Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!

Weir / restriction wall design in water flow

Status
Not open for further replies.

steffeman

Civil/Environmental
Aug 5, 2008
4
0
0
NL
Hello there.

I need to design a separation wall in a channel with water flowing through it. The purpose is to increase the velocity and at the same time create some level drop. The purpose is to improve the transport of light substances, floating on the water.

I know from experiments that it works, but I can not find the right equations that describe the effect.

I am thinking of a wall with slots in it which extend from well above the liquid level to way below it. I want to be able to calculate what size the slots should be to achieve a certain level drop.

Any help would be very much appreciated.

Louis
 
Replies continue below

Recommended for you

I believe the basic information you are looking for would be similar to a flume design. I do not have a reference on design with me at the moment, but try the oracle Google!

Mark Hutton


 
I consulted the "oracle" and it gives loads of references to "flume design", most of them to commercial sites that can supply or desing a flume or have software to help you do it.

What I could find in terms of theory behind it and formulas was not usefull because it all referred to specific flume designs (Pallmer, H-shape).

I have attached a sketch of what I have in mind. If I can calculate the level drop for a given flow rate, dependent on the dimentions of the slot, I am home free.

Louis
 
 http://files.engineering.com/getfile.aspx?folder=af3e992a-82a9-47f2-9252-ec39046c477c&file=sketch_slotted_weir.pdf
Louis,
From the sketch what you are looking for is the formula for a V-notch or square wier. I do not have my fluids reference with me at the moment but search on V-notch will get you close.
Cheers

Mark Hutton


 
Hello Hec

Thanks for taking the time to respond.

I have the formula's for the V-notch and even for the rectangular cut out overflow weir. When I got started with this, I also thought I just had to look up the formulas to apply, but it is more complicated.

The formulas for the V-notch and the rectangular weir only say something about the level upstream of the weir, nothing about the level downstream of the weir. In fact, these known formulas only apply if the level down stream of the weir is well below the weir (free fall flow).

Maybe you have to see it as related conditions at three different spots (not 2): upstream of the weir, at the weir and downstream of the weir. Apply Bernouilly from 1 to 2 to 3. I tried to work it out, but it is not really my league.

Louis
 
What you are looking at is probably a flooded rectangular weir. The downstream level is greater than the critical depth of flow going over the weir.

A previous employer of mine had an equation to give the headloss across a flooded weir, however I have not been able to derive it or find it in any technical literature so it may be proprietry information.

Depending upon your upstream/downstream levels the structure could be acting as a sharp crested weir, a flume or a flooded weir.

Hope this points you in the right direction.
 
I've derived something which may work.

Q is the flow rate
h1 is the upstream depth over the weir lip
h2 is the downstream depth over the weir lip
b is weir width
g is gravitational acceleration
v is the velocity of flow over weir

Assume the dynamic head upstream and downstream of the weir are identical.
Assume the weir is acting like an orifice

h1-h2 = 2.7(v^2)/(2g)

v = Q/(h2*b)

hence

Q^2 = (1/2.7)*(h1-h2)*2*g*(h2^2)*(b^2)

Q=0.61*b*h2*SQRT(2*g*(h1-h2))

This looks just like the equation my previous employer used, except there was a further term added on with looked just like the equation for the flow over a sharp edged weir. I can't see where they got this from though.

If you have tested your theory you must have a model. See how close this equation correlates to your model.

Start with your downstream depth

Calculate the upstream depth based on the equation above and the equation for a sharp crested weir.

Use the greater of the two answers.
 
Hello RCooper

Sure appreciate your help. Even before reading your second post, I found something on the internet following your directions and looking for "flooded weir".

There seems to be a "formula of Honma" which takes into account both upstream and downstream levels. For my situation this would be:

Q = w * (3/2)* sqrt(3) * C * h2 * sqrt(h1 - h2)

However, I have used the formula to calculate the level drop for our model but the outcome is too high. I have attached a detailed discussion of the formulas and calculations.

I also calculated the level drop using your formula, which gives an outcome that seems to correspond better with reality. However, I am not sure your assumtions are correct.

You assume dynamic head upstream and downstream of the weir are identical. If I understand it correctly, that means the velocity v before and after the weir would be the same right? But than, given that the flow Q and the width of channel B are the same before and after the weir, would that not mean that levels before and after the weir must also be the same.

Or am I making a mistake.

If you could take the time and have a look at my "report", maybe you can tell me where the mistake is.

Thanks for your help.

Louis
 
 http://files.engineering.com/getfile.aspx?folder=cdba9cb9-1910-4fa0-a987-7bbdc207082d&file=flooded_weir_formulas.pdf
lovasc,

You are right about the dynamic head being different upstream and downstream of the weir. A better way for me to put it would be that the difference between the dynamic head upstream and downstream is small enough to be negligible.

If you are concerned about the difference in the dynamic head then you would need to solve the following equation (the extended Bernoulli equation using the surface of the water as the streamline so pressure is atmospheric)

(h1+v1^2/(2g)+Hp)-(h2+v2^2/(2g)+Hp)=2.7v^2/(2g)

v1 is the upstream velocity
v2 is the downstream velocity
Hp is atmospheric pressure as a head )i.e. metres water rather than Pascals.

Hp cancels out and disappears from the equation.

If v1 and v2 are almost identical, they will cancel out and you get the equation above, otherwise you have a more complex equation to solve.

 
Most hydraulics text books have a section on open channel flow. I also recommend Camerons (google that name on this forum as it has been discussed several times) as it has a good open channel flow section that I have used.

rmw
 
Status
Not open for further replies.
Back
Top