# Solution of The Value of The Moisture When Fully Saturated

Articles > Solution of The Value of The Moisture When Fully Saturated

### Question:

1) Show that at saturation the moisture (water) content is

$w_{sat}=\frac{n \gamma _w}{(\gamma_{sat}-n \gamma_w)}$

2) Show that at saturation the moisture (water) content is

$w_{sat}=\gamma_w (\frac{1}{\gamma_{d}}-\frac{1}{\gamma_s})$

### Solution:

For formulas of soil phase relationships read Soil Phase Relationships article.

(1) In a fully saturated soil the relation, Se=wGs  becomes simply e=wGs because S = 1

$eGs=w_{sat} \mapsto Gs= \frac{e}{w_{sat}}=\frac{n}{w_{sat}(1-n)}$

but

$\gamma_{sat}=\gamma_w[(1-n)Gs+n]$

rearranging

$\frac{\gamma_{sat}}{\gamma_w}=[(1-n)Gs+n]=[(1-n)\frac{n}{w_{sat}(1-n)}+n]=\frac{n}{w_{sat}}+n$

or

$\frac{\gamma_{sat}}{\gamma_w}-n=\frac{n}{w_{sat}}$

therefore

$w_{sat}=\frac{n \gamma _w}{(\gamma_{sat}-n \gamma_w)}$

(2) Again, in a fully saturated soil:

$w_{sat}=\frac{e}{Gs}=\frac{V_v}{V_s}\frac{\gamma_w}{\gamma_s}=\frac{V_v}{V_s}\frac{\gamma_w}{1}\frac{V_s}{W_s}=\frac{\gamma_wV_v}{W_s}$

Thus

$w_{sat}=\frac{\gamma_wV_v}{W_s}=\gamma_w(\frac{V_v}{W_s})=\gamma_w(\frac{V_v+V_s-V_s}{W_s})=\gamma_w(\frac{V_v+V_s}{W_s}-\frac{V_s}{W_s})$

or

$w_{sat}=\gamma_w (\frac{1}{\gamma_{d}}-\frac{1}{\gamma_s})$