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The heat conduction equation is usually written as

$$\frac{\partial T}{\partial t} = \alpha \vec \nabla^2 T$$

But what if a moving air parcel is considered? The standard derivation of the equation above is to consider the energy balance of a small control volume, hereby considering heat flow through the boundaries due to temperature gradient. But if my control volume moves, the partial derivative is recast in the Lagrangian derivative:

$$\frac{D T}{D t} = \frac{\partial T}{\partial t} + \vec v \cdot \vec \nabla T $$

Now I wonder, if the heat diffusion equation in this case is then still like this:

$$\frac{D T}{D t} = \alpha \vec \nabla^2 T$$

Of course I can re-consider energy balance for the moving control volume, but I'm not sure if the gradient heat flow still applies without additional term due to movement as we have in Reynolds Transport Theorem.

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    $\begingroup$ Your convective heat transfer equation is correct. You can also derive it from an energy balance on a stationary control volume. $\endgroup$ Mar 3 at 11:41
  • $\begingroup$ What do you mean by "stationary control volume" ? I had the feeling that is is correct, but how to derive it correctly would be interesting, too. $\endgroup$
    – MichaelW
    Mar 3 at 16:24
  • $\begingroup$ I mean a stationary control volume that is at rest relative to the laboratory from of reference. For such a control volume fluid flows across the buundaries of the control volume, carrying internal energy into and out of the control volume. $\endgroup$ Mar 3 at 22:32
  • $\begingroup$ The flux at any point is given by $q(x,t) = -\alpha\nabla T + Tv$ when you have both diffusion and advection. Applying the standard control volume analysis and the divergence theorem will yield the desired PDE. $\endgroup$
    – whpowell96
    Mar 6 at 18:06

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Consider creating a differential transport equation for the thermal energy density $e_T$ of some fluid. Applying the Reynolds transport theorem, or any of the other usual equivalent processes to create transport equations, would furnish

$$\frac{\partial e_T}{\partial t} + \nabla \cdot \vec{q}_{e_T} = r$$

where $r$ represents the volumetric generation of thermal energy and $\vec{q}_{e_T}$ is the thermal energy flux vector, often called the heat flux vector too.

Modeling thermal energy transport as entirely diffusive implies $\vec{q}_{e_T} = -k \nabla T$, which when inserted into the above produces the usual heat equation with thermal energy generation. However, we can also include an advective term such that

$$\vec{q}_{e_T} = - k\nabla T + e_T\vec{v}$$

where $\vec{v}$ is the traditional fluid/continuum velocity. Noting that the thermal energy density is usually defined as $e_T = c_p \rho T$ ($c_p$ is the specific heat capacity, $\rho$ is the mass density) up to a constant offset, we can insert these two expressions into the transport equation to obtain:

$$\frac{\partial \left(c_p \rho T\right)}{\partial t} + \nabla \cdot \left(- k\nabla T + c_p \rho T\vec{v}\right) = r$$

Assuming constant/unchanging material properties, and noting that the thermal diffusivity is $\alpha = \frac{k}{c_p \rho}$, we can simplify the expression below into something more intelligible:

$$\frac{\partial T}{\partial t} - \alpha \nabla^2 T + \nabla \cdot \left(T\vec{v}\right) = \frac{r}{c_p \rho}$$

If the flow is incompressible ($\nabla \cdot \vec{v} = 0$), it simplifies even further:

$$\frac{\partial T}{\partial t} - \alpha \nabla^2 T + \nabla T \cdot \vec{v}= \frac{r}{c_p \rho}$$

These last two equations are used, after further simplifications, to model thermal transport in boundary layers (and many other such scenarios).

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