Pertemuan 19 - 22
Open Channel 2
VARIED FLOW IN OPEN
CHANNELS
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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.
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Sketch of downhill flow in an open
channel
In chapter before, 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
(there’s 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:
1.
2.
3.
4.
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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).
Specific Energy:
The combination
energy.
is called the specific
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.
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For given Q and E, usually have 2 allowed depths:
Subcritical and supercritical flow.
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”.
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(Examples a and b above have
different specific energy E)
Critical Depth and the Froude Number
At the turning point (the leftmost 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 won’t 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 < 1 thus move slower than gravity waves.
Flows with Fr > 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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Which will it be?
Flow over a Bump
or
As it turns out:
Left = subcritical
Right = supercritical
We’ll 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 (p=0):
For a channel of rectangular cross-section, again we have
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Substitute Q = V yb 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 < 1
supercritical: Fr > 1
Subcritical flow with dh / dx > 1
Supercritical flow with dh / dx > 1
if flow is subcritical
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dy / dx < 1
dy / dx > 1
if flow is supercritical
Example:
May want to know:
1. How does water
depth change?
2. Where does
jump occur?
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The Hydraulic Jump
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
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and
are the pressures at centroids of A1 and A2 )
… and that’s 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.
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