---
title: "Efficient trapezoid with unequal side slopes"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:efficient-trapezoidal-two-side-slopes-critical-depth"
source: https://www.bctengineer.com/q/semester_4/hydraulics/efficient-trapezoidal-two-side-slopes-critical-depth
site: BCT Engineering
---

# Efficient trapezoid with unequal side slopes

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2082 Kartik, 6+2 marks
- Video: [Most economical trapezoid channel section //open channel flow //](https://www.youtube.com/watch?v=9PJhN517foA) — Civil Engineer FBH
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/efficient-trapezoidal-two-side-slopes-critical-depth
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Trapezoidal Canal with two different side slopes, one having 0.5H:1V and another having 1.5 H:1V carry 12 m³/s of flow. Determine the dimensions of most efficient section if Manning's roughness coefficient is 0.014. Also calculate the critical depth of the flow.

## Short answer

Assuming bed slope 0.001, the most efficient section for side slopes 0.5H:1V and 1.5H:1V has bed width b=1.60m, depth y=1.74m, top width 5.09m. The critical depth for the 12m³/s flow is yc=1.35m; normal depth is greater so flow is subcritical.

## Answer

**The bed slope is not given in the question. It is assumed to be $S_0=1$ in 1000 $=0.001$, and the working is shown so that another slope can be substituted (the depth varies as $S_0^{-3/16}$).**

#### Given data
$z_1=0.5$, $z_2=1.5$ (H:V), $Q=12$ m³/s, $n=0.014$.

#### Condition for most efficient section
The most efficient section has the least wetted perimeter for a given area. As for a symmetric trapezoid, this gives the condition that the top width equals the sum of the two sloping sides, and $R=y/2$. Let $s=\sqrt{1+z_1^2}+\sqrt{1+z_2^2}$.

$$
A=by+\frac{(z_1+z_2)}{2}y^2=by+y^2
$$
$$
P=b+y\left(\sqrt{1+z_1^2}+\sqrt{1+z_2^2}\right)=b+y(1.1180+1.8028)=b+2.9208\,y
$$
Writing $P=\dfrac{A}{y}-\dfrac{z_1+z_2}{2}y+s\,y$ and setting $dP/dy=0$ (with $A$ constant) gives
$$
b=y\left(\sqrt{1+z_1^2}+\sqrt{1+z_2^2}-(z_1+z_2)\right)=(2.9208-2)\,y=0.9208\,y
$$
Then
$$
A=1.9208\,y^2,\qquad P=3.8416\,y,\qquad R=\frac{A}{P}=\frac y2
$$

#### Normal depth from Manning's equation
$$
Q=\frac1nAR^{2/3}S_0^{1/2}\;\Rightarrow\;12=\frac{1}{0.014}(1.9208\,y^2)\left(\frac y2\right)^{2/3}(0.001)^{1/2}
$$
$$
y^{8/3}=4.3905\;\Rightarrow\;y=1.742\ \text{m}
$$
$$
b=0.9208\times1.742=1.604\ \text{m},\qquad A=5.826\ \text{m}^2,\qquad P=6.690\ \text{m}
$$
Top width $T=b+(z_1+z_2)y=5.087$ m.

#### Critical depth
Critical flow condition: $\dfrac{Q^2}{g}=\dfrac{A_c^3}{T_c}$ with $A_c=by_c+y_c^2$ and $T_c=b+2y_c$:
$$
\frac{12^2}{9.81}=14.679=\frac{(1.604\,y_c+y_c^2)^3}{1.604+2y_c}
$$
Solving by trial: $y_c=1.349$ m. Check: $A_c=3.982$ m², $T_c=4.301$ m, $V_c=3.014$ m/s, $Fr=1$.

**Answer (for $S_0=0.001$): the most efficient section has $b=1.60$ m and $y=1.74$ m, with side slopes 0.5H:1V and 1.5H:1V, and the critical depth is $y_c=1.35$ m.** Since $y_n>y_c$, the flow is subcritical.

## Check yourself

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

1. For the most efficient section with two different side slopes, what is the hydraulic radius?
   A. R = y
   B. R = y/2
   C. R = y/4
   D. R = 2y
   Answer: B. Minimum perimeter for given area gives R = y/2, as for a symmetric trapezoid. (Key point: Efficient section condition)

2. With z1 = 0.5 and z2 = 1.5, which expression gives the bed width?
   A. b = y(sqrt(1+z1^2) + sqrt(1+z2^2) - (z1+z2))
   B. b = y(z1 + z2)
   C. b = 2y(sqrt(1+z^2) - z) using z = 1 only
   D. b = y(sqrt(1+z1^2) + sqrt(1+z2^2) + (z1+z2))
   Answer: A. b = y(1.1180 + 1.8028 - 2) = 0.9208 y. (Key point: Bed width)

3. Which equation defines the critical depth?
   A. Q^2/g = A^3/T
   B. Q = (1/n) A R^(2/3) S^(1/2)
   C. Q = C A sqrt(R S)
   D. A = T y
   Answer: A. Critical flow occurs when Froude number is 1, i.e. Q^2/g = A^3/T. (Key point: Critical depth condition)

4. Since the normal depth (1.74 m) exceeds the critical depth (1.35 m), the flow is:
   A. Supercritical
   B. Critical
   C. Subcritical
   D. Unsteady
   Answer: C. y_n > y_c means the Froude number is below 1 (subcritical). (Key point: Flow regime)

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