# A Torque and concentric cylinders fluids

1. Apr 7, 2017

### joshmccraney

Hi PF!

If we have two concentric cylinders with newtonian fluid between them, and the small cylinder is at rest and the larger cylinder with radius $R$ rotates at some angular velocity $\Omega$, how would you calculate torque $\vec{T}$ on the outer edge?

My thoughts: $\vec{T} = \vec{r}\times \vec{F}$ where $\vec{r}=R\vec{r}$. To find $\vec{F}$, we'll need the shear stress in the $\theta$ direction. I know in general the shear stress $\bar{\bar\tau}$ is a second order tensor defined for a newtonian incompressible fluid as $\bar{\bar\tau} = \mu \nabla \vec{V}$. So then $\vec{F} = \bar{\bar\tau} \cdot \hat{\theta} = (\mu \nabla \vec{V}) \cdot \hat{\theta}$. Since I don't know $\nabla \vec{V}$ in cylindrical coordinates, I cannot proceed. Please help me out here.

If I knew $(\mu \nabla \vec{V}) \cdot \hat{\theta}$ then $\vec{F} = \iint_S (\mu \nabla \vec{V}) \cdot \hat{\theta} \, dA$ where $S$ is the boundary of the cylinder and $dA$ is an area element.

2. Apr 7, 2017

### Staff: Mentor

This equation is not correct. The shear stress is a vector and del V is a tensor. The rheological equation for a Newtonian fluid is: $$\pmb{\sigma}=-p\mathbf{I}+\mu(\pmb{\nabla}\mathbf{V}+(\pmb{\nabla}\mathbf{V})^T)$$To get the shear stress you are looking for, you would have to dot this with a unit vector in the radial direction and with a unit vector in the tangential direction.
The easiest thing to do is look up the components of the stress tensor in terms of the components of the velocity vector for cylindrical coordinates on line. Or, do you feel compelled to derive the gradient of velocity vector in cylindrical coordinates on your own?

3. Apr 9, 2017

### joshmccraney

I would like to see how the derivation goes, if you know a reference or would walk me through it?

4. Apr 10, 2017

### Staff: Mentor

Section 1.2, BSL

5. Apr 12, 2017

Thanks!!!