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I have a container (assume open on top) with $1m^3$ of water. That water exits this container from an opening that has a cross-section area A e.g. $πR^2$ = $1cm^2$.

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What is the volumetric flow rate I get out of this opening? I have a feeling that it's related to $R^4$ due to Hagen–Poiseuille equation but on the other hand it has the $|DP|$ that might also be related to R in some way. Is there any way to force the volumetric flow rate to be ~$R^2$ (e.g. adjusting the pressure with a mechanical way?)

neverlastn
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  • The container has a cubic meter of water. Will this be a constant? Also, we need to know the shape of the container so that we can calculate the depth of the water. Knowing the depth (or height) of the water we can determine the elevation head. – Armadillo May 20 '15 at 20:47
  • Yes, let's assume it is constant. Let's assume that h = 1m => gauge pressure = 9.81 kPa (I think) – neverlastn May 20 '15 at 23:50
  • Are you saying you want to change the size of the opening, but you don't want the liquid to come out 16x faster when you make the opening 2x bigger? Indeed you would have to make a system that changes the height of the liquid (pressure) as a function of the size of the hole. This is possible in principle - but what application do you have in mind that you want to, in essence, change the normal laws of physics? – Floris May 21 '15 at 12:38
  • @Floris, you are right. Actually it would help me model server traffic accurately with pipes :) By the way - I'm thinking that $r^4$ is (and makes fully sense) in the case of mostly laminar flow. I think in the case of mostly turbulent flow (i.e. if I have something that rotates the liquid really fast) it's approximating $r^2$ (looking at "Orifice plates") – neverlastn May 21 '15 at 12:51
  • Keep in mind that flow through an orifice and flow through a pipe are different. Flow through a pipe is viscosity-limited, so flow rate is proportional to pressure drop. Flow rate through an orifice is proportional to square root of pressure drop. It also depends strongly on the shape of the entrance to the orifice. Google "orifice flow". – Mike Dunlavey May 21 '15 at 12:52
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    @MikeDunlavey - you make a good point. There is some confusion here between "orifice" and "pipe" and the equations are quite different. In a pipe, it is the losses that dominate the flow rate; in an orifice, the question is "how fast can we accelerate the liquid to get through the hole" - like people trying to get into a subway car at rush hour. Then the flow scales with area (so $r^2$ as needed) - and a bunch of other things, but if you keep those constant you do indeed get an $r^2$ relationship. – Floris May 21 '15 at 12:56
  • @Floris: you put it better than I did. – Mike Dunlavey May 21 '15 at 12:58
  • Thanks a lot @Floris. So essentially if the setup is like this we get $r^2$ and also if the flow is sufficiently low, we get this really nice behavior (the utilization of the process/tube e.g. 20% becomes clearly visible?) – neverlastn May 21 '15 at 15:13
  • Yes - as you show it the level of the liquid gives an indication of the flow rate (goes as square root of height). But combined with the orifice plate it would allow a certain amount of visualization. Hard to calibrate but that might not be needed... – Floris May 21 '15 at 16:03

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