Question

A 50 Hz, four pole turbo generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. Mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW. If the acceleration is maintained for 10 cycles, the rotor speed at the end of this period is

• A. 1404.526 rpm
• B. 1608.635 rpm
• C. 1515.625 rpm
• D. 1505.625 rpm

Answer 1

The correct option is D 1505.625 rpm

$M=\frac{GH}{\pi f}=\frac{100\times 8}{50\times 180}=\frac{4}{45}MJs/elec.degree$

$\frac{M{d}^2\delta }{d{t}^2}=(P_m-P_s)$

$\frac{d^2\delta }{d{t}^2}=30\times \frac{45}{4}$

$\frac{d^2\delta }{d{t}^2}=337.5elec\mathrm{deg}/s^2$

$\frac{d\delta }{dt}=337.5t$

$\frac{d\delta }{dt}=337.5\times \frac{10}{50}\times \frac{60}{360}\times \left(\frac{2}{4}\right)=5.625rpm$

${\omega }_m={\omega }_{m_0}+\frac{d\delta }{dt}$

$N_m=\frac{120\times 50}{4}+5.625=1505.625rpm$