In the Navier Stokes' equation:

$\rho_0 \left( \frac{\partial v}{\partial t} + v \cdot \nabla v\right) = -\nabla p + \mu \nabla^2 v + \hat{f}$

I included the temperature variation of density as per the Boussinesq approximation as

$\hat{f} = \rho_0(1-\alpha\Delta T)g\hat{k}$.

Choosing a viscous time scale $t^* = \frac{t}{d^2/\nu}$ where $d$ is a measure of length (in this case, film thickness) and non-dimensionalizing length and velocity with:

$x^*=x/d$ and $v^*=\frac{v}{\nu/d}$, the Navier Stokes' equation is modified to:

$\rho_0\left(\frac{\nu^2}{d^3} \frac{\partial v}{\partial t} + \frac{\nu^2}{d^3} v\cdot\nabla v \right) = -\frac{1}{d}\nabla p + \frac{\mu\nu}{d^3} \nabla^2 v + \rho_0(1 - \alpha \Delta T) g\hat{k}$.

Dividing throughout by $\frac{\rho_0 \nu^2}{d^3}$, while realizing that the non-dimensional pressure falls out as $\frac{\rho_0 \nu^2}{d^2}$, I get:

$\frac{\partial v}{\partial t} + v\cdot\nabla v = -\nabla p + \nabla^2 v + \frac{gd^3}{\nu}(1 - \alpha \Delta T)g\hat{k}$

Further splitting these terms for simplification:

$\frac{\partial v}{\partial t} + v\cdot\nabla v = -\nabla p + \nabla^2 v + \frac{g d^3}{\nu^2}\hat{k} - \frac{gd^3}{\nu^2} \alpha \Delta T \hat{k}$

Recognizing the following non dimensional numbers:

  1. Galileo number, $Ga = \frac{g d^3}{\nu^2}$
  2. Grashoff number, $Gr = \frac{gd^3}{\nu^2} \alpha \Delta T = \frac{Ra}{Pr}$

Where $Ra$ is the Rayleigh number and $Pr$ is the Prandtl number.

I rewrite my now non dimensional Navier Stokes' equation as:

$\frac{\partial v}{\partial t} + v \cdot \nabla v = -\nabla (p - Ga z \hat{k} + Gr z \hat{k}) + \nabla^2 v$

Where $z$ is the z coordinate and $\hat{k}$ is the unit normal in the $z$ direction.

Is this approach flawed since when I compare this with page 2 of this, equation 4 in Bestehorn et al. and equations 5 thru 7 in Ybarra et al., they all have a temperature difference multiplied with the Rayleight number which doesn't make sense.

Did I do something wrong?

  • $\begingroup$ Dropbox links now dead. $\endgroup$
    – Qmechanic
    Commented Mar 1, 2016 at 22:18

1 Answer 1


Your use of $\Delta T$ is confusing you, as you have it, it is not the temperature difference between the top and the bottom of the film, but the difference between the temperature $T$ at a given point in the fluid, and a reference temperature $T_0$ at which the density of the fluid is $\rho_0$. Replace $\Delta T$ with $T-T_0$ first. You then need to non-dimensionalize temperature, which in your first reference is done as $\theta = \frac{T - (T_{top} + T_{bot})/2}{T_{top} - T_{bot}}$, and taking $T_0 = (T_{top} + T_{bot})/2$, so $\theta = \frac{T - T_0}{\Delta T}$, where $\Delta T$ now IS the difference between top and bottom.

You can rewrite the last equation as $T-T_0 = \theta \Delta T$, and there you have your missing temperature...

  • $\begingroup$ Why must I non-dimensionalize temperature? The overall non-dim. parameter, viz., the grashoff number just falls out. Isnt $\Delta T = T_\text{bot} - T_\text{top}$? $\endgroup$
    – dearN
    Commented Nov 6, 2012 at 17:33
  • $\begingroup$ No, $\Delta T = T - T_0$, and $T$ is another variable of your problem which can change from point to point. $\endgroup$
    – Jaime
    Commented Nov 6, 2012 at 18:47
  • $\begingroup$ Yes, I understand what you are saying now. However on futher introspection I am convinced that this temperature scaling as conducted in the references I included are meant for Rayleigh-Benard convection where there is a cold plate and a hot plate. In my case, with a free surface film, I suppose that such a scaling won't stand good. $\endgroup$
    – dearN
    Commented Nov 7, 2012 at 14:17
  • $\begingroup$ Either with a bounding wall or a free surface, you can go with the assumption of perfect conduction, so that $T=T_{top}$. What the free surface has an effect on is the boundary conditions for the fluid velocities. Get yourself a copy of one of Drazin's books (either 'Hydrodynamic Stability' or 'Introduction to Hydrodynamic Stability') and read the chapter on Rayleigh-Benard convection. My memory is fuzzy, but I think the only analytically solvable case of the linearized equations is when free surfaces are considfered both above and below, which is not very realistic. $\endgroup$
    – Jaime
    Commented Nov 8, 2012 at 0:14
  • $\begingroup$ I can go with $T=T_\text{top}$ with a free surface? I'd assume that I'd need to go with $T_\text{top}=T_\text{interface}$ with a free surface/interface. Rayleigh Benard convection, however, isn't applicable to ultra thin films, which is what I am interestedin, where long wave instabilities can manifest. I was under the impression that it would be a rather terrible thing to make the boussinesq approximation for cases susceptible to long wave instabilities. Which is what I was trying to prove via the modified NSE. $\endgroup$
    – dearN
    Commented Nov 8, 2012 at 0:44

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