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 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\(^3\).
The daily volume of the sludge generated will be ________ m\(^3\) per day.
Assume sludge density = 1000 kg/m\(^3\).
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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.
Updated On: Dec 29, 2025
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Solution and Explanation
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: \[ \boxed{48 \, \text{m}^3/\text{day}} \]
- 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: \[ \boxed{48 \, \text{m}^3/\text{day}} \]
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