--- title: "Load sharing: 250 MW and 500 MW, 20%/10% droop" subject: "Power Plant Equipment" chapter: 2 key: "semester_7/power_plant_equipment:load-sharing-250-500-20-10" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-sharing-250-500-20-10 site: BCT Engineering --- # Load sharing: 250 MW and 500 MW, 20%/10% droop Power Plant Equipment · Chapter 2: Power/Frequency Control in Hydro Generator System - Asked in: 2082 Baisakh, 8 marks; 2072 Chaitra, 8 marks (asked 2 times) - Video: [Speed Droop in Power Control Explained](https://www.youtube.com/watch?v=sQ-AT1WdT3M) — RealPars - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-sharing-250-500-20-10 - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2 ## Question Two generating units rated as 250 MW and 500 MW have speed droop regulation of 20% and 10% respectively based on their rating. They are operated in parallel and each is half loaded at a common frequency of 50 Hz. If the load is increased by 60 MW, calculate: i) New frequency at which operates ii) New loading of each generating units ## Short answer Stiffnesses: K1=25 MW/Hz, K2=100 MW/Hz. With a 60 MW load increase, frequency falls by 0.48 Hz to 49.52 Hz; unit loads become 137 MW and 298 MW. ## Answer 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. | Unit | Rating (MW) | Droop | Initial (MW) | Increase (MW) | New load (MW) | |---|---|---|---|---|---| | 1 | 250 | 20 % | 125 | 12 | 137 | | 2 | 500 | 10 % | 250 | 48 | 298 | **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which expression gives the stiffness K (MW/Hz) of a unit with rating Pr, droop R (pu) and rated frequency f0? A. K = Pr / (R f0) B. K = R f0 / Pr C. K = Pr R / f0 D. K = f0 / (Pr R) Answer: A. A droop line gives dP = -(Pr/(R f0)) df, so the stiffness is Pr/(R f0). (Key point: Stiffness formula) 2. For the 250 MW, 20% unit and the 500 MW, 10% unit at 50 Hz, what are K1 and K2? A. 50 and 50 MW/Hz B. 25 and 100 MW/Hz C. 100 and 25 MW/Hz D. 25 and 250 MW/Hz Answer: B. K1 = 250/(0.20x50) = 25 and K2 = 500/(0.10x50) = 100 MW/Hz. (Key point: Stiffness values) 3. With total stiffness 125 MW/Hz and a 60 MW load rise, what is the new common frequency? A. 49.40 Hz B. 50.48 Hz C. 49.52 Hz D. 49.88 Hz Answer: C. df = -60/125 = -0.48 Hz, so f = 50 - 0.48 = 49.52 Hz. (Key point: New frequency) 4. How is the 60 MW increase divided between the units, and why? A. 30 MW each, since both are half loaded B. 48 MW to unit 1 and 12 MW to unit 2, as the larger droop takes more C. 20 MW and 40 MW, in proportion to ratings only D. 12 MW to unit 1 and 48 MW to unit 2, in proportion to stiffness Answer: D. Each unit picks up K x 0.48: 25x0.48 = 12 MW and 100x0.48 = 48 MW; the stiffer unit takes more. 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