# Fluid flow through pipe

1. Jan 22, 2013

### enc08

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.

2. Jan 22, 2013

### pasmith

A must be zero, or else $u(0) = A \ln(0) + B$, which is mathematical and physical nonsense.

3. Jan 22, 2013

### enc08

I see. So the edge of the cylinder is defined as r = 0, and the centre as r = a?

Thanks for clearing it up.

4. Jan 22, 2013

### pasmith

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.

5. Jan 22, 2013

### enc08

I see, thanks.

Last edited: Jan 22, 2013