×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Time required to exhaust air though an oriface
6

Time required to exhaust air though an oriface

Time required to exhaust air though an oriface

(OP)
Does anybody have the equation(s) to solve for time?  I have a container (206 cubic inches) pressurized with air at 600 psi.  I need to calculate the time required to vent to atmosphere through an oriface (dia = 0.228 in).  
Thanks.

RE: Time required to exhaust air though an oriface

This is not a simple problem and there is no simple formula that I know of that will give you the time required.  I would use a simple forward difference scheme to solve your problem.  Use the ideal gas law to determine how much mass you have in the tank (you have volume and pressure, and you probably know the temperature, right?).  Using the pressure in the tank and the area of the oriface you can determine the initial mass flowrate.  If you use a small enough time step, you can assume constant flowrate over that time step, and therefore determine the mass lost during that time step.  Use that mass, along with volume and temperature (assume temp. is unchanged, i.e., an adiabatic process), to determine the new pressure, and then use that pressure to determine flowrate for the next timestep.  Repeat until your pressure approaches atmospheric pressure.  This can easily be accomplished in Excel.

Things can be simplified slightly.  The flow through your oriface will be choked until the pressure falls to about 30 psi.  Thus, you can assume constant flowrate until that point.

Maybe someone in this forum has a simpler approach, but as far as I know, this is your most straightforward option.  Let me know if you need help with any formulas.

Haf

RE: Time required to exhaust air though an oriface

3
Haf's approach is the one that I have also used.  

Until you no longer have chocked flow, flow rate through the orifice is essentially directly proportional to the inlet pressure.  Once you no longer than chocked flow, at about 15 psig for an orifice going to atmosphere, you then have to calculate the flow rate based on the dP across the orifice.

RE: Time required to exhaust air though an oriface

There was a Tech Brief in Machine Design, 5/22/86, pg. 97 by Robert Shirilla, JRS Manufacturing Co. entitled "Analyzing Blowdown Graphically" which plots mass ratio M/M0, and pressure ratio P/P0 against Nondimensional time, RT0W0t/144p0V where R=Ideal gas constant
T0=Initial flash temperature
P0=Initial flash pressure
W0=Initial mass flowrate
V=Flask volume
From the text, it sounds like you need to do considerable calculation to get the inputs to the nondimensional time expression as indicated above so it may not worth the trouble to track down. If you want a copy (2pages) post a business address and I'll mail it.

RE: Time required to exhaust air though an oriface


Just to repeat something that "TD2K" said is that the flow is (nearly) directly proportional to the upstream pressure.  It is NOT a constant flowrate, as "Haf" had inadvertently said.  

I otherwise agree:  Use a forward difference scheme; writing a rate equation and integrating for the total elapsed time wouldn't be worth it.

RE: Time required to exhaust air though an oriface

Watch out for your properties.  The deviation from ideal behavior gets to be a real issue above about 60 psig.  If your calculation is critical or $-intensive, you might want to use a simulator like HYSYS to get your physical and transport properties.  Unless, of course, you're like TD2K, who had nothing better to do one rainy Sunday afternoon, so he programmed PR and SRK into an Excel spreadsheet...  

Thanks!
Pete

RE: Time required to exhaust air though an oriface

Hey!!!!  What was I supposed to do, she had a headache

RE: Time required to exhaust air though an oriface

I ran the numbers through my spreadsheet at home.  Air at even 600 psig (assuming it's at room temperatures or higher) pretty much follows ideal gas behaviour because it has such a low critical temperature.

206 cubic inches at 600 psig and 60F (assumed) is about 0.38 lbs of air.  Through a 0.228" orifice, I got the pressure down to 15 psig (at which point, you no longer have chocked flow) is about 2.3 seconds.  My spreadsheet doesn't handle non-chocked flows as I was concerned about blowing down some vessels to 100 psig or less in which case, chocked flow still occurs.

RE: Time required to exhaust air though an oriface

Thanks poetix99, you are right.  What I meant is the flow velocity is constant (at Mach 1).  The mass flow rate is directly proportional to pressure, as pointed out above.  Fliegner's formula can be used to calculate mass flow rate.  The flow velocity will remain choked until the ratio of back pressure to tank pressure is greater than 0.52828.  If the back pressure is simply atmospheric pressure, this corresponds to a pressure of about 27.8 psig.

When the flow is no longer choked, you will have to calculate the new Mach number for every time step.  This can be determined using pressure and the isentropic flow relations (be careful to use the correct root).  Then simply use one of the mass flow functions to determine mass flow rate and continue in that fashion.

Haf

RE: Time required to exhaust air though an oriface

2
A good estimate of growth or decay can always be otained using the (good old) standard Naperian logarithmic forms

Qt = Qo *(1 - e^(-k*t))
and
Qt = Qo * e^(-k*t)

where
Qt = quantity at t
Qo = original quantity
e = naperian base (2.7182)
t = time
k = time constant

I'm sure that you will recognise these formulae but if you don't, just plot them out quickly and things will become clearer.
To get an estimate of a realistic time constant k differentiate the equation for time t = 0 and you get

dQ/dt = -k * Qo

dQ/dt is the starting rate of flow when you know all the conditions.  You can calculate this from an orifice formula.

You need to be a little careful choosing the units for the variable Q becuse that's where the errors get involved if the characteristics of Q are non-linear with the other variable in the equation for flow using the orifice formula.
Remember that it's probably better to use moles rather than volumetric measure and the Qo value should be calculated as difference from the end state rather than as an absolute.

The equation never reaches an end point but you can estimate an approximate end when (k*t) = somewhere between 5 (99.33% complete) and 10 (99.996% complete)

In your specific case, there is a slight complication because the equation you use to determine the initial flow rate is based on a choked jet and the P/V relationship is linear.  Once you get below the critical pressure inside, the relationship begins to transition to square root which would be an issue if you had a critically important calculation.  In that case you'd have to start a new calculation from critical down to zero. It's my guess however that you just want to know whether you're looking at hours or minutes, in which case you might be able to ignore the error.

This is a basic transient function and you can use this approach for any situation where decay of a fixed capacitance is involved.  Electrical engineers do it all the time to find the transient conditions in circuits with capacitors and it's right there in damped mechanical vibrations for all the Mech engrs!

RE: Time required to exhaust air though an oriface

I quickly made up a spreadsheet and ran the same numbers TD2K ran (i.e., assumed temp. of 60 deg. F).  According to my calculations, the flow is choked for 1.15 seconds.  This is exactly 1/2 the time TD2K reported, so one of us is off by a factor of 2.

According to my spreadsheet, blowdown is 99% complete (i.e., the atmospheric pressure divided by tank pressure is 0.99) after 2.2 s, 99.9% complete after 2.8 s, and 99.99% complete after 3.4 s.  I assumed atmospheric pressure of 101.3 kPa or 14.69 psi.

Haf

RE: Time required to exhaust air though an oriface

Haf, the ratio of 0.528 for choked flow looks right but isn't it based on dP/P1' where P1 is the inlet pressure in psia?  With an outlet pressure of atmospheric or 14.7 psia, I get the limit of choked flow as 31.1 psia/16.4 psig?

What did you calculate for an initial flow rate through the orifice at 600 psig?  I don't have my spreadsheet here so I'll have to take a look at it when I get home tonight.

RE: Time required to exhaust air though an oriface


A "sidebar" question to Misters TD2K and Haf:

I am not familiar with "Fliegner's formula" (I'll do a web search in the meantime), but I wish to question the accuracy of the flow being proportional to pressure for choked flow.

If we look at an orifice equation, mass flow for compressible fluids may be written as

  q = C * d^2 * Y * [ delta-p/v ]^0.5

  for an ideal gas pv = RT; v = RT/p

  therefore,  q ~ [delta-p / (RT/p)]^0.5

  if (delta-p) is nearly equal to p, then

  q ~ p

But the "error" implied by this analysis is approx. 8%, even when p is 100 psi (with 15 psia downstream).  

What have I missed?

I agree that the analysis is a first order decay, as Flareman has pointed out, and I think that the accumulation of this error, when looking at the entire venting process, might have a significant effect.

RE: Time required to exhaust air though an oriface

Sorry guys but I made a slight math error in the previous memo.  The solution should have included 1/k not k.

However, I'm confused by your discussions.  The key is certainly, as Haf says, to "choose a small enough time step".  That's probably where your difference is and also where the naperian logs come in.  Ultimately, the time step gets so small you need to integrate the solution in order to do the sum and that throws up the log function.

As far as I can see here the  solution resolves to
Air at 600 psia, and 68 degF = 3.0684 lb/cu.ft
206 cu ins contains 0.3658 lb initially and 0.01696 lb at 27.818 psia (critical) and 0.00896 lb at ambient (14.696 psia)
The initial rate of discharge through a 0.228 inch, sharp edged orifice (assuming 0.6 coeff) is 0.302 lb/sec.

If the discharge rate were constant, you'd get down to critical in (0.3658-0.01696)/0.302 = 1.155 secs, but it's not and at the critial condition, the discharge rate has gone down to 0.014 lb/sec.
The time constant of this part of the depressuring is actually (0.3658-0.00896)/0.302 = 1.1816 secs

The decay needed to get to 0.01696 from 3.0684 is 99.445% complete so the exponent of exp is -5.198. 5.198 time constants gives us 6.142 secs to get from 600 psia down to the critical condition. After that we need a new calculation. But hey! were already 99.445% complete.  How much do we care?
If we do care, the formula is a bit more complicated but, very roughly, the time constant for the next bit works out to roughly 3 seconds and it takes about 15 seconds to get down the rest of the way to (almost) atmospheric pressure.

I don't think I've seriously missed anything but I'm open to erudite correction because I use this procedure for a number of different problems and I really need to know if I'm stumbling down the wrong path.

RE: Time required to exhaust air though an oriface

Let me start with the disclaimer that I am a compressible fluid dynamics guy, so you may not like my approach to this problem.  Also, my apologies to TD2K as I found a mistake in my spreadsheet.  I accidentally used an incorrect mass flow function (one with static pressure and static temperature instead of stagnation pressure and stagnation temperature).  With the corrected fass flow function, I calculate that the flow is choked for 2.0 seconds and reaches P/P0 = 0.999 at 3.7 seconds.  The initial mass flow rate is 0.259 kg/s or 0.571 lb/s (I work in SI units).  I put together this spreadsheet in about 7 minutes so I make no guarantees.

The ratio of back pressure (in this case atmospheric pressure) to tank stagnation pressure must be checked to determine if the flow is choked.  When the ratio is less than 0.52828, the flow is choked.  For an atmospheric pressure of 14.7 psia, this occurs at stagnation pressures above 27.8 psia (not psig as I mistakenly typed above).

Fliegners formulae are just simplified versions of the mass flow functions.  Given a Mach number of 1, gamma (ratio of specific heats) of 1.4 for air, and gas constant R of 287 J/(kg*K), Fliegners formula is simply

Mdot = 0.04042*P0*A/(T0)^0.5

where

Mdot is mass flow rate in kg/s
P0 is stagnation pressure in Pa
A is area in square meters
T0 is stagnation temperature in Kelvin

You can determine similar relations with static pres. and temp. instead of stagnation pres. and temp. by simplifying the corresponding mass flow function.  This formula is only valid for choked flow.  More complicated mass flow functions that include Mach number must be used for unchoked flow.

Haf

RE: Time required to exhaust air though an oriface

Haf, I'd love to sit down with you and go over this stuff.  I have a couple of articles that I suspect I just need to sit down and work through a few times but it's an area I'll confess I'm weak in.

My spreadsheet is pretty much the sledge hammer approach.  Calculate the initial mass of air, calculate the initial flow rate through the orifice, calculate how air passes in a time step, calculate the remaining mass of air and calculate a new pressure (for this one, i assumed the temperature in the vessel is constant).

Then go back and calculate the new flow rate and repeat.  I just adjust the time steps until I'm getting reasonably small drops in pressure.

Poetix99, if you have access to a Crane manual, you'll see essentially the same formula as you have.  However, when you have choked flow across an orifice, you no longer use the dP, rather, the 'effective' pressure drop is calculated and you use that which is essentially a constant * inlet pressure.  Thus, flow becomes proportional to pressure during choked flow.

RE: Time required to exhaust air though an oriface

In looking up "Fliegner's formula" in Google, I stumbled onto this website:

  http://www.ijee.dit.ie/articles/999982/experime.htm

Entitled "Experiments to Study the Gaseous Discharge and Filling of Vessels"

Hey, "cbuck", if you're still out there, and we're not merely talking amongst ourselves...

RE: Time required to exhaust air though an oriface

Haf and TD2K - when you guys sit down to go over this, please invite me.  I love this stuff.  How come I didn't get it in college?  How 'bout we meet at Borders in Sacramento next Friday, over hot chocolate?  The folks around here already think I'm a nerd; I'd hate to disappoint them...

Thanks!
Pete

RE: Time required to exhaust air though an oriface

Accoring to Miller on page 13-24 of Flow measurement engineering handbook, a standard thin square edged will not choke at pressure ratios down to 0.2, discharge co-efficient reported as 0.9.
athomas236

RE: Time required to exhaust air though an oriface

TD2K,

I'm using the same approach you are.  In effect, it's a first order numerical solution of the differential equation.  I check convergence by using smaller and smaller time steps until the results using one time step agree with those of the next smallest time step.  Using this approach you can determine the biggest time step you can use and not waste processor time/memory (if you're doing this in Excel you know that these spreadsheets can get pretty large).

It's of course always useful to examine results to see if they make sense.  Keep in mind that 206 cubic inches is a relatively small volume.  If for example, the tank were spherical, it would have a radius of only 3.7 in.  For a tank that small, a hole of 0.228 in is relatively large.  I doubt it would take more than a few seconds to vent to the atmosphere.

As a side note, I'd like to mention that I recently earned my MS in ME (in 2001).  I spent two years studying compressible fluid dynamics and took several advanced compressible fluids and computation fluid dynamics courses.  So, needless to say, I love this stuff too.  I don't use it too much in my current job, so I'd be happy to discuss things before I start to loose it!  Let me know if you have any questions (I'm talking to TD2K and 74Elsinore, self-professed compressible fluid dynamics nerds).

Haf  

RE: Time required to exhaust air though an oriface

Wow, poetix99, it's a small world.  You're link above took me to a paper by J. Craig Dutton.  Turns out he was my grad school advisor.  He's definitely regarded as a world-wide expert in compressible flow, especially on the experimental side of things (although I took two classes from him and can tell you that he is rock solid on the theoretical and computational side of things, too).

RE: Time required to exhaust air though an oriface

Hey Haf, this is slightly off-topic (maybe not?) but sometime I'd love for you to explain the method of characteristics to me.  I never got it.  I had George Emanuel (AIAA stud) for gas dynamics, and while he was extremely intelligent, and a pretty decent guy all-around, communicating the fundamentals of compressible flow to undergrads just wasn't Dr. Emanuel's forté, bless his heart.  I got the feeling that the MOC was a powerful tool but I never got the hang of it.  The text wasn't a lot of help either.  The MOC is also useful in the solution of transient pressure problems in incompressible flow, e.g. waterhammer.

Thanks!
Pete

RE: Time required to exhaust air though an oriface

74Elsinore,

I’m surprised the method of characteristics was even mentioned in an undergrad-level fluids course.  Most undergrad fluids courses spend very little time on compressible gas dynamics at all, much less an advanced concept like MOC.  In fact, the intermediate gas dynamics class I took (graduate level compressible gas dynamics) didn’t even cover it.  I wasn’t exposed to MOC until I took an advanced gas dynamics class.  We spent about 3 weeks on the topic, so, needless to say, I won’t be able to explain everything here.

So here goes with a high-level explanation: It turns out that steady supersonic flow is governed by nonlinear hyperbolic partial differential equations that in general have no closed-form solutions.  As a result, one is forced to use numerical methods.  MOC is a nearly exact method that can be used to solve hyperbolic PDE’s.  The method is quite old, and has essentially been replaced by finite difference methods that are much easier to implement.  In some cases, however, MOC is the best method to use (actually, I can think of only one case, which I will mention later).

Physically, characteristics are paths along which disturbances are propagated in steady supersonic flow.  Mathematically, characteristics are curves along which the governing PDE’s can be manipulated into total differential equations.  If you want to get rigorous, characteristics are curves along which the flow properties are continuous, the flow property derivatives are indeterminate, and across which the flow property derivatives may be discontinuous.  Using these conditions, one can do a bunch of (ugly) calculus and algebra to get compatibility equations that can be used to solve for flow properties.

For supersonic (2D) flow, there are two real characteristics that pass through every point in the flow.  These characteristic curves are inclined at +/- the Mach angle of the flow.  In other words, the characteristics are the Mach waves that pass through the point.  Thus, in 2D, the curves form a V that expands at +/- the Mach angle.  The downstream points that are inside this V are in the zone of influence, i.e., the flow properties at the vertex of the V have an influence on every downstream point inside the V.

Now, let’s assume you have a 2D supersonic flow and an inlet condition (the most convenient would be a velocity profile).  Choose any two points on the velocity profile.  For each point you know the position, (x, y), and the velocity, (u, v).  If you cast a right-running characteristic (as you look downstream the characteristic runs from left to right, i.e., the characteristic has a negative slope) from the “top” point and a left-running characteristic from the “bottom” point, the curves will intersect downstream.  If you solve the characteristic equations you get the position of this point (x, y).  Then, if you solve the compatibility equations, you get the velocity of this point (u, v).  This is how the technique works.

Unfortunately, MOC is extremely tedious, and therefore is usually implemented numerically.  Even then, the programming is tedious when compared to finite difference methods.

As far as I know, there is only one remaining problem in supersonic gas dynamics where MOC is the method of choice (and maybe the only choice).  If you know anything about converging-diverging nozzles, you know that the diverging profile of the nozzle plays a large role in determining flow uniformity.  It turns out you can use MOC to determine the optimum contour.  In grad school, my thesis project involved performing shock boundary layer interaction studies.  I needed a Mach 1.4 flow with a normal shock, and my tunnel was set up for Mach 2.5 flow with an oblique shock.  I used an MOC-based code to design a new nozzle block that would provide uniform Mach 1.4 flow.  Afterwards, I used laser Doppler velocimetry (a non-intrusive laser diagnostic technique that determines instantaneous velocity at discrete points in a flow field) to measure velocity profiles.  It was absolutely amazing how uniform the freestream flow was!  And it was even more beautiful when I saw the normal shock sitting at the center of the windtunnel optical viewport (just as I designed it).  After a little practice, you could actually see the shock in the tunnel with the naked eye (ok, so I spent a lot of time in the lab).  It was almost like looking at one of those Magic Eye 3D pictures.

Anyway, I hope this explanation helps.  And I can’t forget to reference Prof. Craig Dutton.  Much of this explanation came from notes I took in his class.

Haf

RE: Time required to exhaust air though an oriface

Good stuff guys, let have some fun and do this as simple as possible and see how close we get to the real answer:

We will just use the perfect gas law and the equation for the flow through an orifice (see Crane Manual or other text).  Using 600 psig, 60 Deg F, MW =29, Vol = 206 in3 for ATmos press of 14.7 psia; that gives 0.381 lbs in the tank, with a density of 3.197 lbs/ft3.  Using these initial conditions for the orifice flow (with an assumed flow coeff of 0.61)we get 1369 lbs/hr for an expansion factor Y = 0.72 and a critical pressure ratio of 0.531.

Then since we are trying to do this on the back of our hand fast like; instead of doing numerical intergration we will just determine the average flow for time calcs:  avg flow = (1369 + 0)/2 = 685 pph.  Then, 0.381 lbs/ 685 pph is equivalent to 2 seconds.

Son of a gun not too far off!!  

The more you learn, the less you are certain of.

RE: Time required to exhaust air though an oriface

(OP)
WOW!  My sincere thanks and appreciation for sharing your obvious expertise. In particular I would like to thank Haf and TD2K.

I apologize for disappearing on the conversation - I have been lost in the Twilight Zone of a particularly difficult machine installation.

I hope you guys will see this after so much time.  If any of you ever make it to Richmond, VA, I owe you a beer.

Thanks,

cbuck@jewettautomation.com

RE: Time required to exhaust air though an oriface

Don't forget about the temperature of the gas remaining in the vessel.  As gas escapes through the orifice the remaining gas is expanding, and its temperature will be dropping, which also affects the exit flow conditions.  Equations which work for me in the typical spreadsheet, small time-step calculation approach are:

f(k) = sqrt((k(2/k+1))^((k+1)/(k-1)))

Pn+1=Pn(1-(A*sqrt(RTn/V0)*f(k)dt)^k

Tn+1=Tn(1-(A*sqrt(RTn/V0)*f(k)dt)^(k-1)

with
k= ratio of specific heats, 1.4 for air
Pn+1 = pressure in the vessel at n+1 time step
Pn = pressure in the vessel at n time step
A = area of the opening
R = gas constant, 297.1 (m/s)^2*(1/degrees K) for air
Tn+1 = temperature of gas in the vessel at n+1 time step
Tn = temeprature of gas in the vessel at n time step
V0 = volume of the vessel
dt = time step

I haven't tried this formulation on the problem which originally prompted this thread, but I will just as soon as I get time, and let you know what I get.

Of course this doesn't account for the shape of the orifice, sharpness of the edge, etc.  You could do that by appropriately modifying the area, A, with a CvA, with Cv appropriately chosen.

Now let me add another wrinkle.  Stick a pipe of length, L, on the orifice.  How can you account for the pressure drop down the pipe?  You need to calculate the flow speed/Mach number, M, at the entrance to the pipe to be able to calculate the amount of gas which escapes during the time interval, dt.  The only equation I know of which gives M as a function of L is transcendental in M, and can't be solved explicitly for M in terms of L.  The solution requires some sort of iteration or table look-up (Fanno Line tables), which doesn't work too well in my simple spreadsheet implementation.

RE: Time required to exhaust air though an oriface

As promised, I put your parameters (600 psia, .228 in dia opening, 206 cu in vessel) in my equations, and here's what I got:
1. My formulation doesn't account for unchoked flow, so I'm only good to down around 28 psia in the vessel.
2. My calculations go below 28 psia in the vessel in about 1.7 seconds.  Pretty quick, but it's a pretty small vessel, about 6 in^3.
3. As I said in my first post, I'm accounting for the temperature drop of the gas in the vessel.  After 1.7 seconds I calculate the gas temperature as -154 C, or -245 F.  At this point you're probably getting close to the triple-point temperature for oxygen, or something, which is another story.  My equations certainly don't account for that.

RE: Time required to exhaust air though an oriface

Umm jparks1994, I'll have to look at this but your numbers for the gas temperature inside the vessel don't seem right to me.  I've seen blowdown systems on petrochemical facilities in operation.  The piping downstream of the blowdown orifice or valve gets iced up and pretty cold but the piping and equipment upstream does not.  Certainly not to the temperatures you calculate.

How are you calculating the upstream gas temperature drop?  Isenthalphic expansion?

RE: Time required to exhaust air though an oriface

Thanks for the input.  Hands-on experience can't be beat sometimes.  Just a few comments:
1. I posted my equations in an earlier post, same day.  Have a look and see if you can find anything wrong.  Send me an address and I'll email you my derivation.

2. Yes, I did the pressure and temperature drop in the vessel as an isenthalpic expansion.

3. As the gas cools, it will be taking up heat from the vessel and surrounding structure, or whatever, which of course is not isenthalpic.

4.  If there's moisture involved, or other condensing (and freezing) gases, then that will also blow the isenthalpic assumption.

5.  Bottom line is that isenthalpic is an approximation, and the more the temperature drops, the worse it gets.  Hard to tell where it breaks down.  Later, when I get time, I'll take a look at it isothermally, and see how much difference it makes.  Let you know.

RE: Time required to exhaust air though an oriface

In my last post of Oct 7, I seem to have baffled myself with big words (or maybe b.s.).  In response to a question from TD2K, I said that my assumption was an "isenthalpic" expansion.  Later it occurred to me that I meant "adiabatic" expansion, not "isenthalpic."  Maybe they are the same thing, but at least I know what "adiabatic" means.  I can't find "isenthalpic" in any of my references.

RE: Time required to exhaust air though an oriface

I'm confused.  If you assume an adiabatic expansion, that means no heat is transfered to or from the vessel.  This means that stagnation temperature stays constant through the expansion.  Static temperature, of course, changes, but only in areas where the gas is in bulk motion.

RE: Time required to exhaust air though an oriface

Correct, no heat is transferred.  You squirt out a bit of gas over a small time step.  Then the remaining gas in the vessel expands to replace the lost gas, and the temperature and pressure drop according to the adiabatic expansion laws.  Repeat 50 or 100 (or whatever makes you happy) times, and you asymptotically approach zero pressure and zero temperature and zero gas left in the vessel.

RE: Time required to exhaust air though an oriface

What are the basic theory behind the decision to say the flow is sonic until Patm/Ptank < .528?  At what point does this assumption break down?  Is it only for an orifice plate or any kind of converging-diverging nozzle?

The reason I ask is that I have a calculation to make where this assumption would come in handy.  I'm just concerned about applying it without understanding the physics of it.

I have a graduate fluids background, including compressible flow, but it's 11 years old and I'm only now starting to use it again.

Thanks,

Vidaman

RE: Time required to exhaust air though an oriface

Pardon my poor grammatical and typing skills.  I meant "What IS the basic theory behind the decision to say the flow is sonic until Patm/Ptank > (not <) .528?"

Sorry for the mix-up.

RE: Time required to exhaust air though an oriface

Vidaman -
You'll find this relationship derived in most any text which treats flow in constant area ducts, or you can see it in Isentropic Flow Tables, the most famous of which is "Gas Tables", by J.H. Keenan and J. Kaye, published in 1948.  The text I use is "Introduction to Gas Dynamics", by Rotty, published in 1962.  Rotty doesn't derive the value of .528 specifically, that I can find, but you can get it if you plug M=1 and k=1.4 (for air) into his equation for the ratio of pressure in the duct to stagnation pressure (equation 5.45, page 86).  Or, if you find M=1 in the Isentropic Flow Table (Table A.2 in Rotty) for k=1.4, you'll see that pressure ratio is .52828.

RE: Time required to exhaust air though an oriface

jparks1994,

Thanks.  That's exactly what I was looking for.  I'm slowly scraping the rust off my subject knowledge.  It's been too long!

Thanks again.

RE: Time required to exhaust air though an oriface

Vidaman,

The assumption that Patm/Ptank < 0.52828 results in sonic flow DOES NOT apply to a C-D nozzle.  With a CD nozzle, flow will be sonic at Patm/Ptank = 0.582828 and supersonic at Patm/Ptank < 0.52828.  

The assumption applies best to "well-designed" converging nozzles and roughly applies to orifices, etc.

Haf  

RE: Time required to exhaust air though an oriface

how would one apply the above analysis for pressure testing a DN600 -200m long pipeline header at 18 ata  with nitrogen available in bottles at 150ata each of 300L

RE: Time required to exhaust air though an oriface

TD2K and JPARKS:

Remember that the vessel relieving the gas is only 2.06 in3 in volume for the example given; therefore, I would expect the vessel to decrease in temperature in a similar way to the gas expanding to atmosphere since the total expansion only takes about 2 seconds before the vessel is emptied.

TD2K, I agree with your observation and have observed this myself; but I believe this usually occurs when the upstream piping is a header that is continually re-supplied with fluid as the fluid is let-down in pressure to the atmosphere.  Then in the latter case, the expanded gas cools and develops frosting on the surface of any discharge piping, but the upstream piping does not frost.

Right, or am I mising something here?
Just using Boyles and Charles's perfect gas laws, the temperature in the vessel has got to be pretty cool.

The more you learn, the less you are certain of.

RE: Time required to exhaust air though an oriface

TD2K and JPARKS:

Remember that the vessel relieving the gas is only 2.06 in3 in volume for the example given; therefore, I would expect the vessel to decrease in temperature in a similar way to the gas expanding to atmosphere since the total expansion only takes about 2 seconds before the vessel is emptied.

TD2K, I agree with your observation and have observed this myself; but I believe this usually occurs when the upstream piping is a header that is continually re-supplied with fluid as the fluid is let-down in pressure to the atmosphere.  Then in the latter case, the expanded gas cools and develops frosting on the surface of any discharge piping, but the upstream piping does not frost.

Right, or am I missing something here?
Just using Boyles and Charles's perfect gas laws, the temperature in the vessel has got to be pretty cool.

The more you learn, the less you are certain of.

RE: Time required to exhaust air though an oriface

CHD01, the example I was thinking about was a high pressure (5000 psig) injection gas compressor that was depressured on an ESD.  The piping downstream of the piping significantly frosts up (in face, it chilled off close to the calculated gas temperature which was well below the allowable temperature for that piping but that's another story) while the piping upstream of the orifice did not.

RE: Time required to exhaust air though an oriface

A NASA colleague recently showed me that the differential equation for pressure and temperature versus time for gas venting from a vessel can sometimes be separated and integrated to closed form solutions for pressure or temperature versus time.  I haven't worked through all the details yet, but when I get to it, I'll post the equations for your review.  Also, they may be a bit complicated for posting in this text format.  Is there a better way available to post equations with exponents and square roots and exponents inside square roots?

RE: Time required to exhaust air though an oriface

x2

For this you have to write like this. (/sup)x(sup)2

Put square braces[] for the words /sup and sup so that you will get the required result. For more details go to preview post button below the space where you make your posts. Go to editpost option and there you will find lot of useful tags.

Regards,

Repetition is the foundation of technology

RE: Time required to exhaust air though an oriface

jparks,

The paper I referenced above by J.C. Dutton goes into the differential equations and solutions, if available.  There's a link to the paper in one of my above replies.

Haf

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources