Hi everybody, I've been going around in circles with this problem and not sure if I'm approaching it correctly. I have a vessel filled with a fluid and connected to a pipe (circular cross section) with a tap on the end. The pressure on the vessel is not atmospheric, it is about 1 bar absolute (adjustable). The pressure on the pipe is atmospheric and the tap opens to atmospheric. I need a flow of about 2l/s from the fully open tap. How do I go about finding the exact pressure required on the vessel in order to get this flow rate? I was thinking of bernoulli's, but that is only good for inviscid, laminar flow. I think that my flow will have changing viscosities with temperature changes and possibly be in the turbulent region. Am I right in assuming, that if I apply a pressure of 1 bar absolute (so the pressure on the outside of the vessel is 1 bar greater than the pressure in the vessel) to the vessel, that the pressure drop experienced from the vessel through the pipe, to atmospheric, will be 1 bar? Say for example the pressure in the vessel was 10 bar absolute and I applied a pressure of 11 bar absolute to the outside of the vessel. The pressure difference over the vessel wall would be 1 bar, right? then if I attach a pipe to that vessel and the pipe is open to atmospheric, then the pressure drop to atmosphere would be 1 bar, right? Man, for some reason I can't get this through my head. I'm sure the pressure drop through the pipe to atmospheric would actually be 10 bar. any help to ease my fundamental ignorance would be much appreciated. One step at a time... Thanks.
First things first: Bernoulli does not stipulate laminar vs. turbulent flow. That is not one of the restrictions of its derivation. The three main assumptions of the Bernoulli equation are steady state, incompressible, inviscid (no friction) flow along a streamline. I can think of many problems in which the turbulent regime has to be considered. For instance, have you never done a pipe friction factor calculation? If Bernoulli were limited to the laminar regime only, we could always use [tex]f=\frac{Re}{64}[/tex] and go about our business. We would never need a Moody diagram. If your fluid is a liquid, then at lower pressures, the incompressible assumption is quite valid. The frictionless assumption can be acceptable as well, depending on the layout of your system. If you're not going through a lot of valves, orifices and long line lengths, the frictionless assumption may not be too bad also. The steady state assumption is sometimes a tough one to validate because you will, most likely, have an ever changing system. However, it can be a very good starting point to calculate this scenario under that assumption. That is an issue if you have changing temperatures and viscosities. If that is the case then I would first break the system up into multiple small segments and solve for the end conditions at a segment and then fold those into the input for the next segment in line. It's analogous to calculating the area under a curve by calculating the areas of a bunch of rectangles. Close. If you apply a pressure of 1 bar GAUGE to the vessel it will be 1 bar above atmospheric.If you apply 1 bar absolute to the tank, you are actually a bit lower that 1 standard atmosphere in pressure. A vessel pressure of 1 bar gauge would indeed result in an overall delta P of 1 bar.
You didn't say how big the pipe is. 1 L/s would be a lot of air for a small pipe and though you can use Bernoulli's as a good approximation, you may get a velocity big enough to choke the flow. So first calculate the velocity you need, based on the pipe size. From that we can decide what assumptions are valid (ie, things like temperature drop due to pressure drop).
What is the vessel made of (i.e. is it rigid)? Is the vessel sealed (other than the pipe)? Is the fluid liquid or gas? Is fluid entering the system or is it a fixed amount? Do you have a schematic for clarity? CS
First off, thanks Fred, I was going insane trying to get guage and absolute right in my head, I haven't done this stuff since varsity, so I'm a little rusty. I think I had been staring at the problem so long that I just wasn't understanding it any more. This morning I'm feeling a little fresher and ready to keep on tackling it.
Thanks for your input Russ, the fluid is actually a liquid and I have a feeling the pipe inner diameter will be in the range of 2mm to 5mm. I did some velocity calcs yesterday and looking to try a fresh approach today after clarity of a few fundamentals from Fred (above)
Stewart, most of the stuff is confidential information as it pertains to a competitive design, that's why I have to be a little vague. Also, not everything has been specified because it is still in a concept phase. I can't attach a schematic (as an engineer I am a believer of free information and improvement of systems/designs by any means necessary for the benefit of all, but my company sees the world in a different light, so I will have to conform on this one), but I can try to offer some clarity. I'm not asking for a complete solution, just checking my fundamentals to insure I don't go down the wrong path for a week and waste everybody's time and money (oh, and look like a buffoon along the way). I think I can get enough information with general questions without pissing people on the forum off (I personally loathe problems with vagueness) and without breaking any contractual requirements. Like I said, I don't need a total solution from the forum (the solution part is what I get paid to do ) just a few pointers to approve my throries. The vessel is sealed except for the pipe (which has a tap/valve at the end). The vessel is not rigid and is expected to reduce it's volume with the fluid inside. The fluid is a liquid. The fulid is a fixed amount (obviously decreasing in volume with the vessel as the valve/tap at the end of the pipe is opened) I wanted to keep the flow laminar to keep things simple (as an aero engineer, I know turbulence is complicated and never behaves nicely ) but management decided that turbulence (no, they have never seen a moody diagram or the implicit colebrook equation and they think D'Arcy is that chick that played guitar for the smashing pumpkins) would be a great way to slow the flow down. So a thin, rough pipe is what I get to work with and I'm expecting higher than laminar Re numbers (probably Re>8000). An aero has to do what everyone expects an aero can do. Thanks for all the help so far, I found this forum the other day and totally dig it ! My family and friends are going to see a lot less of me as i explore all the interesting problems and solutions.
That sounds basically like a bladder type accumulator except that the fluid and gas are swapped (i.e. the gas is not inside the bladder, the fluid is). Is that correct? CS
Yip, that's it. We apply a pressure to the bladder/vessel. This causes the fluid in the vessel to flow through the connected pipe. Then we model this flow in order to calculate what sort of pressure we need to obtain a certain flowrate. How would you go about modelling this? I am planning to use D'Arcy-Weisbach in the turbulent region (so using a friction factor calculated using the colebrook equation) to calculate the pressure drop over the pipe. Correlate this pressure drop to the flowrate. Could I use Hagen Poiseuille or are there restrictions?
I'd model it like an accumulator, since that is what you have (just inverted which doesn't make any difference), up to the outlet in order to get the flow rate at that point. Then I'd use Bernoulli with Darcy (and the Swamee-Jain equation for the friction factor) from that point on. Caveat: For a rapid discharge make sure to use real gas equations of state (or NIST data) for gas properties. API Specification 16D has the equations for all of this if you need them. CS
the next step Ok, so did some work on this problem and modelled it (basically, to get the fundamentals correct) according to what is in the attached pdf file. I'm not sure if this is correct or not, it just doesn't look right and I can't figure out why. Any help or ideas are much appreciated. Thanks.
nobody have any ideas? after thinking a little more about it, the pressure that the inner chamber "sees" is it's own internal pressure plus the pressure applied to it by the chamber around it, right? So if I had 4 bar in the outer chamber and 2 bar in the inner chamber and atmosphere was 1 bar, then when I opened the valve the pressure would drop from 6 bar to 1 bar atmospheric. right? I really feel like i'm not getting something here.
Re: the next step Just took a quick look at your calcs and didn't check the math but.... Based on what you have drawn, you seem to be on the right track with a few exceptions. In your drawing if P1 is the gas pressure then (until the valve is opened) P2 (as you have shown it) will equal P1 since it is in equilibrium. So you don't need to have P2 represented there. Represent P2 at the outlet of the accumulator instead. Since you've stated that the outlet is atmospheric then P2 = Patm. Your equation therefore accounts for it already. Typically accumulators do not have the gas on the bottom and the fluid on top. Is there a particular reason for that? I would make gravity work on my side personally. Also, the calculation you have (once corrected) will give you the initial velocity at the exit unless the charge pressure (P1) is constant. If it is constant you'll have a constant flow rate at the outlet. If the charge pressure is not constant, but rather "precharged" to a certain value, then you'll have to account for the polytropic expansion of the gas as the fluid exits. Also, I wouldn't necessarily exclude v1. Just relate it in terms of areas and v2 and then substitute it into your equation. Finally, I would break the problem down into two parts. Model the flow rate out of the accumulator as discussed above (just lose the Darcy term and then include a discharge coefficient to correct the outlet flow rate). Then you can find the pressure drop in the pipe using Darcy and the normal flow equations. Hope that helps. CS
That's great and it helps, thanks CS, I'm glad I'm on the right track. So you say P2 will equal P1 if it is in equilibrium before the valve is opened. What if the liquid in the inner vessel is stored with an internal pressure, like if it was soda water or any pressurised liquid? I agree that gravity would be a big help if the whole system was upside down. I suppose one could gain the same benefits of gravity by lowering the pipe exit below the liquid level or lower. Not including v1 would make things slightly easier, but like you say it is just a ratio of the areas of v2, so easy to apply later if it is needed, or even right at the start. I'm going to try the two part approach, will try to find a suitable discharge coefficient. Seeing that dimensions haven't been specified yet (still in the concept phase) it's difficult to get coefficients that apply, but I'll just put a Cd variable in there for now. Thanks for your time.
Referring to the drawing you provided, the vessel is sealed and rigid with a moveable partition (piston or whatever) separating the gas at pressure P1 from the liquid at pressure P2 with the valve at the outlet of the liquid section closed. If that is a true representation of your design, then the pressures P1 and P2 must be in equilibrium since nothing is moving (the valve is closed initially). Just like in a can of soda, until you open the top, the carbonated gas and liquid pressures are in equilibrium. Hope that helps. CS