Question

A bed of uniform sand having particle size 0.66 mm diameter and specific gravity of 2.65, porosity 0.40 and depth 60 cm is to be washed hydraulically. Considering laminar flow and Stoke's law valid, what is the value of backwash velocity?
$\text{(Expansion of bed is 50\%,}\nu \text{= 1.31}\times {10}^{-2}c{m}^2\text{/sec)}$

• A. 2.21 cm/sec
• B. 3.81 cm/sec
• C. 2.933 cm/sec
• D. 3.56 cm/sec

Answer 1

The correct option is C 2.933 cm/sec

Given, $L_e=1.5L$

$\frac{L_e}{L}=\frac{1-\eta }{1-{\eta }_e}(\eta =Porosity)$

$1.5=\frac{1-0.4}{1-{\eta }_e}$

${\eta }_e=0.6$

$V_s=\frac{(G-1)g{d}^2}{18v}=\frac{(2.65-1)\times 9.81\times 100\times (0.066{)}^2}{18\times 1.31\times {10}^{-2}}$

$V_s=29.9cm/sec$

${\eta }_e={\left(\frac{{\overline{V}}_s}{V_s}\right)}^{0.22}$

$({\overline{V}}_s=Backwashvelocity)$

$0.6={\left(\frac{{\overline{V}}_s}{29.9}\right)}^{0.22}$

${\overline{V}}_s=2.933cm/sec$