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Time required to exhaust air though an oriface 6
- Thread starter cbuck
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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).
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
Pete
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 MOC is a nearly exact method that can be used to solve hyperbolic PDE’s. The method is quite old, and has essentially been replaced by finite difference methods that are much easier to implement.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
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
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.
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.
- Thread starter
- #25
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
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
jparks1994
Aerospace
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.
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.
jparks1994
Aerospace
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.
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.
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?
How are you calculating the upstream gas temperature drop? Isenthalphic expansion?
jparks1994
Aerospace
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.
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.
jparks1994
Aerospace
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.
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.
jparks1994
Aerospace
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.
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
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
jparks1994
Aerospace
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.
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.
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
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
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.
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.
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.
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.
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