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Canal head regulator design, Q 200

Short answer

Crest level 301.20 m; waterway 25 m clear (5 bays of 5 m), overall 31.0 m; cut-offs u/s 2.5 m, d/s 3.6 m; impervious floor length 38 m.

Design a canal Head Regulator for the data given below. Crest level of under sluice = 300 m; Pond level = 304 m; Silt factor (f) = 1; U/S HFL = 307 m; Full Supply discharge = 200 m³/s; Full Supply level = 303 m; Bed level of canal = 299.50 m. A silt excluder is provided in the under sluice; Take GE = 1/6.

Data and assumptions

  • Discharge Q=200Q = 200 m³/s; pond level =304= 304 m; full supply level of canal =303= 303 m; canal bed =299.5= 299.5 m; u/s HFL =307= 307 m; Lacey's f=1f = 1; safe exit gradient GE=1/6G_E = 1/6.
  • Crest of the head regulator is kept 1.2 m above the crest of the under-sluice (300.00 m): 300.00+1.2=301.20300.00 + 1.2 = 301.20 m, so bed load is excluded.
  • The regulator is designed as a broad-crested weir for the free-flow condition, Q=1.705 L H3/2Q = 1.705\,L\,H^{3/2}.

1. Crest level

Crest level=301.20 m\text{Crest level} = 301.20\ \text{m}

Head over the crest at pond level: H=304−301.20=2.80H = 304 - 301.20 = 2.80 m. Depth of canal water over the crest: FSL−crest=1.80FSL - \text{crest} = 1.80 m, which is less than 23H\tfrac{2}{3}H, so the weir is not drowned and the free-flow formula applies.

2. Waterway

L=Q1.705 H3/2=2001.705×2.803/2=25.04 mL = \frac{Q}{1.705\,H^{3/2}} = \frac{200}{1.705\times2.80^{3/2}} = 25.04\ \text{m}

Provide 5 bays of 5.0 m clear width (25 m) with 4 piers 1.5 m thick; overall width =31.0= 31.0 m. Check: Q=1.705×25×2.803/2=199.7Q = 1.705\times25\times2.80^{3/2} = 199.7 m³/s, which is accepted as ≈200\approx 200 m³/s.

3. Impervious floor (Khosla's exit gradient)

Discharge intensity q=Q/width=200/31.0=6.45q = Q/\text{width} = 200/31.0 = 6.45 m²/s. Lacey's scour depth:

R=1.35(q2f)1/3=1.35(6.4521)1/3=4.68 mR = 1.35\left(\frac{q^2}{f}\right)^{1/3} = 1.35\left(\frac{6.45^2}{1}\right)^{1/3} = 4.68\ \text{m}

The d/s cut-off is taken down to 1.5R1.5R below the d/s water level: bottom level =303−1.5×4.68=295.98= 303 - 1.5\times4.68 = 295.98 m, so the d/s cut-off depth below the floor (canal bed) is d=299.5−295.98d = 299.5 - 295.98, adopt d=3.6d = 3.6 m. U/s cut-off d1=2.5d_1 = 2.5 m. Maximum static head (canal dry, water at pond level): Hs=304−299.5=4.5H_s = 304 - 299.5 = 4.5 m. The gates are closed in floods; the HFL (307 m) fixes the top of the piers and gates. Exit gradient:

GE=Hsd 1πλ=16⇒1πλ=3.66×4.5=0.1333G_E = \frac{H_s}{d}\,\frac{1}{\pi\sqrt{\lambda}} = \frac{1}{6} \Rightarrow \frac{1}{\pi\sqrt{\lambda}} = \frac{3.6}{6\times4.5} = 0.1333 λ=5.70⇒α=bd=10.35⇒b=10.35×3.6=37.3 m\lambda = 5.70 \Rightarrow \alpha = \frac{b}{d} = 10.35 \Rightarrow b = 10.35\times3.6 = 37.3\ \text{m}

(Using λ=1+1+α22\lambda = \frac{1+\sqrt{1+\alpha^2}}{2}.) Check: GE=0.1667=1/6.00G_E = 0.1667 = 1/6.00. Provide a floor of 38 m: about 1/3 on the u/s side and 2/3 downstream, thickness 1.5 m under the gates tapering to 0.8 m.

4. Other features

  • Top of piers and gate: HFL + 1.0 m free-board. Vertical lift steel gates with a hoist deck.
  • D/s of the floor: inverted filter and loose stone apron (length about 2D2D), and wing walls splayed 45°.
  • The silt excluder in the under-sluice keeps bed load away, so the regulator crest only needs to be 1.2 m above the under-sluice crest.

Summary

ItemValue
Crest level301.20 m
Waterway25 m clear (5 bays of 5 m), overall 31.0 m
Cut-offs u/s / d/s2.5 m / 3.6 m
Floor length38 m

Conceptual sketch (longitudinal section):

 HFL / pond ~~~~~~~~~~|gates|~~~~~~ canal FSL ~~~~~
 river        ________|piers|________
 bed ________|  crest  (1.2 m above   |_____ canal bed
     |d1 u/s floor     U/S sluice crest) d/s floor | d2
       (impervious)                     inverted filter
       <------------ b (Khosla, GE = 1/6) --------->

✓ Answer checked by Claude Opus 5.5 (high effort)

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