# Why use different specific heat capacities in the energy equations for incompressible and compressible fluid flow?

Background:

I notice that there is quite a bit of similarity between the standard heat equation and the energy equation (from the navier stokes equations). The heat equation is given as

$$\frac{\partial\left(\rho h\right)}{\partial t} + \nabla\cdot\left(u\rho h \right) - \nabla\cdot\left(k\nabla T \right)=0$$

with $\rho,h,u,k,T,$ are the density, total specific enthalpy, fluid velocity, thermal conductivity, and temperature, respectively. Note that for many applications, it suffices to write the total specific enthalpy as $h=c_pT$, where $c_p$ is the constant pressure specific heat capacity.

On the other hand, the energy equation (from the navier stokes system) is given as

$$\frac{\partial\left(\rho E\right)}{\partial t}+\nabla\cdot\left(u\rho E\right)-\nabla\cdot(k\nabla T)+\nabla\cdot(\sigma u)=0$$

where $E,\sigma$ are the total energy and stress tensor, respectively. Here, the total energy E is the sum of internal and kinetic energy $E=e+K$, with $=e=c_vT$ and $K\frac{1}{2}|u|^2$.

Since both equations are derived from the conservation of energy, it is no surprise that both equations share the following terms:

1. Rate of change in Enthalpy (time derivative term)
2. Advection of internal heat (divergence term)
3. Heat diffusion (laplacian term)

The energy equation has additional terms which are not found in the heat equation. These additional terms are specific to compressible fluid flow and account for

1. Rate of change in & advection of kinetic energy (K)
2. Stress contributions to energy ($\sigma$)

In many situations, the kinetic energy and stress contributions are very small when the fluid flow is incompressible, and thus can be neglected. This results in the energy equation and heat equation being almost identical except for one subtle difference:

A subtle difference
The heat equation is formulated with the constant pressure specific heat capacity $c_p$ while the energy equation uses the constant volume heat capacity $c_v$. Yet, I assume that they are derived from the same principle of conservation of energy. If so, I would expect the two expressions to use the same specific heat capacity.

This becomes more important when dealing with different types of fluid flows. I have seen models for incompressible flow where the energy equation is simply formulated in terms of the standard heat equation (as above) with the constant pressure heat capacity $c_p$. Since the temperature does not affect incompressible fluid flow, one can think of this as a "scalar transport" which depends on the resolved fluid velocity $u$.

Application: Solidification & Melting
In my particular case, I want to model phase change (solidification and melting). For incompressible flow, it suffices to rewrite the total specific enthalpy as the sum of internal and latent heat ($h=c_pT+h_L$) and substitute h in the equation. For compressible flows, the equation is written in terms of internal energy $c_vT$, not enthalpy. So, I'm not 100% sure that it would suffice to simply add the latent heat term as if it were a part of the total internal energy.

The only way to know for sure is if I understand why the incompressible energy equation is written in terms of $c_p$ while the compressible energy equation is written in terms of $c_v$.

My Question

If the energy equation for incompressible flow is formulated in terms of constant pressure heat capacity $c_p$, why isn't the energy equation compressible flow also formulated in terms of constant pressure heat capacity $c_p$. Why is the heat capacity different for incompressible and compressible flow? What accounts for the difference?

Furthermore, for my specific application, how do these differences affect how to add latent heat effects for solidification and melting of a compressible fluid? Can I simply add the latent heat to internal energy term $e$? Or do I need to completely reformulate the compressible energy equation in terms of enthalpy in order to add latent heat effects?

• You can find a variety of different forms of the energy equation. Some are approximations of the others (including reasons beyond enthalpy and internal energy), and unfortunately this point is not often made clear. Some also may be more useful than others. This combustion textbook derives several of them. As for the latent heat question, I don't know what standard practice is, so I'll leave that to someone else. – Ben Trettel Feb 17 '16 at 23:05

Regardless of context, when you do an energy balance on a thermodynamic system, you're usually interested in (among other things) the change in internal energy of the system and the work done on or by the system by compressing or expanding it. Enthalpy $(dh=c_p dT)$ is more-or-less just a convenience function for the combination of internal energy and work: $dh=du+p \cdot dV$. Any derivation that uses enthalpy (or constant-pressure heat capacity), can also be performed using constant-volume heat capacity instead. The only difference should be whether or not expansion/compression work appears explicitly in the solution or not.
When you say that your system is incompressible, you're basically saying that any changes in volume are negligible, i.e. $dV \approx 0$. The $p \cdot dV$ term then disappears from the definition of enthalpy, so $dh \approx du$. That is, for incompressible materials, there isn't much difference between the heat capacity at constant pressure and at constant volume.
$$E_{tot} = \alpha \cdot u_l + (\alpha -1) u_s +KE+PE+...$$
where $E_{tot}$ is the total energy, $KE$ and $PE$ are kinetic and potential energy, $\alpha$ is the fraction of material in the liquid phase, and $u_l$ and $u_s$ are the internal energy of the liquid and solid phases. As the material melts, $\alpha$ increases towards 1.0, and the change in energy is captured in the first two terms.
• The "logical place to include latent heat" is in the enthalpy - hence terms like "enthalpy of vaporization", etc! You might argue the opposite of what you actually wrote, i.e. "$c_pdt$ is just a convenience function for $H$, when there are no phase changes" ;) – alephzero Dec 1 '16 at 0:55