7 Replies Latest reply on Dec 13, 2013 2:34 PM by Jared Conway

Boundary Condition set up for Rotor help!

Jason Lee Dec 12, 2013 2:00 PM

Hello,

I was wondering if someone could help me with an issue i've come across concerning the proper set up of boundary conditions for a flow simulation of a rotor/propeller I designed.

I attached an image that shows the scenario I am trying to create for my rotor.

So in this scenario, there are two pistons (striped black boxes) that will contract to the striped red boxes position. When the pistons are expanded, there is 6L of a liquid between the pistons and after it is contracted, there will be 2L of liquid remaining in the space seen in the diagram. In total, there should be 4L of the liquid that should go through the center pipe outlet. The pressure at the pistons expanded state will be 5 mm Hg, the pressure of the contracted state is 90 mm Hg and the pressure at the outlet is 100 mm Hg. What I would like to do is add some type of rotor/propeller rotating at 5,000 rpm and placed at the start of the outlet (as seen in the diagram) that will pump the remaining 2 L during the contracted state towards the outlet so that the total volume that reaches the outlet will be 6 L rather than only 4L. The inlet and outlet regions I defined in the flow simulation are seen in the diagram as well (the blue circles). Also assume that the height of the fluid does not affect pressure.

How should I set up my boundary conditions for this scenario?

Some of the appraoches i have tried are:

1. The inlet lid with a total pressure (environmental pressure) of 90 mm Hg and the outlet lid with a static pressure (environmental pressure) of 100 mm Hg.

2. The inlet lid with a volume flow rate of 4 L/min and the oulet lid with a static pressure of 100 mm Hg.

Some additional questions I have are:

1. Will the 4L volume that is contracted by the piston be affected by the rotor?

2. How can we create head flow curve (HQ curve) for this rotor?

Any help will be much appreciated!

Thank you

Jason

Re: Boundary Condition set up for Rotor help!

Jared Conway Dec 12, 2013 2:17 PM (in response to Jason Lee)

unless you can find a way to simplify your pistons moving down to a changing boundary condition in a transient analysis, this likely isn't doable in flow simulation.

but maybe you can break this problem down to just analyzing the rotor with a rotating region. for example if you were to test the rotor on its own, you would put it in a pipe and change the flow rate on one end and leave it open on the other. then you could measure the performance.

i would recommend that before you go much further you go through the technical reference and solving engineering problems docs and also the solidworks kb to get a good handle on how rotating regions work. you've got a pretty complicated problem that i'd start by breaking this down into a bunch of smaller ones before you try going right at the big one.

is this a school project or similar? someone else posted something almost exactly the same yesterday.
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- Re: Boundary Condition set up for Rotor help!
  
  Jason Lee Dec 12, 2013 2:51 PM (in response to Jared Conway)
  
  Hi Jared,
  
  Thanks for your response,
  
  So I did simulate with just the rotor alone in the rotating region. I attached a screenshot of my set up. In this image, the inlet would be at the top and the outlet is at the bottom.
  Could you please elaborate a little more on your second paragraph?
  I believe I tried what you are suggesting here, but there are some assumptions I am confused about. Taking into account the scenario I described above, should I make the inlet lid have a volume of 2 L/min or 6 L/min? while the outlet lid has a static pressure of 100 mm Hg.
  The pistons in the scenario will be responsible to contract 4 L of liquid to the outlet (without the help of the rotor). I would like to have the rotor pump out the remaining 2 L of liquid when the pressure is 90 mm Hg at the inlet. So there is a 10 mm Hg pressure gradient the rotor must go against between the inlet and outlet.
  Would it make sense to have the inlet lid have a boundary condition of 2 L/min or 6 L/min and not take into account the 10 mm Hg pressure gradient? How would i account for the difference in the pressure when the oulet is 100 mm Hg and the inlet should be 90 mm Hg while there is also a volume flow rate that is coming into the inlet lid?
  
  Thank you
  
  Jason
  
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  - Re: Boundary Condition set up for Rotor help!
    
    Jared Conway Dec 12, 2013 2:58 PM (in response to Jason Lee)
    
    you have to choose one set of conditions or another, a pressure difference or a volume flow and pressure, you can't have both.
    
    in the first case, your pressure difference will tell you what the flow rate is through the system
    in the second case, you'll find out what the pressure is at the volume flow rate side
    
    i'd suggest you experiment with both and see which one makes the most sense from a results perspective because you're having to eliminate the piston part and come up with a good assumption. i don't know which one is best, i would have to try them all. in the end it isn't realy exactly the same because you have a "pushing" of the fluid from the pistons and around the corner and you're simplifying that as a volume flow rate. there are assumptions inherent in that as well.
    
    if you choose the flow rate method, it sounds like if you're taking 4L of fluid out over the time of 1 min, the flow rate through the rotor should be 4L/min. but again, assumptions here. is it truly 4l/min the whole time? does it ramp up?
    
    thats why i'm suggesting breaking yoru problem down to a simpler problem. if you were to just TEST the rotor in a pipe, what conditions would you use. most common is to change the pressure drop and find out the flow rate based on the rotation or put in a flow rate and pressure to find the curve. once you have the curve, it doesn't matter what the actual conditions are, you look it up on the curve.
    
    for this test, i would also recommend making your pipe/comp domain larger so that your boundary conditions do not affect the area of interest. 3 dimensions in length is probably a good start.
    
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    - Re: Boundary Condition set up for Rotor help!
      
      Jason Lee Dec 13, 2013 2:10 PM (in response to Jared Conway)
      
      Hi Jared,
      
      Thanks for the advice. Much appreciated.
      I have an unrelated question regarding the torque surface goal value that gets quanitfied. When the simulation quantifies torque, does it take into account the density of the material? So if I used titanium rather than a plastic, i expect the torque to be higher for titanium, but will this be accounted for?
      
      Thank you
      
      Jason
      
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      - Re: Boundary Condition set up for Rotor help!
        
        Jared Conway Dec 13, 2013 2:14 PM (in response to Jason Lee)
        
        the torque is the fluid acting on the solid surface, it doesn't know what the solid is. (ie, you don't enter E for the solid so it doesnt' know how stiff it is)
        
        but i don't think i understand your question. are you thinking the torque changes because it deforms?
        
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        Re: Boundary Condition set up for Rotor help!
        
        Jason Lee Dec 13, 2013 2:26 PM (in response to Jared Conway)
        
        Hi Jared,
        
        I am actually referring to the rotor itself and not the fluid. So I defined a surface goal for the the rotor so that I can figure out what the necessary torque is to rotate the rotor along the y-axis. So if I change the rotor material from plastic to titanium, i would expect the overall torque to increase because titanium is more dense and heavier than a plastic material.
        So does the quanitfied surface goal torque value take into account the change in material?
        
        Thanks
        
        Jason
        
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        Re: Boundary Condition set up for Rotor help!
        
        Jared Conway Dec 13, 2013 2:34 PM (in response to Jason Lee)
        
        you don't enter material strength information into flow > therefore the software doesn't know how it deforms > the torque will not change based on the material.
        
        i'd recommend taking a look at the reference documentation on how torque is calculated to get a better handle of how it works. it is basically just an integration of the flluid flow over the area. it has no concept of the material.
        
        if you export the fluid flow to simulation, then you could see how the model deforms based on the fluid flow.
        
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