are 0.17m and 21.8 * 10−4m2 respectively..Calculate the time required for dad to drop from 0.25m to 0.10m. The area of cross section of stand pipe is $2*10^{-4}m^2$. The sample has three layers with permeability $3*10^{-5}$m/s for first 0.06m. $4*10^{-5}$m/s for second 0.06m and $6*10^{-5}$m/s for the third 0.05m thickness.Assume the floor is taking place perpendicular to the bedding plane.

$K_v = \frac{H}{\frac{H_1}{K_1}+\frac{H_2}{K_2}+\frac{H_3}{K_3}}$ $K_v = \frac{0.17}{\frac{0.06}{3\times10^{-5}}+\frac{0.06}{4\times10^{-5}}+\frac{0.05}{6\times10^{-5}}}$ $k_v = 3.92 \times 10^{-5}$ $K_v = 2.303 \frac{aL}{At} \log_{10} \frac{h_1}{h_2}$ $t = 364.63 sec$