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Time required to exhaust air though an oriface 6
- Thread starter cbuck
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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
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
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- #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.
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.
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.
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.
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.
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
Pete
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.
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.
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
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
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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!
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!
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
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
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.
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.
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.
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.
![[peace]](/data/assets/smilies/peace.gif "[peace] [peace]")
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.
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
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
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.
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.
In looking up "Fliegner's formula" in Google, I stumbled onto this website:
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...
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...
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
Pete
athomas236
Mechanical
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
athomas236
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
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
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