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Topic: Number of turns of E.magnet(Math lovers please) (Read 352 times) previous topic - next topic

tlhsglm

Oct 24, 2019, 08:04 pm Last Edit: Oct 24, 2019, 08:06 pm by tlhsglm
Hi fellows,

I'am going to make an electromagnet.But first everything must be clear.

My power supply 24v and 2 amp max.Thus i want my electromagnet work at 1.5 amps at max.
 24v/1.5amp =16 ohms.Awg 18 is 1mm diameter wire and it's max amperage is 2.3 amps.

For 16 ohms i will buy 763 meters of it.

My core is 50mm diameter and length is 50mm.Now i have to calculate how many turns approximately.And here we go;

i Have to do this with layer to layer because of each layer radius is going bigger. So how can we do that?

Grumpy_Mike

For an electromagnet just use the length of wire you calculated. The number of turns is irrelevant if this is going to be powered by D.C.

tlhsglm

Hi Mike ,

But the every formulas want number of turns on the coil?
As this link

herbschwarz

According to my copper wire spec. chart, #18 AWG enamel coated (magnet wire)
is rated at 23.9 linear turns per inch, 6.386 ohms per 1000 feet @ 25 degrees C.
The chart shows that it can carry 10 Amps continuously in conduit or a bundle.
[That seems too high, to me. But, your spec of max. current seems much too low.]

Magnetic flux is proportional to ampere turns. Since you are "flying blind" here, you
must see how many turns you come up with and if that serves your purpose.
Herb

JCA34F

What will the final diameter of this 50mm thick coil be?

tlhsglm

herbschwarz 
Hmm good point i looked at this link 18AWG 2.3 amps for maximum but this link says diferent.It's interenting..

 JCA34F
Same question with mine.

JCA34F

#6
Oct 25, 2019, 01:39 am Last Edit: Oct 25, 2019, 01:48 am by JCA34F
http://production-solution.com/coil-calculator.htm

For bare wire, enameled will be larger.

MarkT

#7
Oct 25, 2019, 03:57 am Last Edit: Oct 25, 2019, 04:03 am by MarkT
What will the final diameter of this 50mm thick coil be?
About 15cm.  About 2500 turns, linear turns density of 50000 turns/meter, therefore 75000 A/m, and an MMF of 3750A

Calculate the volume of the winding (assume square cross section of wire for packing density), = 7.63e-4
The calculate the area of the coil face = 7.63e-4 / 5e-2 = 1.53e-2.

rest is simple geometry.

You realize this is 5kg of copper? Aluminium wire might save a lot of money. What are you actually trying to do?
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

Grumpy_Mike

#8
Oct 25, 2019, 06:19 am Last Edit: Oct 25, 2019, 06:22 am by Grumpy_Mike
But the every formulas want number of turns on the coil?
That is because they are calculating what magnetic field you will get. You are not designing for that, or so you said. Your design criteria is simply the current the coil will take at the voltage you want to give it. This produces a resistance and then depending on the gauge of wire you use you get a length of wire.

You then wind that into a coil, and as to the magnetic field, you get what you get. Once you have wound this you can use the number of turns you got and then plug that in to calculate the size of magnetic field you should get.

Quote
i Have to do this with layer to layer because of each layer radius is going bigger.
You are over thinking this, it is your length of wire that is fixed by your criteria.

What are you actually trying to do?

JohnRob

When you wind the coil you will not get 100% fill.  Most transformer mfg assume somewhere around 80 - 85%.   This means for your 763 meters of wire will not get you the turns you think.

Another consideration is the power dissipated in the copper wire is approximately 35 watts (aluminum will be much worse).  This means if you end up making your coil it will only survive intermittent duty.

You should also use enamel coated wire for its higher temperature rating.

John

Please do not PM me with thread based messages.  If your thoughts are worth responding,  the group should benefit from your insight.

tlhsglm

#10
Oct 25, 2019, 11:03 am Last Edit: Oct 25, 2019, 11:08 am by tlhsglm
JCA34F
Thank you very much i didnt know that site can do it.

Grumpy_Mike and MarkT
The goal is making 100kg lifting electromagnet.

Another consideration is the power dissipated in the copper wire is approximately 35 watts (aluminum will be much worse).  This means if you end up making your coil it will only survive intermittent duty.
Hi John, i dont understand the" intermittent duty", do you telling me it will get so hot, so when its reach that hot you have to cut the power for cooling down?

I readed some articles that telling if J=6A/mm2 the coil will not get high tempretures.
  J=I/S
Where
J=Charge Density
I=Amperage
S=Cross sectional area(mm2)
Example: If a bobbin that wil take 6amperes the conductor's cross sectional area have to be 1mm2.
J=6/1=6(A/mm2)

Grumpy_Mike

Quote
i dont understand the" intermittent duty", do you telling me it will get so hot, so when its reach that hot you have to cut the power for cooling down?
Basically yes.
You are restrained by the melting point of the enamel coating the wire, but I would not run it at anything higher than 150oC.

MarkT

JCA34F
Thank you very much i didnt know that site can do it.

Grumpy_Mike and MarkT
The goal is making 100kg lifting electromagnet.
For lifting random shaped bits of iron, or do you get to choose the pole-piece shape?  With accurately ground
flat pole-pieces you can get 1000N force with a very small winding.  With random shapes much stronger fields are needed.

You need to first figure out the magnetic circuit, then determine how many amp-turns are needed, then design the winding.  You seem to be going in reverse and hoping.

For efficient lifting a winding plus a permanent magnet is a good design - the coil works against the
permanent magnet to release, and with it for extra lifting force.  And even better is the electropermanent magnet
https://en.wikipedia.org/wiki/Electropermanent_magnet
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

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