Introduction to the Use of MicroTran and other EMTP Versions
R
c 1998 by
Hermann W. Dommel
Department of Electrical and Computer Engineering
The University of British Columbia
2356 Main Mall
Vancouver, B. C.
Canada, V6T 1Z4
Tel.: +1 (604) 822-2793, Fax: +1 (604) 822-5949
Email: hermannd@ece.ubc.ca, WWW: http://www.ece.ubc.ca
These notes were rst used in 1997/98 in a Graduate Course ELEC 553 "Advanced Power Systems
Analysis". The help of the following graduate students in that course is gratefully acknowledged: S.
Bibian, B. D. Bonatto, J. D. Bull, J. Calvino-Fraga, N. Dai, Y. Duan, J. A. Hollman, F. A.
Moreira, R. A. Rivas, T.-C. Yu.
Last revision: May 7, 1999.
Web page created by: Daniel Lindenmeyer and Benedito Donizeti Bonatto.
Contents
1 Computer Programs for Electromagnetic Transients in Power Systems
3
2 Contributions from Research at UBC to Various Versions
4
3 Information for EMTP Users
5
4 Major Applications of MicroTran
6
5 Major Dierences to Short-Circuit, Power Flow (Load Flow) and Stability Programs
13
5.1 Dierences to Short-Circuit Programs . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
5.2 Dierences to Power Flow Programs . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
5.3 Dierences to Stability Programs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
6 MicroTran Input Data
16
7 Case Studies: Short-Circuit in Single-Phase Network
17
7.1
7.2
7.3
7.4
7.5
7.6
7.7
Steady-state Solution Similar to Short-Circuit Programs . . . . . . . . . . . . . . . .
Transient Solution with Simple Single-Phase R-L Circuit . . . . . . . . . . . . . . . .
Transient Solution with Single-Phase R-L Circuit for Line . . . . . . . . . . . . . . .
Transient Solution with Single-Phase -Circuit for Line . . . . . . . . . . . . . . . . .
Transient Solution with Single-Phase Distributed Parameter Line . . . . . . . . . . .
Comparison between R-L, -Circuit, and Distributed Parameter Line Representations
Frequency Scan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 Case Studies: Short-Circuit in Three-Phase Network
18
20
22
23
24
25
26
28
8.1 Steady-State Solution Similar to Short-Circuit Programs . . . . . . . . . . . . . . . . 29
1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
Transient Solution with Three-Phase R-L Circuit for Line . . . . . . . . . . . . . . . .
Transient Solution with Three-Phase -Circuit for Line . . . . . . . . . . . . . . . . .
Transient Solution with Three-Phase Distributed Parameter Line . . . . . . . . . . .
Comparison between R-L, -Circuit, and Distributed Parameter Line Representations
Transient Solution with Detailed Generator Model . . . . . . . . . . . . . . . . . . . .
Transient Solution with Actual Units . . . . . . . . . . . . . . . . . . . . . . . . . . .
Transient Solution for Simultaneous Three-Phase Short-Circuit . . . . . . . . . . . . .
Short-Circuits on HVDC Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A Relationship between Phase Quantities and Sequence Quantities
30
31
33
36
36
38
39
41
42
Chapter 1
Computer Programs for Electromagnetic
Transients in Power Systems
EMTP versions
Technical University Munich 1963 (rst publication May 1964).
BPA EMTP (probably no longer used by Bonneville Power Administration; may use ATP now);
non-commercial. Was distributed for free because of Freedom of Information Act in U.S.A..
R ; commercial. Most of it owned by University of British Columbia; disUBC MicroTran tributed by Microtran Power System Analysis Corporation in Vancouver, Canada.
DCG/EPRI EMTP (DCG = Development Coordination Group, EPRI = Electric Power Research Institute in U.S.A.); commercial. Ontario Hydro is the commercializer.
ATP (Alternative Transients Program of W. S. Meyer); free, but requires a license, which is
not available to everybody.
NETOMAC (Siemens); commercial.
Morgat and Arene (Electricite de France); commercial.
EMTDC (Manitoba HVDC Research Centre); commercial.
PSIM (H. Jin); commercial. Primarily for power electronics studies.
SABER; commercial. For power electronics studies.
SPICE, PSPICE, ...; commercial. Occasionally used for power electronics studies.
?
3
Chapter 2
Contributions from Research at UBC to
Various Versions
Contributions to BPA EMTP (and indirectly to DCG/EPRI EMTP and ATP), and in
some cases to UBC version MicroTran:
TACS of Laurent Dube (Transient Analysis of Control Systems).
Multiphase untransposed transmission line with constant parameters of K. C. Lee (\CP line
model").
Frequency dependent transmission line of J. R. Mart.
Three-phase transformer models of H. W. Dommel and I. I. Dommel (\BCTRAN" in BPA
EMTP, DCG/EPRI EMTP, and ATP; part of input preprocessor MTD in MicroTran).
Synchronous machine model of V. Brandwajn (type 59 in DCG/EPRI EMTP and ATP).
Synchronous machine data conversion of H. W. Dommel (earlier versions in BPA EMTP,
DCG/EPRI EMTP, and ATP; latest version in MicroTran).
Contributions to DCG/EPRI EMTP:
Underground cable models of L. Mart.
New line constants program and improved frequency dependent line model of J. R. Mart.
4
Chapter 3
Information for EMTP Users
EMTP Newsletter from 1979 to 1987, edited by H. W. Dommel (co-editor W. S. Meyer to 1984,
D. Van Dommelen 1985 to 1987), with help from I. I. Dommel and G. Empereur.
EMTP News from 1988 to 1993, edited by D. Van Dommelen, with help from G. Empereur
and A. Laeremans.
EMTP Review from 1987 to 1990, for users of DCG/EPRI EMTP, edited by W. F. Long.
Harmonics and Transients Tech Notes, for members of PATH Users Group, Electrotek Concepts, Inc..
Various User Groups set up or sanctioned by W. S. Meyer.
CAN/AM EMTP News from 1994 (?) to now for ATP users, edited by W. S. Meyer and
Tsu-huei Liu.
R Power System Analysis Corp.,
H. W. Dommel, EMTP Theory Book, 2nd edition. MicroTran Vancouver, Canada, 1992. Latest update: April 1996.
J. A. Martinez-Velasco, editor, Computer Analysis of Electric Power System Transients. IEEE
Press, Piscataway, NJ, U.S.A., 1997.
Specialized Conference IPST (International Power System Transients Conference), co-chaired
by M. T. Correia de Barros and H. W. Dommel:
{ IPST'95 in August 1995 in Lisbon, Portugal.
{ IPST'97 in June 1997 in Seattle, Washington, U.S.A.
{ IPST'99 in June 1999 in Budapest, Hungary.
5
Chapter 4
Major Applications of MicroTran
Steady-state coupling eects between adjacent power lines, or to adjacent communication lines.
Table 4.1: Induced voltages and grounding currents
line phase simulation
Voltages on lines in operation L1
360 kV
L2
550 kV
Induced voltages on open line L3 A
28.20 kV
B
14.19 kV
C
8.04 kV
Grounding currents when
L3 A
10.78 A
line is grounded
B
3.26 A
C
1.53 A
measurement
372 kV
535 kV
30 kV
15 kV
10 kV
11 A
5A
1A
Figure 4.1: Electrostatic coupling at power frequency
6
Harmonics, power quality.
Figure 4.2: Current on ac side of controlled three-phase bridge rectier
Frequency scans to obtain frequency dependent impedances and other frequency responses
(e.g., transfer functions).
Figure 4.3: System impedance in subsynchronous frequency region [1]
7
Detailed waveforms of fault currents in ac and dc systems (shown later in these notes).
Switching surges (from switching transmission lines, capacitors, transformers, and reactors).
Figure 4.4: Measured and calculated overvoltages at receiving end of line [2]
Fault surges.
Figure 4.5: Measured and calculated overvoltage on unfaulted phase (single-line-to-ground fault) [3]
8
Lightning surges (surge arrester protection).
Figure 4.6: Measured and simulated propagation of voltage surge from impulse generator through
substation [4]
Table 4.2: Propagation of surge from impulse generator through substation [4]
Type of
measurement
Mode of Oscillation
Amplitudes
injection frequencies
1st peak
3rd peak
Tests EMTP Dev. Tests EMTP Dev. Tests EMTP Deviation
[kHz] [kHz] [%] [V]
[V]
[%] [V]
[V]
[%]
Measurements homoon the set
polar
230
235
2
65
65
0
43
47
8
of busbars
two-wire 283
300
5.6 70
70
0
45
50
10
Measurements homoat the transpolar
250
230
8
63
69
8
41
48
14
former terminals two-wire 300
285
5
70
65
7
45
49
8
9
Ferroresonance.
Figure 4.7: Measured ferroresonance on 1100 kV test line [5]
Figure 4.8: Calculated ferroresonance on 1100 kV test line [5]
10
Subsynchronous resonance.
Figure 4.9: Measured and calculated shaft torque during out-of-phase synchronization [6]
Inrush current.
Figure 4.10: Measured and calculated inrush current [7]
11
Power electronics.
v
i
Electrical Cable
I.M.
v
i
Figure 4.11: Measured terminal voltage of PWM variable speed drive [8]
2500
2000
1500
1000
V
500
0
-500
-1000
-1500
-2000
-2500
310
315
320
ms
325
330
Figure 4.12: Calculated terminal voltage of PWM variable speed drive [8]
12
Chapter 5
Major Dierences to Short-Circuit,
Power Flow (Load Flow) and Stability
Programs
Phasor equations replaced by dierential equations. For example, the phasor equation V =
j!LI for an inductance is replaced by the dierential equation v = L dtdi ; similar for capacitance,
etc..
5.1 Dierences to Short-Circuit Programs
Generator in short-circuit programs is E 00 behind Xd00; in MicroTran it can be as simple as that,
and as detailed as needed on electrical and mechanical side.
For series capacitors, net reactance Xtransmissionline , Xcapacitor is used in short-circuit programs;
in MicroTran, it becomes an L-C circuit with voltage drop described by
Zt
1
di
vline+capacitor = L dt + C idu + vcapacitor (0)
0
Short-circuit programs solve three single-phase networks, one with positive, one with negative,
and one with zero sequence parameters (negative sequence solution usually omitted by assuming
Znegative = Zpositive); MicroTran uses M-phase representations (e.g., three-phase equations for
single-circuit lines, or six-phase equations for double-circuit lines, etc.).
In short-circuit programs, phase shift through delta-wye transformers is usually ignored, which
can lead to wrong currents on delta side if not corrected for; in MicroTran, phase shift through
delta-wye transformers is automatically correct.
Fault conditions can become complicated in short-circuit programs, particularly for simultaneous faults. In MicroTran, short-circuit conditions are simple because everything is in phase
13
quantities (e.g., Va = 0 for a short from phase a to ground); simultaneous faults are easy in
MicroTran. Example: Va,high = 0 and Vb,low = Vc,low for short \a" to ground on high side
and short between \b" and \c" on low side; high side fault represented by a closed switch from
\a" to ground, and low side fault by a switch between \b-low" and \c-low".
MicroTran produces the following eects, which cannot be obtained from short-circuit programs:
{
{
{
{
decaying dc oset in fault current,
decaying ac amplitude with detailed generator model,
more than one frequency with series capacitors,
dierent X/R ratios automatically accounted for.
5.2 Dierences to Power Flow Programs
EMTP programs use steady-state solutions for the system of linear node equations , either
{ to get initial conditions for transient simulations, or
{ to get steady-state solutions at one frequency, or at many frequencies in so-called frequency
scans.
You can dene voltage and/or current sources for these solutions, by specifying the sources
with respect to magnitude and angle.
You cannot specify real and reactive power, or real power and voltage magnitude, in these
linear solutions.
The incentive for power ow solutions with EMTP-type programs came from the fact that
these programs allow users to set up cases in great detail, such as
{ untransposed lines (single-circuit, double-circuit, etc.),
{ single-phase distribution lines connected to three-phase substations,
{ etc.
Such cases cannot be solved with classical power ow programs, which assume that the power
system is completely "balanced" (all voltages and currents symmetrical) and represented by
the single-phase positive sequence network only.
In the ATP version of the EMTP, a power ow option was added which uses a Gauss-Seidel
type iteration method. This method is unreliable with respect to convergence, and only works
for some cases.
In the DCG/EPRI version of the EMTP, a power ow option was added which uses Newton's iteration method. This method is much more reliable. This option also takes physical
constraints into account, such as [9]:
14
{ specifying the total three-phase power for generators, rather than the power for each phase,
to make sure that the generator, seen from the network side, has the correct negative and
zero sequence impedance in the power ow solution, etc.,
{ specifying the total three-phase power for loads, rather than the power for each phase, if
the loads have known impedance ratios Zneg =Zpos and Zzero=Zpos.
MicroTran does not have a power ow option at this time. If one were to be added, it might
be a completely separate program that just uses EMTP input data.
5.3 Dierences to Stability Programs
Stability programs solve the electrical network with phasor equations of the type [Y ] [V ] = [I ]
(with iterations for nonlinear loads), and use dierential equations only for the mechanical part
of the generator with the \swing equation" !J d!
dt = Pmech , Pelec. Dierential equations are
also used for ux decay in generators, for excitation and governor systems, etc.
Stability programs usually solve the single-phase positive sequence network only. Single-line-toground faults are included in the positive sequence network by adding the impedances Znegative
and Zzero seen from the fault location as a shunt impedance Znegative +Zzero at the fault location.
EMTP programs use M-phase representations.
Stability programs are faster, because EMTP-type programs use three-phase representations
and solve everything with dierential equations.
There are special cases where EMTP-type programs are used for stability studies:
{ cases where non-power-frequency oscillations in the electric network are important, such
as studies of subsynchronous resonance,
{ cases where the \backswing" is important ( = generator angle decreasing immediately
after a fault for a few milliseconds, before it increases),
{ cases where negative sequence currents in generators become important, and
{ other special cases.
15
Chapter 6
MicroTran Input Data
Rarely p.u., usually actual quantities. See SSR fact sheet for comparison between two ap-
proaches. Careful if nonlinearities!
t related to fmax of interest. Unfortunately not automatically chosen.
Basic data:
{ type 0 branch: R-L-C
{ type 1,2,3 ... branches: M-phase symmetric -circuit (type 51, 52, 53 ... for input with
{
{
{
{
higher accuracy)
type -1,-2,-3 ... branches: M-phase distributed parameter line
nonlinear R,L
switches, diodes, thyristors
sources
16
Chapter 7
Case Studies: Short-Circuit in
Single-Phase Network
For simultaneous three-phase short-circuits, we can use a single-phase representation with pos-
itive sequence parameters.
As an example, we use the \First IEEE Subsynchronous Resonance Benchmark Model" (IEEE
Task Force, \First benchmark model for computer simulation of subsynchronous resonance",
IEEE Trans. Power App. Syst., vol. PAS-96, pp. 1565-1572, Sept./Oct. 1977).
The connection to the innite bus through an impedance is left o here, to simplify the case.
Figure 7.1: Single-Phase Network
Per unit values are used at this stage because
{ those familiar with short-circuit and power ow studies may be more used to per unit
values,
{ numbers used here are easier to compare with the numbers in the publication.
17
In general, actual units are better for transient studies, and we will switch to actual units later
on.
With XOPT = 60, input for inductive branches will be as reactance !L in at 60 Hz.
The generator is represented as a voltage source E 00 behind Xd00. Assume Vgen = 1:0 p:u: (RMS)
for E 00.
The values for this single-phase case are the positive sequence values.
The fault resistance is assumed to be 0.04 p.u. In the IEEE benchmark case, it is inductive,
but a resistive fault impedance is more realistic. Maybe the impedance in the IEEE benchmark
case represented a short line from the bus to the fault location.
7.1 Steady-state Solution Similar to Short-Circuit Programs
Data le: 1 STEADY.DAT
This steady-state solution shows what you would obtain
from a conventional short-circuit
V
prefault
program with phasor solutions, namely Ifault = Rtotal+j!Ltotal = 0:06+1:j00:404 p:u:, or jIfault j =
2:448 p:u:.
Note that the series capacitor is represented as a negative reactance in such phasor solutions.
Leave t and tmax empty (or set them zero). Then only a steady-state solution will be performed.
Use tstart = ,1:0 (any negative number) in the source data to indicate that source is already
present in steady state before transient simulation begins at t = 0.
Use tclose = ,1:0 in the data for the switch simulating the fault, to indicate that the switch is
already closed in steady state before transient simulation begins at t = 0.
Ask for fault current with \1" in column 80 of series capacitor data line (do not ask for switch
current because switch current from steady-state solution is not available at this time).
If you want a complete steady-state output, insert \1" in column 38 of time data line.
* File "1_STEADY.DAT".
* To demonstrate the various ways of calculating short-circuit currents, the
* "First IEEE Subsynchronous Resonance Benchmark Model" is used as an example
* (IEEE Task Force, "First benchmark model for computer simulation of sub* synchronous resonance", IEEE Trans. Power App. Syst., vol. PAS-96,
* pp. 1565-1572, Sept./Oct. 1977). The connection to the infinite bus through
18
* an impedance is left off here, so that the case can also be run with the
* student version. Per unit values are used, to make it easier to compare the
* numbers used here with the numbers in the publication.
*
* In this case, the generator is represented as a voltage source E" behind X"d.
* The values for this single-phase case are the positive sequence values. The
* fault resistance is assumed to be 0.04 p.u. (in the IEEE benchmark case, it
* is inductive, but a resistive fault impedance is more realistic).
*
* This steady-state solution shows what you would obtain from a conventional
* short-circuit program with phasor solutions. Note that the series capacitor
* is represented as a negative reactance in such phasor solutions.
*
*
*
.
.
.
.
.
.
. Case identification card
Steady-state solution, short circuit in single-phase circuit
60
*
*
.
.
.
.
0.
.
.
. Time card
0.
1
*
*
$
.
.
.
.
gen
a
a
cap
cap
b
= =
.
.
. Lumped RLC branch
0.275
0.02
0.50
-0.371
1
End of level 1: Linear and nonlinear elements = = = = = = = = = = = =
*
*
.
.
.
.
.
b
.
. Time-controlled switch
-1.
$ = = =
1.0
0.04
End of level 2: Switches and piecewise linear elements = = = = = = = =
*
*
14
.
gen
$ = = =
.
.
.
1.4142
.
.
. Voltage or current sources
60.
End of level 3: Sources
-90.
-1.
= = = = = = = = = = = = = = = = = = = = = = =
cap
$ = = =
End of level 4: User-defined voltage output
$ = = =
Level 5: End of data case
= = = = = = = = = = = = =
= = = = = = = = = = = = = = = = = = = = = =
19
7.2 Transient Solution with Simple Single-Phase R-L Circuit
Data le: 1 SIMPLE.DAT
Change t to 50 s, and tmax to 0.1 s. Step size t is related to maximum frequency which
you expect or want to see in the results: fmax = 81 t , if you want 8 points per cycle at the
highest frequency.
Change tclose of switch to 0, so that fault is initiated at t = 0 (topen = 1:0 will keep switch
closed during transient solution to tmax = 0:1s).
Ask for fault current output with \1" in column 80 of switch data line.
Ask for capacitor current and voltage with \3" in column 80 of capacitor data line.
If it is a simple series connection of resistances and inductances (assuming that the series
capacitor simply decreases the line inductance, which is not correct as shown in the next case),
then there is an exact solution for the rst-order dierential equation of the R-L circuit.
With a voltage source of v(t) = Vmax sin(!t + ), we obtain:
i
h
, Tt ;
i(t) = q 2 Vmax
sin(
!t
+
,
)
,
sin(
,
)
e
Rtotal + (!Ltotal )2
Ltotal
total
where = tan,1 !L
Rtotal , T = Rtotal , Rtotal = 0:06 p:u:, !Ltotal = 0:404 p:u:, f = 60 Hz .
For , = 0, there is no dc oset. For , = 90, there is maximum dc oset.
In high voltage systems, R !L. In that case, maximum dc oset occurs when the voltage is
just going through zero, v(t) = Vmax sin(!t).
With maximum dc oset, the peak current is twice as high as without dc oset if the decay is
very slow, or somewhat less than twice as high with faster decay. As seen from above equation,
total .
the decay time constant is !1 XRtotal
Knowing the dc oset is important because it
{ increases the mechanical forces between busbars carrying the current and return current
by a factor of up to 4,
{ increases the power dissipation i2 Rarc in the circuit breaker arc by a factor of up to 4,
and
20
{ may prevent zero crossings for the rst few cycles in three-phase short circuits close to
generator terminals (see notes "Simple Sources and Machine Models"); in reality, the arc
resistance will probably create enough decay to create zero crossings early enough for the
circuit breaker to operate.
The dc oset in the fault current is taken into account in the circuit breaker standards.
{ Older standards provided data for the inuence of the decaying dc oset on circuit breaker
ratings as functions of X/R ratios.
{ Newer standards use symmetrical current ratings (dc oset implicitly assumed), and show
ratios of asymmetrical to symmetrical interrupting capabilities as a function of contact
parting time.
Exercises:
{ Create the data le for this case and run the case. Include the exact solution for an
R-L circuit for comparison purposes. It consists of a sinusoidal component and of a
decaying dc component. As explained on page 9-24 of the Reference Manual (revision
1997), this decaying dc oseth component
can be simulated with the double-exponential
i
t
t
impulse function i(t) = Imax e , e with = , T1 , and = , 500
t .
{ Inject both components into a node \EXACT", which has a resistance branch of 1:0
to
ground. Ask for the current in that branch. In the lines for the source data, use \-1"
in columns 9-10 to signal that the ac and dc components are current sources rather than
voltage sources.
{ Change tclose to 1/4 of a cycle, and comment on the changes which you see in the fault
current.
Figure 7.2: Wrong Transient Solution, Short Circuit
21
7.3 Transient Solution with Single-Phase R-L Circuit for
Line
Data le: 1 RL.DAT
The transient results of the previous case are not correct. For transient simulations, a capac-
itor must be modelled as a capacitance (this is always correct, for transient and steady-state
solutions). A negative reactance can only be used for steady-state solutions.
1 . Therefore, with f = 60Hz : C =
1
The value for !C in per unit is 0:371
0
:
371
120 = 7149:8 ,
6
10 p:u:. With COPT = 0, input is in F or , p:u:
The fault current looks reasonable now, with the second frequency created by the L-C circuit
clearly noticeable.
Figure 7.3: Transient Solution with RLC
In reality, a protective gap will spark over across the capacitor when the voltage becomes too
high, to protect the capacitors. This is discussed in data le 1 RLGAP.DAT.
Exercises:
{ Create the data le for this case and run the case.
{ Modify the case by putting a protective spark gap in parallel with the series capacitor
(use the \voltage- dependent switch" for the gap). Assume that the gap sparks over when
the voltage reaches 1.8 p.u. Observe the fault current and the current in the spark gap.
22
{ Disable the critical damping adjustment scheme (CDA) with "1" in column 68 of the time
data line, and change t to 1ms (larger t shows numerical oscillations clearer). Observe
the numerical oscillations in the gap current and capacitor current (one should be the
negative of the other).
7.4 Transient Solution with Single-Phase -Circuit for Line
Data le: 1 PI.DAT
We need the capacitance for the -circuit, which we can nd as follows:
{ From Sbase = 892:4 MV A and Vbase = 500 kV , we get a positive sequence reactance of
X1 = 140:07
. Based on known, typical per mile values of 0.5 to 0.6 /mile for 500 kV
lines, we can guess the line length as 250 miles.
{ By ignoring the resistance, the wave velocity becomes c = pL10 C 0 . Since we know c =
:5 p.u./mile, C 0 becomes
300 000 km/s (186 411.4 miles/s), f = 60Hz, and L0 = 2500120
5:4244584 10,6 p.u./mile.
A -circuit is branch type \1" (1, 2, ...n for n-phase -circuits, or 51, 52, ... for input with
higher accuracy). Therefore, add \1" in column 2 of the R-L-C data line, and the capacitance
value for the 250 miles in columns 39-44, C = 5:4244584 10,6 250 = 1356:1 10,6p:u:
Figure 7.4: Transient Solution with 23
If we were to move the fault location to the end of the transmission line in node \cap", we
would short-circuit the shunt capacitance at node \cap" of the -circuit. If the fault impedance
were zero, this would theoretically produce an innite current spike of innitely small duration.
Without the critical damping adjustment scheme for the suppression of numerical oscillations
(CDA), numerical oscillations would appear in the fault current. See data le 1 PINUM.DAT
for this case with numerical oscillations in the current.
Exercises:
{ Create the data le for this case and run the case.
{ Move the fault location to node \cap", assume zero fault resistance, and observe the fault
current.
{ Put a \1" in column 68 of the data line for t, etc., to bypass CDA, and observe the fault
current again.
{ Re-run it again with tclose changed to 1/4 cycle. Observe the fault current again.
7.5 Transient Solution with Single-Phase Distributed Parameter Line
Data le: 1 LINE.DAT
It is very easy to change a -circuit into a distributed parameter line. We can dene the line
length in any units we want, as long as we are consistent. By using \250 miles" as one unit of
length, the values for the total R, X, C (which we already have for the -circuit) become the
distributed parameters per unit length.
Simply add a \-" to column 1; -1, -2, ...-n indicates that it is an n-phase line with distributed
parameters. Use 1.0 for the length, in columns 45-50.
24
Figure 7.5: Transient Solution with Distributed Parameters
Without CDA, numerical oscillations can occur in the fault current with the -circuit model.
With the distributed parameter model, there are no numerical oscillations, because node "cap"
does not have a lumped shunt capacitance, but sees travelling waves from the transmission line
instead. See data le 1 LINUM.DAT.
Exercises:
{ Create the data le for this case and run the case.
{ Put a \1" in column 68 of the data line for t, etc., to bypass CDA, and observe the fault
current again.
7.6 Comparison between R-L, -Circuit, and Distributed
Parameter Line Representations
Exercises:
{ Superimpose the fault currents from the three line models on the same plot.
{ Comment on the similarities and dierences.
The fault current is reasonably accurate with all three representations.
The distributed parameter line model is the correct model, but it can only be used as long as
the travel time (1.34 ms in this case) is larger than the step size t (50 s in this case).
25
Figure 7.6: Comparison between R-L, -Circuit, and Distributed Parameter Line Representations
7.7 Frequency Scan
Data le: 1 FREQ.DAT
To understand the dierences between the three line models, we can do a \frequency scan",
which is a number of steady-state solutions from fmin to fmax , in steps of f .
We will set up the case in such a way that we obtain the frequency-dependent network
impedances seen from the fault location.
To obtain the impedance seen from the fault location, we have to
{
{
{
{
{
add fmin = 1:0 Hz, f = 0:2 Hz, fmax = 500 Hz to the data line with t, etc.
remove the fault switch,
short-circuit all sources in the network,
inject 1 A (RMS) current into node "b" where the fault occurs,
and ask for the voltage in \b", which will be equal to the impedance because of V = Z I ,
where I = 1:0.
The following plots show the impedance for all three line models.
The frequency-dependent impedance obtained with the distributed-parameter line model, with
recurrent resonance points, is the correct one. When we say \correct", we assume that the
distributed parameters R0, L0 , C 0 are constant. As we will see later, they are frequencydependent, particularly for the zero sequence.
26
The -circuit reproduces the series resonance caused by the series capacitor very well (at 39.5
Hz instead of 39.7 Hz), as well as the rst resonance caused by the line capacitance (at 125.3
Hz instead of 124.3 Hz).
The R-L line model duplicates the series resonance with a slight error (at 41.5 Hz instead of
39.7 Hz), but is unable to duplicate any of the resonances caused by the line capacitance.
Figure 7.7: Magnitude of Fault Impedance to 500 Hz
Figure 7.8: Angle of Fault Impedance to 500 Hz
27
Chapter 8
Case Studies: Short-Circuit in
Three-Phase Network
In addition to the positive sequence parameters, we also need the zero sequence parameters
now:
{ The zero sequence source reactance is the transformer reactance alone, because the transformer is seen as an internal short on the delta side, and disconnected from the generator,
X0 = 0:14 p:u:
{ The zero sequence line impedance is R0 = 0:5 p:u:, X0 = 1:56 p:u:
We assume a single-line-to-ground fault in phase A of node \b". In high-voltage transmission
systems, 90 % or so of all faults are single-line-to-ground.
We will also look at the overvoltages in the unfaulted phases at node \b". The steady-state
overvoltages do not depend on the \fault-initiation angle" (angle where fault occurs, counted
from zero crossing of the pre-fault sinusoidal voltage). The transient overvoltages depend very
much on the fault initiation angle. They are largest in one of the unfaulted phases if the fault
occurs when the voltage is just at its maximum. We therefore use tclose = 0:0041667 s (fault
initiation angle = 90).
With a fault initiation angle of 90, there is little dc oset in the fault current. The circuit
breaker duty is therefore less than with a fault initiation angle of 0, but the latter produces
higher transient overvoltages.
Transient overvoltages during line energization can be minimized with controlled closing (closing when the voltages are more or less equal on both sides of the circuit breaker contact). The
disadvantage is the maximum dc oset in the fault current if we happen to re-close into a
permanent fault.
28
8.1 Steady-State Solution Similar to Short-Circuit Programs
Data le: 3 STEADY.DAT
This steady-state solution shows what you would obtain from a conventional short-circuit
program with phasor solutions.
While short-circuit programs usually work with positive and zero sequence networks, MicroTran
works with phase quantities. See Appendix A for relationship between sequence and phase
quantities.
Replace the single-phase R-L branch with three coupled branches (use the -circuit branch
types 1, 2, 3, or branch types 51, 52, 53 for higher accuracy input, for the three coupled
branches, and leave the eld for the capacitances blank). When you use the pre-processor
\MTD", there is a \device" entry form for symmetric -circuits which allows you to input
zero and positive sequence parameters. If you have a version of MTD which creates cascade
connections of -circuits, make certain that you specify the number of sections as 1 and the
length as 1. The positive and zero sequence parameters Z1, Z0 are then converted internally
to self and mutual impedances with
Zs = 13 (Z0 + 2Z1) ; Zm = 31 (Z0 , Z1 ) :
(8.1)
Use the positive sequence capacitance value to model the three-phase series capacitor station
as three uncoupled R-L-C branches, with values in C-eld (elds for R, L blank).
The zero sequence capacitance is identical to the positive sequence capacitance for the series
capacitors, and equal to the capacitance of each phase. If the data shows a zero sequence
capacitance dierent from the positive sequence capacitance, then the data is in error.
We use capacitance representations now, and not negative reactances, because we already know
that this is correct for steady-state solutions as well as transient simulations.
With short-circuit programs, the fault current for a single-line-to-ground fault in phase A is
found from
IA = 3Z V+AZ,prefault
(8.2)
1
2 + Z0
It is usually assumed that the negative sequence impedance is equal to the positive sequence
impedance, Z2 = Z1 .
Using phase quantities, we get the same answer from
IA = VA,prefault
(8.3)
Zs
since Zs = 13 (Z0 + 2Z1). The phase quantities formula is easier to understand than the symmetrical component formula.
29
Exercises:
{ Find the fault current from the above formula.
{ With Z0 = 0:54 + j 1:329 p:u: and Z1 = 0:06 + j 0:404 p:u: , we get jIAj = 1:341 p:u:.
{ Create the data le for this case and run the case, and compare the result with above
hand calculation.
8.2 Transient Solution with Three-Phase R-L Circuit for
Line
Data le: 3 RL.DAT
Except for adding t, tmax , and setting tclose = 0 for the switch, the data le should be the
same as in 8.1.
Exercises:
{ Create the data le for this case and run the case.
{ Observe the transient overvoltages on the unfaulted phases.
Figure 8.1: Transient Solution with RLC: Fault Current
30
Figure 8.2: Transient Solution with RLC: Voltages on Unfaulted Phases
8.3 Transient Solution with Three-Phase -Circuit for Line
Data le: 3 PI.DAT
Assume that the wave velocity in zero sequence is 70 % of the wave velocity in positive
sequence. From the velocity of a lossless line, c0 = pL`10C `0 , and with c0 = 210000 km=s
56 p:u:=mile, C 0 becomes 3:5481805 10,6 p:u:=mile.
(130487:95 miles=s) and L00 = 2501:120
0
The zero sequence capacitance value for the 250 miles becomes C = 3:5481805 10,6 250 =
887:045 10,6 p:u:
Figure 8.3: Transient Solution with : Fault Current
31
Figure 8.4: Transient Solution with : Voltages on Unfaulted Phases
Exercises:
{ Create the data le for this case and run the case.
{ Observe the transient overvoltages on the unfaulted phases.
{ If we look at the short-circuit current, we see an initial spike. It is caused by discharging
the -circuit capacitance through the series capacitor and the 0:04 p:u: fault resistance.
The time step of 50 s is too large to simulate this spike well enough. Change t to 1 s,
and reduce tmax to 0:005 s, to see the correct time constant T = RC for the decay of the
spike. Note that this spike is unrealistic because the -circuit is not valid at the high
frequencies contained in this spike.
{ To get the time constant for the decay of the spike, nd the self impedance looking into one
phase of the 3-phase shunt capacitance matrix at the end of the -circuit: From Z1 = j!1C1
2
and Z0 = j!1C0 , we get Zs = 13 (Z0 + 2Z1) = j!1 1734:8 p:u:, which implies a capacitance
2
of 576:4 , p:u: This, in series with C = 7149:8 , p:u: for the series capacitor, produces
a Ctotal = 533:4 , p:u:, or T = R Ctotal = 21:3s. This is what you should be able to
see in the plot with t = 1s.
v , if v is the instantaneous voltage at the instant
{ The peak value of the spike should be Rp
fault
when the fault occurs; with v = 1:35 2 and Rfault = 0:04 p:u:, we get ipeak = 47:7 p:u:.
This is close to 45:7 p:u: on the plot. The dierences may be caused by the fact that high
frequencies do not just see the shunt capacitance of the -circuit, but the series impedance
as well, though the latter is much larger.
The nal steady-state fault current is now slightly lower (1:2071 p:u: RMS), compared to that
from the coupled R-L model for the line (1:3413 p:u: RMS), because of the shunt capacitance
of the line. As we will see later in Section 8.4, the correct answer is actually 1:3106 p:u:
32
8.4 Transient Solution with Three-Phase Distributed Parameter Line
Data le: 3 LINE.DAT
We assume that the line is perfectly transposed (balanced).
Data input is in the form of zero sequence parameters on the data line for the rst phase: R0 =
0:50 p:u:, X0 = 1:56 p:u:, C0 = 887:045 , p:u:, length = 1:0, and positive sequence parameters
on the data line for the second phase: R1 = 0:02 p:u:, X1 = 0:50 p:u:, C1 = 1356:1 , p:u:
Figure 8.5: Transient Solution with Line: Fault Current
Figure 8.6: Transient Solution with Line: Voltages on Unfaulted Phases
33
Exercises:
{ Create the data le for this case and run the case.
{ Observe the transient overvoltages on the unfaulted phases. For the very sharp overvoltage
peak, the step size of 50 s may be too large. Rerun the case with 1 s and tmax = 0:01 s.
Figure 8.7: Transient Solution with Line: Voltages on Unfaulted Phases
{ Obtain the nal steady-state fault current from a steady-state solution.
The fault current contains travelling wave transients now. Because of travelling waves, the
current can jump suddenly at the fault location. This has been veried with eld tests; see the
comparison below between measurements and EMTP simulations in the Hydro-Quebec system:
Figure 8.8: Measured and calculated fault current and overvoltages in Hydro-Quebec system [10]
34
Figure 8.9: Measured and calculated initial part of fault current in an expanded time scale [10]
The nal steady-state fault current (1:3106 p:u: RMS) is higher now than the one obtained
from the -circuit model for the line (1:2071 p:u: RMS).
The dierence is caused by the fact that a \nominal" -circuit, which we used here, is reasonably accurate only up to approx. 100 miles at 60 Hz. For a nominal -circuit, the series impedance is simply the series impedance per unit length, multiplied with the length:
Zseries = Z 0 l.
For 250 miles, an exact equivalent -circuit would be better than a nominal -circuit, but
the exact equivalent -circuit can only be used in steady-state phasor solutions. For the exact
equivalent -circuit, the series impedance is
Zseries = Z 0 l sinhl(l) :
(8.4)
The distributed parameter model is converted to an exact equivalent -circuit for pure steadystate phasor solutions (no transient simulation starting from ac steady state) inside MicroTran.
If the exact equivalent -circuit is approximated with a cascade connection of nominal circuits, the steady-state fault current comes closer to the exact solution. With ve nominal
-circuits, we obtain 1:3065 p:u: RMS (0.3 % error), and with ten nominal -circuits we obtain
1:3096 p:u: RMS (0.08 % error). The preprocessor \MTD" has a device input option for creating
cascade connections of nominal -circuits automatically.
The cases with cascade connections of nominal -circuits are probably too big to t into the
student version of MicroTran.
35
8.5 Comparison between R-L, -Circuit, and Distributed
Parameter Line Representations
The plot below shows the fault current for the three line models.
Figure 8.10: Comparison between R-L, -Circuit, and Distributed Parameter Line Representations
For the transient overvoltages, only the distributed parameter line model gives meaningful
answers. Even that model is not good enough if constant R0 , L0, C 0 is assumed. A frequencydependent line model is needed to get realistic answers for single-line-to-ground faults, where
the (frequency dependent) zero sequence parameters play a large role. The frequency dependent
line model will be discussed later.
8.6 Transient Solution with Detailed Generator Model
Data le: 3 GEN.DAT
The data for the detailed generator model, with Xd , Xd0 , Xd00 , etc., is available from the IEEE
SSR benchmark model [11]:
Xd = 1:79 p:u:, Xd0 = 0:169 p:u:, X 00 d = 0:135 p:u:,
Xq = 1:71 p:u:, Xq0 = 0:228 p:u:, X 00q = 0:200 p:u:,
Td00 = 4:30 s, Td000 = 0:032 s
Tq00 = 0:85 s, Tq000 = 0:050 s
Ra = 0, Xl = 0:13 p:u:, f = 60 Hz, if (0) = 1:0 p:u: or 600 A
36
Vrating = 26 kV , Srating = 892:4 MV A
Connection: wye, solidly grounded
Since we are using p.u. values at this point, set Vrating = 1 V (10,3 kV ) and Srating = 1 V A
(10,6 MV A), to keep the generator data in per unit.
The data for the mechanical part is left o because we assume here that the speed does not
change.
With a detailed generator model, the step-up transformer with its delta-wye connection has to
be modelled, too. Use the inverse reactance matrix representation, and assume that the threephase bank is made up of single-phase transformers. The connection is achieved by simply
using the correct node names on the coupled branches. For example, the rst single-phase
transformer has the rst branch connected from \a-A" to \ground" and the second branch
from \gen-A" to \gen-B". This automatically creates the correct phase shift between the wye
and delta side as well.
Since the voltages on the delta side lag 30 degrees behind the voltages on the wye side, the
generator voltage angle must be changed from ,90 to ,120 for phase \a".
Exercises:
{ Create the data le for this case from 3 LINE.DAT. For the transformer
p bank, you may
want to use \MTD" to create the transformer model. Use Vrating = 3 on the delta side,
Vrating = 1:0 on the wye side, and Srating = 1:0. Why?
{ Run the case, and observe the dierences in the fault current with respect to the results
from Section 8.4 (they are not too large for the single-line-to-ground fault, but become
larger for the three-phase fault in Section 8.8).
{ The charging current in the pre-fault steady-state solution may exceed the Var-absorption
capability of the generator. Older MicroTran versions may stop with a negative eld
current message. Message in newer versions is more detailed. Cure: add shunt reactors,
or connect to innite bus as in original IEEE case.
37
Figure 8.11: Transient Solution with Detailed Generator Model
8.7 Transient Solution with Actual Units
Data le: 3 GENOHM.DAT
Studies with MicroTran are usually made with parameters in actual units, rather than in per
unit.
Parameters in actual units are usually less confusing than p.u. values, particularly if the case
contains nonlinear inductances or nonlinear resistances.
Exercises:
{ Convert the data le 3 GEN.DAT to the data le 3 GENOHM.DAT in actual units
( , F , V , A, etc.). For the transformer, you may want to createpthe model again with
\MTD", by using Vrating = 26 kV on the delta side, Vrating = 500= 3 kV on the wye side,
and Srating,1phase = 892:4=3.
{ Verify that the results are identical with those from Section 8.6. You may want to use the
scaling factor in the plotting program for this comparison.
38
8.8 Transient Solution for Simultaneous Three-Phase
Short-Circuit
Data les: 3 LINE3.DAT and 3 GEN3.DAT
The dierences between the model with E 00 behind Xd00 and the detailed generator model are
more noticeable for the fault currents of a three-phase fault beyond the rst few cycles.
For a three-phase fault, the rotor uxes decay more noticeably. As a consequence, the ac
component of the fault current changes from its initial subtransient value determined by Xd00 ,
to its transient value determined by Xd0 , and nally to its steady-state value determined by Xd .
Exercises:
{ Convert the data les 3 LINE.DAT and 3 GEN.DAT to the data les 3-LINE3.DAT
and 3 GEN3.DAT for the three-phase short-circuit.
{ Run the cases and compare the fault current.
Figure 8.12: Transient Solution with E 00 behind Xd00 Model
39
Figure 8.13: Transient Solution with Detailed Generator Model
40
8.9 Short-Circuits on HVDC Lines
Short-circuit currents on HVDC lines, and related transient overvoltages, can only be calculated
with EMTP-type programs.
Figure 8.14: Measured and calculated transient overvoltage on unfaulted pole of HVDC line [12]
41
Appendix A
Relationship between Phase Quantities
and Sequence Quantities
Assume that a three-phase transmission line is represented with series R-L branches. For steady-state
solutions, the voltage drop along the three phases would then be
2
6
4
3
2
3 2
Va
Zs Zm Zm
Ia
Vb 75 = 64 Zm Zs Zm 75 64 Ib
Vc
Zm Zm Zs
Ic
3
7
5
(A.1)
where Zs is the self impedance of each phase (with ground and ground wires providing the return
path), and Zm is the mutual impedance between phases. In Equation (A.1), we assume that the
transmission line is \balanced" (perfectly transposed). Otherwise, the matrix would be
2
6
4
Zaa Zab Zac
Zba Zbb Zbc
Zca Zcb Zcc
3
7
5
(A.2)
(still symmetric, Zba = Zab, etc.).
Equations A.1 with phase quantities are coupled. They become simpler, namely decoupled, if we
work with \symmetrical components", by transforming phase quantities Ia , Ib , Ic to \sequence"
quantities Izero, Ipositive , Inegative :
2
6
4
3
2
3 2
Vzero
Zzero
0
0
Izero
7
6
7
6
Vpositive 5 = 4 0 Zpositive
0 5 4 Ipositive
Vnegative
0
0
Znegative
Inegative
3
7
5
(A.3)
Since the matrix is diagonal, we can solve the equations separately for zero, positive, and negative
sequence quantities, as if we had three single-phase networks rather than one three-phase network.
This is a transformation based on eigenvalues and eigenvectors. The transformations are
42
and
2
3
Izero
6 I
7
4 positive 5 =
Inegative
2
3
Ia
6 I 7=
4 b 5
Ic
2
3 2
3
1 1 1
Ia
p1 64 1 a a2 75 64 Ib 75 ;
3 1 a2 a
Ic
2
1
p1 64 1
3 1
3 2
1 1
Izero
a2 a 75 64 Ipositive
a a2
Inegative
3
7
5
(A.4)
(A.5)
where a = ej120 , a2 = e,j120 . The transformations for voltages are identical. Above transformations
are \normalized". If unnormalized (usual practice in power industry), the factor in Equation (A.4)
is 13 , and in (A.5) it is 1.
The symmetrical components have a physical meaning:
1. Zero sequence currents are identical in the three phases (as can be seen from Equation (A.5) if
only Izero is present).
2. Positive sequence currents are the symmetrical currents in the three phases during normal
operation (as can be seen from Equation (A.5) if only Ipositive is present). When the phasors
are rotated counterclockwise, the phases appear in the normal ("positive") sequence a, b, c.
3. Negative sequence currents are also symmetrical currents in the three phases (as can be seen
from Equation (A.5) if only Inegative is present), but they appear in reverse (\negative") sequence
when phasors are rotated counterclockwise.
43
Approach in MicroTran:
Symmetrical component equations make no sense for untransposed lines, because the trans-
formed matrix in sequence quantities is no longer diagonal.
All network components must be three phase with symmetrical component equations. If a
single-phase distribution line (one phase conductor and one neutral conductor) were connected
to a three-phase substation, that would be dicult to model with symmetrical components.
For reasons of generality, MicroTran therefore works with phase quantities, even when input
data is given in sequence quantities.
The zero sequence impedance can be obtained from Equation (A.1). With all voltages equal,
we need only the rst equation for Va , and with Ib = Ia , Ic = Ia, we get
Va = (Zs + 2Zm) Ia
(A.6)
The zero sequence impedance (subscript \zero" or \0") is therefore:
Z0 = Zs + 2Zm
(A.7)
For the positive sequence impedance, we use in Equation (A.1) Ib = a2Ia , Ic = aIa , and we
get
Va = (Zs , Zm) Ia
(A.8)
The positive sequence impedance (subscript \positive" or \1") is therefore:
Z1 = Zs , Zm
(A.9)
Similarly, the negative sequence impedance (subscript \negative" or \2") becomes
Z2 = Zs , Zm
44
(A.10)
Bibliography
[1] M. B. Hughes, R. W. Leonard, and T. G. Martinich, \Measurement of power system subsynchronous driving point impedance and comparison with computer simulations," IEEE
Transactions on Power Apparatus and Systems, vol. PAS-103, pp. 619 - 630, March 1984.
[2] C. A. F. Cunha and H. W. Dommel, \Computer simulation of eld tests on the 345 kV JaguaraTaquaril line," Paper BH/GSP/12 (in Portuguese), in II Seminario Nacional de Produc~ao e
Transmissao de Energia Eletrica, Belo Horizonte, Brazil, 1973.
[3] J. R. Mart, \Accurate modelling of frequency-dependent transmission lines in electromagnetic
transient simulations," IEEE Transactions on Power Apparatus and Systems, vol. PAS-101,
pp. 147 - 157, Jan. 1982.
[4] M. Rioual, \Measurements and computer simulation of fast transients through indoor and
outdoor substations," IEEE Transactions on Power Delivery, vol. 5, pp. 117 - 123, Jan. 1990.
[5] H.W. Dommel, Case Studies for Electromagnetic Transients, 2nd Edition, Microtran Power
System Analysis Corporation, Vancouver, British Columbia, Canada, 1993.
[6] D. H. Baker, \Synchronous machine modeling in EMTP," IEEE Course Text: Digital Simulation of Electrical Transient Phenomena, No. 81 EHO173-5-PWR, IEEE Service Center,
Piscataway, N.J., 1980.
[7] Transformer Inrush Current, Test Case 1., Microtran Power System Analysis Corporation,
Vancouver, British Columbia, Canada, 1990.
[8] A. C. S. de Lima, personal communication, July 1998.
[9] W. Xu, J. R. Mart, and H. W. Dommel, \A multiphase harmonic load ow solution technique,"
IEEE Transactions on Power Systems, vol. 6, pp. 174 - 190, Febr. 1991.
[10] R. Malewski, V. N. Narancic, and Y. Robichaud, \Behavior of the Hydro-Quebec 735-kV
system under transient short-circuit conditions and its digital computer simulation," IEEE
Transactions on Power Apparatus and Systems, vol. PAS-94, pp. 425 - 431, March/April 1975.
[11] IEEE Task Force, \First benchmark model for computer simulation of subsynchronous resonance," IEEE Transactions on Power Apparatus and Systems, vol. PAS-96, pp. 1565 - 1572,
Sept./Oct. 1977.
45
[12] D. J. Melvold, P. C. Odam, and J. J. Vithayathil, \Transient overvoltages on an HVDC bipolar
line during monopolar line faults," IEEE Transactions on Power Apparatus and Systems, PAS96, pp. 591 - 601, March/April 1977.
46