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Volume v's Mass airflow ?

  1. Nov 20, 2005 #1
    Volume v's Mass airflow...???

    Can anybody possible help me to understand how the following situation is affected with respect to the Volume and the Mass Airflow/velocity???

    imagine a sealed square box.
    On the left hand side is a round inlet.
    at the bottom are 4 smaller outlets, in a roow from the lefthand side to the right (no.1 is closest to inlet, no.4 is furthest away).

    The air flows in through the inlet and then out via the 4 outlets.

    Would i be correct in assuming that the outlets air would be different?
    That is, that there velocity and hence mass airflow would be different?
    Is this caused by the fact that from the point the air enters the box from the inlet to the time it passes out of the outlet, each individual outlet is of a different length?

    If so, how can this be compensated for?

    Is the answer to taper the box with the smaller end towards no.4?
    Would this then increase the velocity of no.4 outlet thus increase Mass airflow and help to normalise the 4 outlets???

    i really hope somebody may be able to help me figure this out.
    Thanks in advance.
    :smile:

    I can include a quick drawing if it helps visualise the layout......
     
  2. jcsd
  3. Nov 20, 2005 #2

    Astronuc

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    Staff: Mentor

    Assuming that the air is incompressible, then the mass flowing must equal the mass flowing out, otherwise the box would continuous accumulate air, which we know doesn't happen. This is the basis of the mass continuity equation.

    Mass in = mass out for steady-state equilibrium.

    and consequently mass flow rate in = mass flow rate out or [itex]\dot{m_{in}} = \dot{m_{out}} [/itex].

    The mass flowrate is given by [itex]\dot{m_{in}} = \rho\,V\,A [/itex] where [itex]\rho[/itex] is the air density, V = mean velocity (in this case through an opening) and A is the area.

    If there are multiple opening the A = [itex]\Sigma_i{A_i}[/itex]. However, there are dynamic effects such as flow resistance, which are dependent of the size of a hole. Given holes of different size, then the flow will be partitioned among the holes according to size, but also by flow resistance. If holes are similiar in size, then the difference in flow resistance may be negligible.

    The situation with for the four holes in a row is a little more complicated since there is flow resistance as the air traverses left to right. If the box is large (i.e. deep and the inlet is toward the bottom of the box) and the distance from the inlet to the four exit holes is approximately the same, then the flow through each hole (assuming the same area of each hole) would be approximately equal.

    If however the inlet is much closer to the first hole and the other holes are successively further away, then the flow through the first hole will be greater, and the flow will decrease for each successive hole.

    One could do an easy experiment with ping-pong balls, one on each hole, and vary the box geometry until all 4 ping-pong balls float at the same height. On the other hand, it is not clear what flow rates and pressures one is using for this system.
     
  4. Nov 20, 2005 #3
    Wow :-)

    Thankyou for a very good response......

    If i can draw your attention to this post, i have further details of what im trying to achieve.

    If you could have a quick look it would certainly make it easier for you to see what im explaining, then hopefully you'll be able to give me some guidance more specific to the application :-)

    https://www.physicsforums.com/showthread.php?t=97944
     
  5. Nov 20, 2005 #4
    Thats a very good idea:smile:
    However not quite so easy in reality, as it would require some components which could cost considerable amounts of money just to test...

    In relation to the pressure, under maximum inlet pressure it would be around 26psi above atmospheric.

    Thanks again.
     
  6. Nov 28, 2005 #5
    I guess the back pressure of 4 outlets are same. Air is incompressible at room temperature.

    If so, the only reason that can cause uneven mass flow at outlet is uneven pressure distribution. This depends on inlet air velocity, inlet area comparing to the box volume, and locations of 4 holes. If you have fluid software, I would suggest you to do a three dimensional simulation.

    If the inlet is in the center of top of the box, and 4 outlet symetric on the bottom, definitely you will have same mass flow rate. otherwise, try to give some mixing in the box, which can make even distribution of pressure.

    just a guess.

    Rocketa
     
  7. Dec 1, 2005 #6
    Actually, I should write non-uniform instead of uneven..
     
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