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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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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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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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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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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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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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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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=

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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[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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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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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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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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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

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