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

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

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

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

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

therefore

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

(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

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

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

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