Synchronous Generator Introduction
Synchronous Generator or Alternator
A two-pole round rotor generator and exciter.
http://elektro.fs.cvut.cz/en/SSem/2141025/Synchronous_Machine.pdf
Cross Section of a Large Turbo Generator
(Courtesy Westinghouse)
http://elektro.fs.cvut.cz/en/SSem/2141025/Synchronous_Machine.pdf
Round Rotor of a Large Generator
(Courtesy Westinghouse)
http://elektro.fs.cvut.cz/en/SSem/2141025/Synchronous_Machine.pdf
Round Rotor with Conductor Placed
http://elektro.fs.cvut.cz/en/SSem/2141025/Synchronous_Machine.pdf
Four Pole Salient Pole Rotor
http://elektro.fs.cvut.cz/en/SSem/2141025/Synchronous_Machine.pdf
A Large Salient Pole Hydro Generator (1)
rotor
http://elektro.fs.cvut.cz/en/SSem/2141025/Synchronous_Machine.pdf
A Large Salient Pole Hydro Generator (2)
stator
http://elektro.fs.cvut.cz/en/SSem/2141025/Synchronous_Machine.pdf
Two Methods to Provide DC Field Current
1. Supply the DC power from an external DC source to the rotor
by means of slip rings and brushes.
- used in small synchronous generators, cost effective
- brushes need to be checked for wear regularly
- brush voltage drop can be significant power loss
2. Supply the DC power from a special DC power source (exciter)
mounted on the shaft of the synchronous generator.
Rotor with Slip Rings
2 Stack Synchronous Generator System
3-Stack Synchronous Generator System (1)
(completely independent of any external power source)
3-Stack Synchronous Generator System (2)
Aircraft Synchronous Generator
J. F. Gieras, Advancements in Electric Machines, Springer, 2008.
Synchronous Generator Phasor
Diagram and Power Flow
Generator Phasor Diagram – Lagging Power Factor
E A  V  Rs I A  jX s I A
Over excited ( E A  V )
Generator Phasor Diagram – Unity Power Factor
E A  V  Rs I A  jX s I A
Slightly over excited ( E A  V )
Generator Phasor Diagram – Leading Power Factor
E A  V  Rs I A  jX s I A
Maybe under excited ( E A  V )
Generator Power Flow
Pconv  Pout  PCu
Pconv  Tind m  3E A I A cos 
  E A  I A
V   Rs I A  jX s I A  E A
Output Power
Pout  3V I A cos 
E A sin 
I A cos  
Xs
E A  V  Rs I A  jX s I A  V  jX s I A , since Rs  X s
Pout  Pconv 
3V E A sin 
XS
Induced Torque
Pconv 
3V E A sin 
Tind 
XS
 Tind m
3V E A sin 
m X S
Tind  kB net B R sin 
Synchronous Generator
Model Parameter Measurement
Measurement of Model Parameters
1. The relationship between field current and EA
2. The synchronous reactance
3. The armature resistance
Open Circuit Characteristics (1)
(or ET )
Saturation for large field current
E A, rms
Dl
 8 2 f e 0
geff
 Nˆ a Nˆ f

2
P


 I f

Open Circuit Characteristics (2)
Measurement Procedure:
1. Drive rotor at rated synchronous speed.
2. Increase field current If upward from 0
3. Take data of Voc and plot Voc (or EA) vs If
Short Circuit Characteristics (1)
(or IL )
IA 
EA
Rs  jX s
BR  BS
B net  0
V  0
IA 
EA
Rs2  X s2
V  0
No saturation
Short Circuit Characteristics (2)
Take three rms values in average
and get
1
I A , sc   I a  I b  I c 
3
Measurement Procedure:
1. When  = 0 and If = 0, short the machine through
three ammeters
2. When If = 0, drive the machine to rated
synchronous speed
3. Increase If , take data (rms values) from three
ammeters and plot IA,sc vs If
Measurement of Synchronous Reactance
I A , sc 
EA
R s2  X s2

EA
 Xs 
I A , sc
EA
Xs
(1)
Therefore, an approximate method for determining the synchronous reactance Xs
at a given field current is:
1.
2.
3.
Get the internal voltage EA from the OCC at that field current.
Get the short-circuit current flow IA,SC at that field current from the SCC.
Find Xs by applying (1).
Unsaturated Synchronous Reactance
Saturation for large field current
EA
X S ,u
I A , sc
No saturation
Follow equation
X
s

EA
I A ,sc
The unsaturated synchronous reactance Xs,u can be found simply by applying
X s  E A / I A,sc
at any field current in the linear portion (on the airgap line) of the OCC curve.
Short-Circuit Ratio
Short Circuit Ratio: The ratio of the field current required for the
rated voltage at open circuit to the field current required for the rated
armature current at short circuit.
SCR 
I fV |rated voltage at OC
I fI |rated current at SC
EA
V,rated
SCR 
I fV
I fI

Ix
I A,rated

I x V ,rated
1
Zb

V ,rated I A,rated X s,I fV
SCR is inversely
proportional to Xs
Zb 
V ,rated
I A,rated
base impedance
Measurement of Armature Resistance
The armature resistance Rs can be approximately measured by applying a DC voltage
to the windings while the machine is in stationary and measuring the resulting current
flow. Using DC voltage means that the reactance of the windings will be zero during the
measurement process.
This technique is not perfectly accurate, since the AC resistance will be slightly larger
than the DC resistance (as a results of the skin effect at higher frequencies).
EA versus If under Load
Round rotor generator rated at 93.75 MVA
and 0.8 power factor lagging under two load
conditions.
A
Note: EA is not quite proportional to If
under load since the extrapolation results
in an intercept not at origin.
For many practical applications, people
Develop a useful approximate proportional
relationship between EA and If .
Synchronous Generator
Operation
Effect of Generator Loads – Lagging Power Factor
Keep field excitation the same
E A  E A'
E A  V  jX s I A
If lagging loads (+Q or inductive reactive power loads) are added to a generator,
the phase voltage V and the terminal voltage VT decrease.
Effect of Generator Loads – Unity Power Factor
Keep field excitation the same
E A  E A'
E A  V  jX s I A
If unity-power-factor loads (no reactive power) are added to a generator,
the phase voltage V and the terminal voltage VT slightly decrease.
Effect of Generator Loads – Leading Power Factor
Keep field excitation the same
E A  E A'
E A  V  jX s I A
If leading loads (-Q or capacitive reactive power loads) are added to a generator,
the phase voltage V and the terminal voltage VT may increase.
Generator Voltage Regulation
VR 
Vnl  V fl
V fl
 100%
Lagging Load
-> large positive voltage regulation
Unit Power Factor Load -> small positive voltage regulation
Leading load
-> may be negative voltage regulation
Generator V Curves
• The shape is like the letter “V”
• For each fixed real power, plot
armature current vs. field
current.
Example 1 (1)
A 480 V, 60 Hz,  connected, four pole synchronous generator has the OCC curve shown in the figure.
This generator has a synchronous reactance of 0.1  and an armature resistance of 0.015 .
At full load, the machine supplies 1200 A at 0.8 PF lagging. Under full load conditions, the friction
and windage losses are 40 kW, and the core losses are 30 kW. Ignore any field circuit losses.
(a) What is the speed of rotation of this generator?
(b) How much field current must be supplied to the generator to make the terminal voltage 480 V at
no load?
(c) If the generator is now connected to a load and the load draws 1200 A at 0.8 PF lagging, how much
field current is required to keep the terminal voltage equal to 480 V?
(d) How much power is the generator now supplying? How much power is supplied to the generator
by the prime mover? What is the machine’s overall efficiency?
(e) If the generator’s load were suddenly disconnected from the line, what would happen to its
terminal voltage?
(f) Finally, suppose that the generator is connected to a load drawing 1200 A at 0.8 PF leading, how
much field current would be required to keep VT at 480 V?
Example 1 (2)
sg1.m
Example 2 (1)
A 480V, 60 Hz, Y connected, six pole synchronous generator has a synchronous reactance of 1  and
an armature resistance of 0.1 .
At full load, the machine supplies 60 A at 0.8 PF lagging. Under full load conditions, the friction
and windage losses are 1.5 kW, and the core losses are 1.0 kW. Ignore any field circuit losses.
(a) What is the speed of rotation of this generator?
(b) What is the terminal voltage of this generator at full load assuming the field excitation current
keeps the same as no load?
(c) What is the efficiency of this generator at full load?
(d) How much shaft torque must be supplied by the prime mover at full load? How large is the
induced counter torque?
(e) What is the voltage regulation of this generator ?
After the MatLab program can work, please change the load current to be 60 A
at 1.0 PF, and 60 A at 0.8 PF leading and redo the above.
sg2.m
Example 2 (2)
Let the angle of V :  V =0.
Imaginary part of
V   R s I A  jX s I A  E A

becomes:

0   R s I A sin  I A  X s I A cos  I A  E A sin 
 sin  
X s I A cos  I A  R s I A sin  I A
EA
Note: I is negative when current is lagging.
A
   V   I
A
 V  E A cos   R s I A cos  I A  X s I A sin  I A
Synchronous Generator
Capability Curve
Generator Capability Curves (1)
(1) Stator Copper Loss (stator heating):
PSCL  3I R
2
A s
The maximum allowable heating of the stator sets a maximum
phase current IA for the machine. It’s equivalent to set a maximum
apparent power for the machine. (power factor is irrelevant)
(2) Rotor Copper Loss (rotor heating):
PRCL  IF2 RF
The maximum allowable heating of the rotor sets a maximum
field current IF for the machine. It’s equivalent to set a maximum
EA for the machine.
(3) Prime-mover’s Power Limit.
Generator Capability Curves (2)
E A  V  jX s I A
(rotor heating)
(stator heating)
rotor field current sets the rated power factor
Generator Capability Curves (3)
E A  V  jX s I A
Assume V keeps rated value. Multiply the above figure by
P
Q
3V 
X
S
Generator Capability Curves (4)
Q
P
P
Q
flip
capability curve
Generator Capability Curves (5)
Q
P
Add prime mover’s power limit (real power)
Example 3
A 480V, 50 Hz, Y connected, six pole synchronous generator is rated at 50 kVA at 0.8 PF lagging. It
has a synchronous reactance of 1.0  per phase. Assume that this generator is connected to a steam
turbine capable of supplying up to 45 kW. The friction and windage losses are 1.5 kW, and the core
losses are 1.0 kW.
(a)
(b)
(c)
(d)
Sketch the capability curve for this generator, including the prime-mover power limit.
Can this generator supply a line current of 56 A at 0.7 PF lagging? Why or why not?
What is the maximum amount of reactive power this generator can produce?
If the generator supplies 30 kW of real power, what is the maximum amount of reactive power that
can be simultaneously supplied?
sg3.m
Synchronous Motor
Operation
UCF
Synchronous Motor and Generator (1)
motor
generator
Tind  kB R  B net
Tind  kB R B net sin 
UCF
Synchronous Motor and Generator (2)
Motor
V  Rs I A  jX s I A  E A
Generator
V   Rs I A  jX s I A  E A
Torque of Synchronous Motor
UCF
V  E A  Rs I A  jX s I A
P  3V I A cos 
3
V E A
Tind 
Xs
P
m
sin  (for R A  X s )
3
V E A
m X s
sin 
Tind  kB R B net sin 
Pull-out torque: when sin=1, the maximum torque the machine can get.
Tmax  3
V E A
m X s
Tmax  kB R B net
Typically take Tmax  3T full  load (or sin   1 / 3,   19.47 o )
in the design (leave margin).
Effect of Load Change
UCF
P  3V I A cos   3
V E A
Xs
sin 
V : fixed (from electrical course)
E A : fixed (when I F fixed or using permanent magnets)
P  (load increases)  sin    I A  more heat (3 I A2 Rs )
   (leading  lagging)


At full load, typically pick up cos   1 in the design.
V
EA
 Xs 
sin  , E A 
IA
cos 
UCF
Example 1
Details in sm1.m
Effects of Field Current Change
UCF
V  E A  Rs I A  jX s I A
 E A  jX s I A

UCF
Underexcited Synchronous Motor
UCF
Overexcited Synchronous Motor
Behaves like a capacitor: can be used for power factor correction.
Called synchronous capacitor or synchronous condenser.
UCF
Synchronous Motor V Curves
UCF
Example 2
sm2.m
UCF
Power factor Correction Using Overexcited
Synchronous Motor - Example 3
sm3.m