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|
BEGIN NEW DATA CASE
C BENCHMARK DCNEW-22
C Illustrate modeling of unsymmetric series-RL that uses compensation as
C described in the July, 1997, newsletter story. Begin with uncoupled
C branches of 1 ohm resistance. The answer is obvious by inspection. The
C [R] matrix is diagonal matrix equal to the unit matrix, and [X] is zero.
C In terms of symmetrical components, Zo = Z1 = Z2 (a second half of the
C data this identical but alternative representation to produce the same
C known answer). With each phase having 2 ohms resistance split in half,
C and with balanced, sinusoidal sources having 2 volts, the currents are
C equal to half the source voltage: sinusoidal of amplitude 1. Frequency
C of the 3-phase sources at SENDA, SENDB, and SENDC have been reduced to
C nearly zero so the 1st step of phase "a" is very close to the peak. Step
C 1 is very close to answer at time zero: Va = 1 volt, Vb = Vc = 1/2 volt
C A total of 7 subcases are involved.
C UTPF update of 20 June 2007 allows ATP data to override .PL4 file type
C choices that normally are made within STARTUP. Illustrate use here :
C FMTPL4 L4BYTE NEWPL4 Next card FORMAT ( 16X, A6, 2X, 2I8 )
C CONCATENATE. 1 0 <Force .PL4 type> key text anywhere
C For 2 or more stacked subcases, this declaration must be in the first
C and must be in the original data file. Here, the C-like choice is
C mandatory in order that it be in place for subcase 4, which will use
C $DEPOSIT to change NEWPL4 again, to 2 (for Pisa). That only works
C if a C-like or Pisa file then is in effect. One can not use $DEPOSIT
C to change a formatted .PL4 file to Pisa format in this way. So, the
C preceding ensures that the 4th will work; it ensures that some user
C will not rely upon values in STARTUP that might be incompatible.
.0001 .0005
1 -1 1 0 1 -1
5 5 20 20 100 100
SENDA ENDA 1.0 1
SENDB ENDB SENDA ENDA 1
SENDC ENDC SENDA ENDA 1
ENDA ENDB 1.E7
ENDB ENDC ENDA ENDB
ENDC ENDA ENDA ENDB
91ENDA MODEL Z0Z1Z2 1.0
91ENDB MODEL Z0Z1Z2 1.0
91ENDC MODEL Z0Z1Z2 1.0
SENDA RECA 1.0 1
SENDB RECB SENDA RECA 1
SENDC RECC SENDA RECA 1
RECA RECB 1.E7
RECB RECC RECA RECB
RECC RECA RECA RECB
91RECA MODEL [R][L]
91RECB MODEL [R][L]
91RECC MODEL [R][L]
1.0 0.0 0.0
0.0 0.0 0.0
0.0 1.0 0.0
0.0 0.0 0.0
0.0 0.0 1.0
0.0 0.0 0.0
BLANK card follows the last branch card
BLANK line terminates the last (here, nonexistent) switch
14SENDA 2.0 0.1 0.0 { 1st of 3 sources. Note balanced,
14SENDB 2.0 0.1 -120. { three-phase, sinusoidal excitation
14SENDC 2.0 0.1 120. { with no phasor solution.
BLANK card follows the last source card
SENDA RECA ENDA SENDB RECB ENDB SENDC RECC ENDC
C Step Time SENDA RECA ENDA SENDB RECB ENDB SENDC RECC ENDC SENDA
C ENDA
C SENDB SENDC SENDA SENDB SENDC
C ENDB ENDC RECA RECB RECC
C 0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
C 0.0 0.0 0.0 0.0 0.0
C 1 .1E-3 2.0 .999999848 .999999848 -.99989117 -.49994551 -.49994551 -1.0001088 -.50005434 -.50005434 1.00000015
C -.49994566 -.50005449 1.00000015 -.49994566 -.50005449
C 2 .2E-3 1.99999998 .999999842 .999999842 -.99978234 -.49989109 -.49989109 -1.0002176 -.50010875 -.50010875 1.00000014
C -.49989124 -.5001089 1.00000014 -.49989124 -.5001089
BLANK card ending node voltage outputs
BLANK termination to plot cards
BEGIN NEW DATA CASE
C 2nd of 7 subcases. Progress to symmetrical components having
C Z1 = Z2 so the answer can be shown to be identical to that
C using Type-51,52,53 modeling with sequence impedances. In
C fact, there are 2 identical, uncoupled networks driven from
C the same balanced 3-phase sources at SENDA, SENDB, and SENDC.
.0001 .050
1 1 1 0 1 -1
5 5 20 20 100 100
SENDA ENDA 0.3 1.0 1
SENDB ENDB SENDA ENDA 1
SENDC ENDC SENDA ENDA 1
ENDA ENDB 1.E7 { Balanced, interphase leakage gives 3-phase
ENDB ENDC ENDA ENDB { (rather than 3, single-phase) Z-thevenin.
ENDC ENDA ENDA ENDB { 3 coupled phases are required by "Z0Z1Z2"
91ENDA MODEL Z0Z1Z2 0.3 1.0 { Sequence Ro, Lo in [ohms, mHenry]
91ENDB MODEL Z0Z1Z2 0.1 0.5 { Sequence R1, L1 in [ohms, mHenry]
91ENDC MODEL Z0Z1Z2 0.1 0.5 { Note Z2 = Z1 so [Z] is symmetric
C Next, build a copy of this, but using the old (Type-51,52,53) modeling:
SENDA RECA 0.3 1.0 1
SENDB RECB SENDA RECA 1
SENDC RECC SENDA RECA 1
RECA RECB 1.E7
RECB RECC RECA RECB
RECC RECA RECA RECB
51RECA 0.3 1.0 { Ro, Lo in [ohms, mHenry]
52RECB 0.1 0.5 { R1, L1 in [ohms, mHenry]
53RECC
BLANK card follows the last branch card
BLANK line terminates the last (here, nonexistent) switch
14SENDA 2.0 50. 0.0 { 1st of 3 sources. Note balanced,
14SENDB 2.0 50. -120. { three-phase, sinusoidal excitation
14SENDC 2.0 50. 120. { with no phasor solution.
BLANK card follows the last source card
SENDA RECA ENDA
C Step Time SENDA RECA ENDA SENDA SENDB SENDC SENDA SENDB SENDC
C ENDA ENDB ENDC RECA RECB RECC
C 0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
C 1 .1E-3 1.99901312 .664144463 .664144463 .065757077 -.0310889 -.03466818 .065757077 -.0310889 -.03466818
C 2 .2E-3 1.99605346 .658837702 .658837702 .195443292 -.09061195 -.10483134 .195443292 -.09061195 -.10483134
C 3 .3E-3 1.99112393 .652991396 .652991396 .321457199 -.14486663 -.17659057 .321457199 -.14486663 -.17659057
BLANK card ending node voltage outputs
C 500 .05 -2. -.59687593 -.59687593 -2.0936819 3.18312082 -1.0894389 -2.0936819 3.18312082 -1.0894389
C Variable maxima : 2.0 .664144463 .664144463 3.23520786 3.50393608 3.23541237 3.23520786 3.50393608 3.23541237
C Times of maxima : .02 .1E-3 .1E-3 .0428 .0092 .0361 .0428 .0092 .0361
C Variable minima : -2. -.60247492 -.60247492 -3.3079993 -3.2352894 -3.4484065 -3.3079993 -3.2352894 -3.4484065
C Times of minima : .01 .0496 .0496 .0127 .0394 .0059 .0127 .0394 .0059
CALCOMP PLOT
144 5. 0.0 50. -1.0 1. RECA ENDA { These 2 phase-a voltages should be equal
194 5. 0.0 50. -4.0 4.0BRANCH { Following phase-a currents should be equal
SENDA RECA SENDA ENDA
BLANK termination to plot cards
BEGIN NEW DATA CASE
C 3rd of 7 subcases generalizes the preceding. For the Z0Z1Z2 model,
C Z2 is changed so it no longer is equal to Z1. Since Type-51,52,53
C modeling no longer can provide a comparison, this half is replaced by
C the more general [R][L] alternative that requires the user to supply
C full matrices. Although input precision is limited to 8 digits to the
C right of the fixed decimal point, agreement is close (6 or more digits)
.0001 .050
1 1 1 0 1 -1
5 5 20 20 100 100
SENDA ENDA 0.3 1.0 1
SENDB ENDB SENDA ENDA 1
SENDC ENDC SENDA ENDA 1
ENDA ENDB 1.E7
ENDB ENDC ENDA ENDB
ENDC ENDA ENDA ENDB
91ENDA MODEL Z0Z1Z2 0.3 1.0
91ENDB MODEL Z0Z1Z2 0.1 0.5
91ENDC MODEL Z0Z1Z2 .101 0.8
SENDA RECA 0.3 1.0 1
SENDB RECB SENDA RECA 1
SENDC RECC SENDA RECA 1
RECA RECB 1.E7
RECB RECC RECA RECB
RECC RECA RECA RECB
91RECA MODEL [R][L]
91RECB MODEL [R][L]
91RECC MODEL [R][L]
C Correct the data 1 April 2002. Following a change to USERNL, the
C phase-domain X or L has changed. Previously, we were using units
C of mH, which relied on STATFR of STARTUP for the conversion.
C This is improved by the definition of XOPT = 60 Hz (see below),
C which frees the computation from dependence of STARTUP. Also, we
C switch to input of X in ohms (rather than L in mH):
$DISABLE { For the hysterical record, retain the old matrix on comments:
.1670000000 .0665866025 .0664133975 { R(1,1), R(1,2), R(1,3)
.7666666667-.1720084679 .4053418013 { L(1,1), L(1,2), L(1,3)
.0664133975 .1670000000 .0665866025 { Row 2 of [R]
.4053418013 .7666666667-.1720084679 { Row 2 of [L]
.0665866025 .0664133975 .1670000000 { Row 3 of [R]
-.1720084679 .4053418013 .7666666667 { Row 3 of [L]
$ENABLE { Done showing old (and wrong) matrix [L] in mHenry. Begin new:
C Preceding disabled data was from years past. Correct this 1 April 2002.
$UNITS, 60.0, 60.0, { Define frequency XOPT = 60 for the impedance computation
C The .DBG will includes the following information. We use full precision of
C the matrix values, although blanks and leading zeros have been removed:
C USERNL begins with Lo, L1, L2 [H] = 1.00000E-03 5.00000E-04 8.00000E-04
C Converted to Xo, X1, X2 = 3.76991E-01 1.88496E-01 3.01593E-01
C 3x3 phase-domain impedance matrix in ohms follow. For each row I, X(I,J) is below R(I,J). w = 3.769911E+02 rad/sec.
.1670000000 .0991483886 .0338516114 { R(1,1), R(1,2), R(1,3) in ohms
.2890265241 .0436936220 .0442709723 { X(1,1), X(1,2), X(1,3) in ohms
.0338516114 .1670000000 .0991483886 { Row 2 of [R]
.0442709723 .2890265241 .0436936220 { Row 2 of [X]
.0991483886 .0338516114 .1670000000 { Row 3 of [R]
.0436936220 .0442709723 .2890265241 { Row 3 of [X]
C $UNITS, -1.0, -1.0, { Done with ohms and micromhos at 60 Hz, so restore original
C The preceding line is not tolerated because the associated warning
C message is issued during the first time step. If we cancel the XOPT
C definition, there will be no trace during the first time step. So,
C without changing the answer, we leave XOPT = 60 to suppress the
C warning message that will be seen in output of preceding 2 subcases.
BLANK card follows the last branch card
BLANK line terminates the last (here, nonexistent) switch
14SENDA 2.0 50. 0.0 { 1st of 3 sources. Note balanced,
14SENDB 2.0 50. -120. { three-phase, sinusoidal excitation
14SENDC 2.0 50. 120. { with no phasor solution.
BLANK card follows the last source card
SENDA RECA ENDA
C First 3 output variables are electric-network voltage differences (upper voltage minus lower voltage);
C Next 6 output variables are branch currents (flowing from the upper node to the lower node);
C Step Time SENDA RECA ENDA SENDA SENDB SENDC SENDA SENDB SENDC
C ENDA ENDB ENDC RECA RECB RECC
C 0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
C 1 .1E-3 1.99901312 .784095685 .784095685 .05984815 -.02824862 -.03159953 .05984815 -.02824862 -.03159953
C 2 .2E-3 1.99605346 .776335531 .776335531 .178012015 -.08221824 -.09579377 .178012015 -.08221824 -.09579377
C 3 .3E-3 1.99112393 .768107458 .768107458 .29308232 -.13116828 -.16191404 .29308232 -.13116828 -.16191404
BLANK card ending node voltage outputs
C 500 .05 -2. -.57981099 -.57981099 -2.1430566 3.2140066 -1.07095 -2.1430566 3.2140066 -1.07095
C Variable maxima : 2.0 .784095685 .784095685 3.2728624 3.57534262 3.28836555 3.2728624 3.57534261 3.28836555
C Times of maxima : .02 .1E-3 .1E-3 .0427 .0091 .0161 .0427 .0091 .0161
C Variable minima : -2. -.58387034 -.58387034 -3.4132778 -3.2730642 -3.3604773 -3.4132778 -3.2730642 -3.3604773
C Times of minima : .01 .0496 .0496 .0126 .0394 .0059 .0126 .0394 .0059
CALCOMP PLOT
144 5. 0.0 50. -1.0 1. RECA ENDA
194 5. 0.0 50. -4.0 4.0BRANCH
SENDA RECA SENDA ENDA
BLANK termination to plot cards
BEGIN NEW DATA CASE
C 4th of 7 subcases is unrelated to the preceding three. Instead, it
C illustrates op amp (operational amplifier) modeling as first requested
C by Masahiro Kan of Toshiba Corp. in Japan. See October, 1997, newsletter
C As of 30 Aug 97, the op amp (Type-20 source) is ignored during the
C phasor solution.
C 19 March 2001, expand to illustrate Pisa-format .PL4 file for time
C simulation. Use began with HARMONIC FREQUENCY SCAN in the new 15th
C subcase of DCNEW-21. But correct operation for normal time simulation
C also should be confirmed, so modify this existing 4th subcase for such
C a test. Expand dT and T-max so screen plot is half of a recognizable
C sinusoid:
$DEPOSIT, NEWPL4=2 { Use SPY DEPOSIT to override .PL4 file type given in STARTUP
$DEPOSIT, LUNIT4=4 { Use SPY DEPOSIT to override minus sign specified in STARTUP
$DEPOSIT, NOPOST=1 { Use SPY DEPOSIT to override PostScript choice in STARTUP
$DEPOSIT, NOHPGL=1 { Use SPY DEPOSIT to override the HP-GL choice in STARTUP
$DEPOSIT, NOGNU=1 { Use SPY DEPOSIT to override the GNUPLOT choice in STARTUP
C To prove that Pisa-format code is being used, it is easy to turn on debug
C printout. Use here is like that pioneered in DCNEW-21 for HFS. But there
C are differences. Whereas HFS automatically used LUNIT4 = +4 regardless
C of what STARTUP indicated (usually value -4), this is not true for time
C simulation, so the preceding definition of LUNIT4 is necessary in order
C to force ATP to read from the disk-stored .PL4 file during plotting. About
C overlay number, the .PL4 header is created in overlay 15 rather than 11. In
C the .DBG file, look for the names HEADPI and LU4BEG to see Pisa-related
C data values. To minimize size of the .DBG file, PostScript, HP-GL, and
C GNUPLOT output are turned off (the final 3 $DEPOSIT lines above).
C Turn off diagnostic 22 April 2007 as it disfigures .LIS of Mingw32 ATP:
C DIAGNOSTIC { HEADPI in 15 } 9 { LU4BEG in 28 } 9
PRINTED NUMBER WIDTH, 13, 2, { Request maximum precision for 8 output columns
C .000200 .0005 { Original simulation only advanced 3 time steps
.001 .010 { 10 time steps. Large dT is legal since network is resistive
1 1 1 1
GEN SEND 1.0
SEND 1.0 0.0
OPAMP 1.0 1
BLANK card ending branch cards.
BLANK card ending switch cards.
14GEN 2.0 50. 0.0 -1.
C <____> <____><____> Gain occupies columns 11-20
C BUS1 Gain BUSK BUSM By definition, V-1 = Gain * ( V-k - V-m)
20OPAMP 10.GEN SEND { If V-k or V-m is zero, leave name blank)
BLANK card terminating all EMTP source cards.
GEN SEND OPAMP
BLANK card terminating all output requests.
CALCOMP PLOT { To demonstrate use of Pisa-format .PL4 file, try screen plot
144 1. 0.0 10. SEND OPAMP { 2 sinusoidal signals differ by factor 10
BLANK card terminating all plot cards.
BEGIN NEW DATA CASE
C 5th of 7 subcases really should be an extension of the following DCNEW-23
C since these further support the 12th subcase of DCN23.DAT But the
C following disk file already has 15 subcases as the addition is made on
C 5 April 2000. Symmetrical component data of Type-51, 52, 53 branches is
C the subject. Compensation is not being used. Consider a single, lumped,
C 3-phase, series R-L branch having this imbalanced data:
C Zo = 3.5 * Z1 = 10.5 + j 14.0 Ohm
C Z1 = 3.0 + j 4.0 Ohm
C Z2 = 0.5 * Z1 = 1.5 + j 2.0 Ohm
C The far end will be terminated this way: phase "a" will be open whereas
C phases "b" and "c" will be connected together. I.e., this is a normal
C line-to-line fault. Data comes from Orlando Hevia in Santa Fe, Argentina
C as copied by WSM on 3 April 2002. Mr. Hevia drove the line section using
C a balanced, 3-phase voltage source. Although he included a phasor soution
C for initial conditions, WSM drops this because it might be confusing (ATP
C does not yet correctly represent unsymmetric [R] or [L] during the phasor
C solution). Starting from zero is no problem, however, as the solution
C settles into the steady state smoothly within a cycle or two. Just to be
C sure, 5 cycles are simulated, and extrema are limited to the final 1.25 of
C these. This is necessary to ignore the transient of energization. There
C will be 3 identical tests. Each has different data, but gives identically
C the same answers to the 8 or 9 digits of dT-loop output. Orlando Hevia
C wrote a FORTRAN program to produce the solution exactly, in closed form.
C This produced the following results: POLAR Magnitude Degrees
C IB 230.9401 -143.1301
C IC 230.9401 36.8699
C ATP now will confirm this magnitude using simulation.
DIAGNOSTIC { Cancel diagnostic printout of preceding subcase
PRINTED NUMBER WIDTH, 11, 2, { Request maximum precision (for 8 output columns)
BEGIN PEAK VALUE SEARCH, 0.075, { Ignore 1st 3.75 cycles, until in steady state
.000100 .100 { dT is relatively large to speed simulation. Only 1K steps
1 -1 1 0 1 -1
5 5 20 20 100 100
$UNITS, 50.0, 0.0 { Begin inductance data in ohms at XOPT = 50 Hz
C 1st of 3 line sections follows. SEQ at sending end indicates sequence data:
51GENA OPEN1 MODEL Z0Z1Z2 10.5 14.00 { Ro and Xo, both in ohms
52SEQB FLT1 3.0 4.00 { R1 and X1, both in ohms
53SEQC FLT1 1.5 2.00 { R2 and X2, both in ohms
$UNITS, 0.0, 0.0 { Return to inductance data in mHenry; XOPT = 0
C 3x3 phase-domain [R] in ohms and [L] in mHenries follow. For each row I, L(I,J) is below R(I,J). w = 3.14159265E+02 rad/sec.
C 5.0000000000 2.1726497308 3.3273502692
C 21.2206590789 13.0496847320 10.2930402549
C 3.3273502692 5.0000000000 2.1726497308
C 10.2930402549 21.2206590789 13.0496847320
C 2.1726497308 3.3273502692 5.0000000000
C 13.0496847320 10.2930402549 21.2206590789
C The present comments are diagnostic output from use of MODEL Z0Z1Z2. Such
C output always will be found in the .DBG file. From it, produce the following
C equivalent representation of [R] in ohms and [L] in mHenries:
C 2nd of 3 line sections follows. PHS at sending end indicates phase domain:
51GENA OPEN2 MODEL [R][L]
52PHSB FLT2
53PHSC FLT2
5.000000000 2.172649731 3.327350269 { R(1,1), R(1,2), and R(1,3) in ohms
21.22065908 13.04968473 10.29304025 { L(1,1), L(1,2), and L(1,3) in mHenries
3.327350269 5.000000000 2.172649731 { Row 2 of [R]
10.29304025 21.22065908 13.04968473 { Row 2 of [L]
2.172649731 3.327350269 5.000000000 { Row 3 of [R]
13.04968473 10.29304025 21.22065908 { Row 3 of [L]
C Finally, data for 3rd of 3 line sections comes from Orlando. This is _not_
C the same circuit, note, since [R] & [L] are not the same. The diagonal
C elements are the same, but off-diagonals are not. This 3rd circuit differs
C from the preceding 2, but it gives identically the same fault current. It
C was constructed by Mr. Hevia by transferring impedance between Z1 and Z2
C while maintaining the sum fixed. For either a L-L fault of this subcase or
C a 1-L-G fault of the following one, fault current depends on the sum of
C Z1 and Z2 but not on either Z1 or Z2 independently. Consider these
C parameters, which seem exact to the limits of single precision:
C Z of SEQ branch Z of OPH branch
C Z0 parameter 10.5 + j 14.00 10.500 + j 14.0
C Z1 parameter 3.0 + j 4.00 2.325 + j 13.0
C Z2 parameter 1.5 + j 2.00 2.175 - j 7.0
C But voltage of the faulted node will be different. Note the huge difference
C between X1 and X2. Yet, they sum equally: 4 + 2 = 13 - 7.
51GENA OPEN3 MODEL [R][L]
52OPHB FLT3
53OPHC FLT3
5.0000000 -3.0235027 8.5235027
21.2206591 11.8091947 11.5335303
8.5235027 5.0000000 -3.0235027
11.5335303 21.2206591 11.8091947
-3.0235027 8.5235027 5.0000000
11.8091947 11.5335303 21.2206591
BLANK card ending branches
C Following switches measure line currents in the 3 circuits. Since phase "a"
C is open, this is not of enough interest to warrant output. For the record,
C let's document how close to zero such currents are, however. Salford EMTP
C shows (prior to removal of the 3 switches from phase "a":
C Step Time GENA GENA GENA
C SEQA PHSA OPHA
C 0 0.0 0.0 0.0 0.0
C 1 .1E-3 -.367E-15 .3632E-15 .1798E-15
C 2 .2E-3 -.284E-16 .1234E-15 -.271E-15
C Variable max: .1121E-14 .1246E-14 .1845E-14
C Times of max: .098 .0894 .0764
C Variable min: -.122E-14 -.107E-14 -.229E-14
C Times of min: .09 .0789 .0879
C To conclude, no switches for phase "a". There remain 3 switches for phase
C "b" (these 3 are ordered first, followed by the 3 for phase "c". This
C way, corresponding outputs of the dT loop are contiguous, which encourages
C easy comparison.
GENB SEQB MEASURING 1
GENB PHSB MEASURING 1
GENB OPHB MEASURING 1
GENC SEQC MEASURING 1
GENC PHSC MEASURING 1
GENC OPHC MEASURING 1
BLANK card ending switches
14GENA 1000. 50. -0.00 { Balanced, 3-phase excitation that
14GENB 1000. 50. -120. { is _not_ present during the steady
14GENC 1000. 50. -240. { state. There is no phasor solution.
BLANK card follows the last source card
C Column headings for the 9 EMTP output variables follow. These are divided among the 5 possible classes as follows ....
C First 3 output variables are electric-network voltage differences (upper voltage minus lower voltage);
C Next 6 output variables are branch currents (flowing from the upper node to the lower node);
C Step Time FLT1 FLT2 FLT3 GENB GENB GENB GENC GENC GENC
C SEQB PHSB OPHB SEQC PHSC OPHC
C 0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
C 1 .1E-3 -495.9539 -495.9539 -500.178 .14077359 .14077359 .14077359 -.1407736 -.1407736 -.1407736
C 2 .2E-3 -491.6694 -491.6694 -501.478 .55967714 .55967714 .55967714 -.5596771 -.5596771 -.5596771
C 3 .3E-3 -487.1449 -487.1449 -503.861 1.2498171 1.2498171 1.249817 -1.249817 -1.249817 -1.249817
FLT1 FLT2 FLT3 { Show that the fault voltage of OPH (FLT3; 3rd of 3) differs
BLANK card ending node voltage outputs
C 500 .05 333.3359 333.3359 -572.6502 184.74924 184.74924 184.74924 -184.7492 -184.7492 -184.7492
C 600 .06 -333.3373 -333.3373 572.64118 -184.7477 -184.7477 -184.7477 184.7477 184.7477 184.7477
C 700 .07 333.33716 333.33716 -572.642 184.74784 184.74784 184.74784 -184.7478 -184.7478 -184.7478
C 800 .08 -333.3372 -333.3372 572.64195 -184.7478 -184.7478 -184.7478 184.74783 184.74783 184.74783
C 900 .09 333.33717 333.33717 -572.642 184.74783 184.74783 184.74783 -184.7478 -184.7478 -184.7478
C 1000 0.1 -333.3372 -333.3372 572.64196 -184.7478 -184.7478 -184.7478 184.74783 184.74783 184.74783
C Variable max: 333.33717 333.33717 977.26344 230.90147 230.90147 230.90147 230.90147 230.90147 230.90147
C Times of max: .09 .09 .097 .088 .088 .088 .098 .098 .098
C Variable min: -333.3372 -333.3372 -977.2634 -230.9015 -230.9015 -230.9015 -230.9015 -230.9015 -230.9015
C Times of min: .08 .08 .087 .098 .098 .098 .088 .088 .088
C Note about preceding currents. The maximum of 230.90147 and minimum of
C -230.9015 are nearly negatives of each other. This is a sign that the
C steady state is really steady (equality in the 7th decimal digit). But
C Orlando Hevia's exact solution of 230.9401 implies some error? Why? The
C substantial dT (discretization error). As dT is decreased, the ATP
C answer approaches Orlando Hevia's. Using dT = 2 microseconds, one can
C produce exact agreement to all 7 printed digits.
BLANK termination to plot cards
BEGIN NEW DATA CASE
C 6th of 7 subcases is a modification of the preceding 5th. Repeat
C the preceding subcase, except for a different fault at the far end.
C Termination will be as follows: phase "a" will be grounded whereas
C phases "b" and "c" are left open. I.e., this is a normal single-line-
C to-ground fault. Data comes from Orlando Hevia in Santa Fe, Argentina
C Another change is this: since only a single phase carries current,
C there is no shortage of space in the output vector. So, add a 4th
C alternative: MODEL Z0Z1Z2 with inductance in mHenries. About the
C correct answer, Orlando Hevia wrote a FORTRAN program to produce
C the solution exactly, in closed form. This produced the following
C results: POLAR Magnitude Degrees
C IA 120.0000 -53.1301
C Using ATP simulation, we now will confirm the 120-amp peak current.
PRINTED NUMBER WIDTH, 12, 2, { Request maximum precision (for 8 output columns)
POWER FREQUENCY 50.0 { Remove dependence on STARTUP value
BEGIN PEAK VALUE SEARCH, 0.075, { Ignore 1st 3.75 cycles, until in steady state
.000100 .100 { dT is relatively large to speed simulation. Only 1K steps
1 -1 1 0 1 -1
5 5 20 20 100 100
$UNITS, 50.0, 0.0 { Begin inductance data in ohms at XOPT = 50 Hz
C 1st of 4 line sections follows. SEQ at sending end indicates sequence data:
51SEQA MODEL Z0Z1Z2 10.5 14.00 { Ro and Xo, both in ohms
52GENB OPEN1 3.0 4.00 { R1 and X1, both in ohms
53GENC OPEN2 1.5 2.00 { R2 and X2, both in ohms
$UNITS, 0.0, 0.0 { Return to inductance data in mHenry; XOPT = 0
C 3x3 phase-domain [R] in ohms and [L] in mHenries follow. For each row I, L(I,J) is below R(I,J). w = 3.14159265E+02 rad/sec.
C 5.0000000000 2.1726497308 3.3273502692
C 21.2206590789 13.0496847320 10.2930402549
C 3.3273502692 5.0000000000 2.1726497308
C 10.2930402549 21.2206590789 13.0496847320
C 2.1726497308 3.3273502692 5.0000000000
C 13.0496847320 10.2930402549 21.2206590789
C The present comments are diagnostic output from use of MODEL Z0Z1Z2. Such
C output always will be found in the .DBG file. From it, produce the following
C equivalent representation of [R] in ohms and [L] in mHenries:
C 2nd of 3 line sections follows. PHS at sending end indicates phase domain:
51PHSA MODEL [R][L]
52GENB OPEN3
53GENC OPEN4
5.000000000 2.172649731 3.327350269 { R(1,1), R(1,2), and R(1,3) in ohms
21.22065908 13.04968473 10.29304025 { L(1,1), L(1,2), and L(1,3) in mHenries
3.327350269 5.000000000 2.172649731 { Row 2 of [R]
10.29304025 21.22065908 13.04968473 { Row 2 of [L]
2.172649731 3.327350269 5.000000000 { Row 3 of [R]
13.04968473 10.29304025 21.22065908 { Row 3 of [L]
C Finally, data for 3rd of 3 line sections comes from Orlando. See preceding
C subcase for comments about it. Z1 and Z2 differ while the sum is fixed.
C So, the fault current will be the same, but not the fault voltage, note.
51OPHA MODEL [R][L]
52GENB OPEN5
53GENC OPEN6
5.0000000 -3.0235027 8.5235027
21.2206591 11.8091947 11.5335303
8.5235027 5.0000000 -3.0235027
11.5335303 21.2206591 11.8091947
-3.0235027 8.5235027 5.0000000
11.8091947 11.5335303 21.2206591
C 4th of 4 line sections follows. MH at sending end indicates L in mHenries.
C This 4th alternative is the same as the first except X in ohms has been
C converted to L in mHenries by multiplying by 10 / Pi = 3.18309886184 This
C gives Lo, L1, L2 = 44.5633840657 12.7323954474 6.36619772368
$VINTAGE, 1, { Switch to wide format, with R and L read as 2E16.0
C 3456789012345678901234567890123456789012345678901234567890
C RRRRRRRRRRRRRRRRLLLLLLLLLLLLLLLL
51MHA MODEL Z0Z1Z2 10.5 44.5633840657 { Ro in ohms and Lo in mH
52GENB OPEN7 3.0 12.7323954474 { R1 in ohms and L1 in mH
53GENC OPEN8 1.5 6.36619772368 { R2 in ohms and L2 in mH
$VINTAGE, 0, { Done with wide format; return to old (narrow) format
BLANK card ending branches
C Following switches measure line currents in the 3 lines. Since phase "a"
C is the only one connected, ignore switches for phases "b" and "c":
GENA SEQA MEASURING 1
GENA PHSA MEASURING 1
GENA OPHA MEASURING 1
GENA MHA MEASURING 1
BLANK card ending switches
14GENA 1000. 50. -0.00 { Balanced, 3-phase excitation that
14GENB 1000. 50. -120. { is _not_ present during the steady
14GENC 1000. 50. -240. { state. There is no phasor solution.
BLANK card follows the last source card
C OPEN1 OPEN3 OPEN5 OPEN7 { Output fault voltages (unfaulted phase b, anyway)
C Note. Activate the preceding comment card to demonstrate that the solution
C of the OPH circuit really is different from the other 3. I.e., the voltage
C at OPEN5 will differ from the voltages at OPEN1, OPEN3, and OPEN7.
C The fault currents are identical, but fault voltages will differ. This is
C because the sum of Z1 and Z2 agrees, but neither Z1 nor Z2 does.
C Next 4 output variables are branch currents (flowing from the upper node to the lower node);
C Step Time GENA GENA GENA GENA
C SEQA PHSA OPHA MHA
C 0 0.0 0.0 0.0 0.0 0.0
C 1 .1E-3 2.32761034 2.32761034 2.32761033 2.32761034
C 2 .2E-3 6.92518039 6.92518039 6.92518039 6.92518039
C 3 .3E-3 11.4064981 11.4064981 11.4064981 11.4064981
BLANK card ending node voltage outputs
C 600 .06 71.992366 71.992366 71.9923659 71.992366
C 700 .07 -71.992425 -71.992425 -71.992425 -71.992425
C 800 .08 71.9924194 71.9924194 71.9924193 71.9924194
C 900 .09 -71.99242 -71.99242 -71.99242 -71.99242
C 1000 0.1 71.9924199 71.9924199 71.9924198 71.9924199
C Variable maxima: 119.979925 119.979925 119.979925 119.979925
C Times of maxima: .083 .083 .083 .083
C Variable minima: -119.97993 -119.97993 -119.97993 -119.97993
C Times of minima: .093 .093 .093 .093
C Note about preceding extema. The maximum of 119.979925 and minimum of
C -119.97993 are nearly negatives of each other. This is a sign that the
C steady state is really steady (equality in the 8th decimal digit). But
C Orlando Hevia's exact solution of 120.0000 indicates some error. Why?
C The substantial dT (discretization error). As dT is decreased, the ATP
C answer approaches Orlando Hevia's. For all dT, all 4 maxima agree. Thus
C only a single value need be shown, as a function time-step size dT:
C 100 usec ===> 119.979925
C 50 usec ===> 119.998403
C 20 usec ===> 119.999337
C 10 usec ===> 119.999920
C 5 usec ===> 119.999967
C 2 usec ===> 119.999997
BLANK termination to plot cards
BEGIN NEW DATA CASE
C 7th of 7 subcases is the same as the 5th except that the sources have
C been activated during the steady state. This subcase has a phasor
C solution --- now possible as this subcase is added on 5 August 2008.
C Note that the final time steps of the two subcases are identical.
C BEGIN PEAK VALUE SEARCH has been removed since there no longer is any
C need for allowing the solution to "settle" into the steady-state. In
C place of 5 cycles, simulate just 2 --- plenty to demonstrate sinusoidal
C signals (see addition of 2 plot cards after CALCOMP PLOT).
PRINTED NUMBER WIDTH, 11, 2, { Request maximum precision (for 8 output columns)
.000100 .040 { dT is relatively large to speed simulation. Only 400 steps
1 1 1 1 1 -1
5 5 20 20 100 100
$UNITS, 50.0, 0.0 { Begin inductance data in ohms at XOPT = 50 Hz
C 1st of 3 line sections follows. SEQ at sending end indicates sequence data:
51GENA OPEN1 MODEL Z0Z1Z2 10.5 14.00 { Ro and Xo, both in ohms
52SEQB FLT1 3.0 4.00 { R1 and X1, both in ohms
53SEQC FLT1 1.5 2.00 { R2 and X2, both in ohms
$UNITS, 0.0, 0.0 { Return to inductance data in mHenry; XOPT = 0
51GENA OPEN2 MODEL [R][L]
52PHSB FLT2
53PHSC FLT2
5.000000000 2.172649731 3.327350269 { R(1,1), R(1,2), and R(1,3) in ohms
21.22065908 13.04968473 10.29304025 { L(1,1), L(1,2), and L(1,3) in mHenries
3.327350269 5.000000000 2.172649731 { Row 2 of [R]
10.29304025 21.22065908 13.04968473 { Row 2 of [L]
2.172649731 3.327350269 5.000000000 { Row 3 of [R]
13.04968473 10.29304025 21.22065908 { Row 3 of [L]
51GENA OPEN3 MODEL [R][L]
52OPHB FLT3
53OPHC FLT3
5.0000000 -3.0235027 8.5235027
21.2206591 11.8091947 11.5335303
8.5235027 5.0000000 -3.0235027
11.5335303 21.2206591 11.8091947
-3.0235027 8.5235027 5.0000000
11.8091947 11.5335303 21.2206591
BLANK card ending branches
GENB SEQB MEASURING 1
GENB PHSB MEASURING 1
GENB OPHB MEASURING 1
GENC SEQC MEASURING 1
GENC PHSC MEASURING 1
GENC OPHC MEASURING 1
BLANK card ending switches
14GENA 1000. 50. -0.00 -1.
14GENB 1000. 50. -120. -1.
14GENC 1000. 50. -240. -1.
BLANK card follows the last source card
FLT1 FLT2 FLT3 { Show that the fault voltage of OPH (FLT3; 3rd of 3) differs
C Column headings for the 9 EMTP output variables follow. These are divided among the 5 possible classes as follows ....
C First 3 output variables are electric-network voltage differences (upper voltage minus lower voltage);
C Next 6 output variables are branch currents (flowing from the upper node to the lower node);
C Step Time FLT1 FLT2 FLT3 GENB GENB GENB GENC GENC GENC
C SEQB PHSB OPHB SEQC PHSC OPHC
C *** Phasor I(0) = -1.8475209E+02 Switch "GENB " to "SEQB " closed in the steady-state.
C *** Phasor I(0) = -1.8475209E+02 Switch "GENB " to "PHSB " closed in the steady-state.
C *** Phasor I(0) = -1.8475209E+02 Switch "GENB " to "OPHB " closed in the steady-state.
C *** Phasor I(0) = 1.8475209E+02 Switch "GENC " to "SEQC " closed in the steady-state.
C *** Phasor I(0) = 1.8475209E+02 Switch "GENC " to "PHSC " closed in the steady-state.
C *** Phasor I(0) = 1.8475209E+02 Switch "GENC " to "OPHC " closed in the steady-state.
C 0 0.0 -333.3333 -333.3333 572.66667 -184.7521 -184.7521 -184.7521 184.75209 184.75209 184.75209
C 1 .1E-3 -333.1685 -333.1685 547.50887 -180.3089 -180.3089 -180.3089 180.30888 180.30888 180.30888
C 2 .2E-3 -332.6749 -332.6749 521.81078 -175.6877 -175.6877 -175.6877 175.68774 175.68774 175.68774
C 3 .3E-3 -331.853 -331.853 495.59776 -170.8932 -170.8932 -170.8932 170.89322 170.89322 170.89322
BLANK card ending node voltage outputs
C 400 .04 -333.3372 -333.3372 572.64196 -184.7478 -184.7478 -184.7478 184.74783 184.74783 184.74783
C Variable maxima : 333.33754 333.33754 977.26389 230.90147 230.90147 230.90147 230.90154 230.90153 230.90153
C Times of maxima : .01 .01 .017 .028 .028 .028 .018 .018 .018
C Variable minima : -333.3372 -333.3372 -977.2634 -230.9015 -230.9015 -230.9015 -230.9015 -230.9015 -230.9015
C Times of minima : .04 .04 .027 .018 .018 .018 .028 .028 .028
CALCOMP PLOT { Display some waveforms to show that all is smooth & sinusoidal
144 4. 0.0 40.-1.E31.E3FLT1 FLT2 FLT3
194 4. 0.0 40.-250.250.GENB SEQB GENC PHSC
BLANK termination to plot cards
BEGIN NEW DATA CASE
BLANK
$EOF
In order that it not be lost, let's append Orlando Hevia's FORTRAN
program. This confirms the preceding 5th and 6th subcases as well as
other fault types not illustrated by preceding ATP data:
COMPLEX A,A2,Z0,Z1,Z2,AIR,AIS,AIT,E,J
PI=4.0*ATAN(1.0)
A= CMPLX(-0.5,SQRT(3.0)/2.0)
A2=CMPLX(-0.5,-SQRT(3.0)/2.0)
J=CSQRT(CMPLX(-1.0,0.0))
Z0=CMPLX(10.5,14.0)
Z1=CMPLX(3.0,4.0)
Z2=CMPLX(1.5,2.0)
E=CMPLX(1000.0,0.0)
AIS= -J*SQRT(3.0)*E*(((1.0,0.0)+A2)*Z2+Z0)/
1 (Z1*Z2+Z2*Z0+Z0*Z1)
AIT= J*SQRT(3.0)*E*(((1.0,0.0)+A )*Z2+Z0)/
1 (Z1*Z2+Z2*Z0+Z0*Z1)
AIR=AIS+AIT
AS=CABS(AIS)
AT=CABS(AIT)
AR=CABS(AIR)
FS=ATAN2(AIMAG(AIS),REAL(AIS))*180.0/PI
FT=ATAN2(AIMAG(AIT),REAL(AIT))*180.0/PI
FR=ATAN2(AIMAG(AIR),REAL(AIR))*180.0/PI
C WRITE(*,*)Z0
C WRITE(*,*)Z1
C WRITE(*,*)Z2
C WRITE(*,*)A,A2,J
WRITE(*,*)'PHASE-PHASE-GROUND FAULT'
100 FORMAT('IB ',2F12.4)
101 FORMAT('IC ',2F12.4)
102 FORMAT('IRES ',2F12.4)
WRITE(*,*)'RECTANGULAR'
WRITE(*,100)AIS
WRITE(*,101)AIT
WRITE(*,102)AIR
WRITE(*,*)'POLAR'
WRITE(*,100)AS,FS
WRITE(*,101)AT,FT
WRITE(*,102)AR,FR
C
AIS= -J*SQRT(3.0)*E/(Z1+Z2)
AIT= J*SQRT(3.0)*E/(Z1+Z2)
AIR=AIS+AIT
AS=CABS(AIS)
AR=CABS(AIR)
AT=CABS(AIT)
FS=ATAN2(AIMAG(AIS),REAL(AIS))*180.0/PI
FT=ATAN2(AIMAG(AIT),REAL(AIT))*180.0/PI
FR=ATAN2(AIMAG(AIR),REAL(AIR))*180.0/PI
WRITE(*,*)' '
WRITE(*,*)'PHASE-PHASE FAULT'
WRITE(*,*)'RECTANGULAR'
WRITE(*,100)AIS
WRITE(*,101)AIT
WRITE(*,102)AIR
WRITE(*,*)'POLAR'
WRITE(*,100)AS,FS
WRITE(*,101)AT,FT
WRITE(*,102)AR,FR
C
AIR= E/Z1
AR=CABS(AIR)
FR=ATAN2(AIMAG(AIR),REAL(AIR))*180.0/PI
C
WRITE(*,*)' '
WRITE(*,*)'PHASE-PHASE-PHASE-GROUND FAULT'
WRITE(*,*)'RECTANGULAR'
WRITE(*,202)AIR
WRITE(*,*)'POLAR'
WRITE(*,202)AR,FR
202 FORMAT('IA ',2F12.4)
C
AIR=3.0*E/(Z0+Z1+Z2)
AR=CABS(AIR)
FR=ATAN2(AIMAG(AIR),REAL(AIR))*180.0/PI
C
WRITE(*,*)' '
WRITE(*,*)'PHASE-GROUND FAULT'
WRITE(*,*)'RECTANGULAR'
WRITE(*,302)AIR
302 FORMAT('IA ',2F12.4)
WRITE(*,*)'POLAR'
WRITE(*,302)AR,FR
STOP
END
Output, as produced using GNU FORTRAN compilation, linking, and execution,
is as follows:
PHASE-PHASE-GROUND FAULT
RECTANGULAR
IB -196.3879 -114.6822
IC 165.0836 156.4214
IRES -31.3043 41.7391
POLAR
IB 227.4208 -149.7169
IC 227.4208 43.4567
IRES 52.1739 126.8699
PHASE-PHASE FAULT
RECTANGULAR
IB -184.7521 -138.5641
IC 184.7521 138.5641
IRES 0.0000 0.0000
POLAR
IB 230.9401 -143.1301
IC 230.9401 36.8699
IRES 0.0000 36.8699
PHASE-PHASE-PHASE-GROUND FAULT
RECTANGULAR
IA 120.0000 -160.0000
POLAR
IA 200.0000 -53.1301
PHASE-GROUND FAULT
RECTANGULAR
IA 72.0000 -96.0000
POLAR
IA 120.0000 -53.1301
|
