# Fluid Flow Problem-Viscous Flow

1. Aug 3, 2013

### RGNEM

I am working on a fluid mechanics problem modeling something I've come into contact with at work.

I have an open to atmosphere reservoir of a fluid with properties as follows

SG=1.48
Viscosity= 47000 Cp
Height of fluid column, 3 inches or so. i.e. Neglible

I then have a pressure vessel with air inside of it at

190 Torr = 3.67 psi

If I immerse a plastic tube in the fluid, connect it to the pressure vessel, and open the clamp, I will establish some flow.

Inner Diameter of Tube is .19 in

I have been trying to determine the flow speed and pressure in the flow.

I have been using the equation

$$\frac{P_1}{\gamma} + \frac{V_1^2}{2*g} + Z_1= \frac{P_2}{\gamma} + \frac{V_2^2}{2*g} + Z_2$$

Considering change in height to be negligible (only a few inches), my equation reduces to

$$\frac{P_1}{\gamma} + \frac{V_1^2}{2*g} = \frac{P_2}{\gamma} + \frac{V_2^2}{2*g}$$

Taking initial fluid velocity to be zero, as it is coming from the reservoir of fluid, my equation further simplifies to

$$\frac{P_1}{\gamma}= \frac{P_2}{\gamma} + \frac{V_2^2}{2*g}$$

Plugging in values, I find my V_2 to be about 399 in/s

This is too high, but it must be due to the incorrect assumption of inviscid flow.

$$\frac{P_1}{\gamma} = \frac{P_2}{\gamma} + \frac{V_2^2}{2*g} + H_L$$

To determine this, I first calculate my Reynolds Number using the equation

$$R_e=\frac{ρVD}{\mu}$$

then Friction Factor using the equation

$$f=\frac{64}{R_e}$$

Using this value I find my Head Loss using the equation

$$HL=f\frac{l}{D}\frac{v^2}{2g}$$

My problem is that, after using an assumed initial velocity to begin the iterative process,
my Reynold's Number comes out very low, my Friction Factor very high and my Head Loss astronomical.

For example, for an assumed V1 of 20 in/s
Re= .077
f=828.99
HL=4043058523 in

Plugging into the next iteration of my flow equation, the inside of the eventual square root required for solving for V_2 comes out to be negative, which is just a small problem....

I hope I'm missing something obvious, and after spending many hours just to get to this point, I'd appreciate any ideas/comments/help that any of you could provide.

2. Aug 4, 2013

### SteamKing

Staff Emeritus
Your fluid is so thick, it is probably going to clog such a small tube (ID = 0.19 in). Use a bigger tube or see if the viscosity can be reduced. When highly viscous products like residual fuel oil or asphalt have to be pumped, they must be heated to reduce viscosity sufficiently to allow pumping (residual FO approx. 200 F, asphalt approx. 400 F).
If neither approach is acceptable, you'll probably have to slice your fluid up into chunks and insert it into whatever manually.

3. Aug 4, 2013

### RGNEM

The problem is not making it work.
It flows quite well in this small tube, due to the high pressure gradient.

I am simply trying to describe the process in a quantifiable way.

I've been looking at the Stokes equation and Hagen–Poiseuille equation this morning.
These seem to be the correct equations for flow with a Reynold's number below 1.

4. Aug 4, 2013

### Aero51

If you are running at a low RE bernoullis equation will not be valid as it is only a good approximation for the limit as Re goes to infinity.

5. Aug 4, 2013

### RGNEM

Yes, I've come to realize that.
I was trying to compensate for that problem by incorporating HL from friction, but that was incorrect.

I'm now using the Poiseuille equation:

$$\upsilon \pi R^2 = \frac{\pi R^4}{8\eta} (\frac{|\Delta P|}{\Delta x})$$

6. Aug 4, 2013

### Aero51

I would recommend setting up a full NS equation in the direction of fluid motion and make some simplifying assumptions. I do not have a PC with internet access so I cannot provide an example. The good news is that you know your boundary conditions in the tube and the net pressure gradient, dP/dz.

As another hint, solve the NS 3d polar equation the the z direction. Off the top of my head, you know:

-The flow is constant in time. d()/dt=0
-V_theta = 0 (irrotational)
-Viscosity is only significant in the radial direction
-You have a pressure gradient in 1 direction (dP/dz =/= 0)
-no slip condition at walls

Last edited: Aug 4, 2013