CHEMICAL ENGINEERING

FLOW IN PIPES AND CHANNELS 43

wherekD 4 .0Ns^0.^4 /m^2 andnD 0 .4, calculate the volumetricflowrate per unit width if
thefluidfilm is 10 mm thick.

Solution

Flow with a free surface is discussed in Section 3.6 and the particular case of laminar
flow down an inclined surface in Section 3.6.1. For aflow of liquid of depthυ, widthw
and density down a surface inclined at an angle 4 to the horizontal, a force balance in
thexdirection (parallel to the surface) may be written. The weight offluidflowing down
the plane at a distanceyfrom the free surface is balanced by the shear stress at the plane.
For unit width and unit height:

RyxDgsin4y

RyxDkdx/dynandkdux/dynDgsin4y

SubstitutingkD 4 .0Ns^0.^4 /m^2 ,nD 0 .4, D1200 kg/m^3 and 4 D 15 °:

4. 0 dux/dy^0.^4 D 1200 ð 9. 81 ðsin 15°y

or:dux/dyD 762 yand dux/dyD 1. 60 ð 107 y^2.^5

uxD 4. 57 ð 106 y^3.^5 Cconstant

When thefilm thicknessyDυD 0 .01 m,uxD0. Hence 0D 0. 457 CcandcD 0 .457.

∴ uxD 4. 57 ð 106 y^3.^5  0. 457

The volumetricflowrate down the surface is then:

∫Q

0

dQD

∫w

0

∫ 0

0. 01

uxdwdy

or, for unit width:Q/WD

∫ 0

0. 01 ^4.^57 ð^10

(^6) y 3. (^5)  0. 457 dyD 0. 00357 m (^3) /s/m
PROBLEM 3.32
Afluid with afinite yield stress is sheared between two concentric cylinders, 50 mm long.
The inner cylinder is 30 mm diameter and the gap is 20 mm. The outer cylinder is held
stationary while a torque is applied to the inner. The moment required just to produce
motion is 0.01 Nm. Calculate the torque needed to ensure all thefluid isflowing under
shear if the plastic viscosity is 0.1 Ns/m^2.
Solution
Concentric-cylinder viscometers are in widespread use. Figure 3d represents a partial
section through such an instrument in which liquid is contained and sheared between the
stationary inner and rotating outer cylinders. Either may be driven, but theflow regime