Calculating Cable Resistance for Internal Temperature

[greg997]

Hi. I am a bit confused about the formula for cable resistance.
It is quite a long question but I got stuck at this point.
I have found the resistance Rdc@20 degrees but I need to adjust it for "internal working temp" of 60 degrees.
The formula I have found gives compensation for changes in ambient temp which is R = Rdc ( 1+ alfa (Tambient - 20degrees)).
I believe that ambient temp and internal temp of the cable are not the same so I guess that would be wrong to just stick that 60 degrees into that formula. Is that right?

So the question is, what should I do with that 60 degrees internal temp if its resistance at 20 is let's say 30 Ohm, and I am given temp coefficient?

Thanks

 

[rude man]

What is alfa, really? Isn't it the fractional increase in resistance per degree C above a certain reference temperature, in this case 20C? So if the actual wire temperature is 60 C, what do you think alfa should be multiplied by?

 

[greg997]

I am given alfa = temperature coefficient for given material = 0.004 and in tutorial questions the formula above was used to calculate resistance if the ambient temp was for example -10 or + 45 degrees. Here it is internal temp of 60.
So are you saying it is the same, R (1+alfa( 60-20))?
What if I was given internal temp of the cable and ambient temp, for example 70 cable and 30 ambient, and the reference resistance is at 20? Would that make it ( 70-30-20)?

 

[rude man]

I am given alfa = temperature coefficient for given material = 0.004 and in tutorial questions the formula above was used to calculate resistance if the ambient temp was for example -10 or + 45 degrees. Here it is internal temp of 60.
So are you saying it is the same, R (1+alfa( 60-20))?

Yes.

What if I was given internal temp of the cable and ambient temp, for example 70 cable and 30 ambient, and the reference resistance is at 20? Would that make it ( 70-30-20)?

If the internal temp. is 70 and the reference temp. is 20 then you would multiply alfa by (70 - 20).
The ambient is unimportant. What's important is the effective cable temperature.

Now, a nasty thermodynamics teacher might say 'the ambient is 30 and the axis temp. is 70, what then is the effective resistance?'. In that case you'd have to compute the temperature profile of the cable as a function of radius and integrate the differential shell conductances etc. etc.

We are assuming in your case the cable temperature is uniformly 70.

 

[greg997]

Great. Thanks for good explanations. Really helpful.