Cross drainage (flumed): canal 30, drainage 200

The canal ($Q$ = 30 m³/s) crosses a drainage of $Q$ = 200 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: Canal side slope 1.5:1 (not given); levels are downstream levels, so the work is solved backwards from the downstream bed 200.00 m; ground level taken equal to canal bed; $n$ = 0.014 for the concrete trough as given; Lacey's $f$ = 1.

1. Type of structure

HFL of the drainage (201.00 m) is above the canal bed (200.00 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 = 201.00 − 198.00 = 3.00 m.

2. Drainage waterway

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

$$P = 4.75\sqrt{Q} = 4.75\sqrt{200} = 67.2\ \text{m}$$

Provide 9 spans of 8 m with piers 1.2 m thick: clear waterway = 9 × 8 = 72 m (> 67.2 m). Overall length = 72 + 8 × 1.2 = 81.6 m.

Check: mean velocity = Q/(L·d) = 200/(72 × 3.00) = 0.93 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 2.75 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 4.1 m below HFL, with cut-offs and aprons for scour.

3. Canal waterway (flumed rectangular trough)

Normal canal: area $A_1=(B+zD)D$ = (20 + 1.5 × 2) × 2 = 46.00 m², $V_1$ = 30/46.00 = 0.652 m/s, $h_{v1}=V_1^2/2g$ = 0.0217 m. Top width $T=B+2zD$ = 26.00 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 8 m, with the same full supply depth 2 m (the trough is as long as the drainage waterway, 81.6 m).

• Trough area = 8 × 2 = 16.00 m²; $V_t$ = 30/16.00 = 1.875 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1792 m
• Froude number = 0.42 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 16.00/12.00 = 1.333 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000470 (Manning's n = 0.014); friction loss in trough $h_f=S\,L$ = 0.000470 × 81.6 = 0.038 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 = 26.00 − 8 = 18.0 m
• Outlet length = 3 × (T − B_t)/2 = 27.0 m

Bed level at the downstream canal is fixed (200.00 m). Working backwards with Bernoulli's equation (loss in contraction = 0.3 Δh_v, in expansion = 0.5 Δh_v):

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

$W_1$ = 202.164 m, so $W_2$ = 201.960 m, $W_3$ = 201.921 m, $W_4$ = 202.000 m. Trough bed = water level − 2 m.

Total loss of head = 202.164 − 202.000 = 0.164 m, so the canal bed downstream of the work is 0.164 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 199.32 m, i.e. 1.32 m above the drainage bed; HFL stands 1.68 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 197.82 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Answer: drainage waterway 72 m clear (9 × 8 m, overall 81.6 m); canal flumed to 8 m wide trough; bed/FSL at the four sections = 200.16/202.16, 199.96/201.96, 199.92/201.92, 200.00/202.00 m.