Activated sludge example problems

Lesson 8:

The Activated Sludge Process Calculations

Objective

In this part of the lesson we will learn how to calculate the following:

• Sludge Volume Index (SVI)
• Pounds of solids in the aeration basin
• Pounds of BOD applied to the aeration basin each day
• Sludge age
• Return Activated Sludge (RAS) pumping rate
• Waste Activated Sludge (WAS) puming rate
• Food/Microorganism (F/M) ratio
• Mean Cell Residence Time (MCRT)

Reading Assignment

Read the online lecture.

Lecture

Introduction

The biological reactor is where the microbial population reproduces and feeds on the incoming waste. The environment in the basin must be properly maintained to achieve the desired treatment results. The total solids concentration of the sludge being sent to the aerobic digester plays a key role in the process performance. Thickened sludge results in better volatile solids reduction in the digester. However, if the sludge is too thick, it can cause decanting problems. Typical organic loading is in the range of 0.02 - 0.14 lb VSS/day/ft 3 . If the loading is too high, it could cause anaerobic conditions due to insufficient oxygen transfer capacity, resulting in foul odors. Let's take a look at some of the parameters that need to be monitored during the activated sludge process.

Pounds of Solids in the Aeration Basin

In the activated sludge process it is important to control the amount of solids under aeration. The suspended solids in an aeration tank are called mixed liquor suspended solids (MLSS). To calculate the pounds of solids in the aeration tank, we need to know the MLSS concentration and the aeration tank volume. Pounds of MLSS can be calculated using the following equation:

MLSS, lb = MLSS, mg/L x Volume, MG x 8.34 lb/gal

In a conventional activated sludge treatment plant the mixed liquor suspended solids were 2,200 mg/L. If the aeration basin is 130 feet long by 40 feet wide by 9 feet deep, how many pounds of solids are under aeration?

First, determine the volume of the aeration basin:

Now we can determine the pounds of solids under aeration:

lb = Volume, MG x Concentration, mg/L x 8.34 lb/gal
lb = 0.35 MG x 2,200 mg/L x 8.34 lb/gal
lb = 6,422 lb of solids in the aeration basin

Pounds of BOD Applied to the Basin per Day

When determining BOD (biochemical oxygen demand), it is necessary to have a population of microbes that can oxidize, or consume, the biodegradable organic matter present in the waste stream. BOD refers to the amount of oxygen that the microbes need to break down the organics.

When calculating the BOD loading on the aeration basin it is usually calculated as lb/day using the "pounds" formula we have used before.

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

The BOD concentration of the wastewater entering the aeration basin is 180 mg/L. If the flow to the basin is 2.25 MGD, what is the BOD loading in lb/day?

BOD, lb/day = Flow, MGD x BOD concentration, mg/L x 8.34 lb/gal
BOD, lb/day = 2.25 MGD x 180 mg/L x 8.34 lb/gal
BOD, lb/day = 3,378 lb/day

Sludge Volume Index (SVI)

Sludge volume index (SVI) calculations will tell you whether the mixed liquor suspended solids (MLSS) in the aeration tank are settling at the right rate, or if they are hindering the performance of your facility. This is determiend by how well the sludge blanket in the secondary clarifier is compacting.

To calculate the SVI, you must first take a sample from the aeration tank. Let the sample settle for 30 minutes before beginning analysis. Analyze the sample and determine the concentration of suspended solids. This will be your MLSS concentration, represented in grams per liter (g/L). Divide the wet volume of the settled sludge (represented in mL/L) by the MLSS value from the last step. This calculation will give you your SVI value (represented in mL/g).

For this test, the settled solids are determined by doing a 30-minutes settleability test.

The typical SVI for a system that is operating as it should will be between 50 and 150 mL/g. If your SVI is outside this range, you may need to adjust the levels in your system. A rising SVI is an indication of sludge bulking, usually due to an increase in filamentous microbes.

Looking at the characteristics of different samples will give you some clues as to what you can expect from your own system.

To increase the SVI, you will need to increase the waste sludge rate. This will result in a slower rate of settling, which in turn will trap more of the suspended solids in the mixed liquor, leading to a clearer effluent. If you need to decrease the SVI, do the reverse: reduce the waste rate. This results in a thicker sludge with heavier particles. As the density increases, so does the rate of settling, making the process more efficient.

The sludge volume index can be determined with the following equation:

Lab results for a wastewater treatment plant are listed below. The aeration tank volume is 0.4 MG. Determine the sludge volume index.

There are only two of the values given above that will be needed for SVI determination: MLSS and the 30 minute settleable solids measurement.

With an SVI value of 119 mL/g, this indicates the process is operating under normal conditions with good settling and low effluent turbidity.

Sludge Age

The sludge age is the amount of time, in days, that solids or bacteria are under aeration. It is also known as the mean cell residence time (MCRT). Sludge age is used to maintain the proper amount of activated sludge in the aeration tank. To calculate the sludge age, it is necessary to know the amount of suspended solids (pounds) that are in the aeration tank and the amount of suspended solids (pounds) that enter the aeration tank daily. The sludge age can be calculated by dividing the pounds of suspended solids, or MLSS, already in the aeration basin by the pounds of suspended solids that enter the aeration basin.

Lab results for a wastewater treatment plant are listed below. The aeration tank volume is 0.4 MG.

We will need to do some pre-processing before we have the values needed to determine the sludge age. We can determine the suspended solids already in the aeration basin by:

Solids in the basin = Aeration basin volume, MG x MLSS, mg/L x 8.34 lb/gal
Solids in the basin = 0.4 MG x 1600 mg/L x 8.34 lb/gal
Solids in the basin = 5338 lb

Next we need to determine the solids that are entering the aeration basin from the primary clarifier:

Solids entering the basin = Flow, MGD x Suspended solids, mg/L x 8.34 lb/gal
Solids entering the basin = 2.1 MGD x 100 mg/L x 8.34 lb/gal
Solids entering the basin = 1751 lb/day

Now that we have all the information, we can plug those values into the sludge age formula:

Mean Cell Residence Time (MCRT)

The mean cell residence time (MCRT) is an average measure of how long the microbes remain in contact with the food source. MCRT is not a mass balance, but rather a measure of how many days the microbes are kept in the activated sludge process before being wasted. The treatment system is constantly generating new solids, producing approximately 0.5 lbs of new solids per lb of BOD removed. For instance, if you remove 100 lbs of BOD, you will produce approximately 50 lbs of new solids.

If you fail to remove the new solids produced, your treatment system can suffer in performance. A longer MCRT age yields less sludge production than a younger MCRT. This is because BOD (food) is ued for both staying alive and growing. The MCRT can be calculated using the following formula:

Calculate the mean cell residence time (MCRT) for an activated sludge plant given the following information:

• Aerator basin volume: 0.325 MG
• Final clarifier volume: 0.145 MG
• Flow to aerator: 2.1 MGD
• WAS flow: 0.02 MGD
• MLSS: 2,400 mg/L
• WAS SS: 6,300 mg/L
• Final effluent SS: 12 mg/L

You will need to do a lot of preprocessing before you are able to plug values into the MCRT formula. Let's start with the pounds of solids already in the aeration basin.

First calculate the lb of MLSS under aeration:

MLSS, lb = Volume, MG x Concentration, mg/L x 8.34 lb/gal
MLSS, lb = (0.325 MG + 0.145 MG) x 2,400 mg/L x 8.34 lb/gal
MLSS, lb = 0.47 MG x 2,400 mg/L x 8.34 lb/gal
MLSS, lb = 9,408 lb

Next, calculate the lb of solids leaving aeration by being wasted (WAS) per day:

WAS, lb/day = Flow, MGD x Concentration, mg/L x 8.34 lb/gal
WAS, lb/day = 0.02 MGD x 6,300 mg/L x 8.43 lb/gal
WAS, lb/day = 1,051 lb/day

Now calculate the lb of solids leaving over the effluent weir per day:

Effluent, lb/day = Flow, MGD x Concentration, mg/L x 8.34 lb/gal
Effluent, lb/day = 2.1 MGD x 12 mg/L x 8.34 lb/gal
Effluent, lb/day = 210 lb/day

Now that all the preprocessing is done, we can plug the values into the MCRT formula to determine how long the microbes need to stay in the process:

Microscopic Examination

By examining the mixed liquor (MLSS) under a microscope, an operator can determine the distribution of filamentous bacteria, protozoa and rotifers, which will indicate the condition of the MLSS under aeration.

A predominance of protozoans and rotifers means good quality since they eat the bacteria present and help clean the effluent.

Food/Microorganism (F/M) Ratio

The number of microbes which are used to seed the aeration basin is carefully controlled and is based on the food to microorganism ratio (F/M ratio). If the microbes are present in the proper proportion, they will break down the organic matter present more efficiently. The food value in the F/M ratio can be calculated using the BOD or COD (chemical oxygen demand). The COD analysis is more accurate and doesn't take as long to process as the BOD test. The reason for biosolids production is to convert BOD to bacteria

If the appropriate food to microorganism ratio is followed, then there will be efficient BOD removal in the aeration basin. One manufacturer suggests that the best F/M ratio is around 0.6. Different activated sludge variations will have different F/M ratios, as shown below:

To calculate the proper amount of microbes that need to be added to the aeration basin per day, you will use the following formula:

The BOD represents the amount of BOD going into the aeration basin per day. The MLVSS (mixed liquor volatile suspended solids) represents the amount of microbes already present in the basin that will consume the incoming BOD. The F/M ratio determines how many pounds per day of BOD is available for each pound of microbe in the basin.

Determine the amount of MLVSS in lb as well as mg/L to be maintained in a conventional activated sludge plant. Assume a F/M of 0.5 lb of COD per day for each lb of MLVSS under aeration. The average COD concentration of the primary effluent is 155 mg/L and the average daily flow is 2.0 MGD. The aeration basin volume is 0.3 MG.

A lot of information is given for this problem, so let's break it out to view it easier. We are asked to determine the MLVSS in lb and mg/L. We are given the following information:

• F/M ratio = 0.5
• Primary effluent COD = 155 mg/L
• Average Flow = 2.0 MGD
• Basin volume = 0.3 MG

Given the information provided, we must first calculate the BOD in lb/day:

COD, lb/day = Flow, MGD x Concentration, mg/L x 8.34 lb/gal
COD, lb/day = 2.0 MGD x 155 mg/L x 8.34 lb/gal
COD, lb/day = 2585 lb/day

We are given the F/M ratio and we just calculated the COD in lb/day. We can now use the F/M ratio formula to determine amount of MLVSS that needs to be added. (Remember that BOD and COD are interchangeable)

We need to rearrange the formula to solve for MLVSS, lb:

We have calculated part of what is asked for. Now we need to convert the lb of MLVSS to its concentration. This is done through unit conversion and using the aeration tank volume:

Let's look at another example for F/M calculation.

The aeration tank influent BOD is 155 mg/L and the influent flowrate is 2.28 MGD. What is the F/M ratio if the MLVSS is 2250 mg/L and the aeration tank volume is 1.5 MG?

First, let's determine the BOD in lb/day:

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

BOD, lb/day = 155 mg/L x 2.28 MGD x 8.34 lb/gal

BOD, lb/day = 2,947.36 lb/day

That's one portion of the equation: BOD, lb/day. Now we need to determine the pounds of MLVSS in the tank:

MLVSS, lb = MLVSS, mg/L x Volume, MG x 8.34 lb/gal

MLVSS, lb = 2250 mg/L x 1.5 MG x 8.34 lb/gal

MLVSS, lb = 28,147.5 lb

Now that you have all the components, determine the F/M ratio:

Return Activated Sludge (RAS) Pumping Rate

You may be required to calculate the amount of microbes that need to be added to the aeration basin daily to ensure the proper microbial population for treatment. This mixture of sludge and microbes from the clarifier is known as return activated sludge (RAS). To determine the amount of sludge to add back to the basin (RAS) the BOD and flow is considered for proper food/microbe ratio.

The return sludge is only made up of approximately two percent microbes, so it can take quite a bit of the plant's sludge to seed the aeration basin. Let's determine how many pounds of RAS (MLVSS) need to be added to the aeration basin to achieve an optimum F/M ratio of 0.6.

The plant has a flow of 1.2 MGD and a BOD concentration of 220 mg/L. Determine the amount of microbes needed per day to achieve a F/M ratio of 0.6.

First, determine the BOD loading on the plant per day:

BOD, lb/day = Flow, MGD x BOD, mg/L x 8.34 lb/gal
BOD, lb/day = 1.2 MGD x 220 mg/L x 8.34 lb/gal
BOD, lb/day = 2,202 lb/day

Now you know two components of the F/M ratio formula. Let's rearrange it so we can solve for MLVSS, lb, which indicates the amount of microbes needed to achieve the F/M ratio.

You will need to add 3,670 lb of returned sludge (RAS) to properly seed the aeration basin.

What if you needed to know the solution amount in gallons? Simply convert lbs to gallons:

3,670 lb x (1 gal/8.34 lb) = 440 gal

The question asked for gallons per day, so you have solved the problem with 440 gpd needed. If you needed to know how much to feed per minute, do one more time conversion:

(440 gal/day) x (1 day/1440 min) = 0.31 gpm

Waste Activated Sludge (WAS) Pumping Rate

Waste activated sludge (WAS) is the amount of microbes that must be removed from the process to keep the biological system in balance. You can control the WAS rate by various means. Many plants operate efficiently by keepting their MLSS concentration within a certain range. Other plants keep track of how old the microbes are, and increase the wasting if they are getting too old. Another method uses the ratio of food coming in with the settled solids as compared to the number of microbes available to eat that food. Finally, some plants use the microscope to keep an inventory of the microbial population available.

Each of these methods has been used effectively and is dependent upon how the plant uses the gathered data.

Calculate the desired change in the WAS pumping rate as well as the new pumping rate if the current waste rate is 95 gpm. The MLSS in the aerator is 6,000 lb and the desired MLSS is 5,000 lb. WAS concentration is 6,400 mg/L. The aerator volume is 0.315 MG and the MLSS concentration is 2,300 mg/L.

Wow! That's a lot of information to try to process. Let's start with laying out what we know and we don't know.

Desired change in pumping rate, gpm and the new pumping rate, gpm

Current WAS rate = 95 gpm
Current MLSS in the aerator = 6,000 lb
Desired MLSS in the aerator = 5,000 lb
WAS concentration = 6,400 mg/L
Aeration basin volume = 0.315 MG
MLSS concentration = 2,300 mg/L

The change in WAS pumping rate can be calculated using the formula:

Since the current pumping rate is in gpm, you must convert MGD to gpm:

The means the pumping rate will change by 13 gpm.

To determine the new WAS pumping rate:

New WAS pumping rate = 95 gpm + 13 gpm = 108 gpm

Solids Balance in the Bioreactor (Aeration basin)

To determine if your system is balanced properly you must take into consideration the MLSS, RAS and the flow to the basin. The balance can be determined using the formula below:

Even though the formula has mg/L for MLSS, this needs to be converted to MGD as well, so everything will be in the same unit, allowing you to calculate the return sludge rate, in MGD, needed to keep the solids in the process in balance.

A plant that has a flow of 2.5 MGD has a MLSS of 2850 mg/L. The current RAS is 6,580 mg/L. Determine the return sludge rate-solids balance needed for this plant.

Let's do the pre-processing to get the mg/L measurements into MGD:

MLSS, lb/day = MLSS, mg/L x Flow, MGD x 8.34 lb/gal
MLSS, lb/day = 2850 mg/L x 2.5 MGD x 8.34 lb/gal
MLSS, lb/day = 59,422.5 lb/day

Now convert lb/day to gpd, then MGD:

(59,422.5 lb/day) x (8.34 gal/lb) x (1 MG/1,000,000 gal) = 0.496 MGD

Next we need to do the pre-processing for the return activated sludge:

RAS, lb/day = RAS, mg/L x Flow, MGD x 8.34 lb/gal
RAS, lb/day = 6580 mg/L x 2.5 MGD x 8.34 lb/gal
RAS, lb/day = 137,193 lb/day

Now convert lb/day to gpd, then MGD:

(137, 193 lb/day) x (8.34 gal/lb) x (1 MG/1,000,000 gal) = 1.14 MGD

Now you can plug the values into the equation, exchanging the mg/L for MGD:

This particular basin would need a return sludge rate of 1.92 MGD to keep the solids in the process in balance.

Summary

• Autotrophs use inorganic carbon as a food source.
• Heterotrophs use organic carbon as a food source.
• Microorganism selection in the activated sludge process is accomplished by controlling dissolved oxygen (DO) levels, nutrient levels, mean cell resident time (MCRT) and hydraulic retention time (HRT).
• Proper design and operation of the process can result in the removal or reduction of Carbonaceous BOD (CBOD), ammonia, nitrogen and phosphorus.
• Complete nitrification of ammonia (90% removal) will require more oxygen than a non-nitrifying conventional plant designed to remove CBOD only.
• Nitrification removes alkalinity from the waste stream.
• Denitrification adds alkalinity to the waste stream.
• The MCRT and F:M (Food to Microorganism ratio) are inversely related . As MCRT goes up, F:M goes down.
• The oxidation ditch process is normally operated as an extended aeration system that has an MCRT greater than 10 days.
• High purity oxygen processes are generally used when the waste stream has high levels of biochemical oxygen demand (BOD).
• Raw wastewater flows into an oxidation ditch without grit removal or primary settling.
• The F:M ratio is used as a primary process control parameter for activated sludge plants.
• Inadequate DO levels in the aeration basin can lead to the dominance of filamentous bacteria, which can result in poor sludge settling..
• Solids should not remain in the secondary clarifier longer than necessary since they deteriorate as long as they are in the clarifier.
• Activated sludge is used to remove dissolved and finely divided suspended solids from wastewater.
• Microorganisms in the mixed liquor reproduce at a rapid rate requiring a portion to be wasted (WAS) to maintain a constant population in the aeration basin.
• A brown leathery foam on the surface of an aeration tank indicates a high solids concentration that can be corrected by increasing the wasting rate.
• Causes of foaming in an aeration basin include a high solids retention time, over aeration, or excessive sludge wasting.
• The rotor in an oxidation ditch is used to keep the water flowing, the solids mixed, and to transfer oxygen to the mixed liquor.
• The secondary clarifier receives flow from the aeration basin and is used to separate the solids from the liquid.
• The activated sludge that is removed from the system is called waste activated sludge.
• The mixture of liquid, microorganisms, and solids in the aeration basin is called mixed liquor.
• Under normal operating conditions, activated sludge will have a light to medium brown appearance.
• Chemical Oxygen Demand (COD) is a relatively quick test that can be used to determine organic loading.
• Most of the “work” is performed by the bacteria in an activated sludge treatment plant.
• A term used to describe the combined liquid and solids of the activated sludge process is mixed liquor.
• Primary treatment is not usually part of an extended aeration oxidation ditch process.
• DO in the aeration process of a small activated sludge package plant should be approximately 2 mg/L.
• For optimum water quality during the warm months of summer, the operator should waste approximately 5% of the solids in an extended aeration package plant.
• The MLVSS concentration is used to represent the microorganisms in the aeration basin available to treat incoming waste.
• Most algae are classified in the Protista kingdom.
• Bacteria are classified in the Monera kingdom.
• Acids and Alkali are corrosive and may interfere with biological treatment of wastewater.
• Essential nutrients for wastewater microorganism reproduction include nitrogen, phosphorus, and organic carbon.
• A good ratio of BOD to nitrogen to phosphorus is 100:5:1.
• Secondary sludges typically contain about 20-25 percent of non-volatile inorganic matter.
• Sludge bulking in secondary clarifiers is caused by the excessive growth of filamentous bacteria.
• The sludge age ranges from 0.5 to 2 days in a high rate activated sludge plant.
• The final effluent BOD from a conventional activated sludge plant ranges between 10 mg/L and 20 mg/L.
• It is best to waste excess activated sludge continuously rather than in batches or sporadically.
• The most likely cause of the secondary clarifier sludge blanket uniformly flowing over the effluent weir is hydraulic overload.
• The development of a dark tan foam on the aeration basin surface is most likely caused by a high MCRT.
• If the mixed liquor turns dark, th emost likely cause is inadequate aeration, causing the activated sludge to go septic.
• One likely cause of a low RAS concentration is excess filamentous bacteria growth, resulting in poor sludge compaction (sludge bulking).
• Changes in the wasting rate should be limited to no more than 10% - 15% from one day to the next.
• A conventional activated sludge plant will produce approximately 0.55 lb of sludge volatile matter for every 1 lb of BOD removed in the process.
• A high rate process will produce approximately 0.75 lb of sludge volatile matter for every 1 lb of BOD removed.
• An extended aeration process will produce approximately 0.15 lb of sludge volatile matter for every 1 lb of BOD removed.
• An increase in flow the treatment plant cuased by a storm event will cause the percent volatile matter to decrease in the aeration basin.
• Rising sludge in a secondary clarifier can be caused by septicity or denitrification.
• If sludge is floating in the secondary clarifier due to filamentous organisms, the operator should increase the RAS rate.
• If a toxic shock results in the formation of pin floc that carries over the secondary clarifier weir, the operator should increase the return sludge rate as high as possible.
• If a dark tan foam forms on the surface of the aeration basin due to the MCRT being too long, the operator should slowly increase sludge wasting to reduce the MCRT.
• If a thick sudsy foam forms on the aeration basin surface due to a low MLSS concentration, the operator should reduce sludge wasting to increase the MLSS and MCRT.
• There are approximately 7.2 lb of alkalinity consumed for every 1 lb of ammonia nitrogen oxidized during the nitrification process.
• Nitrification occurs best when the water temperature is at a minimum of 60°F and a maximum of 95°F.

Assignment

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Quiz

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