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Copyright 2011 Cneyt Arslan, All rights reserved.

VISCOSITY of LIQUIDS

Einstein suggested that : B D B T where : B= mobility of liquid particle

D= diffusion coefficient

Since diffusion is an activated process, G* (minimum activation energy)

must be supplied to the system. Fluidity is thermally activated. Moreover,

fluidity is the inverse of viscosity :

*

expG

AR T

where :

: viscosity (Poise)

T : temperature (K)

A : constant

R : gas constant (cal deg1 mol1)

G* : viscosity activation energy (cal mol1)

Albert Einstein1879-1955


The most realistic approach for determining the A is the

Eyring Theory:

No : Avogadro number (6.023 1023 g mol1)

Vm : molar volume

h : Planck constant (6.624 1027 g cm2 s1)

Since G* cannot be determined directly:

m

o

V

hNA

G* = 3.8 R Tb.p. may be used for rough approximations.

After some re-arrangements

T

T

V

pb

m

..3 8.3

exp104

For molecular liquids where bonding forces are Van der Waals type :

G* 0.41 EevaporationHowever, these two equations are not valid for liquid metals, polymers,and chain-like molecules.

Amadeo Avogadro1776-1856Max Ludwig Planck

1858-1947
http://enlarge%28%27planck_7.jpeg%27%29/

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H2O

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LAMINAR FLOW

We have defined the termfluid flowas the flow of parallel plates, creating a

velocity profile, caused by the shear stress applied to the fluid.

This velocity profile is defined as THE REASON or THE POTENTIAL of the

momentum transfer from one plate to another.

In this section, we will derive ordinary differential equations (ODEs) for the

momentum transfer in specific flows, such as: flow between parallel plates, flow

in an open channel, flow on an inclined plane, flow in a tube, etc.

These derivations will utilize the concepts of viscosity and momentum balance

to aid you to derive even more complex formulae.


MOMENTUM BALANCE

Momentum balance is applied to a small control volumeof the fluid in order to

develop a differential equation, describing the whole flow system.

Solutions of these differential equations, when fitted to the physical restrictions(boundary conditions), turn into algebraic relations which can be used to determine

the engineering characteristics of the system.

These solutions present the velocity distributions, from which the characteristics

can be developed such as, the shear stress on the fluid-solid interface.

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For a steady-state flow:

0

systemthe

onactingForces

OUT

MomentumofRate

IN

MomentumofRate

Momentum enters into a system either due to the

viscosity, or

general fluid motion.

Forces affecting the balance are pressure and/or gravitational forces.

Momentum balance is in fact, a balance of forces.

We will be interested in RATE of MOMENTUM (both getting into and coming outof system).

The units ofmomentum:ML/t, rate of momentum:ML/t2

(M: mass, L: distance, t: time).


FLOW BETWEEN PARALLEL PLATES

fully developed flow region

Vx

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Due to the Viscosity:Rate of momentum IN............................(L W)yx|y(through the surface at y)

Rate of momentum OUT.........................(L W) yx|y+y(through the surface at y+y)

Due to the Fluid Motion:

Rate of momentum IN...........................(W y Vx) Vx|x=0(through the surface at x = 0)

Rate of momentum OUT.........................(W y Vx) Vx|x=L(through the surface at x = L)

Pressure force acting on the fluid at x = 0 ........... y W Po

x = L ........... y W PL

MOMENTUM BALANCE


Since the momentum rates, in & out of the system, due to the fluidmotion, are equal to each other;

Dividing both sides by (L W y) and taking the limit when y goes to zero,gives:

L

PP

dy

dLo yx

differential equation.

Boundary conditions can be determined on the axis line (y=0) and on the

plate walls (y = ):

B.C. # 1 at y = 0 yx = 0B.C. # 2 at y = Vx = 0

yx y yx y+y 0 L( ) | ( ) | ( ) 0 L W L W W P P

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Shear stress distribution is therefore;

yL

PP Loyx

L

PPyV LoX

22

2

1

and the velocity profile equation (parabolic) for a Newtonian fluid is:


L

PPV LoX

2

2max

L

PPdyVV Lo

y

XX

3

1

2

0=

Other properties of the system can be easily derived as:

Maximum velocity:

Average velocity:

Volumetric Flowrate:

3

0 L2

3

W P PQ

L

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FLOW in an OPEN CHANNEL

fully developed flow region


22 xa

ag

x

P

Similar to the case of parallel plates:

For the most general case, where: x >> a:

x

g a

dy

Vd

2

2

Boundary conditions:

# 1: Vy = 0 when y = 0

# 2: dV/dy = 0 when y = a

0 < y < a

22 yayx

aVx

2

g

meaning that the velocity profile is parabolic and the maximum velocity

being reached at the liquid surface.

Taking the integral gives:

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FLOW on an INCLINED PLANE


Due to the Viscosity:

Rate of momentum IN.............................(L W)xz|x(through the surface at x)

Rate of momentum OUT.........................(L W)

xz|x+x(through the surface at x+x)

Due to the Fluid Motion:

Rate of momentum IN.............................(W x Vz) Vz|z=0(through the surface at z = 0)

Rate of momentum OUT.........................(W x Vz) Vz|z=L(through the surface at z = L)

Gravitational force acting on the fluid..........(L x W) g cos

MOMENTUM BALANCE

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Substituting all these terms in Momentum Balance equation:

L W xz|x L W xz|x+x + (W x ) ( Vz2|z=0 Vz

2|z=L )

+ L x W g cos = 0

Since we are in a region whereVz is not affected by the entrance & exit

point disturbances, it is independent of z. Thus, 3rd and 4th terms cancel

each other.

Dividing both sides by (L W x) and taking the limit when x goes to zero,gives:

coslim0

gx

xxzxxxz

x

cosgdx

d xz


1cos Cxgxz Taking the integral gives:

Momentum Flux Equation(defining shear stress distribution)

The integral constant C1 can be eliminated by solving at the

boundary condition # 1:xz = 0 at x = 0. cosxgxz

If the fluid is Newtonian,then:

dx

zdV

xz

solving simultaneously:

xg

dx

dVz cos

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Taking the integral gives:2

2

2 Cx

g

Vz

cos

The integral constant C2 can be eliminated by solving at another

boundary condition # 2: Vz = 0 at x = , which gives:

2

2

cos

2

gC

substituting this back:

22

2

xgVz

1

cos

Velocity Profile Equation


Other properties of the system can be easily derived as:

Maximum velocity:

Average velocity:

Volumetric Flowrate:

cos

2

2max gVz

3cos1

0x

2

gdxVV z

3

cos

3WgWVQv

Coming up next... Problem... (about Copper Smelting ?!...)

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Problem: In copper smelting operations, slag is passed over matte in

order to recover all the copper it contains. Process is carried out in

reverberatory furnaces of 80 ft long and 30 ft wide.If the slag layer flows with a volumetric rate of 80 ft3/h over a stationary

matte phase (a major ! simplification) calculate:

a) the velocity distribution profile within the 2 ft high slag layer, and

b) percentage of the slag that remains in the furnace twice the mean

residence time or longer.

Solution: Since the flow is laminar,

3

cos

3WgWVQv

gives the volumetric flow rate. Arranging this equation and multiplying

both sides with 3/2 results in:

2

3

3

cos

2

3 2

g

W

Q


Thus,

2

2

3

x

W

Q

Vz 1

Substituting numerical values:

213

41

ft2ft302

hft803 xVz

2

2

412

22 xxVz

finally,

where, x is the distance (ft) from the top andVz is the velocity of that point.

Right side of this equation

22

2

xgVz 1

cosequals to the first term of

2

3

3

cos

2

3 2

g

W

Q

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This portion remains in the

furnace 2.

SLAG

Matte

x = 0

x =

x = x' Vz = Vave/2

b)Mean residence time

where L is the furnace length (ft), and is themean residence time(h).Thus, the portion of the slag that remains in the furnace for 2,

corresponds to the slag layer between x = x' and x = .x' is the point where Vz = Vave/2 .

aveV

L

W

QVaveSince

1hft3

2

230

80

2

1

zV at x = x'


2

2

2xVz

Substituting this Vz value back into the velocity profile equation:

gives x = 1.63 feet (from the top)

therefore,

The ratio is (2 1.63) / 2 = 18.5 %

SLAG

Matte

x = 0

x = 2

x=1.632 1.63

2

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