Transport Phenomena.pdf

244 Chapter 8 Polymeric Liquids
EXAMPLE 8.3-3
Rework Example 3.6-3 for a power law fluid.
Tangential Annular
f
SOLUTION
r1
,^4J Equations 3.6-20 and 3.6-22 remain unchanged for a non-Newtonian fluid, but in lieu of Eq.
Fluid 4 ' 5
3.6-21 we write the ^-component of the equation of motion in terms of the shear stress by
using Table B.5:
0= -\4-(r 2 r r0
) (8.3-15)
r 2
dr
For the postulated velocity profile, we get for the power law model (with the help of Table B.I)
_ - ... d (
( d he\\ n ~ l d ho
(8.3-16)
Combining Eqs. 8.3-15 and 16 we get
(8.3-17)
Integration gives
(8.3-18)
Dividing by r 2
and taking the nth root gives a first-order differential equation for the angular
velocity
л / 71
\ -i (C \\ln
(8.3-19)
dr\rj r\ r
2 /
This may be integrated with the boundary conditions in Eqs. 3.6-27 and 28 to give
v e
1 - ( K
R/r) 2/n
(8.3-20)
2/п
_
The (z-component of the) torque needed
Ц/
on
"
the
1
outer
- к
cylinder to maintain the motion is then
2TTRL
• R
(8.3-21)
Combining Eqs. 8.3-20 and 21 then gives
T 7
=
(2/n)
(8.3-22)
The Newtonian result can be recovered by setting n = 1 and m = /JL. Equation 8.3-22 can be used
along with torque versus angular velocity data to determine the power law parameters m and n.
§8.4 ELASTICITY AND THE LINEAR VISCOELASTIC MODELS
Just after Eq. 1.2-3, in the discussion about generalizing Newton's "law of viscosity/' we
specifically excluded time derivatives and time integrals in the construction of a linear
expression for the stress tensor in terms of the velocity gradients. In this section, we