cylindrical iron-clad solenoid magnet

Thread Starter

Jess_88

Joined Apr 29, 2011
174
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 :)
 

t_n_k

Joined Mar 6, 2009
5,447
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.
 

Thread Starter

Jess_88

Joined Apr 29, 2011
174
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?
 

t_n_k

Joined Mar 6, 2009
5,447
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 ...

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

Thread Starter

Jess_88

Joined Apr 29, 2011
174
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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Thread Starter

Jess_88

Joined Apr 29, 2011
174
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?

 

t_n_k

Joined Mar 6, 2009
5,447
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?
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.
 
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Thread Starter

Jess_88

Joined Apr 29, 2011
174
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?
 

t_n_k

Joined Mar 6, 2009
5,447
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?
Exactly - you are correct.
 

Thread Starter

Jess_88

Joined Apr 29, 2011
174
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
 

t_n_k

Joined Mar 6, 2009
5,447
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
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

\(A=\pi R^2\)
 
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