SME
Physical Geography · Problem type
Bulk density, porosity and water
Use this for “A 100 cm³ core weighs 132.5 g dry and holds 30 cm³ of water. Find the bulk density, porosity and the share of pores filled.”
Diagram
e
=
Vᵥ
Vₛ
=
50
50
=
1;
S
=
Vw
Vᵥ
=
0.6.
Mₛ
=
M
1 + w
=
163
1.23
=
132
g;
Mw
=
30
g.
The solids fill 50 of the 100 cm³.
This is an example. Type your own values to start a new problem.
ρb (g/cm³)
=
Md (g)
÷
V (cm³)
Md is the oven-dry mass, the picture’s solids Mₛ, and ρb spreads it over the whole core volume V.
Vw
Water volume
example · cm³
ρₛ
Particle density
example · g/cm³
n
Porosity
calculated
θ
Water content by volume
calculated
S
Pores filled
calculated · %
w
Water content by mass
calculated · %
M
Wet mass
worked out (not typed) · g
Formulas
ρb
=
Md
V
1.325
=
132.5
100
n
=
ρₛ − ρb
ρₛ
0.5
=
2.65 − 1.325
2.65
θ
=
Vw
V
0.3
=
30
100
S
=
100
×
θ
n
60
=
100
×
0.3
0.5
w
=
100
×
Vw
Md
22.6415
=
100
×
30
132.5
M
=
Md
+
Vw
×
1
162.5 = 132.5 + 30 × 1
Assumptions
•
The core is taken without packing it down, then oven dried at 105 °C: the dry mass is the solids alone.
•
Particle density ρₛ = 2.65 g/cm³ suits most mineral soils (quartz and feldspars); organic soils are lighter.
•
Water is 1 g/cm³, so the water’s volume in cm³ is its mass in grams.
Step-by-step
WE KNOW
V = 100 cm³, Md = 132.5 g, Vw = 30 cm³, ρₛ = 2.65 g/cm³
FIND
bulk density (ρb), porosity (n), water content by volume (θ), pores filled (S), water content by mass (w), wet mass (M)
Step 1 · Find bulk density (ρb)
Use ρb = Md ÷ V
Bulk density spreads the dry solids over the whole core, pores and all.
ρb
=
132.5
100
ρb = 1.325 g/cm³
Step 2 · Find porosity (n)
Use n = (ρₛ − ρb) ÷ ρₛ
Solids alone would be ρₛ dense; the core is only ρb, so the shortfall is the pores’ share.
n
=
2.65 − 1.325
2.65
n
=
1.325
2.65
n
=
0.5
Step 3 · Find water content by volume (θ)
Use θ = Vw ÷ V
The water’s share of the core’s whole volume.
θ
=
30
100
θ
=
0.3
Step 4 · Find pores filled (S)
Use S = 100 × θ ÷ n
Water sits only in the pores, so its share of the core over the pores’ share is how full they are.
S
=
100
×
0.3
0.5
S
=
30
0.5
S
=
60%
Step 5 · Find water content by mass (w)
Use w = 100 × Vw ÷ Md
Water of 1 g/cm³ weighs Vw grams; by mass, it is that over the dry solids.
w
=
100
×
30
132.5
w
=
3,000
132.5
w
=
22.6415%
Step 6 · Find wet mass (M)
Use M = Md + Vw × 1
The wet core weighs its solids plus its water, 1 g for each cm³.
M
=
132.5
+
30
×
1
M
=
162.5
g
Check
Put the numbers back in. Both sides should match.
1.325
=
132.5
100
✓
0.5
=
2.65 − 1.325
2.65
✓
0.3
=
30
100
✓
60
=
100
×
0.3
0.5
✓
22.6415
=
100
×
30
132.5
✓
162.5 = 132.5 + 30 × 1 ✓
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