22/12/01: This is a work in progress. I've redrafted the first half of this note to keep track of the contribution of work, under constant pressure, to heat, but this needs to be repeated in the second half. Spatial: Change in rate of heat across surfaces $$ \begin{flalign} \frac{dQ}{dt} &= -k A \frac{dT}{dx} &\text{ [W]} &\\ \frac{ \partial Q }{ \partial t } &= -k A \frac{ \partial^2 T }{ \partial x^2 } &\text{ [W]} &\\ \end{flalign} $$ Note that \(\frac{dQ}{dt} = \frac{Q}{\Delta{t}}\) since this only represents rate of heat in a steady-state condition. To account for transient-state where rate of heat changes across two dx, we take the second derivative, and representing the second line as a PDE Temporal: Change in rate of heat within volume over time $$ \begin{flalign} \frac{ \partial Q }{ \partial t } &= \frac{ \partial (dU - W) }{ \partial t } &\text{ [W]} &\\ &= \frac{ \partial \left(dU - (-PdV)\right) }{ \partial t } \text{, assume constant pressure} &\text{ [W]} &\\ &= \frac{ \partial H}{ \partial t} &\text{ [W]} &\\ &= C_p \rho V \frac{ \partial T}{ \partial t } \text{, } C_p \text{ is constant pressure heat capacity} &\text{ [W]} &\\ \end{flalign} $$ Equating these two with an energy balance illustrates that the rate of heat accumulated in an infinitesimal volume, is equal to the derivative of its rate of heat through its surfaces (or the second derivative of heat through a surface), at an infinitesimal unit of time. Energy Balance $$ \begin{flalign} C_p \rho V \frac{ \partial T}{ \partial t } &= -k A \frac{ \partial^2 T }{ \partial x^2 } &\text{ [W]} &\\ C_p \rho (A \Delta{x}) \frac{ \partial T}{ \partial t } &= -k A \frac{ \partial^2 T }{ \partial x^2 } &\text{ [W]} &\\ \frac{ \partial T}{ \partial t } &= \left( \frac{-k }{C_p \rho} \right ) \cdot \frac{ \partial^2 T }{ \partial x^2 } &\text{ [W]} &\\ \end{flalign} $$ Assume the element's spatial temperature distribution is uniform, meaning the core of the element will always be the same temperature as it's surface, the mass \(\rho V \text{ [kg]}\) can be removed from eqn (2). TODO: derive out the extra dx. Lumped Assumptions $$ \begin{flalign} &C \rho V \frac{dT}{dt} = d\left( k A \frac{dT}{dx} \right) &\text{ [W]} &\\ \\ &C \rho V \frac{dT}{dt} = k A \frac{dT}{dx} &\text{ No net change in } dx &\\ \\ &C \frac{dT}{dt} = k \frac{dT}{dx} &\text{ Remove mass } \rho V \text{ and } A &\\ \end{flalign} $$ Set T(t) as y(t), solve calculus $$ \begin{flalign} \frac{dT}{dt} &= \frac{k}{C} \frac{dT}{dx} &\text{ Rearranged } &\\ \\ \frac{dy}{dt} &= ky &\text{ More intuitive y form} &\\ \\ &= k e^{kt} \text{ iff } y(t) = e^{kt} &\\ \end{flalign} $$ Set y(t) back to T(t) for final form $$ \begin{flalign} \frac{dT}{dt} &= kdT &\\ &= k e^{kt} & &\\ \\ dT(t) &= e^{kt} &\text{delta T function} &\\ \end{flalign} $$ Account for dT at t=0 $$ \begin{flalign} &\text{Solve }\frac{dT}{dt} \text{ with chain rule } \mapsto \frac{d}{dx}f(g(x)) = \frac{df}{dg} \cdot \frac{dg}{dx} = \frac{df}{dx} &\\ \\ &\frac{dT}{dt} = \frac{dC}{de^{kt}} \cdot \frac{de^{kt}}{dt} = C k e^{kt} &\\ \\ &T(t) = C e^{kt} &\\ \\ &\text{where } C = T(0) \text{ and } k = k A/C &\\ \end{flalign} $$