---
title: "Minimum hump height with energy loss"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:flat-hump-minimum-height-loss-velocity-head"
source: https://www.bctengineer.com/q/semester_4/hydraulics/flat-hump-minimum-height-loss-velocity-head
site: BCT Engineering
---

# Minimum hump height with energy loss

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

- Asked in: 2069 Poush, 4 marks
- Video: [Critical flow condition in open channel hydraulics](https://www.youtube.com/watch?v=VF4CAetGaGc) — HydroMohsen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/flat-hump-minimum-height-loss-velocity-head
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

A uniform flow of 12 m³/s occurs in a long rectangular channel of 5 m width and depth flow of 1.5 m. A flat hump is to be built at a certain section. Assuming a loss of head equal to upstream velocity head, compute the minimum height of the hump to provide the critical flow.

## Short answer

Head loss equals upstream velocity head, V1^2/2g = 0.131 m. Upstream specific energy E1 = 1.63 m, critical energy Ec = 1.5 yc = 1.26 m (yc = 0.84 m). Minimum hump height Δz = E1 − Ec − hL = 0.24 m.

## Answer

**Given:** $Q = 12\ \text{m}^3/\text{s}$, $b = 5\ \text{m}$, $y_1 = 1.5\ \text{m}$. Head loss over the hump $h_L = V_1^2/2g$. For the minimum hump height the flow is just critical on the crest.

**Upstream:**

$$
V_1 = \frac{12}{5 \times 1.5} = 1.6\ \text{m/s},\quad \frac{V_1^2}{2g} = 0.1305\ \text{m},\quad E_1 = 1.5 + 0.1305 = 1.6305\ \text{m}
$$

**Critical conditions on the crest:**

$$
q = \frac{12}{5} = 2.4\ \text{m}^2/\text{s},\quad y_c = \left(\frac{2.4^2}{9.81}\right)^{1/3} = 0.837\ \text{m},\quad E_c = 1.5\,y_c = 1.256\ \text{m}
$$

**Energy equation between upstream and the crest:**

$$
E_1 = \Delta z + E_c + h_L
$$

$$
\begin{aligned} \Delta z &= E_1 - E_c - h_L \\ &= 1.6305 - 1.2561 - 0.1305 = 0.244\ \text{m} \end{aligned}
$$

**Answer: Minimum hump height $\approx 0.24\ \text{m}$.**

## Check yourself

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

1. For the minimum hump height giving critical flow, the flow on the crest is:
   A. Exactly at critical depth, E = Ec = 1.5 yc
   B. Supercritical with Fr = 2
   C. At the upstream depth 1.5 m
   D. Uniform at normal depth
   Answer: A. The smallest hump that forces critical flow makes the crest energy just equal to Ec. (Key point: Condition for minimum hump)

2. How is yc on the crest found?
   A. yc = V1^2/g
   B. yc = (q^2/g)^(1/3) with q = 12/5 = 2.4 m^2/s, giving 0.837 m
   C. yc = 2/3 y1 = 1.0 m
   D. yc = Q/b = 2.4 m
   Answer: B. Width is unchanged, q = 2.4 m^2/s, and yc = (5.76/9.81)^(1/3) = 0.837 m. (Key point: Critical depth on crest)

3. Which energy equation gives the hump height?
   A. E1 = dz + Ec
   B. E1 = dz + Ec - hL
   C. E1 = dz + Ec + hL, with hL = V1^2/2g
   D. E1 + dz = Ec + hL
   Answer: C. Upstream energy is spent on lifting by dz, the crest energy Ec and the head loss hL. (Key point: Energy equation with loss)

4. With E1 = 1.6305 m, Ec = 1.256 m and hL = 0.1305 m, the minimum hump height is about:
   A. 0.38 m
   B. 0.14 m
   C. 0.50 m
   D. 0.24 m
   Answer: D. dz = 1.6305 - 1.2561 - 0.1305 = 0.244 m. (Key point: Result)

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