6

Context: I am trying to derive an equation given in a Journal of Fluid Mechanics paper (2.2). It deals with the analysis of an axisymmetric turbulent wake where cylindrical coordinate system has been used (which to me is a little hard to understand as I typically deals in Cartesian system).

We start with what is called as a momentum equation (simplified version after certain assumptions):

$$U_\infty \frac{\partial}{\partial x} (U - U_\infty) = - \frac{1}{r} \frac{\partial}{\partial r} (r \ \overline{uv})$$

Here, $r$ is the radial direction. The axial ($x$) velocity is defined as $U$ and $U_\infty$ is a constant freestream velocity. The term $\overline{uv}$ is called as a turbulent stress which tends to zero if $r$ tends to infinity (i.e. restricted within a finite radial distance).

The authors integrate this equation over a cross-section to yield: $$U_\infty \int_0^\infty (U_\infty - U) r \ dr \approx \theta^2U_\infty^2$$ where $\theta$ is the momentum thickness. I think I can tweak the variables to get the $\theta$ in the equation, but I am not able to understand how exactly would we take a cross-section and then integrate over it? Any leads would be appreciated.

PS: Another equation that hasn't been mentioned is a continuity equation that is written as: $$\frac{\partial U}{\partial x} + \frac{1}{r} \frac{\partial}{\partial x} (r V) = 0$$

Tanmay Agrawal
  • 159
  • I haven't seen the paper as it is pay-walled. The co-ordinates are simplified symmetric cylindrical coordinates, that is without any radial angular dependence. Why does the radial distance $r$ extend to infinity in the integration over $r$? This doesn't seem to relate to a practical problem such as fluid flow in a pipe. – James Arathoon May 16 '21 at 12:35
  • Can you please clarify some issues - what is $V$? And also you say $U_\infty$ is a constant, but then the integral $$\int_0^\infty(U_\infty-U)r\mathrm{d}r$$ Surely diverges. – K.defaoite May 16 '21 at 13:52
  • @JamesArathoon Radial distance is taken upto infinity perhaps to include the fact that the turbulent region extends radially along $x$ i.e. radius of that turbulent fluid is an increasing function of $x$. So that implies integrating over all possible values of $r$ in my opinion. – Tanmay Agrawal May 17 '21 at 14:34
  • @K.defaoite $V$ is the radial component of velocity whereas $U$ is the axial ($x$) component. – Tanmay Agrawal May 17 '21 at 14:35

1 Answers1

1

HINT

We have deal with the boundary layer in the axisymmetrical stream, which behavior is invariant to the Reinnolds number $$\text{Re} = \dfrac{U_\infty l}{\nu},$$ where $\,\nu\,$ is the fluid viscosity and $\,l\,$ is the physical layer thickness.

The stress in the turbulent case is quadratic function of the Reinolds number.

We have the radial coordinate $\,r,\,$ which is scaled to the parameter $\,l\,$ and has the lower limit $\,0\,$ (at the surface) and the upper limit $\,\infty\,$ (freestream, or "local infinity"),

The linear scaling of the axial coordinate $\,x\,$ requires the factor $\,R\,$ (physical radial size), and the integration by this coordiate leads to the average value of the square of the "axial" Reinolds number, wherein the factor $\,R^2\,$ is hidden in the momentum thickness.

Yuri Negometyanov
  • 28,835