Applied Fluid Mechanics MACE 20002
Lecture Notes
2.
Lecture Notes
Open Channel
flow
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1.
Introduction
2
2.
Governing Equations
3
3.
Flow over bumps
4
4.
Surface (Gravity) Waves
8
5.
Modes of Flow
11
6.
Energy Analysis
13
7.
Hydraulic jumps
18
8.
9.
7.1.
Introduction
18
7.2.
Jump Analysis
19
7.3.
Energy loss across a jump
21
Flow over Weirs and under Sluice gates
23
8.1.
Introduction
23
8.2.
Broad-crested weir
23
8.3.
Sharp-crested weir
26
8.4.
Sluice gates
28
Summary
31
2.1. Introduction
In this chapter, we are concerned with the important topic of Open-channel
flow. We shall see in the following chapter that this is closely related to
compressible flow. In both cases the behaviour is intimately related to the
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waves. In the case of open-channel flow the waves are gravity waves (see
section 2.4) whilst in the case of compressible flow the waves are sound
waves.
We begin by considering the important features of the flow. Since we are
concerned with the flow of water (which is incompressible), this implies that,
for steady flow, the volume flow rate will be conserved. Thus, for constant
width channels, changes in the channel height lead to changes in the flow
velocity. However, hydrostatics implies that changes in depth also results in a
1
2
change in the pressure force ( gby , b being the channel width, y the flow
2
depth and and g being the fluid density and acceleration due to gravity,
respectively). Hence changes in depth gives rise to a coupling between
velocity changes and pressure changes. This coupling between the velocity
and pressure results in surface (sometimes called gravity) waves. Thus the
shape of the channel can affect the flow. In the next section we consider the
equations of motion in more detail.
2.2. Governing Equations
control volume
Pressure force
1!gby2
y0
0
2
pressur
e consta
nt on su
2
!U0
U0
rface Pressure force
1!gby2
2
y
!U
"w
U
L
Figure 2.1 The Governing equations of open-channel flow
Figure 2.1 shows a general flow (with varying depth) along a channel of
uniform width. Since the the surface is a free surface the pressure there is
constant and equal to atmospheric pressure. The motion of the water is then
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governed by the equations of continuity (conservation of mass), momentum
and energy:
Continuity: AU = A0 U0 = constant
Momentum: ⇢Q(U
(2.1)
1
U0 ) = ⇢gb y02
2
y2
⌧w Lb
1
1
Energy: gy0 + U02 = gy + U 2 + ghf
2
2
(2.2)
(2.3)
Equations (2.2) and (2.3) include a contribution due to viscous losses
( w L and ghf respectively). They are included here for completeness. In
practice, however, there are many situations (although not all) in which the
viscous effects can be ignored. As an example of this, in the next section we
consider the response of the surface of an open-channel flow to a bump on
the base of the channel. As we shall see, the response is non-trivial and
depends upon the speed of the flow.
2.3. Flow over bumps
z(x)
U
d(x)
h(x)
Figure 2.2 Flow over a bump in deep flow.
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d(x) z(x)
U
h(x)
Figure 2.3 Flow over a bump in shallow flow.
Consider the flow over a bump in otherwise uniform flow, shown in
figures 2.2 and 2.3. In figure 2.2 the surface is seen to drop as a result of a
rise in the position of the base (a 'bump'). In figure 2.3 the reverse is seen to
happen with the surface rising in response to the rise in the height of the
base. The essential question, which we intend to answer here, is which of
these two situation arises. As we shall see, the answer is not straightforward
and depends upon both the flow speed and the depth of the flow. In the
equations below, h is the height of the surface and d is the depth (= z h). In
order to solve this problem we must consider both the momentum and
continuity equations. If we treat the flow as inviscid (a fair assumption
provided the depth is not too shallow), then we saw in the previous section
that the momentum equation along a streamline can be simplified (since the
flow is also steady and incompressible) as Bernoulli's equation. It is helpful to
choose a streamline in which the pressure is known everywhere, so we
choose the surface streamline where the pressure is constant (and
atmospheric).
Bernoulli’s equation along surface: pa +
Differentiating wrt x:
1 2
v + gz = constant
2
⇥pa
⇥v
⇥z
⇥v
⇥z
+ v
+ g
=0=v
+g
⇥x
⇥x
⇥x
⇥x
⇥x
(2.4)
(2.5)
Equation (2.5) is a differential equation relating the variation in the flow
velocity to the height of the surface. A second equation relating the flow
velocity to the depth of the flow can now be found using the continuity
equation:
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Continuity: vd = constant
d
+d
x
d
z
h⇥
=
x
x
Differentiating wrt x: v
But d = z
(2.6)
v
v
=0⇥
=
x
x
h
x
v d
d x
(2.7)
(2.8)
Equation (2.7) is a differential equation relating the flow velocity to the
depth of the flow, whilst equation (2.8) relates the variation in depth to the
variation in surface height (unknown) and base (bump) height (known).
Substituting for
v
x
from equation (2.5) and for the depth ( d) from equation (2.8) into equation
(2.7) we have:
⇥
v2
z
h
z
(2.9)
+g
=0
d
x
x
x
Rearranging:
z
⇥
=
x
z
g
x
v2
d
⇥
v2
(gd v 2 )
v2 h
d x
=
⇥
h
x
(2.10)
(2.11)
Equation (2.11) relates the surface height to the height of the base
(bump). In particular whether the surface rises in response to rises in the
base (as in figure 2.3) or dips in response to rises in the base height (as in
figure 2.2) depends upon the sign of the right hand side of equation (2.11).
This in turn depends upon the relative values of v 2 and gd. Hence for deep,
2
slow flow, v < gd so the denominator is positive and the flow dips as it goes
2
over the bump.
For shallow, fast flow, in contrast, v > gd and the
denominator is negative and the level rises as the flow goes over the bump.
When v = gd, then the level experiences a discontinuity and a sudden
rise in height is observed. This is called a hydraulic jump (see section 2.7).
We can explain this rather odd behaviour in a number of different waves. We
begin by considering momentum.
2
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When flow passes over the bump, its momentum changes. There are
two possibilities. If the flow is to move upwards in response to the bump,
then it has to overcome gravity and thus the momentum drops. Since the
mass flow rate is constant, this can only be achieved if its depth increases
which in turn results in a rise in the pressure-area force (the pressure
integrated from the surface to the base). This rise in the pressure-area force
creates the horizontal force needed to reduce the x-component of the
momentum. If the flow is initially shallow and fast moving, then it starts with
high momentum and low pressure-area force. In this situation it has sufficient
momentum to lose and create a higher pressure-area force. If, in contrast,
the flow is initially deep and slow-moving then it has very little momentum and
a large amount of pressure force. In this situation it would need to lose more
momentum than it has and so, instead, the level drops resulting in a
shallower flow which in turn means greater speed, and hence momentum, but
lower pressure force. Next we consider this in terms of energy.
Before the bump, the flow has both potential and kinetic energy, with the
relative quantities depending upon how high it is and how fast and deep it is.
In order to move over the bump there will be an exchange between kinetic
and potential energy. Notice, however, that it cannot simply displace over the
bump as this would result in equal kinetic energy but an increase in potential
energy, which would violate the first law of thermodynamics (conservation of
energy). It can, instead, rise up by more than the size of the bump; resulting
in a rise in potential energy but a drop in kinetic energy (since the velocity
drops for deeper flow); or the surface can drop; resulting in a reduction in
potential energy, but a rise in kinetic energy. For initial flow which is deep and
slow-moving, it does not have sufficient kinetic energy to rise up over the
bump, so the surface level drops with the corresponding rise in kinetic energy
and drop in potential energy. When, in contrast, the initial flow is shallow and
fast-moving, it has plenty of kinetic energy, but insufficient potential energy.
As as result the surface level rises with the associated rise in potential energy
but drop in kinetic energy. Whilst both momentum and energy considerations
explain why the level rises in one situation, but falls in another, the reason for
2
the sudden rise when v = gd is more subtle. Mathematically, it occurs at the
point in which any change in level can occur without a bump being present at
the bottom of the channel. To understand this physically, it is helpful to
consider the surface waves (sometimes called gravity waves).
Surface waves move at a speed of gd in undisturbed flow (see section
2.4). With flow they are convected by the flow and so move at a speed of
v + gd downstream and gd v upstream. Hence surface waves can only
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2
travel upstream in the flow in regions in which v < gd . Suppose, therefore,
2
we have a flow in which v > gd (called a super-critical flow) which is slowed
as a result of an obstruction or other reason. Now suppose that the slowing
is sufficient for there to be a region downstream in which the flow is sub2
critical (v < gd ). In this downstream region, surface waves are free to travel
upstream, whereas in the upstream region they are not. The question then
arises: what happens to these upstream-travelling waves when they get to
the upstream region where they are prevented from travelling upstream? At
2
the intersection, the flow speed passes through the point at which v = gd. At
this point the upstream-travelling waves must stop. In a real flow, there will
be waves caused by slight flow disturbances, noise etc. All such waves, must
2
stop at the point where v = gd giving rise to the jump. In the next section we
discuss these surface (gravity) waves in more detail.
2.4. Surface (Gravity) Waves
In the previous section, we saw that the simple problem of flow over a
bump resulted in two completely different types of behaviour depending upon
the starting conditions. Furthermore a singularity occurred at the boundary
between the two flows (at a speed gd) suggesting that there is an important
phenomenon associated with the speed gd. It turns out, that, as mentioned
gd is the speed of surface waves in
above (but not proved), speed
undisturbed flow. Since they are related to gravitational effects, these waves
are termed gravity waves by some.
In this section we consider the
implications of these waves. We begin by presenting a simplified proof of
there propagation speed. In particular, we assume that the waves are steady.
For a more rigorous proof see Aero-acoustics course.
Figure 2.4 shows a surface wave propagating across a stationary fluid at
speed c . The fluid below the wave is disturbed, so we assume that it moves
at speed V . In order to turn this into a steady problem, it is convenient to
use a relative frame moving at speed c. In this frame, the surface wave
remains in a fixed position, the flow moves at speed c and the fluid below the
V . This is shown in figure 2.5. In both figure
wave moves at a speed c
2.4 and figure 2.5, the width of the channel is taken to be b.
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c
y
!y
still
water
!V
Figure 2.4 A wave disturbance on still fluid in the stationary frame.
!y
y
c
c!!V
Figure 2.5 A wave disturbance on still fluid in the relative frame.
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Since we now have a steady problem, we can analyse the flow, as we
did before for the flow over a bump, using the continuity and momentum
equations. Beginning with continuity:
⇥cyb = ⇥(c
V =c
V )(y + y)b
y
y+ y
(2.12)
⇥
(2.13)
Equation (2.13) gives us an equation relating the motion of the fluid
below the disturbance to the height of the disturbance and the propagation
speed of the disturbance. Another equation relating these quantities can be
obtained by considering the momentum equation. Neglecting viscous effects,
the only force results from the pressure-area force so we have:
1
⇥gb y 2
2
⇥
(y + y)2 = ⇥cby ((c
⇥
y
c V =g 1+
y
2y
V)
c)
(2.14)
(2.15)
Equation (2.15) gives us a second equation for V . Eliminating V using
equation (2.13) then leads to:
y
c2 = gy 1 +
y
⇥
y
1+
2y
⇥
(2.16)
Equation (2.16) tells us the waves speed for a given depth. We note the
following observations:
•
The flow speed depends upon the size of the oscillation, with large
waves (large y ) having higher speeds than small waves.
•
•
The minimum wave speed occurs when y
0
For small waves this minimum speed is c0 =
gy
As discussed previously, the surface waves, which for low amplitude
waves travel at a speed of gy , controls the flow. Furthermore, the speed
V
relative to the surface wave speed ( gd ) is an important quantity called the
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Froude number. When the Froude number is less than 1, the flow is termed
sub-critical, and the surface level drops as it passes over a bump on the
base. When the Froude number is greater than 1 the flow is termed supercritical, and the level rises in response to bumps on the base. Hydraulic
jumps occur at a Froude number of 1 and separates a supercritical region of
flow from a sub-critical region. This behaviour has many parallels with
compressible flow (see next chapter) in which the flow is controlled by the
Mach number.
In the next section which look at the implications of this for disturbances
in a channel with flow.
2.5. Modes of Flow
a) Fr << 1
b) Fr < 1
Figure 2.6 Sub-critical (Froude number (Fr) < 1) pipe modes.
Figure 2.6 shows the propagation of a surface disturbance in a channel
with subcritical flow. In figure 2.6a, the Froude number is much less than 1.
As a consequence, the flow speed is essentially irrelevant to the propagation
and the wavefronts travel away from the centre at equal speeds in all
directions. In figure 2.6b, the Froude number is significant but less than 1.
Here the disturbances are convected by the flow and so have a propagation
speed which is greater in the direction of the flow and smaller in directions
opposed to the flow.
Figure 2.7 shows the propagation of a surface disturbance in a channel
with critical and supercritical flows. Figure 2.7a is for F r = 1. Here the
convection is strong with much greater propagation speeds in the direction of
the flow. In addition the disturbance is unable to propagate at all in the -xLecture Notes
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direction because the propagation speed would be zero. However, the waves
are able to propagate in all other directions. Figure 2.7b is for a supercritical
flow with F r > 1 . Now the waves cannot propagate in any-direction outside a
wave cone, as shown. The angle of the cone is a simple function of the
Froude number. To see this, consider figure 2.8.
a) Fr = 1
b) Fr > 1
Figure 2.7 Critical and super-critical (Froude number (Fr)
modes
1) pipe
c ! Usin(!)
!
U
Figure 2.8 Cone angle for super-critical flow.
The cone side does not move so the component of its velocity normal to
the cone side must be zero. The actual speed is given by c U sin . For
c
c
this to equal zero we require sin = U . Hence the cone angle is 2 sin 1 U .
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This mode behaviour is summarised in the table below.
Open channel flow
Compressible flow
Surface waves
Sound waves
Fr < 1 (subcritical)
M < 1 (subsonic)
Fr = 1 (critical)
M = 1 (sonic)
Fr >1 (supercritical)
M > 1 (supersonic)
Wave cone
Mach cone
Undisturbed region outside cone for
Fr > 1
Zone of silence outside cone for
M>1
2.6. Energy Analysis
In discussing the flow over bumps, we found that the behaviour could be
explained in terms of the distribution between potential and kinetic energy. In
this section we look at this distribution, and the overall energy in more detail.
In our analysis, we will consider the overall energy per unit weight (e
U2
q2
e=y+
=y+
2g
2gy 2
E
= g ):
(2.17)
Equation (2.17) describes the energy in the flow in terms of the depth
and the volume flow rate. It is useful here to consider flows with the same
volume flow rate but different depths.
Figure 2.9 describes the energy in the flow for fixed volume flow rate q
and with varying depth, y . At low y , the flow velocity is large and so the flow
has large energy, almost entirely kinetic.
As the depth increases, the
potential energy increases but the drop in kinetic energy is more dramatic
resulting in an overall energy drop. This behaviour continues until the depth
reaches a critical depth, y , at which point the energy is minimum.
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*
8
yy*
e
y
e
e
0
y2
y
y11
ee
min
8
y2
y
8
y
Figure 2.9 A plot of the total energy in a flow as a function of height, for
fixed flow rate, q
As the depth increases further, the rise in potential energy exceeds the
drop in kinetic energy, resulting in an overall rise in the energy of the flow. It
is worth noting, that at a given energy level there are two possible depths, y1
and y2 . These depths are called conjugate depths. As we shall see later, one
of these depths corresponds to subcritical flow and the other to supercritical
flow. In order to understand the significance of the critical depth, y , and
minimum, energy, emin , we differentiate equation (2.17) to find at what depth
this occurs:
q2
gy 3
(2.18)
q2
U 2 y2
y =
=
g
g
(2.19)
e
=0=1
y
3
U2
g
3y⇤
= e⇤ =
2
y =
emin
(2.20)
(2.21)
It is clear, then, that the minimum occurs when the Froude number is unity
and the flow is critical.
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2
U
2g
y2
Figure 2.10 An unrestricted sluice gate
In practice, unrestricted flow will always take the minimum depth (i.e. y1)
and thus will be supercritical (see figure 2.10). However, if the flow is
restricted, the flow will slow down and the depth will increase to give
subcritical flow (as in figure 2.11).
Let us now consider how this graph relates to the flow
obstruction.
past an
Figure 2.12 shows the flow of water past an obstacle compared with flow of
air through a convergent-divergent nozzle. For clarity the energy graph is
also included on this figure. Initially the flow is deep with a large amount of
potential energy, but very little kinetic energy. As the flow passes over the
obstacle, the level drops and the flow speed increases (consistent with
(2.11)). For this flow, however, the flow increases sufficiently for it to become
critical. It reaches the critical point at the brow of the obstacle (where the
gradient is zero).
At this point is has reached its minimum energy.
Downstream of this point, it continues to accelerate as the level drops (again
consistent with equation (2.11) for supercritical flow moving past a drop in the
base level) and the flow becomes supercritical, as shown. The behaviour is
virtually identical to air being accelerated through a convergent-divergent
nozzle, as seen in the bottom of the diagram. Here the flow is initially
subsonic (M < 1), reaches the sonic point (M=1) at the throat of the nozzle,
and then continues to accelerate, becoming supersonic (M > 1) downstream.
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We will consider this problem in more detail in the next chapter
(Compressible flow).
2
U
2g
y2
Figure 2.11 A restricted sluice gate
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y
Upstream
Downstream
e
critical
Reservoir
Fr < 1
Fr=1
Fr >
1
sonic
M>1
M<1
M=1
Figure 2.12 Variation in flow conditions for flow past an obstruction
compared with compressible flow through a nozzle
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2.7. Hydraulic jumps
2.7.1.
Introduction
Figure 2.13 Hydraulic jump
In the section on bumps (section 2.3), we found that the coupling of the
momentum equation with the continuity equation, allows the possibility of a
sudden rise in depth. Such a rise is common and called a hydraulic jump. It
occurs in a sink (or basin) when the tap is running and is the phenomenon
seen at bores such as the Thames bore or Severn bore. An example of this
is seen in figure 2.13 above. In section 2.3, however, we considered only the
differential forms of the equations. Here we considered the macro form of the
equations.
Since the flow must satisfy both continuity and momentum
simultaneously, this will be sufficient to determine the ratio of downstream to
upstream depth and the ratio of Froude numbers upstream and downstream
of the jump. By also considering the energy equation, we can also
determine the energy lost across the jump to turbulence.
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2.7.2.
Jump Analysis
Pressure
force
2
!gy2
2
y2
Pressure force
!gy12
2
y1
U2
U1
Figure 2.14 A hydraulic jump
Figure 2.14 shows a simple hydraulic jump with upstream depth y1 and
downstream depth y2 . The upstream velocity is taken to be U1 . This can be
related to the downstream velocity (using the depths), using the continuity
equation:
U1 y1 = U2 y2
(2.22)
If we assume that the wall friction is negligible, then the only force acting on
the control body is that due to the pressure difference and is given by
gy12 /2 gy22 /2 per unit mass. Applying the momentum equation we then
have:
⇥
y12 y22
g
= U1 y1 (U2 U1 )
(2.23)
2
2
If we now substitute for U2 from equation (2.22) into equation (2.23) we
obtain:
⇤
⌅
⇥
y1
g 2
2
U1 y1
1 =
y1 y22
(2.24)
y2
2
Dividing through by gy1 this becomes:
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U12
gy1
⇥
y1
y2
⇥
1
1 =
2
y22
y12
which contains a difference of two squares
can cancel
y1
y2
⇥
1
⇥
y12
y22
y22
y12
⇥
y1
y2
1=
from both sides. Noting also that
1
F r12 =
2
⇥
1
y12
y22
y1
+1
y2
⇥
1
U12
= F r12
gy1
(2.25)
y1
+1
y2
⇥
. Thus we
we obtain:
⇥
(2.26)
Equation (2.26) relates the upstream Froude number to the jump ratio
y2
y1 .
However, in reality, the flow (and the jump ratio) is driven by the upstream
Froude number, so we need the inverse relationship. This can be obtained
y2
by rearranging equation (2.26) to give a quadratic equation for y1 and solving
in the usual way. First, rearranging to obtain the quadratic equation:
1
2
y2
y1
⇥2
1
+
2
y2
y1
⇥
F r12 = 0
Equation (2.27) is a quadratic equation for
y2
y1
(2.27)
in terms of the upstream
Froude number, F r1 . This is solved in the usual way. The result is:
y2
=
y1
1
±
2
⇥
⇤
2
1 + 8F r1
2
(2.28)
Equation (2.28) is the general solution of quadratic equation (2.27).
However, note that one solution is unphysical because it is negative. Ignoring
this unphysical solution, the jump ratio is given by:
⇥
⇤
2
1 + 8F r1
y2
1
=
+
(2.29)
y1
2
2
Equation (2.29) tells us the jump ratio. It represents the only physical
jump ratio which satisfies both continuity and momentum. Notice that the
ratio is greater than unity provided F r1 1. In fact, as we shall see, this is
the only physical solution. However, the downstream velocity (and hence the
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downstream Froude number) can now be determined directly using the
continuity equation. First we relate the two Froude numbers:
F r22 = F r12
Next we substitute for
U2
U1
U2
U1
y1
y2
⇥
(2.30)
from equation (2.22)
F r22 = F r12
Finally, we substitute for
⇥2
y2
y1
y1
y2
⇥3
(2.31)
from equation (2.29):
F r22
= ⇤
8F r12
1+
8F r12
1
⇥3
(2.32)
We have now obtained solutions for the jump and Froude number ratios
which satisfy both the continuity and momentum equations. We shall now
see that the singularity necessarily leads to energy loss across the jump.
2.7.3.
Energy loss across a jump
In applying the momentum equation above, we neglected any viscous
forces on the walls. However, it is still possible to have dissipation of energy,
such as from turbulence. As before, we will consider the energy per unit
weight. The energy loss is:
⇥
⇥
U12
U22
hf = e1 e2 = y1 +
y2 +
(2.33)
2g
2g
As before this can be simplified using the continuity equation (2.22)
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U2 =
y1
y2
⇥
U1
(2.34)
⇥
U12
y12
⇥ hf = (y1 y2 ) +
1
2g
y22
⇥
⇥
U12 1
2
2
= (y1 y2 ) +
(y2 y1 )
2g y22
2
The difference of two squares (y2
take y2 y1 out as a factor:
hf = (y2
y12 = (y2
U12
2g
y1 )
1
y22
⇥
(2.35)
(2.36)
y1 )(y2 + y1 )) enables us to
(y2 + y1 )
⇥
1
(2.37)
We can now replace
U12
gy1
with F r1 :
= (y1
y2 ) +
✓
F r12
2
✓
y1
y22
◆
(y2 + y1 )
1
◆
F r12 can be replaced using equation (2.26):
✓
◆✓ 2
◆
y2 y1
y2 + y12 2y1 y2
=
4
y1 y2
Putting this over a common denominator y1 y2 and factorizing:
⇤
⌅
⇥
y 2 y1
hf =
y22 2y1 y2 + y12
4y1 y2
⇥
y 2 y1
2
=
(y2 y1 )
4y1 y2
⇥
(y2 y1 )3
=
4y1 y2
Lecture Notes
MACE 20121/2. Open Channel flow
(2.38)
(2.39)
(2.40)
(2.41)
(2.42)
22/32
Thus there energy is lost provided y2 > y1 which is always the case for
F r1 > 1. Thus we can only have a hydraulic jump and not a 'hydraulic drop'
as this would violate the 1st law of thermodynamics.
2.8. Flow over Weirs and under Sluice gates
2.8.1.
Introduction
In section 2.3 we found two types of behaviour as flow passes over a
bump, depending upon the upstream Froude number. However, provided the
acceleration is sufficient the flow can go from subcritical to supercritical. In
this case the critical point must occur at the crest of the bump, creating a
point in the flow with a known Froude number. This known Froude number is
exploited by weirs giving a known volume flow rate. There are two types of
weirs: broad and sharp-crested weirs. We shall look at both of these
separately.
2.8.2.
Broad-crested weir
Figure 2.15 A broad crested weir
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When we considered flow over bumps we considered the two contrasting
conditions in which the flow was subcritical everywhere (in which case the
level drops above the bump before increasing again) or supercritical
everywhere (in which case the level rises over the bump before dropping
again). Since the drop in height in sub-critical flow accelerates the flow, there
is another possibility, illustrated from the laboratory in figure 2.15, in which the
flow is initially sub-critical and so the level drops, but becomes critical over
the bump and thus continues to drop at the back of the bump. This is the
principle behind a broad-crested weir. Broad-crested weirs are used to both
measure the flow rate and to control it. This is seen in figure 2.16 which
shows a broad-crested weir in Wilmslow, Cheshire.
Figure 2.16 A broad-crested weir in Wilmslow Cheshire.
Figure 2.17 shows a schematic of a broad-crested weir. A subcritical
flow enters from the left and accelerates over the weir as expected for a
subcritical flow (see equation (2.11)). The flow becomes critical above the
weir and thus continues to accelerate (see equation (2.11) for supercritical
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flows). The depth of the weir is y and the depth of the flow upstream is
y + H . The depth of the flow above the weir is yc.
The depth above the weir can be obtained by applying Bernoulli's
equation along the surface from upstream to just above the weir.
U12
Uc2
+ (H + y) =
+ (yc + y)
2g
2g
U12
Uc2
+H =
+ yc
2g
2g
(2.43)
(2.44)
H
yc
U1
y
Figure 2.17 A broad-crested weir.
2
Since the Froude number upstream is small, U1 is small and can be ignored.
2
Furthermore the flow is critical above the weir so Uc = gyc. Thus equation
(2.44) becomes:
H=
yc
3
+ yc = yc
2
2
2
yc = H
3
(2.45)
(2.46)
Having determined the depth above the weir, we simply multiply by the known
critical flow speed (Uc = gyc ) to obtain the volume flow rate per unit width:
q = Uc yc =
p
gyc yc =
p
3
gyc2
(2.47)
Substituting for yc :
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q=3
1
2
2
3
⇥
1
3
(2g) 2 H 2
(2.48)
3
from which we see that q is proportional to H 2 . We can write the relationship
as:
q = Cw
where Cw = 3
1
2
2
3
⇥
1
3
(2g) 2 H 2
(2.49)
= 0.577
(2.50)
In practice, a more accurate relationship is obtained by replacing Cw with the
empirical relationship:
Cw =
2.8.3.
0.65
1+
(2.51)
H
Y
Sharp-crested weir
H!yc
H
Nappe
2 h
U1
y
1 y1
Figure 2.18 A sharp-crested weir
The situation is a bit more complicated when there is a sharp-crested
weir. The situation is illustrated in figure 2.18. As before the height of the
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weir is y and the depth of the flow upstream is y + H . The upstream flow is
assumed to be uniform and horizontal. Shown on figure 2.18 is a streamline
which passes from the upstream region to a point above the weir where the
flow is critical (point 2). The critical point is taken as being a distance h below
the original surface of the flow.
Applying Bernoulli's equation from point 1 to point 2 we have
(pa + g(H + y
y1 )) +
1 2
1
U1 + gy1 = (p2 ) + U22 + g(H + y
2
2
h)
(2.52)
⇥ pa +
1 2
U + g(H + y
2 2
h)
U12
U22
+ (H + y) ⇥
+ (H + y
2g
2g
(2.53)
h)
(2.54)
which simplifies to
U22 = U12 + 2gh
(2.55)
2
If we further assume that U1 is small (as for the broad-crested weir) and
that the depth of the Nappe above the weir is 23 H (as before), we can obtain
the volume flow rate per unit width as:
q=
⇤H
H
3
⇥
⇤H ⌅
2
U2 dh =
U1 + 2gh dh
(2.56)
H
3
2
Since U1 is small this can be simplified to:
⇤
⇥
⇥ 32 ⌅
1
3
2
H
q=
(2g) 2 H 2
3
3
⇥
1
3
2
= 0.81
(2g) 2 H 2
3
(2.57)
(2.58)
So, as for a broad-crested weir, the volume flow rate per unit width is
3
proportional to H 2 :
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q = Cw
2
3
⇥
1
3
(2g) 2 H 2
where Cw = 0.81
(2.59)
(2.60)
Again, for greater accuracy we use the empirical formula:
Cw = 0.611 + 0.075
H
Y
(2.61)
in practice.
We have shown, therefore, that weirs can be use as methods to calculate the
volume flow rate by simply measuring the upstream depth of the flow.
2.8.4.
Sluice gates
Figure 2.19: Sluice gate
We now turn our attention to sluice gates. Like weirs they can be used
as a method of measuring the flow as well as a method of controlling the flow.
An example of a sluice gate, taken from the Hydraulics laboratory is shown in
figure 2.19 above. Figure 2.20 shows a schematic of a typical sluice gate.
Here the depth upstream is h1 and the depth downstream of the gate is h2 .
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As before we analyse this by applying Bernoulli's equation along the surface
streamline:
1 2
1 2
U1 + gh1 =
U + gh2
2
2 2
(2.62)
Since the flow is not completely inviscid, equation (2.62) is an approximation.
We now relate the downstream velocity to the upstream velocity using the
continuity equation and then substitute into equation (2.62):
q = U1 h1 = U2 h2
⇥ 2g(h1
h2 ) = q
2
1
h22
1
h21
⇥
=q
2
(2.63)
h21 h22
h21 h22
2
Once again we have a difference of two squares h1
so the factor h1 h2 cancels to give:
⇥
(2.64)
h22 = (h1 + h2 )(h1
2gh21 h22
q =
h1 + h2
⇥
h2 2gh1
q=
1 + hh21
2
h2 )
(2.65)
(2.66)
In practice, h2 = 0.61d so the volume flow rate per unit width is given by:
q=
0.61d 2gh1
1 + 0.61 hd1
(2.67)
In deriving equation (2.67), we ignored viscous effects. Whist this is a
relatively small approximation for the situation depicted in figure 2.20, it is a
poor assumption in situations where there is a blockage downstream (as
shown in figure 2.21) because there will be viscous energy loss across the
blockage.
Hence equation (2.67) only applies when there is no such
blockage.
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h1
d
h2
Figure 2.20 A sluice gate
Figure 2.21 A sluice gate with a blockage
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2.9. Summary
Open channel flow
Compressible flow
Surface waves
Sound waves
Fr < 1 (subcritical)
M < 1 (subsonic)
Fr = 1 (critical)
M = 1 (sonic)
Fr >1 (supercritical)
M > 1 (supersonic)
Wave cone
Mach cone
Undisturbed region outside cone for
Fr > 1
Zone of silence outside cone for
M>1
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When relevant
Equation
z
=
x
Flow over a bump
Surface wave speed
Height rise across a hydraulic jump
Froude number downstream of a
hydraulic jump
Broad-crested weir:
Broad-crested weir:
v2
(gd v 2 )
y
c = gy 1 +
y
2
y2
=
y1
⇥
⇥
y
1+
2y
8F r12
F r22 = ⇤
1+
q = Cw
2
3
8F r12
⇥
1
0.65
Cw =
1+
2
3
⇥
H
Y
1
q = Cw
Sharp-crested weir
Cw = 0.611 + 0.075
Lecture Notes
1
⇥3
(2g) 2 H 3 2
Sharp-crested weir
Sluice gate
⇥
⇥
⇤
2
1 + 8F r1
2
1
+
2
q=
h
x
3
(2g) 2 H 2
H
Y
0.61d 2gh1
1 + 0.61 hd1
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