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:
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top