--- title: "Cross drainage: Q 50, HFD 550" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:cd-q50-bed30-fsd21-hfd550" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/cd-q50-bed30-fsd21-hfd550 site: BCT Engineering --- # Cross drainage: Q 50, HFD 550 Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2079 Jestha, 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-bed30-fsd21-hfd550 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question The cross-drainage structure across an irrigation channel has the following data: Discharge of canal = 50 cumec; Bed width of canal = 30 m; Full supply depth of canal = 2.1 m; Bed level of the canal = 160.00 m; Side slope of canal = 1.5:1 (H:V); High flood discharge of drainage = 550 cumec; High flood level of drainage = 161.00 m; Bed level of drainage = 158.00 m; General ground level = 160.00 m. Design the drainage waterway, and canal waterway and find the bed levels and FSL at four different sections of the canal trough. ## Short answer A siphon aqueduct is designed with 12 spans of 10 m (120 m clear waterway) for 550 m³/s drainage. Canal flumed to 13 m wide rectangular trough (2.1 m depth) over 136.5 m. Four section levels: 1) Bed 160.000, FSL 162.100; 2) Bed 159.812, FSL 161.912; 3) Bed 159.756, FSL 161.856; 4) Bed 159.829, FSL 161.929. Total head loss 0.171 m. ## Answer The canal ($Q$ = 50 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:** Side slope 1.5:1 (H:V), Lacey's $f$ = 1, concrete trough with Manning's $n$ = 0.015, piers 1.2–1.5 m thick, slab thickness 0.6 m. #### 1. Type of structure HFL of the drainage (161.00 m) is above the canal bed (160.00 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 = 161.00 − 158.00 = 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, $V_1$ = 50/69.61 = 0.718 m/s, $h_{v1}=V_1^2/2g$ = 0.0263 m. Top width $T=B+2zD$ = 36.30 m. To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **13 m**, with the same full supply depth 2.1 m (the trough is as long as the drainage waterway, 136.5 m). - Trough area = 13 × 2.1 = 27.30 m²; $V_t$ = 50/27.30 = 1.832 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1710 m - Froude number = 0.40 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 27.30/17.20 = 1.587 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000408 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000408 × 136.5 = 0.056 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 − 13 = **23.3 m** - Outlet length = 3 × (T − B_t)/2 = **34.9 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): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 162.100 m, so $W_2$ = 161.912 m, $W_3$ = 161.856 m, $W_4$ = 161.929 m. Trough bed = water level − 2.1 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 160.000 | 162.100 | | 2 | End of inlet transition = start of trough | 159.812 | 161.912 | | 3 | End of trough = start of outlet transition | 159.756 | 161.856 | | 4 | End of outlet transition = normal canal | 159.829 | 161.929 | Total loss of head = 162.100 − 161.929 = 0.171 m, so the canal bed downstream of the work is 0.171 m lower than upstream. Check: with a trough slab 0.6 m thick, the soffit of the barrels = 159.756 − 0.6 = 159.16 m, only 1.16 m above the drainage bed, while HFL stands 1.84 m above the soffit, so the barrels run full under pressure. With 1.16 m clear height the barrel velocity would be 550/(120 × 1.16) = 3.95 m/s, which is too high. The drainage bed under the barrels is therefore depressed to RL 157.36 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 13 m wide trough; bed/FSL at the four sections = 160.00/162.10, 159.81/161.91, 159.76/161.86, 159.83/161.93 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 HFL is 1.0 m above the canal bed? A. The trough cannot be kept above HFL economically, so the drain is passed under pressure through barrels B. The drain bed is above the canal C. The canal is flowing below the drain D. The drain discharge is very small Answer: A. Raising the trough above HFL would be costly, so the drain runs under the trough under pressure. (Key point: Type of structure) 2. How is the 120 m clear waterway justified? A. It is 12 x 10 m spans, more than Lacey's regime perimeter 4.75 sqrt(550) = 111.4 m B. It equals the canal bed width C. It is half the regime perimeter D. It equals the trough length minus piers Answer: A. The provided waterway must at least equal the regime perimeter 111.4 m. (Key point: Drainage waterway) 3. What is the effect of flumed trough width 13 m with the same depth 2.1 m? A. Velocity rises to 1.832 m/s, still below 2.5 m/s, and the flow stays subcritical B. Velocity falls below the normal canal C. A hydraulic jump forms in the trough D. Flow becomes supercritical Answer: A. A narrower trough increases velocity (50/27.3), but Froude number 0.40 < 1 so no jump. (Key point: Flumed trough) 4. How are the water levels at the four sections obtained? A. By applying Bernoulli's equation with transition losses (0.3 for contraction, 0.5 for expansion) and friction loss in the trough B. By assuming all levels equal C. By adding the afflux to HFL D. By subtracting 2.1 m from the ground level Answer: A. Levels follow from energy balance between sections plus friction loss h_f = S L, giving 162.100, 161.912, 161.856, 161.929 m. (Key point: Levels along the trough) ## Close variants set in other exams - [Cross drainage: Q 50, HFD 450](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acd-q50-bed30-fsd21-hfd450) — Irrigation and Drainage Engineering, chapter 8, asked 2× ## Related in Irrigation and Drainage Engineering - [Cross drainage: Q 50, bed 35 m, FSD 1.6](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acd-q50-bed35-fsd16-hfd400) — Irrigation and Drainage Engineering, chapter 8, asked 2× - [Cross drainage: Q 25, HFD 360](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acd-q25-bed30-fsd2-hfd360) — Irrigation and Drainage Engineering, chapter 8, asked 2× - [Cross drainage: Q 50, HFD 400, d 1.55 m](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acd-q50-fsd155-hfd400) — Irrigation and Drainage Engineering, chapter 8, asked 2× - [Cross drainage: Q 50, HFD 450, uplift on roof](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acd-q50-fsd15-hfd450-uplift-roof) — Irrigation and Drainage Engineering, chapter 8, asked 2× - [Cross drainage: Q 30, HFL 311 m](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acd-q30-canal-bed310-hfl311) — Irrigation and Drainage Engineering, chapter 8 - [Cross drainage: canal 125, drainage 64](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acd-table-q125-drain64) — Irrigation and Drainage Engineering, chapter 8 ## Related in other subjects - [Irrigation channel earthwork, bed 5 m, 1 in 3000](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Airrigation-channel-j) — Estimating and Costing, chapter 6 - [Irrigation canal earthwork with vertical drop](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Acanal-a-drop-0-400) — Estimating and Costing, chapter 6 - [Irrigation canal earthwork, 0-600 m](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Airrigation-canal-k-drop) — Estimating and Costing, chapter 6 - [Canal earthwork, bed 3 m, free board 0.44 m](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Achannel-q-portion) — Estimating and Costing, chapter 6