Water Resources Engineering

Due: please use give example formulas for Part b and part c

b$)$ If the headloss over the spillway face and the end sill are ignored, the total energy loss between sections $(1)$ and $(4)$ will be only due to hydraulic jump

EGI

In the tailwater:

$q=\frac{400}{25}=16\frac{m^3}{s}\frac{?}{m},y_4=5m,u_4=\frac{16}{5}=3\mathrm{.}2\frac{m}{s}$

Since subcritical flow conditions exist in the tailwater, a hydraulic jump occurs.

$h_{a4}=0\mathrm{.}52m.$

$h_{24}=(3\mathrm{.}2{)}^2$longrightarrow Velocity head $at$ point $(4)$

Elevation of the total energy line at the tailwater:

$EG{L}_4=111+0\mathrm{.}52=111\mathrm{.}52m$

$E=EG{L}_1-EG{L}_4=116\mathrm{.}04-111\mathrm{.}52=4\mathrm{.}52m.$

$E=4\mathrm{.}52m$

$($:$E=\frac{(y_3-y_2{)}^3}{4y_3{y}_2},$ and $,y_3\}$

$8\frac{u_2^2}{g{y}_2}=8\frac{q^2}{g{v}_2^3}=\frac{208\mathrm{.}77}{y_2^3}$

$\frac{[\frac{y_2}{2}(\sqrt[\phantom{2}]{1+\frac{208\mathrm{.}77}{y_2^3}}-1)-y_2{]}^3}{2y_2^2(\sqrt[\phantom{2}]{1+\frac{208\mathrm{.}77}{y_2^3}}-1)}=4\mathrm{.}52m$

By trial and error, the value of $y_2$ is determined as $1\mathrm{.}15m.$ Then $y_3=6\mathrm{.}18m.$

Since $y_3>y_4,$ the river bed should be excavated by a certain depth of $$ whose value can be determined from the energy equation between sections $(3)$ and $(4)$

$y_3+\frac{q^2}{2g{y}_3^2}=+y_4+\frac{q^2}{2g{y}_4^2}$

from which one obtains $=1\mathrm{.}0m.$ Using $=1\mathrm{.}0m,$ the bottom elevation of the stilling basin is obtained as $106-1=105m$ :

c$)$ Type and dimensions of the stilling basin:

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$u_2=\frac{16}{1\mathrm{.}15}=13\mathrm{.}91\frac{m}{s}$

$F_{r2}=\frac{13\mathrm{.}91}{\sqrt[\phantom{2}]{9\mathrm{.}81**1\mathrm{.}15}}=4\mathrm{.}14$

Since and USBR type $4$ basin will $be$ designed. Characteristic dimensions $of$ USBR type $4$ basins are determined using the information given $in$ Figure $4\mathrm{.}29.$

Length $of$ basin:

$L_4=6\mathrm{.}1y_3={6\mathrm{.}1}^{**}6\mathrm{.}18=37\mathrm{.}7m.$

Height $of$ chute blocks above the stilling basin:

$h_1=2y_2=2\mathrm{.}30m.$

Height $of$ the end sill for USBR type $4$ basin $is1\mathrm{.}68m.$ However, the minimum value $of$ required $to$ confine the jump $at$ the spillway toe $is1\mathrm{.}0mas$ determined before. $So,=1\mathrm{.}0mis$ considered $tobe$ satisfactory for economic reasons.

Consider the overflow spillway shown in figure. The design spillway discharge is computed as $Q=180\frac{m^3}{s}.$ The total crest length of the spillway is $10m.$ The approach channel stilling basin and downstream section are of rectangular cross$-$section having $10m.$ There are two rounded nosed piers, each one$-$meter thick. Headwalls of rounded abutments makes angle less than $45$ with the flow direction. Ignoring the head losses over the spillway and over a possible downstream end sill, determine:

a$)$ The design spillway head;

b$)$ The required lowering of the tiver bed for the stilling basin; and

c$)$ The type and dimensions of the stilling basin.