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Plating rack electrical circuit dimensioning

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Luigi C.

Mechanical
Feb 28, 2024
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Hi guys,
I'm a material engineer working in a faclity that build up plating racks.
Often I 've to take into account with the following question: "Is that rack adequate for the flowing current?"
In fact the only one issue is the possibility of overheating and pvc coating burning due to joule effect.

I'm indagating if there is a method to solve the problem. I think i need to work on the ampacity of the plating rack circuit but reading on the internet is more complicated that i can imagine.

Could somebody indicate me the rigth way to calculate the ampacity or give me some ideas?

Thanks

Luigi
20240306_084813_nf1d3e.jpg
 
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Step one; Determine the specific resistivity of the part in question.
Step two; Determine the length of the part in question.
Step three; Determine the cross sectional area of the part in question.
Step four; From the above information, determine the resistance in Ohms of the part in question.
Step five; Determine the current through the part in question.
Step six; Use current squared times resistance to determine the heat developed in Watts in the part in question.
Evaluate the Watts per unit length and the temperature that may be reached.
You are on your own on this. If the part is immersed in a liquid that will efficiently conduct heat away from the part, a higher current density may be allowed.
In the parts shown, do an evaluation for the fingers and an evaluation for the section of the main support that carries the highest current, that is from the connection to the first branch, which will carry the highest current.
Your current limit will be determined by the maximum temperature that your coating will withstand.
Alternately, you may use a test current and measure the temperature that the various parts reach.
Allowable currents of conductors are ultimately determined by temperature rise.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
Hi @waross,
Thanks replying me.
Now I'm trying to use the Neher-McGrath Equation to find out the Ampacity of the spline.
I'm focusing on the idea that the rack is, in first approssimation, a cable insulated. After that I will take into account the tips as the bottleneck of the possible heating zone due to Joule effect and more in depth I 'll try to find out what happnens in the criticest zone that is the coupling zone between the tip and the spline.

I don't know if it could be a rigth path to follow, but another thing I would say it's that We do not require a super specific model, We try to understand in first approximation theoretically the situation with a model that, combined with our experience, is able to fit a "general rack".

My problem now is that I don't have any kind of confidence with Neher-McGrath equation. Especially I understood that there are a series of approximation factors in function of the cases (if the cable is in the air, in the soil, if coupled in the same region with another cable etc.), so I don't know which factor can I use.

I hope I was clear

thanks reading me

Luigi

Neher_McGrath_equation_d3pakc.png
 
What material is used for the rack?
Take the cross sectional area and scale it to a copper area with similar resistance per unit length.
Convert the equivalent cross sectional area to AWG or mm[sup]2[/sup].
Look in any electrical code for a current rating at your choice of ambient temperature.
If the rack is immersed in liquid, it may be satisfactory to consider only the section in free air above the liquid.
Alternately, do actual testing. Increase the current through a sample and monitor the temperature of the part.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
@waross
Thanks replying me,
The core of the rack could be made in brass, aluminum, titanium, carbon steel etc.
tips generally are in Stainless steel, generally AISI 304 or AISI 316.

I'll try to translate the situation to the most similar in the standard tabs.

I really appreciate the piece of advice.

Have a source in a link when i can check the tabs?

Thanks

Luigi
 
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