synchronous machines - montefiore institute · 2020-06-12 · synchronous machines 1 synchronous...

Synchronous machines 1

Synchronous machines

Turbo-alternator Saliant poles

p/w=q!Rotor (inductor) : 2p poles with excitationwindings carrying DC current; non-laminated magnetic material

Stator: polyphase (e.g. 3-phase) windingin slots; laminated magnetic material

Synchronous generator (alternator): transforms mechanical energy into electric energy;designed to generate sinusoidal voltages and currents; used in most power plants, for caralternators, etc.Synchronous motor: transforms electric energy into mechanical energy; used for high-power applications (ships, original TGV…)

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28/05/2020

2

No-load characteristic

Evolution of the voltage Ev in a stator phasevs. intensity of the excitation current Ie, for agiven rotation speed and with no generatedstator current

îíì=

=q=

0Iconstantevitesse

avec)I(fE ev!

)I(kE evEv Fq= !

Magnetic flux producedby the inductor and seen by the stator winding

Non linearity with hysteresis

due due remanent magnetization

• The rotor winding, carrying the DC current Ie and rotatingat speed w/p, produces in the airgap a sliding m.m.f. Fe

(as seen from the stator).• Fe generates a magnetic flux density Br (with the same

phase) in the airgap, which induces sinusoidal e.m.f.s Evin the stator windings, with a phase lag of p/2.


average characteristicaverage characteristic

withconstant speed

3

Vector diagram with loadDiagram of magnetomotive forces and magnetic flux densities

resulting m.m.f. (3) = (1)+(2)

m.m.f. from Ie(1)

m.m.f. from I(2)

(1) The rotor winding, carrying the DC current Ie androtating at speed w/p, produces in the airgap a slidingm.m.f. Fe (as seen from the stator)..

(2) The polyphase current I in the stator winding produces asliding m.m.mf. FI (in phase with I).

(3) The resulting m.m.f. is Fr = Fe+FI.

(4) Fr generates a magnetic flux density Br (with the samephase) in the airgap,, which induces sinusoidal e.m.f.s inthe stator windings, with a phase lag of p/2.

(4b)

rB (4a)

ee IF g= IFI d=and


4

Vector diagram with loadStator leakage flux

1tm

3tm2tm1ti)(

iii)t(e¶l-l-=

¶l-¶l-¶l-=l

IXjE f-=l )(X mf l-lw=

Slot leakage flux

End-winding leakage flux IXjEE frt -=

IRIXjEU fr --=

0iii 321 =++

(seen by the stator but not coming from the rotor)

e.m.f. induced by leakage flux

because

with

Stator leakage reactance

total e.m.f.

Resistance of stator winding

ee IF g= IFI d=and

IIF er d+g=

IIFI er

defr g

d+=

g=

Equivalent excitation current

Diagrams

Er º no-load e.m.f. produced by IrSynchronous machines

5

‘Which excitation current Ie should one impose in the synchronousmachine to reach the functioning point corresponding to a givenvoltage U and current I in the stator, with a phase shift of jbetween U and I?’

Potier diagram

IXjIRUE fr ++=

(1)(0)

(2)(3)

(4)

(5)

(2)

(3)III re gd

-=

Er º no-load e.m.f produced by theequivalent current Ir


6

Reaction

Demagnetizing reaction Magnetizing reaction

The m.m.f. is smaller than the no-load m.m.f. (Ir < Ie)The m.m.f. is larger thanthe no-load m.m.f. (Ir > Ie)

Inductive behaviour of the load(I lagging behind U)

Capacitive behaviour of the load(I in front of U)


7

Zero power factor characteristic

Evolution of the stator voltage U as a function of theexcitation current Ie, for a given rotation speed and statorcurrent, with a zero power factor

ïî

ïí

ì

p±=j=j=

=q=

)2/(0cosconstanteI

constantevitesseavec)I(fU e

!

Example: j = p/2

IXEU fr -»

III re gd

+»


with

speed constant constant

8

Short-circuit characteristic

Evolution of the stator current as a function of the excitationcurrent Ie, for a given rotation speed and with the statorwindings in short-circuit

îíì

==q

=0U

constantevitesseavec)I(fI e

!

IXjIRE fr +=

IXjIRUE fr ++=

0U =with

rr IE b=

IXE fr »

III re gd

+»

ef

IX

I

gdb+

b»

Non saturatedmachine


with speed constant

9

When the magnetic materials are not saturated, the combined effectof the reaction and of stator leakage fluxes can be taken intoaccount thanks to a single parameter: the synchronous reactance

Simplified vector diagramBehn-Eschenburg’s method – Synchronous reactance Xs

constanteIE

IE

r

r

e

v ==

Synchronousreactance

IXjIRUE sv ++=


constant

Equal angles OAB and A’OB’

Similar triangles OAB and OA’B’

A, B and C colinear

10

Experimental determination of Xs

Behn-Eschenburg’s method – Synchronous reactance Xs

)I(I)I(EXecc

evs »

( ) ccsv IXjRE +=

cc

vs IEXjR =+ sXR <<

0U = Þ

Þ with

Approximation when magnetic materialsare saturated!


11

Exterior characteristic

Evolution of the voltage U on a given stator phase as a function of the current Iin thisphase, when the alternator drives a load characterized by a constant power factor, atconstant speed and excitation

ïî

ïí

ì

=j=

=q=

constantecosconstanteI

constantevitesseavec)I(fU e

!

Alternator exterior characteristic


with constant

constant constant

speed

12

Economical organization of power production + Stability of thenetwork despite local defects

Network connectionNeed for interconnection of electric power plants

Synchronization of an alternaltor on an ideal (infinitely powerful) AC network

Large number of production units in parallelÞ constant voltage and frequency

s

vXjRUEI

+-

=

The current should be zero when the connection is made ® 4 conditions

1. same pulsation w (correct rotation speed)2. same amplitudes for Ev and U (adjusting Ie)3. no phase shift between Ev and U4. identical phase ordering (in a 3-phase system)

Daily peak

Daily load


13

Behaviour with load

j=» cosIU3PPelmj

w=

w= cosIUp3

p/PC

intv

s

sinEcosIX

d=j

intvs

sinEUXp3C d

w=

Internal angle

Electromagnetic power

Torque

The variations of the rotormechanical angle Ddmec areproportional to the variations of theinternal (electric) angle Ddint

After network synchronization, there is no exchanged current.Then :

• If mechanical power is provided to the alternator,Ev gets ahead of UÞ dint increases (until the equilibrium ofthe electromagnetic and mechanical torques)

• If a braking torque is applied to the alternator, Ev gets behind UÞ dint decreases (negative torque)

pint

mecdD

=dD

Active electric power

Internal angle


Stat

ic in

stab

ility

St

atic

stab

ility

14

Behaviour with loadStatic stability

0int >d0int <d

Increasing the mechanical (breaking) torque leads toan increase of dint and thus of CelmÞ stable.The equilibrium is reached when the two torques areequal.

Increasing the mechanical(breaking) torque leads to anincrease of dint, and thus to adecrease of CelmÞ unstable

Increasing the mechanical (breaking) torque leads toan increase of the absolute value of dint and thus ofCelmÞ stable.The equilibrium is reached when the two torques areequal.

Increasing the mechanical(breaking) torque leads to anincrease of the absolute valueof dint, and thus to a decreasein the absolute value of CelmÞ unstable

Internal angle


Uns

tabl

e

Uns

tabl

e

synchronous motor alternator

static stability

15

Behaviour with loadPower diagram

elmsss

s PU3XP

U3XcosIU3

U3XcosIXMB ==j=j=

QU3XsinIU3

U3XsinIXB'O ss

s =j=j=

Active power P

Reactive power Q

Alternator

Motor


16

V-curves (Mordey curves)Evolution of the stator current I as afunction of the excitation current Ieof a synchronous machine connectedto an ideal network, at constantactive power

Constant active power

Over-excitedmachine

under-excited machine

"V"

Network = inductive loadMachine = capacitif system

j > 0

j < 0 ...


static stability limit

synchronous machines - montefiore institute · 2020-06-12 · synchronous machines 1 synchronous...

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