10.2 Transport Processes
455
traveled in 1 s. Since the circumference of a shell of radius r is equal to 2πr and since
its thickness is dr,
(contribution of the shell to dV/dt) 2πrdr |u z (r)|
The total volume rate of flow is the integral over all such shells:
dV
dt
2π
R
0
r |u z (r)| dr 2π
P 2 − P 1
4Lη
R
0
R
2 r − r
3
dr
dV
dt
(P 2 − P 1 ) πR 4
8Lη
(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 radius R. 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 of R as coming from the dependence of the
cross-sectional area on the radius and two powers of R as coming from the dependence
of the mean speed on the radius. The mean speed of the fluid is equal to dV /dt divided
by the cross-sectional area of the tube:
||u z |
1
πR 2
dV
dt
(P 2 − P 1 )R 2
8Lη
(10.2-25)
Poiseuille’s equation is named for Jean
Leonard Marie Poiseuille, 1797–1869, a
French physician who studied the
circulation of blood.
E X A M P L E 10.6
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 of P 2 − P 1 is equal to the hydrostatic pressure:
P 2 − P 1 hgρ (0.5700 m)
9.80 m s −2
998.2 kg m −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
||u z |
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., McGrawHill, New York, 1996, p. 126ff.
455
traveled in 1 s. Since the circumference of a shell of radius r is equal to 2πr and since
its thickness is dr,
(contribution of the shell to dV/dt) 2πrdr |u z (r)|
The total volume rate of flow is the integral over all such shells:
dV
dt
2π
R
0
r |u z (r)| dr 2π
P 2 − P 1
4Lη
R
0
R
2 r − r
3
dr
dV
dt
(P 2 − P 1 ) πR 4
8Lη
(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 radius R. 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 of R as coming from the dependence of the
cross-sectional area on the radius and two powers of R as coming from the dependence
of the mean speed on the radius. The mean speed of the fluid is equal to dV /dt divided
by the cross-sectional area of the tube:
||u z |
1
πR 2
dV
dt
(P 2 − P 1 )R 2
8Lη
(10.2-25)
Poiseuille’s equation is named for Jean
Leonard Marie Poiseuille, 1797–1869, a
French physician who studied the
circulation of blood.
E X A M P L E 10.6
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 of P 2 − P 1 is equal to the hydrostatic pressure:
P 2 − P 1 hgρ (0.5700 m)
9.80 m s −2
998.2 kg m −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
||u z |
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., McGrawHill, New York, 1996, p. 126ff.