# Transient heat conductivity with variable heat flux in 1D

I have a metallic block heated by a heat flux variable in time on one of its faces (let's assume that it is infinite in the other dimensions). I would like to compute the temperature over time at a definite point of this block.

I found the equation, $$T(x, t) = T_l + \frac{2 \phi \sqrt{\frac{\alpha \cdot t}{\pi}}}{\lambda} \exp{\left( {\frac{-x^2}{4 \cdot \alpha \cdot t}} \right)} - \frac{\phi \cdot x}{\lambda} \cdot \text{erfc} (\frac{x}{2 \sqrt{\alpha \cdot t}}),$$ which gives the temperature in a given point of a solid at a given time in 1D, where $$\phi := \text{Heat flux density} \\ \alpha := \lambda / (\rho.C_p) : \text{Thermal diffusivity} \\ \lambda := \text{Thermal conductivity}, \; \text{and} \\ C_p := \text{Heat capacity}.$$ However, this equation works only for a constant heat flux during the duration $$t$$.

I would like to find a similar equation that would take into account a variable heat flux. Meaning, it would need to take into account the history of the heat flux that the given point has seen before the current instant of time.

• If there is no exact solution to this problem (the heat flux is arbitrary), I'll try with the differential equation as long as the numerical application is doable easily in Excel or similar. Commented Mar 14, 2020 at 18:36

Since the heat conduction equation is linear, this can be done as a convolution integral. Let $$\Gamma(x,t)=\frac{T^*(x,t)-T_1)}{\phi^*}$$ where $$T^*(x,t)$$ is your existing solution with constant heat flux $$\phi^*$$. Then, for variable heat flux, the solution is $$T(x,t)=T_1+\int_0^t{\Gamma(x,t-\tau)\frac{d\phi}{d\tau}d\tau}$$where $$\tau$$ is a dummy variable of integration. You can also integrate by parts, and arrive at an even simpler integral in terms of the partial derivative of $$\Gamma$$ with respect to $$\tau$$, and $$\phi(t-\tau)$$.
• Are you sure about the $d\phi/d\tau$, shouldn't it be simply $\phi(\tau)$? I had trouble to code your proposal in Excel. But with $\phi(\tau)$, the results seems to make physical sense. Commented Mar 17, 2020 at 21:15
• You can get it in terms of $\phi(t)$ or $\phi(t-\tau)$ by integrating by parts. But, the form I gave is also correct as it stands. Commented Mar 17, 2020 at 22:17
• After integrating by parts, the alternate form is $$T(x,t)=T_1+\int_0^t{\phi(t-\tau)\frac{\partial \Gamma(x,\tau)}{\partial \tau}d\tau}$$ Commented Mar 18, 2020 at 1:44
• Could you explain me how you get to this equation? I am really not sure about the $d\phi/d\tau$. If the flux is contant, then T(x,t) never increases. I coded in Excel T*, T (as you wrote it) and T with $\phi$ instead of $d\phi/d\tau$. However, none of these two match with T* (see this graph) Commented Mar 22, 2020 at 14:38
• If the flux is constant starting at time = 0, then phi is discontinuous, and phi/dt is a Dirac delta function $\phi_0\delta(\tau)$. Then the integral becomes, as required, \phi_0 \Gamma(x,t). Are you familiar with the characteristics of the Dirac delta function? Commented Mar 22, 2020 at 17:35