--- title: "Siphon aqueduct: Q 30, bed 24 m" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:siphon-aqueduct-q30-bed24-hfd500" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/siphon-aqueduct-q30-bed24-hfd500 site: BCT Engineering --- # Siphon aqueduct: Q 30, bed 24 m Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2074 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-q30-bed24-hfd500 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question Design a Siphon aqueduct with the data given below. Full supply discharge of canal = 30 m³/s; Bed width of canal = 24 m; Full supply depth = 1.25 m; Side slope of canal section 1½:1 (H:V); Bed level of the canal = 100.00 m; Max. flood discharge of drain = 500 m³/s; High flood level = 100.50; Bed level of drainage = 98.00 m; Normal ground level = 100.00 m; Lacey's silt factor = 1.0; Rugosity coefficient (n) = 0.016. Make suitable assumptions where necessary. ## Short answer Siphon aqueduct chosen as HFL (100.50m) exceeds canal bed (100.00m). Drainage waterway: 11 spans of 10m (clear 110m) via Lacey's regime perimeter. Canal flumed to 13m wide rectangular trough (1.25m depth). Transitions: inlet 2:1 (14.8m), outlet 3:1 (22.1m). Bed levels/FSL: Sec1 (100.000/101.250), Sec2 (99.831/101.081), Sec3 (99.729/100.979), Sec4 (99.794/101.044). Total head loss 0.206m. ## Answer The canal ($Q$ = 30 m³/s) crosses a drainage of $Q$ = 500 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; $n$ = 0.016 as given is used for the trough; piers 1.5 m thick; slab 0.6 m. #### 1. Type of structure HFL of the drainage (100.50 m) is above the canal bed (100.00 m) by 0.50 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 = 100.50 − 98.00 = 2.50 m. #### 2. Drainage waterway Lacey's regime perimeter (for a natural stream): $$ P = 4.75\sqrt{Q} = 4.75\sqrt{500} = 106.2\ \text{m} $$ Provide **11 spans of 10 m** with piers 1.5 m thick: clear waterway = 11 × 10 = **110 m** (> 106.2 m). Overall length = 110 + 10 × 1.5 = **125.0 m**. Check: mean velocity = Q/(L·d) = 500/(110 × 2.50) = 1.82 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.73 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.6 m below HFL, with cut-offs and aprons for scour. #### 3. Canal waterway (flumed rectangular trough) Normal canal: area $A_1=(B+zD)D$ = (24 + 1.5 × 1.25) × 1.25 = 32.34 m², $V_1$ = 30/32.34 = 0.928 m/s, $h_{v1}=V_1^2/2g$ = 0.0438 m. Top width $T=B+2zD$ = 27.75 m. To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **13 m**, with the same full supply depth 1.25 m (the trough is as long as the drainage waterway, 125.0 m). - Trough area = 13 × 1.25 = 16.25 m²; $V_t$ = 30/16.25 = 1.846 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1737 m - Froude number = 0.53 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 16.25/15.50 = 1.048 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000819 (Manning's n = 0.016); friction loss in trough $h_f=S\,L$ = 0.000819 × 125.0 = 0.102 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 = 27.75 − 13 = **14.8 m** - Outlet length = 3 × (T − B_t)/2 = **22.1 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): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 101.250 m, so $W_2$ = 101.081 m, $W_3$ = 100.979 m, $W_4$ = 101.044 m. Trough bed = water level − 1.25 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 100.000 | 101.250 | | 2 | End of inlet transition = start of trough | 99.831 | 101.081 | | 3 | End of trough = start of outlet transition | 99.729 | 100.979 | | 4 | End of outlet transition = normal canal | 99.794 | 101.044 | Total loss of head = 101.250 − 101.044 = 0.206 m, so the canal bed downstream of the work is 0.206 m lower than upstream. Check: with trough slab 0.6 m thick, soffit of the barrels = 99.13 m, i.e. 1.13 m above the drainage bed; HFL stands 1.37 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 97.63 m (1.5 m clear height) with a 1 in 4 slope at both ends. **Answer:** drainage waterway 110 m clear (11 × 10 m, overall 125.0 m); canal flumed to 13 m wide trough; bed/FSL at the four sections = 100.00/101.25, 99.83/101.08, 99.73/100.98, 99.79/101.04 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. HFL (100.50 m) is above canal bed (100.00 m). Which structure? A. Aqueduct B. Siphon aqueduct C. Canal siphon D. Super passage Answer: B. HFL above the canal bed means the drain flows under pressure through barrels below the trough. (Key point: Structure choice) 2. Q = 500 m3/s: which is correct? A. P = 106.2 m; 11 x 10 m = 110 m clear; overall 125.0 m B. P = 106.2 m; 100 m clear C. P = 53 m D. P = 106.2 m; overall 110 m Answer: A. Clear waterway 110 m > 106.2 m, plus 10 piers of 1.5 m = 125.0 m. (Key point: Drainage waterway) 3. Trough 13 m wide, depth 1.25 m, Q = 30. What is the velocity and flow state? A. 1.85 m/s, subcritical (Fr 0.53) B. 0.93 m/s, supercritical C. 3 m/s, supercritical D. 1.85 m/s, supercritical Answer: A. V = 30/16.25 = 1.846 m/s, Froude 0.53 < 1 so no jump. (Key point: Flumed trough) 4. Top width 27.75 m, trough 13 m. What are the transitions? A. Inlet 2:1 = 14.8 m; outlet 3:1 = 22.1 m B. Inlet 3:1 = 22.1 m; outlet 2:1 = 14.8 m C. Both 14.8 m D. Inlet 27.75 m Answer: A. Inlet = T - Bt = 14.75 m; outlet = 1.5 times that = 22.1 m. 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