- Units
- Soil bulk density
- Apparent density according to texture
- Apparent density according to depth
- How to measure the apparent density?
- Exercise resolved
- Solution to
- Solution b
- Solution c
- Solution d
- Solution e
- Solution f
- References
The apparent density of a sample is defined as the quotient between its mass and the unaltered volume, which includes all the spaces or pores that it contains. If there is air in these spaces, the apparent density ρ b, or bulk density is:
ρ b = Mass / Volume = Mass of particles + Mass of air / Volume of particles + Volume of air
Figure 1. Bulk density is very important to characterize soils. Source: Wikimedia Commons.
When calculating the bulk density of a soil sample, it must be pre-dried in an oven at 105ºC until the mass is constant, indicating that all the air has evaporated.
According to this definition, the apparent density of soils or dry density, is calculated in this way:
ρ s = Weight of solid elements / Solid volume + Pore volume
Denoting as M s the dry weight or mass and V t = V s + V p as the total volume, the formula is:
ρ s = M s / V t
Units
The units of bulk density in the International System of Units are kg / m 3. However, other units such as g / cm 3 and megagrams / cubic meter: Mg / m 3 are also widely used.
The concept of apparent density is very useful when it comes to heterogeneous and porous materials such as soils, as it is indicative of their drainage and aeration capacity, among other qualities.
For example, low-porous soils have high bulk densities, are compact, and tend to water easily, unlike porous soils.
When there is water or another fluid in the pores of the sample, the volume after drying decreases, therefore, when making the calculations, it is necessary to know the proportion of original water (see resolved example).
Soil bulk density
The apparent density of materials in general, including soil, is highly variable, since there are factors such as the degree of compaction, the presence of organic matter, its texture, structure, depth and others, which affect the shape and shape. amount of pore spaces.
Soils are defined as a heterogeneous mixture of inorganic substances, organic substances, air and water. They can be fine, medium or coarse in texture to the touch, while the component particles can be arranged in various ways, a parameter known as structure.
Fine, well-structured soils with a high percentage of organic matter tend to have low values of apparent density. On the contrary, thick soils, with less organic matter and little structure, tend to have higher values.
Apparent density according to texture
According to its texture, the apparent density has the following values:
| Texture | Apparent Density (g / cm 3) |
|---|---|
| Fine | 1.00 - 1.30 |
| Median | 1.30 - 1.50 |
| Gross | 1.50 - 1.70 |
These values serve as a general reference. In peaty soils, abundant in plant residues, the apparent density can be as low as 0.25 g / cm 3, if it is a volcanic mineral soil it is around 0.85 g / cm 3, while in very compacted soils it reaches 1.90 g / cm 3.
Apparent density according to depth
The apparent density value also increases with depth, since the soil is generally more compacted and has a lower percentage of organic matter.
The interior of the terrain is composed of horizontal layers or strata, called horizons. Horizons have different textures, composition, and compaction. Therefore they present variation in terms of apparent density.
Figure 2. A soil profile showing the different horizons. Source: Wikimedia Commons.
A study of the soil is based on its profile, which consists of various horizons that follow one another in an orderly vertical manner.
How to measure the apparent density?
Since the variability in bulk density is very large, it often has to be measured directly by various procedures.
The simplest method is to extract a sample from the soil, inserting a bit with a space metal cylinder of known volume into it and making sure not to compact the soil. The extracted sample is sealed, to avoid loss of humidity or alteration of the characteristics.
Then in the laboratory the sample is extracted, weighed and then placed in an oven at 105ºC to dry for 24 hours.
Although it is the simplest way to find the dry density of the soil, it is not the most recommended for soils with very loose textures or full of stones.
For these, the method of digging a hole and saving the extracted earth is preferable, which will be the sample to dry. The volume of the sample is determined by pouring dry sand or water into the dug hole.
In any case, from the sample it is possible to determine very interesting properties of the soil to characterize it. The following solved exercise describes how to do it.
Exercise resolved
A clay sample of length 100 mm is drawn from the sample cylinder, whose internal diameter is also 100 mm. When weighed, a mass of 1531 g was obtained, which once dry was reduced to 1178 g. The specific gravity of the particles is 2.75. It is asked to calculate:
a) The bulk density of the sample
b) Moisture content
c) The void ratio
d) Dry density
e) The degree of saturation
f) Air content
Solution to
The unaltered volume V t is the original volume of the sample. For a cylinder of diameter D and height h, the volume is:
V cylinder = V t = Area of base x height = πD 2 /4 = π x (100 x 10 -3 m) 2 x 100 x 10 -3 m / 4 = 0.000785 m 3
The statement states that the mass of the sample is M s = 1531 g, therefore according to the equation given at the beginning:
ρ b = M s / V t = 1531 g / 0.000785 m 3 = 1950 319 g / m 3 = 1.95 Mg / m 3
Solution b
Since we have the original mass and the dry mass, the mass of the water contained in the sample is the difference of these two:
M water = 1531 g - 1178 g = 353 g
The percentage of moisture in the sample is calculated as follows:
% Moisture = (Mass of water / Ms) x 100% = (353 g / 1178 g) = 29. 97%
Solution c
To find the void ratio, the total volume of the sample V t must be broken down into:
V t = V particles + pore volume
The volume occupied by the particles is obtained from the dry mass and the specific gravity, data obtained from the statement. The specific gravity s g is the quotient between the density of the material and the density of water under standard conditions, therefore the density of the material is:
ρ = s g x ρ water = 2.75 x 1 g / cm 3 = 2.75 g / cm 3
ρ = M s / V s → V s = 1.178 g / 2.75 g / cm 3 = 0.428 cm 3 = 0.000428 m 3
The volume of voids in the sample is V v = V t - V s = 0.000785 m 3 - 0.000428 m 3 = 0.000357 m 3.
The void ratio e is:
e = V v / V s = 0.000357 m 3 / 0.000428 m 3 = 0.83
Solution d
The dry density of the sample is calculated as indicated in the introduction:
ρ s = Weight of the solid elements / Volume solids + Volume pores = 1178 g / 0.000785 m 3 = 1.5 Mg / m 3
Solution e
The degree of saturation is S = (V water / V v) x 100%. Since we know the mass of water in the sample, calculated in item b) and its density, the calculation of its volume is immediate:
ρ water = M water / V water → V water = 353 g / 1 g / cm 3 = 353 cm 3 = 0.000353 m 3
On the other hand, the volume of voids was calculated in item c)
S = (0.000353 m 3 / 0.000357 m 3) x 100% = 98.9%
Solution f
Finally, the percentage content of air is A = (V air / V t) x 100%. The air volume corresponds to:
V v - V water = 0.000357 m 3 - 0.000353 m 3 = 0.000004 m 3
A = (V air / V t) x 100% = (0.000004 m 3 / 0.000785 m 3) x100% = 0.51%
References
- Berry, P. Soil Mechanics. McGraw Hill.
- Constrummatics. Apparent density. Recovered from: construmatica.com.
- NRCS. Soil Bulk Density. Recovered from: nrcs.usda.gov.
- UNAM. Department of Edaphology. Soil Physics Analytical Procedures Manual. Recovered from: geologia.unam.mx.
- Wikipedia. Bulk Density. Recovered from: en.wikipedia.org.
- Wikipedia. Floor. Recovered from: en.wikipedia.org.