Load sharing: 250 MW and 500 MW, 20%/10% droop

When load rises, both units move down their droop lines to a common lower frequency; each picks up load in proportion to its stiffness $P_r/R$.

Method: with droop $R_i$ (pu, on its own rating $P_{ri}$) and rated frequency $f_0$, the droop line of each unit gives

$$\Delta P_i = -\frac{P_{ri}}{R_i\, f_0}\,\Delta f = -K_i\,\Delta f$$

where $K_i = P_{ri}/(R_i f_0)$ is the stiffness of the unit in MW/Hz. For the units together, $\Delta P_L = -\Delta f\,\sum K_i$.

Data: $P_{r1} = 250$ MW, $R_1 = 0.20$; $P_{r2} = 500$ MW, $R_2 = 0.10$; $f_0 = 50$ Hz. Initial loading: 125 MW and 250 MW. $\Delta P_L = +60$ MW.

Stiffness of each unit:

$$\begin{aligned} K_1 &= \frac{250}{0.20 \times 50} = 25\ \text{MW/Hz} \\ K_2 &= \frac{500}{0.10 \times 50} = 100\ \text{MW/Hz} \\ K_1 + K_2 &= 125\ \text{MW/Hz} \end{aligned}$$

i) New frequency

$$\begin{aligned} \Delta f &= -\frac{\Delta P_L}{K_1 + K_2} = -\frac{60}{125} = -0.48\ \text{Hz} \\ f_{new} &= 50 - 0.48 = 49.52\ \text{Hz} \end{aligned}$$

ii) New loading of each unit

$$\begin{aligned} \Delta P_1 &= K_1 \times 0.48 = 25 \times 0.48 = 12\ \text{MW} \\ \Delta P_2 &= K_2 \times 0.48 = 100 \times 0.48 = 48\ \text{MW} \end{aligned}$$

Check: $12 + 48 = 60$ MW.

Answer: New frequency = 49.52 Hz; unit 1 carries 137 MW and unit 2 carries 298 MW. The 500 MW unit, with the smaller droop (stiffer), takes 80 % of the increase.