new fluids lecture 05 - pages.mtu.edu

9/13/2010

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Microscopic Balances

We have been doing microscopic h ll b l th t ifi t

dzshell balances that are specific to whatever problem we are solving.

dx

dy

We seek equations for microscopic mass, momentum (and energy) balances that are general.

© Faith A. Morrison, Michigan Tech U.

•equations must not depend on the choice of the control volume,

•equations must capture the appropriate balance

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Continuity Equation

n̂dSS

microscopic mass balance written on an arbitrarily shaped volume, V, enclosed by a surface, S

V

vvv

vvv zyx

zyxz

vy

vx

vt zyx

vvt

Gibbs notation:

© Faith A. Morrison, Michigan Tech U.2

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Equation of Motion

n̂dSS

microscopic momentum balance written on an arbitrarily shaped volume, V, enclosed by a surface, S

V

gPvvt

v

Gibbs notation: general fluid

gvPvvt

v

2Gibbs notation:

Newtonian fluid

Navier‐Stokes Equation


Problem‐Solving Procedure ‐ fluid‐mechanics problems

1. sketch system

amended: when using the microscopic balances

2. choose coordinate system

3. simplify the continuity equation (mass balance)

4. simplify the 3 components of the equation of motion (momentum balance) (note that for a Newtonian fluid, the equation of motion is the Navier‐Stokes equation)

5. solve the differential equations for velocity and pressure (if applicable)

6. apply boundary conditions

7. calculate any engineering values of interest (flow rate, average velocity, force on wall)


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EXAMPLE I: Flow of a Newtonian fluid down an inclined plane x

z

xz

fluid

xvz

air

singgx

cosggz

g

cos

0

sin

g

g

g

g

g

g

z

y

x

z



6

www.chem.mtu.edu/~fmorriso/cm310/Navier.pdf

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7


cross-section A:A

r z

r

z

EXAMPLE II: Pressure‐driven flow of a Newtonian fluid in a tube: Poiseuille flow

z

L vz(r)

•steady state•well developed•long tube

Rfluid

g


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9



10


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Poiseuille flow of a Newtonian fluid:

PP

22

0

2)(

)(

rPPgLR

rL

PPr

PzL

PPzP

L

Lorz

Lo

1

4 R

r

L

PPgLRrv Lo

z


2

2

0 0R

R

z

z

drdr

drdrv

vaverage velocity

Engineering Quantities of Interest(tube flow)

0 0

drdr

volumetric flow rate z

R

z vRdrdrvQ 22

0 0

z‐component of force on the wall dzRdF

L

Rrrzz

0

2

0


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A

zdA

dAvn

v

ˆ

average velocity

Engineering Quantities of Interest(any flow)

A

volumetric flow rate

A

dAvnQ ˆ

z‐component of force on the wall dAIpneF

surfaceatA

zz ˆˆ



Pz

L

PPzP Lo)( 0

R

r

L

PPgLRrv

rL

PPr

L

Loz

Lorz

14

2)(

22

L

PPgLRQ Lo

8

4


Hagen‐Poiseuille Equation**

14

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2 R

zav rdrdrvv

14

2

2

0 0

22

0 0

Lo

RLo

PPgLR

rdrdR

r

L

PPgLR

2

8

max,zv

L



-1

-0.5

0

0 0 25 0 5 0 75 1z

Lpp

pp

0

0

1

1.5

2

av

z

v

v

0 0.25 0.5 0.75 1L

z


0

0.5

0 0.25 0.5 0.75 1

R

r

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vz/<v>

EXAMPLE II: Pressure‐driven flow of a Newtonian fluid in a tube: Poiseuille flow


Reynolds Number

Dv velocityaveragev

densityFlow rate

Data may be organized in terms of two dimensionless parameters:

Fanning Friction Factor

DvzRe

lengthpipeL

droppressurePP

viscosity

diameterpipeD

velocityaveragev

L

z

0

1


Pressure Drop

2

0

21

41

z

L

vDL

PPf

18

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Data correlation for friction factor (P) versus Re (flow rate) in a pipe

Re

16

(Geankoplis)© Faith A. Morrison, Michigan Tech U.

Moody Chart

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What is the Fanning Friction Factor for Laminar Flow?

Hagen-Poiseuille E i

PPRvRQ L 0

42

EquationL

vRQ z

8

Re

16161

41

0

DL

PPf

L

TRUE


Re21 2

Dv

vDL z

z

20

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Turbulent flow generates much more friction than themuch more friction than the (unrealizable) laminar flow at

the same flow rate.

(Geankoplis)© Faith A. Morrison, Michigan Tech U.

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Long‐chain polymers can “laminarize” the flow and

reduce drag.


Bird, Armstrong, Hassager, Dynamics of Polymeric Liquids, Wiley, New York 1987. 22

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The hose connecting the city water supply to the washing machine in a home burst while the homeowner was away Water spilled out of the ½ in pipe for 48 hours before the problem was noticed by a neighbor and the water was cut off.

A problem from real life: REDUX

How much water sprayed into the house over the 2‐day period?

The water utility reports that the water pressure supplied to the house was approximately 60 psig.


Home flood: the cold‐water feed to a washing machine burst and was unattended for two days

REDUX


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Discussion:

How do we calculate the total amount of water spilled?

REDUX


Discussion:


What determines flow rate through a pipe?

REDUX


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Discussion:


What determines flow rate through a pipe?

REDUX

What information do we need about the system to calculate the amount of water spilled over two days?


House flood problem

REDUX

Solution Strategy:

•Apply the laws of physics to the situation•Calculate the velocity field in the pipe

(will be function of pressure)•Calculate the flow rate from the velocity field


•Calculate the flow rate from the velocity field (as a function of pressure)

•Calculate the total amount of water spilled

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cross-section A:A

r z

r

z

The Situation:

Steady flow of

REDUX

z

L vz(r)

water in a pipe

Rfluid


Next step: perform balances on flow in a tube

Because flow in a tube is a bit complicated to do as a first problem (because of the curves), let’s consider a somewhat simpler problem first.

REDUX


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Non‐Newtonian Fluids – An Introduction


31

The Weissenberg effect is when a viscoelastic, non‐Newtonian fluid will climb a rotating shaft.

Newtonian Fluids

121

dv

Newton’s Law of Viscosity

Newtonian

slope = viscosity

21

2

21 dx

2

1

x

v


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Non‐Newtonian Fluids

shear‐thinning or pseudoplastic

slope = o

21

shear‐thickening or dilatant

Bingham plastic

Newtonian

2

1

x

v

o


What is the definition of viscosity for Non‐Newtonian Fluids?

1v

2x

)( 21 xv

2

1

x

v

1x

)( 21 xv

21(NOTE on coordinate


system: Viscosity definition is written for shear flow in x1 direction and gradient in x2direction)

34

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loglog

typical polymer behavior

Viscosity (Greek letter eta)The changes in

viscosity with shear t l th

oozero‐shear viscosity (Newtonian plateau)

shear thinning

rate are so large they must be plotted log‐log

loglog


shear rate (gamma dot)

2

1

x

v

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In addition, for many polymers there are shear‐induced

NORMAL (perpendicular) forces.

Morrison, M

ichigan

Tech U.

A

dA21

force on y‐surface in z‐direction

force on 2‐surface in2‐direction

A

dAP22 © Faith A. M

x1

x2H

)( 21 xv

36

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Power‐Law Model (Ostwald‐deWaele Model)

1

1

1 dvdvn

(does notmodel

2

1

2

121 dxdx

m

m or K = consistency index (m = for Newtonian)n = power‐law index (n = 1 for Newtonian)

normal stresses)

rateshearx

v

2

1


What does the power‐law model predict for viscosity?

1

12121

n

d

dvm

d

2

1

1

2

121 dx

dv

dx

dvm

n

2

2

1dx

dxdv

On a log‐log plot, this would give a straight line:

d


2

1log1loglogdx

dvnm

Y B M X

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Power‐Law Fluid

1v

21

Non‐Newtonian viscosity

2x

log

Newtonian

shear thickening

1,1 nm n

1,1 nm n


shear thinning 1,1 nm n

log39

Where do we use the power‐law expression?

e.g., Poiseuille flow in a tube:

d

r

L

PPgL Lorz

2

dr

dvzrz

Newtonian

non‐Newtonian, power‐law

dr

dv

dr

dvm z

n

zrz

1

solve for vz(r)


1‐direction = r2‐direction = z

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EXAMPLE III: Pressure‐driven flow of a Power‐law fluid in a tube

z cross-section A: A

r

r

•steady state•well developed•long tube

z

L vz(r)

Calculate velocity and stress profiles

g R

fluid


Calculate the velocity field forPoiseuille flow of a power‐law fluid:

r

PPgLdvdvm Loz

n

zrz

1

Ldrdrmrz

2

rr

vz


rdr

dvm

dr

dv

dr

dvm

n

zz

n

z

1

Solve for vz(r)

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Boundary Conditions: ?


Velocity fieldPoiseuille flow of a power‐law fluid:

1

11

11

12

nnLo

z R

r

n

R

Lm

PPgLRrv


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Solution to Poiseuille flow in a tubeincompressible, power‐law fluid

2.0n=1.0

0.8

1.0

1.5

vz/v

z,a

v

0.0

0.2

0.4

0.6

0.8

0.0

0.5

0 0.2 0.4 0.6 0.8 1 1.2

r / R© Faith A. Morrison, Michigan Tech U.

45

Solution to Poiseuille flow in a tubeincompressible, power‐law fluid

1

0.4

0.6

0.8

v z/v

max

n= 0.2

n= 0.4

n= 0 6

0

0.2

0 0.2 0.4 0.6 0.8 1 1.2

r/R

n= 0.6

n= 0.8

n= 1


new fluids lecture 05 - pages.mtu.edu

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