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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).
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.
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,
Ia = (V- Eb) / Ra
...... for a d.c. motor
where
Eb = ΦPNZ / 60A
.........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.
or
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.
Fig. 1 Magnetic locking under no load condition
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 alongwith
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 magnitudewise 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.
Fig. 3 Magnetic locking under practical condition
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
alongwith 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).
Fig. 4
So from the above discussion it is clear that on no load, current drawn by the motor is very
small. This is because the stator and the rotor magnetic axes are almost matching transformer
each other i.e. load angle δ 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.
Fig. 5 Torque angle characteristic
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, changes 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.
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.
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 Vcurves 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
1.1 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 top 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.
Observation Table :
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.
Fig. 1 Phasor diagram for under excited condition
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)
where ERph = Iaph x Zs
Applying sine rule to Δ OAB,
Ebph/sinx = ERph/sinδ
......(1)
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.
Fig.2 Phasor diagram for overexcited condition
...
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.
Fig. 3 Phasor diagram for unity p.f. condition
Applying cosine rule to OAB,
(Ebph)2 = (Vph)2 + (ERph)2 - 2Vph ERph cos θ
Applying sine rule to OAB,
Ebph/sinθ = ERph/sinδ
where ERph = Iaph x Zs V
............(5)
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 ([er 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
Consider the phasor diagram of a synchronous motor running on leading power factor cosΦ as
shown in the Fig. 1.
Fig. 1
The line CD is drawn at an angle θ to AB.
The lines AC and DE are perpendicular to CD and AE.
and angle between AB = Ebph and Iaph is also ψ.
The mechanical per phase power developed is given by,
In triangle OBD,
BD = OB cosψ = Ia Zs cosψ
OD = OB sin ψ = Ia Zs sin
Now BD = CD - BC = AE - BC
Substituting in (2),
Ia Zs cosψ = Vph cos (θ-δ) - Eb cosθ
All values are per phase values
Substituting (3) in (1),
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,
Note : Thus when Ra is negligible, θ = 90o for maximum power developed. The corresponding
torque is called pull out torque.
1.1 The Value of Maximum Power Developed
The value of maximum power developed can be obtained by substituting θ =δ in the
equation of Pm.
When Ra is negligible,
...
θ = 90o and cos (θ) = 0 hence,
Ra = Zs cosθ and Xs = Zs sinθ
Substituting cosθ = Ra/Zs in equation (6b) we get,
Solving the above quadratic in Eb we get,
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,
Assume load constant hence δ constant.
but
θ = δ for Pm = (Pm)max
Substitute
cosθ = Ra/Zs
This is the required condition of excitation.
Note : Note that this is not maximum value of but this is the value of foe which power developed
is maximum.
The corresponding value of maximum power is,
Hunting in Synchronous Motor
It is seen that, when synchronous motor is on no load, the stator and rotor pole axes almost
coincide with each other.
When motor is loaded, the rotor axis falls back with respect to stator. The angle by which rotor
retards is called load angle or angle of retardation δ.
If the load connected to the motor is suddenly changed by a large amount, then rotor tries to
retard to take its new equilibrium position.
But due to inertia of the rotor, it can not achieve its final position instantaneously. While
achieving its new position due to inertia it passes beyond its final position corresponding to new
load. This will produce more torque than what is demanded. This will try reduce the load angle
and rotor swings in other direction. So there is periodic swinging of the rotor on both sides of the
new equilibrium position, corresponding to the load. Such a swing is shown in the Fig. 1.
Fig. 1 Hunting in synchronous motor
Such oscillations of the rotor about its new equilibrium position, due to sudden application or
removal of load is called swinging or hunting in synchronous motor.
Due to such hunting, the load angle changes its value about its final value δ. As changes, for
same excitation i.e. Ebph the current drawn by the motor also changes. Hence during hunting
there are changes in the current drawn by the motor which may cause problem to the other
appliances connected to the same line. The changes in armature current due to hunting is shown
in the Fig. 2.
Fig. 2 Current variations during hunting
If such oscillations continue for longer period, there are large fluctuations in the current. If
such variations synchronous with the natural period of oscillation of the rotor, the amplitude of the
swing may become so great that motor may come out of synchronism. At this instant mechanical
stresses on the rotor are sever and current drawn by the motor is also very large. So motor gets
subjected to large mechanical and electrical stresses.
Note : Hence hunting is not desirable phenomenon from motor point of view and must be
prevented.
1.1 Use of Damper Winding to Prevent Hunting
It is mentioned earlier that in the slots provided in the pole faces, a short circuited winding is
placed. This is called damper winding.
When rotor starts oscillating i.e. when hunting starts a relative motion between damper
winding and the rotating magnetic field is created. Due to this relative motion, e.m.f. gets induced
in the damper winding. According to Lenz's law, the direction of induced e.m.f. is always so as to
oppose the cause producing it. The cause is the hunting. So such induced e.m.f. oppose the
hunting. The induced e.m.f. tries to damp the oscillations as quickly as possible. Thus hunting is
minimised due to damper winding.
The time required by the rotor to take its final equilibrium position after hunting is called as
setting time of the rotor. If the load angle is plotted against time, the schematic representation of
hunting can be obtained as shown in the Fig. 3. It is shown in the diagram that due to damper
winding the setting time of the rotor reduces considreably.
Fig. Effect of damper winding on hunting
Synchronous Condensers
When synchronous motor is over excited it takes leading p.f. current. If synchronous motor is on
no load, where load angle δ is very small and it is over excited (Eb > V) then power factor angle
increases almost upto 90o. And motor runs with almost zero leading power factor condition. This
is shown in the phasor diagram Fig. 1.
Fig. 1 Synchronous condenser
This characteristics is similar to a normal capacitor which takes leading power factor current.
Hence over excited synchronous motor operating on no load condition is called as synchronous
condenser or synchronous capacitor. This is the property due to which synchronous motor is
used as a phase advancer or as power improvement device.
1.1 Disadvantage of Low Power Factor
In various industries, many machines are of induction motor type. The lighting and heating
loads are supplied through transformers. The induction motors and transformers draw lagging
current from the supply. Hence the overall power factor is very low and lagging in nature.
The power is given by,
P = VI cosΦ
.............. single phase
...
I = P/(VcosΦ)
The supply voltage is constant and hence for supplying a fixed power P, the current is
inversely proportional to the p.f. cosΦ. Let P = KW is to be supplied with a voltage of 230 V then,
Case i) cosΦ = 0.8,
I = (5 x103)/(230 x 0.8) = 27.17 A
Case ii) cos = 0.6,
I = (5 x103)/(230 x 0.6) = 36.23 A
Thus as p.f. decreases, becomes low, the current drawn from the supply increases to supply
same power to the load. But if p.f. maintained high, the current drawn from supply is less.
The high current due to low p.f. has following disadvantages :
1. For higher current, conductor size required is more which increases the cost.
2. The p.f. is given by
cosΦ = Active power/ Apparent = (P in KW)/ (S i.e. KVA rating)
Thus for fixed active power P, low p.f. demands large KVA rating alternators and
transformers. This increases the cost.
3. Large current means more copper losses and poor efficiency.
4. Large current causes large voltage drops in transmission lines, alternators and other
equipments. This results into poor regulation. To compensate such drop extra equipments is
necessary which further increases the cost.
Note : Hence power factor improvement is must practice. Hence the supply authorities
encourage consumers to improve the p.f.
1.1 Use of Synchronous Condenser in Power Factor Improvement
The low power factor increases the cost of generation, distribution and transmission of the
electrical energy. Hence such low power factor needs to be corrected. Such power factor
correction is possible by connecting synchronous motor across the supply and operating it on no
load with over excitation.
Now let Vph is the voltage applied and I1ph is the current lagging Vph by angle Φ1. This power
factor Φ1 is very low, lagging.
The synchronous motor acting as a synchronous condenser is now connected across the
same supply. This draws a leading current of I2ph.
The total current drawn from the supply is now phasor of Iph and I2ph. This total current IT now
lags Vph by smaller angle Φ due to which effective power factor gets improved. This is shown in
the Fig. 2.
Fig. 2 Power factor correction by synchronous condenser
This is how the synchronous motor as a synchronous condenser is used to improve power
factor of the combined load.