# 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.

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Shopping Cart