SURFACE RESISTANCE

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Chapter 9 SURFACE RESISTANCE
Fluid Mechanics, Spring Term 2007
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So far in this course, we have used
1.) Conservation of mass 2.) Conservation of
momentum 3.) Conservation of energy These
laws are valid for any material. We have not yet
specified that were dealing with a fluid rather
than a solid.
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Shear strain (in one dimension)
y
r2
P2
P2
Dy
Particles P1 and P2 are moved in the x-direction
by displacements r1 and r2.
r1
P1
P1
x
The shear strain e is defined as
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Shear strain in 3 dimensions
y
r2
P2
P2
Dy
r1
P1
P1
x
The displacements are in x The difference in
displacement is taken in y.
(same as before, but clearer what we mean)
In 3-D, there are 9 strains. We limit ourselves
to 1-D.
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In fluids, the strain rate is more useful than
the strain
y
Vx
DVx
Fluid flow
Dy
y
x
Slope at any point is equal to the strain rate at
that point.
Example Flow down an incline.
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Any material deforms when subjected to surface
stresses
Solid (elastic)
Fluid (viscous)
.
Strain rate e
Shear stress t
An elastic material deforms (instantaneously) in
response to a change in stress and then keeps
that shape. A fluid continues to deform as long
as the stress is applied.
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For a fluid in 1-D, we can write
This is what makes a fluid a fluid ! The
parameter m is called the viscosity. The
viscosity depends on the material. It may also
depend on temperature or t (or other
parameters). The above equation is based only on
observations, not on fundamental physics.
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Viscous stresses are surface forces per unit area.
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From the stress-velocity relation, it is clear
that viscous stresses are due to differences in
velocity (i.e., internal deformation of the
fluid).
Viscous stresses are always related to viscous
dissipation. Deforming fluids always experience
viscous resistance, and thus convert mechanical
energy to internal energy (heat).
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Weve seen viscous stresses in previous examples
Shear stresses on fluid in a pipe
We could pick a control volume within the fluid.
The outside fluid then exerts a shear stress on
the cv.
The shear stresses are part of the complete
momentum force balance.
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From last lecture Energy grade line (EGL) and
hydraulic grade line (HGL).
EGL and HGL drop along pipe due to viscous
dissipation.
So far, we were always given the head loss, e.g.
This can be related to heat generation in the
energy equation.
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Another example of viscous dissipation
t t1
t t2
Flow eventually stops.
Initially turbulent flow (e.g. after mixing).
We know that kinetic energy is converted to
heat. But if the control volume is equal to the
entire tank, we have no information about the
heat generated. (Notice that no energy or
momentum enters or leaves the tank).
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Recall in chapter 5 we transformed the continuity
equation from an integral equation to a
differential equation that applies at every
point. Rather than do the same for the momentum
equation, we will derive a differential momentum
equation for a specific application
Steady flow down an inclined plane. We use a
control volume but shrink it to small size.
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For steady flow, inflow and outflow of momentum
are equal
(force balance in s-direction)
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(No Transcript)
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The force balance then becomes
Using the shear stress - velocity relation, we
get
This is the differential form of the momentum
equation for steady flow in one dimension (for
spacial variables s along flow and y across flow).
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This equation can be applied to various flow
problems
e.g., stationary bottom plate and moving top
plate
There is no pressure gradient and no gravity
acting in the flow direction, so the equation
becomes
Boundary conditions
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Example 2 Flow down an inclined plane
(where )
Again, no pressure gradient in flow direction
but gravity!
Integrate twice
Boundary conditions
(after some algebra)
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Example 3 Flow between stationary parallel
plates
(both pressure gradient and gravity in flow
direction)
Again, integrate twice
Boundary conditions
Parabolic velocity profile (very similar to flow
in pipes!)
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I cannot teach a course in fluid mechanics
without at least showing the Navier-Stokes
equation, so here it is
It is quite similar to the equation we derived,
but more general (3-D includes accelerations).
The Navier-Stokes equation is the differential
form of the momentum equation with viscous
stresses included. We wont use it because it is
too difficult to solve
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Boundary Layers
Lets have another look at the Navier-Stokes
equation
a) b) c) d)
e)
When v is very small, term b) can be neglected
(it is quadratic in v). The dominant balance is
then hydrostatic (c against e).

• Local acceleration
• Advective acceleration
• Pressure gradient
• Viscous stresses
• Body forces

If v is small and the flow is steady (term a
0), then the channel flow solutions we just
calculated are valid.
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However, when v is larger, term b) can no longer
be left out. Term b) is the term that leads to
turbulence.
Near the boundaries, v is small and term b) may
be neglected (laminar flow).
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• A bit of scaling (we skipped this in chapter 8)
• Introduce typical scales for the variables, such
  as
• Length scale L pipe diameter (or radius)
• Distance over which velocity changes.
• Velocity scale V mean flow velocity (or
  maximum)

Concentrating on terms b) and d), we find
b)
d)
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To compare the terms, we take the ratio of b) to
d)
This estimate of the relative importance of the
two terms has great relevance and has its own
name
Reynolds number
When Re is very small, term b) can be neglected
Laminar flow! When Re is very large,
term d) can be neglected Fully
turbulent flow! When Re is around 1, both terms
are important.
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Experiments show that Re lt 1000 laminar
flow Re gt 1000 turbulent flow The value of
1000 is approximate and depends on the problem to
be solved and on the choice of scales.
Small systems are less turbulent than large ones
(e.g., flow in very thin pipes is laminar). Slow
flows are laminar, fast flows turbulent. More
viscous materials are less turbulent (e.g., oil
in a pipeline is less turbulent than water in the
same pipeline).
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So how does the Reynolds number relate to
boundary layers?
Recall that the length scale L comes from the
spatial derivatives. Observations show that the
distance over which v goes from 0 to v0 is the
width of the boundary layer d, which is thus a
better choice for L for this problem.
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With a bit of arm-waving, it can be shown that
for this specific problem the boundary layer
thickness can be estimated from
to be
where the system changes from laminar to
turbulent at about Recritical 5 (not 1000)
Dont worry about the details of this every
problem seems to have its own scaling, and I
think the book does not explain things very well.
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The important result is that flow at low Reynolds
number tends to be laminar, at large Re
turbulent. The effective Re increases as one
moves away from the boundaries.
Logarithmic scales
Linear scales