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I wanted to know how would I go about calculating the flow rate of a syringe with a metal tip that is dispensing water using a pressurized air.

I have had a look on internet about this and on this form and I though that laminar flow equation might be the solution for this. $$\text{Flowrate}=\frac{\pi r^4(P-P_0)}{8\eta l}$$

From this equation the variable I can obtain are as follows:

  • $r = \text{radius of the metal tip}$
  • $\eta = \text{viscosity of water}$
  • $L = \text{length of the metal tip}$

However, I don't quite understand how can I get $P$ and $P_0$

Since I'm supplying a certain pressure from the top of the syringe, this could be $P$ and the pressure that comes out of the syringe metal tip could be $P_0$. If what I said is right, how would I calculate the pressure at the tip?

Thank You

I have a follow up question for this:

If I have $P$(compressed air) - $P_0$(atmospheric pressure) and the $P_0$ is higher than $P$, then the value will be negative. So in this case would I do $P_0$ - $P$? I'm asking this as my pressure will range from 0Bar to 5bar and when it's 0.04Bar, atmospheric pressure is higher than the compressed air. what do I do in this case?

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  • $\begingroup$ Assuming your volumetric flow rate for laminar flow is correct (you didn't provide a reference) then $P_0$ is simply atmospheric pressure and $P$ would be the pressure of the compressed air. Make sure you check the Reynolds Number though... $\endgroup$ – Gert Nov 26 '15 at 15:44
  • $\begingroup$ the reference is physics.stackexchange.com/questions/22978/… - would you say it's fine $\endgroup$ – Satvir Singh Nov 26 '15 at 15:47
  • $\begingroup$ Yes, the formula is indeed correct: it's a reworked Darcy-Weisbach equation. $\endgroup$ – Gert Nov 26 '15 at 15:49
  • $\begingroup$ Just another thing, would the amount of liquid in the syringe left affect results. I mean for example if I have xL in syringe and I dispense at 5Bar and amount goes lower and then I dispense again at 5Bar. Would the be affected. $\endgroup$ – Satvir Singh Nov 26 '15 at 15:53
  • $\begingroup$ No. Water is basically incompressible. $\endgroup$ – Gert Nov 26 '15 at 15:54
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I that equation of flow rate the term (P-P0) is the change in the pressure of the column of liquid in the tube. In this case P0 is the atmospheric pressure that is experienced by the fluid on the other end, that is, at the other end of the metal tube.

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  • $\begingroup$ I understand that P is the pressure I apply and PO is the atmosphere pressure. However, what I don't understand is how do I handle the P-PO when PO is bigger than P. If the change in pressure is negative what do I do then? I was going to just find the difference between them, but then I would be changing the formula, since I will have to do PO-P because I cannot have a negative flowrate. $\endgroup$ – Satvir Singh Dec 7 '15 at 9:59
  • $\begingroup$ If the pressure at the end is greater than the pressure at the beginning of the tube, then the flow will reverse as shown by the equation. The value (P-P0) is negative indicating the flow to be in the opposite direction. Flow rate can be negative. Negative flow rate implies flow in opposite direction. You can take an analogy from current in electrical circuits. $\endgroup$ – lattitude Dec 7 '15 at 10:02
  • $\begingroup$ so from you comment I understand that the flow will reverse meaning the direction of water will reverse, meaning water is going into the syringe instead of out of the syringe. If this is the case then I believe I'm missing something, because if I apply air at 0.08Bar to a syringe hanged vertical, I see water is dripping out of the syringe, meaning flow of water is in the direction I want it to be. But if I put 1.01325bar (atmosphere) and supplied will be 0.08bar into the equation , it will give a negative value. Meaning direction is reversed. This is part I'm confused about. $\endgroup$ – Satvir Singh Dec 7 '15 at 10:07
  • $\begingroup$ The value P in the equation takes into account the total pressure applied at that end. It means the pressure you applied and the atmospheric pressure together. $\endgroup$ – lattitude Dec 7 '15 at 10:12
  • $\begingroup$ well if this is the case, then for me P-PO will just be P, since atmosphere pressure remains constant. Is this correct? $\endgroup$ – Satvir Singh Dec 7 '15 at 10:13

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