Ideal interruption of an HV terminal fault

The circuit of Fig. 2.19 models a terminal fault, namely a short-circuit occurred at the terminals of a CB. There is no other component in the circuit between the switch and the fault. This and the circuit used for modeling switch-on (Fig. 2.2) are dissimilar in two points. First, the current model includes the resultant parallel capacitance C of the network, which plays a major role in determining the TRV. During calculation of the fault current, this C could be neglected, because – as 1/ω C>>ω L – the current through it has been very small. The other difference is that there is a resistance r parallel to the contacts of the CB in the model of Fig. 2.19. This r takes into account the resistance of the post-arc.

As a first approximation, we neglect the serial and parallel resistances in the model of Fig. 2.19, that is R=0, and r=∞. This damping - free model is shown in Fig. 2.20.a. It is worthy of notice, that the condition of R=0 is close to the reality of fault interruption in HV circuits, as cosυ≈0.1 there.

Fig. 2.20. Ideal switch-off of a fault in a solely inductive HV circuit

The moment of contact opening (t=0) coincides with the natural current zero of the steady-state prospective short-circuit current i. We assume that all switch-on transients has diminished by this time. We seek the voltage across the switch terminals after the time instant of switch-off. This voltage equals to the capacitor voltage v C(t).

Provided that 1/ω C>>ω L, after a long time, v C becomes identical to the supply voltage that is, v Cst(t)=v(t)=V

mcosω t. This voltage, the steady-state recovery voltage together with the prospective short-circuit current i lasted until its zero (t = 0) is plotted in Fig. 2.20.b. Since the circuit is solely inductive, this current lags behind the voltage by an angle of π/2. The picture clearly indicates that at t=0, v Cst(0)=V m. The capacitor voltage is the sum of two components in this case too:

(2-13)

We still do not know the transient time-function v Ctr(t), but we know that the capacitor voltage has been zero prior to switch-off, since the closed switch shunted out C. This voltage cannot jump, therefore its initial value is zero after current interruption, u C(0)=0. From these assumptions, the equation

(2-14)

follows, which results in v Ctr(0)=- V m. This is the initial value of the transient component, which is followed by a periodic oscillation caused by the current i LC in the L-C circuit of the network. There is no damping in the oscillation (R=0!), and its natural frequency

(2-15)

which is significantly higher (of the order of krad/s) than the angular frequency ω of the power supply. The time-function of the transient component (in other words oscillating component):

(2-16)

Switching transients

which starts with zero steepness. It is easy to verify that i LC(0)=0, as no current has been flowing in the L-C circuit before switch-off, and the current has been interrupted exactly in its natural zero. From this assumption:

(2-17)

consequently, just like v Cst, the initial slope of the resultant v C is zero (see Fig. 2.20c). It follows from this that the initial steepness of v Ctr is also zero.

Fig. 2.20.d shows the time-functions v Ctr(t) and v C (t), the latter one as the sum of v Ctr(t) and v Cst(t). The time-function v C (t) is the voltage appearing across the terminals of the switching device after the current interrupted, and it is called transient recovery voltage (TRV). In this case, the oscillation of the TRV has only one frequency: f 0=ω 0/2π.

The time-function of the TRV:

(2-18)

and its peak factor:

(2-19)

The peak factor is only slightly smaller than kp=2 – we can say that the difference is negligible – since the steady-state recovery voltage v Cst(t)=V mcosω t remains practically constant (V m) during a half cycle of the oscillating transient.

If we take into account the serial resistance R neglected previously, the phase shift between the fault current and the supply voltage will be less than π /2, although in HV circuits the difference from π /2 is usually negligible.

Not negligible however, is the damping of the oscillating part by a serial damping factor of

(2-20)

Besides, the natural frequency also changes, it will be slightly less than in the un-damped case:

(2-21)

Because of the lagging short-circuit current, of the reduced oscillation frequency, and above all, because of the damping, the peak factor of the TRV will be definitely less than 2.

Fig. 2.21. Ideal interruption of fault in a HV circuit with serial damping

Based on the previous assumptions, we can use a simplified, approximate formula to calculate the damped TRV after a fault interruption in an HV circuit:

TRV changes. On the one hand, its damping becomes stronger, since the resultant damping factor δ is the sum of a serial δ s and a parallel δ p part:

(2-23)

On the other hand, if the transient voltage is periodic, the difference between the damped and un-damped natural frequencies will not be negligible:

(2-24)

It is clear from this relation that the damped and un-damped oscillation frequencies can theoretically be equal, if δ p=δ s. In practice however, ω s<ω 0, since δ p>>δ s.

Again in favor of safety, we can use a simplified, approximate equation to calculate the TRV after a fault in a HV circuit with serial and parallel damping:

(2-25)

the plot of which is very similar to that of the serially damped case in Fig. 2.21.