--- title: "Optimizing plant capacity of a RoR project" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:ror-optimum-capacity-discussion" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-optimum-capacity-discussion site: BCT Engineering --- # Optimizing plant capacity of a RoR project Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2080 Bhadra, 4 marks - Video: [How to optimize hydropower](https://www.youtube.com/watch?v=pGfgzHYDEyo) — hec-ressim - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-optimum-capacity-discussion - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question How do you optimize the plant capacity of a RoR project? Discuss. ## Short answer Prepare the flow duration curve, compute power and annual energy for trial design discharges, then calculate annual benefit (energy × price) and annualized cost (fixed + variable). Optimum is maximum net benefit or where marginal benefit equals marginal annual cost, so the last added kW pays for itself. ## Answer The installed capacity of a RoR plant is optimised by comparing the extra cost and the extra benefit of each increase in capacity. Since flow varies throughout the year, a bigger plant uses the large flow only for a short time, so there is an optimum point. #### Procedure 1. Prepare the flow duration curve (FDC) from the long-term flow series (after the environmental release). 2. Select a number of trial design discharges $Q_d$ (e.g. $Q_{30}$, $Q_{40}$, $Q_{50}$ ...). For each, compute the power $P = 9.81\,\eta\,H_n\,Q_d$ and the energy: $$ E = \sum 9.81\,\eta\,H_n\,\min(Q_i,Q_d)\,t_i $$ 3. Compute annual benefit = energy × price (firm and secondary energy may have different prices). 4. Compute annual cost = (fixed cost + variable cost × P) × capital recovery factor + O&M. Fixed cost includes the headworks, civil structures, access and so on; variable cost is proportional to the capacity (turbine, generator, penstock). 5. Plot the net benefit (benefit − cost), the benefit/cost ratio and the marginal values against the capacity. #### Criterion - **Maximum net benefit** (NPV), or - **Marginal benefit = marginal cost:** increase capacity until the extra benefit from the last kW (= price × energy produced by it during the time it is used) equals its extra annual cost. Beyond this point the machines run only a few weeks per year and do not pay back. Other considerations: the cost per kW, the plant factor (usually 50-70% for RoR), the grid demand, the risk and the possibility of staged development. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does an optimum capacity exist for a RoR plant? A. Flow is constant B. A bigger plant uses large flows only for a short time, so extra capacity pays less C. Cost falls with capacity D. Efficiency is zero at large capacity Answer: B. Large flows occur briefly, so extra capacity yields diminishing benefit. (Key point: Why an optimum exists) 2. Which curve is used to select trial design discharges? A. Mass curve B. Rating curve C. Flow duration curve D. Load curve Answer: C. Trial Q_d values such as Q30, Q40, Q50 come from the FDC. (Key point: Starting curve) 3. What is the marginal criterion for optimum capacity? A. Total cost equals total benefit B. Marginal benefit equals marginal cost of the last kW C. Capacity equals mean flow capacity D. Plant factor is 100% Answer: B. Capacity is increased until the extra benefit from the last kW equals its extra annual cost. (Key point: Marginal criterion) 4. Which cost is proportional to capacity? A. Variable cost (turbine, generator, penstock) B. Fixed cost of headworks C. Access road D. Environmental release Answer: A. Fixed cost does not depend on capacity; variable cost does. (Key point: Cost components) ## Related in Hydropower Engineering - [Optimize capacity by marginal cost-benefit](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Amarginal-cost-benefit-installed-capacity) — Hydropower Engineering, chapter 3 - [RoR best installed capacity (USD 2795 million)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-best-capacity-usd-2795) — Hydropower Engineering, chapter 3 - [Load duration curve; diesel plant; ROR flow duration](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aload-duration-diesel-import-fdc) — Hydropower Engineering, chapter 3 - [RoR installed capacity without pondage and with peaking](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-min-flow-pondage-peaking-70m) — Hydropower Engineering, chapter 3 - [RoR best installed capacity (NRs, 45 years)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-best-capacity-nrs-10-percent) — Hydropower Engineering, chapter 3 - [RoR best installed capacity (USD 1200 million)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-best-capacity-usd-1200) — Hydropower Engineering, chapter 3 ## Related in other subjects - [Intake site selection for RoR plant](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aintake-site-selection-ror) — Hydro Power, chapter 5 - [ROR versus PROR hydropower plants](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Aror-vs-pror) — Power Plant Design, chapter 3 - [Design: 360 m head plant, 132 kV](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2078-360m-132kv) — Power Plant Design, chapter 4, asked 5× - [Design: 260 m head plant, generator lead](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2078-260m-generator-lead) — Power Plant Design, chapter 4, asked 5×