Siphon aqueduct: Q 60, bed 35 m

The canal ($Q$ = 60 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: Lacey's $f$ = 1; trough $n$ = 0.015; slab 0.6 m thick.

1. Type of structure

HFL of the drainage (300.80 m) is above the canal bed (300.00 m) by 0.80 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 = 300.80 − 298.20 = 2.60 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: mean velocity = Q/(L·d) = 500/(110 × 2.60) = 1.75 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.73 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.6 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (35 + 1.5 × 2) × 2 = 76.00 m², $V_1$ = 60/76.00 = 0.789 m/s, $h_{v1}=V_1^2/2g$ = 0.0318 m. Top width $T=B+2zD$ = 41.00 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 17 m, with the same full supply depth 2 m (the trough is as long as the drainage waterway, 125.0 m).

• Trough area = 17 × 2 = 34.00 m²; $V_t$ = 60/34.00 = 1.765 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1587 m
• Froude number = 0.40 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 34.00/21.00 = 1.619 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000369 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000369 × 125.0 = 0.046 m

4. Transitions and levels

  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 = 41.00 − 17 = 24.0 m
• Outlet length = 3 × (T − B_t)/2 = 36.0 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):

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

$W_1$ = 302.000 m, so $W_2$ = 301.835 m, $W_3$ = 301.789 m, $W_4$ = 301.852 m. Trough bed = water level − 2 m.

Total loss of head = 302.000 − 301.852 = 0.148 m, so the canal bed downstream of the work is 0.148 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 299.19 m, i.e. 0.99 m above the drainage bed; HFL stands 1.61 m above the soffit, which gives the siphonic head. Since the clear height is less than 1.5 m, the drainage bed is depressed under the barrels to RL 297.69 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 110 m clear (11 × 10 m, overall 125.0 m); canal flumed to 17 m wide trough; bed/FSL at the four sections = 300.00/302.00, 299.83/301.83, 299.79/301.79, 299.85/301.85 m.