Non-dimensionalisation of Navier-Stokes Equations [closed]

Question

I am trying to numerically simulate a flow in which a fluid with kinematic viscosity $\nu=0.00001167$ m$^2$/s is injected into a cylindrical tube of diameter 0.00745m at a velocity of 45.9 m/s. Thus, the studied system has Reynolds number $$ \text{Re} = \frac{0.00745 \cdot 45.9}{0.00001167} \approx 29,300.$$

In order to facilitate the resolution of the Navier-Stokes equations, I am nondimensionalizing the following variables:

• spatial variables $(x,y,z)$, taking \begin{equation*} x^* = \frac{x}{L}, \, \, y^{*} = \frac{y}{L}, \, \, z^{*} = \frac{z}{L}. \end{equation*} with $L=0.3921$m.
• velocities $(u,v,w)$ and time ($t$), taking: $$ u^{*} = \frac{u}{U}, \, \, v^{*} = \frac{v}{U}, \, \, w^{*} = \frac{w}{U}, \, \, t^{*} = \frac{ t}{(L/U)},$$ with U=7.65 m/s
• density ($\rho$) and pressure ($p$), taking $$\rho^* = \frac{\rho}{1.184 \, kg/m^3}, \, \, p^{*} = \frac{p}{100000 \, N/m^2}.$$

Now, after the nondimensionalization procedure, I have some doubts:

1. Can I continue using $\nu=0.00001167$ m$^2$/s in the new form of the Navier-Stokes equations? Or do I need to use $\nu^*=\frac{\left(\frac{0.00745}{L}\right) \cdot \left( \frac{45.9}{U} \right)}{29,300} = 0.0000039$ so that $\text{Re}=29,300$ remains valid?
2. I have $t=t^* \cdot (L/U) = t^* \cdot 0.05125$. Does this means that for me to obtain the solutions of the Navier-Stokes equations at $t=1$ s, it is necessary to solve the dimensionless equations for $t^{*}=\frac{1}{0.05125}=19.5122$? Thank you in advance for your attention and patience!

Answer 1

During the process of non-dimensionalization, the term associated with viscous diffusion

$$ \mu\nabla^2\vec{u} $$

is transformed to

$$ \frac{\mu}{\rho{}LU} \nabla^*\vec{u}^* \equiv \frac1{Re}\vec{u}^*. $$

Hence, the parameter for your simulations should no longer be viscosity, but the Reynolds number. The latter is, in fact, never expressed in terms of dimensionless quantities but rather in terms of scaling dimensional quantities:

$$ Re = \frac{\rho{}LU}{\mu} = \frac{LU}{\nu}. $$

Regarding the simulation time, indeed, to simulate up to $t = 1 \,\text{s}$, it is necessary to integrate the dimensionless equations up to

$$ t^* = \frac{1 \,\text{s}}{L/U}. $$

Furthermore, it is important to note that in the context of flows dominated by viscosity (low Reynolds numbers), the following dimensionless scaling is commonly employed for the pressure:

$$ p^* = \frac{pL}{\mu{}U}$$

In this context, the equation of conservation of momentum is formulated as

$$ Re\left(\frac{\partial\vec{u}^*}{\partial{}t^*} + (\vec{u}^*\cdot\nabla^*)\vec{u}^*\right) = - \nabla^*p^* + \nabla^{*2}\vec{u}^* $$

or, in the case where pressure is normalized by $p^* = \frac{p}{\rho{}U^2}$, as

$$ \frac{\partial\vec{u}^*}{\partial{}t^*} + (\vec{u}^*\cdot\nabla^*)\vec{u}^* = - \nabla^*p^* + \frac1{Re}\nabla^{*2}\vec{u}^*. $$