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What is the correct form of the heat diffusion equation in 1D if we take into account the temperature dependency of specific heat capacity?

$$ \rho\frac{d(cT)}{dt} = \frac{d}{dx}\bigg(k\frac{dT}{dx}\bigg)$$

or

$$ \rho c\frac{dT}{dt} = \frac{d}{dx}\bigg(k\frac{dT}{dx}\bigg)$$

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  • $\begingroup$ First decide what you want to ask. Your title and equations suggest a time-dependent heat capacity, which is rather unusual (does your material decay into something else without any other change worthy of being modelled?). Your text mentions a temperature dependency, which makes sense to model, but that would require a $\frac{\mathrm{d}}{\mathrm{d}T} c$ term, which none of your equations allow for. $\endgroup$ – pyramids Mar 27 '15 at 0:50
  • $\begingroup$ @pyramids what I wanted to ask was while calculating the rate of change of energy, should the 'c' (which is temperature dependent) be kept with 'T' in the derivative (eqn1) or can be pulled out of the derivative (eqn2) $\endgroup$ – nole Mar 27 '15 at 0:57
  • $\begingroup$ To pull it out, you will have to use the product rule. If your heat capacity is not time-dependent, that will simplify to your second equation. If it is time-dependent, it will not. $\endgroup$ – pyramids Mar 27 '15 at 1:00
  • $\begingroup$ @pyramids so equation 2 will hold even in case of the specific heat being temperature dependent? I was thinking to take energy e = rho*(cT - c_refT_ref) and then calculate the rate of change of e as rho*d(cT)/dt. Is this incorrect? $\endgroup$ – nole Mar 27 '15 at 1:03
  • $\begingroup$ pyramids, let me re-phrase my question -> does the second equation take into account the temperature dependency of the specific heat capacity? $\endgroup$ – nole Mar 27 '15 at 2:01
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The differential equation for the conduction of heat is: $$\mathbf{h} = -\kappa\mathbf{\nabla} T$$( This relationship is an approximate one, but holds good for many substances). Also, the equation of continuity for local conservation of heat flow is: $$ - \dfrac{dq}{dt} = \nabla\mathbf{h} \implies \dfrac{dq}{dt} = \kappa{\nabla}^2 T$$ where $q$ the amount of heat in a unit volume & $$\mathbf{\nabla}\cdot\mathbf{\nabla} = {\nabla}^2 = \text{Laplacian operator}$$ Now, we'll assume that the temperature of the material is proportional to the heat content per unit volume - that is, the body has a definite specific heat. So, we can write $$\Delta q = c_v\Delta T \implies \dfrac{dq}{dt} = c_v \dfrac{dT}{dt}$$. The rate of change of heat is proportional to the rate of change of temperature. The constant of proportionality $c_v$ is thd specific heat per unit volume of the material. Using this, we get $$\dfrac{dT}{dt} = \dfrac{\kappa}{c_v} {\nabla}^2 T$$. We find the time rate of change of $T$ at every point as proportional to Laplacian of T. We have a differential equation now for the temperature $T$ using specific heat. So the final equation is $$\dfrac{dT}{dt} = D{\nabla}^2 T$$, where $D$ is the diffusion constant , & is equal to $\dfrac{\kappa}{c_v}$.

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  • $\begingroup$ Does this equation takes into account blackbody radiation? $\endgroup$ – Antonios Sarikas Jun 7 '20 at 16:16
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It always will help you to go back to the derivation of the heat equation. The heat in a region $D$ at a time $t$ is given by the integral, $$ H(t)=\int_Dc(t)\rho T(x,t)\,dx $$ where we've added your assumption of $c$ being time-dependent.

Thus, the change in the heat would be $$ \frac{dH(t)}{dt}=\frac{d}{dt}\int_Dc(t)\rho T(x,t)\,dt\equiv\int_D\rho\frac{\partial}{\partial t}cT\,dx $$ Matching this with Fourier's law would give us something similar to your first equation (the difference being the use of partial derivatives, rather than total derivatives): $$ \rho\frac{\partial}{\partial t}cT=\kappa\nabla^2T $$

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  • $\begingroup$ if the time dependency of specific heat comes from temperature, i.e. c(t) = c(T(t)) would your derivation still hold correct? $\endgroup$ – nole Mar 27 '15 at 1:09
  • $\begingroup$ I'd argue that $c(t)$ and $c(T(t))$ are different things: one is a function of time, the other a function of temperature which itself is a function of time. What do you think the answer is to the question, $\partial_tc(T(t))=?$ $\endgroup$ – Kyle Kanos Mar 27 '15 at 1:36
  • $\begingroup$ mathematically that could be evaluated using chain rule but I don't know if there would be a physically viable situation where that might apply. Nonetheless, I believe that the formulation given by the second equation is correct since by definition $$ de = cdT $$ and so $$ de/dt $$ would be $$ cdT/dt$$. Correct me if I am wrong... $\endgroup$ – nole Mar 27 '15 at 1:46
  • $\begingroup$ Well in that case, it really should be $\delta q$ to denote the path function, rather than a derivative that you're trying to use. If $c$ is a function of temperature, my last question holds. $\endgroup$ – Kyle Kanos Mar 27 '15 at 1:53
  • $\begingroup$ Ok, I think we are deviating from the topic perhaps because by original question wasn't properly framed. To re-frame it, I ask if the second equation of my original post takes into account the temperature dependency of specific heat capacity? $\endgroup$ – nole Mar 27 '15 at 1:59
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This is an old question, yet, it seems the answer is incorrect as it has little to do with the question.

Short answer: your first equation takes into account dependence of heat capacity on temperature, the second one does not. Longer answer: Derivation of heat equation from energy balance (and making use of the Fourier law) leads to the following differential equation:

$\rho(t,x)f(t,x) = \frac{\partial}{\partial t}\left[\rho(t,x)c(t,x)T(t,x)\right] - \nabla\cdot\left[\alpha(t,x)\nabla T(t,x)\right]$,

where $\rho(t,x)$ is material density, $f(t,x)$ is heat generation within the material (e.g. radioactive decay), $c(t,x)$ is heat capacity, $\alpha(t,x)$ is heat conductivity. Now, for well behaved functions, we can write $c(t,x) = c(T(t,x))$, $\alpha(t,x)=\alpha(T(t,x))$. To explicitly write out the derivatives, use the chain rule (e.g. $\frac{\partial c}{\partial x} = \frac{\partial c}{\partial T}\frac{\partial T}{\partial x}$). The second equation comes from the assumption that the parameters are temperature independent, but both capacity and conductivity are temperature dependent but as temperature is space-time dependent, the are too space-time dependent.

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  • $\begingroup$ Change in energy is $c(T)dT$ and so rate of change of energy is $c(T)\frac{dT}{dt}$. Equation 2 is hence the correct form of the diffusion equation. It takes into account temperature dependence of heat capacity as well. $\endgroup$ – nole Jan 11 '17 at 21:03

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