Calculating Thermal Coefficient of a Resistor: Tips and Advice

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Discussion Overview

The discussion revolves around calculating the thermal coefficient of a resistor, specifically how much thermal energy is transmitted to the air from its aluminum outer casing. Participants explore methods to relate surface temperature measurements to power output, considering the context of an engineering student's practical experiment.

Discussion Character

  • Technical explanation
  • Experimental/applied
  • Debate/contested

Main Points Raised

  • One participant describes an experiment involving a resistor placed in free space with recorded temperatures from multiple thermocouples and seeks a method to convert these surface temperatures to power.
  • Another participant asserts that the power radiated from the resistor should equal the electrical power input, suggesting that the thermal coefficient is typically expressed in terms of temperature difference per power.
  • A later reply acknowledges the ideal scenario but highlights real-world inefficiencies, noting that losses such as sound, vibration, and thermal loss in wires complicate the calculation.
  • Another participant reiterates that the thermal coefficient can be expressed as power over temperature difference, while emphasizing that no system is 100% efficient, raising questions about the applicability of this to heaters.

Areas of Agreement / Disagreement

Participants generally agree on the relationship between power and temperature difference but express differing views on the implications of real-world inefficiencies and the complexities involved in applying this to practical scenarios.

Contextual Notes

Participants note that the thermal coefficient's definition may depend on various factors, including efficiency losses and the specific conditions of the experiment, which remain unresolved.

Who May Find This Useful

This discussion may be useful for engineering students, professionals working with thermal systems, and anyone interested in the practical applications of thermal coefficients in resistive heating scenarios.

DJJB
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So I am an engineering student working on co-op and I ran into a problem.

805F%20SERIES.jpg


I have a resistor like the image above. I placed this resistor in free space, leads in vertical position (Ambient: 296K ) and I placed 10W of electrical power into it. What I need to find is the thermal coefficient (How much thermal energy is transmitted to the air from the aluminium outer casing)

I placed 6 thermocouples, one on each side and recorded the following temperatures:

Front: 442K
Back: 486K
Sides x 2: 462K
Bottom: 453K

Is anyone aware of a method to convert surface temperature of a given solid to power? So far I have only come across thermal conductivity which is dependent on thickness. Hopefully there is an easier solution which one of you may know. Thanks.
 
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You know the power radiated from it - it has to equal the electrical power you put into it.

The thermal coeff is normally given in terms of temperature difference per power (or the other way around)
 
NobodySpecial said:
You know the power radiated from it - it has to equal the electrical power you put into it.

The thermal coeff is normally given in terms of temperature difference per power (or the other way around)

Ideally, Yes. But in the real world there are a lot of losses associated with this such as sound, vibration, thermal loss in the wires and leads and so on. Nothing is 100% efficient.
 
All of which must end up as thermal energy anyway - but that doesn't matter

The thermal coef of a resistor is power/temp-difference (or the other way up)

Nothing is 100% efficient.
No machine is 100% - whether this applies to a heater is a surprisingly tricky question
 

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