Load sharing: 900 MW and 450 MW generators

Both generators move down their droop lines to a common lower frequency; the 150 MW is shared 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: 900 MW, 2 %, supplying 750 MW; G2: 450 MW, 3 %, supplying 450 MW; $f_0 = 60$ Hz; $\Delta P_L = 150$ MW.

$$\begin{aligned} K_1 &= \frac{900}{0.02 \times 60} = 750\ \text{MW/Hz} \\ K_2 &= \frac{450}{0.03 \times 60} = 250\ \text{MW/Hz} \\ \Delta f &= -\frac{150}{750 + 250} = -0.15\ \text{Hz} \\ f_{new} &= 60 - 0.15 = 59.85\ \text{Hz} \\ \Delta P_1 &= 750 \times 0.15 = 112.5\ \text{MW} \\ \Delta P_2 &= 250 \times 0.15 = 37.5\ \text{MW} \end{aligned}$$

Answer: New frequency = 59.85 Hz; G1 generates 112.5 MW extra and G2 37.5 MW extra.

Note: G2 is already at its 450 MW rating, so 487.5 MW is an overload. If G2 is held at its limit (gate fully open), G1 must take all 150 MW: $\Delta f = -150/750 = -0.2$ Hz, giving 59.8 Hz and G1 = 900 MW.