3 m clay, cv 0.025 cm²/min, times

Given: clay is drained at the top only (permeable above, impervious rock below), so $H_{dr} = 3$ m $= 300$ cm. $c_v = 0.025\ \text{cm}^2/\text{min}$, $S_{final} = 8$ cm.

(i) Time for 80% of settlement

For $U = 80\%$: $T_v = 1.781 - 0.933\log_{10}(100-80) = 0.567$.

$$t_{80} = \frac{T_v H_{dr}^2}{c_v} = \frac{0.567 \times 300^2}{0.025} = 2.04\times10^6\ \text{min}$$ $$= 1418\ \text{days} \approx 3.9\ \text{years}$$

(ii) Time for 2.5 cm settlement

$$U = \frac{2.5}{8} = 31.25\% \;(<60\%)\;\Rightarrow\; T_v = \frac{\pi}{4}(0.3125)^2 = 0.0767$$ $$t = \frac{0.0767 \times 300^2}{0.025} = 2.76\times10^5\ \text{min} = 192\ \text{days} \approx 0.53\ \text{year}$$

(iii) Settlement in 1 year

$t = 365 \times 1440 = 525{,}600$ min.

$$T_v = \frac{0.025 \times 525600}{300^2} = 0.146 \;\Rightarrow\; U = \sqrt{\frac{4T_v}{\pi}} = 0.431$$ $$S = 0.431 \times 8 = 3.45\ \text{cm}$$

Answer: (i) about 3.9 years; (ii) about 192 days; (iii) about 3.45 cm.