--- title: "Cross drainage: canal 300, drainage 500" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:cd-q300-canal-n0016-drain500" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/cd-q300-canal-n0016-drain500 site: BCT Engineering --- # Cross drainage: canal 300, drainage 500 Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2076 Baisakh, 12 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-q300-canal-n0016-drain500 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8 ## Question Design a suitable cross drainage structure for the following data. Canal Data: Discharge = 300 m³/s; Bed width = 30 m; FSL = 209 m; Bed level = 202.50 m; Side slope = ½:1 (H:V); Rugosity Coefficient (N) = 0.016. Drainage data: Discharge = 500 m³/s; Bed level = 198 m; HFL = 207.70 m; Silt factor (f) = 1.0. ## Short answer Siphon aqueduct chosen since drainage HFL (207.7 m) is 5.2 m above canal bed (202.5 m). Drainage waterway: 11 spans of 10 m (clear 110 m). Canal flumed to 26 m wide trough, depth 6.5 m. Inlet transition 10.5 m (2:1), outlet 15.8 m (3:1). Trough bed levels: start 202.419 m, end 202.391 m. Total head loss 0.064 m. ## Answer The canal ($Q$ = 300 m³/s) crosses a drainage of $Q$ = 500 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 depth = FSL − bed = 209 − 202.5 = 6.5 m; side slope ½H:1V; ground level taken equal to canal bed; Lacey's $f$ = 1 and $n$ = 0.016 (as given) used for the trough; slab 0.6 m. #### 1. Type of structure HFL of the drainage (207.70 m) is above the canal bed (202.50 m) by 5.20 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 = 207.70 − 198.00 = 9.70 m. #### 2. Drainage waterway Lacey's regime perimeter (for a natural stream): $$ P = 4.75\sqrt{Q} = 4.75\sqrt{500} = 106.2\ \text{m} $$ Provide **11 spans of 10 m** with piers 1.5 m thick: clear waterway = 11 × 10 = **110 m** (> 106.2 m). Overall length = 110 + 10 × 1.5 = **125.0 m**. Check: velocity through the barrels (clear height 3.80 m, see section 4) = 500/(110 × 3.80) = 1.20 m/s, which is acceptable. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.73 m (with $f$ = 1), so the scour level 1.5R = 5.6 m below HFL is at RL 202.1 m, which is above the drainage bed (198.00 m). Scour does not govern, so the foundations of piers and abutments are taken to a nominal depth of about 2 m below the drainage bed (about RL 196.0 m), with cut-offs and aprons. #### 3. Canal waterway (flumed rectangular trough) Normal canal: area $A_1=(B+zD)D$ = (30 + 0.5 × 6.5) × 6.5 = 216.12 m², $V_1$ = 300/216.12 = 1.388 m/s, $h_{v1}=V_1^2/2g$ = 0.0982 m. Top width $T=B+2zD$ = 36.50 m. To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **26 m**, with the same full supply depth 6.5 m (the trough is as long as the drainage waterway, 125.0 m). - Trough area = 26 × 6.5 = 169.00 m²; $V_t$ = 300/169.00 = 1.775 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1606 m - Froude number = 0.22 < 1, flow is subcritical, so there is no hydraulic jump. - Hydraulic radius $R=A/P$ = 169.00/39.00 = 4.333 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000114 (Manning's n = 0.016); friction loss in trough $h_f=S\,L$ = 0.000114 × 125.0 = 0.014 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 = 36.50 − 26 = **10.5 m** - Outlet length = 3 × (T − B_t)/2 = **15.8 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): $$ \begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned} $$ $W_1$ = 209.000 m, so $W_2$ = 208.919 m, $W_3$ = 208.905 m, $W_4$ = 208.936 m. Trough bed = water level − 6.5 m. | Section | Location | Bed level (m) | FSL (m) | |---|---|---|---| | 1 | Normal canal, start of inlet transition | 202.500 | 209.000 | | 2 | End of inlet transition = start of trough | 202.419 | 208.919 | | 3 | End of trough = start of outlet transition | 202.405 | 208.905 | | 4 | End of outlet transition = normal canal | 202.436 | 208.936 | Total loss of head = 209.000 − 208.936 = 0.064 m, so the canal bed downstream of the work is 0.064 m lower than upstream. Check: with trough slab 0.6 m thick, soffit of the barrels = 201.80 m, i.e. 3.80 m above the drainage bed; HFL stands 5.90 m above the soffit, which gives the siphonic head. **Answer:** drainage waterway 110 m clear (11 × 10 m, overall 125.0 m); canal flumed to 26 m wide trough; bed/FSL at the four sections = 202.50/209.00, 202.42/208.92, 202.40/208.90, 202.44/208.94 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the canal depth obtained from the data given? A. Depth = FSL - bed = 209 - 202.5 = 6.5 m B. Depth = HFL - drain bed C. Depth = bed width / 5 D. Depth = 1.5 m assumed Answer: A. The full supply depth is the difference between FSL and canal bed level. (Key point: Canal depth assumption) 2. HFL (207.70 m) is 5.20 m above the canal bed (202.50 m). Which work is chosen? A. Aqueduct B. Siphon aqueduct, with the drainage flowing under pressure through barrels C. Super passage D. Level crossing Answer: B. The trough cannot be above HFL economically so barrels under pressure are used. (Key point: Structure type) 3. Which is the correct drainage waterway? A. P = 106.2 m; 11 spans of 10 m, 110 m clear, overall 125.0 m B. P = 4.75 sqrt(500)=53 m C. 10 spans of 10 m with no piers D. P = 106.2 m; 106.2 m clear exactly Answer: A. Clear waterway of 110 m > 106.2 m with 10 piers of 1.5 m gives 125.0 m. (Key point: Drainage waterway) 4. Trough 26 m wide, depth 6.5 m, Q = 300 m3/s. What is true? A. V = 1.78 m/s and Froude number 0.22, so flow is subcritical and no jump forms B. V = 1.39 m/s, supercritical C. V = 5 m/s, jump forms D. Velocity cannot be found without HFL Answer: A. V = 300/169 = 1.775 m/s; Fr = 0.22 < 1. 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