Torque in circular fluid motion

[da_willem]

I'm reading a book (intro to, by Davidson) about MHD now, but found I'm a bit rusty on tensors and curvilinear coordinates. It is written that for a circular flow the azimuthal component of the NS equations in the steady state gives (with F some body force)

$$\tau _{r \theta} r^2 =-\int _0 ^r r^2 F_{\theta} dr$$

Shouldn't this read, for axisymmetric flow, without the square on both r's? I would argue that the remaining terms in the NS equations

$$\sigma _{ij,i}+F_j=0$$

would yield for the azimuthal ($\theta$) component (any suggestions welcome if the notation is obscure):

$$F_{\theta} = -\nabla \cdot \overline{\overline{\sigma}}_{\theta}=-\frac{1}{r} \frac{d}{dr}(r\sigma_{r \theta})$$



Now the author continues,

$$\tau_{r \theta}=\mu r \frac{d}{dr}(\frac{u_{\theta}}{r})$$

In which he says he used Newtons law of viscosity, which I think one can write

$$\tau_{ij}=\mu u_{i,j}$$

(Is it by the way ok to write this as $$\overline{\overline{\tau}}=\mu \nabla \vec{u}$$?)

But how does one come from that to the (r, theta) component?

 

[da_willem]

I managed to get the r,theta component of the stress tensor as it is written in the book now. But I still can't see why there would be squares on the r's in the first formula I posted; anyone?

 

[charng]

Thank you for sharing your thoughts on torque in circular fluid motion. I would like to provide a response to your questions and comments.

Firstly, regarding the equation \tau _{r \theta} r^2 =-\int _0 ^r r^2 F_{\theta} dr, I agree with your observation that the square on both r's may not be necessary for axisymmetric flow. However, it is possible that the author is considering a more general case of non-axisymmetric flow, which would require the square on both r's. It would be helpful to refer to the context and any assumptions made by the author in this case.

Moving on to your suggested equation for the azimuthal component of the Navier-Stokes equations, F_{\theta} = -\nabla \cdot \overline{\overline{\sigma}}_{\theta}=-\frac{1}{r} \frac{d}{dr}(r\sigma_{r \theta}), I believe it is a valid expression for the force in the azimuthal direction. However, it would be helpful to know the specific definitions of \overline{\overline{\sigma}}_{\theta} and \sigma_{r \theta} used by the author.

Regarding the use of Newton's law of viscosity, \tau_{ij}=\mu u_{i,j}, I agree with your suggestion to write it as \overline{\overline{\tau}}=\mu \nabla \vec{u}. This notation is commonly used in fluid mechanics and is a shorthand for the tensor equation.

Finally, to arrive at the (r, theta) component of the torque, \tau_{r \theta}=\mu r \frac{d}{dr}(\frac{u_{\theta}}{r}), the author may have used the fact that for axisymmetric flow, the velocity component in the radial direction, u_r, is zero. This simplifies the equation to \tau_{r \theta}=\mu r \frac{d}{dr}u_{\theta}, which can then be integrated to get the final expression.

I hope this helps clarify your doubts. If you have any further questions or suggestions, please feel free to share them. As scientists, it is important to constantly question and improve our understanding of concepts. Thank you for your contribution to the discussion.