--- title: "Cross drainage: canal 125, drainage 64" subject: "Irrigation and Drainage Engineering" chapter: 8 key: "semester_6/irrigation_and_drainage_engineering:cd-table-q125-drain64" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/cd-table-q125-drain64 site: BCT Engineering --- # Cross drainage: canal 125, drainage 64 Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures - Asked in: 2068 Baisakh (old course), 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-table-q125-drain64 - 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 following data: | Parameters | Canal | Drainage | |---|---|---| | Discharge (m³/s) | 125.0 | 64.0 | | FSL/HFL (downstream) m | 298.00 | 304.00 | | Bed Level (downstream) m | 294.00 | 300.00 | | Bed width (m) | 40.0 | 80.0 | | Side slope (H:V) | 1:1 | 0:1 | Fix the waterway of the canal and drainage designing a suitable type of cross drainage structure and find out the bed level of the drainage designing structure at the upstream end. ## Short answer Super passage, since drainage bed is above canal FSL. Canal waterway: 4 spans of 8 m, 32 m clear, overall 35.6 m. Drainage flumed to a 16 m RCC trough, depth about 3.98 m; trough bed at upstream end 300.00 m, upstream drainage bed 300.04 m. ## Answer The drainage bed (300.00 m) is above the canal FSL (298.00 m) by 2.0 m, and the canal is much larger than the drainage (125 vs 64 m³/s). So the drainage is carried **over** the canal in an RCC trough: a **super passage**. The canal flows beneath as an open channel with a free surface. **Assumptions:** RCC trough and canal waterway with $n$ = 0.015; spans 8 m with 1.2 m piers; slab 0.6 m; losses of $0.3\,\Delta h_v$ at the inlet and $0.5\,\Delta h_v$ at the outlet; downstream drainage section: bed 300.00 m, HFL 304.00 m, width 80 m. #### 1. Canal waterway Normal canal: $A=(40+1\times4)\times4$ = 176 m², $V$ = 125/176 = 0.710 m/s. Through the structure the canal is fitted with a rectangular lined section of width 32 m (depth 4 m): $A$ = 128 m², $V$ = 125/128 = 0.98 m/s (< 2 m/s), Froude number 0.16 (subcritical). Provide **4 spans of 8 m** with 3 piers of 1.2 m: clear canal waterway = **32 m**, overall = 32 + 3 × 1.2 = **35.6 m**. This is also the length of the drainage trough. Clearance: soffit of trough slab = 300.00 − 0.6 = 299.40 m, which is 1.40 m above canal FSL (298.00 m). Adequate (> 0.6 m free board). #### 2. Drainage waterway (flumed trough) Lacey's perimeter $P=4.75\sqrt{64}$ = 38.0 m. The natural channel is 80 m wide with $V$ = 64/(80 × 4) = 0.20 m/s, so it is flumed to a rectangular RCC trough of width $B_t$ = **16 m**, which keeps the velocity near 1 m/s. The depth at the downstream end of the trough ($d_3$) follows from Bernoulli's equation with the expansion loss, taking the trough bed at the downstream bed level 300.00 m: $$ \begin{aligned} d_3 + h_{v3} &= (304.00-300.00) + h_{v4} + 0.5(h_{v3}-h_{v4}) \\ d_3 &= 4.00 + 0.5\,(h_{v4} - h_{v3}) \end{aligned} $$ By trial: $d_3$ = 3.975 m, $V_3$ = 64/(16 × 3.975) = 1.006 m/s, $h_{v3}$ = 0.0516 m. Water level at the end of trough = 303.975 m. Friction in the trough: $R$ = 2.656 m, $S=(nV/R^{2/3})^2$ = 0.000062, $h_f$ = S × 35.6 = 0.0022 m. #### 3. Levels at the upstream end - Water level at the start of trough = 303.975 + 0.0022 = 303.977 m - Keeping the same depth 3.975 m in the trough, the **trough bed at the upstream end = 300.002 m**, practically 300.00 m (2 mm above the downstream end). - Water level in the drainage just upstream of the inlet transition = 303.977 + 0.0516 + 0.3(0.0516 − 0.0020) − 0.0020 = 304.042 m, an afflux of about 0.04 m. Taking the normal 80 m drainage section (depth 4.00 m) upstream, the drainage bed just upstream of the inlet transition = 304.04 − 4.00 = **300.04 m**. - Transitions: inlet (2:1) = 80 − 16 = 64 m; outlet (3:1) = 1.5 × 64 = 96 m (splay measured on the water-surface width), with the drainage bed continuing at 300.00 m. ```text HFL 304 ~~~~~~~\__________________/~~~~~~~ drain bed 300 ---|== RCC trough ==|--- clearance 1.4 m | 4 x 8 m spans | canal FSL 298 ---+~~~~~~~~~~~~~~~~+--- canal bed 294 ---+----------------+--- ``` **Answer:** canal waterway 32 m clear (4 × 8 m spans, overall 35.6 m); drainage flumed to a 16 m wide RCC trough (3.98 m deep); trough bed at the upstream end of the structure = **300.00 m** (water level 303.98 m at the trough start); drainage bed upstream of the inlet transition = 300.04 m (water level 304.04 m). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The drainage bed (300.00 m) is 2.0 m above the canal FSL (298.00 m) and the canal is much larger. Which structure is adopted? A. Siphon aqueduct B. Canal siphon C. Super passage, where the drainage is carried over the canal in an RCC trough and the canal flows beneath as an open channel D. Level crossing Answer: C. Drainage bed above canal FSL means the drainage goes over the canal: a super passage. (Key point: Choice of super passage) 2. The canal is contracted to a 32 m wide lined section carrying 125 m3/s at 4 m depth. What waterway results? A. V = 0.98 m/s; 4 spans of 8 m = 32 m clear, overall 35.6 m with 3 piers of 1.2 m B. V = 0.71 m/s; 40 m clear C. V = 3.9 m/s; 2 spans D. V = 0.98 m/s; 4 spans overall 32 m Answer: A. A = 32 x 4 = 128 m2, V = 125/128 = 0.98 m/s; the 4 spans plus 3 piers give 35.6 m overall. (Key point: Canal waterway) 3. The trough slab soffit is 299.40 m and canal FSL is 298.00 m. What does this indicate? A. Only 0.6 m clearance, which is inadequate B. 1.40 m clearance, adequate since it exceeds the 0.6 m free board C. The canal overtops the slab D. The soffit equals FSL Answer: B. 299.40 - 298.00 = 1.40 m, more than the 0.6 m free board required. (Key point: Clearance above canal) 4. The natural drainage is 80 m wide with V = 0.20 m/s. Why is it flumed to a 16 m RCC trough? A. To lower the velocity further B. To raise HFL by 16 m C. Because the Lacey perimeter equals 80 m D. To keep velocity near 1 m/s and reduce the width, and cost, of the trough Answer: D. A narrow trough raises velocity from 0.20 to about 1.0 m/s and makes the RCC trough much cheaper than 80 m wide. 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