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#1




cylindrical ironclad solenoid magnet
Hey guys
I'm having some trouble withe the following problem. 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
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Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. Archimedes 
#2




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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Jess_88 (09062011) 
#3




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?
__________________
Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. Archimedes 
#4




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 ... 
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Jess_88 (09062011) 
#5




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? Would this be the magnetic equivalent circuit? thanks
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Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. Archimedes Last edited by Jess_88; 09062011 at 06:01 AM. 
#6




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?
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Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. Archimedes 
#7




Quote:
Using the general formula for reluctance 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 where r is the radius as a variable d is the cylinder height μ0 is permeability of free space After integration ... Note that 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 by t_n_k; 09062011 at 07:36 AM. 
#8




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?
__________________
Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. Archimedes 
#9




Exactly  you are correct.

#10




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
__________________
Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. Archimedes 
Tags 
cylindrical, ironclad, magnet, solenoid 
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