Dry sand cylinder saturation numerical

Given

Container: $D = 0.2$ m, $H = 0.2$ m, so $V = \frac{\pi}{4}(0.2)^2(0.2) = 6.283 \times 10^{-3}$ m³. Dry sand $M_s = 10$ kg, $G = 2.65$, $e = 0.54$ (constant). $\rho_w = 1000$ kg/m³.

Assumption: the stated void ratio of 0.54 is used directly with the given mass of sand; the mass of solids fixes $V_s$, and the void volume is $V_v = eV_s$. (Using the container volume with $e = 0.54$ would give a slightly different mass of 10.8 kg, so the data are only approximately consistent.)

Volume of solids

$$V_s = \frac{M_s}{G\rho_w} = \frac{10}{2.65 \times 1000} = 3.774 \times 10^{-3}\ \text{m}^3$$

(i) Water to saturate the sand

$$V_v = eV_s = 0.54 \times 3.774 \times 10^{-3} = 2.038 \times 10^{-3}\ \text{m}^3$$

Water needed $= 2.038 \times 10^{-3} \times 1000 = 2.04$ kg (litres).

Water content at saturation:

$$w = \frac{M_w}{M_s} = \frac{2.038}{10} = 20.4\% \quad (\text{also } w = e/G = 0.54/2.65 = 20.4\%)$$

(ii) Water for 80% saturation

$$M_w = S\,e\,V_s\rho_w = 0.80 \times 2.038 = 1.63\ \text{kg}$$

So 1.63 litres of water must be added to the dry sand.

Answer: (i) 2.04 kg of water, $w = 20.4\%$; (ii) 1.63 kg (litres) of water.