ELEC0047 - Power system dynamics, control and stability
A simple example of
electromagnetic transients in the synchronous machine
Thierry Van Cutsem
t.vancutsem@ulg.ac.be
www.montefiore.ulg.ac.be/~vct
October 2015
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A simple example of electromagnetic transients in the synchronous machine
System modelling
System modelling
ia, ib, ic
synchronous
machine
∼
Le
va, vb, vc
Re
ideal voltage
source
+
−
ea, eb, ec
Network represented by a simple Thévenin equivalent :
resistance Re and inductance Le in each phase
no magnetic coupling between phases, for simplicity
Machine :
only field winding f represented, for simplicity
rotor speed θ̇ assumed constant
the focus is on short-lasting electromagnetic transients, such as fault currents
constant excitation voltage Vf
it is assumed that the automatic voltage regulator hasn’t time to react
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A simple example of electromagnetic transients in the synchronous machine
System modelling
Network equations
va − e a
=
vb − e b
=
vc − e c
=
dia
dt
dib
Re ib + Le
dt
dic
Re ic + Le
dt
Re ia + Le
with
with
with
ea =
√
2E cos(ωN t + ψ)
√
(1)
2π
) (2)
3
√
4π
eb = 2E cos(ωN t + ψ −
) (3)
3
eb =
2E cos(ωN t + ψ −
Park transformation


vd
 vq 
vo
with P


va
= P  vb 
vc
r  cos θ
2
sin θ
=
3
√1
2
where




id
ia
 iq  = P  ib 
io
ic
cos(θ − 2π
3 )
sin(θ − 2π
3 )
√1
2

cos(θ − 4π
3 )

sin(θ − 4π
3 )
(4)
(5)
√1
2
θ = θo + ωN t
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A simple example of electromagnetic transients in the synchronous machine
System modelling
Machine equations
ψd
= Ldd id + Ldf if
(6)
ψq
= Lqq iq
(7)
ψo
= Loo io
(8)
ψf
= Lff if + Ldf id
(9)
vd
=
vq
=
vo
=
Vf
=
dψd
−Ra id − θ̇ψq −
dt
dψq
−Ra iq + θ̇ψd −
dt
dψo
−Ra io −
dt
dψf
Rf if +
dt
(10)
(11)
(12)
(13)
4 / 16
A simple example of electromagnetic transients in the synchronous machine
System modelling
Variable - equations balance
17 variables : va , vb , vc , ia , ib , ic , vd , vq , vo , id , iq , io , ψd , ψq , ψo , ψf , if
17 equations : (1 - 3), 6 eqs. in (4), (6 - 13)
Remarks
The model is made up of Differential-Algebraic Equations (DAEs)
some of the variables and some of the equations could be eliminated but the
additional computational effort of keeping all of them is negligible1
θ being known, the equations are linear with respect to the unknowns
some coefficients of this relationship and some independent terms vary with
time.
1 not
to mention the risk of introducing mistakes in analytical manipulations !
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A simple example of electromagnetic transients in the synchronous machine
System modelling
Passing the equations in per unit
At the stator (a, b, c, d, q, o):
we choose the base voltage VB = nominal RMS phase-to-neutral voltage (kV)
we choose the base power SB = single-phase apparent power (MVA)
the base current is given by IB = SB /VB
the base magnetic flux is given by ψB = VB /ωN
va , vb , vc , vd , vq , vo are divided by VB
ia , ib , ic , id , iq , io are divided by IB
ψd , ψq and ψo are devided by ψB , etc.
In the field winding f :
we use the reciprocal EMFL per unit system
the latter is such that Ldd = Ldf + L` in per unit
After passing in per unit:
θ̇ = 1 pu in Eqs. (10,11)
the equations are unchanged, except that each time derivative is multiplied
by 1/ωN , since we keep the time in second (not in pu)
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A simple example of electromagnetic transients in the synchronous machine
System modelling
Equations converted in per unit and rearranged
1
ωN
1
ωN
1
ωN
dia
dt
dib
dt
dic
dt
Re
1
1
ia + va − ea
Le
Le
Le
Re
1
1
− ib + vb − eb
Le
Le
Le
Re
1
1
− ic + vc − ec
Le
Le
Le
r 2
2π
4π
cos(θ)va + cos(θ −
)vb + cos(θ −
)vc − vd
3
3
3
r 2
2π
4π
sin(θ)va + sin(θ −
)vb + sin(θ −
)vc − vq
3
3
3
1
√ (va + vb + vc ) − vo
3
r 2
2π
4π
cos(θ)ia + cos(θ −
)ib + cos(θ −
)ic − id
3
3
3
r 2
2π
4π
sin(θ)ia + sin(θ −
)ib + sin(θ −
)ic − iq
3
3
3
= −
(14)
=
(15)
=
0
=
0
=
0
=
0
=
0
=
(16)
(17)
(18)
(19)
(20)
(21)
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A simple example of electromagnetic transients in the synchronous machine
0
0
0
0
0
1 dψd
ωN dt
1 dψq
ωN dt
1 dψf
ωN dt
1 dψo
ωN dt
System modelling
1
√ (ia + ib + ic ) − io
3
= Ldd id + Ldf if − ψd
=
(22)
(23)
= Lqq iq − ψq
(24)
= Lff if + Ldf id − ψf
(25)
= Loo io − ψo
(26)
= −Ra id − ψq − vd
(27)
= −Ra iq + ψd − vq
(28)
= −Rf if + Vf
(29)
= −Ra io − vo
(30)
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A simple example of electromagnetic transients in the synchronous machine
System modelling
Model in compact form
With a proper reordering of equations and states, the model can be rewritten in
compact form as:
(1/ωN ) ẋ
=
0 =
Axx x + Axy y + u
Ayx x + Ayy y
(31)
(32)
where:
x
y
u
T
=
[ ia ib ic ψd ψq ψf ψo ]
=
[ va vb vc vd vq vo id iq io if ]
ea
eb
ec
[−
−
−
0 0 Vf 0 ]T
Le
Le
Le
=
T
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A simple example of electromagnetic transients in the synchronous machine
Numerical solution of the DAEs
Numerical solution of the DAEs
Let k denote the discrete time (k = 0, 1, 2, . . .), and h the time step size.
A popular numerical integration formula is the Trapezoidal Method :
xk+1 = xk + h2 (ẋk+1 + ẋk )
Replacing
ẋk+1 by its expression (31) :
xk+1 = xk + h2 ωN Axx xk+1 + h2 ωN Axy yk+1 + h2 ωN uk+1 + h2 ẋk
hωN
and rearranging the various terms :
2
Axx − hω2 I xk+1 + Axy yk+1 = − hω2 xk − ω1 ẋk − uk+1
N
N
N
Dividing by
where
I
(33)
is the unit matrix of same dimension as x .
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A simple example of electromagnetic transients in the synchronous machine
Numerical solution of the DAEs
On the other hand, from Eq. (32) we have :
Ayx xk+1 + Ayy yk+1 = 0
(34)
Grouping Eqs. (33) and (34), the linear system to solve at each time step is :




2
2
1
x
A
−
I
A
−
x
−
ẋ
−
u
k+1
xy 
k+1 
 xx hωN
 hωN k ωN k
(35)
yk+1 =
Ayx
Ayy
0
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A simple example of electromagnetic transients in the synchronous machine
Numerical example
Numerical example
Network and machine data
fN = 50 Hz
Xe = ωN Le = 0.15 pu
Re = 0.015 pu
Ra = 0.005 pu
Xd = 2.0 pu
Xq = 2.0 pu
X` = ωN (Ldd − Ldf ) = 0.2 pu
Xd0 = 0.3 pu
0
Tdo
= 8.0 s
Loo = 0.1 pu
Initial operating point
P = 0.5 pu (active power in one phase in pu on the single-phase base power)
Q = 0.1 pu (reactive power in one phase in pu on the single-phase base power)
V̄a = 1.000 pu ∠0
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A simple example of electromagnetic transients in the synchronous machine
Numerical example
Exercise
A MATLAB script to simulate this system is available in emt.m and linmodel.m.
Analyze that code and explain :
1
0
how the parameters Ldf , Tdo
, Lff and Rf are derived from the data
2
how the initial value (at t = 0) of the states in x and y are determined
3
how the value of E , ψ, θo (see slide # 3) and Vf are determined.
(all values in pu, except θo and ψ in radian)
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A simple example of electromagnetic transients in the synchronous machine
Numerical example
Simulation results
A three-phase short-circuit is simulated by setting E to zero at t = 0.05 s.
It is eliminated after 0.20 s (i.e. at t = 0.25 s) by restoring E to its initial value.
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A simple example of electromagnetic transients in the synchronous machine
Numerical example
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A simple example of electromagnetic transients in the synchronous machine
Numerical example
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