--- title: "Gross and net potential between two sections" subject: "Hydropower Engineering" chapter: 3 key: "semester_7/hydropower_engineering:gross-net-potential-two-sections" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gross-net-potential-two-sections site: BCT Engineering --- # Gross and net potential between two sections Hydropower Engineering · Chapter 3: Power and Energy Potential study - Asked in: 2080 Baishakh, 4 marks - Video: [What is Gross head, Net head, Operating head and Rated Head in Hydropower?](https://www.youtube.com/watch?v=pG1oOjBpawI) — Hydropower - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gross-net-potential-two-sections - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3 ## Question How do you estimate the gross and net hydropower potential between two sections of a river? ## Short answer Gross potential is computed by summing 9.81 Q_i ΔH_i for sub‑reaches using mean flows and head drops. Net potential uses design discharge from the flow duration curve after environmental release, net head (gross minus hydraulic losses) and overall turbine‑generator efficiency (∼0.8–0.9): P_net = 9.81 η Q_d H_n (kW). ## Answer **Gross potential** is the theoretical power of the river between two sections, using all the flow and the full head difference with 100% efficiency. **Net potential** is the power that can actually be developed after deducting head loss, efficiency and the part of the flow that cannot be used. #### Gross potential Divide the reach into small sub-reaches where the flow is nearly constant (the flow increases downstream with tributaries and catchment area). For each sub-reach: $$ P_{gross} = \sum \rho g Q_i \Delta H_i = \sum 9.81\, Q_i\, \Delta H_i \ \ (\text{kW}) $$ where $Q_i$ is the mean flow (m³/s) and $\Delta H_i$ is the drop in elevation (m) of the sub-reach. Annual energy = $P \times 8760$ kWh. Flow comes from the flow series at the two sections (area-ratio method for ungauged sites), and elevations from topographic maps or survey. #### Net (technical/economic) potential - Choose the design discharge $Q_d$ from the flow duration curve (e.g. $Q_{40}$ for RoR), after leaving the environmental flow. - Net head $H_n = H_{gross} - h_f$, where $h_f$ includes the intake, desander, waterway and penstock losses (about 5-10% of gross head). - Overall efficiency $\eta$ of the turbine, generator and transformer (about 80-90%). $$ P_{net} = 9.81\, \eta\, Q_d\, H_n \ \ (\text{kW}) $$ - Annual energy is found from the monthly flows up to $Q_d$. Economic potential is the part for which benefit-cost ratio is greater than one. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is gross hydropower potential between two river sections? A. Theoretical power using all the flow and full head difference at 100% efficiency B. Power after deducting losses C. Power of the design discharge only D. Installed capacity of existing plants Answer: A. Gross potential ignores losses, efficiency and unusable flow. (Key point: Gross potential definition) 2. Why is the reach divided into sub-reaches in the gross estimate? A. To reduce the head B. Because flow increases downstream with tributaries, so flow is nearly constant in each sub-reach C. To avoid using elevations D. Because efficiency changes Answer: B. Flow increases with catchment area, so P = sum of 9.81 Q_i ΔH_i is applied per sub-reach. (Key point: Sub-reaches) 3. How is net head found? A. Gross head plus losses B. Gross head times efficiency C. Gross head minus head losses in intake, desander, waterway and penstock D. Same as gross head Answer: C. H_n = H_gross - h_f, with losses about 5-10% of gross head. (Key point: Net head) 4. Which formula gives net potential? A. P = 9.81 Q H B. P = 9.81 η Q_d H_n C. P = ρ Q D. P = η Q_d / H Answer: B. Net power uses design discharge, net head and overall efficiency. 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