Understanding Reversible Fluid Flow and Its Relationship to Reynolds Number

[member 428835]

Can someone explain what it means for a fluid flow to be reversible and how this relates to Reynolds number, both intuitively and mathematically?

Thanks!

 

[Chestermiller]

What they are referring to is definitely not thermodynamic reversibility. I think what they mean is that, if you have a boundary between two portions of colored liquid and you subject the fluid to a low Re flow, neglecting diffusion, if you reverse the flow, the original boundary shape will be recovered. So the reversibility being referred to is kinematic reversibility. The version I always think of is that, if you write a secret message (with colored fluid) in the fluid and then deform it, the message will apparently disappear as the fluid deforms, but, if you reverse the flow, the secret message reappears.

 

[boneh3ad]

Low Reynolds number flows do represent the theoretical minimum for viscous dissipation, so they are about as close as your could truly get to thermodynamic reversibility. If you took the limit as $Re\to 0$, then the Navier-Stokes equations reduce to a set of equations based on the vector Laplacian of the velocity field and are linear in both velocity and pressure, so they are much easier to solve and use, but dissipation is still not going to be zero.

This is one of my favorite demonstrations illustrating the (near) reversibility of low Reynolds number (i.e. laminar) flows:

 

[Chestermiller]

Low Reynolds number flows do represent the theoretical minimum for viscous dissipation, so they are about as close as your could truly get to thermodynamic reversibility.

I couldn't disagree more. In polymer processing, we routinely dealt with very high viscosity fluids (> 1000 Poise) such as polymer melts which exhibited very significant viscous dissipation at typical deformation rates in screw extruders and pumps, spinning capillaries, transfer lines, spinning packs, film dies, etc. For all these applications, the Reynolds number was always < 1. Pressure drops were on the order of hundreds to thousands of psi.

 

[boneh3ad]

I couldn't disagree more. In polymer processing, we routinely dealt with very high viscosity fluids (> 1000 Poise) such as polymer melts which exhibited very significant viscous dissipation at typical deformation rates in screw extruders and pumps, spinning capillaries, transfer lines, spinning packs, film dies, etc. For all these applications, the Reynolds number was always < 1. Pressure drops were on the order of hundreds to thousands of psi.

Sure, the flow of any fluid with a very large viscosity is going to involve a large amount of dissipation. After all, dissipation is directly proportional to viscosity. However, consider that any fluid can obviously have essentially an infinite number of possible velocity fields depending on the values of its various driving forces. If you limit our view to all of the possible incompressible flows of that fluid (and low Re flows are pretty much exclusively incompressible given they typically feature very low velocities), then Stokes flow ($Re=0$) represents the minimum dissipation rate for that fluid. So I am sure your polymer flow featured large amounts of dissipation, increasing the Reynolds number would have only made it worse.

Source:
https://www.amazon.com/dp/0486442195/?tag=pfamazon01-20by Kim and Karrila

 

[Chestermiller]

Sure, the flow of any fluid with a very large viscosity is going to involve a large amount of dissipation. After all, dissipation is directly proportional to viscosity. However, consider that any fluid can obviously have essentially an infinite number of possible velocity fields depending on the values of its various driving forces. If you limit our view to all of the possible incompressible flows of that fluid (and low Re flows are pretty much exclusively incompressible given they typically feature very low velocities), then Stokes flow ($Re=0$) represents the minimum dissipation rate for that fluid. So I am sure your polymer flow featured large amounts of dissipation, increasing the Reynolds number would have only made it worse.

Source:
https://www.amazon.com/dp/0486442195/?tag=pfamazon01-20https://www.amazon.com/dp/0486442195/?tag=pfamazon01-20

I guess that's not what I got out of what you were saying in post #3, especially the words "they are about as close as your could truly get to thermodynamic reversibility." You can see how I might have interpreted this as "low Reynolds number flows are generally close to being thermodynamically reversible." Of course this isn't correct, but it probably is not what you meant.

 

[boneh3ad]

I guess that's not what I got out of what you were saying in post #3, especially the words "they are about as close as your could truly get to thermodynamic reversibility." You can see how I might have interpreted this as "low Reynolds number flows are generally close to being thermodynamically reversible." Of course this isn't correct, but it probably is not what you meant.

No I get it. My choice of words implied that Stokes flow necessarily has very low viscous dissipation, which is not true. That is not what I was meaning to convey according to the exact meaning of my statement, but it was certainly inadvertently there in the implication.

 

[member 428835]

Thank you both for your insight! So what qualities of the flow would the above-posted video hold? It seems very high viscous flows would be reversible. Is there a reason why?

 

[Chestermiller]

Thank you both for your insight! So what qualities of the flow would the above-posted video hold? It seems very high viscous flows would be reversible. Is there a reason why?

Like I said, if diffusion is negligible, if you reverse the flow, the velocities are exactly reversed (as described by the equations of motion and the continuity equation), the deformation is reversed, and the initial confituration of the system prior to deformation is recovered. For low Re, the equations are linear, and so, -v satisfies the equations as well as +v. This won't happen if the inertial terms are present in the equations of motion.

 

[member 428835]

Thanks!