--- title: "Cross drainage (flumed): canal 30, drainage 200" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:cd-flumed-q30-drain200" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/cd-flumed-q30-drain200 site: BCT Engineering --- # Cross drainage (flumed): canal 30, drainage 200 Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2066 Bhadra (old course), 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-flumed-q30-drain200 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question Find out the waterway, bed level and FSL of a suitable cross drainage structure for the following data given below. The structure should be flumed to achieve economy. | Parameters | Canal | Drainage | |---|---|---| | Qmax (m³/s) | 30 | 200 | | B (m) | 20 | 80 | | Bed level at d/s | 200 | 198 | | FSL / HFL | 202 | 201 | Assume Manning's rugosity coefficient as 0.014 for concrete. ## Short answer Siphon aqueduct. Drainage waterway: 9 spans of 8 m, 72 m clear, overall 81.6 m. Canal flumed to an 8 m trough; transitions 18 m inlet and 27 m outlet. Working back from downstream bed 200.00 m, upstream canal bed 200.164 m, trough bed 199.960 m at start and 199.921 m at end; FSL upstream 202.164 m. ## Answer The canal ($Q$ = 30 m³/s) crosses a drainage of $Q$ = 200 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:** Canal side slope 1.5:1 (not given); levels are downstream levels, so the work is solved backwards from the downstream bed 200.00 m; ground level taken equal to canal bed; $n$ = 0.014 for the concrete trough as given; Lacey's $f$ = 1. #### 1. Type of structure HFL of the drainage (201.00 m) is above the canal bed (200.00 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 = 201.00 − 198.00 = 3.00 m. #### 2. Drainage waterway Lacey's regime perimeter (for a natural stream): $$ P = 4.75\sqrt{Q} = 4.75\sqrt{200} = 67.2\ \text{m} $$ Provide **9 spans of 8 m** with piers 1.2 m thick: clear waterway = 9 × 8 = **72 m** (> 67.2 m). Overall length = 72 + 8 × 1.2 = **81.6 m**. Check: mean velocity = Q/(L·d) = 200/(72 × 3.00) = 0.93 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 2.75 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 4.1 m below HFL, with cut-offs and aprons for scour. #### 3. Canal waterway (flumed rectangular trough) Normal canal: area $A_1=(B+zD)D$ = (20 + 1.5 × 2) × 2 = 46.00 m², $V_1$ = 30/46.00 = 0.652 m/s, $h_{v1}=V_1^2/2g$ = 0.0217 m. Top width $T=B+2zD$ = 26.00 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 m (the trough is as long as the drainage waterway, 81.6 m). - Trough area = 8 × 2 = 16.00 m²; $V_t$ = 30/16.00 = 1.875 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1792 m - Froude number = 0.42 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 16.00/12.00 = 1.333 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000470 (Manning's n = 0.014); friction loss in trough $h_f=S\,L$ = 0.000470 × 81.6 = 0.038 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 = 26.00 − 8 = **18.0 m** - Outlet length = 3 × (T − B_t)/2 = **27.0 m** Bed level at the downstream canal is fixed (200.00 m). Working backwards with Bernoulli's equation (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 202.164 m, so $W_2$ = 201.960 m, $W_3$ = 201.921 m, $W_4$ = 202.000 m. Trough bed = water level − 2 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 200.164 | 202.164 | | 2 | End of inlet transition = start of trough | 199.960 | 201.960 | | 3 | End of trough = start of outlet transition | 199.921 | 201.921 | | 4 | End of outlet transition = normal canal | 200.000 | 202.000 | Total loss of head = 202.164 − 202.000 = 0.164 m, so the canal bed downstream of the work is 0.164 m lower than upstream. Check: with trough slab 0.6 m thick, soffit of the barrels = 199.32 m, i.e. 1.32 m above the drainage bed; HFL stands 1.68 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 197.82 m (1.5 m clear height) with a 1 in 4 slope at both ends. **Answer:** drainage waterway 72 m clear (9 × 8 m, overall 81.6 m); canal flumed to 8 m wide trough; bed/FSL at the four sections = 200.16/202.16, 199.96/201.96, 199.92/201.92, 200.00/202.00 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The given levels are downstream levels. How is the work solved? A. Starting from the upstream canal bed, forward B. Backwards from the downstream bed (200.00 m), giving W4 = 202.000 m and working up to W1 = 202.164 m C. By assuming the upstream FSL equals 202 m D. By ignoring the losses Answer: B. Section 4 is fixed at the downstream level, and Bernoulli's equation is applied in reverse to get upstream levels, so the upstream bed is 0.164 m higher. (Key point: Solving backwards from downstream levels) 2. Canal Q = 30 m3/s flumed to 8 m wide with depth 2 m. What is the trough velocity? A. 0.652 m/s B. 3.75 m/s C. 1.875 m/s, which is under 2.5 m/s and subcritical (Fr 0.42) D. 1.875 m/s, which is supercritical Answer: C. Vt = 30/(8 x 2) = 1.875 m/s; Fr = 0.42 so subcritical. (Key point: Flumed width and velocity) 3. HFL is 201.00 m and the canal bed 200.00 m. What follows? A. The canal passes over HFL freely B. A super passage is needed C. HFL is 1.00 m above the canal bed, so a siphon aqueduct with barrels flowing under pressure is used D. No crossing structure is needed Answer: C. HFL above canal bed prevents a trough above HFL being economical, so the barrels are siphonic. 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