Chapter 7, Problem 7.1CTP

(a)

To determine

Find the quantity of water flowing through the sample per hour.

Answer to Problem 7.1CTP

The quantity of water flowing through the sample per hour is $\underset{\_}{536.0{\text{cm}}^{\text{3}}\text{/hr}}$.

Explanation of Solution

Given information:

The length of each soil layer $\left(H_1,H_2,H_3\right)$ is 200 mm.

The total length of the soil layer H is 600 mm.

The diameter of the cylindrical tube d is 150 mm.

The constant head difference $\Delta h$ is 470 mm.

The porosity of the soil layer I $\left(n_1\right)$ is 0.5.

The porosity of the soil layer II $\left(n_2\right)$ is 0.6.

The porosity of the soil layer III $\left(n_3\right)$ is 0.33.

The hydraulic conductivity of soil layer I $\left(k_1\right)$ is $5\times {10}^{-3}\text{cm/sec}$.

The hydraulic conductivity of soil layer II $\left(k_2\right)$ is $4.2\times {10}^{-2}\text{cm/sec}$.

The hydraulic conductivity of soil layer III $\left(k_3\right)$ is $3.9\times {10}^{-4}\text{cm/sec}$.

Calculation:

Determine the hydraulic conductivity in the vertical direction using the relation.

$k_{V\left(\text{eq}\right)}=\frac{H}{\frac{H_1}{k_1}+\frac{H_2}{k_2}+\frac{H_3}{k_3}}$

Substitute 600 mm for H, 200 mm for $H_1$, $5\times {10}^{-3}\text{cm/sec}$ for $k_1$, 200 mm for $H_2$, $4.2\times {10}^{-2}\text{cm/sec}$ for $k_2$, 200 mm for $H_3$, and $3.9\times {10}^{-4}\text{cm/sec}$ for $k_3$.

$\begin{array}{c}{k}_{V\left(\text{eq}\right)}=\frac{600\text{mm}\times \frac{1\text{cm}}{10\text{mm}}}{\frac{200\text{mm}\times \frac{1\text{cm}}{10\text{mm}}}{5\times {10}^{-3}}+\frac{200\text{mm}\times \frac{1\text{cm}}{10\text{mm}}}{4.2\times {10}^{-2}}+\frac{200\text{mm}\times \frac{1\text{cm}}{10\text{mm}}}{3.9\times {10}^{-4}}}\\ =\frac{60}{4,000+476.19+51,282.05}\\ =1.076\times {10}^{-3}\text{cm/sec}\end{array}$

Determine the hydraulic gradient using the relation.

$i=\frac{\Delta h}{L}$

Here, L is the total length of the soil layer.

Substitute 470 mm for $\Delta h$ and 600 mm for L.

$\begin{array}{c}i=\frac{470}{600}\\ =0.783\end{array}$

Determine the area of the cylindrical tube using the relation.

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

Substitute 150 mm for d.

$\begin{array}{c}A=\frac{\pi {\left(150\text{mm}\times \frac{1\text{cm}}{10\text{mm}}\right)}^2}{4}\\ =176.71{\text{cm}}^{\text{2}}\end{array}$

Determine the rate of seepage per unit length of the dam using the relation.

$q=k_{V\left(\text{eq}\right)}iA$

Substitute $1.076\times {10}^{-3}\text{cm/sec}$ for $k_{V\left(\text{eq}\right)}$, 0.783 for i, and $176.71{\text{cm}}^{\text{2}}$ for A.

$\begin{array}{c}q=1.076\times {10}^{-3}\times 0.783\times 176.71\\ =0.149{\text{cm}}^{\text{3}}\text{/sec}\times \frac{3,600\sec }{1\text{hr}}\\ =536.0{\text{cm}}^{\text{3}}\text{/hr}\end{array}$

Therefore, the quantity of water flowing through the sample per hour is $\underset{\_}{536.0{\text{cm}}^{\text{3}}\text{/hr}}$.

(b)

To determine

Find the elevation head (Z), pressure head $\left(u/{\gamma }_w\right)$, and the total head (h) at the entrance and exit of the each soil layer.

Answer to Problem 7.1CTP

The elevation head ($Z_0$) at x is 0 mm is $\underset{\_}{-220\text{mm}}$.

The pressure head ${\left(u/{\gamma }_w\right)}_0$ at x is 0 mm is $\underset{\_}{690\text{mm}}$.

The total head $h_0$ at x is 0 mm is $\underset{\_}{470\text{mm}}$.

The elevation head ($Z_{200}$) at x is 200 mm is $\underset{\_}{-220\text{mm}}$.

The pressure head ${\left(u/{\gamma }_w\right)}_{200}$ at x is 200 mm is $\underset{\_}{656.3\text{mm}}$.

The total head $h_{200}$ at x is 200 mm is $\underset{\_}{436.3\text{mm}}$.

The elevation head ($Z_{400}$) at x is 400 mm is $\underset{\_}{-220\text{mm}}$.

The pressure head ${\left(u/{\gamma }_w\right)}_{400}$ at x is 400 mm is $\underset{\_}{652.3\text{mm}}$.

The total head $h_{400}$ at x is 400 mm is $\underset{\_}{432.3\text{mm}}$.

The elevation head ($Z_{600}$) at x is 600 mm is $\underset{\_}{-220\text{mm}}$.

The pressure head ${\left(u/{\gamma }_w\right)}_{600}$ at x is 600 mm is $\underset{\_}{220\text{mm}}$.

The total head $h_{600}$ at x is 600 mm is $\underset{\_}{0\text{mm}}$.

Explanation of Solution

Given information:

The length of each soil layer $\left(H_1,H_2,H_3\right)$ is 200 mm.

The total length of the soil layer H is 600 mm.

The diameter of the cylindrical tube d is 150 mm.

The constant head difference $\Delta h$ is 470 mm.

The porosity of the soil layer I $\left(n_1\right)$ is 0.5.

The porosity of the soil layer II $\left(n_2\right)$ is 0.6.

The porosity of the soil layer III $\left(n_3\right)$ is 0.33.

The hydraulic conductivity of soil layer I $\left(k_1\right)$ is $5\times {10}^{-3}\text{cm/sec}$.

The hydraulic conductivity of soil layer II $\left(k_2\right)$ is $4.2\times {10}^{-2}\text{cm/sec}$.

The hydraulic conductivity of soil layer III $\left(k_3\right)$ is $3.9\times {10}^{-4}\text{cm/sec}$.

Calculation:

Determine the elevation head ($Z_0$) at x is 0 mm.

$Z_0=-h_{X-Y}$

Here, $h_{X-Y}$ is the depth of water from the Y point to the middle point of X.

Substitute 220 mm for $h_{X-Y}$.

$Z_0=-220\text{mm}$

Therefore, the elevation head ($Z_0$) at x is 0 mm is $\underset{\_}{-220\text{mm}}$.

Determine the pressure head ${\left(u/{\gamma }_w\right)}_0$ at x is 0 mm.

${\left(u/{\gamma }_w\right)}_0=\Delta h+h_{X-Y}$

Substitute 470 mm for $\Delta h$ and 220 mm for $h_{X-Y}$.

$\begin{array}{c}{\left(u/{\gamma }_w\right)}_0=470+220\\ =690\text{mm}\end{array}$

Therefore, the pressure head ${\left(u/{\gamma }_w\right)}_0$ at x is 0 mm is $\underset{\_}{690\text{mm}}$.

Determine the total head $h_0$ at x is 0 mm.

$h_0={\left(u/{\gamma }_w\right)}_0+Z_0$

Substitute 690 mm for ${\left(u/{\gamma }_w\right)}_0$ and –220 mm for $Z_0$.

$\begin{array}{c}{h}_0=690-220\\ =470\text{mm}\end{array}$

Therefore, the total head $h_0$ at x is 0 mm is $\underset{\_}{470\text{mm}}$.

Determine the elevation head ($Z_{200}$) at x is 200 mm.

$Z_{200}=-h_{X-Y}$

Substitute 220 mm for $h_{X-Y}$.

$Z_{200}=-220\text{mm}$

Therefore, the elevation head ($Z_{200}$) at x is 200 mm is $\underset{\_}{-220\text{mm}}$.

Determine the value of $\Delta H$ value using the relation.

$\begin{array}{l}{k}_{V\left(\text{eq}\right)}i=k_1{i}_1\\ k_{V\left(\text{eq}\right)}i=k_1\frac{\Delta H}{H_1}\end{array}$ (1)

Substitute $1.076\times {10}^{-3}\text{cm/sec}$ for $k_{V\left(\text{eq}\right)}$, 0.783 for i, $5\times {10}^{-3}\text{cm/sec}$ for $k_1$, and 200 mm for $H_1$ in Equation (1).

$\begin{array}{l}1.076\times {10}^{-3}\times 0.783=5\times {10}^{-3}\times \frac{\Delta H}{200\text{mm}\times \frac{1\text{cm}}{10\text{mm}}}\\ \frac{8.425\times {10}^{-4}}{2.5\times {10}^{-4}}=\Delta H\\ \Delta H=3.37\text{cm}\times \frac{10\text{mm}}{1\text{cm}}\\ \Delta H=33.70\text{mm}\end{array}$

Determine the total head $h_{200}$ at x is 200 mm.

$h_{200}=\Delta h-\Delta H$

Substitute 470 mm for $\Delta h$ and 33.70 mm for $\Delta H$.

$\begin{array}{c}{h}_{200}=470-33.70\\ =436.3\text{mm}\end{array}$

Therefore, the total head $h_{200}$ at x is 200 mm is $\underset{\_}{436.3\text{mm}}$.

Determine the pressure head ${\left(u/{\gamma }_w\right)}_{200}$ at x is 200 mm.

${\left(u/{\gamma }_w\right)}_{200}=h_{200}-\left(Z_{200}\right)$

Substitute 436.3 mm for $h_{200}$ and –220 mm for $Z_{200}$.

$\begin{array}{c}{\left(u/{\gamma }_w\right)}_{200}=436.3+220\\ =656.3\text{mm}\end{array}$

Therefore, the pressure head ${\left(u/{\gamma }_w\right)}_{200}$ at x is 200 mm is $\underset{\_}{656.3\text{mm}}$.

Determine the elevation head $\left(Z_{400}\right)$ at x is 400 mm.

$Z_{400}=-h_{X-Y}$

Substitute 220 mm for $h_{X-Y}$.

$Z_{400}=-220\text{mm}$

Therefore, the elevation head ($Z_{400}$) at x is 400 mm is $\underset{\_}{-220\text{mm}}$.

Determine the value of $\Delta H$ value using the relation.

$\begin{array}{l}{k}_{V\left(\text{eq}\right)}i=k_2{i}_2\\ k_{V\left(\text{eq}\right)}i=k_2\frac{\Delta H}{H_2}\end{array}$ (2)

Substitute $1.076\times {10}^{-3}\text{cm/sec}$ for $k_{V\left(\text{eq}\right)}$, 0.783 for i, $4.2\times {10}^{-2}\text{cm/sec}$ for $k_2$, and 200 mm for $H_2$ in Equation (2).

$\begin{array}{l}1.076\times {10}^{-3}\times 0.783=4.2\times {10}^{-2}\times \frac{\Delta H}{200\text{mm}\times \frac{1\text{cm}}{10\text{mm}}}\\ \frac{8.425\times {10}^{-4}}{2.1\times {10}^{-3}}=\Delta H\\ \Delta H=0.40\text{cm}\times \frac{10\text{mm}}{1\text{cm}}\\ \Delta H=4.0\text{mm}\end{array}$

Determine the total head $h_{400}$ at x is 400 mm.

$h_{400}=h_{200}-\Delta H$

Substitute 436.3 mm for $h_{200}$ and 4.0 mm for $\Delta H$.

$\begin{array}{c}{h}_{400}=436.3-4.0\\ =432.3\text{mm}\end{array}$

The total head $h_{400}$ at x is 400 mm is $\underset{\_}{432.3\text{mm}}$.

Determine the pressure head ${\left(u/{\gamma }_w\right)}_{400}$ at x is 400 mm.

${\left(u/{\gamma }_w\right)}_{400}=h_{400}-\left(Z_{400}\right)$

Substitute 432.3 mm for $h_{400}$ and –220 mm for $Z_{400}$.

$\begin{array}{c}{\left(u/{\gamma }_w\right)}_{400}=432.3+220\\ =652.3\text{mm}\end{array}$

Therefore, The pressure head ${\left(u/{\gamma }_w\right)}_{400}$ at x is 400 mm is $\underset{\_}{652.3\text{mm}}$.

Determine the elevation head $\left(Z_{600}\right)$ at x is 600 mm.

$Z_{600}=-h_{X-Y}$

Substitute 220 mm for $h_{X-Y}$.

$Z_{600}=-220\text{mm}$

Therefore, the elevation head ($Z_{600}$) at x is 600 mm is $\underset{\_}{-220\text{mm}}$.

Determine the value of $\Delta H$ value using the relation.

$\begin{array}{l}{k}_{V\left(\text{eq}\right)}i=k_3{i}_3\\ k_{V\left(\text{eq}\right)}i=k_3\frac{\Delta H}{H_3}\end{array}$ (3)

Substitute $1.076\times {10}^{-3}\text{cm/sec}$ for $k_{V\left(\text{eq}\right)}$, 0.783 for i, $3.9\times {10}^{-4}\text{cm/sec}$ for $k_3$, and 200 mm for $H_3$ in Equation (3).

$\begin{array}{l}1.076\times {10}^{-3}\times 0.783=3.9\times {10}^{-4}\times \frac{\Delta H}{200\text{mm}\times \frac{1\text{cm}}{10\text{mm}}}\\ \frac{8.425\times {10}^{-4}}{1.95\times {10}^{-5}}=\Delta H\\ \Delta H=43.21\text{cm}\times \frac{10\text{mm}}{1\text{cm}}\\ \Delta H=432.10\text{mm}\end{array}$

Determine the total head $h_{600}$ at x is 600 mm.

$h_{600}=h_{400}-\Delta H$

Substitute 432.3 mm for $h_{400}$ and 432.10 mm for $\Delta H$.

$\begin{array}{c}{h}_{600}=432.3-432.10\\ =0.2\text{mm}\\ \simeq 0\text{mm}\end{array}$

Therefore, the total head $h_{600}$ at x is 600 mm is $\underset{\_}{0\text{mm}}$.

Determine the pressure head ${\left(u/{\gamma }_w\right)}_{600}$ at x is 600 mm.

${\left(u/{\gamma }_w\right)}_{600}=h_{600}-\left(Z_{600}\right)$

Substitute 432.3 mm for $h_{600}$ and –220 mm for $Z_{600}$.

$\begin{array}{c}{\left(u/{\gamma }_w\right)}_{600}=0+220\\ =220\text{mm}\end{array}$

Therefore, the pressure head ${\left(u/{\gamma }_w\right)}_{600}$ at x is 600 mm is $\underset{\_}{220\text{mm}}$.

(c)

To determine

Plot the variation of the elevation head, pressure head, and the total head with the horizontal distance along the sample axis (X–X).

Explanation of Solution

Given information:

The length of each soil layer $\left(H_1,H_2,H_3\right)$ is 200 mm.

The total length of the soil layer H is 600 mm.

The diameter of the cylindrical tube d is 150 mm.

The constant head difference $\Delta h$ is 470 mm.

The porosity of the soil layer I $\left(n_1\right)$ is 0.5.

The porosity of the soil layer II $\left(n_2\right)$ is 0.6.

The porosity of the soil layer III $\left(n_3\right)$ is 0.33.

The hydraulic conductivity of soil layer I $\left(k_1\right)$ is $5\times {10}^{-3}\text{cm/sec}$.

The hydraulic conductivity of soil layer II $\left(k_2\right)$ is $4.2\times {10}^{-2}\text{cm/sec}$.

The hydraulic conductivity of soil layer III $\left(k_3\right)$ is $3.9\times {10}^{-4}\text{cm/sec}$.

Calculation:

Refer Part b)

Draw the graph between the elevation head pressure head, and the total head with the horizontal distance along the sample axis (X–X) as in Figure (1).

(d)

To determine

Plot the variation of the discharge velocity and the seepage velocity along the sample axis (X–X).

Explanation of Solution

Given information:

The length of each soil layer $\left(H_1,H_2,H_3\right)$ is 200 mm.

The total length of the soil layer H is 600 mm.

The diameter of the cylindrical tube d is 150 mm.

The constant head difference $\Delta h$ is 470 mm.

The porosity of the soil layer I $\left(n_1\right)$ is 0.5.

The porosity of the soil layer II $\left(n_2\right)$ is 0.6.

The porosity of the soil layer III $\left(n_3\right)$ is 0.33.

The hydraulic conductivity of soil layer I $\left(k_1\right)$ is $5\times {10}^{-3}\text{cm/sec}$.

The hydraulic conductivity of soil layer II $\left(k_2\right)$ is $4.2\times {10}^{-2}\text{cm/sec}$.

The hydraulic conductivity of soil layer III $\left(k_3\right)$ is $3.9\times {10}^{-4}\text{cm/sec}$.

Calculation:

Determine the discharge velocity v using the relation.

$v=k_{V\left(\text{eq}\right)}i$

Substitute $1.076\times {10}^{-3}\text{cm/sec}$ for $k_{V\left(\text{eq}\right)}$ and 0.783 for i.

$\begin{array}{c}v=1.076\times {10}^{-3}\times 0.783\\ =0.000843\text{cm/sec}\end{array}$

Determine the seepage velocity of soil I using the relation.

$v_{s1}=\frac{v}{n_1}$

Here, $n_1$ is the porosity of the soil I.

Substitute 0.000843 cm/sec for v and 0.5 for $n_1$.

$\begin{array}{c}{v}_{s1}=\frac{0.000843}{0.5}\\ =0.00168\text{cm/sec}\end{array}$

Determine the seepage velocity of soil II using the relation.

$v_{s2}=\frac{v}{n_2}$

Here, $n_2$ is the porosity of the soil II.

Substitute 0.000843 cm/sec for v and 0.6 for $n_2$.

$\begin{array}{c}{v}_{s2}=\frac{0.000843}{0.6}\\ =0.0014\text{cm/sec}\end{array}$

Determine the seepage velocity of soil III using the relation.

$v_{s3}=\frac{v}{n_3}$

Here, $n_3$ is the porosity of the soil III.

Substitute 0.000843 cm/sec for v and 0.33 for $n_3$.

$\begin{array}{c}{v}_{s3}=\frac{0.000843}{0.33}\\ =0.00255\text{cm/sec}\end{array}$

Draw graph of variation of the discharge velocity and the seepage velocity along the sample axis (X–X).

Refer Figure (1) in Part (c).

(e)

To determine

Find the height of the vertical columns of water inside piezometers A and B installed on the sample axis.

Answer to Problem 7.1CTP

The height of the vertical columns of water at point A is $\underset{\_}{656.3\text{mm}}$.

The height of the vertical columns of water at point B is $\underset{\_}{652.3\text{mm}}$.

Explanation of Solution

Given information:

The length of each soil layer $\left(H_1,H_2,H_3\right)$ is 200 mm.

The total length of the soil layer H is 600 mm.

The diameter of the cylindrical tube d is 150 mm.

The constant head difference $\Delta h$ is 470 mm.

The porosity of the soil layer I $\left(n_1\right)$ is 0.5.

The porosity of the soil layer II $\left(n_2\right)$ is 0.6.

The porosity of the soil layer III $\left(n_3\right)$ is 0.33.

The hydraulic conductivity of soil layer I $\left(k_1\right)$ is $5\times {10}^{-3}\text{cm/sec}$.

The hydraulic conductivity of soil layer II $\left(k_2\right)$ is $4.2\times {10}^{-2}\text{cm/sec}$.

The hydraulic conductivity of soil layer III $\left(k_3\right)$ is $3.9\times {10}^{-4}\text{cm/sec}$.

Calculation:

The height of water column is equal to the Piezometric or pressure head at a point.

Determine the height of water in point A.

$H_A={\left(u/{\gamma }_w\right)}_{200}$

Substitute 656.3 mm for ${\left(u/{\gamma }_w\right)}_{200}$.

$H_A=656.3\text{mm}$

Therefore, the height of the vertical columns of water at point A is $\underset{\_}{656.3\text{mm}}$.

Determine the height of water in point B.

$H_B={\left(u/{\gamma }_w\right)}_{400}$

Substitute 652.3 mm for ${\left(u/{\gamma }_w\right)}_{400}$.

$H_B=652.3\text{mm}$

Therefore, the height of the vertical columns of water at point B is $\underset{\_}{652.3\text{mm}}$.