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Simple fluids question

  1. Jan 8, 2006 #1
    Well, simple may be relative. I've got a question that i'm looking for a simple equation for. Maybe someone can help me out... here it is;

    You have a pressure vessel with known volume that you want to keep at a known positive pressure (if it helps, assume that the atmosphere is at vacuum). If you have a known inlet flow of oxygen (in liters/min and is a set value, or non-variable), how can you find the maximum hole (leak) in the pressure vessel to stay at the desired pressure. Lets say that the pressure starts at what you want, and then you develop a hole or leak at the exact time the inlet flow begins, and friction/thermal effects are negligible.

    Thanks for your smarts!
     
  2. jcsd
  3. Jan 8, 2006 #2

    tbc

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    oh man....this is all ahead of me...(studying aero eng.):frown:
     
  4. Jan 8, 2006 #3

    Clausius2

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    Use Conservation of Energy eliminating the unsteady term (as pressure and volume remains constant, thermal energy of the system is also constant):

    [tex] \dot m (h+u^2/2)_{in}=\dot m (h+u^2/2)_{out}[/tex]

    (h=static enthalpy).
     
  5. Jan 8, 2006 #4

    Q_Goest

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    Hi crvz . . . (you need a vowel in there) Do you live anywhere near LaPorte?

    Clausius quoted you a derivation of the first law of thermodynamics, but unfortunately although it is (kinda) applicable, it's not needed here. (sorry Clausius. besides which for a steady state system enthalpy in = enthalpy out. you can ditch the internal energy terms).

    crve, if your point is to assume there's a mass flow rate in and if we assume we know the state of the fluid inside this pressure vessel, then we need to calculate the area of the hole that will result in an equal mass flow rate out. Is that what you're looking for? I'm assuming it is.

    I lifted this straight from a program I have:

    if Po/Pi>Pcr

    mdot = Cd*Ao*rho1*(Po/Pi)^(1/k1)/144*(2*Gc*R*T*k1/(k1-1)*(1-Po/Pi)^((k1-1)/k1)))^.5

    Else

    mdot = Cd*Ao*Pi*(k1*Gc*(2/(k1+1)/(k1-1))/(T*R*Z))^.5)

    where
    mdot = mass flow (lbm/s)
    Cd = discharge coefficient for the orifice (hole)
    Ao = area of orifice/hole (in^2)
    Pi = inlet pressure (psia)
    Po = outlet pressure (psia)
    Pcr = critical pressure ratio (= 0.528 for oxygen)
    rho1 = inlet density (lbm/ft3)
    k1 = ratio of specific heats
    Gc = constant: 32.174 ft lb / (lbm s^2)
    R = gas constant: 48.25 ft lb / (lbm R)
    T = temp (R)
    Z = compressibility factor

    (rewrite for Ao to determine the area (or diameter) of the hole in the tank)

    Note that the two equations are applied depending on if the flow is "choked" or "unchoked". Choked means there's a shock wave present at the hole, and this can be determined readily by checking the difference in pressure across the hole. If the absolute pressure inside the tank is more than roughly twice the absolute pressure outside the tank, it is above the critical pressure and the flow is "choked" so you use the first equation. Otherwise the flow is unchoked and you use the second equation. Note the if/then rules. For O2, Critial Pressure (Pcr) = 0.528.

    This assumes we know what the fluid state is inside the pressure vessel. If for example, you had a flow into the vessel and you knew what the fluid state was before it entered the vessel but not inside the vessel, you have a slightly more complex analysis. If the flow into the tank is simply from a higher pressure to a lower one, you can assume that flow is isenthalpic as Clausius was pointing out, and calculate the state inside the tank from the initial one. If you'd like more detail on that, feel free to ask.

    Please also explain what you know about the oxygen flowing into the tank, (ie: flow rate, pressure & temperature, is it coming from a compressor, etc…) and if this is an example or a real situation you're trying to understand. ty
     
    Last edited: Jan 8, 2006
  6. Jan 8, 2006 #5
    Q_Goest, thanks for the response (and everyone else!). I live in lsouth east houston, which isnt too far from La Porte. and crvz has no vowels as it is my initials!
    anyways, this is a real life situation that i'm trying to figure out. i'll start plugging away at work tomorrow with this stuff and let you know what i find out. i'm sure i'll have more questions.
    but this system in question has a bottle supply of oxygen (no compressor) starting at pressure of around 400 atmospheres (bottle volume is 2 liters) and bleeds down to 0 with a flow rate of 21 Liters/min (if memory serves, pressure drop vs time is a linear relation). the vessel pressure starts at 0.3 atmospheres and outside pressure is vacuum. bottle temperature and vessel temperature are held constant at about 70 degrees F (system has independant cooling system). obviously the supply of oxygen is limited (at that flow rate, assume you have about 800 standard liters of oxygen, you would get about 38 minutes before you run out of oxygen, but less to hold pressure).
    let me know if i left something out, or if some of my memory has failed me. many moons may have past since i spent time in academia.
     
  7. Jan 9, 2006 #6

    Q_Goest

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    Hi crvz, thanks for the additional details. It sounds like you have a BX cylinder of oxygen or similar high pressure cylinder that is supplying a constant flow of oxygen to your tank that is remaining at 0.3 atmospheres. In order to maintain constant flow into this tank, I'm assuming the cylinder has a regulator on the outlet. Is that right?

    The tank at 0.3 atmospheres is then being maintained at a constant temperature (70 F) using an 'independent cooling system'.

    You then want to control the flow out of this 0.3 atmosphere tank using an orifice. Also, this orifice is passing the oxygen into a vacuum. Did I get that all correct? If so, you can apply the second equation I gave you. (Note: I made a mistake in my other post and said the first equation was the choked flow one, that should be the second equation.) The flow through the orifice is choked with the inlet pressure of 0.3 atmosphere. Make sure you use absolute pressure in the equation. Also, if you use a drilled orifice (ie: a flat plate that has a drilled hole in it) I'd suggest using a discharge coefficient of 0.80. You may need to play around with this a bit, or you might simply put a metering valve on the outlet. Seems like a metering valve would be much easier. If you want a valve recommendation, let me know what tubing you're using.

    Clausius, regarding the thermodynamics of this, the thermo isn't needed to determine the orifice size. Also, I'm unclear on what you're suggesting. I believe the first law for this would reduce as follows.

    Since we have constant pressure and constant temperature inside the tank (the one at 0.3 atmospheres) the internal energy change inside the tank is zero (dU = 0). Thus the first law applied to the 0.3 atm tank would reduce to:

    dU = 0 = Hi - Ho + Qi

    Note that for no change in pressure or temperature, the density also remains constant and thus the mass flow in must equal mass flow out. (I know - I was drunk as a skunk yesterday and said enthalpy in equals enthalpy out, my bad.) So now we have:

    Qi = ho mdot - hi mdot = mdot (ho - hi)

    If I plug some values in, the first thing I do is look at the source. Let's say the bottle pressure is high (~ 5000 psig) and also assume there is little heat transfer between the regulator outlet and this 0.3 atm tank. As the oxygen expands across the regulator (from 5000 psi) and enters the tank, it cools to roughly -55 F. Internal energy is converted to PV energy as the molecules rearrange themselves. Once inside the 0.3 atm tank, heat must be added to get the oxygen back up to 70 F. So actually, assuming I'm corrrect in how this system is set up, the "independent cooling system" is actually warming the gas inside the tank in order to maintain a constant temperature and pressure, not cooling it.

    What is happening also, is that as the pressure inside the high pressure cylinder drops, the oxygen is cooling and heat from the atmosphere is going to warm the contents. It won't cool too much at these flow rates because the thermal mass of the bottle is relatively large and heat transfer is sufficient to maintain a roughly constant temperature.

    Also, as the high pressure cylinder drops in pressure, the enthalpy of the oxygen increases (ie: it picks up heat from the atmosphere/thermal mass of the cylinder) so that there is less and less heat needing to be added to the 0.3 atm tank. At first, the heat flux is relatively high but as the cylinder pressure decays, the amount of heat that must be added to the 0.3 atm tank also drops.
     
    Last edited: Jan 9, 2006
  8. Jan 9, 2006 #7

    Clausius2

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    I've deleted my message here. You said you were drunk, but last times I am stupid and in bad mental fitness, in addition I have got a cold. You're right, and my post was filled with errors.
     
  9. Jan 10, 2006 #8

    FredGarvin

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    Come on Clausius....ease up on yourself. It's all good :approve:
     
  10. Jan 10, 2006 #9
    Thanks again for all the info. I do have a few questions, though...

    While looking through some old textbooks, I've come up with a Z of 1 and k1 of 1.4 (and as these are factors and coefficients, I'm assuming they lack units). But speaking of units, there seems to be a discrepency in the second equation which has an extra ")". This may be the cause of the problem, but if I assume the last ")" is unnecessary, I get an Area in terms of (distance)*(time)^3. I'll do the math again, but Q_Goest, do you mind verifying that equation? Much obliged.

    And to answer the other questions, there is a regulator in the system to maintain the flow. And, if it helps, I actually don't want any leak in the system. Picture a pressure vessel in space. For simplicity, lets make it a cylinder that has a person in it. That person can live at a minimum pressure of about .3 atm. The vessel has an emergency supply of oxgyen for contingency scenarios. So, if a micrometeroid, travelling at 20,000 mph, slaps into the vessel (which will create a near perfect circular hole), I'm just curious to see the physics behind determining what size hole can be made and still maintain the pressure (assuming the emergency supply of oxygen kicks in at the same time the meteroid hits, as well as ignoring the fact that the meteroid will probably travel through the body of the individual enjoying his trip in the space vessel). Anyways, i thought it was a neat problem and just wanted to understand it a little better.
     
  11. Jan 10, 2006 #10

    Q_Goest

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    Yep, I messed up when rewriting that equation, thanks for pointing that out. It gets confusing with all those parenthesis so I've changed a few of them to make it easier to see where they're supposed to be and verified it. I had also left out a (k1+1) term. I also checked the first equation and verified it was correct and changed the parenthesis to make sure there's no missunderstanding. (One of these days I'm going to have to invest some time learning that Latex math thingy). Here's the corrected equations:

    If Po/Pi>Pcr
    mdot = Cd*Ao*rho1*[(Po/Pi)^(1/k1)]/144*{2*Gc*R*T*k1/(k1-1)*([1-Po/Pi]^[(k1-1)/k1])}^.5

    Else
    mdot = Cd*Ao*Pi*{k1*Gc*[2/(k1+1)]^[(k1+1)/(k1-1)]/(T*R*Z)}^.5

    Regarding Z and k1, yes - they are both dimensionless and the values you've mentioned are correct.
     
  12. Jan 16, 2006 #11
    I have a question for Q Goest. Why are you using compressibility for non choked conditions but that is not taken into consideration for choked conditions? Or, are you considering that rho1 takes into consideration compressibility for choked conditions? www.air-dispersion.com/usource.html would also be helpful to this thread.
     
  13. Jan 17, 2006 #12

    Q_Goest

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    (These equations are fairly common, but one can find other equations to calculate choked or unchoked flow. The equations quoted are the one's I've used in the past.) Good question. I've always assumed that compressibility was already taken into account as you say, because rho1 appears in that equation.
     
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