Load sharing: 600 MVA and 500 MVA, 4%/6% droop

Both units share the load increase in proportion to $P_r/R$, and the frequency falls to a common new value.

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: Unit 1: 600 MVA, $R_1 = 0.04$; Unit 2: 500 MVA, $R_2 = 0.06$; $f_0 = 60$ Hz. Initial: $P_1 = 500$ MW, $P_2 = 400$ MW. $\Delta P_L = 90$ MW. (MVA ratings are used as the MW base, i.e. unity power factor rating.)

Stiffness:

$$\begin{aligned} K_1 &= \frac{600}{0.04 \times 60} = 250\ \text{MW/Hz} \\ K_2 &= \frac{500}{0.06 \times 60} = 138.89\ \text{MW/Hz} \\ K_1 + K_2 &= 388.89\ \text{MW/Hz} \end{aligned}$$

New frequency:

$$\begin{aligned} \Delta f &= -\frac{90}{388.89} = -0.2314\ \text{Hz} \\ f_{new} &= 60 - 0.2314 = 59.769\ \text{Hz} \end{aligned}$$

(In per unit, $\Delta f = -0.003857$ pu.)

New generation:

$$\begin{aligned} \Delta P_1 &= 250 \times 0.2314 = 57.86\ \text{MW} \\ \Delta P_2 &= 138.89 \times 0.2314 = 32.14\ \text{MW} \end{aligned}$$

Check: $57.86 + 32.14 = 90$ MW.

Answer: New frequency ≈ 59.77 Hz (drop of 0.231 Hz); unit 1 generates 557.86 MW and unit 2 generates 432.14 MW.