Aqueduct: canal 35, drainage 350

The canal ($Q$ = 35 m³/s) crosses a drainage of $Q$ = 350 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: Flumed width of canal 12 m and depth in trough = canal depth 1.5 m (as given); Lacey's $f$ = 1; trough $n$ = 0.015; slab 0.6 m.

1. Type of structure

Canal bed level (198.50 m) is higher than HFL of the drainage (196.50 m) by 2.00 m, enough for free board and slab depth. The canal can be carried in a trough over the drain with free flow beneath, so an aqueduct is suitable.

2. Drainage waterway

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

$$P = 4.75\sqrt{Q} = 4.75\sqrt{350} = 88.9\ \text{m}$$

Provide 12 spans of 8 m with piers 1.2 m thick: clear waterway = 12 × 8 = 96 m (> 88.9 m). Overall length = 96 + 11 × 1.2 = 109.2 m.

Check: mean velocity = Q/(L·d) = 350/(96 × 3.00) = 1.22 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.31 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 $A_1=(B+zD)D$ = (22 + 1.5 × 1.5) × 1.5 = 36.38 m², $V_1$ = 35/36.38 = 0.962 m/s, $h_{v1}=V_1^2/2g$ = 0.0472 m. Top width $T=B+2zD$ = 26.50 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 12 m, with the same full supply depth 1.5 m (the trough is as long as the drainage waterway, 109.2 m).

• Trough area = 12 × 1.5 = 18.00 m²; $V_t$ = 35/18.00 = 1.944 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1927 m
• Froude number = 0.51 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 18.00/15.00 = 1.200 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000667 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000667 × 109.2 = 0.073 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 = 26.50 − 12 = 14.5 m
• Outlet length = 3 × (T − B_t)/2 = 21.8 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):

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

$W_1$ = 200.000 m, so $W_2$ = 199.811 m, $W_3$ = 199.738 m, $W_4$ = 199.811 m. Trough bed = water level − 1.5 m.

Total loss of head = 200.000 − 199.811 = 0.189 m, so the canal bed downstream of the work is 0.189 m lower than upstream.

Check: soffit of trough slab (0.6 m thick) = 197.64 m; HFL + free board 0.6 m = 197.10 m. Clearance adequate.

Sketch of designed aqueduct

 Longitudinal section along canal
 FSL 200.00 ~~~~\                    /~~~~ 199.81
 bed 198.50  ----\_ trough bed 198.31 _/----
        inlet  |  12 spans x 8 m, piers 1.2 m  |  outlet
     (wing walls)                    (wing walls)
 HFL 196.50 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 bed 193.50 -------------------------------
 Plan:  22 m -> 12 m (2:1 inlet) ... trough 109.2 m ... 12 m -> 22 m (3:1)

Answer: drainage waterway 96 m clear (12 × 8 m, overall 109.2 m); canal flumed to 12 m wide trough; bed/FSL at the four sections = 198.50/200.00, 198.31/199.81, 198.24/199.74, 198.31/199.81 m.