---
title: "RoR installed capacity without pondage and with peaking"
subject: "Hydropower Engineering"
chapter: 3
key: "semester_7/hydropower_engineering:ror-min-flow-pondage-peaking-70m"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-min-flow-pondage-peaking-70m
site: BCT Engineering
---

# RoR installed capacity without pondage and with peaking

Hydropower Engineering · Chapter 3: Power and Energy Potential study

- Asked in: 2075 Ashwin, 6 marks
- Video: [Calculate Power,  pondage requirement  || Hydropower Numerical](https://www.youtube.com/watch?v=Dt3SBc-UOuY) — OPengineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-min-flow-pondage-peaking-70m
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3

## Question

A RoR plant has a minimum flow of 30 m³/s and net head of 70 m. The overall efficiency of the plant is 85%. Calculate the installed capacity of the plant (i) without pondage (designed for pure RoR plant) and (ii) if the plant is designed for a peaking plant with 6 hours peaking. The plant has two sets of units such that one unit at full capacity is operating during off-peak hours. Total evaporation and other losses is 5% of the stored water.

## Short answer

For pure run-of-river, installed capacity = 17.51 MW. With pondage and 6 h peaking using two units, peak discharge is 47.63 m³/s, giving installed capacity = 27.80 MW.

## Answer

#### (i) Pure RoR plant (no pondage)
Only the minimum flow can be used continuously:

$$
P_1 = 9.81\,\eta\,Q\,H = 9.81\times0.85\times30\times70 = 17,511\ \text{kW} = 17.51\ \text{MW}
$$

#### (ii) Peaking plant, 6 h peak, two units
Let $Q_u$ be the discharge of one unit at full load. Off-peak (18 h) only one unit runs, using $Q_u$. In peak hours (6 h) both units run, using $2Q_u$.

Water stored in the 18 off-peak hours (inflow 30 m³/s) after 5% loss:

$$
V_s = 0.95\,(30-Q_u)\times18\times3600
$$

During the 6 peak hours the extra flow required is $(2Q_u-30)$, supplied from storage:

$$
(2Q_u-30)\times6\times3600 = 0.95\,(30-Q_u)\times18\times3600
$$

$$
12Q_u - 180 = 513 - 17.1\,Q_u \Rightarrow Q_u = \frac{693}{29.1} = 23.814\ \text{m}^3/\text{s}
$$

Peak discharge $=2Q_u=47.629$ m³/s.

$$
P_2 = 9.81\times0.85\times47.629\times70 = 27,801\ \text{kW} = 27.80\ \text{MW}
$$

Check: off-peak, one unit uses 23.81 m³/s (less than 30), so the surplus goes to the pond. Water put into the pond $=(30-23.814)\times18\times3600 = 400,825$ m³ (the pondage capacity, about 4.01 × 10⁵ m³). After the 5% loss, $0.95\times400,825 = 380,784$ m³ is usable, which equals the peak draw $(47.629-30)\times6\times3600 = 380,784$ m³.

**Answer: (i) installed capacity = 17.51 MW (pure RoR); (ii) installed capacity = 27.80 MW with 6-hour peaking (2 units of 13.90 MW each), a gain of 1.59 times.**

## Check yourself

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

1. What flow decides the capacity of a pure RoR plant without pondage?
   A. Average flow
   B. The minimum flow of 30 m3/s
   C. Twice the minimum flow
   D. Peak flow
   Answer: B. Only the minimum flow is certain all the time, so P = 9.81 x 0.85 x 30 x 70 = 17.5 MW. (Key point: Pure RoR capacity)

2. In the peaking design with two units, how do the units operate?
   A. Both units run all day
   B. One unit at full capacity runs in the 18 off-peak hours and both units run in the 6 peak hours
   C. Both units run only off-peak
   D. One unit runs only in the peak
   Answer: B. The off-peak flow is Qu and peak flow is 2Qu. (Key point: Off-peak and peak operation)

3. What equation gives Qu?
   A. (2Qu - 30) x 6 = 0.95 (30 - Qu) x 18, peak deficit equals stored water after 5% loss
   B. 2Qu = 30
   C. Qu x 24 = 30 x 24 x 0.95
   D. (2Qu - 30) x 18 = (30 - Qu) x 6
   Answer: A. Stored water in the 18 off-peak hours is reduced by 5% and must equal the extra water needed in the 6 peak hours. (Key point: Water balance with loss)

4. The result Qu = 23.81 m3/s leads to an installed capacity of
   A. 17.51 MW
   B. About 27.8 MW, using the peak discharge 2Qu = 47.63 m3/s
   C. 13.9 MW, from one unit
   D. 55.6 MW
   Answer: B. P = 9.81 x 0.85 x 47.63 x 70 = 27.8 MW. (Key point: Peak capacity)

## Related in Hydropower Engineering

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- [RoR pondage: capacity and pondage volume](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-pondage-capacity-load-factor-65) — Hydropower Engineering, chapter 3
- [Installed capacity, energy and daily peaking reservoir (2079)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-capacity-plant-factor-18cumec-peaking-79) — Hydropower Engineering, chapter 3, asked 2×
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