Chapter 7, Problem 7.9P

(a)

To determine

Find the absolute permeability of the soil.

Answer to Problem 7.9P

The absolute permeability of the soil is $\underset{\_}{8.4\times {10}^{-13}{\text{m}}^{\text{2}}}$.

Explanation of Solution

Given information:

The length of the soil sample L is 400 mm.

The area of the sample A is $7,854{\text{mm}}^{\text{2}}$.

The diameter of the standpipe (d) is 11 mm.

The head difference $\left(h_1\right)$ at $t=0$ is 450 mm.

The head difference  $\left(h_2\right)$ at $t=8\min$ is 200 mm.

The unit weight of water $\left({\gamma }_w\right)$ is $9.789{\text{kN/m}}^{\text{3}}$.

The dynamic viscosity of water $\left(\eta \right)$ is $1.005\times {10}^{-3}\text{N}\cdot {\text{s/m}}^{\text{2}}$.

Calculation:

Determine the area of the standpipe a using the relation.

$a=\frac{\pi d^2}{4}$

Substitute 11 mm for d.

$\begin{array}{c}a=\frac{\pi {\left(11\right)}^2}{4}\\ =95.03{\text{mm}}^2\end{array}$

Determine the hydraulic conductivity k using the relation.

$k=2.303\left(\frac{aL}{At}\right){\log }_{10}\left(\frac{h_1}{h_2}\right)$

Substitute $95.03{\text{mm}}^{\text{2}}$ for a, 400 mm for L, $7,854{\text{mm}}^{\text{2}}$ for A, 8 min for t, 450 mm for $h_1$, and 200 mm for $h_2$.

$\begin{array}{c}k=2.303\left(\frac{95.03\times 400}{7,854\times 8\min \times \frac{60\sec }{1\min }}\right){\log }_{10}\left(\frac{450}{200}\right)\\ =8.18\times {10}^{-3}\text{mm/s}\times \frac{1\text{cm}}{10\text{mm}}\\ =8.18\times {10}^{-4}\text{cm/sec}\end{array}$

Determine the absolute permeability of the soil using the relation.

$k=\frac{{\gamma }_w}{\eta }\overline{K}$

Substitute $8.18\times {10}^{-4}\text{cm/sec}$ for k, $9.789{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, and $1.005\times {10}^{-3}\text{N}\cdot {\text{s/m}}^{\text{2}}$ for $\eta$.

$\begin{array}{l}8.18\times {10}^{-4}\text{cm/sec}\times \frac{1\text{m}}{100\text{cm}}=\frac{9.789\times {10}^3}{1.005\times {10}^{-3}}\times \overline{K}\\ \frac{8.18\times {10}^{-6}\times 1.005\times {10}^{-3}}{9.789\times {10}^3}=\overline{K}\\ \overline{K}=8.4\times {10}^{-13}{\text{m}}^{\text{2}}\end{array}$

Therefore, the absolute permeability of the soil is $\underset{\_}{8.4\times {10}^{-13}{\text{m}}^{\text{2}}}$.

(b)

To determine

Find the head difference at 4 min time duration.

Answer to Problem 7.9P

The head difference at 4 min time duration is $\underset{\_}{30\text{cm}}$.

Explanation of Solution

Given information:

The length of the soil sample L is 400 mm.

The area of the sample A is $7,854{\text{mm}}^{\text{2}}$.

The diameter of the standpipe (d) is 11 mm.

The head difference $\left(h_1\right)$ at $t=0$ is 450 mm.

The head difference $\left(h_2\right)$ at $t=8\min$ is 200 mm.

The unit weight of water $\left({\gamma }_w\right)$ is $9.789{\text{kN/m}}^{\text{3}}$.

The dynamic viscosity of water $\left(\eta \right)$ is $1.005\times {10}^{-3}\text{N}\cdot {\text{s/m}}^{\text{2}}$.

Calculation:

Determine the head difference at 4 min time duration using the relation.

$k=2.303\left(\frac{aL}{At}\right){\log }_{10}\left(\frac{h_1}{h_2}\right)$

Substitute $8.18\times {10}^{-4}\text{cm/sec}$ for k, $95.03{\text{mm}}^{\text{2}}$ for a, 400 mm for L, $7,854{\text{mm}}^{\text{2}}$ for A, 8 min for t, and 450 mm for $h_1$.

$\begin{array}{l}8.18\times {10}^{-4}\text{cm/sec}\times \frac{10\text{mm}}{1\text{cm}}=2.303\left(\frac{95.03\times 400}{7,854\times 4\min \times \frac{60\sec }{1\min }}\right){\log }_{10}\left(\frac{450}{h_2}\right)\\ \frac{8.18\times {10}^{-3}}{0.0464}={\log }_{10}\left(\frac{450}{h_2}\right)\\ {10}^{\left(0.176\right)}=\left(\frac{450}{h_2}\right)\\ h_2=\frac{450}{1.5}\end{array}$

$\begin{array}{l}{h}_2=300\text{mm}\times \frac{1\text{cm}}{10\text{mm}}\\ h_2=30\text{cm}\end{array}$

Therefore, the head difference at 4 min time duration is $\underset{\_}{30\text{cm}}$.