Document Text Contents
Page 1
Lesson 13:
Sedimentation Basin Design and Problems
Designing a Rectangular Sedimentation Tank
Introduction
Designing a rectangular sedimentation tank is similar in many ways to designing a flocculation
chamber.However, water in a sedimentation basin is not agitated, so the velocity gradient is not
a factor in the calculations. Instead, two additional characteristics are important in designing a
sedimentation basin.
The overflow rate (also known as the surface loading or the surface overflow rate) is equal to
the settling velocity of the smallest particle which the basin will remove. Surface loading is
calculated by dividing the flow by the surface area of the tank.Overflow rate should usually be
less than 1,000 gal/day-ft.2
The weir loading is another important factor in sedimentation basin efficiency.Weir loading,
also known as weir overflow rate, is the number of gallons of water passing over a foot of weir
per day. The standard weir overflow rate is 10,000 to 14,000 gpd/ft and should be less than
20,000 gpd/ft.Longer weirs allow more water to flow out of the sedimentation basin without
exceeding the recommended water velocity.
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Specifications
The sedimentation basin we will design in this lesson will be a rectangular sedimentation basin
with the following specifications:
• Rectangular basin
• Depth: 7-16 ft
• Width: 10-50 ft
• Length: 4 × width
• Influent baffle to reduce flow momentum
• Slope of bottom toward sludge hopper >1%
• Continuous sludge removal with a scraper velocity <15 ft/min
• Detention time: 4-8 hours
• Flow through velocity: <0.5 ft/min
• Overflow rate: 500-1,000 gal/day-ft2
• Weir loading: 15,000-20,000 gal/day-ft
Overview of Calculations
We will determine the surface area, dimensions, and volume of the sedimentation tank as well as
the weir length. The calculations are as follows:
1. Divide flow into at least two tanks.
2. Calculate the required surface area.
3. Calculate the required volume.
4. Calculate the tank depth.
5. Calculate the tank width and length.
6. Check flow through velocity.
7. If velocity is too high, repeat calculations with more tanks.
8. Calculate the weir length.
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1. Divide the Flow
The flow should be divided into at least two tanks and the flow through each tank should be
calculated using the formula shown below:
Qc = Q / n
Where:
Qc = flow in one tank
Q = total flow
n = number of tanks
We will consider a treatment plant with a flow of 1.5 MGD.We will divide the flow into three
tanks, so the flow in one tank will be:
Qc = (1.5 MGD) / 3
Qc = 0.5 MGD
2. Surface Area
Next, the required tank surface area is calculated.We will base this surface area on an overflow
rate of 500 gal/day-ft2 in order to design the most efficient sedimentation basin.
The surface area is calculated using the following formula:
A = Qc / O.R.
Where:
A = surface area, ft2
Qc = flow, gal/day
O.R. = overflow rate, gal/day-ft2
In our example, the surface area of one tank is calculated as follows:
A = (500,000 gal/day) / (500 gal/day-ft2)
A = 1,000 ft2
Page 4
(Notice that we converted the flow from 0.5 MGD to 500,000 gal/day before beginning our
calculations.)
3. Volume
The tank volume is calculated just as it was for flocculation basins and flash mix chambers, by
multiplying flow by detention time. The optimal detention time for sedimentation basins
depends on whether sludge removal is automatic or manual.When sludge removal is manual,
detention time should be 6 hours.We will consider a tank with automatic sludge removal, so the
detention time should be 4 hours.
The volume of one of our tanks is calculated as follows:
V = Q t
V = (500,000 gal/day) (4 hr) (1 day/24 hr) (1 ft3/7.48 gal)
V = 11,141 ft3
(Notice the conversions between days and hours and between cubic feet and gallons.)
4. Depth
The tank's depth is calculated as follows:
d = V / A
Where:
d = depth, ft
V = volume, ft3
A = surface area, ft2
For our example, the depth is calculated to be:
d = (11,141 ft3) / (1,000 ft2)
Page 5
d = 11.1 ft
The specifications note that the depth should be between 7 and 16 feet.Our calculated depth is
within therecommended range.If the depth was too great, we would begin our calculations
again, using a larger number of tanks.If the depth was too shallow, we would use a smaller
number of tanks.
5. Width and Length
You will remember that the volume of a rectangular solid is calculated as follows:
V = L W d
Where:
V = volume
L = length
W = width
d = depth
For our tank, the length has been defined as follows:
L = 4 W
Combining these two formulas, we get the following formula used to calculate the width of our
tank:
In the case of our example, the tank width is calculated as follows:
W = 15.8 ft
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The length is calculated as:
L = 4 (15.8 ft)
L = 63.2 ft
6 and 7. Flow Through Velocity
Checking the flow through velocity is done just as it was for the flocculation basin.First, the
cross-sectional area of the tank is calculated:
Ax = W d
Ax = (15.8 ft) (11.1 ft)
Ax = 175.4 ft
2
Then the flow through velocity of the tank is calculated (with a conversion from gallons to cubic
feet and from days to minutes):
V = Qc / Ax
V = (0.0000928 ft3-day/gal-min) (500,000 gal/day) / (175.4 ft2)
V = 0.26 ft/min
The velocity for our example is less than 0.5 ft/min, so it is acceptable.As a result, we do not
need to repeat our calculations.
8. Weir Length
The final step is to calculate the required length of weir.We will assume a weir loading of
15,000 gal/day-ft and use the following equation to calculate the weir length:
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Lw = Qc / W.L
Where:
Lw = weir length, ft
Qc = flow in one tank, gal/day
W.L. = weir loading, gal/day-ft
So, in our example, the weir length is calculated as follows:
Lw = (500,000 gal/day) / (15,000 gal/day-ft)
Lw = 33.3 ft
The weir length should be 33.3 ft.
Conclusions
Our plant should build a sedimentation tank which is 11.1 feet deep, 15.8 feet wide, and 63.2 feet
long. This tank will have a surface area of 1,000 ft2 and a volume of 11,141 ft3. The flow
through velocity will be 0.26 ft/min. Theweir length will be 33.3 ft.
Review
Sedimentation basin efficiency is influenced by floc characteristics, water temperature, short-
circuiting, gases in the water, algal growth on tank walls, intermittent tank operation, surface
loading, and weir loading. To ensure optimal performance, the operator should test turbidity and
temperature of the water and should visually survey the basin
Design of a sedimentation basin involves the following steps:
· Divide flow into at least two tanks.
· Calculate the required surface area.
· Calculate the required volume.
· Calculate the tank depth.
· Calculate the tank width and length.
· Check flow through velocity.
· If velocity is too high, repeat calculations with more tanks.
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· Calculate the weir length.
New Formulas Used
To calculate tank surface area:
A = Qc / O.R.
To calculate tank depth:
d = V / A
To calculate width of a rectangular tank where length is four times the width:
To calculate length of a rectangular tank where length is four times the width:
L = 4 W
To calculate flow through velocity:
V = Qc / Ax
To calculate weir length:
Lw = Qc / W.L
Lesson 13:
Sedimentation Basin Design and Problems
Designing a Rectangular Sedimentation Tank
Introduction
Designing a rectangular sedimentation tank is similar in many ways to designing a flocculation
chamber.However, water in a sedimentation basin is not agitated, so the velocity gradient is not
a factor in the calculations. Instead, two additional characteristics are important in designing a
sedimentation basin.
The overflow rate (also known as the surface loading or the surface overflow rate) is equal to
the settling velocity of the smallest particle which the basin will remove. Surface loading is
calculated by dividing the flow by the surface area of the tank.Overflow rate should usually be
less than 1,000 gal/day-ft.2
The weir loading is another important factor in sedimentation basin efficiency.Weir loading,
also known as weir overflow rate, is the number of gallons of water passing over a foot of weir
per day. The standard weir overflow rate is 10,000 to 14,000 gpd/ft and should be less than
20,000 gpd/ft.Longer weirs allow more water to flow out of the sedimentation basin without
exceeding the recommended water velocity.
Page 2
Specifications
The sedimentation basin we will design in this lesson will be a rectangular sedimentation basin
with the following specifications:
• Rectangular basin
• Depth: 7-16 ft
• Width: 10-50 ft
• Length: 4 × width
• Influent baffle to reduce flow momentum
• Slope of bottom toward sludge hopper >1%
• Continuous sludge removal with a scraper velocity <15 ft/min
• Detention time: 4-8 hours
• Flow through velocity: <0.5 ft/min
• Overflow rate: 500-1,000 gal/day-ft2
• Weir loading: 15,000-20,000 gal/day-ft
Overview of Calculations
We will determine the surface area, dimensions, and volume of the sedimentation tank as well as
the weir length. The calculations are as follows:
1. Divide flow into at least two tanks.
2. Calculate the required surface area.
3. Calculate the required volume.
4. Calculate the tank depth.
5. Calculate the tank width and length.
6. Check flow through velocity.
7. If velocity is too high, repeat calculations with more tanks.
8. Calculate the weir length.
Page 3
1. Divide the Flow
The flow should be divided into at least two tanks and the flow through each tank should be
calculated using the formula shown below:
Qc = Q / n
Where:
Qc = flow in one tank
Q = total flow
n = number of tanks
We will consider a treatment plant with a flow of 1.5 MGD.We will divide the flow into three
tanks, so the flow in one tank will be:
Qc = (1.5 MGD) / 3
Qc = 0.5 MGD
2. Surface Area
Next, the required tank surface area is calculated.We will base this surface area on an overflow
rate of 500 gal/day-ft2 in order to design the most efficient sedimentation basin.
The surface area is calculated using the following formula:
A = Qc / O.R.
Where:
A = surface area, ft2
Qc = flow, gal/day
O.R. = overflow rate, gal/day-ft2
In our example, the surface area of one tank is calculated as follows:
A = (500,000 gal/day) / (500 gal/day-ft2)
A = 1,000 ft2
Page 4
(Notice that we converted the flow from 0.5 MGD to 500,000 gal/day before beginning our
calculations.)
3. Volume
The tank volume is calculated just as it was for flocculation basins and flash mix chambers, by
multiplying flow by detention time. The optimal detention time for sedimentation basins
depends on whether sludge removal is automatic or manual.When sludge removal is manual,
detention time should be 6 hours.We will consider a tank with automatic sludge removal, so the
detention time should be 4 hours.
The volume of one of our tanks is calculated as follows:
V = Q t
V = (500,000 gal/day) (4 hr) (1 day/24 hr) (1 ft3/7.48 gal)
V = 11,141 ft3
(Notice the conversions between days and hours and between cubic feet and gallons.)
4. Depth
The tank's depth is calculated as follows:
d = V / A
Where:
d = depth, ft
V = volume, ft3
A = surface area, ft2
For our example, the depth is calculated to be:
d = (11,141 ft3) / (1,000 ft2)
Page 5
d = 11.1 ft
The specifications note that the depth should be between 7 and 16 feet.Our calculated depth is
within therecommended range.If the depth was too great, we would begin our calculations
again, using a larger number of tanks.If the depth was too shallow, we would use a smaller
number of tanks.
5. Width and Length
You will remember that the volume of a rectangular solid is calculated as follows:
V = L W d
Where:
V = volume
L = length
W = width
d = depth
For our tank, the length has been defined as follows:
L = 4 W
Combining these two formulas, we get the following formula used to calculate the width of our
tank:
In the case of our example, the tank width is calculated as follows:
W = 15.8 ft
Page 6
The length is calculated as:
L = 4 (15.8 ft)
L = 63.2 ft
6 and 7. Flow Through Velocity
Checking the flow through velocity is done just as it was for the flocculation basin.First, the
cross-sectional area of the tank is calculated:
Ax = W d
Ax = (15.8 ft) (11.1 ft)
Ax = 175.4 ft
2
Then the flow through velocity of the tank is calculated (with a conversion from gallons to cubic
feet and from days to minutes):
V = Qc / Ax
V = (0.0000928 ft3-day/gal-min) (500,000 gal/day) / (175.4 ft2)
V = 0.26 ft/min
The velocity for our example is less than 0.5 ft/min, so it is acceptable.As a result, we do not
need to repeat our calculations.
8. Weir Length
The final step is to calculate the required length of weir.We will assume a weir loading of
15,000 gal/day-ft and use the following equation to calculate the weir length:
Page 7
Lw = Qc / W.L
Where:
Lw = weir length, ft
Qc = flow in one tank, gal/day
W.L. = weir loading, gal/day-ft
So, in our example, the weir length is calculated as follows:
Lw = (500,000 gal/day) / (15,000 gal/day-ft)
Lw = 33.3 ft
The weir length should be 33.3 ft.
Conclusions
Our plant should build a sedimentation tank which is 11.1 feet deep, 15.8 feet wide, and 63.2 feet
long. This tank will have a surface area of 1,000 ft2 and a volume of 11,141 ft3. The flow
through velocity will be 0.26 ft/min. Theweir length will be 33.3 ft.
Review
Sedimentation basin efficiency is influenced by floc characteristics, water temperature, short-
circuiting, gases in the water, algal growth on tank walls, intermittent tank operation, surface
loading, and weir loading. To ensure optimal performance, the operator should test turbidity and
temperature of the water and should visually survey the basin
Design of a sedimentation basin involves the following steps:
· Divide flow into at least two tanks.
· Calculate the required surface area.
· Calculate the required volume.
· Calculate the tank depth.
· Calculate the tank width and length.
· Check flow through velocity.
· If velocity is too high, repeat calculations with more tanks.
Page 8
· Calculate the weir length.
New Formulas Used
To calculate tank surface area:
A = Qc / O.R.
To calculate tank depth:
d = V / A
To calculate width of a rectangular tank where length is four times the width:
To calculate length of a rectangular tank where length is four times the width:
L = 4 W
To calculate flow through velocity:
V = Qc / Ax
To calculate weir length:
Lw = Qc / W.L
