Fish belly flap gate –
- This gate is show in Fig. 14.26.
- This gate is also know as the Bascule type of gate.
- This is use at the top of the weir crest to store extra stormwater.
- The gate is fitt on the crest with the help of a hinged joint. It is operated with the help of a lever rod.
Example.1
- The Head of water over the crest of the ogee spillway is 3 m and the coefficient of discharge 2.5.
- Weir is 100 m long and the height of the crest above the base of the approach channel is 10 m.
- The width of the approach channel is equal to the length of the weir.
- Find out the discharge passing over the spillway.
Solution.
\[Q= CLH^{3/2}\]
\[Q= 2.5\times 100\times 3^{3/2}= 1300 cumec.\]
Velocity of approach
\[V_{a}= \frac{Q}{Head\times width.of.channel}\]
\[= \frac{1300}{(10+3)100}= 1m\]
\[H_{a}= \frac{V_{a}^{2}}{2g}= \frac{(1)^{2}}{2\times 9.81}= 0.05m\]
\[H= h+H_{a}= 3+0.05= 3.05m.\]
\[Modified- dischargeQ= 2.5\times 100\times (3.05)^{3/2}\]
=1330 cumecs.
Example. 2
- Find out the discharge of a siphon spillway from following data:
- Number of siphon units = 4 Area at throat in m2 = 3
- Full reservoir level = 150 m R.L. of centre of outlet = 128
- Tailwater level on D/S side during rains = 130 m
- Tailwater level during winter = 125 m
- Discharge coefficient = 0.60 Solution.
Case 1.
- In rainy days outlet remains submerged and hence discharge depends upon the Tailwater level.
Working head = R.L. of reservoir – R.L. or T.W.L.
= 150 – 130 = 20 m
\[Q= CA\sqrt{2gh}\]
\[0.60\times 3\sqrt{2g\times 20}\]
Discharge of Four units = 4 × 35.75 = 143 cumecs.
Case II.
In winter T.W.L. falls down and spillways discharge free in the air.
Available head = Reservoir level – R.L. of centre of outlet
= 150 – 128 = 22 m.
\[Q= CA\sqrt{2gh}\]
\[= 0.60\times 3\times \sqrt{2\times 9.81\times 22}\]
= 37.5 cumecs.
Discharge of four units = 4 × 37.5 = 152 cumecs.
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