--- title: "Weir waterway and crest levels, 16000 m3/s" subject: "Irrigation and Drainage Engineering" chapter: 5 key: "semester_6/irrigation_and_drainage_engineering:weir-waterway-crest-q16000" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/weir-waterway-crest-q16000 site: BCT Engineering --- # Weir waterway and crest levels, 16000 m3/s Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks - Asked in: 2073 Bhadra, 6 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/weir-waterway-crest-q16000 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5 ## Question A river carries a high flood discharge of 16000 m³/s with its average bed level at 200.0 m. A canal carrying 200 m³/s is to take off from the headworks. The full supply level of the canal at its head is 203.0 m. The high flood level before construction is 205.7 m and Lacey's silt factor is equal to unity. Fix suitable values for the waterway and crest levels of weir, undersluices and canal head regulator. Assume suitably any other data if required. ## Short answer Total waterway 600 m (undersluices 46 m, weir 554 m). Weir crest ~200.4 m (gates pond level 203.5 m); undersluice crest 200.0 m; head regulator crest 201.5 m with clear waterway ~42 m. ## Answer #### Assumptions Afflux $= 1.0$ m, loss through the head regulator $= 0.5$ m, weir coefficient $C = 1.7$ ($Q = CLH^{3/2}$), average bed level $= 200.0$ m, undersluice discharge intensity about 10 m³/s/m. Velocity of approach is neglected. #### 1. Waterway (Lacey) $$ P = 4.75\sqrt{16000} = 600.8\ \text{m} \Rightarrow L = 600\ \text{m} $$ Discharge intensity $q = 16000/600 = 26.67$ m³/s/m; scour depth $R = 1.35(q^2/f)^{1/3} = 12.05$ m. #### 2. Levels - HFL before construction $= 205.7$ m; HFL after construction $= 205.7 + 1.0 = 206.7$ m. - Pond level $=$ FSL $+$ regulator loss $= 203.0 + 0.5 = 203.5$ m. - Head needed to pass the flood over the full waterway: $H = (16000/(1.7\times 600))^{2/3} = 6.27$ m. - Weir (open-flood) crest level $= 206.7 - 6.27 = 200.43$ m, say **200.4 m** (only about 0.4 m above the bed). The structure is therefore a gated barrage: gates (3.1 m high) hold the pond at 203.5 m. - Undersluice crest level $=$ river bed $=$ **200.0 m**. - Head regulator crest level $=$ undersluice crest $+$ 1.5 m $=$ **201.5 m** (the canal FSL of 203.0 m gives 1.5 m depth over the crest). #### 3. Waterway distribution Undersluice discharge $= 2\times$ canal discharge $= 400$ m³/s, so the width $= 400/10 = 40$ m, taken as 5 bays of 8 m with 4 piers of 1.5 m: $40 + 6 = 46$ m. Weir bays: $600 - 46 = 554$ m. Head regulator: head over its crest $=$ pond level $-$ crest $= 203.5 - 201.5 = 2.0$ m, so $$ L = \frac{200}{1.7\times 2.0^{3/2}} = 41.6\ \text{m} $$ Provide about 42 m clear waterway (more if the partly drowned flow is allowed for). **Answer:** Total waterway 600 m (undersluices 46 m, weir 554 m); weir crest $\approx 200.4$ m (gates raise the pond to 203.5 m); undersluice crest 200.0 m; head regulator crest 201.5 m with about 42 m clear waterway. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The total waterway for Q = 16000 m3/s is: A. Lacey's 4.75 sqrt(Q) = 600.8, say 600 m B. About 300 m C. About 1200 m D. About 126 m Answer: A. 4.75 x sqrt(16000) = 600.8 m. (Key point: Waterway) 2. Weir crest level is found from: A. HFL after afflux (206.7) - H, H = 6.27 m, giving about 200.4 m B. Canal FSL - 1.5 m C. Bed level + 3 m D. Pond level + H Answer: A. H = (Q/(1.7L))^(2/3) = 6.27 m; 206.7 - 6.27 = 200.43 m. (Key point: Weir crest) 3. Why is the structure a gated barrage? A. Crest (~200.4 m) is only slightly above the bed, so gates hold the 203.5 m pond B. Because canal is large C. Afflux is zero D. Because the undersluice is high Answer: A. A crest this low cannot hold the pond itself, so gates about 3.1 m high retain it. (Key point: Gated structure) 4. Pond level is obtained as: A. Canal FSL + regulator loss = 203.5 m B. FSL - loss C. HFL - afflux D. Bed + R Answer: A. The pond must be above FSL by the head loss through the regulator. (Key point: Pond level) 5. Undersluice width is based on: A. Discharge twice the canal discharge at about 10 m3/s/m B. Whole flood C. One-tenth of waterway D. Canal width Answer: A. 400/10 = 40 m of waterway, plus piers giving 46 m. 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