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Weir design by Bligh's theory, 3200 m3/s

Short answer

Using Bligh's theory: waterway 270 m from Lacey's perimeter; crest level 182.35 m with shutters to pond level 184.5 m; upstream pile depth 2.5 m, downstream pile depth 6.5 m from scour depth; impervious floor length 66 m between piles (total creep 84 m); exit gradient 0.145 < 1/6, safe.

Determine the length of water way, crest level, depth of upstream and downstream pile and length of impervious floor for a vertical drop weir using Bligh's theory for the following data. And also check the exit gradient. Given data: maximum flood discharge = 3200 m³/sec; HFL before construction = 185.0 m; downstream bed level = 178.0 m; canal FSL = 184.0 m; allowable afflux = 1 m; coefficient of creep = 12; permissible exit gradient = 1/6; Assume head loss through head regulator is 0.5 m, weir coefficient is 1.7, depth of retrogression is 0.5 m and silt factor is 1.0.

Data and assumptions

Q=3200Q = 3200 m³/s, HFL before construction =185.0= 185.0 m, d/s bed =178.0= 178.0 m, canal FSL =184.0= 184.0 m, afflux =1.0= 1.0 m, C=12C = 12, f=1.0f = 1.0, safe exit gradient 1/6, regulator loss =0.5= 0.5 m, weir coefficient 1.71.7, retrogression 0.50.5 m. The u/s bed is assumed equal to the d/s bed (178.0 m).

1. Length of waterway (Lacey)

P=4.75Q=4.753200=268.7 m⇒adopt L=270 mP = 4.75\sqrt{Q} = 4.75\sqrt{3200} = 268.7\ \text{m} \Rightarrow \text{adopt } L = 270\ \text{m}

2. Levels and crest

  • Pond level == FSL ++ regulator loss =184.0+0.5=184.5= 184.0 + 0.5 = 184.5 m.
  • HFL after construction =185.0+1.0=186.0= 185.0 + 1.0 = 186.0 m.
  • Discharge per metre q=3200/270=11.85q = 3200/270 = 11.85 m³/s/m.
  • Head over the crest from Q=1.7 L H3/2Q = 1.7\,L\,H^{3/2}:
H=(32001.7×270)2/3=3.65 mH = \left(\frac{3200}{1.7\times 270}\right)^{2/3} = 3.65\ \text{m}
  • Crest level (open flood) =186.0−3.65=182.35= 186.0 - 3.65 = 182.35 m. The pond level of 184.5 m is held by shutters or gates of height 184.5−182.35=2.15184.5 - 182.35 = 2.15 m above this crest.

3. Scour depth and pile depths

R=1.35(q2f)1/3=1.35(11.8521.0)1/3=7.02 mR = 1.35\left(\frac{q^2}{f}\right)^{1/3} = 1.35\left(\frac{11.85^2}{1.0}\right)^{1/3} = 7.02\ \text{m}
  • U/s pile bottom =186.0−1.5R=175.47= 186.0 - 1.5R = 175.47 m, so depth below bed d1=178.0−175.47=2.53d_1 = 178.0 - 175.47 = 2.53 m, adopt 2.5 m.
  • D/s pile bottom =185.0−2R=170.96= 185.0 - 2R = 170.96 m. The bed after retrogression is 178.0−0.5=177.5178.0 - 0.5 = 177.5 m, so d2=177.5−170.96=6.54d_2 = 177.5 - 170.96 = 6.54 m, adopt 6.5 m.

4. Length of impervious floor (Bligh)

Maximum static head Hs=184.5−177.5=7.0H_s = 184.5 - 177.5 = 7.0 m. Total creep length

Lc=CHs=12×7.0=84 mL_c = C H_s = 12\times 7.0 = 84\ \text{m} Lc=2d1+2d2+b⇒b=84−2(2.53+6.54)=65.9 mL_c = 2d_1 + 2d_2 + b \Rightarrow b = 84 - 2(2.53 + 6.54) = 65.9\ \text{m}

Adopt a horizontal floor length b≈66b \approx 66 m between the piles.

5. Exit gradient check

α=bd2=65.96.54=10.08,λ=5.565,GE=Hsd2⋅1πλ=7.06.54×1π5.565=0.145\alpha = \frac{b}{d_2} = \frac{65.9}{6.54} = 10.08,\quad \lambda = 5.565,\quad G_E = \frac{H_s}{d_2}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{7.0}{6.54}\times\frac{1}{\pi\sqrt{5.565}} = 0.145

GE=0.145<1/6=0.167G_E = 0.145 < 1/6 = 0.167, so the floor is safe against piping.

Answer: Waterway ≈270\approx 270 m; crest level ≈182.35\approx 182.35 m (shutters to 184.5 m); u/s pile ≈2.5\approx 2.5 m deep; d/s pile ≈6.5\approx 6.5 m deep; impervious floor ≈66\approx 66 m between piles (total creep 84 m); exit gradient =0.145= 0.145 (safe).

✓ Answer checked by Claude Opus 5.5 (high effort)

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