module g1 electric power generation and machine controls james d. mccalley

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

http://powerlearn.ee.iastate.edu/library/html/isupp39.html

Click on the below for some pictures of power plantsand synchronous generators

http://powerlearn.ee.iastate.edu/library/html/amespp1.html



ISU Power Plant

ISU Power Plant synchronous generator

Ames Power Plant

Ames Power Plant synchronous generator




















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

A Two Pole Machine (p=2)Salient Pole Structure

Synchronous Machine Structure

N

S

+

+

DCVoltage

Phase A

Phase B

Phase C

STATOR(armaturewinding)

ROTOR(field winding)

The negative terminalfor each phase is180 degrees fromthe correspondingpositive terminal.

+

Salient PoleConstruction

Smooth rotorConstruction

A Four Pole Machine (p=4)(Salient Pole Structure)


N

S

N

S

Generation Operation

• The generator is classified as a synchronous machine because it is only at synchronous speed that it can develop electromagnetic torque

• • = frequency in rad/sec

• = machine speed in RPM

• p = number of poles on the rotor of the machine

m ep2

e f2

Np

fs 120

For 60 Hz operation (f=60)

• Synchronous generator

Rotor construction

Round Rotor Salient Pole

Two pole s = 3600 rpm

Four Poles = 1800 rpm

Eight Poles = 900 rpm

No. of Poles (p) Synchronous speed (Ns)------------------- ----------------------------- 2 3600 4 1800 6 1200 8 900 10 720 12 600 14 514 16 450 18 400 20 360

For 60 Hz operation (f=60)

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


N

S

+

+

DCVoltage

Phase A

Phase B

Phase C+

• voltage induced in phase wdgs by flux from field wdg• current in phase wdgs produces flux that also induces voltage in phase wdgs.

Stator

Statorwinding Slip

rings

Brushes

Rotorwinding

+-

Generation Operation: The revolving magnetic field

• = 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.

f

dlBuec

A )(

B

u

eB field

Direction of rotation+

_

Voltage induced in the conductor = Blu

Magneticfield lines

Stator

RotorAirgap

0

2

B, flux densityin the air gap

Voltage induced in one turn ( 2 conductors)is e = 2Blu.

If there are N turns per winding, the voltageinduced is e = 2NBlu.

If the speed of rotation of the rotor is , andthe radius of the rotor is r, u = rand e = 2NBlr

If the flux density changes sinusoidally in the airgap, B = Bmax sin t, and the induced voltage is

e = 2BmaxNlrsin t = Emax sin t

Generation Operation: The revolving magnetic field (cont’d)

• = 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 and are 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 = +

ar

f ar

r ar f

a cb

If each of the three armature windings are connected across equal impedances, balanced three phase currents will flow in themproducing their own magnetic fields = =

Generation Operation:The phasor diagram

• From Faraday’s Law of Induction, a voltage is induced in each of the three armature windings:

where N = number of winding turns

• All voltages, ,lag their corresponding fluxes, ,by 90 degrees

• The current winding a, denoted by , is in phase with the flux it produces

dt

dNe r

Ia

ar

E E Er ar f, ,

r ar f, ,

Generation Operation: The phasor diagram (cont’d)

• Note: E E Er f ar

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

E Nddtar

ar ( )X Iar a90 jX Iar a

E E jX Ir f ar a

V E jX I E j X X It r l a f l ar a ( ) E jX If s a

Generation Operation: The equivalent circuit model (cont’d)


Ef

jXs

ZloadVt

Ia


In per-phase, Ef and Vt are both line to neutral voltages,Ia is the line current, and Z is the impedance of theequivalent 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.

You can perform per-phase equivalent analysisor you can perform per-unit analysis.

Leading and Lagging Generator Operation

LetVtt VVZZ || ,||

From the equivalent circuit,

)(||||

||||

V

tVtta Z

VZ

VZV

I

So here we see thatiVVi


Assign Vt as the reference: 0V

Then,

)(||||

)(||||

ZV

ZV

ZV

I tV

tta

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.


Inductive load Capacitive load----------------------------------------------------------------

ggingleading/la iscurrent

0?

0?

0?

Q ppliesabsorbs/sugen

Q ppliesabsorbs/su load

i

X

ggingleading/la iscurrent

0?

0?

0?

Q bsorbssupplies/agen

Q ppliesabsorbs/su load

i

X

Circle the correct answer in each column

Phasor Diagram for Equivalent Circuit

Ef

jXs

ZloadVt

Ia

From KVL: astf IjXVE

Phasor Diagram for Equivalent Circuitastf IjXVE

This equation gives directions for constructing the phasor diagram.1. Draw Vt phasor2. Draw Ia phasor3. Scale Ia phasor magnitude by Xs and rotate it by 90 degrees.4. Add scaled and rotated vector to Vt Vt

Try it for lagging case. Ia

XsIa

jXsIa

jXsIa

Ef


astf IjXVE

This equation gives directions for constructing the phasor diagram.1. Draw Vt phasor2. Draw Ia phasor3. 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.


astf IjXVE

Let’s define the angle that Ef makes with Vt as

|| ff EE

For generator operation (power supplied by machine),this angle is always positive.

For motor operation, this angle is negative.

module g1 electric power generation and machine controls james d. mccalley

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