ESLAVA, RITCHELLE M.

1. Symmetrical Faults on 3 Phase System:

Most of the Symmetrical Faults on 3 Phase System on the power system lead to a short-
circuit condition. When such a condition occurs, a heavy current (called short circuit current)
flows through the equipment causing considerable damage to the equipment and interruption of
service to the consumers. There is probably no other subject of greater importance to an
electrical engineer than the question of determination of short circuit currents under fault
conditions. The choice of apparatus and the design and arrangement of practically every
equipment in the power system depends upon short-circuit current considerations.

That fault on the power system which gives rise to symmetrical fault currents (i.e. equal fault
currents in the lines with 120° displacement) is called a symmetrical fault.

The symmetrical fault occurs when all the three conductors of a 3-phase line are brought
together simultaneously into a short-circuit condition as shown in Fig. 17.1. This type of fault
gives rise to symmetrical currents L e. equal fault currents with 120° displacement. Thus
referring to Fig. 17.1, fault currents IR, ly and IB will be equal in magnitude with 120°
displacement among them. Because of balanced nature of fault, only one phase need be
considered in calculations since condition in the other two phases will also be similar.

The following points may be particularly noted:

1. The symmetrical fault rarely occurs in practice as majority of the faults are of
unsymmetrical nature. However, symmetrical fault calculations are being discussed in
this chapter to enable the reader to understand the problems that short circuit conditions
present to the power system.
2. The symmetrical fault is the most severe and imposes more heavy duty on the circuit
breaker.
 ESLAVA, RITCHELLE M.

2. Limitation of Fault Current:

When a short circuit occurs at any point in a system, the short-circuit current is limited by
the impedance of the system up to the point of fault. Thus referring to Fig. 17.2, if a Symmetrical
Faults on 3 Phase System occurs on the feeder at point F, then the short circuit current from the
generating station will have a value limited by the impedance of generator and transformer and
the impedance of the line between the generator and the point of Symmetrical Faults on 3
Phase System. This shows that the knowledge of the impedance of various equipment and
circuits in the line of the system is very important for the determination of short-circuit currents.

In many situations, the impedances limiting the Symmetrical Faults on 3 Phase System current
are largely reactive, such as transformers, reactors and generators. Cables and lines are mostly
resistive, but where the total reactance in calculations exceeds 3 times the resistance, the latter
is usually neglected. The error introduced by this assumption will not exceed 5%.

3. Percentage Reactance in Power System:

The Percentage Reactance in Power System of generators, transformers, reactors etc.
is usually expressed in percentage reactance to permit rapid short circuit calculations.

The percentage reactance of a circuit is defined as under:

It is the percentage of the total phase-voltage dropped in the circuit when full-load
current is flowing.

i.e.,

Where
I = full-load current
V = phase voltage
X = reactance in ohms per phase

Alternatively, percentage reactance (%X) can also be expressed in terms of kVA and kV as
under:
 ESLAVA, RITCHELLE M.

Where X is the reactance in ohms.

If X is the only reactance element in the circuit, then short-circuit current is given by;

i.e. short circuit current is obtained by multiplying the full-load current by 100 / % X.

For instance, if the Percentage Reactance in Power System of an element is 20% and the full-
load current is 50 A, then short-circuit current will be 50 x 100/20 = 250 A when only that
element is in the circuit.

It may be worthwhile to mention here the advantage of using Percentage Reactance in Power
System instead of ohmic reactance in short-circuit calculations. Percentage reactance values
remain unchanged as they are referred though transformers, unlike ohmic reactances which
become multiplied or divided by the square of transformation ratio. This makes the procedure
simple and permits quick calculations.

4. Percentage Reactance and Base WA:

It is clear from exp. (ii) above that percentage reactance of equipment depends upon its
kVA rating. Generally, the various equipment used in the power system had different kVA
ratings. Therefore, it is necessary to find the Percentage Reactance in Power System of all the
elements on a common kVA rating. This common kVA rating is known as base kVA. The value
of this base kVA is quite unimportant and may be:

 equal to that of the largest plant

 equal to the total plant capacity
 any arbitrary value

The conversion can be effected by using the following relation:

Thus, a 1000 kVA transformer with 5% reactance will have a reactance of 10% at 2000 kVA
base.
 ESLAVA, RITCHELLE M.

The fact that the value of base kVA does not affect the short circuit current needs
illustration. Consider a 3-phase transmission line operating at 66 kV and connected through a
1000 kVA transformer with 5% reactance to a generating station bus-bar. The generator is of
2500 kVA with 10% reactance. The single line diagram of the system is shown in Fig. 17.3.
Suppose a short-circuit fault between three phases occurs at the high voltage terminals of
transformer. It will be shown that whatever value of base kVA we may choose, the value of
short-circuit current will be the same.

(i) Suppose we choose 2500 kVA as the common base kVA. On this base value, the reactance of
the various elements in the system will be:

Reactance of transformer at 2500 kVA base

Reactance of generator at 2500 kVA base

Total percentage reactance on the common base kVA

The full load current corresponding to 2500 kVA base at 66 kV is given by;

(ii) Now, suppose we choose 5000 kVA as the common base value.
Reactance of transformer at 5000 kVA base
 ESLAVA, RITCHELLE M.

Reactance of generator at 5000 kVA base

Total percentage reactance on the common base kVA

Full-load current corresponding to 5000 kVA at 66 kV is

which is the same as in the previous case.

From the above illustration, it is clear that whatever may be the value of base kVA, short-circuit
current is the same: However, in the interest of simplicity, numerically convenient value for the
base kVA should be chosen.

5. Short-Circuit kVA
Although the potential at the point of fault is zero, it is a normal practice to express the short-
circuit current in terms of Short Circuit kVA based on the normal system voltage at the point of
fault.

The product of normal system voltage and short-circuit current at the point of fault expressed in
kVA is known as Short Circuit kVA.

Let
V = normal phase voltage in volts
I = full-load current in amperes at base kVA
%X = percentage reactance of the system on base kVA up to the fault point

Short-circuit kVA for 3-phase circuit

 ESLAVA, RITCHELLE M.

i.e. Short Circuit kVA is obtained by multiplying the base kVA by 100/% X.

6. Reactor Control of Short Circuit

With the fast expanding power system, the fault level (i.e. the power available to flow into
a fault) is also rising. The circuit breakers connected in the power system must be capable of
dealing with maximum possible short-circuit currents that can occur at their points of connection.
Generally, the reactance of the system under fault conditions is low and fault currents may rise
to a dangerously high value. If no steps are taken to limit the value of these short-circuit
currents, not only will the duty required of circuit breakers be excessively heavy, but also
damage to lines and other equipment will almost certainly occur.

In order to limit the short-circuit currents to a value which the circuit breakers can handle,
additional reactance known as reactors are connected in series with the system at suitable
points. A reactor is a coil of number of turns designed to have a large inductance as compared
to its ohmic resistance. The forces on the turns of these reactors under short-circuit conditions
are considerable and, therefore, the windings must be solidly braced. It may be added that due
to very small resistance of reactors, there is very little change in the efficiency of the system.

Advantages
 Reactors limit the flow of short-circuit current and thus protect the equipment from over-
heating as well as from failure due to destructive mechanical forces.
 Troubles are localized or isolated at the point where they originate without
communicating their disturbing effects to other parts of the power system. This increases
the chances of continuity of supply.
 They permit the installation of circuit breakers of lower rating.

7. Location of Reactors
Reactors are located at different location in a power system for reducing the short circuit
current. These reactors may be connected in series with the generators, feeders or in
bus-bars as explained below.

a. Generators Reactors
Generator reactors are inserted between the generator and the generator bus. Such
reactors protect the machines individually. In power station generator, reactors are
installed along with the generators. The magnitude of reactors is approximately about
0.05 per unit. The main disadvantages of such type of reactors are that if the fault occurs
on one feeder, then the whole of the system will be adversely affected by it.
ESLAVA, RITCHELLE M.

b. Feeders Reactors
Reactors, which is connected in series with the feeder is called feeders
reactor. When the fault occurs on any one feeder, then the voltage drops occur only in
its reactors and the bus bar is not affected much. Hence the machines continue to
supply the load. The other advantage is that the fault occurs on a feeder will not affect
the others feeders, and thus the effects of fault are localized.

The disadvantage of such type of reactors is that it does not provide any protection to
the generators against short circuit faults occurs across the bus bars. Also, there is a
constant voltage drop and constant power loss in reactors during normal operating
conditions.
c. Bus-Bar Reactor
When the reactors are inserted in the bus bar, then it is called bus-bar reactors.
The constant voltage drop and constant power loss in reactors may be avoided by
inserting the reactors in the bus bars. The bus bar reactor for ring system and the tie
system are explained below.

 Bus-Bar Reactors (Ring System)

Bus-bar reactors are used to tie together the separate bus sections. In this system sections
are made of generators and feeders and these sections are connected to each other to a
common bus bar. In such type of system normally one feeder is fed from one generator. In
ESLAVA, RITCHELLE M.

normal operating conditions a small amount of power flows through the reactors. Therefore
voltage drop and the power loss in the reactor is low. The bus bar reactor, therefore, made with
high ohmic resistance so that there is not much voltage drop across it.

When the fault occurs on any one feeders, only one generator feeds the fault while the
current of the other generator is limited because of the presence of the bus-bar reactors. The
heavy current and voltage disturbances caused by a short circuit on a bus section are reduced
and restricted to that faulty section only. The only drawback of such type of reactor is that it
does not protect the generators connected to the faulty sections.

 Bus-bar Reactors (Tie-Bus System)

This is the modification of the above system. In tie-bus system, the generator is
connected to the common bus-bar through the reactors, and the feeder is fed from generator
side.

The operation of the system is similar to the ring system, but it has got additional advantages. In
this system, if the number of sections is increased, the fault current will not exceed a certain
value, which is fixed by the size of the individual reactors.
 ESLAVA, RITCHELLE M.

8. Steps for Symmetrical Fault Calculation

Symmetrical Fault in Power System refers to kind of fault which leads to short circuit of
three phases. This may be either short circuit of three phases or three phases to ground fault. In
symmetrical fault, fault currents in phases are symmetrical in the sense that their magnitudes
are equal and they are equally displaced by an angle of 120°. Thus, a symmetrical fault may be
assumed as a normal case but with high value of current in phases.

Let us consider a symmetrical fault as shown in figure below.

In the figure above, all the three phases i.e. R, Y and B are shorted. This can be assumed as
the three lines are connected to a delta connected load with zero impedance in each phase.
Taking this assumption, the current in lines will be symmetrical but with higher value. The value
of fault current i.e. IR, IY and IB in such fault will only be limited by the source impedance as well
as the impedance of line up to the point of fault.

Well, we can analyze such fault by assuming single phase as the condition in other two phases
are similar.

Symmetrical Fault in power system is very rare and they rarely take place. But still the study of
Symmetrical Fault is important as it is the severe fault and imposes high duty on breaker and
equipment.

Calculation of Symmetrical Fault Current

When a short-circuit take place at any point in the system, the short-circuit current is
limited by the system impedance up to the point of fault. Let us assume a fault as shown below.
ESLAVA, RITCHELLE M.

In the above figure, the impedance up to the point of fault is the sum of impedance of
Generator, Transformer and Line. Thus we need to calculate the impedance of each element of
system up to the location of fault. As different element of system has different kVA rating
therefore per unit or percentage impedance is calculated assuming appropriate kVA rating as
base value.

If the value of per unit or percentage impedance up to the point of fault is X % then short circuit
current Isc is given as

Isc = (V / X %)

But percentage impedance X % = 100x( IX / V)

Therefore,

Isc = I (100 / % X)

where I = Full load current

For example, if the percentage impedance up to point of fault is 10% and full load current is 40
A then short circuit current will be [40 x (100 / 10)] = 400 A.

Example:

Consider the figure below and calculate the symmetrical fault current considering three
phase short circuit at the HV terminal of Transformer. The impedance values are mentioned in
the figure.

Solution:

As we need to calculate the short circuit current at the HV terminal of the transformer, therefore
first of all we need to calculate the total impedance of the system up to the HV terminal of
transformer. But for finding the percentage impedance we need to select a base kVA rating.
Well, we choose 1000 kVA as the Base value as calculation with 1000 is quite easier.

% impedance of Generator at Base kVA = (10 / 2500)x1000

=4%
ESLAVA, RITCHELLE M.

% impedance of transformer at base kVA = (5 / 1000)x1000

=5%

Thus total impedance up to the point of fault = (5 + 4) %

=9%

Now we need to find the full load current at the base kVA at the HV side of transformer.

Full load current I = 1000 / (1.732 x66)

= 8.75 A

Therefore,

Short Circuit current Isc = 8.75x(100 / 9)

= 97.2 A