Select type of CD structure: Q 32, HFD 303

The canal ($Q$ = 32 m³/s) crosses a drainage of $Q$ = 303 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); drainage bed = HFL − 2.5 m = 207.5 m; Lacey's $f$ = 1; trough $n$ = 0.015. Depth of canal = FSL − bed = 1.5 m.

1. Type of structure

Canal bed level (212.00 m) is higher than HFL of the drainage (210.00 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):

$$P = 4.75\sqrt{Q} = 4.75\sqrt{303} = 82.7\ \text{m}$$

Provide 11 spans of 8 m with piers 1.2 m thick: clear waterway = 11 × 8 = 88 m (> 82.7 m). Overall length = 88 + 10 × 1.2 = 100.0 m.

Check: mean velocity = Q/(L·d) = 303/(88 × 2.50) = 1.38 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.16 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 4.7 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 × 1.5) × 1.5 = 33.38 m², $V_1$ = 32/33.38 = 0.959 m/s, $h_{v1}=V_1^2/2g$ = 0.0469 m. Top width $T=B+2zD$ = 24.50 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, 100.0 m).

• Trough area = 12 × 1.5 = 18.00 m²; $V_t$ = 32/18.00 = 1.778 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1611 m
• Froude number = 0.46 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 18.00/15.00 = 1.200 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000558 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000558 × 100.0 = 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 = 24.50 − 12 = 12.5 m
• Outlet length = 3 × (T − B_t)/2 = 18.8 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):

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

$W_1$ = 213.500 m, so $W_2$ = 213.352 m, $W_3$ = 213.296 m, $W_4$ = 213.353 m. Trough bed = water level − 1.5 m.

Total loss of head = 213.500 − 213.353 = 0.147 m, so the canal bed downstream of the work is 0.147 m lower than upstream.

Check: soffit of trough slab (0.6 m thick) = 211.20 m; HFL + free board 0.6 m = 210.60 m. Clearance adequate.

Sketch (section along the drainage)

   FSL 213.50  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
   bed 212.00  _____________________________   canal trough
               |_____________________________|  slab 0.6 m
               ||  ||  ||  ||  ||  ||  ||  ||  piers
   HFL 210.00  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  drainage flow
   bed 207.50  _______________________________
               <-- 11 spans of 8 m, 100 m -->

Answer: drainage waterway 88 m clear (11 × 8 m, overall 100.0 m); canal flumed to 12 m wide trough; bed/FSL at the four sections = 212.00/213.50, 211.85/213.35, 211.80/213.30, 211.85/213.35 m.