--- title: "Cross drainage: Q 50, HFD 450, uplift on roof" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:cd-q50-fsd15-hfd450-uplift-roof" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/cd-q50-fsd15-hfd450-uplift-roof site: BCT Engineering --- # Cross drainage: Q 50, HFD 450, uplift on roof Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2073 Magh, 10 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-fsd15-hfd450-uplift-roof - 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 = 1.5 m, Bed level of canal = 100.0 m, High flood discharge of drain = 450 m³/s, High flood level of drainage = 100.50 m, Bed level of drainage = 98.8 m, General ground level = 100.0 m. (i) Design of drainage water-way (ii) Design of canal water way (iii) Design of transition and (iv) Uplift pressure on the roof. ## Short answer Drainage waterway: 11 spans of 10 m, clear waterway 110 m. Canal waterway: flumed trough 19 m wide, 1.5 m deep. Inlet transition 2:1 splay 15.5 m; outlet 3:1 splay 23.2 m. Bed levels: 100.000, 99.867, 99.806, 99.857 m. Uplift pressure on roof = (100.50 − 99.21) × 9.81 ≈ 12.7 kN/m². ## Answer The canal ($Q$ = 50 m³/s) crosses a drainage of $Q$ = 450 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; Lacey's $f$ = 1; trough $n$ = 0.015; roof (trough) slab 0.6 m thick, unit weight of concrete 24 kN/m³; barrel $n$ = 0.015; barrel length ≈ trough width + walls ≈ 20 m. #### 1. Type of structure HFL of the drainage (100.50 m) is above the canal bed (100.00 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 = 100.50 − 98.80 = 1.70 m. #### 2. Drainage waterway Lacey's regime perimeter (for a natural stream): $$ P = 4.75\sqrt{Q} = 4.75\sqrt{450} = 100.8\ \text{m} $$ Provide **11 spans of 10 m** with piers 1.5 m thick: clear waterway = 11 × 10 = **110 m** (> 100.8 m). Overall length = 110 + 10 × 1.5 = **125.0 m**. Check: mean velocity = Q/(L·d) = 450/(110 × 1.70) = 2.41 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.60 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.4 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 × 1.5) × 1.5 = 48.38 m², $V_1$ = 50/48.38 = 1.034 m/s, $h_{v1}=V_1^2/2g$ = 0.0545 m. Top width $T=B+2zD$ = 34.50 m. To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **19 m**, with the same full supply depth 1.5 m (the trough is as long as the drainage waterway, 125.0 m). - Trough area = 19 × 1.5 = 28.50 m²; $V_t$ = 50/28.50 = 1.754 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1569 m - Froude number = 0.46 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 28.50/22.00 = 1.295 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000490 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000490 × 125.0 = 0.061 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 = 34.50 − 19 = **15.5 m** - Outlet length = 3 × (T − B_t)/2 = **23.2 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): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 101.500 m, so $W_2$ = 101.367 m, $W_3$ = 101.306 m, $W_4$ = 101.357 m. Trough bed = water level − 1.5 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 100.000 | 101.500 | | 2 | End of inlet transition = start of trough | 99.867 | 101.367 | | 3 | End of trough = start of outlet transition | 99.806 | 101.306 | | 4 | End of outlet transition = normal canal | 99.857 | 101.357 | Total loss of head = 101.500 − 101.357 = 0.143 m, so the canal bed downstream of the work is 0.143 m lower than upstream. Check: with trough slab 0.6 m thick, soffit of the barrels = 99.21 m, i.e. 0.41 m above the drainage bed; HFL stands 1.29 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 97.71 m (1.5 m clear height) with a 1 in 4 slope at both ends. #### Uplift on the roof (trough slab) The critical case is the canal dry with the drainage at HFL. Barrels: 11 openings of 10 m × 1.5 m (bed depressed), area 165 m², $V_b$ = 450/165 = 2.73 m/s, $h_{vb}$ = 0.379 m; $R$ = 15/23 = 0.652 m. $$ h_{fb} = \frac{n^2V_b^2L}{R^{4/3}} = \frac{0.015^2 \times 2.73^2 \times 20}{0.652^{4/3}} = 0.059\ \text{m} $$ Upstream HFL = d/s HFL + entry loss ($0.5h_{vb}$) + exit loss ($1.0h_{vb}$) + $h_{fb}$ = 100.50 + 0.569 + 0.059 = 101.13 m (afflux 0.63 m). The maximum uplift is just inside the entry (upstream end of the roof): pressure head = 101.128 − 1.5 × 0.379 − 99.206 (soffit) = 1.35 m, so the uplift pressure = 9.81 × 1.353 = **13.3 kN/m²**. Downward load = slab weight 24 × 0.6 = 14.4 kN/m² plus canal water 9.81 × 1.5 = 14.7 kN/m², total 29.1 kN/m². The uplift (13.3) is less than the slab weight alone (14.4), so the roof does not lift even with the canal dry; the net downward load with the canal running = 15.8 kN/m². The slab is reinforced for bending under these loads. **Answer:** drainage waterway 110 m clear (11 × 10 m, overall 125.0 m); canal flumed to 19 m wide trough; bed/FSL at the four sections = 100.00/101.50, 99.87/101.37, 99.81/101.31, 99.86/101.36 m; maximum uplift on the roof 13.3 kN/m² (less than the 14.4 kN/m² slab weight). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the upward pressure on the trough slab (roof) found when the barrel runs full? A. gamma_w x (HFL - soffit level) = 9.81 x 1.29 = 12.7 kN/m2 B. gamma_w x canal depth = 14.7 kN/m2 C. Slab weight 24 x 0.6 = 14.4 kN/m2 D. gamma_w x drainage depth at HFL Answer: A. Pressure under the slab is the water head above the soffit times unit weight of water. (Key point: Uplift on roof) 2. Uplift is 12.7 kN/m2 and slab weight is 14.4 kN/m2. What is concluded? A. The roof floats when the canal is dry B. The roof is safe against uplift even with the canal dry; net downward load with canal running is 16.4 kN/m2 C. The uplift equals canal water load D. The roof needs anchoring because uplift exceeds total load Answer: B. Uplift is less than the slab weight alone, so no flotation; with canal water (29.1 kN/m2 downward) net is 29.1 - 12.7 = 16.4 kN/m2. (Key point: Stability against uplift) 3. Why is a siphon aqueduct chosen here? A. Drain bed is above the canal bed B. Drain HFL (100.50 m) is above the canal bed (100.00 m), so the drain flows under pressure through barrels below the trough C. Canal FSL is below drain bed D. The drain discharge is smaller than canal discharge Answer: B. HFL above canal bed means the trough cannot be kept above HFL economically. (Key point: Type of structure) 4. For Q = 450 m3/s the Lacey perimeter is 100.8 m. What is provided? A. 11 spans of 10 m, 110 m clear, overall 125.0 m B. 10 spans of 10 m, overall 100 m C. 11 spans of 10 m, overall 110 m D. 100.8 m of single span Answer: A. 110 m clear exceeds 100.8 m; adding 10 piers of 1.5 m gives 125.0 m overall. (Key point: Drainage waterway) ## Close variants set in other exams - [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× ## Related in Irrigation and Drainage Engineering - [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× - [Cross drainage: Q 25, uplift on roof of barrel](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acd-q25-uplift-roof-barrel-800) — Irrigation and Drainage Engineering, chapter 8, asked 2× - [Cross drainage: Q 50, HFD 550](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acd-q50-bed30-fsd21-hfd550) — Irrigation and Drainage Engineering, chapter 8, asked 2× - [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 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 - [Siphon aqueduct: Q 30, bed 24 m](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asiphon-aqueduct-q30-bed24-hfd500) — Irrigation and Drainage Engineering, chapter 8 ## Related in other subjects - [Gravity dam stresses with drainage gallery uplift](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-stresses-2080-baishakh) — Hydropower Engineering, chapter 4 - [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 - [Uplift force with and without drainage gallery](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Auplift-force-drainage-gallery) — Hydro Power, chapter 4 - [Hydrostatic thrust and uplift with drainage gallery](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Adam-thrust-uplift-gallery) — Hydro Power, chapter 4