Physical Chemistry Third Edition

10.2 Transport Processes 455

traveled in 1 s. Since the circumference of a shell of radiusris equal to 2πrand since

its thickness isdr,

(contribution of the shell todV/dt) 2 πrdr|uz(r)|

The total volume rate of flow is the integral over all such shells:

dV

dt

 2 π

∫R

0

r|uz(r)|dr 2 π

P 2 −P 1

4 Lη

∫R

0

(

R^2 r−r^3

)

dr

dV

dt



(P 2 −P 1 )πR^4

8 Lη

(Poiseuille’s equation) (10.2-24)

This is Poiseuille’s equation for the rate of flow of an incompressible liquid undergoing

laminar flow in a tube of radiusR. A different version holds for the flow of a gas.^3 Some

people are surprised that the rate of flow is proportional to the fourth power of the radius

of the pipe. You can think of two powers ofRas coming from the dependence of the

cross-sectional area on the radius and two powers ofRas coming from the dependence

of the mean speed on the radius. The mean speed of the fluid is equal todV /dtdivided

by the cross-sectional area of the tube:

|〈uz〉|

1

πR^2

dV

dt



(P 2 −P 1 )R^2

8 Lη

(10.2-25)

Poiseuille’s equation is named for Jean
Leonard Marie Poiseuille, 1797–1869, a
French physician who studied the
circulation of blood.

EXAMPLE10.6

Assume that the lower portion of a buret (including the stopcock) consists of a tube of length

7.00 cm and a radius of 0.500 mm. The upper portion is a tube of uniform diameter that is

large enough so that the upper tube does not inhibit the flow. The distance from the 0.00-mL

mark to the constriction of the lower part is 57.00 cm and the distance from the 50.00-mL

mark to the constriction is 4.50 cm. If the buret is filled with water at 20.0◦C, find the volume

rate of flow and the mean speed of flow when the meniscus is at the 0.00-mL mark. Assume

laminar flow.

Solution

The viscosity of water at 20◦C is equal to 0.001002 kg m−^1 s−^1. The density of water at

20 ◦C is 998.23 kg m−^3. The value ofP 2 −P 1 is equal to the hydrostatic pressure:

P 2 −P 1 hgρ(0.5700 m)

(

9 .80 m s−^2

)(

998 .2kgm−^3

)

5576 Pa

dV

dt



(5576 Pa)(π)( 0 .000500 m)^4

( 8 )( 0 .0700 m)

(

0 .001002 kg m−^1 s−^1

) 1. 95 × 10 −^6 m^3 s−^1

 1 .95 mL s−^1

|〈uz〉|

1

πR^2

dV

dt



1. 95 × 10 −^6 m^3 s−^1

π( 0 .000500 m)^2

 2 .48 m s−^1

(^3) D. P. Shoemaker, C. W. Garland, and J. W. Nibler,Experiments in Physical Chemistry, 6th ed., McGraw-
Hill, New York, 1996, p. 126ff.