A vertical underflow sluice gate $5\mathrm{.}5$m wide discharges $18\mathrm{.}2$ m$3$

$/$s into a

channel of the same width. The gate is set $0\mathrm{.}90$m above the bed $($Y$),$ Cc is

assumed to be $0\mathrm{.}6$ and the normal depth in the channel is DN $=1\mathrm{.}5$ m$.$

$($a$)$ Determine if the flow condition that gate is operating $($i$.$e$.$ submerged or

free flow$).$

$($b$)$ Determine the upstream depth behind the sluice gate by ignoring minor

and friction head losses due to the sluice gate and $\backslash$alpha $=1\mathrm{.}0.$ I have attched the anwser to this question. I dont understand the trail and error part and when to stop doing the trail and error. A vertical underflow sluice gate $5\mathrm{.}5m$ wide discharges $18\mathrm{.}2\frac{m^3}{s}$ into a

channel of the same width. The gate is set $0\mathrm{.}90m$ above the bed $($Y$),C_c$ is

assumed to be $0\mathrm{.}6$ and the normal depth in the channel is $D_N=1\mathrm{.}5m.$

$($a$)$ Determine if the flow condition that gate is operating $($i$.$e$.$ submerged or

free flow$).$

$($b$)$ Determine the upstream depth behind the sluice gate by ignoring minor

and friction head losses due to the sluice gate and $=1\mathrm{.}0.$

Solution of Problem $2$

$($a$)$ Assume $D_N=1\mathrm{.}5m=$ the depth after a hydraulic jump.

$V_N=\frac{18\mathrm{.}2}{1\mathrm{.}55\mathrm{.}5}=2\mathrm{.}21\frac{m}{s}so$

$F_N=\frac{V_2}{(g{D}_2{)}^{\frac{1}{2}}}=\frac{2\mathrm{.}21}{(9\mathrm{.}811\mathrm{.}5{)}^{\frac{1}{2}}}=0\mathrm{.}575$

The sequent depth to $D_N$ is $D_1$ where from hydraulic jump equation

$D_1=\frac{D_2}{2}(\sqrt[\phantom{2}]{1+8F_2^2}-1)$

$D_1==(\frac{1\mathrm{.}5}{2})[(1+8{0\mathrm{.}575}^2{)}^{\frac{1}{2}}-1]$

$=0\mathrm{.}682m$

Thus, a jump can form only if the initial depth $(D_1)$ is $0\mathrm{.}682m($or less$).$ Since the depth

downstream of the gate is $0\mathrm{.}90\mathrm{.}6=0\mathrm{.}54m$ a jump is clearly possible and the gate is free flow.

$($b$)H_1+{}_1\frac{V_1^2}{2}g=H_2+{}_2\frac{V_2^2}{2}g+0\frac{V_2^2}{2}g$

$H_2=0\mathrm{.}54m$ then:

$V_2=\frac{18\mathrm{.}2}{5\mathrm{.}50\mathrm{.}54}=6\mathrm{.}13\frac{m}{s}$

We use trial and error for $V_1$ and $H_1$ and start with $V_1=0$

$V_1=0$ then $H_1=2\mathrm{.}454$

$V_1=\frac{18\mathrm{.}2}{5\mathrm{.}52\mathrm{.}45}=1\mathrm{.}35$ then $H_1=2\mathrm{.}36$

$V_1=\frac{18\mathrm{.}2}{5\mathrm{.}52\mathrm{.}36}=1\mathrm{.}41$ then $H_1=2\mathrm{.}35$

After a few trials of $H_1$ finally for $H_1=2\mathrm{.}36$ we have