--- title: "Combined sewer section, 85 ha area" subject: "Sanitary Engineering" chapter: 3 key: "semester_6/sanitary_engineering:combined-sewer-85ha" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/combined-sewer-85ha site: BCT Engineering --- # Combined sewer section, 85 ha area Sanitary Engineering · Chapter 3: Design and Construction of Sewers - Asked in: 2074 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/combined-sewer-85ha - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-3 ## Question What would be your preferable combined sewer section for an 85-hectare residential area having average runoff coefficient of 0.45 for serving altogether 1500 population? Average rainfall duration is 26 min. Self-cleansing velocity is 0.98 m/sec. Residential area have average elevation difference of 22 m in horizontal 5 km longitudinal distance. Assume any other appropriate data if required. ## Short answer Total flow 2.249 m³/s. With n=0.013, slope 22/5000=1/227, Manning's gives D=1.139 m; adopt 1.2 m circular section, full-flow velocity 2.29 m/s (above self-cleansing 0.98 m/s, below scouring limit). A low-flow channel is advisable. ## Answer #### Assumptions Sewage 135 lpcd, of which 80% reaches the sewer; peak factor 3; $n=0.013$ for concrete; circular section running full at peak flow; intensity from the assumed formula $i=\dfrac{760}{t+10}$ mm/hr with storm duration 26 min. #### Step 1: Discharge $$ \begin{aligned} Q_{sew}&=\frac{1500\times135\times0.8\times3}{86400\times1000}=0.0056\ \text{m}^3/\text{s}\\ i&=\frac{760}{26+10}=21.11\ \text{mm/hr}\\ Q_{storm}&=\frac{0.45\times21.11\times85}{360}=2.243\ \text{m}^3/\text{s}\\ Q&=Q_{sew}+Q_{storm}=2.249\ \text{m}^3/\text{s} \end{aligned} $$ #### Step 2: Slope $$ S=\frac{\text{fall}}{\text{length}}=\frac{22}{5000}=0.00440\approx\frac{1}{227} $$ #### Step 3: Size $$ D=\left(\frac{Q\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{2.249\times0.013}{0.3117\times0.0663}\right)^{3/8}=1.139\ \text{m} $$ Adopt $D=1.200$ m. Then $v_f=\dfrac{0.3969}{0.013}(1.200)^{2/3}S^{1/2}=2.29$ m/s and $Q_f=2.586$ m³/s. #### Step 4: Check The full-flow velocity of 2.29 m/s is greater than the required self-cleansing velocity of 0.98 m/s and below the maximum of 3 m/s, so the section is satisfactory. At the very small dry weather flow (about 0.0056 m³/s) the velocity in this big pipe will be lower, so a small low-flow channel (cunette) in the invert or an egg-shaped section is advisable. **Answer: A circular combined sewer of about 1200 mm (1.2 m) diameter at a slope of 22/5000 (1 in 227), carrying 2.249 m³/s at a velocity of 2.29 m/s.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which flow dominates the design of this combined sewer? A. Storm water flow (2.243 m3/s) far exceeds sewage flow (0.0056 m3/s) B. Sewage flow, since it is dry weather flow C. Both equally D. Neither; infiltration controls Answer: A. Q_storm = CiA/360 with A = 85 ha dwarfs the sewage from 1500 people. (Key point: Dominant flow) 2. How is the slope obtained from the data? A. S = length / fall = 5000/22 B. S = fall / length = 22/5000 = 1 in 227 C. S = 22 x 5 D. Slope from Manning with v = 0.98 m/s only Answer: B. The ground elevation difference over the length gives the gradient. (Key point: Slope of the sewer) 3. With sewer running full, what gives D? A. D from the sewage flow only B. D = Q / v with v = 0.98 C. D = (Q n / (0.3117 S^(1/2)))^(3/8) = 1.139 m, adopt 1.2 m D. D = (Q n / S)^(3/8) with no constant Answer: C. Full-pipe Manning with Q = 2.249 m3/s gives 1.139 m. (Key point: Size from Manning) 4. Why may a cunette or egg-shaped section be advisable? A. Because storm flow is too small B. Because slope is too steep C. Because velocity at full flow is above 3 m/s D. At very small dry weather flow the velocity in a large circular pipe falls below self-cleansing Answer: D. A narrow low-flow channel keeps depth and velocity adequate for the small sewage flow. 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