--- title: "Silt excluder design, Q 200 cumec" subject: "Irrigation and Drainage Engineering" chapter: 5 key: "semester_6/irrigation_and_drainage_engineering:silt-excluder-design-q200" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/silt-excluder-design-q200 site: BCT Engineering --- # Silt excluder design, Q 200 cumec Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks - Asked in: 2078 Chaitra, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/silt-excluder-design-q200 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5 ## Question Design a silt excluder for the diversion head works having following data: Full supply discharge of canal = 200 m³/sec; Crest level of undersluices = 225.25 m; Crest level of head regulator = 227.35 m; Bay width of undersluices = 10.2 m. ## Short answer Assume tunnel discharge 40 m³/s (20% of 200). Height = (227.35−225.25)−0.30 = 1.80 m. Area = 40/2.5 = 16 m², width = 8.89 m. Provide 3 tunnels each 3.0 m wide with two 0.6 m piers to fit 10.2 m bay. Velocity = 2.47 m/s. Result: 3 tunnels 3.0 m×1.80 m, piers 0.6 m, carrying 40 m³/s at 2.47 m/s. ## Answer #### Given and assumptions Canal discharge $Q = 200$ m³/s; undersluice crest level (tunnel floor) $= 225.25$ m; head regulator crest level $= 227.35$ m (tunnel roof level); bay width of the undersluices $= 10.2$ m. Assumed: discharge through the tunnels = 20% of the canal discharge; roof slab thickness 0.3 m; tunnel velocity about 2.5 m/s (non-silting); pier thickness 0.6 m. #### Design 1. **Discharge in tunnels** $$ Q_t = 0.20\times 200 = 40\ \text{m}^3/\text{s} $$ 2. **Height of tunnels** Difference of crest levels $= 227.35 - 225.25 = 2.10$ m. Subtracting the roof slab of $0.30$ m: $$ h = 2.10 - 0.30 = 1.80\ \text{m} $$ 3. **Area of tunnels** $$ A = \frac{Q_t}{V} = \frac{40}{2.5} = 16.0\ \text{m}^2 $$ 4. **Total width of tunnels** $$ W = \frac{A}{h} = \frac{16.0}{1.80} = 8.89\ \text{m} $$ 5. **Number of tunnels.** Provide 3 tunnels of 3.0 m each and 2 piers of 0.6 m: $$ \text{Total width} = 3\times 3.0 + 2\times 0.6 = 10.2\ \text{m} $$ which fits exactly in the bay width of 10.2 m. Area provided $= 3\times 3.0\times 1.80 = 16.20$ m². 6. **Velocity check** $$ V = \frac{40}{16.20} = 2.47\ \text{m/s} $$ This is very slightly below 2.5 m/s (about 1% less), which is practically acceptable; it lies within the usual tunnel velocity range of 2-3 m/s, so the silt will be carried through the tunnels. #### Section ```text regulator crest 227.35 ----------------------- roof slab (0.30) =================== tunnel height 1.80 | | | | | | undersluice crest 225.25 ----------------------- 3 tunnels x 3.0 m + 2 piers x 0.6 m = 10.2 m ``` **Answer:** 3 tunnels, each 3.0 m wide and 1.80 m high (roof 227.35 m, floor 225.25 m), piers 0.6 m, carrying 40 m³/s at about 2.47 m/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the tunnel discharge Q_t? A. 200 m3/s B. 10 m3/s C. 20 m3/s D. 40 m3/s Answer: D. 20% of 200 = 40 m3/s. (Key point: Tunnel discharge) 2. What is the tunnel height? A. 1.80 m (2.10 m level difference minus 0.30 m roof slab) B. 2.10 m C. 2.40 m D. 0.30 m Answer: A. 227.35 - 225.25 = 2.10 m; minus 0.30 m gives 1.80 m. (Key point: Tunnel height) 3. What is the required total tunnel width? A. 16.0 m B. 8.89 m (A/h = 16.0/1.80) C. 10.2 m D. 4.44 m Answer: B. A = 40/2.5 = 16 m2; W = 16/1.8 = 8.89 m. (Key point: Total width) 4. What layout fits the 10.2 m bay? A. 2 tunnels of 5 m B. 4 tunnels of 2.5 m C. 3 tunnels of 3.0 m with 2 piers of 0.6 m D. 1 tunnel of 10.2 m Answer: C. 3 x 3.0 + 2 x 0.6 = 10.2 m. 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