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SHORT-CIRCUIT CALCULATIONSHORT-CIRCUIT CALCULATION

INTRODUCTION

Designing an electrical system is easy and simple,if only the normal operation of the network istaken into consideration. However, abnormalconditions which are likely to occur anytime mustbe foreseen and should be taken seriously duringthe design stage. A good design must not only bemade simple, but most importantly, safe andmade simple, but most importantly, safe andreliable. An Electrical system must operatecontinuously during normal and healthy overloadsituations; its protective device must also tripexpeditiously to isolate the affected parts of thesystem during fault conditions.

Short-Circuit Calculation

Why Fault Occur?

Fault in power system occur because of insulation failure in plant

which may be caused by a system over-voltage such as switching

surge or a lightning stroke, or maybe due to broken insulators or

conductors and various causes in the system.

The following table shows approximate percentage wise the

various causes of faults:various causes of faults:

(i) Lightning 6%

(ii) Sleet, wind, Mechanical (jumping conductors) 10%

(iii) Apparatus failure 10%

(iv) Switching to a fault 10%

(v) Miscellaneous (tree falling on lines, sabotage) 10%


These faults can take one of the following forms:

Type Probability of failures

1. Single phase to ground faults 10%

2. Phase to phase faults 15%

3. Two-phases to ground faults 10%

4. Three-phase faults 5%

Such faults cause heavy currents called, short-circuit currents, to flow

in the system.in the system.

The determination of the values of such currents enables us to make

proper selection of circuit breakers, protective relays and also helps to

ensure that the associated apparatus e.g. bus bars, connections,

current transformers will withstand the forces which arise due to the

fault currents during the period prior to the interrupting device

clearing the fault.


Single phase to ground faults are the most common whereas the threephase short-circuit fault are the most severe faults and also the mostamenable to calculations since these involve symmetrical conditionsonly.For unsymmetrical ground faults, line-to-ground fault on a solidlygrounded system is used to size ground cable as well as to coordinateground fault protective devices.

Sources of Fault Power.

� All the generating and trandformers which under normalconditions take power from the system.conditions take power from the system.

� Synchronous Machines. Under short-circuit conditions, a drop infrequency or voltage is common and in this event, synchronousmachines will feed back into the system for a short period.

� Large Induction Motor. large induction motor’s flywheel effect willact as generators in the event of reduces frequency. Wheremachines such as these are connected, and they are of size as tohave effect particularly where they are connected to a point close tothat for which short-circuit values are being calculated, they shouldbe calculated.

� Frequency chargers


FAULT CURRENTS LIMITERS:

Cables

Busbars

Circuit breakers,

Reactors

Transformer

x

P Q

F

x

P Q

G F1

x

G .F1

G F3x

x F2

Fig. 1.2 - short-circuit fed from a source

having more then one generator

Fig. 1.1 - short-circuit fed from a

single generator


In Fig. 1.1, let P and Q represents the bus bars at the power station and

the sub-station respectively. Also F1 represents a feeder outgoing from

the sub-station bus-bars and equipped with circuit breaker at X.

If there is a single feeder connecting P and Q and the generating station

also consists of a single generator as in Fig 1.1, in the event of fault

occurring at any point in on feeder F1, the short-circuit current from the

generating station will be a certain value limited by the impedance of

the generator and the impedance of the feeder up to the point of fault.

Now supposed that an increased load is required to be fed from the

sub-station bus-bars. In that case, it may become necessary to increase

the generating units at the power station and also to install a second

feeder between P and Q to carry the increased load. ( It is assumed that

the original generating unit and feeder between P and Q were fully loaded).

With the increased load, which may be due to additional consuming

apparatus, it may be similarly necessary to have more outgoing feeders

F1 and F2 etc. as shown in Fig. 1.2.


The fault current from the generating station due to a fault occurring

on, say F1 will be greater than that in the original system due to

(i) large kVA of the generating plant,

(ii) smaller impedance of the generating plant. ( the equivalent

impedance of two generators in parallel is smaller than the

individual impedance of each.)

(iii) the smaller impedance of the two feeders connecting P and Q

since they are now in parallel.

The value of fault current which can flow in any system under short-

circuit is limited only by the impedance in that system. Therefore, it is

necessary in any calculation to have knowledge of these impedances.


Reactance. Often the resistance is so small that in most cases

reactance alone is considered for calculating the fault currents. In

general it is to be noted that if the reactance exceeds 3 times the

resistance, the resistance may be neglected. The error in the

assumption will not exceed 5%.

The reactance of synchronous machines, transformers, reactors, is

usually expressed as a percentage.usually expressed as a percentage.


In electrical systems, power flows from the sources to

loads through circuit components which has specific

impedances. The power flow is principally determined by

the load impedance.

A FAULT can be described as a load with very low

impedance. Depending upon the type of fault, Phase orimpedance. Depending upon the type of fault, Phase or

ground faults can be regarded as Source Limited (phase

fault) or Fault limited (ground faults).

All the fault calculating methods are based upon this

principle.


FAULTS CALCULATING METHODS:

Ohms Method

The basic formula for all current flow is Ohms Law,

Volts = Ampere x Ohms.

This is easily applied to single phase circuits operating at

only one voltage level. However, when several differentonly one voltage level. However, when several different

voltages occur, the situation becomes more complicated.

Each location in the circuit where faults must be

calculated has an specific operating voltage. All circuit

impedances must be referred to the fault voltage before

calculating the short-circuit current.


Formula 1 - is based for referring impedance to a new voltage at which

the fault is being calculated.

When all the impedances have been referred to the fault voltage, they

can be added.

Ohms at new V = Ohms at old V x ( Formula 1)

Example: Calculate the two faults indicated on the following

diagram.

New Voltage

Old Voltage

2

diagram.


Fault 1 - ISC1 = = 96 Amps

Fault 2 - refer 5 ohms from 480V to 240V

Ohms @ 240V = 5 x = 5 x 0.25

480 volts

5 ohms

240 V

480V

2

Ohms @ 240V = 1.25

Assume perfect transformer (Z = 0%)

Total Impedance = 1.25 + 3 = 4.25 ohms

Fault 2 - ISC2 = = 56.5 Amps

480V

240 volts

4.25 ohms


This conversion is necessary when a transformer must be included or

motor contribution is being considered. The formula is derived from the

fact that the impedance of the transformer or motor is the only

limitation to the circulation of current with a winding shorted or rotor

locked. The rating of voltage required to circulate full rated current to

the rated voltage is the Per Unit Impedance of the particular transformer

of motor.

Consider the following system consisting of a utility in-feed through the

cable to a transformer. The step by step solution is shown below.cable to a transformer. The step by step solution is shown below.

Example:


2


(Formula 5)

(Formula 6)

(Formula 9)


The circuit used to illustrate the Ohmic calculations is used below to

illustrate Per Unit calculations.


So far only three phase faults have been considered. Inall cases, the fault impedance was assumed to be zero.However, three phase faults do not often occur. Groundfaults are by far the most common and they generallyhave arc and contact impedance at the point of the fault.Impedance of the ground return path may also be animportant factor. The fact that the ground faults arerarely bolted (near zero fault impedance), does notmean the case may be ignored. Bolted phase to groundfaults are generally the theoretical worst case. For thisfaults are generally the theoretical worst case. For thisreason it is necessary to calculate them and understandthe theory behind the calculations.

The study of ground faults necessarily leads us tosymmetrical component analysis. The theory presentedhere is brief and no mathematical proof is included.


The bolted 3-phase faults which have been consideredthus far were calculated with the assumption that allvoltages and impedances were balanced, i.e., asymmetrical fault. When it becomes necessary to analyzean unsymmetrical fault (phase to phase, phase to ground,etc.) the calculations become quite complicated.

To simplify the analytical process, the unbalanced systemof vectors is reduced into three balanced vector systemswhich are known as POSITIVE, NEGATIVE, and ZEROwhich are known as POSITIVE, NEGATIVE, and ZEROphase sequence components. The positive sequencecomponents consist of three vectors equal in magnitude,120 degrees displaced and rotating ABC. The negativesequence components are three equal vectors, 120degrees displaced and rotating ACB. The zero sequencecomponents consists of three vectors equal in magnitudeand in phase. Each of these sets of voltage vectors isshown in the following diagram.


Now that three sets of sequence voltages have been assumed, It

follows that they are the result of sequence currents flowingfollows that they are the result of sequence currents flowing

through sequence impedances. Positive sequence currents only

flow through positive sequence impedance, etc. It is then

possible to draw an impedance diagram for each sequence in

order to calculate faults. The Theory and generation of these

diagrams is best described in applied Protective relaying

references. It will simply be stated here that the impedance

diagram is drawn and connected as shown in the following

sketches.


IMPEDANCE DIAGRAM


Since delta-delta and delta-wye groundedtransformers are most common, the equivalentdiagram of each is shown below:


For most circuits, the positive and negativesequence impedances can be assumed equal. Thezero sequence impedance will depend upon thelocation of delta windings in the circuit. The term Znrefers to impedance in the neutral or return pathfrom the fault to the transformer neutral connection.Zf refers to the impedance of the fault arc. Theexample assumes Zn and Zf are zero, i.e., a boltedphase to ground fault.phase to ground fault.


Step 1. Draw the impedance diagram. Note that Z0

only includes the transformer. Z0 isassumed equal to Z1.

Step 2. The impedance of each system element hasalready been calculated in the 3-phaseexample. All of the following impedanceshave been referred to the 480 volts level.

Utility = 0.0023 ohms (slide 16)

Cable = 0.0016 ohms (slide 17)

Transformer = 0.0115 ohms (slide 18)


Example: The basic form of the Ohmic impedancediagram is suitable for this example.

Step 1. Draw the impedance diagram.

Step 2. The impedance calculated for the 3-phasefault are valid for this condition.

Utility = 0.1 0/1 ohm

Cable = 0.0694 0/1 ohm

Transformer = 0.5 0/1 ohmTransformer = 0.5 0/1 ohm

Step 3. Calculate the total impedance.

Zt1 = 0.6694 0/1 ohm = Zt2

Zt2 = 0.6694 0/1 ohm

Zt0 = 0.5 0/1 ohm

ZT = 1.834 0/1 ohm

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