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Consider the flow over a circular cylinder at a high Reynolds number shown here.

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For the region outside the boundary layer $(y>\delta)$, derive a relationship for the normal pressure gradient $\frac{\partial p}{\partial y}$ at $y=\delta$. Using this, derive another relationship for the normal gradient of the azimuthal velocity component $\frac{\partial u}{\partial y}$ at $y=\delta$ in terms of the edge velocity $u_e(x)\equiv u(x,\delta)$.

I have made the following assumptions to help analyze this problem.

  1. The flow is incompressible $\frac{1}{\rho}\frac{D\rho}{Dt}=\nabla\cdot \overrightarrow{V}=0$.
  2. The flow is steady $\frac{\partial \overrightarrow{V}}{\partial t}=0$.
  3. The cylinder is very long $R\ll L$.
  4. The streamlines close to the cylinder are roughly circular.
  5. Body forces are negligible.

In cylindrical coordinates, the continuity equation is, $$\nabla\cdot \overrightarrow{V}=\frac{1}{r}\frac{\partial}{\partial r}(rv_r)+\frac{1}{r}\frac{\partial}{\partial \theta}(v_\theta)+\frac{\partial}{\partial z}(v_z)=0$$ applying the 3rd, and 4th assumptions gives $v_r=0$ and $\frac{\partial}{\partial z}(\overrightarrow{V})=0$. Therefore the the azimuthal velocity is not a function $\theta$ and can only depend on $r$ since nothing depends on $z$.

The radial component of the Navier-Stokes equations is, $$\frac{\partial v_r}{\partial t}+v_r\frac{\partial v_r}{\partial r}+\frac{v_\theta}{r}\frac{\partial v_r}{\partial\theta}+v_z\frac{\partial v_r}{\partial z}-\frac{v_\theta^2}{r}=-\frac{1}{\rho}\frac{\partial p}{\partial r}+b_r+\nu\left(\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial v_r}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 v_r}{\partial\theta^2}+\frac{\partial^2 v_r}{\partial z^2}-\frac{v_r}{r^2}-\frac{2}{r^2}\frac{\partial v_\theta}{\partial\theta}\right)$$ which, upon applying the result of continuity and the other assumptions simplifies to, $$\frac{v_\theta^2}{r}=\frac{1}{\rho}\frac{\partial p}{\partial r}.$$ The azimuthal component of the Navier-Stokes equations is, $$\frac{\partial v_\theta}{\partial t}+v_r\frac{\partial v_\theta}{\partial r}+\frac{v_\theta}{r}\frac{\partial v_\theta}{\partial\theta}+v_z\frac{\partial v_\theta}{\partial z}-\frac{v_r v_\theta}{r}=-\frac{1}{\rho r}\frac{\partial p}{\partial \theta}+b_\theta+\nu\left(\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial v_\theta}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 v_\theta}{\partial\theta^2}+\frac{\partial^2 v_\theta}{\partial z^2}-\frac{v_\theta}{r^2}-\frac{2}{r^2}\frac{\partial v_r}{\partial\theta}\right)$$ and becomes $$\frac{1}{r}\frac{\partial p}{\partial \theta}=\mu\left(\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial v_\theta}{\partial r}\right)-\frac{v_\theta}{r^2}\right)=\mu\frac{\partial}{\partial r}\left(\frac{1}{r}\frac{\partial}{\partial r}(rv_\theta)\right).$$ The first portion seems pretty straight forward, $$\frac{\partial p_e}{\partial r}=\frac{\rho v_{\theta,e}^2}{\delta+R},$$ but for the next part I am confused, how can I relate my normal pressure gradient to my result from the azimuthul component? I know that in general $\frac{\partial p}{\partial\theta}\ne0$ due to the curvature of the cylinder, but does that mean I can simply say that the entire term on the right with $r v_\theta$ is a constant since this is a PDE? Specifically, $$\frac{d}{d r}\left(\frac{1}{r}\frac{d}{d r}(rv_\theta)\right)=k?$$ I'm also thinking I might be able to do something with potential flow theory far from $y=\delta$, but closer to the edge I'm not sure if this will work. Thanks for any suggestions on how to proceed further.

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  • $\begingroup$ Are you allowed to solve this using the stream function? $\endgroup$ – Chet Miller Jan 17 '18 at 23:56
  • $\begingroup$ @ChesterMiller, the problem doesn't specify how to arrive at a final equation so I don't see why not. $\endgroup$ – WnGatRC456 Jan 19 '18 at 14:22
  • $\begingroup$ This is basically asking you to determine the characteristics of the potential flow (modeled as an inviscid flow) outside the boundary layer. So what do the 2D equations look like in terms of the pressure and stream function, with the viscosity set to zero? $\endgroup$ – Chet Miller Jan 19 '18 at 14:51
  • $\begingroup$ That seems to imply that OUTSIDE the boundary layer that $\frac{dp}{d\theta}=0$, interesting. $\endgroup$ – WnGatRC456 Jan 19 '18 at 16:05
  • $\begingroup$ I don’tthink so. Certainly not at the leading edge. $\endgroup$ – Chet Miller Jan 19 '18 at 16:18
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The solution for potential flow past a cylinder of radius a (in cylindrical coordinates) is as follows: $$v_r=U\cos{\theta}\left[1-\frac{a^2}{r^2}\right]$$ and $$v_{\theta}=U\sin{\theta}\left[1+\frac{a^2}{r^2}\right]$$where $\theta$ is the angle measured relative to the negative x axis and $v_{\theta}$ is the $\theta$ velocity in the clockwise direction. The rest of the details are for you to work out.

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