Is Bernoulli's law valid for turbulent flow for a non-viscous, non-compressible, non-rotating fluid with isoentropic flow? An ideal fluid has coeffieicent of viscosity=0,which implies that it's Reynold number is infinte,which in turn means that even an ideal fluid is turbulent!?
Firstly, there is a misconception in your question: turbulent flow cannot be inviscid. Turbulence comes from vorticity in the boundary layer that somehow finds its way into the bulk flow of the fluid. This can happen due to separation from a sharp edge or an adverse pressure gradient, or the vorticity can be stripped from the walls by a flow instability (e.g. in the case of turbulent flow through a pipe). That vorticity in the boundary layer is due to the velocity gradient that is created by viscosity. Turbulence is both created and destroyed by viscosity. Without viscosity, you cannot have turbulence.
Now, the requirements for Bernoulli's Equation to be valid are as follows:
- flow must be steady
- flow must be incompressible
- flow must be inviscid
- flow is reversible
- the equation is applied along a streamline
A turbulent flow violates several of these requirements.
Firstly, as mentioned above, turbulent flows are always viscous. Secondly, turbulent flows are inherently unsteady, and thirdly, it is not possible to identify streamlines in a turbulent flow, because they all get tangled up in the highly complex mixing eddies.
So, no, you cannot use Bernoulli's Equation for a turbulent flow.
To address your point regarding the Reynolds Number going to infinity as viscosity goes to zero:
It seems you are having some difficulty accepting that viscosity is a necessary requirement for turbulence to exist. Yes, it is true that $Re$ tends to infinity as viscosity goes to zero. However, Reynolds Number is a simple concept, based on dimensional analysis, which is intended to be used for comparing realistic, viscous flows. You are trying to extend it to a theoretical, unrealistic, inviscid fluid, where it doesn't really apply.
Even a vanishingly tiny amount of viscosity imposes a 'no-slip' condition, which means the velocity of the fluid is zero at the wall. However, if the viscosity vanishes completely, the behavior changes dramatically, because there is no longer a no-slip condition at the wall and you can have an arbitrary velocity there. That change in behavior is discontinuous and is not captured by the simple Reynolds Number concept.
So, no, it is still the case that an inviscid fluid cannot be turbulent.
No it is not valid Bernoulli equation is all about energy conservation, let say we have equation is this, $p+kgh+1/2kv^2=l$ (where $k$ is liquid density and where $l$ is constant). Multiply both side by $m$ (where $m$ is mass of fluid) $pm+kghm+1/2kmv^2=lm$ ($=C$ new constant), here only translation kinetic energy is included, there is no mention about energy which losses due to viscosity and rotational kinetic energy.
Bernoulli's law (equation) is not valid for turbulent flow for a non-viscous, non-compressible, non-rotating, isoentropic fluid, because this can't manipulate our supposition about deriving Its relationship, i.e when we take sum of work done on the fluid by fluid behind it
and also when we suppose to take equation of continuity to find the Volume...