## Hydrotechnical calculation for Outlet piping

Outlet pipe:
It is used to drain excess water from the dam, if necessary, to empty the dam. The outlet pipe is at the deepest point. The drain pipe they lead outside the dam. Or the drain pipe they lead directly into the base of the dam.
In the case of a pressure pipe, a strong pressure drop can result in a sudden or gradual expansion of the cross section (in the narrower cross section) and in the curved parts of the pipeline of the most curved water fibers. With the gradual expansion of the pipe, the absolute pressure in the narrower section of the cross section is closer to zero, and thus there is a greater risk of cavitation than in the suddenly expanded pipe. In practice, however, zero pressure in the pipeline cannot arise because water is excreted out of the water when the pressure drops, the gases contained in the water and forming water vapor.
Each outlet has at least two valves: operational and revision. To achieve complete safety, an emergency valve is placed, which is permanently in the open position and is designed to quickly close the piping in the most critical cases. Trying to construct an operating valve is to achieve the best mechanical and hydraulic properties not only in the fully open position but also in all positions. Machine requirements are mainly: a simple construction of a fully secure valve control. Hydraulically, the water flow at all valve positions should be smooth and undisturbed with the least pressure drop, and to avoid cavitation phenomena and associated unpleasant shocks and shocks to make the shape of the outlet water stream contribute to the destruction of the energy flowing out of the water stream and achieved a practical watertight valve in the closed position.

Pipeline flow:

$Q=\mu F\sqrt{2gh_0};\ d\le d_p$
where:
 $Q$ pipeline flow $\mathrm{m^3/s}$ $\mu$ output coefficient $\mathrm{-}$ $F$ the flow area of the output hole $\mathrm{m^2}$ $g$ gravitational acceleration $\mathrm{m/s^2}$ $h_0$ rated net head (hydrostatic pressure) $\mathrm{m}$ $d$ diameter of the output hole $\mathrm{m}$ $d_p$ pipe diameter $\mathrm{m}$

Output coefficient:

$\mu=\cfrac{1}{\sqrt{1+\sum\xi}}$
where:
 $\mu$ output coefficient $\mathrm{-}$ $\sum\xi$ sum of partial pressure losses $\mathrm{-}$

The flow area of the output hole:

$F=\cfrac{\pi d^2}{4}$
where:
 $F$ the flow area of the output hole $\mathrm{m^2}$ $d$ diameter of the output hole $\mathrm{m}$

Sum of partial pressure losses:
It is the sum of all losses in the outlet, from the trash rack, in the inlet to the pipeline, the friction of water from the wall, the water passage through the valve etc.

Example 1:
We have to determine the capacity of the outlet steel pipe with the following parameters:
rated net head (hydrostatic pressure) $h_0=24\ \mathrm{m}$; diameter of the output hole (operational valve diameter) $d=2\ \mathrm{m}$; steel pipe length $l=22\ \mathrm{m}$; loss in trash rack $ξ1=0,1$; loss in the inlet $ξ2=0,25$; loss in the revision valve $ξ3=0,24$; loss in operational valve $ξ4=0,67$.

Loss of water friction on pipe walls:

$\xi_5=\lambda\cdot\cfrac{l}{d}=0,026\cdot\cfrac{22}{2}=0,286$

Sum of partial pressure losses:

$\sum\xi=\xi_1+\xi_2+\xi_3+\xi_4+\xi_5=0,1+0,25+0,24+0,67+0,286=1,546$

The flow area of the output hole:

$F=\cfrac{\pi d^2}{4}=\cfrac{\pi2^2}{4}=3,142\ \mathrm{m^2}$

Output coefficient

$\mu=\cfrac{1}{\sqrt{1+\sum\xi}}=\cfrac{1}{\sqrt{1+1,5468}}=0,6278$

Pipeline flow:

$Q=\mu F\sqrt{2gh_0}=0,627\cdot3,142\cdot\sqrt{2\cdot9,81\cdot24}=42,724\ \mathrm{m^3/s}$
$d\le d_p; 2\le2$

$\rightarrow$ does suit

Example 2:
We have to determine the capacity of the outlet steel pipe with the following parameters:
rated net head (hydrostatic pressure) $h_0=24\ \mathrm{m}$; diameter of the output hole (operational valve diameter) $d=2\ \mathrm{m}$; steel pipe length $l=22\ \mathrm{m}$; loss in trash rack $ξ1=0,1$; loss in the inlet $ξ2=0,25$; loss in the revision valve $ξ3=0,24$; loss in operational valve, which is open to 50% stroke $ξ4=1,38$.

Loss of water friction on pipe walls:

$\xi_5=\lambda\cdot\cfrac{l}{d}=0,026\cdot\cfrac{22}{2}=0,286$

Sum of partial pressure losses:

$\sum\xi=\xi_1+\xi_2+\xi_3+\xi_4+\xi_5=0,1+0,25+0,24+1,38+0,286=2,256$

The flow area of the output hole:

$F=\cfrac{\pi d^2}{4}=\cfrac{\pi2^2}{4}=3,142\ \mathrm{m^2}$

Output coefficient:

$\mu=\cfrac{1}{\sqrt{1+\sum\xi}}=\cfrac{1}{\sqrt{1+2,256}}=0,554$

Pipeline flow:

$Q=\mu F\sqrt{2gh_0}=0,554\cdot3,142\cdot\sqrt{2\cdot9,81\cdot24}=37,78\ \mathrm{m^3/s}$
$d\le d_p; 2\le2$

$\rightarrow$ does suit

Example 3:
We have to determine the capacity of the outlet steel pipe with the following parameters:rated net head (hydrostatic pressure) $h_0=24\ \mathrm{m}$; diameter of the output hole (operational valve diameter) $d=2\ \mathrm{m}$; steel pipe length $l=22\ \mathrm{m}$ and diameter $d_p=3\ \mathrm{m}$; loss in trash rack $ξ1=0,1$; loss in the inlet $ξ2=0,25$; loss in the revision valve $ξ3=0,24$; loss in operational valve $ξ4=0,67$. The revision and operating valve is located at the end of the outlet about diameter $2\ \mathrm{m}$. before the revision valve, there is a tapered pipe that has a loss $ξ5=0,15$.

Loss of water friction on pipe walls:

$\xi_6=\lambda\cdot\cfrac{l}{d_p}=0,026\cdot\cfrac{22}{3}=0,191$

Sum of partial pressure losses:

$\sum\xi=\xi_1\left(\cfrac{d^2}{d_p^2}\right)^2+\xi_2\left(\cfrac{d^2}{d_p^2}\right)^2+\xi_3+\xi_4+\xi_5\left(\cfrac{d^2}{d_p^2}\right)^2+\xi_6\left(\cfrac{d^2}{d_p^2}\right)^2=0,1\left(\cfrac{2^2}{3^2}\right)^2+0,25\left(\cfrac{2^2}{3^2}\right)^2+0,24+0,67+0,15\left(\cfrac{2^2}{3^2}\right)^2+0,191\left(\cfrac{2^2}{3^2}\right)^2=1,046$

The flow area of the output hole:

$F=\cfrac{\pi d^2}{4}=\cfrac{\pi2^2}{4}=3,142\ \mathrm{m^2}$

Output coefficient:

$\mu=\cfrac{1}{\sqrt{1+\sum\xi}}=\cfrac{1}{\sqrt{1+1,046}}=0,699$

Pipeline flow:

$Q=\mu F\sqrt{2gh_0}=0,699\cdot3,142\cdot\sqrt{2\cdot9,81\cdot24}=47,654\ \mathrm{m^3/s}$
$d\le d_p;2\le3$

$\rightarrow$ does suit

Literature: