--- title: "Siphon aqueduct: Q 20, FSL 251.5" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:siphon-aqueduct-q20-hfd200-251" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/siphon-aqueduct-q20-hfd200-251 site: BCT Engineering --- # Siphon aqueduct: Q 20, FSL 251.5 Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2064 Jestha (old course), 16 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/siphon-aqueduct-q20-hfd200-251 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question Canal Data: Discharge = 20 m³/s, Depth of water 1.5 m and FSL = 251.5 m. Drainage data: Discharge = 200 m³/s, HFL = 250.7 m, Bed Level = 248.5 m and Ground level = 250.0 m. From above data, design following components of siphon aqueduct. a) canal waterway b) drainage waterway c) afflux and head losses through siphon barrel d) uplift on drainage slab. ## Short answer Drainage waterway: 9 spans of 8 m, clear 72 m (Lacey's P=67.2 m). Canal flumed to 7 m wide trough, depth 1.5 m; velocity 1.905 m/s, friction loss 0.062 m. Total head loss (trough + transitions) 0.175 m. Uplift on drainage slab is the difference between HFL and soffit level, reduced by slab self-weight. ## Answer The canal ($Q$ = 20 m³/s) crosses a drainage of $Q$ = 200 m³/s. The design covers the type of work, the drainage waterway, the canal waterway (flumed trough) with the bed and water levels of the trough, the afflux and head loss through the siphon barrels, and the uplift on the roof slab. **Assumptions:** Canal bed width is not given and is taken as 12 m with side slope 1.5:1; bed level = 251.5 − 1.5 = 250.0 m; Lacey's $f$ = 1; trough $n$ = 0.015; slab 0.6 m thick, concrete 24 kN/m³. #### 1. Type of structure The HFL of the drainage (250.70 m) is 0.70 m above the canal bed (250.00 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), so a **siphon aqueduct** is suitable. Drainage depth at HFL = 250.70 − 248.50 = 2.20 m. #### 2. (b) 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); the foundations of piers and abutments are taken about 1.5R = 4.1 m below HFL, with cut-offs and aprons for scour. #### 3. (a) 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 of 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, so the flow is subcritical and 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 the 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): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 251.500 m, so $W_2$ = 251.318 m, $W_3$ = 251.255 m, $W_4$ = 251.325 m. Trough bed = water level − 1.5 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 250.000 | 251.500 | | 2 | End of inlet transition = start of trough | 249.818 | 251.318 | | 3 | End of trough = start of outlet transition | 249.755 | 251.255 | | 4 | End of outlet transition = normal canal | 249.825 | 251.325 | Total loss of head = 251.500 − 251.325 = 0.175 m, so the canal bed downstream of the work is 0.175 m lower than upstream. Check: with a trough slab 0.6 m thick, the soffit of the barrels = 249.16 m, only 0.66 m above the drainage bed, while 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 (see below). #### c) Afflux and head loss through the siphon barrels Soffit of barrel = trough bed − slab = 249.16 m. Provide barrels of clear height 1.5 m (barrel floor at 247.66 m, i.e. the drainage bed is depressed by 0.84 m under the structure, with 1 in 4 slopes at the ends): 9 barrels of 8 m, total clear width 72 m, barrel length (along the canal) $L$ = 7 + 1.2 = 8.2 m. - Area = 72 × 1.5 = 108 m²; $V$ = 200/108 = 1.852 m/s; $V^2/2g$ = 0.1748 m - Hydraulic radius of one barrel $R$ = (8 × 1.5)/(2(8+1.5)) = 0.632 m - Unwin's formula: $h_L=\left(1+f_1+f_2\dfrac{L}{R}\right)\dfrac{V^2}{2g}$ with $f_1$ = 0.505 and $f_2=0.00316\,(1+0.1/R)$ = 0.00366 $$ \begin{aligned} h_L &= \left(1+0.505+0.00366\times\frac{8.2}{0.632}\right)\times0.1748 \\ &= 1.5525\times0.1748 = 0.271\ \text{m} \end{aligned} $$ This head is lost between upstream and downstream, so the **afflux** = 0.27 m. Upstream HFL = 250.70 + 0.27 = **250.97 m**, and the upstream flood banks/guide bunds are raised accordingly. #### d) Uplift on the drainage (roof) slab Taking the full static head above the soffit (the severe case): $p$ = 9.81 × (250.70 − 249.16) = 15.2 kN/m². Downward load = slab 24 × 0.6 = 14.4 kN/m² + canal water 9.81 × 1.5 = 14.7 kN/m², total 29.1 kN/m². With the canal running, the net downward load = 29.1 − 15.2 = 13.9 kN/m² (safe against uplift). With the canal empty, the net uplift = 15.2 − 14.4 = 0.8 kN/m², which is resisted by reinforcement and by anchoring the slab to the piers. **Answer:** drainage waterway 72 m clear (9 × 8 m, overall 81.6 m); canal flumed to a 7 m wide trough; bed/FSL at the four sections = 250.00/251.50, 249.82/251.32, 249.76/251.26, 249.83/251.33 m; head loss through the barrels (afflux) ≈ 0.27 m; uplift on the roof slab ≈ 15.2 kN/m² (net uplift 0.8 kN/m² with the canal empty). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is a siphon aqueduct chosen for this crossing? A. The drainage HFL (250.70 m) is above the canal bed (250.00 m), so the trough cannot stay above HFL without a very costly structure B. The canal discharge is larger than the drainage discharge C. The drainage bed is higher than the canal bed D. The canal must be carried in an open channel over the drain with no trough Answer: A. HFL is 0.70 m above the canal bed, so drainage water is passed under the trough through barrels flowing under pressure (siphonic action). (Key point: Choice of siphon aqueduct) 2. How is the clear waterway of the drainage fixed, and what is adopted here? A. P = 2.67 sqrt(Q); adopt 6 spans of 8 m B. P = 4.75 sqrt(Q) = 67.2 m; adopt 9 spans of 8 m = 72 m clear (more than 67.2 m) C. P = 4.75 sqrt(Q) = 67.2 m; adopt exactly 67.2 m with no piers D. Waterway equals the canal bed width of 12 m Answer: B. Lacey's regime perimeter 4.75 sqrt(200) = 67.2 m is the minimum; 9 x 8 = 72 m clear is provided, giving an overall length of 81.6 m with piers. (Key point: Drainage waterway by Lacey's perimeter) 3. How is the water level at the end of the inlet transition (section 2) found, giving 251.318 m? A. W2 = W1 minus the friction loss in the trough only B. W2 = W1 plus the expansion loss 0.5(hvt - hv1) C. Bernoulli's equation: W2 = W1 + hv1 - hvt - 0.3(hvt - hv1), allowing for the velocity-head increase and contraction loss D. W2 is equal to W1 since the depth stays the same Answer: C. Velocity rises in the 7 m trough, so water level drops by the velocity head difference plus 0.3 times that difference as contraction loss. (Key point: Flumed trough and level calculation) 4. How are head loss through the barrels and afflux obtained? A. Using Manning's formula only, giving zero afflux B. The afflux equals the canal head loss 0.175 m C. Afflux = V^2/2g only = 0.1748 m, subtracted from HFL D. Unwin's formula hL = (1 + f1 + f2 L/R) V^2/2g = 0.271 m; this loss is the afflux, so upstream HFL = 250.97 m Answer: D. Unwin's formula includes entry, exit and friction losses; the head lost is the afflux, which raises upstream HFL to 250.97 m. 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