--- title: "Separate sewer, 0.70 full, 100000 persons" subject: "Sanitary Engineering" chapter: 3 key: "semester_6/sanitary_engineering:sewer-0p7-full-100000" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sewer-0p7-full-100000 site: BCT Engineering --- # Separate sewer, 0.70 full, 100000 persons Sanitary Engineering · Chapter 3: Design and Construction of Sewers - Asked in: 2069 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/sewer-0p7-full-100000 - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-3 ## Question Design a circular sewer running 0.70 full at maximum discharge for a town provided with the separate system serving a population of 100000 persons. Water is supplied from the water works at a rate of 200 liters per capita per day. Take a constant value of n = 0.013 at all depths of flow. The permissible slope is 1 in 600. Take a peak factor of 2.25. ## Short answer Peak flow = 0.4167 m³/s. Using Q/Q_f=0.8372 at 0.7 depth, Q_f=0.4977 m³/s; from Manning's, D=0.776 m (adopt 825 mm). At design flow, the sewer runs about 0.62 full with v ≈ 1.19 m/s. ## Answer #### Design flow Assume 80% of the water supplied reaches the sewer. Peak factor = 2.25. $$ Q_{max}=\frac{100000\times200\times0.8\times2.25}{86400\times1000}=0.4167\ \text{m}^3/\text{s} $$ Data: $S=1/600$, $n=0.013$ (constant at all depths), depth ratio $d/D=0.70$. 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: $$ Q_f=\frac{Q}{Q/Q_f}=\frac{0.4167}{0.8372}=0.4977\ \text{m}^3/\text{s} $$ Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00167$ and $n=0.013$: $$ D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.4977\times0.013}{0.3117\times0.04082}\right)^{3/8}=0.776\ \text{m} $$ Adopt the next standard size, **D = 825 mm**. #### Check of velocity and capacity For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=1.096$ m/s and $Q_f=0.5861$ m³/s. $$ \begin{aligned} v_{0.7}&=v_f\times(v/v_f)=1.096\times1.1198=1.23\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=0.5861\times0.8372=0.4907\ \text{m}^3/\text{s}\ (\ge 0.4167) \end{aligned} $$ At the design flow, $Q/Q_f=0.4167/0.5861=0.711$, which gives $d/D\approx0.62$ and $v/v_f\approx1.086$, so $v\approx1.19$ m/s. The velocity at maximum flow, about 1.2 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 D = 825 mm at slope 1 in 600** (computed diameter 0.776 m runs exactly 0.7 full). The adopted pipe can carry 0.4907 m³/s at 0.7 depth; at Q = 0.4167 m³/s it runs about 0.62 full with v ≈ 1.19 m/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is Q_max for 100,000 people at 200 lpcd, 80% to sewer, peak factor 2.25? A. 0.1852 m3/s B. 0.4167 m3/s C. 0.5208 m3/s D. 1.0417 m3/s Answer: B. 100000 x 200 x 0.8 x 2.25 / 86400000 = 0.4167 m3/s. (Key point: Design flow) 2. With constant n at d/D = 0.7, the ratios are A. A/A_f = 0.7, v/v_f = 1.0, Q/Q_f = 0.7 B. A/A_f = 0.8372, v/v_f = 0.7477, Q/Q_f = 1.1198 C. A/A_f = 0.7477, v/v_f = 1.1198, Q/Q_f = 0.8372 D. A/A_f = 0.7477, v/v_f = 0.7477, Q/Q_f = 0.559 Answer: C. Q/Q_f is the product of A/A_f and v/v_f; velocity at 0.7 depth exceeds full velocity because R is larger. (Key point: Proportionate values at 0.7 depth) 3. Why is velocity at 0.7 depth greater than at full flow (v/v_f = 1.12)? A. The hydraulic radius is larger than at full flow (R/R_f = 1.185) B. The area is larger than full area C. The slope increases in a partly full sewer D. Manning's n decreases at partial depth Answer: A. R = A/P peaks near 0.81 depth, so velocity at partial depth can exceed full velocity when n is constant. (Key point: Why v exceeds v_f) 4. The computed D is 0.776 m. What is provided and why? A. 700 mm, the nearest smaller standard size B. 900 mm, the next standard size; capacity 0.619 m3/s exceeds 0.4167 C. 776 mm exactly D. 1200 mm since peak factor is high Answer: B. Adopt the next larger standard size, then verify velocity (1.30 m/s) and capacity. 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