Navier – Stokes Equations
These are one of the hardest PDEs that had never solved exactly and it
is one of the one million USD equations. These equations along with the
conservation of mass and energy equations are the corner stone of all of
the fluid flow and heat transfer problems due to their wide range of
applications. As there is not exact solutions, an approximate solutions
using CFD had been developed for various problems and using of different
models. Since the present work concentrates on the turbulent flow,
various turbulence models will be discussed in full – details later.
Let us first obtained the Navier – Stokes Equation in x – direction;
\(\left(\frac{\partial\sigma_{\text{xx}}}{\partial x}+\frac{\partial\tau_{\text{yx}}}{\partial y}+\frac{\partial\tau_{\text{zx}}}{\partial z}\right)+\rho g_{x}=\rho\left(\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)\)
\(\frac{\partial}{\partial x}\left(-P+2\mu\frac{\partial u}{\partial x}\right)+\frac{\partial}{\partial y}\left[\mu\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\right]+\frac{\partial}{\partial z}\left[\mu\left(\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}\right)\right]+\rho g_{x}=\rho\left(\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)\)
Now let us develop the N – S equation in x – direction for a special
case study which involves for incompressible, Inviscid flow;
\(-\frac{\partial P}{\partial x}+2\mu\frac{\partial^{2}u}{\partial x^{2}}+\mu\frac{\partial}{\partial y}\frac{\partial v}{\partial x}+\mu\frac{\partial}{\partial y}\frac{\partial u}{\partial y}+\mu\frac{\partial}{\partial z}\frac{\partial w}{\partial x}+\mu\frac{\partial}{\partial z}\frac{\partial u}{\partial z}+\rho g_{x}=\rho\left(\frac{\partial u}{\partial t}+u\frac{\partial
u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)\)
Now, since\(\mu\frac{\partial}{\partial y}\frac{\partial v}{\partial x}=\mu\frac{\partial}{\partial x}\frac{\partial v}{\partial y}\)and\(\mu\frac{\partial}{\partial z}\frac{\partial w}{\partial x}=\mu\frac{\partial}{\partial x}\frac{\partial w}{\partial z}\)
and by put\(2\mu\frac{\partial^{2}u}{\partial x^{2}}=\mu\frac{\partial^{2}u}{\partial x^{2}}+\mu\frac{\partial^{2}u}{\partial x^{2}}\)
Then,
\(-\frac{\partial P}{\partial x}+\mu\frac{\partial^{2}u}{\partial x^{2}}+\mu\frac{\partial^{2}u}{\partial x^{2}}+\mu\frac{\partial}{\partial x}\frac{\partial v}{\partial y}+\mu\frac{\partial}{\partial y}\frac{\partial u}{\partial y}+\mu\frac{\partial}{\partial x}\frac{\partial w}{\partial z}+\mu\frac{\partial}{\partial z}\frac{\partial u}{\partial z}+\rho g_{x}=\rho\left(\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)\)
\(-\frac{\partial P}{\partial x}+\mu\frac{\partial^{2}u}{\partial x^{2}}+\mu\frac{\partial}{\partial x}\frac{\partial u}{\partial x}+\mu\frac{\partial}{\partial x}\frac{\partial v}{\partial y}+\mu\frac{\partial^{2}u}{\partial y^{2}}+\mu\frac{\partial}{\partial x}\frac{\partial w}{\partial z}+\mu\frac{\partial^{2}u}{\partial z^{2}}+\rho g_{x}=\rho\left(\frac{\partial u}{\partial
t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)\)
Let us re-arrange them so that we can use continuity equation;
\(-\frac{\partial P}{\partial x}+\mu\frac{\partial^{2}u}{\partial x^{2}}+\mu\frac{\partial}{\partial x}\left[\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}\right]+\mu\frac{\partial^{2}u}{\partial y^{2}}+\mu\frac{\partial^{2}u}{\partial z^{2}}+\rho g_{x}=\rho\left(\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)\)
\(\therefore\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0\)
\(-\frac{\partial P}{\partial x}+\mu\frac{\partial^{2}u}{\partial x^{2}}+\mu\frac{\partial^{2}u}{\partial y^{2}}+\mu\frac{\partial^{2}u}{\partial z^{2}}+\rho g_{x}=\rho\left(\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial
x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)\)
\(-\frac{\partial P}{\partial x}+\mu\left(\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}+\frac{\partial^{2}u}{\partial z^{2}}\right)+\rho g_{x}=\rho\left(\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)\)
In this way, the N – S Equation in x – direction will be
\(\rho\left(\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)=-\frac{\partial P}{\partial x}+\rho g_{x}+\mu\left(\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}+\frac{\partial^{2}u}{\partial z^{2}}\right)\text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\left(3.34\right)\)
Similarly and using of the same procedure, we can obtain the N – S
equations in
y – direction
\(\rho\left(\frac{\partial v}{\partial t}+u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}\right)=-\frac{\partial P}{\partial y}+\rho g_{y}+\mu\left(\frac{\partial^{2}v}{\partial x^{2}}+\frac{\partial^{2}v}{\partial y^{2}}+\frac{\partial^{2}v}{\partial z^{2}}\right)\text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\left(3.35\right)\)
z – direction
\(\rho\left(\frac{\partial w}{\partial t}+u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}\right)=-\frac{\partial P}{\partial z}+\rho g_{z}+\mu\left(\frac{\partial^{2}w}{\partial x^{2}}+\frac{\partial^{2}w}{\partial y^{2}}+\frac{\partial^{2}w}{\partial z^{2}}\right)\text{\ \ \ \ \ \ \ \ \ }\left(3.36\right)\)
Energy Equation
Figure 6 demonstrates the two components of stresses which they are the
normal and shear stresses on a fluid particle. The normal stress is
denoted as”\(\sigma_{\text{xx}}\)” and the shear stress is denoted by
the symbol”\(\tau_{\text{xy}}\)”.
Firstly, we consider the x – component of the forces due to pressure,
normal and shear stresses components as illustrated below in Fig. .