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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
AF .X!
Where F is in Newtons (N), A is area in
metres (m) and Xis the shear stress
(N/m2).
Large velocity gradients lead to larger
shear stresses, hence
y
u
x
xwX
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Viscosity
y
u
xx! QX
Ifthe variation is linear, then a constant coefficient may
be introduced, called the viscosity.
By re-arrangement, the units of viscosity are Ns/m2,
which inSt
andard units is kg/ms, and is of
ten 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
X
y
u
x
x
Newtonian
Dilatant
Plastic
Pseudo-plastic
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Viscosity (con
t)
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 beconstant.
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Flow in a viscous fluidConsidertwo infinitely wide parallel plates
y=0
y=c
x
y z
pp+dp
X+ dX
X
Pdydz - (p+dp)dydz + (X+ dX)dxdz - Xdxdz = 0
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Differential equation of motion
of a fluid
pdy - (p+dp)dy + (X+ dX)dx - Xdx = 0
-dpdy + dXdx = 0
dXdx = dpdy
dx
dp
dy
d
We already know that for a Newtonian Fluidyux
x!
dx
dp
dy
ud
2
2
!Substituting
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Solution of flow between two flat
plates (Couette flow)The differential equation may be solved by integration
dydx
dp1dydy
ud2
2
!Hence
y
dx
dp1
dy
du!
nd a further integration wrt y yields
By2
y
dx
dp1u
2
!
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Boundary conditions
Due to molecular bonding between the fluid and the wall
it may be assumed thatthe fluid velocity on the wall is zerou=0 at y=0
u=0 at
y=cThis is known as the no-slip condition.
To satisfy the first boundary condition, B=0
Then the second b.c. givesAc
2
c
dx
dp
10
2
!
2
c
dx
dp
1A !
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Quadratic velocity profile for
flow in a channelSubstituting the values for A and B into the previous
equation gives the quadratic equation:
ycydx
dp
2
1u 2 !
For a long, straight channel, of length l, p decreases
with length at a constant rate, so
l
p
dx
dp !
ycyp2
1u 2
(!
l
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Graph of velocity profile
0
0.2
0.4
0.6
0.8
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
y
u
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Volume flow rate
To calculate the volume flow rate, integrate from y=0 to y=c
y=0
y=c
dy
udydq ! dyyyc2p
q
c
0
2
(
!l
l12
pcq
3
! per unit width (z direction)
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Maximum and mean velocity
Max velocity occurs at y=c/2, the centre ofthe channel
lu p
8c
2
max !
Mean velocity is gained by dividing the flow rate by the
channel width
q/cu !
!! max
2
u3
2
12
pcu
l
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Eulers Equation
Flow along a streamtube of area A with no viscosity
p
p+dpds
mg
u
Forces along streamline: 0dt
duAdssgcos()AdAdp !
pressure + gravity + inertia (or F=ma)
U
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Eulers equation (con
t)
Dividing through by VAds:
0
dt
du
ds
dzg
ds
dp
1!
By the chain rule, the time derivative of u, which is a function
of both s and t, may be expressed as:
t
s
s
u
t
u
dt
du
x
x
x
x
x
x
!
s
uu
t
u
dt
du
x
x
x
x!
(We will come
backto this
equation later)
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Euler and BernoulliEulers equation is independent oftime, so for 0!
x
x
t
u
0ds
duu
ds
dzg
ds
dp
1!
For an incompressible fluid, integrating along the streamline,
const
z2g
u
g
p 2!
Eulers equation
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 andviscosity
Need three dimensions and all forces in
orderto provide a full solu
tion for anygeneral flow problem
The following is not rigorous- see Batchelor
for a rigorous derivation
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Eulers equation (reminder)
0
dt
du
ds
dzg
ds
dp
1!
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Navier-St
okes equationsLooking backto Eulers equation with unsteadiness, the
gravity term is simply the component of gravity, gs.
Introducing viscosity as well gives 3 similar equations:
dtdu
zu
yu
xu
xpg
2
2
2
2
2
2
x !
xx
xx
xx
xx
dt
dv
z
v
y
v
x
v
y
pg
2
2
2
2
2
2
y !
x
x
x
x
x
x
x
x
dt
dw
z
w
y
w
x
w
z
pg
2
2
2
2
2
2
z !
x
x
x
x
x
x
x
x
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Navier-St
okes 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:
dt
udupg
2!
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Channel flow
dt
du
z
u
y
u
x
u
x
pg
2
2
2
2
2
2
x !
x
x
x
x
x
x
x
x
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 0!x
x
x
u
3. No velocity variation in z direction - walls infinitely far away
4. Steady flow, so 0!x
x
t
u
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Channel flowThe n.s. equations therefore reduce to:
dxdp
dyud 2
2
!
Which may be solved as before.
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Laminar pipe flow 1
xr U
For flow in a pipe, cylindrical polar coordinates x, r, U, are
most
useful.
Consider steady laminar flow in a horizontal circular pipe
of radius, a.
a
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Laminar pipe flow 2The Navier-Stokes equation:
dt
udupg
2 !
may be expressed as three components in x, r, U, but for steady
flow in the x-direction, we only need one component:
0r
ur
rr
1
x
pg
x
!
x
x
x
x
x
x
(= 0 for steady flow
in a straight pipe)
And we will drop the gravitational term for a horizontal pipe.
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Laminar pipe flow 3
x
p
r
ur
rr
1
x
x!
x
x
x
x
Integrating twice wrt r, gives:
BrAx
p
4
1u 2
x
x! lnr
The constantA must be zero, because atthe centre ofthe pipe
ln(0) is infinite. The bc atthe wall, u=0 at r=a gives a value for B.
22x
p
4
1u ar
x
x!Poisseuille
equation
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Laminar pipe flow 4The discharge may be found by integrating annuli of width dr
from the centre to the edge ofthe pipe, with the flow through
each annulus being dq=u(r)2Trdr.
l8
paq
4
!l
p
x
pwhere
!
x
x
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Laminar flow 5
Velocity at centre of pipe (r=0):
l8
pau
2
!Mean velocity in pipe:
u24
pau
2
max !! l
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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:
4d
q128p
l
!
Expressing this in terms of head loss,g
ph !f
du16
2gdu4
gdu32h
2
2V
QV
llf !!
g2
u
d
4h
2
flf !DArcys equation where, for laminar flow, f=16/Re
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