Synchronous machines 1

Synchronous machines

Turbo-alternator Saliant poles

p/ω=θ!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…)

...

01/03/2018

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

⎩⎨⎧

=

=θ=

0Iconstantevitesse

avec)I(fE ev!

)I(kE evEv Φθ= !

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

Synchronous machines

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 ω/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 π/2.

(4b)

rB (4a)

ee IF γ= IFI δ=and

Synchronous machines

4

Vector diagram with loadStator leakage flux

1tm

3tm2tm1ti)(

iii)t(e∂λ−λ−=

∂λ−∂λ−∂λ−=λ

IXjE f−=λ )(X mf λ−λω=

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 γ= IFI δ=and

IIF er δ+γ=

IIFI er

defr γ

δ+=

γ=

Equivalent excitation current

Diagrams

Er ≡ 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 ϕbetween U and I?’

Potier diagram

IXjIRUE fr ++=

(1)(0)

(2)(3)

(4)

(5)

(2)

(3)III re γδ

−=

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

Synchronous machines

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)

Synchronous machines

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

⎪⎩

⎪⎨

⎧

π±=ϕ=ϕ

=

=θ

=

)2/(0cosconstanteI

constantevitesseavec)I(fU e

!

Example: ϕ = π/2

IXEU fr −≈

III re γδ

+≈

Synchronous machines

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

⎩⎨⎧

=

=θ=

0Uconstantevitesse

avec)I(fI e!

IXjIRE fr +=

IXjIRUE fr ++=

0U =with

rr IE β=

IXE fr ≈

III re γδ

+≈

ef

IX

I

γδ

β+

β≈

Non saturatedmachine

Synchronous machines

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 ++=

Synchronous machines

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 = ⇒

⇒ with

Approximation when magnetic materialsare saturated!

Synchronous machines

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

⎪⎩

⎪⎨

⎧

=ϕ

=

=θ

=

constantecosconstanteI

constantevitesseavec)I(fU e

!

Alternator exterior characteristic

Synchronous machines

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 parallel⇒ constant voltage and frequency

s

vXjRUEI

+−

=

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

1. same pulsation ω (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

Synchronous machines

13

Behaviour with load

ϕ=≈ cosIU3PPelmϕ

ω=

ω= cosIUp3

p/PC

intv

ssinEcosIX

δ=

ϕ

intvs

sinEUXp3C δ

ω=

Internal angle

Electromagnetic power

Torque

The variations of the rotormechanical angle Δδmec areproportional to the variations of theinternal (electric) angle Δδint

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

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

• If a braking torque is applied to the alternator, Ev gets behind U⇒ δint decreases (negative torque)

pint

mecδΔ

=δΔ

Active electric power

Internal angle

Synchronous machines

Stat

ic in

stabi

lity

St

atic

stab

ility

14

Behaviour with loadStatic stability

0int >δ0int <δ

Increasing the mechanical (breaking) torque leads toan increase of δint and Celm ⇒ stable.The equilibrium is reached when the two torques areequal.

Increasing the mechanical(breaking) torque leads to anincrease of δint, and adecrease of Celm ⇒ unstable

Increasing the mechanical (breaking) torque leads toan increase of the absolute value of δint et of Celm ⇒stable.The equilibrium is reached when the two torques areequal.

Increasing the mechanical(breaking) torque leads to anincrease of the absolute valueof δint, and a decrease in theabsolute value of Celm ⇒unstable

Internal angle

Synchronous machines

Uns

tabl

e

Uns

tabl

e

synchronous motor alternator

static stability

15

Behaviour with loadPower diagram

elmsss

s PU3XP

U3XcosIU3

U3XcosIXMB ==ϕ=ϕ=

QU3XsinIU3

U3XsinIXB'O ss

s =ϕ=ϕ=

Active power P

Reactive power Q

Alternator

Motor

Synchronous machines

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

ϕ > 0

ϕ > 0 ...

Synchronous machines

static stability limit