Raw primary and waste activated sludge containing 4% solids is to be anaerobically digested at a loading of 3 kg/(m³.day). The total sludge produced in the plant is 1500 kg of dry solids per day. Assume 1 L of sludge weighs about 1 kg. The approximate hydraulic retention time required is :
- A13·5 daysCorrect
- B15·5 days
- C10·5 days
- D12·5 days
Solution
Concept
Digester volume and hydraulic retention time
Formula
V = solids load / loading rate; HRT = V / Qsludge
Step-by-step solution
1. Digester volume = 1500 / 3 = 500 m³. 2. Sludge flow = 1500 / 0.04 = 37 500 kg/day = 37.5 m³/day. 3. HRT = 500 / 37.5 = 13.3 days ≈ 13.5 days.
Explanation
The digester volume follows from the solids loading rate. The hydraulic retention time is the digester volume divided by the daily volume of wet sludge fed.
Why the other options are wrong
15.5, 10.5 and 12.5 days do not agree with 500/37.5 = 13.3.
Shortcut
HRT = (Solids/loading) ÷ (Solids/concentration).
Common mistake
Using the dry solids mass as the volume.
Correct answer: A) 13·5 days