Calculating Cable Resistance for Internal Temperature

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Discussion Overview

The discussion revolves around calculating the resistance of a cable at an internal temperature of 60 degrees Celsius, starting from a known resistance at 20 degrees Celsius. Participants explore the implications of using a temperature coefficient in the resistance formula and the differences between internal and ambient temperatures.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions whether it is appropriate to use the formula R = Rdc (1 + alfa (Tambient - 20 degrees)) for an internal temperature of 60 degrees, suggesting that internal and ambient temperatures are not the same.
  • Another participant clarifies that alfa represents the fractional increase in resistance per degree Celsius above a reference temperature, which is 20 degrees Celsius in this case.
  • There is a proposal to use the formula R = Rdc (1 + alfa (60 - 20)) for calculating resistance at the internal temperature of 60 degrees.
  • A hypothetical scenario is presented where both internal and ambient temperatures are given (70 degrees internal and 30 degrees ambient), prompting a discussion about how to adjust the formula accordingly.
  • One participant asserts that the ambient temperature is unimportant for calculating effective resistance, emphasizing the need to focus on the effective cable temperature instead.
  • Another participant introduces a more complex consideration involving thermodynamics, suggesting that if the ambient temperature is 30 degrees and the internal temperature is 70 degrees, one would need to compute the temperature profile of the cable for accurate resistance calculation.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of ambient temperature in resistance calculations, with some agreeing on the use of the formula for internal temperature while others raise concerns about its applicability. The discussion remains unresolved regarding the best approach to incorporate both internal and ambient temperatures.

Contextual Notes

There are limitations regarding the assumptions made about temperature uniformity within the cable and the potential complexities introduced by varying ambient conditions. The discussion does not resolve how to handle these complexities in practical scenarios.

greg997
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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
 
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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?
 
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)?
 
greg997 said:
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.
 
Great. Thanks for good explanations. Really helpful.
 

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