---
title: "RoR pondage: capacity and pondage volume"
subject: "Hydropower Engineering"
chapter: 3
key: "semester_7/hydropower_engineering:ror-pondage-capacity-load-factor-65"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-pondage-capacity-load-factor-65
site: BCT Engineering
---

# RoR pondage: capacity and pondage volume

Hydropower Engineering · Chapter 3: Power and Energy Potential study

- Asked in: 2075 Chaitra, 4+4 marks
- Video: [Pondage Requirement Type3- Hydropower Engineering #dipakdahal #engineering](https://www.youtube.com/watch?v=qDp0cinS8us) — 𝐄𝐫. 𝐃𝐢𝐩𝐚𝐤 𝐃𝐚𝐡𝐚𝐥
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-pondage-capacity-load-factor-65
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3

## Question

A run-of-river hydroelectric power station is proposed across a river at a site where a net head of 30 m is available on the turbine. The river carries a sustained minimum flow of 25 m³/s in dry weather and behind the power station sufficient pondage is provided to supply the daily peak load demand with a load factor of 65%. Assuming the plant efficiency of 60%, determine the maximum generating capacity of the generator to be installed at the power house. If the daily load pattern indicates 20 hours average load and 4 hours of peak load, determine the volume of pondage to be provided to supply the daily demand.

## Short answer

Generator capacity is 6.79 MW. Pondage volume needed is about 193,846 m³, calculated from the peak flow deficit over 4 hours.

## Answer

#### Maximum generating capacity
The minimum flow gives the average power available throughout the day:

$$
P_{avg} = \gamma Q H \eta = 9.81\times25\times30\times0.60 = 4,414.5\ \text{kW}
$$

With pondage the plant can follow the daily load, and the load factor is

$$
\text{Load factor} = \frac{P_{avg}}{P_{max}} = 0.65 \Rightarrow P_{max} = \frac{4,414.5}{0.65} = 6,791.5\ \text{kW} \approx 6.79\ \text{MW}
$$

#### Volume of pondage (20 h average load + 4 h peak load)
Peak load: $P_{max}=6,791.5$ kW for 4 h. The average over the day must equal $P_{avg}$, so the off-peak load for 20 h is

$$
P_{off} = \frac{24\times4,414.5 - 4\times6,791.5}{20} = 3,939.1\ \text{kW}
$$

Flows needed at peak and off-peak (with $9.81\times30\times0.6 = 176.58$ kW per m³/s):

$$
Q_{peak} = \frac{6,791.5}{176.58} = 38.46\ \text{m}^3/\text{s},\qquad Q_{off} = \frac{3,939.1}{176.58} = 22.31\ \text{m}^3/\text{s}
$$

The river supplies 25 m³/s, so the deficit of $(38.46-25)$ m³/s for 4 hours must come from the pond:

$$
V = (38.46-25)\times4\times3600 = 193,846\ \text{m}^3 \approx 1.94\times10^5\ \text{m}^3
$$

(During the 20 off-peak hours the pond fills by $(25-22.31)\times20\times3600 = 193,846$ m³, the same volume.)

**Answer: generator capacity = 6.79 MW; pondage volume ≈ 193,846 m³ (≈ 1.94 × 10⁵ m³).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does the sustained minimum flow of 25 m3/s give?
   A. The maximum capacity directly
   B. The average power available through the day, 9.81 x 25 x 30 x 0.6 = 4,414.5 kW
   C. The peak power only
   D. The off-peak power only
   Answer: B. The minimum flow gives continuous average power; pondage then shapes it into peak and off-peak. (Key point: Average power from minimum flow)

2. How is the maximum generator capacity obtained?
   A. P_max = P_avg x 0.65
   B. P_max = P_avg / load factor = 4,414.5 / 0.65 = 6.79 MW
   C. P_max = P_avg + 0.65
   D. P_max = P_avg / (1 - 0.65)
   Answer: B. Load factor = average load / maximum load, so P_max = P_avg / 0.65. (Key point: Load factor use)

3. What flow is needed at peak load?
   A. 25 m3/s
   B. Q = P_max / (9.81 x H x eta) = 38.46 m3/s
   C. 50 m3/s
   D. Q = P_avg / (9.81 x H x eta)
   Answer: B. 6,791.5 / 176.58 = 38.46 m3/s. (Key point: Peak flow)

4. Why is the pondage (38.46 - 25) x 4 x 3600?
   A. The pond supplies the part of the peak flow that the river does not provide, over 4 hours
   B. It is the full peak flow stored for 24 h
   C. It is the minimum flow stored for 20 h
   D. It is the load factor times daily flow
   Answer: A. The river contributes 25 m3/s; the deficit of 13.46 m3/s for 4 h is taken from the pond, about 1.94 x 10^5 m3. (Key point: Pondage volume)

## Related in Hydropower Engineering

- [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
- [Peaking RoR: max power and wet/dry energy ratio](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apeaking-ror-reservoir-250m-q25) — 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 capacity, peaking capacity and benefit](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-three-sets-peaking-benefit) — Hydropower Engineering, chapter 3
- [Pondage surface area and weekly output](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apondage-weekly-operation-40cumec) — 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

- [Daily pondage ROR with 150 m head](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adaily-pondage-150m-head) — Power Plant Design, chapter 3
- [Daily pondage run-of-river components](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adaily-pondage-ror-components) — Power Plant Design, chapter 3
- [Micro hydro net head, power and energy](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Amicro-hydro-power-output-numerical) — Energy Resources, chapter 5
- [Primary/secondary energy and pondage volume](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aenergy-and-pondage-volume) — Hydro Power, chapter 3