Symmetrical Fault Analysis in Power Systems

Choose a base of 50 MVA, 20 kV in the circuit of the generator. Then the motor synchronous reactance is given by \[X_{m}^{''}=0.2\times\dfrac{50}{25}=0.4~pu\]

Base impedance of the transmission line is \[Z_{base}=\dfrac{66^{2}}{50}=87.12~\Omega\]

Impedance of the transmission line is \[X_{line}=j\dfrac{10}{87.12}=j0.1148~pu\]

In impedance diagram, switch S indicates the fault

Motor draws a load current at rated voltage and rated MVA with 0.9 lagging power factor. Therefore \[i_{L}=1\angle-cos^{-1}\left(0.9\right)=0.9-j0.4359~pu\]

Sub-transient voltages of the motor and the generator are \[\begin{aligned} E_{m}^{''} & =1.0-j0.4\times i_{L}=0.8256-j0.36~pu\\ E_{g}^{''} & =1.0+j0.5148\times i_{L}=1.2244+j0.4633~pu \end{aligned}\]

Sub-transient fault currents fed by the motor and the generator are \[\begin{aligned} I_{m}^{''} & =\dfrac{E_{m}^{''}}{j0.4}=-0.9-j2.0641~pu\\ I_{g}^{''} & =\dfrac{E_{g}^{''}}{j0.5148}=0.9-j2.3784~pu \end{aligned}\]

Total current flowing to the fault is \[I_{f}^{''}=I_{g}^{''}+I_{m}^{''}=-j4.4425~pu\]

Note that the base current in the circuit of the motor is \[I_{base}=\dfrac{50\times10^{3}}{\sqrt{3}\times18}=1603.8~A\]

Therefore while load current was 1603.8 A, fault current is 7124.7 A.

The Thevenin impedance at the circuit between the terminals A and B \[Z_{th}=j\dfrac{0.4\times0.5148}{0.4+0.5148}=j0.2251~pu\]

Since voltage at the motor terminals before the fault is 1.0 per unit, the fault current is \[I_{f}^{''}=\dfrac{1.0}{Z_{th}}=-j4.4425pu\]

If we neglect the pre-fault current flowing through the circuit, then fault current fed by the motor and the generator can be determined using the current divider principle, i.e., \[\begin{aligned} I_{m0}^{''} & =\dfrac{I_{f}^{''}}{j0.9148}\times j0.5148=-j2.5~pu\\ I_{g0}^{''} & =\dfrac{I_{f}^{''}}{j0.9148}\times j0.4=-j1.9425~pu \end{aligned}\]

If, on the other hand, the pre-fault current is not neglected, then the fault current supplied by the motor and the generator are \[\begin{aligned} I_{m}^{''} & =I_{m0}^{''}-I_{L}=-0.9-j2.0641pu\\ I_{g}^{''} & =I_{g0}^{''}+I_{L}=0.9-j2.3784pu \end{aligned}\]

Assume a symmetrical fault has occurred in bus-4 such that it is now connected to the reference bus

Let the pre-fault voltage at this bus is \(V_f\)

To denote that bus-4 is short circuit, we add two voltage sources \(V_f\) and -\(V_f\) together in series between bus-4 and the reference bus

Note that the sub-transient fault current \(I_{f}^{''}\) flows from bus-4 to the reference bus

This implies that a current that is equal to -\(I_{f}^{''}\) is injected into bus-4

This current, which is due to the source -\(V_f\) will flow through the various branches of the network and will cause a change in the bus voltages

Assuming that the two sources and \(V_f\) are short circuited

Then -\(V_f\) is the only source left in the network that injects a current -\(I_{f}^{''}\) into bus-4.

The voltages of the different nodes that are caused by the voltage -\(V_f\) and the current -\(I_{f}^{''}\) are then given by \[\left[\begin{array}{c} \varDelta V_{1}\\ \varDelta V_{2}\\ \varDelta V_{3}\\ -V_{f} \end{array}\right]=Z_{bus}\left[\begin{array}{c} 0\\ 0\\ 0\\ -I_{f}^{''} \end{array}\right]\] where \(\varDelta\) indicates the changes in the bus voltages due to -\(I_{f}^{''}\)

Now, we can write

\[\begin{aligned} V_{f} & =Z_{44}I_{f}^{''}\\ \varDelta V_{i} & =-Z_{i4}I_{f}^{''}=-\dfrac{Z_{i4}}{Z_{44}}V_{f},~i=1,2,3 \end{aligned}\]

We further assume that the system is unloaded before the fault occurs and that the magnitude and phase angles of all the generator internal emfs are the same.

Then there will be no current circulating anywhere in the network and the bus voltages of all the nodes before the fault will be same and equal to \(V_f\).

Then the new altered bus voltages due to the fault will be given by \[V_{i}=V_{f}+\varDelta V_{i}=\left(1-\dfrac{Z_{i4}}{Z_{44}}\right)V_{f},~i=1,\cdots,4\]

Once the fault occurs, the protective devices get activated.

A certain amount of time elapses before the protective relays determine that there is over-current in the circuit and initiate trip command, called the .

The contacts of the CBs are held together by spring mechanism and, with the trip command, the spring mechanism releases the contacts.

When two current carrying contacts part, a voltage instantly appears at the contacts and a large voltage gradient appears in the medium between the two contacts.

This voltage gradient ionizes the medium thereby maintaining the flow of current.

This current generates extreme heat and light that is called .

Different mechanisms are used for elongating the arc such that it can be cooled and extinguished.

Therefore the circuit breaker has to withstand fault current from the instant of initiation of the fault to the time the arc is extinguished.

Important factors for the selection of circuit breakers are:

• The maximum instantaneous current that a breaker must withstand

• The total current when the breaker contacts part

In this chapter we have discussed the calculation of symmetrical sub-transient fault current in a network.

However the instantaneous current following a fault will also contain the dc component.

In a high power CB selection, the sub-transient current is multiplied by a factor of 1.6 to determine the rms value of the current the CB must withstand.

This current is called the .

The interrupting current of a circuit breaker is lower than the momentary current and will depend upon the speed of the CB.

The interrupting current may be asymmetrical since some dc component may still continue to decay.

Breakers are usually classified by their

• nominal voltage

• continuous current rating

• rated maximum voltage

• K -factor which is the voltage range factor

• rated short circuit current at maximum voltage

• operating time.

The K -factor is the ratio of rated maximum voltage to the lower limit of the range of the operating voltage.

The maximum symmetrical interrupting current of a CB is given by K times the rated short circuit current.

Short-circuit interrupting MVA can be estimated from the knowledge of pre fault voltage and short-circuit interrupting current \[\begin{aligned} \text{Short circuit interrupting MVA} & = \sqrt{3}\cdot |V_{pfL}|\cdot |I_{fL}| \end{aligned}\]

where

\[ \begin{aligned} |V_{pfL}| & = \text{mag. of prefault line voltage at the fault point in kV}\\ |I_{fL}| & = \text{mag. of line value of short-circuit interrupting current at the fault in kA} \end{aligned} \]

Short circuit interrupting MVA = \(|V_{pfL,pu}| \times |I_{fL,pu}| \times MVA_{b}\)

where \[ \begin{aligned} |V_{pfL,pu}| & = \text{mag. of prefault voltage at the fault point in p.u.}\\ |I_{fL,pu}| & = \text{mag. of short-circuit interrupting current at the fault in p.u.} \end{aligned} \]

To compute interrupting KVA in normal working condition, instead of fault voltage/current normal working voltage/current is used