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I have a conductor that's being heated with current. I want to know the temperature of the conductor

Steady state soultion is:

$$P_{\text{generated}}=W_{\text{conduction}}+W_{\text{convection}}+W_{radiation}$$

Where generated heat is equal to dissipated heat. I got those equations to have only $T_{\text{conductor}}$ to be unknown from $\Delta T=\left(T_{\text{conductor}}-T_{\text{ambient}}\right)$ and with that I got result for $T_{conductor}$ when it's in steady state.

My question is how do I get this all time dependent so I can see how long it would take the conductor to get to steady state and what is the temperature of conductor at any time?

Thanks

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1 Answer 1

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In general we could use Newton's cooling law for this problem.

An object with surface area $A$ and heat transfer coefficient $h$ will lose thermal energy acc.:

$$\frac{dQ}{dt}=-hA(T(t)-T_0),$$

with $\frac{dQ}{dt}$ the heat flux (loss, in $\mathrm{W}$), $T(t)$ temperature evolution of the object and $T_0$ the ambient temperature.

But you're also providing power as heat energy $P$.

The heat balance is thus:

$$P-hA(T(t)-T_0)$$

Assume now that the object has mass $m$ and specific heat capacity $C_p$, then for each infinitesimal increment in heat content $dQ$ there is an increment in temperature $dT(t)$:

$$dQ=mC_pdT(t)$$

Or divided by $dt$:

$$\frac{dQ}{dt}=mC_p\frac{dT(t)}{dt}$$

We can now create the following identity:

$$P-hA(T(t)-T_0)=mC_p\frac{dT(t)}{dt}$$

Which is a first order differential equation with separable variables.

Substitute: $u=P-hA(T-T_0)$, then $du=-hAdT$, so:

$$u=-\frac{mC_p}{hA}\frac{dT}{dt}$$

For ease of notation set $\alpha=\frac{hA}{mC_p}$, so:

$$u=-\frac{1}{\alpha}\frac{du}{dt}$$

Or $\ln{u}=-\alpha t + C$, with $C$ an integration constant.

The determined integral is obtained by substituting back and applying boundary conditions $0,T_0$ and $t,T$:

$$\ln[{\frac{P-hA(T-T_0)}{P}}]=-\alpha t$$

$$\frac{P-hA(T-T_0)}{P}=e^{-\alpha t}$$

$$\large{T=T_0+\frac{P}{hA}(1-e^{-\alpha t})}$$

Note that for $t \to \infty$, $T=T_0+\frac{P}{hA}$, which is the point where heat losses precisely match power input $P$.

No matter how you've calculated your value of $T_{conductor}$, this principle should be applicable to your problem.

This of course assumes the object is homogeneous in temperature, a reasonable assumption for small objects. For larger objects, temperature gradients $\frac{dT}{dr}$ would arise and Fourier's Law would have to be used, as basis for the derivation.

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  • $\begingroup$ With this, can i put put my current to vary with time, let's say it pulses, and see how much would my conductor would heat up and cool down between pulses? $\endgroup$
    – SRC90
    Commented Dec 8, 2015 at 15:35
  • $\begingroup$ @SRC90: absolutely. During parts of the cycle where current is 'off', set $P=0$ and reintegrate the fourth equation (DE) with appropriate boundaries. $\endgroup$
    – Gert
    Commented Dec 8, 2015 at 15:50

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