# Silicone tube with three holes, flow rate, pressure

I have a silicone tube -- a saline solution flows in from one end, and then flows out of three holes of equal diameter and equal distance from each other that are along the side of the tube. What can I do to make the pressure at the three holes equal?

• Do you want equal pressure as such, or is your final goal equal flow rates of solution from the 3 holes? Jul 7, 2015 at 12:41
• In principle, it's not possible to get equal pressures at all 3 holes. Flow is caused by pressure drop. The pressure at the last hole must be lower than the pressure at the first hole if there is flow in the line. If you want the same FLOW RATE through each hole, that is certainly achievable, as per the recommendations of @docscience. Aug 20, 2016 at 23:58

If injected from a single end, and assuming a constant diameter pipe, the pressure decreases linearly along the length of the pipe due to friction. Therefore equal diameter holes will not yield equal flow rates or pressures.

## For equal flow rates:

To control the flow rates, you must control the injection pressure (ex. reservoir with regulator valve). Specify a desired flow rate (per hole or total) where $$Q_\text{total}=3Q_\text{hole}$$. For each hole location along the length of the pipe, $$\ell$$, calculate the pressure.
$$P = P_\text{injection}-\frac{8\mu\ell Q_\text{total}}{\pi r_\text{pipe}^4}$$ Next, apply a loss coefficient depending on the cross sectional geometry of the holes (sharp-edged, rounded, etc.) to obtain the pressure at the holes. $$P_{hole} = P*K_{L}$$

Next, for the desired flowrate, $$Q_{hole}$$, and known exit pressure, solve for the radius required at each hole: $$P_\text{hole}-P_\text{exit} = \frac{8\mu\ell Q_\text{hole}}{\pi r_\text{hole}^4} \rightarrow r_\text{hole} = \left(\frac{8\mu\ell Q_\text{hole}}{\pi(P_\text{hole}-P_\text{exit})} \right)^{1/4}$$

## For equal pressures:

The only practical way I can think to accomplish this is adding regulator valves to each hole, with an injection pressure high enough to overcome frictional losses to the furthest hole.

• This looks like a good answer, but I'm having trouble understanding it because at least two of the variables are not explicitly defined: what are $\mu$ and $Q$? Aug 20, 2016 at 17:34
• I'm also not sure what $P_\text{exit}$ is. Could you go through and try to clear up the discussion? It looks like you know the right way to solve this so I'm interested in getting it all fixed up. Aug 20, 2016 at 17:42

I'm guessing the flow rates in your system are slow so the flow is laminar. In that case the flow is described by the Hagen-Poiseuille equation:

$$\Delta P = \frac{8\mu\ell V}{\pi r^4}$$

where $\Delta P$ is the pressure drop, $\mu$ is the viscosity of your saline solution, $\ell$ is distance along the pipe, $r$ is the pipe radius and $V$ is the volume flow rate. To get the pressure difference between the first and last hole set $\ell$ to the distance between them.

You can't do much about the viscosity of the saline, and I'm guessing the distance between the holes is fixed so you can't change $\ell$. Likewise the volume flow rate $V$ is probably dictated by the experiment. So the only variable under your control is the pipe radius $r$. The good news is that the pressure difference is very sensitive to $r$ because it varies as $r^{-4}$. So if you double $r$ you reduce the pressure difference by a factor of 16.

If the equal flow rate is really critical you could change the pipe geometry to reduce the difference in length between the holes. Something like this:

• Dear John, thank you very much for your prompt response. I am actually working on an artificial tear duct, so the pipe geometry needs to remain the same, however I can change the radii and the length between the holes. Feb 4, 2015 at 8:55

If your objective is to obtain precise and equal flow rates through the three holes however constrained by serial arrangement you sketched then you need to progressively increase the radii of each duct that are downstream of one another.