Question

In a 100 kV, 60 Hz system, the reactance and capacitance up to the location of circuit breaker are $4\Omega and0.02\mu F$ respectively. The maximum value of RRRV will be

• A. $6.255kV/\mu sec$
• B. $5.607kV/\mu sec$
• C. $2.720kV/\mu sec$
• D. $9.010kV/\mu sec$

Answer 1

The correct option is B $5.607kV/\mu sec$

We know that, $X_L=4\Omega ,C=0.02\mu F$
$L=\frac{X_L}{2\pi f}=\frac{4}{2\pi \times 60}=\frac{1}{30\pi }=0.0106H$
The frequency of transient oscillation
$f_n=\frac{1}{2\pi \sqrt{LC}}$
$=\frac{1}{2\pi \sqrt{0.0106\times 0.02\times {10}^{-6}}}=10.93kHZ$
The rate of rise of restriking voltage

$\frac{d}{dt}\left[V_m\right(1-cos{\omega }_{n}t\left)\right]$
$RRRV=V_m{\omega }_{n}sin{\omega }_{n}t$
$\therefore MaximumvalueofRRRV=V_m{\omega }_n$
$=2\pi f_n\times V_m$
$V_m=\frac{100}{\sqrt{3}}\sqrt{2}\times 1000=81649.66V$
$(RRRV{)}_{max}=2\pi \times 10.93\times {10}^3\times \frac{100\sqrt{2}}{\sqrt{3}}\times {10}^3$
$=5607.308\times {10}^{6}V/sor5.607kV/\mu sec$