Cross drainage: Q 50, HFD 400, d 1.55 m

The canal ($Q$ = 50 m³/s) crosses a drainage of $Q$ = 400 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: Side slope not given, taken 1.5:1; Lacey's $f$ = 1; trough $n$ = 0.015.

1. Type of structure

HFL of the drainage (300.50 m) is above the canal bed (300.00 m) by 0.50 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 = 300.50 − 298.80 = 1.70 m.

2. Drainage waterway

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

$$P = 4.75\sqrt{Q} = 4.75\sqrt{400} = 95.0\ \text{m}$$

Provide 12 spans of 8 m with piers 1.2 m thick: clear waterway = 12 × 8 = 96 m (> 95.0 m). Overall length = 96 + 11 × 1.2 = 109.2 m.

Check: mean velocity = Q/(L·d) = 400/(96 × 1.70) = 2.45 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.46 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.2 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 × 1.55) × 1.55 = 50.10 m², $V_1$ = 50/50.10 = 0.998 m/s, $h_{v1}=V_1^2/2g$ = 0.0508 m. Top width $T=B+2zD$ = 34.65 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 18 m, with the same full supply depth 1.55 m (the trough is as long as the drainage waterway, 109.2 m).

• Trough area = 18 × 1.55 = 27.90 m²; $V_t$ = 50/27.90 = 1.792 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1637 m
• Froude number = 0.46 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 27.90/21.10 = 1.322 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000498 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000498 × 109.2 = 0.054 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 = 34.65 − 18 = 16.6 m
• Outlet length = 3 × (T − B_t)/2 = 25.0 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):

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

$W_1$ = 301.550 m, so $W_2$ = 301.403 m, $W_3$ = 301.349 m, $W_4$ = 301.405 m. Trough bed = water level − 1.55 m.

Total loss of head = 301.550 − 301.405 = 0.145 m, so the canal bed downstream of the work is 0.145 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 299.20 m, i.e. 0.40 m above the drainage bed; HFL stands 1.30 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 297.70 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 18 m wide trough; bed/FSL at the four sections = 300.00/301.55, 299.85/301.40, 299.80/301.35, 299.86/301.41 m.