Question

A $3-\varphi$, 2 pole, 50 Hz, synchronous generator has rating of 200 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 840 MJ. The machine is operating at synchronous speed and delivering 50 MW power at a power angle of 10° electrical degree. If load is suddenly removed, assuming acceleration is constant for 10 cycles, the value of power angle after 5 cycles will be (in electrical degree)

• A. 11.725°
• B. 10.50°
• C. 12.679°
• D. 14.72°

Answer 1

The correct option is C 12.679°

We know, $\delta ={\delta }_0+\frac{Pa}{M}\frac{t^2}{2}$...(1)

Also, 5 cycles of 50 Hz frequency

$=20ms\times 5=100ms=0.1sec$

$M=\frac{GH}{180f}=\frac{840}{180\times 50}(MJ-s/elec-deg)$

Accelerating power,
$P_a=P_m-P_e$

Using equation (1) as load in removed,
$P_a=P_m$

$\delta =10°+\frac{50\times 180\times 50\times (0.1{)}^2}{2\times 840}=12.679°elec-deg$