Question

A 3 m thick soil starta has coefficient of permeability $4\times {10}^{-3}cm/secat{25}^{\circ }C$. A separate test gave porosity 30% and unit weight of 2.1 gm/cc at moisture content of 25%. If the temperature rises to ${40}^{\circ }C$, the increase in discharge per $c{m}^2$ to maintain critical condition is $3.92\times {10}^{-3}c{m}^3/sec/m^2$.
[ $\mu =10\times {10}^{-3}poise@{25}^{\circ }Cand6\times {10}^{-3}poise@{40}^{\circ }C$]

• A. 2.62
• B. 1.38
• C. 4.60
• D. 6.15

Answer 1

The correct option is A 2.62

Given

$L=3m$

$At{25}^{\circ }C,K_1=4\times {10}^{-3}cm/sec$

$n=30\%$

$\gamma =21g/cc$

$w=25\%$

$e=\frac{n}{1-n}=\frac{0.3}{1-0.3}=0.43$

To maintain critical condition

$i=i_c=(G-1)(1-n)$

$\gamma =\left(\frac{G+e{s}_r}{1+e}\right){\gamma }_w=\left(\frac{G+wG}{1+e}\right){\gamma }_w$

$\therefore \gamma =\frac{G{\gamma }_w}{1+e}(1+w)$

$2.1=\frac{G\times (1+0.25)}{1+0.43}\times 1\Rightarrow G=2.4$

$\therefore i_c=(2.4-1)(1-0.3)=0.98@{25}^{\circ }C$

$Discharge/c{m}^2\Rightarrow \frac{Q}{A}=V=k\times i=4\times {10}^{-3}\times 0.98$

$=3.92\times {10}^{-3}c{m}^3/s/c{m}^2$

$K\propto \frac{1}{\mu }$

$\frac{K_{{40}^{\circ }C}}{K_{{25}^{\circ }C}}=\frac{{\mu }_{{25}^{\circ }C}}{{\mu }_{{40}^{\circ }C}}=\frac{10\times {10}^{-3}}{6\times {10}^{-3}}=1.67$

$K_{{40}^{\circ }C}=1.67\times 4\times {10}^{-3}=6.67\times {10}^{-3}cm/sec$

$NewDischarge/c{m}^2$

$Q_2=6.67\times {10}^{-3}\times 0.98\times 1$

$=6.54\times {10}^{-3}c{m}^3/s/c{m}^2$

Increase in discharge

$=6.54\times {10}^{-3}-3.92\times {10}^{-3}$

$=2.62\times {10}^{-3}c{m}^3/s/c{m}^2$