Calculating Power & Temperature Loss from Cable

Click For Summary

Discussion Overview

The discussion revolves around calculating the power and temperature loss from a cable, focusing on thermodynamic calculations related to heat transfer. Participants explore the relationship between electrical power in the cable and the heat transferred to a neighboring material, considering both theoretical and experimental approaches.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Experimental/applied

Main Points Raised

  • Finch calculates power using the formula P = VI and temperature loss as I^2 x R, expressing uncertainty about whether this represents the power transmitted to the surroundings.
  • Berkeman clarifies that the power calculated is indeed the heat generated in the cable, which leads to a temperature rise until equilibrium is reached with the environment.
  • Berkeman notes the lack of simple equations for determining the cable's temperature rise based on external conditions, suggesting experimental methods or simulations instead.
  • Finch specifies the use of UK AC current and a transformer to reduce voltage, aiming for a low-cost material that can achieve high temperatures.
  • Another participant suggests that while Finch's method may yield an average temperature, it will not determine the peak temperature, which would require differential equations and experimental data.
  • Discussion includes considerations of thermal resistance and the geometry of the wire, with references to external resources for further information.

Areas of Agreement / Disagreement

Participants express differing views on the adequacy of Finch's calculations for determining temperature loss, with some suggesting that more complex methods are necessary for accurate peak temperature assessments. There is no consensus on the best approach to take.

Contextual Notes

Participants acknowledge limitations in their calculations, particularly regarding the assumptions made about thermal resistance and the need for experimental validation to obtain accurate results.

HelpFinch
Messages
2
Reaction score
0
Hi guys,

I am trying to calculate the power and temperature lost from a cable. So I can do some crude thermodynamic calculations on heat transfer.

I am assuming in this case that the object next to the cable is touching, and I am trying to calculate how much power and temperature is transferred from the cable to the neighbouring material.

I have presently been calculating it with the Power = VI and the temperature loss is equal to I^2 x R.

I am not that great at the electronics but I think I am actually calculating just the power in the cable and I am being stupid in assuming this is the actual power transmitted to the surroundings I would just like some clarrification or correction if I am misleading myself.

Thanks for any help in advance,
Finch
 
Engineering news on Phys.org
HelpFinch said:
Hi guys,

I am trying to calculate the power and temperature lost from a cable. So I can do some crude thermodynamic calculations on heat transfer.

I am assuming in this case that the object next to the cable is touching, and I am trying to calculate how much power and temperature is transferred from the cable to the neighbouring material.

I have presently been calculating it with the Power = VI and the temperature loss is equal to I^2 x R.

I am not that great at the electronics but I think I am actually calculating just the power in the cable and I am being stupid in assuming this is the actual power transmitted to the surroundings I would just like some clarrification or correction if I am misleading myself.

Thanks for any help in advance,
Finch

Welcome to the PF.

The power converted to heat in the cable is P = VI = V^2/R = I^2 R

Where V is the end-to-end voltage drop across the cable, R is the resistance of the cable, and I is the current flowing through the cable. This power goes into heating the cable materials. That heat will cause a temperature rise in the cable, until it reaches some equilibrium with the excess heat flowing out into the enviroment.

AFAIK, there are no simple equations for calculating what temperature the cable rises to, given the external environment's characteristics. You can determine it experimentally, or you can use simulation software to calculate what it will probably be.

Do you have a particular application in mind? What kind of cable are you thinking of? Is the power transmission down the cable at DC or 50/60Hz AC, or some other frequency?
 
Thanks for answering Berkeman I appreciate it.

I intend on using transmforming the UK AC current and use a transformer to reduce the standard to a low voltage input most likely below 40. I believe in the UK it is 220V and 50Hz.

The desired output here is the temperature and the purpose is to find a material that gives out high temperature while being low cost (a balance will obviously met).


I only need to do some crude calculations to prove concepts so that experimental data can be made. Thus I hope I do not have to go through long differential equations in order to find accurate data!

Thanks and Regards,
Finch
 
If all you need is an average temp of the wire this technique will work.

This will not find the peak temp in the wire. For that you need differential equations and information you can probably only get experimentally so you might as well just run the experiment.

http://en.wikipedia.org/wiki/Thermal_conductivity
http://en.wikipedia.org/wiki/Thermal_resistance_in_electronics

The trick is how you define the thermal resistance of the wire's insulation, assuming it has some, and the connection to ambient. Probably somebody already solved it for the geometry of a wire and some standard materials, with the assumption that the wire is suspended in air. I wouldn't be surprised to find out it was a typical physics homework question or something. Try googling with the keywords in the wikipedia article.
 

Similar threads

Replies
2
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 4 ·
Replies
4
Views
10K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
22
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 8 ·
Replies
8
Views
15K