Siphon aqueduct: Q 60, Manning 0.025

The canal ($Q$ = 60 m³/s) crosses a drainage of $Q$ = 550 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: Manning's $n$ = 0.025 (as given) is used for the friction loss in the trough; Lacey's $f$ = 1; piers 1.5 m thick.

1. Type of structure

HFL of the drainage (117.00 m) is above the canal bed (116.00 m) by 1.00 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). A siphon aqueduct is suitable. Drainage depth at HFL = 117.00 − 114.00 = 3.00 m.

2. Drainage waterway

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

$$P = 4.75\sqrt{Q} = 4.75\sqrt{550} = 111.4\ \text{m}$$

Provide 12 spans of 10 m with piers 1.5 m thick: clear waterway = 12 × 10 = 120 m (> 111.4 m). Overall length = 120 + 11 × 1.5 = 136.5 m.

Check: mean velocity = Q/(L·d) = 550/(120 × 3.00) = 1.53 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.85 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.8 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (30 + 1.5 × 1.4) × 1.4 = 44.94 m², $V_1$ = 60/44.94 = 1.335 m/s, $h_{v1}=V_1^2/2g$ = 0.0909 m. Top width $T=B+2zD$ = 34.20 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 24 m, with the same full supply depth 1.4 m (the trough is as long as the drainage waterway, 136.5 m).

• Trough area = 24 × 1.4 = 33.60 m²; $V_t$ = 60/33.60 = 1.786 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1625 m
• Froude number = 0.48 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 33.60/26.80 = 1.254 m; friction slope $S=(nV/R^{2/3})^2$ = 0.001474 (Manning's n = 0.025); friction loss in trough $h_f=S\,L$ = 0.001474 × 136.5 = 0.201 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 = 34.20 − 24 = 10.2 m
• Outlet length = 3 × (T − B_t)/2 = 15.3 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):

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

$W_1$ = 117.400 m, so $W_2$ = 117.307 m, $W_3$ = 117.106 m, $W_4$ = 117.141 m. Trough bed = water level − 1.4 m.

Total loss of head = 117.400 − 117.141 = 0.259 m, so the canal bed downstream of the work is 0.259 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 115.11 m, i.e. 1.11 m above the drainage bed; HFL stands 1.89 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 113.61 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 120 m clear (12 × 10 m, overall 136.5 m); canal flumed to 24 m wide trough; bed/FSL at the four sections = 116.00/117.40, 115.91/117.31, 115.71/117.11, 115.74/117.14 m.