In a fully developed turbulent flow of an in-compressible fluid inside a pipe of radius $R$, velocity at the center is $U_m$. If we define $U^*=\sqrt{\tau_0/\rho}$, where $\tau_0$ is the wall shear stress and $\rho$ is the density, then find the velocity distribution as a function of $y=R-r$ distance from the wall. Consider the $l=k \frac{du/dy}{d^2u/dy^2}$ as Von Karman mixing length.

Now, if we write $\tau \approx \tau_0=-\overline{\rho u' v'}= \rho l^2 (du/dy)^2$, then we get $$(U^*)^2=k^2 \left(\frac{du/dy}{d^2u/dy^2}\right)^2 (du/dy)^2$$ and $$U^*=k\frac{(du/dy)^2}{d^2u/dy^2}$$ Now let $p=u'$ to get $p'/p^2=k/U^*$. Integrating twice gives $$-1/p=\frac{k}{U^*} y+C_1$$and $$u=-\frac{U^*}{k} \ln \left( \frac{k}{U^*} y +C_1\right)+C_2. $$

Now, one of the conditions for finding $C_1$ and $C_2$ is $u(y=R)=U_m$. What will be the other condition? This is the problem I encountered solving a similar problem:

In a pipe with diameter $0.8 \ m$ water is flowing (turbulent) and velocity at $y=0.2 \ m$ is $2 \ m/s$. If the relation $u/U^*= C_1 \ln(y/R)+ C_2$ is true, then find $C_1$, $C_2$, and wall shear $\tau_0$ (notation is as same as above).

Should we relate this to viscous sub-layer somehow?

Note: I have posted this in the physics site also. If I get my answer in one SE site I'll delete the question in the other one.


1 Answer 1


I am not entirely sure I follow your derivations, specifically I'm not sure where you get the equation for the Von Karman mixing length, and why the equivalence of tau and $\tau_0$ give you the Prandtl turbulence model that you write afterwards (apologies since I can't get the equations to format properly, but I am referring to the equations you write immediately after stating $\tau = \tau_0$).

Regardless, in the context of the last question, you typically don't need to solve any differential equation once they have told you that the logarithmic overlap law desrcibes the flow, which is: $u/U^*=C_1 \ln(y/R)+C_2$. In Fluid Mechanics by White on page 364 he says as much "For turbulent pipe flow, we need not solve a differential equation, but instead proceed with the logarithmic law..."

In response to whether you need to relate the viscous sub-layer, I believe the answer is no. Once the question has told you the flow is determined by the logarithmic relationship, then any viscous sub-layer equation will not be valid for that specific flow. In this case, you can assume the Von Karman constant $k=0.41$ (for pipe flow) and $C_1 = 1/k$ and $C_2 = 5.0$ (also for pipe flow). Admittedly this has just been pulled from tables, so maybe this is not what you're looking for. Nonetheless that's how we were asked to solve these problems. After using these values, you then just make a simple substitution using the data given, and calculate $U^*$ from the equation $U^* = \sqrt{\tau_0/\rho}$. (see edit below)

In response to the question of the boundary conditions, my understanding is that the equation supplied to you in the question is the logarithmic overlap law. These constants were not solved by using boundary conditions; they were solved by experiment. You can see this here https://en.wikipedia.org/wiki/Law_of_the_wall#General_logarithmic_formulation.

If you are familiar with the Reynold's stresses equation, it can be reduced in 2D to yield:

$\tau + \tau_{turbulence} = \tau_{wall}$ (or in your notation $\tau_0$)

From here it can be non-dimensionalised and then boundary conditions are used with THIS equation in order to generate the logarithmic overlap law, and other valid solutions to describe the flow. In other words, the logarithmic overlap law is the outcome from the Reynold's stresses equation, and the constants in the logarithmic overlap law have been determined experimentally. I hope this helps, and sorry again for the lousy formatting.

EDIT: I made a significant error in the third paragraph. The question asks you to find the shear stress, but "make a simple substitution using the data given" won't help you do that, since you know neither U* nor $\tau_o$.

Specifically, for the equation to be the logarithmic overlap law, then

$ y/R = (y*U^*)/\nu$, where $\nu$ is the kinematic viscosity. --> If you're not sure where this comes from read this: https://en.wikipedia.org/wiki/Law_of_the_wall. It is simply using the normalised equations.

From this, you can calculate $\tau_o$ in terms of R, as per the question. To do this, equate U* in terms of R, and then use $U^* = \sqrt{\tau_o/\rho}$. This will allow you to solve for tau in terms of R.

  • 1
    $\begingroup$ Thank you very much :) I'll look at this and inform you :) $\endgroup$
    – Ghartal
    Feb 28, 2018 at 19:20
  • $\begingroup$ No worries, I hope it helps! $\endgroup$
    – masiewpao
    Feb 28, 2018 at 19:26

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