GATE Civil Engineering (CE) 2019 Shift 1 Solved Paper

Question : 61 of 65

Marks: +1, -0

Sedimentation basin in a water treatment plant is designed for a flow rate of

0.2 {m}^{3} {/} {s}

. The basin is rectangular with a length of

32 {m}

, width of

8 {m}

and depth of

4 {m}

. Assume that the settling velocity of these particles is governed by the Stokes' law. Given: density of the particles

=2.5 {g}{/} {cm}^{3} ;

density of water

=1 {g} \/ {cm} 3 ;

dynamic viscosity of water

=0.01 {g} {/}( {cm} . {s})

gravitational acceleration

=980 {cm} {/} {s}^{2} .

If the incoming water contains particles of diameter 25

µ {m}

(spherical and uniform) the removal efficiency of these particles is

Solution:

We know that
Settling velocity

( {V}_{ {s}})=\;{ {g}(ρ_{ {s}}-ρ_{ {m}}) {d}^{2}}/{18 µ}

{V}_{ {s}}=\;{980 ×(2.51-)(25 × 10^{-4})^{2}}/{18 × 0.01}=0.051 {cm} {/} {sec}

Given flow rate

( {Q})=0.2 {m}^{2}{/} {sec}

We know that surface overflow rate

( {V}_{0})=\;{\;\text " Volume "\; {/} \;\text " time "\;}/{\;\text " surface area "\;}

{V}_{0}=\;{ {Q}}/{\;\text " surface area "\;}

surface area

= {B} × {L}=8 × 32=256 {m}^{2}

{V}_{0}=\;{ {Q}}/{ {BL}}=\;{0.2 {m}^{3}{/} {sec}}/{256}=7.8125 × 10^{-4} {m} {/}{sec}

{V}_{0}=7.8125 × 10^{-2} {cm} {/} {sec}

Particle removal efficiency

(η)=\;{ {V}_{ {s}}}/{ {V}_{0}} × 100

η=\;{0.051}/{0.078} × 100=65.38 \%

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