Syphonic aqueduct levels, Q 36

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: A drainage bed width of 14 m is far too narrow to carry 400 m³/s at a depth of 2.4 m, so the given 14 m is taken as the flumed width of the canal trough; trough $n$ = 0.015; Lacey's waterway is used only to find the trough length.

Normal canal

$A_1=(B+zD)D=(28+1.5\times1.5)\times1.5$ = 45.38 m², $V_1$ = 36/45.38 = 0.793 m/s, $h_{v1}$ = 0.0321 m, FSL $W_1$ = 210.40 + 1.5 = 211.90 m.

Trough

Drainage waterway by Lacey: $P=4.75\sqrt{400}$ = 95.0 m, so the trough is about 109.2 m long (12 spans of 8 m, piers 1.2 m). Canal flumed to $B_t$ = 14 m, depth 1.5 m:

• $V_t$ = 36/(14 × 1.5) = 1.714 m/s, $h_{vt}$ = 0.1498 m, Froude number 0.45 (subcritical)
• $R$ = 1.235 m, $S=(nV/R^{2/3})^2$ = 0.000499 (n = 0.015), $h_f$ = 0.000499 × 109.2 = 0.054 m
• Inlet length (2:1 splay) = T − B_t = 32.50 − 14 = 18.5 m; outlet length (3:1 splay) = 27.8 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 = 211.900 − 211.751 = 0.149 m. Trough bed = water level − 1.5 m.

Check: with a 0.6 m slab the barrel soffit is at 209.59 m, below the HFL of 211.00 m by 1.41 m, so the barrels run full (syphonic).

Answer: bed/water levels = 210.40/211.90, 210.25/211.75, 210.19/211.69, 210.25/211.75 m.