9/25/13 An Overview Of Short Circuit Current (part 1) | EEP

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An Overview Of Short Circuit Current (part 1)

An Overview Of Short Circuit Current (part 1)

Basic concept

There are essentially four types of faults: three-phase, single line-to-ground, double line-to-ground, and line-to-line.

Each of these types of faults can result in different magnitudes of fault current.

In all types, however, there is a common element: an abnormally low-impedance path or shorted path for

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current to flow, hence the name short circuit current. Such a condition can lead to extremely high currents.

By Ohm’s Law, voltage equals current times impedance (resistance) . Therefore, when the impedancebecomes very low and the voltage does not change, the current becomes very high. Large electrical currentsproduce a lot of heat transfer, which increases the temperature of cables, transformers, etc.

The increase in temperature can cause insulation damage. These currents also produce high magneticforces, which can actually bend buses in switchgear.

H igh fault currents cause magnetic forces that are proportional to the square of the fault current.

Mathematical background, X/R ratio and type of fault current

The treatment of electrical faults should be carried out as a function of time, from the start of the event attime t = 0+ until stable conditions are reached, and therefore it is necessary to use differential equationswhen calculating these currents.

In order to illustrate the transient nature of the current, consider an RL circuit as a simplifiedequivalent of the circuits in electricity distribution networks.

This simplification is important because all the system equipment must be modeled in some way in orderto quantify the transient values which can occur during the fault condition.

For the circuit shown in Figure, the mathematica l expression which defines the behavior of thecurrent is:

e(t) = L di + Ri(t)

RL circuit as a simplified equivalentof the circuits in electricity-distribution networks

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This is a differentia l equation with constant coefficients, of which the solution is in two parts:

ia (t): ih (t) + ip(t)

Where:

ih(t) is the solution of the homogeneous equation corresponding to the transient period.ip(t) is the solution to the particular equation corresponding to the steady-state period.

By the use of differential equation theory, the complete solution can be determined and expressed in thefollowing form:

Where:

α – the closing angle which defines thepoint on the source sinusoidal voltagewhen the fault occurs

Ø = tan-1(ωL/R) or Ø = tan-

1(X/R)

The second term in the equation for fault current is recognized as the DC component of the current, andhas an initial maximum value when:

α - Φ = ± π / 2, and zero value when α = Φ.

Notes:

Here we introduce the concept of X/R ratio. We can very well see that since ωL = XL or simply X

hence DC component of fault current to large extent depends upon Ø = tan-1(X/R) or simply X/Rratio.The X/R ratio is important because it determines the peak asymmetrica l fault current.In X/R ratio when X equals zero, there is only symmetrical current with no DC component. With Requals zero, the DC component would never decay. One can say there will always be bothresistance and reactive components in the system.The resistance and reactance of a circuit establishes a power factor.The power factor (p.f.) is given by the following equation: p.f. = cos(tan-1(X/R)) this equationmeans that the power factor and X/R ratio are related.

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Therefore, system power factor and system X/R ratio are different ways of saying the samething . Please note that as power factor decreases, the X/R ratio increases.

It is impossible to predict a that is at what point the fault will be applied or take place on the sinusoidalcycle and therefore it is not possible to determine exactly what magnitude the DC component will reach.

Symmetrical fault current

If in a circuit mainly containing reactance a short circuit occurs at the peak of the voltage wave, theshort-circuit current would start at zero and trace a sine wave which would be symmetrical about the zeroaxis.

This is known as a symmetrica l short circuit current.

Asymmetrical fault current

Right after a fault occurs, the current waveform is no longer a sine wave.

Instead, it can be represented by the sum of a sine wave and a decaying exponentia l. Figure belowillustrates this phenomenon. Please note that the decaying exponential added to the sine wave causes thecurrent to reach a much larger value than that of the sine wave alone.

The waveform that equals the sum of the sine wave and the decaying exponential is called theasymmetrica l current because the waveform does not have symmetry above and below the time axis.

The sine wave alone is called the symmetrical current because it does have symmetry above and below thetime axis.

Sine wave, decaying exponential and their sum

Hence we can define asymmetrical fault current in the following way: If, in a circuit containing onlyreactance, the short circuit occurs at any point at the peak of the voltage wave, there will be some offsetof the current.

The amount of offset depends upon the point on the voltage wave at which the short circuit occurs.

This is known as asymmetrica l short circuit current. Maximum asymmetry occurs when short circuittakes place when voltage is zero.

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Asymmetrical fault remains only for few cycles after which it becomes symmetrica l fault. Decay ofasymmetrical component depends on the value of X/R. More the value of R, faster is the decay ofasymmetrical fault current.

Magnitude of asymmetrica l fault current is more than that of symmetrica l fault current.

If the short circuit current does not include DC component it is called symmetrica l short circuit current.If the short circuit current contains DC component it is called as asymmetrica l component.

Figure above represents the short circuit current with and without DC component.

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An Overview Of Short Circuit Current (part 2)

An Overview Of Short Circuit Current - Part 2 (on photo Main Switchboard by jayreynoldsisreal@Flickr)

Reactance

Sub transient reactance Xd” is the apparent reactance of the stator winding at the instant short circuitoccurs, and it determines the current flow during the first few cycles of a short circuit.

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Transient reactance Xd’ is the apparent initial reactance of the stator winding, if the effect of allamortisseur windings is ignored and only the field winding considered. This reactance determines thecurrent following the period when subtransient reactance is the controlling value.

Transient reactance is effective up to 1/2 sec. i.e. 30 cycles or longer, depending upon the design ofthe machine.

Synchronous reactance Xd is the apparent reactance that determines the current flow when a steady statecondition is reached.

It is not effective until several seconds after the short circuit occurs, consequently it has no value in shortcircuit calculations for the application of circuit breakers, fuses and contactors selection but is useful forrelay setting studies.

Below figure gives simplified representation of Asymmetrica l and symmetrica l fault current as well asdifferent reactance:

Asymmetrical and symmetrical fault current

During first few cycles reactance of system/synchronous machine is least and short circuit current ishighest. This stage is called subtransient reactance. This reactance is denoted by X”. After first fewcycles decrement in RMS value of short circuit current is less. This state is called transient reactance andis denoted by X’.

Finally transient dies out and current reaches the steady sinusoidal state. Reactance in this state is calledsteady state reactance and is denoted by Xd.

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Here we can introduce the concept of short circuit making and breaking current. During the first few cyclesof fault current the reactance is least and magnitude of short circuit current is highest. Current increases tomaximum value at the peak of first current loop.

All switching device are subjected to high electro-magnetic forces. To ensure that switching devicelike circuit breakers withstand safely this high magnitude of short circuit current, switching device is testedfor short circuit making current. Hence we can also define the short circuit making current as peakvalue of first current loop of short circuit current.

Short circuit making current= Peak value of steady state SC current + doubling effect caused by first peak containing DC component= 1.8 x peak value of steady state short circuit current (considering doubling effect)= 1.8 x √2 x RMS value of steady state short circuit current= 2.5 x RMS value of steady state short circuit current

Since RMS value of steady state short circuit current is called breaking current so short circuit makingcurrent can be written as:

Short circuit making current = 2.5 x short circuit breaking current

Above expression for calculating the making current is also given by Indian standard 10118, part-2 forselection, installation and maintenance of switchgear and controlgear.

However as per Indian standard 8623-part-1 for low voltage switchgear and controlgear assembly:

Relationship between peak and RMS. values of short-circuit current The value of peak short-circuit current(peak value of the first loop of the short-circuit current including DC. component) for determining theelectrodynamic stresses shall be obtained by multiplying the RMS. value of the short-circuit current bythe factor n.

Standard values for the factor n and the corresponding power factor are given in below table:

Table for selecting asymmetrical peak value

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RMS Value of Short circuit current cosΦ n

I ≤ 5 kA 0.7 1.5

5kA <I ≤ 10 kA 0.5 1.7

10kA <I ≤ 20 kA 0.3 2

20kA <I ≤ 50 kA 0.25 2.1

50kA <I 0.2 2.2

One can observe the difference in selection of multiplying factor n in case of two different IS. As per IS10118 part-2 multiplying factor should be 2.5 and as per 8623 part-1 multiplying factor should be n timesand n should be selected as per above table.

Since IS 8623 latest edition is 1998 and IS 10118 has been published before 8623 hence generalized valueof n in IS 10118 must have been elaborated in IS 8623. Also note the fact that IS 10118 considers thedoubling effect as 1.8 times which may vary depending upon amount of DC component which in turndepends upon the X/R ratio.

Exact and accurate knowledge of system X/R ratio is difficult to obtain, only power system engineerswho are exclusively involved in system studies can throw some light on it. Hence different values ofmultiplying factor at different power factor (in other words X/R) in IS 8623 is more reliable and is used byall switchgear manufacturers.

All the switchgears are type tested as per IS 8623 part-1 Readers are advised to refer the equivalent IEC439 part-1 for technical comparison and analysis of multiplying factor n.

Sources and limiters of short circuit current

When determining the magnitude of short-circuits currents, it is extremely important that a ll sources ofshort circuit current be considered and that the reactance characteristics of these sources are known.

Electrica l loads are either static (such as lighting) or dynamic loads (like motors). Dynamic loads haveresidual voltage and voltage of fault point is zero (if it is ground fault) or very less than line voltage hencecurrent starts flowing from dynamic loads to fault points.

During a short circuit condition the system voltage will decay. A stable voltage supply no longer exists.The rotating magnetic field in the rotor will attempt to support the reduced voltage condition by becominga power source. The motor is now providing additional current into the faulted electrical system.

This phenomenon is ca lled “motor contribution”.

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The amount of current is dependent on the motor impedance. At first there is an asymmetrica l currentcontaining both AC and DC components.

Lack of a stable voltage supply causes the AC component to decay when the rotor flux begins to drop.Without a stable voltage supply, the transient DC component also decays. Induction motor contributiontypically lasts from one to four cycles from time equal zero during a short circuit condition.

However, synchronous motors’ short circuit contribution can last from six to eight cycles. The maindifference is the induction motor does not have an excitation capability of a synchronous motor; therefore,it cannot maintain voltage for the same amount of time.

In either case the motor contribution is present during the first cycle.

There are three basic sources of short circuit current:

1. Generators2. Synchronous motors and synchronous condensers3. Induction motor

Due to residual flux in the rotor of the induction motor, it contributes fault current for 1-4 cycles.Normally, induction motor current contribution is considered for fault calculations.

ANSI standard C37.010 [1] offers guidance when calculating motor contribution for a group of lowvoltage motors if detail motor data are not available. Assuming a motor contribution of four times ratedfull load current is acceptable. The standard arrived at this value by assuming the motor contribution of 3.6times rated current came from 75% induction motors and 4.8 times rated current from 25% synchronousmotors.

A circuit element where voltage is induced by changing current in it is inductor and property is calledinductive property. As per Lenz Law rate of change of current is positive and induced voltage is negative.

Thus inductance acts in negative direction around the circuit to oppose change in current and hence canalso limit the short circuit current.

There are three basic limiters of short circuit current:

1. Transformer impedance2. Cable impedance3. Series reactor impedance (if any)

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An Overview Of Short Circuit Current (part 3)

Sample calculation fo r small LT system

Fault ca lculations are carried out to find the magnitude of fault current at various voltage levels ofelectrical system.

Short circuit calculations are actually just an elaborate version of Ohm’s Law. One of the key componentsin the calculation process is to determine the total impedance of the circuit from the utility / source,through the transmission system, transformers, and conductors, down to the point in question such as a

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panel or switchboard location. The impedances of the various circuit elements have both resistance andreactance and are often referred to as the “complex impedance” or “polar notation”.

Fault current values and time helps in deciding equipment short time withstand capacity and derivingsettings of protection relays. Interrupting capacity of protection equipment should be high enough to opensafely the maximum short circuit current which the power system can cause to flow through thatequipment.

One sample calculation for calculating the short circuit current at downstream of transformer is shownbelow.

Purpose and intent of this calculation is to ca lculate the short term current rating of a marshallingkiosk to be fed by AC distribution board (ACDB). ACDB being fed by a source of 630kVAtransformer.

Calculation basis

1/ The busbar and switchgear of Marshalling kiosk is sized for short time rating as per contribution fromMV source through LT transformer.

2/ For circuits connected by transformer PU system is particularly suitable. By selecting suitable base kVfor circuits the per unit reactance and resistance remains same, referred to either side (HV or LV ) oftransformer.

3/ For circuits connected by transformer same base kVA is selected for both the circuits (HV and LV )because power remains constant throughout so same base kVA should be considered throughout.

4/ As a rule only two bases should be selected first and from these two the remaining bases should becalculated. This is so because kV, kVA, I and Z are interrelated. They must obey ohms law. If we selectbase kVA and base kV than other base like base I and base Z are calculated from base kV and base kVA.

Vice-versa will be inconvenient that is selecting base I and Z and calculating other bases like kV and kVAwill make calculation difficult.

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Input data to be collected:

1. Transformer Rating = 0.63 MVA

2. Transformer Voltage ratio = 11/0.433 kV

3. Frequency = 50Hz

4. Transformer Impedance = 5% = 0.05 PU

5. MV System fault level (Maximum) = 40 kA

6. MV System fault MVA = √3 x 40 x 11 = 762 MVA

Calculation

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Actual Fault Current available at AC distribution board

Base MVA = 0.63

Base kV = 11

Base Current in kA =Base MVA/(√3 x Base kV) = 0.63/(√3 x 11)= 0.033

Base Impedance = (Base kV)2 / Base MVA = 192.1

Source Impedance = MV System fault MVA / BaseMVA

= 0.0008

LT Transformer impedance at 0.63MVA & 11kVBase

= 0.05

Total MV system impedance (MV System + LTTransformer)

= 0.0508

Fault MVA contributed by Source through LTTransformer

= Base MVA / Total Impedance

= 0.63 / 0.0508 = 12.40

Fault current contribution in kiloAmpers from MV system at LV side through (Switchyard) LT Transformer:

=Fault MVA x 1000 x 1000 / (√3 x 0.433 x1000 x 1000)

=12.40 x 1000 x 1000 / (√3 x 0.433 x 1000 x1000)

= 17.245 kA

Actual Fault Current available at marshalling kioskbusbar

Busbars and switchgear components of marshalling kiosk shall be braced for the peak value of thefaultcurrent contribution from the MV system through 630kVA rated source transformer.

Hereafter Marshalling kiosk to be referred as BMK and AC distribution board to be referred as ACDB.

Actual Fault Current available at marshalling kioskbusbar

Base kVA = same as above, since this parameter remains constant throughout the circuit

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Base kV = 0.415V Base kV at LV circuitDistance in meters of transformer fromACDB = 20Distance in meters of BMK from ACDB =50Size of connecting cable in Sq mm fromtransformer to ACDB = 3.5C x 300 Sqmm Al, XLPEResistance in Ohms/kM of connectingcable from transformer to BMK = 0.128

Total resistance over route length =20×0.128 /1000 = 0.003PU resistance = Actual Resistance x BasekVA/ (BasekV2 x 1000)= 0.003 x 0.63 x 1000 / (0.415 x 0.415 x1000) = 0.009

Reactance in Ohms/kM of connectingcable from transformer to BMK = 0.0705Total reactance over route length =0.0705×20 /1000 = 0.001

PU reactance = Actual Reactance x BasekVA/(Base kV2x 1000)= 0.001 x 0.63 x 1000 / (0.415 x 0.415 x

1000) = 0.0052

PU impedance of cable from LT transformer to ACDB= √((PU resistance)2+(PU reactance)2)= √(0.0092+ 0.07052) = 0.011

Size of conecting cable in Sq mm from ACDB to BMK = 3.5C x 35 Al, XLPEResistance in Ohms/kM of connecting cable from ACDB to BMK = 0.671Total resistance over route length = 0.671 x 50 /1000 = 0.034PU resistance = Actual Resistance x Base kVA/(Base kV2 x 1000)= 0.034 x 0.63 x 1000 / (0.415 x 0.415 x 1000) = 0.12

Reactance in Ohms/kM of connecting cable from ACDB to BMK = 0.0783Total reactance over route length = 0.0783 x 50 /1000 = 0.004

PU reactance = Actual Reactance x Base kVA/(Base kV2x 1000)= 0.004 x 0.63 x 1000 / (0.415 x 0.415 x 1000) = 0.14

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PU impedance of cable from ACDB to BMK = √((PU resistance)2+(PU reactance)2) = √((0.12)2+(0.14)2) = 0.124

Total PU Impeadnce of connecting cable from LT transformer to BMK = 0.011 + 0.124 = 0.134Total PU Impedance from LT Transformer to BMK = PU Impedance of Transformer + Total PUImpedance of connecting cable from LT transformer to BMK = 0.05 + 0.134 = 0.1842

Fault MVA at BMK busbar = Base MVA/Total Impedance = 0.63 / 0.1842 = 3.42

Fault current in kiloAmps at BMK busbar = Fault MVA x 1000x 1000 / (√3 x 0.415 x 1000 x 1000)= 3.42 x 1000 x 1000 / (√3 x 0.415 x 1000 x 1000) = 4.757 kA

Hence selection of 10kA busbar and switchgear components like MCB is safe and appropriate as per theactual fault level existing at BMK main busbar.

Si. No Equipment

CURRENT RATING

CALCULATED SHORT TERM CURRENT

RATING IN kA

OPTIMUM SELECTION OF SHORT TIME

CURRENT RATING IN kA

RMSSymmetrical

Assymmetricalpeak value =

nxRMSSymmetrical

RMSSymmetrical

Assymmetricalpeak value =

nxRMSSymmetrical

1 Main LT board 17.24 34.5 (n=2) 35 73.5 (n=2.1)

2 Marshalling kiosk 4.75 7.1 (n=1.5) 10 17 (n= 1.7)

References:

1. Indian Standard 8623, part-1-SPECIFICATION FOR LOW-VOLTAGE SWITCHGEAR ANDCONTROLGEAR ASSEMBLIES

2. Indian Standard 10118, part-2-CODE OF PRACTICE FOR THE SELECTION, INSTALLATIONAND MAINTENANCE OF SWITCHGEAR AND CONTROLGEAR

3. The Importance of the X/R Ratio in Low-Voltage Short Circuit Studies- Research paper DATE:November 17, 1999 REVISION: 0 by AUTHOR: John Merrell

4. Short-circuit-current Calculating Procedures by Donald Beeman, Alan Graeme Darling, and R. H.Kaufmann

5. Industrial Power Engineering and Applications Handbook by K.C. Agrawal