# SM2- 1 SYNCHRONOUS MACHINES OPERATION AND ```SYNCHRONOUS MACHINES OPERATION AND CHARACTERISTICS
Ef = f(If) for n=nN (const.)
open-circuit characteristic (occ)
Short-circuit state.
Isc = f(If) for n=nN (const.)
short-circuit characteristic (scc)
Per-unit values: Ir=I/IN Ur=U/UN (Er=E/UN)
IfoN – nominal field current at
IfscN – nominal field current at
short circuit (Isc=IN)
IfoN is usually applied as a
base for relative values of
field current
Ifr = If/IfoN
Short-circuit ratio is an important parameter of synchronous generator:
K sc =
IfoN
1
=
IfscN IfscNr
≈ 0.3 − for cylindrical mach. (turbogene rator)

≈ 1.5 - for salient - pole m. (hydrogene rator)
Assume SM excited by IfoN. At no-load: Ef=UN.
At short circuit: Isc=CD and the state of machine can be shown by the
following phasor diagram:
 BD [V ]
CD [A ] = X d [Ω] for saturated machine
Ef

Xd =
⇒
Isc
 AD [V ] = X
dun [Ω ] for unsaturate d machine
 CD [A ]
BD [p.u.]
= X dr [p.u.]
CD [p.u.]
X dr =
Xd
Xd
=
ZN U N I N
From the triangles:
BD [p.u.] IfscNr
=
CD [p.u.]
1
⇒
X dr =
1
K sc
SM2- 1
a)
External characteristics; voltage-current characteristics; constant
excitation characteristics
U = f(I)
for
If = const
cosϕ = const
n = const
b)
Regulation characteristics; constant voltage characteristics
If = f(I)
for
U = const
cosϕ = const
n = const
‘ lagging character of load; Ef &gt; U ; machine is
overexcited
underexcited
Characteristics for a) and b):
Voltage regulation
∆U r =
Ef − U
U
[p.u.]
for nominal load current, nominal excitation and nominal power factor
∆U rN =
EfN − UN
UN
[p.u.]
∆Ur for lagging power factor can be of significant value; it can be of zero
value for leading power factor or even be negative.
Usually – in power stations – synchronous generators operate on
lagging loads (p.f. = 0.8 – 0.9 lag) and have a large positive voltage
regulation (voltage drop). To keep the voltage constant a wide range
control of field current is required.
SM2- 2
c)
V-curves of synchronous generator (Mordey’s curves)
I = f(If)
for
SM2- 3
P = const
U = const
n = const
For example for P = 0
U s = jX d I
Us
E −U
I = −j
= −j f
Xd
Xd
I =
E
U
− f
Xd
Xd
⇒
If we neglect saturation of magnetic core
Ef = cIf and V-curves for P = 0 and other
values of active power are shown ⇒
d)
Active power – load angle characteristic P = f(ϑL) or Te = f(ϑL)
In cylindrical generator (Xd = Xq) and with assumption R1 ≅ 0
P = mUI cos ϕ
and due to phasor
diagram
X d I cos ϕ = Ef sinϑL
I cos ϕ =
Ef
sin ϑL
Xd
P =m
UEf
sinϑL
Xd
hence
and due to the relation T =
1
P to so
Ω
described active power corresponds electromagnetic torque
Te =
1 UEf
sinϑL
m
Ω
Xd
Both relations are sinusoidal
functions.
Tb – breakdown torque (maximum
torque) for given Ef (i.e. for given
excitation); - it appears for ϑL = 90o.
Tb
P
= b ≥2
TN
PN
(stability margin);
therefore, usually ϑLN ≤ 30o
For ϑL &lt; 0 Te &lt; 0 - negative value of
torque means DRIVING TORQUE.
MOTOR MODE OF OPERATION
These characteristics are
named “power angle
characteristics” in US and
“angle characteristics” in
Poland.
In salient-pole generator (Xd ≠ Xq) and with assumption R1 ≅ 0
I is resolved into d- and qcomponents.
SM2- 4
This phasor diagram is an
example of Blondel’s diagram.
Voltage drops corresponding to Id &amp; Iq
currents are:
Ud = XqIq
Uq = XdId
From the diagram
Ef = U cos ϑL + X d Id from where
U sin ϑL = X q Iq ⇒
Id =
Ef
U
−
cos ϑL and
Xd
Xd
Iq =
U
cos ϑL
Xq
Substituting these relations (forId and Iq) to the following expression for
power
P = mUI cos ϕ = mU (Id sin ϑL + Iq cos ϑL )
yields
UEf
U 2  1
1 
−
P =m
sin ϑL + m
sin 2ϑL
Xd
2  X q X d 
144244
3 14444
4244444
3
1
(T=P/Ω)
2
1 – power (torque) due to field
excitation (synchronous
torque)
2 – power (torque) due to
saliency (Xd ≠ Xq); it doesn’t
depend on Ef (or If) !
RELUCTANCE TORQUE !
RELUCTANCE MACHINES
(with no excitation)
3 – total power (torque).
Rotors of synchronous reluctance motors
1 – ferromagnetic core
2 – starting &amp; damping cage
3 – magnetic flux barriers
Xd
= 1. 5 &divide; 2
Xq
Xd
= 2. 5 &divide; 3
Xq
Xd
=5
Xq
SM2- 5
OPERATION OF SYNCHRONOUS GENERATOR WITH A POWER
SYSTEM
SYNCHRONIZATION OF GENERATOR – connection of synchronous
generator on to the system
busbars.
CB – circuit-breaker
Condition of correct synchronization:
To avoid heavy currents flow after
switching on, the voltages (potentials)
of generator terminals and system
terminals should be of equal value
before and after connection – equal
potentials at CB terminals:
uug = uusy
uvg = uvsy
uwg = uwsy
instantaneous values for t o−+ must be the same.
to – time of CB closing
Practically this one general conditions is satisfied when the following
“more practical” conditions are fulfilled:
1.
rms values Ug = Usy (≈5% difference allowed);
2.
frequencies fg = fsy (0.2 – 0.4 Hz difference allowed);
3.
phase sequences are the same (of generator &amp; system)
4.
instantaneous values are practically equal at the moment of
connection
α &lt; 5o
co-phasal position
of generator &amp;
system voltage
stars;
5.
voltage curves (waveforms) of generator &amp; system are sinusoidal.
In power stations – synchronization by means of synchroscope or
automatic.
In laboratory: voltmeter + lamps connected in so called “rotating light
arrangement with one dark bulb”:
Three bulbs are located
symmetrically at circumference
of a circle.
In case of the same phase
sequences of GEN and SYS an
effect of rotating light appears.
The speed of rotation is
proportional to the difference of
frequencies.
In case of different sequences
all bulbs pulsate
simultaneously.
OPERATION OF GENERATOR AT SYSTEM BUSBARS
(infinite busbars)
After synchronization with infinite busbars: Ug = Usy = const; f = const
a) Field current regulation ( = reactive power regulation)
ϑL = 0 P = 0
By means of excitation regulation we can control
ONLY REACTIVE POWER:
when If &gt; IfoN – reactive (inductive) power is delivered to the system;
when If &lt; IfoN – reactive (capacitive) power is delivered to the system or
reactive (inductive) power is absorbed from the system
and delivered to generator.
b) Torque (mechanical power, active power) regulation
point 0 – after “ideal”
synchronization Efo = Us1 and I = 0
point 1 – If ↑ ; Ef ↑ and Ef1 &gt; U ⇒
Us1 ⇒ I1 (overexcitation)
P1 = mUI1 cos ϕ1 = 0
Q1 = mUI1 sin ϕ1 = mU1I1
point 2 – driving torque T ↑
(mechanical power ↑)
Ef2 = Ef1 but rotor accelerates and
ϑL = ϑL2 ⇒ Us2 ⇒ I2
P2 = mUI2 cos ϕ 2
(b )
Q2 = mUI2 sin ϕ 2
Te 2 = cmUI2 cos ϕ 2 = m
UEf 2
sin ϑL 2 = T
Xd
steady-state operation, active power P2 delivered to the system.
Point 3 - If ↑ ; Ef ↑ Ef3 ⇒ Us3 ⇒ I3 only transiently! It is not steady state
because
P3 = mUI3 cos ϕ 3
Te3 = cmUI3 cos ϕ 3 f T and generator is slowing down to point 4
Point 4 – due to Te3 &gt; T angle ϑL ↓ ϑL4 ⇒ Us3 ⇒ I4
P4 = mUI 4 cos ϕ 4 = P2 (b ) Te 4 = T
Q4 = mUI 4 sin ϕ 4 f Q2
steady-state operation with the same P but with Q increased. Lines a
and b are hodographs of constant active power (torque), active currents.
SM2- 6
Conclusion: During operation of synchronous generator with U = const
(connection to infinite busbars):
1. By means of field current the control of only reactive power is
possible;
2. Control of active power delivered to the system is possible only by
means of mechanical power driving the generator (regulation of
prime mover – turbine);
3. Both above regulations practically don’t affect each other (are
independent).
SM2- 7
SYNCHRONOUS MOTOR
Consider 3 steps of synchronous machine (generator) operation:
1. synchronization,
2. overexcitation,
3. application of breaking torque (mechanical load) to the machine’s
shaft:
What is the result?
Te =
1 UE f
cos ϕ &lt; 0 (!)
Ω Xd
For the machine considered as generator the negative torque
(electromagnetic torque produced in the machine) means driving
character of this torque (not breaking as it is in generator).
The machine becomes a motor. Active power delivered to the system
P = mUI cos ϕ &lt; 0
i.e. the active power is delivered to the machine
(to the motor) from the mains.
All possible positions of armature current phasor:
Overexcited synchronous machine
can operate as synchronous motor
(converting electrical active power into
mechanical) and simultaneously can
deliver reactive inductive power to the
supply (to the system).
How to start such a motor?
Compare this situation to
induction motor. Induction
motor always absorbs reactive
inductive power from the mains.
Synchronous motor seems to
be much better from this point
of view.
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