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3 Synchronous

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Synchronous machines
Synchronous generator (alternator): transforms mechanical energy into electric energy;
designed to generate sinusoidal voltages and currents; used in most power plants, for car
alternators, etc.
Synchronous motor: transforms electric energy into mechanical energy; used for highpower applications (ships, original TGV…)
Saliant poles
Turbo-alternator
Rotor (inductor) : 2p poles with excitation
windings carrying DC current; nonlaminated magnetic material
θ! = ω / p
Stator: polyphase (e.g. 3-phase) winding
in slots; laminated magnetic material
...
01/03/2018
Synchronous machines
1
No-load characteristic
Evolution of the voltage Ev in a stator phase
vs. intensity of the excitation current Ie, for a
given rotation speed and with no generated
stator current
• The rotor winding, carrying the DC current Ie and rotating
at 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 same
phase) in the airgap, which induces sinusoidal e.m.f.s Ev
in the stator windings, with a phase lag of π/2.
speed θ! = constante
constant
⎧vitesse
E v = f (Ie ) with
avec ⎨
⎩I = 0
average characteristic
average characteristic
E v = k E θ! Φ v (Ie )
Magnetic flux produced
by the inductor and seen by the
stator winding
due due remanent magnetization
Non linearity with hysteresis
Synchronous machines
2
Vector diagram with load
Diagram of magnetomotive forces and magnetic flux densities
Br (4a)
resulting m.m.f.
(3) = (1)+(2)
m.m.f. from I
(2)
(4b)
m.m.f. from Ie
(1)
Fe = γ Ie
and
FI = δ I
(1) The rotor winding, carrying the DC current Ie and
rotating at speed ω/p, produces in the airgap a sliding
m.m.f. F e (as seen from the stator)..
(2) The polyphase current I in the stator winding produces a
sliding m.m.mf. F I (in phase with I).
(3) The resulting m.m.f. is F r = F e+F I.
(4) F r generates a magnetic flux density Br (with the same
phase) in the airgap,, which induces sinusoidal e.m.f.s in
the stator windings, with a phase lag of π/2.
Synchronous machines
3
Vector diagram with load
Stator leakage flux
(seen by the stator but not
coming from the rotor)
e.m.f. induced by leakage flux
eλ ( t ) = −λ ∂ t i1 − λ m ∂ t i 2 − λ m ∂ t i3
Slot leakage
flux
= −(λ − λ m ) ∂ t i1
because
i1 + i 2 + i3 = 0
Eλ = − j X f I
with
X f = ω (λ − λ m )
Stator leakage reactance
End-winding
leakage flux
E t = Er − j Xf I
Diagrams
Fe = γ Ie
U = Er − j Xf I − R I
total e.m.f.
and
FI = δ I
Fr = γ Ie + δ I
Resistance of
stator winding
def
Ir =
Fr
δ
= Ie + I
γ
γ
Equivalent
excitation current
Synchronous machines
E r ≡ no-load e.m.f. produced by Ir
4
Potier diagram
‘Which excitation current Ie should one impose in the synchronous
machine to reach the functioning point corresponding to a given
voltage U and current I in the stator, with a phase shift of ϕ
between U and I?’
E r ≡ no-load e.m.f produced by the
equivalent current Ir
(2)
(4)
Er = U + R I + j Xf I
(5)
Ie = I r −
(2)
(3)
δ
I
γ
(3)
(0)
(1)
Synchronous machines
5
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 than
the no-load m.m.f. (Ir > Ie)
Capacitive behaviour of the load
(I in front of U)
Inductive behaviour of the load
(I lagging behind U)
Synchronous machines
6
Zero power factor characteristic
Evolution of the stator voltage U as a function of the
excitation current Ie, for a given rotation speed and stator
current, with a zero power factor
⎧vitesse
speed θ! = constante
constant
⎪
constant
U = f (Ie ) with
avec ⎨I = constante
⎪cos ϕ = 0 (ϕ = ± π / 2)
⎩
Example: ϕ = π/2
δ
Ie ≈ I r + I
γ
U ≈ E r − Xf I
Synchronous machines
7
Short-circuit characteristic
Evolution of the stator current as a function of the excitation
current Ie, for a given rotation speed and with the stator
windings in short-circuit
Er = U + R I + j Xf I
with
⎧vitesse
speed θ! = constante
constant
I = f (Ie ) with
avec ⎨
⎩U = 0
U=0
Er = R I + j Xf I
E r ≈ Xf I
I≈
β
δ
Xf + β
γ
Ie
δ
Ie ≈ I r + I
γ
Non saturated
machine
E r = β Ir
Synchronous machines
8
Simplified vector diagram
Behn-Eschenburg’s method – Synchronous reactance Xs
When the magnetic materials are not saturated, the combined effect
of the reaction and of stator leakage fluxes can be taken into
account thanks to a single parameter: the synchronous reactance
E v Er
constant
=
= constante
Ie I r
Equal angles OAB and A’OB’
Similar triangles OAB and OA’B’
A, B and C colinear
E v = U + R I + j Xs I
Synchronous
reactance
Synchronous machines
9
Experimental determination of Xs
Behn-Eschenburg’s method – Synchronous reactance Xs
U=0
⇒
⇒
E v = (R + j Xs ) Icc
R + j Xs =
Ev
Icc
with
Xs ≈
R << Xs
E v ( Ie )
Icc (Ie )
Approximation when magnetic materials
are saturated!
Synchronous machines
10
Exterior characteristic
Alternator exterior characteristic
Evolution of the voltage U on a given stator phase as a function of the current Iin this
phase, when the alternator drives a load characterized by a constant power factor, at
constant speed and excitation
⎧vitesse
speed θ! = constante
constant
⎪
constant
U = f (I) with
avec ⎨Ie = constante
⎪cos ϕ = constante
constant
⎩
Synchronous machines
11
Network connection
Need for interconnection of electric power plants
Daily peak
Economical organization of power production + Stability of the
network despite local defects
Daily load
Synchronization of an alternaltor on
an ideal (infinitely powerful) AC network
Large number of production units in parallel
⇒ constant voltage and frequency
The current should be zero when the connection is made → 4 conditions
I=
Ev − U
R + j Xs
1.
2.
3.
4.
same pulsation ω (correct rotation speed)
same amplitudes for E v and U (adjusting Ie)
no phase shift between E v and U
identical phase ordering (in a 3-phase system)
Synchronous machines
12
Behaviour with load
Active electric power
Static stability
Static instability
Electromagnetic power
Internal angle
Torque
Pelm ≈ P = 3 U I cos ϕ
C=
P
3p
=
U I cos ϕ
ω/ p ω
C=
3p
U E v sin δint
ω Xs
Xs I cos ϕ
= E v sin δint
Internal angle
After network synchronization, there is no exchanged current.
Then :
• If mechanical power is provided to the alternator,
E v gets ahead of U ⇒ δint increases (until the equilibrium of
the electromagnetic and mechanical torques)
• If a braking torque is applied to the alternator, E v gets
behind U ⇒ δint decreases (negative torque)
Synchronous machines
The variations of the rotor
mechanical
angle
Δδmec
are
proportional to the variations of the
internal (electric) angle Δδint
Δδmec =
Δδint
p
13
Behaviour with load
Static stability
Internal angle
δint < 0
δint > 0
Unstable
Increasing the mechanical
(breaking) torque leads to an
increase of the absolute value
of δint , and a decrease in the
absolute value of Celm ⇒
unstable
alternator
Unstable
synchronous motor
Increasing the mechanical
(breaking) torque leads to an
increase of δint , and a
decrease of Celm ⇒ unstable
static stability
Increasing the mechanical (breaking) torque leads to
an increase of the absolute value of δint et of Celm ⇒
stable.
The equilibrium is reached when the two torques are
equal.
Increasing the mechanical (breaking) torque leads to
an increase of δint and Celm ⇒ stable.
The equilibrium is reached when the two torques are
equal.
Synchronous machines
14
Behaviour with load
Power diagram
MB = Xs I cos ϕ =
Xs
X
X
3 U I cos ϕ = s P = s Pelm
3U
3U
3U
Active power P
Alternator
O' B = Xs I sin ϕ =
Xs
X
3 U I sin ϕ = s Q
3U
3U
Motor
Reactive power Q
Synchronous machines
15
V-curves (Mordey curves)
under-excited
machine
Evolution of the stator current I as a
function of the excitation current Ie
of a synchronous machine connected
to an ideal network, at constant
active power
Over-excited
machine
"V"
Constant active power
ϕ > 0 ...
ϕ>0
Network= inductive load
Machine = capacitif system
static stability limit
Synchronous machines
16
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