Transcript IES PE Short Course

Module G1
Electric Power Generation and
Machine Controls
James D. McCalley
Overview
• Energy transformation into electrical form
• Generation operation
–
–
–
–
–
Revolving magnetic field
Phasor diagram
Equivalent Circuit
Power relationships
Generator pull-out power
• Excitation control
• Turbine speed control
Energy Transformation
• Transformation processes:
– Chemical
– photovoltaic
– electromechanical
• Electromechanical: conversion of energy from
coal, petroleum, natural gas, uranium, water flow,
geothermal, and wind into electrical energy
• Turbine-synchronous AC generator conversion
process most common in industry today
Click on the below for some pictures of power plants
and synchronous generators
ISU Power Plant
http://powerlearn.ee.iastate.edu/library/html/isupp39.html
ISU Power Plant synchronous generator
http://powerlearn.ee.iastate.edu/library/html/isupp1.html
Ames Power Plant
http://powerlearn.ee.iastate.edu/library/html/amespp34.html
Ames Power Plant synchronous generator
http://powerlearn.ee.iastate.edu/library/html/amespp1.html
Feedback Control Systems for
Synchronous Generators
• Turbine-generator basic form
• Governor and excitation systems are known as
feedback control systems; they control the speed
and voltage respectively
Synchronous Machine Structure
ROTOR
(field
winding)
+
N
+
DC
Voltage
The negative terminal
for each phase is
180 degrees from
the corresponding
positive terminal.
STATOR
(armature
winding)
S
+
A Two Pole Machine (p=2)
Salient Pole Structure
Salient Pole
Construction
Smooth rotor
Construction
Synchronous Machine Structure
N
S
S
N
A Four Pole Machine (p=4)
(Salient Pole Structure)
Generation Operation
• The generator is classified as a synchronous
machine because it is only at synchronous speed
that it can develop electromagnetic torque
2
•  m  p  e 
• e  2f = frequency in rad/sec
• Ns 
120
f = machine speed in RPM
p
• p = number of poles on the rotor of the machine
For 60 Hz operation (f=60)
• Synchronous generator
Rotor construction
Round Rotor
Two pole
Four Pole
Eight Pole
Salient Pole
s = 3600 rpm
s = 1800 rpm
s = 900 rpm
For 60 Hz operation (f=60)
No. of Poles (p)
------------------2
4
6
8
10
12
14
16
18
20
Synchronous speed (Ns)
----------------------------3600
1800
1200
900
720
600
514
450
400
360
Fact: hydro turbines are slow speed,
steam turbines are high speed.
Do hydro-turbine generators have few poles or many?
Do steam-turbine generators have few poles or many?
Fact: salient pole incurs significant
mechanical stress at high speed.
Do steam-turbine generators have salient poles or smooth?
Fact: Salient pole rotors are cheaper to build than smooth.
Do hydro-turbine generators have salient poles or smooth?
Generation Operation
• A magnetic field is provided by the DC-current
carrying field winding which induces the desired
AC voltage in the armature winding
• Field winding is always located on the rotor where
it is connected to an external DC source via slip
rings and brushes or to a revolving DC source via
a special brushless configuration
• Armature winding is located on the stator where
there is no rotation
• The armature consists of three windings all of
which are wound on the stator, physically
displaced from each other by 120 degrees
Synchronous Machine Structure
• voltage induced in phase
wdgs by flux from
field wdg
+
• current in phase wdgs
produces flux that
also induces
voltage in phase
wdgs.
N
+
DC
Voltage
S
+
Rotor
Stator
winding
Brushes
+Stator
winding
Slip
rings
Generation Operation:
The revolving magnetic field
•  f = flux associated with the revolving magnetic field which
links the armature windings. It will have a flux density of B.
• At any given time t, the B-field will be constant along the coil.
• By Faraday’s Law of Induction, the rotating magnetic field
will induce voltages phase displaced in time by 120 degrees
(for a two pole machine) in the three armature windings
• Let’s consider just the A-phase.
eA   (u  B)  dl
c
_
u
B
B field
+
Direction of rotation
Voltage induced in the
conductor = Blu
e
Stator
t
0
2t
Air
B, flux density gap
in the air gap
Rotor
Magnetic
field lines
Voltage induced in one turn ( 2 conductors)
is e = 2Blu.
If there are N turns per winding, the voltage
induced is e = 2NBlu.
If the speed of rotation of the rotor is , and
the radius of the rotor is r, u = r, and e = 2NBlr
If the flux density changes sinusoidally in the air
gap, B = Bmax sin t, and the induced voltage is
e = 2BmaxNlr sin t = Emax sin t
Generation Operation:
The revolving magnetic field (cont’d)
If each of the three armature windings are connected across equal
impedances, balanced three phase currents will flow in them
producing their own magnetic fields a = b = c
• ar = resultant field with associated flux obtained as the
sum of the three component fluxes is the field of
armature reaction
• The two fields represented by f and arare stationary with
respect to each other
• The armature field has “locked in” with the rotor field
and the two fields are said to be rotating in synchronism
• The total resultant field r = ar +  f
Generation Operation:
The phasor diagram
• From Faraday’s Law of Induction, a voltage is
induced in each of the three armature windings:
d r
where N = number of winding turns
e  N
dt
• All voltages, Er , Ear , E f ,lag their corresponding
fluxes, r , ar ,  f ,by 90 degrees
• The current winding a, denoted by Ia , is in phase
with the flux it produces ar
Generation Operation:
The phasor diagram (cont’d)
• Note: Er  E f  Ear
Generation Operation:
The equivalent circuit model
• Develop equivalent model for winding a only;
same applies to winding b and c with appropriate
120 degree phase shifts in currents and voltages
given a balanced load
Ear   N
dar
dt
 ( X ar   90) Ia   jX ar I a
Er  E f  jX ar Ia
Vt  Er  jX l Ia  E f  j( X l  X ar ) Ia  E f  jX s I a
Generation Operation:
The equivalent circuit model (cont’d)
Generation Operation:
The equivalent circuit model (cont’d)
Ia
jXs
Ef
Vt
Zload
Generation Operation:
The equivalent circuit model (cont’d)
You can perform per-phase equivalent analysis
or you can perform per-unit analysis.
In per-phase, Ef and Vt are both line to neutral voltages,
Ia is the line current, and Z is the impedance of the
equivalent Y-connected load.
In per-unit, Ef and Vt are per-unit voltages,
Ia is the per unit current, and Z is the per unit load
impedance.
Leading and Lagging Generator
Operation
Let Z | Z |  , Vt | Vt | V
From the equivalent circuit,
Vt | Vt | V | Vt |
Ia  

(V   )
Z | Z |  | Z |
So here we see that
 i  V  
   V   i
Leading and Lagging Generator
Operation
Assign Vt as the reference:
V  0
Then,
Vt | Vt |
| Vt |
Ia  
(V   ) 
( )
Z |Z|
|Z|
So here we see that  i  
This gives an easy way to remember the relation
between load, sign of current angle, leading/lagging,
and sign of power angle.
Leading and Lagging Generator
Operation
Circle the correct answer in each column
Inductive load
Capacitive load
----------------------------------------------------------------
load absorbs/supplies Q
load absorbs/supplies Q
gen absorbs/supplies Q
gen supplies/absorbs Q
X ?0
X ?0
 ?0
i  ?  0
 ?0
i  ?  0
current is leading/lagging current is leading/lagging
Phasor Diagram for Equivalent Circuit
Ia
jXs
Ef
From KVL: E f  Vt  jX s I a
Vt
Zload
Phasor Diagram for Equivalent Circuit
E f  Vt  jX s I a
This equation gives directions for
constructing the phasor diagram.
1. Draw Vt phasor
2. Draw Ia phasor
3. Scale Ia phasor magnitude by Xs and
rotate it by 90 degrees.
4. Add scaled and rotated vector
to Vt
Try it for lagging case.
Ia
XsIa
Ef
jXsIa
jXsIa
Vt
Phasor Diagram for Equivalent Circuit
E f  Vt  jX s I a
This equation gives directions for
constructing the phasor diagram.
1. Draw Vt phasor
2. Draw Ia phasor
3. Scale Ia phasor magnitude by Xs and
rotate it by 90 degrees.
4. Add scaled and rotated vector
to Vt
You do it for leading case.
Phasor Diagram for Equivalent Circuit
E f  Vt  jX s I a
Let’s define the angle that Ef makes with Vt as 
E f | E f | 
For generator operation (power supplied by machine),
this angle is always positive.
For motor operation, this angle is negative.