Supply sluices in gravity dams -

Supply sluices in gravity dams –

Some openings in the dam have to be provid so as to pass the excess flow D/S of the dam.
These openings are know outlet sluices or supply sluices.
If water from the reservoir is to be release for irrigation purpose at a controll rate, it is done with the help of these supply sluices.
All the supply sluices are fitted with gates which can be raise or lowere.
The control on the gates is exercise from the top of the dam.
The gates are fitted in the grooves form at the sides of the openings in the dam.
When gates are lowere they stop flow of water and when raised they again start discharging water D/S.
Sluices may be provided at more than one depth.
By this, water can be drawn from different

elevations or depths.
Supply sluices are also sometimes used to scour out the silt deposit in the vicinity of the U/S of the dam.

Example.

Determine the equations for base width of an elementary profile of gravity dam so that resultant passes through the outer middle third points.
Consider earthquake force, hydrostatic pressure and uplift pressure for computations.

Solution. (a) Vertical forces

\[Force- due -to- self -weight -of -dam-W= \frac{1}{2}bhw_{\rho }\]

This force acts downwards.

II. Force due to vertical acceleration of earthquake

\[F= \alpha W= \alpha bhw_{\rho }\]

This force acts upwards.

III. Force due to uplift u= 1/2 bwh

This force also acts upwards.

Hence, total vertical force = ΣV = W – α W – U

\[= \frac{1}{2}bhw_{\rho }-\alpha bhw-\frac{1}{2}bhw\]

\[= \frac{1}{2}bhw[(1-\alpha )p-1]\]

where

(b) Horizontal forces

\[Force-due- to -self- weight -of- dam- W =\frac{1}{2}wh^{2}\]

II. Force due to hydrodynamic pressure of water at base

\[C_{m}= 0.735\frac{\theta }{90}= 0.735 since \theta = 90^{0}\]

\[P_{e}= C_{m}\alpha wh= 0.735\times \alpha wh\]

\[P_{e}= 0.726p_{e}h\]

\[M_{e}= 0.299p_{e}h^{2}= 0.299\times 0.725\alpha wh^{3}= 0.2205\alpha wh^{3}\]

III. Inertia force horizontal = αw = 1 /2 α bhwp

If the resultant of all the forces have to pass through the outer third point M2, the moment of all these forces at this point must be zero

\[\Sigma V\times \frac{B}{3}= \frac{1}{2}WH^{2}\times \frac{H}{3}+0.2205\alpha WH^{3}+\frac{1}{2}\alpha BHW_{\rho }\times \frac{h}{3}\]

\[\frac{B^{2}hw}{6}[(1-\alpha )\rho -1]= \frac{bh^{2}w\rho \alpha }{6}+\frac{wh^{3}}{6}[1+\alpha (1.323)]\]

\[b^{2}[(1-\alpha )p-1]= bhp.\alpha +h^{2}[1+1.323\alpha ]\]

Solving this equation for b we get

\[b= h\frac{p\alpha \pm \sqrt{p^{2}\alpha ^{2}+4(1+1.323\alpha )(1-\alpha )p-1}}{2{(1-\alpha )-1}}\]

This is the required general equation for b.

When there is no earthquake, value of α = 0 and the equation for b reduces to the form

\[b= \frac{h}{\sqrt{p-1}}\]

Er. Banwari Lal