ECU0406,
In steady-state conditions, the specific heat or the mass of the wire has no effect on its temperature.
In steady-state, the power absorbed by the wire is totally dissipated in the environment.
Only during the warm-up transient will the specific heat and density of the wire play some role.
Highed heat capacity will slow-down the heat-up process, for example.
If the temperature is high, the dominant mode of cooling of the wire will be radiation.
If the wire is straigth and not isolated, then in first approximation, the "black-body" radiation law can be applied.
Usually an additional factor is used called "emissivity" that takes the quality of the radiating surface into account.
The best radiatiors have an emissivity = 1 .
The "black-body" radiation law reads:
(emitted power) = Pemit = eps sig (T^4-To^4) Surf
where
eps < 1 is the emissivity (close to 1 for good 'radiatiors')
sig = 5.67E-9 W/m²/K^4 is the Stefan-Boltzmann Constant
T is the temperature of the emitting body in K
To is the temperature of the surrounding in K
Surf is the (apparent) surface of the emitting body (wire)
As an example, let us assume that your wire had a length of 1 meter.
Then you have these likely values:
eps = 1 (assumption with a slight effect when in the usual range)
To = 300 K (assumption with negligible impact)
P = 3720 W
Surf = 0.001 m²
Solving the balance equation Pemit = P for the the wire temperature gives T = 2842 K.
These calculations will not be correct for another geometry (I assumed a straight wire far fom any object).
For example, when the wire is torsaded, then its "apparent surface" is decreased, and its temperature will increase.
Similarly, if the wire is enclosed within a glass cylinder, its temperature will increase a little bit. For such a geometry, the calculations are similar, but the (radiative) properties of the glass have to be taken into account, and the temperature of the glass must be solved simultaneously to the temperature of the wire.
As you can guess from the high temperatures above, the radiation cooling process will always dominate. However, if you introduce a strong forced convective cooling, you could still calculate the wire temperature in a similar way. All you need to do, is to include an additional term for the additional cooling mode (and solve numerically the equation). Usually the corresponding heat tranfer coefficient will depend on the speed of the cooling fluid and its temperature. These are given by empirical correlations. The same applies for natural convection, but this will be even more negligible. Usually the heat transferred is calculated by a formula like
powerTransfered [W] = h S (T-Tfluid)
where
h is the heat exchange coefficient [W/m²/K], given by empirical correlations
S is the exposed surface
T is the temperature of the wire
Tfluid is the temperature of the fluid
There are many variants and all the details can be found in thick books on heat transfer.
Additional note:
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Keeping in mind how the radiative heat transfer works and increases with temperature, it should be clear that measuring the temperature by contact would modifiy the temperature quite a lot. Therefore, a non-contact method should be used. Google about hot-wire techniques for measuring flame temperatures, this might suggest you some way of measuring your wire temperature. There are many other techniques and strategies available.
