Page 260 - Bird R.B. Transport phenomena

244 Chapter 8 Polymeric Liquids

EXAMPLE 8.3-3 Rework Example 3.6-3 for a power law fluid.
Tangential Annular SOLUTION
f
,^ J Equations 3.6-20 and 3.6-22 remain unchanged for a non-Newtonian fluid, but in lieu of Eq.
r1 4 5 4
Fluid ' 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 r ) (8.3-15)
2
r0
r 2 dr
For the postulated velocity profile, we get for the power law model (with the help of Table B.I)
_ - ... d (

n
( d he\\ ~ 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 / 7 1 \ -i (C \\ln
(8.3-19)
dr\rj r\ 2
r /
This may be integrated with the boundary conditions in Eqs. 3.6-27 and 28 to give
v 1 - ( R/r) 2/n
e K (8.3-20)
_ 2/п
Ц/ " 1 - к
The (z-component of the) torque needed on the outer cylinder to maintain the motion is then

2TTRL • R (8.3-21)
Combining Eqs. 8.3-20 and 21 then gives
(2/n)
T = (8.3-22)
7
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