# Non-dimensionalization of Navier-Stokes equations multiphase flows

I am currently dealing with multiphase flows and have to use the non-dimensional form of the Navier-Stokes equations (NSE). In the scientific literature I found various formulations (and almost no explaining o how they get them) but I wanted to get to the point by myself as well. Although the procedure should be simple, the results do not lign up and I cannot wrap around my head on why. NSE can be written for incompressible fluids as: $$\frac{\mathrm{D}\mathbb{u}}{\mathrm{D}t}=-\frac{1}{\rho}\mathbb{\nabla}p+\mathbb{g}+\frac{\mu}{\rho}\Delta\mathbb{u}$$ where the ratio $\frac{\mathrm{D}}{\mathrm{D}t}$ represent the material derivative, $\mathbb{u}$ is the fluid velocity, $\rho$ the density, $\mathbb{g}$ the gravitational acceleration and $\mu$ the dynamic viscosity. Also, $\Delta$ is the Laplacian operator.

With surface tension, the equation becomes (CSF formulation of the surface tension) $$\frac{\mathrm{D}\mathbb{u}}{\mathrm{D}t}=-\frac{1}{\rho}\mathbb{\nabla}p+\mathbb{g}+\frac{\mu}{\rho}\Delta\mathbb{u}+\frac{\sigma\kappa\hat{\mathbb{n}}\delta_{\varepsilon}}{\rho}$$ where $\sigma$ is the coefficient of surface tension, $\kappa$ is the curvature, $\hat{\mathbb{n}}$ is the normal unit vector to the interface surface and $\delta_{\varepsilon}$ a Dirac delta function (smeared out version).

Introducing the following dimensionless variables (marked with $'$): $$x'L=\mathbb{x} \hspace{5mm} u'U=\mathbb{u} \hspace{5mm} g'g=\mathbb{g} \hspace{5mm} \nabla'=\mathbb{\nabla}L \hspace{5mm} \Delta'=\Delta L^2 \hspace{5mm} p=p'\rho U^2 \hspace{5mm} t'=tU/L$$ I get $$\frac{U^2}{L} \frac{\mathrm{D}u'}{\mathrm{D}t'}=-\frac{U^2}{\rho}\frac{\nabla'}{L}p'+g'g+\frac{\mu}{\rho}\frac{U}{L^2}\Delta' u'+\frac{\sigma\kappa\hat{\mathbb{n}}\delta_{\varepsilon}}{\rho}$$ and, after multiplying both sides of the equation for $L/U^2$, $$\frac{\mathrm{D}u'}{\mathrm{D}t'}=-\frac{1}{\rho}\nabla'p'+\frac{g'gL}{U^2}+\frac{\mu}{\rho}\frac{1}{LU}\Delta' u'+\frac{\sigma\kappa\hat{\mathbb{n}}\delta_{\varepsilon}}{\rho}\frac{L}{U^2}$$ Now, after introducing the well-known Reynolds and Froude numbers $$\mathrm{Re}=\frac{\rho LU}{\mu} \hspace{10mm} \mathrm{Fr}=\frac{U}{\sqrt{gL}}$$ it follows that: $$\frac{\mathrm{D}u'}{\mathrm{D}t'}=-\frac{1}{\rho}\nabla'p'+\frac{1}{\mathrm{Fr}^2}g'+\frac{1}{\mathrm{Re}}\Delta' u'+\frac{\sigma\kappa\hat{\mathbb{n}}\delta_{\varepsilon}}{\rho}\frac{L}{U^2}$$ Until here no problem.

However, for the manipulation of the surface tension (SF) form, some authors (e.g. "An improved level set method for incompressible two-phase flows". Sussman et al., 1998) use the Weber number: $$\mathrm{We}=\frac{\rho LU^2}{\sigma}$$ Plugging this latter in the SF term though, I get: $$SF'=\frac{\kappa\hat{\mathbb{n}}\delta_{\varepsilon}L^2}{\mathrm{We}}$$ instead of $$SF'=\frac{\kappa\hat{\mathbb{n}}\delta_{\varepsilon}}{\mathrm{We}}$$

I dare to think that this latter formulation is incorrect due to the incontrovertible principle of dimensional homogeneity.

$\kappa$ is measured (I hereinafter use the S.I. units for simplicity) in [1/m], $\hat{\mathbb{n}}$ is dimensionless [-] (unit vector) and the delta function has as unit of measure the inverse of its argument (which is a distance function) [1/m]. Thus $$\mathrm{dimensionless \hspace{3mm}number}[-] \neq \frac{[1/m][-][1/m]}{\mathrm{dimensionless \hspace{3mm}number}[-]}=[1/m^2]$$

If someone can explain to me if I made some mistake I would greatly appreciate it.

Note: $$\hat{\mathbb{n}}=\frac{\mathbb{\nabla}{d}}{|\mathbb{\nabla}{d}|}=[-]$$ $$\kappa=\mathbb{\nabla}\cdot\hat{\mathbb{n}}=[1/m]$$

• Since when is a surface tension term included in the Navier Stokes equations? Surface tension occurs at a free surface, and is treated as a boundary condition rather than a term in the momentum balance. – Chet Miller Sep 3 '18 at 13:07
• As I wrote in the question title, I'm dealing with multiphase flows; to model the interfaces between the phases, surface tension must be taken into account. – M.Giacchello Sep 3 '18 at 13:13
• I'm voting to close this question as off-topic because check-my-work questions are generally not appropriate for physics SE. See physics.meta.stackexchange.com/questions/6093/… – user191954 Sep 7 '18 at 7:37

As you pointed out, your $\kappa$ and $\delta_\varepsilon$ are still dimensional quantities, both in the units of [1/m]. To make them dimensionless, you just need to multiply by characteristic length $L$:
$\kappa' = \kappa L$, $\delta'_\varepsilon=\delta_\varepsilon L$
Then, $SF' = \frac{\kappa'\hat{n}\delta'_\varepsilon}{\rm We}$ is a dimensionless quantity.