Siphon aqueduct: Q 20, FSL 251.5

The canal ($Q$ = 20 m³/s) crosses a drainage of $Q$ = 200 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 afflux and head loss through the siphon barrels, and the uplift on the roof slab.

Assumptions: Canal bed width is not given and is taken as 12 m with side slope 1.5:1; bed level = 251.5 − 1.5 = 250.0 m; 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 (250.70 m) is 0.70 m above the canal bed (250.00 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 = 250.70 − 248.50 = 2.20 m.

2. (b) 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); the foundations of piers and abutments are taken about 1.5R = 4.1 m below HFL, with cut-offs and aprons for scour.

3. (a) 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 of 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, so the flow is subcritical and 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 the 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):

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

$W_1$ = 251.500 m, so $W_2$ = 251.318 m, $W_3$ = 251.255 m, $W_4$ = 251.325 m. Trough bed = water level − 1.5 m.

Total loss of head = 251.500 − 251.325 = 0.175 m, so the canal bed downstream of the work is 0.175 m lower than upstream.

Check: with a trough slab 0.6 m thick, the soffit of the barrels = 249.16 m, only 0.66 m above the drainage bed, while 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 (see below).

c) Afflux and head loss through the siphon barrels

Soffit of barrel = trough bed − slab = 249.16 m. Provide barrels of clear height 1.5 m (barrel floor at 247.66 m, i.e. the drainage bed is depressed by 0.84 m under the structure, with 1 in 4 slopes at the ends): 9 barrels of 8 m, total clear width 72 m, barrel length (along the canal) $L$ = 7 + 1.2 = 8.2 m.

• Area = 72 × 1.5 = 108 m²; $V$ = 200/108 = 1.852 m/s; $V^2/2g$ = 0.1748 m
• Hydraulic radius of one barrel $R$ = (8 × 1.5)/(2(8+1.5)) = 0.632 m
• Unwin's formula: $h_L=\left(1+f_1+f_2\dfrac{L}{R}\right)\dfrac{V^2}{2g}$ with $f_1$ = 0.505 and $f_2=0.00316\,(1+0.1/R)$ = 0.00366

$$\begin{aligned} h_L &= \left(1+0.505+0.00366\times\frac{8.2}{0.632}\right)\times0.1748 \\ &= 1.5525\times0.1748 = 0.271\ \text{m} \end{aligned}$$

This head is lost between upstream and downstream, so the afflux = 0.27 m. Upstream HFL = 250.70 + 0.27 = 250.97 m, and the upstream flood banks/guide bunds are raised accordingly.

d) Uplift on the drainage (roof) slab

Taking the full static head above the soffit (the severe case): $p$ = 9.81 × (250.70 − 249.16) = 15.2 kN/m².

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

With the canal running, the net downward load = 29.1 − 15.2 = 13.9 kN/m² (safe against uplift). With the canal empty, the net uplift = 15.2 − 14.4 = 0.8 kN/m², which is resisted by reinforcement and by anchoring the slab to the piers.

Answer: drainage waterway 72 m clear (9 × 8 m, overall 81.6 m); canal flumed to a 7 m wide trough; bed/FSL at the four sections = 250.00/251.50, 249.82/251.32, 249.76/251.26, 249.83/251.33 m; head loss through the barrels (afflux) ≈ 0.27 m; uplift on the roof slab ≈ 15.2 kN/m² (net uplift 0.8 kN/m² with the canal empty).