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## Q. 3.18

An ogee spillway of large height is to be designed to evacuate a flood discharge of 200 m³/s under a head of 2 m. The spillway is spanned by piers to support a bridge deck above. The clear span between piers is limited to 6 m. Determine the number of spans required in order to pass the flood discharge with the head not exceeding 2 m. Assume the pier contraction coefficient $k_{p}$ = 0.01 and the abutment contraction coefficient $k_{a}$ = 0.10.

## Verified Solution

The flow between the piers and abutments is contracted, thus reducing the spillway width for the flow to $B_{e}$. Each pier has two end contractions and one abutment, and hence the effective width is given by

$B_{e} = B − 2(nk_{p} + k_{a})H_{e}$
n being the number of piers.

The pier contraction coefficient depends on the shape of its nose ($k_{p}$ = 0 for a pointed nose and $k_{p}$ = 0.02 for a square nose), whereas the abutment contraction coefficient may be as high as 0.2 for a square abutment, reducing to zero for a rounded abutment. If the velocity of approach $V_{a}$ is not negligible, a trial-and-error procedure is to be used for the discharge computations; for large heights (P), $V_{a}$ ≈ 0 and hence $H_{e}$ ≈ H. Here, assuming $V_{a}$ ≈ 0, we can write Equation 3.43 as

$Q=\frac{2}{3}C_{d0}B\sqrt{2g}H^{3/2}_{de}$             (3.43)

$Q=200=\frac{2}{3}C_{d0}\sqrt{(2g)}\left[6(n+1)-2(n\times 0.01+0.10)2.0\right]2^{3/2}$

which gives n = 4.36 with $C_{d0}$ = 0.75 (P/H > 3). Therefore round up to n = 5, and so provide five piers. Thus the clear span of the spillway (for flow) = 36 m. From the discharge equation we can now compute the corresponding head for this flow. In fact the spillway is capable of discharging a larger flood flow at the specified design head of 2 m. A stage (head)–discharge relationship can be established by using appropriate discharge coefficients (refer Figure 3.20).