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.
Thanks.





RE: Time required to exhaust air though an oriface
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
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
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
Thanks!
Pete
RE: Time required to exhaust air though an oriface
RE: Time required to exhaust air though an oriface
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
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
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
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
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
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
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
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
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
Thanks!
Pete
RE: Time required to exhaust air though an oriface
athomas236
RE: Time required to exhaust air though an oriface
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
RE: Time required to exhaust air though an oriface
Thanks!
Pete
RE: Time required to exhaust air though an oriface
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
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
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
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
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
How are you calculating the upstream gas temperature drop? Isenthalphic expansion?
RE: Time required to exhaust air though an oriface
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
RE: Time required to exhaust air though an oriface
RE: Time required to exhaust air though an oriface
RE: Time required to exhaust air though an oriface
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
Sorry for the mix-up.
RE: Time required to exhaust air though an oriface
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
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
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
RE: Time required to exhaust air though an oriface
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
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
RE: Time required to exhaust air though an oriface
RE: Time required to exhaust air though an oriface
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
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