Length of bottom intake rack

Given: intercepted flow $Q=8$ m³/s; rack width $B=10$ m; rack inclination $\beta=30^\circ$; bar thickness $s=10$ mm; clear spacing $a=15$ mm; contraction factor $\mu=0.82$.

The rack length is found from Frank's formula for a bottom (Tyrolean) intake, in which the depth of flow falls along the rack as the water is drawn through the bars:

$$L=\frac{2.561\,q}{\lambda\sqrt{h}},\qquad \lambda=\psi\,\mu\sqrt{2g\cos\beta},\qquad \psi=\frac{a}{a+s}$$

where $q$ is the discharge per metre width and $h$ the depth of flow at the upstream end of the rack. The flow is assumed to arrive at the rack at critical depth (over the weir crest), so $h=h_c$.

Step 1: Unit discharge and approach depth

$$q=\frac{Q}{B}=\frac{8}{10}=0.8\ \text{m}^2/\text{s},\qquad h=h_c=\left(\frac{q^2}{g}\right)^{1/3}=\left(\frac{0.8^2}{9.81}\right)^{1/3}=0.403\ \text{m}$$

Step 2: Opening ratio

$$\psi=\frac{a}{a+s}=\frac{15}{15+10}=0.60$$

Step 3: Rack coefficient

$$\lambda=\psi\mu\sqrt{2g\cos\beta}=0.60\times0.82\times\sqrt{2\times9.81\times\cos30^\circ}=0.492\times4.122=2.028\ \text{m}^{1/2}/\text{s}$$

Step 4: Length of the rack

$$L=\frac{2.561\times0.8}{2.028\times\sqrt{0.403}}=\frac{2.049}{1.287}=1.59\ \text{m}$$

Allow about 25% extra for partial clogging by gravel and leaves: $L=1.25\times1.59=1.99$ m. Adopt about 2.0 m.

Answer: required rack length $\approx1.6$ m; adopt $2.0$ m (rack 10 m wide, inclined at $30^\circ$).