# Bernoulli's law for turbulent flow

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!?

• There will be no turbulence in a non-viscous flow. So, no, Bernoulli does not apply to turbulent flows. Besides. turbulent flows are unsteady, which also rules out Bernoulli. – Time4Tea Sep 14 at 12:47
• What is an isentropic fluid? I've heard of isentropic flows, but never isentropic fluids. – Chet Miller Sep 14 at 13:22
• I meant that.Will edit.Thanks for the input – Schwarz Kugelblitz Sep 14 at 13:36
• Is turbulent flow even possible when all of your other assumptions are true? Address this fundamental issue first. Otherwise, the question seems to ask whether the Bernoulli equation is valid for turbulent flow but sets conditions that are the hallmarks of laminar flow. – Jeffrey J Weimer Sep 14 at 13:58

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 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.

Edit:

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.

• Hold on...an ideal fluid has coeffieicent of viscosity=0 , so it follows that it's Reynold number is infite,so by this logic even an ideal fluid is turbulent? – Schwarz Kugelblitz Sep 15 at 19:40
• What is a no-slip condition? – Schwarz Kugelblitz Sep 15 at 20:48
• @SchwarzKugelblitz https://en.m.wikipedia.org/wiki/No-slip_condition – Time4Tea Sep 15 at 21:36
• So u mean to say we can apply the Bernoulli's law on inviscous fluid which is rotating?Since,according to your condition of turbulent flow this flow is not turbulent as it is not viscous? – Schwarz Kugelblitz Sep 20 at 23:16
• @SchwarzKugelblitz I believe that is true: Bernoulli's equation can be applied to rotating flows that are inviscid (e.g. inviscid flow around a pipe bend), because there are clear streamlines present that are not disrupted by turbulence. – Time4Tea Sep 23 at 2:37

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.

• But it is not necessary that turbulent flow has loss of energy! – Schwarz Kugelblitz Sep 14 at 12:50
• As i mention in my answer there will be viscous force in turbulent flow. – yuvraj singh Sep 14 at 12:55
• Multiplying both sides with $m$ doesn't solve anything. Also, badly formatted answer. – Gert Sep 14 at 13:14
• I see no reason as to why it is necessary that viscous force be present in turbulent flow...besides Bernoulli isn't valid for streamline flow which is viscous.. – Schwarz Kugelblitz Sep 14 at 13:15

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

W=F delta(x)

and also when we suppose to take equation of continuity to find the Volume...

V=Avt

• But if energy is being conserved in turbulent flow,why can't we apply berniulli's equation.It is certainly not necessary that a turbulent flow dissipates energy – Schwarz Kugelblitz Sep 14 at 13:51
• In turbulent flow energy energy dissipation is must, due to the fact that fluid does not obey the conditions of constant Volume. – Sallo Sep 14 at 13:55