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Pressure drop and exit velocity

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Old   August 1, 2013, 04:50
Default Pressure drop and exit velocity
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I am simulating the blood flow inside the coronary arteries.But before going further I have the need the advice of CFD experts.

My question is(attached) , if we have the same pressure at the one side of two pipes like the same head , and at the other side on pipe constricts at the end compared to the other constant area . What would be the exit velocity and why.
According to the Bernoulli'se equation , the exit velocity is not a function of the exit area , then we should have the same velocity at the exit , but the second pipe with constricted exit has lower mass.
But according to the experiment , the constricted exit pipe has higher velocity .
There is another theory that expresses , on the constricted exit channel , the resistance to the flow will increase then there will be mass flow decrease , but nothing about the exit velocity.

I would appreciate if the experts help me solve this problem,
Thank you in advance
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File Type: jpg uexit.jpg (21.6 KB, 35 views)
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Old   August 1, 2013, 06:14
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This is basic fluid mechanics. I wonder what you are doing with CFD if you cannot answer this question.....

On the top plot the only resistance to flow is the friction in the tube. So the flow rate in the top pipe will be high.

On the bottom plot you have pressure being converted to velocity (Bernoulli), and conservation of mass/volume. So the inlet pressure is working against friction and the pressure of accelerating the flow in the nozzle. So the flow rate will be lower.

So no, the Bernoulli equation does not have area in it, but conservation of volume/mass does. That makes the bottom pipe squirt a small fast jet and the top pipe flow a large volume of lower velocity fluid.
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Old   August 1, 2013, 17:06
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Thank for your reply,

I was not clear on explaining the problem.

Suppose two pipes are connected to a huge tank place at the height of h with large surface area that we can consider the velocity of the fluid on the surface exposed to the atmosphere , zero.
Then on the B.Eq. we have three terms of static pressure , dynamic and head pressure . Because the jet pressure is atmospheric pressure then if we write the B.Eq between one point on the top surface of the tank an another at the exit jet , the static pressure is the same and omitted , then it is the head pressure that cause the velocity .
For two pipes there is noting to distinguish on the B.Eq then the exit velocity according to the equation would be the same. But the pipe with nuzzle has lower mass flow rate because of the lower exit area.

My question is that how can we justify the higher velocity at the second pipe with the fluid dynamic equations?

Thanks
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Old   August 1, 2013, 18:02
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In this case, if you ignore flow losses then both pipes will squirt a jet of equal velocity. But the nozzle will have lower flow rate due to the smaller cross section area.

But if you include losses there will be more losses in the open pipe due to the higher velocity, so this velocity will reduce more than the nozzle pipe which has less losses. So fluid losses will cause the nozzle to squirt a faster jet.
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Old   August 2, 2013, 01:16
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The pressure and friction loss goes to one term added to the B.Eq , because it is not for viscous flow.
Then obviously the constricted channel has lower mass flow rate but not the same velocity because of lower pressure drop.
But I have seen a paper that discussed a numerical investigation on the channel with different degree of stenosis and the same pressure drop for all cases.
The results show that mass flow decreases with degree of stenosis but the velocity remains almost the same without any significant change.

Although the area and mass flow rate both will affect the exit velocity , but we expect to have higher exit velocity at higher degree of stenosis.

I do not know how to justify these controversies.

Thanks .
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Old   August 3, 2013, 00:59
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Hi,
I think you have simplified the problem for us with the tank explanation. If you could tell us about the actual case, some people might be able to understand your problem better. It might be that you might have missed some important detail.
Also
1) If you are referring a paper, see what are the assumptions and what are the boundary conditions.


Quote:
Originally Posted by mejahan View Post
The pressure and friction loss goes to one term added to the B.Eq , because it is not for viscous flow.
Then obviously the constricted channel has lower mass flow rate but not the same velocity because of lower pressure drop.
But I have seen a paper that discussed a numerical investigation on the channel with different degree of stenosis and the same pressure drop for all cases.
The results show that mass flow decreases with degree of stenosis but the velocity remains almost the same without any significant change.
.
2) what is the 'almost zero change in velocity, it is easier to compare the total pressure drop in both the systems

3) The observation that the velocity is same for same pressure drops doesnt fit the logic especially since the exit areas are different, I think the simulation in the paper you are referring to might be an inviscid assumption or it is something to do with boundary condition. I think the answer lies somwhere in them.
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Old   August 4, 2013, 16:57
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The paper used the same pressure inlet and outlet for all of the cases with the different degree of stenosis .
You are right in the viscous flow like blood , there should be noticeable difference between them.
Anyway I received my answer here , thank you
I will take an exact look at the paper about the details
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