What is fault in power system-Symmetrical and unsymmetrical fault

Today in this article we will get some information from fault in Power System. So let’s see what we will get to know from this article.

What is a fault in power system?

A fault indicates abnormal conditions in the circuit. Most faults in a power system cause a short-circuit condition. because when such a condition occurs, a heavy current (called short circuit current) flows through the equipment, causing significant damage to the equipment and interruption of service to consumers.

There is no other subject of greater importance to an electrical engineer than the question of the determination of short circuit currents under fault conditions.

So the choice of equipment and design and arrangement of practically every piece of equipment in a power system depends on short circuit current considerations.

See also this : What is electrical short circuit-Definition, Effects

Define fault in power system

A fault occurs when two or more conductors that normally operate with a potential difference come in contact with each other. These faults may be caused by sudden failure of a piece of equipment, accidental damage or insulation failure as a result of a short-circuit or lightning surge to overhead lines.

Regardless of the cause, faults in a 3-phase system can be classified into two main categories. Hence the fault is (1) Symmetrical faults and (2) Un symmetrical fault.

Symmetrical fault

That fault which gives rise to symmetrical fault currents is called a symmetrical fault. The most common example of a symmetrical fault is when all three conductors of a three-phase line are simultaneously brought into a short-circuit condition.

So you can see the method of calculating fault currents for symmetrical faults is discuss below.

Unsymmetrical fault

Those faults which give rise to unsymmetrical currents (That is unequal line currents with unequal displacement) are called unsymmetrical faults, which may take any one of the following forms :

(a) Single line-to-ground fault (b) Line-to-line fault (c) Double line-to-ground fault. Since most faults on a power system are of unsymmetrical nature, and the most common type being a short-circuit from a line to ground. So such fault currents are calculated by the symmetrical components method.

Faults on the power system which give rise to symmetrical fault currents (that is equal fault currents in the lines with 120° displacement) are called symmetrical faults.

Symmetrical fault occurs when all three conductors of a three-phase line are brought together in short-circuit condition. So this type of fault gives rise to symmetrical currents, You can see in the figure below.

What is fault in power system

So now the fault current I_R , I_Y and I_B will be equal in magnitude and 120 of them will be displacement. Due to the balanced nature of the fault, only one phase needs to be considered in the calculation as the condition will be same in the other two phases also. Special attention may be given to the following points :

(1) Symmetrical fault rarely occurs in practice as most of the faults are unsymmetrical in nature. However, symmetrical fault calculations are being discuss in this chapter so that the reader can understand the problems that exist under short circuit conditions present in a power system.

(2) The symmetrical fault is the most severe and imposes more heavy duty on the circuit breaker.

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What are the 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 to the point of fault.

What is fault in power system

Since if a fault occurs at point F on the feeder, the value of the short circuit current from the generating station will be limited by the impedance of the generator and transformer and the impedance of the line between the generator and the point of fault.

So this shows that knowledge of the impedances of the various devices and circuits in the systems line is very important for the determination of the short-circuit current.

Percentage reactance

So the reactance of generators, transformers, reactors is usually express in % (percentage) reactance for short circuit calculation. The % reactance of a circuit is defined as the % of total phase voltage dropped in the circuit when full load current is flowing.

$X_{pu} = \frac{X_{actual}}{X_{base}}$

$X_{pu} = \frac{IX}{V}$

%X = X_pu_{× 100}

$%X = \frac{IX}{V} \times 100$

Where, I = Full load current

V = Phase voltage

X = Reactance in ohms per phase

$X_{pu} = \frac{IX}{V}$

$= \frac{VIX}{V^{2}}$

$%X = \frac{\frac{VA}{10^{3}}\times X \times 100}{\frac{V^{2}}{10^{6}}}$

$%X = \frac{kVA \times X \times 100}{(kV)^{2}\times 1000}$

Therefore, $%X = \frac{(kVA) X}{10(kV)^{2}} - What is fault in power system$

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

I_SC = V/X

Since we know that, %X = $\frac{IX}{V}\times 100$

So we can write, I_SC = $\frac{V \times I\times 100}{%X \times V}$

Therefore, I_SC = $\frac{I\times 100}{%X} - What is fault in power system$

So now we will see one example.

Question : If % reactance of an element is 20% and full load current is 50A then find the short circuit current.

Solution :

Given data : Full load current = 50A, % Reactance = 20%

So we know the formula, I_SC = $\frac{I\times 100}{%X} - What is fault in power system$

I_SC = $\frac{50 \times 100}{20}$

So the answer is, I_SC = 250A

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What is short circuit kva

Although the potential at the point of fault is zero, it is a common practice to express the short circuit current as short kVA based on the normal system voltage at the point of fault.

So the product of normal system voltage and short circuit current at the point of fault expressed in kVA is know as short circuit kVA.

Let, V = Normal phase voltage in volt

I = Full load current in ampere at base kVA

% X = % Reactance of the system on the base kVA up to the fault point

Since we know, Short circuit current (I_SC ) = $\frac{I \times 100 }{%X}$

Short circuit kVA for 3-Φ circuit = $\frac{3 V I_{SC} }{1000}$

= $\frac{3 V I \times 100}{%X \times 1000}$

= $\frac{Base kVA \times 100}{%X}$

Therefore, Short circuit kVA = $\frac{Base kVA \times 100}{%X} - What is fault in power system$

Therefore, Short circuit MVA = $\frac{Base MVA}{X_{_{pu}}} - What is fault in power system$

Steps for symmetrical fault calculation

So First of all Draw the single line diagram of the complete network indicating the rating, voltage and percentage reactance of each element of the network.
After than Choose a numerical convenient value of base kVA and convert all the percentage reactance to this base value.
Corresponding to the single line diagram of the network draw the reactance diagram showing one phase of the system and neutral. Indicate the percentage reactance on the base kVA in the reactance diagram. So the transformer in the system should be represented by a reactance in series.
Find the total % reactance of the network up to the point of fault, Let it be X%.
Find the full load current corresponding to the selected base kVA and at the normal system voltage at the fault point, let it be I.
Then various short circuit calculation are :

So, Short circuit current, I_SC = $\frac{I \times 100}{%X}$

Short circuit kVA, I_SC kVA = $\frac{Base kVA \times 100}{%X} - What is fault in power system$

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So now we will see one more example,

Q1. A three-phase, 20 MVA, 10kV alternator has internal reactance of 5% and negligible. Find the external reactance per phase to be connect in series with the alternator so that steady current on short circuit does not exceed 8 times the full load current.

Solution :

Given data : Base MVA = 20 MVA, Base kV = 10 kV and Reactance (X) = 5%

Voltage per phase, V_ph = (10×1000) / √3 = 5773.5 V

Full load current, √3×V_L ×I_L = 20×10⁶

I_L = (20×10⁶ ) / (√3×10×10³ ) = 1154.7 A

As the short circuit current is to be 8 times the full load current,

We know the formula, I_SC = I×100 / %X

So, I_SC / I = 8

Therefore, Total percentage reactance, %X = I×100 / I_SC

= (1 / 8) × 100 = 12.5 %

Therefore, External percentage reactance = 12.5 − 5 = 7.5 %

So we know that, %X = $\frac{IX}{V}\times 100$

$X = \frac{%X \times V}{I \times 100} - What is fault in power system$

$= \frac{7.5 \times 5773.5}{1154.7\times 100}$

Therefore, X = 0.375 Ω

Causes of faults in power system?

Faults in power system can occur due to many reasons like (1) lightning (2) high speed winds (3) snow (4) insulation failure (5) falling of tree on a line (6) bird shorting the line etc.

Effect of fault in power system?

Faults in a power system have the following harmful effects on the system, So that is : (1) Faults generally give rise to large currents which may damage the equipment in the line of the system.(2) Large fault current overheats the system equipment.(3) Some faults block the flow of power.(4) Faults can cause the system to become unstable.(5) Faults can cause the 3-phase system to become unbalanced and this affects the operation of 3-phase.

What is fault calculation?

Fault calculations are performed to determine the fault currents for different types of faults at different locations in the power system. So the data obtained from the fault calculation also serves to determine the setting of the relay that controls the circuit breaker.

What is symmetrical fault?

A symmetrical fault occurs when all three conductors of a 3-phase line are brought together under short circuit condition. Symmetrical faults rarely occur in power system. Experience shows that hardly 5% of overhead line faults are symmetrical faults.

What is unsymmetrical fault?

Unsymmetrical faults involve only one or two phases. When an unsymmetrical fault occurs, the fault currents become unequal in the lines with unequal phase displacements. In other words, unsymmetrical fault causes an unbalance among the phases. For this reason, each phase must be considered separately for fault calculations.

So friends, this was some information about Faults in Power System. Hope you will get a lot of information from this article, Thank you.