Load sharing: 600 and 300 MW, 51 Hz no-load

Both generators start from the same no-load frequency (51 Hz) and move down their droop lines until their outputs add up to the load. Rated frequency is taken as $f_0 = 50$ Hz.

Method: a unit with droop $R$ (pu on its own rating $P_r$) and rated frequency $f_0$ changes its output by

$$\Delta P = -\frac{P_r}{R\,f_0}\,\Delta f = -K\,\Delta f$$

where $K = P_r/(R f_0)$ is its stiffness in MW/Hz. All units run at the same frequency, so $\Delta P_L = -\Delta f \sum K$.

Stiffness:

$$\begin{aligned} K_1 &= \frac{600}{0.03 \times 50} = 400\ \text{MW/Hz} \\ K_2 &= \frac{300}{0.06 \times 50} = 100\ \text{MW/Hz} \\ K_1 + K_2 &= 500\ \text{MW/Hz} \end{aligned}$$

(Full-load frequency drop: G1 $= 0.03 \times 50 = 1.5$ Hz, G2 $= 0.06 \times 50 = 3$ Hz.)

Sharing of 600 MW

$$\begin{aligned} \Delta f &= \frac{600}{500} = 1.2\ \text{Hz}, \quad f = 51 - 1.2 = 49.8\ \text{Hz} \\ P_1 &= 400 \times 1.2 = 480\ \text{MW} \\ P_2 &= 100 \times 1.2 = 120\ \text{MW} \end{aligned}$$

Load increased by 200 MW (total 800 MW)

$$\begin{aligned} \Delta f &= \frac{800}{500} = 1.6\ \text{Hz}, \quad f = 51 - 1.6 = 49.4\ \text{Hz} \\ P_1 &= 400 \times 1.6 = 640\ \text{MW} \\ P_2 &= 100 \times 1.6 = 160\ \text{MW} \end{aligned}$$

(Equivalently, the extra 200 MW is shared as $400 \times 0.4 = 160$ MW and $100 \times 0.4 = 40$ MW.)

Answer: At 600 MW: 49.8 Hz, G1 = 480 MW, G2 = 120 MW. At 800 MW: 49.4 Hz, G1 = 640 MW, G2 = 160 MW.

Note: 640 MW is above G1's 600 MW rating. In practice G1's governor would stop at full gate (600 MW) and G2 would carry the remaining 200 MW, so the frequency would settle on G2's line at $51 - 200/100 = 49$ Hz. Load reference settings should be changed to avoid this overload.