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Tube Rupture 2
- Thread starter hnaik
- Start date
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1
- #2
Harshal,
Crane TP 410 is one of the world's best references on basic fluid flow. It does not contain an equation or example of a tube rupture scenario. What you have to do is evaluate your tube rupture scenario, use good engineering judgement, and put the pertinent parameters of that specific case into the form of a basic fluid flow problem so you might use Crane TP 410 to solve the problem.
The key is good engineering judgement. That comes from experience. The best way I know to get experience is roll up your sleeves and dig into the problem. Think! Don't sit back and expect the answer to be handed to you.
Since you did not give us any data, I will infer from your wording that MAWPtube side > MAWPshell side. What's the highest expected or allowed pressure on the tube side for your scenario? MAWPtube side x 1.1? What's the highest expected or allowed pressure on the shell side for your scenario? MAWPshell side x 1.1? By golly, I think we have a delta P now! dP = 1.1 x (MAWPtube side - MAWPshell side). Now, how do you expect the rupture to occur? Is the operating pressure of the tube side slightly or moderately greater than the operating pressure of the shell side? Or, is this difference in operating pressure huge? Is there a corrosive environment on either side? Do you expect large pressure transients? These all play a part in trying to predict the most likely failure mode, i.e. will a pin hole leak develop, or will the tube shear in two pieces in a catastrophic failure?
If operating pressure differences are not so big and there is no corrosion on either side, I'd lean towards a pin hole leak, say like a 1/8” diameter hole in the tube. But if mechanisms are in place such that a catastrophic failure is possible, like stress corrosion cracking in chloride service, then you have to plan on flow coming out both ends of the severed tube into the shell.
Don’t forget to look at your conditions to see if two-phase flow will occur or not. Without any information on your application, I don’t know if it is or not, but you need to know.
Good luck,
Latexman
Crane TP 410 is one of the world's best references on basic fluid flow. It does not contain an equation or example of a tube rupture scenario. What you have to do is evaluate your tube rupture scenario, use good engineering judgement, and put the pertinent parameters of that specific case into the form of a basic fluid flow problem so you might use Crane TP 410 to solve the problem.
The key is good engineering judgement. That comes from experience. The best way I know to get experience is roll up your sleeves and dig into the problem. Think! Don't sit back and expect the answer to be handed to you.
Since you did not give us any data, I will infer from your wording that MAWPtube side > MAWPshell side. What's the highest expected or allowed pressure on the tube side for your scenario? MAWPtube side x 1.1? What's the highest expected or allowed pressure on the shell side for your scenario? MAWPshell side x 1.1? By golly, I think we have a delta P now! dP = 1.1 x (MAWPtube side - MAWPshell side). Now, how do you expect the rupture to occur? Is the operating pressure of the tube side slightly or moderately greater than the operating pressure of the shell side? Or, is this difference in operating pressure huge? Is there a corrosive environment on either side? Do you expect large pressure transients? These all play a part in trying to predict the most likely failure mode, i.e. will a pin hole leak develop, or will the tube shear in two pieces in a catastrophic failure?
If operating pressure differences are not so big and there is no corrosion on either side, I'd lean towards a pin hole leak, say like a 1/8” diameter hole in the tube. But if mechanisms are in place such that a catastrophic failure is possible, like stress corrosion cracking in chloride service, then you have to plan on flow coming out both ends of the severed tube into the shell.
Don’t forget to look at your conditions to see if two-phase flow will occur or not. Without any information on your application, I don’t know if it is or not, but you need to know.
Good luck,
Latexman
- Thread starter
- #3
Thank you Latexman for your quick reply and sorry for not providing much information.
MAWP Tube side (cooling water): 150 psig and operating 135 psig
MAWP Shell Side: 65 psig and operating around 52 psig
So, i think catastrophic failure is a possible scenario.
If i am not wrong Equation 3-21 (Crane, page 3-5 and i am sure you have one) are applicable to find flow during tube rupture case (for compressible flow), but i dont know how to find flow coefficient "C". TUBE size is 1" with 14 BWG.
So can you plz guide me how to find flow?
To find out weather two-phase flow occur or not, i just need to flash it at adiabatic condition, am i right?
Thank you
MAWP Tube side (cooling water): 150 psig and operating 135 psig
MAWP Shell Side: 65 psig and operating around 52 psig
So, i think catastrophic failure is a possible scenario.
If i am not wrong Equation 3-21 (Crane, page 3-5 and i am sure you have one) are applicable to find flow during tube rupture case (for compressible flow), but i dont know how to find flow coefficient "C". TUBE size is 1" with 14 BWG.
So can you plz guide me how to find flow?
To find out weather two-phase flow occur or not, i just need to flash it at adiabatic condition, am i right?
Thank you
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1
- #5
If you have predetermined that a heat exchanger tube failure is a credible scenario then the more common thinking, which is short and sweet, is to consider the tube failure as nothing more than an orifice flow exercise. You take the flow through two orifices, one for each end of tube break, using the diameter as the tube ID. For the orifice coefficient, you will probably end up somewhere around 0.6 but check against Crane TP410, Page A-20.
But then the calculation becomes tricky because you have to check for two-phase flow. You can perform an adiabtic flash across the tube break to determine if this is indeed your scenario. Or, this may not be two-phase exiting the tubes but you can still have a two-phase in your relief device after your cooling water mixes with whatever is in your shell.
But then the calculation becomes tricky because you have to check for two-phase flow. You can perform an adiabtic flash across the tube break to determine if this is indeed your scenario. Or, this may not be two-phase exiting the tubes but you can still have a two-phase in your relief device after your cooling water mixes with whatever is in your shell.
If the cooling water remains liquid once in the shell side there will not be much energy ( [∫]PdV ) available to tear/shear the tube further open/apart immediately following the loss of containment. This is the same reason that hydrotests are inherently safer than pnuematic pressure tests, all else being equal. I have seen some cases where a little more effort saved a lot of money over the life of a relief system. This may not be one of them, we don't even know if this is the controlling scenario, but all that is needed are the temperatures (in and out) on both sides and a knowledge of the fluid on the shell side. Since this data should be readily available anyway, why not consider it?
Good luck,
Latexman
Good luck,
Latexman
- Thread starter
- #7
Thank you Pleckner and Latexman,
PLECKNER,
My tube ID is .782" and Shell ID is 58". But i am not able to figure it out how to get C value. Either it has some trick or i am not able to figure out simple way to read graph. If you can guide me i will really appreciate.
Latexman,
Shell sibe: Benzene and Toluene, Overhead vapor from Column, Tin=349 °F and P=15.76 psig. Tout=135.45 °F
Tube side: Cooling water: 135 psig and as inlet temp is not available i have assumed it as 80 °F.
I think, tube rupture will not be a controlling case (or may not be a valid bz of fill out time, but i am not able to figure out the way to calculate flow), as reflux failure is controlling.
Thank you
PLECKNER,
My tube ID is .782" and Shell ID is 58". But i am not able to figure it out how to get C value. Either it has some trick or i am not able to figure out simple way to read graph. If you can guide me i will really appreciate.
Latexman,
Shell sibe: Benzene and Toluene, Overhead vapor from Column, Tin=349 °F and P=15.76 psig. Tout=135.45 °F
Tube side: Cooling water: 135 psig and as inlet temp is not available i have assumed it as 80 °F.
I think, tube rupture will not be a controlling case (or may not be a valid bz of fill out time, but i am not able to figure out the way to calculate flow), as reflux failure is controlling.
Thank you
- Thread starter
- #10
- Thread starter
- #13
Thank you Pleckner and Latexman,
Pleckner,
If i m not misunderstanding it, than beta should be high (in our case isn't it should be equal to 1 (d2=d1), technically its not orifice but pipe exit).
let me know if i m getting it wrong and sorry if i m going in wrong direction.
Latexman,
I will look at it when ever i get perry.
Thank you
Pleckner,
If i m not misunderstanding it, than beta should be high (in our case isn't it should be equal to 1 (d2=d1), technically its not orifice but pipe exit).
let me know if i m getting it wrong and sorry if i m going in wrong direction.
Latexman,
I will look at it when ever i get perry.
Thank you
For an orifice calculation to apply the "plate thickness" < dorifice/8. In pleckner's model, I believe "plate thickness" = tubesheet thickness. So, is the tubesheet thickness < 0.1" (0.782"/8)? I doubt it is.
In this scenario, I normally assume the break occurs at one tubesheet. Two simultaneous breaks seems unlikely to me. I'd model it as two pipes or tubes, same thing. One is the length of the tubes - tubesheet thickness with an inlet loss and an outlet loss. The other is the length of the tubesheet thickness with an inlet loss and an outlet loss.
Good luck,
Latexman
In this scenario, I normally assume the break occurs at one tubesheet. Two simultaneous breaks seems unlikely to me. I'd model it as two pipes or tubes, same thing. One is the length of the tubes - tubesheet thickness with an inlet loss and an outlet loss. The other is the length of the tubesheet thickness with an inlet loss and an outlet loss.
Good luck,
Latexman
The model is not for two simultaneous breaks. It is one break so the flow comes out of (or into) two ends of the tubes.
I found a quide from a previous employer I worked for and a couple of articles. There is not 100% consenses on how to calculate the flow. One gives the orifice equation (see CRANE TP410 equation 3-21, Page 3-5) with a C of 0.61. I have one source that says to use a C of 0.7. Another source says to use the Darcy equation, see CRANE TP410 equation 3-19 , Page 3-4 (times 2 for both ends of the broken tube). This equation uses the entrance/exit loss rather than an orifice coefficient.
When working for the other empolyer, I used their guide (CRANE equation 3-19). Since then and in the very rare times I've needed to do this calculation, I've used the orifice equation with a C of 0.61. Basically it doesn't matter because this scenario has never been controlling in anything I've ever done. And besides, we make good and sure this calculation is never needed by designing our heat exchangers to eliminate this as a credible scenario in the first place.
One thing to note is that the two equations don't differ except between the value of 'K' and 'C'. In equation 3-19, 'K' is in the denominator of the radical. In equation 3-21, the orifice equation, 'C' is outside the radical. One can easily convert from one to the other to see the impact of using 'C' or 'K' and use the most conservative.
I found a quide from a previous employer I worked for and a couple of articles. There is not 100% consenses on how to calculate the flow. One gives the orifice equation (see CRANE TP410 equation 3-21, Page 3-5) with a C of 0.61. I have one source that says to use a C of 0.7. Another source says to use the Darcy equation, see CRANE TP410 equation 3-19 , Page 3-4 (times 2 for both ends of the broken tube). This equation uses the entrance/exit loss rather than an orifice coefficient.
When working for the other empolyer, I used their guide (CRANE equation 3-19). Since then and in the very rare times I've needed to do this calculation, I've used the orifice equation with a C of 0.61. Basically it doesn't matter because this scenario has never been controlling in anything I've ever done. And besides, we make good and sure this calculation is never needed by designing our heat exchangers to eliminate this as a credible scenario in the first place.
One thing to note is that the two equations don't differ except between the value of 'K' and 'C'. In equation 3-19, 'K' is in the denominator of the radical. In equation 3-21, the orifice equation, 'C' is outside the radical. One can easily convert from one to the other to see the impact of using 'C' or 'K' and use the most conservative.
Phil,
I totally misunderstood the tube break in your method. Thanks for the clarification!
It's comforting to see my company's method is used by other companies you worked for. The cases where I found tube rupture to be controlling were on really old heat exchangers. Today we usually take care that the design mitigates this scenario, unless that gets cost prohibitive.
Good luck,
Latexman
I totally misunderstood the tube break in your method. Thanks for the clarification!
It's comforting to see my company's method is used by other companies you worked for. The cases where I found tube rupture to be controlling were on really old heat exchangers. Today we usually take care that the design mitigates this scenario, unless that gets cost prohibitive.
Good luck,
Latexman
- Thread starter
- #18
you should also ask yourself if this is a double jeopardy issue.
since we essentially never close cooling water unless the equipment is being isolated for service, we would say that having the cooling water flow closed at the same time as a tube rupture is unlikely.
if the process path can relieve the cooling water flow based on a tube rupture, that also makes it unlikely.
for the relief case to be valid, you would need the following type of sequence of events:
1. process side isolated
2. cooling water isolated
3. cooling water tube rupture
this would appear to be a rather remote possibility, imho.
since we essentially never close cooling water unless the equipment is being isolated for service, we would say that having the cooling water flow closed at the same time as a tube rupture is unlikely.
if the process path can relieve the cooling water flow based on a tube rupture, that also makes it unlikely.
for the relief case to be valid, you would need the following type of sequence of events:
1. process side isolated
2. cooling water isolated
3. cooling water tube rupture
this would appear to be a rather remote possibility, imho.
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