---
title: "Water surface over hump with contraction"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:hump-contraction-surface-elevations-1-6m"
source: https://www.bctengineer.com/q/semester_4/hydraulics/hump-contraction-surface-elevations-1-6m
site: BCT Engineering
---

# Water surface over hump with contraction

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

- Asked in: 2081 Chaitra, 8 marks
- Video: [Open channel flow-18 | channel transition hump | shubham sarathe](https://www.youtube.com/watch?v=SXy9LMLzviE) — EIQ
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/hump-contraction-surface-elevations-1-6m
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

Water flows at a depth of 1.6 m and velocity of 1.10 m/s in an open channel of rectangular cross section of width 4.0 m. At a certain section the width is reduced to 3.5 m and the bed is raised by 0.35 m through a smooth flat top hump. Calculate the water surface elevations at the contracted section as well as at a section upstream of it. Assume energy losses to be negligible. Show the results in specific energy diagram.

## Short answer

Upstream water surface elevation remains 1.600 m. Over the 0.35 m hump and 3.5 m width, the depth is 1.158 m, giving a surface elevation of 1.508 m above the upstream bed.

## Answer

#### Given data
$y_1=1.6$ m, $V_1=1.10$ m/s, $b_1=4.0$ m; contracted width $b_2=3.5$ m with a hump of $\Delta z=0.35$ m; no losses.
$$
Q=b_1y_1V_1=4\times1.6\times1.10=7.04\ \text{m}^3/\text{s}
$$
$$
E_1=y_1+\frac{V_1^2}{2g}=1.6+\frac{1.21}{19.62}=1.6617\ \text{m},\qquad Fr_1=\frac{1.10}{\sqrt{9.81\times1.6}}=0.278
$$

#### Check for critical flow at the contraction
$$
q_2=\frac{7.04}{3.5}=2.0114\ \text{m}^2/\text{s},\quad y_{c2}=\sqrt[3]{\frac{q_2^2}{g}}=0.7444\ \text{m},\quad E_{min2}=1.1165\ \text{m}
$$
Energy available over the hump: $E_2=E_1-\Delta z=1.6617-0.35=1.3117$ m $>E_{min2}$. So the flow passes without choking, and the upstream depth is unchanged.

#### Depth over the hump (subcritical)
$$
y_2+\frac{q_2^2}{2gy_2^2}=y_2+\frac{0.2062}{y_2^2}=1.3117\;\Rightarrow\;y_2=1.1579\ \text{m}
$$
(here $q_2^2/2g=0.2062$). Velocity $V_2=q_2/y_2=1.737$ m/s.

#### Water surface elevations (datum: upstream bed)
- At a section upstream: $1.600$ m.
- At the contracted section: $0.35+1.1579=1.5079$ m.

Drop in water surface $=1.600-1.508=0.092$ m.

#### Specific energy diagram
```text
 y (m)
1.700|                                  .:
1.600|                              1:         (1) upstream, E1 = 1.662 m
1.500|                           .:
1.400|                       .:
1.300|                    .:
1.158|               .2                        (2) over hump, E2 = 1.312 m
1.100|             . :
1.000|          . :
0.900|        . :
0.800|      .  :
0.700|     .   :                               noses: yc1 = 0.681 m, yc2 = 0.744 m
0.600|      .    :
0.500|         .       :
0.400|                   .            :
     +-----+---+------+-------------+---------> E (m)
 ticks, left to right: E_min1 = 1.021, E_min2 = 1.117, E2 = 1.312, E1 = 1.662
 "." = curve for q1 = 1.76 m²/s,   ":" = curve for q2 = 2.011 m²/s
```
Point (1) is on the upper (subcritical) limb of the curve for $q_1=1.76$ m²/s; point (2) is on the upper limb of the curve for $q_2=2.011$ m²/s, shifted left by $\Delta z=0.35$ m.

**Answer: depth over the hump $=1.158$ m, water surface elevation there $=1.508$ m (above the upstream bed); upstream elevation $=1.600$ m (unchanged).**

## Check yourself

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

1. What is the first step in the solution?
   A. Find Q = b1 y1 V1 and E1 = y1 + V1^2/2g
   B. Assume critical flow at the hump
   C. Compute the hydraulic jump height
   D. Find the normal depth
   Answer: A. Q = 4 x 1.6 x 1.1 = 7.04 m^3/s and E1 = 1.6617 m give the energy used for the contraction. (Key point: Discharge and energy)

2. How is it confirmed that the upstream depth is unaffected?
   A. Froude number is below 1
   B. E2 = E1 - Δz = 1.312 m exceeds Emin2 = 1.1165 m for q2 = 2.011 m^2/s
   C. Velocity is below 2 m/s
   D. The hump is smooth
   Answer: B. Available energy over the hump is greater than the minimum needed for q2, so no choking. (Key point: Choking check)

3. The depth over the hump is found by
   A. y = yc
   B. y = y1 - Δz
   C. y = E2/2
   D. solving y + q2^2/(2g y^2) = E2 for the subcritical root
   Answer: D. Using q2 = 2.011 and E2 = 1.3117 gives y2 = 1.158 m (above yc2 = 0.744 m). (Key point: Depth over the hump)

4. The water surface elevation over the hump above the upstream bed is
   A. 1.158 m
   B. 0.35 + 1.158 = 1.508 m
   C. 1.600 m
   D. 1.158 - 0.35 = 0.808 m
   Answer: B. Surface = hump height + depth = 0.35 + 1.1579 = 1.508 m, a drop of 0.092 m from 1.600 m. (Key point: Surface elevation)

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