Siphon aqueduct: Q 20, drainage 200

The canal ($Q$ = 20 m³/s) crosses a drainage of $Q$ = 200 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: Bed level of canal = FSL − depth = 151.50 − 1.50 = 150.00 m; Lacey's $f$ = 1; trough $n$ = 0.015; slab 0.6 m, concrete 24 kN/m³.

1. Type of structure

HFL of the drainage (150.70 m) is above the canal bed (150.00 m) by 0.70 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 = 150.70 − 148.50 = 2.20 m.

2. Drainage waterway

Lacey's regime perimeter (for a natural stream):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{200} = 67.2\ \text{m}$$

Provide 9 spans of 8 m with piers 1.2 m thick: clear waterway = 9 × 8 = 72 m (> 67.2 m). Overall length = 72 + 8 × 1.2 = 81.6 m.

Check: mean velocity = Q/(L·d) = 200/(72 × 2.20) = 1.26 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 2.75 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 4.1 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (12 + 1.5 × 1.5) × 1.5 = 21.38 m², $V_1$ = 20/21.38 = 0.936 m/s, $h_{v1}=V_1^2/2g$ = 0.0446 m. Top width $T=B+2zD$ = 16.50 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 7 m, with the same full supply depth 1.5 m (the trough is as long as the drainage waterway, 81.6 m).

• Trough area = 7 × 1.5 = 10.50 m²; $V_t$ = 20/10.50 = 1.905 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1849 m
• Froude number = 0.50 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 10.50/10.00 = 1.050 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000765 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000765 × 81.6 = 0.062 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 = 16.50 − 7 = 9.5 m
• Outlet length = 3 × (T − B_t)/2 = 14.2 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):

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

$W_1$ = 151.500 m, so $W_2$ = 151.318 m, $W_3$ = 151.255 m, $W_4$ = 151.325 m. Trough bed = water level − 1.5 m.

Total loss of head = 151.500 − 151.325 = 0.175 m, so the canal bed downstream of the work is 0.175 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 149.16 m, i.e. 0.66 m above the drainage bed; HFL stands 1.54 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 147.66 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Uplift on the roof (trough slab)

Upward pressure under the slab (barrel running full) = γ_w (HFL − soffit) = 9.81 × (150.70 − 149.16) = 15.2 kN/m².

Downward load = slab weight 24 × 0.6 = 14.4 kN/m² plus canal water 9.81 × 1.5 = 14.7 kN/m², total 29.1 kN/m².

Uplift exceeds the slab weight but not the total load: net downward = 14.0 kN/m² with canal running. With the canal dry the net uplift is 0.8 kN/m², which the slab and its anchorage to the piers must resist.

Answer: drainage waterway 72 m clear (9 × 8 m, overall 81.6 m); canal flumed to 7 m wide trough; bed/FSL at the four sections = 150.00/151.50, 149.82/151.32, 149.76/151.26, 149.83/151.33 m.