Design of orifice type intake

Given: $Q=5$ m³/s; weir crest is 2.5 m above the river bed; canal water level is 0.3 m below the crest; intake velocity $\le0.8$ m/s; $C_d=0.6$; bar-contraction coefficient 0.03; trash-rack bar shape factor $\beta=1.83$.

Step 1: Net opening area from the velocity limit

$$A_{net}=\frac{Q}{V}=\frac{5}{0.8}=6.25\ \text{m}^2$$

Step 2: Gross area allowing for the bars

The bars reduce the opening by the contraction coefficient 0.03:

$$A_{gross}=\frac{A_{net}}{1-0.03}=\frac{6.25}{0.97}=6.44\ \text{m}^2$$

Step 3: Dimensions of the opening

Sill level 1.0 m above the river bed (to keep bed load out). Take the height of the orifice $h=1.2$ m so that its top is at 2.2 m above the bed, i.e. at the canal water level (0.3 m below the crest), so the opening stays submerged.

$$b=\frac{A_{gross}}{h}=\frac{6.44}{1.2}=5.37\ \text{m}\ \Rightarrow\ \text{adopt }5.4\ \text{m (2 bays of 2.7 m)}$$

Adopted opening area $A=1.2\times5.4=6.48$ m².

Step 4: Head required across the orifice (check)

$$Q=C_dA\sqrt{2g\,\Delta h}\ \Rightarrow\ \Delta h=\frac{1}{2g}\left(\frac{Q}{C_dA}\right)^2=\frac{1}{19.62}\left(\frac{5}{0.6\times6.48}\right)^2=0.084\ \text{m}$$

Available head difference $=0.3$ m $>0.084$ m, so the opening can pass the design flow; the extra head is absorbed by throttling the gate.

For comparison, the area that would pass 5 m³/s under the full 0.3 m head is $A=\dfrac{5}{0.6\sqrt{2g(0.3)}}=3.43$ m², so the velocity limit (6.25 m²) controls the size.

Step 5: Trash rack loss (Kirschmer)

Assume bars 10 mm thick at 50 mm clear spacing, inclined at $75^\circ$:

$$h_t=\beta\left(\frac{s}{b}\right)^{4/3}\frac{V^2}{2g}\sin\theta=1.83\left(\frac{10}{50}\right)^{4/3}\frac{0.8^2}{19.62}\sin75^\circ=0.0067\ \text{m}$$

 SECTION                         PLAN
 crest 2.5 ~~~~~~~~~~~~          river ==>
        |   |gate|  2.2         |rack|gate|--> canal
  rack  |## |    |~~~~ canal    | 5.4 m opening |
 sill   |___|____|              2 bays of 2.7 m
 1.0 m above bed

Answer: orifice opening $5.4\ \text{m}\times1.2$ m with sill 1.0 m above the bed (2 bays of 2.7 m); required head $0.084$ m < 0.3 m available; rack loss $\approx0.0067$ m.