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Week 10: Non-Uniform Flow: University of Nueva Caceres College of Engineering and Architecture

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University of Nueva Caceres

College of Engineering and Architecture

Week 10: Non-Uniform Flow

I. Introduction
A hydraulic jump is a fluid shockwave created at the transition between laminar and turbulent
flow. One common example of a hydraulic jump can be seen in the water radiating outward
when the stream of tap water strikes the horizontal surface of a sink. The water initially flows in
a smooth sheet with consistent current patterns. In addition to the kitchen sink example,
hydraulic jumps are also typical features of river rapids where the water swirls and foams
around rocks and logs.

In this chapter, you will learn different applications of hydraulic jump and how to solve problems
related to it.

II. Objectives
After completion of the course the student should be able to:

1. Analyze applications of Hydraulic Jump

III. Hydraulic Jump


Non- uniform flow occurs where the stream enters and leaves the channel at obstruction
such as dams, weirs, bridges, or piers.

The hydraulic jump is defined as the rise of water level, which takes place due to
transformation of the unstable shooting flow (super-critical) to the stable streaming flow (sub-critical).
When hydraulic jump occurs, a loss of energy due to eddy formation and turbulence flow occurs.

Figure 15-1: Froude Numbers and Fluid Depths across a Hydraulic


Jump

HYDRAULICS V.1.0 BY:


University of Nueva Caceres
College of Engineering and Architecture

Applications of Hydraulic Jump

 Usually, hydraulic jump reverses the flow of water. This phenomenon can be used to mix
chemicals for water purification.
 Hydraulic jump usually maintains the high-water level on the downstream side. This high-water
level can be used for irrigation purposes.
 Hydraulic jump can be used to remove the air from water supply and sewage lines to prevent
the air locking.
 It prevents the scouring action on the downstream side of the dam structure.

Figure 15-2: Flow Under a sluice gate accelerates from subcritical to


critical to supercritical and then jump back to subcritical flow

Formulas for Hydraulic Jump


q2 ( d 1+ d 2 ) (d 1 d 2)
=
g 2
Energy Head loss due to the jump
3
( d 2−d 1 )
HL =
4 d1 d2
(For rectangular channel unit)
Height of Jump = d2 – d1
Hydraulic Jump of Non-Rectangular Sections

h1 A 1−h2 A 2
Q2
= 1 1
g −
A2 A1

Where:

HYDRAULICS V.1.0 BY: LOSARES 2


University of Nueva Caceres
College of Engineering and Architecture
Q = rate of flow
A1 = cross sectional area of section 1
A2 = cross sectional area of section 2
h1 = centroid of section 1 below the water surface

h2 = centroid of section 1 below the water surface

g = gravitational constant is 9.81 m/s2

Example 1:

A rectangular channel 2 m. wide carries 2.2 m 3/s of water in sub critical uniform flow at a depth of
1.0 m
a) Compute the specific energy
b) Compute the minimum specific energy
c) What is the lowest transverse hump in the bottom such that a critical depth is
attained at the peak?
Solution:
a) Specific Energy b) Min. Specific Energy
V2 2.2
E= +d q=
2g 2
Q = AV q = 1.1 m3/s
2.2 = 1 (2) V
V = 1.1 m/s 2
dc = 3 q
E=
V2
2g
+d √ g
dc = 0.4978 m
(1.1)2 3
E= + 1.0 Emin = d
2(9.81) 2 c
E = 1.0617 m Emin = 0.7467 m

c) Height of Jump
h = E - Ec
h= 1.0617- 0.7467
h = 0.315 m

Example 2:

A vertical sluice gate with an opening of 0.67 m produces a downstream jet depth of 0.4
m when installed in a long rectangular channel 5m wide conveying a steady discharge of 20
m3/s. Assuming a hydraulic jump occurs and the flow downstream of the gate eventually returns
to the uniform flow depth of 2.5 m.
a) Compute the initial depth before hydraulic jump occurs.

HYDRAULICS V.1.0 BY: LOSARES 3


University of Nueva Caceres
College of Engineering and Architecture
b) Calculate the head loss in the jump

HYDRAULICS V.1.0 BY: LOSARES 4


University of Nueva Caceres
College of Engineering and Architecture

a) Initial depth before the jump occurs


20
q=
5
q = 4 m3/s/m
q2 ( d 1+ d 2 ) (d 1 d 2)
=
g 2
(4)2
( d + 2.5 ) (d 1∗2.5)
= 1
9.81 2
d12 + 2.5d1 – 1.305 = 0
d1 = 0.443 m

b) Head loss at the jump


V 21 V2
+ d1 = HL + 2 + d2
2g 2g
V1A1 = q
V1(0.443)(1) =4
V1 = 9.03 m/s

V2A2 = q
V2(2.5)(1) = 4
V2 = 1.6 m/s
( 9.03 )2 (1.6 )2
+ 0.443= HL + + 2.5
2(9.81) 2(9.81)
HL = 1.97 m

HYDRAULICS V.1.0 BY: LOSARES 5

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