The problem provides the following information about a soil mass: moisture content ($w = 21\% = 0.21$), void ratio ($e = 1.1$), specific gravity of soil solids ($G_s = 2.71$), and total volume ($V = 15 m^3$). We need to determine the degree of saturation ($S$), total unit weight ($\gamma_t$), dry unit weight ($\gamma_d$), and the weight of water ($W_w$) in the soil.

1. Problem Description

The problem provides the following information about a soil mass: moisture content ($w = 21\% = 0.21$), void ratio ($e = 1.1$), specific gravity of soil solids ($G_s = 2.71$), and total volume ($V = 15 m^3$). We need to determine the degree of saturation ($S$), total unit weight ($\gamma_t$), dry unit weight ($\gamma_d$), and the weight of water ($W_w$) in the soil.

2. Solution Steps

First, let's calculate the degree of saturation $S$ using the formula:

$Se = wG_s$

$S = \frac{wG_s}{e}$

Substitute the given values:

$S = \frac{0.21 \times 2.71}{1.1} = \frac{0.5691}{1.1} = 0.51736$

$S = 51.74\%$

Next, let's calculate the total unit weight $\gamma_t$ using the formula:

$\gamma_t = \frac{(G_s + Se)\gamma_w}{1 + e}$

where $\gamma_w$ is the unit weight of water, which is approximately $9.81 kN/m^3$.

$\gamma_t = \frac{(2.71 + 0.51736)(9.81)}{1 + 1.1} = \frac{(3.22736)(9.81)}{2.1} = \frac{31.6604}{2.1} = 15.0764 kN/m^3$

Now, let's calculate the dry unit weight $\gamma_d$ using the formula:

$\gamma_d = \frac{G_s \gamma_w}{1 + e}$

$\gamma_d = \frac{2.71 \times 9.81}{1 + 1.1} = \frac{26.5851}{2.1} = 12.6596 kN/m^3$

Finally, let's calculate the weight of water $W_w$ in the soil. We can use the moisture content and dry unit weight.

We know that the dry weight of the soil solids is given by:

$W_s = \gamma_d \times V = 12.6596 \frac{kN}{m^3} \times 15 m^3 = 189.894 kN$

The weight of water is related to the weight of solids and moisture content as:

$w = \frac{W_w}{W_s}$

Therefore,

$W_w = w \times W_s$

$W_w = 0.21 \times 189.894 = 39.87774 kN$

3. Final Answer

$S = 51.74\%$

$\gamma_t = 15.08 kN/m^3$

$\gamma_d = 12.66 kN/m^3$

$W_w = 39.88 kN$