transient r-l circuit example- updated: see page 9, 10-13 and 25 · 2020. 2. 1. · ece 524:...

ECE 524: Transients in Power Systems

Session 6; Page 1/26 Spring 2018

Transient R-L Circuit Example-Updated: see page 9, 10-13 and 25

Given a 230kV:34kV, D-Ygrounded, 100MVA transformer supplied supplied by a sourcewith a short circuit MVA of 2000. The transformer has a per unit reactance of 0.1pu on itsratings base. Both the transformer and the source impedance have an X/R ratio of 12.Calculate the complete phase A fault current if a fault occurs on the LV sideof the transformer at 70 degrees past the voltage peak.

Define Units:

MVA 1000kW:= SBASE 100MVA:=

pu 1:=

Transformer parameters:

Srated 100MVA:= XoverR 12:=

Vhv 230kV:= Vlv 34.5kV:=

Xtrans 0.1pu:=

RtransXtrans

XoverR:= Rtrans 0.0083 pu=

Find the transformer impedances referred to the LV side since the fault occurs on that side.

ZBLVVlv

2

Srated:= ZBLV 11.9025 Ω=

RtransLV Rtrans ZBLV:= RtransLV 0.0992Ω=

XtransLV Xtrans ZBLV:= XtransLV 1.19Ω=

LtransLVXtransLV

2 π 60 Hz:= LtransLV 3.157 mH=



Source impedance

MVAsc 2000MVA:= Vpu 1.0pu:=

MVAsc_puMVAsc

SBASE:= MVAsc_pu 20 pu=

mZsourceVpu

2

MVAsc_pu:= mZsource 0.05 pu=

η atan XoverR( ):= η 85.2364 deg=

Zsource mZsource ej η

:= Zsource 0.0042 0.0498i+( ) pu=

Source impedance in Ohms referred to LV side:

Zsource_Ohm Zsource ZBLV:= Zsource_Ohm 0.049 0.593i+( ) Ω=

Lsource_LVIm Zsource_Ohm( )

2 π 60 Hz( ):= Lsource_LV 1.5732 mH=

If we wanted to find source impedance referred to LV side in 1 step:

Vlv( )2

MVAsc

ej atan XoverR( )

0.049 0.593i+( ) Ω=

Equivalent circuit (since this is a three phase fault we can use per phase analysis:

RsrcLsrc

VLN_LVFault

RtransLtrans



Transient Current Solution:

Requiv Re Zsource_Ohm( ) RtransLV+:= Requiv 0.1486 Ω=

Lequiv Lsource_LV LtransLV+:= Lequiv 4.7304 mH=

Decay constant:

αRequiv-

Lequiv:= α 31.4159- 1

s=

τ 1α

:= τ 0.032- s=

Driving point voltage:

Vm2

3Vlv:= Vm 28.1691 kV=

Fault inception angle:

If we define our driving voltage as:

v t( ) Vm sin ω t ϕ+( )=

ω 2 π 60 Hz:= ω 376.9911 rads

=

Since the fault occurs 70 degrees phase the peak of the voltage (for analytical solutionlet fault occur at time t = 0):

ϕ 90deg 70deg+:= ϕ 160 deg=

Initial condition:

i0 0A:= Switch is open

Equiavlent impedance:

Zequiv_LV Requiv2 ω Lequiv( )2+:= Zequiv_LV 1.7895Ω=



Create time vector:

v t( ) Vm sin ω t ϕ+( ):=t 0sec 1 10 4- sec, 460Hz

..:=

iss t( )Vm

Zequiv_LV

sin ω t ϕ+ η-( ):=

itrans t( )Vm

Zequiv_LV

sin ϕ η-( ) eα t:=

0 0.02 0.04 0.064- 10

4

2- 104

0

2 104

4 104

v t( )

iss t( )

itrans t( )

t

0 0.02 0.04 0.060

5 103

1 104

1.5 104

2 104

itrans t( )

t



i t( ) iss t( ) itrans t( )-:=

0 0.02 0.04 0.063- 10

4

2- 104

1- 104

0

1 104

i t( )

t

Other items of interest: peak current:

Peak current occurs when the derivative of the current equals 0 (the second timein this case).

ti

d

d

Vm

Zequiv_LV

ω cos ω t ϕ+ η-( ) α sin ϕ η-( ) eα t( )+ =

Plot this for comparison.....

di t( )Vm

Zequiv_LV

ω cos ω t ϕ+ η-( ) α sin ϕ η-( ) eα t( )+ :=

0 0.01 0.02 0.031- 10

7

5- 106

0

5 106

1 107

300 i t( )

di t( )

0

t

Note that the secondzero crossing is at the peak in this case.



Now we want to find the values of t where the derivative is zero.

tzero 0.01sec:= millisecondsec

1000:=

Given

Vm

Zequiv_LV

ω cos ω tzero ϕ+ η-( ) α sin ϕ η-( ) eα tzero( )+ 0=

tpeak Find tzero( ):= tpeak 9.1986 millisecond=

di tpeak( ) 3.3555 10 10-A

s=

i tpeak( ) 27.09- kA=

ATP Simulation results:·

34.5 kV_LL

VS

0.149 ohm 4.73 mH

LV

0.02407 sec

I

V V

Settings for simulation:

Δt 5 10 5- sec:= tmax 0.1sec:=

Vm 28169 V= Requiv 0.149 Ω= Lequiv 4.73 mH=

Note, ATP uses peak voltage, PSCAD/EMTDC uses RMS line-to-line

Since the source is a cosine, set phase angle for sine:

θ 90- deg:=

Time for fault to occur (let run 1 full cycle and then 70 degrees past positive peak):

tfault1

60Hz

1160

360+

:= tfault 0.02407 s= This is time that the switch closes.



(f ile Example.pl4; x-v ar t) c:LV -

0.00 0.02 0.04 0.06 0.08 0.10-30

-20

-10

0

10

20

*103



EMTDC Simulation results:·

VLN_RMS34.5kV

3:= VLN_RMS 19.919 kV=

0.149 [ohm] 0.00473 [H]

R=

0

TimedFaultLogic

Fa

ult

Fault

I_Fault

Main : Graphs

0.000 0.010 0.020 0.030 0.040 0.050 0.060 0.070 0.080 0.090 0.100 ... ... ...

-30.0

-25.0

-20.0

-15.0

-10.0

-5.0

0.0

5.0

10.0

15.0

y (k

A)

I_Fault

Ipk = 27.122kA goes up a little with smaller time step



EMTP-RV Simulation results:·

+34.5kVRMSLL /_-90 ?v

AC1

+0.149

R1+

4.73mH

L1

+

0.0

24

07

|10

|0?

vi

Fa

ult

Vs Load

Ipk 27197.7A=



Finding the Worst Case DC Offset

vs t( ) Vm cos ω t θ+( )=

if t( )Vm

Zequiv_LVcos ω t θ+ η-( ) eα t cos θ η-( )-( )=

We want to find the worst case peak fault current, which occurs when the 60Hzterm and the dc offset term are maximum, so we want

θ η- 0deg=Note that this maximum should also occur at or close to t = 0, so

cos ω t θ+ η-( ) 1-= (or +1) as well

θb 0deg η+:= θb 85.2364 deg=

Therefore:

vs t θ, ( ) Vm cos ω t θ+( ):=

if t θ, ( )Vm

Zequiv_LVcos ω t θ+ η-( ) eα t cos θ η-( )-( ):=

Find time for the maximum value:

0tif t( )

d

d= the key part is: 0 ω- sin ω t θ+ η-( ) α eα t cos θ η-( )( )-=

Use solve block:

Initial guess:

tg 0.01sec:=

Given

0 ω- sin ω tg θb+ η-( ) α eα tg cos θb η-( )( )-=Set constraints on t: 1

2 60 Hztg 0sec

tmax Find tg( ):= tmax 8.1622 ms=



0 0.02 0.043- 10

4

2- 104

1- 104

0

1 104

if t θb, ( )

t

if tmax θb, ( ) 27.8895- kA=

However, the above calculation doesn't account for the variation in the 60Hz term. Instead weneed to find the optimum combination of t and θ to get the most negative solution.

ifm β θm, ( )Vm

Zequiv_LVcos β θm+ η-( ) e

αβ

376.99rad

s

cos θm η-( )-

:=

β π:=

θm 0:=

Given

π β> 0>

0 θm 300deg

Sol Minimize ifm β, θm, ( ):=

Sol171.5261

90

deg= tmaxaltSol0

2π 60 Hz:= tmaxalt 7.94102 ms=

θmax Sol1:= θmax 90 deg=

if tmaxalt θmax, ( ) 27.9318- kA= Slightly larger than the answer above·based only on the dc offset.



Differenceif tmaxalt θmax, ( ) if tmax θb, ( )-

if tmaxalt θmax, ( ):= Difference 0.1515 %=

Note that the worst case usually occurs with the fault happing at 90 degrees past·the peak for most circuits (zero crossing of the voltage)

Find time to use in simulation cases:·

tfault901

60Hz

1180

360+

:= tfault90 0.025 s=

ATP Results with new fault initiation time

(file SeriesRL.pl4; x-var t) c:LV -

0.00 0.02 0.04 0.06 0.08 0.10[s]-30

-20

-10

0

10

20

[A]

27924- A



EMTDC Results with new fault initiation time

Main : Graphs

0.000 0.020 0.040 0.060 0.080 0.100

-30.0

-25.0

-20.0

-15.0

-10.0

-5.0

0.0

5.0

10.0

15.0

Ipk 27924- A=

EMTP-RV Results with new fault initiation time

Ipk 27.932- kA=



Series LC Example

t 0 0.000001sec, 2

60sec..:= MVAR MVA:=

For the system below find the following:

Worst case inrush current when closing into C1 with C2 open1.

jXs

C2C1

Assume source and capacitorsare Y connected.

VLL 34.5kV:= RMS

Capacitor Banks:

Q1 18MVAR:= at 34.5kV

Q2 10MVAR:= at 34.5kV

L1 19.2μH:= between caps

Source Isc 25kA:=

Determine circuit parameters:

Find capacitance using:Q

VLL( )2

Xc=

Xc1VLL

2

Q1:= Xc1 66.125 Ω= Xc2

VLL2

Q2:= Xc2 119.025 Ω=

C11

ω Xc1:= C1 40.1147 μF= C2

1

ω Xc2:= C2 22.2859 μF=

Find source impedance using: IscVln

Xs=

Xs

VLL

3

Isc:= Xs 0.797 Ω= Ls

Xs

ω:= Ls 2.1134 mH=



Rs 0ohm:= no source resistance

Equivalent circuit:

Ls

C2C1

L1

Load

Neglect load for now.

1. Worst case inrush current when closing into C1 with C2 open

Vm VLL2

3:= Vm 28.1691 kV= ϕ 0deg:=

vs t( ) Vm cos ω t ϕ+( ):=

Z0Ls

C1:= Z0 7.2584 Ω=

ωn1

Ls C1:= fn

ωn2 π

:= fn 546.6066 Hz=

Initial conditons: Vc1_0 0V:= capacitor is discharged

iLs_0 0A:= no load current

Resonant frequency

fn1

2 π1

Ls C1:= fn 546.6066 Hz= approx. 9 times source frequency

ωn 2 π fn:=



Use biggestpossible voltagedifference for givenVc1(0)

Homogeneous solution:ih_1 t( )

Vm Vc1_0-

Z0

sin ωn t( ):=

peak current at resonant frequency Vm Vc1_0-

Z0

3880.8905A=

Rs 1010-

ohm:= Set to a very small number more numerically stable

Particular solution: η atanω Ls

1

ω C1-

Rs

:= η 90- deg=

ip_1 t( )

2

3VLL

1

ω C1ω Ls-

cos ω t ϕ+ η-( ):=

IPrms

1

3VLL

1

ω C1ω Ls-

:= IPrms 304.9A=

i1 t( ) ip_1 t( ) ih_1 t( )+:=

Worst possible current will be when both the 60Hz current and the resonant current peak at thesame time. Since there is no resistance, the resonant response will continue forever. So at somepoint in time they will peak at the same time. Realistically, we would look at the peak in the first60Hz cycle.

Imax1Vm Vc1_0-

Z0

2 IPrms+:= Imax1 4312.0842A=



Capacitor voltage (assume switch at voltage peak)

vc1 t( )1

C1

0sec

t

ti1 t( )

d Vc1_0+:=

Lets look at this analytically and using the fact that the circuit is linear::

1

C1

0sec

t

ti1 t( )

d

Vc1_0+1

C1

0sec

t

tip_1 t( )

d

0sec

t

tih_1 t( )

d+

Vc1_0+=

1

C1 0sec

t

tip_1 t( )

d1

C1 0sec

t

t2 IPrms( ) cos ω t ϕ+ η-( )

d=

Note that:cos ω t ϕ+ η-( ) = -sin ω t ϕ+( ) since η 90- deg= , so this derivation could be done interms of sine instead, but I will stick with starting from cosine here.

Vp_c1 t( )1

C1 0sec

t

t2 IPrms( )- cos ω t ϕ+ η-( )

d=

Evaluating the integral symbolically results in: ·

t

0Vp_c1 t( )

1

ω C1

2 IPrms( ) sin ω t ϕ+ η-( )=

Simplifying and taking into accout that η 90- deg= results in:

sin ω t ϕ+ η-( ) cos ω t ϕ+( )=So now:

t

0Vp_c1 t( )

1

ω C1

2 IPrms( ) cos ω t ϕ+( )( )=

Plugging in the limits of integratin results in:

Vp_c1 t( ) 2 IPrms( ) Xc1 cos ω t ϕ+( ) cos ω 0 ϕ+( )-( )=which reduces to:

Vp_c1 t( ) 2 IPrms( ) Xc1 cos ω t ϕ+( ) 1-( ):=Note that this will be bigger than Vm due to the source inductance.·



Now for the transient part of the response:

1

C1 0sec

t

tih_1 t( )

d Vc1_0+1

C10sec

t

tVm Vc1_0-

Z0

sin ωn t( )

d

Vc1_0+=

Evaluting the integral and plugging in the limits of integration:

Vh_c1 t( )1

C1

Vm Vc1_0-

ωn Z0

cos ωn t( )- cos ωn 0( )+( ) Vc1_0+=

Note that: cos ωn 0( ) 1=

and, in this case: C1 ωn Z0( ) C11

Ls C1

Ls

C1=

C1 Ls

C1 Ls= 1=

Vh_c1 t( ) Vm Vc1_0-( )- cos ωn t( ) cos ωn 0( )-( ) Vc1_0+ :=

Note that: Vp_c1 0( ) 0 kV= Vh_c1 0( ) 0V=

vc1 t( ) Vp_c1 t( ) Vh_c1 t( )+:=

We can also write this out as: ·

vcap1 t( ) 2 IPrms( ) Xc1 cos ω t ϕ+( ) 1-( ) Vm Vc1_0-( ) cos ωn t( ) 1-( ) Vc1_0+ - :=

If Ls is small such that: 2 IPrms( ) Xc1 = Vm Vc1_0- we can say:

vcap1_approx t( ) 2 IPrms cos ω t ϕ+( ) Xc1 Vm Vc1_0-( ) cos ωn t( )-:=

In this case, we there is an error in this result since:

2 IPrms( ) Xc1 Vm Vc1_0-( )- 343.5507V=

Or 2 IPrms( ) Xc1 Vm Vc1_0-( )-

Vm1.2196 %=

VcapPK 2 IPrms( ) Xc1 Vm Vc1_0-( )+:= VcapPK 56.6818 kV=



Plotting these out we get:

0 0.01 0.02 0.031- 10

5

5- 104

0

5 104

1 105

vc1 t( )

vcap1 t( )

vcap1_approx t( )

t

0 1 104-

2 104-

3 104-

4 104-

0

1 104

2 104

3 104

vc1 t( )

vcap1 t( )

vcap1_approx t( )

t

Zooming in:

Note that the appoximate curve (green) is higher than the other two which are aligned).



Note the angle on the current (account for multiplying times a capacitive reactance)

0 0.01 0.02 0.034- 10

3

2- 103

0

2 103

4 103

ip_1 t( )

ih_1 t( )

t

0 1 103-

2 103-

3 103-

4 103-

6- 103

2- 103

2 103

6 103

i1 t( )

t

Maximum value 4312A

First positive peak:

I1stPos 3807.1A:=

First negative peak:

I1stNeg 4094.3-:=

0 2 103-

4 103-

6 103-

8 103-

1- 105

5- 104

0

5 104

1 105

vc1 t( )

t


V1stPos 54.679kV:=


V1stNeg 6.572- kV:=



ATP Model:

set time step near 5% of period of resonant freq· μsec 10 6- sec:=

can use deltaT=100microseconds

tn1

fn:= tn 1829.4693 μsec=

tn

2091.4735 μsec=

Vpk=28.169kV

VS

2.11mH 19.2uF B2B1

C1 R2

40.11uF

22.29uF

C2

CSW1

T

ATPDraw schematic

Peak and RMS steady-state current (C1 inserted for steady-state initialization)

(f ile Exampl1.pl4; x-v ar t) c:B1 -C1 t: IC1RMS

0.00 0.02 0.04 0.06 0.08 0.10[s]-500

-375

-250

-125

0

125

250

375

500

Irms 304.89A:=

Ipk 431.14A:=

from above·

IPrms 304.9A=

ATP results for energizing C1:

Breaker current ·

( f i le E x a m p l1 R . p l4 ; x -v a r t ) c : B 1 -C 1

0 .0 9 5 0 .1 0 0 0 .1 0 5 0 .1 1 0 0 .1 1 5 0 .1 2 0 0 .1 2 5 0 .1 3 0[s ]- 5 0 0 0

- 3 7 5 0

- 2 5 0 0

- 1 2 5 0

0

1 2 5 0

2 5 0 0

3 7 5 0

5 0 0 0

[A ]


I1stPos 3843.5A:=


I1stNeg 4144.2-:=



Capacitor voltage·

(f ile Exampl1R.pl4; x-v ar t) v :C1 -

0.095 0.100 0.105 0.110 0.115 0.120 0.125 0.130[s]-60

-40

-20

0

20

40

60

[kV]

First positive peak: Some differences·from analytical

Overall positive peak:V1stPos 55.343kV:=56.876kV

Overall negative peak: First negative peak:

57.016- kV V1stNeg 6.592- kV:=

Repeat if capacitor is storing charge:

Vc1_0 Vm-:= Charged to negative peak voltage

Imax1Vm Vc1_0-

Z0

2 IPrms+:= Imax1 8192.9748A=

VcapPK 2 IPrms Xc1 Vm Vc1_0-( )+:= VcapPK 84.8509 kV=



ATP resultsUse the added circuit to precharge the capacitor. The resistor has a very small resistance·and is used to prevent a loop of switches.

Vpk=28.169kV

VS

2.11mH 19.2uF B2B1

C1 R2

40.11uF

22.29uF

C2

CSW1

T

(f ile Exampl1.pl4; x-v ar t) c:B1 -CSW1 c:B1 -C1

30 35 40 45 50 55 60 65 70[ms]-9000

-6000

-3000

0

3000

6000

9000

[A]

Current response:

Ipk 8172A:=

(f ile Exampl1.pl4; x-v ar t) v :C1 - v :VS

30 35 40 45 50 55 60 65 70[ms]-90

-60

-30

0

30

60

90

[kV]

Voltage response:

Vpk 84.462kV:=



EMTDC Results

40

.11

[uF

]

22

.29

[uF

]

R=

0 BR

K1

BRK1

TimedBreaker

LogicOpen@t0

IL1

BRK2

BRK2

TimedBreaker

LogicOpen@t0

VS

VC2

0.0000192 [H]

ILS

IL1

VC1

VC2

VC1

VS

RMSILS

0.002113 [H]

Main : Graphs

0.0900 0.0950 0.1000 0.1050 0.1100 0.1150 0.1200 0.1250 0.1300 ... ... ...

-5.0

-4.0

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

4.0

5.0

y (k

A)

ILS



Main : Voltages

0.0900 0.0950 0.1000 0.1050 0.1100 0.1150 0.1200 0.1250 0.1300 0.1350 0.1400 ... ... ...

-60

-40

-20

0

20

40

60

y (k

V)

VS VC1 VC2

EMTP-RV Results

+ AC1

?v34.5kVRMSLL /_0

+L1

2.11mH

+L2

19.2uH

+S

W2?i

10

|10

|0+S

W1?i

0.0

5|1

0|0

+ C1?v

40.11uF + C2?v

22.29uF

VS B2B1

transient r-l circuit example- updated: see page 9, 10-13 and 25 · 2020. 2. 1. · ece 524:...

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