Question

A sewage treatment plant (STP) receives sewage at a flow rate of 20000 m\(^3\) per day. The sewage has 200 mg/L of suspended solids. Assume 60% suspended solids are removed in the primary clarifier. The underflow (i.e. sludge) removed from the clarifier contains 5% solids (by weight).
The daily volume of the sludge generated will be ________ m\(^3\) per day.
Assume sludge density = 1000 kg/m\(^3\).

• A. 48
• B. 80
• C. 480
• D. 800

Hint:

To calculate the volume of sludge generated, calculate the mass of suspended solids removed, then divide by the fraction of solids in the sludge and the density of the sludge.

Answer 1

The Correct Option is A

We are given the following data:
- Flow rate of sewage = 20000 m\(^3\) per day.
- Suspended solids concentration = 200 mg/L.
- 60% of suspended solids are removed in the primary clarifier.
- The sludge contains 5% solids (by weight).
- Sludge density = 1000 kg/m\(^3\).
Step 1: Calculate the mass of suspended solids in the sewage.
Mass of suspended solids per day is given by: \[ \text{Mass of suspended solids} = \text{Flow rate} \times \text{Concentration of solids} = 20000 \, \text{m}^3/\text{day} \times 200 \, \text{mg/L}. \] Convert mg/L to kg/m\(^3\): \[ 200 \, \text{mg/L} = 0.2 \, \text{kg/m}^3. \] Thus: \[ \text{Mass of suspended solids} = 20000 \times 0.2 = 4000 \, \text{kg/day}. \] Step 2: Calculate the amount of solids removed by the primary clarifier.
Since 60% of the solids are removed, the mass of solids removed is: \[ \text{Solids removed} = 4000 \times 0.6 = 2400 \, \text{kg/day}. \] Step 3: Calculate the volume of the sludge.
The sludge contains 5% solids by weight, so the mass of solids in the sludge is 5% of the total mass of the sludge: \[ \text{Total mass of sludge} = \frac{\text{Mass of solids}}{\text{Fraction of solids in sludge}} = \frac{2400}{0.05} = 48000 \, \text{kg}. \] Step 4: Calculate the volume of sludge.
The volume of sludge is given by: \[ \text{Volume of sludge} = \frac{\text{Mass of sludge}}{\text{Density of sludge}} = \frac{48000}{1000} = 48 \, \text{m}^3/\text{day}. \] Final Answer: \[ \boxed{48 \, \text{m}^3/\text{day}} \]