A rectangular concrete channel $(n=0\mathrm{.}013)$ has a slope of $0\mathrm{.}15\%.$ The width of the channel is $9\mathrm{.}00m$ wide and is

carrying a discharge of $17m\frac{3}{s}($see notes below to determine individual discharge value$).$ Water in the channel

flows over a broad crest weir where at the upstream end of the weir the water depth is $4\mathrm{.}80m.$

The discharge in the channel is $(17+4)m\frac{3}{s}.$

Show calculation of the following and submit along with the summary Table:

Calculate the critical depth.

Calculate the normal depth.

Determine the water surface profile type $($backwater or tailwater$)$ and state whether it is a $$1,$$$2,$

S$3,$ M$1$ M$2$ or M$3$ profile $($Lecture $=4$ from Week $10).$

Calculate the number of segments to be used.

Calculate the upstream and downstream depth of each segment.

Determine the length of the backwater curve.

Draw a sketch depicting the weir and the backwater curve.

Procedure:

First determine whether the flow is subcritical or supercritical by computing the normal depth

and critical depth. Compare the upstream depth $(4\mathrm{.}80m)$ with the normal depth $(D_n)$ and critical

depth $(D_c)$ to determine the water surface profile in question $3$ above.

Column $1$: Assuming each segment should not be more than $0\mathrm{.}50m$ in length.

i$)$ Number of segments $=($upstream depth $\frac{\text{:}(4\mathrm{.}80)-D_n}{0\mathrm{.}5}$

ii$)$ Difference in depth between segments $=($upstream depth $(4\mathrm{.}80)-D_n)/$ number of segments

Column $2$: Cross$-$sectional area at that depth.

Column $3$: Velocity.

Column $4$: Energy at each of the location $=$ depth $$ velocity head $(\frac{V^2}{2g})$

Column $5$: Difference in energy between downstream and upstream energy.

Column $6$: Hydraulic radius.

Column $7$: Slope of the Energy Grade Line: $s_f=(\frac{nQ}{A{R}^{\frac{2}{3}}}{)}^2$

Column $8$: Average $S_f($e$.$g$.,$ slope of the EGL at the midpoint$).$

Column $9$: Difference between the actual channel slope and calculated EGL.

Column $10$: Segment length $($d$1).$ The total segment length is the length of the backwater curve

before the flow depth gets back to the normal depth.

$dl=\frac{E}{s_0-s_f}$

All answers must be provided in the Table below. Supporting calculations must also be

submitted.

BONUS $2$ marks: Why is the segment lengths and total length of the backwater curve negative?

Submission Table

Flow Rate, $Q,\frac{m^3}{s})$

$($Please solve them all step by step $)$