Math Solutions: Waste Activated Sludge (WAS) Calculations

By Daniel Theobald, "Wastewater Dan"

Water Online’s “Math Solutions” series, presented by wastewater consultant and trainer Dan Theobald (“Wastewater Dan”), provides an understanding of the sometimes difficult calculations involved in achieving wastewater operator certification.

Waste activated sludge (WAS) calculations in your plant may be based on data similar to the following:

An activated sludge process has three aeration basins. Each basin is 90 feet long, 45 feet wide, and 12 feet deep. The mixed liquor suspended solids (MLSS) concentration is 2,750 mg/L. There are two clarifiers. Each clarifier is 36 feet in diameter and 11 feet deep. A clarifier core sample indicates that the solids concentration in the clarifier is equivalent to 500 mg/L. If the return activated sludge (RAS)/WAS concentration is 5,950 mg/L, find the WAS pumping rate in gallons per minute (GPM) if the pump operates 17 minutes out of each hour to maintain a sludge age of 15 days.

Math Solutions to solve WAS pumping rate in GPM include calculations of tank volumes, pounds in activated sludge process, WAS pounds, and WAS pumping rate in GPM.

Calculate Volumes In Million Gallons (MG)

Aeration basins:

Clarifiers:

Calculate Pounds (Lbs.) In Activated Sludge Process

Aeration basins:

Clarifiers:

Calculate WAS

WAS pounds (lbs.) per day:

WAS GPM:

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