---
title: "Siphon aqueduct: Q 30, drainage 800"
subject: "Irrigation and Drainage Engineering"
chapter: 8
key: "semester_6/irrigation_and_drainage_engineering:siphon-aqueduct-q30-bed23-hfd800"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/siphon-aqueduct-q30-bed23-hfd800
site: BCT Engineering
---

# Siphon aqueduct: Q 30, drainage 800

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

- Asked in: 2062 Kartik (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/siphon-aqueduct-q30-bed23-hfd800
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8

## Question

Design a syphon-aqueduct with the following data: Canal: discharge 30 m³/s; bed width 23 m; depth of water 1.7 m; bed level 230.00; side slope 1.5:1. Drainage: high flood discharge 800 m³/s; high flood level 231.25; bed level 227.60; general ground level 230.00.

## Short answer

Drainage waterway: 14 spans of 10 m, clear 140 m (Lacey's P=134.4 m). Canal flumed to 10 m wide trough, depth 1.7 m; velocity 1.765 m/s, friction loss 0.081 m. Total head loss 0.189 m. Uplift on barrel roof equals HFL minus soffit elevation minus slab weight.

## Answer

The canal ($Q$ = 30 m³/s) crosses a drainage of $Q$ = 800 m³/s. The design covers the type of work, the drainage waterway, the canal waterway (flumed trough) with the bed and water levels of the trough, the siphon barrels with the afflux, and the uplift on the barrel roof.

**Assumptions:** Side slope 1.5:1 as given; Lacey's $f$ = 1; trough $n$ = 0.015; slab 0.6 m thick, concrete 24 kN/m³.

#### 1. Type of structure
The HFL of the drainage (231.25 m) is 1.25 m above the canal bed (230.00 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), so a **siphon aqueduct** is suitable. Drainage depth at HFL = 231.25 − 227.60 = 3.65 m.

#### 2. Drainage waterway
Lacey's regime perimeter (for a natural stream):

$$
P = 4.75\sqrt{Q} = 4.75\sqrt{800} = 134.4\ \text{m}
$$

Provide **14 spans of 10 m** with piers 1.5 m thick: clear waterway = 14 × 10 = **140 m** (> 134.4 m). Overall length = 140 + 13 × 1.5 = **159.5 m**.

Check: mean velocity = Q/(L·d) = 800/(140 × 3.65) = 1.57 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 4.36 m (with $f$ = 1); the foundations of piers and abutments are taken about 1.5R = 6.5 m below HFL, with cut-offs and aprons for scour.

#### 3. Canal waterway (flumed rectangular trough)
Normal canal: area $A_1=(B+zD)D$ = (23 + 1.5 × 1.7) × 1.7 = 43.44 m², $V_1$ = 30/43.44 = 0.691 m/s, $h_{v1}=V_1^2/2g$ = 0.0243 m. Top width $T=B+2zD$ = 28.10 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = **10 m**, with the same full supply depth of 1.7 m (the trough is as long as the drainage waterway, 159.5 m).
- Trough area = 10 × 1.7 = 17.00 m²; $V_t$ = 30/17.00 = 1.765 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1587 m
- Froude number = 0.43 < 1, so the flow is subcritical and there is no hydraulic jump.
- Hydraulic radius $R=A/P$ = 17.00/13.40 = 1.269 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000510 (Manning's n = 0.015); friction loss in the trough $h_f=S\,L$ = 0.000510 × 159.5 = 0.081 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 = 28.10 − 10 = **18.1 m**
- Outlet length = 3 × (T − B_t)/2 = **27.2 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):

$$
\begin{aligned} W_2 &= W_1 + h_{v1} - h_{vt} - 0.3(h_{vt}-h_{v1}) \\ W_3 &= W_2 - h_f \\ W_4 &= W_3 + h_{vt} - h_{v1} - 0.5(h_{vt}-h_{v1}) \end{aligned}
$$

$W_1$ = 231.700 m, so $W_2$ = 231.525 m, $W_3$ = 231.444 m, $W_4$ = 231.511 m. Trough bed = water level − 1.7 m.

| Section | Location | Bed level (m) | FSL (m) |
|---|---|---|---|
| 1 | Normal canal, start of inlet transition | 230.000 | 231.700 |
| 2 | End of inlet transition = start of trough | 229.825 | 231.525 |
| 3 | End of trough = start of outlet transition | 229.744 | 231.444 |
| 4 | End of outlet transition = normal canal | 229.811 | 231.511 |

Total loss of head = 231.700 − 231.511 = 0.189 m, so the canal bed downstream of the work is 0.189 m lower than upstream.

Check: with a trough slab 0.6 m thick, the soffit of the barrels = 229.14 m, i.e. 1.54 m above the drainage bed; HFL stands 2.11 m above the soffit, which gives the siphonic head.

#### 5. Siphon barrels and afflux
With the natural bed (227.60 m), the clear height of the barrels would be only 1.54 m and the barrel velocity 800/(140 × 1.54) = 3.7 m/s, which is too high. The barrel floor is therefore depressed to 226.84 m (0.76 m below the drainage bed, with 1 in 4 slopes at both ends), giving a clear height of 2.3 m: 14 barrels of 10 m × 2.3 m, barrel length (along the canal) $L$ = 10 + 1.2 = 11.2 m (trough width plus side walls).

- Area = 140 × 2.3 = 322 m²; $V$ = 800/322 = 2.48 m/s; $V^2/2g$ = 0.315 m
- Hydraulic radius of one barrel $R$ = (10 × 2.3)/(2(10 + 2.3)) = 0.935 m
- Unwin's formula with $f_1$ = 0.505 and $f_2=0.00316\,(1+0.1/R)$ = 0.00350:

$$
h_L=\left(1+0.505+0.00350\times\frac{11.2}{0.935}\right)\times0.315 = 1.547\times0.315 = 0.49\ \text{m}
$$

**Afflux** ≈ 0.49 m, so the upstream HFL = 231.25 + 0.49 = **231.74 m**; the upstream flood banks and the canal banks near the work are raised to this level plus free board.

#### 6. Uplift on the barrel roof (trough floor)
Severe case, full static head at the upstream end: $p$ = 9.81 × (231.74 − 229.14) = 25.5 kN/m².
- Slab weight = 24 × 0.6 = 14.4 kN/m²; with the canal running, add 9.81 × 1.7 = 16.7 kN/m², total 31.1 kN/m² > 25.5 kN/m² (safe).
- With the canal empty, net uplift = 25.5 − 14.4 = 11.1 kN/m². The roof slab is designed as a reinforced slab spanning between the piers for this uplift and is anchored to them.

**Answer:** drainage waterway 140 m clear (14 × 10 m, overall 159.5 m); canal flumed to a 10 m wide trough; bed/FSL at the four sections = 230.00/231.70, 229.83/231.53, 229.74/231.44, 229.81/231.51 m; barrels 10 m × 2.3 m with afflux ≈ 0.49 m; net uplift on the roof ≈ 11.1 kN/m² with the canal empty.

## Check yourself

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

1. Why is a siphon aqueduct selected here?
   A. HFL (231.25 m) is 1.25 m above the canal bed (230.00 m), so the drain is passed under the trough through pressure barrels
   B. The canal bed is well above HFL
   C. The drainage discharge is small
   D. The canal has to be siphoned over a hill
   Answer: A. With HFL above the canal bed, a trough above HFL would be too high and costly, hence barrels flowing under pressure. (Key point: Type of structure)

2. What waterway is adopted for the 800 m^3/s drainage?
   A. P = 134.4 m; 10 spans of 10 m = 100 m
   B. P = 4.75 sqrt(800) = 134.4 m; 14 spans of 10 m = 140 m clear, overall 159.5 m
   C. P = 134.4 m; 14 spans of 10 m with no piers, overall 140 m
   D. P = 4.75 x 800 = 3800 m
   Answer: B. Clear waterway must exceed Lacey's perimeter 134.4 m; 14 x 10 = 140 m plus 13 piers of 1.5 m gives 159.5 m. (Key point: Drainage waterway)

3. In the 10 m trough with depth 1.7 m, what is the velocity and is it acceptable?
   A. V = 0.69 m/s, too low
   B. V = 3.0 m/s, requiring a jump
   C. V = 30/17 = 1.765 m/s, acceptable (< 2.5 m/s) and subcritical
   D. V = 1.765 m/s but supercritical
   Answer: C. Trough area = 10 x 1.7 = 17 m^2, giving 1.765 m/s with Froude number 0.43, so flow is subcritical. (Key point: Trough hydraulics)

4. What is the total loss of head through the work and its effect?
   A. 0.081 m, equal to friction only
   B. Zero, as levels return to the start
   C. 0.189 m, so the downstream bed is raised
   D. 231.700 - 231.511 = 0.189 m, so the downstream canal bed is 0.189 m lower
   Answer: D. W4 = 231.511 m is lower than W1 = 231.700 m, so the downstream canal bed is lowered by 0.189 m. (Key point: Level changes and head loss)

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