ENGNG 2024 Electrical Engineering
E Levi, 2000
5
One observes that in terms of spatial dependence, all the three individual phase magneto-
motive forces are stationary and they act along the defined magnetic axis of the winding. From
(3) one notices that each of the three m.m.f.’s is varying in time. The values of the three phase
m.m.f.’s in the given instant of time correspond to those met in any three phase system.
The resultant magneto-motive force that stems from the three phase system of currents
flowing through spatially displaced windings is the sum of the individual contributions of the
three phases. The summation is done in the cross-section of the machine, and it is necessary to
observe the spatial displacement between the three m.m.f.’s. One may regard the cross-section
of the machine as a Cartesian co-ordinate system in which phase a magnetic axis corresponds
to x-axis, while y-axis is perpendicular to it. For the purposes of calculation this co-ordinate
system may be treated as a complex plane, with x-axis corresponding to the real axis, and y-
axis corresponding to the imaginary axis. In terms of the complex plane, spatial displacement
of the m.m.f. by 120 degrees corresponds to the so-called ‘vector rotator’,
()
aj= exp /23
π
.
Hence the resultant magneto-motive force can be written as
()
()()
()
FFFF e e
FNI t t t
res a b c
jj
res m
=++ = =
=+−+−
aa a a
aa
2
2
3
2
4
3
2
23 43
π
π
ωωπ ωπ
,
cos cos / cos /
(4)
The expression for the resultant magneto-motive force is most easily found if one recalls the
well-known correlation
()
cos . exp( ) exp( )
δδδ
=+−05 jj. Hence
()() () ()
( )
()
()
FNIee ae ae ae ae
FNIeeaaeaaeaaeaae
aa
aa a a a aa
FNIe aaaae
res
jt jt jt jt jt jt
res m
jt jt jt jt jt jt
res m
jt jt
=+++ + + =
++ + + +
++ =
====
=++++
−−−− −−−
−− −
−
1
2
1
2
10
1
1
2
11
23 23
2
43
2
43
22 22
2
22 3 4
22
ω ω ωπ ωπ ωπ ωπ
ωω ω ω ω ω
ωω
/// /
**
**
=
()
()
() ()
()
aa
FNIe e
FNIe
res m
jt jt
res m
jt
2
1
2
30
3
2
+=
=+
=
−
ωω
ω
(5)
Symbol * in (5) stands for complex conjugate. The result obtained in (5) is an important one.
The equation
FNIe
res m
jt
=
3
2
ω
(6)
describes a circular trajectory in the complex plane. This means that the resulting magneto-
motive force, produced by three stationary, time-varying m.m.f.’s (often called pulsating
m.m.f.’s) is a time-independent and rotating magneto-motive force. Thus it follows that the
three-phase system creates a rotating field (called also Tesla’s field) in the machine. Figure 4
illustrates the rotating field in different instants of time. The speed at which the rotating field
travels equals the angular frequency of the stator three-phase supply.
ENGNG 2024 Electrical Engineering
E Levi, 2000
6
Im
ω
t=90
°
ω
t=135° F
res
ωt=45°
ω
t=0
°
Re (a)
1.5NI
m
Fig. 4 - Resultant field in the three-phase machine for sinusoidal supply conditions.
Since the rotor winding carries excitation current, a field is produced by this current.
This field is stationary with respect to rotor. Since the rotor rotates at synchronous speed,
then, looking in from stationary stator, this rotor field rotates at synchronous speed. This is
always the case in any multi-phase AC machine: regardless of whether the rotor rotates
synchronously or asynchronously, all the fields in the machine rotate at synchronous speed.
Since the resulting m.m.f. is responsible for the resulting flux density and ultimately
resulting flux, this means that apart from the rotating m.m.f., there is a rotating flux density
wave and a rotating flux in the machine as well. The term rotating field in general denotes any
of the three.
In other to show how a synchronous machine can produce reactive power, consider the
operation of a synchronous motor. The motor will always consume real power; however, it can
either generate or absorb reactive power. Let us further consider a couple of characteristic
situations. Let us assume at first that a synchronous motor operates under ideal no-load
conditions, so that all the losses in the motor can be neglected. This means that the input real
power and the output mechanical power are both zero. Such an ideal situation is useful in
explaining the reactive power production and consumption.
Suppose at first that the rotor current is zero. Therefore the rotor does not produce any
rotating field. Rotating field of the stator is generated due to current flow in stator and the
stator rotating field equals the total rotating field of the machine (total field is a vectorial sum
of the field generated by the stator currents and the rotor field; it is fixed since the supply
voltage is fixed, and does not depend on individual values of the rotor and stator fields). Since
rotor current is zero, there is not an induced emf in the stator winding due to rotor flux. Under
these conditions the machine behaves like a pure reactance and consumes reactive power. The
situation is illustrated in Fig. 5a, where total field and stator rotating field are shown for one
instant in time. If the reactance of a stator phase winding is X
s
, then the current in one phase is
I=V/X
s
, where V is the rms of the phase to neutral voltage. Hence the reactive power
consumedbythemachineisQ=3VI=3V
2
/X
s
.
Suppose next that rotor current is now increased to a certain value, say, such that the
emf in the stator winding equals one half of the applied phase to neutral voltage. The situation
is illustrated in Fig. 5b. The stator field is now just one half of the value for zero rotor current,
since one half of the total field is provided by rotor. The current in stator is I = (V - E) /X
s
,and
ENGNG 2024 Electrical Engineering
E Levi, 2000
7
since E is 0.5V, then I = 0.5V/X
s
. Consequently, reactive power consumed by the machine
becomesQ=3VI=1.5V
2
/X
s
. The consumed reactive power is just one half of what it was
originally.
Let the rotor current be now exactly such that the induced emf in stator due to rotor
flux equals applied voltage. This means that the total field exactly equals rotor field and, since
there is not any current in stator, stator field is zero (Fig. 5c). As the stator current is zero, the
machine now neither consumes nor delivers the reactive power.
Finally, let the rotor current be further increased, so that the induced emf becomes
1.5V. The corresponding rotor filed is now larger than the total field and consequently stator
field has to change the direction (Fig. 5d). This means that the stator current now flows from
the machine towards the grid, i.e. I = (V - E) /X
s
= -0.5V/X
s
. Hence the reactive power is now
equal to Q = - 1.5V
2
/X
s
, where negative sign indicates that the reactive power is delivered to
the grid rather than being absorbed.
In all the cases discussed so far stator current was purely reactive and the reactive
power was either taken from or delivered to the grid. If a synchronous machine absorbs
reactive power it is said that it operates in under-excited mode. If a synchronous machine
delivers reactive power to the grid, it is said that it operates in over-excited mode. These two
terms are exclusively related to the reactive power balance of the machine. One notes that for
all these cases the angle between the total rotating field and the rotor rotating field in the
machine is zero.
3. MOTORING MODE OF OPERATION
Suppose now that the synchronous motor drives a load torque. Hence it delivers
mechanical power at the shaft and consequently consumes real (active) power at stator winding
terminals. Let the rotor current be such that the machine is in over-excited mode (Fig. 5d).
What now happens is that an angular displacement occurs between the rotor field and total
field. This angle is called load angle and it is illustrated in Fig. 6, where motoring operation in
over-excited mode is depicted. This load angle will appear in the phasor diagram later on. As
will be shown, power delivered by a synchronous motor directly depends on the value of this
angle.
On the basis of considerations in the previous section, it is easy to derive an equivalent
per-phase representation of the synchronous motor. The applied stator phase to neutral voltage
(note that the value given is always line to line voltage and that stator winding of a
synchronous motor is always connected in star) can be represented with a corresponding
phasor. This voltage phasor corresponds to the total field in Fig. 6. Rotor current causes rotor
flux, that rotates at synchronous speed and therefore induces an emf in the stationary stator
winding. This emf corresponds to the rotor field in the Fig. 6. Hence, in phasor diagram, the
angle between the grid voltage and the induced emf will be equal to the load angle of Fig. 6.
Thus equivalent circuit contains an internal voltage source (induced emf) and the external
applied voltage. In between is the stator reactance of the machine, called synchronous
reactance. Stator winding resistance can usually be neglected and it is omitted from all the
further considerations. Furthermore, iron loss (that takes place in stator core) and mechanical
loss will be neglected as well, so that under these idealised conditions input active power
equals output mechanical power. Equivalent circuit of a synchronous motor with excitation
winding on rotor and uniform air-gap is shown in Fig. 7.
ENGNG 2024 Electrical Engineering
E Levi, 2000
8
F
total
F
s
(= F
res
)
ω
Phase a axis
a.
F
total
F
s
(= F
res
)
F
r
ω
Phase a axis
b.
F
total
F
r
ω
Phase a axis
c.
F
s
F
total
F
r
ω
Phase
a
axis
d.
Fig. 5 - Illustration of rotating fields in a synchronous motor (for ideal no-load conditions)
for four different values of the rotor current.
F
s
F
total
F
r
δ
ω
Phase
a
axis
Fig. 6 - Rotating fields in motoring operation in over-excited mode.
jX
s
I+
VE
Fig. 7 - Equivalent circuit of a synchronous motor.
ENGNG 2024 Electrical Engineering
E Levi, 2000
9
The following phasor equation follows directly from Fig. 7:
EjXI
s
=+ (7)
The phasor diagram, that this equation describes, is drawn in Fig. 8 for two cases: operation in
the under-excited mode when the power factor is lagging (since the machine consumes reactive
power) and operation in the overexcited mode when the power factor is leading (since the
machine consumes active power but simultaneously delivers reactive power to the grid).
jX
s
I
VjX
s
I
VE
δ
E
φδ
φ
II
lagging power factor leading power factor
Fig. 8 - Phasor diagram for motoring operation in under-excited and overexcited modes.
Phasor diagrams of Fig. 8 are used most frequently to determine the unknown load
angle and the induced emf on the basis of the known motor loading and stator terminal
conditions. They simultaneously represent the starting point in deriving the so-called load angle
characteristic of a synchronous motor. Consider the phasor diagram of Fig. 8 for overexcited
mode, that is for convenience re-drawn in Fig. 9. Due to the correlation that exists, one
immediately recognises that the angle BCA is equal to the power factor angle (angle between
current and voltage phasors). On the other hand, angles ODA and ABC are right angles. When
solving the phasor diagram of Fig. 9, it is much easier to use one of the two triangles (triangle
ODC or triangle OBC) and project the phasors on the sides of these two triangles, than to
directly solve the complex phasor equation (7).
In order to obtain the load angle characteristic of a synchronous motor one considers
the triangle OBC in Fig. 9. By projecting the phasors on sides OB and CB one has
C
B
φ
AjX
s
I
D
VE
φδ
I
O
Fig. 9 - Phasor diagram for derivation of load angle characteristic.
ENGNG 2024 Electrical Engineering
E Levi, 2000
10
XI E
XI E
s
s
+=
=
sin cos
cos sin
φ
δ
φδ
(8)
When terminal voltage, stator current and the power factor are known, these two equations
enable simple calculation of the load angle and the induced emf. However, for the purpose of
deriving the load angle characteristic one expresses the active stator current component from
the second equation of (8) as
I
E
X
s
cos
sin
φ
δ
= (9)
Next, it is necessary to recall that the phasor diagram represents one phase of the machine and
that the machine is three-phase. Hence the input active power and the reactive power (that is
either absorbed or delivered, as in Fig. 9) are given in terms of phase to neutral voltage and
phase current as
PVI QVI==33cos sin
φ
φ
(10)
Load angle characteristic of a synchronous machine relates the active power with the load
angle, δ. Note that, since the stator resistance loss and the rotational losses are neglected, input
active power equals active power transferred from stator to rotor and it ultimately equals the
output mechanical power. Substitution of (9) into (10) yields
PVI V
E
X
VE
X
ss
== =33 3cos
sin
sin
φ
δ
δ
(11)
The correlation between real (or output) power and load angle is called load angle
characteristic of a synchronous motor. One observes from (11) that, for the given stator
voltage and rotor current (that is, emf) power depends only on the sine of the load angle. This
means that, higher the load is, higher the value of the load angle will be, and vice versa. In
other words, rotor field in Fig. 6 will lag the total field more and more as the loading of the
machine increases.
The maximum power that a synchronous motor can deliver is met when load angle
equals 90 degrees. Hence, for each rotor current setting (that is, for each value of the emf)
there is a maximum power that the motor can deliver. The maximum power is
P
VE
X
s
max
= 3 (12)
If the power required by the load increases above the maximum power, load angle would go
over 90 degrees and the so-called loss of synchronism would take place. Since increase of the
load angle above 90 degrees leads to decrease of the motor power (compared to maximum
power) rather than an increase, power demanded by the load cannot be met and the motor
becomes unstable (it looses synchronism, that is, the speed starts decreasing). The stable
operation of the motor is therefore possible only for load angle values between zero and 90
degrees.
Torque of the motor can be directly obtained from the load angle characteristic by
dividing the power with the mechanical synchronous angular speed of rotation:
T
P
V
E
XfP
P
f
VE
X
TP
f
VE
X
e
ss s
e
s
== =
=
ω
δ
ππ
δ
π
3
1
2
3
1
2
3
1
2
sin
sin
max
(13)
ENGNG 2024 Electrical Engineering
E Levi, 2000
11
Hence the motor torque dependence on load angle is the same as for the power. The scaling
factor is the constant, synchronous mechanical angular speed of rotation.
Load angle characteristics of a synchronous motor are illustrated in Fig. 10, for a
couple of values of the rotor current (i.e. emf). An increase in the rotor current causes an
increase in the induced emf. Consequently, for the given load, the load angle decreases. An
increase of the rotor current leads to an increase in the maximum power (maximum torque)
that the motor can develop.
P
max1
,T
emax1
P, T
e
E
1
P
max2
,T
emax2
E
2
P
max3
,T
emax3
E
3
STABLE UNSTABLE
E
1
>E
2
>E
3
090
°
180
°δ
Fig. 10 - Load angle characteristic of a synchronous motor.
Reactive power characteristic of the motor can be derived in exactly the same way as
the active power characteristic. Consider the phasor diagram for under-excited operation, that
is for convenience re-drawn in Fig. 11. The following two equations now correspond to (8):
XI E
XI E
s
s
−=
=
sin cos
cos sin
φ
δ
φδ
(14)
They are obtained by using projections of phasor in triangle ABC in Fig. 11. From (14) one
expresses the reactive component of the stator current and then substitutes this expression into
the equation for reactive power, (10). Thus
QVI= 3sin
φ
(10a)
and further
I
VE
X
QVI V
VE
X
VVE
X
Q
V
X
VE
X
s
ss
s
s
sin
cos
sin
cos cos
cos
φ
δ
φ
δδ
δ
=
−
==
−
=
−
=−
33 3
33
2
2
(15)
The first term in this equation is the reactive power that the machine needs for its own
magnetisation. The second term is the reactive power produced by the machine. It follows
from (15) that reactive power is positive (i.e. absorbed) as long as V>Ecosδ. The reactive
power is zero when V=Ecosδ and it becomes negative (i.e. the motor delivers reactive power
to the grid) when V<Ecosδ. Reactive power characteristic is illustrated in Fig. 12.
ENGNG 2024 Electrical Engineering
E Levi, 2000
12
A
jX
s
I
V
φ
BC
δ E
φ
I
Fig. 11 - Phasor diagram of a synchronous motor for under-excited operation.
It is easy to observe in Fig. 12 that the reactive power may be both positive and
negative for the given value of the induced electro-motive force (i.e. rotor current). Whether
the reactive power is delivered or absorbed will depend on the loading of the motor, since the
load at the shaft determines how much the load angle is. The internal reactive power
production of a synchronous motor is at maximum value when load angle is zero. This
situation corresponds to the no-load operation, that is not normally met when the machine is
used as a motor. However, this fact is made use of in so-called synchronous condensers that
are exclusively used for reactive power production and delivery to the grid. A synchronous
condenser is in essence a synchronous motor that operates at all times under no load
conditions. The load angle, although not exactly zero since there are losses in the machine (that
were neglected throughout here) is very small, so that the machine produces maximum possible
amount of reactive power. Synchronous condensers are essentially reactive power generators
and they are installed at various points in a power system to provide the necessary reactive
power support.
Generating mode of operation is discussed in the subsequent section. In principle, all
the characteristics remain valid, but one has to take into account that the active power will now
be produced and delivered to the grid, rather than absorbed. This fact does cause some
inevitable differences,, especially in terms of the phasor diagrams.
Q
3V(V
−
Ecos
δ
)/X
s
3V
2
/X
s
090° 180°δ
−
3VE cos
δ
/X
s
Fig. 12 - Reactive power characteristic of a synchronous motor.
ENGNG 2024 Electrical Engineering
E Levi, 2000
13
EXAMPLES
Example 1:
Calculate the number of poles on the rotor of the synchronous motor whose frequency
is 60 Hz and rated speed of rotation is 200 rpm.
Solution:
Since the frequency is 60 Hz and the synchronous speed is 200 rpm, then from (1) one calculates the
number of pole pairs as P =60xf / n
s
, that is 60 x 60 / 200 = 18. The machine thus has 36 poles, 18
north and 18 south poles.
Example 2:
Calculate the synchronous speeds of two series of synchronous motors designed for 50
Hz and 60 Hz operation, respectively, if the number of pole pairs is from 1 to 10.
Solution:
Synchronous mechanical speed is given with (1),
n
fP
s
=
6
0
/
. Hence
f =50Hz
P 12345678910
n
s
3000 1500 1000 750 600 500 428.5 375 333 300
f =60Hz
P 12345678910
n
s
3600 1800 1200 900 720 600 514 450 400 360
Example 3:
An induction motor drives a load and takes 350 kW from the mains at 0.707 power
factor lagging. An over-excited synchronous motor is connected in parallel with the
induction motor and it takes 150 kW from the grid. If the overall power factor of the
two motors combined becomes 0.9 lagging, calculate the reactive power of the
synchronous motor and its kVA rating.
Solution:
An induction motor always consumes reactive power. The synchronous motor in this example however
produces reactive power. Active powers that the motors take are (1 stands for induction motor, 2
stands for the synchronous motor):
P1 = 350 kW P2 = 150 kW
so that the total power taken from the grid is P = P1 + P2 = 350 + 150 = 500 kW. Since the combined
power factor of the two motors is 0.9 lagging, total reactive power that the two motors take from the
grid is equal to
Q=Q1+Q2=Ptanφ = P tan (arccos0.9) = 500 tan (arccos 0.9) = 242.16 kVAr
Since the power factor of the induction motor is known, one can determine further reactive power
consumed by the induction motor as
Q1 = P1 tan φ1 = 350 tan (arccos 0.707) = 350 kVAr
Hence the synchronous motor delivers the reactive power of
Q2 = Q - Q1 = 242.16 - 350 = - 107.95 kVAr
ENGNG 2024 Electrical Engineering
E Levi, 2000
14
where the negative sign indicates that the synchronous motor delivers reactive power.
Power factor of the synchronous motor is
cos φ2=P2/√(P2
2
+Q2
2
) = 0.81 leading
andthekVAratingis
S2 = √(P2
2
+Q2
2
) = 184.8 kVA
Situation is illustrated in accompanying Figure.
IM SM
350 kW 350 kVAr 150 kW 108 kVAr
500 kW 242 kVAr
Example 4:
A synchronous motor operates on mains of 6.3 kV phase to neutral voltage.
Synchronous reactance is 14 Ohms and the stator resistance can be neglected. The
machine operates under ideal no-load conditions. For induced electromotive forces
equal to 6000 V, 6300 V and 7850 V construct the phasor diagram, and calculate
stator current and reactive power.
Solution:
In this example the machine operates as a synchronous condenser with zero intake of active power.
Hence it either delivers or consumes reactive power. With respect to rotating fields, this operation
corresponds to the one analysed in conjunction with Fig. 5. As noted there, load angle is zero for such
an operation. Hence the induced emf and the applied voltage are in all the three cases under
consideration in phase. Since the voltage is 6300 V phase to neutral, the induced emf of 6000 V means
that the machine consumes reactive power (E cosδ =E,E<V).Statorcurrentis
I=(V-E)/X
s
= (6300 - 6000) / 14 = 21.4 A
and the consumed reactive power is
Q = 3 V I = 3 x 6300 x 21.4 = 404.46 kVAr
In the second case the induced emf and the applied voltage are the same, 6300 V. There is
consequently no current flow in stator and the reactive power is zero. The machine neither consumes
nor delivers reactive power.
In the third case induced emf is 7850 V. It is greater than applied voltage and the machine now
delivers reactive power to the grid. Stator current is
I=(V-E)/X
s
= (6300 - 7850) / 14 = - 110.7 A
and the delivered (negative sign) reactive power is
Q = 3 V I = - 3 x 6300 x 110.7 = - 2.092 MVAr
Corresponding phasor diagrams are given in Figure.
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