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Rod moving inside a cylinder with fluid of certain viscosity 

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April 11, 2016, 20:03 
Rod moving inside a cylinder with fluid of certain viscosity

#1 
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Jack Alderson Taggart Penny
Join Date: Apr 2016
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I am trying to solve a problem and I don't know where to start. I have a rod with radius R1. It is moving inside a long concentric cylinder of radius R2. The rod is moving under the influence of a fluid (with viscosity "nu") inside the cylinder. The rod is moving along its long axis (let's say the xdirection). My given task is to find the corresponding flow field V(r) between cylinder and rod. (Yes, this IS a homework problem). Further, what force is required to pull rod with constant speed U (neglect end effects)?
Now, I need some help with this. But if no one can actually help, can they tell me what keywords I can enter into Google search bar. I know that this is NOT a piston problem, so all the hundreds of online pdf files on piston dynamics is useless. This is also NOT Couette's flow. I mean, what the hell is this? 

April 11, 2016, 22:47 

#2 
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Troy Snyder
Join Date: Jul 2009
Location: Akron, OH
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What you described is 'cylindrical Couette flow'. Consult google and you should be on your way.


April 12, 2016, 17:12 
Cylindrical Couette

#3 
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Jack Alderson Taggart Penny
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I posted a question yesterday and I was helpfully informed that my question was actually a cylindrical Couette problem. I looked it up, but I am still having difficulties. A rod is inside a cylinder. Fluid is present between the rod and the cylinder. Fluid velocity at the wall of the cylinder is zero, due to no slip condition. The rod moves in an axial direction (zaxis, or the long axis) because the fluid molecules move in that direction. How fast the fluid particles move depend ONLY on the radius. Speed of fluid is not dependent on the zaxis or on theta. The velocity is given by "v". So, all the derivatives (in Navier Stokes) which are "dv/d(theta)" or dv/dz can be completely ignored. Only dv/dr matters. And lets say we don't care about the pressure gradient either, so the grad(P) which is the first term on te RHS of Navier Stokes can also be ignored. Also ignore gravity term. And the advection term on the LHS can ALSO be ignored. So, ALL you have left is the Laplacian on the RHS. Assuming that "mu" is the dynamic viscosity, I found out from an online document that the solution to the Couette problem is:
0 = 0 + 0 + (mu/r)*d/dr(r*dv/dr) I just don't understand the above. WHY is there a "1/r" on the outside, and WHY is there an "r" inside the bracket. I know what the gradient operator in cylindrical coordinates is, but I still don't understand. Could someone please kindly help? 

April 12, 2016, 17:48 

#4  
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Filippo Maria Denaro
Join Date: Jul 2010
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Quote:
that is just from the definition of Laplacian operator in cyilindrical system 

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constant velocity, cylinder, rod, viscosity 
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