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I Time to equalization

  1. Feb 24, 2017 #1
    Hi,
    I have a pipe that is unpressurized initially at atm internally with air. Inside the pipe there is a welded container that is vented at the top. The vent is made to have air pass through so that the internal pressure of the container eventually equals the pressure of the internal pressure of the pipe when it is pressurized to high pressures. The vents are there so that there isn't a large pressure difference. The walls of the container would fail if the vent wasn't there.

    T=266F
    Ppipe,initial=14psi
    Ppipe,final= 1400psi
    Pcontainer, initial= 14psi
    Vcontainer= 106in^3

    The pipe is then pressuriazed at a rate of 95psi/sec is this design good enough to prevent collapse of the container walls? In other words does the container equalize pressure fast enough to prevent failure? Assume a pressure difference of 200 psi or more between the walls of the container will fail it.

    The vent comes off the container in the picture below:
    https://ibb.co/hvz9qa
     
    Last edited: Feb 24, 2017
  2. jcsd
  3. Feb 24, 2017 #2

    BvU

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    Hi,
    You write about a vent. Then it changes to plural. Where's the other ?
    And you tell us nothing about the vents !?
     
  4. Feb 24, 2017 #3
    Only singular. 1 vent. The vent is shown in the picture with the diameter of the opening.

    Edit: Also assume the pipe is a closed container
     
  5. Feb 25, 2017 #4

    BvU

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    So, just to make sure I get the right idea about he problem to be solved:
    The red box is the pipe and it's closed for all our purposes (how the pressure increases can be left outside our considerations).

    upload_2017-2-25_12-56-28.png

    If I treat this as a mathematician I'd set up the following problem statement:
    $$P_{\rm pipe} = 14 {\rm\ psi } + 95 {\rm \ psi /s}\ \times t \quad {\rm for\ \ } t = \left [0, {1400-14\over 95 } \right ]$$ ##P_{\rm \ ves} (0) = 14 {\rm\ psi }## (this 14 psi surprisingly turns out to be atmospheric :smile:) and we want to check that $$\Delta P = P_{\rm \ pipe} - P_{\rm \ ves} < 200 {\rm \ psi }\ \ \forall t $$With that, the problem statement isn't complete: we need a flow into the vessel as a function of ##\Delta P## and we need an equation for ## P_{\rm \ ves} ## as a function of the holdup. The latter is easy : ideal gas law is good enough ##PV = nRT_{\rm abs}##. Brings us to a next level of detail: what happens to the temperature. Here I could enter only 211 F as the highest temperature, but it does allow entry in your funny units (sorry, couldn't reSIst :rolleyes: -- [edit] too hasty: in fact it does, sorry TLV!. Only 5/8" has to be converted to .625" and you don't get cubic inches out) and shows the equations used. Add them into your problem statement and get an altogether hideous differential equation for which I am way too lazy -- but you are welcome.

    As a physicist I'd leave out the temperature business detail and still have a differential equation.

    As an engineer I would be happy with treating this differential equation as a series of pseudo-steady states to get an impression of an estimated worst case:
    The TLV link above tells me that if I wait 1 second before opening the vent, a pressure difference of 95 psi at 211 F though 0.625 " gives a flow of 2.6 Nm3/s;
    (sorry, no cubic inch/fortnight available, but:) some work shows that's over 160000 N in3. In other words: the vessel catches up in no time.

    As a blue collar mechanic (I'm not) I'd say: 5/8" is more than wide enough for a 2.75" diam vessel to keep up :smile: , even if it were five times as long; no worry at all.

    As a PF HHelper I'd consult with @Chestermiller , to see if i don't type nonsense.

    Note that in the end you are responsible, so you have to convince yourself :smile:.
     
  6. Feb 25, 2017 #5
    The key to this is knowing the pressure drop-flow rate relationship for the 5/8" neck/vent. Without that, this doesn't seem doable. We need to get a blown-up view of the neck to determine it analytically, or it needs to be measured experimentally on a test rig. Of course, some experimental measurements would be preferable.
     
  7. Feb 27, 2017 #6
    The diagram you made is correct for the problem I am talking about.Thank you all for your help so far! cubic inch/fortnight had me rolling on the floor.

    Chestermiller, without measuring the flow rate into the vessel experimentally, you said we can determine it analytically with a blown up view?
    http://imgur.com/a/9cBDP
     
  8. Feb 27, 2017 #7

    BvU

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    yea yea, but if you would have checked my antics you'd have hit the ceiling: the 2.6 Nm3/s is for 1400 to 14 psi and 95 to 14 psi only yields some 0.3 Nm3/s ! Depending on the volume of your vessel that may still be just fine, but the margin isn't so huge as I made it look.

    TLV calculation is for flow through an orifice -- your geometry is more favourable.

    (*) the unit furlongs/fortnight for velocity came from some murphy poster or - booklet I think. We used it to tease our poor UK colleagues
     
  9. Feb 27, 2017 #8
    What I'm thinking is this: Suppose, rather than having a short section of 5/8" opening, you had a 100' long section of hose. At the far end of the 100' section of hose, the pressure is rising at 95 psi/sec. Certainly, if this didn't lead to a 200 psi difference, then a short section would not either. And, we know how to calculate the results for the case of a 100' section of hose. So, the 100' section bounds the result for the actual situation. How would this grab you?
     
  10. Feb 27, 2017 #9
    I actually simplified the opening to a straight tube coming off the vessel. In reality it looks like this:
    http://imgur.com/a/Kd1tz
    The domecap has slots cut in to allow air to flow. The air travels to the top of an internal tube and shoots down into the larger volume portion of the vessel.

    Do you think it would be good enough to simplify it as a 100' tube coming off the top? I know turbulence may affect it with the actual design.
     
  11. Feb 27, 2017 #10
    There will be turbulence in the tube. In my judgment, it would be a very very conservative calculation.
     
  12. Feb 27, 2017 #11

    BvU

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    100' of 5/8" is a hefty resistance.
    But the dome cap in #9 sure looks a lot different (and less favourable) from what we saw in post #6 ! Is it true to scale ?

    (an alternative to the 100' hose: Make a conservative estimate of ##\sum k## (or http://file:///C:/Users/aa844/Downloads/CraneNuclear-USletter-Apr2013-Section8.pdf [Broken]) and do ##\Delta p = \sum k \;{1\over 2} \rho v^2 ## )
     
    Last edited by a moderator: May 8, 2017
  13. Feb 27, 2017 #12
    I'm having trouble doing the math on this. Tell me if I'm doing this right. For the 100' assumption, I should calculcate the flow rate for a 95psi pressure difference (the worst case flow rate). Once I calculate the flow rate into the vessel do I calculate the time it takes to fill the total volume of the container? If that time is less than 14.56 sec. than theoretically it would be a good design?

    BvU,
    I tried to draw it as close to scale as possible. The rectanglular shaped vents go around the whole circumference of the domecap. I have areas for these if we need them. I will look into this alternative approach. Thanks.
     
  14. Feb 27, 2017 #13

    Nidum

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    I find this description of your problem worrying . Vague statements of equalising pressures so that the container doesn't fail are not adequate when dealing with pressure vessels charged to 1400 psi .

    What are you actually trying to do ? Do you have a sufficient level of technical knowledge to do this work safely ?
     
  15. Feb 27, 2017 #14
    The pressure variation in the hose for turbulent flow would be given by:$$\frac{dp}{dz}=-\frac{4}{D}\left(\frac{1}{2}\rho v^2\right)\frac{0.0791}{Re^{0.8}}\tag{1}$$
    where v is the local gas velocity, D is the hose diameter, ##\rho## is the local gas density, z is axial position, and Re is the Reynolds number for the flow:
    $$Re=\frac{\rho v D}{\mu}=\frac{4\dot{m}}{\pi D \mu}\tag{2}$$where ##\mu## is the gas viscosity and ##\dot{m}## is the mass flow rate. The term ##\frac{1}{2}\rho v^2## can be rewritten as: $$\frac{1}{2}\rho v^2=\frac{1}{2\rho}(\rho v)^2=\frac{8}{\rho \pi^2 D^4}\dot{m}^2\tag{3}$$Substituting Eqn. 3 into Eqn. 1 then yields:$$p\frac{dp}{dz}=-\frac{32}{\pi^2D^5}\left(\frac{RT}{M}\right)\frac{0.0791}{Re^{0.8}}\dot{m}^2\tag{4}$$where used has been made here of the ideal gas law:$$\rho=\frac{pM}{RT}$$with M being the molecular weight and T being the temperature. For a given mass flow rate, the entire right hand side of Eqn. 4 is constant, so the equation can be immediately integrated to yield:$$p^2(0,t)=p^2(L,t)-\frac{64L}{\pi^2D^5}\left(\frac{RT}{M}\right)\frac{0.0791}{Re^{0.8}}\dot{m}^2\tag{5}$$where ##p(0,t)## is the pressure in the inner chamber, ##p(L,t)## is the pressure at the far end of the hose (14.7+95 t psi), and L is the length of the hose. The pressure in the inner chamber is determined by the ideal gas law:$$p(0,t)=\frac{mRT}{MV}\tag{6}$$, where V is the volume of the inner chamber. So, $$\frac{64L}{\pi^2D^5}\left(\frac{RT}{M}\right)\frac{0.0791}{Re^{0.8}}\dot{m}^2=p^2(L,t)-\left(\frac{RT}{MV}\right)^2m^2\tag{7}$$
    Eqn. 7 provides a first order differential equation for ##\dot{m}=dm/dt## as a function of the cumulative mass m in the inner chamber, and the pressure at time t ##p(L,t)## that can be readily integrated to get the mass m in the inner chamber as a function of time, and also, from the ideal gas law, the pressure in the inner chamber as a function of time.
     
  16. Feb 27, 2017 #15
    Wow this is impressive. It's been awhile since I've done any differential equations. I really appreciate what you've done.

    Edit: Someone who enjoys doing differential equations, please help. I am trying to relearn right now
    I'd like to find the pressure inside the container at t=14.56sec
     
    Last edited: Feb 27, 2017
  17. Mar 2, 2017 #16
    Do I assume a Reynolds number? Cause I do not know the velocity of the fluid.
     
  18. Mar 2, 2017 #17
    You have an algebraic equation involving ##\dot{m}## to the 0.8 power inside the Reynolds number, and 2 outside the Reynolds number. So the left hand side of the equation involves ##\dot{m}## to the 2 - 0.8 = 1.2 power. So solve for ##\dot{m}=dm/dt##. What do you get?
     
  19. Mar 3, 2017 #18
  20. Mar 3, 2017 #19
    Your Reynolds number in the denominator was to the 0.8 power. That's missing. You need to do two things now:

    1. Combine the groups of parameters that are constant into single constants so you don't keep writing them out, and the equation looks simpler.

    2. Solve the Eqn. algebraically for ##\dot{m}=dm/dt##
     
  21. Mar 6, 2017 #20
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