cylindrical iron-clad solenoid magnet

Discussion in 'Homework Help' started by Jess_88, Sep 4, 2011.

  1. Jess_88

    Jess_88 Thread Starter Member

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    Hey guys :)

    I'm having some trouble withe the following problem.

    [​IMG]

    For the above circuit I need to
    a) Draw the magnetic equivalent circuit.
    b) Compute the flux density in the working air gap for x=10 mm.
    c) Compute the value of the energy stored in Wfld (for x=10 mm).
    d) Compute the value of the inductance L (for x=10 mm).
    e) For a force ffld of 1000 N determine for x=10 mm the current I=I0 required.

    given
    grad = 1 mm, R=20 mm, and d=10 mm. The exciting coil has N=1000 turns and carries a constant current of I=10A.

    My main problem at this point is actually part 'a'. This circuit is quite different to ones analysed in my lecturers and my text book hasn't got anything on it either. Is this a common configuration?

    I can't really see how the windings work in this case. Could someone explain what is going on with that? If its not to much trouble, a diagram would be much appreciated.

    thanks a lot guys :)
  2. t_n_k

    t_n_k AAC Fanatic!

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    The total mmf (NxI) is driving flux through two air gap reluctances in series.

    There is a central air gap created by a "solid" cylinder gap of length 'X' sitting atop the infinitely permeable plunger of radius 'R' - use area of a circle of radius 'R' to compute the reluctance of gap length 'X' with permeability μ0.

    There is also a thin cylindrical air gap of length 'grad' and cross sectional area equal to the total surface area of a cylinder of radius 'R' and height 'd'. Again gap permeability is μ0.
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  3. Jess_88

    Jess_88 Thread Starter Member

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    oh ok, I see. For the air gap of length 'grad' would that be times 2, because there is a gap on ether side?

    for the magnetic equivalent circuit, would that just be the resistance of the air gaps in series?
  4. t_n_k

    t_n_k AAC Fanatic!

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    I believe the grad gap is a continuous cylindrical air gap. What the picture shows is (I think) a section through a fully cylindrical apparatus.

    My guess is that calculating the cylindrical reluctance value would be similar in nature to a cylindrical capacitor problem - with some obvious differences.

    So I think for that part of the magnetic circuit ...

    Reluctance=\frac{1}{2\pi d \mu_0}ln(1+\frac{g_{rad}}{R})
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  5. Jess_88

    Jess_88 Thread Starter Member

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    I think your right.
    Can you show me how you got to the formula?
    mainly the 'ln(1 + grad/R)'
    I why wouldn't it be Reluctance= length/μ*Area ?

    Also, is this kinda what the system looks like in 3d?
    [​IMG]

    Would this be the magnetic equivalent circuit?
    [​IMG]

    thanks :)
    Last edited: Sep 6, 2011
  6. Jess_88

    Jess_88 Thread Starter Member

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    Another Question in relation calculating the Reluctance.
    When calculating the thin cylindrical air gap of length 'grad' (see my diagram), do I use the area of the green shaded surface or the red shaded surface?

    [​IMG]
  7. t_n_k

    t_n_k AAC Fanatic!

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    The red shaded surface

    Using the general formula for reluctance

    R_{el}=l/\mu A

    In the cylindrical case the flux density across the gap [grad] is not constant with varying radius r so we must integrate for small changes over the range of R to (R+grad) using the difference relationship

    ΔRel=Δr/μ0 A

    with

    A=2∏rd

    For a small change in the radial distance in the cylindrical case

    R_{el}=\int_{R}^{(R+g_{rad})}\frac{\delta r}{\mu_02\pi d r}

    where
    r is the radius as a variable
    d is the cylinder height
    μ0 is permeability of free space

    After integration ...

    R_{el}=[\frac{1}{2\pi \mu_0 d} log_e(r)]_R^{(R+g_{rad})}

    R_{el}=\frac{1}{2\pi \mu_0 d} \{ log_e(R+g_{rad})-log_e(R)\}

    R_{el}=\frac{1}{2\pi \mu_0 d}  log_e[\frac{(R+g_{rad})}{R}]

    R_{el}=\frac{1}{2\pi \mu_0 d}  log_e[1+\frac{g_{rad}}{R}]

    Note that ln()==log_e()

    Because the gap is small in relation to the radius R the error in taking a simple linear gap length of 1mm with the same cross sectional area gives only a slight difference in reluctance value between the true and linear approximation values.
    Last edited: Sep 6, 2011
  8. Jess_88

    Jess_88 Thread Starter Member

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    Ah thats great. Very nice explanation.
    One thing that is confusing me about this is, that the plunger is guided so that it can move in vertical direction only. So wouldn't Rgrad remain the same (constant), and the distance 'x' change?
  9. t_n_k

    t_n_k AAC Fanatic!

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    Exactly - you are correct.
  10. Jess_88

    Jess_88 Thread Starter Member

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    so the flux density across the gap [grad] is constant but not across the x gap.
    then I use Re = length/μ*Area for the grad gap?
    Would I need to use ΔRel=Δr/μ0 A for the x gap?
    how would I do that?

    thanks
  11. t_n_k

    t_n_k AAC Fanatic!

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    For a given mag circuit flux value, the flux density across the 'x' gap will be constant.

    The 'x' gap reluctance is then easy to calculate. Just apply the standard formula.

    R_{el}=\frac{x}{\mu_0 A}

    Where

    A=\pi R^2
    Last edited: Sep 6, 2011

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