279 39 Solutions Instructor Manual Chapter 10 Symmetrical Components Unsymmetrical Fault Analyses

8/19/2019 279 39 Solutions Instructor Manual Chapter 10 Symmetrical Components Unsymmetrical Fault Analyses

1/28

Chapter 10

10.1 Input:

» a=exp(i*2*pi/3);

» A=[1 1 1;1 a^2 a;1 a a^2];

» I=[15*exp(0);20*exp(i*120*pi/180);30*exp(i*(-120)*pi/180)];

The zero, positive and negative sequence currents are given by

» IS=inv(A)*I

I ao = -3.3333 - 2.8868i = 4.4096 ∠-139.11 0 A

I a1 = -3.3333 + 2.8868i = 4.4096∠

139.110

AI a2 = 21.6667 + 0.0000i = 21.6667 ∠00 A

10.2 Input:

» a=exp(i*2*pi/3);

» A=[1 1 1;1 a^2 a;1 a a^2];

» VS=[50*exp(i*180*pi/180);100*exp(i*90*pi/180);50*exp(i*0)];

Computation of phase to neutral voltages

» VP=A*VS

Van = 0.0000 + 100.00i = 100 ∠900 V

V bn = 11.6000 – 6.7000i = 13.3975 ∠ –30 0 V

Vcn = -161.600 – 93.300i = 186.6025 ∠ –150 0 V

10.3 Assume that the neutral is un-earthed. Hence, there is no path for the zero sequence

currents to flow. Therefore, there is no zero sequence component of voltages in the line

voltages.


2/28

If V c a is assumed to have an angle of 180 0, the phase angles of the other line voltages are

calculated using the cosine law which yields the following

V a b = 200 ∠70.53 0 V, V b c = 220 ∠-58.99 0 V, and V c a = 180 ∠180 0 V

Input data:

» a=exp(i*2*pi/3);

» Vab=200*exp(i*70.53*pi/180);Vbc=220*exp(i*(-58.99)*pi/180);

» Vca=180*exp(i*180*pi/180);

Computation of symmetrical components of line voltages

» Vab1=(Vab+a*Vbc+a^2*Vca)/3 = (8.7762e+001 +1.7896e+002i) V= 199.3220 ∠63.8767 0 V

» Vab2=(Vab+a^2*Vbc+a*Vca)/3 = (-21.1011 + 9.6002i) V

= 23.1823 ∠155.5361 0 V

Computation of sequence components of voltages to neutral

» Van1=i*Vab1 = (-1.7896e+002 +8.7762e+001i) V

» Van2=-i*Vab2 = (9.6002 +21.1011i) V

» Van0=0;

Computation phase ‘a’ to neutral voltage

» Van=Van1+Van2+Van0 = (-1.6936e+002 +1.0886e+002i)

= 201.3313 ∠147.2675 0 V

Computation of phase current in line ‘a’

» Ian=Van/10 = -16.9361 +10.8863i = 20.1331 ∠147.2675 0 A

Computation phase ‘b’ to neutral voltage

» Vbn=a^2*Van1+a*Van2+Van0 = (1.4241e+002 +1.0887e+002i)


3/28

= 179.2564 ∠37.3964 0 V

Computation of phase current in line ‘b’

» Ibn=Vbn/10 = 14.2411 +10.8867i = 17.9256 ∠37.3964 0 A

Computation phase ‘c’ to neutral voltage

» Vcn=a*Van1+a*2*Van2+Van0 = (-3.2672e+001 -2.0334e+002i)

= 205.9470 ∠-99.1282 0 V

Computation of phase current in line ‘c’

» Icn=Vcn/10 = -3.2672 -20.3339i = 20.5947 ∠-99.1282 0 A

10.4 Computation of phase sequence currents» Ian1=Van1/10 = (-17.8961 + 8.7762i) A

» Ian2=Van2/10 = (0.9600 + 2.1101i) A

» Ian0=Van0/10 = 0 A

Computation of power from sequence voltages and currents

» S=3*(Van0*conj(Ian0)+Van1*conj(Ian1)+Van2*conj(Ian2))

= 1.2080e+004 W

Verification by computing power fro line currents and resistance

» (abs(Ian)^2+abs(Ibn)^2+abs(Icn)^2)*10 = 1.1508e+004

The slight difference is due to round off error in computations

10.5 When the neutral is grounded, it provides a return path for the line currents to flow.

Thus if I n is the neutral and I a, I b, and I c are the line currents,

I n = I a + I b + I c

Also, I a0 =31

(I a + I b + I c).

That is, I n = 3 I a0


4/28

10.6 Computation of generator 1 reactance

» Xg1=0.12*(10.6/11)^2*(100/50) = 0.2229 pu

Computation of generator 2 reactance

» Xg2=0.10*(6.6/6.6)^2*(100/30) = 0.3333 pu

Computation of transformer 1 reactance

» Xt1=0.15*(220/220)^2*(100/75) = 0.2000 pu

Computation of transformer 2 reactance

» Xt2=0.15*(220/220)^2*(100/50) = 0.3000 pu

Computation of transformer 3 reactance» Xt3=0.15*(220/220)^2*(100/30) = 0.5000 pu

Computation of motor reactance

» Xm=0.08*(10/11)^2*(100/25) = 0.2645 pu

» Zbase=(220^2)/100 = 484 Ω

Computation of line 1 impedance

» Zl1=(50+i*100)/Zbase = (0.1033 + 0.2066i) pu


» Zl2=(25+i*50)/Zbase = (0.0517 + 0.1033i) pu


» Zl3=(45+i*60)/Zbase = (0.0930 + 0.1240i) pu

Representation of the load as a parallel R- X combination

» S=30*(0.8+i*0.6) = (24.0000 +18.0000i) MVA

» Rload=(11/11)^2*100/real(S) = 4.1667 pu

» Xload=(11/11)^2*100/imag(S) = 5.5556 pu


5/28

The per unit single line diagram of the network is shown below:


6/28

10.7 Here 0.Also.and,, b ====== cabcabcbbaa ZZZZZZZZZ Hence, the

sequence impedance is given by

[Z 012 ] =⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

012

201

120

sss

sss

sss

ZZZ

ZZZ

ZZZ

(I)

where

( ) ( )

( ) [ ]( ) ( )

( ) [ ]( ) ( )babacbas

bababbas

babbas

ZZZaaZaZZaZZ

ZZZaaZZaaZZZ

ZZZZZZ

−=++=++=

−=++=++=

+=++=

31

31

31

31

31

31

231

31

222

221

0

Substitution for the sequence impedances in (I) leads to

[Z 012 ] =

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

+−−−+−−−+

=⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

bababa

bababa

bababa

ZZZZZZ

ZZZZZZ

ZZZZZZ

Z

Z

Z

2

2

2

31

2

1

0

10.8 Connection of the neutral point to ground provides a path for the zero sequence

currents to flow. Staring from first principle, the sequence impedance matrix is written as

[ ] [ ]( )

( )( )

[ ]AZZ

ZZ

ZZ

AZ⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

+

+

+

= −

gb

gb

ga

00

00

001

012

On performing the indicated matrix multiplications leads to the final result

[Z 012 ] =( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

++−−

−++−−−++

=⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

gbababa

bagbaba

babagba

ZZZZZZZ

ZZZZZZZZZZZZZZ

Z

ZZ

32

3232

31

2

1

0

The expression for the neutral voltage is given by


7/28

gan ZIV 30 ×=

10.9 Assume a system base of 100 MVA. The voltage base for the various parts of the

network is as per the transformation ratios of the transformers.

Generator 1 per unit reactance: 5 .0;25.0 021 === X X X

Generator 2 per unit reactance: 3333.0;25.0 021 === X X X

Transformer 1-2 per unit reactance: 2 .0021 === X X X

Transformer 1-5 per unit reactance: 1143.0021 === X X X



Transmission line 2-3 per unit impedance:

( )( )

5165.01722.0132

10090302

j j

Z +=×+=

Transmission line 5-6 per unit impedance:

( )( )

1653.01033.0220

10080502

j j

Z +=×+=

Series representation of load

( ) ( )( )( ) ( )( ) ( )( )( ) ( ) pu2.110301040

1030101000.1

pu6.110301040

1040101000.1

0.300.40)6.08.0(50

2626

662

2626

662

=×+×

××=

=×+×

××=

+=+×=

X

R

j jS

pu05.0= f Z

Assembly of positive and negative sequence networks:

The positive sequence network is shown in the figure below.


8/28

Thus, )3937.01133.0(21 j Z Z +==

Computation of zero sequence network:

Based on the types of transformer connections, the zero sequence network is

shown below.


9/28

From the diagram, it is seen that )4506.01595.0(0 j Z +=


10/28

The network below depicts the inter-connection of the sequence networks for a

SLG fault.

Take the node 4 voltage (11 kV) as the fault voltage pu 0.1= f V (reference

phasor) for fault calculations.

( ) pu j j j j Z Z Z Z

V I

f

f a

°−∠=−=

++++++=

+++=

5856.667412.06802.02946.0

15.0)4506.01595.0()3937.01133.0()3937.01133.0(0.1

30211

Base current kA I Base 2486.510113

101003

6=

×××=

021 aaa III ==

Fault current pu pu I I aa °−∠=−∠×== 5856.662236.25856.667412.0330

1

kA°−∠=−∠×= 5856.666708.115856.662236.22486.5 0

Computation of sequence voltages:

» Va0=-Ia0*Z0 = (–0.3535 – 0.0242i) pu = 0.3542 ∠ –176.0837 ° pu

» Va1=Vf-Ia1*Z1 = (0.6988 – 0.0389i) pu = 0.6999 ∠ –3.1862 ° pu

» Va2=-Ia2*Z2 = (–0.3012 – 0.0389i) pu = 0.3037 ∠ –17.6410 ° pu

Computation of line voltages:

» A=[1 1 1;1 a^2 a;1 a a^2];


11/28

» VS=[Va0;Va1;Va2];

» Vabc=A*VS

Vabc = =⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

c

b

a

VV

V

⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

°∠°−∠

°∠

=⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

+ 122.09321.0395973.1221.0148

66.5603-0.1112

0.8807i0.5523-0.8514i-0.5523-

0.1020i-0.0442

Line voltage aV is also given by

°−∠×=×= 5856.667412.015.01a f a I Z V = 0.1112 ∠ –66.5856 °

The above verifies the computations.

10.10 (i) 2LL fault:

The interconnection of the sequence networks is drawn below:


12/28


13/28

» Zf=0.05;

Computation of sequence currents:

» Ia1=Vf/(Z1+Z2+Zf) = (0.3971 - 1.1305i) = 1.1982 ∠ –70.6445 ° pu

» Ia2=-Ia1 = (–0.3971 + 1.1305i) = 1.1982 ∠109.3555 ° pu

Computation of sequence voltages

» Va0=0;

» Va1=Vf–Ia1*Z1 = (0.5099 – 0.0283i) pu

» Va2=-Ia2*Z2 = (0.4901 + 0.0283i) pu

Computation of phase voltages:


» Vabc=A*VS

Vabc = =⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

c

b

a

VV

V

⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

°∠°−∠

°∠

=⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

+−−

177.81670.4514178.20580.5492

01.0000

0.0172i0.45100.0172i-0.5490

1.0000

Computation of line voltages

» Vab=Vabc(1) –Vabc(2) = (1.5490 + 0.0172i) = 1.5490 ∠0.6361 ° pu

» Vbc=Vabc(2) –Vabc(3) = (–0.0979 – 0.0344i) = 0.1038 ∠ –160.6445 ° pu

» Vca=Vabc(3) –Vabc(1) = (–1.4510 + 0.0172i) = 1.4511 ∠179.3210 ° pu

Computation of line currents

» Ib=–i*sqrt(3)*Ia1 = (–1.9581 – 0.6878i) = 2.0754 ∠ –160.6445 ° pu

» Ic=-Ib = (1.9581 + 0.6878i) = 2.0754 ∠19.3555 ° pu


14/28

Verification of computations

» Vbc=Zf*Ib = (–0.0979 – 0.0344i) = 0.1038 ∠ –160.6445 ° pu

Computation of actual line currents

» IBase*abs(Ib) = 10.8929 kA

(ii) 2LG fault

The inter-connected sequence networks are shown below:

Computation of sequence currents

» Ia1=Vf/((Z1+Zf)+(Z2+Zf)*(Z0+Zf)/(Z2+Z0+2*Zf)) = (0.5939 –1.4048i) pu

» Ia2=–((Z0+Zf)/(Z2+Z0+2*Zf))*Ia1 = (–0.3050 + 0.7624i) pu

» Ia0=–((Z2+Zf)/(Z2+Z0+2*Zf))*Ia1 = (–0.2889 + 0.6424i) pu

» Ia0+Ia1+Ia2 = 0; which verifies the computations

» IS=[Ia0;Ia1;Ia2];

» Iabc=A*IS

Iabc = =⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

c

b

a

I

I

I

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

++−

−

1.7420i1.4435

0.1851i2.3101

0.0000i0.0000

» 3*Ia0 = (–0.8666 + 1.9271i) pu

» Iabc(2)+Iabc(3) = (–0.8666 + 1.9271i)

Since 03 acb III =+ , the computations are again verified.

Computation of sequence voltages

» Va0=–Ia0*Z0 = (0.3355 + 0.0277i) pu

» Va1=1.0–Ia1*Z1 = (0.3797 – 0.0747i) pu


15/28

» Va2=-Ia2*Z2 = (0.3347 + 0.0337i) pu

Computation of phase voltages


» Vabc=A*VS

Vabc = =⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

c

b

a

V

V

V

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

°∠°∠°−∠

=⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

++−

−

3532.500.1131

4195.1750.1159

7226.01.0500

0.0871i0.0722

0.0093i0.1155

0.0132i1.0499

pu


» Vab=Vabc(1) –Vabc(2) = (1.1654 – 0.0225i) = 1.1656 ∠ –1.1058 ° pu

» Vbc=Vabc(2) –Vabc(3) = (–0.1877 – 0.0778i) = 0.2032 ∠ –157.4722 ° pu

» Vca=Vabc(3) –Vabc(1) = (–0.9777 + 0.1003i) = 0.9829 ∠174.1404 ° pu

Computation of actual line currents

» IBase*abs(Iabc)

kA

I

I

I

c

b

a

⎥⎥

⎥

⎦

⎤

⎢⎢

⎢

⎣

⎡

°∠°∠°−∠

=⎥⎥

⎥

⎦

⎤

⎢⎢

⎢

⎣

⎡

3532.5011.8743

4195.17512.1637

9638.750.0000

(iii) Three phase fault

The sequence networks for a three-phase fault are drawn below.

Computation of sequence currents

» Ia1=Vf/(Z1+Zf) = (0.8989 – 2.1672i) = 2.3462 ∠ –67.4722 ° pu

» Ia=Ia1;

Computation of actual current

» IBase*abs(Ia) = 12.3142 ∠ –67.4722 ° kA

» angleIb=angleIa+(angle(a^2)*180/pi) = ∠ –187.4722 °


16/28

» angleIc=angleIa+(angle(a)*180/pi) = ∠52.5278 °

Computation of voltages

» Va=Zf*Ia = (0.0449 – 0.1084i) = 0.1173 ∠ –67.4722 ° pu

» angleVb=angleVa+angle(a^2)*180/pi = ∠ –187.4722 °

» angleVc=angleVa+angle(a)*180/pi = ∠52.5278 °

10.11 Input data from problem 10.9

» Ia1=0.2946–i*0.6802;Ia2=Ia1;Ia0=Ia1;

» Va0=-0.3535–i*0.0242;

» Va1=0.6988–i*0.0389;

» Va2=-0.3012–i*0.0389;

» a=exp(i*2*pi/3);

» A=[1 1 1;1 a^2 a;1 a a^2];

Computation of positive and negative sequence currents flowing in the circuit on the

generator side

» Ia1g=((1.6+i*1.2)/(i*0.125+(0.0646+i*0.3551)+1.6+i*1.2))*Ia1

= (0.1625 – 0.6054i) pu

» Ia2g=Ia1g;

Computation of zero sequence current flowing from node 5

» Ia05=((1.6+i*1.2)/(i*0.1143+(0.1033+i*0.1653)+i*0.3+1.6+i*1.2))*Ia0

= (0.1459 – 0.5839i) pu

Computation of positive and negative sequence currents flowing from node 5.


17/28

Ia15=((0.1722+i*0.5165+i*0.4)/(i*0.1143+0.1033+i*0.1653+i*0.3+0.1722+i*0.5165

+i*0.4))*Ia1g = (0.0983 – 0.3715i) pu

» Ia25=Ia15;

Computation of phase currents flowing from node 5

» IS5=[Ia05;Ia15;Ia25];

» Iabc5=A*IS5

Iabc5 =

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

°−∠

°−∠°−∠

=

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

−

−−

=

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

3481.772177.0

3481.772177.0

5286.753703.1

0.2124i0.0477

0.2124i0.0477

1.3268i0.3424

c

b

a

I

I

I

pu

Computation of sequence voltages at node 5

» Va15=1.0+Ia15*(0.1033+i*0.1653+i*0.3) = (1.1830 + 0.0073i) pu

» Va25=Ia25*(0.1033+i*0.1653+i*0.3) = (0.1830 + 0.0073i) pu

» Va05=(0.1033+i*0.1653+i*0.3)*Ia05 = (0.2867 + 0.0076i)

Computation of phase voltages at node 5» VS5=[Va05;Va15;Va25];

» Vabc5=A*VS5

Vabc5 =⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

°∠°−∠

°∠=⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

+−−−

+=⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

5805.1149526.0

5927.1149522.0

7721.06529.1

0.8663i0.3963

0.8658i0.3963

0.0223i1.6527

c

b

a

V

V

V

pu

Computation of line voltages at node 5

» Vab=Vabc5(1) –Vabc5(2) = (2.0490 + 0.8881i) = 2.2332 ∠23.4324 ° pu

» Vbc=Vabc5(2) –Vabc5(3) = (–0.0000 – 1.7321i) = 1.7321 ∠ –90.0000 ° pu

» Vca=Vabc5(3) –Vabc5(1) = (–2.0490 + 0.8440i) = 2.2160 ∠157.6129 ° pu


18/28

10.12 Computation of positive sequence impedance

» Zp=(0.1722+i*0.9165)*(0.1033+i*0.5796)/(0.1722+i*0.9165+0.1033+i*0.5796)

= (0.0646 + 0.3551i) pu

» Z1=i*0.125*(Zp+1.6+i*1.2)/(i*0.125+Zp+1.6+i*1.2)

= (0.0046 + 0.1203i) pu

The positive sequence network for a 3-phase fault at node-1 is shown in the

figure below.


19/28

Computation of base current

» IBase=100*10^6/(sqrt(3)*6.6*10^3) = 8.7477 kA

» Zf=0.05;

Computation of voltage (pre-fault) at node-1

» Vnode=1.0+(1.0/(1.6+i*1.2))*Zp = (1.1324 + 0.1226i) = 1.1390 ∠6.1815 °

Set the fault voltage Vf equal to the magnitude of voltage at node-1, that is,

Vf = 1.1390 ∠0°

» Vf=abs(Vnode);

Computation of positive sequence current

» Ia1=Vf/(Z1+Zf) = (3.5650 – 7.8479i) = 8.6197 ∠ –65.5699 ° pu

» Ia=Ia1;

Computation of actual current

» IBase*abs(Ia) = 75.403 kA

» angleIb=angleIa+(angle(a^2)*180/pi) = ∠ –185.5699 °

» angleIc=angleIa+(angle(a)*180/pi) = ∠ 54.4301 °

Computation of voltages

» Va=Zf*Ia = (0.1782 – 0.3924i) = 0.4310 ∠ –65.5699 ° pu

» angleVb=angleVa+angle(a^2)*180/pi = ∠ –185.5699 °

» angleVc=angleVa+angle(a)*180/pi = ∠54.4301 °

10.13 Input data:

» Zl2=i*0.2+0.1722+i*0.5165+i*0.2;


20/28

» Zl1=i*0.1143+0.1033+i*0.1653+i*0.3;

» Zlo=1.6+i*1.2;Zg=i*0.125;

» IBase=8.7477;Zf=0.05; Va=0.1782–i*0.3924;

Ia = (3.5650 – 7.8480i);

Parallel combination of lines 1 and 2

» Zp=Zl1*Zl2/(Zl1+Zl2) = (0.0646 + 0.3551i);

Component of current in the load side of the network

» Ialo=(Zg/(Zg+Zp+Zlo))*Ia = (0.4258 – 0.1620i) pu

= 3.9853 ∠ –20.8347 ° KAComponent of current in the generator side

» Iag=((Zp+Zlo)/(Zg+Zp+Zlo))*Ia = (3.1392 – 7.6860i) pu

= 72.6266 ∠ –67.7833 0 KA

Voltage at node-4

» Van4=Va+Ialo*Zp = (0.2632 – 0.2517i) = 0.3642 ∠ –43.7142 ° pu

Computation of current in line-1

» Ial1=(Zl2/(Zl1+Zl2))*Ialo = (0.2606 – 0.1003i) pu

= 2.4430 ∠ –21.0420 ° KA

Computation of current in line-2

» Ial2=(Zl1/(Zl1+Zl2))*Ialo = (0.1651 – 0.0618i) pu

= 1.5423 ∠ –20.5063 KA

Computation of voltage at node-2

» Van2=Va+Ial2*(i*0.2) = (0.1906 – 0.3594i) = 0.4068 ∠ –62.0658 ° pu



21/28

» Van3=Va+Ial2*(i*0.2+0.1722+i*0.5165) = (0.2509 – 0.2847i) pu

= 0.3795 ∠ –48.6134 ° pu

The computations are verified by computing the voltage at node-4 via the line-2 circuit as

under

» Van4=Va+Ial2*Zl2 = (0.2632 – 0.2517i) = 0.3642 ∠ –43.7142 ° pu


» Van5=Va+Ial1*(i*0.1143) =(0.1897 – 0.3626i) = 0.4092 ∠ –62.3883 ° pu


» Van6=Va+Ial1*(i*0.1143+0.1033+i*0.1653) = (0.2332 – 0.3299i)

= 0.4040 ∠-54.7471 ° pu

10.14 Computation of positive sequence network (reference figs.(a) and (b))

» Z14=0.1722+i*0.9165;

» Z4F=0.0517+i*0.3826;

» ZF1=0.0517+i*0.1970;

» ZFs=Z4F*ZF1/(Z14+Z4F+ZF1) = (0.0107 + 0.0506i) pu

» Z1s=ZF1*Z14/(Z14+Z4F+ZF1) = (0.0321 + 0.1206i) pu

» Z4s=Z14*Z4F/(Z14+Z4F+ZF1) = (0.0325 + 0.2344i) pu

» Z1=ZFs+(Z1s+i*0.125)*(Z4s+1.6+i*1.2)/(Z1s+Z4s+i*0.125+1.6+i*1.2)

= (0.0557 + 0.2737i)

» Z2=Z1;

Computation of zero sequence network

» Z0=(Z4F+1.6+i*1.2)*(ZF1)/(Z4F+ZF1+1.6+i*1.2) = (0.0559 + 0.1807i)

Computation of fault and ground impedances


22/28

» Zf=5*100/(220^2) = 0.0103 pu

» Zg=10*100/(220^2) = 0.0207 pu

The interconnection of the sequence networks is shown below


23/28


24/28

Computation of sequence fault currents

» Vf=1.0;

» Ia1=Vf/((Z1+Zf)+(Z2+Zf)*(Z0+Zf+3*Zg)/(Z2+Z0+2*Zf+3*Zg))

= (0.7345 – 2.3521i) pu

» Ia2=-((Z0+Zf+3*Zg)/(Z2+Z0+2*Zf+3*Zg))*Ia1 = (–0.0989 + 1.1002i) pu

» Ia0=-((Z2+Zf)/(Z2+Z0+2*Zf+3*Zg))*Ia1 = (–0.6356 + 1.2519i)Verification of computations

» Ia0+Ia1+Ia2 = 0

Computation of line fault currents

» IS=[Ia0;Ia1;Ia2];

» Iabc=A*IS

Iabc =⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

°∠°∠

°∠=⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

++

+=⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

9268.513.3023

6605.1634.1093

00.0000

2.5996i2.0364

1.1560i3.9433-

0.0000i0.0000

c

b

a

I

I

I

Computation of fault sequence voltages

» Va0=-Z0*Ia0 = (0.2617 + 0.0449i) pu

» Va1=Ea-Z1*Ia1 = (0.3153 – 0.0699i) pu

» Va2=-Z2*Ia2 = (0.3067 – 0.0343i) pu

Computation of voltages at fault point


» Vabc=A*VS


25/28

Vabc =⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

∠

∠

−∠=⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

++

−=⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

0

0

0

9696.990.1061

8273.1310.1202

8380.30.8856

1045001840

0895008010

0593088360

i..-

i..-

i..

V

V

V

c

b

a


» Vab=Vabc(1) –Vabc(2) = (0.9638 – 0.1488i) = 0.9752 ∠ –8.7779 ° pu

» Vbc=Vabc(2) –Vabc(3) = (–0.0618 – 0.0149i) = 0.0635 ∠ –166.4276 ° pu

» Vca=Vabc(3) –Vabc(1) = (–0.9020 + 0.1637i) = 0.9167 ∠169.7115 ° pu

If Z g is set equal to ∞ , it is seen from the interconnected sequence network that the zero

sequence network is on open circuit. Therefore, there will be no zero sequence currents

and the boundary conditions are similar to a double line fault.

10.15 Computation of positive sequence network (reference figs.(a) and (b))

» Z1F=i*0.2+(0.1722+i*0.5165)/4 = (0.0430 + 0.3291i) pu

» Z4F=i*0.2+3*(0.1722+i*0.5165)/4 = (0.1291 + 0.5874i) pu

» Z14=i*0.1143+0.1033+i*0.1653+i*0.3 = (0.1033 + 0.5796i) pu

» Z1s=Z14*Z1F/(Z1F+Z4F+Z14) = (0.0159 + 0.1275i) pu

» ZFs=Z1F*Z4F/(Z1F+Z4F+Z14) = (0.0215 + 0.1295i) pu

» Z4s=Z4F*Z14/(Z1F+Z4F+Z14) = (0.0487 + 0.2276i) pu

» Z1=ZFs+(i*0.125+Z1s)*(1.6+i*1.2+Z4s)/(i*0.125+Z1s+1.6+i*1.2+Z4s)

= (0.0539 + 0.3605i) pu

» Z2=Z1;

Computation of fault impedance

» Zf=5*100/(132^2) = 0.0287 pu

The inter-connection of the sequence networks is as under


26/28

» Vf=1.0;

Computation of sequence fault currents


27/28


28/28

» Ic=–Ib = (2.3195 + 0.4390i) = 2.3606 ∠10.7185 ° pu

» Zf*Ib = (–0.0666 – 0.0126i) (II)

Comparing (I) and (II) shows that Vbc = Zf*Ib which verifies the computations.

In case Zf is set equal to zero, it is a dead short circuit between two lines. The fault

currents and voltages at the fault point can be determined by following the procedure

outlined above by setting Zf = 0.

A couple of problems in the book for this chapter had a few errors. Herewith, pleasefind the errata:

Problem As printed in book Correction______________

10.3 V ab = 200∠0° V ab = 200V bc = 220∠120 ° V bc= 220V ca = 180∠ –60 ° V bc= 180

Add the text at the end of the problem:[Note: Assume V ca = V ca ∠120 °, use thecosine law for the triangle to calculate the

phase angles of the other line voltages]

10.9 SLG occurs at node 3 SLG occurs at node 4___________________________________________________________________-

Documents

279 39 Solutions Instructor Manual Chapter 10 Symmetrical Components Unsymmetrical Fault Analyses