Load sharing: 1000 MW and 500 MW units

When frequency rises, both units move up their droop lines and reduce output in proportion to $P_r/R$. 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 $\Delta P_L = -\Delta f \sum K$.

Data: G1: 1000 MW, 2 %; G2: 500 MW, 4 %; fully loaded at 49.5 Hz; frequency rises to 50 Hz, so $\Delta f = +0.5$ Hz.

$$\begin{aligned} K_1 &= \frac{1000}{0.02 \times 50} = 1000\ \text{MW/Hz} \\ K_2 &= \frac{500}{0.04 \times 50} = 250\ \text{MW/Hz} \\ \Delta P_1 &= -1000 \times 0.5 = -500\ \text{MW} \\ \Delta P_2 &= -250 \times 0.5 = -125\ \text{MW} \end{aligned}$$

Answer: At 50 Hz, G1 supplies 500 MW and G2 supplies 375 MW; the system load is 875 MW (a reduction of 625 MW).