Problem $2(30$ points$)$

Rework example $5\mathrm{.}4$ below that we solved during class time if the length of the river reach is $L=700m$

instead of $L=500m.$ Also included below are some of the formulas and the Table used in class to solve

the problem for $L=500m.$ When solving for $L=700m,$ assume the free surface elevation at the upstream

section is $21\mathrm{.}2$ m $($first guess$).$

Example $5-4.$ Determine the change in the river stage in a $500-$it$-$long river reach for a

flow of $3000\frac{m^3}{s}.$ The cross section of the river in this reach consists of two subsections, a

main$-$charnel section and an overbank$-$channel section, both of which are approsimately

rectangular in section. The properties of each subsection at the downstream end of the reach

are $B_{n1}=150m$;${}_{a1}=15\mathrm{.}0m$;$B_z=300m$; and $z_z=18\mathrm{.}0m$; and at the upstream end of the

reach are $B_n=170m,t_m=15\mathrm{.}2m,B_n=250m,$ and $z_n=18\mathrm{.}5m,$ where the subscripts $m$ and

o tefer to the main channel and overbank channel, respectively. The values of the Manring

n for the main and overbank channels are $0\mathrm{.}03$ and $0\mathrm{.}05.$ respectively. The river Mage at the

downstrean section is $20\mathrm{.}5$ inl.

The compututions are summarized in Tables$-3.$

Explomory Remarks on Table $5-3$

The flow parameter are computed separitely for ench subsection, A trial stage

$h=21\mathrm{.}00m$ in column $3$ is assumed of $L=-500m.$ Le$.,$ in section $2.$ The $K$ values in

column $7$ are computed from Eq $.(2-20).$ The energy coefticient $$ in column $9$ is oblained

from Eq$,(3-41).$ The values of tolal head $H$ in section $2$ are computed using Eqs. $(5-17)$ and

$(5-18)$ und are listed in colamns $11$ und $16,$ respectively. For the assumed value of

$h=21,00m$ the two values of the total head in section $2,21\mathrm{.}26m$ fin column $11$ and

$21\mathrm{.}13$ m in column $16,$ are not the same. Another value of $h=20\mathrm{.}87m,$ which is defermined

from Eq$.(5-20),$ is wied, and it yields the sume value of the rotal head in section $2.$ The new

trial value is computed as follows.

For $\{$:$h_2^2=5\mathrm{.}5(m),ft{y}_2)$ from $Eq.(5-19)$ is

$f(y_2^{*})=H_a(y_2)-H_{a1}\frac{+}{b}ar(S{)}_{j}x=21\mathrm{.}26-20\mathrm{.}80-0\mathrm{.}33=0\mathrm{.}13m$

$f(y_2^{*})$ from Eq$.(5-22)$ is

$f^{\text{'}}(p_0^{*})=1-1\mathrm{.}5\frac{{1\mathrm{.}80}^2}{9\mathrm{.}813\mathrm{.}733}-\frac{30\mathrm{.}572{10}^{-3}(-300)}{23\mathrm{.}733}=0\mathrm{.}973$

The value of $R_2$ in Eq$.(5\mathrm{.}22)$ is obtained by dividing the total area $A$ in column $4$ by the

total wetted perimeter $P$ in column $5.$ The value of $h_2$ from Eq$-15-201$ is

$h_2=21\mathrm{.}001-\frac{0\mathrm{.}13}{0\mathrm{.}973}=20\mathrm{.}87m$