Load sharing: 600 MW and 300 MW, 7%/8% droop

Both units drop along their droop lines to a common new frequency, each taking a share of the extra 150 MW in proportion to $P_r/R$.

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: 600 MW, $R_1 = 0.07$, initial 300 MW; G2: 300 MW, $R_2 = 0.08$, initial 150 MW; $f_0 = 50$ Hz; $\Delta P_L = 150$ MW.

Stiffness:

$$\begin{aligned} K_1 &= \frac{600}{0.07 \times 50} = 171.43\ \text{MW/Hz} \\ K_2 &= \frac{300}{0.08 \times 50} = 75\ \text{MW/Hz} \\ K_1 + K_2 &= 246.43\ \text{MW/Hz} \end{aligned}$$

a) New frequency

$$\begin{aligned} \Delta f &= -\frac{150}{246.43} = -0.6087\ \text{Hz} \\ f_{new} &= 50 - 0.6087 = 49.39\ \text{Hz} \end{aligned}$$

b) Power supplied by each generator

$$\begin{aligned} \Delta P_1 &= 171.43 \times 0.6087 = 104.35\ \text{MW} \\ \Delta P_2 &= 75 \times 0.6087 = 45.65\ \text{MW} \end{aligned}$$

Answer: New frequency ≈ 49.39 Hz; G1 supplies 404.35 MW and G2 supplies 195.65 MW.