Solving for the Constants in Fluid Flow through a Pipe

enc08
Messages
40
Reaction score
0
Hi,

I'm looking at the solution to a question on fluid flow through a rigid pipe.

Original equation: \mu u = 0.25r^{2} dp/dx + Aln(r) + B
After applying boundary conditions: \mu u = 0.25dp/dx (r^{2} - a^{2})

I don't understand how the constants have been solved for. Below is as far as I get:
Starting with
\mu u = 0.25r^{2} dp/dx + Aln(r) + B
Assume a no-slip boundary condition, so
u(r = a) = 0: 0 = 0.25a^{2} dp/dx + Aln(a) + B

The notes somehow end up with Aln(a) = 0.

Thanks for any input.
 
Physics news on Phys.org
A must be zero, or else u(0) = A \ln(0) + B, which is mathematical and physical nonsense.
 
I see. So the edge of the cylinder is defined as r = 0, and the centre as r = a?

Thanks for clearing it up.
 
enc08 said:
I see. So the edge of the cylinder is defined as r = 0, and the centre as r = a?

Thanks for clearing it up.

No, the center of the cylinder is r = 0.

You have a 2nd order ODE for u(r); it has two boundary conditions. One is that u(a) = 0, the other is that u(0) is finite. It is this condition that requires you to reject the ln(r) complimentary function.
 
I see, thanks.
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top