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Optimize the design of a magnetic circuit actuator

  1. Oct 2, 2015 #1
    1. The problem statement, all variables and given/known data

    The aim is to optimize this actuator to get the maximum force from it. Here is the circuit diagram below:
    Circuit act.PNG model.PNG

    2. Relevant equations


    3. The attempt at a solution

    Length of air gap top (Ragt in diagram) = 0.5 mm
    Length of air gap centre (Ragc) = 0.25 mm
    Current in coil = 0.8 A (I calculated it), current in coil must be less than 1A

    I'm not going to put too many calculations here because it might be too long and people reading it will get a headache, but anyway I have some questions

    Will the flux divide equally ?, my guess is no because the air gap reluctance's are different

    To get the drop in RcoreBCDA, I use F = Hl, now my question is will the mean path length(l) run from ABCD ?

    Lastly, any guidelines on how to solve for the magnetic fluxes and mmf drops ?, I did attempt it but not sure it it's correct

    Any help will be greatly appreciated, this my design project and it's due very soon
     
    Last edited by a moderator: Oct 2, 2015
  2. jcsd
  3. Oct 3, 2015 #2

    rude man

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    Right.
    Use mmf = Σ RΦ (which = H l but I don't think you need it). R = reluctance.
    You drew a good equivalent circuit; use it.
    BTW for mmf you need more than the current. what is mmf for the coil?
     
  4. Oct 3, 2015 #3
    What's the relative permeability for "Hot rolled silicon steel" ?, I can't find it online
     
  5. Oct 3, 2015 #4

    rude man

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    Since this is a lab project, a 'real-world' problem, I suggest assuming all the mmf drops be considered across the gaps.
     
  6. Oct 3, 2015 #5
    Yeah, which is why I was given a BH curve for "hot rolled silicon steel", i'm guessing I will have to use that

    Here's what I did:
    Since magnetic circuits are analogus to electrical circuits, I used KVL and made two equations like this,

    Top loop:

    Ftotal = (phi1*RcoreAB) + (phi1*Rgc) + (phi2*Rgt) + (phi2*RcoreBCDA)

    Bottom loop:

    Ftotal = (phi1*RcoreAB) + (phi1*Rgc) + (phi3*RcoreBEFA)

    If I simplify by equating them and saying that RcoreBEFA = RcoreBCDA, I get this expression

    phi3/phi2 = (Rgt/RcoreBEFA) + 1

    I'm not sure if that's right ?, also will RcoreBCDA = RcoreBEFA ?, thanks
     
  7. Oct 3, 2015 #6

    rude man

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    Well, that wasn't my suggestion.
    My suggestion amounted to RcoreBCDA = RcoreBEFA = RcoreAB = RAGB = 0. That's because the permeability of rolled steel is >> permeability of air.
    OK, so look at your equivalent circuit in this (totally justifiably) simplified way. What do you think the flux in gap AGT will be?
     
  8. Oct 4, 2015 #7
    Will it be zero ?, because if their's no reluctance in the bottom then all the flux would flow their
     
  9. Oct 4, 2015 #8

    rude man

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    Right! But this assumes the bottom gap reluctance RAGB really is zero which it really isn't, but I guess your teacher said to assume it's zero so it's zero! :smile:
     
  10. Oct 4, 2015 #9
    So how does that help me get the mmf drops and fluxes of the circuit ?
     
  11. Oct 4, 2015 #10

    rude man

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    Your equivalent circuit answers that question. The mmf drop will be entirely across the middle gap. Flux = mmf/reluctance of gap.
    As I hinted before, you need to know the number of turns in your coil, not just the current.
    Your next question should be: at a given setting of the middle air gap, what's the force pulling the actuator arm into the actuator?
     
  12. Oct 4, 2015 #11
    The number of turns is 850 and the current is 0.89A (I recalculated it)

    So F = NI = 850*0.89 = 753.63 = Which is the drop across the center gap ?

    To calculate force, this is the equation I was given

    F = -1/2(N*I/R)^2 * dR/dx

    I don't know how to get dR/dx which is the change in reluctance with respect to a change in air gap
     
  13. Oct 4, 2015 #12

    rude man

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    right
    Can you compute the reluctance of the air gap for a given opening, say a given angle the arm makes with the body?
     
  14. Oct 4, 2015 #13
    For the top air gap I chose randomly 1.2 mm, then by using similar triangles I got the center gap to be 0.6 mm, like this

    lgt/40 = lgc/20..............where lgt = 1.2

    Solving for lgc, I get 0.6 mm

    The reluctance = L/u*a
    = 0.6/(4pi*10^-7)*10*16, where 10*16 is the area
    = 2.98*10^6 At/Wb

    Oh and these are the dimensions:

    Lam details.PNG
    Dim.PNG
     
  15. Oct 4, 2015 #14

    rude man

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    OK, so you have a gap with width x, what is R as a function of x? If the gap width x is changed by an amount dx, what is the change in R?
     
  16. Oct 4, 2015 #15
    R = L/u*A
    So,
    dR = dx/u*A , Is that the formula ?
     
  17. Oct 4, 2015 #16

    rude man

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    Good chance! :smile:
     
  18. Oct 4, 2015 #17
    So dR/dx = 1/u*A ?, but it's just a constant
     
  19. Oct 4, 2015 #18

    rude man

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    you don't like constants? :smile:
     
  20. Oct 4, 2015 #19
    I love them :)

    Then would the area be the area of the center gap ?

    And in the formula, F = -1/2(N*I/R)^2 * dR/dx

    Is R the total reluctance of the circuit ?
     
  21. Oct 4, 2015 #20

    rude man

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    right
    well, you were given the formula - I have not derived it but I would go with R of the middle gap. You might check with the instructor.
    It's too bad you weren't asked to derive the formula as it would have introduced you to a very important concept, that of "virtual work".
     
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