A Presentation On

Short Circuit Analysis

BY

KALYAN RANJAN

RAHUL RANJAN

ISHAQ KHAN

M.Tech - EPSM

1ST SEMESTER

How short-circuit fault takes place ?

A short-circuit fault takes place when two or more conductors at different potentials are connected or a live conductor touching the ground either with zero resistance or with a substance of low resistance.

Effects of short circuit fault ……

A fault will cause currents of high value to flow through the network to the faulted point. The amount of current may be much greater than the designed thermal ability of the conductors in the power lines or machines feeding the fault. As a result, temperature rise may cause damage by annealing of conductors and insulation failure. In addition, the low voltage in the healthy phase of the faulty system will cause equipment to malfunction.

Fault analysis required because …….

The fault analysis of a power system provide information for the design of switchgear like isolaters, Circuit breakers, short circuit current limitting reactors and the design of settings of protective relays.

Cause for Faults 1. Insulation failure 3. flashover 2. human error 4. physical damage

Faults – Types

1. Unsymmetrical Three Phase Fault

* Occur frequently.

* Fault calculations are more difficult.

(A) Phase-to-earth fault

(B) Phase-to-phase fault

(C) Phase-to-phase-to-earth fault

2. Symmetrical Three Phase Fault

* Occur infrequently, but severe due to heavy fault level.

* Estimate the breaking capacity of the CB.

(D) Three phase fault

(E) Three phase-to-earth fault

Symmetrical Three Phase Fault Analysis

A three phase fault is a condition where either

(a) all 3-ph are short circuited to each other, or

(b) all 3-ph are earthed.

During these fault, symmetry (or balanced condition) may be observed in all the 3-phases. Due to symmetry, knowledge of the condition of one phase in a 3-phase system is sufficient to estimate the condition of other phases.This simplifies short circuit calculations to a great extent as calculations are made on a single-phase, rather than on 3-phase basis.

Per Unit quantity --Any electrical quantity (voltage, current, power, impedance)

in the per-unit system is define as,

Note that the numerator terms in these equations are in general complex while the base values are real-valued.

* Once any two of the four base values are defined, the remaining two base values can be determined according their fundamental circuit relationships.

* Usually the base values of power and voltage are selected and the base values of current and impedance are determined according to

In 1-ph power system : voltages and power are usually expressed in kV and MVA, thus it is usual to select an MVAbase and a kVbase and to express them as

In 3-ph systems :

The line voltage and the total power are usually used rather than the 1-ph quantities. It is thus usual to express base quantities in terms of these.

If MVA3φbase and KVLLbase are the base three-phase power and line-to-line voltage respectively then…

.In three phase, the calculations of per unit quantities becomes….

Change of bases ……..

Per unit impedance is directly proportional to the base kVA and inversely proportional to the square of the base voltage. Therefore, to change from per unit impedance on a given base to per unit impedance on a new base, the following equation applies:

Symmetrical Fault Calculation

Two tasks 1. Converting electrical quantities into per unit values and

2. Converting per unit single-line diagram into Thevenin’s equivalent circuit across the faulted point terminal & the ZPB terminal.

Let a 3-ph fault occur at i-th bus in power system. The Thevenin’s equivalent circuit is shown.

Where,

E = rated voltage(= Base voltageV)

I = Full load/ Rated current

(= Base current)

Isc = Short circuit current (V / Xth )

Xth = Thevenin’s equivalent reactance

E= 1 0

Xth ith

Isc

.

Base impedance Zb = V/ I Ω

Thevenin’s equivalent reactance in p.u. is given by

Xp.u. = Xth / Zb = (Xth I) / V p.u. = I / Isc

Percentage reactance , % X = Xp.u. * 100

= I / Isc * 100

or Isc = I * ( 100 / %X )

Multiplying both side with Rated voltage , we get,

V Isc = VI * ( 100 / %X )

or Short circuit KVA = Rated KVA * ( 100 / %X )

Example: A 3-ph 10 MVA, 11 KV alternator has 10% sub-transient reactance. Find short circuit MVA and current, if a symmetical fault occurs at its terminals.

Solution:Take the alternator rating as a base values.

Base MVA = 10 MVA,

Base KV = 11KV.

Base current = (10*10⁶)/(√3*11000)

= 524.86 A

Short circuit MVA = Base MVA*(100/%X)

= 10*(100/10) = 100 MVA

Fault current in amps = (Short circuit MVA*1000)/(√3*rated KV)

= 100*10⁶/(√3*11*1000 ) = 5248.6 A p.u. fault current = Fault current in amps / Base current

= 5248.6/524.86

= 10 p.u.

Unsymmetrical Three Phase Fault Analysis

Unbalanced three phase systems can be split into three balanced components

namely Positive Sequence, Negative Sequence and Zero Sequence.

These are known as the Symmetrical Components or the

Sequence Components and are shown in figure.

The phase components are the addition of the symmetrical components and can be written

as follows. a = a₂ + a₁ + a₀ b = b₂ + b₁ + b₀ c = c₂ + c₁ + c₀

Let us again examine the sequence components of the unbalanced

quantity, with each of the components written in terms of phase a

components, and the operator α, as in figure below.

Thus, This can be written in matrix form.

a = a0 + a1 + a2 b = a0 + α a1 + α a2

c = a0 + α a1 + α a2

2

2

Where,

[A] =

and [A] =

Sequence Impedances

The sequence impedance matrix in term of phase components.

[Vp] = [Zp] [Ip]

or [A] [Vs] = [Zp] [A] [Is]

or [Vs] = [A] [Zp] [A] [Is]

Thus, [Zs] = [A] [Zp] [A]

-1

-1

-1

Since,

So, [Zs] = 1/3 [A] [Zp] [A]

=

Thus, [Vs] = [Zs] [Is] …… From this equation it is clear that in symmetrical circuits with or without mutual coupling, currents of a given sequence produce voltage drops of the same sequence.The sequence imped –ances are said to be uncoupled.

-1

Definition of the operator α

we define a new complex operator αwhich has a magnitude of unity and when operated on any complex number rotates it anti-clockwise by an angle of 120⁰.

i.e. α= 1 120∠ ⁰ = - 0.500 + j 0.866

Some Properties of α

α2 = 1 240∠ ⁰; α3 = 1 360∠ ⁰ = 1 ; 1 + α + α2 = 0

The phase current can be wirtten in terms of sequence components as -

Iₐ = Iₐₒ + Iₐ₁ + Iₐ₂ = Ia0 + Ia1 + Ia2 ….. (1)

Ib = Ib₀ + Ib₁ + Ib₂ = Ia0 + α2 Ia1 + αIa2 ..… (2)

Ic = Ic₀ + Ic₁ + Ic₂ = Ia0 + αIa1 + α2 Ia2 ….. (3)

Adding equations (1),(2) & (3), we have

Iₐ + Ib + Ic = Iₐₒ + Ib₀ + Ic₀ ….. Since, Iₐₒ = Ib₀ = Ic₀ = 3Iₐₒ

Basic Voltage Current Network equations in Sequence Components

The generated voltages in the transmission system are assumed balanced prior to the fault, so that they consist only of the positive sequence component Ef (pre-fault voltage). This is in fact the Thevenin’s equivalent at the point of the fault prior to the occurrence of the fault.

Va0 = 0 – Z0 Ia0 This may be written in matrix form as

Va1 = Ef – Z1 Ia1

Va2 = 0 – Z2 Ia2

Single-Line-to-Ground FaultLet a 1-LG fault has occurred in a unloaded network such that phase-a

touched the ground through an impedance Zf .

At the fault

If = Ia, Ib = 0, Ic = 0

Vf = Va = IaZf = IfZf .

These can be converted to equivalent

conditions in symmetrical components as

follows.

So we have, Ia0 = Ia1 = Ia2 = Ia/3 = If/3 .

Since, Va0 = 0 – Z0 Ia0 Va1 = Ef – Z1 Ia1 Va2 = 0 – Z2 Ia2

So, fault voltage Va = Va0 + Va1 + Va2 = If.Zf = 3 Ia1.Zf

or, 0 – Z0 Ia0 + Ef – Z1 Ia1 + 0 – Z2 Ia2 = 3 Ia1.Zf

Or, 0 – Z0 If + Ef – Z1 If + 0 – Z2 If = 3 If.Zf

Thus,

Connection of Sequence Networks for L-G fault with Zf :

Example A 25 MVA ,13.2 KV alternator with solidly grounded neutral has a subtransient reactance of 0.25 p.u. The negative and zero sequence reactances are 0.35 and 0.1 p.u respectively. A L-G fault occurs at the terminal of an unloaded alternator, determine the fault current and L-L voltages. Neglect resistance.

Solution: Let the L-N voltage at the fault point before the fault is 1+j0.0 p.u.

For a L-G fault the fault impedance is j0.25+j0.35+j0.1=j0.7

Ia1= Ea /(Z1+Z2+Z0 )

=(1+j0.0)/j0.7

=-j1.428

Now we know that for a L-G fault Ia1 = Ia2 = Ia0 =-j1.428

The p.u fault current Ia = Ia1+Ia2+Ia0 =-j4.285

Let the base quantity be 25 MVA,13.2 KV and hence

The base current = 25000/(√3 ×13.2) = 1093 A

The fault current in amperes = 1093 × 4.285 =4685 ANow Va1 = Ea – Ia1 Z1

=1+j 0.0 – (-j1.428)(j0.25)

=0.643

Va2 = -Ia2 Z2

= - (-j1.428)(j0.35) = - 0.4998Similarly , Va0 = - Ia0 Z0

= - (-j1.428)(j0.1) = - 0.1428Now Va = Va1+Va2+Va0 = 0.643-0.4998-0.1428 = 0.0 Vb = Vb1+Vb2+Vb0

= α²Va1 + αVa2 + Va0

= (-0.5-j0.866)(0.643) + (-0.5+j0.866)(-0.4998) + (- 0.1428) = - 0.2143-j0.9898And Vc = Vc1+Vc2+Vc0

= αVa1+α²Va2+Va0 = (-0.5+j0.866)(0.643) + (-0.5-j0.866)(-0.4998) + (-0.1428) = (-0.2143+j0.9898)

Line to line voltage: Vab = Va – Vb

= 0.0 - (-0.2143 - j0.9898) = 0.2143 + j0.9898 Vbc = Vb – Vc = (- 0.2143 - j0.9898) - (- 0.2143 + j0.9898) = - j1.9796 Vac = Va – Vc = 0.0 – (- 0.2143 + j 0.9898) = 0.2143 - j0.9898 Vab = √(0.2143² + 0.9898²) =1.0127 p.u.

The line to line voltage (in volt) will be …… Vab = 1.0127 × 13.2/√3 = 7.717 kv Vbc = 1.9796 × 13.2/√3 = 15.08 kv Vac = 1.0127 × 13/ √3 = 7.717 kv

Line-to-Line FaultThe faulted segment for an L-L fault is shown below, where it is assumed that

the fault has occurred at node k of the network. In this the phases b and c got shorted through the impedance Zf . Since the system is unloaded before the occurrence of the fault we have

Ia = 0, If = Ib = - Ic

Vb = Vc + Zf If

=

Iao = 0 , Ia1 = - Ia2 = (αIf)/3

Now, Vb = Vc + Zf If

or, Va0 + α 2Va1 + α Va2 = Va0 + α Va1 + α 2 Va2 + Zf If

or, (α 2- α) Va 1 - (α 2- α) Va2 = Zf (Ia0 + α 2Ia1 + α Ia2)

or, (a2- a) (Va 1 - Va2) = (a2- a) Ia1Zf or, (Va 1 - Va2) = Ia1 Zf

Since, Va0 = 0 – Z0 Ia0, Va1 = Ef – Z1 Ia1, Va2 = 0 – Z2 Ia2 So, Ef – Z1 Ia1 – (– Z2 Ia2 ) = Ia1 Zf

or, Ef – Z1 Ia1 – Z2 Ia1 = Ia1 Zf

or,

So fault current,

If = Ib = Ia0 + a2Ia1 + aIa2 = (a2- a) Ia1 = -j√3 Ia1 =

Example: Determine the fault current and L-L Voltage at the fault when a L-L fault occurs at the terminals of the alternator described in the previous example.

Solution: Ia1 = Ea / (Z1 +Z2 ) = 1+j0.0 / (j0.25 + j 0.35 ) = - j1.667Now for a L-L fault Ia1 = - Ia2 = -j1.667 Ia2 = j1.667and Ia0 = 0.0 Ia = Ia1 + Ia2 + Ia0 = 0.0 Ib = Ib1 + Ib2 + Ib0

= α²Ia1 + α Ia2 + Ia0

= (-0.5-j0.866)(-j1.667) + (0.5+j0.866)(j1.667) + 0.0 = -2.8872 p.uThe base current is 1093 A Hence fault current = 1093 × 2.8872 = 3155.71 A

Now , Va1 = Ea – Ia1Z1

= 1+j0.0 – (-j1.667)(j0.25) = 0.5833 Va2 = -Ia2 Z2 = (-j1.667)(j0.35) = 0.5834 Va0 = 0.0 Va = Va1 + Va2 + Va0

= 0.5833 + 0.5834 + 0.0 = 1.1666 p.u Vb = α²Va1 + α Va2 + Va0

= (-0.5-j0.866)(0.5833) + (-0.5+j0.866)(0.5833) = - 0.5833 p.u. Vc = - 0.5833 p.u.

Hence Line Voltages are …..

Vab = Va – Vb =1.1666 – ( - 0.5833) =1.7499 p.uVbc = 0.0 p.uVac = Va – Vc =1.1666 – ( - 0.5833) = 1.7499 p.u

Now L-L Voltages are …….Vab = 1.7499 × 13.2 / √3 = 13.33 kvVbc = 0.0

Vac = 13.33 kv

Double- Line -to Ground FaultThe faulted segment for a 2LG fault is shown in Fig below where it is assumed

that the fault has occurred at node k of the network. In this the phases b and c got shorted through the impedance Zf to the ground. Let the system is unloaded before the occurrence of the fault. Therefore

………….. (1)

Also voltages of phases b and c are given by

…. (2)

Therefore,

We thus get the following two equations from above equation,

… (3) & …. (4)

Substituting (2) and (3) in (4) and rearranging we get (5)

……….. (5)

Also since I fa = 0 we have

Thevenin equivalent circuit for 2LG fault

From this figure we get

The zero and negative sequence currents can be obtained using the current divider principle as

And

Example: Determine the fault current and Line to Line Voltages at the fault when a L-L-G fault occurs at the terminals of the alternator described in the previous example.

Solution:Assuming the pre fault voltage as 1 + j 0.0 p.u Ia1 = Ea / (Z1+(Z0 Z2 / (Z0 + Z2 ) ) = (1 + j 0.0) / (j0.25+(j0.1*j0.35)/j0.45) = - j 3.0506 p.u.

Now for L-L-G Fault

Va1 = Va2 = Va0

Since Va1 = Ea – Ia1 Z1 = (1+j0.0) – (-j3.0506)(j0.25) = 0.2374 p.u

Va2 = Va0 = 0.2374 p.u Ia2 = - Va2/Z2

= -0.2374 /j 0.35 = j0.678 p.u Ia0 = -Va0 / Z0

= - 0.2374 / - j0.1 = j 2.374 p.u. Ia1 = - (Ia2 + Ia0) = - ( j0.678 + j2.374 ) = - j3.05 p.u

Now fault current,

If = Ib + Ic = 3 Ia0 = 3*j2.374 = j7.122So fault current in amperes = Base current * Fault current in p.u

=1093*7.122 = 7784.3 AVa = Va1 + Va2 + Va0 = 3 Va1 = 3*0.2374 = 0.7122 AVb = Vc = 0 A

And Line to line voltages , Vab = Va – Vb = 0.7122 A

Vbc = Vb – Vc = 0 A

Vac = Va - Vc = 0.7122 A

THANK YOU