CivilSolve

Environmental

Sedimentation Tank (Overflow Rate)

Clarifier/sedimentation-basin performance numbers: surface overflow rate (surface loading) SOR = Q/A — the settling-velocity benchmark — plus detention time V/Q when the depth is given, weir loading rate Q/L_weir, and horizontal flow-through velocity. Use for: 'determine the overflow rate and detention time', clarifier loading checks, weir loading. Not for: solids flux/thickening analysis.

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Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

overflow rate + detention (the motivating exam problem)

Given

flowrate
10000 m^3/day
length
20 m
width
10 m
depth
3 m

Assumptions

  • Ideal horizontal-flow settling: the overflow rate equals the critical settling velocity that is just fully removed.

Solution steps

  1. Surface overflow rate

    Flow divided by plan area — dimensionally a velocity: particles settling faster than this are captured.

    SOR = Q / A

    SOR = 50 m per day (= 0.000579 m/s)

    = 50

  2. Detention time

    Tank volume over flow — how long a parcel of water stays inside.

    td = V/Q = A·d/Q

    V = 600 m^3; td = 1.44 h

    = 1.44 h

Results

Surface overflow rate (m³/m²·day = m/day)

50

Surface area

200m^2

Tank volume

600m^3

Detention time

1.44h

Horizontal flow-through velocity

0.003858m/s

Where this answer was checked
source
SOR = Q/A; td = A·d/Q — 10,000 m³/day tank, 20×10×3 m
verified by
hand-recomputed
derivation
A = 200 m²; SOR = 10000/200 = 50 m/day. V = 600 m³; td = 600/10000 day = 0.06 day = 1.44 h.

Example 2

US clarifier: 2 MGD over 4000 ft²

Given

flowrate
2 mgd
surface area
4000 ft^2

Assumptions

  • Ideal horizontal-flow settling: the overflow rate equals the critical settling velocity that is just fully removed.

Solution steps

  1. Surface overflow rate

    Flow divided by plan area — dimensionally a velocity: particles settling faster than this are captured.

    SOR = Q / A

    SOR = 20.37 m per day (= 0.000774 ft/s)

    = 20.37

Results

Surface overflow rate (m³/m²·day = m/day)

20.37

Surface area

4000ft^2

Where this answer was checked
source
SOR = Q/A with unit conversion — MGD and square feet
verified by
hand-recomputed
derivation
Q = 7570.82 m³/day; A = 4000·0.092903 = 371.61 m²; SOR = 20.37 m/day.

Example 3

weir loading rate

Given

flowrate
5000 m^3/day
surface area
150 m^2
weir length
30 m

Assumptions

  • Ideal horizontal-flow settling: the overflow rate equals the critical settling velocity that is just fully removed.

Solution steps

  1. Surface overflow rate

    Flow divided by plan area — dimensionally a velocity: particles settling faster than this are captured.

    SOR = Q / A

    SOR = 33.33 m per day (= 0.000386 m/s)

    = 33.33

  2. Weir loading rate

    Flow per length of effluent weir — controls outlet currents.

    WLR = 166.7 m³ per m of weir per day

Results

Surface overflow rate (m³/m²·day = m/day)

33.33

Surface area

150m^2

Weir loading rate (m³/m·day)

166.7

Where this answer was checked
source
WLR = Q/L_weir — 5000 m³/day over a 30 m weir
verified by
hand-recomputed
derivation
WLR = 5000/30 = 166.7 m³/(m·day).