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MAVERICK [17]
3 years ago
14

Consider the function f(n) = n

Engineering
1 answer:
alexira [117]3 years ago
4 0

get off this

Explanation:

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A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.
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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  = \frac{V_v}{V_s} =  \frac{\left\begin{array}{ccc}volume&of&void\end{array}\right}{\left\begin{array}{ccc}volume&of&solid\end{array}\right}------ (1)

Specific gravity G_{s} = \frac{P_s}{P_w} =  \frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)

% Saturation S = \frac{V_w}{Vv} × \frac{100}{1} =  \frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right} × \frac{100}{1}--------(3)

water content w =  \frac{M_w}{M_s} = \frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)

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 =  \frac{M_w}{Ms}

Recall; mass = Density × volume

w = \frac{V_wP_w}{V_sP_s} ------(5)

From eqn. (2)  G_{s} = \frac{P_s}{P_w}

∴ \frac{1}{G_s} = \frac{P_w}{P_s} ------(6)

putting eqn. (6) into (5)

w = \frac{V_w}{V_sG_s} -----(7)

Again, from eqn (1)

V_s = \frac{V_v}{e}

substituting into eqn. (7)

w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v}  = S

∴ w = \frac{Se}{G_s} -----(8)

With eqn. (7), we can calculate

upper limit of water content

when S = 100% = 1

Given, G_{s} = 2.7, e= 0.9

∴w= \frac{0.9*1}{2.7} = 0.333

∴ %w = 33.3%

Lower limit of water content

when S = 1% = 0.01

w= \frac{0.01*0.9 }{2.7} = 0.0033

∴ % w = 0.33%

b) Calculating mass of water in 100 cm³ sample of soil (P_w=\frac{1_g}{cm^{3} } )

Given, V_{s} = 100 cm^{3 }, S = 50% = 0.5

%S = \frac{V_w}{V_v} × \frac{100}{1} = \frac{V_w}{eV_s} × \frac{100}{1}

0.50 = \frac{V_w}{0.9* 100}  = 45cm^{3}

mass of water = P_wV_w= 1 * 45 = 45_{g}

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