A very long trapezoidal channel has a B $=20$ ft$,$ t$=2\mathrm{.}0$ H:V$,$ So $=0\mathrm{.}0016,$ and

Manning$$s n $=0\mathrm{.}025,$ it carries Q $=400$ cfs$.$ A velocity energy correction coefficient of $=1\mathrm{.}1$ is to $\backslash$table$[[$Station$,$WSEL,y$,$A$,$V$,$V$^(2)/2$g$,$H$,$R$,$R$^(4/3),$S$\_($f$),$Ave S$\_($f$),\backslash$Delta x$,$h$\_($f$),$H$],[0\mathrm{.}0,605\mathrm{.}00,5\mathrm{.}00,,2\mathrm{.}667,0\mathrm{.}1217,605\mathrm{.}122,3\mathrm{.}54,,\mathrm{.}000370,$dots,dots,.....$+-,605\mathrm{.}122],[155\mathrm{.}0,605\mathrm{.}048,,142\mathrm{.}08,2\mathrm{.}819,,,3\mathrm{.}43,,,0\mathrm{.}000402,155,0\mathrm{.}062,605\mathrm{.}184],[318\mathrm{.}0,,4\mathrm{.}60,,2\mathrm{.}979,0\mathrm{.}1517,,3\mathrm{.}31,4\mathrm{.}92,\mathrm{.}000507,,163,,],[491\mathrm{.}0,605\mathrm{.}186,,126\mathrm{.}72,,,605\mathrm{.}357,3\mathrm{.}19,,,\mathrm{.}000598,,0\mathrm{.}096,]]$be assumed.

Compute the backwater profile created by a dam which backs up water to a depth of $5$ ft immediately behind the dam.

Calculate the water surface profile in this channel using the standard step method by filling the bolded cell in the table below.

Show your calculations $($iterations$)$ on another sheet by using the standard step table.

The table to fill in is on the next sheet for you to complete. The elevation or datum at the dam site is $600$ ft$.$

Yn $=\_\_\_\_\_\_\_\_\_$ ft yc $=\_\_\_\_\_\_\_$ ft Profile type: $\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$