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Thursday, 26 February 2015

MATLAB coding to find out post fault Current and Voltages

In a power system, faults occur because of insulation failure, or because of a damaged insulator, or a broken conductor. Various other reasons such as improper operating habits may also lead to a fault; for example, loading a distribution transformer beyond its normal rated capacity. 

Nearly one half of the faults occur on power lines which are widely branched, have greater length, operate under variable weather conditions and are more exposed to atmospheric disturbances. Faults give rise to abnormal operating conditions. 

When a fault occurs at any point in the power system large currents, large forces and or abnormal voltages are developed. The excessive current because of the fault is determined by the internal e.m.f.s of the machines in the network, their impedances, and the impedance in the network between the machines and the fault.

Faults currents, also called short circuit currents, are many times greater than the normal currents. Large voltage stresses the insulation of the various equipments which are way beyond their breakdown value causing the failure. Sometimes faults lower the system voltage below the permissible voltage limit causing unwanted and teasing interruption of various equipments and components. Faults can also cause a three-phase system to become unbalance.

To obtain proper setting of the protective relays and the interrupting capacities of circuit breakers, the values of these fault currents and voltages should be known with great accuracy. Short circuit studies and calculations provide currents and voltages on a power system during fault conditions. 

For Unsymmetrical faults

The majority of faults that occur in a power system are unsymmetrical faults involving only one or two phases. The most common type of unsymmetrical fault is a short circuit between a phase and the earth. In case of unsymmetrical faults, voltages and currents in the network become unbalanced and each phase is to be treated individually for computational purpose.

The magnitude of fault currents in the three lines is different having unequal phase displacements. The calculation procedure called as “method of symmetrical components” is used to find the currents and voltages in this type of fault.      

In this blog we are going to find out how to write the MATLAB code so that the post fault currents and voltages, in case of occurrence of an unsymmetrical fault, can be determined. 

Lets us assume that a 25 MVA alternator is working without load. A single line to ground fault occurs at one of the terminals of the alternator. The alternator has positive sequence impedance (Z1) of 0.25 p.u., negative sequence impedance (Z2) of 0.35 p.u. and zero sequence impedance (Z0) of 0.1 p.u. Now we have to find the fault current and line to line voltages.

Let the Line to neutral voltage at the fault point before the fault, ‘Ea’ be 1+ 0i p.u. 

The MATLAB coding is as follows:
>> Ea= 1+0i;
>> Z1= 0.25i; Z2= 0.35i; Z0= 0.1i;
Assuming that fault occurs at phase ‘a’, the positive sequence component of current in the ‘a’ phase (for a single line to ground fault without impedance),
>> Ia1= (Ea/ (Z1+Z2+Z0));
% Also, for a single line to ground fault, Ia1=Ia2=Ia0
>> Ia2= Ia1;
>> Ia0 =Ia1;
% Also, Fault current in phase ‘a’, Ia = Ia1+ Ia2 + Ia0
>> Ia = Ia1+ Ia2+ Ia0;
% From the positive sequence network
>> Va1= Ea-(Ia1*Z1);
% From the negative sequence network
>> Va2 = -Ia2*Z2;
% From the zero sequence network
>> Va0= -Ia0*Z0;
% For operator ‘a’ i.e. an operator which causes a rotation of 120 degrees in the anti-clockwise direction.
>> a=pol2 rect(1,((pi/180)*120));
% If Va1, Vb1 and Vc1 are the positive sequence component of the unbalanced voltages,
>> Vb1=a^2*Va1;
% If Va2, Vb2 and Vc2 are the negative sequence component of the unbalanced voltages,
>> Vb2=a*Va2;
% If Va0, Vb0 and Vc0 are the negative sequence component of the unbalanced voltages,
>> Vb0=Va0;
>> Vc0= Va0;
>> Vc1=a*Va1;
>> Vc2=a^2*Va2;
>> Vb= Vb1+Vb2+Vb0;
>> Vc= Vc1+Vc2+Vc0;
>> Va=0;
>> Vab= Va-Vb;
>> Vab_mag=abs(Vab);

 Similarly we can find out the values of line voltages Vbc and Vca. The values of fault current and post fault voltages are in p.u. values, we can convert them into actual values by assuming proper base values.