# Top 3 Equations for Activated Sludge Process Control

Experiencing a plant upset? Here are three go-to equations to help you pinpoint major activated sludge issues.

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Activated sludge process control is a daily consideration for producing high-quality effluent. Typically, a plant superintendent or chief operator makes decisions for daily plant achievement goals. However, the entire plant staff is responsible for providing safe effluent to a community. During times of plant upset, the stress of noncompliance or being offline can trickle down to all shifts and personnel.

Here are three math equations that will help you pinpoint major activated sludge issues.

1. Food to microorganism (F/M ratio)

As basic as this formula is, the key to the activated sludge process is a balance between food and organisms. For microorganisms, food is the organic content of raw wastewater as reflected in the biochemical oxygen demand. BOD is known as the strength of the wastewater because the greater the need for oxygen by bacteria and microorganisms, the more organic content is available in the influent.

Here is the calculation:

BOD, mg/L x 8.34 x plant flow

MLVSS, mg/L x 8.34 x aerbay capacity in mgd

Situation: A treatment plant is experiencing an increase in bar screen runtime with more frequent cleanings. During a cleaning, the night operator notices what appears to be increasing fats, oils and grease (FOG) on the screening and lift station. Two days later, the treatment process is just not settling correctly, and the filter media is backwashing often. It’s time to check the BOD loading and the available mass to treat the presumed excess loading.

Data:

• Plant flow = 10.5 mgd
• Aerbay capacity = 15 mgd
• Five-day BOD average = 325 mg/L
• Five-day mixed liqueur volatile suspended solids (MLVSS) = 1,500 mg/L

An extended aeration plant should have 0.05-0.1 pounds of BOD/day per pound MLVSS, therefore this plant is organically overloaded (Kerri, Dendy, et al. 1991 Operations of Wastewater Treatment Plants Volume II p. 19). The plant superintendent should discontinue wasting and consider building the solids inventory.

Hydraulic loading refers to the physical capacity of the process tank to take the flow being delivered to the tank. The process tank can have problems if the loading is too little or too great due to the detention time of organisms in the flow. Understanding the concept of hydraulic loading can shorten the time of an upset.

Here is the calculation:

Hydraulic loading, gpd/sq. ft = flow, gal/day divided by surface area, square feet

Situation: Each day for a week, the secondary clarifier has shown high effluent TSS, and the sludge blanket is far less than average for the time of year. The effluent water quality has deteriorated to the point operators must divert the flow to a deep-well injection instead of a reclamation pond. Searching for an explanation, the plant manager decides to do a hydraulic loading calculation. (Plant design characteristics states the hydraulic loading is 2.225 gpd/sq. ft.)

Plant data:

• Plant flow = 1,000,000
• Clarifier square foot = 350,000

1,000,000/350,000 = 2.857 gpd/sq. ft.

The data indicates there is a hydraulic overloading for the treatment plant. Therefore, the sludge is not getting enough time and condition to settle in the clarifier. The plant manager should consider putting another clarifier online to reduce the scouring effect and add detention time.

3. Alkalinity

A big indicator for nitrifying plants is alkalinity (the capacity of wastewater to neutralize an acid). An acidic condition is created as the nitrification process converts ammonia-nitrogen to nitrite. This activity requires an adequate supply of alkalinity to keep the pH level at optimum range. It takes 7.14 mg of alkalinity as expressed as calcium carbonate (CaCO3) to convert 1 mg of ammonia-nitrogen.

Situation: A plant manager notices tiny bubbles in the settling tank that brings up clumps of sludge from time to time. When the oxidation ditch effluent is sampled, results reveal nitrates are low, but the nitrites are much higher than usual. It appears the process is stuck in mid-nitrification because the operator also gets an ammonia-nitrogen concentration in the effluent sample. He decides to test the alkalinity.

Oxidation ditch plant data:

• Carrousel diameter 46 ft
• Carrousel volume 12,466 ft3

Alkalinity required for nitrification if a stable alkalinity for pH control is 72 mg/L calcium carbonate (CaCO3). The process has 130 mg/L of alkalinity present in the process carrousel.

Alkalinity depletion from nitrification= 7.14 mg CaCO3/1 mg N oxidized

Alkalinity required= 7.14 x 24= 171.36 mg/L CaCO3

Alkalinity needed= (alkalinity goal) + (alkalinity required) – (alkalinity present)

Alkalinity needed= (72) + (171.36 – 130)

Alkalinity needed= 113.36

To increase the alkalinity of the process, the addition of hydrated lime (Ca(OH)2 is typical. There is an estimated 1.33 pounds alkalinity/1 pound lime (Water Environment Federation, 2005).

To find the bags of lime needed for 113.36 mg/L of alkalinity:

• Convert concentration into pounds
• Pounds of alkalinity needed = 113.36 x 8.34 x 2.35
• Pounds of alkalinity needed = 2,221.74
• Pounds of lime needed = 2,221.74/1.33
• Pounds of lime needed =1,670.48
• Number of 50-pound bags needed = 1,670.48/50
• 33.41 bags of lime needed to give this oxidation ditch enough alkalinity to nitrify.

Conclusion
Though several other considerations should be made when troubleshooting process control, these three equations will prove invaluable. In addition to these equations, operators should use their sense of smell, sight and perception to get a complete view of the process control problem. There is no substitute for a well-trained and observant operator.