Load sharing between two stations in parallel

Each station follows a linear speed (frequency) droop: frequency falls from the no-load value (50 Hz) by its regulation percentage at its full load.

Droop of each station

$$\begin{aligned} \text{A: drop at full load} &= 0.03\times50 = 1.5\ \text{Hz for } 50\ \text{MW} \Rightarrow 0.03\ \text{Hz/MW} \\ \text{B: drop at full load} &= 0.035\times50 = 1.75\ \text{Hz for } 25\ \text{MW} \Rightarrow 0.07\ \text{Hz/MW} \end{aligned}$$

Load sharing

Both run at the same frequency $f$:

$$\begin{aligned} f &= 50 - 0.03P_A = 50 - 0.07P_B \\ P_A + P_B &= 50 \\ 0.03P_A &= 0.07(50 - P_A) \Rightarrow P_A = \frac{3.5}{0.10} = 35\ \text{MW} \\ P_B &= 15\ \text{MW} \\ f &= 50 - 0.03\times35 = 48.95\ \text{Hz} \end{aligned}$$

Comment

• A supplies 35 MW (70% of its capacity) and B supplies 15 MW (60% of its capacity).
• The load is not shared in proportion to the ratings (which would be 33.3 : 16.7 MW). B has the steeper droop (larger Hz/MW), and A, with the smaller speed regulation, takes a larger share relative to its rating.
• For load sharing in proportion to ratings, the per-unit droops must be equal.
• The frequency falls to 48.95 Hz; the governor speed settings must be raised (secondary control) to restore 50 Hz.

Answer: P_A = 35 MW, P_B = 15 MW, common frequency = 48.95 Hz.