1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Transfer function of flow measurement system HELP

  1. Dec 15, 2011 #1
    Transfer function of flow measurement system HELP!!

    1.

    I have attached the question, which is to derive the transfer function of a volumetric flow measurement system....

    2.

    I know of the following relevant equations:

    F = Ma = MD^2.x

    Flow f = area x velocity

    and f through a restriction f = C(p2-p1)

    Also Force F = pressure x Area

    3.

    When considering the Mass alone I have

    Sum of forces F = F - kx - BDx = MD^2.x

    therefore x/F = 1 / K + BD + MD^2

    and finally, x/F = (1/k) / 1 + (B/k)D + (M/k)D^2




    However I am struggling to break the problem down further. Which parts of the system do I also need to consider and how do I go about manipulation of the equations to derive the transfer function....??

    Your help is very much appreciated.
     

    Attached Files:

  2. jcsd
  3. Dec 16, 2011 #2
    Re: Transfer function of flow measurement system HELP!!

    You have an equation relating pressure difference to volumetric flow. The pressure on each side of the transducer would have a contribution to force on the transducer plate.
     
  4. Dec 19, 2011 #3
    Re: Transfer function of flow measurement system HELP!!

    I am still struggling with this one im afraid. I cant figure out where the 2Mf/k . A^2/a^2 term originates from...

    Some guidance would be very much appreciated...
     
  5. Dec 21, 2011 #4
    Re: Transfer function of flow measurement system HELP!!

    Ok this is where I have solved to thus far:

    I have the equation x/F = 1 / (k + BD + MD^2)

    but F = Pressure x Area

    so Pressure difference on transducer mass P2 - P3 x Area = Force (F)

    therefore,

    x / A (P2-P3) = 1 / (k + BD + MD^2)

    and then,

    x / (P2 - P3) = A / (k + BD + MD^2)

    then finally divide rhs of equation by k to give,

    x / (P2 - P3) = A/k / ((1 + B/k(D) + M/k(D^2))

    My next equation of (P2 - P3)C = f

    so substituting this in gives,

    x/flow = A/CK / ((1 + B/k(D) + M/k(D^2))

    BUT where does the A^2 / a^2 term come in....??

    Any pointers?

    Thanks
     
  6. Dec 21, 2011 #5
    Re: Transfer function of flow measurement system HELP!!

    I would guess this.
    You may consider the smaller diameter pipes to be obstructions. Whenever the plate would be in motion [itex]\left( \frac{dx}{dt} \neq 0 \right)[/itex] then fluid would be flowing through these pipes. Thus p1 would not be equal to p2 and p3 would not be equal to p4.

    Even though fluid would not pass through the plate, you may consider the path through both smaller pipes as one flow.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Transfer function of flow measurement system HELP
  1. Transfer Function help (Replies: 1)

Loading...