Reynolds Number We know that the Reynolds Number is $R_e=\frac{\rho vd}{\eta}$ ($\rho$ density, $v$ velocity, $d$ diameter and $\eta$ the viscosity of fluid). 
We also know that an ideal fluid has no viscosity which means that $\eta=0$ and what we get is $R_e=\frac{\rho vd}{0}$. Does this mean that the ideal fluid doesn't have a Reynolds Number?
In my book it says that every fluid has it's own Reynolds Number.
Can anybody help me?
Thank you!
 A: At the moment the answer to your question is still matter of active research. In many aspects liquid helium at very low temperatures is an ideal fluid but the Reynolds number is still finite (Barenghi, 2008).
So it seems that every fluid has a finite Reynolds number and only in a gedanken experiment one can have an infinite Reynolds number.

Barenghi, C., "Is the Reynolds number infinite in superfluid turbulence?", Physica D: Nonlinear Phenomena,
V. 237, 15 Aug. 2008, pp 2195–2202.
A: So, I guess, the Reynolds number is infinite for an ideal fluid. Or, if you insist that infinity is not a number, then an ideal fluid does not have a Reynolds number. However, the author of your book can say that, if we want to be pedantic, there is no such thing in reality as an ideal fluid, so each fluid does have a Reynolds number:-)
A: 1) The Reynolds number is a combination of fluid properties, $\eta/\rho$, and flow properties, $vd$. So yes, for a given flow profile, every fluid has its own Reynolds number.
2) Ideal fluids are, as the name suggests, an idealization. Real fluids have finite viscosity. There are reasons to believe that there is a fundamental bound to how small viscosity can get, see for example http://arxiv.org/abs/0904.3107. 
A: "We also know that an ideal fluid has no viscosity" - this statement is incorrect. We know that there is no such a thing "ideal flow" - it's a mathematical abstraction created in order to solve the equations that are otherwise impossible or at least very difficult to solve. Therefore, the dimensionless parameter defined following the research of Reynolds cannot be applied to the non-existing matter. 
