--- title: "Grit chamber for 180 l/s, SOR 1.5" subject: "Sanitary Engineering" chapter: 7 key: "semester_6/sanitary_engineering:grit-180-ls-sor-1p5" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/grit-180-ls-sor-1p5 site: BCT Engineering --- # Grit chamber for 180 l/s, SOR 1.5 Sanitary Engineering · Chapter 7: Wastewater Treatment - Asked in: 2071 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/grit-180-ls-sor-1p5 - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7 ## Question Design a grit chamber for a wastewater flow of 180 l/s with SOR = 1.5 cm/second and detention period of 50 seconds. Specific gravity of organic and inorganic particles are 1.2 and 2.65 respectively. Assume the size of both organic and inorganic particles as 0.25 mm. Take k = 0.06 and f = 0.03. ## Short answer For180 L\/ s, SOR=1.5 cm\/ s, detention 50 s: surface area=12 m\u00b2, volume=9 m\u00b3, depth=0.75 m. Scour velocities: organic 0.089 m\/ s, grit 0.254 m\/ s, adopt horizontal velocity0.25 m\/ s. Length=12.5 m, width=1.0 m. Chamber: L=12.5 m, B=1.0 m, water depth0.75 m (total ~1.30 m), velocity0.25 m\/ s; two units. ## Answer $Q=180$ l/s $=0.18$ m³/s, SOR $=1.5$ cm/s $=0.015$ m/s, detention $=50$ s, particle size 0.25 mm, $k=0.06$, $f=0.03$. #### Step 1: Surface area and depth $$ \begin{aligned} Q &= 0.180\ \text{m}^3/\text{s}\\ \text{Surface area } A &= \frac{Q}{SOR} = \frac{0.180}{0.015} = 12.00\ \text{m}^2\\ \text{Volume } V &= Q\,t = 0.180\times 50 = 9.00\ \text{m}^3\\ \text{Depth } H &= \frac{V}{A} = 0.75\ \text{m} \end{aligned} $$ #### Step 2: Scour velocity for the two kinds of particles $$ v_c=\sqrt{\frac{8k}{f}(G-1)\,g\,d} $$ | Particle | $G$ | $d$ (mm) | $v_c$ (m/s) | |---|---|---|---| | Organic | 1.2 | 0.25 | 0.089 | | Inorganic (grit) | 2.65 | 0.25 | 0.254 | The horizontal velocity must be **above** the organic scour velocity (organic matter stays suspended and is carried through) and **below** the grit scour velocity (settled grit is not re-suspended). Adopt $v_h = 0.25$ m/s, which satisfies 0.089 < 0.25 < 0.254 m/s. (The flow velocity should also be near 0.2-0.3 m/s in practice.) #### Step 3: Dimensions $$ \begin{aligned} L &= v_h\,t = 0.25\times 50 = 12.50\ \text{m}\\ B &= \frac{A}{L} = \frac{12.00}{12.50} = 0.96\ \text{m}\\ \text{Check: } A_c &= B\,H = 0.96\times 0.75 = 0.720\ \text{m}^2 \approx \frac{Q}{v_h} = 0.720\ \text{m}^2 \end{aligned} $$ Adopt $L = 12.5$ m, $B = 1.0$ m, water depth $H = 0.75$ m, plus 0.25 m grit storage and 0.30 m free board (total depth about 1.30 m). Provide two units in parallel. **Answer: grit chamber $L\times B\times H = 12.5\times 1.0\times 0.75$ m, flow velocity 0.25 m/s, detention 50 s, surface overflow rate 1.5 cm/s.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the surface area of the grit chamber found from the given flow and SOR? A. A = Q x SOR = 0.18 x 0.015 B. A = Q / SOR = 0.18 / 0.015 = 12 m^2 C. A = Q x t = 0.18 x 50 D. A = SOR / Q = 0.015 / 0.18 Answer: B. SOR is flow per unit surface area, so A = Q/SOR = 0.18/0.015 = 12 m^2. (Key point: Surface area from SOR) 2. Which route gives the water depth H = 0.75 m? A. H = A / V = 12 / 9 B. H = Q / A = 0.18 / 12 C. H = V / A = (Q x t) / A = 9 / 12 D. H = SOR x t = 0.015 x 50 x 12 Answer: C. Volume V = Q t = 9 m^3 and depth = volume divided by surface area = 0.75 m. (Key point: Depth from detention) 3. Why is the horizontal velocity chosen between the organic scour velocity (0.089 m/s) and the grit scour velocity (0.254 m/s)? A. Below 0.089 m/s organic matter is carried off; above 0.254 m/s grit settles faster B. It must exceed 0.254 m/s so that all particles are scoured out C. It must be below 0.089 m/s so organic matter can settle with grit D. So organic matter stays moving (not settling) while settled grit is not re-suspended Answer: D. Velocity above the organic scour value keeps organics moving through; below the grit scour value stops settled grit being picked up again. (Key point: Choice of horizontal velocity) 4. After adopting v_h = 0.25 m/s, how are L and B obtained? A. L = v_h x t = 12.5 m, then B = A / L = 0.96 m B. L = A / H, then B = v_h x t C. L = Q / v_h, then B = H x t D. L = v_h / t, then B = A x L Answer: A. Length is the distance travelled in the detention time, and width is the surface area divided by length. 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