Question

A $0.80m$ deep bed of sand filter (length $4m$ and width $3m$) is made of uniform particles (diameter = $0.40mm$, specific gravity $=2.65$, shape factor $=0.85$) with bed porosity of $0.4$. The bed has to be backwashed at a flow rate of $3.60m^3/min$ . During backwashing, if the terminal settling velocity of sand particles is $0.05m/s$, the expanded bed depth (in m, round off to 2 decimal places) is

• A. 1.2075

Answer 1

The correct option is A 1.2075
$n_{ex}={\left(\frac{V_B}{V_t}\right)}^{0.22}$

$V_B=\frac{3.6}{4\times 3\times 60}=5\times {10}^{-3}m/sec$

$n_{ex}={\left(\frac{5\times {10}^{-3}}{0.05}\right)}^{0.22}$

$n_{ex}=0.6025$

then,

$L_{ex}(1-n_{ex})=L(1-n)$

$\Rightarrow L_{ex}(1-0.6025)=0.8\times (1-0.4)$

$L_{ex}=1.2075m$