---
title: "Siphon design: canal 18, drainage 60"
subject: "Irrigation and Drainage Engineering"
chapter: 8
key: "semester_6/irrigation_and_drainage_engineering:siphon-q18-drain60-lacey"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/siphon-q18-drain60-lacey
site: BCT Engineering
---

# Siphon design: canal 18, drainage 60

Irrigation and Drainage Engineering · Chapter 8: Cross-Drainage structures

- Asked in: 2075 Bhadra, 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/siphon-q18-drain60-lacey
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8

## Question

Design a siphon for the data given below:

| | Canal | Drainage |
|---|---|---|
| Discharge (m³/s) | 18.0 | 60 |
| Bed level (m) | 100.0 | 101.5 |
| Side slope (V:H) | 1:1.5 | |
| Bed width (m) | 6.0 | |
| FSL/high flood level (m) | 103.0 | 103.5 |

Lacey's silt factor = 1.0, Rugosity coefficient N = 0.016, Normal ground level = 102.0 m.

## Short answer

Canal siphon adopted since drainage bed (101.5 m) and HFL (103.5 m) exceed canal bed (100 m) and FSL (103 m). Drainage crossing 37 m wide. Two barrels 2 m × 1.8 m at velocity 2.5 m/s. Barrel floor at 99.20 m. Inlet transition 9.2 m (2:1), outlet 13.8 m (3:1). Total head loss 0.419 m; downstream FSL 102.581 m, bed 99.581 m.

## Answer

The drainage bed (101.50 m) is above the canal bed (100.00 m) and the drainage HFL (103.50 m) is above the canal FSL (103.00 m), so the drainage cannot be passed under the canal. The canal is carried **under** the drainage through closed barrels flowing full under pressure: a **canal siphon** (the drainage flows over the barrels as an open channel).

**Assumptions:** two rectangular RCC barrels; roof slab 0.5 m with its top flush with the drainage bed (101.50 m); entry loss 0.5 $V^2/2g$ and exit loss $0.5(V_b-V_1)^2/2g$; barrel length = width of drainage crossing; permissible velocity in barrels 2–3 m/s.

#### 1. Drainage waterway

$$
P = 4.75\sqrt{60} = 36.8\ \text{m}
$$

Provide a pitched drainage crossing of bottom width 37 m (> 36.8 m) over the barrels. Depth at HFL = 103.50 − 101.50 = 2.0 m, velocity = 60/(37 × 2.0) = 0.81 m/s, so a pitched floor with end aprons and cut-offs is enough. The length of the barrels is therefore about 38 m (crossing width plus flanking walls).

#### 2. Canal waterway (barrels)
Normal canal: $A_1=(6+1.5\times3)\times3$ = 31.5 m², $V_1$ = 18/31.5 = 0.571 m/s, $h_{v1}$ = 0.0166 m, top width $T$ = 15 m.

Adopting $V_b$ = 2.5 m/s, required area = 18/2.5 = 7.2 m². Provide **2 barrels, each 2 m wide × 1.8 m high**, separated by a 0.6 m pier: area = 2 × 2 × 1.8 = 7.2 m², $V_b$ = 18/7.2 = 2.50 m/s, $h_v$ = 0.3186 m. Overall width = 2×2 + 0.6 + 2×0.6 = 5.8 m.

#### 3. Bed levels and transitions
- Top of roof = 101.50 m; soffit = 101.50 − 0.5 = 101.00 m; barrel floor = 101.00 − 1.8 = **99.20 m** (canal bed depressed by 0.80 m, with a slope of 1 in 4 at the inlet and outlet).
- Transitions in plan: inlet (2:1) = T − barrel width = 15 − 5.8 = 9.2 m; outlet (3:1) = 1.5 × 9.2 = 13.8 m.

Losses of head (N = 0.016, $R$ = 0.474 m for one barrel):

$$
\begin{aligned} h_{f} &= \frac{N^2V_b^2L}{R^{4/3}} = \frac{0.016^2 \times 2.50^2 \times 38}{0.474^{4/3}} = 0.165\ \text{m} \\ h_e &= 0.5\,\frac{V_b^2}{2g} = 0.159\ \text{m} \\ h_x &= 0.5\,\frac{(V_b-V_1)^2}{2g} = 0.095\ \text{m} \end{aligned}
$$

Total loss = 0.159 + 0.165 + 0.095 = **0.419 m**.

| Section | Bed level (m) | Water level (m) |
|---|---|---|
| U/s canal (normal) | 100.00 | 103.000 |
| Barrel (floor level) | 99.20 | HGL falls from 103.00 to 102.58 |
| D/s canal (normal) | 99.581 | 102.581 |

```text
  HFL 103.50 ~~~~~~~~~~~~~~~~~~~~~~~~~
  drain bed 101.50 ============ roof (0.5 m)
  canal FSL 103.00 --.  barrels 2 x (2.0 x 1.8)  .-- 102.58
  canal bed 100.00 ---'__________________________'--- 99.58
                       floor 99.20
```

**Answer:** drainage crossing 37 m wide; 2 barrels of 2.0 m × 1.8 m (V = 2.50 m/s); barrel floor 99.20 m; total head loss 0.419 m; downstream canal FSL 102.581 m and bed 99.581 m.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Drainage bed (101.50 m) is above canal bed (100.00 m) and drain HFL (103.50 m) above canal FSL (103.00 m). Which structure is suitable?
   A. Aqueduct
   B. Siphon aqueduct
   C. Canal siphon, with the canal under the drainage in closed barrels flowing full
   D. Super passage
   Answer: C. The drainage cannot pass under the canal, so the canal goes under it through pressure barrels. (Key point: Choosing a canal siphon)

2. How are the barrels sized?
   A. Required area = Q/Vb = 18/2.5 = 7.2 m2, giving 2 barrels of 2 m x 1.8 m
   B. Area = canal normal area 31.5 m2
   C. Area = 60/2.5
   D. Barrel width equals canal bed width
   Answer: A. Adopt Vb = 2.5 m/s (within 2-3 m/s) and divide Q by it. (Key point: Barrel sizing)

3. How is the barrel floor level fixed?
   A. Roof top at drainage bed 101.50, soffit 101.00 (roof 0.5 m), floor = 101.00 - 1.8 = 99.20 m
   B. Equal to canal bed 100.00 m
   C. Equal to drainage bed 101.50 m
   D. HFL - 1.8 m
   Answer: A. Working down from roof top flush with the drain bed gives floor 99.20 m, so canal bed is depressed 0.80 m. (Key point: Barrel floor level)

4. Which losses make up the total 0.419 m?
   A. Entry 0.5 V^2/2g, friction in barrel, exit 0.5 (Vb - V1)^2/2g
   B. Only friction loss
   C. Only entry and exit
   D. Uplift and seepage
   Answer: A. Entry 0.159 + friction 0.165 + exit 0.095 = 0.419 m. (Key point: Head losses)

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