Question

A $3-\varphi$, 50 Hz, Y-connected, 100 MVA, 50 Hz alternator has inertia constant H = 5 MJ/MVA. Initially it delivers 60 MW at equilibrium. If the prime mover input is suddenly increased to 70 MW, then the load angle after 6 cycles is______ electrical degrees.

• A. ${10.23}^{\circ }$
  
• B. ${31.30}^{\circ }$
• C. ${20.46}^{\circ }$
• D. ${15.46}^{\circ }$

Answer 1

The correct option is B ${31.30}^{\circ }$

According to swing equation

$\frac{d^2\delta }{d{t}^2}=\frac{P_a}{M}$

$M=\frac{GH}{180\times f}=\frac{100\times 5}{180\times 50}=\frac{1}{18}$

$P_a=10MW$

$\frac{d^2\delta }{d{t}^2}=\frac{10}{\left(1/18\right)}=180\text{electrical degrees}/se{c}^2$

Integrating the above expression twice,

$\Delta \delta =\frac{180\times t^2}{2}=90\times t^2$

Where t for 6 cycles,

$=\frac{6}{50}=0.12s$

$\Delta \delta =90\times (0.12{)}^2={1.296}^{\circ }$

$\delta ={\delta }_0+\Delta \delta ={31.296}^{\circ }$