Reluctance of EN1A Steel. 2

[Lakey]

Hi Forum,

Can anyone provide me (or point me in the right direction) with a ball park figure for the Reluctance of ENIA steel?

I have contacted and questioned our materials supplier, unfortunately they don’t have the information I require, but they have supplied me with a magnetic properties table which cross references “magnetic induction B (gauss) for magnetizing field H (Oersted). – Unfortunately I don’t have the expertise to know if it’s useful or not!

EN1A is leaded mild steel that is easily machined.

I would appreciate your comments.

Thanks.

 

[jbartos]

Suggestion: Check with

http://www.westyorkssteel.com/

 

[electricpete]

Do you have B and H?

Then find a parir of (B,H) in the linear range (assuming that's where you'll be using it).

mu=B/H

Now reluctance can be calculated from mu.

Reluctance = Lm/(mu*A)
where A is cross-sectional area (perpendicular to the flux) and Lm is mean path length of flux

Sorry if I have misunderstood the quesiton.

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[aolalde]

Reluctance is a parameter of a defined magnetic circuit. It is a function of the configuration and dimensions you have as well as the magnetic saturation.

Electricpete gave the expression to calculate it, but first you should know the flux intensity, then the flux density then find “mu= B/H” on the curve provided for your material.

This is a kind of equivalent to “Resistance” on electric circuits; you need a defined conductor length and section to calculate its resistance, before you will have only conductivity of the material (equivalent to “mu” permeability of the material, which changes with flux density or magnetic saturation).

 

[Lakey]

Thank you for your input guys.

I now have an answer to the original post.

But more information is required I think: I’m making a small homemade electromagnet / solenoid and I’m curious about the formula, which calculates the force developed by an electromagnet: -

Example 1 – Ref: Relay Handbook.

F = 2*Pi(N)**2 / A (R + x/A)**2

F = Force (N)
N = ampere-turns (At/m)
A = pole piece area (sqmetres?)
x = distance between moving core and its stop (sqmetres?)
R = reluctance of magnetic circuit
**2 = exponent 2


[1] I’m using a 416At coil, which has 3200 turns, and where the current is set at 130mA. H = 416 (At/m)

[2] Therefore, using data provided for ENIA, the flux density would be approximately 0.3T.

Note:

[a] The moving core, frame and stop are made from mild steel (ENIA), with the coil it is configured as an ‘open D-frame solenoid’.

The moving core and stop is flat ended and provides a surface area of 100.53sqmm or 0.10053sqm

[c] The air gap between moving core and stop is 0.65mm or 0.00065m

[3] Re: The reluctance ‘R’ = Lm/(mu*A) [Thx Pete] of the magnetic frame is thought to be approximately = 0.00065m / ((0.30T/416H)*0.10053sqm) = 8.96 – do my peers concur?

That is, if Lm (mean path length of flux) is equal to the air gap between the moving core and its stop.

Question 1: Is Lm (mean path length of flux) is equal to the air gap between the moving core and its stop?
Question 2: Or is Lm equal to the length of the air gap plus the length of the magnetic frame that surrounds the coil?

Because when F = 2*Pi(N)**2 / A (R + x/A)**2
F = 2*Pi(416)**2 / 0.10053 (8.96 +(0.00065 / 0.10053))**2
F = 134532.8N

Which is not correct because I’ve taken the readings from the electromagnet and they don’t come even close to the calculated values above. The measured values were approximately 10N each time!!

Question 3: What am I missing – Is it the units or is there something else !!

Question 4: would the shape of the moving core and its stop affect the original formula? I.e. should there be any consideration for the geometry of the pole pieces. I know those conical moving cores and their stops produce a different Force/Distance plots to that of the flat-ended type that I’m using).

Guys, I’m an amateur in this field and I appreciate your patience.

P.S I can assure you that it is not a school project. It’s for personal achievement only.

 

[Lakey]

Sorry about the format of the text - it looked OK in the Preview!

 

[jbartos]

Suggestion: The equation needs to be revisited. Please notice that:
L(x)=muo * N^2 * l * d * (1 - x/d)/(2 * g)
from
Page 122, "Electric Machinery" 6th Edition by A.E. Fitzgerald, Charles Kingsley, Jr., Stephen D. Umans.
Permeability for free space muo=4 x pi x 10^-7 H/m

 

[electricpete]

The magnetic circuit has series combinatio of ReluctanceAir and ReluctanceIron. Since MuAir << MuIron, then ReluctanceAir >> ReluctanceIron, it will neglect ReluctanceIron. Do the whole problem only looking at the airgap.

A small error in my reluctance formula (Mu goes in denominator, not numerator)
Rair = g / (A*Mu0)

PhiAir = NI/Rair = NI * A*Mu0 / g

Airgap Energy W = A * g * 0.5 * B^2 / Mu0
= A* g * 0.5 * (Phi/A)^2 / Mu0
= g * 0.5 * (Phi^2/A) / Mu0
substitute in PhiAIR = NI * A*Mu0 / (g)
Airgap Energy W = g * 0.5 * ( NI * A*Mu0 / g )^2 / (A* Mu0 )
=

Force = -dw/dg = 0.5 * N^2I^2 A *Mu0 / g^2

Mu0=4PiE-7 * H/m = 4PiE-7 Newton/A^2

Plug in and make sure the units work out right. Also double check me, I am going from memory.


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[electricpete]

One dangling = sign was supposed to be simplificaiton of previous expression:

[W] = 0.5*N^2*I^2*A*Mu0/g

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[electricpete]

Force = 0.5 * N^2I^2 A *Mu0 / g^2
= 0.5*3200^2*[0.13^2 Amp^2] *[ 0.10053m^2] * [4*PIE-7 Newton/(Amp^2) ]/ (0.00065^2*m^2)
Amp^2 and m^2 cancel from numerator and denominator and result is in Newton.

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[electricpete]

4PIE-7 of course should be 4*Pi E-7
A humourous &quot;correction&quot; from my friendly word processor

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[Lakey]

Thank you for that guys - I will re-evaluate.

Can you tell me what the Lm is? The air gap or the whole length of the magnetic circuit, which surrounds the coil.

Do you know if geometry of the pole pieces will effect the formula?

Best regards,

 

[electricpete]

My simple approach as outlined above is to neglect the reluctance of the iron. It is somewhat similar to neglecting the resistance of copper wires connecting resistors in an electric circuit.

With that assumption, the only Lm of importance is air-gap length (I called it g).

If you are wanting to add effect of iron the Lm is the length that the flux has to travel through pole pieces.

Recognize there is bound to be a lot of error due to inability to calculate things like fringing. Based on that I think neglecting iron is fine.

Your equation I have not seen and not too sure exactly what they're doing. Your term (R + x/A) I take it is supposed to represent the series combination of magnetic reluctance (R) and air gap reluctance (x/A). I think in that case there is an assumption that relative permeability is used. I don't see anywhere in the whole thing the factor on the order of 1E-6 associated with mu0 (although I see the 4 Pi). It may be correct but I can't figure it out. Is there a derivation provided?

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[electricpete]

Your solution:
F = 2*Pi(N)**2 / A (R + x/A)**2
Assume R~0
F = 2*Pi(NI)**2 / A (x/A)**2
= 2*Pi(NI)**2 * A / x^^2

Compare to my solution:
F = 0.5 * N^2I^2 A *Mu0 / g^2
Substitue Mu0 = 4*Pi E - 7
F = 0.5 * N^2I^2 A *4*Pi E – 7 / g^2
F = 2*Pi E-7 * N^2I^2 A / g^2

It looks like they are the same except for minor detail factor of 1E-7

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[electricpete]

If I were you I would trust the solution that allows you to plug in the units and verify correct units on final result (mine).

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[jbartos]

Suggestion: The Lm should be pertinent to area A only. If there is a fringing flux creating the area A bigger, then the Lm should be restricted to the bigger area only, and called Lma or so.

 

[Lakey]

Guys,

Thank you for you patience.

I accept the reluctance of the gap/gaps far exceeds that of the magnetic circuit and will focus on that. Also, at this stage I will ignore any fringing that is occurring.

But I do need to clarify one thing:

The gap and area between the moving core (armature) and its stop is known – no problem.

But should allowance be made for the air gap between the moving core and the metal insulated tube, in which, the core travels?


! winding
! -------------------!
[stop] g [moving Core]
! ------------------!
! winding !
!--------------------------!


---- & !!! - Indicates the metal parts/frame

The coil is wound around the tube. The tube is connected to a frame, which surrounds the coil and is connected to the metal stop that limits the travel of the core. The stop has a hole in it for force measurements.

I agree. E-pete’s formula was better than my original, but I have another one for the pot:

F = Muo (NI)^2 A / 4 g^2

What I can’t see, in any of the methods, is an allowance for the winding’s mean diameter; I think this should have a bearing on things!

The harder I dig the more interesting my problem becomes, and I won’t cease until the calculated value is within 5% of the measured value.

Thanks & regards,

 

[electricpete]

My assumption was that the area A applied to the air gap.

Now to model the effects of the two air gaps, you need to treat as 2 reluctances in series.
R1 = g1/(A1*mu0) can be reluctance of gap by the stop..
R2 = g2/(A2*mu0) can be the reluctance of gap between core and plunger.
(where A2 is the area of the unfolded cylinder = 2*Pi*Length*Radius )
Rtotal = R1+R2 = g1/(A1*mu0) + g2/(A2*mu0) = (A2*g1 + A1*g2 )/ (A1*A2*mu0)

Use this in place of previous Rair = g/(Amu0) to derive the expression.

Ie. Go back to
PhiAir = NI/Rtotal = (NI * A1*A2*Mu0 )/ (A2*g1 + A1*g2 )

Airgap Energy W = A * g * 0.5 * B^2 / Mu0
= A* g * 0.5 * (Phi/A)^2 / Mu0
= g * 0.5 * (Phi^2/A) / Mu0
substitute in Phi = (NI * A1*A2*Mu0 )/ (A2*g1 + A1*g2 )

Do the algebra to simplify W (I will do it later if I have the chance)

Force = -dW/dg

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[Lakey]

Aaah. Thank you Pete.

That mostly makes sense.

All I need to do now is get a grip of:

Force (N) = -dW/dg

I guess, F = minus delta Watts (air gap energy)/ delta gap.

Once sussed, solenoid force (Newtons) = xyz (design parameters) etc.

Please define or correct me if i'm wrong.

I'm most obliged,
Lakey

 

[jbartos]

Suggestion: Reference:
A.E. Fitzgerald, Charles Kingsley, Jr., Stephen D. Umans &quot;Electric Machinery,&quot; Fifth Edition, McGraw-Hill, Inc., 1990,
page 110 Example 3-3

Ffld = - muo * pi * r * (N * i)^2 / (4g)

or

Ffld = - (g * lambda^2)/{muo * pi * r * [N * d * (1-x/d)]^2}

The example 3-3 uses rectangular area under the South & North poles around the plunger. This was modified to half-cylindrical area under the North & South poles around the plunger.