SME
Environmental Engineering
Water treatment
Use this for “0.1 m³/s flows through a settling basin 30 m long, 10 m wide and 3 m deep. Find the overflow rate, the detention time and the share of 15 μm sand removed.”
Diagram
v₀
=
Q
LW
=
0.1
30 × 10
=
3.33 ×
10
-4
m/s;
detention
t
=
LWD
Q
=
2.5
h.
vₛ
<
v₀:
a
particle
entering
at
the
surface
reaches
the
outlet
still
1.18
m
above
the
floor;
those
entering
in
the
lowest
vₛ
v₀
=
60.7%
of
the
depth
settle.
The depth is drawn 3 times its scale.
This is an example. Type your own values to start a new problem.
Q
Flow
example
L
Length
example
W
Width
example · m
D
Depth
example · m
v₀
Overflow rate
calculated · m/s
t
Detention time
calculated
d
Grain diameter
example
vₛ
Settling speed
calculated · m/s
R
Share removed
calculated · %
Formulas
v₀
=
Q
L
×
W
3.3333 ×
10
-4
=
0.1
30 × 10
t
=
L
×
W
×
D
3,600 ×
Q
2.5
=
30
×
10
×
3
3,600 × 0.1
vₛ
=
9.81
×
(2,650
−
1,000)
×
d
2
18 × 0.001
2.0233 ×
10
-4
=
9.81
×
(2,650
−
1,000)
×
(1.5 ×
10
-5
)
2
÷
(18
×
0.001)
R
=
min(100,
100
×
vₛ
v₀
)
60.7
=
min(100,
100
×
2.0233 ×
10
-4
3.3333 ×
10
-4
)
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
An ideal (Camp) basin: the water moves evenly from inlet to outlet, and grains reaching the floor stay.
•
Discrete grains that don’t flocculate: sand of 2650 kg/m³, finer than about 0.1 mm so Stokes’ law holds.
•
Water at 20 °C: ρw = 1000 kg/m³ and μ = 0.001 Pa·s; g = 9.81 m/s².
Step-by-step
WE KNOW
Q = 0.1 m³/s, L = 30 m, W = 10 m, D = 3 m, d = 15 μm
FIND
overflow rate (v₀), detention time (t), settling speed (vₛ), share removed (R)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m).
d = 15 μm = 1.5 × 10⁻⁵ m (1 m = 10⁶ μm)
Step 2 · Find overflow rate (v₀)
Use v₀ = Q ÷ (L × W)
The overflow rate is the flow spread over the basin’s surface, the speed the water rises past.
v₀
=
0.1
30 × 10
v₀
=
0.1
300
v₀
=
3.3333 ×
10
-4
m/s
Step 3 · Find detention time (t)
Use t = L × W × D ÷ (3,600 × Q)
The basin’s volume over the flow is how long the water stays, in seconds; 3600 s make an hour.
t
=
30
×
10
×
3
3,600 × 0.1
t
=
30
×
10
×
3
360
t
=
300
×
3
360
t
=
900
360
t
=
2.5
h
Step 4 · Find settling speed (vₛ)
Use vₛ = 9.81 × (2,650 − 1,000) × d² ÷ (18 × 0.001)
Stokes’
law:
the
grain’s
weight
less
its
buoyancy
balances
the
drag,
g
(ρₚ
−
ρw)
d
2
18μ
.
vₛ
=
9.81
×
(2,650
−
1,000)
×
(1.5 ×
10
-5
)
2
÷
(18
×
0.001)
vₛ
=
9.81
×
1,650
×
(1.5 ×
10
-5
)
2
÷
0.018
vₛ
=
9.81
×
1,650
×
2.25 ×
10
-10
0.018
vₛ
=
16,186.5
×
2.25 ×
10
-10
0.018
vₛ
=
3.642 ×
10
-6
0.018
vₛ
=
2.0233 ×
10
-4
m/s
Step 5 · Find share removed (R)
Use R = min(100, 100 × vₛ ÷ v₀)
Grains
entering
below
the
share
vₛ
v₀
of
the
depth
reach
the
floor;
once
vₛ
≥
v₀
they
all
do.
R
=
min(100,
100
×
2.0233 ×
10
-4
3.3333 ×
10
-4
)
R
=
60.7%
Check
Put the numbers back in. Both sides should match.
3.3333 ×
10
-4
=
0.1
30 × 10
✓
2.5
=
30
×
10
×
3
3,600 × 0.1
✓
2.0233 ×
10
-4
=
9.81
×
(2,650
−
1,000)
×
(1.5 ×
10
-5
)
2
÷
(18
×
0.001)
✓
60.7
=
min(100,
100
×
2.0233 ×
10
-4
3.3333 ×
10
-4
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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