FUNDAMENTALS OF
FLUID MECHANICS
Chapter 10
Flow in Open Channels
Jyh-Cherng Shieh
Department of Bio-Industrial Mechatronics Engineering
National Taiwan University
1
MAIN TOPICS
General Characteristics of Open-Channel Flow
Surface Waves
Energy Considerations
Uniform Depth Channel Flow
Gradually Varies Flow
Rapidly Varies Flow
2
Introduction
Open channel flow involves the flows of a liquid in a channel or
conduit that is not completely filled.
There exists a free surface between the flowing fluid (usually water)
and fluid above it (usually the atmosphere).
The main deriving force is the fluid weight-gravity forces the fluid
to flow downhill.
Under steady, fully developed flow conditions, the component if the
weight force in the direction of flow is balanced by the equal and
opposite shear force between the fluid and the channel surface.
3
Open Channel Flow vs. Pipe Flow
There can be no pressure force driving the fluid through the channel
or conduit.
For steady, fully developed channel flow, the pressure distribution
within the fluid is merely hydrostatic.
4
Examples of Open Channel Flow
The natural drainage of water through the numerous creek and river
systems.
The flow of rainwater in the gutters of our houses.
The flow in canals, drainage ditches, sewers, and gutters along roads.
The flow of small rivulets, and sheets of water across fields or
parking lots.
The flow in the chutes of water rides.
5
Variables in Open-Channel Flow
Cross-sectional shape.
Bends.
Bottom slope variation.
Character of its bounding surface.
Most open-channel flow results are based on correlation
obtained from model and full-scale experiments.
Additional information can be gained from various analytical
and numerical efforts.
6
General Characteristics of
Open-Channel Flow
7
Classification of Open-Channel Flow
For open-channel flow, the existence of a free surface allows
additional types of flow.
The extra freedom that allows the fluid to select its free-surface
location and configuration allows important phenomena in openchannel flow that cannot occur in pipe flow.
The fluid depth, y, varies with time, t, and distance along the
channel, x, are used to classify open-channel flow:
8
Classification - Type I
Uniform flow (UF): The depth of flow does not vary along the
channel (dy/dx=0).
Nonuniform flows:
►Rapidly varying flows (RVF):
The flow depth changes
considerably over a relatively
short distance dy/dx~1.
►Gradually varying flows (GVF):
The flow depth changes slowly
with distance dy/dx <<1.
9
Classification - Type II
R e = ρVR h / μ
V is the average velocity of the fluid.
Rh is the hydraulic radius of the channel.
Laminar flow: Re < 500.
Transitional flow:
Turbulent flow: Re > 12,500.
Most open-channel flows involve water (which has a fairly small
viscosity) and have relatively large characteristic lengths, it is
uncommon to have laminar open-channel flows.
10
Classification - Type III
Fr = V / gl
Critical Flow: Froude number Fr =1.
Subcritical Flow: Froude number Fr <1.
Supercritical Flow: Froude number Fr >1.
11
Surface Wave
12
Surface Wave
The distinguishing feature of flows involve a free surface (as in
open-channel flows) is the opportunity for the free surface to distort
into various shapes.
The surface of a lake or the ocean is usually distorted into everchanging patterns associated with surface waves.
13
Kinds of Surface Wave
Some of the surface waves are very high, some barely ripple the
surface; some waves are very long, some are short; some are
breaking wave that form white caps, others are quite smooth.
Dsmall amplitude
DFinite-sized solitary
DContinuous sinusoidal shape
14
small amplitude
Wave Speed 1/5
Consider a single elementary wave of small height, by δy, is
produced on the surface of a channel by suddenly moving the
initially stationary end wall with speed δV.
15
small amplitude Wave Speed 2/5
The water in the channel was stationary at the initial time, t=0.
A stationary observer will observe a single wave move down the
channel with a wave speed c, with no fluid motion ahead of the
wave and a fluid velocity of δV behind the wave.
The motion is unsteady.
For a observer moving along the channel with speed c, the flow will
appear steady.
v
v
To this observer,
the fluid
v velocity will be V = -c i on the observer’s
v
right and V = (-c + δ V) i to the left of the observer.
Momentum Equation + Continuity Equation
16
small amplitude
Wave Speed 3/5
With the assumption of uniform one-dimensional flow, the
continuity equation becomes
− cyb = ( −c + δV )( y + δy )b
( y + δy )δV
δV
(1)
=y
⇒c=
δy
δy
δy << y
Similarly, the momentum equation
1 2
1
F2 = γy c 2 A 2 = γy b
F1 = γy c1A1 = γ ( y + δy )2 b
2
2
1 2
1
F2 − F1 = γy b − γ ( y + δy ) 2 b = ρbcy[(c − δV ) − c]
2
2
17
small amplitude
Wave Speed 4/5
(δy ) 2 << yδy
⇒
δV g
=
δy c
(1)+(2)
(2)
c = gy
(3)
Energy Equation + Continuity Equation
18
small amplitude
Wave Speed 5/5
The single wave on the surface is seen by an observer moving with
the wave speed, c.
Since the pressure is constant at any point on the free surface, the
Bernoulli equation for this frictionless flow is
V2
VδV
+ y = cons tan t →
+ δy = 0
2g
g
The continuity equation
Vy = cons tan t → yδV + Vδy = 0
Combining these two equations and using the fact V=c
c = gy
19
finite-sized solitary
Wave Speed
More advanced analysis and experiments show that the wave speed
for finite-sized solitary wave
1/ 2
⎛ δy ⎞
c = gy ⎜⎜1 + ⎟⎟
y⎠
⎝
1/ 2
⎛ δy ⎞
c = gy ⎜⎜1 + ⎟⎟
y⎠
⎝
> gy
The larger the amplitude, the faster the wave travel.
20
Continuous sinusoidal shape
Wave Speed 1/2
A more general description of wave motion can be obtained by
considering continuous (not solitary) wave of sinusoidal shape.
DBy combining waves of various wavelengths, λ, and amplitudes,
δy.
DThe wave speed varies with both the wavelength and fluid depth
as
1/ 2
⎡ gλ
⎛ 2πy ⎞⎤
c =& ⎢ tanh ⎜
⎟⎥ (4)
⎝ λ ⎠⎦
⎣ 2π
21
Continuous sinusoidal shape
Wave Speed 2/2
1/ 2
⎡ gλ
⎛ 2πy ⎞⎤
c =& ⎢ tanh ⎜
⎟⎥
⎝ λ ⎠⎦
⎣ 2π
y
⎛ 2 πy ⎞
→ ∞ ⇒ tanh ⎜
⎟ =1
λ
⎝ λ ⎠
⇒c=
gλ
2π
Deep layer
y
⎛ 2 πy ⎞ 2 πy
→ 0 ⇒ tanh ⎜
⎟=
λ
λ
⎝ λ ⎠
⇒ c = gy
Shallow layer
Wave speed as a function of wavelength.22
Froude Number Effects 1/3
Consider an elementary wave travelling on the surface of a fluid.
If the fluid layer is stationary, the wave moves to the right with
speed c relative to the fluid and stationary observer.
When the fluid is flowing to the left with velocity V.
DIf V<c, the wave will travel to the right with a speed of c-V.
DIf V=c, the wave will remain stationary.
DIf V>c, the wave will be washed to the left with a speed of V-c.
Froude Number
F r = V / gl = V / c
Is the ratio of the fluid
velocity to the wave speed.
23
Froude Number Effects 2/3
When a wave is produced on the surface of a moving stream, as
happens when a rock is thrown into a river.
DIf V=0, the wave speeds equally in all directions.
DIf V<c, the wave can move upstream. Upstream locations are
said to be in hydraulic communication with the downstream
locations. Such flow conditions, V<c, or Fr<1, are termed
subcritical.
DIf V>c, no upstream communication with downstream locations.
Any disturbance on the surface downstream from the observer
will be washed farther downstream. Such conditions, V>c, or
Fr>1, are termed supercritical.
24
Froude Number Effects 3/3
DIf V=c or Fr=1, the upstream propagating wave remains
stationary and the flow is termed critical.
25
Energy Considerations
26
Energy Considerations 1/3
z1 − z 2
S0 =
l
The slope of the channel bottom or bottom
slope is constant over the segment
Very small for most open-channel flows.
x and y are taken as the
distance along the
channel bottom and the
depth normal to the
bottom.
27
Energy Considerations 2/3
With the assumption of a uniform velocity profile across any section
of the channel, the one-dimensional energy equation become
p1 V12
p 2 V22
+
+ z1 =
+
+ z2 + hL
2g
2g
γ
γ
(5)
hL is the head loss due to viscous effects between sections (1) and (2).
(5)
V12
V22
y1 +
+ Sol = y 2 +
+ hL
2g
2g
(6)
z1 − z 2 = S o l
p1 / γ = y1
p2 / γ = y2
28
Energy Considerations 3/3
(6)
V22 − V12
y1 − y 2 =
+ (S f
2g
− So )l
(7)
Sf = h L / l
(7)
V22 − V12
y1 − y 2 =
2g
For a horizontal channel bottom (S0=0)
and negligible head loss (Sf=0)
29
Specific Energy 1/4
Define specific energy, E
V2
E=y+
2g
(8)
E1 = E 2 + (Sf − So )l
(9)
(9)
E1 + z1 = E 2 + z 2 The sum of the specific energy and
Head losses are negligible, Sf=0
−So l = z 2 − z1
the elevation of the channel bottom
remains constant.
This a statement of the Bernoulli
equation.
30
Specific Energy 2/4
If the cross-sectional shape is a rectangular of width b
E=y+
q2
2gy
2
(10)
Where q is the flowrate per unit width, q=Q/b=Vyb/b=Vy
For a given channel
b= constant
q = constant
E = E (y) Specific energy diagram
31
Specific Energy 3/4
E=y+
q2
2gy 2
(10)
For a given q and E, equation (10) is a
cubic equation with three solutions , ysup,
ysub, and yneg.
If E >Emin, two solutions are positive
and yneg is negative (has no physical
meaning and can be ignored).
These two depths are term alternative
depths.
32
Specific Energy 4/4
Approach y=E
Very deep and very slowly
y sup < y sub
Vsup > Vsub
E > Emin
Two possible depths of flow,
one subcritical and the other
supercritical
Approach y=0
Very shallow and
very high speed
33
Determine Emin
To determine the value of Emin
1/ 3
⎛q ⎞
dE
dE
q
=0⇒
= 1 − 3 = 0 ⇒ yc = ⎜ ⎟
⎜ g ⎟
dy
dy
gy
⎝ ⎠
2
Sub. (11) into (10)
2
E min
(11)
3y c
q
=
⇒ Vc =
= gy c ⇒ Frc = 1
2
yc
1. The critical conditions (Fr=1) occur at the location of Emin.
2. Flows for the upper part of the specific energy diagram
are subcritical (Fr<1)
3. Flows for the lower part of the specific energy diagram are
supercritical (Fr>1)
34
Example 10.1 Specific Energy diagram Qualitative
Water flows under the sluice gate in a constant width rectangular
channel as shown in Fig. E10.1a. Describe this flow in terms of the
specific energy diagram. Assume inviscid flow.
35
Example 10.1 Solution1/2
Inviscid flow Sf=0
Channel bottom is horizontal z1=z2 (or So=0)
E1=E2 q1=q2
The specific energy diagram for this flow is as shown in Fig. E10.1b.
The flowrate can remain the same for this channel even if the
upstream depth is increased. This is indicated by depths y1’ and y2’
in Fig E10.1c.
To remain same flowrate, the distance between the bottom of the
gate and the channel bottom must be decreased to give a smaller
flow area (y2’ < y2 ), and the upstream depth must be increased to
give a bigger head (y 1’ > y1 ).
36
Example 10.1 Solution2/2
On the other hand, if the gate remains fixed so that the downstream
depth remain fixed (y2’’ = y2 ), the flowrate will increase as the
upstream depth increases to y 1’’ > y1.
q”>q0
37
Example 10.2 Specific Energy diagram –
Quantitative
Water flows up a 0.5-ft-tall ramp in a constant width rectangular
channel at a rate of q = 5.75 ft2/s as shown in Fig. E10.2a. (For now
disregard the “bump”) If the upstream depth is 2.3 ft, determine the
elevation of the water surface downstream of the ramp, y2 + z2.
Neglect viscous effects.
38
Example 10.2 Solution1/4
With S0 l =z1-z2 and hL=0, conservation of energy requires that
p1 V12
p 2 V22
+
+ z1 =
+
+ z2
γ
γ
2g
2g
V22
⇒ 1.90 = y 2 +
64.4
(10.2-1)
The continuity equation
y 32 − 1.90 y 22 + 0.513 = 0
y 2 V2 = y1V1
y 2 = 1.72ft
⇒ y 2 V2 = 5.75ft 2 / s
(10.2-2)
0.638ft
y 2 = −0.466ft
39
Example 10.2 Solution2/4
The corresponding elevations of the free surface are either
y 2 + z 2 = 1.72ft + 0.50ft = 2.22ft
y 2 + z 2 = 0.638ft + 0.50ft = 1.14ft
Which of these flows is to be expected?
This can be answered by use of the specific energy diagram
obtained from Eq.(10), which for this problem is
E=y+
0.513
y2
The diagram is shown in Fig.E10.2(b).
40
Example 10.2 Solution3/4
41
Example 10.2 Solution4/4
The upstream condition corresponds to subcritical flow; the
downstream condition is either subcritical or supercritical,
corresponding to points 2 or 2’.
Note that since E1=E2+(z2-z1)=E2+0.5 ft, it follows that the
downstream conditions are located to 0.5 ft to the left of the
upstream conditions on the diagram.
….. The surface elevation is
y 2 + z 2 = 1.72ft + 0.50ft = 2.22ft
42
Channel Depth Variations 1/3
Consider gradually varying flows.
For such flows, dy/dx<<1, and it is reasonable to impose the onedimensional velocity assumption.
At an section the total head
V2
H=
+y+z
2g
and the energy equation H1 = H 2 + h L
The slop of the energy line
dH dh L
=
= Sf
dx
dx
dz
= So
dx
⎞ V dV dy dz
dH
d ⎛⎜ V 2
+
+
=
+ y + z⎟ =
⎟
⎜
dx dx ⎝ 2g
⎠ g dx dx dx
43
Channel Depth Variations 2/3
dh L V dV dy
=
+
+ S0
dx
g dx dx
V dV dy
+
= Sf − S o
(12)
g dx dx
For a given flowrate per unit width, q, in a rectangular channel of
constant width b, we have
V=
(12)
q
q dy
dV
V dy
⇒
=−
=−
y
dx
y dx
y 2 dx
V dV
V 2 dy
dy
=−
= − Fr2
g dx
gy dx
dx
(13)
44
Channel Depth Variations 3/3
Sub. (13) into (12)
dy Sf − S o
=
dx (1 − Fr2 )
(14)
Depends on the local slope of the channel bottom, the
slope of the energy line, and the Froude number.
45
Uniform Depth Channel Flow
46
Uniform Depth Channel Flow 1/3
Many channels are designed to carry fluid at a uniform depth all
along their length.
DIrrigation canals.
DNature channels such as rivers and creeks.
Uniform depth flow (dy/dx=0) can be accomplished by adjusting the
bottom slope, S0, so that it precisely equal the slope of the energy
line, Sf.
A balance between the potential energy lost by the fluid as it coasts
downhill and the energy that is dissipated by viscous effects (head
loss) associated with shear stress throughout the fluid.
47
Uniform Depth Channel Flow 2/3
Uniform flow in an open channel.
48
Uniform Depth Channel Flow 3/3
Typical velocity and shear stress distributions in an open channel:
(a) velocity distribution throughout the cross section. (b) shear
stress distribution on the wetted perimeter.
49
The Chezy & Manning Equation 1/6
Control volume for uniform flow in an open channel.
50
The Chezy & Manning Equation 2/6
Under the assumption of steady uniform flow, the x component of
the momentum equation
∑F
x
= ρQ( V2 − V1 ) = 0
F1 − F2 − τ w Pl + W sin θ = 0
(15)
where F1 and F2 are the hydrostatic pressure forces across either end
of the control volume.
P is wetted perimeter.
51
The Chezy & Manning Equation 3/6
y1=y2
F1=F2
(15)
− τ w Pl + W sin θ = 0
τw =
W sin θ
Pl
W = γAl
γAlSo
τw =
= rR h So
Pl
Rh =
A
P
(16)
Wall shear stress is proportional to the dynamic pressure
(Chapter 8)
V2
τ w = Kρ
2
V2
τw ∝ ρ
2
K is a constant dependent upon the roughness of the pipe
52
The Chezy & Manning Equation 4/6
(16)
V2
Kρ
= rR h S o
2
V = C R hSo
C is termed the Chezy coefficient
(17)
Chezy equation
Was developed in 1768 by A. Chezy (1718-1798), a French
engineer who designed a canal for the Paris water supply.
(17)
V ∝ S1o/ 2
V ∝ Rh
Reasonable
V ∝ R 2h / 3
Manning Equation
53
The Chezy & Manning Equation 5/6
In 1889, R. manning (1816-1897), an Irish engineer, developed the
following somewhat modified equation for open-channel flow to
more accurately describe the Rh dependence:
R 2h / 3S1o/ 2
V=
n
(18)
Manning equation
n is the Manning resistance coefficient.
Its value is dependent on the surface material
of the channel’s wetted perimeter and is
obtained from experiments.
It has the units of s/m1/3 or s./ft1/3
54
The Chezy & Manning Equation 6/6
κ 2 / 3 1/ 2
V = R h So
n
κ
Q = AR 2h / 3S1o/ 2
n
(18)
(19)
(20)
Where κ=1 if SI units are
used, κ=1.49 if BG units
are used.
The best hydraulic cross section is defined as the section of minimum
area for a given flowrate Q, slope, So, and the roughness coefficient, n.
κ ⎛A⎞
Q = A⎜ ⎟
n ⎝P⎠
Rh =
A
P
2/3
S1o/ 2
κ
=
n
A 5 / 3S1o/ 2
2/3
P
⎡ nQ ⎤
⇒ A = ⎢ 1/ 2 ⎥
⎢⎣ kSo ⎥⎦
3/ 5
P2 / 5
constant
A channel with minimum A is one with a minimum P.
55
Value of the Manning Coefficient, n
56
Uniform Depth Examples
57
Example 10.3 Uniform Flow, Determine
Flow Rate
Water flows in the canal of trapezoidal cross section shown in Fig.
E10.3a. The bottom drops 1.4 ft per 1000 ft of length. Determine the
flowrate if the canal is lined with new smooth concrete. Determine
the Froude number for this flow.
58
Example 10.3 Solution1/2
(20)
κ
Q = AR 2h / 3S1o/ 2
n
κ=1.49 if BG units are used.
⎛ 5
⎞
A = 12 ft (5ft ) + 5ft ⎜
ft ⎟ = 89.8ft 2
⎝ tan 40° ⎠
P = 12 ft + 2(5 / sin 40°ft ) = 27.6ft
R h = A / P = 3.25ft
From Table 10.1, n=0.012
10.98
1.49
2
2/3
1/ 2
Q=
(89.8ft )( 3.25ft ) ( 0.0014 )
=
= 915cfs
n
n
V = Q / A = 10.2ft / s
Fr =
V
gy
= ... = 0.804
59
Example 10.3 Solution2/2
60
Example 10.4 Uniform Flow, Determine
Flow Depth
Water flows in the channel shown in Fig. E10.3 at a rate o Q = 10.0
m3/s. If the canal lining is weedy, determine the depth of the flow.
61
Example 10.4 Solution
A = 1.19 y 2 + 3.66 y
⎛ y ⎞
P = 3.66 + 2⎜
⎟ = 3.11y + 3.66
⎝ sin 40° ⎠
A 1.19 y 2 + 3.66 y
Rh = =
P
3.11y + 3.66
From Table 10.1, n=0.030
κ
1 .0
2 / 3 1/ 2
Q = 10 = AR h S o =
...
n
0.03
(1.19 y 2 + 3.66 y ) 5 − 515( 3.11y + 3.66) 2 = 0
y=1.50 m
62
Example 10.5 Uniform Flow, Maximum
Flow Rate
Water flows in a round pipe of diameter D at a depth of 0 ≤ y ≤ D,
as shown in Fig. E10.5a. The pipe is laid on a constant slope of S0,
and the Manning coefficient is n. At what depth does the maximum
flowrate occur? Show that for certain flowrate there are two depths
possible with the same flowrate. Explain this behavior.
63
Example 10.5 Solution1/2
(20)
κ
Q = AR 2h / 3S1o/ 2
n
κ=1.49 if BG units are used.
D2
A=
( θ − sin θ)
8
Dθ
A D( θ − sin θ)
Rh = =
P=
2
P
4θ
κ 1 / 2 D 8 / 3 ⎡ ( θ − sin θ) 5 / 3 ⎤
Q = So
⎢
⎥
2
/
3
2
/
3
n
θ
8( 4)
⎢⎣
⎥⎦
This can be written in terms of the flow depth by using
D
y = [1 − cos( θ / 2)]
2
64
Example 10.5 Solution2/2
A graph of flowrate versus flow depth, Q = Q(y), has the
characteristic indicated in Fig. E10.5(b).
The maximum flowrate occurs when y=0.938D, or θ=303º
Q = Q max when y = 0.938 D
65
Example 10.6 Uniform Flow, Effect of
Bottom Slope
Water flows in a rectangular channel of width b = 10 m that has a
Manning coefficient of n = 0.025. Plot a graph of flowrate, Q, as a
function of slope S0, indicating lines of constant depth and lines of
constant Froude number.
66
Example 10.6 Solution1/2
A = by = 10 y
Rh =
V=
(19)
A
by
=
P ( b + 2 y)
κ 2 / 3 1/ 2
1.0 ⎛ 10 y ⎞
⎜⎜
⎟⎟
R h So =
n
0.025 ⎝ 10 + 2 y ⎠
1.0 ⎛ 10 y ⎞
⎟⎟
⎜⎜
Fr =
0.025 ⎝ 10 + 2 y ⎠
2/3
⎛5+ y⎞
⎟⎟
⇒ S o = 0.00613 Fr y⎜⎜
⎝ 5y ⎠
4/3
1/ 2
( gy )
2
2/3
S1o/ 2
(10.6-1)
S1o/ 2
(10.6-2)
67
Example 10.6 Solution2/2
For given value of Fr, we pick
various value of y, determine the
corresponding value of So from Eq
(10.6-2), and then calculate Q=VA,
with V from either Eq (10.6-1) or
V=(gy)1/2Fr.
The results are indicated in
Fig. E10.6
68
Example 10.7 Uniform Flow, Variable
Roughness
Water flows along the drainage canal having the properties shown in
Fig. E10.7a. If the bottom slope is S0 = 1 ft/500 ft=0.002, estimate
the flowrate when the depth is y = 0.8 ft + 0.6 ft = 1.4 ft.
69
Example 10.7 Solution
Q = Q1 + Q 2 + Q 3
1.49
Qi =
A i R 2h / 3S1o/ 2
i
ni
Q = ... = 16.8ft 3 / s
70
Example 10.8 Uniform Flow, Best
Hydraulic Cross Section
Water flows uniformly in a
rectangular channel of width
b and depth y. Determine the
aspect ratio, b/y, for the best
hydraulic cross section.
71
Example 10.8 Solution1/3
(20)
A = by
κ
Q = AR 2h / 3S1o/ 2
n
P = b + 2y
κ=1.49 if BG units are used.
A
by
A
Ay
Rh = =
=
=
P ( b + 2 y) ( b + 2 y) (2 y 2 + A)
⎞
Ay
κ ⎛⎜
⎟
Q= A
n ⎜⎝ ( 2 y 2 + A ) ⎟⎠
A
5/ 2
2/3
2
S1o/ 2
y = K(2y + A)
⎛ nQ
K = ⎜ 1/ 2
⎜ κS
⎝ o
constant
⎞
⎟
⎟
⎠
3/ 2
72
Example 10.8 Solution2/3
The best hydraulic section is the one that gives the minimum A
for all y. That is, dA/dy = 0.
⎛
dA
5 3 / 2 dA
dA ⎞
5/ 2
⎟⎟
y+A
=0⇒ A
= K ⎜⎜ 4 y +
dy
2
dy
dy ⎠
⎝
⇒ A 5 / 2 = 4 ky
2 y 2 = by
The rectangular with the best hydraulic cross section twice
as wide as it is deep, or
b/y = 2
73
Example 10.8 Solution3/3
The best hydraulic cross section for other shapes
74
Gradually Varied Flow
dy
<< 1
dx
75
Gradually Varied Flow 1/2
Open channel flows are classified as uniform depth, gradually
varying or rapidly varying.
If the channel bottom slope is equal to the slope of the energy line,
So=Sf, the flow depth is constant, dy/dx=0.
DThe loss in potential energy of the fluid as it flows downhill is
exactly balanced by the dissipation of energy through viscous
effects.
If the bottom slope and the energy line slope are not equal, the flow
depth will vary along the channel.
76
Gradually Varied Flow 2/2
dy Sf − S o
=
dx (1 − Fr2 )
(14)
The sign of dy/dx, that is, whether the flow depth
increase or decrease with distance along the channel
depend on Sf -So ad 1-Fr2
77
Classification of Surface Shapes 1/3
The character of a gradually varying flow is often classified in terms
of the actual channel slope, So, compared with the slope required to
produce uniform critical flow, Soc.
The character of a gradually varying flow depends on whether the
fluid depth is less than or greater than the uniform normal depth, yn.
12 possible surface configurations
78
Classification of Surface Shapes 2/3
Fr<1 : y>yc
Fr>1 : y<yc
79
Examples of Gradually Varies Flows 1/5
Backwater curve
Drop-down profile
Typical surface configurations for nonuniform depth flow with a
mild slope. S0 < S0c.
80
Examples of Gradually Varies Flows 2/5
Typical surface configurations for nonuniform depth flow with a
critical slope. S0 = S0c.
81
Examples of Gradually Varies Flows 3/5
Typical surface
configurations for
nonuniform depth
flow with a steep
slope. S0 > S0c.
82
Examples of Gradually Varies Flows 4/5
Typical surface configurations for nonuniform depth flow with a
horizontal slope. S0 =0.
83
Examples of Gradually Varies Flows 5/5
Typical surface configurations for nonuniform depth flow with a
adverse slope. S0 <0.
84
Classification of Surface Shapes 3/3
The free surface is relatively free to conform to the shape that
satisfies the governing mass, momentum, and energy equations.
The actual shape of the surface is often very important in the design
of open-channel devices or in the prediction of flood levels in
natural channels.
☯ The surface shape, y=y(x), can be calculated by solving
the governing differential equation obtained from a
combination of the Manning equation (20) and the
energy equation (14).
Numerical techniques have been developed and
used to predict open-channel surface shapes.
85
Rapidly Varied Flow
dy
~1
dx
86
Rapidly Varied Flow
Rapidly varied flow: flow depth changes occur over a relatively
short distance.
DQuite complex and difficult to analyze in a precise fashion.
DMany approximate results can be obtained by using a simple
one-dimensional model along with appropriate experimentally
determined coefficients when necessary.
87
Occurrence of Rapidly Varied Flow 1/2
Flow depth changes significantly un a short distance: The flow
changes from a relatively shallow, high speed condition into a
relatively deep, low speed condition within a horizontal distance of
just a few channel depths.
Hydraulic Jump
88
Occurrence of Rapidly Varied Flow 2/2
Sudden change in the channel geometry such as the flow in an
expansion or contraction section of a channel.
Rapidly varied flow may occur in a
channel transition section.
89
Example of Rapidly Varied Flow 1/2
The scouring of a river bottom in the neighborhood of a bridge pier.
Responsible for
the erosion near
the foot of the
bridge pier.
The complex three-dimensional flow structure around a bridge pier.
90
Example of Rapidly Varied Flow 2/2
Many open-channel flow-measuring devices are based on
principles associated with rapidly varied flows.
DBroad-crested weirs.
DSharp-crested weirs.
DCritical flow flumes.
DSluice gates.
91
Hydraulic Jump
92
The Hydraulic Jump 1/6
Under certain conditions it is possible that the fluid depth will
change very rapidly over a short length of the channel without any
change in the channel configuration.
DSuch changes in depth can be approximated as a discontinuity in
the free surface elevation (dy/dx=∞).
DThis near discontinuity is called a hydraulic jump.
.
93
The Hydraulic Jump 2/6
A simplest type of hydraulic jump in a horizontal, rectangular channel.
Assume that the flow at sections (1) and (2) is nearly uniform, steady,
and one-dimensional.
94
The Hydraulic Jump 3/6
The x component of the momentum equation
F1 − F2 = ρQ( V2 − V ) = ρV1y1b( V2 − V1 )
F1 = p c1A1 = γy12 b / 2
F2 = p c 2 A 2 =
γy 22 b / 2
y12 y 22 V1y1
−
=
( V2 − V )
2
2
g
The conservation of mass equation y1bV1 = y 2 bV2 = Q
The energy equation
V12
V22
y1 +
= y2 +
+ hL
2g
2g
(21)
(22)
(23)
The head loss is due to the violent turbulent mixing and dissipation.
95
The Hydraulic Jump 4/6
Nonlinear equations
(21)+(22)+(23)
One solution is y1=y2, V1=V2, hL=0
Other solutions?
(21)+(22)
⎞ V12 y1
y12 y 22 V1y1 ⎛ V1y1
⎜⎜
( y1 − y 2 )
−
=
− V1 ⎟⎟ =
2
2
g ⎝ y2
⎠ gy 2
2
⎛ y2 ⎞ ⎛ y2 ⎞
⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ − 2 Fr21 = 0
⎝ y1 ⎠ ⎝ y1 ⎠
Solutions
y2 1 ⎛
= ⎜ − 1 ± 1 + 8Fr21 ⎞⎟
y1 2 ⎝
⎠
Fr1 =
V1
gy1
y2 1 ⎛
= ⎜ − 1 + 1 + 8Fr21 ⎞⎟ (24)
y1 2 ⎝
⎠
96
The Hydraulic Jump 5/6
(23)
(24)+(25)
2⎤
⎡
⎛ ⎞
y
hL
⎢1 − ⎜ y1 ⎟ ⎥
=1− 2 +
y1
2 ⎢ ⎜⎝ y 2 ⎟⎠ ⎥
y1
⎣
⎦
Fr21
(25)
Depth ratio and dimensionless head loss across a
hydraulic jump as a function of upstream Froude
number.
97
The Hydraulic Jump 6/6
The head loss is negative if Fr1<1.
Violate the second law of
thermodynamics
Not possible to produce a
hydraulic jump with Fr1<1.
98
Classification of Hydraulic Jump 1/2
The actual structure of a hydraulic jump is a complex function of Fr1,
even though the depth ratio and head loss are given quite accurately
by a simple one-dimensional flow analysis.
A detailed investigation of the flow indicates that there are
essentially five type of surface and jump conditions.
99
Classification of Hydraulic Jump 2/2
100
Hydraulic Jump Variations 1/2
Hydraulic jumps can occur in a variety o channel flow
configurations, not just in horizontal, rectangular channels as
discussed above.
Other common types of hydraulic jumps include those that occur in
sloping channels and the submerged hydraulic jumps that can occur
just downstream of a sluice gate.
101
Hydraulic Jump Variations 2/2
Hydraulic jump variations:
(a) jump caused by a change
in channel slope,
(b) submerged jump
102
Example 10.9 Hydraulic Jump
Water on the horizontal apron of the 100-ft-wide spillway shown in
Fig. E10.9a has a depth o 0.60 ft and a velocity of 18 ft/s. Determine
the depth, y2, after the jump, the Froude numbers before and after
the jump, Fr1 and Fr2, and the power dissipated, Pd, with the jump.
103
Example 10.9 Solution1/3
Conditions across the jump are determined by the upstream
Froude number
V1
18ft / s
=
= 4.10
Fr1 =
gy1
(32.2ft / s 2 )(0.60ft )
(24)
y2 1 ⎛
= ⎜ − 1 + 1 + 8Fr21 ⎞⎟ = ... = 5.32
y1 2 ⎝
⎠
y 2 = 5.32(0.60ft ) = 3.19ft
Since Q1=Q2, or V2=(y1V1)/y2=3.39ft/s
104
Example 10.9 Solution2/3
Fr2 =
V2
gy1
=
3.39ft / s
(32.2ft / s 2 )(3.19ft )
= 0.334
The poser dissipated, Pd, by viscous effects within the jump can
be determined from the head loss
Pd = γQh L = γby1V1h L
(23)
2⎞ ⎛
2⎞
⎛
V
V
1
2
⎟ = ... = 2.26ft
⎟ − ⎜ y2 +
h L = ⎜ y1 +
⎜
2g ⎟⎠ ⎜⎝
2g ⎟⎠
⎝
Pd = γQh L = γby1V1h L = ...1.52 × 105 ft ⋅ lb / s = 277hp
105
Example 10.9 Solution3/3
q = q1 = q 2 =
E=y+
q2
2gy
2
Q
= y1V1 = 10.8ft 2 / s
b
=y+
1.81
y2
Various upstream depth
106
Weirs and Gate
107
Weir
A weir is an obstruction on a channel bottom over which the fluid
must flow.
Weir provides a convenient method of determining the flowrate in
an open channel in terms of a single depth measurement.
108
Sharp-Crested Weir 1/4
A sharp-Crested weir is essentially a vertical-edged flat plate placed
across the channel.
The fluid must flow across the sharp edge and drop into the pool
downstream of the weir plate.
109
Sharp-Crested Weir - Geometry 2/4
Sharp-crested weir plate geometry: (a) rectangular, (b) triangular, (c)
trapezoidal.
110
Sharp-Crested Weir – Flowrate 3/4
Assume that the velocity profile upstream of the weir plate
is uniform and that the pressure within the nappe is
atmosphere.
Assume that the
fluid flows
horizontally over the
weir plate with a
nonuniform velocity
profile.
111
Sharp-Crested Weir – Flowrate 4/4
With PB=0, the Bernoulli equation for flow along the arbitrary
streamline A-B indicated can be written as
p A V12
u 22
+
+ z A = ( H + Pw − h ) +
2g
2g
γ
(26)
Since the total head for any particle along the vertical section (1) is
the same
p A V12
V12
+
= H + pw +
zA +
γ
2g
2g
(26)
2⎞
⎛
V
1
⎟
u 2 = 2g⎜ h +
⎜
2g ⎟⎠
⎝
∫
∫
h=H
Q = u 2 dA = u 2 ldh
(27)
h =0
112
Rectangular Weir – Flowrate 1/2
For a rectangular weir, l=b
Q = 2g b
(28)
∫
H⎛
0
3/ 2
3/ 2 ⎫
⎧⎛
2⎞
2⎞
⎛
V
V
2
⎪
⎪
dh =
2g b ⎨⎜ H + 1 ⎟
−⎜ 1 ⎟ ⎬
⎜
⎜ 2g ⎟
3
2g ⎟⎠
⎪⎩⎝
⎝
⎠ ⎪⎭
2g bH 3 / 2 (29)
1/ 2
⎞
⎜h +
⎟
⎜
⎟
2
g
⎝
⎠
Q=
V12
2
3
(28)
V12
<< H
2g
Because of the numerous approximations made to obtain Eq. (29)
Q = C wr
2
3
2g bH 3 / 2
(30)
113
Rectangular Weir – Flowrate 2/2
Cwr is the rectangular weir coefficient.
Cwr is function of Reynolds number (viscous effects), Weber
number (surface tension effects), H/Pw (geometry effects).
In most practical situations, the Reynolds and Weber number
effects are negligible, and the following correction can be used.
C wr
⎛ H
= 0.611 + 0.075⎜⎜
⎝ Pw
⎞
⎟⎟
⎠
(31)
114
Triangular Weir – Flowrate 1/2
For a triangular weir l = 2( H − h ) tan θ2
Q=
8
θ
tan
2g H 5 / 2
15
2
V12
<< H
2g
An experimentally determined triangular weir coefficient, Cwt, is
used to account for the real world effects neglected in the analysis
so that
8
θ
Q = C wt tan
2g H 5 / 2
15
2
(32)
115
Triangular Weir – Flowrate 2/2
Weir coefficient for
triangular sharp-crested
weirs
116
About Nappe
Flowrate over a weir depends on whether the napple is free or
submerged.
Flowrate will be different for these situations
than that give by Eq. (30) and (32).
Flow conditions over a weir without a free nappe: (a) plunging nappe,
(b) submerged nappe.
117
Broad-Crested Weir 1/3
A broad-crested weir is a structure in an open channel that has a
horizontal crest above which the fluid pressure may be considered
hydrostatic.
118
Broad-Crested Weir 2/3
Generally, these weirs are restricted to the range 0.08 < H/Lw < 0.50.
For long weir block (H/Lw < 0.08), head losses across the weir
cannot be neglected.
For short weir block (H/Lw > 0.50), the streamlines of the flow over
the weir are not horizontal.
Apply the Bernoulli equation
Vc2
V12
= y c + Pw +
H + Pw +
2g
2g
If the upstream velocity head is negligible
Vc2 − V12 Vc2
=
H − yc =
2g
2g
119
Broad-Crested Weir 3/3
Since V2 = Vc = gy c
yc
H − yc =
2
2
yc = H
3
The flowrate is Q = by 2 V2 = by c Vc = b g y 3c / 2
⎛2⎞
= b g⎜ ⎟
⎝ 3⎠
3/ 2
H3/ 2
Again an empirical broad-crested weir coefficient, Cwb, is used to
account for the real world effects neglected in the analysis so that
⎛2⎞
Q = C wb b g ⎜ ⎟
⎝ 3⎠
3/ 2
H3/ 2
(33)
C wb =
0.65
1/ 2
⎛
H ⎞
⎜⎜1 +
⎟⎟
Pw ⎠
⎝
(34)
120
Example 10.10 Sharp-Crested and broadCrested Weirs
Water flows in a rectangular channel of width b = 2 m with flowrate
between Qmin = 0.02 m3/s and Qmax = 0.60 m3/s. This flowrate is to
be measured by using (a) a rectangular sharp-crested weir, (b) a
triangular sharp-crested weir with θ=90º, or (c) a broad-crested
weir. In all cases the bottom of the flow area over the weir is a
distance Pw = 1 m above the channel bottom. Plot a graph of Q=
Q(H) for each weir and comment on which weir would be best for
this application.
121
Example 10.10 Solution1/3
For the rectangular weir with Pw=1.
(30)+(31)
Q = C wr
2
3
⎛
H
2 g bH 3 / 2 = ⎜⎜ 0.611 + 0.075
Pw
⎝
⎞2
3/ 2
⎟⎟
2
g
bH
3
⎠
Q = 5.91( 0.611 + 0.075 H ) H 3 / 2
For the triangular weir
8
θ
Q = C wt tan
2g H 5 / 2 = 2.36C wt H 5 / 2
(32)
15
2
122
Example 10.10 Solution2/3
For the broad-crested weir
(33)+(34)
⎛2⎞
Q = C wb b g ⎜ ⎟
⎝ 3⎠
Q=
3/ 2
2.22
(1 + H )1 / 2
H
3/ 2
H3/ 2
=
0.65
1/ 2
⎛
H ⎞
⎟⎟
⎜⎜1 +
Pw ⎠
⎝
⎛2⎞
b g⎜ ⎟
⎝ 3⎠
3/ 2
H3/ 2
123
Example 10.10 Solution3/3
124
Underflow Gates 1/4
A variety of underflow gate structure is available for flowrate
control at the crest of an overflow spillway, or at the entrance of an
irrigation canal or river from a lake.
Three variations of underflow gates: (a) vertical gate, (b) radial gate, (c) drum gate.
125
Underflow Gates 2/4
The flow under the gate is said to be free outflow when the fluid
issues as a jet of supercritical flow with a free surface open to the
atmosphere.
In such cases it is customary to write this flowrate as
q = C d a 2gy1
(35)
Where q is the flowrate per unit width.
The discharge coefficient, Cd, is a function of contraction coefficient,
Cc = y2/a, and the depth ration y1/a.
126
Underflow Gates 3/4
Typical discharge coefficients for underflow gates
127
Underflow Gates 4/4
The depth downstream of the gate is controlled by some downstream
obstacle and the jet of water issuing from under the gate is overlaid
by a mass of water that is quite turbulent.
Drowned outflow from a sluice gate.
128
Example 10.11 Sluice gate
Water flows under the sluice gate shown in Fig. E10.11. The
channel width is b = 20 ft, the upstream depth is y1= 6 ft, and the
gate is a = 1.0 ft off the channel bottom. Plot a graph of flowrate, Q,
as a function of y3.
129
Example 10.11 Solution1/2
(35)
q = bq = baC d 2gy1 = 393C d cfs
(Figure 10.29)
Along the vertical line y1/a=6.
For y3=6 ft, Cd=0
The value of Cd increases as y3/a decreases, reading a maximum of
Cd=0.56 when y3/a=3.2. Thus with y3=3.2a=3.2ft
Q = 393(0.56) cfs = 220 cfs
For y3 < 3.2 ft the flowrate is independent of y3, and the outflow
is a free outflow.
130
Example 10.11 Solution2/2
The flowrate for
3.2ft ≤ y3 ≤ 6ft
131