Calculating Thermal Coefficient of a Resistor: Tips and Advice

AI Thread Summary
The discussion focuses on calculating the thermal coefficient of a resistor subjected to a 10W power input in free space. The user recorded temperatures from six thermocouples placed on the resistor's surface, noting significant temperature differences. They seek a method to convert these surface temperatures into a power output, acknowledging that thermal conductivity is typically dependent on thickness. The conversation highlights that the thermal coefficient is defined as power per temperature difference, while also noting that real-world factors like losses from sound and vibration complicate efficiency calculations. Ultimately, the user aims to understand how to accurately determine the thermal energy transmitted to the air from the resistor's casing.
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 ALOT 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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