- #1
member 428835
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.
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.