Silt excluder design, Q 200 cumec

Given and assumptions

Canal discharge $Q = 200$ m³/s; undersluice crest level (tunnel floor) $= 225.25$ m; head regulator crest level $= 227.35$ m (tunnel roof level); bay width of the undersluices $= 10.2$ m.

Assumed: discharge through the tunnels = 20% of the canal discharge; roof slab thickness 0.3 m; tunnel velocity about 2.5 m/s (non-silting); pier thickness 0.6 m.

Design

1. Discharge in tunnels

$$Q_t = 0.20\times 200 = 40\ \text{m}^3/\text{s}$$

2. Height of tunnels

Difference of crest levels $= 227.35 - 225.25 = 2.10$ m. Subtracting the roof slab of $0.30$ m:

$$h = 2.10 - 0.30 = 1.80\ \text{m}$$

3. Area of tunnels

$$A = \frac{Q_t}{V} = \frac{40}{2.5} = 16.0\ \text{m}^2$$

4. Total width of tunnels

$$W = \frac{A}{h} = \frac{16.0}{1.80} = 8.89\ \text{m}$$

5. Number of tunnels. Provide 3 tunnels of 3.0 m each and 2 piers of 0.6 m:

$$\text{Total width} = 3\times 3.0 + 2\times 0.6 = 10.2\ \text{m}$$

which fits exactly in the bay width of 10.2 m. Area provided $= 3\times 3.0\times 1.80 = 16.20$ m².

6. Velocity check

$$V = \frac{40}{16.20} = 2.47\ \text{m/s}$$

This is very slightly below 2.5 m/s (about 1% less), which is practically acceptable; it lies within the usual tunnel velocity range of 2-3 m/s, so the silt will be carried through the tunnels.

Section

 regulator crest 227.35 -----------------------
        roof slab (0.30)  ===================
        tunnel height 1.80 |  |  |  |  |  |
 undersluice crest 225.25 -----------------------
 3 tunnels x 3.0 m + 2 piers x 0.6 m = 10.2 m

Answer: 3 tunnels, each 3.0 m wide and 1.80 m high (roof 227.35 m, floor 225.25 m), piers 0.6 m, carrying 40 m³/s at about 2.47 m/s.