Skip to main content

Load decrease: 250 and 500 MW at 49 Hz

Short answer

Each unit's stiffness is K = P_r/(R f0). G1: 250/(0.02×50)=250 MW/Hz; G2: 500/(0.04×50)=250 MW/Hz. With Δf=+1 Hz, each sheds 250 MW, so total load decrease is 500 MW; G1 falls to 0 MW, G2 to 250 MW.

Two generators of 250 MW and 500 MW capacities respectively are operating in parallel and supplying power to a common load. When each generator is fully loaded, they operate at a common frequency of 49 Hz. Their droop characteristics are 2% and 4% respectively based on their respective ratings. When the system load is decreased, the frequency increases to 50Hz. calculate the decrease in system load shared by each generator.

Both units are fully loaded at 49 Hz. As the load falls, frequency rises to 50 Hz and each unit sheds load equal to its stiffness times the rise (f0=50f_0 = 50 Hz).

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.

K1=2500.02×50=250 MW/HzK2=5000.04×50=250 MW/HzΔf=50−49=+1 HzΔP1=−250×1=−250 MWΔP2=−250×1=−250 MW\begin{aligned} K_1 &= \frac{250}{0.02 \times 50} = 250\ \text{MW/Hz} \\ K_2 &= \frac{500}{0.04 \times 50} = 250\ \text{MW/Hz} \\ \Delta f &= 50 - 49 = +1\ \text{Hz} \\ \Delta P_1 &= -250 \times 1 = -250\ \text{MW} \\ \Delta P_2 &= -250 \times 1 = -250\ \text{MW} \end{aligned}
GeneratorAt 49 Hz (MW)Decrease (MW)At 50 Hz (MW)
G1 (250 MW, 2 %)2502500
G2 (500 MW, 4 %)500250250
Total750500250

Answer: The system load decreases by 500 MW, shared equally: 250 MW by G1 (which falls to no load) and 250 MW by G2 (which now supplies 250 MW).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Power Plant Equipment chapter 2 questions →