I Assumptions for Navier Stokes equations in this system

[dor040101]

TL;DR Summary

Why is it true to use cartesian coordinates for NS equations in the small gaps? How to find $P_2$?

There is a cylinder which is held by a rope, inside a pipe. Fluid is flowing (laminar) in the direction of Q. I'm trying to calculate the velocity profile in the gaps between the pipe and cylinder, using Navier Stokes.

First question is, if $\frac{R_1-R}{R_1}<<1$, which means the gap width is very small compared to the cylinder, is it okay to use cartesian coordinates in NS equations? why?

Also need to find $\frac{dP}{dz}$, assuming the pressure is constant along every horizontal plane $P=P(z)$ ($z$ is up).

I know that $\frac{dP}{dz}=const$ if the flow is steady state, fully developed and constant $\mu$. So $\frac{dP}{dz}=\frac{P_{atm}-P_{L_2}}{L_1}$

Given everything in the diagram, how to find $P_{L_2}$? (pressure at entrance to the gap).

 

[Chestermiller]

TL;DR Summary: Why is it true to use cartesian coordinates for NS equations in the small gaps? How to find $P_2$?

There is a cylinder which is held by a rope, inside a pipe. Fluid is flowing (laminar) in the direction of Q. I'm trying to calculate the velocity profile in the gaps between the pipe and cylinder, using Navier Stokes.

First question is, if $\frac{R_1-R}{R_1}<<1$, which means the gap width is very small compared to the cylinder, is it okay to use cartesian coordinates in NS equations? why?

To an observer at the scale of the gap, it appears like the flow is between two infinite parallel plates. The curvature of the cylinders has little effect on the flow and pressure variation.

Also need to find $\frac{dP}{dz}$, assuming the pressure is constant along every horizontal plane $P=P(z)$ ($z$ is up).

I know that $\frac{dP}{dz}=const$ if the flow is steady state, fully developed and constant $\mu$. So $\frac{dP}{dz}=\frac{P_{atm}-P_{L_2}}{L_1}$

Given everything in the diagram, how to find $P_{L_2}$? (pressure at entrance to the gap).

View attachment 358256

Inertial effects can be neglected (and probably gravitational effects also, if the gap is very small and the pressure drop significantly exceeds $\rho g h$).

So what you are dealing with is a linear combination of drag flow between parallel plates and a superimposed pressure flow parallel to the plates. Do you know how to solve this kind of steady flow problem?

 

[dor040101]

To an observer at the scale of the gap, it appears like the flow is between two infinite parallel plates. The curvature of the cylinders has little effect on the flow and pressure variation.

Inertial effects can be neglected (and probably gravitational effects also, if the gap is very small and the pressure drop significantly exceeds $\rho g h$).

So what you are dealing with is a linear combination of drag flow between parallel plates and a superimposed pressure flow parallel to the plates. Do you know how to solve this kind of steady flow problem?

I'm not sure because everyone makes different assumptions in fluid mechanics in this level... this is a question from an exam, the professor didn't neglect gravity. The NS equation is:
$0=- \frac{dP}{dz} + \mu \frac{d^2u}{d^2y} -\rho g$
where $u$ is the velocity at $z$ direction.
For the pressure difference, he actually assumed that the pressure drop in the entrance to the gap, is a combination of hydrostatic and pipe flow pressure drop:

$P_2=P_{in}-\rho g L_2 - \frac{64}{Re} \frac{L_2}{2R} \frac{\rho U_{avg}^2}{2}$
Where we use Darcy–Weisbach equation for the pipe pressure drop.

Maybe your solution is above this level?

 

[Chestermiller]

I'm not sure because everyone makes different assumptions in fluid mechanics in this level... this is a question from an exam, the professor didn't neglect gravity. The NS equation is:
$0=- \frac{dP}{dz} + \mu \frac{d^2u}{d^2y} -\rho g$
where $u$ is the velocity at $z$ direction.as
For the pressure difference, he actually assumed that the pressure drop in the entrance to the gap, is a combination of hydrostatic and pipe flow pressure drop:

$P_2=P_{in}-\rho g L_2 - \frac{64}{Re} \frac{L_2}{2R} \frac{\rho U_{avg}^2}{2}$
Where we use Darcy–Weisbach equation for the pipe pressure drop.

Maybe your solution is above this level?

I assumed that the lower section was not supported mechanically from below, and, instead was sliding downward relative to the top plug. This would give it a downward velocity, and would result in a drag flow component downward. Otherwise, I would have gotten the same result as your professor. Basically, the flow contribution to the pressure change can be gotten as follows:

If Q is the volumetric flow rate in the gap, then the mean upward velocity is $$U_{ave}=\frac{Q}{2\pi R \delta}$$where $\delta$ is the gap opening. For flow between parallel plates, the shear rate at the wall is $$\gamma=\frac{6U_{ave}}{\delta}$$The shear stress at the wall is then equal to the viscosity times the shear rate: $$\tau=\mu \gamma$$ And the pressure difference solely due to the flow plus the pressure difference due to gravity is $$\Delta P=\rho g (L_1+L_2)+2\tau \frac{L_1}{\delta}$$