--- title: "Siphon: canal 50, FSL 201.80, RCC flumed" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:siphon-aqueduct-q50-hfl2515" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/siphon-aqueduct-q50-hfl2515 site: BCT Engineering --- # Siphon: canal 50, FSL 201.80, RCC flumed Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2064 Kartik (old course), 12 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-q50-hfl2515 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question Design syphon of a syphon aqueduct for the following data: Canal: Q = 50 m³/s; FSL = 201.80 m; CBL = 200.00 m; B = 36 m; z (side slope) = 1.5. Drainage: Qmax = 450 m³/s; HFL = 200.50 m; DBL = 198.00 m; GL = 200.00 m. Assume that the aqueduct will be made of RCC having flumed width of 18 m. Assume other data suitably, if necessary. ## Short answer Siphon aqueduct. Drainage waterway: 11 spans of 10 m, 110 m clear, overall 125 m. Canal flumed to an 18 m rectangular trough; transitions 23.4 m inlet and 35.1 m outlet. Trough bed levels 199.876 m and 199.837 m; downstream canal bed 199.885 m, FSL 201.685 m, upstream bed 200.000 m. ## Answer The canal ($Q$ = 50 m³/s) crosses a drainage of $Q$ = 450 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:** Flumed (RCC) trough width 18 m as given; $n$ = 0.015 for RCC; Lacey's $f$ = 1; slab 0.6 m; barrel $n$ = 0.015 and barrel length ≈ trough width + walls ≈ 19 m. Depth of canal = 201.80 − 200.00 = 1.80 m. #### 1. Type of structure HFL of the drainage (200.50 m) is above the canal bed (200.00 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 = 200.50 − 198.00 = 2.50 m. #### 2. Drainage waterway Lacey's regime perimeter (for a natural stream): $$ P = 4.75\sqrt{Q} = 4.75\sqrt{450} = 100.8\ \text{m} $$ Provide **11 spans of 10 m** with piers 1.5 m thick: clear waterway = 11 × 10 = **110 m** (> 100.8 m). Overall length = 110 + 10 × 1.5 = **125.0 m**. Check: mean velocity = Q/(L·d) = 450/(110 × 2.50) = 1.64 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.60 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.4 m below HFL, with cut-offs and aprons for scour. #### 3. Canal waterway (flumed rectangular trough) Normal canal: area $A_1=(B+zD)D$ = (36 + 1.5 × 1.8) × 1.8 = 69.66 m², $V_1$ = 50/69.66 = 0.718 m/s, $h_{v1}=V_1^2/2g$ = 0.0263 m. Top width $T=B+2zD$ = 41.40 m. To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **18 m**, with the same full supply depth 1.8 m (the trough is as long as the drainage waterway, 125.0 m). - Trough area = 18 × 1.8 = 32.40 m²; $V_t$ = 50/32.40 = 1.543 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1214 m - Froude number = 0.37 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 32.40/21.60 = 1.500 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000312 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000312 × 125.0 = 0.039 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 = 41.40 − 18 = **23.4 m** - Outlet length = 3 × (T − B_t)/2 = **35.1 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): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 201.800 m, so $W_2$ = 201.676 m, $W_3$ = 201.637 m, $W_4$ = 201.685 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 | 199.876 | 201.676 | | 3 | End of trough = start of outlet transition | 199.837 | 201.637 | | 4 | End of outlet transition = normal canal | 199.885 | 201.685 | Total loss of head = 201.800 − 201.685 = 0.115 m, so the canal bed downstream of the work is 0.115 m lower than upstream. Check: with trough slab 0.6 m thick, soffit of the barrels = 199.24 m, i.e. 1.24 m above the drainage bed; HFL stands 1.26 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 197.74 m (1.5 m clear height) with a 1 in 4 slope at both ends. #### 5. Design of the siphon (drainage barrels) - Barrels: 11 openings of 10 m clear width × 1.5 m clear height (drainage floor depressed to 197.74 m), area = 11 × 10 × 1.5 = 165 m². - Velocity in barrels $V_b$ = 450/165 = 2.73 m/s (within the usual 2–3 m/s); $h_{vb}$ = 2.73²/19.62 = 0.379 m. - One barrel: $R$ = 15/23 = 0.652 m; friction loss $h_{fb}=n^2V_b^2L/R^{4/3} = 0.015^2 \times 2.73^2 \times 19/0.652^{4/3}$ = 0.056 m. - Head loss through the siphon = entry ($0.5h_{vb}$) + exit ($1.0h_{vb}$) + friction = 0.190 + 0.379 + 0.056 = **0.625 m** (approach velocity neglected). - Upstream HFL = 200.50 + 0.63 = **201.13 m**, so the drainage banks and guide banks upstream are raised above this level, with free board. - Uplift on the roof (canal dry, drainage at HFL): the maximum is just inside the entry, with pressure head = 201.125 − 1.5 × 0.379 − 199.237 (soffit) = 1.32 m, so the uplift = 9.81 × 1.32 = **12.9 kN/m²**. This is less than the slab weight 24 × 0.6 = 14.4 kN/m², so the roof is safe against uplift. **Answer:** drainage waterway 110 m clear (11 × 10 m, overall 125.0 m); canal flumed to 18 m wide trough; bed/FSL at the four sections = 200.00/201.80, 199.88/201.68, 199.84/201.64, 199.88/201.68 m; siphon barrels 11 × (10 m × 1.5 m), V = 2.73 m/s, head loss 0.63 m (upstream HFL 201.13 m), maximum uplift on the roof 12.9 kN/m². ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. FSL is 201.80 m and CBL is 200.00 m. What is the full supply depth, and area of the normal canal for B = 36 m, z = 1.5? A. D = 1.8 m; A = 69.66 m2 B. D = 1.8 m; A = 64.8 m2 C. D = 201.8 m D. D = 1.5 m; A = 69.66 m2 Answer: A. D = 201.80 - 200.00 = 1.8 m and A = (36 + 1.5 x 1.8) x 1.8 = 69.66 m2. (Key point: Canal depth from data) 2. For Q = 450 m3/s, P = 4.75 sqrt(450) = 100.8 m. Which provision is made? A. 10 spans of 10 m = 100 m clear B. 11 spans of 10 m = 110 m clear (greater than P), overall 125.0 m with 10 piers of 1.5 m C. A single 18 m span D. 11 spans with overall 110 m Answer: B. The clear waterway must exceed 100.8 m; 110 m is chosen, and the piers add 15 m. (Key point: Drainage waterway) 3. The trough width is given as 18 m. What are its velocity and flow state? A. Vt = 0.718 m/s, supercritical B. Vt = 2.78 m/s, supercritical C. Vt = 1.543 m/s, Fr 0.37, subcritical, so no jump D. Vt = 1.543 m/s, Fr 1.2 Answer: C. Vt = 50/(18 x 1.8) = 1.543 m/s and Fr = 0.37, subcritical. (Key point: Trough sizing) 4. W1 = 201.800 m and W4 = 201.685 m. What is concluded? A. The downstream canal bed is 0.115 m lower because of the head lost in the work B. The downstream bed is raised by 0.115 m C. There is no head loss D. The downstream bed is 0.039 m lower Answer: A. Total loss = 201.800 - 201.685 = 0.115 m, so the downstream canal bed is set 0.115 m lower. 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