Not the expected pressure drop through resistance material

[Anonymous]

Hello,

I have defined a resistance material with the following parameters:

Trough-flow K: Constant: 40

Normal direction 1 and 2: Constant: 1000  (the idea is that air shouldn't flow in this direction)

Permeability: 0

The fluid surrounding it is air.

The pressure drop I get using this material in a simulation is about 40 Pa when a 2.1 m/s flow goes through it.

This is the pressure drop formula provided in the documentation: Creating and Editing Resistances | CFD | Autodesk Knowledge Network

Using it: DeltaP = 40*1.2*(2.1^2)/2 = about 100 Pa.

So I am getting less than half the expected pressure drop.

Have I understood this material/formula correctly? Or is the problem in my simulation?

Thank you in advance,

Pol

[marwan_azzam]

Hello,

Please zip the support share file (*.cfz) of your model and share with us.

Marwan

[Anonymous]

Hello,

Thank you for taking the time to answer me.

I am attaching the cfz file (sadly without the results as it would be too heavy to post). In it, you will find the filter I was mentioning in the previous post as solid number 43.

In the results I got for this simulation, there were about 7.5 Pa of pressure drop across the filter at about 2.5 m/s. Using the formula provided in the documentation I am obtaining 150 Pa of theoretical pressure drop at this speed.

Have I understood correctly the formula in the documentation? (See original post). Or is there a problem in my simulation?

Thanks again.

Pol

[marwan_azzam · Accepted solution]

Hello Pol,

I would use a uniform mesh on the resistance material with 4 to 5 elements through the flow direction (Z-direction).

Please try that and let us know if you get better results.

Marwan

[Anonymous]

Hello Marwan,

Thank you for your insight! I will test it and let you know my results as soon as possible.

Pol

[Anonymous]

Hello,

I have run some tests and you were right Marwan: it was a meshing problem. Here are my results:

Resistance defined with "constant" method:

 - 1 element in flow direction: 3 Pa pressure drop

 - 4 elements in flow direction: 130 Pa pressure drop

 - 7 elements in flow direction: 125 Pa pressure drop

The last two results are coherent with the formula stated in the first post of the topic.

Other resistance defined with the "free area ratio" method: (different part, results shouldn't be the same as above)

 - 1 element in the flow direction: 3 Pa pressure drop

 - 2 elements in the flow direction: 5 Pa pressure drop

 - 4 elements in the flow direction: 5 Pa pressure drop

So two elements are needed to achieve mesh independence in this second case.

Also, using the "Parts" result calculator provides a  "Resistance Pressure:" value that has nothing to do with the pressure drop result through the part. I don't know what that number means.

Thank you for pointing me in the right direction,

Pol