---
title: "Flow rate from drop in surface over hump"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:contraction-3m-to-2m-floor-raised-drop-0-2"
source: https://www.bctengineer.com/q/semester_4/hydraulics/contraction-3m-to-2m-floor-raised-drop-0-2
site: BCT Engineering
---

# Flow rate from drop in surface over hump

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

- Asked in: 2073 Magh, 7 marks
- Video: [Channel Transitions | Open Channel Flow | Hydraulics and Fluid Mechanics](https://www.youtube.com/watch?v=al6BM6xvtnU) — 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/contraction-3m-to-2m-floor-raised-drop-0-2
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

The width of a rectangular channel is reduced gradually from 3 m to 2 m and the floor is raised by 0.3 m at a given section. When the approaching depth of flow is 2.05 m, what rate of flow will be indicated by a drop of 0.2 m in the water surface elevation at the contracted section?

## Short answer

With a 0.2 m water surface drop and 0.3 m bed rise, y2 = 1.55 m. Equating upstream and downstream energies yields Q²(1/3.10² − 1/6.15²)/19.62 = 0.2, so Q = 7.11 m³/s.

## Answer

#### Given data
$b_1=3$ m, $y_1=2.05$ m; $b_2=2$ m; floor raised $\Delta z=0.30$ m; surface drops $0.20$ m at the contracted section. No loss assumed.

#### Depth at the contracted section
Water surface level at the contraction $=2.05-0.20=1.85$ m (above the approach bed). The floor is 0.30 m higher, so
$$
y_2=1.85-0.30=1.55\ \text{m}
$$

#### Energy equation (datum: approach bed)
$$
y_1+\frac{V_1^2}{2g}=\Delta z+y_2+\frac{V_2^2}{2g},\qquad V_1=\frac{Q}{3\times2.05}=\frac{Q}{6.15},\quad V_2=\frac{Q}{2\times1.55}=\frac{Q}{3.10}
$$
$$
2.05+\frac{Q^2}{2g(6.15)^2}=0.30+1.55+\frac{Q^2}{2g(3.10)^2}
$$
$$
\frac{Q^2}{2g}\left(\frac{1}{3.10^2}-\frac{1}{6.15^2}\right)=2.05-1.85=0.20
$$
$$
\frac{Q^2}{19.62}\left(0.10406-0.026439\right)=0.20\;\Rightarrow\;Q^2=\frac{0.20\times19.62}{0.07762}=50.555
$$
$$
Q=7.110\ \text{m}^3/\text{s}
$$
Check: $V_1=1.156$ m/s, $V_2=2.294$ m/s; $E_1=2.05+0.0681=2.1181$ m and $E_2$ (with datum) $=0.30+1.55+0.2681=2.1181$ m ✓. Flow is subcritical at the contraction ($Fr_2=0.588$).

**Answer: discharge $Q\approx7.11$ m³/s.**

## Check yourself

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

1. If the surface drops 0.20 m from 2.05 m and the floor is raised 0.30 m, the depth at the contraction is
   A. 1.85 m
   B. 1.55 m
   C. 1.75 m
   D. 2.05 m
   Answer: B. Surface level = 1.85 m above the approach bed; subtract the floor rise 0.30 m: y2 = 1.55 m. (Key point: Depth at contraction)

2. The energy equation used (datum approach bed) is
   A. y1 + V1^2/2g = y2 + V2^2/2g
   B. y1 + V1^2/2g = Δz + y2 + V2^2/2g
   C. y1 = Δz + y2
   D. V1 = V2
   Answer: B. Energy at the contraction includes the 0.30 m elevation of the floor. (Key point: Energy equation)

3. V1 and V2 are expressed as
   A. Q/3 and Q/2
   B. Q/(3 x 2.05) and Q/(2 x 1.55)
   C. Q/2.05 and Q/1.55
   D. Q/6.15 and Q/6.15
   Answer: B. V = Q/(b y): V1 = Q/6.15 and V2 = Q/3.10. (Key point: Velocities in terms of Q)

4. The resulting discharge is about
   A. 3.55 m^3/s
   B. 7.11 m^3/s
   C. 14.2 m^3/s
   D. 50.6 m^3/s
   Answer: B. Q^2 = 0.20 x 19.62/0.07762 = 50.56, so Q = 7.11 m^3/s. (Key point: Result)

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