---
title: "Maximising discharge with same lining"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:rectangular-channel-max-discharge"
source: https://www.bctengineer.com/q/semester_6/hydro_power/rectangular-channel-max-discharge
site: BCT Engineering
---

# Maximising discharge with same lining

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2072 Asoj, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/rectangular-channel-max-discharge
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

A rectangular channel, 8 m wide and 1.5 m deep, has a slope of 0.001 in 1 and is lined with smooth concrete plaster. It is desired to enhance the discharge to a maximum by changing the dimensions of the channel but keeping the same amount of lining. Find the new dimensions and percentage increase in discharge. For smooth concrete plaster, take Manning's constant = 0.015.

## Short answer

Keeping the same wetted perimeter (11 m), the best rectangular section (b=2y) gives y=2.75 m, b=5.5 m. Discharge increases from 26.81 to 39.43 m³/s, a 47% increase.

## Answer

Keeping the same amount of lining means the **wetted perimeter stays the same**. For a given perimeter, discharge is maximum when the section is the best (most economical) rectangle: $b = 2y$.

#### Existing channel
$$
\begin{aligned}
A_1 &= 8 \times 1.5 = 12\ \text{m}^2, \quad P = 8 + 2 \times 1.5 = 11\ \text{m} \\
R_1 &= 12/11 = 1.091\ \text{m} \\
Q_1 &= \frac{1}{n}A_1R_1^{2/3}S^{1/2} = \frac{1}{0.015} \times 12 \times 1.091^{2/3} \times 0.001^{1/2} \\
&= 66.67 \times 12 \times 1.0597 \times 0.03162 = 26.81\ \text{m}^3/\text{s}
\end{aligned}
$$

#### New channel (same P = 11 m, $b = 2y$)
$$
\begin{aligned}
P &= b + 2y = 4y = 11 \;\Rightarrow\; y = 2.75\ \text{m}, \quad b = 5.5\ \text{m} \\
A_2 &= 5.5 \times 2.75 = 15.125\ \text{m}^2, \quad R_2 = 15.125/11 = 1.375\ \text{m} \\
Q_2 &= 66.67 \times 15.125 \times 1.375^{2/3} \times 0.03162 = 39.43\ \text{m}^3/\text{s}
\end{aligned}
$$

#### Percentage increase
$$
\frac{Q_2 - Q_1}{Q_1} \times 100 = \frac{39.43 - 26.81}{26.81} \times 100 = 47.1\%
$$

| Item | Existing | New |
|---|---|---|
| Width (m) | 8.0 | 5.5 |
| Depth (m) | 1.5 | 2.75 |
| Perimeter (m) | 11 | 11 |
| R (m) | 1.091 | 1.375 |
| Q (m³/s) | 26.81 | 39.43 |

**Answer:** New width = 5.5 m, depth = 2.75 m; discharge increases by about 47%.

## Check yourself

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

1. What does keeping the same amount of lining mean?
   A. Same area
   B. Same wetted perimeter
   C. Same width
   D. Same depth
   Answer: B. Lining covers the wetted perimeter, so P = 11 m stays fixed. (Key point: Same lining)

2. For fixed perimeter, which rectangle gives maximum discharge?
   A. b = y
   B. b = 2y
   C. b = 3y
   D. b = y/2
   Answer: B. The best rectangular section has b = 2y. (Key point: Maximum discharge shape)

3. With P = b + 2y = 4y = 11, the new dimensions are:
   A. b = 5.5 m, y = 2.75 m
   B. b = 4 m, y = 3.5 m
   C. b = 2.75 m, y = 5.5 m
   D. b = 8 m, y = 1.5 m
   Answer: A. y = 11/4 = 2.75 m and b = 2y = 5.5 m. (Key point: New dimensions)

4. With Q1 = 26.81 and Q2 = 39.43 m^3/s, the increase is about:
   A. 32%
   B. 47%
   C. 12.6%
   D. 68%
   Answer: B. (39.43 - 26.81)/26.81 x 100 = 47.1%, driven by larger R (1.375 vs 1.091) and area. (Key point: Percentage increase)

## Related in Hydro Power

- [Flood discharge in rectangular channel](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aflood-discharge-slope-area) — Hydro Power, chapter 2
- [Uniform flow in rectangular channel (Chezy)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Arectangular-channel-chezy-flow) — Hydro Power, chapter 1
- [Economical rectangular condition and trapezoidal bed slope](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atrapezoidal-channel-bed-slope) — Hydro Power, chapter 1
- [Efficient rectangular section and triangular channel depth](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-rect) — Hydro Power, chapter 1, asked 2×
- [Most economical rectangular channel section](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Amost-economical-rectangular-section) — Hydro Power, chapter 1, asked 2×
- [Slope area method of discharge](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aslope-area-method) — Hydro Power, chapter 2

## Related in other subjects

- [Rectangle with same lining for maximum discharge](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Arectangular-channel-same-lining-max-discharge) — Hydraulics, chapter 6
- [Economical channel: unlined versus concrete lined](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aeconomical-trapezoidal-unlined-vs-lined) — Hydraulics, chapter 6, asked 2×
- [Concrete-lined canal: Q 70, slope 1 in 4000](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Alined-canal-slope-14000-q70) — Irrigation and Drainage Engineering, chapter 4