--- title: "Siphon aqueduct: Q 50, bed 32 m" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:siphon-aqueduct-q50-bed32-hfd400" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/siphon-aqueduct-q50-bed32-hfd400 site: BCT Engineering --- # Siphon aqueduct: Q 50, bed 32 m Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2069 Bhadra, 10 marks - Video: [Siphon Aqueduct | Cross Drainage Structure | Irrigation Engineering| IOE, TU](https://www.youtube.com/watch?v=VKfu13VWL2Y) — Parash Joshi - Civil Construction and Tutor - 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/siphon-aqueduct-q50-bed32-hfd400 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question Design a syphon aqueduct (Drainage water way, Canal water way, Bed levels and Transitions) if the following data at the crossing of canal and drainage are given. Discharge of canal = 50 cumecs; Bed width of canal = 32 m; Full supply depth of canal = 1.80 m; Canal bed level = 200.0 m; Side slopes of canal = (1.5H:1V); High flood discharge of drainage = 400 cumecs; High flood level of drainage = 200.60 m; Bed level of drainage = 198.0 m; General ground level = 200.20 m. ## Short answer Siphon aqueduct. Drainage waterway: 12 spans of 8 m (96 m clear), overall 109.2 m. Canal flumed to a 15 m rectangular trough; transitions 22.4 m inlet and 33.6 m outlet. Trough bed levels 199.815 m and 199.764 m; downstream canal bed 199.835 m, FSL 201.635 m. ## Answer The canal ($Q$ = 50 m³/s) crosses a drainage of $Q$ = 400 m³/s. Design is done in four steps: type of work, drainage waterway, canal waterway (flumed trough) and bed/water levels of the trough. **Assumptions:** Lacey's $f$ = 1; trough $n$ = 0.015; slab 0.6 m. #### 1. Type of structure HFL of the drainage (200.60 m) is above the canal bed (200.00 m) by 0.60 m, so the trough cannot be kept above HFL without a very high, costly structure. The drainage is passed under the trough through barrels flowing under pressure (siphonic action). A **siphon aqueduct** is suitable. Drainage depth at HFL = 200.60 − 198.00 = 2.60 m. #### 2. Drainage waterway Lacey's regime perimeter (for a natural stream): $$ P = 4.75\sqrt{Q} = 4.75\sqrt{400} = 95.0\ \text{m} $$ Provide **12 spans of 8 m** with piers 1.2 m thick: clear waterway = 12 × 8 = **96 m** (> 95.0 m). Overall length = 96 + 11 × 1.2 = **109.2 m**. Check: mean velocity = Q/(L·d) = 400/(96 × 2.60) = 1.60 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.46 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.2 m below HFL, with cut-offs and aprons for scour. #### 3. Canal waterway (flumed rectangular trough) Normal canal: area $A_1=(B+zD)D$ = (32 + 1.5 × 1.8) × 1.8 = 62.46 m², $V_1$ = 50/62.46 = 0.801 m/s, $h_{v1}=V_1^2/2g$ = 0.0327 m. Top width $T=B+2zD$ = 37.40 m. To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **15 m**, with the same full supply depth 1.8 m (the trough is as long as the drainage waterway, 109.2 m). - Trough area = 15 × 1.8 = 27.00 m²; $V_t$ = 50/27.00 = 1.852 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1748 m - Froude number = 0.44 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 27.00/18.60 = 1.452 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000469 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000469 × 109.2 = 0.051 m #### 4. Transitions and levels ```text FSL ---. .--- FSL '--.______________________.--' bed ---. .--- bed '--.______________________.--' normal inlet flumed trough outlet normal canal (2:1) (3:1) canal ``` Wing-wall (straight) transitions: inlet splay 2:1 (contraction) and outlet splay 3:1 (expansion), measured on each side: - Inlet length = 2 × (T − B_t)/2 = 37.40 − 15 = **22.4 m** - Outlet length = 3 × (T − B_t)/2 = **33.6 m** Bernoulli's equation between sections (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v; Δh_v = difference of velocity heads): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 201.800 m, so $W_2$ = 201.615 m, $W_3$ = 201.564 m, $W_4$ = 201.635 m. Trough bed = water level − 1.8 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 200.000 | 201.800 | | 2 | End of inlet transition = start of trough | 199.815 | 201.615 | | 3 | End of trough = start of outlet transition | 199.764 | 201.564 | | 4 | End of outlet transition = normal canal | 199.835 | 201.635 | Total loss of head = 201.800 − 201.635 = 0.165 m, so the canal bed downstream of the work is 0.165 m lower than upstream. Check: with trough slab 0.6 m thick, soffit of the barrels = 199.16 m, i.e. 1.16 m above the drainage bed; HFL stands 1.44 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 197.66 m (1.5 m clear height) with a 1 in 4 slope at both ends. **Answer:** drainage waterway 96 m clear (12 × 8 m, overall 109.2 m); canal flumed to 15 m wide trough; bed/FSL at the four sections = 200.00/201.80, 199.82/201.62, 199.76/201.56, 199.84/201.64 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. HFL is 200.60 m, canal bed is 200.00 m. Which statement justifies a siphon aqueduct? A. HFL is below the canal bed, so the drainage flows freely under an ordinary aqueduct B. HFL is above the canal bed, so a trough above HFL would be costly; barrels run full under the trough C. The canal discharge exceeds the drainage discharge D. The ground level is above the canal bed Answer: B. HFL above the bed makes an ordinary aqueduct uneconomic, so the barrels pass the flood under pressure. (Key point: Why siphon aqueduct) 2. The trough (15 m wide, 1.8 m deep) carries 50 m3/s. What are Vt and the flow condition? A. Vt = 0.801 m/s, supercritical B. Vt = 1.852 m/s, below 2.5 m/s, with Froude number 0.44 (subcritical) C. Vt = 3.7 m/s, which needs a jump D. Vt = 1.852 m/s, Froude number 1.4 Answer: B. Vt = 50/27 = 1.852 m/s; Fr = 0.44, so flow is subcritical and the velocity is acceptable. (Key point: Trough flow check) 3. How is friction loss hf in the trough found? A. hf = S x L, where S = (nV/R^(2/3))^2 from Manning and L is trough length (109.2 m) B. hf = n x V C. hf = 0.5 x velocity head difference D. hf = V^2/2g Answer: A. Manning's equation gives the friction slope S, and multiplying by trough length gives hf = 0.051 m. (Key point: Friction loss in trough) 4. W1 = 201.800 m. After the inlet transition W2 = 201.615 m. What is the trough bed at section 2? A. 201.615 m B. 199.815 m, i.e. water level - 1.8 m C. 200.000 m D. 203.415 m Answer: B. The trough depth is 1.8 m, so bed = 201.615 - 1.8 = 199.815 m. 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