Solids Loading Rate

S L = \frac{M}{A}

Mass of solids applied per unit clarifier surface area per day. M = Q X (plus recycle solids if you load the clarifier with Q(1+R) and mixed liquor X). Typical secondary-clarifier SL is on the order of 100–150 kg/m²·d; primary tanks are much lower because influent TSS is lower. Distinct from overflow rate vo = Q/A.

$S L$ (kg/m²·d): Solids loading
$M$ (kg/d): Solids mass rate
$A$ (m²): Clarifier surface area
$X$ (mg/L = g/m³): Mixed-liquor (or influent) TSS

Step-by-step solved example

Q = 10,000 m³/d, X = 2500 mg/L, A = 250 m². SL? (ignore recycle)

Solids loading = mass flux / plan area.

1. Mass rate
X = 2500 mg/L = 2.50 kg/m³. M = Q X = 10,000 × 2.50 = 25,000 kg/d.
2. Loading
SL = M/A = 25,000/250 = 100 kg/m²·d.
$S L = 100 kg/m^{2} \cdot d$

Answer: SL = 100 kg/m²·d

10 practice questions

0/10 correct

1.Q = 6000 m³/d, X = 3000 mg/L, A = 300 m². SL =

2.Overflow rate vo = Q/A has units of

3.Including recycle, the solids load on a secondary clarifier uses

4.A high SVI (bulking) means the allowable SL

5.2500 mg/L equals

6.Primary clarifiers run at much lower SL than secondary because

7.M = 12,000 kg/d, A = 200 m². SL =

8.If area is halved at the same M, SL

9.Solids flux theory plots

10.Q = 8000 m³/d, X = 2500 mg/L, A = 400 m², no recycle. SL =