--- title: "Separate sewer for 0.5 m³/s, 0.7 depth" subject: "Sanitary Engineering" chapter: 3 key: "semester_6/sanitary_engineering:separate-sewer-0p5-slope-1-in-1000" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/separate-sewer-0p5-slope-1-in-1000 site: BCT Engineering --- # Separate sewer for 0.5 m³/s, 0.7 depth Sanitary Engineering · Chapter 3: Design and Construction of Sewers - Asked in: 2068 Magh (old course), 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/separate-sewer-0p5-slope-1-in-1000 - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-3 ## Question Design a sewer for separate system to carry a maximum flow of 0.5 m³/s at a slope of 1 in 1000 (10 in 10000). Sewer should run 0.7 times depth at maximum flow. Assume n = 0.012. ## Short answer Using hydraulic elements at 0.7 depth, full-flow discharge Q_f = 0.5/0.8372 = 0.5972 m³/s. From Manning's formula, D = 0.887 m, so adopt 900 mm. At this size, the sewer runs 0.7 full with velocity 1.09 m/s. ## Answer #### Given Maximum flow $Q=0.5$ m³/s, slope $S=1/1000=0.001$, $n=0.012$, depth of flow $d/D=0.7$ at maximum flow. #### Hydraulic elements For $d/D=0.7$: $\theta = 2\cos^{-1}(1-2d/D)=227.16^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.7477$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.1849$, so $v/v_f=(R/R_f)^{2/3}=1.1198$ and $Q/Q_f=(A/A_f)(v/v_f)=0.8372$ (n taken constant). #### Size of the sewer Discharge when running full: $$ Q_f=\frac{Q}{Q/Q_f}=\frac{0.5000}{0.8372}=0.5972\ \text{m}^3/\text{s} $$ Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00100$ and $n=0.012$: $$ D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.5972\times0.012}{0.3117\times0.03162}\right)^{3/8}=0.887\ \text{m} $$ Adopt the next standard size, **D = 900 mm**. #### Check of velocity and capacity For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=0.975$ m/s and $Q_f=0.6202$ m³/s. $$ \begin{aligned} v&=v_f\times(v/v_f)=0.975\times1.1198=1.09\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=0.6202\times0.8372=0.5193\ \text{m}^3/\text{s}\ (\ge 0.5000) \end{aligned} $$ The velocity at maximum flow, 1.09 m/s, lies between the minimum (0.6 m/s) and the maximum non-scouring limit (3.0 m/s), so it is satisfactory. **Answer: Provide a circular sewer of diameter 900 mm at slope 1 in 1000; it runs 0.7 full with velocity 1.09 m/s at 0.5 m³/s.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At d/D = 0.7 (constant n), what is Q/Q_f and what does it mean? A. 0.7: the discharge is 70% of full B. 0.8372: the sewer carries about 84% of its full-pipe discharge C. 1.0: the sewer is running full D. 0.7477: equal to the area ratio only Answer: B. Q/Q_f = (A/A_f)(v/v_f) = 0.7477 x 1.1198 = 0.8372. (Key point: Q/Q_f at 0.7 depth) 2. Which step gives Q_f for the pipe design? A. Q_f = 0.5 x 0.8372 B. Q_f = 0.5 / 0.7 C. Q_f = Q / (Q/Q_f) = 0.5 / 0.8372 = 0.597 m3/s D. Q_f = 0.5 m3/s itself Answer: C. Dividing the design flow by the discharge ratio gives the full-flow capacity needed. (Key point: Full discharge needed) 3. Which formula gives D from Q_f, n and S for a full circular pipe? A. D = (Q_f n / (0.3117 S^(1/2)))^(3/8) B. D = (Q_f n / (0.3117 S^(1/2)))^(8/3) C. D = (Q_f S / n)^(1/2) D. D = Q_f / (v A) Answer: A. From Q_f = (0.3117/n) D^(8/3) S^(1/2), D is raised to the power 3/8. (Key point: Diameter formula) 4. The adopted 900 mm sewer gives 1.09 m/s at maximum flow. Why is this acceptable? A. It is below 0.6 m/s so no scour occurs B. It exceeds 3.0 m/s so it self-cleanses C. Any velocity is acceptable if capacity exceeds Q D. It lies between the minimum self-cleansing 0.6 m/s and the maximum non-scouring 3.0 m/s Answer: D. Velocity must be at least 0.6 m/s to avoid silting and at most about 3.0 m/s to avoid scouring. 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