--- title: "Cross drainage: Q 30, HFL 311 m" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:cd-q30-canal-bed310-hfl311" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/cd-q30-canal-bed310-hfl311 site: BCT Engineering --- # Cross drainage: Q 30, HFL 311 m Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2079 Asoj, 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-q30-canal-bed310-hfl311 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question A canal carrying 30 m³/sec of discharge with a bed width of 30 m and bed level at downstream is 310.0 m is crossing the natural drainage having bed level 308.0 m and the high flood level of 311.0 m which carries the 50 year return period flood as 550 cumecs. The full supply depth of canal is 2.1 m and side slope is 1.5:1. Design appropriate cross drainage structure with determining the drainage waterway, canal waterway and bed and full supply level at four different sections of the canal trough. ## Short answer A siphon aqueduct with 12 spans of 10 m (120 m clear waterway) passes 550 m³/s drainage. Canal flumed to 8 m wide rectangular trough (2.1 m depth) over 136.5 m. Four section levels (downstream bed fixed at 310.000 m): 1) Bed 310.186, FSL 312.286; 2) Bed 309.987, FSL 312.087; 3) Bed 309.923, FSL 312.023; 4) Bed 310.000, FSL 312.100. Total head loss 0.186 m. ## Answer The canal ($Q$ = 30 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:** Bed level 310.0 m is the downstream canal bed; general ground level = canal bed = 310.0 m; Lacey's $f$ = 1; concrete trough $n$ = 0.015. The 50-year flood (550 m³/s) is the design flood. #### 1. Type of structure HFL of the drainage (311.00 m) is above the canal bed (310.00 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 = 311.00 − 308.00 = 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**. Approach velocity at HFL = Q/(L·d) = 550/(120 × 3.00) = 1.53 m/s (the velocity through the barrels is checked at the end). 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 × 2.1) × 2.1 = 69.61 m², $V_1$ = 30/69.61 = 0.431 m/s, $h_{v1}=V_1^2/2g$ = 0.0095 m. Top width $T=B+2zD$ = 36.30 m. To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **8 m**, with the same full supply depth 2.1 m (the trough is as long as the drainage waterway, 136.5 m). - Trough area = 8 × 2.1 = 16.80 m²; $V_t$ = 30/16.80 = 1.786 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1625 m - Froude number = 0.39 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 16.80/12.20 = 1.377 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000468 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000468 × 136.5 = 0.064 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 = 36.30 − 8 = **28.3 m** - Outlet length = 3 × (T − B_t)/2 = **42.4 m** Bed level at the downstream canal is fixed (310.00 m). Working backwards with Bernoulli's equation (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 312.286 m, so $W_2$ = 312.087 m, $W_3$ = 312.023 m, $W_4$ = 312.100 m. Trough bed = water level − 2.1 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 310.186 | 312.286 | | 2 | End of inlet transition = start of trough | 309.987 | 312.087 | | 3 | End of trough = start of outlet transition | 309.923 | 312.023 | | 4 | End of outlet transition = normal canal | 310.000 | 312.100 | Total loss of head = 312.286 − 312.100 = 0.186 m, so the canal bed downstream of the work is 0.186 m lower than upstream. Check: with a trough slab 0.6 m thick, the soffit of the barrels = 309.923 − 0.6 = 309.32 m, only 1.32 m above the drainage bed, while HFL stands 1.68 m above the soffit, so the barrels run full under pressure. With 1.32 m clear height the barrel velocity would be 550/(120 × 1.32) = 3.47 m/s, which is too high. The drainage bed under the barrels is therefore depressed to RL 307.52 m (1.8 m clear height) with 1 in 4 slopes at both ends, giving a barrel velocity of 550/(120 × 1.8) = 2.55 m/s, which is acceptable. **Answer:** drainage waterway 120 m clear (12 × 10 m, overall 136.5 m); canal flumed to 8 m wide trough; bed/FSL at the four sections = 310.19/312.29, 309.99/312.09, 309.92/312.02, 310.00/312.10 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What structure suits the 30 m3/s canal over the drain with HFL 311.00 m and canal bed 310.00 m? A. Siphon aqueduct B. Super-passage C. Level crossing D. Plain aqueduct Answer: A. HFL is above the canal bed, so the drain flows under pressure through barrels beneath the trough. (Key point: Type of structure) 2. What controls the drainage waterway chosen for the 550 m3/s flood? A. The Lacey perimeter 4.75 sqrt(Q) = 111.4 m, so 12 spans of 10 m (120 m) are given B. The canal discharge 30 m3/s C. The trough width of 8 m D. The general ground level Answer: A. The waterway must be at least the regime perimeter, so 120 m clear is provided. (Key point: Drainage waterway) 3. Why are levels worked backwards from the downstream end? A. Because the downstream canal bed level (310.00 m) is fixed B. Because the upstream level is unknown by nature C. Because friction is zero D. Because the trough is dry Answer: A. The d/s bed is given, so Bernoulli's equation with losses is applied upstream from section 4 to get W1 = 312.286 m. (Key point: Working backwards) 4. What is the net result of the transitions and friction on the canal bed? A. Canal bed downstream is lowered by the total loss of head, 0.186 m B. The downstream bed is raised by 0.186 m C. No change in level D. The bed is lowered by the afflux Answer: A. Losses of 0.186 m must be provided by dropping the d/s canal bed and FSL. 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