CHEMICAL ENGINEERING

56 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

PROBLEM 3.40

Afluid which exhibits non-Newtonian behaviour isflowing in a pipe of diameter 70 mm
and the pressure drop over a 2 m length of pipe is 4ð 104 N/m^2. When theflowrate is
doubled, the pressure drop increases by a factor of 1.5. A pitot tube is used to measure
the velocity profile over the cross-section. Confirm that the information given below is
consistent with the laminarflow of apower-lawfluid.
Any equations used must be derived from the basic relation between shear stressRand
shear rate:

RDk,Pn

radial distance from centre velocity

of pipe (smm) (m/s)

00.80

10 0.77

20 0.62

30 0.27

Solution

For a power-lawfluid:

At the initialflowrate: 4 ð 104 Dkun

With aflow of: 6 ð 104 Dk 2 un

Dividing: 1. 5 D 2 n

and hence: nD 0. 585

For the power-lawfluid:

RDk

∣

∣

∣

∣

dux

dy

∣

∣

∣

∣

n

A force balance on afluid core of radiussin pipe of radiusrgives:

Rs 2 slDPs^2

or: RsDk

(



dux

ds

)n

D

Ps

2 l



dux

ds

D

(



P

2 kl

) 1 /n

s^1 /n

Integrating: uxD

(



P

2 ks

) 1 /n(

n

nC 1

)

s

nC 1

n Cconstant