5- Basics Of Fault Calculation

Power System Protection Date: 6th March 2010 Venue: L&T, Manapakkam, Chennai

Steady State Fault Analysis Dr. G. Pradeep Kumar

What is Fault Analysis? Fault Analysis 

Analysis of a power system 



during and after a fault

Involves computation of voltages and currents under the fault conditions.

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Why Fault Analysis? Fault Analysis 

Important to understand what happens in the system as a result of a fault

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Why Fault Analysis? Study the effects of fault 

How high will the current go?



Will the voltage rise above a limit or fall below a limit?

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Why Fault Analysis? Ensure Ratings 

Use calculated voltage and current values 

Ensure that load and short circuit ratings of the plant are not exceeded



For proper selection of switchgear

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Why Fault Analysis? Selection of Switchgear 

Rating of a breaker 

Current it has to carry under normal conditions 



Continuous rating

Current flowing immediately after the fault occurs 



Short time withstand

Current which the breaker must interrupt 

Breaking capacity

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Why Fault Analysis? Selection of Switchgear 

Calculated fault voltages and currents 

Verify that the breaking capacity is not exceeded



Short time withstand is not exceeded

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Why Fault Analysis? Protection System Design 

Fault detection 

In which system parameters will the change due to fault be reflected more?

combination of both ??

Voltage

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Current

8

Why Fault Analysis? Protection System Design 

Fault detection 

Fault voltages and currents calculated at various system locations 

Select most suitable relay characteristics for fault detection

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Why Fault Analysis? Software Packages 

Software packages can do it - why bother? 

Understanding what is happening is important 

Eg. In a Δ-Y transformer for a single phase fault on the Y-side, why fault current in two phases on the Δ- side.



Spot check on the computer results 



Catch data entry errors

Easier for simpler systems

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How to Perform Fault Analysis? Computations 

Steady state analysis 

Solved by algebraic equations



Does not mean system is static



System may be fast changing



This is like taking a set of photographs of the system behavior under certain specified conditions 

Algebraic equations are much easier to solve!

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Per-unit Systems Definition

A is any quantity 

A per unit quantity is a “normalized” quantity with respect to a chosen base value.



Any quantity can be converted to a selected base quantity of the same dimension. 

Per unit quantities are dimensionless

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Per-unit Systems Per-unit Applied to Power Systems 



Power system quantities that are represented in per unit are: 

Voltage



Current



Impedance



Apparent power

All of these are completely described by only two of them 

Sufficient to choose only two base quantities

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Per-unit Systems Per-unit Applied to Power Systems 

Generally the two chosen base quantities are, 

Apparent power (SB)



Voltage (VB)



Equipment ratings are usually given in MVA and kV



Other two base quantities (ZB, IB) can be fixed as follows:

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Per-unit Systems Per-unit Calculations 

Converting from actual values to per unit

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Per-unit Systems Per-unit Calculations 

Change of Base: 

Often require to convert per unit values from one base to another 

Most equipment impedances are given in its own MVA and kV base



Need to be converted to a common base when solving a network

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Per-unit Systems Per-unit Calculations

A is any quantity 

Basically we, 

Multiply the old per unit value by the old base to get back the actual value



Then divide by new base to get the new per unit value

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Per-unit Systems Per-unit Calculations – Example Problem T1 G

50 + j100 ohms

ΔY 

Generator: 15 MVA, 13.8 kV, x = 0.15 pu



Load 1: 10 MVA at 0.9 pf lead



T1: 25 MVA, 13.2/161 kV, x = 0.10 pu



T2: 15 MVA, 161/13.8 kV, x = 0.10 pu



Load 2: 4 MVA at 0.8 pf lag

T2 YΔ

Load 1 Load 2

If the load bus voltage is 15.18 kV, find the voltage at the generator terminals, Vg.

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Per-unit Systems Calculations – using per unit system 

Choose an MVA base

25 MVA T1

15 MVA T2

10 MVA Load 1

G

Load 2

15 MVA

4 MVA Choose an MVA base of 15

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Per-unit Systems Calculations – using per unit system 

Choose a kV base at any part of the system and fix the kV bases at different sections of the system

13.2/161 kV

161/13.8 kV

T1

T2

13.2 kV

G

Load 1 Load 2

13.8 kV 13.2 kV

161 kV

13.8 kV

Choose a kV base of 161 kV at the transmission line

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Per-unit Systems Calculations – using per unit system 

MVAB = 15

Chosen bases T1

T2

G

Load 1 Load 2

kVB = 13.2

kVB = 161

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kVB = 13.8

32

Per-unit Systems Calculations – using per unit system 

Convert all impedances to pu in the chosen bases

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Per-unit Systems Calculations – using per unit system 

Solve the network to obtain required voltages & currents in pu Load bus voltage

Load 1 current

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Per-unit Systems Calculations – using per unit system 

Solve the network to obtain required voltages & currents in pu Load 2 current

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Per-unit Systems Calculations – using per unit system 

Solve the network to obtain required voltages & currents in pu Total load current

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Per-unit Systems Calculations – using per unit system 

Solve the network to obtain required voltages & currents in pu Generator terminal voltage

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Per-unit Systems Calculations – using per unit system 

Convert the pu values back to actual values

The generator terminal voltage is 14.623 ∠8.540 kV

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Per-unit Systems Advantages 

Name plate details 

Manufacturers usually specify impedance of the equipment in per unit or percent

pu ?

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Per-unit Systems Advantages 

Relative strength of different parts of the system on the same scale 

E.g. Currents in two parts of the system with a few transformers in between cannot be compared directly



Per unit gives the relative values directly

Actual values

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PU values

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Classification of Faults Power System Faults Symmetrical

Shunt Faults

Unsymmetrical

Shunt Faults

Series Faults

Three-phase

Single-line to gnd

Unequal Series

3-phase to gnd

phase-to-phase

One line open

2-phase to gnd

Two lines open

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Classification of Faults Balanced or Symmetrical Faults 

Three phase faults



System remains balanced even after the fault



Per-phase analysis can be done 



the easiest fault to analyze

Generally the most severe fault a b c

a b c 3-phase

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3 phase - ground

53

Classification of Faults Un-balanced or Un-symmetrical Faults 



Un-balanced situation 

Unbalanced load



Any other unsymmetrical condition

System becomes unbalanced 

Single phase representation is not valid 



Per phase analysis CANNOT be done

Further classified into 

Shunt faults



Series faults

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Classification of Faults Shunt Faults 

Unbalance is between phases or between phase and neutral



Causes 

Insulation breakdown



Lightning discharges and other over-voltages



Mechanical damage

a b c

a b c

a b c e

e Single-line to ground

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Line to line

Double-line to ground

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Classification of Faults Series Faults 

Unbalance in the line impedances



Doesn’t involve ground or inter-connection between the phases



Causes 

Broken conductor



Operation of fuses



Maloperation of single phase circuit breakers

a b c

za zb zc

Unequal series Impedances

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a b c

a b c One line open

Two lines open

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Classification of Faults Other Faults 

Cross-country faults

a

a'

b

b'

c

c'

e

e

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Classification of Faults Other Faults 

Faults between adjacent parallel lines

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Classification of Faults Other Faults 

Open-circuit + Single-line to ground fault

a b c e

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Classification of Faults Other Faults 

Changing fault in cable - evolving faults

a

a c

b

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a c

b

c b

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Reactance of Synch. Machines Transients Caused by Synchronous Generator 

Consider only this ac component 

The amplitude of the ac component is given by,

Where E – Internal voltage or no-load voltage Xd – steady-state reactance, Xd’ – transient reactance, Xd” – sub-transient reactance Td’ – transient decay time constant, Td” – sub-transient decay time constant

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Reactance of Synch. Machines Transients Caused by Synchronous Generator 

The ac component of the current waveform 

Has three components

Steady-state component

Sub-transient component

Transient component

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Reactance of Synch. Machines Transients Caused by Synchronous Generator 

Transient and sub-transient currents 

decay at different rates over the sub-transient and transient period



Reduces to negligible values after a few cycles 

The sub-transient component first



The transient component next

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Reactance of Synch. Machines Sub Heading here

i(t)

t (sec) © Protection Engineering And Research Laboratories

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Reactance of Synch. Machines Sub Heading here

Steady-state component

i(t)

t (sec) © Protection Engineering And Research Laboratories

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Reactance of Synch. Machines Sub Heading here

Transient component

i(t)

t (sec) © Protection Engineering And Research Laboratories

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Reactance of Synch. Machines Sub Heading here Sub-transient component

i(t)

t (sec) © Protection Engineering And Research Laboratories

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Reactance of Synch. Machines Transients Caused by Synchronous Generator 

The time after a fault occurs can be divided into 3 periods 

Sub-transient – 2 to 3 cycles after the fault 



Transient – 10 to 15 cycles after the fault 



Characterized by the transient reactance Xd’

Steady-state – > 20 cycles after the fault 



Characterized by the sub-transient reactance Xd”

Characterized by the steady-state reactance Xd

Xd” < Xd’ < Xd

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Reactance of Synch. Machines Transients Caused by Synchronous Generator 

rms value of the steady current I 



rms value of the transient current I’ 



I’rms = Erms/Xd’

rms value of the sub-transient current I” 



Irms = Erms/Xd

I”rms= Erms/Xd”

I”rms > I’rms > Irms 

Since Xd” < Xd’ < Xd

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Transients due to Faults Symmetrical and Un-Symmetrical Fault Currents 

When to consider unsymmetrical ? 



Circuit breaker rating

Unsymmetrical currents usually ignored 

Assumed that it will decay fast

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Transmission Lines & Cables Representation of Transmission Lines & Cables 



Practical transmission lines 

Not spaced equilaterally



May not be transposed

The errors due to this are considered minimal 

Lines are taken to be symmetrical

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Transmission Lines & Cables Representation of Transmission Lines & Cables 

Distributed model of lines

L

R

L

R

C

L C

G

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R

L C

G

R C

G

G

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Transmission Lines & Cables Representation of Transmission Lines & Cables 

Shunt branches are ignored for short circuit calculations

L

R

L

R

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L

R

L

R

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Transmission Lines & Cables Representation of Transmission Lines & Cables 

A lumped parameter model is used 

Entire line lumped into a single impedance XL

RL

RL = the total resistance of the line XL = 2ᴨf×LLine, where LLine - is the total inductance of the line

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Transmission Lines & Cables Representation of Transmission Lines & Cables 

Represented by a series impedance 

Resistance equal to the total resistance of the line/cable



Reactance equal to the total series reactance of the line/cable

ZL

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Transformers Two Winding Transformer Equivalent Circuit I1

XL1

R1

XL2

I’2= I2/n

R2

I2

n = N1/N2

I0

V1

Xm

Rc

E2=E1/n

E1

N1



Winding resistance & Leakage reactance

Secondary side impedance 



N2

Primary side impedance 



V2

Winding resistance & Leakage reactance

Magnetizing branch

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Transformers Two Winding Transformer Equivalent Circuit I1

XL1

n2XL2

R1

Xm

Rc

V’2=nV2

N1



n = N1/N2

n2R2

I0 V1

I’2= I2/n

N2

Referring the secondary impedances to the primary side

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Transformers Two Winding Transformer Equivalent Circuit I1

XL1

R1

n2XL2

I’2= I2/n

n = N1/N2

n2R2 V’2=nV2

V1

N1



N2

Magnetizing impedance is very large compared to the leakage reactance 

Can be ignored without much error

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Transformers Two Winding Transformer Equivalent Circuit I1

XL1

n2XL2

I’2= I2/n

V’2=nV2

V1

N1



n = N1/N2

N2

The winding resistances are very small 

Can be safely neglected

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Transformers Two Winding Transformer Equivalent Circuit n = N1/N2

XT = XL1 + n2XL2

N1



N2

The primary & secondary leakage reactances can be lumped together 

Single reactance, XT



Obtained from the short circuit test of the transformer

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Transformers Two Winding Transformer Equivalent Circuit XTpu = XL1pu + XL2pu



n = N1/N2

Per unit equivalent diagram 

Total reactance is sum of the per unit reactance of the two windings

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Transformers Representation of Two Winding Transformers 

Represented by a series reactance 

Equal to the leakage impedance of the transformer



Obtained from the short circuit testing of the transformer

P

S

P

XTpu

S

Transformer

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Transformers Representation of Three Winding Transformers 

Each winding is represented by its own leakage reactance 



XP, XS, XT

Magnetizing impedance represented by the parallel branch ZM

Transformer

P

S

T

P

XP-pu

ZM-pu

XS-pu S XT-pu T N

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Transformers Representation of Three Winding Transformers 

Magnetizing impedance is very large 

Usually ignored

Transformer

P

S

P

XP-pu

XS-pu S XT-pu

T

T N

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Transformers Representation of Autotransformers 

The HV & LV windings are represented by the leakage reactances XH & XL



Magnetizing impedance represented by the parallel branch ZM

Autotransformer

H

L

H

XH-pu

XL-pu L

ZM-pu N

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Transformers Representation of Autotransformers 

Magnetizing impedance is very large 

Usually ignored

Autotransformer

H

L

H

XH-pu

XL-pu L

N

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Transformers Representation of Autotransformers 

Represented by a single leakage reactance 



XH & XL lumped together

Same as two winding transformer

Autotransformer

H

L

H

XHL-pu

L

N

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Transformers Representation of Autotransformers with Tertiary Delta Winding 

The HV & LV windings are represented by the leakage reactances XH & XL



The tertiary delta winding is represented by its leakage reactance XT



Magnetizing impedance is ignored

Autotransformer

H

L

∆ T

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H

XH-pu

XL-pu L XT-pu T N

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Synchronous Machines Representation of a Synchronous Generator 

An unloaded generator is represented by the no-load phase to neutral voltage in series with the appropriate reactance

Xg +

Eg -

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Synchronous Machines Representation of a Synchronous Generator 

Appropriate reactance is considered 

depends on time-period is of interest

Xg = Xd” or Xd’ or Xd +

Eg -

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Synchronous Machines Representation of a Synchronous Generator 

For fault currents immediately after the fault (typically 2-5 cycles) 

Use sub-transient reactance 



Eg: Instantaneous protection setting calculations

For fault currents few cycles after the fault (typically 25 cycles) 

Use transient reactance 

Eg: Time delayed protections such as generator back-up impedance protection



For sustained fault currents 

Use steady-state reactance 

Eg: Power swing calculations

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Synchronous Machines Representation of a Synchronous Motor 

When short-circuited, 

No more electric energy supplied to it



Still keeps rotating for some time 

Inertia of its rotor and connected load



Field remains energized



Starts acting like a generator

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Synchronous Machines Representation of a Synchronous Motor 

Synchronous motor behaves like a generator 



Especially larger capacity motors 



Contributes current to the fault

Contribution to fault current is significant

This current decays with time 

Stored energy is dissipated



Rotor starts slowing down

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Synchronous Machines Representation of a Synchronous Motor 

Contribution from small capacity motors 



Neglected – not included in fault calculations

Contribution from large motors 

Usually considered only if the sub-transient fault current is calculated

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Synchronous Machines Representation of a Synchronous Motor 

Usually modeled as a voltage behind Xd”

Xm = Xd”

Typical values from IEEE standard 399 



+



Em

2 - 6 poles

8 - 14 poles 

-



Xd” = 20%

≥ 16 poles 

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Xd” = 15%

Xd” = 28%

95

Transients due to Faults Representation of a Induction Motor 

When fault occurs 

Input power to motor is cut



Energy stored in the motor contributes to the fault



Contribution decays with time 



Typical decay rate - 100 to 150 ms

Modeled like the synchronous motor 

Voltage behind an impedance

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Transients due to Faults Representation of a Induction Motor 

Small Motors 

Contribution from small motors are negligible



Not considered for fault calculations



Motors of ratings < 35kW considered as small motors

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Transients due to Faults Representation of a Induction Motor Modeled as a voltage behind an impedance





Xm = Xd”

Large Motors > 35kW 

Xd” is approximately equal to the locked rotor reactance

+

Em -



Typical value can be used when exact values not known 

Typical value Xd” = 16.7% (IEEE Standard 399)

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Summary Synchronous Motors

Xm = Xd”

Xg = Xd” or Xd’ or Xd

+ Em

Induction Motors

Synchronous Generators

Xm = Xd”

+

+

-

Eg

Em

-

Typical Xd” = 16.7%

-

Trans. Lines & Cables ZL Autotransformer with ∆ tertiary

Three-winding transformer

P

XP-pu

2-wdg txfr & Auto-txfr XT-pu

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XS-pu S

H

XH-pu

XL-pu L XT-pu

XT-pu T

T

N

N

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Analysis of Symmetrical Faults

Introduction Symmetrical Faults 

As we saw earlier 

These are three-phase faults



System remains balanced



Per-phase analysis can be done

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Introduction Procedure 

In the given network replace … 

Transformers with their leakage reactances



Transmission lines with their series impedance  All shunt connections are neglected



Load impedances are neglected  Much larger than the network components



Synchronous machines represented by  No load-voltage behind Xd” or Xd’ 

Depends on time period of interest

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Introduction Procedure (contd.) 

Express voltage and impedances in per unit



Reduce the network



Solve for 

Fault current



Current through each branch



Bus voltages

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Single Source System Problem 1 T1

G

50 + j100 ohms

ΔY



Generator: 15 MVA, 13.8 kV, xd” = 0.15 pu



Load: 10 MVA at 0.9 pf lead



T1: 25 MVA, 13.2/161 kV, x = 0.10 pu



T2: 15 MVA, 161/13.8 kV, x = 0.10 pu

T2

Load

YΔ

For a three-phase fault at the load bus, find the symmetrical current fed into the fault

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Single Source System Problem 1 - Per-unit equivalent

MVA base = 15 kV base of 161 kV at the transmission line

0.0289 + j0.0579

G j0.06

j0.1639

j0.06

j0.1

0.0289 + j0.0579

j0.1

j0.1639 +

IF”

1.045 © Protection Engineering And Research Laboratories

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Single Source System Problem 1 - Reduced equivalent circuit 0.0289 + j0.0579

G j0.1639

j0.06

j0.1

0.0289 + j0.3818 + 1.045

IF” -

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Single Source System Problem 1 - Fault current calculation 0.0289 + j0.0579

G j0.1639

j0.06

j0.1

0.0289 + j0.3818

IF" =

+ 1.045

IF” -

1.045 = 0.206 " j2.7214 pu 0.0289 + j0.3818

IF" = 2.7292# " 85.67 0 pu

! © Protection Engineering And Research Laboratories

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Single Source System Problem 1 - Fault current calculation 0.0289 + j0.0579

G j0.1639

j0.06

j0.1

0.0289 + j0.3818

IBase =

+ 1.045

IF” -

MVA B 15 = kA = 627.55 A 3 " kVB 3 "13.8

IF" = 2.7292 " 627.55 = 1712.72# $ 85.67 0 A

! © Protection Engineering And Research Laboratories

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Single Source System Problem 1 - Solution T1

T2

ΔY

YΔ

Load

G

Fault current = 1712.72 ∠-85.670 A

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Three-phase Systems Unbalanced 3 Phase system

Va

Vc

Vb Its Sequence Components POSITIVE SEQUENCE

NEGATIVE SEQUENCE

Vb2

Vc1 1200

1200 1200

Vb1

ZERO SEQUENCE

1200

Va1

Vc2

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Va0

1200

Vb0

1200

Va2

Vc0

126

Three-phase Systems Points to note Vc1

Vb2 1200

Va1

1200

Vc2

Vb1



1200

1200

1200 Positive Sequence

1200

Va0 Vb0 Vc0

Va2 Negative Sequence

Zero Sequence

There is no direct relationship between the magnitude or angles of Va0, Va1 and Va2.



Similarly no direct relationship between the magnitude or angles of Vb0, Vb1 and Vb2

or Vc0, Vc1 and Vc2.

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Three-phase Systems Points to note Vb2

Vc1

Va0 1200

1200

Va1

1200 1200

Vc2

Vb1 Positive Sequence



1200

1200

Va2

Negative Sequence

Vb0 Vc0 Zero Sequence

The magnitude and the angles of Va0, Va1 and Va2 are determined by the system conditions.

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Symmetrical Transmission Line Sequence Impedances

a

Z0

+ Van0

I0

a’

a +

_ n

Zero

+

Va’n’0

_ n’

Z1

Van1

a’

I1

a +

_ n

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Positive

+

Va’n’1

_ n’

Z2

Van2

a’

I2

+ Va’n’2

_

_ n

Negative

n’

142

Star/star - Both Neutrals Grounded A

⦁

IA

⦁

N1:N2

N C B I0

P

3ZN

IC

⦁

a

Ia

n

ZN

⦁ ⦁

IN

⦁

Zn

In

IB

Ib

b

Ic

ZT

3Zn

V0’

Q

V0

P

I1

ZT

V1’

Reference Bus

Zero © Protection Engineering And Research Laboratories

I1

Q

V1 Reference Bus

Positive

P

I2

c ZT

I2

V2’

Q

V2 Reference Bus

Negative 143

Star/star - Ungrounded Neutral A

⦁

IA

N C B I0

P

3ZN

IC

⦁

⦁

N1:N2

a

Ia

n

ZN

⦁ ⦁

IN

⦁

Zn= ∞

IB

Ib

b

Ic

ZT

V0’

Q

V0

P

I1

ZT

V1’

Reference Bus

Zero © Protection Engineering And Research Laboratories

I1

Q

V1 Reference Bus

Positive

P

I2

c ZT

I2

V2’

Q

V2 Reference Bus

Negative 144

Delta/delta - Pu Sequence Circuit N1:N2

A

Ia

IA

∆

P

∆

Q

C

IC

⦁

⦁

Transformer

I0

ZT

V0’

⦁

⦁ ⦁

⦁

Vab

Ib Ic

IB

B

P

VAB

Q

V0

P

I1

ZT

V1’

Reference Bus

Zero © Protection Engineering And Research Laboratories

I1

Q

V1 Reference Bus

Positive

a

P

I2

ZT

b c I2

V2’

Q

V2 Reference Bus

Negative 145

Star/delta - Neutral Grounded A

∆

P

Per-unit Sequence Ckt. P V0’

Q

V0

B

P

IC

⦁

⦁

⦁

N C

ZT

Ia

VAN

Q

Transformer

I0 3ZN

N1:N2

⦁

IA

VN

ZN

Vab

⦁

⦁

IN

Ib Ic

IB

IA1

ZT

1∠30˚:1

VA1

Ia1

Q

Va1

P

IA1

ZT

1∠-30˚:1

Ia1

VA1

© Protection Engineering And Research Laboratories

Positive

b c

Q

Va1

Reference Bus

Zero

a

Negative 146

Star/delta - Neutral Ungrounded A

∆

P

Per-unit Sequence Ckt. P V0’

Q

V0

B

P

IC

⦁

⦁

N C

ZT

Ia

VAN

Q

Transformer

I0

N1:N2

⦁

IA

⦁

Vab

⦁

⦁

Ib Ic

IB

IA1

ZT

1∠30˚:1

VA1

Ia1

Q

Va1

P

IA1

ZT

1∠-30˚:1

Ia1

VA1

© Protection Engineering And Research Laboratories

Positive

b c

Q

Va1

Reference Bus

Zero

a

Negative 147

Synchronous Machines 

Positive & Negative sequence circuits 



n

Grounding has no effect

Zero sequence impedance 

Grounding impedance is included



Ungrounded generators - no path for zero sequence

XG1

I1

n

a +

XG2

I2

V1 _

Positive

n

n

© Protection Engineering And Research Laboratories

Negative

a +

V1 _

n

3ZN

gnd

XG0

I0

a +

V0 _

Zero 148

Sequence Networks Positive Sequence Network T1

T2

Load

G ZE

(Dyn1) ΔY

YY ZF

Equivalent Positive Sequence Network ZT1(1)

ZLine(1)

ZT2(1)

F1

1:1∠-300

ZG(1) + Eg

1:1∠-300 -

© Protection Engineering And Research Laboratories

Reference bus

N1 149

Sequence Networks Negative Sequence Network T1

T2

Load

G ZE

(Dyn1) ΔY

YY ZF

Equivalent Negative Sequence Network ZT1(2)

ZG(2)

ZLine(2)

ZT2(2)

F2

1:1∠300

1:1∠300

N2 © Protection Engineering And Research Laboratories

150

Sequence Networks Zero Sequence Network T1

T2

Load

G ZE

ΔY

YY ZF

Equivalent Zero Sequence Network ZT1(0)

ZLine(0)

ZT2(0)

F0

ZG(0)

3ZE

G0 © Protection Engineering And Research Laboratories

151

Inter-connection of seq. networks Power System Network – Before Fault T1

T2

G

ZE

YY

ΔY

Positive Sequence Network

Load

Negative Sequence Network

Z1

Z2

Zero Sequence Network Z0

+ Vf © Protection Engineering And Research Laboratories

152

Inter-connection of seq. networks Power System Network – Before Fault 

Positive Sequence 



Voltage source is present  Positive sequence currents flow

Negative and Zero Sequence 

No negative and zero sequence sources 

No Negative and zero sequence currents

Positive Sequence Network

Negative Sequence Network

Z1

Z2

i1

i2=0

Zero Sequence Network Z0

+ Vf

i0=0

© Protection Engineering And Research Laboratories

153

Inter-connection of seq. networks Faulted Power System T1

T2

Load

G

ZE

YY

ΔY

ZF

Sequence circuits for a fault at the load bus Z1

F1

Z2

F2

Z0

F0

+ Vf -

N1

© Protection Engineering And Research Laboratories

N2

N0 154

Inter-connection of seq. networks Faulted Power System– Redrawn sequence circuits Z1 Vf

+

-

+

I1

V1 -

Z2 I2

+

N1 F2

V2 -

Z0 +

I0

F1

N2 F0

V0 -

N0

© Protection Engineering And Research Laboratories

155

Single Line to Ground Fault Inter-connection of Sequence Networks Z1 Vf

+

-

+

I1

F1

V1 N1

Z2 I2

F2 +

V2 -

Z0 +

I0

3 × ZF N2 F0

V0 N0

© Protection Engineering And Research Laboratories

156

Three Phase Fault Inter-connection of sequence networks Z1 Vf

+

-

+

I1

F1 ZF

V1 N1

Z2 I2

F2 +

V2 -

Z0 +

I0

ZF

 Fault impedance to ground Zg  No effect on the fault

N2

current

F0 ZF + 3Zg

V0 N0

© Protection Engineering And Research Laboratories

170

Analysis of Un-Symmetrical Faults Three Phase vs. Single Phase Fault Level 

Three phase faults are generally the most severe ones 



Three phase fault level is the highest

However, sometimes 

Single phase fault levels can be higher than 3-phase 

Solidly grounded systems



On the Y-grounded side of Δ-Y-grounded transformer banks

© Protection Engineering And Research Laboratories

171

Analysis of Un-Symmetrical Faults Three Phase vs. Single Phase Fault Level 

Consider the following system with fault as shown

T G ΔY

© Protection Engineering And Research Laboratories

ZF

172

Analysis of Un-Symmetrical Faults Three Phase vs. Single Phase Fault Level 

Hence for the system shown …

G

T1 ΔY

Single-phase fault current

ZF

If Z0 < Z1, then Three-phase fault current

© Protection Engineering And Research Laboratories

1-φ fault level > 3-

φ fault level

173