--- title: "RoR at 40% exceedence; storage regulation" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:ror-40-percent-exceedence-storage-regulation" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-40-percent-exceedence-storage-regulation site: BCT Engineering --- # RoR at 40% exceedence; storage regulation Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2070 Chaitra, 2+2+2+2+2+2 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-40-percent-exceedence-storage-regulation - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question A hydropower project is planned to develop in a river having net head of 100 m and overall efficiency of 85% with the monthly hydrograph as shown below. | Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Discharge (m³/s) | 100 | 120 | 140 | 300 | 320 | 1800 | 2000 | 2500 | 2100 | 900 | 500 | 300 | i) Calculate the installed capacity, annual spill energy and firm energy if the RoR project is designed based on the 40% probability of exceedence flow. ii) If the storage project is developed with full regulation of the annual hydrograph (design discharge is equal to the average monthly flow), calculate the storage requirements. iii) Calculate the installed capacity and annual energy generation from the storage project as mentioned in the above case. ## Short answer (i) P = 683.8 MW, spill energy ≈ 3165 GWh, firm energy = 730.5 GWh. (ii) Storage required = 12,369 Mm³. (iii) P = 769.9 MW, annual energy = 6745 GWh. ## Answer #### Data $H=100$ m, $\eta=0.85$, so $9.81\,\eta\,H = 833.85$ kW per m³/s. One month $=730$ h $=2.628\times10^6$ s. #### i) RoR designed for $Q_{40}$ Flows in descending order: 2500, 2100, 2000, 1800, 900, 500, 320, 300, 300, 140, 120, 100. By the Weibull formula $p=\frac{m}{N+1}\times100$ with $N=12$, $p=40\%$ gives $m=0.4\times13=5.2$, which lies between 900 (m = 5) and 500 (m = 6): $$ Q_{40} = 900 - 0.2\,(900-500) = 820\ \text{m}^3/\text{s} $$ $$ P = 833.85\times820 = 683,757\ \text{kW} = 683.8\ \text{MW} $$ - **Spill (flow above $Q_{40}$)** by month (m³/s): Jun 980, Jul 1180, Aug 1680, Sep 1280, Oct 80, total $=5,200$ m³/s-months. $$ E_{spill} = 833.85\times5,200\times730 = 3,165\ \text{GWh} $$ - **Firm energy** (minimum flow 100 m³/s available all year): $$ E_{firm} = 833.85\times100\times8760 = 730.5\ \text{GWh} $$ - (Total energy of the RoR plant $= 833.85\times5,880\times730 = 3,579$ GWh, so the non-firm energy is 2,849 GWh.) #### ii) Storage with full regulation The design discharge equals the mean flow: $$ \bar Q = \frac{11,080}{12} = 923.33\ \text{m}^3/\text{s} $$ Deficits ($\bar Q-Q$) in the months with flow below the mean, from Oct to May (m³/s-months): Oct 23.33, Nov 423.33, Dec 623.33, Jan 823.33, Feb 803.33, Mar 783.33, Apr 623.33, May 603.33. Sum $=4,706.67$ (this equals the surplus of Jun-Sep, which refills the reservoir). $$ V_s = 4,706.67\times2.628\times10^6 = 12,369\times10^6\ \text{m}^3 \approx 12.37\ \text{billion m}^3 $$ #### iii) Installed capacity and energy of the storage project $$ P = 833.85\times923.33 = 769,922\ \text{kW} = 769.9\ \text{MW} $$ $$ E = P\times8760 = 6,745\ \text{GWh/year} $$ (The head is taken as constant at 100 m.) **Answer: (i) P = 683.8 MW at $Q_{40}=820$ m³/s; spill energy ≈ 3,165 GWh; firm energy = 730.5 GWh. (ii) storage = 12,369 Mm³. (iii) P = 769.9 MW; E = 6,745 GWh.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is Q40 found? A. Rank flows in descending order and interpolate at m = 0.4 x 13 = 5.2 between 900 and 500, giving 820 m3/s B. Average of all flows C. The 4th flow in the list D. 40% of the maximum flow Answer: A. m = 5.2 lies between the 5th (900) and 6th (500) flows, so Q40 = 900 - 0.2 x 400 = 820 m3/s. (Key point: Q40 by Weibull) 2. On what flow is firm energy based? A. Q40 = 820 B. The minimum flow, 100 m3/s, available all year C. The mean flow D. The Q90 Answer: B. E_firm = 833.85 x 100 x 8760 = 730.5 GWh. (Key point: Firm energy) 3. What is the spill energy in the RoR project? A. Energy of the flow above Q40, which cannot pass through the turbines B. Energy of the flow below Q40 C. Firm energy D. Zero Answer: A. The flow above the design discharge (total 5,200 m3/s-months) is spilled. (Key point: Spill) 4. How is the storage needed for full regulation obtained? A. Sum of the deficits (mean flow - Q) of months below the mean, 4,706.67 m3/s-months x 2.628 x 10^6 s B. Sum of all flows C. Mean flow x 12 D. Q40 x one month Answer: A. The reservoir must supply the deficits of the low-flow months; this equals the surplus stored in Jun-Sep, about 12,369 Mm3. (Key point: Storage requirement) 5. Why is the storage plant larger than the RoR plant (769.9 MW vs 683.8 MW) and why is its energy higher? A. Its design discharge is the mean flow 923 m3/s, used all year without spill B. Its head is higher C. Its efficiency is higher D. It uses Q90 Answer: A. With full regulation Qd = mean flow and the plant runs at constant Qd all year: E = 769.9 MW x 8760 h = 6,745 GWh. 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