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?

  • $\begingroup$ The only way im aware of is by graphing your function $\endgroup$
    – Bardeen
    Commented Feb 28, 2018 at 7:54
  • $\begingroup$ @001 how do you solve the 2nd question? What are the values of $C_1$, $C_2$ and $U^*$? $\endgroup$
    – user115350
    Commented Feb 28, 2018 at 16:20
  • $\begingroup$ @user115350 C1 is the inverse of karmann´s constant C2 is when U+ = y/R very close to the wall $\endgroup$
    – Bardeen
    Commented Feb 28, 2018 at 16:34
  • $\begingroup$ @user115350 C2 should give a value close to 5.5 and C1 2.43 $\endgroup$
    – Bardeen
    Commented Feb 28, 2018 at 16:42

1 Answer 1


try to write $$ y = k(R-r)(1-\frac{R-r}{R})$$ now find when y is maximum then you graph this function :$$u^+ = \frac{1}{\kappa} \ln\, ((R-r)y_m) + C^+$$ you sould get something like this 1) REICHARDT 2) PRESENT 3) UNIVERSALenter image description here

  • $\begingroup$ @JohnRennie hey john are you familiar with this topic? $\endgroup$
    – Bardeen
    Commented Feb 28, 2018 at 8:27
  • $\begingroup$ Thanks for the time you put on this :) I'll look at this and inform you :) $\endgroup$
    – Ghartal
    Commented Feb 28, 2018 at 19:24

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