To model an inviscid flow analytically, we can simply let the viscosity be zero (𝜇=0). It is more difficult to create an inviscid flow experimentally, because all fluids of interest (such as water and air) have viscosity.

The question then become; are there flows of interest in which the viscous effects are negligibly small? The answer is ,Yes, if the shear stresses in the flow are small

𝜏=𝜇(𝑑𝑢/𝑑𝑦).

So, all fluids have aviscosity and can not assume zero.

Then the invisced flow means du/dy=0 which leads to zero shear stress

Hello,

It would be infinite for a perfect fluid, indeed.

However, there is no "perfect fluid". What comes closest to a perfect fluid is a superfluid (fluid whose viscosity is almost zero). In this case, one can calculate the Reynold number, and this one would be very large.

Hoping to have answered your question,

BAUDRIMONT Roman

for real fluids viscosity will not be zero, so Reynolds number will bot be infinity.

This question often arises in the context of the analysis of the perfect fluids, which are characterized by the vanishing viscosity. The traditional definition of the Reynolds number in such a case leads to the unphysical values going to infinity. The problem can be avoided by recognizing that even in a superfluid, when the flow is sufficiently fast, the transition to turbulent flow can be observed. One of the best discussions about such a situation is made by Reeves et al. (2015), (T. Reeves, M & Billam, Thomas & P. Anderson, B & Bradley, Ashton. (2015). Identifying a Superfluid Reynolds Number via Dynamical Similarity. Physical Review Letters. 114. 155302. 10.1103/PhysRevLett.114.155302.)

The authors suggest a generalized Reynolds number that follows the ideas of Onsager. The explicit formula for Re in the case of a superfluid is: Re = uLm / h, where h is the Planck constant and m is the atomic mass.