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Forebay design with spillway length (14 m³/s)

Short answer

Volume = 2520 m³, area 70 m², depth 4.8 m; forebay 36 m×15 m×4.8 m. Spillway length L = Q/(C h^1.5) = 14/(1.7×0.5^1.5) ≈24 m for 0.5 m head.

Design a forebay for a power plant with a design discharge of 14 m³/s. The system uses two penstock pipes, each with a length of 500 m and diameter of 2 m, to convey water. Assume a detention time of 3 minutes and a flow velocity of 0.2 m/s in the forebay. Additionally, design an appropriate spillway length for the forebay.

Given: Q=14.0Q=14.0 m³/s; 2 penstocks of diameter 2.0 m and length 500 m; retention (detention) time 3 min; limiting velocity in the forebay 0.20 m/s.

Step 1: Volume of the forebay

The forebay stores the water for the retention time:

V=Q t=14.0×(3×60)=2520 m3V=Q\,t=14.0\times(3\times60)=2520\ \text{m}^3

Step 2: Cross-sectional area from the velocity limit

A=Qv=14.00.20=70.0 m2A=\frac{Q}{v}=\frac{14.0}{0.20}=70.0\ \text{m}^2

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

Velocity in each penstock: Vp=Q2×π4D2=2.228V_p=\dfrac{Q}{2\times\frac{\pi}{4}D^2}=2.228 m/s. Minimum submergence to avoid vortices (Gordon, with C=0.7245C=0.7245 for an asymmetrical approach, which is on the safe side):

S=0.7245 VpD=0.7245×2.228×2.0=2.28 mS=0.7245\,V_p\sqrt{D}=0.7245\times2.228\times\sqrt{2.0}=2.28\ \text{m}

Water depth = S+D+S+D+ bottom clearance of 0.5 m =2.28+2.0+0.5=4.78=2.28+2.0+0.5=4.78 m, adopt H=4.8H=4.8 m.

Step 4: Width and length

B=AH=70.04.8=14.58 m,L=VA=252070.0=36.0 mB=\frac{A}{H}=\frac{70.0}{4.8}=14.58\ \text{m},\qquad L=\frac{V}{A}=\frac{2520}{70.0}=36.0\ \text{m}

Adopt B=15.0B=15.0 m and L=36L=36 m (the volume is then B L H=2592B\,L\,H=2592 m³ ≥V\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.

Step 5: Spillway of the forebay

If the turbines trip, the whole discharge must spill over the forebay spillway (side-channel type) while the water level rises by h=0.5h=0.5 m:

Q=C Ls h3/2 ⇒ Ls=QC h3/2=14.01.7×0.53/2=23.3 mQ=C\,L_s\,h^{3/2}\ \Rightarrow\ L_s=\frac{Q}{C\,h^{3/2}}=\frac{14.0}{1.7\times0.5^{3/2}}=23.3\ \text{m}

(C=1.7C=1.7 for a broad-crested weir.) Adopt Ls=24L_s=24 m, placed along the side wall (or folded), with a spill channel to the river.

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
ItemValue
Storage volume2520 m³
Water depth4.8 m (+0.5 m free board)
Width15.0 m
Length36 m
Submergence of penstock crown2.28 m
Spillway length24 m (head 0.5 m)

Answer: forebay about 36 m long x 15.0 m wide x 4.8 m deep (water), with a volume of 2520 m³, and a spillway 24 m long.

✓ Answer checked by Claude Opus 5.5 (high effort)

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