fault calculation

FAULT CALCULATION

Sompol C

Power System Fault Analysis

All protection Engineers should have and understanding to :- Calculate power system currents and voltages during fault condition Check the breaking capacity of switchgear is not exceeded

Determine the quantities which can be used by relays to distinguish between healthy (i.e. loaded ) and fault conditions

Appreciate the effect of the method of earthing on the detection of earth faults

Selected the best relay characteristics for fault detection

Ensure that load and short circuit ratings of plant are not exceeded Select relay settings for fault detection and discrimination

Understand principles of relay operation

Conduct post fault analysis

Power System Fault Analysis

Power System Fault Analysis also used to :- Consider stability conditions

Required fault clearance times

Need for 1 phase or 3 phase auto

reclose

Computer Fault Calculation Programmes

Widely available, particularly in large power utilities

Powerful for large power systems

Sometimes overcomplex for simple circuits

Not always use friendly

Sometimes operated by other department and not directly available to protection engineers

Programme calculation methods :-understanding is important Need for ‘by hand’spot checks of calculation

Per-Unit System Per-unit impedance of transformer is the same regardless of

whether it is determined from ohmic values referred to the high or low side of transformer .

Only one MVA Base is used for all equipments in the network . The impedance of each equipment can be merged directly . Manufacturer usually specify the impedance in per-unit on the

base of the nameplate rating .

~Y 11kV 20MVA 11/132kV 50MVA Z G = 0.3PU Z T = 10%Z L = 40?Y 132/33kV 50MVA

Z T = 10%

O/H Line Z L = 8?

Feeder

Different voltage

How do we Analyses ?

Referring Impedance R 1X 1R 2X 2

N :1Ideal

Transformer

R 1+N 2R 2X 1+N 2X 2R 1/N 2+R 2X 1/N 2+X 2

Primary Secondary

Consider the equivalent CCT refer to :-

Base Quantities and Per Unit Values Particularly useful analyzing large systems with several voltage

levels .

All system parameters referred to common base quantities .

Base quantities fixed in on part of system .

Base quantities at other parts at different voltage levels depend

on ratio of intervening transformers .

~Y 11kV 20MVA 11/132kV 50MVA Z G = 0.3PU Z T = 10%Z L = 40?Y 132/33kV 50MVA

Z T = 10%

O/H Line Z L = 8?

Feeder

Base Quantities and Per Unit Values Base quantities normal used :- Base MVA =MVA b =3?MVA Constant all voltage levels Value –MVA rating of largest item of plant or 100 MVA

Base voltage =kV b =?/?voltage in kV Fixed in one part of system

This value is referred through transformers to obtain base voltage on other parts of system

Base voltage on each side of transformer are in same ratio as voltage ratio.

Base Quantities and Per Unit Values Other base quantities :-

Base Impedance =Z b =

(kV b )2 in Ohms

MVA b

Base Current =I b =

MVA b in kA

√3 x kV b

Per Unit System

Use to simplify calculation on system with more than two voltage Definition :

P.U.Value =Actual Value

Base Value

Base Quantities and Per Unit Values Per unit values=Actual Value

Base Value

Per unit MVA =MVA p.u.= MVA a

MVA

b

Per unit voltage =kV p.u.=kV a

kV

b

Per unit impedance=Z p.u.= Z a=Z a MVA b

Z b (kV

b

)2

Per unit current=I p.u.= I a

I

b

Transformer percentage impedance If Z T = 5%with secondary short circuit

5% V (rated)produces I (rated)in secondary

V (rated)produce 100 x I (rated)

5

If Source impedance Z S = 0

Fault current =20 x I (rated)

Fault power =20 x kVA (rated)

Z T is base on I (rated)& V (rated)

i.e. Base on MVA (rated)& kV (rated)

Is same value viewed from either side of transformer Conversion of Per Unit from one Set Quantities to Another

Z b1Z b2

MVA b1 MVA b2

kV b1kV b2

Actual Z = Z a

Z P.U.1=Z a

Z b1

Z P.U.2=Z a = Z P.U.1 x Z b1

Z b2Z b2

= Z P.U.1x (kV b1)2x MVA b2MVA b1 (kV b2)2= Z P.U.1x MVA b2 x (kV b1)2

MVA b1 (kV b2)2

Z P.U.1Z P.U.2

Example 0

Incorrect selection of kV b

11.8kV 132kV 11kV

Correct selection of kV b

132x11.8 132kV 11kV

141

=10.05kV

Alternative correct selection

11.8kV 141kV 141x11 = 11.75kV 132

11.8/141kV OHL Distribution System 11.8kV 132/11kV ~Base voltage on each side of transformer must be in the same ratio as voltage ratio of transformer Example 1

kV b 1113233MVA b 505050Z b = 2.42 ?348.5 ?21.8 ?(kV b )2

MVA b

I b =2624 A 219 A 875 A MVA b

√3kV b

Z P.U.O.3 x 50408

20348.521.8

= 0.75 P.U.= 0.115 P.U.= 0.367 P.U.

~Y 11kV 20MVA 11/132kV 50MVA Z G = 0.3PU Z T = 10%Z L = 40?Y 132/33kV 50MVA

Z T = 10%

O/H Line Z L = 8?

Feeder

3-?Fault

I 11kV =0.698 x 2624=1832 A I 132kV =0.698 x 219=153 A I 33kV =0.698 x 875=611 A I F = 1/1.432 = 0.698 P.U.

~ 1.432 p.u.

1 p.u.Example 2

Generator Transformer Line Motor

50 MVA 11kV 50MVA X1=X2= 1.3225 ohm 20 MVA 115 kV Xd”= 20% 11/115kV Ynd1X0= 6.6125 ohm Xd”= 20%X2 = 20% Imp voltage =12.5%X2 = 20%X0 = 10% at 50 MVA Base X0 = 10%Neutral resister = 12.1 ohm

~Y 11kV Bus Transformer Xd”M O/H Line

Rn 115kV Bus 115kV Bus

Xd”

Convert Impedance to Per-Unit Generator

MVAb= 100 MVA, kVb= 11 kV

Zb= kVb2/MVAb = 112/100 = 1.21 ohm

Xd”= j0.2 x 100/50 = j0.4 p.u.

X2 = j0.2 x 100/50 = j0.4 p.u.

X0 = j0.1 x 100/50 = j0.2 p.u.

Rn= 12.1/1.21 = 10 p.u.

Transformer

MVAb= 100 MVA, kVb= 11&115 kV

Imp voltage = 0.125 x 100/50 = 0.25 p.u.

X1 = X2 = X0 = j0.25 p.u.

Convert Impedance to Per-Unit Line

MVAb= 100 MVA, kVb= 115 kV

Zb= kVb2/MVAb = 1152/100 = 132.25 ohm

X1 = X2 = j1.3225/132.5 = j0.01 p.u.

X0 = j6.6125/132.5 = j0.05 p.u.

Motor

MVAb= 100 MVA, kVb= 115 kV

Zb= kVb2/MVAb = 1152/100 = 132.25 ohm

Xd”= j0.2 x 100/20 = j1.0 p.u.

X2 = j0.2 x 100/20 = j1.0 p.u.

X0 = j0.1 x 100/20 = j0.5 p.u.

Symmetrical Components Vectors

Vector notation can be used to represent phase relationship between 2 electrical quantities

~I

V Z

v = V sin ωt = V/0°

i = I sin (ωt-θ)=I/θ°

θLagging = -

Leading = +

“j ”operator : j = 1/90°

Rotates vector by 90°Anti-clockwise :-

90°

90°90°90°

Used to express vectors in terms of ‘real’and ‘imaginary’parts 1/0°

j 2= 1/180°= -1j = 1/90°

j 3= 1/270°= -j ‘a ’operator : a = 1/120° Rotates vectors by 120°Anti-clockwise

Used extensively in ‘Symmetrical Component Analysis’

120°

120°120°

a = 1/120°= -?+j √3/2

a 2= 1/240°= -?-j √3/2

1 , a 3

Balance 3?voltages :-

120°

120°120°

V C = a V A V A

V B = a 2V A

a 2 + a + 1 = 0

‘a ’operator : a = 1/120°Symmetrical Components Property of unbalanced phasors ‘n’phasors may be resolved into :-

(n-1) sets of balanced n-phase systems of phasors, each set having a different phase sequence, plus

1set of zero phase sequence or unidirectional phasors

Symmetrical Components V A = V A1+ V A2+ V A3 + V A4 + ………. +V A(n-1)+ V An

V B = V B1+ V B2+ V B3 + V B4 + ………. +V B(n-1)+ V Bn

V C = V C1+ V C2+ V C3 + V C4 + ………. +V C(n-1)+ V Cn

V D = V D1+ V D2+ V D3 + V D4 + ………. +V D(n-1)+ V Dn

………………………………………………………….

V N = V N1+ V N2+ V N3 + V N4 + ………. +V N(n-1)+ V Nn

(n-1) x Balanced 1 x Zero Sequence Consider 5-phase unbalanced

System V a1V b1First Set of Balanced Phasors V c1V

d1V e1V a2

V d2

Second Set of Balanced Phasors

V b2

V

e2V c2V a3V c3Third Set of Balanced Phasors V e3V

b3V d3V a4

V e4Fourth Set of Balanced Phasors

V d4

V

c4V b4V a0V b0V c0V d0V e0

Fifth Set of Balanced Phasors

( 360°/ 5 )

Unbalanced 3 Phase System V A = V A1+ V A2+ V A0

V B = V B1+ V B2+ V B0V C = V C1+ V C2+ V C0

V A1V B1V C1Positive Sequence V A2

V C2Negative Sequence V B2V A0V B0V C0

Zero Sequence ( 360°/ 3 )Symmetrical Components

V A1V B1V C1+V A0V B0V C0

+V A2

V C2V B2

V A

V B V C

V A = V A1+ V A2+ V A0

V B = V B1+ V B2+ V B0

V C = V C1+ V C2+ V C0

Converting from sequence Component to Phase Values V A = V A1+ V A2+ V A0

V B = V B1+ V B2+ V B0=a 2V A1+ aV A2+ V A0

V C = V C1+ V C2+ V C0=aV A1+ a 2V A2+ V A0

V A1

V B1

V C1V A2V B2

V B0V C2V A0

V C0V A V B

V C

V A1V B1V C1V A2V C2

V

B2

V A0V B0V C0

Converting from Phase Values to Sequence Component

V

A1 = 1/3 { V A + aV B + a 2V C }

V A2= 1/3 { V A + a 2V B + aV

C }

V A0= 1/3 { V A + V B + V C }

V B

V A V

C

V A0

V C0

= 3V 0

V B0

Summary

V A = V

A1

+ V

A2

+ V

A0

V B = a2V

A1

+ aV

A2

+ V

A0

V C = aV

A1

+ a2V

A2

+ V

A0

I A = I

A1

+ I

A2

+ I

A0

I B = a2I

A1

+ aI

A2

+ I

A0

I C = aI

A1

+ a2I

A2

+ I

A0

V

A1

= 1/3 { V

A

+ aV

B

+ a2V

C

}

V

A2

= 1/3 { V

A

+ a2V

B

+ aV

C

}

V

A0

= 1/3 { V

A

+ V

B

+ V

C

}

I

A1

= 1/3 { I

A

+ aI

B

+ a2I

C

}

I

A2

= 1/3 { I

A

+ a2I

B

+ aI

C

}

I

A0

= 1/3 { I

A

+ I

B

+ I

C

}

Residual Current

I RESIDUAL is Zero for :-

Balanced Load

3?Faults

??Faults Residual Current = I

A

+ I

B

+I

C

=3I

I RESIDUAL is present for :-

?/E Faults

??/E Faults

Open circuits (with current

remaining phases )

Use to detect earth faults

E/F I A

I B

I C

Residual Voltage RESIDUAL Voltage is measured

from Open Delta or Broken Delta

VT Secondary windings.

V RESIDUAL is Zero for :-

Healthy Unfaulted System

3?Faults

??Faults

V Residual = V A + V B + V C

=3V 0 V RESIDUAL is present for :-

?/E Fault

??/E Fault

Open circuits (on supply side

of VT)

Use to detect earth faults R g = n

Ground Plane A B C A

Vb Vc Va n B

C Van

Vbn Vcn

Normal Balanced Condition

Primary voltage

Secondary voltage Va+Vb+Vc = 0

g = n Ground Plane A B C A

Vb Vc Va n B

C Van Vbn

Vcn Fault Condition

3V0Va+Vb+Vc = 3V0

3V0Secondary voltage Primary voltage

Example 3

Evaluate the positive,

negative and Zero Sequence

components for the unbalanced phase vectors

V B

V A

V C

V A = 1 /0°

V B = 1.5/-90°

V C = 0.5/120°

Solution

V

A1

=

1/3

{V

A

+ aV

B

+ a2V

C

}

=1/3 {1 + (1 /120°)(1.5/-90°)+(1 /240°)(0.5/120°}

= 0.965

/15°

V

A2

=

1/3

{V

A

+ a2V

B

+ aV

C

}

=1/3 {1 + (1 /240°)(1.5/-90°)+(1 /120°)(0.5/120°} =0.211 /150°

V A0= 1/3 {VA + VB + VC}

=1/3 {1 + (1.5/-90°) + (0.5/120°} =0.434 /-55°

Positive Sequence Voltage

V

A1

= 0.965/15°

15°

V B1= a2V

A1

V C1 = aV

A1