Question

A soil sample is partially saturated. It's natural moisture content was found to be 22% and bulk density 2g/cm³. If G-2.65 and density of water is 1gm/cm³. Find out degree of saturation and void ratio​

Answer 1

Given : -

Partially saturated, so degree of saturation ( S ) < 1 .

Natural moisture content ( ω ) = 22% = 0.22 .

Bulk density ( γ ) = 2 g/cm³.

Specific gravity ( G ) = 2.65.

density of water ( γ$_{w}$ ) = 1 gm/cm³.

Need to find : -

Degree of saturation ( S ) and void ratio ( e )

Solution : -

we have the relation,

$S*e =G *w$

⇒ $S * e = 2.65 * 0.22$  

⇒$Se = 0.583$    _____(1)

Also we have for bulk density ( γ )

⇒ γ = $\frac{(G+Se)}{1+e} *$γ$_{w}$

⇒ $2 = \frac{(2.65+Se) }{(1+e)} * 1$

⇒ $2 = \frac{(2.65+0.583)}{(1+e)}$    ( ∵ from equation (1) )

⇒ $1+e = \frac{3.233}{2}$

 ∴ e = 0.6165

from the equation (1),

⇒ $S*e = 0.583$

⇒ $S = \frac{0.583}{0.6165} = 0.9456$

 ∴ S = 94.56%

Degree of saturation and void ratio​ are 94.56 % and 0.6165 respectively