--- title: "Siphon aqueduct: Q 20, drainage 200" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:siphon-aqueduct-q20-hfd200-uplift" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/siphon-aqueduct-q20-hfd200-uplift site: BCT Engineering --- # Siphon aqueduct: Q 20, drainage 200 Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2070 Bhadra, 6 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-q20-hfd200-uplift - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question Following data are obtained at the crossing of a canal and a drainage. Canal data: Q = 20 m³/s, depth of water = 1.5 m and FSL = 151.50 m, Bed width = 12 m, side slope (H:V) = (1.5:1). Drainage data: Q = 200 m³/s, HFL = 150.7 m, Bed level = 148.5 m and Ground level = 150.0 m. Design the following components of siphon aqueduct. i) Drainage waterway ii) Canal waterway iii) Transition iv) Uplift. ## Short answer Drainage waterway 72 m clear (9 spans @8 m, overall 81.6 m). Canal trough width 7 m, depth 1.5 m, velocity 1.905 m/s. Inlet transition 9.5 m (2:1), outlet 14.2 m (3:1). Uplift resisted by 0.6 m thick slab with soffit at 149.16 m; cut-offs and aprons provided. ## Answer The canal ($Q$ = 20 m³/s) crosses a drainage of $Q$ = 200 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:** Bed level of canal = FSL − depth = 151.50 − 1.50 = 150.00 m; Lacey's $f$ = 1; trough $n$ = 0.015; slab 0.6 m, concrete 24 kN/m³. #### 1. Type of structure HFL of the drainage (150.70 m) is above the canal bed (150.00 m) by 0.70 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 = 150.70 − 148.50 = 2.20 m. #### 2. Drainage waterway Lacey's regime perimeter (for a natural stream): $$ P = 4.75\sqrt{Q} = 4.75\sqrt{200} = 67.2\ \text{m} $$ Provide **9 spans of 8 m** with piers 1.2 m thick: clear waterway = 9 × 8 = **72 m** (> 67.2 m). Overall length = 72 + 8 × 1.2 = **81.6 m**. Check: mean velocity = Q/(L·d) = 200/(72 × 2.20) = 1.26 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 2.75 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 4.1 m below HFL, with cut-offs and aprons for scour. #### 3. Canal waterway (flumed rectangular trough) Normal canal: area $A_1=(B+zD)D$ = (12 + 1.5 × 1.5) × 1.5 = 21.38 m², $V_1$ = 20/21.38 = 0.936 m/s, $h_{v1}=V_1^2/2g$ = 0.0446 m. Top width $T=B+2zD$ = 16.50 m. To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **7 m**, with the same full supply depth 1.5 m (the trough is as long as the drainage waterway, 81.6 m). - Trough area = 7 × 1.5 = 10.50 m²; $V_t$ = 20/10.50 = 1.905 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1849 m - Froude number = 0.50 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 10.50/10.00 = 1.050 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000765 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000765 × 81.6 = 0.062 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 = 16.50 − 7 = **9.5 m** - Outlet length = 3 × (T − B_t)/2 = **14.2 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): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 151.500 m, so $W_2$ = 151.318 m, $W_3$ = 151.255 m, $W_4$ = 151.325 m. Trough bed = water level − 1.5 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 150.000 | 151.500 | | 2 | End of inlet transition = start of trough | 149.818 | 151.318 | | 3 | End of trough = start of outlet transition | 149.755 | 151.255 | | 4 | End of outlet transition = normal canal | 149.825 | 151.325 | Total loss of head = 151.500 − 151.325 = 0.175 m, so the canal bed downstream of the work is 0.175 m lower than upstream. Check: with trough slab 0.6 m thick, soffit of the barrels = 149.16 m, i.e. 0.66 m above the drainage bed; HFL stands 1.54 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 147.66 m (1.5 m clear height) with a 1 in 4 slope at both ends. #### Uplift on the roof (trough slab) Upward pressure under the slab (barrel running full) = γ_w (HFL − soffit) = 9.81 × (150.70 − 149.16) = 15.2 kN/m². Downward load = slab weight 24 × 0.6 = 14.4 kN/m² plus canal water 9.81 × 1.5 = 14.7 kN/m², total 29.1 kN/m². Uplift exceeds the slab weight but not the total load: net downward = 14.0 kN/m² with canal running. With the canal dry the net uplift is 0.8 kN/m², which the slab and its anchorage to the piers must resist. **Answer:** drainage waterway 72 m clear (9 × 8 m, overall 81.6 m); canal flumed to 7 m wide trough; bed/FSL at the four sections = 150.00/151.50, 149.82/151.32, 149.76/151.26, 149.83/151.33 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is uplift on the trough slab (barrel running full) estimated? A. gamma_w x slab thickness B. gamma_w x (HFL - soffit) = 9.81 x (150.70 - 149.16) = 15.2 kN/m2 C. gamma_w x canal depth = 14.7 kN/m2 D. Weight of concrete 24 x 0.6 = 14.4 kN/m2 Answer: B. Water pressure under the slab equals the head of HFL above the soffit times the unit weight of water. (Key point: Upward pressure on the roof) 2. T = 16.5 m and trough width 7 m. What are the inlet (2:1) and outlet (3:1) lengths? A. 14.2 m inlet and 9.5 m outlet B. 9.5 m inlet and 14.2 m outlet C. 19.0 m inlet and 28.5 m outlet D. 4.75 m inlet and 7.1 m outlet Answer: B. Inlet = 2 x (16.5 - 7)/2 = 9.5 m; outlet = 3 x 4.75 = 14.2 m. (Key point: Transition lengths) 3. For Q = 200 m3/s, which waterway is provided? A. P = 4.75 sqrt(200) = 67.2 m; 9 spans of 8 m = 72 m clear with 1.2 m piers (overall 81.6 m) B. P = 67.2 m; exactly 67.2 m with no piers C. P = 4.75 x 200 = 950 m D. 9 spans of 8 m but overall length 72 m Answer: A. Lacey perimeter 67.2 m is rounded up to 72 m clear; 8 piers of 1.2 m make the overall length 81.6 m. 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