--- title: "Siphon aqueduct: Q 60, Manning 0.025" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:siphon-aqueduct-q60-bed30-fsd14" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/siphon-aqueduct-q60-bed30-fsd14 site: BCT Engineering --- # Siphon aqueduct: Q 60, Manning 0.025 Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2079 Chaitra, 10 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-q60-bed30-fsd14 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question Design following components of Siphon aqueduct for the following data: (i) canal water way, ii) Bed level at different sections of canal. Discharge of canal = 60 Cumecs; Bed width of canal = 30 m; Full supply depth of canal = 1.4 m; Bed level = 116 m; Side slope of canal = 1.5 H:1 V; High Flood discharge of drainage = 550 Cumecs; High flood level of drainage = 117 m; Bed level of drainage = 114 m; General ground level = 116 m; Manning's n = 0.025. ## Short answer Canal waterway is a flumed rectangular trough 24 m wide, 1.4 m deep, velocity 1.786 m/s. Bed levels at sections: 116.000 m (start), 115.907 m (trough start), 115.706 m (trough end), 115.741 m (end), total head loss 0.259 m. ## Answer The canal ($Q$ = 60 m³/s) crosses a drainage of $Q$ = 550 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:** Manning's $n$ = 0.025 (as given) is used for the friction loss in the trough; Lacey's $f$ = 1; piers 1.5 m thick. #### 1. Type of structure HFL of the drainage (117.00 m) is above the canal bed (116.00 m) by 1.00 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). A **siphon aqueduct** is suitable. Drainage depth at HFL = 117.00 − 114.00 = 3.00 m. #### 2. Drainage waterway Lacey's regime perimeter (for a natural stream): $$ P = 4.75\sqrt{Q} = 4.75\sqrt{550} = 111.4\ \text{m} $$ Provide **12 spans of 10 m** with piers 1.5 m thick: clear waterway = 12 × 10 = **120 m** (> 111.4 m). Overall length = 120 + 11 × 1.5 = **136.5 m**. Check: mean velocity = Q/(L·d) = 550/(120 × 3.00) = 1.53 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.85 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.8 m below HFL, with cut-offs and aprons for scour. #### 3. Canal waterway (flumed rectangular trough) Normal canal: area $A_1=(B+zD)D$ = (30 + 1.5 × 1.4) × 1.4 = 44.94 m², $V_1$ = 60/44.94 = 1.335 m/s, $h_{v1}=V_1^2/2g$ = 0.0909 m. Top width $T=B+2zD$ = 34.20 m. To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **24 m**, with the same full supply depth 1.4 m (the trough is as long as the drainage waterway, 136.5 m). - Trough area = 24 × 1.4 = 33.60 m²; $V_t$ = 60/33.60 = 1.786 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1625 m - Froude number = 0.48 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 33.60/26.80 = 1.254 m; friction slope $S=(nV/R^{2/3})^2$ = 0.001474 (Manning's n = 0.025); friction loss in trough $h_f=S\,L$ = 0.001474 × 136.5 = 0.201 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 = 34.20 − 24 = **10.2 m** - Outlet length = 3 × (T − B_t)/2 = **15.3 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): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 117.400 m, so $W_2$ = 117.307 m, $W_3$ = 117.106 m, $W_4$ = 117.141 m. Trough bed = water level − 1.4 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 116.000 | 117.400 | | 2 | End of inlet transition = start of trough | 115.907 | 117.307 | | 3 | End of trough = start of outlet transition | 115.706 | 117.106 | | 4 | End of outlet transition = normal canal | 115.741 | 117.141 | Total loss of head = 117.400 − 117.141 = 0.259 m, so the canal bed downstream of the work is 0.259 m lower than upstream. Check: with trough slab 0.6 m thick, soffit of the barrels = 115.11 m, i.e. 1.11 m above the drainage bed; HFL stands 1.89 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 113.61 m (1.5 m clear height) with a 1 in 4 slope at both ends. **Answer:** drainage waterway 120 m clear (12 × 10 m, overall 136.5 m); canal flumed to 24 m wide trough; bed/FSL at the four sections = 116.00/117.40, 115.91/117.31, 115.71/117.11, 115.74/117.14 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Here the drainage HFL (117.00 m) is 1.00 m above the canal bed (116.00 m). Which structure is chosen and why? A. Aqueduct, because the canal can easily be kept above HFL B. Siphon aqueduct, because the trough cannot economically be kept above HFL so the drainage runs under pressure through barrels below the trough C. Canal siphon, because the canal is carried under the drainage D. Level crossing, because the two levels are almost the same Answer: B. HFL is above the canal bed, so an aqueduct would need a very high costly structure; the drainage is passed under the trough through barrels flowing under pressure. (Key point: Choice of structure) 2. How is the minimum clear drainage waterway found, and what is provided here? A. P = 2.75 sqrt(Q) = 64.5 m, so 7 spans B. P = 4.75 sqrt(Q) = 111.4 m, so 12 spans of 10 m (120 m clear) is provided C. P = 4.75 sqrt(Q) = 111.4 m, so exactly 111.4 m is provided with no piers D. Waterway equals the canal bed width, 30 m Answer: B. Lacey's regime perimeter P = 4.75 sqrt(550) = 111.4 m; the clear waterway of 12 x 10 = 120 m exceeds it. (Key point: Drainage waterway (Lacey)) 3. Why is the canal flumed to a 24 m rectangular trough with the same full supply depth? A. To reduce the length and cost of the trough while velocity (1.79 m/s) stays acceptable and subcritical B. To make the flow supercritical so a hydraulic jump forms C. To increase the canal discharge D. To reduce the drainage waterway Answer: A. A narrower trough saves cost; velocity stays below 2.5 m/s and Froude number is below 1, so there is no hydraulic jump. (Key point: Reason for fluming) 4. In the Bernoulli calculation, which losses are used and how does the water level behave across the trough? A. Contraction loss 0.5 delta hv, expansion loss 0.3 delta hv; level rises in the trough B. Contraction loss 0.3 delta hv, expansion loss 0.5 delta hv; level falls by hf along the trough (W3 = W2 - hf) C. No losses are considered in transitions; only friction D. Contraction and expansion both 1.0 delta hv Answer: B. The answer uses 0.3 delta hv for contraction and 0.5 for expansion, and friction loss hf = S L lowers the level from W2 to W3. 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