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I've recently come across Fourier's Heat Equation: $k \cdot \nabla^2 \; T = \rho c \cdot \dot{T}$ and the solution to Newton's Law of Cooling: $T(t)=\left(T_\text{initial} - T_\text{cool}\right) \cdot \mathrm e^{-\alpha t} + T_\text{cool}$ for a constant cooling temperature. Does this equation still hold when $T_\text{cool}$ itself is a function of time?

In other words, does $T(t)=\left[T_\text{initial} - T_\text{cool}(t)\right] \cdot\mathrm e^{-\alpha t} + T_\text{cool}(t)$ accurately represent a cooling process when the cooling temperature is for example $T_\text{cool}(t)=T_\text{start} - \frac{t}{5}$?

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Firstly:

$$k \cdot \nabla^2 \; T = \rho c \cdot \dot{T}\tag{1}$$

is NOT the "lumped capacity heat conduction equation" and:

$$T(t)=\left(T_{initial} - T_{cool}\right) \cdot e^{-\alpha t} + T_{cool}\tag{2}$$

is NOT "its one-dimensional solution".

Rather, $(1)$ is the Fourier Heat Equation and $(2)$ is the solution of Newton's Law of Cooling, for $T_{cool}=\text{constant}$.

Lumped thermal analysis in fact uses Newton's Law of Cooling by assuming the temperature of the cooling body is uniform in space (that is what 'lumped' really means here). Fourier's equation does not require this.

Does this equation still hold when $T_{cool}$ itself is a function of time?

NO. But Newton's Cooling Law can be written, taken that in mind, as:

$$\frac{\text{d}Q}{\text{d}t}=-hA[T(t)-T_{cool}(t)]\tag{3}$$

where $T_{cool}(t)$ is a function of time.

With: $$\text{d}Q=mc_p\text{d}T(t)$$

we have:

$$\frac{\text{d}T(t)}{\text{d}t}=-k[T(t)-T_{cool}(t)]\tag{4}$$ where $k=\frac{hA}{mc_p}$ (not to be confused with the same symbol in $(1)$)

$(4)$ is a 1st order ordinary differential equation. Whether it has an analytical solution will depend strongly on the shape of $T_{cool}(t)$. $(4)$ can be rewritten as:

$$T'+kT=f(t)$$

Again, depending on $f(t)$, this can often be tackled using an integrating factor.

The OP suggested that:

$$f(t)=k(a-bt)$$

So we have:

$$T'+kT=k(a-bt)$$

The integrating factor is:

$$I=e^{\int k \text{d}t}=e^{kt}$$

$$IT'+kIT=k(a-bt)I$$

$$\int(IT'+kIT)\text{d}t=\int k(a-bt)I\text{d}t$$

$$e^{kt}T=\int k(a-bt)I\text{d}t=\int k(a-bt)e^{kt}\text{d}t$$

$$e^{kt}T=ka\int e^{kt}\text{d}t-kb\int t e^{kt}\text{d}t$$

$$e^{kt}T=a e^{kt}-e^{kt}\frac{b(kt-1)}{k}+C$$

So that:

$$T(t)=a-\frac{b(kt-1)}{k}+Ce^{-kt}\tag{5}$$

Apply the initial condition to determine $C$.

You can see how it's fundamentally different from $(2)$ because it contains a linear term.

That Eq. $(5)$ is in all likelihood correct can be seen for $t \gg 0$ because then the exponential term vanishes and we get:

$$T(t)=a-bt=T_{cool}(t)$$

So $T(t)$ tends asymptotically to $T_{cool}$.

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  • $\begingroup$ I have corrected my original post to use the right names for those formulas. I will give this solution a try, thanks! $\endgroup$ Apr 10, 2020 at 10:26
  • $\begingroup$ You're welcome. $\endgroup$
    – Gert
    Apr 10, 2020 at 10:27
  • $\begingroup$ I didn't really see your last line so I'll now include an answer to that. $\endgroup$
    – Gert
    Apr 10, 2020 at 10:36
  • $\begingroup$ I leave the numerical side to you. $\endgroup$
    – Gert
    Apr 10, 2020 at 11:19
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    $\begingroup$ Yes, that's correct. But Mathjax works in the comments too, e.g.: $T_c(t) = p - q.t$ $\endgroup$
    – Gert
    Apr 10, 2020 at 11:25

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