Skip to main content

Load sharing: 250 MW and 500 MW, 4%/3% droop

Short answer

Common frequency is approximately 49.78 Hz. G1 generates 152.27 MW and G2 generates 322.73 MW.

Two generator of 250 MW and 500 MW capacities respectively are operating in parallel and connected to common bus. When each generator is half loaded, they operate at a common frequency of 50 Hz. Their drop characteristics are 4% and 3% respectively based on respective ratings. (i) What will be the common frequency if the common load is increased by 100 MW and calculate the new generation of each units.

Both units move down their droop lines to a common lower frequency, sharing the extra 100 MW in proportion to Pr/RP_r/R.

Method: a unit with droop RR (pu on its own rating PrP_r) and rated frequency f0f_0 changes its output by

ΔP=−PrR f0 Δf=−K Δf\Delta P = -\frac{P_r}{R\,f_0}\,\Delta f = -K\,\Delta f

where K=Pr/(Rf0)K = P_r/(R f_0) is its stiffness in MW/Hz. All units run at the same frequency, so ΔPL=−Δf∑K\Delta P_L = -\Delta f \sum K.

Data: G1: 250 MW, 4 %, initial 125 MW; G2: 500 MW, 3 %, initial 250 MW; f0=50f_0 = 50 Hz; ΔPL=100\Delta P_L = 100 MW.

K1=2500.04×50=125 MW/HzK2=5000.03×50=333.33 MW/HzΔf=−100458.33=−0.2182 Hzfnew=50−0.2182=49.78 HzΔP1=125×0.2182=27.27 MWΔP2=333.33×0.2182=72.73 MW\begin{aligned} K_1 &= \frac{250}{0.04 \times 50} = 125\ \text{MW/Hz} \\ K_2 &= \frac{500}{0.03 \times 50} = 333.33\ \text{MW/Hz} \\ \Delta f &= -\frac{100}{458.33} = -0.2182\ \text{Hz} \\ f_{new} &= 50 - 0.2182 = 49.78\ \text{Hz} \\ \Delta P_1 &= 125 \times 0.2182 = 27.27\ \text{MW} \\ \Delta P_2 &= 333.33 \times 0.2182 = 72.73\ \text{MW} \end{aligned}
GeneratorInitial (MW)Increase (MW)New (MW)
G1 (250 MW, 4 %)12527.27152.27
G2 (500 MW, 3 %)25072.73322.73
Total375100475

Answer: Common frequency ≈ 49.78 Hz; G1 generates 152.27 MW and G2 322.73 MW.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Power Plant Equipment chapter 2 questions →