I have a trapezoidal channel like this:

Trapezoidal channel

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:

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


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

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    $\begingroup$ Math seems correct to me $\endgroup$ – Michiel Feb 4 '14 at 16:38
  • $\begingroup$ @Michiel I verified that everything related to math is correct before I ask here. $\endgroup$ – Mhmd Feb 4 '14 at 16:39
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    $\begingroup$ $m/m$ 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 $\endgroup$ – Michiel Feb 6 '14 at 17:24
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    $\begingroup$ 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. $\endgroup$ – tpg2114 Feb 6 '14 at 19:17
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    $\begingroup$ @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. $\endgroup$ – tpg2114 Feb 6 '14 at 19:40

I agree with this result. First of all, the math is correct, though

\begin{equation} R = \frac{1000}{76 + \frac{20}{5/12}} \approx 8.06 \ m \end{equation}

but the change to your result is negligible. Second, it is consistent with the Manning Formula

\begin{equation} v = \frac{1}{n} R^{2/3} S^{1/2} \end{equation}

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).

The reason your slope is so small is because the mean velocity is very small relative to the desired Chézy coefficient, though this is not something one could point out from a first-glance examination of the problem.


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