---
title: "Pondage surface area and weekly output"
subject: "Hydropower Engineering"
chapter: 3
key: "semester_7/hydropower_engineering:pondage-weekly-operation-40cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pondage-weekly-operation-40cumec
site: BCT Engineering
---

# Pondage surface area and weekly output

Hydropower Engineering · Chapter 3: Power and Energy Potential study

- Asked in: 2076 Ashwin, 5+5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pondage-weekly-operation-40cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3

## Question

During a low water week a river has an average daily flow of 40 m³/s. Fluctuation during the day required a pondage capacity of approximately 30% of the daily discharge. A hydroelectric plant is to be located on the river which will operate 6 days a week, 24 hours a day, but will supply power at a varying rate such that the daily load factor is 50%, corresponding to which the pondage required is equal to 0.2 times the mean flow to the turbine. On Saturday all the flow is ponded for use on the rest of the days. If the effective head on the turbines when the pond is full is to be 25 m and the maximum allowable fluctuation in pond level is 1 m, find (i) the surface area of the pond to satisfy all the operating conditions, (ii) the weekly output at the switch board in kWh. Assume turbine efficiency 80% and generator efficiency 90%.

## Short answer

Pond surface area is about 4.49 km². Weekly switchboard output is about 1.16 GWh, using average head of 24.5 m.

## Answer

#### Data
Inflow $Q=40$ m³/s; daily inflow volume $=40\times86400 = 3,456,000$ m³. The plant works 6 days (Sunday-Friday) and on Saturday the whole flow is stored. $H_{full}=25$ m, maximum fluctuation $=1$ m, $\eta_t=0.80$, $\eta_g=0.90$.

#### (i) Surface area of the pond
**1. Weekly pondage (Saturday flow):** all of Saturday's inflow is stored for use on the other days.

$$
V_1 = 40\times86400 = 3,456,000\ \text{m}^3
$$

**2. Daily pondage:** two requirements are given.
- To take care of the daily flow fluctuation: $0.30\times3,456,000 = 1,036,800$ m³
- For the 50% daily load factor: mean flow to the turbines on working days $=40\times\frac{7}{6}=46.67$ m³/s, so the pondage $=0.2\times46.67\times86400 = 806,400$ m³

The pond must satisfy both, so the larger value is taken: $V_2=1,036,800$ m³.

**3. Total live storage:**

$$
V = V_1+V_2 = 3,456,000+1,036,800 = 4,492,800\ \text{m}^3
$$

The allowed draw-down is 1 m, so

$$
A = \frac{V}{\Delta h} = \frac{4,492,800}{1} = 4,492,800\ \text{m}^2 \approx 4.49\ \text{km}^2
$$

#### (ii) Weekly output at the switchboard
The whole weekly inflow is used: $40\times7\times24 = 6,720$ m³/s·h. The pond level goes down by 1 m, so the average head is $25-0.5 = 24.5$ m.

$$
E = 9.81\,\eta_t\eta_g\,H\,\sum Q\,t = 9.81\times0.80\times0.90\times24.5\times6,720 = 1,162,885\ \text{kWh}
$$

(With the full-pond head of 25 m, the figure would be 1,186,618 kWh.)

**Answer: pond surface area ≈ 4.49 km² (4,492,800 m²); weekly output ≈ 1.16 GWh (1,162,885 kWh).**

## Check yourself

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

1. What is the weekly pondage in this problem?
   A. 30% of daily flow
   B. Saturday's whole inflow, 40 x 86400 = 3,456,000 m3, stored for use on the other days
   C. 0.2 times mean flow
   D. The 1 m fluctuation volume
   Answer: B. Since the plant does not run on Saturday, all the flow of that day is ponded for the other days. (Key point: Weekly pondage)

2. Two daily pondage requirements are found (1,036,800 m3 and 806,400 m3). Which is used?
   A. The smaller
   B. The average
   C. The larger, 1,036,800 m3, since the pond must satisfy both conditions
   D. Their sum
   Answer: C. To satisfy all operating conditions the larger of the two is adopted. (Key point: Daily pondage choice)

3. How is the surface area of the pond found?
   A. Total live storage / allowable draw-down: 4,492,800 / 1 m
   B. Storage x 1 m
   C. Weekly pondage only / 25 m
   D. Daily flow / head
   Answer: A. A = V / Δh with V = weekly + daily pondage and Δh = 1 m. (Key point: Surface area)

4. Which head is used for the weekly energy and why?
   A. Full head of 25 m always
   B. 24.5 m, the average head, since the pond falls 1 m
   C. 24 m, the minimum head
   D. 26 m
   Answer: B. The level falls by 1 m, so the average head is 25 - 0.5 = 24.5 m. (Key point: Head for weekly energy)

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