CHEMICAL ENGINEERING

54 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

From which, whenaD 0 .0025 m and D 1 ð 10 ^3 Ns/m^2 ,

kD 6 .24 Nsn/m^2

PROBLEM 3.39

Two liquids of equal densities, the one Newtonian and the other a non-Newtonian“power-
law”fluid,flow at equal volumetric rates down two wide vertical surfaces of the same
widths. The non-Newtonianfluid has a power-law index of 0.5 and has the same apparent
viscosity, in SI units, as the Newtonianfluid when its shear rate is 0.01 s^1. Show that,
for equal surface velocities of the twofluids, thefilm thickness for the Newtonianfluid
is 1.125 times that of the non-Newtonianfluid.

Solution

For a power-lawfluid:

RDk

(

du

dy

)n

(equation 3.121)

Dk

(

du

dy

)n 1 (

dux

dy

)

D (^) adux/dy (equation 3.122)
where (^) ais the apparent velocityDkdux/dyn^1
For a Newtonianfluid:
RD

(

dux

dy

)

(equation 3.3)

WhennD 0 .5anddux/dyD 0 .01, D (^) aand:
(^) aD Dkdux/dyn^1 Dk 0. 01 ^0.^5 D 10 kD andkD 0. 1 .
The equation of state of the power-lawfluid is therefore:
RD 0. 1 dux/dy^0.^5
For afluidflowing down a vertical surface, lengthland widthwandfilm thicknessS,
at a distanceyfrom the solid surface, a force balance gives:
SywlgDRwlDkdux/dynwl
or:
dux
dy

D

(g

k

) 1 /n

Sy^1 /n

and: uxD

(g

k

) 1 /n

Sy

nC 1

n

(



n

nC 1

)

Cconst.

WhenyD0,uxD0 and the constantD

(g

K

) 1 /n

S

nC 1

n

n

nC 1

and: uxD

(g

k

) 1 /n n

nC 1

[

S

nC 1

n Sy

nC 1

n

]