Sedimentation
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/dayft.^{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.
Specifications
The sedimentation basin we will design in this lesson will be a rectangular sedimentation basin with the following specifications:

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:
 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.
 Calculate the weir length.
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:
Q_{c} = Q / n
Where:
Q_{c} = flow in one tank
Q = total flow
n = number of tanks
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:
Q_{c} = (1.5 MGD) / 3
Q_{c} = 0.5 MGD
Q_{c} = 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/dayft^{2} in order to design the most efficient sedimentation basin.
The surface area is calculated using the following formula:
A = Q_{c} / O.R.
Where:
Where:
A = surface area, ft^{2}
Q_{c} = flow, gal/day
O.R. = overflow rate, gal/dayft^{2}
Q_{c} = flow, gal/day
O.R. = overflow rate, gal/dayft^{2}
In our example, the surface area of one tank is calculated as follows:
A = (500,000 gal/day) / (500 gal/dayft^{2})
A = 1,000 ft^{2}
(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 ft^{3}/7.48 gal)
V = 11,141 ft^{3}
V = (500,000 gal/day) (4 hr) (1 day/24 hr) (1 ft^{3}/7.48 gal)
V = 11,141 ft^{3}
(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, ft^{3}
A = surface area, ft^{2}
V = volume, ft^{3}
A = surface area, ft^{2}
For our example, the depth is calculated to be:
d = (11,141 ft^{3}) / (1,000 ft^{2})
d = 11.1 ft
The specifications note that the depth should be between 7 and 16 feet. Our calculated depth is within the recommended 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:
Where:
V = volume
L = length
W = width
d = depth
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
The length is calculated as:
L = 4 (15.8 ft)
L = 63.2 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 crosssectional area of the tank is calculated:
A_{x} = W d
A_{x} = (15.8 ft) (11.1 ft)
A_{x} = 175.4 ft^{2}
A_{x} = (15.8 ft) (11.1 ft)
A_{x} = 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 = Q_{c} / A_{x}
V = (0.0000928 ft^{3}day/galmin) (500,000 gal/day) / (175.4 ft^{2})
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/dayft and use the following equation to calculate the weir length:
L_{w} = Q_{c} / W.L
Where:
Where:
L_{w} = weir length, ft
Q_{c} = flow in one tank, gal/day
W.L. = weir loading, gal/dayft
Q_{c} = flow in one tank, gal/day
W.L. = weir loading, gal/dayft
So, in our example, the weir length is calculated as follows:
L_{w} = (500,000 gal/day) / (15,000 gal/dayft)
L_{w} = 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 ft^{2} and a volume of 11,141 ft^{3}. The flow through velocity will be 0.26 ft/min. The weir length will be 33.3 ft.
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