frequency of the stator (the primary) voltage
Frequency
• The frequency of the voltage induced in the
rotor is given by
Pn
fr 
120
Where fr = the rotor frequency (Hz)
P = number of stator poles
n = slip speed (rpm)
P  (ns  nm )
fr 
120
P  sns

 sf e
120
Frequency
• What would be the frequency of the rotor’s
induced voltage at any speed nm?
fr  s fe
• When the rotor is blocked (s=1) , the frequency
of the induced voltage is equal to the supply
frequency
• On the other hand, if the rotor runs at
synchronous speed (s = 0), the frequency will be
zero
Torque
• While the input to the induction motor is electrical
power, its output is mechanical power and for that
we should know some terms and quantities
related to mechanical power
• Any mechanical load applied to the motor shaft
will introduce a Torque on the motor shaft. This
torque is related to the motor output power and
the rotor speed
 load 
Pout
m
N .m
and
2 nm
m 
60
rad / s
Horse power
• Another unit used to measure mechanical
power is the horse power
• It is used to refer to the mechanical output
power of the motor
• Since we, as an electrical engineers, deal with
watts as a unit to measure electrical power,
there is a relation between horse power and
watts
hp  746 watts
Example
A 208-V, 10hp, four pole, 60 Hz, Y-connected
induction motor has a full-load slip of 5
percent
1. What is the synchronous speed of this motor?
2. What is the rotor speed of this motor at rated
load?
3. What is the rotor frequency of this motor at
rated load?
4. What is the shaft torque of this motor at rated
load?
Solution
1.
2.
nsync
120 f e 120(60)


 1800 rpm
P
4
nm  (1  s)ns
 (1  0.05) 1800  1710 rpm
f r  sfe  0.05  60  3Hz
3.
 load
4.
Pout
Pout


m 2 nm
60
10 hp  746 watt / hp

 41.7 N .m
1710  2  (1/ 60)
Why Synchronous Motor is not self
starting
• Consider the rotating magnetic field as
equivalent to physical rotation of two stator
poles N1 and S1.
• Consider an instant when two poles are at
such a position where stator magnetic axis is
vertical, along A-B as shown in the Fig. 1(a).
Why Synchronous Motor is not self
starting
•
At this instant, rotor poles are arbitrarily positioned as shown in the
Fig. 1.
•
At this instant, rotor is stationary and unlike poles will try to attract
each other. Due to this rotor will be subjected to an instantaneous torque
in anticlockwise direction as shown in the Fig. 1(a).
•
Now stator poles are rotating very fast i.e. at a speed Ns r.p.m. Due to
inertia, before rotor hardly rotates in the direction of
anticlockwise
torque, to which it is subjected, the stator poles change their positions.
Consider an instant half a period latter where stator poles are exactly
reversed but due to inertia rotor is unable to rotate from its initial
position. This is shown in the Fig. 1(b)
At this instant, due to the unlike poles trying to
attract each other, the rotor will be subjected to
a torque in clockwise direction. This will tend to
rotate rotor in the direction of rotating magnetic
field.
But before this happen, stator poles again
change their position reversing the direction of
the torque exerted on the rotor.
Key Point : As a result, the average torque
exerted on the rotor is zero. And hence the
synchronous motor is not self starting.
Note : The question is obvious that will happen if by
chance the rotor position is in such a way that the
unlike rotor and stator poles are facing each other ?
But owing to the large inertia of the rotor, the rotor
fails to rotate along with the stator poles. Hence
again the difference of position of magnetic axes
gets created and rotor gets subjected to quickly
reversing torque. This is because the speed with
which rotating magnetic field is rotating is so high
that it is unable to rotate the rotor from its initial
position, due to the inertia of the rotor. So under
any case, whatever may be the starting position of
the rotor, synchronous motor is not self starting.
Procedure to Start a Synchronous Motor
• Now suppose the rotor is rotated by some external means at a speed
almost equal to synchronous speed. And then the rotor is excited to
produce its poles. At a certain instant now, the stator and rotor unlike
poles will face each other such that their magnetic axes are near each
other. Then the force of attraction between the two, pulls both of them
into the magnetic locking condition.
•
Once magnetic locking is established, the rotor and stator poles
continue to occupy the same relative positions. Due to this, rotor
continuously experiences a unidirectional torque in the direction of the
rotating magnetic field. Hence rotor rotates at synchronous speed and
said to be in synchronism with rotating magnetic field. The external device
used to rotate rotor near synchronous speed can be removed once
synchronism is established. The rotor then continues its rotation at Ns due
to magnetic locking. This is the reason why synchronous motor runs only
at synchronous speed and does not rotate at any speed other than the
synchronous. This operation is shown in the Fig 1(a) and (b).
It is necessary to keep field winding i.e. rotor excited
from d.c. supply to maintain the magnetic locking,
as long as motor is operating.So a general procedure
to start a synchronous motor can be stated as :
1. Give a three a.c. supply to a three phase winding.
This will produce rotating magnetic field rotating at
synchronous speed Ns r.p.m.
2. Then drive the rotor by some external means like diesel
engine in the direction of rotating magnetic field, at a speed
very near or equal to synchronous speed.
3. Switch on the d.c. supply given to the rotor which will
produce rotor poles. now there are two fields one is rotating
magnetic field produced by stator while the other is produced
by rotor which is physically rotated almost at the same speed
as that of rotating magnetic field.
4. At a particular instant, both the fields get magnetically
locked. The stator field pulls rotor field into synchronism. Then
the external device used to rotate rotor can be removed. But
rotor will continue to rotate at the same speed as that of
rotating magnetic field i.e. Ns due to magnetic locking.
Key Point : So the essence of the discussion is that to start the
synchronous motor, it needs some device to rotate the rotor
at a speed very near or equal to the synchronous speed.
Methods of Starting Synchronous Motor
As seen earlier, synchronous motor is not self starting. It is necessary
to rotate the rotor at a speed very near to synchronous speed. This is
possible by various method in practice. The various methods to start
the synchronous motor are,
1. Using pony motors
2. Using damper winding
3. As a slip ring induction motor
4. Using small d.c. machine coupled to it.
1.
Using pony motors
In this method, the rotor is brought to the synchronous speed
with the help of some external device like small induction motor.
Such an external device is called 'pony motor'.
Once the rotor attains the synchronous speed, the d.c. excitation
to the rotor is switched on. Once the synchronism is established pony
motor is decoupled. The motor then continues to rotate as
synchronous motor.
2. Using Damper Winding:
In a synchronous motor, in addition to the
normal field winding, the additional winding
consisting of copper bars placed in the slots in
the pole faces. The bars are short circuited with
the help of end rings. Such an additional
winding on the rotor is called damper winding.
This winding as short circuited, acts as a squirrel
cage rotor winding of an induction motor. The
schematic representation of such damper
winding is shown in the Fig.1.
• Once the rotor is excited by a three phase supply, the motors starts
rotating as an induction motor at sub synchronous speed. Then d.c. supply
is given to the field winding. At a particular instant motor gets pulled into
synchronism and starts rotating at a synchronous speed. As rotor rotates
at synchronous speed, the relative motion between damper winding and
the rotating magnetic field is zero. Hence when motor is running as
synchronous motor, there can not be any induced e.m.f. in the damper
winding. So damper winding is active only at start, to run the motor as an
induction motor at start. Afterwards it is out of the circuit. As damper
winding is short circuited and motor gets started as induction motor, it
draws high current at start so induction motor starters like star-delta,
autotransformer etc. used to start the synchronous motor as an induction
motor.
3. As a Slip Ring Induction Motor
The above method of starting synchronous motor as a
squirrel cage induction motor does not provide high starting
torque. So to achieve this, instead of shorting the damper
winding, it is designed to a form a three phase star or delta
connected winding. The three ends of this winding are
brought out through slip rings. An external rheostat then can
be introduced in series with the rotor circuit. So when stator
is excited, the motor starts as a slip ring induction motor and
due to resistance added in the rotor provides high starting
torque. The resistance is then gradually cut off, as motor
gathers speed. When motor attains speed near synchronous.
d.c. excitation is provided to the rotor, then motors gets pulled
into synchronism and starts rotating at synchronous speed.
The damper winding is shorted by shorting the slip rings. The
initial resistance added in the rotor not only provides high
starting torque but also limits high inrush of starting current.
Hence it acts as a motor resistance starter.
The synchronous motor started by this method is called
a slip ring induction motor is shown in the Fig.1(b).
It can be observed from the Fig. 1(b) that the same
three phase rotor winding acts as a normal rotor winding
by shorting two of the phases. From the positive
terminal, current 'I' flows in one of the phases, which
divides into two other phases at start point as 1/2
through each, when switch is thrown on d.c. supply side.
4. Using Small D.C. Machine
Many a times, a large synchronous motor are
provided with a coupled d.c. machine. This machine is
used as a d.c. motor to rotate the synchronous motor at a
synchronous speed. Then the excitation to the rotor is
provided. Once motor starts running as a synchronous
motor, the same d.c. machine acts as a d.c. generator
called exciter. The field of the synchronous motor is then
excited by this exciter itself.
Behaviour of Synchronous Motor on Loading
• When a d.c. motor or an induction motor is loaded, the speed of
the motors drops. This is because the load torque demand
increases then the torque produced by the motor. Hence motor
draws more current to produce more torque to satisfy the load but
its speed reduces. In case of synchronous motor speed always
remains constant equal to the synchronous speed, irrespective of
load condition. It is interesting to study how synchronous motor
reacts to changes in the load condition.
•
In a d.c. motor, armature develops an e.m.f. after motoring
action starts, which opposes supply voltage, called back e.m.f. Eb.
•
Hence if Ra the armature resistance and V is the supply voltage,
we have established the relation for the armature current as,
•
• where
Ia = (V- Eb) / Ra
Eb = ΦPNZ / 60A
...... for a d.c. motor
.........for a d.c. motor
•
In case of synchronous motor also, once rotor starts
rotating at synchronous speed, the stationary stator
(armature) conductors cut the flux produced by rotor. The
only difference is conductors are stationary and flux is
rotating. Due to this there is an induced e.m.f. in the stator
which according to Lenz's law opposes the supply voltage.
This induced e.m.f. is called back e.m.f. in case of
synchronous motor. It is obtained as Ebph i.e. back e.m.f.
per phase. This gets generated as the principle of alternator
and hence alternating in nature and its magnitude can be
calculated by the equation,
Ebph α Φ
•
As speed is always synchronous, the frequency is
constant and hence magnitude of such back e.m.f. can be
controlled by changing the flux Φ produced by the rotor.
•
Keypoint: So back e.m.f. in case of synchronous motor
depends on the excitation given to the field winding and
not on the speed, as speed is always constant.
•
As stator construction is similar to the armature
of a three phase alternator, the impedance of the
stator is called synchronous impedance of
synchronous motor consisting of Ra as the stator
winding resistance and Xs as the synchronous
reactance. All the values are generally expressed on
per phase basis.
Zs = Ra + jXs Ω per phase
•
So similar to the d.c. motor, we can write voltage
equation for a synchronous motor as,
• The difference is that this equation is vector
equation as each quantity is alternating and
has different phase. So addition is to be
performed vectorially to obtain the result.
• where Vph is the supply voltage per phase. The
magnitude of Ebph is adjusted almost equal to
Vph, on no load by controlling flux produced by
rotor i.e. field winding.
1.1 Ideal Condition on No Load
•
The ideal condition on no load can be
assumed by neglecting various losses in the
motor.
•
And
Vph = Ebph
•
Under this condition, the magnetic locking
between stator and rotor is in such a way that
the magnetic axes of both, coincide with each
other as shown in the Fig.1. As this is possible
only under no losses condition, is said to be
ideal in case of synchronous motor.
• As magnitude of Ebph and Vph is same and opposes the
phasor diagram for this condition can be shown as in the
Fig. 2.
• In practice this is impossible. Motor has to supply
mechanical losses and iron losses along with small copper
losses. Let us see how it can be explained in case of
synchronous motor.
1.2 Synchronous Motor on No Load (With Losses)
We have seen that Ebph and Vph are
magnitude wise same, which is adjusted by
controlling field current, in turn controlling the flux.
Now due to the various losses practically
present on no load, the magnetic locking exists
between stator and rotor but in such a way that
there exists a small angle difference between the
axes of two magnetic fields as shown in the Fig.3.
• So the rotor axis falls back with respect to stator
axis by angle 'δ' as shown in the Fig.3 This angle
decides the amount of current required to
produce the torque to supply various losses.
•
Hence this angle is called load angle, power
angle, coupling angle, torque angle or angle of
retardation and denoted as δ as mentioned
earlier.
•
The magnetic locking still exists between the
two and rotor rotates at synchronous speed along
with rotating magnetic field maintaining angle
difference between the axes of two fields, as
shown in the Fig. 3(b). The flux lines between the
two get stretched due to such retardation of
rotor axis with respect to stator.
Now though │Ebph │ = │ Vph │, Ebph will not be
located in exact opposition with Vph , but will get
displaced from its initial position by angle'δ' as
shown in the Fig. 4(a).
Fig. 4(a) Phasor diagram for no load condition with losses
Hence the vector difference between the two, Ebph and
Vph is not zero but give rise to a phasor 'OB' as shown.
• This resultant decides the amount of current Iaph to be
drawn to produce the torque which meets the various
losses present in the synchronous motor. Under no
load condition, δ is very small and hence ERph is also
very small.
•
So current drawn by the motor is also very small on
no load which is the case in all the various type of
motors.
• 1.3 Synchronous Motor on Load
•
As the load on the synchronous motor increases, there is no change in
its speed. But what gets affected is the load angle 'δ' i.e. the angle by
which rotor axis retards with respect to stator axis.
•
Hence as load increases, δ increases but speed remains synchronous.
•
As δ increases, though Ebph and Vph magnitudes are same,
displacement of Ebph from its ideal position increases.
•
As synchronous impedance is constant, the magnitude of Iaph drawn by
the motor increases as load increases. This current produces the
necessary torque which satisfied the increased load demand. The
magnetic locking still exists between the rotor and stator.
•
The phasor diagrams showing ERph increases as load increases are
shown in the Fig. 4(b) and (c).
• So from the above discussion it is clear that on no load,
current drawn by the motor is very small.
• As load increases, rotor magnetic axis starts retarding
with respect to stator axis i.e. load angle δ increases
maintaining the magnetic locking condition.
• And hence in case of the synchronous motor load
affects the angle δ without affecting the speed.
• As δ increases, the magnitude of ERph increases which
shows that motor draws more current from the supply.
This satisfies the increased load torque demand.
•
Key point: So torque produced in the synchronous
motor depends on the load angle 'δ' for small values of
and to be precise depends on 'sinδ'. The load angle 'δ'
is measured in degrees electrical.
• As angle δ increases, the magnetic flux lines producing the force of
attraction between the two get more and more stretched. This
weakens the force maintaining the magnetic locking, though torque
produced by the motor increases. As δ reaches upto 90o electrical
i.e. half a pole pitch, the stretched flux lines get broken and hence
magnetic locking between the stator and rotor no longer exists.
The motor comes out of synchronism. So torque produced at δ
equal to 90oelectrical is the maximum torque, a synchronous motor
can produce, maintaining magnetic locking i.e. synchronism. Such s
torque is called pull out torque. The relationship between torque
produced and load angle is shown in the Fig 5
Torques in synchronous motor
•
•
•
•
Starting torque: at starting
Running torque: running condition
Pull in torque: at synchronism
Pullout torque: after load
• Analysis of phasor diagram – Refer Book.
Operation of S.M. at constant Load
Variable Excitation
• We have seen previously that when load
changes, for constant excitation, current
drawn by the motor increases. But if
excitation i.e. field current is changed
keeping load constant, the synchronous
motor reacts by by changing its power factor
of operation. This is most interesting feature
of synchronous motor. Let us see the details of
such operation.
• Consider a synchronous motor operating at a certain load. The
corresponding load angle is δ.
•
At start, consider normal behaviour of the synchronous motor, where
excitation is adjusted to get Eb = V i.e. induced e.m.f. is equal to applied
voltage. Such an excitation is called Normal Excitation of the motor.
Motor is drawing certain current from the supply and power input to the
motor is say Pin. The power factor of the motor is lagging in nature as
shown in the Fig. 1(a).
• Now when excitation is changed, but there is hardly any change in the
losses of the motor. So the power input also remains same for constant
load demanding same power output.
•
Now
Pin = √3 VL IL cos Φ = 3 (Vph Iph cos Φ)
•
Most of the times, the voltage applied to the motor is constant. Hence
for constant power input as Vph is constant, 'Iph cos Φ' remains constant.
•
Note : So far this entire operation of variable excitation it is necessary
to remember that the cosine component of armature current, Ia cosΦ
remains constant.
•
So motor adjusts its cos Φ i.e. p.f. nature and value so that Ia cos Φ
remains constant when excitation of the motor is changed keeping load
constant. This is the reason why synchronous motor reacts by changing
its power factor to variable excitation conditions.
• 1.1 Under Excitation
•
When the excitation is adjusted in such a way that
the magnitude of induced e.m.f. is less than the
applied voltage (Eb < V) the excitation is called Under
Excitation.
•
Due to this, ER increases in magnitude. This means
for constant Zs, current drawn by the motor increases.
But ER phase shifts in such a way that, phasor Ia also
shifts (as ER ^ Ia = θ) to keep Ia cos Φ component
constant. This is shown in the Fig. 1(b). So in under
excited condition, current drawn by the motor
increases. The p.f. cos Φ decreases and becomes more
and more lagging in nature.
• θ: internal machine angle
• 1.2 Over Excitation
•
The excitation to the field winding for which the
induced e.m.f. becomes greater than applied voltage
(Eb < V), is called over excitation.
•
Due to increased magnitude of Eb, ER also increases
in magnitude. But the phase of ER also changes. Now =
ER ^ Ia = θ is constant, hence Ia also changes its phase.
So Φ changes. The Ia increases to keep Ia cos Φ
constant as shown in Fig.1(c). The phase of ER changes
so that Ia becomes leading with respect to Vph in over
excited condition. So power factor of the motor
becomes leading in nature. So overexcited
synchronous motor works on leading power factor. So
power factor decreases as over excitation increases but
it becomes more and more leading in nature.
1.3 Critical Excitation
•
When the excitation is changed, the power factor
changes. The excitation for which the power factor of
the motor is unity (cos Φ = 1) is called critical
excitation. Then Iaph is in phase with Vph. Now Ia cos Φ
must be constant, cos Φ = 1 is at its maximum hence
motor has to draw minimum current from supply for
unity power factor condition.
•
So for critical excitation, cos Φ = 1 and current
drawn by the motor is minimum compared to current
drawn by the motor for various excitation conditions.
This is shown in the Fig. 1(d).
Fig. 1 Constant load variable excitation operation
V-Curves and Inverted V-Curves
• From the previous article, it is clear that if excitation is
varied from very low (under excitation) to very high (over
excitation) value, then current Ia decreases, becomes
minimum at unity p.f. and then again increases. But initial
lagging current becomes unity and then becomes leading in
nature. This can be shown as in the Fig. 1.
•
Excitation can be increased by increasing the field
current passing through the field winding of
synchronous motor. If graph of armature current
drawn by the motor (Ia) against field current (If) is
plotted, then its shape looks like an english alphabet V.
If such graphs are obtained at various load conditions
we get family of curves, all looking like V. Such curves
are called V-curves of synchronous motor. These are
shown in the Fig. 2a).
•
As against this, if the power factor (cos Φ) is
plotted against field current (If), then the shape of the
graph looks like an inverted V. Such curves obtained by
plotting p.f. against If, at various load conditions are
called Inverted V-curves of synchronous motor. These
curves are shown in the Fig. 2(b).
Fig. 2 V-curves and Inverted V-curves
Experimental Setup to Obtain V-Curves
•
Fig. 3 shows an experimental setup to
obtain V-curves and Inverted V-curves of
synchronous motor.
•
Stator is connected to three phase supply
through wattmeters and ammeter. The two
wattmeter method is used to measure input
power of motor. The ammeter is reading line
current which is same as armature (stator)
current. Voltmeter is reading line voltage.
Fig. 3 Experimental setup for V-curves
•
A rheostat in a potential divider arrangement
is used in the field circuit. By controlling the
voltage by rheostat, the field current can be
changed. Hence motor can be subjected to
variable excitation condition to note down the
readings.
•
Now IL = Ia, per phase value can be
determined, from the stator winding
connections.
•
IL = Iaph for stator connection
•
IL/√3 = Iaph for delta connection
•
The power factor can be obtained as
The result table can be prepared as :
•
The graph can be plotted from this result table.
• 1) Ia Vs If → V-curve
• 2) cosΦ Vs If → Inverted V-curve
•
The entire procedure can be repeated for various load
conditions to obtain family of V-curves and Inverted V-curves.
Expression for Back E.M.F or Induced
E.M.F. per Phase in S.M.
• Case i) Under excitation, Ebph < Vph .
•
Zs = Ra + j Xs = | Zs | ∟θ Ω
•
θ = tan-1(Xs/Ra)
•
ERph ^ Iaph = θ, Ia lags always by angle θ.
•
Vph = Phase voltage applied
•
ERph = Back e.m.f. induced per phase
•
ERph = Ia x Zs V
... per phase
•
Let p.f. be cosΦ, lagging as under excited,
•
Vph ^ Iaph = Φ
•
Phasor diagram is shown in the Fig. 1.
• Applying cosine rule to Δ OAB,
•
(Ebph)2 = (Vph)2 + (ERph)2 - 2Vph ERph x (Vph ^ ERph)
•
but Vph ^ ERph = x = θ - Φ
•
(Ebph)2 = (Vph)2 + (ERph)2 - 2Vph ERph x (θ - Φ)
......(1)
•
where ERph = Iaph x Zs
•
Applying sine rule to Δ OAB,
•
Ebph/sinx = ERph/sinδ
• So once Ebph is calculated, load angle δ can be determined by using sine
rule.
• Case ii) Over excitation, Ebph > Vph
•
p.f. is leading in nature.
•
ERph ^ Iaph = θ
•
Vph ^ Iaph = Φ
•
The phasor diagram is shown in the Fig. 2.
•
Applying cosine rule to Δ OAB,
•
(Ebph)2 = (Vph)2 + (ERph)2 - 2Vph ERph x cos(Vph ^ ERph)
•
Vph ^ ERph = θ + Φ
• ... (Ebph)2 = (Vph)2 + (ERph)2 - 2 Vph ERph cos(θ + Φ)
.......(3)
•
But θ + Φ is generally greater than 90o
• ... cos (θ + Φ) becomes negative, hence for leading
p.f., Ebph > Vph .
•
Applying sine rule to Δ OAB,
•
Ebph/sin( ERph ^ Vph) = ERph/sinδ
• Hence load angle δ can be calculated once Ebph is
known.
• Case iii) Critical excitation
•
In this case Ebph ≈ Vph, but p.f. of synchronous
motor is unity.
• ...
cos = 1 ... Φ = 0o
•
i.e. Vph and Iaph are in phase
•
and ERph ^ Iaph = θ
•
Phasor diagram is shown in the Fig. 3.
• Applying cosine rule to OAB,
•
(Ebph)2 = (Vph)2 + (ERph)2 - 2Vph ERph cosθ
•
Applying sine rule to OAB,
•
Ebph/sinθ = ERph/sinδ
• where ERph = Iaph x Zs V
Power Flow in Synchronous Motor
• Net input to the synchronous motor is the three phase
input to the stator.
• ...
Pin = √3 VL IL cosΦ
W
•
where
VL = Applied Line Voltage
•
IL = Line current drawn by the motor
•
cosΦ = operating p.f. of synchronous motor
•
or
Pin = 3 (per phase power)
•
= 3 x Vph Iaph cosΦ W
•
Now in stator, due to its resistance Ra per phase there
are stator copper losses.
Total stator copper losses = 3 x (Iaph)2 x Ra W
• ... The remaining power is converted to the mechanical
power, called gross mechanical power developed by the
motor denoted as Pm.
• ...
Pm = Pin - Stator copper losses
Now P = T x ω
...
Pm = Tg x (2πNs/60) as speed is always Ns
•
This is the gross mechanical torque developed. In d.c.
motor, electrical equivalent of gross mechanical power
developed is Eb x Ia, similar in synchronous motor the
electrical equivalent of gross mechanical power developed
is given by,
•
Pm = 3 Ebph x Iaph x cos (Ebph ^ Iaph)
i) For lagging p.f.,
Ebph ^ Iaph = Φ - δ
ii) For leading p.f.,
Ebph ^ Iaph = Φ + δ
iii) For unity p.f.,
Ebph ^ Iaph = δ
• Note : While calculating angle between Ebph and Iaph from
phasor diagram, it is necessary to reverse Ebph phasor. After
reversing Ebph, as it is in opposition to Vph, angle between
Ebph and Iaph must be determined.
•
In general,
• Positive sign for leading p.f.
Neglecting sign for lagging p.f.
•
Net output of the motor then can be obtained by
subtracting friction and windage i.e. mechanical losses
from gross mechanical power developed.
• ...
Pout = Pm - mechanical losses.
•
where
Tshaft = Shaft torque available to load.
Pout = Power available to load
... Overall efficiency = Pout/Pin
Alternative Expression for Power
Developed by a Synchronous Motor
Refer Book
• This is the expression for the mechanical
power developed interms of the load
angle δ and the internal machine angle θ, for
constant voltage Vph and constant Eph i.e.
excitation.
Condition for Maximum Power Developed
• The value of δ for which the mechanical power
developed is maximum can be obtained as,
The Value of Maximum Power Developed
The value of maximum power developed can be
obtained by substituting θ =δ in the equation of Pm.
• As Eb is completely dependent on excitation, the
equation (8) gives the excitation limits for any
load for a synchronous motor. If the excitation
exceeds this limit, the motor falls out of step.
1.2 Condition for Excitation When Motor
Develops (Pm ) Rmax
• Let us find excitation condition for maximum
power developed. The excitation controls Eb.
Hence the condition of excitation can be
obtained as,