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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