Soils 205 Lecture 3

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 DENSITY AND POROSITY

Particle density (D_p or r _p)

1. mass (wt.) of soil solids per unit volume of soil solids:

D_p or r _p = Weight soil(dry)/Volume particles

= W_soil/V_particles

2. Units and values

Mg/m³ (megagrams per cubic meter)

(old) g/cm³ (grams per cubic centimeter)

same absolute value

2.6 Mg/m³ = 2.6 g/cm³

for most mineral soils 

Dp = 2.6 - 2.75 Mg/m³ (assume 2.65, unless told otherwise)

3. Determination

pycnometer or water displacement

W_soil = weight of soil used

V_particles = weight (volume)of water
displaced by particles

Bulk Density (D_b or r _b)

1. Mass (wt.) of soil solids per unit volume of soil

D_b or r _b = Weight soil(dry)/Volume(particles + pores)

= W_soil/(V_particles + V_pores)

= W_soil/V_total

2. Units - same as D_p

3. Measurement

(a) clod method

- dry wt of clod = g soil

- weigh clod after coated with a sealant (saran or paraffin are generally used) while suspended in water

- clod wt. in air - clod wt. in H₂O = wt. water displaced

since the density of H₂O = 1 g/cm³

g H₂O displaced = cm³ H₂O and clod

- g/cm³ = Mg/m³ = D_b

(b) core sampler

- dimensions of core

V = p r²h

V = volume (cm³)

r = radius (cm)

h = height or length (cm) of core

- dry weight of soil in core = g soil

- g/cm³ = D_b = Mg/m³

Example A: A soil core 5 cm long and 8 cm in diameter weighed 402 grams at sampling and 309 grams when oven-dried. What is the bulk density of the soil?

core volume = p r²h = p (4)²(5) = 251 cm³

D_b = soil weight/total volume

= 309 g/251 cm³ = 1.23 g/cm³ = 1.23 Mg/m³

Example B: A clod of soil weighed 62 grams when oven dry and 19 grams when suspended in water. What is the bulk density of the soil?

volume = water displaced = 62 g - 19 g = 43 g = 43 cm³

D_b = soil weight/total volume

= 62 g/43 cm³ = 1.44 g/cm³ = 1.44 Mg/m³

4. Values

(a) relate to compaction and pore space

1.2 Mg/m³ = more pore space

vs

1.6 Mg/m³ = more compacted

(b) wide range

0.8 Mg/m³ ¬ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ® 2.0⁺ Mg/m³

organic and volcanic ash influenced

high clay and low O.M. subsoils (compacted)

(c) can be altered by management

 

 

 

 

 

 

 

 

Timber Harvest
Volcanic Ash Soil

(d) convert to "field" values

Mg ¾ ® kg, lb

m³ ¾ ® ft³, yd³, ha-15 cm, acre-6 inches, acre-ft

medium D_b yields about:

· 1300 kg/m³

· 2150 lb/yd³

· 2 x 10⁶ kg/ha-15 cm

4 x 10⁶ lb/A-ft

example: if 50 tons per acre of soil were lost from a watershed each year,

how many years to erode the surface foot of soil?

(4 x 10⁶ lbs/acre ft)(acre year/50 T)(T/2000 lbs)

= 40 years/foot

C. Pore space

1. Related to D_p and D_b:

solve D_b & D_p for W_soil: W_soil = D_b x V_total

W_soil = D_p x V_particles

equate and rearrange: D_b/D_p = V_particles/V_total = particle fraction

{V_particles/V_total} x 100 = % particle space = (D_b/D_p) x 100

% particle space + % pore space = 100 %

% pore space = 100 - % particle space

= 100 - [(D_b/D_p) x 100]

or

= [1 - (D_b/D_p)] x 100

2. What is % pore space if D_b = 1.3 Mg/m³?

(1 - 1.3/2.6) x 100 = 50 %

3. Compaction of soil:

ü higher D_b

ü lower pore space (no change in D_p)

4. Pore size

macropores - aeration and drainage

micropores - hold water against gravity

 

soil aggregates

macropores

micropores

Example: A 98 cm³ core of soil weighed 132 grams when oven-dried and 180 grams when saturated with water. What is the pore space of the soil?

Answer A:

180 g - 132 g = grams water in pores at saturation

= cm³ of pores = 48 cm³

% pore space = (V_pores/V_total) x 100

= (48/98) x 100 = 49 %

Answer B:

D_b = W_soil/V_total = 132 g/98 cm³ = 1.35 g/cm³

V_particles = total volume - water volume at saturation

= 98 cm³ - 48 cm³ = 50 cm³

D_p = W_soil/V_particle = 132 g/50 cm³ = 2.64 g/cm³

% pore space = (1 - D_b/D_p) x 100

= (1 - 1.35/2.64) x 100 = 49 %

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