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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

or

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})

or

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


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