--- title: "Siphon aqueduct: Q 60, bed 35 m" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:siphon-aqueduct-q60-bed35-hfd500" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/siphon-aqueduct-q60-bed35-hfd500 site: BCT Engineering --- # Siphon aqueduct: Q 60, bed 35 m Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2077 Chaitra, 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-q60-bed35-hfd500 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question Design siphon aqueduct (Drainage waterway, Canal waterway, Bed levels and transitions) if the following data at crossing of canal and drainage are given: a) Discharge of canal = 60 cumecs b) Bed width of canal = 35 m c) Full supply depth of canal = 2 m d) Canal bed level = 300 m e) Side slopes of canal = 1.5H:1V f) High flood discharge of drainage = 500 cumecs g) High flood level of drainage = 300.8 m h) Bed level of drainage = 298.2 m i) General ground level = 300.2 m. ## Short answer Siphon aqueduct selected as drainage HFL (300.8 m) exceeds canal bed (300 m). Drainage waterway: 11 spans of 10 m (clear 110 m). Canal flumed to 17 m wide rectangular trough, depth 2 m. Inlet transition 24 m (2:1), outlet 36 m (3:1). Trough bed levels: start 299.835 m, end 299.789 m. Total head loss 0.148 m; downstream bed 0.148 m lower. ## Answer The canal ($Q$ = 60 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; trough $n$ = 0.015; slab 0.6 m thick. #### 1. Type of structure HFL of the drainage (300.80 m) is above the canal bed (300.00 m) by 0.80 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 = 300.80 − 298.20 = 2.60 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 × 2.60) = 1.75 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 $A_1=(B+zD)D$ = (35 + 1.5 × 2) × 2 = 76.00 m², $V_1$ = 60/76.00 = 0.789 m/s, $h_{v1}=V_1^2/2g$ = 0.0318 m. Top width $T=B+2zD$ = 41.00 m. To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **17 m**, with the same full supply depth 2 m (the trough is as long as the drainage waterway, 125.0 m). - Trough area = 17 × 2 = 34.00 m²; $V_t$ = 60/34.00 = 1.765 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1587 m - Froude number = 0.40 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 34.00/21.00 = 1.619 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000369 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000369 × 125.0 = 0.046 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 = 41.00 − 17 = **24.0 m** - Outlet length = 3 × (T − B_t)/2 = **36.0 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): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 302.000 m, so $W_2$ = 301.835 m, $W_3$ = 301.789 m, $W_4$ = 301.852 m. Trough bed = water level − 2 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 300.000 | 302.000 | | 2 | End of inlet transition = start of trough | 299.835 | 301.835 | | 3 | End of trough = start of outlet transition | 299.789 | 301.789 | | 4 | End of outlet transition = normal canal | 299.852 | 301.852 | Total loss of head = 302.000 − 301.852 = 0.148 m, so the canal bed downstream of the work is 0.148 m lower than upstream. Check: with trough slab 0.6 m thick, soffit of the barrels = 299.19 m, i.e. 0.99 m above the drainage bed; HFL stands 1.61 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 297.69 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 17 m wide trough; bed/FSL at the four sections = 300.00/302.00, 299.83/301.83, 299.79/301.79, 299.85/301.85 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. HFL (300.80 m) is 0.80 m above canal bed (300.00 m). What does this indicate? A. An aqueduct is possible B. A siphon aqueduct is needed since the trough cannot be kept above HFL economically C. A canal siphon is needed D. No structure is needed Answer: B. With HFL above the canal bed, drainage must flow in barrels under pressure beneath the trough. (Key point: Choice of structure) 2. For Q = 500 m3/s, which result is correct? A. P = 4.75 sqrt(500) = 106.2 m; provide 11 x 10 m = 110 m clear, overall 125.0 m B. P = 4.75 x 500 = 2375 m C. P = 106.2 m; provide 100 m clear D. P = 106.2 m; overall length 110 m Answer: A. Clear waterway must exceed the Lacey perimeter; piers add 10 x 1.5 m. (Key point: Drainage waterway) 3. Trough 17 m wide, depth 2 m, Q = 60 m3/s. What is the velocity and is it acceptable? A. 0.79 m/s, too low B. 1.76 m/s, acceptable (below 2.5 m/s) and subcritical (Fr = 0.40) C. 3.5 m/s, too high D. 2.5 m/s, supercritical Answer: B. V = 60/34 = 1.765 m/s, below 2.5 m/s with Froude number 0.40 so no jump. (Key point: Flumed trough check) 4. What is the net effect of the structure on the canal level downstream? A. Downstream bed is 0.148 m higher B. Unchanged C. Downstream bed is 0.148 m lower because of the total head loss D. Downstream bed is 2 m lower Answer: C. Total loss 302.000 - 301.852 = 0.148 m, so the downstream canal bed is lowered by that amount. 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