Mechanical Treatment of Storm Water. Thomas Soerens University of Arkansas. Outline. Fundamentals of Settling Catch basin sizing examples Alternative mechanical treatment technologies. Settling. Example Regulation Storm water treatment should remove 80% of Total Suspended Solids (TSS).
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University of Arkansas
Fundamentals of Settling
Catch basin sizing examples
Alternative mechanical treatment technologies
Storm water treatment should remove 80% of Total Suspended Solids (TSS).
vague: what size solids?
System 1: Removes 80% solids with d50 of 50 microns
System 2: Removes 80% solids with d50 of 100 microns
System 2 not remove 80% of solids with d50 of 50 microns
In comparing systems, must see data side by side and compare apples to apples
Example Basin (next slide)
A rectangular settling tank processes 48,000 m3/day, is 6 m wide, 36 m long, and 4 m deep.
What is the average hydraulic retention time in the tank (hr)?
t0 = Vol/Q = (6m x 4m x 36m) / 48000 m3/day = 0.018 day = 0.432 hr = 26 min
Assuming horizontal flow, what is the flow (approach) velocity (m/d)?
vx = Q/(w x h) = Q / (6m x 4m) = 2000 m/day = 1.4 m/min
What is the overflow rate for the tank (m/d)?
v0 = Q/(w x L) = Q / (6m x 36m) = 222 m/day = 0.15 m/min note: vo = 4 m / 0.018 day = depth / retention time
If it enters 4 m above bottom, it has to drop 4 m in 26 min to hit bottom
If particle has a settling velocity greater than the overflow rate (0.15 m/min), it will settle out.
example: vs = 0.20 m/min
in 26 minutes, it drops 0.20 x 26 = 5.2 m > depth
to drop 4 m, it takes 4/0.20 = 20 min < t0
in 20 minutes, it travels 20 x 1.4 m/min = 28 m < L
If the settling velocity is less than the overflow rate, it doesn’t hit bottom
example: vs = 0.10 m/min
in 26 minutes, it drops 0.10 x 26 = 2.6 m < depth
to drop 4 m, it takes 4/0.10 = 40 min > t0
in 40 minutes, it travels 40 x 1.4 m/min = 56 m > L
Approximately vs/vo fraction of particles will settle out
example: vs = 0.10 m/min
Removal =~ 0.10/0.15 = 0.65 = 65% removal
note: this is for horizontal clarifiers
note: turbulence happens
Stoke’s law for settling velocity of spheres:
vs = [(rp – rw)d2g]/18m
rp , rw = density of particle, water
d = diameter of particle
g = gravity
m = viscosity
A 100 micron particle will have a settling velocity 4 times that of a 50 micron particle
side note for water or wastewater treatment:
In Stoke’s Law, what can be changed?
Do you see why we coagulate and flocculate
Set overflow rate of basin at design flow equal to d50 of a grain-size analysis of dirt you want to remove.
Can have v0 up to 1/0.8 = 1.25 of settling velocity
100 micron particle
choose aspect ratio: Length = 4 x width
set vo = Q/Asurface = Q/(w x 4w) = 0.015 m/sec
w = 1.7 m (5.5 ft) , L = 6.7 m (22 ft)
will a 5 ft x 20 ft basin work?
vo = Q/wL = 0.018 m/sec
vs/vo = 0.015/0.018 = 0.82 82% removal
okay for 80% removal
disclaimer: the above process is a principle, not a regulation or a standard.
depth not involved in calculation
choose depth based on practical considerations of separating clean water from dirt.
1 inch deep?
1.7 second retention time - solids only have to fall 1 in to reach bottom
100 feet deep?
34 min retention time - solids fall 100 feet in 34 minutes
4 feet deep?
1.4 min ret time, velocity = 16 ft/min, might be good
at 24 min, 45% of particles have hit bottom (7m)
60% of particles have settled to 2 m; 75% to 0.6m
avg settling velocity of 15% of particles between 45% and 60% contours is about 3.4 m in 24 min; for next interval it’s 1.3m/24min.
removal rate = vs/vo
overall removal = 45% + 15% x (3.4m/24min)/(7m/24min) + 15% x (1.3/7) + …
= 45% + 7.3% + 2.8% =~ 55%
note: could also take this approach with grain size analysis data
next: examples of mechanical storm water treatment systems
2 units in series
Water Quality Unit (WQU)
series of weirs from 60-in diameter HDPE pipe.
two manholes for maintenance
Detention/Infiltration Unit (DIU)
three 40-ft sections of 48 in perforated HDPE pipe
top and sides of excavation are wrapped in geotextile
1 cfs or less though WQU then DIU
> 1 cfs bypass WQU and go into DIU
WQU size: 5 ft x 20 ft
catchment area: 1 acre
peak flow 1 cfs
treatment volume 3264 cf
$50k per acre
requires high maintenance
University of Arkansas
Identify existing and emerging mechanical storm water treatment technologies and describe design and decision parameters.
Santa Monica Urban Runoff Recycling Facility
Joint Santa Monica-Los Angeles Project
Rotating Drum Screens