---
title: "Rectangle with same lining for maximum discharge"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:rectangular-channel-same-lining-max-discharge"
source: https://www.bctengineer.com/q/semester_4/hydraulics/rectangular-channel-same-lining-max-discharge
site: BCT Engineering
---

# Rectangle with same lining for maximum discharge

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2069 Bhadra, 6 marks
- Video: [Hydraulics - Open Channels (Past Board Exam Problem)](https://www.youtube.com/watch?v=cPSYpOmRYY8) — Engr. Jed
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/rectangular-channel-same-lining-max-discharge
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

A rectangular channel 8 m wide and 1.5 m deep has a slope of 0.001 and is lined with smooth plaster. It is desired to enhance the discharge to a maximum by changing the dimension of the channel, but keeping the same amount of lining. Work out the new dimension and the percentage increase in discharge. Take roughness coefficient n = 0.015.

## Short answer

Existing channel (b=8 m, y=1.5 m) has P=11 m, Q₁=26.81 m³/s. With same lining (constant P=11 m), maximum discharge occurs for b=2y: y=2.75 m, b=5.5 m, Q₂=39.43 m³/s. Discharge increases by 47%.

## Answer

Lining is on the wetted perimeter, so "same amount of lining" means the wetted perimeter is kept constant. For a fixed $P$ and $n$, $S$, discharge is maximum when the area $A$ is maximum (since $Q\propto A^{5/3}P^{-2/3}$).

#### Existing channel
$b=8$ m, $y=1.5$ m, $S=0.001$, $n=0.015$:
$$
A=8\times1.5=12\ \text{m}^2,\quad P=8+2(1.5)=11\ \text{m},\quad R=\frac{12}{11}=1.0909\ \text{m}
$$
$$
Q_1=\frac1nAR^{2/3}\sqrt S=\frac{1}{0.015}(12)(1.0909)^{2/3}\sqrt{0.001}=26.81\ \text{m}^3/\text{s}
$$

#### New dimensions for maximum discharge at $P=11$ m
$$
A=by=(P-2y)y\;\Rightarrow\;\frac{dA}{dy}=P-4y=0\;\Rightarrow\;y=\frac P4=2.75\ \text{m}
$$
$$
b=P-2y=5.5\ \text{m}=2y\quad(\text{most economical rectangle})
$$
$$
A=5.5\times2.75=15.125\ \text{m}^2,\quad R=\frac y2=1.375\ \text{m}
$$
$$
Q_2=\frac{1}{0.015}(15.125)(1.375)^{2/3}\sqrt{0.001}=39.43\ \text{m}^3/\text{s}
$$

#### Increase in discharge
$$
\frac{Q_2-Q_1}{Q_1}\times100=\frac{39.43-26.81}{26.81}\times100=47.1\%
$$
**Answer: new dimensions $b=5.5$ m and $y=2.75$ m (i.e. $b=2y$); discharge increases from 26.81 to 39.43 m³/s, an increase of about 47%.**

## Check yourself

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

1. 'Keeping the same amount of lining' means keeping constant:
   A. The cross-sectional area
   B. The bed width
   C. The wetted perimeter
   D. The depth
   Answer: C. Lining covers the wetted perimeter, so P = 8 + 2(1.5) = 11 m stays fixed. (Key point: Meaning of same lining)

2. For fixed P, n and S, Q is proportional to A^(5/3)P^(-2/3), so Q is greatest when:
   A. R is smallest
   B. Top width is smallest
   C. The area A is greatest
   D. Depth is smallest
   Answer: C. With P constant, Q increases with A, so maximise A. (Key point: Maximising discharge at fixed P)

3. With A = (P - 2y)y, setting dA/dy = P - 4y = 0 gives:
   A. y = P/2 = 5.5 m
   B. y = P/3 = 3.67 m
   C. y = P/4 = 2.75 m with b = 5.5 m (b = 2y)
   D. y = 1.5 m with b = 8 m
   Answer: C. y = 11/4 = 2.75 m and b = 11 - 5.5 = 5.5 m = 2y, the most economical rectangle. (Key point: New dimensions)

4. Existing Q1 = 26.81 m3/s and new Q2 = 39.43 m3/s. The increase is about:
   A. 12.6%
   B. 32%
   C. 47%
   D. 68%
   Answer: C. (39.43 - 26.81)/26.81 = 0.471, about 47%. (Key point: Percentage increase)

## Related in Hydraulics

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- [Direct step, 16 m wide channel, 40 m³/s](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agvf-profile-40m3s-16m-wide-direct-step) — Hydraulics, chapter 8
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- [94% depth for maximum discharge in circular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acircular-channel-94-percent-depth-max-discharge) — Hydraulics, chapter 6

## Related in other subjects

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