Hello. This is a good question and the answer is not so straightforward.

1- The Hagen-Poiseuille solution is obtained under the asumptions that the velocity field is stationary, and independent of the axial and tangential coordinates. Then, the analytical solution of the Navier-Stokes equations leads to a velocity profile u(r)=v(1-(r/R)^2). This solution is only observed after a certain distance from the entrance of the pipe.

2- When the flow becomes turbulent, the velocity is time dependent and 3D: there is no more analytical solution. The time-averaged velocity is still stationary, axisymmetric and the radial profile is "flat": for convenience, it might be well described in the bulk by a profile like \overbar{u}=v(1-r/R)^(1/n), with n between 6 and 10.

3- All the question is to know or decide whether the flow is turbulent or not. The base flow is linearly stable whatever the Reynolds number (for a cylindrical pipe). It is possible, with extreme care, to maintain laminar pipe flows until Re=10^5. So the common thinking "Re>2300 in a pipe is equivalent to turbulent flow" is incorrect. It is however true that in industrial configurations, turbulent flows are observed, due to unavoidable perturbations.

Here are some references:

http://naca.central.cranfield.ac.uk/reports/1954/naca-report-1174.pdf

MULLIN, T. & PEIXINHO, J. 2006. Transition to Turbulence in Pipe Flow. Journal

of Low Temperature Physics, 145, 75-88.

HOF, B., JUEL, A. & MULLIN, T. 2003. Scaling of the turbulence transition

threshold in a pipe. Phys. Rev. Lett., 91,244502

Hello. This is a good question and the answer is not so straightforward.

1- The Hagen-Poiseuille solution is obtained under the asumptions that the velocity field is stationary, and independent of the axial and tangential coordinates. Then, the analytical solution of the Navier-Stokes equations leads to a velocity profile u(r)=v(1-(r/R)^2). This solution is only observed after a certain distance from the entrance of the pipe.

2- When the flow becomes turbulent, the velocity is time dependent and 3D: there is no more analytical solution. The time-averaged velocity is still stationary, axisymmetric and the radial profile is "flat": for convenience, it might be well described in the bulk by a profile like \overbar{u}=v(1-r/R)^(1/n), with n between 6 and 10.

3- All the question is to know or decide whether the flow is turbulent or not. The base flow is linearly stable whatever the Reynolds number (for a cylindrical pipe). It is possible, with extreme care, to maintain laminar pipe flows until Re=10^5. So the common thinking "Re>2300 in a pipe is equivalent to turbulent flow" is incorrect. It is however true that in industrial configurations, turbulent flows are observed, due to unavoidable perturbations.

Here are some references:

http://naca.central.cranfield.ac.uk/reports/1954/naca-report-1174.pdf

MULLIN, T. & PEIXINHO, J. 2006. Transition to Turbulence in Pipe Flow. Journal

of Low Temperature Physics, 145, 75-88.

HOF, B., JUEL, A. & MULLIN, T. 2003. Scaling of the turbulence transition

threshold in a pipe. Phys. Rev. Lett., 91,244502

thank you Florent so much for your detailed answer. I can understand more now! I have here please some other questions related to this subject:

1. Do you think that the time averaged velocity in a turbulent flow is axisymmetric? Why do we observe so coherent structures (streaks and vortices) giving rise to radial and tangential components of velocity.

2. Is there any approach that lead to derive the flat profile describing a turbulent pipe folw?

3. I would be pleased if you send me the second paper published in Phys. Rev. Lett. I read a similar paper published in nature by Hof et al. and I would like to deepen my knowledge in transition in pipe flows.

Thank you!

How can i send the paper to you ?

I agree with Florent.

However, what can derived analytically from the balance equation is the stresses only (shear stress and static pressure).

The behaviour law, which is not a balance equation, is necessary to obtain the velocity field. The velocity profile is parabolic if and only if the Reynold stress is zero and the behaviour law is the Newton's law of viscosity.

Please check the work of Jirkovsky and Muriel mentioned in ResearchGate.

The Hagen-Poiseuille solution is a solution for the incompressible stationary Navier-Stokes equations (that means without turbulence model). Now if the flow that you look for is turbulent, it is definitely not stationary. Now you can talk a lot about averaging and artificially removing fluctuations. But if you do this you need to introduce new terms in the equations to account for the correction that you would need. So the answer is no.

@Amador : thank you for the information. Very interesting. So the answer maybe 'yes' with a modified 'Hagen–Poiseuille' solution ? :)

Have you considered the wall roughness in your approach (how the probability sigma can be affected by this roughness ?)

In the strictest sense, the answer is "no." However, you didn't state how you wanted to apply the flow solution.

If you wanted to look at the flow profile, Florent is right that the profile will no longer be parabolic; rather, it becomes flatter in the turbulent center with greater shear at the boundary, approaching "plug" flow. Of course I am speaking about the velocity that is averaged in some way, not the instantaneous velocity.

If instead you are just interested in the amount of fluid passing through a pipe for a given loss in static pressure (or vice versa), that is another matter. The answer is strictly speaking still "no," but you can fudge a little. The best place to start with this problem is a Moody diagram. This illustrates the friction factors (head loss) for pipes as a function of Reynolds number and pipe roughness.

It is true that strictly speaking no linear instability for pipe flow has been found, but for any practical engineering problem it is pretty much a given that flow with Re in excess of a few thousand is probably turbulent, unless one has gone to extreme measures to ensure a smooth perturbation-free inflow, etc. Convention is 2300 IIRC for the transition, although that exact number should be taken with a hefty dose of skepticism.

The interesting thing about the Moody diagram is that you'll see that for sufficiently high Re, the friction factor becomes independent of Re and depends only on surface roughness, and only weakly at that, whereas for laminar flow the friction factor is 64/Re (Darcy-Weisbach, not Fanning). Invert the laminar relation and treat it as defining the "effective" Reynolds number due to some effective turbulent velocity. Write this turbulent velocity as the mean velocity in the pipe V and the pipe diameter D times some proportionality constant C. You find that C is of order 1/few for practical problems, and a very-weakly varying function of surface roughness only (for "completely turbulent" flows). This is in line with what you would have expected the turbulent viscosity to be, just using rough mixing-length arguments. You *could* then plug this effective viscosity back into the H-P solution and interpret this as using the normal laminar Poiseuille solution but with an effective turbulent viscosity, although people don't normally do that. So In that sense, you can kinda say that you are using the H-P solution for turbulent flow, but again, that's a bit of a backwards way to think of it.

If fact, likewise, you *could* say that the non-parabolic flow profile in turbulent flow was due to some non-uniform effective turbulent viscosity, say where the turbulence behaves like a shear-thinning fluid like ketchup. Shear-thinning fluids have flatter velocity profiles than ordinary viscous fluids, in pipe flow.

If you look at your turbulent pipe flow and apply the Reynolds-averaged Navier Stokes equations and interpret the Reynolds stress as a turbulent viscous stress, you will get just such a radially-varying effective viscosity, because ultimately you still have to satisfy momentum balance, and the only additional term to do this (assuming Mach number is low enough) is velocity-velocity correlations.

Consider popular turbulence models like k-epsilon that have a turbulent viscosity in them. In k-epsilon, the turbulent viscosity at any given point in your pipe will be a function of k, the turbulent kinetic energy, and epsilon, the dissipation rate. In turn, k and epsilon are found by solving coupled transport equations, so that for a pipe, k and epsilon will both be functions of radius, and so the turbulent viscosity will be a function of radius. Then the mean flow profile will be non-parabolic, but will match what you would have gotten for a laminar flow with the same radial profile in viscosity. Normal H-P flow can be thought of as the limiting case of this system in which the viscosity is chosen to be independent of radius. Again, normally people don't think of it this way, but you could.

Dear Abdallah

No ! You can NOT use it for turbulent flow. NEVER !

Prof. A.A. Ranjbar

The answer as mentioned above is: NO. The Hagen-Poiseuile flow solution was obtained subject to the following explicit assumptions that simplify the governing equations:

1. Incompressible flow, i.e. the density is constant, rho=const.

2. Steady state, i.e. the velocity vector V and the pressure p are independent of time.

3. Developed flow, i.e. ∂V/∂x =0, the velocity does not change in the axial direction.

4. Two-dimensional flow, i.e. there are no changes of velocity or pressure in the angular direction. There is no velocity component in the angular direction.

All these assumptions (except assumption 1) violate the typical behavior of turbulent flow which is three dimensional and time dependent.

However, one may use the Hagen-Poisseuille velocity profile to investigate its stability and the transition point to turbulence.