Fluid Mechanics 4

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Fluid Mechanics 4

• Laminar Flows
• Professor William J Easson

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Shear Stress in Fluids

• In liquids, molecular bonds provide forces
  between layers
• In gases, the interaction between layers is due
  to collisions

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Shear Stress
Where F is in Newtons (N), A is area in metres
(m) and ? is the shear stress (N/m2). Large
velocity gradients lead to larger shear stresses,
hence
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Viscosity
If the variation is linear, then a constant
coefficient may be introduced, called the
viscosity.
By re-arrangement, the units of viscosity are
Ns/m2, which in Standard units is kg/ms, and is
often shown as Pa s.
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Viscosity values

• Water 1.005 x 10-3 Pa s
• Air 1.815 x 10-5 Pa s
• Lubricating oil 0.26 Pa s

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Non-Newtonian Fluids
Plastic
Newtonian
Dilatant
Pseudo-plastic
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Viscosity (cont)

• Viscosity varies with temperature
• in a gas increases with increasing temperature
• in a liquid decreases with increasing temperature
• We will only deal with Newtonian fluids in this
  course, so viscosity will always be constant.

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Flow in a viscous fluid
Consider two infinitely wide parallel plates
yc
? d?
pdp
p
?
y0
Pdydz - (pdp)dydz (? d?)dxdz - ?dxdz 0
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Differential equation of motion of a fluid
pdy - (pdp)dy (? d?)dx - ?dx 0
-dpdy d?dx 0
d?dx dpdy
We already know that for a Newtonian Fluid
Substituting
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Solution of flow between two flat plates (Couette
flow)
The differential equation may be solved by
integration
Hence
And a further integration wrt y yields
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Boundary conditions
Due to molecular bonding between the fluid and
the wall it may be assumed that the fluid
velocity on the wall is zero
u0 at y0 u0 at yc
This is known as the no-slip condition.
To satisfy the first boundary condition, B0
Then the second b.c. gives
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Quadratic velocity profile for flow in a channel
Substituting the values for A and B into the
previous equation gives the quadratic equation
For a long, straight channel, of length l, p
decreases with length at a constant rate, so
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Graph of velocity profile
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Volume flow rate
To calculate the volume flow rate, integrate from
y0 to yc
yc
dy
y0
per unit width (z direction)
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Maximum and mean velocity
Max velocity occurs at yc/2, the centre of the
channel
Mean velocity is gained by dividing the flow rate
by the channel width
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Eulers Equation
Flow along a streamtube of area A with no
viscosity
?
Forces along streamline
pressure gravity inertia (or Fma)
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Eulers equation (cont)
Dividing through by ?Ads
(We will come back to this equation later)
By the chain rule, the time derivative of u,
which is a function of both s and t, may be
expressed as
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Euler and Bernoulli
Eulers equation is independent of time, so for
Eulers equation
For an incompressible fluid, integrating along
the streamline,
Bernoullis equation
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Navier-Stokes equations

• So far we have separately considered flow
• in one dimension affected by pressure and gravity
• in one dimension affected by pressure and
  viscosity
• Need three dimensions and all forces in order to
  provide a full solution for any general flow
  problem
• The following is not rigorous- see Batchelor for
  a rigorous derivation

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Eulers equation (reminder)
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Navier-Stokes equations
Looking back to Eulers equation with
unsteadiness, the gravity term is simply the
component of gravity, gs. Introducing viscosity
as well gives 3 similar equations
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Navier-Stokes equations

• There is no general solution to the N-S equations
• Some analytical solutions may be obtained by
  simplification
• The equations may be written in vector (div/grad)
  notation

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Channel flow
Flow between two horizontal flat plates, as in
1st lecture
1. Horizontal - no g component
2. Parallel plates - u is constant along the
flow, so
3. No velocity variation in z direction - walls
infinitely far away
4. Steady flow, so
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Channel flow
The n.s. equations therefore reduce to
Which may be solved as before.
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Laminar pipe flow 1
Consider steady laminar flow in a horizontal
circular pipe of radius, a.
a
For flow in a pipe, cylindrical polar coordinates
x, r, ?, are most useful.
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Laminar pipe flow 2
The Navier-Stokes equation
( 0 for steady flow in a straight pipe)
may be expressed as three components in x, r, ?,
but for steady flow in the x-direction, we only
need one component
And we will drop the gravitational term for a
horizontal pipe.
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Laminar pipe flow 3
Integrating twice wrt r, gives
The constant A must be zero, because at the
centre of the pipe ln(0) is infinite. The bc at
the wall, u0 at ra gives a value for B.
Poisseuille equation
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Laminar pipe flow 4
The discharge may be found by integrating annuli
of width dr from the centre to the edge of the
pipe, with the flow through each annulus being
dqu(r)2?rdr.
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Laminar flow 5
Mean velocity in pipe
Velocity at centre of pipe (r0)
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Laminar flow 6 - Friction
The pressure loss in a straight pipe is due to
friction, and we can re-arrange the discharge
equation
Expressing this in terms of head loss,
DArcys equation
where, for laminar flow, f16/Re