1. Raw wastewaters contain suspended solid materials, or simply suspended solids, that must be removed. The main method used to remove these solids is sedimentation.
[Picture: A beaker filled with liquid and particles labeled "Solids."]
2. Sedimentation tanks, or clarifiers, provide a large space, or volume, where the suspended solids can be settled and removed from the wastewater. There are typically two different types of sedimentation tanks - circular and rectangular.
[Pictures: Picture 1 is an animation showing a tank in the ground with water flowing into it. There are brown particles that settle at the bottom of this tank, and then they are sucked through a pipe at the bottom. Picture 2 shows a circular structure, and then the top is removed so show that there is water in the structure with a walkway across it to the center. Picture 3 is of a rectangular structure with beams separating it into eight sections, and it is filled with water.]
3. As the flow of wastewater enters the large volume of the tank, it slows down and the suspended solids begin to settle. The solids that reach the bottom of the tank before they exit with the wastewater will be removed. The solids removed from the bottom of the tank are called sludge.
[Picture: A rectangular tank is shown from the side. There is a pipe on the left, right, and on the bottom. There is an arrow inside the tank pointing to the right, and there are smaller arrows in the tank pointing down. There are arrows pointing right from the pipe on the right, and arrows pointing down from the pipe at the bottom. Animations: 1. water moves into influent (at left) and out of effluent (at right); relatively fast. Water in tank moves relatively slow. Fade in text: 1st- "IN" to "INFLUENT" 2nd- "OUT" to "EFFLUENT." 2. Solids "settling" are added. Use greater density near front of tank, less at rear. Settling is relatively slow. 3. Bring in sludge pipe. Sludge flows from pipe in brief spurts. Fade in text: "Sludge Removal."]
4. A mass balance analysis of the sedimentation process can be used to estimate the quantity of sludge that is removed.
[Picture: A rectangular tank is shown, the same drawing as in step 3.]
First, we need to know something about the wastewater flowing through the tank and the concentration of suspended solids in the wastewater. All wastewater treatment facilities will measure these parameters.
[Picture: Transition from animated graphic to static representation. A drawing of a box with an arrow pointing to it from the left, and arrow pointing down from the bottom, and an arrow pointing right from the right.]
5. We will know the flow rate of the wastewater entering the tank and the concentration of its suspended solids.
[Drawing: The same box with the arrows. Text fade in: "Q in" and "SS in" next to the arrow on the left.]
We will also know the suspended solids concentration leaving the tank but not the wastewater flow rate because it is typically not measured here.
[Drawing: Text fade in: "Q out = ?" and "SS out" next to the arrow on the right.]
We will, however, measure the volume of sludge pumped so a flow rate value for the sludge removal is also known, but...
[Drawing: Text fade in: "Q Sludge" and "SS Sludge = ?" next to the bottom arrow. Text fade in: *What is the concentration of the sludge solids? We need to know this in order to calculate the quantity of sludge removed.]
6. In this example lets say that our influent wastewater flow rate is one million gallons per day, or simply, 1 MGD.
[Drawing: Same drawing with labeled arrows. Text fade in: "Q In = 1 MGD."]
Let's also say our influent suspended solids concentration is two hundred fifty milligrams per liter and our effluent suspended solid concentration is one hundred milligrams per liter.
[Drawing: Text fade in: "SS In = 250 mg/L" and "SS Out = 100 mg/L."]
These are both very typical values for solids concentration entering and leaving a primary clarifier.
[Drawing: A red arrow is now pointing at the "SS In = 250 mg/L" and an red arrow is pointing at "SS Out = 100 mg/L."]
7. To complete what we know about this problem lets also say that we pump two thousand three hundred gallons of sludge from the tank each day. This is expressed as a sludge flowrate of 2,300 gpd.
[Drawing: Red arrows are taken away. Text fade in: "Q Sludge = 2300 gpd."]
Remember that since mass is neither created nor destroyed, we must be able to account for all of the suspended solids either entering or leaving the tank. We must also be able to account for, or balance, all of the flows entering or leaving the tank. Basically what goes IN must equal what goes OUT.
[Drawing: Text fade in below the drawing: "IN = OUT."]
8. Balancing the flows requires that the total flow incoming is equal to the total flow outgoing.
[Drawing: Text fade in: "IN = OUT" arrow to the right "Qin = Qout + Qsludge"]
Before we balance the flow, let's make sure the flow rate units are the same. Use the quick converter below to do this. It doesn't matter whether you convert MGD to gpd or gpd to MGD, it just matters that the units expressed are the same.
[Drawing: The QUICK CONVERTER has a box next to MGD with "Input Value" inside. There is a circle with an arrow pointing to the right to a box next to gpd. Below this, there are the same two boxes. Now, the box next to gpd has "Input Value" inside, and the arrow is pointing to the left.]
9. Now that we have expressed our flows in the same units we can balance the flow system by solving for our tank's effluent flow rate. Enter what you think this flow rate is and then check it to see if you're correct.
[Drawing: "Qout = Qin - Qsludge" and "Qout =" a blank box next to gpd and a button below that says "CHECK IT."]
10. Once all of the flows are known, a mass balance analysis can be done to determine the solids concentration of the sludge. As before, what goes in must equal what goes out. Since solids entering or leaving the tank are influenced by the flows, we use the mass flow rate relationship for this analysis.
[Drawing: The same box with the arrows. Text fade in: "Q Out = 997700 gpd" and "IN = OUT."]
In this relationship, the rates of solids mass entering the tank must equal the rate of solids mass leaving the tank. mass rate is calculated by multiplying the flow rate by the solids concentration. The result is the solid mass flow per unit time.
[Drawing: Text fade in: "Solids Mass Rate" arrow to right "Q x S."]
The overall mass rate balance for our tank is based on the mass rate of solids entering the tank being equal to the sum of the solids mass rates leaving the tank.
[Drawing: Text fade in: "Q In x SS In = Q Out x Ss Out + Q Sludge x SS Sludge."]
11. Rearranging this equation we can solve for the solids concentration of the sludge pumped from the tank.
[Drawing: Text fade in: "Q In x Ss In - Q Out x SS Out" over "Q Sludge" equals "SS Sludge." Text fade in: *Besides being set up to solve for the sludge solids concentration we have to appreciate how the "gpd" units will all cancel out leaving only mg/L, which is what we want.]
12. Okay, let's calculate our sludge solids concentration.
[Drawing: The above equation now has an arrow pointing down to an equation box. The box goes as follows: (gpd x mg/L - gpd x mg/L)/gpd with a button that has "=" on it to the right. Caption below it: *Enter all of our known values into their correct position and hit the equal button to complete the calculation.]
Use script to check that each entry field is correct. If true, complete calculation and display. If false, it will display "Try again. You must have entered something wrong." If true, it will display "Correct. The concentration of solids in the sludge pumped from the tank is [blank] milligrams per liter." If it is false, it will void all entry fields.
13. Now that we have used mass balance to estimate the concentration of sludge solids, we can now estimate the total quantity of sludge that is removed. Let's do that for one day.
[Drawing: The rectangular tank with the three pipes and the arrows indicating the direction of flow.]
14. In the U.S.A., sludge quantities are still calculated in pounds. But let's do it using kilograms. So what we will be calculating is the kilograms of sludge per day. These units express our mass of sludge per unit of time.
[Drawing: Text fade in: "Q Sludge = 2300 gpd" and "SS Sludge = 65,317 mg/L" *Notice that the concentration has been rounded.]
15. To express this mass rate quantity, we simply multiply the sludge flow rate by the sludge solids concentration. But in doing so, we must make sure to convert our units correctly.
[Drawing: Text fade in next to previous two equations "Q Sludge x SS Sludge = mass/day."]
16. We will do this by converting gallons to liters and milligrams to kilograms.
[Drawing: Text fade in: "gallon [arrow] liter" and "mg [arrow] kg."]
Using a units conversion table you should find that one gallon equals 3.785 liters. You should also find that it takes 1,000,000 milligrams to equal 1 kilogram.
[Drawing: Text fade in: "1 gallon = 3.785 L" and "1,000,000 mg = 1 kg."]
17. Now that we have our conversion factors, we can calculate the kilograms of sludge that is removed each day by this sedimentation tank.
[Drawing: Text fade in: "Q Sludge x Conversion Factor" arrow pointing down to "2,3000 gallons/day x 3.785 Liters/gallons"]
We convert the sludge flow rate by multiplying by 3.785 L. The gallon units cancel out and we are left with the unit expression of liters per day. Our result rounded up is 8,706 liters per day.
[Drawing: Text fade in: "Q Sludge x Conversion Factor" arrow pointing down to "= 8,706 Lpd."]
18. To convert milligrams per liter to kilograms per liter, we simply divide by 1,000,000. Thus, the concentration of sludge solids expressed as kilograms per liter is as shown.
[Drawing: Text fade in: "65,317 mg/L [divided by] 1,000,000 mg/kg" arrow pointing down "= 0.065317 kg/L."]
19. Multiplying our converted sludge flow and sludge concentration values together now gives us the mass quantity of sludge removed per day.
[Drawing: Text fade in: "Q Sludge x SS Sludge = 568.6 kg/day."]
20. You still want to know what this is in pounds per day? To do this let's multiply our kilograms per day by the conversion factor of 2.205 lbs per kilogram. Rounding up again, we have 1,254 pounds of sludge being removed from the sedimentation tank each day.
[Drawing: Text fade in: "568.6 kg sludge/day" arrow down to "568.6 kg/day x 2.205 lbs/kg" with the "kg" on each side crossed out, and an arrow down to "= 1,254 lbs/day."]
Since 1 ton is 2,000 pounds, that is well over a half-ton of sludge per day!
[Drawing: Text fade in: "or 0.627 tons of sludge per day."]
21. Now that we have used mass balance analysis to determine all of our flow rates and solids concentrations, and the quantity of sludge removed from the wastewater each day, what else is there?
[Drawing: The rectangular tank with the pipes, and text that says "SLUDGE REMOVED = 568.6 kg/day" and "what else?"]
As engineers, one of the questions we want to know is "How well does the sedimentation tank perform?" To calculate the efficiency of the tank for removing solids, we will once again use mass balance relationships.
[Drawing: Text fade in: "Efficiency of Sludge Removal?"]
22. The efficiency of sludge removal is expressed as the percentage of incoming solids that are removed from the wastewater as sludge.
[Drawing: The box with the three arrows. The arrow on the left is labeled "Influent Solids," the bottom arrow is labeled "Sludge Solids," the right arrow is labeled "Effluent Solids," and the inside of the box is labeled "% Removal?"]
Since we have determined all the flow rates and solids concentrations, we can calculate removal efficiency using either one of two ways.
1. One way is so simply divide the sludge solids removed by the total incoming solids and then multiply that by 100 to get the percent.
[Drawing: Text fade in: Sludge Solids/Influent Solids x 100 = % Removed]
2. The other way is to subtract the solids leaving the tank in the effluent flow from the total incoming solids, then divide that by the total incoming solids and multiply by 100.
[Drawing: Text fade in: (Influent Solids - Effluent Solids/Influent Solids) x 100 = % Removed]
23. Using our original data and the results from our mass balance analysis, choose the method you would like to use for calculating the solids removal efficiency.
[Drawing: The box with the arrows labeled. The left arrow is labeled "Q In = 1 MGD" and "SS In = 250 mg/L." The bottom arrow is labeled "Q Sludge = 2300 gpd" and "SS Sludge = 65317 mg/L." The right arrow is labeled "Q Out = 997700 gpd" and SS Out = 100 mg/L."]
Click to solve for one of the two methods shown below. Enter the values indicated and then hit the calculate button. Be sure that ll of your flow rate units are the same. Note: 2nd method format. Qin x SSin - Qout x SSout /Qin xSSin.
[Drawing: 1) Sludge Removed/Total Solids x 100 [button labeled solve] 2) Total Solids - Solids Out/Total Solids x 100 [button labeled solve] Equation One: "Q Sludge x SS Sludge/ Q in x SS in x 100 =" Equation 2: "(Q in x SS in) - (Q out x SS out)/ Q in x SS in x 100 =."]
If "Correct" display "Calculate using other method? Yes/No." If "Incorrect" display "Oops. Try again [continue button] "hit blanks fields for another try."]
24. Great, you have completed the mass balance analysis of a sedimentation tank. You can use the same basic methods demonstrated in this exercise to solve any material flow system where the material of concern is conserved.
[Drawing: The rectangular sedimentation tank with the three pipes and the direction of flow.]
When the system is more complex that the sedimentation example presented here, simply break down into smaller sub-systems that can be easily solved. By solving each sub-system separately, the larger more complex system can be solved.
[Drawing: An arrow pointing to a circle labeled "TANK 1" with an arrow pointing down to a box formed by an arrow pointing to a box labeled "TANK 3" which has an arrow pointing down. This points to an arrow pointing to the right from a circle labeled "TANK 2," which has the arrow pointing to "TANK 3." The junction of the arrows from "TANK 2" and "TANK 3" is circled and labeled "Sub-System Boundary." There is an arrow pointing down from this to an enlargement of the junction, which shows an arrow pointing down to make a perpendicular crossing with an arrow pointing right. The arrow pointing down is labeled "Q2" and "C2," the left end of the horizontal arrow is labeled "Q1" and "C1," and the right end is labeled "Q3" and "C3."]