# Biological Fluid Dynamics - Approximate Navier Stokes Equations Solns

1. Dec 11, 2013

### ghughes

1. The problem statement, all variables and given/known data

Good evening. First post on this forum! The problem I wish to state would take too long to write by hand so I thought it best to do so via attachment. The question I am stuck on is part d and, in fact, part e also.

2. Relevant equations

All relevant equations are given within the attachment

3. The attempt at a solution

To be honest, I am unsure how to even start this problem but I know the appropriate BCs would be non slip conditions which are dimensionless. I.e. URL^-1u_r = 0 => u_r = 0. From here I am unsure whether to proceed. I have imposed that u remains finite for the other condition but I am at a loss of how this can be done.

Due in Friday so really panicking!

Any helps, tips or relevant information would be very much appreciated. Thank you very much to anybody for their time! My expertise lies in group theory so would be happy to help anybody in that field!

ghughes

#### Attached Files:

• ###### Physics.docx
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2. Dec 11, 2013

### Staff: Mentor

Hi GHughes. Welcome to Physics Forums.

I can see where this problem is trying to get you headed, but I don't know how to get you there without working the problem myself. It is trying to get you to the point where you determine that the axial flow can be approximated locally by a parabolic velocity distribution, so it is locally the same as axial flow in a tube of constant radius. You can see that from the f4 in the equation. Don't worry about the boundary conditions for now. What they want you to do is make the substitutions into the continuity and momentum equations that they are recommending, and combine parameters so you then have the equations in terms of the dimensionless groups and the dimensionless velocities, times, distances, etc. You then make the determination the the Reynolds number is very small, and set it equal to zero to get rid of the inertial terms. Just try doing what they said. Let us know how far you get, and, maybe we can help you further.

Chet

3. Dec 11, 2013

### Staff: Mentor

Incidentally, if I were doing this analysis, I wouldn't have done it the way they recommended. I would have defined the dimensionless radius as r/f(t,z), and transformed coordinates so that the dimensionless radial coordinate was a new independent variable in place of r. That way, the non-slip boundary condition would always be applied at dimensionless radius equal to 1. This method would remove all the moving-boundary and undulating-boundary complexity from the boundary conditions and put it into the differential equations. This would be much easier to deal with, and it would allow you to get to the final results that you are seeking in a more structured way.

Chet

4. Dec 12, 2013

### ghughes

Well I've done a and b and c. I've non dimensionalised the equations and arrived at what they say for c.

I've taken that second equation, multiplied through by r and differentiated both sides w.r.t. r.

I ended up with u_z = (dp/dz)((r^2)/4) - (dp/dz)((f^2)/4) after applying the fact that u must remain finite and the no slip condtion that u_z(r = f) = 0

That u_z sorted I think. I don't know what to do for u_r though. I tried a similar approach. Multiplying that first equation through by r and integrating between 0 and f but it didn't really work.

Hope that's all clear!

5. Dec 12, 2013

### Staff: Mentor

Your z velocity looks like it's missing a viscosity. If you've otherwise solved for the axial velocity correctly, I can see what to do next. It's suggested by the form of the equation you are trying to arrive at. You substitute uz into the continuity equation, and integrate with respect to r from r = 0 to r = f. At r = 0, urr = 0, and, at r = f, urr = -f∂f/∂t. Don't forget to multiply both terms by r before you start integrating, and don't forget to move the r into the partial derivative with respect to z.

Chet

6. Dec 12, 2013

### ghughes

Where would the viscosity term come from. So integration ru_r between 0 and f gives -f(df/dt) and that equals the integral between 0 and f of (d/dz)ru_z where u_z which is what I have worked out. Kind of figured that was the right approach. Thanks Chestermiller. Just confused regarding the viscosity for u_z!

7. Dec 12, 2013

### ghughes

So I'd integrate integral between 0 and f of (d/dz)(ru_z) = (d/dz)((dp/dz)(r^3)/4 - (dp/dz)((rf^2)/4) dr

?

ghughes

8. Dec 12, 2013

### Staff: Mentor

Yes. Oh, I think the viscosity got absorbed into the dimensionless parameters. So, yes, if you integrate this way, you end up with the result you were looking for (I think). If not, figure out what adjustment you would have to make to uz to get the result, and then figure out how to modify your derivation of uz to get what you want.

Chet

9. Dec 12, 2013

### ghughes

Yeah, I was going to say that regarding the viscosity. We have an extra term that's different from the entire answer coming from the u_z bit. If for u_z we got ((r^2)/4)(dp/dz) instead of ((r^2)/4)(dp/dz) - ((f^2)/4)(dp/dz) (i.e. our constant of integration was 0) then we would get the desired result. Almost there I guess!

Not too sure why that constant would be 0 though.

10. Dec 12, 2013

### ghughes

The no slip condition u_z(r=f) = 0 gives a value for B :s

11. Dec 12, 2013

### ghughes

Just a thought...

If the boundary condition were u_z(r=0) = 0 then we would get that B=0. But at r=0 we have already imposed the velocity must be finite and that eliminates the other constant.

12. Dec 12, 2013

### Staff: Mentor

I don't see where there's an extra term. I get f4/2-f4/4=f4/4.

13. Dec 12, 2013

### ghughes

Well (dp/dz) = (1/r)(d/dr)(r(du_z/dr)) then multiplying through by r:

r(dp/dz) = (d/dr)(r(du_z/dr)). Intgrate both wrt r:

((r^2)/2)(dp/dz) + A = r(du_z/dr). Divide by r and integrate again:

((r^2)/4)(dp/dz) + Aln(r) + B = u_z.

Imposing that its finite implies A = 0. So: ((r^2)/4)(dp/dz) + B = u_z.

Using u_z(r=f) gives ((f^2)/4)(dp/dz) + B = 0 which implies B = -((f^2)/4)(dp/dz)

Therefore u_z = ((r^2)/4)(dp/dz) - ((f^2)/4)(dp/dz). Is that wrong?

14. Dec 12, 2013

### Staff: Mentor

No, it's correct. But when you multiply by rdr and integrate between r = 0 and r = f, the two terms combine into a single term.

15. Dec 13, 2013

### ghughes

Thank you for all the help Chestermiller, handed it in now.