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Glacis fall: crest, cistern, HGL; Q 118

Short answer

Flumed crest width 46.5 m, crest level 208.43 m, head 1.304 m, crest width 3.0 m. Cistern depth 0.25 m, floor level 205.75 m, length 8.28 m. Total impervious floor length 38 m. HGL drops from 209.70 m at u/s end to 206.85 m at d/s pile, then to 206.00 m exit.

Design crest elements, cistern elements and draw HGL line of a designed glacis fall on a canal waterway with the following data: Full supply discharge = 118 cumec; Full supply level of canal (U/S) = 209.7 m; Full supply level of canal (D/S) = 208.2 m; Safe exit gradient of canal material = 1/6; Canal bed level (U/S) = 207.5 m; Canal bed level (D/S) = 206.0 m; Canal bed width (U/S & D/S) = 62 m; Fluming ratio = 75%.

A glacis fall passes the flow over a raised crest, down a sloping glacis, and into a cistern in which a hydraulic jump forms. The canal is flumed (contracted) at the crest to reduce the cost and to increase the discharge intensity, so that the jump is more stable. Here the canal width is 62 m and the fluming ratio is 75 %.

Data

Q=118Q = 118 m³/s, FSL u/s 209.70209.70 m and d/s 208.20208.20 m, bed u/s 207.50207.50 m and d/s 206.00206.00 m, so Du=Dd=2.2D_u = D_d = 2.2 m, bed width 62 m, fluming ratio 75 %, safe exit gradient 1/6.

1. Crest elements

  • Fluming: throat width L=0.75×62=46.5L = 0.75\times62 = 46.5 m.
  • Approach velocity: v=11862×2.2=0.865v = \dfrac{118}{62\times2.2} = 0.865 m/s, ha=0.0381h_a = 0.0381 m.
  • Head over a broad crest: Q=1.705 L H3/2Q = 1.705\,L\,H^{3/2}
H=(1181.705×46.5)2/3=1.304 mH = \left(\frac{118}{1.705\times46.5}\right)^{2/3} = 1.304\ \text{m}
  • Crest level =FSLu+ha−H=209.70+0.038−1.304=208.43= FSL_u + h_a - H = 209.70 + 0.038 - 1.304 = 208.43 m, which is 0.93 m above the u/s bed.
  • Crest width: about 2H2H to 3H3H for a broad crest; adopt 3.0 m. U/s approach slope 1:1 (ramp), glacis slope 3:1 from the crest to the cistern.

2. Cistern elements

Discharge per metre of flumed crest q=118/46.5=2.538q = 118/46.5 = 2.538 m²/s. Energy above the d/s bed E1=209.70+0.038−206.00=3.738E_1 = 209.70 + 0.038 - 206.00 = 3.738 m. The conjugate depth without depression is 1.911 m, which is less than the tail water 2.20 m, so the jump is submerged and a nominal depression is adopted:

  • Cistern depth X=0.25X = 0.25 m; cistern floor level =206.00−0.25=205.75= 206.00 - 0.25 = 205.75 m.
  • y1=0.298y_1 = 0.298 m, F1=4.97F_1 = 4.97, y2=1.954y_2 = 1.954 m.
  • Cistern length Lc=5(y2−y1)=8.28L_c = 5(y_2 - y_1) = 8.28 m.
  • Fall height from crest to cistern floor =208.43−205.75=2.68= 208.43 - 205.75 = 2.68 m, so the horizontal length of the 3:1 glacis =3×2.68=8.05= 3\times2.68 = 8.05 m.

3. Floor length and HGL

Maximum static head Hs=209.70−206.00=3.70H_s = 209.70 - 206.00 = 3.70 m. Lacey's scour (f = 1): R=1.35(q2/f)1/3=2.51R = 1.35(q^2/f)^{1/3} = 2.51 m, so the d/s cut-off need only reach 204.43 m; adopt d=2.5d = 2.5 m (practical minimum) below the floor.

1πλ=16dHs=0.1126⇒λ=7.99, α=14.95, b=37.4 m\frac{1}{\pi\sqrt{\lambda}} = \frac{1}{6}\frac{d}{H_s} = 0.1126 \Rightarrow \lambda = 7.99,\ \alpha = 14.95,\ b = 37.4\ \text{m}

Provide a total impervious floor of 38 m.

HGL (Khosla) at the d/s pile: ϕE=23.0 %\phi_E = 23.0\,\% and ϕD=16.1 %\phi_D = 16.1\,\% of HsH_s. Pressure head levels:

PointResidual headHGL level (m)
U/s end of floor100 % of 3.70 m209.70
Junction at d/s pile23.0 %206.85
Bottom of d/s pile16.1 %206.60
D/s exit0206.00

The HGL is a straight line from 209.70 m at the u/s end to 206.85 m at the d/s pile, then falls to the exit.

 209.7 ~~~~~~~~~~
        \__ crest 208.43
 207.5 ___\___ glacis 3:1
              \_cistern_______ 206.0
          HGL: 209.7 ------------> 206.85
                                 |  d/s pile

Answer: crest level 208.43 m, crest width 3.0 m, flumed length 46.5 m; cistern depth 0.25 m, length 8.28 m; floor length 38 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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