Inlet/Outlet Loss due to sharp edged contraction 2

[peglor]

This is a question that was put to me for which I can't find a satisfactory answer.

Any fluids book will tell you that the pressure drop due to a sudden contraction or expansion is found by:

PL = KL*rho*u^2/2

PL=Pressure Drop
KL=Loss coefficient
rho=fluid density
u=fluid velocity

KL is found from a table for the inlet loss and calculated as follows for outlet loss:

KL=(1-A1/A2)^2
A1=Initial CSA
A2=Expanded CSA

The question is: What velocity goes in as the u value in the PL equation?
I'm assuming u is the initial mean velocity over the cross section prior to the contraction for the contraction losses and the initial mean velocity over the cross section prior to the expansion for the expansion losses.

The problem is no books I can find are totally clear on it. Some say to use the average velocity, which I take to mean the averaged velocity profile prior to the area change, though it could be interpreted to mean the average of the mean velocities before and after the area change.

This seems like a simple question, but I can't find a clear answer. Hopefully someone here will know.

 

[hacksaw]

in this case u is the velocity in the pipe prior to the area change.

 

[Latexman]

peglor,

I believe the convention is u = the average velocity in the smaller section in an expansion or contraction. Otherwise, if the smaller pipe terminated to an open area with essentially infinite area and no velocity, what to do?


Good luck,
Latexman

 

[hacksaw]

whether you use an upsteam or downstream basis, assuming that you choose a non-trivial section, is up to you. You just need to be consistent.

 

[peglor]

Hacksaw:
What do you mean by using an upstream or downstream basis?

Latexman makes a very valid point in saying the formula won't work in the case of an inlet going from an infinite area zero velocity zone to a finite area zone with non-zero velocity since u for this case would be zero. Does this mean there is no pressure drop due to this type of inlet or that there is a flaw in the formula?

The convention of always using the velocity in the smaller cross section should work in all cases, but choosing to apply it is not as obvious.

Even the fact there are two different answers given to my question here shows that this is seems to be something that isn't covered carefully enough by fluids books.

 

[Latexman]

The K needs to be consistent with the reference conduit and the corresponding u. Most K's I've seen are correlated based on choosing the smaller conduit as the reference. Choosing the larger conduit requires converting the K, more chance for error and more time spent. Both ways work. One is more efficient.


Good luck,
Latexman

 

[peglor]

The data I have for an inlet loss is as follows (From Engineering Tables and Data by Howatson, Lund and Todd):
A2/A1 KL
0.1 0.37
0.2 0.35
0.4 0.27
0.6 0.17
0.8 0.06
0.9 0.02

Does this look like it uses the velocity in the A2 (Smaller area) or A1 (Bigger initial area) as u in the PL = KL*rho*u^2/2 formula?

At least now I'm reasonably confident that for the expansion losses the value of u used is the velocity in the smaller channel prior to the expansion.

 

[hacksaw]

peglor,

I believe that you are free to choose you reference section, under the proviso that it not produce a trivial result. Latexman's comment about infinite expansion is a case in point. The velocity may be small, but conservation of mass still applies. A second example is seen in flow metering, where the metering can be performed on an upstream or downstream basis.

 

[Latexman]

peglor,

The data you presented *looks* like KL corresponds to A2 (smaller area) to me. I compared it to the following contraction equation using your nomenclature:

K2 = 0.5 * (1 - A2/A1)

This compares to the following contraction equation out of Crane TP410 where A1 is the smaller area:

K1 = 0.5 * (1 - A1/A2)

Therefore, it appears you need to use KL with u2 in A2.


Good luck,
Latexman

 

[peglor]

Thank you very much for that posting. I'm finally happy with my calculations. Turns out I was doing them correctly all along .

My peace of mind has earned you a star anyway.