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Forebay design considerations; design for 12 m³/s

Short answer

Required volume = 12 × 180 = 2160 m³; limiting velocity 0.2 m/s gives area 60 m². Submergence 1.96 m, water depth 4.5 m. Forebay 36 m long × 13.5 m wide × 4.5 m deep.

What are the design considerations of a forebay? Design a forebay with turbine discharge 12 m³/sec, water is conveyed from the forebay to the powerhouse by two penstocks of 2 m diameter each. Take retention time 3 minutes and limiting velocity 0.2 m/sec.

Design considerations of a forebay

  • Storage: enough volume for the retention time (2-3 min) to meet a sudden load increase until the headrace flow adjusts.
  • Low velocity (0.2-0.3 m/s), so that the remaining sediment settles and the flow at the intake is calm.
  • Adequate submergence of the penstock intake below the minimum water level to avoid vortices and air entry (Gordon's rule), with a trash rack and a smooth bell-mouth.
  • Spillway able to pass the design flow on load rejection, with a safe discharge channel.
  • Flushing/desilting arrangement at the bottom (flushing gate and bed slope) and a bottom clearance of about 0.5 m.
  • Stable foundation, water-tightness (lining), 0.5 m free board and access for maintenance; located on firm ground close to the powerhouse and protected from slides.

Design for the given data

Given: Q=12.0Q=12.0 m³/s; 2 penstocks of diameter 2.0 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=12.0×(3×60)=2160 m3V=Q\,t=12.0\times(3\times60)=2160\ \text{m}^3

Step 2: Cross-sectional area from the velocity limit

A=Qv=12.00.20=60.0 m2A=\frac{Q}{v}=\frac{12.0}{0.20}=60.0\ \text{m}^2

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

Velocity in each penstock: Vp=Q2×π4D2=1.910V_p=\dfrac{Q}{2\times\frac{\pi}{4}D^2}=1.910 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=0.7245×1.910×2.0=1.96 mS=0.7245\,V_p\sqrt{D}=0.7245\times1.910\times\sqrt{2.0}=1.96\ \text{m}

Water depth = S+D+S+D+ bottom clearance of 0.5 m =1.96+2.0+0.5=4.46=1.96+2.0+0.5=4.46 m, adopt H=4.5H=4.5 m.

Step 4: Width and length

B=AH=60.04.5=13.33 m,L=VA=216060.0=36.0 mB=\frac{A}{H}=\frac{60.0}{4.5}=13.33\ \text{m},\qquad L=\frac{V}{A}=\frac{2160}{60.0}=36.0\ \text{m}

Adopt B=13.5B=13.5 m and L=36L=36 m (the volume is then B L H=2187B\,L\,H=2187 m³ ≥V\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
ItemValue
Storage volume2160 m³
Water depth4.5 m (+0.5 m free board)
Width13.5 m
Length36 m
Submergence of penstock crown1.96 m

Answer: forebay about 36 m long x 13.5 m wide x 4.5 m deep (water), with a volume of 2160 m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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