How to exactly calculate the required wire diameter?

Hi!

I would like to know how to exactly calculate required wire diameter? For example, here I found this calculator on internet: https://www.solar-wind.co.uk/CST.html .

It takes acceptable loss, volts, amps, and cable length for arguments. It specifies Vdc. Would it be different formula with AC? In this case I am interested in DC.

This site uses the following code:
JavaScript:
Result = Math.round(((localobject.crun.value * localobject.amps.value * 0.04) / ((localobject.volts.value * localobject.perc_loss.value) / 100 )) * 100) / 100;
Which in formulas looks like this:
g.latex?\dfrac{(\dfrac{Length\cdot%20A%20\cdot%200.png




Which doesnt look too complex to understand, but can anyone explain what that 0.04 does?
I am interested to know, if it's correct calculator at all and if it is then what needs to be done in formulas to make it work with other materials? I assume this one works for copper. But if for example I want to make wire out of aluminium or gold or any other periodic table element that conducts electricity, how would that formula look like then?
 

anorlunda

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DC and RMS AC are equivalent for wire size purposes.

Note that wire size can be limited by voltage drop, by power loss, or by maximum temperature. It is up to you to decide which applies. I would not use the word exact for any of them.

The term ampacity refers to the maximum current in a conductor, usually based on temperature.

Yes it does depend on the material. That calculator you found does not say what material it applies to. Tsk tsk, I wouldn't trust it.

Here's another calculator that gives answers for copper and for aluminum based on voltage drop.


You can find other calculators for ampacity.

Also be aware that local electrical codes may set wire sizes by law. It would not be smart to invest a lot of time and money without checking the local codes first. They may tell you the codes don't apply to free standing wind generators.
 

rbelli1

Gold Member
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It would not be smart to invest a lot of time and money without checking the local codes first. They may tell you the codes don't apply to free standing wind generators.
"The Codes" are not there to make your life difficult. They are there to make your life not end. Even if you can prove definitively that the codes do not apply it is a good idea to follow them anyway.

Also if you don't follow best practices and an insurance company gets involved then any sub standard work will cause them to deny the claim. Remember that the insurance company is not there to pay to fix your house when it burns down. It's there to make money.

BoB
 
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I'm pretty confused about the nature of your question. A bit of context would help get a more targeted answer.
However, I will assume this question is practical in nature. As others have said, look at the building codes for guidance.

What most people outside of the field don't understand about wire sizing is it is really about how hot you can allow the wires to get.
There are three main considerations here:
1) How much heat is generated by I2R losses. This is the part everyone focuses on. Remember that hot wire has more resistance than cold wire.
2) What is the local thermal environment like; i.e. are there other wires in a bundle also generating heat? What sort of ambient operating temperatures are expected? How will the heat flow away from the wire? Any ampacity table that doesn't distinguish between single wires and cables with multiple conductors is probably rubbish. Many tables assume you are dealing with AC power distribution and will only have entries for a couple of common configurations.
3) What materials are subjected to the heat generated. The big one here is the wire insulation which comes in different temperature grades (and costs). Ampacity tables should refer to the insulation type (or temperature rating) to be useful. These grades can be cryptic to people not in the trades (UF, THHN, etc.). Find a table that looks complex and read ALL of the footnotes if you want really useful data.

So, as you can see it can get pretty complicated. I don't think you will find an exact calculation, there are too many different variables involved. That is why no one ever seems to have a great answer to people who ask this question. When people used to ask me expecting a quick simple answer, all they really got back was a lot of questions from me in return. In practice, what equipment designers actually have to do is design based on tables and experience, and then verify proper temperatures by measurement in worse case conditions. This is what is required to get UL, TUV, CSA, CE... safety approvals.

In my experience, you probably need a larger wire gauge than you initially think, primarily because of thermal conditions/requirements around the wire, not just from the current flowing through it. It can be a better trade-off sometimes to pay more for high temperature insulation than to pay for a larger wire with low temperature insulation.

On the other hand, if you have a standard application, like house wiring, for example, then you can just follow the rules and not worry too much.
 
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First I agree with anorlunda and DaveE.
Voltage drop =R*I[A]=Vdc*loss/100
R=2*length*ρ/scu [there are two wire:+ and -] ρ=resistivity[specific resistance] in Ω*mm^2/m
Roughly this could be copper conductor at 60oC [maximum temperature]
ρ=1/58*(234.5+60)/(234.5+20)≈0.02 then:
scu=2*length*ρ/R=0.04*length/(V*loss%/100)=4*length/(V.loss)
In order to calculate the conductor cross section area and ampacity-current carrying capacity-we need also permissible temperature, insulation thermal resistance, the distance between the two wires [+ and -], the ambient [air, underground],medium temperature ,rho etc.
For a.c. we need frequency, conductor diameter, and distance between wires in order to calculate skin and proximity effect.
 

Baluncore

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I would like to know how to exactly calculate required wire diameter?
Remember that “perfection is the enemy of progress”.

It largely depends on your application. You have not identified the application.
Are you winding a solenoid, or hanging bare wires from poles?

https://en.wikipedia.org/wiki/American_wire_gauge
https://en.wikipedia.org/wiki/IEC_60228

Conductor material is usually copper. You can change to another material by adjusting for the change in resistivity. Multiply by the factor = (resistivity_of_conductor) / (resistivity_of_copper). For DC adjust the cross-sectional area, for higher frequency AC, change the diameter.
 
Hi!

Well, actually my question was both practical and theoretical. Practical part was that if I want to power n watt led that is m meters away from x volt DC PSU and I want to keep wires in healthy temperature like 40C max at room temperature of 25C then by what formula would I calculate the wire diameter.

Theoretical part of question was that if I wanted to calculate it for other materials then how would I do that? For example if wires were made of gold instead of copper how much would then the required diameter be? Not that I actually planned to get gold wires but just wanted to calculate the result.

Roughly this could be copper conductor at 60oC [maximum temperature]
ρ=1/58*(234.5+60)/(234.5+20)≈0.02 then:
Could you explain where did 1/58 come from and what exactly is scu in your formulas?
 
379
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Hi!

Well, actually my question was both practical and theoretical. Practical part was that if I want to power n watt led that is m meters away from x volt DC PSU and I want to keep wires in healthy temperature like 40C max at room temperature of 25C then by what formula would I calculate the wire diameter.

Theoretical part of question was that if I wanted to calculate it for other materials then how would I do that? For example if wires were made of gold instead of copper how much would then the required diameter be? Not that I actually planned to get gold wires but just wanted to calculate the result.


Could you explain where did 1/58 come from and what exactly is scu in your formulas?
The basic equation is R=ρ*L/A (Ohms).

ρ = resistivity of the conductor, copper is ~1.6e-8, aluminium is ~2.7e-8, (Note these are 20C values, they should be adjusted for actual conductor temperature using the temperature coefficient of resistivity)
L is the length of the conductor in m
A is the cross section area of the conductor (in m2) normal to direction of current.

So to calculate the required wire size, first calculate the current draw of the load (LED in this case), then take your supply voltage, subtract the LED operating voltage, this is the maximum you can lose over the wire, divide this voltage by the load current to get the maximum total resistance you can have to ensure LED gets the right voltage. Then re arrange that equation to solve for your unknown (Area), with max resistance, length (there and back) and material, the equation will spit out the minimum x section area needed to achieve this, then use pi()r^2 equation to get your minimum wire diameter.

As far as different materials, its a linear thing, so area scales by ratio of resistivity, which means for example you'd need 2.7/1.6= 1.7x area if you wanted to have the same resistance using aluminium as you had with copper.
 

Baluncore

Science Advisor
6,662
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Insulation is very important.
Electrical insulation is also thermal insulation, and so will limit the rate that I2R heat can be lost. That will increases the temperature of an insulation blanketed wire. The maximum wire temperature permitted is usually limited by the thermal properties of the insulation.
 

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