VARIED FLOW IN OPEN CHANNELS

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Chapter 15 VARIED FLOW IN OPEN CHANNELS
Fluid Mechanics, Spring Term 2009
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Open Channels vs. Closed Conduits

• In a closed conduit there can be a pressure
  gradient that drives the flow.
• An open channel has atmospheric pressure at the
  surface.
• The HGL is thus the same as the fluid surface.

Sketch of downhill flow in an open channel
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In chapter 10, we looked at some uniform open
channel flows. Now we deal with varied flow
which is steady but nonuniform. (Flow is
constant in time, but velocity and depth may vary
along the flow). We will only deal with two very
simple cases here (theres much more in chapter
15), but these do illustrate the main points of
open channel flow.
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Energy equation applied to open channel
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• We make the following simplifications
• Assume turbulent flow (a 1).
• Assume the slope is zero locally, so that z1
  z2.
• Write pressure in terms of depth (y p / g).
• Assume friction is negligible (hL 0).

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Specific Energy
The combination is
called the specific energy.
For our example (no slope, turbulent, )
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The specific energy can be written in terms of
discharge Q V A (from continuity)
For a channel with rectangular cross-section, A
b y, (where b is the width)
For a given Q, we now have E in terms of y alone.
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Thus, for flat slope ( other assumptions) we
can graph y against E (Recall for given flow, E1
E2 )
Curve for different, higher Q.
For given Q and E, usually have 2 allowed
depths Subcritical and supercritical flow.
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Subcritical vs. Supercritical Flow
These 2 different types of flow are in fact
observed Example Flow past a sluice
gate. Subcritical Calm, tranquil
flow. Supercritical Rapid flow, whitewater.
(Examples a and b above have different specific
energy E)
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Critical Depth and the Froude Number
At the turning point (the left-most point of the
blue curve), there is just one value of
y(E). This point can be found from
It can easily be shown (but we wont do it here)
that at
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Define the Froude number
(Recall that the Reynolds number is the ratio of
acceleration to viscous forces). The Froude
number is the ratio of acceleration to
gravity. Perhaps more illustrative is the fact
that surface (gravity) waves move at a speed
of Flows with Fr lt 1 thus move slower than
gravity waves. Flows with Fr gt 1 move faster than
gravity waves. Flows with Fr 1 move at the same
speed as gravity waves.
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Flows sometimes switch from supercritical to
subcritical
(The switch depends on upstream and downstream
velocities our theory is not sufficient to
determine which type of flow the fluid
chooses) Gravity waves If you throw a rock
into the water, the entire circular wave will
travel downstream in supercritical flow. In
subcritical flow, the part of the wave trying to
travel upstream will in fact move upstream
(against the flow of the current).
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Flow over a Bump
Which will it be?
or
As it turns out Left subcritical Right
supercritical
Well derive this using the Bernoulli equation
for frictionless, steady, incompressible flow
along a streamline
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Apply Bernoulli equation along free surface
streamline (p0)
For a channel of rectangular cross-section, again
we have
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Substitute Q V z b into Bernoulli equation
To find the shape of the free surface, take the
x-derivative
Solve for dz / dx
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(from last page)
Since subcritical Fr lt 1
supercritical Fr gt 1
Subcritical flow with dh / dx gt 1
dy / dx lt 1 Supercritical flow with dh / dx gt 1
dy / dx gt 1
if flow is subcritical if flow is supercritical
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The Hydraulic Jump

• Example
• May want to know
• How does water depth change?
• Where does jump occur?

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A look at the hydraulic jump in greater detail
Note that there is a lot of viscous dissipation (
head loss ) within the hydraulic jump.
So our previous analysis does not apply to the
jump (and unless we know V1, V2, y1, y2, and Q,
we cannot determine hL ).
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It turns out that it is more useful to apply the
momentum eqn.
Why? Because there is an unknown loss of energy
(where mechanical energy is converted to
heat). But as long as there is no friction along
the base of the flow, there is no loss of
momentum involved.
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Momentum balance
The forces are hydrostatic forces on each end
(where and are the pressures at
centroids of A1 and A2 )
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and thats actually all for this problem For
example, if y1 and Q are given, then for
rectangular channel
is the pressure at mid-depth.
So entire left-hand side is known, and we also
know the first term on the right-hand side. So we
can find V2.
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There are obviously many more applications. For
example, now that we have V2 we could find hL (by
using the energy equation) But this is enough
for this course.
The End