# Flow through a glaucoma tube

1. Oct 13, 2009

### spal123

Hi,

I'm trying to improve my understanding of flow rates through fluid filled tubes.

The backround is that I am an eye doctor specialising in glaucoma. We implant small plastic tubes in to patients' eyes to relieve raised pressure.

The rate of fluid production in an eyeball is approx 2.5 microliters per minute which hopefully will provide info as to the head of pressure.
The tubes are currently approx 10 millimeters long and empty into a fluid filled surgically constructed second chamber outside the eyeball. From this second chamber the fluid is absorbed into veins and removed from the system. There should be no air/gas in this system.

A big problem is overdrainage, causing low post-operative pressure and with it a significant risk to sight. At the moment we try to guard against this overdrainiage by blocking the lumen of the tube with a piece of stitch material and sometimes tying a stich aroung the outside of the tube to block flow through it. This then makes the post-operative pressure high but these stitches can be removed later once the second chamber has had a chance to become established (i.e. scarred up a bit) and itself can offer some resistance at the tail end of the system.

I'm trying to work out if the tube parameters could be targeted to improve predictability of fluid drainage from the eye. Options include choosing a smaller internal diameter tube of using a longer length of tube. My google reading suggests that length is not that important a factor which is a shame as it would be relatively easy just to loop the tube around and around in the eye if more length meant more resistance.

Thanks in advance for any thoughts.

2. Oct 13, 2009

### spal123

And just to show I am trying I have found this on google which might help me work out the optimum diameter of a 10mm tube.
Problem is I can't do the mathematics involved for cubic feet per second, etc.
Thanks again.

Formula For Diameter:
When the length of a pipe, the hydrostatic head, and the quantity of water required to be delivered per second are. known, the diameter of pipe that will safely take care of that quantity can be found by the formula:

d=0.234 5√(q2L)/h

In which d=diameter of pipe in feet q=cubic feet per second to be delivered L=length of pipe in feet h=head in feet
Example - What diameter of pipe will be required to deliver .5 cubic foot of water per second through a pipe 2,000 feet long with a head of 400 feet?

3. Oct 13, 2009

### Andy Resnick

You haven't mentioned what the tube diameter is- that's the critical dimension. For your application, the flow is laminar and the length of the tube does not affect the result, with the assumption that the pressure difference between inner and outer chamber does not change as you increase the length of the tube.

Poiseuille flow (or Hagen-Poiseuille flow) can be recast into a volumetric flux calculation (microliters/minute) and then you do not need to calculate the pressure change; I'll PM you the details. One question to consider- you mention the volumetric flow rate is 2.5 microliters/minute; how much can that vary without causing over (or under-) drainage?

4. Oct 14, 2009

### Andy Resnick

I'll just post it here:

The flow profile, in terms of the volumetric flow, is

U(r) = $$\frac{2Q}{\pi R^{4}} (R^{2}-r^{2})$$

Where Q is the volumetric flow (volume/time), and R the radius of the tube. Again, the relevant point is whether you are trying to control the pressure difference between chambers, or control the flow rate.

If you maintain a constant pressure jump, as the diameter of the tube changes the fluid velocity will change, altering the volumetric flow. Alternatively, maintaining a constant flow rate means the pressure jump and tube radius are connected.