CHEMICAL ENGINEERING

FLOW IN PIPES AND CHANNELS 45

As motion just initiates under the action of this torque, this shear stress must equal the
yield stress and:

RYD 141 .5N/m^2

If all thefluid is to be in motion, the shear stress at the surface of the outer cylinder must
be at least this value and the shear stress at the inner cylinder will be higher, and will be
given by:
RiDRoro/ri^2 D[141. 5  0. 035 / 0. 015 ^2 ]D770 N/m^2

The required torque is then:

TDRið 2 ri^2 hD 770 ð 2 ð 0. 0152 ð 0. 05 D 0 .054 Nm

PROBLEM 3.33

Experiments with a capillary viscometer of length 100 mm and diameter 2 mm gave the
following results:

Applied pressure (N/m^2 ) Volumetricflowrate (m^3 /s)

1 ð 103 1 ð 10 ^7

2 ð 103 2. 8 ð 10 ^7

5 ð 103 1. 1 ð 10 ^6

1 ð 104 3 ð 10 ^6

2 ð 104 9 ð 10 ^6

5 ð 104 3. 5 ð 10 ^5

1 ð 105 1 ð 10 ^4

Suggest a suitable model to describe thefluid properties.

Solution

Inspection of the data shows that the pressure difference increases less rapidly than the
flowrate. Taking thefirst and the last entries in the table, it is seen that when theflowrate
increases from 1ð 10 ^7 to 1ð 10 ^4 m^3 /s, that is by a factor of 1000, the pressure differ-
ence increases from 1ð 103 to 1ð 105 N/m^2 that is by a factor of only 100. In this way,
thefluid appears to be shear-thinning and the simplest model, the power-law model, will
be tried.
From equation 3.136:

QD/ 4 d^2 uDP/ 4 kl^1 /n[n/ 6 nC 2 ]/ 4 d^3 nC^1 /n

Using the last set of data:

1. 0 ð 10 ^4 D[ 1 ð 105 / 4 kð 0. 1 ]^1 /n/ 8 n/ 3 nC 1  2 ð 10 ^3 ^3 nC^1 /n

or: QDKP^1 /n