Simulation Analysis of Lighting Switching ImpulseVoltages

Ahmad Shaiful Amiruddin Taher(1071118187)Irwan Ramli(1081116195)Ariff Musa(1051105968)

December 27, 2010

Abstract

This report serve the purpose of the simulation and analysis of lighting and switch-ing impulse voltages as part of our High Voltage Engineering subject demand. Theobjectives is to simulate switching and lightning impulse voltages, to study the be-havior of circuit parameters on the impulse wave shape (tf&tt) and to be able tosimulate and compute the required impulse peak voltage for given front and tailtimes by varying circuit parameters.

Contents

1 Introduction 2

2 Question 1 32.1 Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 MATLAB® simulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3 Question 2 53.1 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.2 Matlab® simulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.3 the Matlab program to compute for lightning impulse wave form . . . . . . 103.4 Analysis and verification of the result . . . . . . . . . . . . . . . . . . . . . 12

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Figure 1: This is the caption for the picture.

1 Introduction

Power systems equipment must withstand not only the rated voltage (Vm), which corre-sponds to the highest voltage of a particular system, but also overvoltages. Accordingly,it is necessary to test high voltage equipment during its development stage and prior tocommissioning. The magnitude and type of test voltage varies with the rated voltage ofa particular apparatus. The standard method of measurement of high voltage and thebasic techniques for application to all types of apparatus for alternating voltages, directvoltages, switching impulse voltages and lightning impulse voltages are laid down in therelevant national and international standards

• Testing with lightning impulse voltages. Lightning strokes terminating on trans-mission lines will induce steep rising voltages in the line and set up travelling wavesalong the line and may damage the systems insulation. The magnitude of theseovervoltages may reach several thousand kilovolts, depending upon insulation. Ex-haustive measurements and long experience have shown that lightning overvoltagesare characterized by short front duration, ranging from a fraction of a microsecondto several tens of microseconds and then slowly decreasing to zero. The standard im-pulse voltage has been accepted as an aperiodic impulse that reaches its peak valuein 1.2s and then decreases slowly (in about 50s) to half its peak value. In additionto testing equipment, impulse voltages are extensively used in research laboratoriesin the fundamental studies of electrical discharge mechanism, notably when the timeto breakdown is of interest.

• Testing with switching impulse voltages. Transient overvoltages accompanyingsudden changes in the state of power systems, e.g. switching operations of faults,are known as switching impulse voltages. It has become generally recognized thatswitching impulse voltages are usually the dominant factor affecting the design ofinsulation in high voltage power system for rated voltages of about 300kV and above.Accordingly, the various international standards recommend that equipment designedfor voltages above 300kV be tested for switching impulses. Although the waveshapeof switching overvoltages occurring in the system may vary widely, experience hasshown that for flashovevr distances in atmospheric air of practical interest the lowestwithstand values are obtained with surges with front times between 100 and 300s.Hence, the recommended switching surge voltage has been designated to have a fronttime of about 250s and half value time of 2500s.

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2 Question 1

A Cockcroft Walton type voltage tripler circuit has C1 = C2 = 0.025 µF and C3 = 0.15µF.The supply voltage is 105 sin ωt kV, where ω= 314. If the load current is 25 mA,determine:

(a) ripple voltage

(b) voltage drop

(c) average output voltage

(d) ripple factor

(e) simulate the circuit using MATLAB simulink to verify above calculations

2.1 Calculations

• ripple voltage

∆V =

(I1

f

) [2

C1

+1

C3

]

=(

25mA

50

) [2

0.025µF+

1

0.15µF

]= 43.33kV (1)

• Voltage drop

δV +∆V

2=

(I1

f

) [1

C2

+1

C2

+1

2C3

]

=

(25mA

3142π

)[1

0.025µF+

1

0.025µF+

1

2(0.01µF )

]= 41.6878kV (2)

• Average output voltage

Vav = 3Vs(max) −(δV +

∆V

2

)= 3(105kV − 41.6878kV )

= 273.3122kV (3)

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• Ripple factor

RF =

(δV + ∆V

2

)3Vs(max)

=41.6878kV

3(105kV )

= 0.1323 (4)

2.2 MATLAB® simulations

• circuit diagram

4

• graph plotted

Figure 2: Graph plotted from Matlab® simulink

3 Question 2

Figure 3 shows the equivalent circuit of a single stage impulse generator with input voltageof 300 kV. The value of each component used in this circuit is C1= 7 nF, C2= 300 pF,R1=1.6 kΩ, R2= 6.5 kΩ.

(a) Obtain the Vout(t) and its efficiencies.

(b) Using the MATLAB® simulink, plot the waveform.

(c) Using the file in the Energy Systems Lab (name of the file is IGA), modify the Matlabprogram to compute for lightning impulse wave form

(d) Analyse and verify the results

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Figure 3: Single stage impulse generator with input voltage of 300 kV

3.1 Solutions

• Obtain the Vout(t) and its efficiencies.

K =Vc

R1C2

=300kV

(1.6kΩ)(200pF )

= 6.25× 1011

α =R1C2 +R2C2 +R2C1

R1R2C1C2

=(1.6kΩ)(200pF ) + (6.5kΩ)(300pF ) + (6.5kΩ)(7nF )

(1.6kΩ× 6.5kΩ× 7nF × 300pF )

= 2.1946× 106

β =1

R1R2C1C2

=1

(1.6kΩ× 6.5kΩ× 7nF × 300pF )

= 4.5788× 1010

Vout =K

s2 + αs+ β

Vout =6.25× 1011

[s2 + 2.1946× 106s+ 4.5788× 1010]

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B(s) = [s2 + 2.1946× 106s+ 4.5788× 1010] = 0

a = 1

b = 2.1946× 106

c = 4.5788× 1010

S =−b±

√b2 − 4ac

2a

S =−b±

√b2 − 4ac

2a

S =−2.1946× 106 ±

√(2.1946× 106)2 − 4(1)(4.5788× 1010)

2(1)= −21066,−2173534

Vout = 6.25× 1011[

A

s+ 21066+

B

s+ 2173534

]

by comparison with:

Vout =6.25× 1011

[s2 + 2.1946× 106s+ 4.5788× 1010]

we get

A+B = 0 (5)

2173534A+ 21066B = 1 (6)

solve for (5) and (6):

A =1

2152468= 4.6458× 10−7

B =−1

2152468= −4.6458× 10−7

Vout = 6.25× 1011

[4.6458× 10−7

s+ 21066− 4.6458× 10−7

s+ 2173534

]

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by taking Laplace Transform.

Vout = 290.3625(e−21066t − e−2173534t)kV

dVoutdt

=d

dt290.3625(e−21066t − e−2173534t)kV = 0

−21066e−21066t +−2173534e−2173534t = 0e−21066t

e−2173534t=

21066

2173534e−2152468t = 9.6920× 10−3

lne−2152468t = ln9.6920× 10−3

t = 2.1540µs

Vout = 290.3625(e−21066(2.1540µs) − e−2173534(2.1540µs))kV

Vout = 274.78095kV

η =VoutVin× 100%

=274.08795kV

300kV× 100%

= 91.59365%

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3.2 Matlab® simulations

Figure 4: Circuit design with Matlab® Simulink

Figure 5: Wavefrom generated by Matlab® Simulink

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3.3 the Matlab program to compute for lightning impulse waveform

This program Calculates Front time and tail time of Lighting Impulse Generator circuitEET 4106 High Voltage Engineering

r1=input(’Enter wave front resistor, R1 on IG side = ’);

r2=input(’Enter wave tail resistance, R2 = ’);

c1=input(’Enter capacitance of the impulse generator, C1 = ’);

c2=input(’Enter capacitance of the test object, C2 = ’);

Vin=input(’Enter the charging voltage of impulse generator, Vin = ’);

a = (1/(r1*c2))+(1/(r2*c2))+(1/(r1*c1));

b = 1/(r1*r2*c1*c2);

c = r1*c2;

alpha = a/2;

beta = sqrt((a/2)^2 - b);

y1 = alpha - beta;

y2 = alpha + beta;

tf = (1/(y2-y1))*log(y2/y1);

K = 0.7/((y1)*tf);

tt = K*tf;

A=(Vin/(c*(y2-y1)));

step=0.1e-6;

t(1) = 0;

for i = 1 : 1000

e(i) = A*(exp(-y1*t(i))-exp(-y2*t(i)));

t(i+1) = t(i) + step;

end

t = t(1:length(t)-1);

plot(t,e);grid;

title(’LIGHTNING IMPULSE VOLTAGE WAVEFORM’);

xlabel(’Time in seconds’);

ylabel(’Magnitude’);

Vp = max(e);

disp(sprintf(’a = (1/(r1c2) + 1/(r2c2) + 1/(r1c1)) = \%g’,a))

disp(sprintf(’b = (1/r1r2c1c2) = \%g’,b));

disp(sprintf(’y1 = \%g’,y1));

disp(sprintf(’y2 = \%g’,y2));

disp(sprintf(’K = \%g’,K));

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disp(sprintf(’TIME TO FRONT, Tf = \%g sec’,tf));

disp(sprintf(’TIME TO TAIL, Tt = \%g sec’,tt));

disp(sprintf(’Impulse Peak Waveform Voltage, Vp = \%f’, Vp));hold on;

Vtp = 0.5*Vp;

for i = 1 : 1000

hp(i) = Vtp;

end

plot(t,hp,’.r’);

for i = 1 : 1000

if e(i) == Vtp

if t(i) > 2e-6

tt = t(i);

disp(sprintf(’Tail time of the impulse wave, Tt = %g’, tt));

end

end

end

Figure 6: Lightning Impulse Voltage Wavefrom generated by Matlab® Simulink

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3.4 Analysis and verification of the result

α =1

R1C1

+1

R2C1

+1

R1C2

=1

(1600)(7nF )+

1

(6500)(7nF )+

1

(1600)(300pF )

= 2.1946× 106

α =α

2= 1.0973× 106

β =1

R1R2C1C2

=1

2.184× 10−11

= 4.5788× 1010

β =√α2 − b

=√

1.0973× 1062 − 4.5788× 1010

= 1.0762× 106

α1 = α− β= 1.0973× 106 − 1.0762× 106

= 21100

α1 = α− β= 1.0973× 106 − 1.0762× 106

= 21100

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α2 = α + β

= 1.0973× 106 + 1.0762× 106

= 2173500

Vpeak =

[Vo

2R1C2β

]e−α1tf − e−α2tf

= [235.76k]e−21100tf − e−2173500tf

dVpeak(tf )

dtf=

d

dtf[235.76k]e−21100tf − e−2173500tf = 0

−e−21100tf = e−2173500tf

e−2152400tf = 9.7078× 103

lne−2152400tf = ln9.7078× 103

tf = 2.1533× 10−6

Vpeak = [235.76k]e−21100(2.1533×10−6) − e−2173500(2.1533×10−6)= 223.1051kV

η =VoutVin× 100%

=223.1051

300× 100%

= 74.3684%

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