Transcript Synchronous Motors

Synchronous Motors





When a synchronous machine is used as a motor, it is not self-starting.
If the rotor field poles are excited by the field current and the stator
terminals are connected to the ac supply, the motor will not start; instead, it
vibrates.
Let consider two-pole synchronous machine connected to a 3, 60 Hz ac
supply. Stator currents will produce a rotating field that will rotate at 3600
rpm in the air gap.
At start (t=0), Fig. a, the rotor will therefore experience a clockwise torque,
making it rotate in the direction of the stator rotating poles.
At t=t1 , let the stator poles move by half a revolution, shown in Fig (b). The
rotor poles have hardly moved, because of the high inertia of the rotor.
Synchronous Motors




Therefore, at this instant the rotor experiences a counterclockwise
torque tending to make it rotate in the direction opposite to that of
the stator poles.
The net torque on the rotor in one revolution will be zero, and
therefore the motor will not develop any starting torque.
The stator field is rotating so fast that the rotor poles cannot catch
up or lock onto it. The motor will not speed up but will vibrate.
Because it not self-started, two methods are normally used to start a
synchronous motor:
 Use a variable-frequency supply
 Start the machine as an induction motor.
Start with Variable-Frequency Supply





By using a frequency converter, a synchronous motor can be
brought from standstill to its desired speed.
The motor is start with a low-frequency supply.
This will make the stator field rotate slowly so that the rotor poles
can follow the stator poles.
Afterwards, the frequency is gradually increased and the motor
brought to its desired speed.
This method is expensive since the frequency converter is a costly
power conditioning unit.
inverter
Start as an Induction Motor







For this purpose, a damper or amortisseur
winding, which resembles the cage of an
induction motor, is mounted on the rotor.
To start the motor the field winding is left
unexcited.
If the motor terminals are now connected to
the ac supply, the motor will start as an
induction motor.
The motor will speed up and will approach synchronous speed.
The rotor is then closely following the stator field poles, which are
rotating at the synchronous speed.
Now if the rotor poles are excited by a field current from a dc source,
the rotor poles will be locked to them.
The rotor will then run at synchronous speed.
Synchronous Generator
Equivalent Circuit Model






An equivalent circuit model can be used to study the
performance characteristics with sufficient accuracy.
Since the steady-state behavior will be studied, the
circuit time constant of the field and damper windings
need not be considered.
The equivalent circuit will be derived on per-phase basis.
The excitation current If in the field winding produces a
flux f in the air gap.
The current Ia in the stator winding produces flux a.
Flux a consists of leakage flux al (link with stator
winding only) and armature reaction flux ar (link with
field winding)
Equivalent Circuit Model
Er - Ear = Ef
a1= leakage flux
ar= armature reaction flux
Ef induced by f,
Ra=effective resistance=1.5 Rdc
Ear induced by ar
Equivalent Circuit Model
Xs=Xar+Xa1 = synchronous reactance
Zs = Ra + jXs = synchronous impedance
Nre = effective field winding resistance
Nse = effective stator phase winding resistance
Generally
Ra << Xs,
in most
cases Ra is
neglected
Determination of The Synchronous
Reactance


The synchronous reactance is an important parameter in
the equivalent circuit of the synchronous machine.
It can be determined by performing two tests, opencircuit test and short circuit test.
Determination of The Synchronous
Reactance
Explain OCC & SCC graphs
Determination of The Synchronous
Reactance
Determination of The Synchronous
Reactance - Saturated
Neglect the drop Ra and Xa1
Zs
• Machine connected to infinite bus bar
•If If chance, Ef changes along the line oc (modified air gap line), not on OCC line.
•For calculation of voltage regulation, Zs is determine at Isc = 2 x full load
Phasor Diagram with Ra
• Indicate relationship between voltage and currents
• Power angle= 
generator
motor
 - magnetic angle
between stator and
rotor axis
Without Ra
Ef
IajXs


vt
 
Ia
sin( 90   )
IajXs
Ia
generating
Ef = Vt + Ia.jXs
Ef
vt

Or I a cos  
Ia X s
Xs
V t I a cos   Pe 
Ef = Vt - Ia.jXs
Using sine rules
sin 
Ef
motoring
sin  ; multiplyin g by V t
Vt E f
Xs
sin 
Pe  Iacos
Effect on generating voltage with
power factor


If the generator drive lagging load power factor, the main air gap flux
(Fr) is reduced since the flux from armature current (Fa) is opposing
the excitation flux (Ff) . This reduced the generated voltage, Er.
If the generator drive leading power factor, the main air gap flux (Fr)
is increased since the flux from armature current (Fa) is aiding the
excitation flux (Ff) . This increased the generated voltage. The better
the power factor, the higher the generated voltage, Er.
Fa
Ff
Ef
Fr
IaXar
Fr
Fa
Er


Ia
vt
IaX1
IaRa
Lagging p factor, Er<Ef
Ef
Ff
Ia


IaXar
Er
IaX1
IaRa
vt
Leading p factor, Er>Ef
Voltage Regulation

When a synchronous generator is connected to a load,
the terminal voltage may vary to a certain extent. The
amount of variation is known as the voltage regulation.

The voltage regulation can be determined by
E f  V  0  I a   x Zs
o
 V  0  I a ( R a  jX s )(cos   j sin  )
o
where  is phase different
% PV 
No load voltage
between phase current and phase voltage
- loaded voltage
No load voltage

E f - Vt
Ef
x 100 %

Based on phasor diagram, at lagging power factor, the
voltage Ef also can be calculated as
Ef 

V
cos   I a R a   V sin   I a X s 
2
2
Based on phasor diagram also, at leading power factor,
the voltage Ef also can be calculated as
Ef 
V
cos   I a R a   V sin   I a X s 
2
2
Example 6.2

The following data are obtained for 3-phase, 10 MVA,14 kV star
connected synchronous machine. Ra is given as 0.07 /phase
If
OC (kV) line to line Air gap voltage(kV) line to line SC current (A)
100
9
150
12
200
14
18
490
250
15.3
300
15.9
350
16.4
a) Find the unsaturated and saturated value of the syn. reactance (in ohm
and p.u)
b) Find the field current required if syn. Generator is connected to infinite
bus and delivers rated MVA at 0.8 lagging power factor
c) If generator in (b) is disconnected from infinite bus without changing the
field current, find the terminal voltage
Sol-Pg17SM
Sen pg. 312
Power and Torque Characteristics



A synchronous machine is normally connected to a
fixed-voltage bus and operates at constant speed.
There is a limit on the power a synchronous generator
can deliver to the infinite bus and on the torque can be
applied to the synchronous motor without losing
synchronism.
Analytical expressions for the steady-state power
transfer between the machine and the constant-voltage
bus or the torque developed by the machine will be
derived.
Power and Torque
Per phase
equivalent
circuit
Complex power S
Complex
phasor
diagram
Power and Torque
(VA)
(Watt)
(VAr)
Power and Torque
Zs=Xs, s = 90o
Power angle /
torque angle
•Both power and torque vary sinusoidal with power angle, 
•Machine can be loaded gradually up to Pmax (static stability limit)
•Beyond 90 degree, machine loss synchronism
•Field current (Ef) need to be increase if the machine loss synchronism
Example

A 3, 5 kVA, 208 V, 4-pole, 60Hz star connected
synchronous generator has negligible stator winding
resistance and a synchronous reactance of 8
ohms/phase at rated terminal voltage.

a) Determine the excitation voltage and power angle
when machine delivers rated kVA at 0.8 lagging PF.
Draw the phasor diagram
b) If the field current is increased by 20% (fixed prime
Mover), find the stator current, power factor, and
reactive kVA supplied by the machine.
c) With field current as in (a), the prime mover is slowly
increased. What is steady state stability limit ?


Sol_pg22_SM
Sen 319
Example 1

The synchronous machine in example 6.3 is operated
as a synchronous motor from the 3, 208 V, 60Hz
power supply. The field excitation is adjusted so that
the power factor is unity when the machine draws 3 kW
from the supply.
a. Find the excitation voltage and the power angle.
Draw the phasor diagram for this condition.
b. If the field excitation is held constant and the
shaft load is slowly increased, determine the
maximum power & torque that the motor can
deliver.
Sen pg. 322
Ans: 137.35< -29o V, 6185 W, 32.8 Nm
The Effect of Changing Excitation Current at
Constant Power Output When connected to
infinite bus bar



Generator output power (W) control by the amount of steam applied.
Pe = VIcos , i.e. Pe proportional to Icos  (fixed)
Assume the generator operates in lagging power factor
Power factor =
unity (1)
under-excited
Ef2
Ef3
Limit of
instability
Over-excited (lagging p f)
Ef1
Ia1jXs
Ia3


Ia1
Constant
power
output
vt
Ia2
Ia = Iacos  + jIasin 
Excitation
voltage(E) slides
on the constant
power line
Under excited
= If small
Over excited
= If big
fixed changed
**Remarks: Reactive Power delivered by the generator can be
controlled by the excitation/field current, in over excitation
machine supplies inductive reactive power**
The Effect of Changing Excitation Current at
Constant Power Output When connected to
infinite bus bar

Example :
The star-connected synchronous machine has
synchronous reactance of 20 ohm/phase. It supplies
load current of 150 A with power factor 0.8 lagging to an
infinite bus bar. The bus bar voltage is 11 kV. If the
supply steam to the machine prime mover is fixed,
calculate the percentage change in excitation voltage if
the load power factor load is 0.8 leading. Neglect power
losses.
Ans: Eo1 = 8497 < 16.41o, Eo2 = 5145<21.8o; -39%
The effect of Changes the Output Power on Infinite
Bus Bar at Fixed Excitation Current
Constant
power
lines
Ia3jXs
Ef3
Ef2
Limit of
stability
of Pmax
Ia2jXs
Ef1

2
1
Ia1
vt
Unity
pf line
Ia1jXs
power
output
varies,
P  Iacos
Excitation
voltage(E)
swings from one
the constant
power line to
another
|Ef1|=|Ef2|=|Ef3|
Ia = Iacos  + jIasin 
changed fixed
**Remarks: Active Power delivered by the generator can be
controlled by the steam supplied into prime mover**
Example 2
 A 3, 250 hp, 2300 V, 60Hz, Y-connected non-salient rotor
synchronous motor has a synchronous reactance of 11  per phase.
When it draws 165.8 kW the power angle is 15 electrical degrees.
Neglect ohmic losses. Determine:
a. The excitation voltage per phase.
b. The supply line current.
c. The supply power factor.
1769   15 V 
o
54 .14  39 .8 A 
(0.77 leading)
d. If the mechanical load is thrown off and all losses become
negligible,
a. Determine the new line current and supply power factor.
 40 .1  90
o
A  40 . 1  90 A ; pf  0
o

a. Draw the phasor diagram for the condition in (i).
b. By what percent should the field current If be changed to
minimize the line current?
(75 %)
Page 371, Q: 6.2
Complex Power Locus/ Capability Curve
•Define the bounds within which generator can work safely
1.
MVA loading < gen. rating (-- stator/armature heating)
2. MW-loading < turbine rating (MVA x power factor)
3. Save from steady state stability limit (  < 90 o)
4. Max. field current < specified by rotor heating
Neglect Ra, multiply each phasor by Vt/Xs
M
P
Ef
IajXs
0’


vt
XsIasin
VtIacos
Xs

0
Ia
Vt E f
XsIacos
N
M
0’


Vt
X
2
s
0
VtIa
VtIasin
Q
0’
Power Factor Control



An outstanding feature of the synchronous machine is
that the power factor of the machine can be controlled by
the field current.
The field current can be adjusted to make the stator (or
line) current lagging or leading as desired.
The power factor characteristic can be explained by
drawing phasor diagrams of machine voltages and
currents.
Power Factor Control
Unity p factor
Leading pf
Lagging pf
Under-excited
Over-excited
Normal -excited
Vt=Ef + jIaXs
Ef =Vt - jIaXs
Utilize field current to stator current leading or lagging
Power Factor Control




The unique feature of power factor control by the field
current can be utilized to improve the power factor of a
plant.
In a plant most of the motors are normally induction
motors, which draw power at lagging power factors.
Synchronous motors can be installed for some drives in
the plant and made to operate in an overexcited mode
so that these motors operate at leading power factors.
This will compensate the lagging power factor of the
induction motors and thereby improving the overall
power factor of the plant.
Example: 2009/10
Q5:
(a) Discuss briefly the effect of the generated voltage when synchronous
generator operates in lagging and leading power factor.
(b) Draw a per phase equivalent circuit of a synchronous machine. Label
all parameters clearly. Derive the real power generated output of that
machine.
(c) A three phase, 1500 kVA, 12 kV, six pole, 50 Hz, star connected
synchronous generator has a synchronous resistance and synchronous
reactance of 2 W/phase and 35 W/phase respectively. If the generator is
connected to a load of 120 kW, determine the voltage regulation if the load
is operated at a power factor of
i. 0.8 lagging.
ii. 0.8 leading.
Sol_pg33-SM
Example 07/08-2
Sol-pg34