Stored energy and rotor acceleration

Data: $G = 100$ MVA, $H = 8$ MJ/MVA, $f = 50$ Hz, 4 poles. Synchronous speed $N_s = 120f/P = 1500$ rpm.

(i) Stored energy at synchronous speed

$$KE = G H = 100 \times 8 = 800\ \text{MJ}$$

(ii) Rotor acceleration

Accelerating power: $P_a = P_m - P_e = 80 - 50 = 30$ MW.

The inertia constant in MJ·s/elec-degree is

$$M = \frac{GH}{180 f} = \frac{800}{180\times 50} = 0.08889\ \text{MJ·s/elec deg}$$

Swing equation: $M\,\dfrac{d^2\delta}{dt^2} = P_a$, so

$$\begin{aligned} \alpha &= \frac{P_a}{M} = \frac{30}{0.08889} = 337.5\ \text{elec deg/s}^2 \\ &= \frac{337.5}{2} = 168.75\ \text{mech deg/s}^2 \quad (\text{4 poles} = 2\ \text{pole pairs}) \\ &= \frac{168.75\times 60}{360} = 28.125\ \text{rpm/s} \end{aligned}$$

(iii) After 10 cycles

Time: $t = 10/50 = 0.2$ s. Starting from synchronous speed with constant $\alpha$:

The change in torque (load) angle is

$$\Delta\delta = \tfrac{1}{2}\alpha t^2 = \tfrac{1}{2}\times 337.5\times 0.2^2 = 6.75\ \text{elec deg}\ (= 3.375\ \text{mech deg})$$

The rotor speed at the end of 10 cycles is

$$\begin{aligned} \Delta N &= 28.125\times 0.2 = 5.625\ \text{rpm} \\ N &= 1500 + 5.625 = 1505.625\ \text{rpm} \end{aligned}$$

"Change in torque" is read here as the change in torque (load) angle, which is the usual form of this problem. For reference, the accelerating torque is $T_a = P_a/\omega_{sm} = 30\times10^6/(2\pi\times1500/60) = 1.91\times10^5$ N·m.

Answer: (i) 800 MJ; (ii) $\alpha = 337.5$ elec deg/s² (28.125 rpm/s); (iii) $\Delta\delta = 6.75$ elec deg, speed $= 1505.625$ rpm.