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.

 

[electricpete]

I plugged it into computer to help on the algebra. It makes is a little tough to read, sorry about that. Please check it for yourself.

Please check it yourself.

PhiAir := N*I_/(g1*A2+g2*A1)*A1*mu0*A2

W:= A1* g1 * 0.5 * (PhiAir/A1)^2 / Mu0;
W := .5*A1*g1*N^2*I_^2/(g1*A2+g2*A1)^2*mu0^2*A2^2/Mu0

Derivative of W with respect to g1 is:
0.500000*A1*N^2*I_^2*mu0^2*A2^2*(g1*A2-1.*g2*A1)/(g1*A2+g2*A1)^3/Mu0


=====================================
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[jbartos]

Suggestion to Lakey (Electrical) Feb 2, 2004
Please, would you elaborate on the shape of magnetic circuit since various forms, leading to different formulae, are possible? My cited reference has another example, where the magnetic circuit surrounds the coil leading to two spaces in the magnetic circuit attracting the plunger. The formulae will be different from the posted ones.

 

[Lakey]

OK Jbartos, here goes….

Cross section of electromagnet:

ZZZZZZZZZZZZZZZZZZZZZZZZZZ
Z

http://WWWWWWWW

Z
Z

http://WWWWWWWW

Z
Z

http://WWWWWWWW

Z
Z ZZZZZZZZZZZZZZZZ
SSSS CCCCCCCCCCCC
------ CCCCCCCCCCCC
SSSS CCCCCCCCCCCC
Z ZZZZZZZZZZZZZZZZZ
Z

http://WWWWWWWWW

Z
Z

http://WWWWWWWWW

Z
Z

http://WWWWWWWWW

Z
ZZZZZZZZZZZZZZZZZZZZZZZZZZZZ


W = winding
Z = metal frame / tube
C = moving core / metal cylinder
S = stop for moving core.
---- = a non magnetic probe which is attached to e.g. force gauge

Air gap between S & C is between 0 and 5mm
Air gap between C & Z is 0.8mm

When winding (W) is activated - C is attracted to S.

I hope this clarifies.

Regards,

 

[jbartos]

Suggestion to the previous posting:
The magnetic circuit is not properly posed. It will not provide a closed path for magnetic flux if the flux path in the air outside the tube is considered unacceptable because of high Reluctance, say Ra.
The my cited reference diferent example 3.7 and figure 3-19 on page 126 would lead to:
zzzzzz zzzzzz
z

http://wwwccccwwwz

z

http://wwwccccwwwz

z

http://wwwccccwwwz

zzzzzzcccczzzzzz
cccc

Let:
g=0.8mm, air gap
a=5mm, distance of travel=frame z thickness
d=?, mean diameter, Distance from Z to Z
Reluctance of the upper gap is:
Rg=g/[muo * pi * x * d]
Reluctance of the lower gap is:
Rg=g/[muo * pi * a * d]
Total Reluctance R=Rg+Ra=[g/(muo*pi*a*d)]*[(a+x)/x]
L(x)=N^2/R=(muo * pi * a * d * N^2 / g)*[x/(a+x)]
L'=(muo * pi * a * d * N^2 / g)
F(x)=(1/2) * i^2 * dL(x)/dx = (1/2) * L' * a * i^2 /(a + x)^2

 

[Lakey]

Dear, Jb & E-pete

Thanks again for you assistance, and I will try to acquire a copy of the reference that has been mentioned.


Jb, with regard to the previous post, please clarify a couple of points for me and forgive me if I’ve misunderstood.

//Reluctance of the upper gap is: Rg=g/[muo * pi * x * d]\
///Reluctance of the lower gap is: Rg=g/[muo * pi * a * d]\\
Rg = gap / (1.256*10^-6)*3.14*area*mean diameter of frame

I am unclear about “x” and “a”.

Also,

//// L(x)=N^2/R=(muo * pi * a * d * N^2 / g)*[x/(a+x)]
L'=(muo * pi * a * d * N^2 / g)
F(x)=(1/2) * i^2 * dL(x)/dx = (1/2) * L' * a * i^2 /(a + x)^2\\\
I am unclear about: -

L
L’
DL

Thanks & regards,

 

[jbartos]

Comment on the previous posting marked ////\\\Jb, with regard to the previous post, please clarify a couple of points for me and forgive me if I’ve misunderstood.

//Reluctance of the upper gap is: Rg=g/[muo * pi * x * d]\////This reluctance in the upper part of the magnetic circuit where the space in the magnetic circuit will be filled with the plunger as it travels distance x to the stopping point.\\\///Reluctance of the lower gap is: Rg=g/[muo * pi * a * d]\\////The lower space in the magnetic circuit (frame) is filled with the plunger iron\\\Rg = gap / (1.256*10^-6)*3.14*area*mean diameter of frame

I am unclear about “x” and “a”.
////x is the distance to be traveled by the plunger in the upper part of the magnetic circuit (frame), namely: your specified "Air gap between S & C is between 0 and 5mm"
a is the thickness of the lower part of the magnetic circuit (frame) already filled with the plunger. It may also be your specified "Air gap between S & C is between 0 and 5mm".\\\
Also,

//// L(x)=N^2/R=(muo * pi * a * d * N^2 / g)*[x/(a+x)]
L'=(muo * pi * a * d * N^2 / g)

////L' is a constant defined for convenience only\\\F(x)=(1/2) * i^2 * dL(x)/dx = (1/2) * L' * a * i^2 /(a + x)^2\\\////F(x) is a force upward the plunger will experience. It will depend among others on x-variable (0 to 5mm), current i-variable, etc.
for x=0
F(0)=(1/2) * i^2 * dL(x)/dx = (1/2) * L' * i^2 /a
and
for x=a
F(a)=(1/2) * i^2 * dL(x)/dx = (1/2) * L' * i^2 /(4a)
\\\I am unclear about: -

L
////L is inductance in the empty gap in the magnetic circuit that will be filled with the plunger iron.\\\L’
////L' is and auxiliary constant defined for convenience\\\DL
dL(x)/dx is a derivative of the inductance L(x) by x, where x is distance variable as the plunger travels upward.\\\////The presentation is on the University Undergraduate Level, equivalent to approximately BSEE\\\