Load sharing: 500 MW and 250 MW, 5%/6% droop

The two units move down their droop lines to a common lower frequency, sharing the 100 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: 500 MW, 5 %, initial 250 MW; G2: 250 MW, 6 %, initial 125 MW; $f_0 = 50$ Hz; $\Delta P_L = 100$ MW.

$$\begin{aligned} K_1 &= \frac{500}{0.05 \times 50} = 200\ \text{MW/Hz} \\ K_2 &= \frac{250}{0.06 \times 50} = 83.33\ \text{MW/Hz} \\ K_1 + K_2 &= 283.33\ \text{MW/Hz} \end{aligned}$$

a) New frequency

$$\Delta f = -\frac{100}{283.33} = -0.3529\ \text{Hz}, \qquad f = 50 - 0.3529 = 49.65\ \text{Hz}$$

b) Power supplied by each generator

$$\begin{aligned} \Delta P_1 &= 200 \times 0.3529 = 70.59\ \text{MW} \\ \Delta P_2 &= 83.33 \times 0.3529 = 29.41\ \text{MW} \end{aligned}$$

Answer: New frequency ≈ 49.65 Hz; G1 supplies 320.59 MW, G2 supplies 154.41 MW.