Symmetrical Fault Analysis in Power Systems

Demonstrative Video

Analysis of Symmetrical Faults in PS

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  • Consider a power system in which a synchronous generator supplies a synchronous motor

  • Motor is operating at rated voltage and rated MVA while drawing a load current at a power factor of 0.9 (lagging)

  • \(3\phi\) symmetrical short circuit occurs at its terminals

  • calculate the fault current that flow from both the generator and the motor?

  • \[X_{m}^{''}=0.2\times\dfrac{50}{25}=0.4~pu\]
    Choose a base of 50 MVA, 20 kV in the circuit of the generator. Then the motor synchronous reactance is given by
  • \[Z_{base}=\dfrac{66^{2}}{50}=87.12~\Omega\]
    Base impedance of the transmission line is
  • \[X_{line}=j\dfrac{10}{87.12}=j0.1148~pu\]
    Impedance of the transmission line is
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  • In impedance diagram, switch S indicates the fault

  • \[i_{L}=1\angle-cos^{-1}\left(0.9\right)=0.9-j0.4359~pu\]
    Motor draws a load current at rated voltage and rated MVA with 0.9 lagging power factor. Therefore
  • \[\begin{aligned} E_{m}^{''} & =1.0-j0.4\times i_{L}=0.8256-j0.36~pu\\ E_{g}^{''} & =1.0+j0.5148\times i_{L}=1.2244+j0.4633~pu \end{aligned}\]
    Sub-transient voltages of the motor and the generator are
  • \[\begin{aligned} I_{m}^{''} & =\dfrac{E_{m}^{''}}{j0.4}=-0.9-j2.0641~pu\\ I_{g}^{''} & =\dfrac{E_{g}^{''}}{j0.5148}=0.9-j2.3784~pu \end{aligned}\]
    Sub-transient fault currents fed by the motor and the generator are
  • \[I_{f}^{''}=I_{g}^{''}+I_{m}^{''}=-j4.4425~pu\]
    Total current flowing to the fault is
  • \[I_{base}=\dfrac{50\times10^{3}}{\sqrt{3}\times18}=1603.8~A\]
    Note that the base current in the circuit of the motor is

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

  • \[Z_{th}=j\dfrac{0.4\times0.5148}{0.4+0.5148}=j0.2251~pu\]
    The Thevenin impedance at the circuit between the terminals A and B
  • \[I_{f}^{''}=\dfrac{1.0}{Z_{th}}=-j4.4425pu\]
    Since voltage at the motor terminals before the fault is 1.0 per unit, the fault current is
  • \[\begin{aligned} I_{m0}^{''} & =\dfrac{I_{f}^{''}}{j0.9148}\times j0.5148=-j2.5~pu\\ I_{g0}^{''} & =\dfrac{I_{f}^{''}}{j0.9148}\times j0.4=-j1.9425~pu \end{aligned}\]
    If 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.,
  • \[\begin{aligned} I_{m}^{''} & =I_{m0}^{''}-I_{L}=-0.9-j2.0641pu\\ I_{g}^{''} & =I_{g0}^{''}+I_{L}=0.9-j2.3784pu \end{aligned}\]
    If, on the other hand, the pre-fault current is not neglected, then the fault current supplied by the motor and the generator are
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  • 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.

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

  • 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 \(V_f\).

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

Selection of Circuit Breakers

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  • 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.

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

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

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

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