--- title: "Cross drainage: Q 50, HFD 400, d 1.55 m" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:cd-q50-fsd155-hfd400" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/cd-q50-fsd155-hfd400 site: BCT Engineering --- # Cross drainage: Q 50, HFD 400, d 1.55 m Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2078 Chaitra, 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-q50-fsd155-hfd400 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question Design the following components of a suitable C/D work for the following data: Discharge of canal = 50 m³/s, Bed width of canal = 30 m, Depth of water in canal = 1.55 m, Bed level of canal = 300.0 m, High flood discharge of drain = 400 m³/s, High flood level of drainage = 300.50 m, Bed level of drainage = 298.8 m, General ground level = 300.0 m. i) Design of drainage water-way ii) Design of canal water way iii) Design of transition iv) Design of bed levels. ## Short answer Drainage waterway: 12 spans of 8 m, clear waterway 96 m. Canal waterway: flumed trough 18 m wide, 1.55 m deep. Inlet transition 2:1 splay 16.6 m; outlet 3:1 splay 25.0 m. Bed levels: 300.000, 299.853, 299.799, 299.855 m; head loss 0.145 m. ## Answer The canal ($Q$ = 50 m³/s) crosses a drainage of $Q$ = 400 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 not given, taken 1.5:1; Lacey's $f$ = 1; trough $n$ = 0.015. #### 1. Type of structure HFL of the drainage (300.50 m) is above the canal bed (300.00 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 = 300.50 − 298.80 = 1.70 m. #### 2. Drainage waterway Lacey's regime perimeter (for a natural stream): $$ P = 4.75\sqrt{Q} = 4.75\sqrt{400} = 95.0\ \text{m} $$ Provide **12 spans of 8 m** with piers 1.2 m thick: clear waterway = 12 × 8 = **96 m** (> 95.0 m). Overall length = 96 + 11 × 1.2 = **109.2 m**. Check: mean velocity = Q/(L·d) = 400/(96 × 1.70) = 2.45 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.46 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.2 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 × 1.55) × 1.55 = 50.10 m², $V_1$ = 50/50.10 = 0.998 m/s, $h_{v1}=V_1^2/2g$ = 0.0508 m. Top width $T=B+2zD$ = 34.65 m. To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **18 m**, with the same full supply depth 1.55 m (the trough is as long as the drainage waterway, 109.2 m). - Trough area = 18 × 1.55 = 27.90 m²; $V_t$ = 50/27.90 = 1.792 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1637 m - Froude number = 0.46 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 27.90/21.10 = 1.322 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000498 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000498 × 109.2 = 0.054 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 = 34.65 − 18 = **16.6 m** - Outlet length = 3 × (T − B_t)/2 = **25.0 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): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 301.550 m, so $W_2$ = 301.403 m, $W_3$ = 301.349 m, $W_4$ = 301.405 m. Trough bed = water level − 1.55 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 300.000 | 301.550 | | 2 | End of inlet transition = start of trough | 299.853 | 301.403 | | 3 | End of trough = start of outlet transition | 299.799 | 301.349 | | 4 | End of outlet transition = normal canal | 299.855 | 301.405 | Total loss of head = 301.550 − 301.405 = 0.145 m, so the canal bed downstream of the work is 0.145 m lower than upstream. Check: with trough slab 0.6 m thick, soffit of the barrels = 299.20 m, i.e. 0.40 m above the drainage bed; HFL stands 1.30 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 297.70 m (1.5 m clear height) with a 1 in 4 slope at both ends. **Answer:** drainage waterway 96 m clear (12 × 8 m, overall 109.2 m); canal flumed to 18 m wide trough; bed/FSL at the four sections = 300.00/301.55, 299.85/301.40, 299.80/301.35, 299.86/301.41 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. HFL of the drain (300.50 m) is above the canal bed (300.00 m). What structure is suitable? A. Aqueduct B. Super passage C. Siphon aqueduct D. Canal siphon Answer: C. HFL is above the canal bed, so the trough cannot be above HFL cheaply; the drainage runs under pressure through barrels, a siphon aqueduct. (Key point: Choice of structure) 2. For Q = 400 m3/s, Lacey's perimeter is 95.0 m. What is provided and what is the overall length? A. 12 spans of 8 m with 1.2 m piers: 96 m clear, overall 109.2 m B. 12 spans of 8 m with 1.2 m piers: 96 m clear, overall 96 m C. 11 spans of 8 m: 88 m clear, overall 100 m D. 95 m clear, no piers Answer: A. Clear waterway 96 m exceeds 95.0 m; overall length adds 11 piers of 1.2 m to give 109.2 m. (Key point: Drainage waterway) 3. With normal top width T = 34.65 m and trough width 18 m, how are the transitions sized? A. Inlet 3:1 and outlet 2:1 B. Inlet 2:1 (16.6 m) and outlet 3:1 (25.0 m), splay measured on each side C. Both transitions 1:1 D. Inlet equals the trough length Answer: B. The inlet (contraction) uses 2:1 and the gentler outlet (expansion) uses 3:1, measured per side, giving 16.6 m and 25.0 m. (Key point: Transition lengths) 4. How is the friction loss in the trough obtained? A. hf = S x L, with S = (nV/R^(2/3))^2 from Manning's equation B. hf = 0.3 x velocity head C. hf = Q/A D. hf = HFL minus canal bed Answer: A. Friction slope from Manning's equation multiplied by the trough length (109.2 m) gives hf = 0.054 m. 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