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Pressure test failure velocity 3
- Thread starter sbatch20
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EdStainless
Materials
It is a function of the stored energy, there is a whole class of work on problems like this.
They require a lot of assumptions.
= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube
They require a lot of assumptions.
= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube
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- #4
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2
- #5
The Bernoulli equation is not any help here.
It is actually a kinematics problem. The fluid will experience explosive decompression (which is unbounded by choked flow) for something like 20-50 mS (0.02-0.05 S). After explosive decompression completes, the fluid velocity is limited by the speed of sound and the plug will have outrun any exiting fluid, so you only need to be concerned with the first 50 mS (to be conservative).
F=ΔP*A
and
F=m*a
so
a=(ΔP&A)/m
and
v=a*t=(ΔP&A)/m * 0.05S
Put in the cross sectional area of the plug and its mass and get the units right and you have the velocity at the end of the explosive acceleration. From that point it is going to fall under the influence of gravity and slow due to wind resistance. This calculation works for any length pneumatic test. For a hydrostatic test the duration of the explosive decompression falls rapidly with decreasing volume of the test so it is only valid for tests over about 100 ft long.
I've got a narrated presentation on my web page that talks about this stuff, takes about a half hour and it may be useful.
[bold]David Simpson, PE[/bold]
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
It is actually a kinematics problem. The fluid will experience explosive decompression (which is unbounded by choked flow) for something like 20-50 mS (0.02-0.05 S). After explosive decompression completes, the fluid velocity is limited by the speed of sound and the plug will have outrun any exiting fluid, so you only need to be concerned with the first 50 mS (to be conservative).
F=ΔP*A
and
F=m*a
so
a=(ΔP&A)/m
and
v=a*t=(ΔP&A)/m * 0.05S
Put in the cross sectional area of the plug and its mass and get the units right and you have the velocity at the end of the explosive acceleration. From that point it is going to fall under the influence of gravity and slow due to wind resistance. This calculation works for any length pneumatic test. For a hydrostatic test the duration of the explosive decompression falls rapidly with decreasing volume of the test so it is only valid for tests over about 100 ft long.
I've got a narrated presentation on my web page that talks about this stuff, takes about a half hour and it may be useful.
[bold]David Simpson, PE[/bold]
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
- Thread starter
- #6
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1
- #7
There were no empirical equations in this calculation so use any units you want, just make sure that they cancel properly (I am not going to presume to dictate units to anyone who uses '2" NPT' and "1650 bar" in the same sentence--you may want velocity in furlong per fortnight for all I know).
A --> Area (length squared)
a --> acceleration (length per unit (time squared))
f --> force
ΔP --> differential pressure (just use gauge pressure here)
m --> mass
v --> velocity (length per unit time)
[bold]David Simpson, PE[/bold]
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
A --> Area (length squared)
a --> acceleration (length per unit (time squared))
f --> force
ΔP --> differential pressure (just use gauge pressure here)
m --> mass
v --> velocity (length per unit time)
[bold]David Simpson, PE[/bold]
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
To simplify and make some assumptions such as no friction being present:
Area of 2-Inch plug = 2-Inch NPT = pi*r2 = 3.14 in2
Force = 3.14 in2 * 1650 bar = 3.14 * 23,931 lbs/in2 = 75,140 lbs = 334,200 Newtons
2-Inch pipe plug weight = 0.36 lbs = 0.617 kg-f
Mass of pipe plug (corrected) = 0.617 kg-f/ 9.8 = 0.063 kg
Since you know that F = ma, F = N and m = mass in kg, a = acceleration in m/s2
334,200 = 0.063 *(a)
a = 5,310,000 m/s2
In physics problems, when given the magnitude of a force, the mass of an object on which that force has been acting, and the time in seconds that has elapsed since the force began acting on the object, which is presumed to be at rest initially. To solve such a problem, one needs to access the basic equations of motion in mathematical physics, specifically, the one that states:
v = v0 + at
where v is velocity at time t.
Substitute the acceleration into the kinematic equation, with the initial velocity v0 being equal to zero and assuming the plug has traveled 1 meter:
v = v0 + at
v = 0 + (5,310,000 m/s2)(0.0005 s)
v = 2,655 m/s
As a point of reference, military bomb particle velocities from TNT explosions are in the range of 2,000 to 8,000 ft/sec.
Area of 2-Inch plug = 2-Inch NPT = pi*r2 = 3.14 in2
Force = 3.14 in2 * 1650 bar = 3.14 * 23,931 lbs/in2 = 75,140 lbs = 334,200 Newtons
2-Inch pipe plug weight = 0.36 lbs = 0.617 kg-f
Mass of pipe plug (corrected) = 0.617 kg-f/ 9.8 = 0.063 kg
Since you know that F = ma, F = N and m = mass in kg, a = acceleration in m/s2
334,200 = 0.063 *(a)
a = 5,310,000 m/s2
In physics problems, when given the magnitude of a force, the mass of an object on which that force has been acting, and the time in seconds that has elapsed since the force began acting on the object, which is presumed to be at rest initially. To solve such a problem, one needs to access the basic equations of motion in mathematical physics, specifically, the one that states:
v = v0 + at
where v is velocity at time t.
Substitute the acceleration into the kinematic equation, with the initial velocity v0 being equal to zero and assuming the plug has traveled 1 meter:
v = v0 + at
v = 0 + (5,310,000 m/s2)(0.0005 s)
v = 2,655 m/s
As a point of reference, military bomb particle velocities from TNT explosions are in the range of 2,000 to 8,000 ft/sec.
bimr,
"lbs" is not a unit. What I hope you meant was "3.14 * 23,931 lbf/in2" and "2-Inch pipe plug weight = 0.36 lbf so, m=0.36 lbf *gc=0.163 kg". You needed to be a lot more careful with your units and conversions. The 10 ft assumption is pretty shaky, I've found a lot of data that puts the duration of an explosive decompression between 20 mS and 50 mS. 4 mS is really short.
Chemical explosions are limited by the burn rate through the explosive material and numbers upwards of 10,000 m/s are reasonably common.
(see Wiki ), but explosive decompression works under a different set of constraints (e.g., no flame front), and I can't find a reference to a maximum projectile velocity.
Running through the math with the correct mass and 50 mS, I get 102,000 m/s at 50 mS, but I was hoping that the OP would do the arithmetic himself.
[bold]David Simpson, PE[/bold]
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
"lbs" is not a unit. What I hope you meant was "3.14 * 23,931 lbf/in2" and "2-Inch pipe plug weight = 0.36 lbf so, m=0.36 lbf *gc=0.163 kg". You needed to be a lot more careful with your units and conversions. The 10 ft assumption is pretty shaky, I've found a lot of data that puts the duration of an explosive decompression between 20 mS and 50 mS. 4 mS is really short.
Chemical explosions are limited by the burn rate through the explosive material and numbers upwards of 10,000 m/s are reasonably common.
(see Wiki ), but explosive decompression works under a different set of constraints (e.g., no flame front), and I can't find a reference to a maximum projectile velocity.
Running through the math with the correct mass and 50 mS, I get 102,000 m/s at 50 mS, but I was hoping that the OP would do the arithmetic himself.
[bold]David Simpson, PE[/bold]
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
zdas04 - I watched the video you posted earlier and I do not understand why on slide 9 the energy released by a failure of the hydrotest increases with the pipeline length that is outside of the explosive decompression zone. Is this because all of the additional test pressure water mass is located within the 50ms zone?
- Thread starter
- #11
LittleInch
Petroleum
sbatch20
whilst it is quite fun to try and work out what is going to happen, I think there needs to be some greater thought here.
I have no idea what sort of vessel you have which has a test pressure of 1650 bar so presumably a MAWP of 1500 bar?? What is a vessel of that pressure doing with a 2 inch screwed plug??
I can't begin to think how deep the plug needs to be to withstand 33 tonnes of force. - That's basically like hanging a large articulated truck off a 2" connection.
Don't neglect the massive force that would be applied to the vessel in the opposite direction caused by high density air at sonic velocity through a 2" hole and also the reaction force of the vessel if this plug ejects out of one side. The jet force will continue for some time, though slowly reducing depending on your volume. This might well be a bigger problem.
If you're still going to do this and essentially wrap the vessel in a further very thick steel sheet 1m away then you might be able to reduce force a little, but these types of calculations above are just to get you into an order of magnitude. You will need to study this further using some sort of transient study on ballistics.
Oh and make sure you tell everyone here so we can be some distance away....
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
whilst it is quite fun to try and work out what is going to happen, I think there needs to be some greater thought here.
I have no idea what sort of vessel you have which has a test pressure of 1650 bar so presumably a MAWP of 1500 bar?? What is a vessel of that pressure doing with a 2 inch screwed plug??
I can't begin to think how deep the plug needs to be to withstand 33 tonnes of force. - That's basically like hanging a large articulated truck off a 2" connection.
Don't neglect the massive force that would be applied to the vessel in the opposite direction caused by high density air at sonic velocity through a 2" hole and also the reaction force of the vessel if this plug ejects out of one side. The jet force will continue for some time, though slowly reducing depending on your volume. This might well be a bigger problem.
If you're still going to do this and essentially wrap the vessel in a further very thick steel sheet 1m away then you might be able to reduce force a little, but these types of calculations above are just to get you into an order of magnitude. You will need to study this further using some sort of transient study on ballistics.
Oh and make sure you tell everyone here so we can be some distance away....
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
- Thread starter
- #13
OK thanks again. I understand your points, this is more of an attempt to get a calculation sheet created and determine some maximum plug sizes and pressures.
If I now know the velocity along with the mass and diameter of the missile is there a simple way of calculating the thickness of containment of say a mild steel sheet 1 meter away?
If I now know the velocity along with the mass and diameter of the missile is there a simple way of calculating the thickness of containment of say a mild steel sheet 1 meter away?
jojoyohan,
Yes, for all but the longest tests, the entire added mass can be expected to be ejected during the explosive decompression period and when it is gone the explosion is over.
sbatch20,
The distance (S) is the integral of the velocity equation.
S=(1/2)a*t2,
so if you put your barrier 1 m from the plug, you can calculate how long it will take the plug to reach that point, then use that time in the above equations to get velocity. Then you can estimate a deceleration duration (that number is simply a guess) to get a force on the barrier.
[bold]David Simpson, PE[/bold]
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
Yes, for all but the longest tests, the entire added mass can be expected to be ejected during the explosive decompression period and when it is gone the explosion is over.
sbatch20,
The distance (S) is the integral of the velocity equation.
S=(1/2)a*t2,
so if you put your barrier 1 m from the plug, you can calculate how long it will take the plug to reach that point, then use that time in the above equations to get velocity. Then you can estimate a deceleration duration (that number is simply a guess) to get a force on the barrier.
[bold]David Simpson, PE[/bold]
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
LittleInch
Petroleum
You are into ballistics issues here - allowable deformation of the plate, impact velocity, mass etc.
A lot of armour plating stuff now is actually a composite material designed to absorb the energy instead of just being x mm of steel.
I'm still very dubious about having any size or type of plug or screwed connection here.
Will the vessel always have this barrier around it in service?
Why are you even looking at a screwed connection for such a high pressure vessel??
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
A lot of armour plating stuff now is actually a composite material designed to absorb the energy instead of just being x mm of steel.
I'm still very dubious about having any size or type of plug or screwed connection here.
Will the vessel always have this barrier around it in service?
Why are you even looking at a screwed connection for such a high pressure vessel??
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
- Thread starter
- #16
We are just looking at test plugs, not service items. In reality the size pressure will not go to the extremes detailed, I am just looking for a way to calculate a containment shield thickness given the velocity, mass and diameter of an object so we can set limitations.
Looking in more detail it seems a hyrdo test chamber is usually a mix of steel, composite and concrete block construction rather than solely a steel plate.
Looking in more detail it seems a hyrdo test chamber is usually a mix of steel, composite and concrete block construction rather than solely a steel plate.
When I have something like a threaded fill pipe in a blind flange I use sand bags instead of any kind of metal. The bags absorb a lot of energy and tend to convert the collision towards inelastic. The more rigid the barrier, the higher the chance of fully elastic collisions that result in shrapnel heading in unpredictable directions.
[bold]David Simpson, PE[/bold]
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
[bold]David Simpson, PE[/bold]
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
Thanks for checking my calculations. However, there appears to be a major error in your calculation. Units maybe?
If something is moving at "I get 102,000 m/s", it should be somewhat obvious that the particle would take just 0.009 mS to move 1 meter.
Rather than wasting your time fixing calculations, you might consider becoming an advisor to NASA or Elon Musk because at the speed of 228,000 miles per hour, you might be able to put that pipe plug into earth orbit.
If something is moving at "I get 102,000 m/s", it should be somewhat obvious that the particle would take just 0.009 mS to move 1 meter.
Rather than wasting your time fixing calculations, you might consider becoming an advisor to NASA or Elon Musk because at the speed of 228,000 miles per hour, you might be able to put that pipe plug into earth orbit.
sbatch20 (Mechanical) (OP) said:
The speed of the pipe plug would have to determined empirically as there is no method to calculate the forces that would be resisting the pipe plug as it is ejected.
The 0.5mS time would mean the plug is traveling at 2,000 m/s. Unless you come up with empirical data, I don't think anyone would object to that velocity as it is in the range of velocities from military bomb explosions.
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