Respuesta :
Answer:
The correct answers are:
a. % w = 33.3%
b. mass of water = 45g
Explanation:
First, let us define the parameters in the question:
void ratio e  = [tex]\frac{V_v}{V_s}[/tex] =  [tex]\frac{\left\begin{array}{ccc}volume&of&void\end{array}\right}{\left\begin{array}{ccc}volume&of&solid\end{array}\right}------ (1)[/tex]
Specific gravity [tex]G_{s}[/tex] = [tex]\frac{P_s}{P_w} =[/tex] Â [tex]\frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)[/tex]
% Saturation S = [tex]\frac{V_w}{Vv}[/tex] × [tex]\frac{100}{1}[/tex] =  [tex]\frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right}[/tex] × [tex]\frac{100}{1}--------(3)[/tex]
water content w = Â [tex]\frac{M_w}{M_s}[/tex] = [tex]\frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)[/tex]
a) To calculate the lower and upper limits of water content:
when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.
when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.
Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.
To get the relationship between water content and saturation, we will manipulate the equations above;
w = Â [tex]\frac{M_w}{Ms}[/tex]
Recall; mass = Density × volume
w = [tex]\frac{V_wP_w}{V_sP_s} ------(5)[/tex]
From eqn. (2) Â [tex]G_{s}[/tex] = [tex]\frac{P_s}{P_w}[/tex]
∴ [tex]\frac{1}{G_s} = \frac{P_w}{P_s} ------(6)[/tex]
putting eqn. (6) into (5)
w = [tex]\frac{V_w}{V_sG_s} -----(7)[/tex]
Again, from eqn (1)
[tex]V_s = \frac{V_v}{e}[/tex]
substituting into eqn. (7)
[tex]w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v} = S[/tex]
∴ [tex]w = \frac{Se}{G_s} -----(8)[/tex]
With eqn. (7), we can calculate
upper limit of water content
when S = 100% = 1
Given, [tex]G_{s} = 2.7, e= 0.9[/tex]
∴[tex]w= \frac{0.9*1}{2.7} = 0.333[/tex]
∴ %w = 33.3%
Lower limit of water content
when S = 1% = 0.01
[tex]w= \frac{0.01*0.9 }{2.7} = 0.0033[/tex]
∴ % w = 0.33%
b) Calculating mass of water in 100 cm³ sample of soil ([tex]P_w=\frac{1_g}{cm^{3} }[/tex] )
Given, [tex]V_{s} = 100 cm^{3 }[/tex], S = 50% = 0.5
%S = [tex]\frac{V_w}{V_v}[/tex] × [tex]\frac{100}{1}[/tex] = [tex]\frac{V_w}{eV_s}[/tex] × [tex]\frac{100}{1}[/tex]
0.50 = [tex]\frac{V_w}{0.9* 100} = 45cm^{3}[/tex]
mass of water = [tex]P_wV_w= 1 * 45 = 45_{g}[/tex]
