Saved Bookmarks
| 1. |
A cylindrical vessel with water is rotated about its vertical axis with a constant angular velocity omega. Find: (a) the shape of the free surface of the water, (b) the water pressure distribution over the bottom of the vessel along its radius provided the pressure at the central point is equal to p_0. |
|
Answer» Solution :The Euler's equation is `rho(dvecv)/(dt)=vecf-vecnablap` in the space fixed frame where `vecf=-rhogveck` downward. We assume incompressible fluid so `rho` is constant. Then `vecf=-vecnabla(rhogz)` where Z is the height vertically upwards from some fixed origin. We go to rotating frame where the equation becomes `rho(dvecv)/(dt)=-vecnabla(p+rhogz)+rhoomega^2+2rho(overset(rarr')vxxvecomega)` the additional terms on the right are the well known coriolis and centrifugal FORCES. In the frame rotating with the liquid `overset(rarr')v=0` so `vecnabla(p+rhogz-1/2rohomega^2r^2)=0` or `p+rhogz-1/2rhoomega^2r^2=consta nt` On the free surface `p=const ant`, thus `z=(omega^2)/(2g)r^2+const ant` If we choose the origin at point `r=0` (i.e. the axis) of the free surface then "constant" =0 and `z=(omega^2)/(2g)r^2` (The PARABOLOID of REVOLUTION) At the bottom `z=const ant` So `p=1/2rhoomega^2r^2+const ant` If `p=p_0` on the axis at the bottom, then `p=p_0+1/2rhoomega^2r^2`. |
|