CHEMICAL ENGINEERING

FLOW IN PIPES AND CHANNELS 41

From Fig. 3.8:ReDud/ D 5 ð 104

∴uD 5 ð 104 ð 1 ð 10 ^3 / 1000 ð 0. 025 D 2 .0m/s

Flowrate per tubeD 2. 0 ð/ 4 ð 0. 0252 D 0. 982 ð 104 m^3 /s

TotalflowrateD 110 ð 1000 / 1000 ð 3600 D 0 .3056 m^3 /s

∴Number of tubes requiredD 0. 3056 / 0. 982 ð 10 ^4 D 31 .1 or 32 tubes

If 10 per cent of the tubes are blocked, velocity offluidD 2. 0 / 0. 9 D 2 .22 m/s

ReD 5 ð 104 / 0. 9 D 5. 5 ð 104 and, from Fig. 3.7,R/u^2 D 0 .00245.

From equation 3.18, pressure drop is:

PD 4 ð 0. 00245 ð 5 / 0. 025 ð 1000 ð 2. 222 D9650 N/m^2 , an increase of 20.6%

PROBLEM 3.29

The effective viscosity of a non-Newtonianfluid may be expressed by the relationship:

(^) aDk^00

(



dux

dr

)

wherek^00 is constant.
Show that the volumetricflowrate of thisfluid in a horizontal pipe of radiusaunder
isothermal laminarflow conditions with a pressure dropP/lper unit length is:

QD

2 

7

a^7 /^2

(

P

2 k^00 l

) 1 / 2

Solution

In Section 3.4.1 of Volume 1 it is shown that for anyfluid, the shear stress,Rr,ata
distancerfrom the centre of the pipe may be found from a force balance for an element
offluid of lengthlacross which the pressure drop isPby:

Pr^2 D 2 rlRror RrD

r

2

(

P

l

)

(equation 3.7)

The viscosity is related to the velocity of thefluid,ux, and the shear stress,Rr, by:

RrD (^) adux/dr (from equation 3.4)
If, for the non-Newtonianfluid, (^) aDk^00 dux/dr
then: RrDk^00 dux/drdux/dr
Dk^00 dux/dr^2
Combining the two equations forRr:
k^00 dux/dr^2 D
r
2

(

P

l

)