Question

Given that V${}_{s1}$ = V${}_{s2}$ = 1 + j0 p.u. +ve sequence impendance are Z${}_{s1}$ = Z${}_{s2}$ = 0.001 + j0.01 p.u and Z${}_L$ = 0.006 + j0.06 p.u, 3-$\varphi$. Base MVA = 100, voltage base = 400 kV(L-L).

Nominal system frequency = 50 Hz. The reference voltage for phase 'a' is deifned as V(t) = V${}_m$ cos($\omega$t). The symmetrical 3$\varphi$ fault occurs at centre of the line i.e, at point 'F'. At the time '$t_0$' the +ve sequence impendance from source S${}_1$ to point 'F' equals (0.004 + j0.04) p.u. The wave form corresponding to phase 'a' fault current from bus X reveals that decaying d.c. offset current is -ve in magnitude at its maximum initial value. Assuming that the negative sequence impedances are equal to +ve sequence impendances and the zero sequence (Z) are 3 times of +ve sequence (Z).

The instant (t${}_0$) of the fault will be

• A. 4.682 ms
• B. 9.667 ms
• C. 14.667 ms
• D. 19.667 ms

Answer 1

The correct option is A 4.682 ms



To find i(t) for t(0${}^{+}$) we can convert the above circuit in following way

$Let,t^{\prime }=t-t_0$

$\Rightarrow$ t = t' + $t_0$

$i\left(t^{\prime }\right)=\frac{\sqrt{2}V}{\left|Z\right|}$ cos ($\omega t^{\prime }+\omega t_0-\alpha$) + Ae${}^{\frac{t^{\prime }}{\pi }}$ (t' $\ge$ t${}_0^{+}$)

symmetrical short circuit current,

$\left|Z\right|=\sqrt{R^2+(\omega L{)}^2}$

$\alpha$ = tan${}^{-1}$$\frac{\omega L}{R}$

$\pi$ = $\frac{L}{R}$

Now, i(t' = 0${}^{-}$) = i(t' = 0${}^{+}$) = 0

$\Rightarrow$ $\frac{\sqrt{2}V}{\left|Z\right|}$ cos($\omega t_0-\alpha$) + A = 0

A = $\frac{\sqrt{2}V}{\left|Z\right|}$ cos($\omega t_0-\alpha$)

Now change,

t' = t - t${}_0$

i(t) = $\frac{\sqrt{2}V}{\left|Z\right|}$ cos($\omega t-\alpha$)

$-\frac{\sqrt{2}V}{\left|Z\right|}$ cos($\omega t_0-\alpha$) e$\frac{(t-t_0)R}{L}$

$t\ge t_0^{+}$

Now, if the initial value of dc offset ( A) is - $\frac{\sqrt{2}V}{\left|Z\right|}$ it requires that
cos ($\omega t_0-\alpha$) = 1

$\Rightarrow$ $\omega t_0-\alpha$ = 0 $\Rightarrow$ t${}_0$ = $\frac{\alpha }{\omega }$

t${}_0$ = $\frac{ta{n}^{-}1\frac{\omega L}{R}}{\omega }$

In given problem, the circuit for 3-$\varphi$ fault can be drawn as:



According to the derivation:

For initial value of dc off set to be - $\frac{\sqrt{2}V}{\left|Z\right|}$
we required

t${}_0$ = $\frac{ta{n}^{-}1\frac{\omega L}{R}}{\omega }=\frac{ta{n}^{-}1\frac{0.004}{0.004}}{2\times 3.14\times 15}$

= 4.68 msec.