Cross drainage: Q 50, HFD 450, uplift on roof

The canal ($Q$ = 50 m³/s) crosses a drainage of $Q$ = 450 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; Lacey's $f$ = 1; trough $n$ = 0.015; roof (trough) slab 0.6 m thick, unit weight of concrete 24 kN/m³; barrel $n$ = 0.015; barrel length ≈ trough width + walls ≈ 20 m.

1. Type of structure

HFL of the drainage (100.50 m) is above the canal bed (100.00 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 = 100.50 − 98.80 = 1.70 m.

2. Drainage waterway

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

$$P = 4.75\sqrt{Q} = 4.75\sqrt{450} = 100.8\ \text{m}$$

Provide 11 spans of 10 m with piers 1.5 m thick: clear waterway = 11 × 10 = 110 m (> 100.8 m). Overall length = 110 + 10 × 1.5 = 125.0 m.

Check: mean velocity = Q/(L·d) = 450/(110 × 1.70) = 2.41 m/s. Lacey's regime depth $R = 0.47(Q/f)^{1/3}$ = 3.60 m (with $f$ = 1); foundations of piers and abutments are taken about 1.5R = 5.4 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 × 1.5) × 1.5 = 48.38 m², $V_1$ = 50/48.38 = 1.034 m/s, $h_{v1}=V_1^2/2g$ = 0.0545 m. Top width $T=B+2zD$ = 34.50 m.

To save cost, the canal is flumed to a rectangular trough of width $B_t$ = 19 m, with the same full supply depth 1.5 m (the trough is as long as the drainage waterway, 125.0 m).

• Trough area = 19 × 1.5 = 28.50 m²; $V_t$ = 50/28.50 = 1.754 m/s (acceptable, < 2.5 m/s); $h_{vt}$ = 0.1569 m
• Froude number = 0.46 < 1, flow is subcritical, so there is no hydraulic jump.
• Hydraulic radius $R=A/P$ = 28.50/22.00 = 1.295 m; friction slope $S=(nV/R^{2/3})^2$ = 0.000490 (Manning's n = 0.015); friction loss in trough $h_f=S\,L$ = 0.000490 × 125.0 = 0.061 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 = 34.50 − 19 = 15.5 m
• Outlet length = 3 × (T − B_t)/2 = 23.2 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):

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

$W_1$ = 101.500 m, so $W_2$ = 101.367 m, $W_3$ = 101.306 m, $W_4$ = 101.357 m. Trough bed = water level − 1.5 m.

Total loss of head = 101.500 − 101.357 = 0.143 m, so the canal bed downstream of the work is 0.143 m lower than upstream.

Check: with trough slab 0.6 m thick, soffit of the barrels = 99.21 m, i.e. 0.41 m above the drainage bed; HFL stands 1.29 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 97.71 m (1.5 m clear height) with a 1 in 4 slope at both ends.

Uplift on the roof (trough slab)

The critical case is the canal dry with the drainage at HFL. Barrels: 11 openings of 10 m × 1.5 m (bed depressed), area 165 m², $V_b$ = 450/165 = 2.73 m/s, $h_{vb}$ = 0.379 m; $R$ = 15/23 = 0.652 m.

$$h_{fb} = \frac{n^2V_b^2L}{R^{4/3}} = \frac{0.015^2 \times 2.73^2 \times 20}{0.652^{4/3}} = 0.059\ \text{m}$$

Upstream HFL = d/s HFL + entry loss ($0.5h_{vb}$) + exit loss ($1.0h_{vb}$) + $h_{fb}$ = 100.50 + 0.569 + 0.059 = 101.13 m (afflux 0.63 m).

The maximum uplift is just inside the entry (upstream end of the roof): pressure head = 101.128 − 1.5 × 0.379 − 99.206 (soffit) = 1.35 m, so the uplift pressure = 9.81 × 1.353 = 13.3 kN/m².

Downward load = slab weight 24 × 0.6 = 14.4 kN/m² plus canal water 9.81 × 1.5 = 14.7 kN/m², total 29.1 kN/m².

The uplift (13.3) is less than the slab weight alone (14.4), so the roof does not lift even with the canal dry; the net downward load with the canal running = 15.8 kN/m². The slab is reinforced for bending under these loads.

Answer: drainage waterway 110 m clear (11 × 10 m, overall 125.0 m); canal flumed to 19 m wide trough; bed/FSL at the four sections = 100.00/101.50, 99.87/101.37, 99.81/101.31, 99.86/101.36 m; maximum uplift on the roof 13.3 kN/m² (less than the 14.4 kN/m² slab weight).