Cross drainage: Q 50, HFD 550

The canal ($Q$ = 50 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: Side slope 1.5:1 (H:V), Lacey's $f$ = 1, concrete trough with Manning's $n$ = 0.015, piers 1.2–1.5 m thick, slab thickness 0.6 m.

1. Type of structure

HFL of the drainage (161.00 m) is above the canal bed (160.00 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 = 161.00 − 158.00 = 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.

Approach velocity at HFL = Q/(L·d) = 550/(120 × 3.00) = 1.53 m/s (the velocity through the barrels is checked at the end). 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 × 2.1) × 2.1 = 69.61 m², $V_1$ = 50/69.61 = 0.718 m/s, $h_{v1}=V_1^2/2g$ = 0.0263 m. Top width $T=B+2zD$ = 36.30 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 13 m, with the same full supply depth 2.1 m (the trough is as long as the drainage waterway, 136.5 m).

• Trough area = 13 × 2.1 = 27.30 m²; $V_t$ = 50/27.30 = 1.832 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1710 m
• Froude number = 0.40 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 27.30/17.20 = 1.587 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000408 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000408 × 136.5 = 0.056 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 = 36.30 − 13 = 23.3 m
• Outlet length = 3 × (T − B_t)/2 = 34.9 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):

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

$W_1$ = 162.100 m, so $W_2$ = 161.912 m, $W_3$ = 161.856 m, $W_4$ = 161.929 m. Trough bed = water level − 2.1 m.

Total loss of head = 162.100 − 161.929 = 0.171 m, so the canal bed downstream of the work is 0.171 m lower than upstream.

Check: with a trough slab 0.6 m thick, the soffit of the barrels = 159.756 − 0.6 = 159.16 m, only 1.16 m above the drainage bed, while HFL stands 1.84 m above the soffit, so the barrels run full under pressure. With 1.16 m clear height the barrel velocity would be 550/(120 × 1.16) = 3.95 m/s, which is too high. The drainage bed under the barrels is therefore depressed to RL 157.36 m (1.8 m clear height) with 1 in 4 slopes at both ends, giving a barrel velocity of 550/(120 × 1.8) = 2.55 m/s, which is acceptable.

Answer: drainage waterway 120 m clear (12 × 10 m, overall 136.5 m); canal flumed to 13 m wide trough; bed/FSL at the four sections = 160.00/162.10, 159.81/161.91, 159.76/161.86, 159.83/161.93 m.