Sewerage and Sewage Treatment
JKSSB Junior Engineer (Civil) · Water Supply and Waste Water Engineering · 22 practice questions with answers and explanations: 15 Easy, 6 Medium, 1 Hard.
Sample questions
Sample Easy question 1
The BOD of sewage is normally measured over:
- 5 days at 20°C
- 5 days at 37°C
- 1 day at 37°C
- 10 days at 20°C
Show answer
Answer: A. BOD₅ at 20°C is the standard test.
Sample Easy question 2
A sewerage system that carries sewage and storm water in the same pipe is called:
- A septic system
- A partially separate system
- A separate system
- A combined system
Show answer
Answer: D. A separate system uses two sets of pipes.
Sample Easy question 3
The minimum velocity in a sewer that prevents the deposition of solids is called the:
- Self-cleansing velocity
- Settling velocity
- Terminal velocity
- Critical velocity
Show answer
Answer: A. A sanitary sewer is designed to run at about 0.6 m/s or more at its design flow.
Sample Medium question 4
The full-flow velocity in a circular sewer of 0.6 m diameter with n = 0.013 and a slope of 0.002 is about:
- 3.88 m/s
- 0.97 m/s
- 1.94 m/s
- 0.49 m/s
Show answer
Answer: B. v = (1 ÷ n) × (D ÷ 4)^(2/3) × S^(1/2) = 0.97 m/s.
Sample Medium question 5
A town of 30,000 people uses 135 litres per head per day, and 80% of it reaches the sewers. The average dry weather flow is:
- 4.05 MLD
- 6.48 MLD
- 0.324 MLD
- 3.24 MLD
Show answer
Answer: D. 30,000 × 135 × 0.8 ÷ 10⁶ = 3.24 MLD.
Sample Medium question 6
After 30 minutes of settling, a sample of activated sludge occupies 300 mL per litre, and the MLSS is 3,000 mg/L. The sludge volume index is:
- 10 mL/g
- 100 mL/g
- 0.1 mL/g
- 1,000 mL/g
Show answer
Answer: B. SVI = settled volume (mL/L) × 1,000 ÷ MLSS (mg/L) = 300 × 1,000 ÷ 3,000 = 100 mL/g.
Sample Hard question 7
A sewage has an ultimate BOD of 300 mg/L and a reaction rate constant k = 0.1 per day (base 10). The 5-day BOD is about:
- 205 mg/L
- 270 mg/L
- 300 mg/L
- 150 mg/L
Show answer
Answer: A. BOD₅ = L₀ (1 − 10^(−5k)) = 300 × (1 − 10^−0.5) = 205 mg/L.