Lesson 16:
Sedimentation Calculations
Objective
In this lesson we will learn the following:
Lecture
Introduction to Sedimentation
Sedimentation is the process of allowing particles in suspension in water to settle out of the suspension under the effect of gravity. The particles that settle out from the suspension become sediment, and is known as sludge. In water treatment, sedimentation might be used to reduce the concentration of particles in suspension before the application of coagulation, to reduce the amount of coagulating chemicals needed, or after coagulation, possibly, flocculation. When sedimentation is applied after coagulation, its purpose is to reduce the concentration of solids in suspension so that filtration can function effectively.
Calculations
The two most common tank shapes of sedimentation basins are rectangular and cylindrical. The equations for calculating the volumes of these shapes were given in lesson 10. If you need to review how to work those, please do.
Tank Volume
Rectangular sedimentation basin:
Volume, gal = Length, ft x Width, ft x Depth, ft x (7.48 gal/ft3)
Cylindrical clarifiers:
Volume, gal = 0.785 x (Diameter)2 x Depth, ft x (7.48 gal/ft3)
Example:
A sedimentation basin is 35 ft wide by 75 ft long and contains water to a depth of 12.5 ft. What is the volume of water in the basin, in gallons?
Volume, gal = Length, ft x Width, ft x Depth, ft x (7.48 gal/ft3)
Volume, gal = 75 ft x 35 ft x 12.5 ft x (7.48 gal/ft3)
Volume, gal = 245,438 gal
Example:
A circular clarifier has a diameter of 40 ft and a depth of water of 25 ft. What is the volume of water in the clarifier?
Volume, gal = 0.785 x (Diameter)2 x Depth, ft x (7.48 gal/ft3)
Volume, gal = 0.785 x (40 ft)2 x 25 ft x (7.48 gal/ft3)
Volume, gal = 0.785 x (1600 ft2) x 25 ft x (7.48 gal/ft3)
Volume, gal = 234,872 gal
Detention Time
Detention time for clarifiers, or sedimentation basins, varies from 1 to 3 hours and can be determined from one of the following equations:
Basic Detention Time
Example:
A sedimentation basin has a volume of 142,000 gallons. If the flow to the tank is 133,00 gph, what is the detention time in the basin, in hours?
Rectangular Sedimentation Basin Detention Time
Example:
A flocculation basin is 55 ft long and 20 ft wide with a depth of water of 8.3 ft. If the flow to the basin is 1,825,000 gpd, what is the flocculation basin detention time?
First, convert the flow from gpd to gph so the units can cancel:
1,825,000 gal/day x (1 day/24 hrs) = 76,041.67 gph
Now calculate the detention time:
Circular Basin Detention Time
Example:
Determine the detention time of a circular clarifier that has a flow rate of 800,000 gpd with a diameter of 40 ft and a water depth of 15 ft.
First convert the flow from gpm to gph:
800,000 gal/min x (1 day/24 hrs) = 33,333 gph
Now determine the detention time of the clarifier:
Let's watch a video that shows you how to calculate detention time of a circular clarifier.
Surface Overflow Rate
The surface loading rate (similar to hydraulic loading rate - flow per unit area) is used to determine loading on sedimentation basins and circular clarifiers. Hydraulic loading rate measures the total water entering the process, whereas surface overflow rate measures only the water overflowing the process (plant flow only). Surface overflow rate calculations do not include recirculated flows and can be determined with the following formula:
A circular clarifier has a diamter of 60 ft. If the flow to the clarifier is 1650 gpm, what is the surface overflow rate, in gpm/ft2.
First determine the area of the clarifier: This can be done one of two ways:
Area = πr2
Area = π(30 ft)2
Area = π(900 ft2)
Area = 2826 ft2
or
Area = 0.785 x (Diameter)2
Area = 0.785 x (60 ft)2
Area = 0.785 x 3600 ft2
Area = 2826 ft2
Once you have the area calculated you can determine the surface overflow rate:
Example:
A sedimentation basin that is 65 ft by 20 ft receives a flow of 1250 gpm. What is the surface overflow rate, in gpm/ft2.
First determine the area of the sedimentation basin:
Area = L x W
Area = 65 ft x 20 ft
Area = 1300 ft2
Now determine the surface overflow rate:
Mean Flow Velocity
The measure of average velocity of the water as it travels through a rectangular sedimentation basin is known as mean flow velocity and is calculated by using the following equation:
Q = A x V
or
Flow, ft3/min = Cross-sectional area, ft2 x Volume, ft/min
*Hint: If the cutting plane is parallel to one of the two set sides, the cross-sectional area is found by L x H or W x H.
Example:
A sedimentation basin that is 45 ft long by 15 ft wide has water to a depth of 8.4 ft. When the flow through the basin is 875,000 gpd, what is the mean flow velocity in the basin, in ft/min?
Because we need the flow in cubic feet per min we must convert the flow from gpd to ft3/min:
875,000 gal/day x (1 ft3/7.48 gal) x (1 day/1440 min) = 81.24 ft3/min
Determine the cross-sectional area of the basin:
Area = L x H
Area = 45 ft x 8.4 ft
Area = 378 ft2
Now plug in the values you have calculated:
Flow, ft3/min = Cross-sectional area, ft2 x Volume, ft/min
81.24 ft3/min = 378 ft2 x Volume, ft/min
Volume, ft/min = 0.21 ft/min
Example:
A rectangular sedimentation basin that is 30 ft long and 18 ft wide with a water depth of 7.7 ft. If the flow to the basin is 1,220,000 gpd, what is the mean flow velocity, in ft/min?
Because we need the flow in cubic feet per min we must convert the flow from gpd to ft3/min:
1,220,000 gal/day x (1 ft3/7.48 gal) x (1 day/1440 min) = 113.27 ft3/min
Determine the cross-sectional area of the basin:
Area = L x H
Area = 30 ft x 7.7 ft
Area = 231 ft2
Now plug in the values you have calculated:
Flow, ft3/min = Cross-sectional area, ft2 x Volume, ft/min
113.27 ft3/min = 231 ft2 x Volume, ft/min
Volume, ft/min = 0.49 ft/min
Weir Loading Rate (Weir Overflow Rate)
The weir loading, or overflow, rate is the amount of water leaving the settling tank per linear foot of weir. The result of this calculation can be compared with design. Normally, weir overflow rates of 10,000 to 20,000 gpd/ft are used in the design of a settling tank. Typically, weir loading rates are measured in gallons per minute (gpm) flow over each foot (ft) of weir and is determined with the following equation:
Example:
A rectangular sedimentation basin has a total of 120 ft of weir. What is the weir loading rate, in gpm/ft, when the flow is 1,220,000 gpd.
We need to first convert the flow from gpd to gpm:
1,220,000 gal/day x (1 day/1440 min) = 847.22 gpm
Determine the weir loading rate:
Example:
A circular clarifier receives a flow of 4.25 MGD. If the diameter of the weir is 100 ft, what is the weir loading rate, in gpm/ft?
To determine the feet of weir, use the formula:
π x Diameter
Weir length, ft = π x Diameter
Weir length, ft = π x 100 ft
Weir length = 314 ft
Convert flow from MGD to gpm:
4.25 MG/day x (1,000,000 gal/1MG) x (1 day/1440 min) = 2951.39 gpm
Now plug in the values to determine the weir loading rate:
Determining Lime Dosage
mg/L
During the alum dosage process, lime is sometimes added to provide adequate alkalinity (HCO3) in the solids contact clarification process for the coagulation and precipitation of the solids. To determine the necessary lime dose, in mg/L, three steps are required.
First, the total alkalinity required is calculated. The total alkalinity required to react with the alum to be added and provide proper precipitation is determined using the following formula:
Total alkalinity required, mg/L = Alkalinity reacting with alum, mg/L + Alkalinity in the water, mg/L
*Hint: 1 mg/L alum reacts with 0.45 mg/L alkalinity.
Example:
Raw water requires an alum dose of 37 mg/L, as determined by jar testing. If a residual 25 mg/L alkalinity must be present in the water to ensure complete precipitation of alum added, what is the total alkalinity required, in mg/L?
First calculate the alkalinity that will react with 37 mg/L alkalinity:
Next, calculate the total alkalinity required:
Total alkalinity required, mg/L = Alkalinity reacting with alum, mg/L + Alkalinity in the water, mg/L
Total alkalinity required, mg/L = 16.65 mg/L + 25 mg/L
Total alkalinity required, mg/L = 41.65 mg/L
Example:
Jar tests indicate that 32 mg/L alum are optimum for a particular raw water. If a residual 20 mg/L alkalinity must be present to promote complete precipitation of alum added, what is the total alkalinity required, in mg/L?
First calculate the alkalinity that will react with 32 mg/L alkalinity:
Next, calculate the total alkalinity required:
Total alkalinity required, mg/L = Alkalinity reacting with alum, mg/L + Residual alkalinity, mg/L
Total alkalinity required, mg/L = 14.4 mg/L + 20 mg/L
Total alkalinity required, mg/L = 34.4 mg/L
In the next step we make a comparison between required alkalinity and alkalinity already in the raw water to determine how many mg/L alkalinity should be added to the water:
Alkalinity to be added, mg/L = Total alkalinity required, mg/L - Alkalinity in the water, mg/L
Example:
A total of 52 mg/L alkalinity is required to react with alum and ensure proper precipitation. If the raw water has an alkalinity of 33 mg/L as bicarbonate, how many mg/L alkalinty should be added to the water?
Alkalinity to be added, mg/L = Total alkalinity required, mg/L - Alkalinity in the water, mg/L
Alkalinity to be added, mg/L = 52 mg/L - 33 mg/L
Alkalinity to be added, mg/L = 19 mg/L
Finally, after determining the amount of alkalinity to be added to the water, we determine how much lime (the source of alkalinity) must be added.
Example:
It has been calculated that 19 mg/L of alkalinity must be added to a raw water. How much mg/L lime will be required to provide this amount of alkalinity?
*Hint: 1 mg/L alum reacts with 0.45 mg/L alkalinity.
1 mg/L alum reacts with 0.35 mg/L lime.
First, determine the mg/L lime required by using a proportion that relates bicarbonate alkalinity to lime. We calculate this through cross-multiplying:
This means you would multiply (0.45 x "x") and (0.35 x 19) and becomes:
0.45 x "x" = 0.35 x 19 You need "x" by itself so you divide both sides by 0.45, cancelling it out on the left
x = (0.35 x 19) / (0.45)
x = 14.78 mg/L lime
Example:
Given the following data, calculate the lime dose required, in mg/L:
To calculate the total alkalinity required, we must first calculate the alkalinity that will react with 48 mg/L: (use cross-multiplying again)
You can now determine the total alkalinity requirement:
Total alkalinity required, mg/L = Alkalinity reacting with alum, mg/L + Residual alkalinity, mg/L
Total alkalinity required, mg/L = 21.6 mg/L + 32 mg/L
Total alkalinity required, mg/L = 53.6 mg/L
Next, calculate how much alkalinity must be added to the water:
Alkalinity to be added, mg/L = Total alkalinity required, mg/L - Alkalinity in the water, mg/L
Alkalinity to be added, mg/L = 53.6 mg/L - 37 mg/L
Alkalinity to be added, mg/L = 16.6 mg/L
Finally, calculate the lime required to provide the additional alkalinity:
This means you would multiply (0.45 x "x") and (0.35 x 16.6) and becomes:
0.45 x "x" = 0.35 x 16.6 You need "x" by itself so you divide both sides by 0.45, cancelling it out on the left
x = (0.35 x 16.6) / (0.45)
x = 12.91 mg/L lime
lb/day
After the lime dose has been determined in terms of mg/L, it is a fairly easy process to calculate the lime dosage in lb/day, which is one of the most common calculations you will be doing. To convert from mg/L to lb/day lime dose, we use the following equation:
Lime, lb/day = Lime, mg/L x Flow, MGD x 8.34 lb/gal
Example:
The lime dose for raw water has been calculated to be 12.91 mg/L. If the flow to be treated is 3.2 MGD, how many lb/day lime will be required?
Lime, lb/day = Lime, mg/L x Flow, MGD x 8.34 lb/gal
Lime, lb/day = 12.91 mg/L x 3.2 MGD x 8.34 lb/gal
Lime, lb/day = 344.54 lb/day
Example:
The flow to a clarifier is 2,800,000 gpd. If the lime dose required is determined to be 16.6 mg/L, how many lb/day lime will be required?
First convert the flow from gpd to MGD:
2,800,000 gal/day x (1 MG/1,000,000 gal) = 2.8 MGD
Now determine the lb/day of lime required:
Lime, lb/day = Lime, mg/L x Flow, MGD x 8.34 lb/gal
Lime, lb/day = 16.6 mg/L x 2.8 MGD x 8.34 lb/gal
Lime, lb/day = 387.64 lb/day
g/min
To convert lime dosage from mg/L to grams per minute (g/min) lime, use the following equation:
Example:
A total of 387.64 lb/day lime will be required to raise the alkalinity of the water passing through a clarification process. How many grams per minute (g/min) lime does this represent?
Example:
A total of 344.54 lb/day lime will be required to raise the alkalinity of the water passing through a clarification process. How many grams per minute (g/min) lime does this represent?
Summary
Sedimentation is the process of allowing particles in suspension in water to settle out of the suspension under the effect of gravity. The particles that settle out from the suspension become sediment, and is known as sludge. In water treatment, sedimentation might be used to reduce the concentration of particles in suspension before the application of coagulation, to reduce the amount of coagulating chemicals needed, or after coagulation, possibly, flocculation. When sedimentation is applied after coagulation, its purpose is to reduce the concentration of solids in suspension so that filtration can function effectively. Detention time for clarifiers, or sedimentation basins, varies from 1 to 3 hours. The surface loading rate (similar to hydraulic loading rate - flow per unit area) is used to determine loading on sedimentation basins and circular clarifiers. Hydraulic loading rate measures the total water entering the process, whereas surface overflow rate measures only the water overflowing the process (plant flow only). Surface overflow rate calculations do not include recirculated flows. The measure of average velocity of the water as it travels through a rectangular sedimentation basin is known as mean flow velocity. The weir loading, or overflow, rate is the amount of water leaving the settling tank per linear foot of weir. The result of this calculation can be compared with design. Normally, weir overflow rates of 10,000 to 20,000 gpd/ft are used in the design of a settling tank. Typically, weir loading rates are measured in gallons per minute (gpm) flow over each foot (ft) of weir. During the alum dosage process, lime is sometimes added to provide adequate alkalinity (HCO3) in the solids contact clarification process for the coagulation and precipitation of the solids. To determine the necessary lime dose, in mg/L, three steps are required. First, the total alkalinity required is calculated. The total alkalinity required to react with the alum to be added and provide proper precipitation is determined.
Assignment
Complete the math worksheet for this lesson. You must be logged into Canvas to submit this assignment. Make sure you choose the appropriate semester.
