Water operator math tutorial
Sludge Age
Divide the solids in the aeration tank by the solids coming in each day.
Sludge age is similar to MCRT but measures how long solids stay based on the solids entering the aeration tank each day.
It is a simpler number to compute when you only know the primary effluent solids, which is why older exam questions favor it.
The formula
sludge age (days) = [MLSS (mg/L) × tank volume (MG) × 8.34] ÷ [influent TSS (mg/L) × flow (MGD) × 8.34]
| Variable | Unit |
|---|---|
| MLSS | mg/L |
| Aeration tank volume | MG |
| Primary effluent TSS | mg/L |
| Flow | MGD |
| Result: Sludge age | days |
Conversion factors used: 8.34 lb per gallon of water (cancels in the ratio).
Applies to: Wastewater Treatment.
Common exam tip: Sludge age uses solids entering the tank. MCRT uses solids leaving. Read which one the question wants.
Calculator
Sludge age
3.33 days
- 1. Solids in tank = 2,000 × 0.5 × 8.34 = 8,340 lb
- 2. Solids added = 150 × 2 × 8.34 = 2,502 lb/day
- 3. Sludge age = 8,340 ÷ 2,502 = 3.33 days
For instructional use. Results assume the standard conversion factors shown on this page. Actual field calculations may need chemical strength, density, temperature, and other adjustments, so check your plant’s procedures.
Worked examples
Example 1
MLSS 2,000 mg/L in a 0.5 MG tank; primary effluent TSS 150 mg/L at 2 MGD.
- Solids in tank = 2,000 × 0.5 × 8.34 = 8,340 lb
- Solids added = 150 × 2 × 8.34 = 2,502 lb/day
- Sludge age = 8,340 ÷ 2,502 = 3.33 days
Answer: 3.33 days
Example 2
MLSS 2,800 mg/L in a 1.2 MG tank; primary effluent TSS 200 mg/L at 4 MGD.
- Solids in tank = 2,800 × 1.2 × 8.34 = 28,022 lb
- Solids added = 200 × 4 × 8.34 = 6,672 lb/day
- Sludge age = 28,022 ÷ 6,672 = 4.2 days
Answer: 4.2 days
Common mistakes
Wrong result: 0.3 days
dividing solids added by solids in the tank (upside down)
Wrong result: 0.14 days
converting days to hours incorrectly
Practice questions
1. MLSS 2,500, tank 0.8 MG, TSS 120 mg/L, flow 2.5 MGD. Sludge age?
- A0.15 days
- B0.28 days
- C6.67 days
- D66.7 days
Show answer and explanation
Correct answer: C. Solids in tank = 2,500 × 0.8 × 8.34 = 16,680 lb Solids added = 120 × 2.5 × 8.34 = 2,502 lb/day Sludge age = 16,680 ÷ 2,502 = 6.67 days Answer: 6.67 days. A common slip is dividing solids added by solids in the tank (upside down), which gives 0.15 days.
2. MLSS 1,800, tank 0.6 MG, TSS 180 mg/L, flow 1.5 MGD. Sludge age?
- A0.25 days
- B0.17 days
- C40 days
- D4 days
Show answer and explanation
Correct answer: D. Solids in tank = 1,800 × 0.6 × 8.34 = 9,007 lb Solids added = 180 × 1.5 × 8.34 = 2,252 lb/day Sludge age = 9,007 ÷ 2,252 = 4 days Answer: 4 days. A common slip is dividing solids added by solids in the tank (upside down), which gives 0.25 days.
3. MLSS 3,200, tank 2.0 MG, TSS 210 mg/L, flow 5.0 MGD. Sludge age?
- A6.1 days
- B0.16 days
- C0.25 days
- D61 days
Show answer and explanation
Correct answer: A. Solids in tank = 3,200 × 2 × 8.34 = 53,376 lb Solids added = 210 × 5 × 8.34 = 8,757 lb/day Sludge age = 53,376 ÷ 8,757 = 6.1 days Answer: 6.1 days. A common slip is dividing solids added by solids in the tank (upside down), which gives 0.16 days.
Where this formula applies
This calculation can appear in the disciplines below. Exact exam coverage depends on your state, certification level, and official exam outline.
Wastewater Treatment
Study guides
Keep going with the full guide
Water Operator Math Made Simple walks through operator math step by step and includes printable references plus a full practice exam.
Water Operator Math Made Simple
A Step-by-Step Guide to the Math Every Water Operator Needs to Know