# I Thin film around circle and ignoring curvature

1. Mar 13, 2017

### joshmccraney

Hi PF!

If we have flow around a curved object that is sufficiently thin, I Have seen many texts assume the surface is linear rather than curved. Can someone help me with what "sufficiently thin" is quantitatively and how this allows us to neglect surface curvature?

As a simple toy problem, perhaps a circle of radius $R$ and a film of thickness $\epsilon$.

2. Mar 13, 2017

### pasmith

If you're near the north pole of a sphere of radius $R$, you can consider a local coordinate system $(s,\phi,z)$ where, in cartesian coordinates, $$\mathbf{r} = (R\sin \frac{s}{R} \cos\phi, R\sin \frac{s}{R} \sin\phi, (R + z)\cos \frac{s}{R}) - (0,0,R)$$ or $$\mathbf{r}(s,\phi,z) = R\left( \sin \frac{s}{R} \cos \phi, \sin \frac{s}{R} \sin \phi, \left(1 + \frac{z}{R}\right)\cos\frac{s}{R} - 1\right).$$
To leading order with $|z/R| < |\epsilon/R| \ll 1$ and $|s/R| < |\epsilon/R| \ll 1$ this is $(s \cos\phi, s \sin\phi, z)$ - which are cylindrical polar coordinates with axis normal to the sphere and origin at the north pole. Of course there is some error, and you can expand $\|\mathbf{r}(s,\phi,z) - (s\cos\phi, s \sin\phi, z)\|$ as a Taylor series in $s$ and $z$ to see how large that might be.

3. Mar 13, 2017

### hilbert2

Do you mean something like what's done here: http://www.itg.cam.ac.uk/people/heh/Paper219.pdf ?

4. Mar 13, 2017

### A.T.

Is it explicitly the thickness that is used to justify neglect surface curvature? Or for example the low acceleration due to curvature, compared to other effects?

5. Mar 13, 2017

### Staff: Mentor

Why don't you consider a simple heat conduction situation in cylindrical coordinates, where the temperature at r = R is $T_0$ and the temperature at $r=(R+\delta)$ is $T_1$ and the thermal conductivity is k. You want to determine the heat flow if (a) you include the curvature and (b) you neglect the curvature. You also want to compare the temperature profiles. A good rule of thumb is that, if $\delta/R<0.1$, you can neglect the curvature.

6. Mar 14, 2017

### 256bits

The first sentence states a condition where the object is sufficiently thin. ( Likely not what you meant, just be picky )
The second sentence states a condition where the film is thin.( in line with the heading )

Relate it to two infinite parallel plate moving relative to one another. A linear velocity profile of the fluid wrt the y=direction perpendicular to the plates can be obtained.
The "infinite" is unobtainable, so we instead assume long enough and wide enough so that the steady state velocity profile has a chance to be set up, and edge effects are minimal and can be neglected for the flow interior.

For two concentric drums with a relative rotational velocity, the edge effect is pretty much the same as above. The "infinite" length problem appears to go away as there is no start end to a circle as far as I know. Except that one "plate" is now just a little bit longer than the other due to the different radii. If ε << R of the inner drum, the outer drum length ( circumference ) is nearly the same as the inner. The velocity profile of the fluid is then assumed to be as in the flat plate case - ie a linear function of y.

Another problem crops up though, and that is centrifugal effects upon the fluid. That also is minimized with a thin film, and by having the outer drum rotate wrt the inner drum.

7. Mar 14, 2017

### joshmccraney

I am using Bird, Stewart, and Lightfoot, and a sample problem they give (section 4.4) they have flow around a cylinder. They say to consider the boundary layer very small, so that we can neglect the curvature. They proceed to give the Navier-Sokes equations in rectangular coordinates.

Yea, this looks great, but can you explain the last term on equation 2.2? Is it from Young-Laplace, because it looks similar to that but different too.

Ok, so in cylindrical coordinates I'm getting $$\partial_r(r\partial_rT)=0 \implies\\ T = \frac{(T_0-T_1)\log(r)+T_1 \log(R) - T_0 \log(R + \delta)}{\log\frac{R}{R-\delta}}$$I notice $\delta \to 0 \implies T \to \infty$ but this doesn't seem right. Setting this up in rectangular coordinates is a pain since the boundary conditions are tough due to the circular boundaries.

pasmith, I don't really understand the coordinate system you've adopted; do you have a reference website you could direct me to?

Could you explain why the velocity is assumed linear when $\epsilon \ll R$?

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