Weir waterway and crest levels, 16000 m3/s

Assumptions

Afflux $= 1.0$ m, loss through the head regulator $= 0.5$ m, weir coefficient $C = 1.7$ ($Q = CLH^{3/2}$), average bed level $= 200.0$ m, undersluice discharge intensity about 10 m³/s/m. Velocity of approach is neglected.

1. Waterway (Lacey)

$$P = 4.75\sqrt{16000} = 600.8\ \text{m} \Rightarrow L = 600\ \text{m}$$

Discharge intensity $q = 16000/600 = 26.67$ m³/s/m; scour depth $R = 1.35(q^2/f)^{1/3} = 12.05$ m.

2. Levels

• HFL before construction $= 205.7$ m; HFL after construction $= 205.7 + 1.0 = 206.7$ m.
• Pond level $=$ FSL $+$ regulator loss $= 203.0 + 0.5 = 203.5$ m.
• Head needed to pass the flood over the full waterway: $H = (16000/(1.7\times 600))^{2/3} = 6.27$ m.
• Weir (open-flood) crest level $= 206.7 - 6.27 = 200.43$ m, say 200.4 m (only about 0.4 m above the bed). The structure is therefore a gated barrage: gates (3.1 m high) hold the pond at 203.5 m.
• Undersluice crest level $=$ river bed $=$ 200.0 m.
• Head regulator crest level $=$ undersluice crest $+$ 1.5 m $=$ 201.5 m (the canal FSL of 203.0 m gives 1.5 m depth over the crest).

3. Waterway distribution

Undersluice discharge $= 2\times$ canal discharge $= 400$ m³/s, so the width $= 400/10 = 40$ m, taken as 5 bays of 8 m with 4 piers of 1.5 m: $40 + 6 = 46$ m. Weir bays: $600 - 46 = 554$ m.

Head regulator: head over its crest $=$ pond level $-$ crest $= 203.5 - 201.5 = 2.0$ m, so

$$L = \frac{200}{1.7\times 2.0^{3/2}} = 41.6\ \text{m}$$

Provide about 42 m clear waterway (more if the partly drowned flow is allowed for).

Answer: Total waterway 600 m (undersluices 46 m, weir 554 m); weir crest $\approx 200.4$ m (gates raise the pond to 203.5 m); undersluice crest 200.0 m; head regulator crest 201.5 m with about 42 m clear waterway.