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EEET 421 lab 3

Name and No. Date:

Section : Score:

Calculation of symmetrical fault current in a power system by calculation and using MATLAB

PERFORMANCE OBJECTIVE

Upon completion of this laboratory exercise, the student can use MATLAB program to solve the

symmetrical fault calculation problems.

3. The reactance data for the power system is shown in figure in

per unit on a common base as shown below.

Item X1 X2 X0

G1 0.10 0.10 0.05

G2 0.10 0.10 0.05

T1 0.25 0.25 0.25

T2 0.25 0.25 0.25

Line 1-2 0.30 0.30 0.50

Calculate the fault current if a symmetrical fault occurs at bus 1.

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Theory:Short circuits and other abnormal conditions often occur on a power system.Short circuits areusually called faults by power system engineers. Some defects, other thanshort circuits are

also termed as faults.Faults are caused either by insulation failures or by conducting path failures.The failureof insulation results in short circuits which are very harmful as they maydamage someequipment of the power system. Most of the faults in transmission anddistribution lines arecaused by over voltages due to lightning or switching surges, or by externalconducting objectsfalling on overhead lines. Overvoltages due to lightning or switching surgescause flashover on

the surface of insulators resulting in short circuits. Short circuits are alsocaused by tree branchesor other conducting objects falling on the overhead lines.The fault impedance being low, the fault currents are relatively high. Thefault currents being excessive, they damage the faulty equipment and thesupply installation. Also, the systemvoltage may reduce to a low level, windings and busbars may suffermechanical damage due tohigh magnetic forces during faults and the individual generators in a powerstation or group ofgenerators in different power stations may loose synchronism

The symmetrical fault occurs when all the three conductors of a three-phaselineare brought together simultaneously into a shortcircuit condition as shownin Figure 1.

This type of fault gives rise to symmetrical currents i.e. equal fault currentswith 1200

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EEET 421 lab 3

displacement. Thus referring to Figure 1, fault currents IA, IB and IC will beequal in magnitudewith 1200 displacement among them. Because of balanced nature of fault,only one phase needs to be considered in calculations since condition in theother two phases will also be similar.

A three-phase short circuit occurs rarely but it is most severe type of faultinvolving largestcurrents. For this reason, the balanced short-circuit calculations areperformed to determine theselarge currents to be used to determine the rating of the circuit breakers.

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Matlab Program

zdata1 = [0 1 0 0.35

0 2 0 0.35

1 2 0 0.3];

Zbus1 = zbuild(zdata1);

disp('(a) Symmetrical three-phase fault')

symfault(zdata1, Zbus1)

Review Questions-

1. What is a fault in power system?

2. What are the effects of fault?

3. What is meant by symmetrical and unsymmetrical fault?

4. Why fault calculation is necessary?

5. Change the fault location to bus number 4 and solve the problem again. What is the new fault current?