Physical Chemistry Third Edition

10.2 Transport Processes 453

exerted on the fluid by the upper plate, but also for the force exerted on one layer of

the fluid by the adjacent layer.

Newton’s law is formulated so that the rate of shear (∂uy/∂z) plays the role of a

driving force while the force per unit area plays the role of a rate variable. This seems

like a role reversal, but defining Newton’s law in this way corresponds to a viscosity

coefficient that is larger for more viscous fluids. Newton’s law is valid only forlaminar

flow, which means flow in layers. Flow that is not laminar is calledturbulent flow, and

Newton’s law does not hold for turbulent flow. There are some liquids, such as blood

and polymer solutions, that do not obey Newton’s law even for laminar flow. These

fluids are callednon-Newtonian fluidsorthixotropic fluidsand can be described by a

viscosity coefficient that depends on the rate of shear.

EXAMPLE10.5

The viscosity coefficient of water at 20◦C equals 0.001002 kg m−^1 s−^1 (0.001002 Pa s). For

an apparatus like that in Figure 10.1, find the force per unit area required to keep the upper

plate moving at a speed of 0.250 m s−^1 if the tank is 0.0500 m deep.

Solution

The rate of shear has an average value of

∆uy

∆z



0 .250 m s−^1

0 .0500 m

 5 .00 s−^1

so that

Pzy

(

0 .001002 kg m−^1 s−^1

)(

5 .00 s−^1

)

 5. 01 × 10 −^3 kg m−^1 s−^2  5. 01 × 10 −^3 Nm−^2

Poiseuille’s Equation for Viscous Flow in a Tube

This equation describes the rate of laminar flow of a liquid through a straight tube or

pipe of lengthLand radiusR. We assume that the tube is parallel to thezaxis and we

assume that the velocity depends only onr, the perpendicular distance from the center

of the tube. Consider a portion of the fluid that is contained in an imaginary cylinder

of radiusrthat is concentric with the tube walls, as depicted in Figure 10.5.

When a steady state has been reached there is no net force on the fluid since it

does not accelerate. The frictional force due to viscosity at the surface of the cylinder

balances the hydrostatic force pushing the liquid through the cylinder. From Newton’s

law of viscous flow,

Ff

A

η

∣

∣

∣

∣

duz

dr

∣

∣

∣

∣ (10.2-22)

whereFfis the magnitude of the frictional force on the liquid in the imaginary cylinder

andA is the area of its surface (excluding the area of its ends). This area is the

circumference times the length of the cylinder:

A 2 πrL