The derivation of the energy equation is based on the following physical facts:
Then the physical meaning of energy equation can be written as
$$\left\{\begin{array}{c}
\textrm{rate of increase of }\\
\textrm{energy of fluid element}
\end{array}
\right\}=\left\{\begin{array}{c}
\textrm{net rate of heat }\\
\textrm{added to the fluid}
\end{array}
\right\}-\left\{\begin{array}{c}
\textrm{net rate of work }\\
\textrm{done on fluid element}
\end{array}
\right\}$$
Applying this fact to the infinitesimally small fluid element.
First find a relation for the rate of increase of energy. There are two types of energy concerning the fluid element in motion
Then the total energy (E) is the sum of these two
$$E=e+\frac{u^{2}}{2}$$
Now the the rate of increase of the energy E is
$$\rho \frac{DE}{Dt}\Delta x\Delta y\Delta z$$
Next we need a relation for the work done by surface forces. There are two types surface forces, pressure force and shear stress. These forces are multiplied by the velocity to find the work done. Considering the $x-$ direction only, the work is
$$
\begin{split}
\left[up-\frac{\partial (up)}{\partial x}\frac{\Delta x}{2}\right]\Delta y\Delta z-&\left[up+\frac{\partial (up)}{\partial x}\frac{\Delta x}{2}\right]\Delta y\Delta z\\
-\left[u\tau_{xx}-\frac{\partial (u\tau_{xx})}{\partial x}\frac{\Delta x}{2}\right]\Delta y\Delta z
+&\left[u\tau_{xx}+\frac{\partial (u\tau_{xx})}{\partial x}\frac{\Delta x}{2}\right]\Delta y\Delta z\\
-\left[u\tau_{xy}-\frac{\partial (u\tau_{xy})}{\partial x}\frac{\Delta x}{2}\right]\Delta y\Delta z+&\left[u\tau_{xy}+\frac{\partial (u\tau_{xy})}{\partial x}\frac{\Delta x}{2}\right]\Delta y\Delta z\\
-\left[u\tau_{zy}-\frac{\partial (u\tau_{xz})}{\partial x}\frac{\Delta x}{2}\right]\Delta y\Delta z+&\left[u\tau_{xz}+\frac{\partial (u\tau_{xz})}{\partial x}\frac{\Delta x}{2}\right]\Delta y\Delta z
\end{split}
$$
Rearranging the above expression gives
$$\left[-\frac{\partial(up)}{\partial x}+\frac{\partial (u\tau_{xx})}{\partial x}+\frac{\partial(u\tau_{xy})}{\partial x}+\frac{\partial (u\tau_{xz})}{\partial x}\right]\Delta x\Delta y\Delta z$$
Similarly for the $y$ and $z$ direction
$$\left[-\frac{\partial(vp)}{\partial y}+\frac{\partial (v\tau_{yx})}{\partial y}+\frac{\partial(v\tau_{yy})}{\partial y}+\frac{\partial (v\tau_{yz})}{\partial y}\right]\Delta x\Delta y\Delta z$$
$$\left[-\frac{\partial(wp)}{\partial z}+\frac{\partial (w\tau_{zx})}{\partial z}+\frac{\partial(w\tau_{zy})}{\partial z}+\frac{\partial (w\tau_{zz})}{\partial z}\right]\Delta x\Delta y\Delta z$$
The work done by the body forces will be
$$\rho\textbf{g}\cdot\textbf{u}\Delta x\Delta y\Delta z$$
Finally, the energy flux due to heat conduction is
$$-\frac{\partial q_{x}}{\partial x}\Delta x\Delta y\Delta z$$
$$-\frac{\partial q_{y}}{\partial y}\Delta x\Delta y\Delta z$$
$$-\frac{\partial q_{z}}{\partial z}\Delta x\Delta y\Delta z$$
for $x$,$y$ and $z-$ direction respectively, where $\textbf{q}=-k\nabla T$. Combining all these relations, the energy equation becomes
$$
\begin{split}
\rho\frac{DE}{Dt}=\frac{\partial}{\partial x}\left(k\frac{\partial T}{\partial x}\right)+\frac{\partial}{\partial y}\left(k\frac{\partial T}{\partial y}\right)+\frac{\partial}{\partial z}\left(k\frac{\partial T}{\partial z}\right)-&\frac{\partial (up)}{\partial x}-\frac{\partial(vp)}{\partial y}-\frac{\partial(wp)}{\partial z}\\
+&\frac{\partial (u\tau_{xx})}{\partial x}+\frac{\partial (u\tau_{xy})}{\partial x}+\frac{\partial (u\tau_{xz})}{\partial z}\\
+&\frac{\partial(v\tau_{yx})}{\partial y}+\frac{\partial (v\tau_{yy})}{\partial y}+\frac{\partial (v\tau_{yz})}{\partial y}\\
+&\frac{\partial(w\tau_{zx})}{\partial z}+\frac{\partial (w\tau_{zy})}{\partial z}+\frac{\partial (w\tau_{zz})}{\partial z}+\rho\textbf{g}\cdot\textbf{u}
\end{split}
$$
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[1] Computational Fluid Dynamics, J.D. Anderson, McGraw-Hill Education, 1995.
[2] An Introduction to Computational Fluid Dynamics, H.K. Versteeg and W. Malalasekera, Pearson Education Limited, 2007.