Chapter 6, Problem 6.12P

To determine

Find the hydraulic conductivity of the sand.

Answer to Problem 6.12P

The hydraulic conductivity of the sand is $\underset{\_}{0.0108\text{cm/sec}}$.

Explanation of Solution

Given information:

The shape factor (SF) is 6.5.

The void ratio of clay e is 0.5.

Calculation:

Find the opening size of the sieve:

Refer Table 2.6, “US sieve sizes with number designation” in the textbook.

For sieve no 30, the opening size of the sieve is 0.06 cm.

For sieve no 40, the opening size of the sieve is 0.0425 cm.

For sieve no 60, the opening size of the sieve is 0.02 cm.

For sieve no 100, the opening size of the sieve is 0.015 cm.

For sieve no 200, the opening size of the sieve is 0.075 cm.

For sieve no 40:

Find the fraction between two consecutive sieves:

$\left(\begin{array}{l}\text{Fraction between two}\\ \text{ consecutive sieves}\end{array}\right)=\left(\begin{array}{l}\text{Percent}\text{passing}\text{of}\text{sieve}\text{number}30\\ -\text{Percent}\text{passing}\text{of}\text{sieve}\text{number}40\end{array}\right)$

Substitute 100 % for percent passing of sieve number 30 and 80 % for percent passing of sieve number 40.

$\begin{array}{c}\left(\begin{array}{l}\text{Fraction between two}\\ \text{ consecutive sieves}\end{array}\right)=100\%-80\%\\ =20\%\end{array}$

For sieve no 60:

Find the fraction between two consecutive sieves:

$\left(\begin{array}{l}\text{Fraction between two}\\ \text{ consecutive sieves}\end{array}\right)=\left(\begin{array}{l}\text{Percent}\text{passing}\text{of}\text{sieve}\text{number}40\\ -\text{Percent}\text{passing}\text{of}\text{sieve}\text{number}60\end{array}\right)$

Substitute 80 % for percent passing of sieve number 40 and 68 % for percent passing of sieve number 60.

$\begin{array}{c}\left(\begin{array}{l}\text{Fraction between two}\\ \text{ consecutive sieves}\end{array}\right)=80\%-68\%\\ =12\%\end{array}$

Similarly calculate the fraction between two consecutive sieves.

Summarize the calculated values of opening size of the sieve and fraction between two consecutive sieves as in Table 1.

Sieve Number	Opening (cm)	Percent passing	Fraction between two consecutive sieves
30	0.06	100
40	0.0425	80	20
60	0.02	68	12
100	0.015	28	40
200	0.0075	0	28

Table 1

Refer Table 1;

For fraction between sieve numbers 30 and 40:

Find the term ${\left(\frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}\right)}_{30-40}$:

Substitute 20% for $f_i$, 0.06 cm for $D_{li}$, and 0.0425 cm for $D_{si}$.

$\begin{array}{c}{\left(\frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}\right)}_{30-40}=\frac{20}{{0.06}^{0.404}\times {0.0425}^{0.595}}\\ =408\end{array}$

For fraction between sieve numbers 40 and 60:

Find the term ${\left(\frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}\right)}_{40-60}$:

Substitute 12% for $f_i$, 0.0425 cm for $D_{li}$, and 0.02 cm for $D_{si}$.

$\begin{array}{c}{\left(\frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}\right)}_{40-60}=\frac{12}{{0.0425}^{0.404}\times {0.02}^{0.595}}\\ =440.75\end{array}$

For fraction between sieve numbers 60 and 100:

Find the term ${\left(\frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}\right)}_{60-100}$:

Substitute 40% for $f_i$, 0.02 cm for $D_{li}$, and 0.015 cm for $D_{si}$.

$\begin{array}{c}{\left(\frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}\right)}_{60-100}=\frac{40}{{0.02}^{0.404}\times {0.015}^{0.595}}\\ =2,364.12\end{array}$

For fraction between sieve numbers 100 and 200:

Find the term ${\left(\frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}\right)}_{100-200}$:

Substitute 28% for $f_i$, 0.015 cm for $D_{li}$, and 0.0075 cm for $D_{si}$.

$\begin{array}{c}{\left(\frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}\right)}_{100-200}=\frac{28}{{0.015}^{0.404}\times {0.0075}^{0.595}}\\ =2,807.73\end{array}$

Find the term $\frac{100\%}{{\displaystyle \sum \frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}}}$:

Substitute 408 for ${\left(\frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}\right)}_{30-40}$, 440.75 for ${\left(\frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}\right)}_{40-60}$, 2,364.12 for ${\left(\frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}\right)}_{60-100}$, and 2,807.73 for ${\left(\frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}\right)}_{100-200}$.

$\begin{array}{c}\frac{100\%}{{\displaystyle \sum \frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}}}=\frac{100}{408+440.75+2,364.12+2,807.73}\\ =0.0166\end{array}$

Find the hydraulic conductivity of the sand at void ratio of 0.5 (k) using the formula:

$k=1.99\times {10}^4{\left[\frac{100\%}{{\displaystyle \sum \frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}}}\right]}^2{\left(\frac{1}{SF}\right)}^2\left(\frac{e^3}{1+e}\right)$

Substitute 0.0166 for $\frac{100\%}{{\displaystyle \sum \frac{f_i}{D_{li}^{0.404}\times D_{si}^{0.595}}}}$, 6.5 for SF, and 0.5 for e.

$\begin{array}{c}k=1.99\times {10}^4{\left[0.0166\right]}^2{\left(\frac{1}{6.5}\right)}^2\left(\frac{{0.5}^3}{1+0.5}\right)\\ =0.0108\text{cm/sec}\end{array}$

Thus, the hydraulic conductivity of the sand is $\underset{\_}{0.0108\text{cm/sec}}$.