emtp simul(8)
TRANSCRIPT
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72 Power systems electromagnetic transients simulation
Table 4.1 Norton components for different integrationformulae
Integration method Req IHistory
Inductor
Backward EulerL
tin1
Trapezoidal2L
tin1 +
t
2Lvn1
Gear 2nd order3L
2t43
in1 13 in2...................................................................................................
Capacitor
Backward Eulert
C C
tvn1
Trapezoidalt
2C 2C
tvn1 in1
Gear 2nd order2t
3C 2C
tvn1
C
2tvn2
Substituting equation 4.17 into equation 4.16 yields:
IHistory =t
2Lv(t t) tR
2Li(t t) + i(t t )
=
1 tR2L
i(t t ) + t
2Lv(t t) (4.18)
The Norton equivalent circuit current source value for the complete RL branch
is simply calculated from the short-circuit terminal current. The short-circuit circuitconsists of a current source feeding into two parallel resistors (R and 2L/t), with
the current in R being the terminal current. This is given by:
Ishort-circuit =(2L/t)IHistory
R + 2L/t
= (2L/t) ((1 tR/(2L)) i(t t) + (t/(2L))v(t t))R + 2L/t
= (2L/t R)(R + 2L/t) i(t t) + 1(R + 2L/t) v(t t)
= (1 tR/(2L))(1 + tR/(2L)) i(t t) +
t/(2L)
(1 + tR/(2L)) v(t t) (4.19)
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Numerical integrator substitution 73
Reff =t
2L
2L
R+
t
IRL History =
IL History
2LIL History = ikm (tt) vL (tt)
t
vR (
t)
vL (t)
R R
m mm
k k k
(2L/t) IL History
R 2L/t
Figure 4.6 Reduction of RL branch
The instantaneous current term is obtained from the current that flows due to an
applied voltage to the terminals (current source open circuited). This current is:
1(R + 2L/t) v(t) = t/(2L)(1 + tR/(2L)) v(t) (4.20)
Hence the complete difference equation expressed in terms of branch voltage is
obtained by adding equations 4.19 and 4.20, which gives:
i(t) = (1 tR/(2L))(1 + tR/(2L)) i(t t ) +
t/(2L)
(1 + tR/(2L)) (v(t t) + v(t)) (4.21)
The corresponding Norton equivalent is shown in Figure 4.6.
The reduction of a tuned filter branch is illustrated in Figure 4.7, which shows the
actual RLCcomponents, their individual Norton equivalents and a single Norton rep-resentation of the complete filter branch. Parallel filter branches can be combined into
one Norton by summing their current sources and conductance values. The reduction,
however, hides the information on voltages across and current through each individual
component. The mathematical implementation of the reduction process is carried out
by first establishing the nodal admittance matrix of the circuit and then performing
Gaussian elimination of the internal nodes.
4.3 Dual Norton model of the transmission line
A detailed description of transmission line modelling is deferred to Chapter 6. The
single-phase lossless line [4] is used as an introduction at this stage, to illustrate the
simplicity of Dommels method.
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74 Power systems electromagnetic transients simulation
R
R
ICHistory IL History
2L= ikm (tt) + vL (tt)
t2C
= ikm (tt) vC(tt)t
Reff = t
2LReff =
t
2C
vR (t) vL (t) vC(t)
k m
IL History IC History
R +2Ct
t
2L+
R +2Ct
2Ct
t
2L+
t
2L+
Figure 4.7 Reduction of RLC branch
x = 0 x = d
ikm imk
v (x, t)
i (x, t)
mk
Figure 4.8 Propagation of a wave on a transmission line
Consider the lossless distributed parameter line depicted in Figure 4.8, where
L is the inductance and C the capacitance per unit length. The wave propagationequations for this line are:
v(x,t)x
= L i(x,t)t
(4.22)
i(x,t)x
= C v(x,t)t
(4.23)
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Numerical integrator substitution 75
and the general solution:
i(x,t) = f1(x t ) + f2(x + t) (4.24)v(x,t) = Z f1(x t ) Z f2(x + t) (4.25)
with f1(x t) and f2(x + t ) being arbitrary functions of(x t) and (x + t)respectively. f1(x t ) represents a wave travelling at velocity in a forward direc-tion (depicted in Figure 4.8) and f2(x+ t ) a wave travelling in a backward direction.ZC , the surge or characteristic impedance and , the phase velocity, are given by:
ZC =
L
C (4.26)
= 1LC
(4.27)
Multiplying equation 4.24 by ZC and adding it to, and subtracting it from,
equation 4.25 leads to:
v(x,t) + ZC i(x,t) = 2ZC f1(x t ) (4.28)v(x,t) ZC i(x,t) = 2ZC f2(x + t) (4.29)
It should be noted that v(x,t)+ZC i(x,t) is constant when (x t) is constant.Ifd is the length of the line, the travelling time from one end (k) to the other end (m)
of the line to observe a constant v(x,t) + ZC i(x,t) is:
= d/ = d
LC (4.30)
Hence
vk (t ) + ZC ikm (t ) = vm(t ) + ZC (imk (t)) (4.31)
Rearranging equation 4.31 gives the simple two-port equation for imk , i.e.
imk(t ) =1
ZCvm(t) + Im(t ) (4.32)
where the current source from past History terms is:
Im(t ) = 1
ZCvk (t ) ikm(t ) (4.33)
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76 Power systems electromagnetic transients simulation
ZC ZCvk(t) vm (t)
ikm (t) imk(t)
Ik(t)
Im (t)
Figure 4.9 Equivalent two-port network for a lossless line
Similarly for the other end
ikm(t ) =1
ZCvk (t) + Ik (t ) (4.34)
where
Ik (t ) = 1
ZCvm(t ) imk(t )
The expressions (x t ) = constant and (x + t) = constant are called thecharacteristic equations of the differential equations.
Figure 4.9 depicts the resulting two-port model. There is no direct connection
between the two terminals and the conditions at one end are seen indirectly and with
time delays (travelling time) at the other through the current sources. The past History
terms are stored in a ring buffer and hence the maximum travelling time that can be
represented is the time step multiplied by the number of locations in the buffer. Since
the time delay is not usually a multiple of the time-step, the past History terms on
either side of the actual travelling time are extracted and interpolated to give the
correct travelling time.
4.4 Network solution
With all the network components represented by Norton equivalents a nodal
formulation is used to perform the system solution.
The nodal equation is:
[G]v(t) = i(t) + IHistory (4.35)
where:
[G] is the conductance matrixv(t) is the vector of nodal voltages
i(t ) is the vector of external current sources
IHistory is the vector current sources representing past history terms.
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Numerical integrator substitution 77
i12
i13
i14i15
i1
Transmission line
5
2
3
41
Figure 4.10 Node 1 of an interconnected circuit
The nodal formulation is illustrated with reference to the circuit in Figure 4.10
[5] where the use of Kirchhoffs current law at node 1 yields:
i12(t ) + i13(t) + i14(t) + i15(t ) = i1(t) (4.36)Expressing each branch current in terms of node voltages gives:
i12(t) =1
R(v1(t) v2(t)) (4.37)
i13(t) =t
2L(v1(t) v3(t)) + I13(t t) (4.38)
i14(t) =2C
t(v1(t) v4(t)) + I14(t t) (4.39)
i15(t) =1
Z v1(t) + I15(t ) (4.40)
Substituting these gives the following equation for node 1:1
R+ t
2L+ 2C
t+ 1
Z
v1(t )
1
Rv2(t )
t
2Lv3(t)
2C
tv4(t)
= I1(t t) I13(t t ) I14(t t ) I15(t t) (4.41)Note that [G] is real and symmetric when incorporating network components. If con-trol equations are incorporated into the same
[G
]matrix, the symmetry is lost;
these are, however, solved separately in many programs. As the elements of [G]are dependent on the time step, by keeping the time step constant [G] is constant andtriangular factorisation can be performed before entering the time step loop. More-
over, each node in a power system is connected to only a few other nodes and therefore
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78 Power systems electromagnetic transients simulation
2C1
t
t
2L1
Is = Vm sin(t)/R1
ISt
2L2
IhL1
IhL2IhC
1
R1
R1
R2
R2
C1Vm sin(t)
1
1
2
2
3
3
(a)
(b)
Figure 4.11 Example using conversion of voltage source to current source
the conductance matrix is sparse. This property is exploited by only storing non-zero
elements and using optimal ordering elimination schemes.
Some of the node voltages will be known due to the presence of voltage sources
in the system, but the majority are unknown. In the presence of series impedance with
each voltage source the combination can be converted to a Norton equivalent and the
algorithm remains unchanged.
Example: Conversion of voltage sources to current sources
To illustrate the incorporation of known voltages the simple network displayed in
Figure 4.11(a) will be considered. The task is to write the matrix equation that must
be solved at each time point.
Converting the components of Figure 4.11(a) to Norton equivalents (companion
circuits) produces the circuit of Figure 4.11(b) and the corresponding nodal equation:
1
R1+ t
2L1 t
2L10
t
2L1
t
2L1+
1
R2+
2C1
t 1
R2
0 1R2
1
R2+ t
2L2
v1
v2v3
=
Vm sin(t)
R1 IhL1
IhL1 IhC1IhL2
(4.42)
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Numerical integrator substitution 79
Equation 4.42 is first solved for the node voltages and from these all the branch
currents are calculated. Time is then advanced and the current sources representing
History terms (previous time step information) are recalculated. The value of the
voltage source is recalculated at the new time point and so is the matrix equation.
The process of solving the matrix equation, calculating all currents in the system,
advancing time and updating History terms is continued until the time range of the
study is completed.
As indicated earlier, the conversion of voltage sources to Norton equivalents
requires some series impedance, i.e. an ideal voltage source cannot be represented
using this simple conductance method. A more general approach is to partition the
nodal equation as follows:
[GU U
] [GU K
][GKU] [GKK ] .vU(t)
vK (t) =
iU(t )
iK (t )+
IUHistory
IKHistory =
IU
IK
(4.43)
where the subscripts U and K represent connections to nodes with unknown and
known voltages, respectively. Using Krons reduction the unknown voltage vector is
obtained from:
[GU U]vU(t ) = iU(t) + IUHistory [GU K ]vK (t ) = IU (4.44)
The current in voltage sources can be calculated using:
[GKU]vU(t) + [GKK ]vK (t ) IKHistory = iK (t ) (4.45)
The process for solving equation 4.44 is depicted in Figure 4.12. Only the right-
hand side of this equation needs to be recalculated at each time step. Triangular
factorisation is performed on the augmented matrix [GU U GU K ] before entering thetime step loop. The same process is then extended to iU(t ) IHistory at each time step(forward solution), followed by back substitution to get VU(t). Once VU(t) has been
found, the History terms for the next time step are calculated.
4.4.1 Network solution with switches
To reflect switching operations or time varying parameters, matrices [GU U] and[GU K ] need to be altered and retriangulated. By placing nodes with switches last, asillustrated in Figure 4.13, the initial triangular factorisation is only carried out for the
nodes without switches [6]. This leaves a small reduced matrix which needs altering
following a change. By placing the nodes with frequently switching elements in the
lowest part the computational burden is further reduced.
Transmission lines using the travelling wave model do not introduce off-diagonal
elements from the sending to the receiving end, and thus result in a block diagonal
structure for [GU U], as shown in Figure 4.14. Each block represents a subsystem(a concept to be described in section 4.6), that can be solved independently of the rest
of the system, as any influence from the rest of the system is represented by the History
terms (i.e. there is no instantaneous term). This allows parallel computation of the
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80 Power systems electromagnetic transients simulation
=
GUU
[GUU GUK]
[GUU ]
0
0
[G]
(1)
'
UUG
GUK
GUK
GBA GKK
V
VU
I
VK
VK
VK
VK
VU
VU
VU
V
IK
IU
IU
IU
IU
IU
GUU GUK
GUK
(3) (2)
=
=
(1) Triangulation of matrix
(2) Forward reduction
(3) Back substitution
Figure 4.12 Network solution with voltage sources
solution, a technique that is used in the RTDS simulator. For non-linear systems, each
non-linearity can be treated separately using the compensation approach provided
that there is only one non-linearity per subsystem. Switching and interpolation are
also performed on a subsystem basis.
In the PSCAD/EMTDC program, triangular factorisation is performed on a sub-
system basis rather than on the entire matrix. Nodes connected to frequently switched
branches (i.e. GTOs, thyristors, diodes and arrestors) are ordered last, but other
switching branches (faults and breakers) are not. Each section is optimally ordered
separately.
A flow chart of the overall solution technique is shown in Figure 4.15.
4.4.2 Example: voltage step applied to RL load
To illustrate the use of Kron reduction to eliminate known voltages the simple circuit
shown in Figure 4.16 will be used.
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Numerical integrator substitution 81
[G] V I
0
Frequently switching
components placed last
=
=
(1) Partial triangulation of matrix (prior to time step loop)
(2) Complete triangulation
(3) Forward reduction of current vector
(4) Back substitution for node voltages
(1)
(2)
(3)(4)
Figure 4.13 Network solution with switches
VU
[GUU]
IU= IU [GUK] VK
0
= + + +
Figure 4.14 Block diagonal structure
Figure 4.17 shows the circuit once the inductor is converted to its Norton
equivalent. The nodal equation for this circuit is:
GSwitch GSwitch 0GSwitch GSwitch + GR GR
0 GR GL_eff + GR
v1v2
v3
=
iv0
IHistory
(4.46)
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82 Power systems electromagnetic transients simulation
Initialisation
Build upper part of
triangular matrix
Is run finished
Check switches for
change
Solve for history
terms
User specified
dynamics file
Network solution
User-specified
output definition file
Interpolation,
switching procedure
and chatter removal
Write output files
Yes Stop
NoSolve for voltages
Update source voltages and
currents
Figure 4.15 Flow chart of EMT algorithm
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Numerical integrator substitution 83
ivR = 100
L = 50mH
1 2
3
R
Figure 4.16 Simple switched RL load
1 2
3
RSwitch
Riv
RL eff
Figure 4.17 Equivalent circuit for simple switched RL load
As v1 is a known voltage the conductance matrix is reordered by placing v1 last in
the column vector of nodal voltages and moving column 1 of [G] to be column 3;
then move row 1 (equation for current in voltage source) to become row 3. This then
gives:
GSwitch + GR GRGR GL_eff + GR
GSwitch0
GSwitch 0 GSwitch
v2v3
v1
=
0IHistory
iv
(4.47)
which is of the form
[GU U] [GU K ][GKU] [GKK ]
vU(t )
vK (t)
=
iU(t)
iK (t)
+
IU_History
IK_History
i.e. GSwitch + GR GR GSwitch
GR GL_eff + GR 0v2v3
v1
=
0
IHistory
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84 Power systems electromagnetic transients simulation
Note the negative IHistory term as the current is leaving the node. Performing Gaussianelimination gives:
GSwitch + GR GR GSwitch
0 GL_eff + GR M(GR ) M(GSwitch)v2v3
v1
= 0IHistory
(4.48)
where
M = GL_effGSwitch + GR
Moving the known voltage v1(t) to the right-hand side gives:
GSwitch + GR GR
0 GL_eff + GR M(GR )
v2v3
= 0IHistory GSwitchM(GSwitch)
v1
(4.49)
Alternatively, the known voltage could be moved to the right-hand side before
performing the Gaussian elimination. i.e.
GSwitch + GR GR
GR GL_eff + GR
v2v3
=
0
IHistory
GSwitch
0
v1 (4.50)
Eliminating the element below the diagonal, and performing the same operation on
the right-hand side will give equation 4.49 again. The implementation of these equa-
tions in FORTRAN is given in Appendix H.2 and MATLAB in Appendix F.3. The
FORTRAN code in H.2 illustrates using a d.c. voltage source and switch, while the
MATLAB version uses an a.c. voltage source and diode. Note that as Gaussian elimi-
nation is equivalent to performing a series of NortonThevenin conversion to produce
one Norton, the RL branch can be modelled as one Norton. This is implemented in
the FORTRAN code in Appendices H.1 and H.3 and MATLAB code in Appendices
F.1 and F.2.Table 4.2 compares the current calculated using various time steps with results
from the analytic solution.
For a step response of an RL branch the analytic solution is given by:
i(t) = VR
1 et R/L
Note that the error becomes larger and a less damped response results as the time
step increases. This information is graphically displayed in Figures 4.18(a)4.19(b).
As a rule of thumb the maximum time step must be one tenth of the smallest time
constant in the system. However, the circuit time constants are not generally known
a priori and therefore performing a second simulation with the time step halved will
give a good indication if the time step is sufficiently small.
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Numerical integrator substitution 85
Table 4.2 Step response of RL circuit to various step lengths
Time (ms) Current (amps)
Exact t = /10 t = t = 5 t = 10
1.0000 0 0 0 0 0
1.0500 63.2121 61.3082 33.3333
1.1000 86.4665 85.7779 77.7778
1.1500 95.0213 94.7724 92.5926
1.2000 98.1684 98.0785 97.5309
1.2500 99.3262 99.2937 99.1770 71.4286
1.3000 99.7521 99.7404 99.7257
1.3500 99.9088 99.9046 99.9086 1.4000 99.9665 99.9649 99.9695
1.4500 99.9877 99.9871 99.9898
1.5000 99.9955 99.9953 99.9966 112.2449 83.3333
1.5500 99.9983 99.9983 99.9989
1.6000 99.9994 99.9994 99.9996
1.6500 99.9998 99.9998 99.9999
1.7000 99.9999 99.9999 100.0000
1.7500 100.0000 100.0000 100.0000 94.7522
1.8000 100.0000 100.0000 100.0000
1.8500 100.0000 100.0000 100.0000 1.9000 100.0000 100.0000 100.0000
1.9500 100.0000 100.0000 100.0000
2.0000 100.0000 100.0000 100.0000 102.2491 111.1111
2.0500 100.0000 100.0000 100.0000
2.1000 100.0000 100.0000 100.0000
2.1500 100.0000 100.0000 100.0000
2.2000 100.0000 100.0000 100.0000
2.2500 100.0000 100.0000 100.0000 99.0361
2.3000 100.0000 100.0000 100.0000
2.3500 100.0000 100.0000 100.0000
2.4000 100.0000 100.0000 100.0000
2.4500 100.0000 100.0000 100.0000
2.5000 100.0000 100.0000 100.0000 100.4131 92.5926
2.5500 100.0000 100.0000 100.0000
2.6000 100.0000 100.0000 100.0000
2.6500 100.0000 100.0000 100.0000
2.7000 100.0000 100.0000 100.0000
2.7500 100.0000 100.0000 100.0000 99.8230
2.8000 100.0000 100.0000 100.0000
2.8500 100.0000 100.0000 100.0000
2.9000 100.0000 100.0000 100.0000
2.9500 100.0000 100.0000 100.0000
3.0000 100.0000 100.0000 100.0000 100.0759 104.9383