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GATE Civil Engineering (CE) 2019 Shift 1 Solved Paper
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© examsnet.com
Question : 61
Total: 65
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
/
c
m
3
;
density of water
=
1
g
∕
c
m
3
;
dynamic viscosity of water
=
0.01
g
/
(
c
m
.
s
)
gravitational acceleration
=
980
c
m
/
s
2
.
If the incoming water contains particles of diameter 25
µ
m
(spherical and uniform) the removal efficiency of these particles is
100
%
65
%
78
%
51
%
Validate
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
c
m
/
s
e
c
Given flow rate
(
Q
)
=
0.2
m
2
/
s
e
c
We know that surface overflow rate
(
V
0
)
=
Volume
/
time
surface area
V
0
=
Q
surface area
surface area
=
B
×
L
=
8
×
32
=
256
m
2
V
0
=
Q
B
L
=
0.2
m
3
/
s
e
c
256
=
7.8125
×
10
−
4
m
/
s
e
c
V
0
=
7.8125
×
10
−
2
c
m
/
s
e
c
Particle removal efficiency
(
η
)
=
V
s
V
0
×
100
η
=
0.051
0.078
×
100
=
65.38
%
© examsnet.com
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