[SOLVED] Calculating the temperature rise of a coil

[Zak28]

An air core coil with with 3Ω dissipates 50W on average. The coil is a donut weights ~817 grams and perfectly layered thus heat transfer wouldn't be obscured with air gaps if it were hand wound. Coppers specific heat is ~0.4 J/g °C would the coils temperature rise be roughly (50 * 0.4) / 817 thus a rise of 0.024°C per second?

Greatly appreciated.

 

[c_mitra]

1. three ohm resistance is not relevant here.

2. On an average, 50J is produced per second.

3. Specific heat is 0.4J/g OR heat capacity is 817*0.4 J/C; say 327J/C

4. Input of 50J/s will have a temp rise of 50/327=0.153C/s

5. after 100s the temp will be 15.3C (over the ambient) and after 1000s the temp will be 153C (over the ambient)

6. There will be heat loss due to conduction convection and radiation. The higher the temp the greater is the cooling (heat loss)

7. Insufficient information to calculate the cooling rate

8. At some temp, the heat loss will become 50W and the temp will not rise above that.

9. You need to determine the steady state temp when the heat loss equals 50W

 

[Zak28]

...the temp will not rise above that...

Thank you very much, I am sure thermal equilibrium temp would be sufficiently high to melt the coils plastic housing. I intend to run this at ~5min durations anyways yielding ~45.9C over ambient which isn't sufficient to melt its housing. I guess the best I can do is make slots/rib patterns to expose the coil to ambient air outside its housing or even have it enclosed in an aluminum block with thermal grease to conduct heat away.

 

[c_mitra]

I doubt. A copper block of 817 gram will not rise above 100C with a input of 50W (assuming the ambient is around 30C). For additional safety, you can paint the coil black.

Just assume, to get some rough idea, you are embedding a 50W soldering iron to a 800g copper block. Copper is heavy, and 800g copper cube will have a volume of 90cm3 and the surface area 25cm2 (rough guess). It can happily sink 50W of power at 100C (30 ambient).

The plastic housing will block cooling but perhaps not melt.

 

[Zak28]

I doubt. A copper block of 817 gram will not rise above 100C with a input of 50W...

The above coil resistance, weight and power was fiction for me to grasp the calculation. I might as well depict the real deal here and that is 1017g @ 270W thus 407 J/°C yielding 0.6°C/second another words 5mins of operation and the coil climbs 180°C over ambient. I am going to glue a 10mm thick aluminum plate to the coil with silicon thermal goop otherwise Im sure the plastic wont last for 5min intact.

 

[FvM]

I intend to run this at ~5min durations

Once per day? Once per hour? Obviously the specification is insufficient to calculate the maximum temperature.

If this is a kind of interval operation with fixed duty cycle, problem reduces to two points
- temperature rise during heating interval according to temperature capacity and energy feed
- steady state overtemperature according to thermal resistance and average power

You have already calculated the first point.

 

[Zak28]

Once per day? Once per hour? Obviously the specification is insufficient to calculate the maximum temperature...

~5mins every 3-5hours - and only on days it will be used, anotherwords a sufficient time will elapse for the coil to drop back down to ambient temp ~25°C The coil will reside in a gizmo Im making and keeping things from melting from this coils heat is the greatest challenge so far.

 

[KlausST]

Hi,

If you want to go the safe way, then measure the temperature.
Either with a temperature sensor, or you measure the DC resistance of the coil.
The higher the resistance, the higher the temperature.

Klaus

 

[c_mitra]

The above coil resistance, weight and power was fiction for me to grasp the calculation...

Smaller objects have greater surface to volume proportions; they cool faster.

Heat production is roughly proportional to the volume but the heat loss is roughly proportional to the surface area.

Two half watt resistors have smaller size compared to a single one watt resistor. It may be cheaper to break down the inductor into several smaller units (it will be finally cheaper too) and use series-parallel combinations.