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In order to maintain a constant water deep in canal, how much water must flow trought the pipe ? enter image description here

As shown on picture, canal have a rectangular shape. I don't know if canal length have an influence.

EDIT : to simplify things let's consider there is no turbulence, no viscosity, and that water falling from pipe do not disturb water in canal.


I tried to solve the problem by myself (i'm a physics beginner so it could be totally wrong, please do not downvote the question if you think this is not correct) :

Area of canal section : $A = w \, h$

If I calculate water velocity $v$ in canal, using this and surface $A$, I can calculate how much water $Q$ will flow :

$$Q = A \, v$$

and solve the problem...

So only thing left is to calculate $v$.

Let's say the canal have no inclination $Z = 0$, I think water velocity for a given water height can be calculated like this (I'm not sure about this) :

$$ v = \sqrt{ 2 \, g \, h } $$

$$ \delta Q = A \, v = w \, \delta h \, \sqrt{ 2 \, g \, h } $$

integrating h from 0 to H and gives :

$$ Q = w \, \sqrt{ 2 \, g } \int_0^H h^{1/2} \, dh$$

so discharge for a given height and width :

$$ Q = \frac{2}{3} \, w \, \sqrt{ 2 \, g } \, H ^{3/2} $$

Could anyone tell me if the above is correct (assuming there is not inclination), and try to answer my initial question ?

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Your expression for v is not correct, as Bernouilli's law is only valid along streamlines. –  Bernhard Jun 13 '12 at 22:10
    
Hi tigrou, and welcome to Physics Stack Exchange! I'm adding the homework tag because your question seems like a homework-like question (not necessarily from an actual homework assignment). Could you clarify whether that's the case, or why you're asking this question? Note that there's nothing wrong with it being a homework question! We're just interested to know what kind of answer will be most useful to you. –  David Z Jun 14 '12 at 2:26
    
Hi, first of all this is not homework. I got interest for fluid dynamics and engineering recently and read some of information about it. This is a question i'm looking for since a long time without finding an answer. I'd like to know if Bernoulli's principle can be applied here and if there is a simple answer to this problem. Maybe some conditions are missing. People trying to answer the question are free to change/add some conditions (or neglect some details) if it makes solving the problem easier. –  tigrou Jun 14 '12 at 11:42
    
@tigrou Why did you make this question community wiki? –  Bernhard Jun 17 '12 at 10:57

1 Answer 1

up vote 0 down vote accepted

I think there is a fundamental difference between the inclined slope and a flat surface.

If you consider the sloped case, you have gravity as a driving force to accelerate the flow downwards. As the fluid accelerates, you know from mass conservation that the value of $h$ along the slope will decrease. But at some point, you got friction at the wall preventing further acceleration. Lets focus at this far downstream point where $$ \frac{dh(x)}{dx}=0$$

Therefore, we also know that $\vec{u}(x,y)$ reduces to $u(y)$, only a velocity component in the direction parallel to the bottom, that is only varying in the direction perpendicular to the bottom.

We know from experience that this kind of laminar flow will give a Poisseuille profile, so we assume that $u(y)=ay^2+by+c$ and boundary conditions.

  • At $y=0$, $u(0)=0$
  • At $y=h$, $\frac{du}{dy}=0$, because there is no slip at the free surface.
  • We now that mass is conserved, thus $\int_0^h u(y)dy=Q$ with $Q$ the flow rate. Note that $Q$ has dimensions $m^2/s$ as this case is two-dimensional.

This give $c=0$, $b=-2ha$ and $a=-\frac{2}{3}\frac{Q}{h^3}$ respectively, and thus $$u(y)=\frac{3Q}{h^3}(yh-\frac{1}{2}y^2)$$

Now we will balance the friction along the bottom with the gravity force on the bulk. This is valid, as there is no pressure gradient acting (the pressure at the top is uniform ambient pressure).

The friction force along the bottom over a length $\Delta x$ equals $\mu\frac{du}{dy}(0)\Delta x$ and the gravity force in that direction equals $\rho h g \sin \alpha$, where $\mu$ is the viscosity, $\rho$ the density and $\alpha$ the angle of the slope with the horizontal.

$$friction = gravity\\ \mu \frac{3Q}{h^2} \Delta x = \rho h \Delta x g \sin \alpha \\ h=\sqrt[3]{\frac{3\mu Q}{\rho g \sin \alpha}}$$

From this expression, you see that for increasing flow rate, the height of the film increases. Also when you decrease the slope angle, the film thickness will increase.

Finally, increasing the viscosity, will increase film height. This makes sense when you realize that when there is more friction, the same area can withstand more mass.

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This has been flagged as "not an answer", and in it's current form I think it teeters on the edge: it's rather more like a long comment. That said, I can see where you are going with this and think that the addition of a little quantitative discussion could improve it considerably. –  dmckee Jun 14 '12 at 15:13
    
@dmckee I had the homework tag and this discussion in mind meta.physics.stackexchange.com/questions/714/… I've provided a lot of steps that will give you at least some answer. Maybe I can extend my answer later, but not without doing the exact derivation myself. Actually I wrote my answer without pen or paper in the neighborhood, making it tough to add more then I did. –  Bernhard Jun 14 '12 at 20:35
    
great post, thanks. i'm reading it. –  tigrou Jun 17 '12 at 13:25
    
@tigrou Does it answer your question, or is anything unclear? –  Bernhard Jun 17 '12 at 18:26
    
There is two things i'm not sure to understand : 1) from v = ax^2 + bx + c how did you find that c = 0, b = -2ha, a = -2/3... ? (you integrate v(x), but then ?) 2) you said gravity force = phg * sin(alpha) where does this formula came from? (i understand why sin is here and what it does but not the rest). –  tigrou Jun 17 '12 at 21:01

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