Outlines
❖Open Channels
❖Types of Open Channels
❖Types of Flow
❖Critical Flow
❖Continuity
Introduction
❖“Occur when free water surface in the channel
is at atmosphere pressure”
❖Example of open channel:
• Rivers and streams
• Drainage
• Ditches
• Irrigation canal
Application
❖Interest to hydraulic engineers
• location of free surface
• velocity distribution
• discharge - stage (depth) relationships
• optimal channel design
Types of channels
1. Man made
• Channel designed and made by
human
• Examples: earth or concrete
lined drainage and irrigation
• Prismatic channel (no change in
geometry with distance)
2. Natural
• Examples: River and streams
• Changes with spatial and
temporal (non prismatic channel)
FLOW IN OPEN CHANNEL
TEMPORAL (Time)
STEADY FLOW UNSTEADY FLOW
UNIFORM FLOW NON-UNIFORM FLOW
RAPIDLY VARIED FLOW
SPATIAL (Space)
GRADUALLY VARIED FLOW
Types of flow
❖Based on temporal (Time, t) and Spatial (Space,x)
❖Time Criteria
• Steady flow (dy/dt = 0). Water depth at one point same all the time. (Flow
constant with time)
• Unsteady flow (dy/dt ≠ 0). Water depth changes all the time. (Flow variation
with time)
❖Space criteria
• Uniform flow (dy/dx = 0). Water depth same along the whole length of flow.
• Non-uniform flow (dy/dx ≠ 0). Water depth changes either rapidly or
gradually flow
Steady and Non-Steady Flow
Flow Rate
Steady
Unsteady
Time
Uniform and Non-Uniform Flow
V1 = V2
A1 = A2
V1
V2
V1 V2 A1 A2
A1 A2
Uniform Flow Non-Uniform Flow
States of flow
❖Flow vary with following forces:
• Viscous
• Inertia
• Gravity
❖De nes by Reynolds number (Re) and Froude numbers (Fr)
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Reynolds Number
❖Refer textbook page 324:
Reynolds Number
❖To determine:
• Laminar ow : Re < 500 (viscous > inertia)
• Transitional ow : 500 < Re < 1300
• Turbulent ow : Re > 1300 (inertia > viscous)
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Froude Number
❖Refer textbook page 683.
Froude Number
❖The Froude Number, Fr describes the following states of ow:
❖Fr < 1 : ow is subcritical
❖Fr = 1 : ow is critical ( inertia < gravity)
❖Fr > 1 : ow is supercritical ( inertia > gravity)
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Froude Number
❖A ow is called critical if the ow velocity is equal to the
velocity of a gravity wave having small amplitude.
❖The ow is called subcritical flow, if the ow velocity is less
than the critical velocity
❖The ow is called supercritical flow if the ow velocity is
greater than the critical velocity.
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Critical Flow 3
y
1
❖ Characteristics
0
0 1 2 3 4
• Unstable surface
E
• Series of standing waves
Difficult to measure depth
❖ Occurrence
• Broad crested weir (and other weirs)
• Channel Controls (rapid changes in cross-section)
• Over falls
• Changes in channel slope from mild to steep
❖ Used for ow measurements
• ___________________________________________
Unique relationship between depth and discharge
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Parameters of Open Channels
❖Wetted Perimeter (P) :The Length of contact between Liquid and sides and
base of Channel
❖Hydraulic Mean Depth or Hydraulic Radius (R): If cross sectional area is A,
then R = A/P
Parameters of Open Channels
❖Depth of flow section (d) : depth of flow normal to the direction of flow.
❖Flow depth (y)
❖Top width (T) : the width of channel section at the free surface.
❖Hydraulic depth (D) : D = A/T
❖Base slope (So) : So = tan θ
Parameters of Open Channels
❖ Freeboard: Vertical distance between the highest water level anticipated in
the design and the top of the retaining banks. It is a safety factor to prevent
the overtopping of structures.
❖ Side Slope (Z): The ratio of the horizontal to vertical distance of the sides
of the channel. Z = e/d = e’/D
General Flow Equation
Q = vA Equation 1
Area of the
cross-section
Avg. velocity of (m2)
flow at a cross-
Flow rate (m3/ section (m/s)
s)
Uniform flow in Open Channel
Energy lines
i
Water Surface
Sw
Flow
yo
So
For uniform flow (in prismatic channel), i = Sw = So
yo= normal depth for uniform flow only
Parameters of Open Channels
Parameters of Open Channels
Character of Flow and Alternate Depth
Water is owing steadily in a 0.4-m-wide rectangular open channel at a rate of 0.2 m3/s (Fig. 13–14). If the ow dep
is 0.15 m, determine the ow velocity and if the ow is subcritical or supercritical. Also determine the alternate ow
depth if the character of ow were to change.
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Continuity Equation
3a
Inflow 3 A
Change in Storage
3b
Outflow
1 A 2
Section AA
Inflow – Outflow = Change in Storage
Resistance Equation
1. Chezy Equation
❖ By Antoine Chezy (France), 1768
2. Manning Equation
❖ By Robert Manning (Irish), 1889
Chezy Equation
❖Introduced by the French engineer Antoine Chezy in 1768 while
designing a canal for the water-supply system of Paris
v = C Ri
❖ Because i = So, Therefore,
v = C RS o
Q = AC RS o
Chezy Equation
❖ where C = Chezy coe cient
= L1/2/T (Unit m1/2/s)
where 60 is for rough and 150 is for smooth
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Manning Equation
❖ Most popular in for open channels around
the world C = R1/6 / n
(SI units!)
Dimensions of n? T /L1/3
(English system)
n = Manning roughness coe cient
= T/L1/3 (Unit s/m1/3)
Bottom slope
very sensitive to n
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Manning roughness coefficient, n
d in ft
d = median size of bed material
d in m
Manning roughness coefficient, n
Uniform Flow
The ow depth in uniform ow is called the normal
depth yn, and the average ow velocity is called the
uniform-flow velocity V0.
The ow remains uniform as long as
the slope,
cross section,
and surface roughness of the channel
remain unchanged
Critical Uniform Flow
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Examples:
Flow Rate in an Open Channel in Uniform Flow
Water is owing in a weedy excavated earth channel of trapezoidal cross section with a bottom width of 0.8 m,
trapezoid angle of 60°, and a bottom slope angle of 0.3°, as shown. If the ow depth is measured to be 0.52 m,
determine the ow rate of water through the channel. What would your answer be if the bottom angle were 1°?
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Examples: Tutorial
The Height of a Rectangular Channel
Water is to be transported in an un nished-concrete rectangular channel with a bottom width of 4 ft at a rate
of 51 ft3/s. The terrain is such that the channel bottom drops 2 ft per 1000 ft length. Determine the minimum
height of the channel under uniform- ow conditions. What would your answer be if the bottom drop is just 1 ft
per 1000 ft length?
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Examples: Tutorial
Channels with Nonuniform Roughness
Water ows in a channel whose bottom slope is 0.003 and whose cross section is shown in. The dimensions
and the Manning coef cients or the surfaces of different subsections are also given on the gure. Determine
the ow rate through the channel and the effective Manning coef cient for the channel.
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Example: Tutorial
❖Trapezoidal channel:
• Bottom width = 3.0 m
• Side slope = 1: 1.5
• Base slope = 0.0016
• Manning coe cient = 0.013
❖Determine Q if yo = 2.6m.
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Determination of yo
❖ If Q, So and n given or known and you need to estimate yo, direct
calculation cannot give you answer. So there are another method can
be use:
1. Try and error
2. Graphical
3. Curves chart
Example
❖ A rectangular channel with n = 0.017 with width 6 meter, base slope
0.0016 and to carry10 m3/s owrate.
Determine yo with:
1. Try and error
2. Graphical
3. Curves chart
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Best Hydraulic Section : Class Activity
The best hydraulic cross section for an open channel is the one with
the maximum hydraulic radius or, equivalently, the one with the
minimum wetted perimeter for a speci ed cross section.
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Rectangular Channels
Best hydraulic cross section (rectangular channel):
Trapezoidal Channels
Best hydraulic cross section (trapezoidal channel):
Hydraulic radius for the best cross section:
Best trapezoid angle:
Example : Tutorial
Best Cross Section of an Open Channel
Water is to be transported at a rate of 2 m3/s in uniform ow in an open channel whose surfaces are asphalt lined.
The bottom slope is 0.001. Determine the dimensions of the best cross section if the shape of the channel is (a)
rectangular and (b) trapezoidal .
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Non-Uniform Flow
Grdually Varied Flow
EXAMPLE: Classification of Channel Slope
Water is owing uniformly in a rectangular open channel with un nished concrete surfaces. The
channel width is 6 m, the ow depth is 2 m, and the bottom slope is 0.004. Determine if the channel
should be classi ed as mild, critical, or steep for this ow.
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RAPIDLY VARIED FLOW & HYDRAULIC JUMP
Rapidly varied ow occurs when there is a sudden change in
ow, such as an abrupt change in cross section.
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Hydraulic Jump
❖ What happens if the Froude
n u m b e r c ro s s e s f ro m F r > 1
(shallow, fast) to Fr < 1 (deep,
slow)? At the transition, the ow
has to suddenly change from one
ow depth to the other. It forms a
jump between one and the other.
The two regions are separated by
a continuously collapsing wall of
water referred to as a hydraulic
jump.
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RAPIDLY VARIED FLOW & HYDRAULIC JUMP
The energy dissipation ratio represents
the fraction of mechanical energy
dissipated during a hydraulic jump.
Energy Dissipation ratio =
The Depth Ratio for a Hydraulic Jump
The ratio of the depths is:
Example: Calculation of a Hydraulic Jump
A sluice gate is constructed across an open channel. Water owing under
it creates a hydraulic jump. Determine the depth just downstream of the
jump (point b) if the depth of ow at point a is 0.0563 meters and the
velocity at point a is 5.33 meters/sec.
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Example Hydraulic Jump - Tutorial
Water discharging into a 10-m-wide rectangular horizontal channel from a sluice gate
is observed to have undergone a hydraulic jump. The ow depth and velocity before
the jump are 0.8 m and 7 m/s, respectively. Determine (a) the ow depth and the
Froude number after the jump, (b) the head loss and the dissipation ratio, and (c) the
wasted power production potential due to the hydraulic jump.
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Example Hydraulic Jump -Tutorial
Example Hydraulic Jump -Tutorial
QUIZ
Draw HGL and EGL for the following situa on
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Specific Energy: Sluice Gate
sluice gate
10
q = 5.5 m2/s
EGL y2 = 0.45 m
y1 7.5
V2 = 12.2 m/s
1
5
E2 = 8 m
y
2.5 2
y2
0
0 2.5 5 7.5 10
E E1 = E 2
Given downstream depth and discharge, find upstream depth.
y1 and y2 are ___Alternate_ depths (same specific energy)
Why not use momentum conservation to find y1?
Specific Energy: Raise the Sluice Gate
sluice gate
4
y1 3 EGL
2 1 2
y
y2 1
E1 = E2
0
0 1 2 3 4 2
E
q
E = y +
2
2gy
as sluice gate is raised y1 approaches y2 and E is
minimized: Maximum discharge for given energy.
Specific Energy: Step Up
Short, smooth step with rise Δy in channel
Given upstream depth and 4
discharge find y2
4 3
3 2
y
2 1
y
1 0 Δy
0 1 2 3 4
E
0
0 1 2 3 4 Increase step
height?
E
E1 = E2 + y D
Table 1: Maximum Canal Side Slopes (Z)
Sand, Soft Clay 3: 1 (Horizontal: Vertical)
Sandy Clay, Silt Loam,
2:1
Sandy Loam
Fine Clay, Clay Loam 1.5:1
Heavy Clay 1:1
Sti Clay with Concrete
0.5 to 1:1
Lining
Lined Canals 1.5:1
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M.Hanif Chaudry, Open Channel Flow 2nd Edition, Springer, 2008
Design of Non-Erodible Channels
When a channel conveying clear water is to be lined, or
the earth used for its construction is non-erodible in the
normal range of canal velocities, Manning's equation is
used. We are not interested about maximum velocity in
design. Manning's equation is:
1 2/3 1/ 2
` Q= AR S .......(1)
n
Q and S are basic requirements of canal determined from crop
water needs. The slope of the channels follows the natural
channel. Manning's n can also be got from Tables or estimated
using the Strickler equation: n = 0.038 d1/6 , d is the particle
size diameter (m)
Design of Non-Erodible Channels Contd.
■ LHS of equation (1) can be calculated in terms of A
R2/3 termed section factor. For a trapezoidal
section:
■ A = b d + Z d2 ; P = b + 2 d (1 + Z)1/2
■ The value of Z is decided (see Table 6.1) and the
value of b is chosen based on the material for the
construction of the channel.
■ The only unknown d is obtained by trial and error
to contain the design flow. Check flow velocity and
add freeboard.
Example - Tutorial
Design a Non-Erodible Channel to convey 10
m3/s flow, the slope is 0.00015 and the mean
particle diameter of the soil is 5 mm. The side
slope is 2 : 1.
■ Solution: Q = 1/n AR 2/3 S 1/2 ….. (1)
■ With particle diameter, d being 5 mm, Using
Strickler Equation, n = 0.038 d 1/6
■ = 0.038 x 0.005 1/6 = 0.016
Solution of Example Contd.
1 2/3 1/ 2
Q= A R 0.00015 = 0.77 A R 2/3
0.016
Z = 2. Choose a value of 1.5 m for 'b‘
For a trapezoidal channel, A = b d + Z d2 = 1.5 d + 2 d2
P = b + 2 d (Z2 + 1)1/2 = 1.5 + 2 d 51/2 = 1.5 + 4.5 d
Try different values of d to contain the design flow of 10 m3/s
Soln of Example Contd.
d(m) A(m2 ) P(m) R(m) R2/3 Q(m3/s) Comment
2.0 11.0 10.5 1.05 1.03 8.74 Small flow
2.5 16.25 12.75 1.27 1.18 14.71 Too big
2.2 12.98 11.40 1.14 1.09 10.90 slightly big
2.1 11.97 10.95 1.09 1.06 9.78 slightly small
2.13 12.27 11.09 1.11 1.07 10.11 O.K.
The design parameters are then d = 2.13 m and b = 1.5 m
Check Velocity : Velocity = Q/A = 10/12.27 = 0.81 m/s
Note: For earth channels, it is advisable that Velocity should be above 0.8 m/s
to inhibit weed growth but this may be impracticable for small channels.
Assuming freeboard of 0.2 d ie. 0.43 m, Final design parameters are:
D = 2.5 m and b = 1.51 m
Final Design Diagram
T = 11.5 m
Z = 2:1 D = 2.5 m
d = 2.13 m
b = 1.5 m
T = b + 2 Z d = 1.5 + 2 x2 x 2.5 = 11.5 m
Design of Erodible Channels Carrying Clean Water
■ The problem here is to find the velocity at which
scour is initiated and to keep safely below it.
Different procedures and thresholds are involved
including maximum permissible velocity and
tractive force criteria.
■ Maximum Permissible Velocities: The
maximum permissible velocities for different earth
materials can be found in text books e.g.
Hudson's Field Engineering, Table 8.2.
Procedure For Design
■ i) Determine the maximum permissible velocity
from tables.
■ ii) With the permissible velocity equal to Q/A,
determine A.
■ iii) With permissible velocity = 1/n S1/2 R2/3
■ Slope, s and n are normally given.
■ iv) R = A/P , so determine P as A/R
■ v) Then A = b d + Z d and
■ P = b+ 2 d (Z2 + 1)1/2 ,
■ Solve and obtain values of b and d
Example - Tutorial
■ From previous example, design the
channel using the maximum
permissible velocity method.
Example:
From previous example, design the channel using the
maximum permissible velocity method.
Solution: Given: Q = 10 m3 /s , Slope = 0.00015 , n = 0.016
, Z = 2:1
i) From permissible velocity table, velocity = 0.75 m/s
ii) A = Q/V = 10/0.75 = 13.33 m
3/ 2 3/ 2
⎡ Vn ⎤ ⎡ 0.75x0.016 ⎤
iii) ` R = ⎢ 1/ 2 ⎥ R=⎢ 1/ 2 ⎥
= 0.97
⎣S ⎦ ⎣ 0.00015 ⎦
iv) P = A/R = 13.33/0.97 = 13.74 m
v) A = b d + Z d2 = b d + 2 d2
P = b + 2 d (Z2 + 1)1/2 = b + 2 d 51/2 = b + 4.5 d
ie. b d + 2 d2 = 13.33 m 2 ........(1)
b + 4.5 d = 13.74 m ........ (2)
Solution of Equation Contd.
From (2), b = 13.74 - 4.5 d .......(3)
Substitute (3) into (1), (13.74 - 4.5 d)d + 2 d2 = 13.33
13.74 d - 4.5 d2 + 2 d = 13.33
13.74 d - 2.5 d2 = 13.33
ie. 2.5 d2 - 13.74 d + 13.33 = 0
Recall the quadratic equation formula:
2
−b ± b − 4 a c 13.74 ± 7.44
x= d= = 1016
. m and 126
. m
2a 5
d = 1.26 m is more practicable
From (3), b = 13.74 - (4.5 x 1.26) = 8.07 m
Adding 20% freeboard, Final Dimensions are depth = 1.5 m and
width = 8.07 m
Final Design Diagram
T = 14.1 m
Z = 2:1 D = 1.5 m
d = 1.26m
b = 8.07 m
T = b + 2 Z d = 8.05 + 2 x2 x 1.5 = 14.1 m
