---
title: "Tidal power output: 3 km × 5.5 km barrage"
subject: "Power Plant Design"
chapter: 1
key: "semester_8/power_plant_design:tidal-barrage-3x5-5km"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/tidal-barrage-3x5-5km
site: BCT Engineering
---

# Tidal power output: 3 km × 5.5 km barrage

Power Plant Design · Chapter 1: Energy Sources and Electric Power Generation

- Asked in: 2078 Chaitra, 4 marks
- Video: [energy in tidal barrages, HEP and pumped storage systems](https://www.youtube.com/watch?v=f3Ymi-9LFyQ) — physicshelp.online
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/power_plant_design/tidal-barrage-3x5-5km
- Whole chapter: https://www.bctengineer.com/md/semester_8/power_plant_design/chapter-1

## Question

Assume that there is barrage of size 3 km × 5.5 km for tidal power project. Average number of tides per day is 2 and the tide height is 10 m. Considering power conversion efficiency of 60%, determine power output capacity of the tidal project. The density of sea water is 1023 kg/m³.

## Short answer

A tidal barrage traps basin water at high tide and releases it through turbines when the sea falls; energy per tide is 0.5 rho g A h^2. Using area 1.65e7 m^2, height 10 m, density 1023 kg/m^3, two tides, and 60% efficiency, average power capacity is about 115 MW.

## Answer

In a tidal barrage, the basin fills at high tide and the trapped water is released through turbines when the sea falls. The potential energy of water of area $A$ and tidal range $h$ (centre of mass at $h/2$) is:
$$
E = \rho g A h \cdot \frac{h}{2} = \frac{1}{2}\rho g A h^2
$$

#### Given
- $A = 3\ \text{km} \times 5.5\ \text{km} = 3000 \times 5500 = 1.65\times 10^7\ \text{m}^2$
- $h = 10$ m, $\rho = 1023$ kg/m³, $g = 9.81$ m/s²
- 2 tides per day, $\eta = 0.60$

#### Step 1: Energy per tide
$$
\begin{aligned}
E &= 0.5 \times 1023 \times 9.81 \times 1.65\times 10^7 \times 10^2 \\
&= 8.279 \times 10^{12}\ \text{J} \\
&= \frac{8.279\times 10^{12}}{3.6\times 10^9} = 2299.8\ \text{MWh}
\end{aligned}
$$

#### Step 2: Energy per day available
$$
E_{day} = 2 \times 8.279\times 10^{12} = 1.656\times 10^{13}\ \text{J}
$$

#### Step 3: Average electrical power
$$
\begin{aligned}
P &= \frac{\eta E_{day}}{86400} = \frac{0.6 \times 1.656\times 10^{13}}{86400} \\
&= 1.150\times 10^{8}\ \text{W} = 115.0\ \text{MW}
\end{aligned}
$$

#### Annual energy (for reference)
$$
E_{year} = 115.0\ \text{MW} \times 8760\ \text{h} \approx 1007\ \text{GWh}
$$

**Answer:** Average power output capacity $\approx 115$ MW (energy per tide $\approx 2300$ MWh before losses).

Assumption: one emptying of the basin per tide (single-effect ebb generation), averaged over 24 hours.

## Check yourself

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

1. Which formula gives the energy of water in the basin per tide?
   A. E = rho g A h
   B. E = 1/2 rho g A h^2
   C. E = rho g A h^2
   D. E = 1/2 rho A h^2
   Answer: B. Mass rho A h falls through the centre of mass h/2, giving 1/2 rho g A h^2. (Key point: Energy per tide)

2. What is the basin area for 3 km x 5.5 km?
   A. 1.65 x 10^7 m^2
   B. 1.65 x 10^6 m^2
   C. 8.5 x 10^3 m^2
   D. 1.65 x 10^4 m^2
   Answer: A. 3000 x 5500 = 1.65 x 10^7 m^2. (Key point: Basin area)

3. How is average power obtained from the energy per tide?
   A. Multiply by 2 tides, apply 60% efficiency, divide by 86400 s
   B. Divide by 3600 only
   C. Multiply by 10 m
   D. Divide by 2 tides and by 60%
   Answer: A. P = eta x (2E) / 86400 s. (Key point: Daily energy and averaging)

4. The average power output is about
   A. 11.5 MW
   B. 230 MW
   C. 1150 MW
   D. 115 MW
   Answer: D. 0.6 x 1.656 x 10^13 / 86400 = 115 MW. (Key point: Result)

## Related in Power Plant Design

- [Tidal power generation and output: 3 km × 3 km](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Atidal-generation-3x3km) — Power Plant Design, chapter 1

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