2. Hydraulic data
2.1 Model A
Model A is represented in Fig.l. Hydraulic data and savings are described next for a value of the wall height H equal to 4 m and are shown in Fig 3.
For other values of H, the specific flows and relevant saving should be multiplied by (H/4) 1.5 , and the depth saving by H/4.
The length of elements is 12 m, the width of an element is 2 x 2.40 = 4.80 m. The ratio N between the total wall length (2 x 12 m + 4.8) and the spillway length 4.8 m is 6.
N = 6, H = 4 m, L = 12m, Width = 4.80 m
For H = 4 m.
The required quantity of reinforced concrete is close to 4 m3 per metre of spillway. The increase of specific flow is 11 m3/s, that is 1,4 H 1.5
The nappe depth saving is 1.7 m, that is, 0.42 H.
These savings are significantly reduced for nappe depths of less than 1 m (or 0.25 H) if this solution is used for multiplying a Creager weir flow by a coefficient R higher than 4 (for N = 6).
It is possible to increase the values N and R if narrowing the elements for the same length and depth. However, the cost increase is roughly proportional to the increase of N and the increase in the saving is progressively reduced. It will usually be advisable to use values of N lower than 10 (for R lower than 7.) It is also possible to increase the width of the elements. For instance, if it is increased from 4.80 to 8 m, N will become 4, the savings and costs will be reduced by about 30 per cent and the maximum value of R is limited to 2.5.
It is thus possible to choose the width according to the required value of R, N being higher than 1.5 R.
For model A, for the same overall width of the inlet and outlet, it is also possible for the same cost to increase inlet width by 20 per cent and to reduce, accordingly, the outlet width: the savings are increased by 5 per cent as compared with the figures above. |