Cross drainage: Q 40, 45 cumec drainage

The canal ($Q$ = 40 m³/s) crosses a drainage of $Q$ = 45 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.

1. Type of structure

HFL of the drainage (207.00 m) is above the canal bed (206.40 m) by 0.60 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 = 207.00 − 204.50 = 2.50 m.

2. Drainage waterway

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

$$P = 4.75\sqrt{Q} = 4.75\sqrt{45} = 31.9\ \text{m}$$

Provide 6 spans of 6 m with piers 1.0 m thick: clear waterway = 6 × 6 = 36 m (> 31.9 m). Overall length = 36 + 5 × 1.0 = 41.0 m.

Check: mean velocity = Q/(L·d) = 45/(36 × 2.50) = 0.50 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 1.67 m (with $f$ = 1); 1.5R = 2.5 m below HFL gives RL 204.5 m, which is only at the natural bed and above the depressed floor under the barrels (204.11 m, see below), so scour does not govern; the foundations of piers and abutments are taken at least 1.5 m below the depressed floor (about RL 202.6 m), with cut-offs and aprons.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (30 + 1.5 × 1.6) × 1.6 = 51.84 m², $V_1$ = 40/51.84 = 0.772 m/s, $h_{v1}=V_1^2/2g$ = 0.0303 m. Top width $T=B+2zD$ = 34.80 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 14 m, with the same full supply depth 1.6 m (the trough is as long as the drainage waterway, 41.0 m).

• Trough area = 14 × 1.6 = 22.40 m²; $V_t$ = 40/22.40 = 1.786 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1625 m
• Froude number = 0.45 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 22.40/17.20 = 1.302 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000504 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000504 × 41.0 = 0.021 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 = 34.80 − 14 = 20.8 m
• Outlet length = 3 × (T − B_t)/2 = 31.2 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):

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

$W_1$ = 208.000 m, so $W_2$ = 207.828 m, $W_3$ = 207.807 m, $W_4$ = 207.874 m. Trough bed = water level − 1.6 m.

Total loss of head = 208.000 − 207.874 = 0.126 m, so the canal bed downstream of the work is 0.126 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 205.61 m, i.e. 1.11 m above the drainage bed; HFL stands 1.39 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 204.11 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 36 m clear (6 × 6 m, overall 41.0 m); canal flumed to 14 m wide trough; bed/FSL at the four sections = 206.40/208.00, 206.23/207.83, 206.21/207.81, 206.27/207.87 m.