Objective

In this lesson we will learn the following calculations:

• Surface loading rate
• Weir overflow rate
• Percent Total Solids
• BOD and SS Removal, lb/day

Lecture

Primary treatment (primary sedimentation and clarification) should remove both settleable organic and floatable solids. Poor solids removal during this step of treatment may cause organic overloading of the biological treatment processes following primary treatment. Normally, each primary clarification unit can be expected to remove 90 to 95% of settleable solids, 40 to 60% of the total suspended solids (TSS), and 25 to 35% of biochemical oxygen demand (BOD).

Let's watch a video showing what occurs during primary wastewater treatment.

Process Control Calculations

As with other treatment plant unit processes, several control calculations may be helpful in evaluating the performance of the primary treatment process. The expected range of hydraulic detention time for a primary clarifier is 1 to 3 hours with an expected range of surface loading/settling rate of 300 to 1200 gpd/ft². The expected range of weir overflow rate for a primary clarifier is 10,000 to 20,000 gpd/ft.

Surface Loading Rate

Surface loading rate is the number of gallons of wastewater passing over 1 square foot of tank per day. Plant designs generally use a surface loading rate of 300 to 1200 gpd/ft². The surface loading rate can be determined by:

Example:

A circular clarifier has a diameter of 150 ft. If the flow to the unit is 4.85 MGD, what is the surface loading rate, in gpd/ft²?

Remember to find the area of a circle use: (0.785)(Diameter)².

Area, ft² = (0.785)(Diameter)²

Area, ft² = (0.785)(150 ft)²

Area, ft² = 17,662.5 ft²

Next convert MGD flow to gpd:

4.85 MGD x (1,000,000 gal/1MG) = 4,850,000 gpd

 

Now determine the surface loading rate of the circular clarifier:

Example:

A sedimentation basin has a length of 120 ft and a width of 50 ft. If the primary effluent flow is 3.25 MGD, what is the surface loading, or overflow rate, in gpd/ft²?

First determine the area of the rectangular clarifier:

Area, ft² = Length, ft x Width, ft

Area, ft² = 120 ft x 50 ft

Area, ft² = 6000 ft²

Next, convert the flow from MGD to gpd:

3.25 MGD x (1,000,000 gal/1MG) = 3,250,000 gpd

Now determine the surface loading rate of the rectangular clarifier:

Weir Overflow Rate

A weir is a device used to measure wastewater flow.

Weir overflow rate (or weir loading rate) is the amount of water leaving the settling tank per linear foot of water. The result of this calculation can be compared with design. Normally, weir overflow rates of 10,000 to 20,000 gpd/ft² are used in the design of a settling tank. The weir overflow rate can be determined by:

*Hint: to calculate weir circumference, use: total feet of weir = 3.14 x weir diameter, ft

Example:

A circular clarifier has a diameter of 100 ft and has a weir along its circumference. The effluent flow rate is 2.22 MGD. What is the weir overflow rate in gallons per day per foot (gpd/ft)?

First determine the length of the weir:

Total feet weir = 3.14 x Diameter, ft

Total feet weir = 3.14 x 100 ft

Total feet weir = 314 ft

Convert the flow from MGD to gpd:

2.22 MGD x (1,000,000 gal/1MG) = 2,220,000 gpd

Now determine the weir overflow rate in gpd/ft:

Example:

A sedimentation basin has a total of 125 ft of weir. What is the weir overflow rate, in gpd/ft, when the flow is 2,055,000 gpd?

Just plug and play on this one, no pre-conversions needed.

Biochemical Oxygen Demand (BOD) and Suspended Solids (SS)

BOD Measurement, mg/L

Domestic wastewater influent, or effluent from biological water treatment plants before disinfection, provide the best source of seed, and give the most reproducible results when measuring for BOD. The term "seed" refers to microbes that consume the biodegradable organic matter in samples for measurement of BOD. Some sources of wastewater may contain toxins that prevent organisms from growing, so seed must be added to the sample when calculating the BOD of the wastewater.

Whatever source of seed is used, it is simply a solution that contains a sufficient population of bacteria. In some cases there are sufficient amounts of microbes in wastewater and using a seed for BOD calculation is not needed. BOD can be determined by the following equations, depending upon if seed needed to be added to the sample or not:

or

 

Example:

A 50 mL sample of wastewater influent is found to have an initial DO reading of 10.2 mg/L and a final DO reading of 7.5 mg/L. The sample does not need to be seeded for BOD calculation. Using the given information, determine the BOD in mg/L for the influent.

Example:

A 100 mL sample of wastewater influent is found to have an initial DO reading of 7.8 mg/L and a final DO reading of 3.2 mg/L. The sample needs a correction factor of 0.69 mg/L. Using the given information, determine the BOD for the influent, in mg/L.

BOD Removed, lb/day

To calculate the pounds of Biochemical Oxygen Demand (BOD) or suspended solids (SS) removed each day, we need to know the mg/L BOD or suspended solids removed and the plant flow, then we can use the mg/L to lb/day formula:

BOD removed, lb/day = BOD, mg/L x Flow, MGD x 8.34 lb/gal

Example:

If 180 mg/L suspended solids is removed by the primary clarifier, how many lb/day suspended solids will be removed when the flow is 4,250,000 gpd?

SS removed, lb/day = SS, mg/L x Flow, MGD x 8.34 lb/gal

SS removed, lb/day = 180 mg/L x 4.25 MGD x 8.34 lb/gal

SS removed, lb/day = 6380.1 lb/day

Example:

The flow to a secondary clarifier is 1.2 MGD. If the influent BOD concentration is 220 mg/L and the effluent BOD concentration is 85 mg/L, how many pounds of BOD are removed daily?

First determine the mg/L of BOD removed per day:

BOD removed, mg/L = 220 mg/L - 85 mg/L

BOD removed, mg/L = 135 mg/L

Now calculate the BOD removed in lb/day:

BOD removed, lb/day = BOD, mg/L x Flow, MGD x 8.34 lb/gal

BOD removed, lb/day = 135 mg/L x 1.2 MGD x 8.34 lb/gal

BOD removed, lb/day = 1351.08 lb/day

Summary

Primary treatment (primary sedimentation and clarification) should remove both settleable organic and floatable solids. Poor solids removal during this step of treatment may cause organic overloading of the biological treatment processes following primary treatment. Normally, each primary clarification unit can be expected to remove 90 to 95% of settleable solids, 40 to 60% of the total suspended solids (TSS), and 25 to 35% of biochemical oxygen demand (BOD). As with other treatment plant unit processes, several control calculations may be helpful in evaluating the performance of the primary treatment process. The expected range of hydraulic detention time for a primary clarifier is 1 to 3 hours with an expected range of surface loading/settling rate of 600 to 1200 gpd/ft². The expected range of weir overflow rate for a primary clarifier is 10,000 to 20,000 gpd/ft.

Assignment

Complete the math worksheet for this lesson and return to instructor via email, fax or mail.