CAN SOMEONE PLEASE EXPLAIN THIS STEP BY STEP. First we must develope the emplying time eqn

$-A_{ws}h=Q_{A}tt=-(\frac{A_{w_s}}{Q_A})h$

for a rectangolar weir $Q_{}=C_0(\frac{2}{3})b\sqrt[\phantom{2}]{2g}{h}^{\frac{3}{2}}$

$=>{}_t=\frac{2A_s}{-\frac{3}{3}\text{c}\text{o}\text{b}\sqrt[\phantom{2}]{2g}}{\displaystyle \underset{H_1}{\overset{H_2}{}}}{h}^{-\frac{1}{2}}{h}^{-\frac{1}{2}}h=\frac{A_{ws}}{\frac{2}{3}{C}_{0b}\sqrt[\phantom{2}]{2g}}[-2h^{-\frac{1}{2}}{]}_{H_1}^{H_2}$

$\frac{2A_{ws}}{\frac{2}{3}{C}_{p}b(2g{)}^{\frac{1}{a}}}[\frac{1}{H_2^{\frac{1}{a}}}-\frac{1}{H_1^{\frac{1}{2}}}]$ discharge time through rectangular weir

Given $T-136s$;$A_{ws}=63=18m$;$b=0\mathrm{.}2m$;$H_1=0\mathrm{.}3m$;$H_2=0\mathrm{.}1m$

$136=\frac{2(18)}{\frac{3}{3}{C}_0(0\mathrm{.}2)(29\mathrm{.}81{)}^{\frac{1}{2}}}[\frac{1}{{0\mathrm{.}1}^{\frac{1}{2}}}-\frac{1}{{0\mathrm{.}3}^{\frac{1}{2}}}]$

$C_0=\frac{2(18)}{\frac{2}{3}(136)(0\mathrm{.}2)(29\mathrm{.}81{)}^{\frac{1}{2}}}[\frac{1}{{61}^{\frac{1}{2}}}-\frac{1}{{6\mathrm{.}3}^{\frac{1}{2}}}]$

Plan

front view