Mathcad - L27 examples - uidaho.edu · 2020. 1. 15. · ECE 524: Transients in Power Systems...

ECE 524: Transients in Power Systems

Session 27; Page 1/28 Spring 2018

Three Phase Travelling Wave Examples

132kV LL

VS VSEND

I

100km line

VRECVVV

R0 0.32ohmkm

R1 0.025ohmkm

L0 3.2mHkm

L1 0.91mHkm

C0 0.008μFkm C1 0.0125

μFkm

Zc0L0C0

Zc0 632.46 Ω Zc1L1C1

Zc1 269.81 Ω

Applied voltage: Vm 107.8kV

Balanced three phase set with close at peak of phase: Vabc0

Vm

Vm

2

Vm

2

Vabc0

107.8

53.9

53.9

kV

Clarke Transform:

Ti1

3

1

1

1

2

1

2

1

2

0

32

32

Te Ti Te TiT

1

0

0

0

1

0

0

0

1



Vmodal0 Te1 Vabc0 Vmodal0

0

132.03

0

kV

Close into open circuit, we expect receiving end voltage to double for each mode, and the convert back to ABC

Voltages when close all three poles

(file lect27.pl4; x-var t) v:VSENDA v:VSENDB v:VSENDC v:VRECA v:VRECB v:VRECC 0.094 0.096 0.098 0.100 0.102 0.104[s]

-200

-100

0

100

200

300

[kV] Note that each individual phase doublesthe initial applied voltageSuggests only one mode excited, asimplied above

Initial modal currentVmodal01

Zc1489.33A Ti

0

Vmodal01Zc1

0

399.53

199.77

199.77

A



(file lect27.pl4; x-var t) c:VSA -VSENDA c:VSB -VSENDB c:VSC -VSENDC 0.096 0.098 0.100 0.102 0.104 0.106 0.108[s]

-500

-350

-200

-50

100

250

400

[A]

Consistent with the resultsabove for the initial transient

Single phase close to open circuited line:

Voltage

Vmodal_single0 Te1

107.8kV

0kV

0kV

Vmodal_single0

62.24

88.02

0

kV Note the cross-coupling, the first term is a ground mode term.

Receiving end: Vrecmodal 2 Vmodal_single0

Vabc_rec Te Vrecmodal Vabc_rec

215.6

0

0

kV



Sending end voltages

( fi le le c t2 7 .p l4 ; x-va r t) v :V S E N D A v :V S E N D B v :V S E N D C 0 .0 9 7 0 .0 9 8 0 .0 9 9 0 .1 0 0 0 .1 0 1 0 .1 0 2 0 .1 0 3 0 .1 0 4 0 .1 0 5[s ]

- 4 0

- 2 0

0

2 0

4 0

6 0

8 0

1 0 0

1 2 0[k V ]

Phases B and C show a zero mode response(same voltage on each).

(fi le le c t27 .p l4 ; x-va r t) v:V RE C A v:V RE C B v:V RE C C 0.095 0.097 0.099 0.101 0.103 0.105 0.107[s]

-100

-50

0

50

100

150

200

250

[kV ]Receiving endvoltages: Note effect of two different

propagation times for ground andline mode



Current

(f ile lec t25.pl4; x -v ar t ) c :VSA -VSEN D A c :VSB -VSEN D B c :VSC -VSEN D C 0.096 0.098 0.100 0.102 0.104 0.106 0.108[s ]

-400

-300

-200

-100

0

100

200

300

[A ]

Close 1 phase, with load 3 phase short.....

(f ile lec t25 .p l4 ; x -v a r t ) c :V S A -V S E N D A c :V S B -V S E N D B c :V S C -V S E N D C 0 .000 0 .026 0 .052 0 .078 0 .104 0 .130[s ]

-2000

-1500

-1000

-500

0

500

1000

1500

2000[A ]

Sending end currents:



(f ile lect25.pl4; x-v ar t) c:VRECA - c:VRECB - c:VRECC - 0.095 0.100 0.105 0.110 0.115 0.120 0.125 0.130[s]

-2000

-1500

-1000

-500

0

500

1000

1500

2000[A]

Receiving end currents



EMTDC ImplementationBreaker set for single pole operation (eachphase can operate independently)

Now need three trip inputs, see in circuit diagram below



R=0

T im e dB re a k e r

L o g icO p e n @ t0B R K A

TL in eT

TER

M


L o g icO p e n @ t0T E R M

VS E N D VR E C

VS E N D

VR E C

IB R E AK A

IB R E AK B

IB R E AK C

Ire c A

Ire c B

Ire c C


L o g icO p e n @ t0B R K B


L o g icO p e n @ t0B R K C Need a seperate control for each

phase of the breaker now

Main : Graphs

0.100

-250 -200 -150 -100 -50

0 50

100 150 200 250

y (k

V)

VSEND VREC

Voltages for 3 pole close, withterminal open



Main : Graphs

0.100

-0.50

0.00

0.50

y (k

A)

IBREAKA IBREAKB IBREAKCCurrents for 3 pole close to opencircuited line

Main : Graphs

0.100

-250 -200 -150 -100

-50 0

50 100 150 200 250

y (k

V)

VSEND VRECSingle phase close:

Voltage (open circuited line)



Main : Graphs

0.100

-0.50

0.00

0.50

y (k

A)

IBREAKA IBREAKB IBREAKC

Current (open circuited line)

Main : Graphs

0.100

-250 -200 -150 -100

-50 0

50 100 150 200 250

y (k

V)

VSEND VRECClose 1 phase, with load 3 phaseshort.....

Sending end voltage (note reflections onsending end)



Main : Graphs

0.100 0.150

-2.00

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

y (k

A)

IBREAKA IBREAKB IBREAKC

Sending end currents:

Main : Graphs

0.100 0.150

-2.00

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

y (k

A)

IrecA IrecB IrecC

Receiving end currents



Trapped Charge and Inrush Current

When clearing a line, breakers open at natural current zeroIf line is unloaded this is a capacitive current, so clear at voltage peakPotential for inrush currentsNote that emt-programs exaggerate this effect since don't allow for self-dischargeMismatch in voltage between source side of breaker and line side of breaker determines:

Inrush current1.Magnitude of worst case overvoltage at receiving end of line if that is open2.

Won't see same on all three phase with 3 pole closing

1 Zc Zc

132kV LL

VS

I

SWITI

10 uOhm

V1

100km line

V2VV



( f i l e R E C L O S E . p l 4 ; x - v a r t ) v : V 1 A v : V 1 B v : V 1 C 0 . 0 0 0 0 . 0 2 6 0 . 0 5 2 0 . 0 7 8 0 . 1 0 4 0 . 1 3 0[ s ]

- 1 2 0

- 8 0

- 4 0

0

4 0

8 0

1 2 0

[ k V ]

Voltages at sending end of the linewhen energize open circuited line:

Energize lineThen clear it Then reclose

The transient voltages in the trappedcharge are largely due to residualeffects of the initial closing transient.

( f i le R E C L O S E .p l4 ; x - v a r t ) v :V 2 A v :V 2 B v :V 2 C 0 . 0 0 0 0 . 0 2 6 0 . 0 5 2 0 . 0 7 8 0 . 1 0 4 0 . 1 3 0[ s ]

- 3 0 0

- 2 0 0

- 1 0 0

0

1 0 0

2 0 0

3 0 0

[ k V ]And a receiving end

Note the larger transientsfor the reclose



Sending end currents (combine the two switches)

( f i le R E C L O S E . p l4 ; x - v a r t ) c : V S A - V 1 A c : V S B - V 1 B c : V S C - V 1 C c : V S A - S W IT A c : V S B - S W IT B c : V S C - S W IT C

0 . 0 0 0 0 . 0 2 6 0 . 0 5 2 0 . 0 7 8 0 . 1 0 4 0 . 1 3 0[ s ]- 6 0 0

- 4 0 0

- 2 0 0

0

2 0 0

4 0 0

6 0 0

[ A ]Bigger current on reclose due to largervoltage across the charactersitic impedance

Repeat with the system in steady-state before open the breakers.

( f i l e R E C L O S E . p l 4 ; x - v a r t ) v : V 1 A v : V 1 B v : V 1 C 0 . 0 0 0 0 . 0 2 6 0 . 0 5 2 0 . 0 7 8 0 . 1 0 4 0 . 1 3 0[ s ]

- 1 2 0

- 6 8

- 1 6

3 6

8 8

1 4 0

[ k V ]

Sending end voltages:note that trappedcharge changedsince voltage atpoint on wave forbreaker clearinghas changed



( f i l e R E C L O S E . p l4 ; x - v a r t ) v : V 2 A v : V 2 B v : V 2 C 0 . 0 0 0 0 . 0 2 6 0 . 0 5 2 0 . 0 7 8 0 . 1 0 4 0 . 1 3 0[ s ]

- 3 0 0

- 2 0 0

- 1 0 0

0

1 0 0

2 0 0

3 0 0

[ k V ]

Receiving end voltages

R=0


L o g icO p e n @ t0B R K 1

B R K 1

R E S IS T


L o g icO p e n @ t0R E S IS T

1 E -5 [o h m ]

1 E -5 [o h m ]

1 E -5 [o h m ]TL in e

T

TER

M


L o g icO p e n @ t0T E R M

VS E N D VR E C

VS E N D

VR E C

IB R E AK A IR A

IB R E AK B

IB R E AK C

IR B

IR C

EMTDC implementation:



Ma in : G ra p h s

0 .0 0 0 0 .0 5 0 0 .1 0 0 0 .1 5 0 0 .2 0 0 0 .2 5 0 0 .3 0 0

-1 2 5 -1 0 0

-7 5 -5 0 -2 5

0 2 5 5 0 7 5

1 0 0 1 2 5

y (k

V)

V S ENDSending end voltageswith two transients

Some differencefrom ATP

Main : Graphs

0 .000 0 .050 0 .100 0 .150 0 .200 0 .250 0 .300

-300

-200

-100

0

100

200

300

y (k

V)

V RECReceiving end voltages



Ma in : G raphs

0 .100 0 .120 0 .140 0 .160 0 .180 0 .200 0 .220 0 .240 0 .260

-0 .80

-0 .60

-0 .40

-0 .20

0 .00

0 .20

0 .40

0 .60

0 .80

y (k

A)

IB REA K A IB REA K B IB REA K C IRA IRB IRC

Sending end currents

Main : Graphs

0.000 0.050 0.100 0.150 0.200 0.250 0.300

-125 -100

-75 -50 -25

0 25 50 75

100 125

y (k

V)

VSENDRepeat without firstenergization transient

Sending end voltage



Main : Graphs

0.000 0.050 0.100 0.150 0.200 0.250 0.300

-250 -200 -150 -100

-50 0

50 100 150 200 250

y (k

V)

VRECReceiving end voltages



Options to Reduce Inrush

Controlled breaker pole closingPre-insertion resistor, bypass after a few travel timesSecondary transient (small) when bypassSingle phase case: size R = ZcNo reflection for return waves

Voltage divider - apply 50% of voltage to the1.input of the lineReduces voltage and current2.

Resistor sized for current loading and dissipationHow about three phase line:

Ideally size for the first line mode,1.Some mismatch if do single pole reclosing2.Many utilities just use one standard size3.resistor, rather than optimize for each line. Not ideal, but eliminates most of the transient4.

2 Zc

1 R

Zc

132kV LL

VS

I

SWITI

632 Ohm

V1

100km line

V2VVV

132kV LL

VSALT

I

SWITAI

269.8 Ohm

V1A

100km line

V2AVVV

Rpre = Zc1

Rpre = Zc0



132kV LL

VS

I

SWITI

632 Ohm

V1

100km line

V2VVV

132kV LL

VSALT

I

SWITAI

269.8 Ohm

V1A

100km line

V2AVVV

Rpre = Zc1

Rpre = Zc0

( f i le P R E IN S . p l4 ; x - v a r t ) v : V 1 A A v : V 2 A A v : V S A 0 . 0 9 8 0 . 0 9 9 0 . 1 0 0 0 . 1 0 1 0 . 1 0 2 0 . 1 0 3 0 . 1 0 4[ s ]

- 2 0

0

2 0

4 0

6 0

8 0

1 0 0

1 2 0

[ k V ]

Initial close with R = Zc1

No overvoltageatthe receiving endSending end voltage increases whenafter 2 travel times



( f i le P R E IN S . p l4 ; x - v a r t ) c : V S A L T A - S W IT A A c : V S A L T B - S W IT A B c : V S A L T C - S W IT A C 0 . 0 9 6 0 . 0 9 8 0 . 1 0 0 0 . 1 0 2 0 . 1 0 4 0 . 1 0 6 0 . 1 0 8 0 . 1 1 0 0 . 1 1 2[ s ]

- 1 5 0

- 1 0 0

- 5 0

0

5 0

1 0 0

1 5 0

2 0 0

[ A ]

Sending end current

No longer have repeatedreflections, transient lasts2 travel times.



( f i le P R E IN S .p l4 ; x - v a r t ) v :V 1 A A v :V 2 A A v :V S A 0 . 1 0 0 0 . 1 0 4 0 . 1 0 8 0 . 1 1 2 0 . 1 1 6 0 . 1 2 0[ s ]

- 1 2 0

- 8 0

- 4 0

0

4 0

8 0

1 2 0

[ k V ]

Bypass the resistor after 1 cycle:



( f i le P R E IN S .p l4 ; x - v a r t ) v :V 1 A v :V 2 A v :V S A 0 . 0 9 5 0 . 0 9 9 0 . 1 0 3 0 . 1 0 7 0 . 1 1 1 0 . 1 1 5[ s ]

- 1 2 0

- 8 0

- 4 0

0

4 0

8 0

1 2 0

[ k V ]

Repeat with R = Zc0

Actually reducevoltages further



( fi le P R E IN S .p l4 ; x -va r t) v :V 2 A v :V 2 B v :V 2 C 0 . 0 0 0 0 .0 2 6 0 . 0 5 2 0 .0 7 8 0 .1 0 4 0 . 1 3 0[s ]

- 2 5 0 .0

- 1 8 7 .5

- 1 2 5 .0

- 6 2 .5

0 .0

6 2 .5

1 2 5 .0

1 8 7 .5

2 5 0 .0[ k V ]Reclose with R=Zc0



R=0

Tim edBreaker

LogicOpen@t0BRK1

BRK1

RESIST 269.8[ohm ]

269.8[ohm ]

269.8[ohm ]TLine

T

TER

M

Tim edBreaker

LogicOpen@t0TERM

VSEND VREC

EMTDC implementation(with Rpre = Zc1)

Main : Graphs

0.2020 0.2030 0.2040 0.2050 0.2060 0.2070 0.2080 0.2090

-150

-100

-50

0

50

100

150

y (k

V)

VSEND VRECPhase sending andreceiving endvoltage with Rpre = Zc1



Main : Graphs

0.2000 0.2020 0.2040 0.2060 0.2080 0.2100 0.2120 0.2140

-0.200

-0.150

-0.100

-0.050

0.000

0.050

0.100

0.150

0.200

y (k

A)

IRA IRB IRC

Sending end currents



Main : Graphs

0.1990 0.2000 0.2010 0.2020 0.2030 0.2040 0.2050 0.2060 0.2070 0.2080 0.2090

-150

-100

-50

0

50

100

150

y (k

V)

VSEND VRECRepeat with Rpre=Zc0

Main : Graphs

0.180 0.190 0.200 0.210 0.220

-0.125 -0.100 -0.075 -0.050 -0.025 0.000 0.025 0.050 0.075 0.100 0.125

y (k

A)

IRA IRB IRC



Main : Graphs

0.000 0.050 0.100 0.150 0.200 0.250 0.300

-250

-200

-150

-100

-50

0

50

100

150

200

y (k

V)

VRECReclose intotrapped charge with Rpre=Zc0

Mathcad - L27 examples - uidaho.edu · 2020. 1. 15. · ECE 524: Transients in Power Systems...

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Transcript of Mathcad - L27 examples - uidaho.edu · 2020. 1. 15. · ECE 524: Transients in Power Systems...