Coil Heat Dissipation

[ElmerFu]

Greetings:

I am new this and I apologize if this is a common question. I have reviewed some previous threads and didn't find what I was looking for.

I am being tasked to design/manufacture a solenoid for a one off application. I can calculate most aspects, but the heat dissipation escapes me. I have read several old (all I could find) books which cite some specifics, but the age, 1909, does not leave me with a good feeling. Also the vernacular of the time period results in techlogical inferences that escape me for today. I have Googled some of the "rules" named for the men who created them looking for a modern day refernce, but to no avail.

I am looking for soem sort of guidance to deterrmine the heat dissiaption for the coil so I can "dial in" a power level.

I have found:

1) 2.0 to 2.5 Sq*Inch per watt lost in the coil to ensure a thermal rise no more than 75F. (Watt lost? Is not that all of the power for a DC coil?)

2) Esson's Rule Degrees F= 100*(Watts lost)/superficial area (Superficial area? OD including ends?

Carhart: Highest Temp = Ambient+0.0000445*(current density^2)* coil depth.

Do these seem right?

I appeciate your help,

Elmer

 

[MJR2]

In most DC coils all power goes to heat.

1) This looks reasonable in my experience.

2) This doesn't look to agree well with (1).

Carhart needs a little more info, units maybe.



Mike

 

[ElmerFu]

Hello, Mike:

Thank you for the response.

(1) seems to be a Practical rule of thumb for limiting the heat rise to 75F above ambient. It obviously ignores convective coefficients, etc. Earlier in the text ii is asserted that " a suposition is generally made that the heat is radiated from the outside or cylindrical surafce of teh coil." It seems to me that this menas the surface area used in the "Practical Rule" is the outside perimeter'Pi*OD*the Axial Length.

Do you agree with this?

(2) I am guessing for Esson's rule the 100 is some general coefficient. The equations seems to calculate the thermal rise. If we plug in 75F and constrain the area or power we can solve for the other vaiable.

If we use 75F as the target and 2 sq*Inch/watt lost the resulting power is 1.5 watts and not 1. I agree that this does not agree well with (1). Probalbly too easy. Also the coefficient may only work as a first order approximation for cotton insulation. (Yes, the book is that old)

Also Carhart (yes, he was busy) mdified Esson's rule to account for the fact that typically the hottest part of the coil is somewhere in the interior. Also only about 65% of the heat is radiated from the outer (cylindrical) surface of the coil. The modification is:

t(F) = 140*Lost watts / Surface area.

Spo if we use the target 75F = 140*(1 watt)/ ? watts our result is 1.86 watts which is much closer to the "Practical Rule" of 2 - 2.5 sq*inch / radiated watt of power.


(3) Carhart_ The books lists this as an empirical formula.

T=t + 0.0000445 d^2 * D

T= Highest temeprature Degree (C)
t= ambient air Temp (C)
d= current density (I would guess a/m^2, but the coil depth "D" is in inches so a/in^2????))
D= Depth of coil (I am infering from other locations in the text this is value is the coil sidewall. I listed this as axial length prviously and I was wrong.)

Even with what is listed belwo and not giving away and proprietary information, is there a FRIST ORDER method?

Thank you again,

Elmer

 

[ElmerFu]

Correction:

(2) The Carhrt modified Esson rule should be 1.86 SqInch not watts.

Sorry

 

[EdStainless]

There are ways to increase the heat dissipation. If this is a DC coil you can use a metal bobbin (Al) and you could add fins. Adding a resin filler in between the windings will also help by improving conduction in the coil.
And of course there is the duty cycle. Not just the percentage, but the max time on also.
I hate to say it but build and test is the best way forward.

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