--- title: "Load sharing: 600 and 300 MW, 51 Hz no-load" subject: "Power Plant Equipment" chapter: 2 key: "semester_7/power_plant_equipment:load-sharing-600-300-51hz" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-sharing-600-300-51hz site: BCT Engineering --- # Load sharing: 600 and 300 MW, 51 Hz no-load Power Plant Equipment · Chapter 2: Power/Frequency Control in Hydro Generator System - Asked in: 2081 Baisakh, 10 marks - Video: [Speed Load characteristics | Two Generators rated 200MW and 400MW | Solved Problem -1 | PSOC | droop](https://www.youtube.com/watch?v=MEGzoyxTT6w) — Simple n Easy Learning - 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-600-300-51hz - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2 ## Question Two generators of capacities 600MW and 300MW are connected in parallel to supply a common load of 600 MW. Their droop regulations are 3% and 6% respectively w.r.t. their respective ratings. At no-load, they operates at a common frequency of 51 Hz. How they will share the common load of 600MW? When the load is increased by 200 MW, at which frequency they will operate and calculate the power supplied by each generator? ## Short answer Stiffness: G1 400 MW/Hz, G2 100 MW/Hz. For 600 MW load, frequency is 49.8 Hz, G1 480 MW, G2 120 MW. After 200 MW extra, frequency 49.4 Hz, G1 640 MW, G2 160 MW (G1 overloaded). ## Answer Both generators start from the same no-load frequency (51 Hz) and move down their droop lines until their outputs add up to the load. Rated frequency is taken as $f_0 = 50$ Hz. **Method:** a unit with droop $R$ (pu on its own rating $P_r$) and rated frequency $f_0$ changes its output by $$ \Delta P = -\frac{P_r}{R\,f_0}\,\Delta f = -K\,\Delta f $$ where $K = P_r/(R f_0)$ is its stiffness in MW/Hz. All units run at the same frequency, so $\Delta P_L = -\Delta f \sum K$. **Stiffness:** $$ \begin{aligned} K_1 &= \frac{600}{0.03 \times 50} = 400\ \text{MW/Hz} \\ K_2 &= \frac{300}{0.06 \times 50} = 100\ \text{MW/Hz} \\ K_1 + K_2 &= 500\ \text{MW/Hz} \end{aligned} $$ (Full-load frequency drop: G1 $= 0.03 \times 50 = 1.5$ Hz, G2 $= 0.06 \times 50 = 3$ Hz.) #### Sharing of 600 MW $$ \begin{aligned} \Delta f &= \frac{600}{500} = 1.2\ \text{Hz}, \quad f = 51 - 1.2 = 49.8\ \text{Hz} \\ P_1 &= 400 \times 1.2 = 480\ \text{MW} \\ P_2 &= 100 \times 1.2 = 120\ \text{MW} \end{aligned} $$ #### Load increased by 200 MW (total 800 MW) $$ \begin{aligned} \Delta f &= \frac{800}{500} = 1.6\ \text{Hz}, \quad f = 51 - 1.6 = 49.4\ \text{Hz} \\ P_1 &= 400 \times 1.6 = 640\ \text{MW} \\ P_2 &= 100 \times 1.6 = 160\ \text{MW} \end{aligned} $$ (Equivalently, the extra 200 MW is shared as $400 \times 0.4 = 160$ MW and $100 \times 0.4 = 40$ MW.) | Case | Frequency | G1 (MW) | G2 (MW) | |---|---|---|---| | Load 600 MW | 49.8 Hz | 480 | 120 | | Load 800 MW | 49.4 Hz | 640 | 160 | **Answer:** At 600 MW: **49.8 Hz, G1 = 480 MW, G2 = 120 MW**. At 800 MW: **49.4 Hz, G1 = 640 MW, G2 = 160 MW**. **Note:** 640 MW is above G1's 600 MW rating. In practice G1's governor would stop at full gate (600 MW) and G2 would carry the remaining 200 MW, so the frequency would settle on G2's line at $51 - 200/100 = 49$ Hz. Load reference settings should be changed to avoid this overload. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are K1 and K2 for 600 MW at 3% and 300 MW at 6% (f0 = 50 Hz)? A. 400 and 100 MW/Hz B. 200 and 50 MW/Hz C. 600 and 300 MW/Hz D. 400 and 200 MW/Hz Answer: A. K1 = 600/(0.03x50) = 400; K2 = 300/(0.06x50) = 100. (Key point: Stiffness) 2. Since both units are at 51 Hz at no load, how is the frequency drop at a given load found? A. From 50 Hz B. From the 51 Hz no-load frequency using total load = K total x drop C. From 49 Hz D. Using the initial load difference Answer: B. Each unit's output is K x (51 - f), so total load = 500 x (51 - f). (Key point: Reference frequency) 3. At 600 MW load, what are f and the shares? A. 49.8 Hz; 400 and 200 MW B. 50 Hz; 300 and 300 MW C. 49.8 Hz; 480 and 120 MW D. 49.4 Hz; 480 and 120 MW Answer: C. drop = 600/500 = 1.2 Hz; P1 = 480, P2 = 120. (Key point: 600 MW case) 4. At 800 MW, what are f and the problem noted? A. 49.4 Hz; no problem B. 49.0 Hz; G2 overloaded C. 49.6 Hz; G1 at 560 MW D. 49.4 Hz; G1 at 640 MW exceeds its 600 MW rating Answer: D. drop = 1.6 Hz, P1 = 640 MW which exceeds 600; G1 would stick at full gate and G2 would take the rest. (Key point: 800 MW case) ## Related in Power Plant Equipment - [Load sharing: 600 MW and 300 MW, 7%/8% droop](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-600-300-7-8) — Power Plant Equipment, chapter 2, asked 2× - [Load sharing: 600 and 300 MW at 49 Hz](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-600-300-49hz) — Power Plant Equipment, chapter 2, asked 4× - [Load sharing: 500 MW and 250 MW, 5%/6% droop](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-500-250-5-6) — Power Plant Equipment, chapter 2, asked 2× - [Load sharing: 250/500 MW, frequency falls to 49.5 Hz](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-250-500-4-2) — Power Plant Equipment, chapter 2, asked 2× - [Droop calculation: 800 MW and 300 MW units](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Adroop-calculation-800-300) — Power Plant Equipment, chapter 2 - [Load sharing: 250 MW and 500 MW, 4%/3% droop](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-250-500-4-3) — Power Plant Equipment, chapter 2 ## Related in other subjects - [Load sharing between two stations in parallel](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aload-sharing-two-stations) — Electric Machines II, chapter 1 - [Load sharing of two transformers in parallel](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aparallel-transformers-load-sharing) — Electrical Machines, chapter 2