FAULT Analysis presentation Armstrong

Presented by Armstrong Okai Ababio

1Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016

Goal of the Presentation-1

To Appreciate the use of Ohmic and MVA Methods for Symmetrical Short Circuit Calculation.

A symmetrical or balanced fault affects each of the three-phases equally (Three phase fault)


Goal of the Presentation-2To appreciate the use of Symmetrical Components

for Asymmetrical Short Circuit Calculation.An asymmetric or unbalanced fault does not

affect each of the three phases equally.Common types of asymmetric faults:

line-to-lineline-to-grounddouble line-to-ground

The analysis of this type of fault is often simplified by using methods such as symmetrical components.


http://en.wikipedia.org/wiki/Symmetrical_components

IntroductionThe operation of a power system departs from normal after

the occurrence of a fault. The most common faults in any electrical installation are short circuits, i.e. a breakdown of insullation between conductive parts which normally are at different potential.

Fault give rise to abnormal operating conditions such as excessive current and voltages at certain points in the network.

The magnitude of fault current depends on the generated power in the system, the distance to these sources, fault resistance and for ground fault it depends also on the neutral point treatment of the system.


Effect of Faults in a 3 phase Network A fault will cause voltage and current disturbances, including a

complete voltage collapse near the fault location.

The voltage gradient in the ground and across the earthing resistance create dangerous step and touch voltages near the fault.

The dynamic forces of the fault current, the electric arc at the fault location and the thermal effects of the current on the network elements in the current path all cause damage to the plant equipment and personnel.

The voltage and current disturbances interrupt the transmission of power and thus also affect customers.


Effect of Faults-2 A large amount of power is dissipated at the fault point, which means lost

of revenue for the utility.

Failure to remove a fault will usually result in rapid expansion of damage

to the system.

Various protection equipment are used in guarding against such faults

condition. An idea of the magnitudes of such fault currents gives the

Engineer the current settings of the various protection devices and

ratings of circuit breakers in the network.


Effect of Faults-3 In the light of the above discussed effects of faults, FAULT or SHORT

CIRCUIT ANALYSIS is highly needed.

Performing short-circuit calculations requires an understanding of various system components and their interaction.

Components of Power System:

Generators, Transformers, Transmission lines, Protection Systems(control and monitoring system), Switch gear, Load.


Types of Fault/ Short Circuits in a 3 phase network

Three-phase fault with or without earth (5%)

Phase-to- phase clear of ground (10 – 15%)

Two-phase-to-earth fault (10 – 20%)

Phase-to-earth fault (65 – 70%) Fault Incident:

85% of faults are overhead line.

50% of these are due to lightning strikes.

L1L2L3

L1L2L3

L1L2L3

L1L2L3


3 PHASE FAULTS-1 A 3-ph fault affects the three-phase network equally (symmetrical fault).

They rarely occur. All three conductors are equally involved and carry the same rms short- circuit current. There is the need to use only one conductor for the calculation.

It is valid because system is maintained in a balanced state during fault. Voltages are equal and 120° apart. Currents are equal and 120° apart. Power system plant symmetrical:- Phase impedance equal Mutual impedance equal Shunt admittances equal Causes: System energization with maintenance earthing clamps still connected. 1ø faults developing into 3ø faults. Etc.


3 PHASE FAULTS-2 A 3-ph fault affects the three-phase network symmetrically. They

rarely occur. All three conductors are equally involved and carry the same rms short- circuit current. There is the need to use only one phase (conductor ) for the calculation.

The Per unit system or MVA method is used for the analysis of Three phase fault.


3 PHASE FAULTS


Over View of Per Unit System Per- Unit Notations. It is used to simplify calculations on system with more

than two voltages. ( The √ are eliminated) It can be seen by inspection of any power system diagram that: a. several voltage levels exist in a system b. it is common practice to refer to plant MVA in terms of per unit or

percentage values c. transmission line and cable constants are given in ohms/km

Before any system calculations can take place, the system parameters must be referred to 'base quantities' and represented as a unified system of impedances in either, percentage, or per unit values.

Per-unit analysis is based on "normalized" representations of the electrical quantities (i.e., voltage, current, impedance, etc.). The per-unit equivalent of any electrical quantity is dimensionless. It is defined as the ratio of the actual quantity in units (i.e., volts, amperes, ohms, etc.) to an appropriate base value of the electrical quantity. Per Unit (p.u.) value = Actual value / Base value


Referring Impedances

14Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana

Can be derived from V1/V2 = I2/I1

Which translates into V1/V2=(V2/R2)/ (V1/R1)

9/9/2016

Formulae used in Per-Unit Calculations The base quantities are: three-phase power in MVA (MVA rating of largest item or 100MVA) It is

constant at all voltages . The line voltage in kV. Fixed at one part and it is transferred through

transformers to obtain base voltages of the other part of the system. The following Equations are used in Per- Unit Calculation: - Ip.u = Iactual / Ibase - Sp.u = Sactual / S base

- Vp.u = Vactual / Vbase - Base Current (kA), Ib = MVAb (3ø)/ {1.732 x kVb (L-L)} - Base Impedance (Ohm) Zb = (kV)2

(L-L) / MVAb 3ø

- Zp.u(new) = Z p.u

(old) x[MVAb (N) / MVAb

(old)] x[kVb(old) / kVb

(new)]2


Transformer Per Unit Impedance


Transformer Percentage Impedance


Transformer Base Voltage Selection


Example


MVA METHOD FOR 3 PHASE SHORT CIRCUIT CALCULATION

The MVA Method is recognized and widely acceptable by industry in calculating power system short circuits where the reactance of all circuit components far exceeds resistance producing a consistently high X/R ratio throughout the system.

Combined MVA of components connected in series and parallel are calculated using the following formulas:

series: MVA1, 2 = MVA1 X MVA2 / (MVA1 + MVA2) parallel: MVA1, 2 = MVA1 + MVA2

As can be seen from the formulas above, series MVA’s are being calculated same as resistances in parallel. Parallel MVA combinations are done same as resistances in series.

MVA diagram undergoes same reduction process as impedance diagram, only that MVA values are used instead of per unit impedances or reactances.


MVA METHOD FOR 3 PHASE SHORT CIRCUIT CALCULATION

The MVA method is a modification of the Ohmic method where the impedance of a circuit equals the sum of the impedances of components constituting the circuit.

In practice, the MVA method is used by separating the circuit into components and calculating each component with its own infinite bus as shown in figures 1 and 2 below:

M15 MVAXd = 0.2

20MVA33 / 11kV X = 10%

T2161 / 33 kVX = 10%

161 kV T1 50MVA

Line 33kV

X = 4.0Ω

Sk’’ = 700MVA

Figure 1-One Line Diagram

700 500 272.25

200

75

3ø F.

700 / 1 50 / 0.1 (33)2 / 4

15 / 0.2

20 / 0.1

3ø F.

Figure 2- MVA Diagram

1 2 34

5

11kV

G

103.8 MVA


MVA METHOD FOR 3 PHASE SHORT CIRCUIT CALCULATION In our example: MVA’s 1& 2 are in series - MVA (1&2) = (700 x 500) / (700 + 500) = 291.67

MVA’s (1&2) & 3 are in series - MVA = (291.67 x 272.25) / (291.67 + 272.25) = 140.81

MVA’s (1,2&3) // MVA 4 - MVA at the point = 140.81 + 75 = 215.81

MVA’s (1,2,3&4) in series with MVA 5 MVA at the Fault point = (251.81 x 200) /(251.81 + 200) = 103.802 Considering Voltage Factor of 1.1, MVA = 103.802 x 1.1 = 114.18

- 3Ø ISC =MVA /(1.732 x Un) =114.18 /(1.732 x 11) = 5.99 kA

Per Unit System is more accurate than the MVA method in the analysis of three (3) phase faults.


D.C. Transient and Offsets


END OF PART 1

WAGE OPENER FOR PART 2

RIDDLE!!!


Goal of the Presentation-2To appreciate the use of Symmetrical Components

for Asymmetrical Short Circuit Calculation.An asymmetric or unbalanced fault does not

affect each of the three phases equally.Common types of asymmetric faults:

line-to-lineline-to-grounddouble line-to-ground

The analysis of this type of fault is often simplified by using methods such as symmetrical components.


http://en.wikipedia.org/wiki/Symmetrical_components

ASYMMETRICAL OR UNBALANCED FAULTS

CALCULATIONS

Developed by Charles LeGeyt Fortescue (1876–1936) in 1918. Fortescue was an Electrical Engineer and he worked for Westinghouse Corporation at East Pittsburgh, Pennsylvania. He developed the method of Symmetrical Components to resolve unbalance fault conditions.


http://en.wikipedia.org/wiki/Westinghouse_Electric_%281886%29

http://en.wikipedia.org/wiki/Pennsylvania

UNBALANCED FAULTS-1Unbalanced Faults may be classified into SHUNT

FAULTS and SERIES FAULTSSHUNT FAULTSLine to GroundLine to LineLine to Line to GroundThe above faults are described as single shunt

faultsbecause they occur at one location and involve aconnection between one phase and another or to

earth.32

Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016

UNBALANCED FAULTS-2Unbalanced Faults may be classified into

SHUNT FAULTS and SERIES FAULTS

OPEN CIRCUIT/ SERIES FAULTSSingle phase open circuitDouble phase open circuit


CAUSES OF UNBALANCE FAULTSCAUSES OF SHUNT FAULTS Insulation BreakdownLightning Discharges Mechanical DamageCAUSES OF OPEN CIRCUIT FAULTSBroken ConductorOperation of FusesMal-operation of single phase CBDuring Unbalanced Faults, symmetry of system

is lost. Hence single phase representation is no longer valid for the fault analysis.


Analysis of Unbalanced FaultsUnbalanced faults are analysed using :-Symmetrical ComponentsEquivalent Sequence Networks of the Power

SystemConnection of Sequence Networks

appropriate to the Type of Fault


Sequence ComponentsThe Symmetrical Components consist of

three subsystems for an unbalanced three-phase system. They are;

Positive Sequence Subsystem-consisting of three phasers of equal magnitude and 120° phase displacement, and having the same phase sequence as the original balanced system of phasers.


Sequence ComponentsThe second subsystem is termed the

'negative sequence' system, consisting of three phasers of equal magnitude and 120° phase displacement, and having a phase sequence which is the reverse of the original balanced system of phasers.

The final subsystem is termed the 'zero sequence' system, consisting of three phasers of equal magnitude and zero phase displacement.


ILLustration of Symmetrical Components

Orientation of the symmetrical components are as follows:V A1

V B1V C1V C2

V A2 VAo

V B2

VBo

120° 240°V Co

Zero Sequence PhasorsNegative Sequence PhasorsPositive Sequence Phasors


Sequence Components-Three Phase Fault


Sequence Components PHASE TO PHASE FAULT

ONLY +VE AND –VE SEQUENCE COMPNENTS EXISTS

E

Z1 Z2

I a11 I a22 ZoI a0 0 = 0= 0

Sequence Network Interconnection for phase to phase fault


Sequence Components-Single Phase-Earth Fault. All sequence components are present


By definition, the magnitude and phase angle of any phasor in an unbalanced three-phase system is equal to the vector addition of the symmetrical components from the respective sequence subsystems.

Symmetrical Components


Symmetrical Components


Symmetrical ComponentsTo provide a mathematical perspective on the

preceding discussion of phasor magnitudes and phase displacements, it is necessary to define a phase displacement operator 'a'.

a = 1<120° = -0.5 + j0.866

Successive applications of operator 'a' to a given vector will result in rotation of that vector through 120°, 240° and 360°, degrees, as shown below.


Symmetrical Componentsa = 1<120° = 1ej2π/3 = -0.5 +j0.866 a2 = 1<240° = 1ej4π/3 = -0.5 -j0.866 a3 = 1<360° = 1e]2π = 1Therefore we have;


Converting from phase value to Converting from phase value to sequence componentssequence components


Converting from phase value to sequence components


Symmetrical Components-Similarly Equations for sequence Currents can be derived

48Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana

It can be deduced thatIn= Ia + Ib + Ic = 3I0

9/9/2016

Phase Sequence Equivalent Circuits

E

a2E

aE

I

a2 I

a I

P Q

P1 Q1

Z1 = E/I

Positive Sequence Impedance

9/9/2016Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 49


E

aE

a2E

I

a2 I

a I P Q

P2 Q2

Z2 = E/I

Negative Sequence Impedance For static non-rotating plant :- Z2= Z1



E

I

I

I P Q

P0 Q0

Z0 = E/I

3I

Zero Sequence Impedance


Sequence Networks

+ve, -ve and zero sequence networks are drawn for a ‘reference’ phase. This is usually taken as the ‘A’ phase.

Faults are selected to be ‘balanced’ relative to the reference ‘A’ phase.

e.g. For Ø/E faults, we consider an A-E fault For Ø/Ø faults, we consider a B-C fault Sequence network interconnection is the simplest

for the reference phase.


1. Start with neutral point N1 -All generator and load neutrals are connected to

N1 2. Include all source voltages:- Phase-neutral

voltage 3. Impedance network:- Positive sequence

impedance per phase 4. Diagram finishes at fault point F1

Positive Sequence Diagram

N1E1

Z1F1


Positive Sequence Diagram G

T Line F

R

N

E

N1

E1 Z G1 Z T1 Z L1 I1 F1

N1

V1

V1=Positive sequence PH-N voltage at fault pointI1=Positive sequence phase current flowing into F1

V1=E1 –I1 (ZG1+ ZT1 + ZL1)

System Single Line Diagram

Positive Sequence Diagram


Negative Sequence Diagram

Start with neutral point N2

-All generator and load neutrals are connected to N2

No voltages included -No negative sequence voltage is generated! Impedance network -Negative sequence impedance per phase Diagram finishes at fault point F2

N2

Z2

F2


Negative Sequence Diagram G

T Line F

R

N

E

N2

Z G2 Z T2 Z L2 I2 F2

N2

V2

V2=Negative sequence PH-N voltage at fault pointI2=Negative sequence phase current flowing into F2

V2= –I2 (ZG2+ ZT2 + ZL2)

System Single Line Diagram

Negative Sequence Diagram


Zero Sequence Diagram

For “In Phase” (Zero Phase Sequence) currents to flow in each phase of the system, there must be a fourth connection (this is typically the neutral or earth connection).

IAo+ IBo+ICo=3IAo

IAo

IBo

ICo

N


Zero Sequence Diagram

N

E

R3IAo

Resistance Earthed System :-

Zero sequence voltage between N & E is given byVo= 3IAo * RZero sequence impedance of neutral to earth pathZo= Vo/ IAo = 3R


Zero Sequence Equivalent “D y” Transformer”


Zero Sequence Equivalent “D y” Transformer

Thus, Equivalent single phase zero sequence diagram is as shown:-

Side terminal Z ToI o

Y sideterminal

N o(Eo)


Zero Sequence Equivalent Circuits


Equations Defining Shunt Fault Conditions

It should be noted that for any type of fault there are three equations that define the fault conditions. They are as follows:

Single Phase-to-earth (A-E)

Ib= 0 (Ib= 0, Ic = 0, Because Phases B and C do not contribute to fault current)Ic = 0

Va = 0 (Va = 0, because voltage at the faulted phase decreases to Zero)



Phase-phase (B-C)

Ia= 0 (Ia= 0, Because Phase A does not contribute to fault current)

Ib =-Ic

Vb = Vc



Phase-phase-to-earth (B-C-E)

Ia= 0 (Ia= 0, Because Phase A does not contribute to fault current)

Ib + Ic = In

Vb = Vc=0 (Because voltages at the faulted phases decrease to Zero)



Three Phase fault (A-B-C or A-B-C-E)

Ia + Ib + Ic = 0 (Because the system is balanced and hence In= Ia + Ib + Ic =0) Va = Vc

Vb = Vc


Single Phase to Earth Fault-1We recall that for Single Phase to ground fault:

Ib=Ic = 0

Va = 0 and therefore we can write;

I1=1/3(Ia + aIb + a2Ic)=1/3Ia

I2=1/3(Ia + a2Ib + aIc)=1/3Ia Therefore I0=I1=I2= 1/3Ia

I0=1/3(Ia + Ib + Ic)=1/3Ia


Single Phase to Earth Fault-2Also

Va = 0 and therefore we can write;

V1 +V2 +V0 =0 (Since Va =V1 +V2 +V0 =0)But

V1 =V-I1 Z1 , V2 =-I2 Z2 and V0=-I0Z0

Substituting we have;

V-I1 Z1 -I2 Z2-I0Z0=0 since I1=I2=I0, we haveV=I1 (Z1 + Z2 + Z0)


Single Phase to Earth Fault-3The above analysis indicate that the

equivalent circuit for the fault is obtained by connecting the sequence networks in series as shown below;


Single Phase to Earth Fault-4It also follows that;

I1 = E/(Z1 + Z2 + Z0)

From the equation described earlier, I0=I1=I2= 1/3Ia and there fore

Ia = 3I1 = 3E/(Z1 + Z2 + Z0)


Single Phase to Earth Fault-5In the more general case, with nonzero fault

resistance, the equality of I1; I2 and I0 is maintained, and 3I1 flows through the fault resistance. Therefore, it is necessary that 3Zf exist in series with the zero sequence subsystem to achieve the required effect. The generalized equation is as follows:

Ia = 3I1 = 3E/(Z1 + Z2 + Z0 + 3Zf)


Phase to Phase Fault (B-C)-1We recall that for a phase-phase fault,

Ia = 0, implying

I1=1/3(Ia + aIb + a2Ic)=1/3(aIb + a2Ic)

I2=1/3(Ia + a2Ib + aIc)=1/3(a2Ib + aIc)

Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 719/9/2016

Phase to Phase Fault (B-C)-2Also,

Ib= -Ic I1=1/3(aIb + a2Ic) = 1/3(alb-a2lb) = l/3Ib (a-a2)

I2=1/3(a2Ib + aIc) = 1/3(a2lb-alb) = l/3Ib (a2-a) The above equations illustrate that;I 1 = - I 2


Phase to Phase Fault (B-C)-3By inspection, the equations for this fault

condition are as follows:

I1=E/(Z1 + Z2)

I2=-E/(Z1 + Z2)

I0 = 0It must be noted that no zero sequence current

exist since there is no connection to ground.Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 739/9/2016

Phase to Phase Fault (B-C)-4From the equation below;

I1=1/3(aIb + a2Ic) = 1/3(alb-a2lb) = l/3Ib (a-a2), Since Ib=-Ic

But

a = 1<120° = 1ej2π/3 = -0.5 +j0.866 a2 = 1<240° = 1ej4π/3 = -0.5 -j0.866

Therefore it follows that, a-a2 = j1.732= j√3 and I1= j√3/3Ib


Phase to Phase Fault (B-C)-5Finally it implies that;

I1=1/3(aIb + a2Ic) = 1/3(alb-a2lb) = l/3Ib (a-a2) Since,

a-a2 = j1.732= j√3I1 = j √3 /3Ib = jIb/ √3 by rationalization.

hence Ib =-j √3 I1


Phase to Phase Fault (B-C)-6Similarly;

I2=1/3(a2Ib + aIc) = 1/3(a2lb-alb) = l/3Ib (a2-a) And since

a2 -a = -j1.732= -j√3, we haveI2 = -j √3 /3Ib = -jIb/ √3 by rationalization.

hence Ib =j √3 I2


Phase to Phase Fault (B-C)-7By inspection, the equations for this fault

condition are as follows:

I1=E/(Z1 + Z2) but Ib =-j √3 I1 Therefore

Ib =-j √3 E/(Z1 + Z2)

Also, Ic =-Ib

Therefore Ic =j √3 E/(Z1 + Z2)


Phase to Phase Fault (B-C)-8In the more general case, with nonzero fault

resistance, the generalized equation, based on the fault impedance of Zf is expressed below:

Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana 78

fcb ZZZ

EjII

21

3

9/9/2016

Double Phase to Ground Fault (B-C-E)We recall that for double Phase to Ground Fault

(B-C-E), the following equations can be written:

Ia= 0 (Ia= 0, Because Phase A does not contribute to fault current)Ib + Ic = In

Vb = Vc=0 (Because voltages at the faulted phases decrease to Zero)


Double Phase to Ground Fault (B-C-E)From the general equation below, we have:

V-I1 Z1 = -I2 Z2 from(V1=V2=V0=Va/3) and Vb = Vc=0

Also, Since Ia= 0 (Ia= 0, Because Phase A does not contribute to fault current)

we have, I1 + I2 + I0 = 0

And it follows that;

I1 = -( I2 + I0)


Double Phase to Ground Fault (B-C-E)Vb = Vc=0 (Since voltages at the faulted phases

decrease to Zero)Also implies, V1 = V2 = V0 = 1/3Va

Given that all three sequence voltages are equal in phase and magnitude, and given that the sequence currents sum vectorially to zero, it is obvious that the three sequence networks are in parallel


Double Phase to Ground Fault (B-C-E)From the fact that V1 = V2 = V0 = 1/3Va, we can write

V2 = V0, and hence I2 Z2 = I0Z0

The negative and zero sequence currents can be derived on the basis of the current divider principle as shown below:

I2=-I1 Z0/(Z2+ Z0), and I0=-I1 Z2/(Z2+ Z0)


Double Phase to Ground Fault (B-C-E)Equating V1 and V2, we have

V-I1 Z1 = -I2 Z2 or V = I1 Z1 - I2 Z2 , where V=E= phase voltage

Substituting I2, I2=-I1 Z0/(Z2+ Z0), into the above equation,

It gives, V = I1 [Z1 + Z0Z2 /(Z0 + Z2)] Therefore;

I1 = E(Z0 + Z2)/(Z1Z0 + Z1Z2 +Z0Z2)


Double Phase to Ground Fault (B-C-E)Finally,

Substituting I2 and I0 into the above equation,

I2=-I1 Z0/(Z2+ Z0) , I0=-I1 Z2/(Z2+ Z0)andI1 = E(Z0 + Z2)/(Z1Z0 + Z1Z2 +Z0Z2)


0212 IaIIaIb

9/9/2016

Double Phase to Ground Fault (B-C-E)We have;


020121

203ZZZZZZ

aZZEIb

9/9/2016

Three Phase Fault (A-B-C or A-B-C-E)We recall that for a three phase fault;

Ia + Ib + Ic = 0 (Because the system is balanced and hence In= Ia + Ib + Ic

=0) Va = Vc

Vb = Vc


Three Phase Fault (A-B-C or A-B-C-E)It should be noted that, because this fault

type is completely balanced, there are no zero- or negative-sequence currents;

By inspection we can write;

I1= E/(Z1+ Zf) also I2 = I0 =0


Three Phase Fault (A-B-C or A-B-C-E)

It finally follows that for a three phase fault;

Ia= Ib = Ic = E/(Z1+ Zf) also I2 = I0 =0


CONCLUSION-1

It must be emphasized that in very large and complex networks, system computer programs are used for short circuit analysis.

In the Electricity Company of Ghana (ECG), ASPEN software is used for short circuit analysis.

Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016 89

CONCLUSION-2 I would like to first of all thank God for His guidance and for

the wisdom and understanding He gave me to put this presentation together.

Secondly, I appreciate the efforts of Ing. Godfred Mensah/ SM/System Planning and Mr. Frank Osei Owusu of Protection Applications whom I understudied when preparing for this presentation.

Finally, I very much appreciate your presence for this presentation

THANK YOU ALL FOR COMING.!!!

Armstrong Okai Ababio/ Electrical Engineer/ Electricity Company of Ghana9/9/2016 90

FAULT Analysis presentation Armstrong

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