Forebay design (14.5 m³/s, two penstocks)

Given: $Q=14.5$ m³/s; 2 penstocks of diameter 1.8 m; retention (detention) time 3 min; limiting velocity in the forebay 0.22 m/s.

Step 1: Volume of the forebay

The forebay stores the water for the retention time:

$$V=Q\,t=14.5\times(3\times60)=2610\ \text{m}^3$$

Step 2: Cross-sectional area from the velocity limit

$$A=\frac{Q}{v}=\frac{14.5}{0.22}=65.9\ \text{m}^2$$

Step 3: Depth of water (submergence of the penstock intake)

Velocity in each penstock: $V_p=\dfrac{Q}{2\times\frac{\pi}{4}D^2}=2.849$ m/s. Minimum submergence to avoid vortices (Gordon, with $C=0.7245$ for an asymmetrical approach, which is on the safe side):

$$S=0.7245\,V_p\sqrt{D}=0.7245\times2.849\times\sqrt{1.8}=2.77\ \text{m}$$

Water depth = $S+D+$ bottom clearance of 0.5 m $=2.77+1.8+0.5=5.07$ m, adopt $H=5.1$ m.

Step 4: Width and length

$$B=\frac{A}{H}=\frac{65.9}{5.1}=12.92\ \text{m},\qquad L=\frac{V}{A}=\frac{2610}{65.9}=39.6\ \text{m}$$

Adopt $B=13.0$ m and $L=40$ m (the volume is then $B\,L\,H=2652$ m³ $\ge V$).

PLAN
 headrace        forebay                 penstocks
 tunnel/canal  +------------------+
 ====> \       |                  |==> penstock 1
        \ trans| trash rack       |
 ====> /       |                  |==> penstock 2
              +--+-spillway-+-flush+
                   |          |
                   v          v  to river (spillway/flushing)

SECTION
 canal bed ____ WL ~~~~~~~~~~~~~~~~~~~~~~
                  |        S (submergence)
                  |   rack  ____ crown
                  |   ####  \   penstock
 flushing  ______ |___####____\_D__
 bottom clearance 0.5 m

Answer: forebay about 40 m long x 13.0 m wide x 5.1 m deep (water), with a required volume of 2610 m³.