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Attracting force of an electromagnet ? 1
- Thread starter Warpspeed
- Start date
To add to my previous post, using the electromagnet dimensions the magntic circuit need to be solved calculating the flux passing in the electromagnet and the piece it arttract including flux leakages.
Are you designing an electromagnet or using one and want to calculate the force from 1.6mm distance?
If you like to study the matter the best is the book "Electomagnetic devices" by Roters 1941.
Are you designing an electromagnet or using one and want to calculate the force from 1.6mm distance?
If you like to study the matter the best is the book "Electomagnetic devices" by Roters 1941.
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1
- Thread starter
- #4
All my literature here deals with electrical and magnetic quantities, but nowhere can I find reference to mechanical force.
The dimensions of the magnetic steel path are such that it is well below saturation, all of the effective reluctance is in the airgap. This airgap is 1.6mm thick with an effective diameter of around 140mm.
The dimensions of the magnetic steel path are such that it is well below saturation, all of the effective reluctance is in the airgap. This airgap is 1.6mm thick with an effective diameter of around 140mm.
The force is the derivetive of the energy in the gap with respect to the distance between the electromagnet and the attracted part. You will find the derivation in the book from Roters. You can also do an FEA analysis. Why not use a consultant to do the analysis for you?
- Thread starter
- #6
I have designed and developed DC electromagnets and electro-pneumatic valves for many years including computer simulation writing based on Roters book to analyse transient and steady state forces and movement. I also did FEA analysis for electromagnets and electro-pneumatic valves and permanent DC motors.
To my best knowledge there is no thumb rule answer to your question without seeing the detailed design of the electromagnet. If it is a standard off the shelf one then you could consult the manufacturer documentation/catalog if you can find one. If it is a new design then you have to do an analysis or a test. To get an answer it will take work to do.
This forum (at least to my understanding) is not meant that someone will invest hours to give an answer but to give guidance in such cases and in simpler cases where the answer doesn't require time investment to give you a direct answer.
To my best knowledge there is no thumb rule answer to your question without seeing the detailed design of the electromagnet. If it is a standard off the shelf one then you could consult the manufacturer documentation/catalog if you can find one. If it is a new design then you have to do an analysis or a test. To get an answer it will take work to do.
This forum (at least to my understanding) is not meant that someone will invest hours to give an answer but to give guidance in such cases and in simpler cases where the answer doesn't require time investment to give you a direct answer.
- Thread starter
- #10
Thank you all, I thought it was a fairly simple question, and I only really needed a fairly rough ballpark figure.
I will simply have to measure it, not too difficult.
Like many electrical problems there is the quick dirty first order estimate, good enough in this case. Actual measurement is probably the best real world solution anyhow.
Cheers......
I will simply have to measure it, not too difficult.
Like many electrical problems there is the quick dirty first order estimate, good enough in this case. Actual measurement is probably the best real world solution anyhow.
Cheers......
Roughly,
F = B x H x Area
For only one gap and and a decently designed magnet, H is approx = NI/L where NI = your amp turns and L = length of air gap. B = u x H. Area is region that the flux is passing through. F in Newtons, B in Tesla, H in amp-turns/mm, Area in meters squared, u = 4 x pi x 10E-07.
F = B x H x Area
For only one gap and and a decently designed magnet, H is approx = NI/L where NI = your amp turns and L = length of air gap. B = u x H. Area is region that the flux is passing through. F in Newtons, B in Tesla, H in amp-turns/mm, Area in meters squared, u = 4 x pi x 10E-07.
BobM2
This is maybe true assuming no sigma(H*L) of the irom parts But:
1. Who know how long the iron parts are and how saturated they are?
2. There maybe not just one air gap but more depends on the magnet design.
3. What about leakage?
4. How about the side force? Once I had to modified/re-design Ledex solenoid that gave 6kg to get 10kg by adding few mm to it's length. FEA and theoretical analysis gave pulliong force of 50kg but the actual pulling force was 10kg. Further investigation revealed that the side force was so high that the friction mutiplied by this side force deducted 40kg (80%).
This is maybe true assuming no sigma(H*L) of the irom parts But:
1. Who know how long the iron parts are and how saturated they are?
2. There maybe not just one air gap but more depends on the magnet design.
3. What about leakage?
4. How about the side force? Once I had to modified/re-design Ledex solenoid that gave 6kg to get 10kg by adding few mm to it's length. FEA and theoretical analysis gave pulliong force of 50kg but the actual pulling force was 10kg. Further investigation revealed that the side force was so high that the friction mutiplied by this side force deducted 40kg (80%).
- Thread starter
- #13
BobM2,
Many thanks for your assistance. While I am very familiar with transformer design, and the application of Faraday's law, dc solenoids are an entirely different beast and quite new to me. I previously attempted a reverse design along the lines you suggest but the various odd assortment of magnetic units finally defeated me. I will attempt it again using your method.
Israelkk,
I appreciate your concern that the devil is in the details, but in this case the magnetic path is both physically large, open, and simple. Fringing, flux leakage, and saturation are not going to significantly effect the result, and neither is the slight superimposed ac ripple on the rectified drive current. I am only looking for a +/-20% ballpark figure to resolve some mechanical mounting and loading issues.
Many thanks for your assistance. While I am very familiar with transformer design, and the application of Faraday's law, dc solenoids are an entirely different beast and quite new to me. I previously attempted a reverse design along the lines you suggest but the various odd assortment of magnetic units finally defeated me. I will attempt it again using your method.
Israelkk,
I appreciate your concern that the devil is in the details, but in this case the magnetic path is both physically large, open, and simple. Fringing, flux leakage, and saturation are not going to significantly effect the result, and neither is the slight superimposed ac ripple on the rectified drive current. I am only looking for a +/-20% ballpark figure to resolve some mechanical mounting and loading issues.
Hi Warpspeed,
A very quick calculation of your application shows that for the parameters that you have described:
NI = 5,700
Diameter = 140mm (5.5 in)
Air-Gap = 1.6mm (0.063 in)
will definitely produce a magnetic field that would drive any common low-carbon steel well into saturation, which was NOT your original assumption. Not only that, but the very large cross-sectional area of your application, coupled with the incredibly small air-gap (1.6mm), and an NI of 5,700 will develop several thousand pounds of pull-in force. Please consider this when testing or attempting to measure the actual force. I would suggest that you start your force testing with a very small amount of current and then SLOWLY and CAREFULLY increase the coil current after you have a better feel for the forces generated with this application.
A very quick calculation of your application shows that for the parameters that you have described:
NI = 5,700
Diameter = 140mm (5.5 in)
Air-Gap = 1.6mm (0.063 in)
will definitely produce a magnetic field that would drive any common low-carbon steel well into saturation, which was NOT your original assumption. Not only that, but the very large cross-sectional area of your application, coupled with the incredibly small air-gap (1.6mm), and an NI of 5,700 will develop several thousand pounds of pull-in force. Please consider this when testing or attempting to measure the actual force. I would suggest that you start your force testing with a very small amount of current and then SLOWLY and CAREFULLY increase the coil current after you have a better feel for the forces generated with this application.
- Thread starter
- #16
Thank you frankeng9 for your concern.
I agree the ampere turns do sound awfully high, and testing will proceed as you suggest starting off in small increments. Saturation should be readily apparent when it does occur.
This device already exists as a commercial product, but has been slightly mechanically redesigned, and operates at a different voltage but with the same ampere turns. The forces were expected to be very high which is why I am trying to get a ballpark figure for the magnetic attraction.
The whole job is on hold at the moment, but should hopefully reach the testing stage in a few more weeks. Once again, thank you all for your help in this.
I agree the ampere turns do sound awfully high, and testing will proceed as you suggest starting off in small increments. Saturation should be readily apparent when it does occur.
This device already exists as a commercial product, but has been slightly mechanically redesigned, and operates at a different voltage but with the same ampere turns. The forces were expected to be very high which is why I am trying to get a ballpark figure for the magnetic attraction.
The whole job is on hold at the moment, but should hopefully reach the testing stage in a few more weeks. Once again, thank you all for your help in this.
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