OPEN CHANNEL FLOW
OPEN CHANNEL FLOW
To be Covered
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Flow classification-Uniform, Rapidly and Gradually
varied ,critical flow
Open Channel cross sections
Design of Open channels (Chezy’s formula and
Manning’s)
Specific energy
The hydraulic jump
Flow measurement (Weirs and Fumes)
Learning Objectives
Students should be able to:
• Calculate Normal Depths in Open Channels using
Manning’s Equation and Chezy Equation.
• Develop Specific Energy Curves and Determine
Critical Depths for Open Channels
• Evaluate Hydraulic Jumps in Open Channels.
• Design Hydraulically Efficient and Stable Open
Channels
• Calculate the flow through weirs and flumes.
OPEN CHANNEL FLOW
• Open-channel flow must have a free surface,
whereas pipe flow has none. A free surface is subject
to atmospheric pressure.
• The flow therefore always takes place due to the fact
that the canal is at a slope and a component of the
weight of the liquid causes the foreward motion .
• Physical conditions in open-channels vary much
more than in pipes.
• Flow conditions in open channels are complicated by
the position of the free surface which will change
with time and space.
Comparison between open channel and pipe
flow
Pipe flow
Open channel flow
Flow driven by
Pressure work
Gravity (potential
energy)
Flow cross section
Known ,fixed
Unknown in advance
because flow depth is
unknown
Characteristics flow
parameters
Velocity deduced from
continuity
Flow depth deduced
from continuity and
momentum equations
Specific boundary
conditions
Atmospheric pressure at
the free surface
Properties of Open Channels
• Artificial channels -These are man made e.g.
irrigation canals, navigation canals, spillways
,drainage ditches, culverts.
• Usually constructed in regular shape throughout,
and has reasonably defined roughness.
• Constructed of concrete ,steel or earth.
• Natural channels- not regular in shape and usually
constructed of earth and the surface roughness
coefficient varies.
• Affected by erosion and deposition of sediments.
Flow Classification
• The flow in an open channel is classified according to
the change in the depth of flow with respect to
space and time.
• Uniform - if the depth of flow remains the same at
every section of the channel.
• Non-uniform flow- the depth changes along the
length of the channel.
• Steady uniform flow -Depth is constant both with
time and distance.
• Steady non uniform flow -Depth varies with
distance, but not with time.
• Unsteady flow -Depth varies with both time and
distance.
Gradually varied and Rapidly Varied Flow
• When the change in the depth occurs abruptly over a short
distance, it is a Rapidly Varied Flow (RVF).
• When the change in depth occurs gradually it is a Gradually
Varied Flow (GVF.
Geometric properties of Open Channels
• To determine the flow in a canal the shape
and size of the canal are important:
• Channel cross sections:
–
–
–
–
Rectangular.
Triangular.
Trapezoidal.
Semi Circular.
• To compare different canal sections, the hydraulic
radius and the hydraulic depth are used.
Geometric Properties of Open Channels
• Depth (y)-the vertical distance of the lowest point of a
channel section from the free surface.
• Stage (h)- the vertical distance of the free surface from an
arbitrary datum.
• Area (A)- the cross sectional area of flow normal to the
direction of flow.
• Wetted perimeter (P)- the length of the wetted surface
measured normal to the direction of flow.
• Surface width (B) -the width of the channel section at the free
surface.
• Hydraulic radius (R)-the ratio of the wetted area to the
wetted perimeter (A/P).
• Hydraulic mean depth (Dm)-the ratio of the area to the
surface width (A/B)
Hydraulic Sections
• To minimize the quantity of material required max
hydraulic radius should be used.
• Semi-circle is the most effective but difficult to
construct freshly placed concrete tend to slide down
the sides.
• Trapezoidal channels are efficient especially for
greater discharges.
• Rectangular channels can be used where space is not
limited.
Rectangular section
Trapezoidal Section
Channel free board
• The channel free board is necessary to prevent overtopping
due to waves or variations in water level.
• Free board variation of 10 -30% of the normal flow depth is
acceptable.
Channel depth (m)
<0,25
0,25-0,4
0,4-0,65
0,65-0,9
>0,9
Free board height (mm)
50
75
100
125
150
Fundamental equations
• Equations which describe the flow of a fluid are
derived from three fundamental laws of physics
– Conservation of matter (mass)
– Conservation of energy
– Conservation of momentum
Laminar and Turbulent flow
• For channels the Reynolds number is
RV
Re 

• Where: V-velocity of flow
µ-dynamic viscosity
ρ-fluid density
R-hydraulic radius
• Re(channel) <500 flow is laminar
• Re(channel)>1000 flow is turbulent
Elements of Channel Section
• The channel bottom should have a slope in the
direction of flow
The Chezy’s formula
• Chezy's formula can be derived by equating the propulsive
force due to the weight of the water in the direction of flow
with the retarding shear force at the channel boundary.
The Chezy equation
• Shear force is proportional to velocity squared
 o  kV
2
• Substitute into
V 
• Then
V  C RS o
• C-Chezy coefficient
g
k
RS o
The Manning’s formula
• In terms of velocity (V)
2
3
R S0
V 
n
1
2
• In terms of discharge
2
3
1
1
Q  AR S o 2
n
•
•
•
•
R-hydraulic radius
So- channel bed slope
n-Manning’s roughness coefficient
A-Cross sectional area
Typical Values of n
Examples
• A trapezoidal concrete lined channel with uniform
flow of water has a normal depth of 2 m. The base
width is 5 m has equal side slopes at 1:2.The channel
bed slope is 0,001 and Manning’s n =0,015. Dynamic
viscosity of water is 1,14 x 10-3 kg/m.s. Calculate the
(a) Discharge
(b) Mean velocity
(c) Reynolds number
• If the discharge in the channel given above is 30
m3/s.Find the normal depth of flow.
Example
• A rectangular channel with a bottom width of 3 m, a bed slope of 1:
1500 and a Chezy C of 30 in SI units. When the flow is 3.5 m3/s
determine:
– The normal depth of flow
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Example
Examples
Problem No. 1
• Water flows at a depth of 1.83 m in a trapezoidal, riprap-lined
channel with a bottom width of 3 m, side slopes of 2:1 (H:V), and a
bottom slope of 0.005. Assuming uniform flow, what is the discharge
in the channel?
Problem No. 2
• Determine normal depth for a 4 m wide trapezoidal channel that is
carrying a discharge of 49.7 m3/s. The excavated earth channel is
well maintained (clean) with a flow slope of 0.2 percent and side
slopes of 4:1 (H:V).
Example
• Determine the hydraulic radius of a trapezoidal canal
with the following dimensions. Bottom width 2,5 m,
sloping sides 2,4m at 45o to the horizontal with a flow
depth of 1,5 m.
Example
• Determine the flow depth and average flow
velocities for a concrete channel with slope 1:2
500 changing to 1: 3 000. Assume a Manning’s
n=0,0017.The channel is rectangular with base
width of 3 m and must be able to handle a flow
rate of 2 m3/s.
Compound Channels Example
• If the channel above was to be designed for
flooding it might have a trapezoidal channel and
then flood plains so that it carries more
discharge.
• During flooding the water level in the channel
given above exceeds the bank full-level of 2,5 m.
The flood banks are 10 m wide and are grassed
with side slopes of 1:3. The estimated Manning’s
n for the flood banks is 0,035. Estimate the
discharge for a maximum flood level of 4 m.
Example on Erodible Channels
• Determine the floor width (b) and safe flow depth
(y) of a trapezoidal spillway with a floor slope of
0.0016 and a flow rate of 7.750 m3/h.The
spillway is built in sandy loam soils.
• The n value of a trapezoidal channel in a sandy
soil weakens from 0.025 to 0.3 as a result of bad
maintenance (no weed control). The channel
was initially designed to handle a flow rate of
2m3/s.Channel slope is 1:2 500. Determine the
reduction in flow rate with the new n-value.
Open Channel Flow
SPECIFIC ENERGY
• If water flows in a canal at a depth y and
average velocity v, the specific energy is
V2
E  y
2g
• This is the energy of the liquid in relation to the
bottom of a canal
Specific energy
• E1= E2= E3
• If the canal width remains constant
2
2
2
V1
V2
V2
y1 
 y2 
 y3 
2g
2g
2g
• Q = AV = By1V1= By2V2
and
Q
V1 
By1
Q
V2 
By 2
Specific energy
• If q = Q/B is the flow rate per unit width
then
q
V1 
y1
q
V2 
y2
q2
q2
y1 
 y2 
2
2 gy2
2 gy1
y 3  Ey 2
q2

0
2g
Specific Energy
• For a certain flow there are two depths at which the
water can flow
• Consider a canal with a sluice gate
• Before the gate water flows slowly with a large depth y1
(Sub critical) and after sluice flow is fast with a small
depth y2 (supercritical) but the specific energy is the
same
Specific Energy
• Energy before sluice is stored as
potential and after sluice is mainly
Kinetic
• Critical depth can be determined by
differentiating
V2
E  y
2g
Froude Number
• A dimensionless ratio of the inertia forces
to the gravitational force
Fr 
V
gyc
• Fr determine the velocity of the surface
wave
• Fr>1 supercritical
• Fr=1 critical
• Fr<1 subcritical.
The Hydraulic Jump
• It is the change from shooting flow to tranquil
flow which occurs abruptly.
• This is due to change in slope from being very
steep (high velocity) to gentle which destroys
the high velocity and water moves slower at a
greater depth.
• It occurs when a supercrtical flow meets a
subcritical flow.
• The resulting flow transition is rapid and involves
large energy loss due to turbulence.
The Hydraulic Jump
The Hydraulic Jump
• The depth of flow before the jump
h2
2
h1  ( 1  8 Fr2  1)
2
• Depth of flow after the jump
h1
2
h2  ( 1  8Fr1  1)
2
• Head or Energy loss due to the jump
(h2  h1 ) 3
E 
4h1h2
Head losses in a hydraulic Jump
• The loss of mechanical energy that takes
place in a hydraulic jump may be readily
determined from the energy equation.
Velocity after a Hydraulic Jump
Example
• Water flows in a nearly horizontal canal at a
velocity of 17m/s and a depth of 300mm.
• (a) Will it be possible for a hydraulic jump to
occur?
• (b) what will be the velocity of flow just after the
jump?
• (c )What percentage of the initial power in the
stream remains after the jump?
Solution
Solution
Example
Water flows in a nearly horizontal rectangular canal at a
velocity of 17m/s and a depth of 300 mm.
a)Show that it is possible for a hydraulic jump to occur.
b)Calculate the depth of flow just after the jump
c)Calculate the velocity of flow just after the jump
d)Calculate the percentage of initial power in the stream
that remains after the jump.
Discharge measurement in Open
channels
• Important for:
 To ensure the maintenance of proper delivery
schedules
 To determine the amount of water delivered for
water pricing, where it is applicable
 To detect the origin of water losses and to
estimate the quantity
 To ensure efficient water distribution
 To conduct applied research
Weir
• Use the depth–discharge relationship to measure Q
based on the specific energy equation.
• Equations have a modification/coefficient to account for
loses of energy.
Weir
Example
• A long channel 1.5m wide terminates in a full
width rectangular weir whose crest height is
300mm above the stream bed. Estimate the
discharge when the measuring station is
recording a depth of 400mm above weir crest
.Take Cd=0.7.
Flumes
• Designed to achieve critical flow at the narrowest section
called the throat
Parshall Flume
Flume
• Discharge through the Flume
Example
• An rectangular open channel is 2 m wide.
A venturi flume having a throat width of 1
m is installed at one point. Estimate the
discharge if the upstream depth is 1.2m
and critical flow occurs in the flume
Examples
• A concrete –lined trapezoidal channel has
a bed width of 3.5m, side slopes at 450 to
the horizontal, a bed slope 1 in 1000 and a
Manning roughness coefficient of 0.015.
Calculate the depth of uniform flow when
the discharge is 20m3/s.
Example
• Water flows at a rate of 1.1m3/s along a
rectangular channel 1.5m wide. The bed
slope of the channel is 1 in 100 and the
Chezy is 30 in SI units. Determine the
critical and normal depth of flow.