1. A 20 g oven-dry soil sample, consisting of two soil particle size classes, is dispersed and
mixed with water. The soil suspension is placed in a 100 cm tall column and allowed to
settle. The coarsest fraction of soil particles reaches the bottom after 2.5 minutes. The
second size fraction of soil particles reaches the bottom of the column after 350 hours.
Calculate the diameter of each of the soil particle size classes.
Note that: particle density is 2.7 g∙cm-3; fluid density is 1.0 g∙cm-3(Note: 1 dyne = 1
g∙cm∙s-2); acceleration due to gravity is 980 cm∙s-2; and fluid viscosity is 0.01
dynes∙sec∙cm-2.
© Luc Bernier [2019]
Solution
1. A 20 g oven-dry soil sample, consisting of two soil particle size classes, is dispersed and
mixed with water. The soil suspension is placed in a 100 cm tall column and allowed to
settle. The coarsest fraction of soil particles reaches the bottom after 2.5 minutes. The
second size fraction of soil particles reaches the bottom of the column after 350 hours.
Calculate the diameter of each of the soil particle size classes. Assume the density of all
particles is 2.7 g cm-3 and the viscosity of water is 0.01 dynes sec cm-2.
Stokes Law:
d 2 g (ρ s − ρ w )
v=
18η
where:
v = terminal velocity (distance / time)
d = particle diameter
g = gravitational force (980 cm s-2)
ρs = particle density
ρw = water density (1 g cm-3)
η = water viscosity
We must arrange the equation so that we can solve for particle diameter.
v ∙ 18 ∙ η = d2∙g ∙( ρs – ρw)
v ⋅ 18 ⋅ η
= d2
g ⋅ (ρ x − ρ w )
a) For the first size fraction of the soil:
We know:
t = 2.5 min x 60 s/min = 150 s
v = 100 cm / 150s ≈ 0.66 cm/s
g = 980 cm/s2
ρs = 2.7 g/cm3
ρw = 1 g/cm3
η = 0.01 dyne∙s∙cm-2
Note: 1 dyne = 1 g∙cm∙s-2
= 0.01 g/cm∙s
d2 =
v ⋅ 18 ⋅ η
g ⋅ (ρ x − ρ w )
© Luc Bernier [2019]
(0.66 cm/s) ·18 · (0.01 g/cm·s)
(980 cm/s 2 ) · (2.7 g/cm 3 1.0 g/cm 3 )
(0.1188 cm/s · g/cm·s)
=
(980 cm/s 2 ) · (1.7 g/cm 3 )
=
=
0.1188 cm/s · g/cm·s
1666 cm/s 2 · g/cm 3
-5
Units
2
= cm/s ∙ g/cm∙s
d = 7.13 x10 −5 cm 2
cm/s2 ∙ g/cm3
= 8.44 x10 −3 cm
= cm g s2 cm3
= 7.13 x 10 cm
= 8.44 x 10 -2 mm
The particle diameter for the grains that settled after 2.5 min is 8.44 x 10-2 mm.
b) For the second size fraction of the soil:
We know:
t = 350 hrs x 60 min x 60 s_ = 1,260,000s
1 hr
1 min
v = 100 cm / 1,260,000s ≈ 7.94 x 10-5 cm/s
g = 980 cm/s2
ρs = 2.7 g/cm3
ρw = 1.0 g/cm3
h = 0.01 dyne∙s∙cm-2 Note: 1 dyne = 1 g∙cm∙s-2
= 0.01 g/cm∙s
© Luc Bernier [2019]
d2 =
v ⋅ 18 ⋅ η
g ⋅ (ρ x − ρ w )
d2 =
(7.94 x 10 -5 cm/s) ·18 · (0.01 g/cm·s)
(980 cm/s 2 ) · (2.7 g/cm 3 1.0 g/cm 3 )
d2 =
(1.43 x 10 -5 cm/s · g/cm·s)
(980 cm/s 2 ) · (1.7 g/cm 3 )
1.43 x 10 -5 cm/s · g/cm·s
d2 =
1666 cm/s 2 · g/cm 3
d 2 = 8.6 x 10 -9 cm 2
d = 8.6 x 10 -9 cm 2
Units
= cm/s ∙ g/cm∙s
cm/s2 ∙ g/cm3
= cm g s2 cm3
d = 9.27 x10 −5 cm
d = 9.27 x10 − 4 mm
The particle diameter for the grains that settled after 350 hrs is 9.27 x 10-4 mm.
© Luc Bernier [2019]
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© Luc Bernier [2019]
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© Luc Bernier [2019]