# Slope of a channel

I have a trapezoidal channel like this: The question is to get the slope of the channel such that we have a Chézy coefficient of 64 $m^{1/2}/s$.

Given: $B= 76\: \mathrm{m}$, $y = 10\: \mathrm{m}$, $\tan(\gamma) = 5/12$ where $\gamma$ is the angle between the horizontal line and the oblique line. $q= 1200\: \mathrm{m^3/s}$.

Here is how I solved it:

we know that $$v = C \sqrt{ R\space i}$$

so $$i = \frac{v^2}{C^2 \times R}$$ $$v = \frac{q}{S} = \frac {1200}{ y \times(B + \frac{y}{\tan(\gamma)})} = 1.2\: \mathrm{m/s}$$

$$R = \frac{1000}{76 + \frac {20}{5/12}}=7.8\: \mathrm{m}$$ and then we have:

$$i = \frac{1.2^2}{64^2 \times 7.8} = 4.5 \times 10^{-5}$$

# Question:

Is this an expected value of a slope? shouldn't this be something like: $4.5$ or $45$?

• Math seems correct to me – Michiel Feb 4 '14 at 16:38
• @Michiel I verified that everything related to math is correct before I ask here. – Mhmd Feb 4 '14 at 16:39
• $m/m$ refers to how many meters in height the channel changes for a given meter of horizontal distance. That means that $45 m/m$ is a change of 45 meters in height for 1 $m$ of distance which is obviously close to 90 degrees. To be exact it is: $Arctan(45/1)=88.7$ degrees – Michiel Feb 6 '14 at 17:24
• Because there is a bounty I can't vote to close the question even though it doesn't conform to our homework policy @Qmechanic linked to above. There are no conceptual questions asked and it appears to just be looking for somebody to check the work of the questioner. – tpg2114 Feb 6 '14 at 19:17
• @user689 That is still not a physics or conceptual question. A slope can be any real number from 0 to $\pi/2$ and you're within those bounds. So there isn't anything about it that's on topic if all you want is to know if your number is reasonable. – tpg2114 Feb 6 '14 at 19:40

for your calculated velocity, slope, and hydraulic radius. Solving for the Manning coefficient gives $n \approx 0.022$, well in the ballpark of most materials, which is between 0.01 and 0.1 (in fact, it seems to be closest to corrugated metal).