Syphonic aqueduct canal waterway levels, Q 40

At a syphonic aqueduct the canal is flumed into a rectangular trough over the drainage barrels. The levels follow from Bernoulli's equation between the normal canal and the trough, allowing losses of 0.3 Δh_v in the contraction (inlet) and 0.5 Δh_v in the expansion (outlet).

Assumptions: The data are inconsistent: FSL 151.8 m − bed 150.00 m = 1.8 m, but the depth is given as 1.5 m. The depth 1.5 m and bed 150.00 m are used, so the FSL is 151.50 m. (If instead the FSL 151.8 m and depth 1.5 m are taken as correct, the bed is at 150.30 m and every bed and water level in the table rises by 0.30 m.) Trough $n$ = 0.015.

Normal canal

$A_1=(B+zD)D=(32+1.5\times1.5)\times1.5$ = 51.38 m², $V_1$ = 40/51.38 = 0.779 m/s, $h_{v1}$ = 0.0309 m, FSL $W_1$ = 150.00 + 1.5 = 151.50 m.

Trough

Drainage waterway by Lacey: $P=4.75\sqrt{520}$ = 108.3 m, so the trough is about 125.0 m long (11 spans of 10 m, piers 1.5 m). Canal flumed to $B_t$ = 15 m, depth 1.5 m:

• $V_t$ = 40/(15 × 1.5) = 1.778 m/s, $h_{vt}$ = 0.1611 m, Froude number 0.46 (subcritical)
• $R$ = 1.250 m, $S=(nV/R^{2/3})^2$ = 0.000528 (n = 0.015), $h_f$ = 0.000528 × 125.0 = 0.066 m
• Inlet length (2:1 splay) = T − B_t = 36.50 − 15 = 21.5 m; outlet length (3:1 splay) = 32.3 m

Levels at the four sections

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

Total head lost in the work = 151.500 − 151.330 = 0.170 m. Trough bed = water level − 1.5 m.

Check: with a 0.6 m slab the barrel soffit is at 149.16 m, below the HFL of 150.60 m by 1.44 m, so the barrels run full (syphonic).

Answer: bed/water levels = 150.00/151.50, 149.83/151.33, 149.76/151.26, 149.83/151.33 m.