Question

A 50 Hz, $3-\Phi$ synchronous generator has an inductance per phase of 1.75 mH and its neutral is grounded. It feeds a line through a circuit breaker. The total stray capacitance to ground of the generator and circuit breaker is 0.0025 $\mu$F. A fault occurs just beyond the circuit breaker which opens when the symmetrical SC current is 7500 A (rms). The value of maximum rate of rise of TRV is

• A. 3876 $V/\mu$S
• B. 2786 $V/\mu$S
• C. 1786 $V/\mu$S
• D. 4786 $V/\mu$S

Answer 1

The correct option is B 2786 $V/\mu$S

Transient recovery voltage,
$TRV=e=E_m(1-\cos \omega t)$

$RRRV=\frac{de}{dt}=E_m\omega \sin \omega t$

$RRR{V}_{max}=E_m\omega =\frac{E_m}{\sqrt{L_C}}$

$E=I\omega L$

$=7500\times 314\times 1.75\times {10}^{-3}$

$=4.121kW$

$RRR{V}_{max}=\frac{4121\sqrt{2}}{\sqrt{1.75\times {10}^{-3}\times 25\times {10}^{-10}}}=2786V/\mu S$