the generalized newtonian fluid - isothermal flows · the generalized newtonian . fluid -...

The Generalized Newtonian Fluid - Isothermal Flows

10.531J/2.341J !Spring 2014!

MIT!Cambridge, MA 02139!

•  Constitutive Equations!•  Viscosity Models!•  Solution of Flow Problems!

1

Generalized Newtonian FluidSimple Shear FlowNewtonian Fluid

yx

dvτ yx = −µ x

dy

Non-Newtonian Fluid

yx

dvτ η xyx = −

dywhere

η = η( )dvx dy

Arbitrary FlowNewtonian Fluid

= −µγ̇ ij

dvτ ij µ j dv

= − + i dxi dx

j

Non-Newtonian Fluid

where

τ ηγij ij= − ˙

η η γ= ( )˙˙ ˙γ γ= 1

2 II

2

Non-Newtonian Viscosity�η�LV�D�VFDODU� ,W�PXVW�GHSHQG�RQO\�RQ�VFDODU�LQYDULDQWV�RI�! ̇� 7KUHH�VFDODU�LQYDULDQWV�DUH�GHILQHG�IRU ! ̇

Iγ̇ = =∑ γi˙ii 2 0( )∇ ⋅v =

IIγ̇ = = γji∑ ∑ γ γ˙ ˙ij ji 2 ˙

IIIγ̇ = =∑ ∑ γ γ˙ ˙ γi jj∑ ij jk ˙ki 0

�ODVW�HTXDOLW\�LV�IRU�VKHDU�IORZ�

� 6LQFH�,,�LV�WKH�RQO\�QRQ�]HUR�LQYDULDQW�LQ�VLPSOH�VKHDU�IORZ��GHILQH

˙ ˙γ definition of = 12 IIγ shear rate 3

Generalized Newtonian Fluid� �������������� �� ������� ����� ������ ������������ �� ������� �������������� ��������� ������� ��������� �� ������� ��� � shear

stresses �� steady shear flow

� ���������� �� η γ̇( )� �������� ����� ������� �������� �������� ����� ������ ����� !������ ����� ���� ����

4

Power-Law Model

η = mγ̇ n−1

�Captures high shear rate behavior normally occuring in processing

n < 1 shear thinning[0.15 < n < 0.6]

n = 1 Newtonian fluidm = µ �Defects

n > 1 shear thickening � No η0

� No λ

slope = n-1η0

η∞

log η

log γ⋅10-2 100 102

m

5

Spriggs "Truncated Power-Law"

η η= 0

η

˙ ˙

˙ n 1 γ

−

= η0 γ̇ 0

γ γ≤ 0

˙ ˙γ > γ 0

�Contains a characteristic time λ

1λ =γ̇ 0

slope = n-1

η0

η∞

log η

log γ⋅10-2 100 102

γ0⋅

6

Carreau-Yasuda Model! !! !

"#$$

= + ( )[ ]%

%

$

0

1

1 ˙ ana

! (P

a·s)

" (s-1)#

7

Bingham Model

slope µ0

- τyx

= ηγ

γ⋅

τ0

⋅

η∞ ≤τ τ ( )0

= τµ 0 0 + ( )τ τ≥ γ̇ 0

�Bingham plastics have a yield stress τ0 and will not flow unless the magnitude of the stress τ exceeds τ0

τ = 12 ( )" ":

�Time constantµ0

τ 0 8

Casson Model

�Useful for chocolate

dv± =τ τyx + µ x0 0 m for τ yx > τdy 0

γ τ˙ yx = <0 for yx τ 0

9

Tube Flow of a Power-Law Fluid� )RUFH�EDODQFH�RQ�F\OLQGULFDO�FRQWURO�YROXPH�RI�OHQJWK�/�DQG�UDGLXV�U

L

p0 pL

r

z

R

control volume formomentum balance

( )p0 − pL rπ τr2 − ⋅z 2 0πrL=

rτ τrz = w R∆pRτw =2L

� &RQVWLWXWLYH�HTXDWLRQ�JLYHV�D�VHFRQG�H[SUHVVLRQ�IRU�τU]

˙ ˙ − dvτ ηrz = − ( )γ γ rz = −mγ̇ n 1 z

dr

τ rz zn

dv= m − dr 10

Tube Flow Results� ��������� ����

� ��������� ���

vm

R

n

rRz

w n n= !"

#$ +

% !"#$

&

'((

)

*++

+,1 1 1

1 11

Q R

nmw n

=+

&'(

)*+

- ,31

1 3

� � ������������˙ ˙. .R a

nn

= +!"

#$

3 14

.̇ azv

D=8

true wall shear rate apparent (Newtonian) wall shear rate

11

Radial Flow between Parallel Disks

z Problem: z Determine Q(p1-p2, B, R1, R2, m, n)

z Assume the fluid viscosity isdescribed by the power-law function

z Solution:z Use the lubrication approximation

to simplify the problem to pressure driven slit flow

z First find Q for this simple flow

12

z Assume

z z-component of the equation of motion (Table B.1)

z Integrate to get the shear stress distribution from conservation of momentum

( ) 0

( )z z x y

ij ij

v v x , v v

x

! !

z z z zx y z xz yz zz z

v v v v pv v v g

t x y z x y z zw w w w§ · ª ºw w w w

� � � � � � � �« »¨ ¸© w w w w ¹ w w w w¬ ¼" ! ! ! "

0 Lxz

p pddx L

� !

01

Lxz

p px c

L�

�!0

13

z For the generalized Newtonian fluid

and the power-law model gives

z In order to avoid problems with the absolute value, consider only the region x > 0, for which

z Then

zxz

dvdx

�! #

11

n

n zdvm m

dx

�� # $�

0z z zdv dv dv

dx dx dx§ ·

� � �¨ ¸© ¹

1

0( )n n

z z zxz L

dv dv dv xm m p p

dx dx dx L

�§ · § ·

� � � �¨ ¸ ¨ ¸© ¹ © ¹!

14

z The differential equation for the velocity is thus

which can be integrated to give

z Finally, the volume flow rate is found as

where

110

nnz Ldv p p

xdx mL

�§ ·� ¨ ¸© ¹

� �1 11

1 101

11

n n

n

Lz

n

p pv B x

mL� ��§ · �¨ ¸© ¹ �

12

10

22

2

nBB

zn

WBQ W v dx

m§ · ¨ ¸© ¹�³

!

0 LB

p p dpB B

L dz�

�!

15

Apply the slit flow results locally to the disk problem

z The corresponding quantities in the two geometries arez pressure gradient

z width

z The volume flow rate expression is adapted as

which can be integrated from r = R1 2the fact that for incompressible fluids Q is independent of r. This gives

to r = R , by taking advantage of

0 Lp p dp dpL dz dr�

� o �

2W r %

� �1

2

2 12 2

n

n Q m dpB r B dr

ª º� �« »

¬ ¼%

16

� � � �1 112 1

1 2 2

2

4 (1 )

n n nn

R RQ mp p

B B n

� ��ª º�� « » �¬ ¼%

z Finally, the above is inverted to give the volume flow rate in terms of the pressure gradient

� �� �

� �

12

1

1 112 1

(1 )42

n

n nn

p p B nBQ

m R R� �

ª º� �« »

� �« »¬ ¼

%

17

Justification for Applying the Lubrication Approximation

Use an order of magnitude analyis to justify the use of the lubrication approximation in adapting the slit flow results to the radial disk flow problem

z For the radial flow problem, assume that

z r-component of the equation of motion

z For the power-law fluid

( ) 0r r zv v r ,z , v v &

� �w w w wª º � � � �« »w w w w¬ ¼

1rr rr zr

v pv r

r r r z r r&&!

" ! !

18

z For the power-law fluid

z From Table B.3 in DPL, we get the rate-of-strain tensor in cylindrical coordinates

from which the shear rate is found

� �1 12; n

ij ijm � � ! !! $ $ $ � �� � � :

2 0

0 2 0

0 0

r r

r

r

v vr z

vr

vz

w w§ ·¨ ¸w w¨ ¸¨ ¸ ¨ ¸¨ ¸w¨ ¸¨ ¸© ¹w

!�

2 2 2

2 2r r rv v vr r z

w w§ · § · § · � �¨ ¸ ¨ ¸ ¨ ¸© ¹ © ¹ © ¹w w$�

19

z Order of magnitude estimates for contributions to the shear rate

for R1 << R2

z For small gap, B << R2, the shear rate is well approximated by

z Next evaluate the order of magnitude of the terms that appear in the equation of motion

14r

Qv V

R Ba

%

2 2

; ; r r rv v vV V Vr R r R z B

w wa a a

w w

rv Vz B

wa a

w$�

2

2

rr

v Vv

r Rww

" "�(Inertial term)

20

(Stress terms)1

2 12 2 2

1; ;

n nn

rr zrn

mV m V m V Vr

r r R B z B B r R B R

�

�

w w § · § ·¨ ¸ ¨ ¸© ¹ © ¹w w

&&!! !� � �

z Comparison of different terms shows that is the largest term on the right side by a factor of

z The ratio of inertial to viscous forces is

z If the Reynolds number is at most (R2/B), then inertial forces can still be neglected

zr zw w!

� �2

2 1B R �

� �

2

2 2

221

2 2

Ren n

VVR B BR

R Rm V m V BB B

�

§ · § · � �¨ ¸ ¨ ¸© ¹ © ¹§ ·

¨ ¸© ¹

""

21

z Neglecting terms of order (B/R2) and smaller gives the equation of motion as

which is locally (in r) the same as the slit flow equation

w w � �

w w0 zr

pr z

!

22

Distributor Design (Power-Law)

z For flow down the circular tube

0 1z

Q QL

ª º �« »¬ ¼

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23

z Equate this to the power-law result for Q(dp/dz) for tube flow

z Integrate to get the pressure drop down the tube

z At any position z, there is p - pa driving force to force fluid through the slit of local length l(z)

z Slit flow for a power law fluid gives

� �� �

1

21 2 2 ( )

nap p BB

Vn ml z

ª º� « »� ¬ ¼

1 13

0 11 23

n nz R R dpQ

L m dzn

%ª º § · § ·� �¨ ¸ ¨ ¸« » © ¹ © ¹¬ ¼ �

0

11

3

321

1

n nnn

a

QmL zp p

R R n L%

��§ · § ·� �¨ ¸¨ ¸ © ¹© ¹ �

24

z Equating the available to needed pressure gradient gives l(z)

� �� �

1

03

1 3( ) 1

( 1) 2 1 2

n n nn QBL B zl z

R n R n V L%

�§ ·� § · § · �¨ ¸¨ ¸¨ ¸ © ¹© ¹� �© ¹

25

Squeezing Flow between Parallel Disks

(DPL Example 4.2-7)

26

z Volume flow rate across surface at rz Mass conservation

z Equation of motion with the lubrication approximation

z Hence

� � � �22Q r r h% � �

Radial flowSlit flow

� �

2 rh

dp dr

Q r

%

�� �0 L

WB

P P L

Q

�

� � � �122 2

1 2

nr h h dp

Q rn m dr%� � § · �¨ ¸© ¹�

27

z Solve for p(r) with the boundary condition that p(R) = pa

z A force balance on the top plate gives

� � 11

2 1

2 11

2 1

nn nn

a n

m h n R rp p

h n n R

� ��

�

ª º� �§ · § ·� �« »¨ ¸ ¨ ¸© ¹ © ¹� « »¬ ¼

� �2

0 0

( )R

a zz z hF t p p r dr d

%

! &

� �³ ³= 0 on any solid surface

� � 3

2 1

2 12 3

nn n

n

h n mRF

h n n%� �

�

� �§ · ¨ ¸© ¹ �

Scott equation

28

z For constant force this can be integrated to give the half time t1/2 for h to go from h0 to (½)h0:

� �1 1121 2

0

nn

n

t R m RK

n F h%

�§ ·§ ·

¨ ¸ ¨ ¸© ¹ © ¹

function of n

29

= De-1

� � � �12

n nm m'

�c c

where1nm# $� �

Viscous response2

1nm( $� �c c

��VRXUFH�XQNQRZQ��$OO�ULJKWV�UHVHUYHG��7KLV�FRQWHQW�LV�H[FOXGHG�IURP�RXU�&UHDWLYH &RPPRQV�OLFHQVH��)RU�PRUH�LQIRUPDWLRQ��VHH�KWWSV���RFZ�PLW�HGX�KHOS�IDT�IDLU�XVH��

30

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2.341J / 10.531J Macromolecular HydrodynamicsSpring 2016

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