Forebay design (15 m³/s, 4 minutes)

Given: $Q=15.0$ m³/s; 1 penstock (assumed) of diameter 2.0 m and length 500 m; storage (retention) time 4 min; limiting velocity in the forebay 0.20 m/s (assumed, the usual limit, as it is not given).

Step 1: Volume of the forebay

The forebay stores the water for the retention time:

$$V=Q\,t=15.0\times(4\times60)=3600\ \text{m}^3$$

Step 2: Cross-sectional area from the velocity limit

$$A=\frac{Q}{v}=\frac{15.0}{0.20}=75.0\ \text{m}^2$$

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

Velocity in the penstock: $V_p=\dfrac{Q}{\frac{\pi}{4}D^2}=4.775$ m/s. Minimum submergence to avoid vortices (Gordon, with $C=0.7245$ for an asymmetrical approach, which is on the safe side):

$$S=0.7245\,V_p\sqrt{D}=0.7245\times4.775\times\sqrt{2.0}=4.89\ \text{m}$$

Water depth = $S+D+$ bottom clearance of 0.5 m $=4.89+2.0+0.5=7.39$ m, adopt $H=7.4$ m.

Step 4: Width and length

$$B=\frac{A}{H}=\frac{75.0}{7.4}=10.14\ \text{m},\qquad L=\frac{V}{A}=\frac{3600}{75.0}=48.0\ \text{m}$$

Adopt $B=10.5$ m and $L=48$ m (the volume is then $B\,L\,H=3730$ m³ $\ge V$).

The penstock length (500 m) is not needed to size the forebay by this method; it matters for the penstock head loss and the water hammer.

PLAN
 headrace        forebay                 penstock
 tunnel/canal  +------------------+
 ====> \       |                  |
        \ trans| trash rack       |==> penstock
 ====> /       |                  |
              +--+-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 48 m long x 10.5 m wide x 7.4 m deep (water), with a volume of 3600 m³.