Calculating temp rise in copper wire

In summary: I'm not sure if this question belongs in the forum or the physics forum, I guess I'll post it there as well. In summary, the temperature rise in a coil of wire (27AWG) wound around a toroid will depend on the loss of the ferrite and the resistance of the wire. It might be easiest to just prototype it and measure it.
  • #1
philopher
3
0
I'm not sure if this question belongs in the forum or the physics forum, I guess I'll post it there as well. I'm trying to calculate the temperature rise in a coil of wire (27AWG) wound around a toroid. The current through the wire is ~0.5A with a calculated voltage drop of 14.64mV across the 18" wire due to the wire resistance. The ambient temperature of the wire is 125C. The wire insulation max temp is ~220C. I'm interested in calculating the temperature rise in the wire. How would I go about this?

Background info: This design is for a switching power supply, 28V to 5V, isolated, Fs=250khz, Tambient=125C.

Thanks for any insight.
 
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  • #2
The temperature rise will depend on the loss of the ferrite as well as the resistance of the wire.It might be easiest to just prototype it and measure it...
 
  • #3
philopher said:
I'm not sure if this question belongs in the forum or the physics forum, I guess I'll post it there as well.

BTW, multi-posting is not allowed here on the PF. I've deleted the two duplicate threads. The EE frorum is probably the best place for this question.
 
  • #4
I'm not really concerned about the ferrite loss. For argument sake, let's ignore it or throw it out of the eqn. Is there a method/approach to solving this problem?

I spoke with EE Phd's and scoured the internet for two days with no solution. I would suspect this was a rather simple question if you were a physics person. I didn't do so well in physics. However, I did find the following approach but am missing some info/parameters:

T=E/k; where E=kinetic energy, k boltzman constant, T=temp

The energy across the wire is 7.32mJ as E=W/sec. The wire diameter is 0.0142". The eqn above seem to be at the atomic level. I get some astronomical temp.

Sorry about the multiple posts. Like I said, this seem like a real physics equation as no EE that I spoke to seem to know.
 
  • #5
philopher said:
I'm not really concerned about the ferrite loss. For argument sake, let's ignore it or throw it out of the eqn. Is there a method/approach to solving this problem?

I spoke with EE Phd's and scoured the internet for two days with no solution. I would suspect this was a rather simple question if you were a physics person. I didn't do so well in physics. However, I did find the following approach but am missing some info/parameters:

T=E/k; where E=kinetic energy, k boltzman constant, T=temp

The energy across the wire is 7.32mJ as E=W/sec. The wire diameter is 0.0142". The eqn above seem to be at the atomic level. I get some astronomical temp.

Sorry about the multiple posts. Like I said, this seem like a real physics equation as no EE that I spoke to seem to know.

Ignoring core losses, the power dissipation is just RI^2. Use the RMS value of the current and the wire resistance, or do a more explicit calculation based on the current waveform if you have a simulation of that.

The variability (and the difficulty in the calculation) comes in when you try to calculate how quickly the heat is dissipated from the assembly. If the thermal transfer rate is low (lots of windings, still air, etc., you will get a higher temperature rise for the same power dissipation. For heat sinks, for example, they are rated with different thermal transfer theta values, in units of degrees C per watt. The theta value depends on the size of the heat sink, and whether it is bolted to a big piece of metal, or is free-standing, and whether it has forced air across it or not.

You might be able to use a FEA program to help you calculate the theta value for your transformer, but I still think it would be easier to just build one ane measure it. You don't need to build the whole supply, just put the expected RMS current through the coil and orient it how it will be in the final product. Attach a thermocouple, and you have your answers.
 
  • #6
Thanks for your reply. I already know what the power consumption is. What I'm after is the rise in temperature due to the power. I don't think this is a hard question. I'm sure there is some smart physics guy/gal on this forum that can answer it quickly. I think the winding interaction and all that has a small affect, which we can ignore. What if the problem was changed such that the wire is a straight piece of copper wire with known volume, no insulation, keep it real simple. It seem like you know all the viriables. The thing that I don't understand is at the atomic level with all the electron colliding into the ions in the copper and giving off a certain amount of heat due to all the collisions.
 
  • #7
The issue isn't the heat generation, it's the conduction/convection/radiation of the heat away. That's the harder part to calculate, and why how thick the winding layers are compared to the outside surface area of the final winding layer matters.

There have been threads here before about the temperature rise of wires by themselves... let me try a little searching...
 
  • #8
  • #9
berkeman posted:
The issue isn't the heat generation, it's the conduction/convection/radiation of the heat away. That's the harder part to calculate, and why how thick the winding layers are compared to the outside surface area of the final winding layer matters.

This is precisely the correct answer and is similar to one posted in a mechanial engineering problem not so long ago: how much heat would be dissipated by concrete footings of a building. Both are complex problems not amenable to theoretical approaches.

You can visualize the importance of ambient conditions if you note that you would have very different coil temperatures if operating in ambient temperature near absolute zero,say, or in a 200F oven...Automobile alternators, for example, require fans to keep them from overheating under a hot hood in summer...

In general, conduction is proportional to temperature difference, but convection is affected by RPM, as well as the factors already noted.
 
  • #10
philopher said:
Thanks for your reply. I already know what the power consumption is. What I'm after is the rise in temperature due to the power. I don't think this is a hard question. I'm sure there is some smart physics guy/gal on this forum that can answer it quickly. I think the winding interaction and all that has a small affect, which we can ignore. What if the problem was changed such that the wire is a straight piece of copper wire with known volume, no insulation, keep it real simple. It seem like you know all the viriables. The thing that I don't understand is at the atomic level with all the electron colliding into the ions in the copper and giving off a certain amount of heat due to all the collisions.

It is not an easy question. Any time you want a real temperature things get very difficult. The trouble is knowing the heat loss. You have a pretty good idea of the heat being generated but really know very little about the heat loss. As said above the best route forward is to build one and take measurements.
 
  • #11
Philopher, since you're in an engineering forum can I assume the question you really want answered is, "will I damage my part?" Accurately predicting the real deltaT, as mentioned, is quite hard. But ball-parking and guard banding it can work.

Here is the technique:
http://en.wikipedia.org/wiki/Thermal_resistance_in_electronics

You can ballpark the thermal resistivity from the geometry spec'ed in the datasheet and the tabulated values in the link below. The datasheet for the wire should have a worst case allowed temp too. You spec'ed the ambient at 125C.
http://en.wikipedia.org/wiki/Thermal_conductivityCompare the Qmax to the power you expect the wire to dissipate. If the ratio is between 1 or 2 be worried, but a prototype will likely work for a little while. One or less and you're definitely broken. :)

More than two and chances are you'll be fine. Definitely confirm by measurement though.
 
  • #12
hello friend's,
thank's to all of you i am facing this problem last 2 month how to calculating temp rise in copper wire.in this forum solve my problem thank's again to all of you.i am a student of physics...

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Last edited by a moderator:

1. How do you calculate the temperature rise in a copper wire?

To calculate the temperature rise in a copper wire, you need to know the current flowing through the wire, the resistance of the wire, and the time the current is flowing. You can then use the formula Q=I^2RT to calculate the heat generated in the wire, where Q is the heat, I is the current, R is the resistance, and T is the time. Once you have the heat, you can use the specific heat capacity of copper to calculate the temperature rise.

2. What is the specific heat capacity of copper?

The specific heat capacity of copper is 0.385 J/g·K. This means that it takes 0.385 joules of energy to raise the temperature of 1 gram of copper by 1 Kelvin (or 1 degree Celsius).

3. How does the current affect the temperature rise in a copper wire?

The current has a direct relationship with the temperature rise in a copper wire. This means that as the current increases, so does the temperature rise. This is because the higher the current, the more heat is generated in the wire, leading to a higher temperature rise.

4. What is the main factor that affects the temperature rise in a copper wire?

The main factor that affects the temperature rise in a copper wire is the resistance of the wire. This is because the higher the resistance, the more heat is generated in the wire for a given current, leading to a higher temperature rise.

5. How can I ensure the accuracy of my temperature rise calculation in a copper wire?

To ensure the accuracy of your temperature rise calculation in a copper wire, it is important to use accurate measurements of the current, resistance, and time. It is also important to use the correct specific heat capacity for copper and to consider any external factors that may affect the temperature rise, such as ambient temperature or heat dissipation. Additionally, performing multiple calculations and comparing the results can help to verify the accuracy of your calculation.

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