---
title: "Minimum hump height for critical depth, 3 m channel"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:minimum-hump-height-critical-flow-3m-channel"
source: https://www.bctengineer.com/q/semester_4/hydraulics/minimum-hump-height-critical-flow-3m-channel
site: BCT Engineering
---

# Minimum hump height for critical depth, 3 m channel

Hydraulics · Chapter 7: Energy and Momentum Principles in Open channel flow

- Asked in: 2075 Baisakh, 8 marks
- Video: [Critical Parameters (Depth, Velocity and Flow) | Open Channel Flow | Hydraulics and Fluid Mechanics](https://www.youtube.com/watch?v=4NqoLy1WBKI) — The Content Paradox
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/minimum-hump-height-critical-flow-3m-channel
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

A 3 m wide rectangular channel carries a discharge of 15 m³/s at a depth of 2 m. What will be the minimum height of hump at which the depth over the hump will be critical? Calculate the height of hump for which upstream water depth will be 2.5 m. What will be the depth of flow on the upstream and on the hump when its height is 0.2 m?

## Short answer

(a) Minimum hump height for critical flow is Δz_min = E1 − Emin = 0.270 m. (b) For upstream depth 2.5 m, hump height is 0.655 m. (c) With a 0.2 m hump, upstream depth stays 2.00 m and depth over hump is 1.65 m.

## Answer

#### Given data
$b=3$ m, $Q=15$ m³/s, $y_1=2$ m. $q=\dfrac{15}{3}=5$ m²/s, $V_1=2.5$ m/s.
$$
E_1=2+\frac{2.5^2}{19.62}=2.3186\ \text{m}
$$
$$
y_c=\sqrt[3]{\frac{q^2}{g}}=\sqrt[3]{\frac{25}{9.81}}=1.3659\ \text{m},\qquad E_{min}=1.5y_c=2.0489\ \text{m}
$$

#### (a) Minimum hump height for critical depth over the hump
$$
\Delta z_{min}=E_1-E_{min}=2.3186-2.0489=0.270\ \text{m}
$$

#### (b) Hump height for upstream depth 2.5 m
The upstream depth is raised, so the hump causes critical flow. Upstream energy:
$$
E_1'=2.5+\frac{(5/2.5)^2}{19.62}=2.5+\frac{4}{19.62}=2.7039\ \text{m}
$$
$$
\Delta z=E_1'-E_{min}=2.7039-2.0489=0.655\ \text{m}
$$

#### (c) Hump of 0.20 m
Since $0.20<\Delta z_{min}=0.270$ m, the flow is not choked: **upstream depth remains 2.0 m**. Depth over the hump (subcritical):
$$
E_2=E_1-0.20=2.1186\ \text{m},\qquad y_2+\frac{25}{19.62\,y_2^2}=y_2+\frac{1.2742}{y_2^2}=2.1186
$$
$$
y_2=1.651\ \text{m}
$$
**Answer: (a) minimum hump height $=0.270$ m; (b) hump height $=0.655$ m; (c) upstream depth 2.00 m, depth over the 0.2 m hump $=1.65$ m.**

## Check yourself

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

1. The minimum hump height giving critical depth over it is
   A. E1 + Emin
   B. E1 - Emin
   C. Emin - y1
   D. yc - y1
   Answer: B. Critical over the hump needs E1 - Δz = Emin, so Δz = 2.3186 - 2.0489 = 0.270 m. (Key point: Minimum hump for critical depth)

2. For an upstream depth of 2.5 m, the upstream energy used is
   A. E1' = 2.5 + V^2/2g with V = 2 m/s, i.e. 2.704 m
   B. E1' = 2.5 m
   C. E1' = 1.5 x 2.5
   D. E1' = 2.5 + yc
   Answer: A. V = 5/2.5 = 2 m/s, so E1' = 2.5 + 4/19.62 = 2.7039 m and Δz = E1' - Emin = 0.655 m. (Key point: Hump for upstream depth 2.5 m)

3. Why is the upstream depth unchanged for a 0.2 m hump?
   A. 0.2 m is less than the minimum critical hump height 0.270 m
   B. 0.2 m exceeds the critical height
   C. The discharge is reduced
   D. The flow is supercritical
   Answer: A. E2 = 2.1186 m > Emin = 2.0489 m, so flow is not choked and y1 stays 2.0 m. (Key point: Hump 0.2 m: is it choked?)

4. The depth over the 0.2 m hump is
   A. 1.366 m (critical)
   B. 2.0 m
   C. 1.65 m, the subcritical root of y + 1.2742/y^2 = 2.1186
   D. 1.8 m
   Answer: C. Not choked, so the depth is the subcritical root, about 1.651 m, above yc = 1.366 m. (Key point: Depth over 0.2 m hump)

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