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I'm trying to write the heat transfer equation in an arbitrary fluid (compressible and viscous). Consider an adiabatic system where the only heat generated is due to the internal friction/viscosity. In a fluid, the dissipation function is:

$$\Phi_v= \check{\tau}:\check{\nabla}^T\check{\nu} \tag{1}$$

$\boldsymbol{\tau}$ is defined as

$$\check{\tau}=\check{\sigma}-P\check{I} \tag{2}$$

Where $\boldsymbol{\sigma}$ is the Cauchy stress tensor and $P$ is the hydrostatic pressure.

Assuming that conduction and radiation are negligible, and convection is the only form of heat transfer, I think I can write the conservation of heat as:

$$\frac{\partial}{\partial t}\left( q \right) +\check{\nabla}\left( q \, \check{\nu}^T \right)= \check{\tau}: \check{\nabla}^T \check{\nu} \tag{3}$$

Now what I do not understand is which one represent heat: enthalpy $h$ or internal energy $e$? For example I'm trying to write the equation for an ideal gas. Can I use enthalpy and write the conservation of heat as:

$$ \frac{\partial}{\partial t}\left( \rho T \right) +\check{\nabla}\left( \rho T \, \check{\nu}^T \right)= \frac{1}{c_P} \check{\tau}:\check{\nabla}^T\check{\nu} \tag{4}$$

Or should I use the internal energy:

$$ \frac{\partial}{\partial t}\left( \rho T \right) +\check{\nabla}\left( \rho T \, \check{\nu}^T \right)= \frac{1}{c_{\nu}} \check{\tau}:\check{\nabla}^T\check{\nu} \tag{5}$$

Which one of the equations 4 or 5 is the correct form? Or maybe they are both wrong?

Notation:

  • $\check{A}$ is the matrix representation of tensor $\boldsymbol{A}$
  • $:$ is the the double dot product of two square matrices (i.e. $\check{A}:\check{B}=a_{ij}b_{ji}$ in Einstein notation form)
  • $\check{a}^T\check{b}$ (Where $\check{a}$ and $\check{b}$ are row matrices) is the dyadic product of first rank tensors $\boldsymbol{a}$ and $\boldsymbol{b}$, usually noted as $\boldsymbol{a}\boldsymbol{b}$. Also known as outer product $ \vec{a} \otimes \vec{b}$ in vector form.
  • $\check{A}\check{B}$ is the matrix multiplication and for row matrices $\check{a}\check{b}^T \equiv\vec{a}.\vec{b} $ dot product in vector form
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  • $\begingroup$ Do you know the difference between $c_p$ and $c_v$? $\endgroup$ – nluigi Feb 21 '18 at 10:47
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    $\begingroup$ Yesterday, I referred you to BSL for the viscous dissipation. BSL also has much more, including the derivation of the equation that you are looking for. They start out with (a) the overall differential energy balance equation (i.e., the open system version of the first law) and (b) the mechanical energy balance equation (which is basically the equation of motion dotted with the velocity vector). When they subtract (b) from (a), they obtain what I like to call the "thermal energy balance equation." This is the equation you are looking for. Table 11.4-1, Eqn. I. $\endgroup$ – Chet Miller Feb 21 '18 at 12:28
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    $\begingroup$ It is the heat flux vector (heat flow per unit area). $\mathbf{q}=-k\nabla T$. You might also consider using the other form of Eqn. I given in BSL (involving Cp); your choice depends on whichever one is more convenient for solving your specific problem. $\endgroup$ – Chet Miller Feb 21 '18 at 12:55
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    $\begingroup$ Is heat conduction really negligible in your situation? $\endgroup$ – Chet Miller Feb 21 '18 at 13:48
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    $\begingroup$ Sorry. Not my "thing." Hopefully, someone else can help. $\endgroup$ – Chet Miller Feb 21 '18 at 15:32
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Ok, after extensive research I think I have found the right answer to this question. Unfortunately there are a lot of different notations and some sources, such as Transport Phenomena by Bird (ed2 2002) which was suggested in the comments, are entirely different.

Firstly, the definition I wrote for viscous stress tensor is wrong. In a perfect fluid:

$$\check{\tau}=\check{\sigma}+P\check{I} \tag{6}$$

Where $\check{\sigma}$ is the cauchy stress tensor.

Secondly internal energy $e$ is heat and enthalpy $h=e+\frac{P}{\rho}$ is the total non-kinetic energy (if there is a such a term! or maybe potential energy?), the total energy is $h+\frac{1}{2}\check{\nu}\check{\nu}^T$. The detailed proof can be found in [1,2] but in the end the result is:

$$ \rho \frac{D e}{D t}=\check{\sigma}:\check{\nabla}\check{\nu} \tag{7}$$

Or in expanded form for an ideal fluid:

$$\rho\left( \frac{\partial e}{\partial t}+\left( \check{\nu}\check{\nabla}^T \right)e \right)=-P\left( \check{\nabla}\check{\nu}^T \right)+ \check{\tau}:\check{\nabla}\check{\nu} \tag{8}$$

Obviously for ideal gas $e=c_\nu T$ and $P=\rho \mathring{R} T$. I have the feeling that 8 can be derived from Eqn. 4 in the OP. But I'm not sure.

References:

  1. Mathematical Modeling in Continuum Mechanics By Roger Temam, Alain Miranville page 94 Eqn. (6.5)
  2. Introduction to continuum mechanics By Michael Lai, page 378 Eqn. (6.18.1)
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