--- title: "Speed droop in sharing common load" subject: "Power Plant Equipment" chapter: 2 key: "semester_7/power_plant_equipment:speed-droop-load-sharing" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/speed-droop-load-sharing site: BCT Engineering --- # Speed droop in sharing common load Power Plant Equipment · Chapter 2: Power/Frequency Control in Hydro Generator System - Asked in: 2076 Chaitra, 5 marks; 2074 Asoj, 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/speed-droop-load-sharing - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2 ## Question Explain the importance of speed droop in sharing common load by two alternator operating in parallel with suitable block diagram and derivation. ## Short answer Speed droop provides each unit a definite speed-load line, ensuring stable load division. The load change share ratio equals (P_r/R) of the units; equal per-unit droop divides load proportionally to ratings, preventing instability. ## Answer **Speed droop** is the drop in speed (frequency) of a governed unit as its load goes from no load to full load, expressed in per unit of rated speed. It is defined by $$ R = \frac{\Delta f / f_0}{\Delta P / P_r} \quad \text{(pu)} $$ A 4 % droop means the frequency falls by 4 % (2 Hz at 50 Hz) from no load to full load. Droop gives every governor a definite speed–load line, so parallel units share a load change in a fixed, stable ratio. Without droop (isochronous governors), parallel units fight each other and the load division is undefined. #### Block diagram (two units on a common bus) ```text +-->[1/R1]--+ | v - | dPref1->(+)->[Gov1]->[Turb1]--dPm1--+ | v | -dPL ------->(S)--+ | ^ | | dPref2->(+)->[Gov2]->[Turb2]--dPm2--+ | | ^ - v +-->[1/R2]--+ [1/(2Hs+D)] | | +--------------- d(omega) -----------------+ ``` S = summing point: $\Delta P_{m1} + \Delta P_{m2} - \Delta P_L$. Both units see the same frequency deviation $\Delta f$ because they are synchronised. #### Derivation In steady state, the governor of unit $i$ obeys its droop line: $$ \Delta P_i = \Delta P_{ref,i} - \frac{1}{R_i}\frac{\Delta f}{f_0}P_{ri} $$ With fixed references ($\Delta P_{ref,i} = 0$): $$ \Delta P_1 = -\frac{P_{r1}}{R_1}\frac{\Delta f}{f_0}, \qquad \Delta P_2 = -\frac{P_{r2}}{R_2}\frac{\Delta f}{f_0} $$ The total change must supply the load change: $$ \begin{aligned} \Delta P_L &= \Delta P_1 + \Delta P_2 = -\frac{\Delta f}{f_0}\left(\frac{P_{r1}}{R_1} + \frac{P_{r2}}{R_2}\right) \\ \frac{\Delta f}{f_0} &= \frac{-\Delta P_L}{\frac{P_{r1}}{R_1} + \frac{P_{r2}}{R_2}} \end{aligned} $$ and the ratio of sharing is $$ \frac{\Delta P_1}{\Delta P_2} = \frac{P_{r1}/R_1}{P_{r2}/R_2} $$ If $R_1 = R_2$ (same per-unit droop), $\Delta P_1/\Delta P_2 = P_{r1}/P_{r2}$, i.e. **load is shared in proportion to ratings**. ```text f f f0|\ unit 1 |\ unit 2 | \ | \ f'|--*-----common f'--|---* | \ | \ +----+---> P1 +-----+---> P2 P1' P2' ``` #### Importance - Gives a unique, stable division of load among parallel units. - A smaller droop makes a unit "stiffer", so it takes a bigger share of any change. - Equal per-unit droop on all units makes each carry load proportional to its rating, avoiding overload of small units. - The frequency drop needed is small (typically 4–5 % droop), and the load reference setting is then used to bring frequency back to nominal. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does a 4% speed droop mean? A. Frequency falls 4% from no load to full load B. Load falls 4% as frequency rises C. Voltage falls 4% at full load D. Frequency falls 4% for every 1% load Answer: A. Droop R = (df/f0)/(dP/Pr); 4% is a 2 Hz fall at 50 Hz from no load to full load. (Key point: Droop definition) 2. In what ratio do two parallel units share a load change? A. R1/R2 B. Pr1/R1 : Pr2/R2 C. Pr1 : Pr2 always D. R2 Pr1 : R1 Pr2 inverted Answer: B. dP1/dP2 = (Pr1/R1)/(Pr2/R2). (Key point: Sharing ratio) 3. If R1 = R2, how is the load shared? A. Equally B. Inversely to ratings C. In proportion to ratings D. Undefined Answer: C. The ratio reduces to Pr1/Pr2. (Key point: Equal per-unit droop) 4. Why are isochronous governors unsuitable for parallel units? A. They are too slow B. They have high droop C. They cannot sense speed D. Each tries to hold its own set frequency, so units fight and load division is undefined Answer: D. Without droop there is no definite speed-load line, so load division is not unique or stable. 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