I have a solenoid electromagnet whose cylindrical steel core protrudes out and has a chamfer at the end. When the core end contacts the mating part, the chamfer reduces the actual contact area by 1/2 (compared to the core diameter). I know pull force is dependent on core diameter, but is it dependent on the diameter of the contact area or of the core diameter within the solenoid? If I remove the chamfer (so that the contact diameter is the same as the core diameter, doubling the contact area) will I nearly double my pull force?
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Electromagnet contact area vs. force
- Thread starter cativo
- Start date
hi
I don't think it will double but the force will increase because you've reduced the air gap the flux will travel through.
I played with this calculator:-
I don't think it will double but the force will increase because you've reduced the air gap the flux will travel through.
I played with this calculator:-
- Thread starter
- #3
Thanks for the reply. Yes, I would think the clamp force would go up since the region that has the air gap (the chamfer area) would be getting closer to the contact plate. However, I'm not sure how the field would change with or without the chamfer, so it may just redistribute and stay almost the same. It's difficult to use those force calculators in this situation since they assume a uniform field within the air gap (I think).
Anyone care to run it through FEMM?
Anyone care to run it through FEMM?
Compositepro
Chemical
If you have any experience with magnets, you would know that if you have a cylindrical bar magnet it can be very difficult to pull it straight off a steel surface, and that you should push it to the side to create an air gap on one side of the contact area. With this air gap it is much easier to to pull the magnet away from the steel. So yes, there will be significantly reduced force with your chamfer and air gap.
- Thread starter
- #5
Intuitively, that would make sense. However, I know the field calculations change when there there is an air gap vs. no air gap, so I'm not sure what happens when both situations exist simultaneously. What might happen is that the magnetic flux lines may redistribute themselves within the smaller contact area, increasing the flux density by a the same amount of area reduction, thus maintaining the same clamping force. But I really don't know!
electricpete
Electrical
Tough question.
One aspect of the system to consider: when the core contacts the mating part, is there still an airgap in another part of the circuit, or does it become a closed magnetic circuit?
=====================================
(2B)+(2B)' ?
One aspect of the system to consider: when the core contacts the mating part, is there still an airgap in another part of the circuit, or does it become a closed magnetic circuit?
=====================================
(2B)+(2B)' ?
bascially F is proptional to square flux density (B^2) and contacting area (S), normally the chamfer is to increase B when the flux is constant. to simplify your case, two situations are considered:
case 1. your core is not saturated: decreasing S by half increases B twice, F = c. (2B)^2 x (1/2 S)= 2 c B^2.S will increase twice.
case 2. core is saturated: decreasing S will not change B, F decreases simply by half
it also can be between case 1 and 2
other condistions, such as close or open circuit, limited size of pore pieces, air gap, can affect B, so affect F.
case 1. your core is not saturated: decreasing S by half increases B twice, F = c. (2B)^2 x (1/2 S)= 2 c B^2.S will increase twice.
case 2. core is saturated: decreasing S will not change B, F decreases simply by half
it also can be between case 1 and 2
other condistions, such as close or open circuit, limited size of pore pieces, air gap, can affect B, so affect F.
- Thread starter
- #8
Excellent explanation, thanks!
Ok, so my coil parameters are as follows:
Coil Length (L): 17mm
Core Diameter (D): 4.75mm
# of Turns (N): 3,700
Current (I): .060 Amps
Core Material: 1110 Low Carbon Steel
Using the equation for Field Strength H = (N*I)/L = (3700*.060)/.017 or H = 13,059 A/m
Using the value of H in the B-H curve for 1010 Steel (don't have one for 1110), I get a B value of about 1.7T. And the point seems to be right at the knee of the curve, so its not saturated yet, but starting to make the turn.
Are my calculations correct?
If so, then my next question is, what alternate material can I use (instead of 1110 Steal) to increase the Flux B, thereby increase my clamping force? For example Carpenter Hiperco50 would produce a flux of over 2.2T and would be more saturated (so removing the chamfer would be more effective). I assume Hiperco50 is really expensive? What other material could give me significant improvement?
Ok, so my coil parameters are as follows:
Coil Length (L): 17mm
Core Diameter (D): 4.75mm
# of Turns (N): 3,700
Current (I): .060 Amps
Core Material: 1110 Low Carbon Steel
Using the equation for Field Strength H = (N*I)/L = (3700*.060)/.017 or H = 13,059 A/m
Using the value of H in the B-H curve for 1010 Steel (don't have one for 1110), I get a B value of about 1.7T. And the point seems to be right at the knee of the curve, so its not saturated yet, but starting to make the turn.
Are my calculations correct?
If so, then my next question is, what alternate material can I use (instead of 1110 Steal) to increase the Flux B, thereby increase my clamping force? For example Carpenter Hiperco50 would produce a flux of over 2.2T and would be more saturated (so removing the chamfer would be more effective). I assume Hiperco50 is really expensive? What other material could give me significant improvement?
Your calculations are incorrect. You assume that all the magnetomotive force (H) go into the iron. However, the magnetic flux flows through the air parts of the magnetic circuit too. Most of the magnetomotive force goes to overcome the air gaps resistance to the magnetic flux flow. Therefore, the force will be due to much less than 1.7T. It is more likely to assume 0.8 to 1 T. Only a full magnetic circle analysis either by FEA analysis or analytic analysis as suggested in the book Electromagnetic Devices by Roter, 1941 will give an intelligent estimate. Actual testing following the analysis is needed.
the calculaiton is reasonable assuming the solenoid is very long. regardless of what core materials used, open or close circuit, the magnetic field created by the soiloid itself is the same.
with an H field of 13,059A/m = 164 Oe, assuming a B of 1.7 T for 1110 steel is reasonable, which is close to saturation. if using annealed 1010 steel, the B could increase to 2.0T. A litle bit of chamfer can help to increase B at the contacting point, 1/2 area of chamfer is too much, sacrified too much of area.
With Hiperco 50, you can easily tget 2.2T at 164Oe. but if you have room to increase current or turn to increase the H field, you may want to go easily with this option.
with an H field of 13,059A/m = 164 Oe, assuming a B of 1.7 T for 1110 steel is reasonable, which is close to saturation. if using annealed 1010 steel, the B could increase to 2.0T. A litle bit of chamfer can help to increase B at the contacting point, 1/2 area of chamfer is too much, sacrified too much of area.
With Hiperco 50, you can easily tget 2.2T at 164Oe. but if you have room to increase current or turn to increase the H field, you may want to go easily with this option.
Of course, the calculated field is at the center of solenoid. The field will decrease drastically outside the solenoid if it is open circuit. For a close circuit with a uniform core (rings or toroid), the field is inversely proportional to the path length. I guess that was what Israelkk concerned.
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- #12
Yes, I would have to agree with israelkk. My calculation would have been more realistic if I were modeling it after a toroid (closed circuit), but I do not have a toroidal solenoid so the calculation is not valid.
I did manage to measure the B field within the empty solenoid with a probe and it came out to be 165 gauss. Not sure if it's correct since adding a steel core would bring it up hundreds of times, and that is too high once again.
In any case, even if I do have a field as low as 0.5 to 1 T, then the core's saturation would be at the lower steap side of the B-H curve. In this region, if I replace the 1110 Steel with 1006 Steel I would get around a 30-60% increase, according to the B-H data I have. If I replace the 1110 core with a Silicon Steel I would double the field.
Does this sound reasonable?
I did manage to measure the B field within the empty solenoid with a probe and it came out to be 165 gauss. Not sure if it's correct since adding a steel core would bring it up hundreds of times, and that is too high once again.
In any case, even if I do have a field as low as 0.5 to 1 T, then the core's saturation would be at the lower steap side of the B-H curve. In this region, if I replace the 1110 Steel with 1006 Steel I would get around a 30-60% increase, according to the B-H data I have. If I replace the 1110 core with a Silicon Steel I would double the field.
Does this sound reasonable?
no, your calculation is realistic and correct, your mesurement (165 G) approved it already.
you just missed the point. The H field inside the solenoid has nothing to do with the circuit, has nothing to do with what kinds of core materials used. when you measured the B field of 165 Gs, it is also H field of 165 Oe! with a steel core in it, the B field will increase to 165 + permeability x 165.
From the B-H curves attached, 165Oe (13200A/M) is close to saturtion for all the core materials. i.e. all three materials will give you a simlimar B field!
you just missed the point. The H field inside the solenoid has nothing to do with the circuit, has nothing to do with what kinds of core materials used. when you measured the B field of 165 Gs, it is also H field of 165 Oe! with a steel core in it, the B field will increase to 165 + permeability x 165.
From the B-H curves attached, 165Oe (13200A/M) is close to saturtion for all the core materials. i.e. all three materials will give you a simlimar B field!
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- #14
Ok, I got it now, my apologies. My problem was I was mixing units. Funny, the answer was staring at me in face when the calculation came out 165Oe and I measured 165G, Ha! Couldn't ask for a better empirical confirmation.
The only part I need clarification is calculating the B field with the steel core in it. You said it would be 165 + Permeability X 165. Is that relative permeability? Relative MUr for 1010 steel is about 200 @ 165G, so .165 + (200 X .165) = 33.165T. Is it really that high inside the core?
The only part I need clarification is calculating the B field with the steel core in it. You said it would be 165 + Permeability X 165. Is that relative permeability? Relative MUr for 1010 steel is about 200 @ 165G, so .165 + (200 X .165) = 33.165T. Is it really that high inside the core?
The unit system in magnetics is complicated. People are very easy to be confused.
It is correct you measured B of 165 Gs using a Hall probe. Since in the air Gs = Oe, the H field generated by the solenoid is also 165 Oe.
at 165 Oe = 13,200 A/m, the steel is close to saturation, no way the relative permeability is 200, i guess it is about 100 @ 165, so the B is 165 Gs + 100x165 Gs = 16,665 Gs = 1.67 T.
It is correct you measured B of 165 Gs using a Hall probe. Since in the air Gs = Oe, the H field generated by the solenoid is also 165 Oe.
at 165 Oe = 13,200 A/m, the steel is close to saturation, no way the relative permeability is 200, i guess it is about 100 @ 165, so the B is 165 Gs + 100x165 Gs = 16,665 Gs = 1.67 T.
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- #16
i didnot see any conflict. permeability of a material keeps changing with the applied field, for carbon steels, it can vary from thousands to tens. or even close to 1. if you apply an H field of 20,000 Oe, and you get a B of 20,000 Gs, your relative perm is only 1 Gs/Oe.
never heard of 1.6T limit
never heard of 1.6T limit
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- #18
The conflict is that the B-H curves for all three of those materials indicate that the MUr at 13,000 A/m should be 200 to 250, yet you're saying it should be closer to 10. That's a huge error. Why is the data off by that much? It makes the data useless.
i took another look at your B-H curve, at 10,000A/m (=136Oe), 1006 and 1010 have a B of about 1.75 T, so the permeability is 17,500 Gs / 136Oe =128 G/Oe. If the H increses to 13,000 A/M, I am sure the permeability will be further decreased to about 100, but will be 10.
I said 100. so 100 x 165 = 16500 =1.65 T! 1T = 10,000 Gs, not 1000Gs
I said 100. so 100 x 165 = 16500 =1.65 T! 1T = 10,000 Gs, not 1000Gs
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