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I have seen many derivations of the heat transfer equation. It always has a form something like the following:

$$\rho C_{P} \frac{\partial T}{\partial t}-\nabla\cdot(k\nabla T)=\dot{q}_{V}$$ No matter how you write it, there is always a $C_{P}$ term, for specific heat capacity at constant pressure even though you are not necessarily considering an example that is at constant pressure. None of the derivations explain why they choose $C_{P}$ specifically either (they just say that one should use heat capacity, but why not $C_{V}$, for instance). Where does the constant pressure part come into play?

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The underlying equation is the entropy equation $$ \frac{\partial s}{\partial t } = \frac{1}{T}\nabla \cdot(\kappa\nabla T) + \ldots $$ where $s$ is the entropy density. In general, $s$ is a function of two variables, e.g. $T$ and $P$, and we have to make some assumption to reduce this result to an equation for $T$. It does not make sense to assume $n$ (this implies $V$) to be constant, because materials expand when heated. It makes sense to assume that $P$ is approximately constant, as long as the system is in mechanical equilibrium and fluid velocities are small.

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When first law is written in terms of enthalpy you get \begin{align} \frac{dh}{dt}-v\frac{dp}{dt} & =\dot{q} \\ \rho C_p \frac{dT}{dt}-v\frac{dp}{dt} & =\dot{q} \\ \rho C_p \frac{dT}{dt}-v\frac{dp}{dt} & =\nabla \cdot(k~\nabla T)+\dot{q}_v \end{align} This form of the equation is preferred because mostly we deal with constant pressure processes occurring in open atmosphere, in which case you get your equation \begin{align} \rho C_p \frac{dT}{dt}-\nabla \cdot(k~\nabla T)=\dot{q}_v \end{align} However if pressure is not constant, a work term, $\dot{w}_v\equiv -v\frac{dp}{dt}$, must appear in the equation: \begin{align} \rho C_p \frac{dT}{dt}-\nabla \cdot(k~\nabla T)=\dot{q}_v-\dot{w}_v \end{align}

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In the derivations you are referring to, the fluid is assumed to be incompressible. in the limit of an incompressible fluid, the heat capacity and internal energy are functions only of temperature. So it doesn't matter whether you call it Cp or Cv or just C.

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  • $\begingroup$ Surely I can have an incompressible fluid that has different value for $C_{P}$ and $C_{V}$, right? So, shouldn't it matter? $\endgroup$ – Argon Jan 7 '17 at 3:41
  • $\begingroup$ @Argon Be careful what you mean by "incompressible" here. "Incompressible flow" and "an incompressible material" are two different concepts. Does an "incompressible material" also have a zero coefficient of thermal expansion $\alpha$? If $\alpha = 0$, then $C_P$ and $C_V$ are necessarily equal. If not, they can be different. $\endgroup$ – alephzero Jan 7 '17 at 7:27

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