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).