Download Distributed Power Systems ELCT 908

Modeling of Synchronous
Generators
99% of the electric power allover the
world is generated by Synchronous
Generators
System Components

The main components of power systems are:
1.
Generators (Synchronous Generators).
2.
Power Transformers.
3.
Transmission lines and cables.
4.
Loads:
- Static (lighting, heating,…)
- Dynamic (Motors).
Generators
Input

The Generator converts mechanical power into
electrical power.

Synchronous generators are constant speed
generators.

The conversion of mechanical power into
electrical power is done through a coupling field
(magnetic field).
Mechanical
Magnetic
Electrical
Output
Construction
Stator
Armature
Electrical
Mechanical
Rotor
Field
Synchronous Generators
Generator
Exciter
View of a two-pole round rotor generator and exciter.
(Westinghouse)
Synchronous Generators
Stator with
laminated iron core
B+
C-
Armature Windings
N
-
A+
Rotor with
dc winding
+
+
+
+
+
A-
S
Field Windings
B-
C+
Slots with
phase
winding
Construction of a two-pole salient pole generator
Salient Pole Generator
Slip
rings
Pole
Fan
DC excitation
winding
Rotor of a four-pole salient pole generator
Operation Concepts
Flux f
nsy
B+
C+
N
+
+
-
A+
A-
+
-
+
-
B-
S
C+
Operating concept of a synchronous generator
Flux Linkage
Maximum flux linkage with phase A
C-
No flux linkage with phase A
B+
-
C-
+
B+
-
+
-
N
N
A+
+
-
+
-
A-
A+
+
S
-
B-
-
S
+
+
-
B-
C+
(a) Flux is perpendicular to phase A
+
C+
(b) Flux is parallel to phase A
Flux linkage variation.
A-
Rotating Flux
rot
link
t
nsy
B+
C+
N
+
+
-
A+
-
30
+
-
S
A-
+
-
B-
C+
Rotating flux linkage to phase A.
EMF Equation
According to Faraday’s law,
the induced emf in the armature coil of Nsta turns is given by:
E s (t )  N sta
d  link (t )
dt
 link (t )   rot cos( t )
E s (t )   N sta  rot  sin(  t )
 N sta  rot  cos( t  90)
  2 f
Esta  Emax cos(t  90)
Emax  N sta  rot 
N sta  rot 
where Erms 
2
Erms  4.44 N sta  rot f
Speed and Frequency

nS is the synchronous speed (r/m)

f is the frequency in Hz.

2P = total number of poles.
Pns
f 
60
Example:
calculate the frequency of a 1800 rpm , 4 pole synchronous generator
ns = 1800 r/m, 2P = 4 Then, f = 2*1800/60 = 60 Hz
Armature Reaction
nsy
B+
C+
N
+
Field flux f
-
A+
A-
+
-
30
+
-
S
+
-
B-
Armature
flux ar
C+
The main field flux (Φf) and the load generated rotating fluxe (Φar)
Armature Reaction
Load current generates a rotating flux reducing the main flux and
induced voltage
I arm (t )  2 I sta cos( t )
 arm (t )   ar cos( t )
Ear (t )  N sta
E arm 
d arm (t )
  N sta  ar  sin ( t )
dt
N sta  ar 
2
Vt  E sta  E arm
Armature Reactance
E ar (t )  Larm
dI
arm
(t )
dt
  Larm 2 I sta
  X arm
X arm 
d
dt
sin(  t )
 Larm
2 I sta sin(  t )
N sta  ar 
2 I sta
X syn  X arm  X leakage
2 I sta cos( t )
Equivalent Circuit
E arm syn  I sta ( j X syn )
Vt  E sta  E arm syn  E sta  I sta j X syn
Single-phase equivalent circuit of a synchronous generator.
Phasor Diagram
Ia

E  E
E  V0
I
jX

V  V0
Power Calculations
The apparent Power S is given by
S  VI
*
S  V I  P  jQ
*
*
Using the current expression derived from the equivalent circuit we get
E  V0
S  P  jQ  V0  (
)
jX
*
VE
V2
EV
V2



  90  j
X90 X90 X
X
EV
P
sin 
X
EV
V2
Q
cos 
X
X
Power Angle
P
Pmax = EV/X

Power – Angle characteristics
Example
A 1,250-kVA, three-phase, Y-connected, 4,160-V , tenpole, 60-Hz generator has an armature resistance of 0.126
ohms per phase and a synchronous reactance of 3 ohms
per phase.
Find the full load generated voltage per phase at a power
factor of 0.8 lagging.
Solution
The magnitude of full load current is obtained as
The terminal voltage per phase is taken as reference
The synchronous impedance is obtained as
The generated voltage per phase is calculated for
a power factor of 0.8 lagging ( φ = -36.87°).