incompressible flow?

Because many calculations are much easier assuming constant volume.

Michael Dahms

Yes!

The reason for making assumptions is to make the calculation easier (or possible).

Although fluids like air are certainly not incompressible, the variation in their density is often unimportant.

Usually, as long as the Mach number is small (

Because assuming a liquid fluid (not gas fluid) incompressible is good enough in 98% of situations. But perhaps you get some kind of enjoyment out of solving PDEs with non-symmetric stress tensors? You can keep the term if you really want or need it.

Maybe good enough in situations without a pressure like open flow channels. Hydropower pressure pipes need a surge tank to regulate elastic energies during valve closing!

I trie to give the answer found in most text books of elasticity. The bulk modulus K becomes K=E/(3*(1-2*n)) . So the denominator vanishes for materials with Poisson number n=0.5 and therefor K is becoming infinite (incompressibel). This singularity backs on the assumption that the elastic modulus E>0 which is not the case for materials with n=0.5 where E=0 and G=0. The bulk modulus for fluids can be found better with K=L*(1+n)/(3*n) where L is the lame constante which tend toward K=L for n=>0.5 !

Peter

Don't you think that the use of elastic modulus equations for solids is a poor beginning to understand the flow of fluids?

Encountering singularities is often a good clue that your simplifying assumptins are inapplicable.

I think that the use of solid elasticity to understand fluids is loaded with inconsistencies. Such as what you describe above.