--- title: "Cross drainage waterway and transitions, Q 56" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:cd-water-way-q56-bed24" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/cd-water-way-q56-bed24 site: BCT Engineering --- # Cross drainage waterway and transitions, Q 56 Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2078 Baisakh, 8 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/cd-water-way-q56-bed24 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question Design drainage waterway, canal waterway and canal transitions of a cross drainage work for a canal having discharge 56 m³/s, bed width 24 m, full supply depth 1.98 m, and bed level 267.00 m. The drainage has a high flood discharge of 425 m³/s, high flood level of 268.20 m, general bed level of low water cross-section of 265.50 m, and general ground level of 267.20 m. ## Short answer Siphon aqueduct selected as HFL (268.20m) exceeds canal bed (267.00m). Drainage waterway: 13 spans of 8m (clear 104m) using Lacey's regime perimeter. Canal flumed to 16m wide rectangular trough (1.98m depth). Transitions: inlet 2:1 (13.9m), outlet 3:1 (20.9m). Bed levels/FSL: Sec1 (267.000/268.980), Sec2 (266.866/268.846), Sec3 (266.821/268.801), Sec4 (266.872/268.852). Head loss 0.128m. ## Answer The canal ($Q$ = 56 m³/s) crosses a drainage of $Q$ = 425 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:** Side slope of canal 1.5:1; Lacey's $f$ = 1; trough $n$ = 0.015. #### 1. Type of structure HFL of the drainage (268.20 m) is above the canal bed (267.00 m) by 1.20 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 = 268.20 − 265.50 = 2.70 m. #### 2. Drainage waterway Lacey's regime perimeter (for a natural stream): $$ P = 4.75\sqrt{Q} = 4.75\sqrt{425} = 97.9\ \text{m} $$ Provide **13 spans of 8 m** with piers 1.2 m thick: clear waterway = 13 × 8 = **104 m** (> 97.9 m). Overall length = 104 + 12 × 1.2 = **118.4 m**. Check: mean velocity = Q/(L·d) = 425/(104 × 2.70) = 1.51 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.53 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.3 m below HFL, with cut-offs and aprons for scour. #### 3. Canal waterway (flumed rectangular trough) Normal canal: area $A_1=(B+zD)D$ = (24 + 1.5 × 1.98) × 1.98 = 53.40 m², $V_1$ = 56/53.40 = 1.049 m/s, $h_{v1}=V_1^2/2g$ = 0.0561 m. Top width $T=B+2zD$ = 29.94 m. To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **16 m**, with the same full supply depth 1.98 m (the trough is as long as the drainage waterway, 118.4 m). - Trough area = 16 × 1.98 = 31.68 m²; $V_t$ = 56/31.68 = 1.768 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1593 m - Froude number = 0.40 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 31.68/19.96 = 1.587 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000380 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000380 × 118.4 = 0.045 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 = 29.94 − 16 = **13.9 m** - Outlet length = 3 × (T − B_t)/2 = **20.9 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): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 268.980 m, so $W_2$ = 268.846 m, $W_3$ = 268.801 m, $W_4$ = 268.852 m. Trough bed = water level − 1.98 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 267.000 | 268.980 | | 2 | End of inlet transition = start of trough | 266.866 | 268.846 | | 3 | End of trough = start of outlet transition | 266.821 | 268.801 | | 4 | End of outlet transition = normal canal | 266.872 | 268.852 | Total loss of head = 268.980 − 268.852 = 0.128 m, so the canal bed downstream of the work is 0.128 m lower than upstream. Check: with trough slab 0.6 m thick, soffit of the barrels = 266.22 m, i.e. 0.72 m above the drainage bed; HFL stands 1.98 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 264.72 m (1.5 m clear height) with a 1 in 4 slope at both ends. **Answer:** drainage waterway 104 m clear (13 × 8 m, overall 118.4 m); canal flumed to 16 m wide trough; bed/FSL at the four sections = 267.00/268.98, 266.87/268.85, 266.82/268.80, 266.87/268.85 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is a siphon aqueduct chosen when the drainage HFL (268.20 m) is 1.20 m above the canal bed (267.00 m)? A. The drainage discharge is smaller than the canal discharge B. The canal can simply be carried over the drainage in an open trough above HFL at no extra cost C. The drainage flows under the trough through barrels under pressure, because keeping the trough above HFL would need a very high, costly structure D. The canal must be taken below the drainage bed in an inverted siphon Answer: C. HFL is above the canal bed, so a trough above HFL is too costly; the drainage is passed under pressure through barrels, which is siphonic action. (Key point: Choice of structure) 2. How is the clear drainage waterway fixed for Q = 425 m3/s? A. P = 4.75 sqrt(Q) = 97.9 m, then 13 spans of 8 m = 104 m clear is provided (more than P) B. P = 4.75 sqrt(Q) = 97.9 m, then exactly 97.9 m is provided with no piers C. P = 0.47 (Q/f)^(1/3) = 3.53 m is used as the waterway D. The waterway equals the canal bed width of 24 m Answer: A. The regime perimeter 4.75 sqrt(Q) gives the minimum waterway; whole spans are chosen to give 104 m clear, which exceeds 97.9 m. (Key point: Drainage waterway by Lacey) 3. The canal is flumed to a 16 m rectangular trough with V = 1.768 m/s and Froude number 0.40. What does this tell us? A. Flow is supercritical, so a hydraulic jump forms at the outlet B. Flow is subcritical, so no hydraulic jump occurs; flume narrowing saves cost on the trough C. Flow is critical, so the trough must be widened D. The velocity is too high (above 2.5 m/s) and the flume must be rejected Answer: B. Fr = 0.40 is below 1 and V is below 2.5 m/s, so the flume is acceptable and no jump occurs; narrowing the trough cuts its cost. (Key point: Flumed trough and subcritical flow) 4. In computing W2 and W4 by Bernoulli's equation, what head losses are allowed? A. 0.5 times the velocity-head difference in contraction and 0.3 times in expansion B. 0.3 times the velocity-head difference in contraction (inlet) and 0.5 times in expansion (outlet) C. Equal losses of 0.3 times at both inlet and outlet D. No losses; only friction in the trough is considered Answer: B. The inlet (contraction) loss is 0.3 delta hv and the outlet (expansion) loss is 0.5 delta hv; friction loss hf is applied separately along the trough. 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