PHY 711 Fall 2015 -- Lecture 32 111/13/2015

PHY 711 Classical Mechanics and Mathematical Methods

10-10:50 AM MWF Olin 103

Plan for Lecture 32

Viscous fluids – Chap. 12 in F & W1. Viscous stress tensor2. Navier-Stokes equation3. Example for incompressible fluid

PHY 711 Fall 2015 -- Lecture 32 211/13/2015 2

PHY 711 Fall 2015 -- Lecture 32 311/13/2015

General formulation of viscosity from Chapter 12 of Fetter and Walecka

33

1

3

For a non-viscous fluid, we can define a stress tensor:

From Euler and Newton equations for fluid:

ideak l kl

kl kl k

lkl

V A Vl

v v p

v

T

d r TdA fd rt

3

1

th component of force acting on surface l klAl

dA k dAT

For an ideal (non-viscous) fluid: kl lkT T

3

1

Differential form of momentum conservation law:

i

ijii

j

Tvt

fx

PHY 711 Fall 2015 -- Lecture 32 411/13/2015

Note that in absence of viscous effect, the stress tensor relations are identical to the Newton-Euler and continuity equations:

3

1

Differential form of momentum conservation law in terms of stress tensor:

ijii

ji

fTv

t x

1

Continuity condition

Newton-Euler equa n

0

tio

pt

t

v v v f

v

PHY 711 Fall 2015 -- Lecture 32 511/13/2015

Effects of viscosity

Formulate viscosity stress tenso

Argue that viscosity is due to s

r with traceless and diagonal t

hear forces in a fluid of the for

e

m

rms

23

:

:

xdrag

v kkl kl kl

l

kl

vdA

y

v vTx x

F

v v

viscosity bulk viscosity

Total stress tensor:

23

kl

k

ideal vkl kl

il kl

v kkl kl kl

dealkl

l

kl

v v p

v v

T

T

T

T T

x x

v v

PHY 711 Fall 2015 -- Lecture 32 611/13/2015

Effects of viscosity -- continued

223 3 3

21 1 1

3

1

13

Continuity condition

0

Vector form (Nav

Incorporating generalized stress tensor into Newton-Euler

ier-S

equations

i j ji ii

j j ji j i j

j

j

j

j

v vv vpf

t x x x x x

tvx

v

2

tokes equation)1 1 1

3p

t

vv v v f v

Continuity condition

0t

v

PHY 711 Fall 2015 -- Lecture 32 711/13/2015

Newton-Euler equations for viscous fluids

2

Navier-Stokes equation1 1 1

3Continuity condition

0

pt

t

v v v f v

v

v

Fluid / (m2/s) (Pa s)Water 1.00 x 10-6 1 x 10-3

Air 14.9 x 10-6 0.018 x 10-3

Ethyl alcohol 1.52 x 10-6 1.2 x 10-3

Glycerine 1183 x 10-6 1490 x 10-3

Typical viscosities at 20o C and 1 atm:

PHY 711 Fall 2015 -- Lecture 32 811/13/2015

Example – steady flow of an incompressible fluid in a long pipe with a circular cross section of radius R

2

Navier-Stokes equation1 1 1

3Continuity condition

0

pt

t

v v v f v

v

v

2

0

Steady flow 0

Irrotational flow 0No applied force

Incompressible flu

=0

Neglect non-linear terms

id

0

t

v

vv

vf

PHY 711 Fall 2015 -- Lecture 32 911/13/2015

Example – steady flow of an incompressible fluid in a long pipe with a circular cross section of radius R -- continued

2

Navier-Stokes equation becomes:10 p

v R

v(r)

2

2

( ) (independent of )

Suppo

ˆAssu

se that

( )

me that , z

z

z

pv r z

zp pz Lpv rL

t v r

v r z L

(uniform pressure gradient)

PHY 711 Fall 2015 -- Lecture 32 1011/13/2015

Example – steady flow of an incompressible fluid in a long pipe with a circular cross section of radius R -- continued

R

v(r)

2

2

1 2

2

1 2

( )

1 ( )

( ) ln( ) C4

0 ( ) 0 C4

z

z

z

z

pv r

Ld dv r pr

r dr dr Lprv r C rL

pRv RL

C

2 2( )4zpv r R rL

L

PHY 711 Fall 2015 -- Lecture 32 1111/13/2015

Example – steady flow of an incompressible fluid in a long pipe with a circular cross section of radius R -- continued

R

v(r)

2 2

4

0

( )4

Mass flow rate through the pipe:

2 ( )8

z

R

z

pv r R rL

dM p Rrdrv rdt L

L

Poiseuille formula; Method for measuring

PHY 711 Fall 2015 -- Lecture 32 1211/13/2015

Example – steady flow of an incompressible fluid in a long tube with a circular cross section of outer radius R and inner radius kR

L

RkR

2

2

1 2

2

1 2

2 2

1 2

( )

1 ( )

( ) ln( ) C4

( ) 0 ln( ) C4

( ) 0 ln( ) C4

z

z

z

z

z

pv r

Ld dv r pr

r dr dr Lprv r C rLpRv R C RLp Rv R C RL

kk k

PHY 711 Fall 2015 -- Lecture 32 1311/13/2015

Example – steady flow of an incompressible fluid in a long tube with a circular cross section of outer radius R and inner radius kR -- continued

L

RkR

1 2

22 2

Solving for

1( ) 1 l

an

n4 ln

d :

z

C

pR r Rv rL R r

C

k k

2244

Mass flow rate through th

1

e pipe:

2 ( ) 1ln8

R

zR

dM p Rrdrv rdt Lk

k k k