--- title: "Load duration curve; diesel plant; ROR flow duration" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:load-duration-diesel-import-fdc" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/load-duration-diesel-import-fdc site: BCT Engineering --- # Load duration curve; diesel plant; ROR flow duration Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2073 Shrawan, 5+5 marks - Video: [Plotting load duration curve](https://www.youtube.com/watch?v=XcXPinRTgC0) — Understanding Electrical Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/load-duration-diesel-import-fdc - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question The power supplied by the state electricity authority throughout the year by a steam power plant is as shown in the table below. | Month | Power supplied (MW) | |---|---| | Jestha | 550 | | Asar | 500 | | Shrawan | 450 | | Bhadra | 380 | | Asoj | 330 | | Kartik | 280 | | Mangsir | 250 | | Poush | 220 | | Magh | 200 | | Falgun | 150 | | Chaitra | 145 | | Baisakh | 100 | But the current demand forced them to have load shedding. To minimize the load shedding by providing at least power equivalent to the Magh month throughout the year, the Authority has decided to import power from a neighbouring country for only 3 months, i.e. Falgun, Chaitra and Baisakh, as 50 MW, 55 MW and 100 MW respectively. a) Despite importing power, the authority felt that they cannot provide uniform power of Magh throughout. So they decide to have a diesel plant for the deficit. Estimate the minimum capacity of the diesel plant. (Use load duration curve for analysis.) b) If instead of the above system (steam plant + import + diesel plant), the Authority has planned to provide the power in the near future by constructing a ROR hydropower plant by its own to substitute the current model, derive the flow duration curve for such a new hydro project to supply the power demand given in the table. Assume power demand is constant in future. ## Short answer The imports of 50 MW, 55 MW, and 100 MW exactly cover the deficits in Falgun, Chaitra, and Baisakh, so no diesel plant is needed (0 MW). A ROR plant would require a flow duration curve given by Q = P/(9.81ηH); with example H=100 m, η=0.85, flows range from about 660 m³/s (8.3% exceedance) to 120 m³/s (100% exceedance). ## Answer Each month is taken as 1/12 of the year (8.33% of the time). The months are already in descending order of power, so the table is the load duration curve (LDC). #### a) Load duration curve and diesel plant | Rank | Month | Power (MW) | % of time exceeded | |---|---|---|---| | 1 | Jestha | 550 | 8.3 | | 2 | Asar | 500 | 16.7 | | 3 | Shrawan | 450 | 25.0 | | 4 | Bhadra | 380 | 33.3 | | 5 | Asoj | 330 | 41.7 | | 6 | Kartik | 280 | 50.0 | | 7 | Mangsir | 250 | 58.3 | | 8 | Poush | 220 | 66.7 | | 9 | Magh | 200 | 75.0 | | 10 | Falgun | 150 | 83.3 | | 11 | Chaitra | 145 | 91.7 | | 12 | Baisakh | 100 | 100.0 | ```text MW 550 |* 500 | * 450 | * 380 | * 330 | * 280 | * 250 | * 220 | * 200 |--------*---- Magh level (target) 150 | * 145 | * 100 | * +------------------------ 0 25 50 75 100 % ``` Target: at least the Magh power, 200 MW, in every month. Months below 200 MW and their deficits: | Month | Steam supply (MW) | Deficit to 200 MW | Import (MW) | Remaining deficit | |---|---|---|---|---| | Falgun | 150 | 50 | 50 | 0 | | Chaitra | 145 | 55 | 55 | 0 | | Baisakh | 100 | 100 | 100 | 0 | The import in these three months is exactly equal to the deficits, so on the LDC the supply is at or above 200 MW for 100% of the time and **no further deficit is left** to be covered by the diesel plant. The minimum diesel capacity needed is therefore **0 MW** for the stated imports. If the imports were not available, or fell short by $x$ MW in a month, the minimum diesel plant would be the largest remaining deficit: with no import at all it would be the largest gap, **100 MW** (in Baisakh), and it would run only for about 25% of the time (energy ≈ (50+55+100) MW × 730 h = 149,650 MWh/year). #### b) Flow duration curve for a RoR hydro plant A constant (future) demand equal to the table has to be met by the hydropower plant, so the required flow follows from $$ Q = \frac{P}{9.81\,\eta\,H} $$ Head and efficiency are not given, so assume $H=100$ m and $\eta=0.85$: $9.81\times0.85\times100 = 833.85$ kW per m³/s, i.e. $Q=P/833.85$ with $P$ in kW. The FDC has the same shape as the LDC. | Rank | Power (MW) | Flow Q (m³/s) | % of time exceeded | |---|---|---|---| | 1 | 550 | 659.6 | 8.3 | | 2 | 500 | 599.6 | 16.7 | | 3 | 450 | 539.7 | 25.0 | | 4 | 380 | 455.7 | 33.3 | | 5 | 330 | 395.8 | 41.7 | | 6 | 280 | 335.8 | 50.0 | | 7 | 250 | 299.8 | 58.3 | | 8 | 220 | 263.8 | 66.7 | | 9 | 200 | 239.9 | 75.0 | | 10 | 150 | 179.9 | 83.3 | | 11 | 145 | 173.9 | 91.7 | | 12 | 100 | 119.9 | 100.0 | So the flow required is 659.6 m³/s for 8.3% of the time, 263.8 m³/s for about 67% of the time and at least 119.9 m³/s all the year (flow scales as $1/(\eta H)$ for other values). The plant must be able to deliver 550 MW; for the minimum months the river must provide at least the flow of the Baisakh row. **Answer: (a) diesel capacity = 0 MW with the stated imports (deficits 50, 55, 100 MW are fully covered); (b) FDC as tabulated, with Q = P/(9.81ηH).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the monthly table be used directly as the load duration curve? A. The months are already in descending order of power, each representing 1/12 (8.33%) of the time B. Because power is constant during each month and the months are in calendar order C. Because the table is converted to energy first D. Because the LDC must always be drawn from the monthly average only Answer: A. An LDC ranks loads in descending order against percent of time exceeded; the table is already ranked, with each month being 8.33% of the year. (Key point: Load duration curve construction) 2. For the target of at least 200 MW (Magh level), which months have a deficit and what are the deficits? A. Poush, Magh and Falgun: 20, 0 and 50 MW B. Falgun, Chaitra, Baisakh: 50, 55 and 100 MW C. Jestha, Asar, Shrawan: 350, 300 and 250 MW D. Only Baisakh: 100 MW Answer: B. Only Falgun (150), Chaitra (145) and Baisakh (100) are below 200 MW, giving deficits of 50, 55 and 100 MW. (Key point: Deficit below target level) 3. How is the flow needed for the RoR plant obtained from each power value? A. Q = P x 9.81 x eta x H B. Q = P/(9.81 eta H), with an assumed head and efficiency, giving an FDC of the same shape as the LDC C. Q = P/H only D. Q = P x eta / 9.81 only Answer: B. From P = 9.81 eta Q H, Q = P/(9.81 eta H); with constant head and efficiency, flow is proportional to power, so the FDC has the LDC's shape. 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