Cross drainage: Q 25, HFD 360

The canal ($Q$ = 25 m³/s) crosses a drainage of $Q$ = 360 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 (211.00 m) is above the canal bed (210.30 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 = 211.00 − 208.50 = 2.50 m.

2. Drainage waterway

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

$$P = 4.75\sqrt{Q} = 4.75\sqrt{360} = 90.1\ \text{m}$$

Provide 12 spans of 8 m with piers 1.2 m thick: clear waterway = 12 × 8 = 96 m (> 90.1 m). Overall length = 96 + 11 × 1.2 = 109.2 m.

Check: mean velocity = Q/(L·d) = 360/(96 × 2.50) = 1.50 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.34 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.0 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 × 2) × 2 = 66.00 m², $V_1$ = 25/66.00 = 0.379 m/s, $h_{v1}=V_1^2/2g$ = 0.0073 m. Top width $T=B+2zD$ = 36.00 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 7 m, with the same full supply depth 2 m (the trough is as long as the drainage waterway, 109.2 m).

• Trough area = 7 × 2 = 14.00 m²; $V_t$ = 25/14.00 = 1.786 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1625 m
• Froude number = 0.40 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 14.00/11.00 = 1.273 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000520 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000520 × 109.2 = 0.057 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 = 36.00 − 7 = 29.0 m
• Outlet length = 3 × (T − B_t)/2 = 43.5 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):

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

$W_1$ = 212.300 m, so $W_2$ = 212.098 m, $W_3$ = 212.041 m, $W_4$ = 212.119 m. Trough bed = water level − 2 m.

Total loss of head = 212.300 − 212.119 = 0.181 m, so the canal bed downstream of the work is 0.181 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 209.44 m, i.e. 0.94 m above the drainage bed; HFL stands 1.56 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 207.94 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 96 m clear (12 × 8 m, overall 109.2 m); canal flumed to 7 m wide trough; bed/FSL at the four sections = 210.30/212.30, 210.10/212.10, 210.04/212.04, 210.12/212.12 m.