5 m clay, Cc 0.25, 50% consolidation

Assumptions: the layer is sandwiched between permeable layers (double drainage), so $H_{dr} = 2.5$ m; $\gamma_w = 9.81$ kN/m³. $0.15\ \text{N/mm}^2 = 150$ kPa and $0.20\ \text{N/mm}^2 = 200$ kPa.

Void ratio at 200 kPa

$$\Delta e = C_c \log_{10}\frac{\sigma_2'}{\sigma_1'} = 0.25\log_{10}\frac{200}{150} = 0.0312$$ $$e_2 = 1.9 - 0.0312 = 1.869$$

Settlement

$$S = \frac{\Delta e}{1+e_0}H = \frac{0.0312}{2.9}\times 5000 = 53.9\ \text{mm}$$

Time for 50% consolidation

Coefficient of compressibility and volume compressibility:

$$a_v = \frac{\Delta e}{\Delta\sigma'} = \frac{0.0312}{50} = 6.25\times10^{-4}\ \text{m}^2/\text{kN}, \quad m_v = \frac{a_v}{1+e_0} = 2.154\times10^{-4}\ \text{m}^2/\text{kN}$$

With $k = 3\times10^{-3}$ mm/s $= 3\times10^{-6}$ m/s:

$$c_v = \frac{k}{m_v\gamma_w} = \frac{3\times10^{-6}}{2.154\times10^{-4}\times 9.81} = 1.42\times10^{-3}\ \text{m}^2/\text{s}$$

For $U = 50\%$, $T_v = 0.197$:

$$t_{50} = \frac{T_v H_{dr}^2}{c_v} = \frac{0.197 \times 2.5^2}{1.42\times10^{-3}} = 867\ \text{s} \approx 14.5\ \text{min}$$

Answer: $e_2 \approx 1.869$; settlement $\approx 54$ mm; $t_{50} \approx 867$ s (about 14.5 min). With single drainage ($H_{dr}=5$ m) the time would be four times longer, about 58 min.