Fault Level Analysis

FAULT LEVEL ANALYSIS

G.K.Thayanithy

Fault level analysis

What is fault level ?

“The apparent power in MVA or KVA, supplied by one or more power generating or supplying sources to a system under fault conditions”

What is fault current ? “ The current flowing through a faulted branch of an Electrical system” Generally expressed in Kilo Ampere(kA)

GKT/FLS/2008

Fault level analysis

Need for fault level analysis Fault level analysis is needed to determine,

 The interruption duty that the CB has to perform and interruption rating required for the breaker

 The current making capacity required for the CB when closed on existing fault-( Also known as momentary duty or first cycle duty or make & latch duty)

 The S.C. withstand capability required for cables, busbars, CTs, etc…. till the fault is cleared by the protection relay

 Suitable relays and relay setting co ordination  Transient behavior of the system under fault conditions, and hence suitable system design and stability co ordination

GKT/FLS/2008

Fault level analysis

Classification of faults Electrical faults are classified as 1. Three phase faults

– A short circuit between all the three phases

2. Single line to earth fault

-- One phase getting connected to earth

3. Line to line fault

-- Short between two phases

4. Double line to earth fault – Two phases shorted and connected to earth

5. Simultaneous fault

-- All three phases shorted and also connected to earth

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Fault level analysis

Types of faults

 Symmetrical -- Arcless bolted type of faults( no voltage drop & zero fault impedance ) are considered for FL study

-- Can be solved easily using single phase concept -- LL voltage is considered as LN voltage

 Unsymmetrical – Can be converted to symmetrical components -- generally not considered

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Fault level analysis

Sources of fault current

 Grid --- No of generators interconnected into a common power grid, to which no. of consumers are also connected

 Generators at site --- One or more of generators installed at site --- Generally used as internal power source --- Some times used to export power to grid for sale to grid or to wheel power to other consumer through grid

 Induction motors --- Sq.cage type: Contributes constant fault current

--- Slip ring type: Fault level reduces with resistance included in the rotor GKT/FLS/2008

Fault level analysis

Assumptions made in fault level analysis  Resistance if any, small compared with reactance , is neglected  Capacitance if any, and of the system, are neglected since the time constant (RC) is too small  Machine reactances are assumed be constant- Saturation effects are neglected  Generator voltages are assumed to be constant

 Generators are replaced by EMF sources in series with their reactances

GKT/FLS/2008

Fault level analysis

Procedure for fault level calculation 1. Draw the single line diagram for the system assuming as single phase 2. Draw the reactance or positive sequence diagram 3. Select a suitable base KV & base KVA 4. Calculate the base quantities for current & impedances 5. Calculate the Thevenin`s equivalent impedances at the point of fault 6. Calculate the current through the faulted branch- this will be the fault current

7. Calculate the fault MVA using the relation Fault MVA = √3 X Voltage X Fault current GKT/FLS/2008

Fault level analysis

Representation of Power System Single line diagram

GENERATOR

GENERATOR TRANSFORMER 1

1

TRANSFORMER 2

3

2 LOAD LOAD

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Fault level analysis

Impedance diagram Load

Generators

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Transformer

Transmission line

Neutral bus

Transformer

Load

Generator

Fault level analysis

Reactance diagram or Positive Sequence diagram

Reference bus

GKT/FLS/2008

Fault level analysis

Selection of base values  As a rule only two bases should be selected first & the remaining bases are calculated from these two - Generally the base KV and base KVA are selected and other quantities are calculated Base current = Base KVA/ Base voltage Base impedance = Base voltage/ Base current

 For circuits connected by transformer, - same KVA base is selected for both circuits - Base KVs are chosen such that the ratio of the base KVs is the ratio of the transformer ( This will give the same p.u reactance of transformer referred to both the circuits)

GKT/FLS/2008

Fault level analysis

Calculation of base current & impedence 1. If base voltage assumed is L-N voltage, Base KVA

= 3phase KVA/3

Base current

= Base KVA/Base voltage

Base Impedance

= (Base KV)² X 1000/ Base KVA

2. If base voltage assumed is L-L voltage, Base current

= Base KVA/√3 X Base KV(L-L)

Base Impedance

= ( Base KV/ √3 )² X 1000/ (Base KVA/3) = (Base KV)² X 1000/ Base KVA

GKT/FLS/2008

Fault level analysis

Thevenin`s Theorem  Used to determine the current in a branch of a network having one or more sources. The current in a branch Ib is given by,

Ib

=

Voc Zth + Zb

Where, Voc

= the voltage across the terminals, with branch disconnected

Zth

= Thevenin`s equivalent impedance ( The impedance of the network between the terminals of the branch with the branch disconnected & voltage sources replaced by short circuits)

Zb

= Impedance of the branch whose current is to be measured

GKT/FLS/2008

Fault level analysis

Per Unit (p.u) method  To simplify calculations, p.u method is used  Electrical quantities such as voltage(V), current(I), KVA & Impedences (Z) are expressed as a “Percentage” or “Per Unit” of their selected base Example – Assume 200 Volt as selected base,then, 100 100 Volt = ------ = 0.5 p.u 200 400 400 Volt= ------ = 2.0 p.u 200 In general Actual value Per unit value = ---------------------Base value GKT/FLS/2008

Fault level analysis

Important relations in fault level calculation Select Base KV & Base KVA Base KV Actual KV given in kV p.u. kV Base current, I Amps Base Impedance, Z Ohms

Base power kW p.u. Impedance Z

= 1 p.u = Actual kV/Base kV = Base KVA/Base voltage = Base kV X 1000/ Base current I = Base kV X Base kV X 1000/Base KVA = (Base kV)² X 1000/Base KVA = Base KVA = Actual Z/Base Z = Actual Z XBase KVA/(Base KV)²X 1000

Change of base If the base is changed, the p.u.Z referred to new base = p.u Z (old base)X ( Base kV old/Base kV new)² X ( Base KVA new/ Base KVA old) GKT/FLS/2008

Fault level analysis

More information on percentage impedance

 The base impedance of the system is that impedance in Ohms which would cause 100% drop of the adopted base voltage when base MVA or base current flows in the circuit

 The percentage impedance of an element in a circuit is it`s actual ohmic impedance in Ohms expressed as a percentage of the base impedance Actual ohms Percentage impedance = X 100 Base ohms

 The p.u. impedance of an element in a circuit is it`s actual ohmic impedance in Ohms expressed as a decimal fraction or as a multiple of base impedance corresponding to the base MVA selected Actual ohms P.u impedance = Base ohms GKT/FLS/2008

Fault level analysis

Passive and Dynamic reactance  In many electrical circuits, elements such as cables, transformer, O/H lines, etc.., the variables are voltage & current  In such circuit elements, the values of resistance & reactance remain the same in all conditions  Reactance of such elements are referred to as “Static Reactance” Or “Passive reactance”( Resistance is normally ignored in FLS)  In motors & generators, the internal reactance vary during the fault duration, due to armature reaction and dynamic nature of magnetic circuit  Reactance of such elements which contribute for varying current at different instances of fault duration are termed as “Dynamic Reactance” GKT/FLS/2008

Fault level analysis

Behavior of Fault current in alternator Sub-transient period  When a three phase short occurs at the terminals of an alternator, a very high current flows for a few cycles, this current is known as “ Sub transient component of current”  The internal impedance at this point is Voltage Impedance = sub transient current  Since the resistance is neglected being very small compared to the inductance, this impedance is assumed to be equal to the reactance  This reactance is known as “Sub transient reactance”- denoted by

the symbol Xd”, and the value will be 0.1 to 0.27 p.u.  The duration of existence of sub transient current is known as “ Sub transient time” and will be in the order of 0.08 to 0.10 Sec. GKT/FLS/2008

Fault level analysis

Behavior of Fault current in alternator Transient period  The internal reactance of the m/c increases due to armature reaction of the sub transient current and dynamic nature of magnetic circuit  The increased reactance is termed as “Transient reactance” and is denoted by X`d, and the value will be 0.15 to 0.35 p.u.  The fault current now decays from the sub transient value to a lesser value, known as “Transient component of current”  The duration of existence of transient current is known as “Transient time” and will be in the order of 0.6 to 0.7 Sec.

GKT/FLS/2008

Fault level analysis

Behavior of Fault current in alternator Sustained condition  The internal reactance of the m/c further increases to the value known as “ Synchronous reactance”, denoted by Xd, & the value will be 0.5 -1.5 p.u.  The fault current further reduces to a steady state value known as “ Sustained SC current”  This current will have to be sustained by the machine until the protective relay senses the fault and isolates the alternator from fault  The fault currents produce, large electro-mechanical forces between the coils, in addition to high temperature rise of the winding Note:  The fault current also contains some amount of DC components & Negative sequence components generated by Zero sequence reactance (X0), & negative sequence reactance (X2) GKT/FLS/2008

Fault level analysis

Role of Fault current in selection of switch gear Besides the voltage, nominal current, & frequency rating, the SWG has three current ratings,namely  The circuit making current capacity- also known as making capacity or momentary rating: The current capacity of the SWG or CB to close on fault and latch on to the fault, against the forces of the total first cycle fault current  The interruption duty current capacity or the interruption rating; The fault current in kA at the time of contact parting of the SWG - about 200ms. Or 10 cycles after inception of the fault when the machine reactance have increased from sub transient value , reducing the first cycle value to interruption cycle value.  The short time rating ( 1 or 3 sec rating ; The fault current in kA.the SWG can withstand for a short period incase the SWG or CB fails to clear the fault, till an upstream SWG clears the fault. GKT/FLS/2008

Fault level analysis

Examples on FL calculation 1. Two 11KV, 3 Ph 3000 KVA generators having sub transient reactance 15% operate in parallel. The grs. Supply power to a transmission line through a 6000 KVA transformer of ratio 11/22 KVA and having a leakage reactance of 5%, Calculate fault current and fault MVA for a three fault on a)HT side, b) LT side of the transformer

2. Two generators rated 11KV, 3000 KVA having 20% reactance are interconnected by a 100 km long transmission line. The reactance of line is 0.10 Ohms/km.The transformers near the generators are rated 6000 KVA 11KV/66 and have leakage reactance of 5%, A three phase fault occurs at a distance of 20km from one end of the line, when the system is on no load but at rated voltage Calculate fault MVA & Fault current

GKT/FLS/2008

Fault level analysis

Behavior of Fault current in alternator

Symmetrical current wave of a three phase synchronous generator subjected to three phase short circuit GKT/FLS/2008