---
title: "Syphonic aqueduct canal waterway levels, Q 40"
subject: "Irrigation and Drainage Engineering"
chapter: 8
key: "semester_6/irrigation_and_drainage_engineering:syphonic-aqueduct-levels-q40-hfd520"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/syphonic-aqueduct-levels-q40-hfd520
site: BCT Engineering
---

# Syphonic aqueduct canal waterway levels, Q 40

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

- Asked in: 2071 Bhadra, 8 marks
- Video: [Siphon Aqueduct | Cross Drainage Structure | Irrigation Engineering| IOE, TU](https://www.youtube.com/watch?v=VKfu13VWL2Y) — Parash Joshi - Civil Construction and Tutor
- 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/syphonic-aqueduct-levels-q40-hfd520
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-8

## Question

Determine bed and water levels at four critical locations of the canal water way at transition of Syphonic Aqueduct designed with the following data. Canal: Full supply discharge = 40 m³/s; Full supply level = 151.8 m; Side slope = 1.5:1; Depth of water = 1.5 m; Bed level = 150.00 m; Bed width = 32 m. Drainage: Maximum flood discharge = 520 m³/s; High flood level = 150.6 m; Bed level = 148.2 m; Normal ground level = 150.00 m.

## Short answer

Bed/water levels: normal canal: 150.00/151.50 m; start of trough: 149.83/151.33 m; end of trough: 149.76/151.26 m; normal canal outlet: 149.83/151.33 m. (Canal flumed to 15 m wide trough, trough length 125 m, friction loss 0.066 m.)

## Answer

At a syphonic aqueduct the canal is flumed into a rectangular trough over the drainage barrels. The levels follow from Bernoulli's equation between the normal canal and the trough, allowing losses of 0.3 Δh_v in the contraction (inlet) and 0.5 Δh_v in the expansion (outlet).

**Assumptions:** The data are inconsistent: FSL 151.8 m − bed 150.00 m = 1.8 m, but the depth is given as 1.5 m. The depth 1.5 m and bed 150.00 m are used, so the FSL is 151.50 m. (If instead the FSL 151.8 m and depth 1.5 m are taken as correct, the bed is at 150.30 m and every bed and water level in the table rises by 0.30 m.) Trough $n$ = 0.015.

#### Normal canal
$A_1=(B+zD)D=(32+1.5\times1.5)\times1.5$ = 51.38 m², $V_1$ = 40/51.38 = 0.779 m/s, $h_{v1}$ = 0.0309 m, FSL $W_1$ = 150.00 + 1.5 = 151.50 m.

#### Trough
Drainage waterway by Lacey: $P=4.75\sqrt{520}$ = 108.3 m, so the trough is about 125.0 m long (11 spans of 10 m, piers 1.5 m). Canal flumed to $B_t$ = 15 m, depth 1.5 m:
- $V_t$ = 40/(15 × 1.5) = 1.778 m/s, $h_{vt}$ = 0.1611 m, Froude number 0.46 (subcritical)
- $R$ = 1.250 m, $S=(nV/R^{2/3})^2$ = 0.000528 (n = 0.015), $h_f$ = 0.000528 × 125.0 = 0.066 m
- Inlet length (2:1 splay) = T − B_t = 36.50 − 15 = 21.5 m; outlet length (3:1 splay) = 32.3 m

#### Levels at the four sections

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

| Section | Location | Bed level (m) | Water level (m) |
|---|---|---|---|
| 1 | Normal canal, start of inlet transition | 150.000 | 151.500 |
| 2 | Start of trough (end of inlet) | 149.831 | 151.331 |
| 3 | End of trough (start of outlet) | 149.765 | 151.265 |
| 4 | Normal canal, end of outlet transition | 149.830 | 151.330 |

Total head lost in the work = 151.500 − 151.330 = 0.170 m. Trough bed = water level − 1.5 m.

Check: with a 0.6 m slab the barrel soffit is at 149.16 m, below the HFL of 150.60 m by 1.44 m, so the barrels run full (syphonic).

**Answer:** bed/water levels = 150.00/151.50, 149.83/151.33, 149.76/151.26, 149.83/151.33 m.

## Check yourself

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

1. The data give FSL 151.8 m, bed 150.00 m and depth 1.5 m, which conflict. How does the answer treat this?
   A. It takes depth 1.5 m as stated (FSL 151.50 m) and notes that using FSL 151.8 m would add 0.3 m to all water levels
   B. It takes FSL 151.8 m and changes the bed levels
   C. It ignores the depth and uses 1.8 m
   D. It rejects the problem as unsolvable
   Answer: A. Depth 1.5 m is used; bed levels are unchanged and a 0.3 m shift to water levels would apply if FSL 151.8 m were used. (Key point: Depth of water assumption)

2. With Bt = 15 m and depth 1.5 m, Q = 40 m3/s. What are Vt and the flow state?
   A. Vt = 1.778 m/s, Froude 0.46, subcritical
   B. Vt = 0.779 m/s, Froude 0.46, subcritical
   C. Vt = 2.67 m/s, supercritical
   D. Vt = 1.778 m/s, supercritical
   Answer: A. Vt = 40/(15 x 1.5) = 1.778 m/s, and Fr = V/sqrt(gD) = 0.46, so flow is subcritical. (Key point: Trough velocity and flow state)

3. Section 1 water level is 151.500 m and section 4 is 151.330 m. What is the total head lost and why?
   A. 0.170 m, lost to contraction, friction in the trough and expansion
   B. 0.331 m, equal to the velocity head in the trough
   C. Nil, since the canal returns to its original section
   D. 0.066 m, the friction loss only
   Answer: A. 151.500 - 151.330 = 0.170 m, the sum of the inlet loss, friction loss hf and the outlet loss net of velocity head recovery. (Key point: Total head lost)

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