Question

"Is the minimum viscosity of a fluid bounded?"

"If Ψ(x,t) is a solution to a wavefunction, why mathematically would AΨ(x,t)∣A∈C also be a solution?"

Had these questions scribbled down in my notes. I'm wondering if anyone could help me with a rigorous, mathematical treatment of them.

Answer 1

Let me dig up my Fluid Mechanics book, and I'll see what it says about the first one.

I'm not much use on the second.

Answer 2

@eashmore I appreciate it. :D

I don't have any of my textbooks. :( I have to sell them to afford the next round.

Answer 3

I wish I still had my Resnick books. They were the best fundamentals.

Answer 4

That name sounds so familiar. I'm sure I have one of his books lying around here somewhere (I live in organized chaos).

It looks like for a Newtonian fluid viscosity is a function of temperature and pressure. Since gasses exhibit an increasing trend in viscosity with increasing temperature, and liquids have the opposite trend. The minimum viscosity would likely occur at the condensation point. That sounds reasonable.

Answer 5

That sounds reasonable, but I feel as if there's a minimum viscosity for fluids before the condensation point. Given the context of my notes.

Answer 6

It's a shame you had to sell your books. I feel that sometimes the book is more valuable than the tuition you pay for the class. I tend to forget the nitty gritty stuff books are so good about. I have nearly all the books I have bought throughout my college career.

My book doesn't offer much than that. It offers up some equations for the effect of temperature.

For gasses. \[\mu = {b \sqrt{T} \over 1 + S/T} \]where b and S are constants. This is the Sutherland empirical correlation. From The US Standard Atmosphere. US Government Print Office. 1976

For liquids\[\mu = A e^{B/(T-C)}\] where A,B, and C are constants. I believe this is also the Sutherland correlation.

Answer 7

So, essentially the minimum\[\frac{B}{T-C}\]and\[A\] should give us the smallest value for \(\mu\). Knowing the general minimum ranges for these constants, I think, we have a \(\mu\) min.

Answer 8

I wish I could give you a medal, but that'd close the question, and then there'd be no help for the quantum. :(

Answer 9

The books gives, for water\[\left(\begin{matrix}A = 2.414 \times 10^5 [N \cdot s/m^2]\\ B = 247.8 [K] \\ C = 140 [K]\end{matrix}\right)\]This is within \(\pm 2.5\)% from \(0^\circ C ~{\rm to}~370^\circ C\) according to Touloukina, Y. S., S. C. Saxena, and P. Hestermans, Thermophysical Properties of Matter, the TPRC Data Series. Vol.11-Viscosity. New York: Plenum Publishing. 1975.

Let me do a university library search and see what I can come up with.

No worries on the medal. I'm not in it for the medals. :-)

Answer 10

Aha, so in it for principles? Thanks, though. You are immeasurably helpful.

Answer 11

Especially since I don't exactly have access to my school's libraries.

Answer 12

I found something close. Same title and author, but five years prior. Let me pick it up and see if the tables reveal anything. I'll grab some scans if I come across anything.

Unfortunately, my books are of the engineering type and typically lack much of the nitty gritty your questions always seems to require. Stop reading through old notes dang it! :-P

Answer 13

:D You can blame my professors, asking questions in class and not providing answers. Then going over a ton of new topics, so we forget to ask. :(

Nah, he was a good teacher. Just a bit too much breadth and not enough depth.

Answer 14

Interesting. I don't think many liquids are significantly less viscous than water, so finding water's minimum is a good start.

Answer 15

I feel your pain. The "jack of all trades, master of none" rings true with me. It seems unsatisfactory to lack the depth. I don't know how much more my brain can handle. It feels awfully saturated as of late. Or maybe that's the sleep depravation that is making me hazy. I forget. :-)

Here are some plots from my book.

Answer 16

good question

Answer 17

@JamesJ Don't mind if I drag you in. :D

Answer 18

With respect to the second question, the wave equation is a linear homogeneous differential equation, LHDE.

If f is a solution of an LHDE, then cf is also a solution of that equation.

If this general answer doesn't help you, make the calculation explicitly: take your solution \( C\Psi \) and substitute it back into the wave equation. You will see that both a necessary and sufficient condition for \( C\Psi \) to be a solution is that \( \Psi \) is also a solution.

Answer 19

Thanks!

Answer 20

I'm giving Eashmore the medal since he isn't level 99 haha