Internal forces acting on a dive system

In summary: There is no calculation for a diver exiting Bell and diving into EL, but the differential pressure between them would presumably be at the flange.
  • #1
Boyles
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TL;DR Summary
trying to find the answer to a query on the force acting on a leak at a bell trunk.

https://www.longstreath.com/community/index.html/articles/stena-workhorse-fire-r371/
Hello please be aware that my understanding of physics is basic, but I want to learn - and I will if explained simply enough please. I looked on here and found the cube in space query from last year, and I just about understood it, but the scalars and vectors did throw me, so hopefully it will be more understandable.

There was a fire many years ago on the dive vessel Stena Workhorse. When they retrieved the diving bell out of the water to lock it on to the dive system chamber the clamp that holds the bell on was distorted by the fire and wouldn't close. I believe the depth was 50 msw (5 barg), and it was a top mating bell (bell door is on bottom and sits on top of system). Apparently the weight (unknown but likely circa 6 t) of the bell was enough for the trunk (tunnel between bell and system) to be pressurised to 50 msw to allow divers to transfer. I think that this must be an incorrect, and the clamp did work because the force acting on the seal at the mating face flange would not be overcome by the bell weight.

It would be an 800 mm diameter trunk, so initially I worked out the pressure acting on the closed bell door would be 5024 cm2 x 1.1 kg x 5 bar = 27 t. Far in excess of the potential weight of a bell therefore it would not stay on. 1. Is that the correct calculation?

Then theoretically I thought if the trunk was at the same pressure as the bell (and system) at 5 barg there is no pressure differential on the 800 mm door therefore the differential is at the flange. So
2. So would it be the area of the internal flange line that I should use to calculate the force?
3. Then I thought if the door is open is it the area of the internal bell that I should use?

I'm sure in all likelihood all 3 suggestions are incorrect (particularly 3.) but an explanation would be most helpful.

Thanks,

John
 
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  • #2
Welcome to PF.

That is certainly an unusual scenario you present. It is good that the divers were able to transfer successfully.

I don't fully understand the nature of your question. Are you trying to figure out why the clamp did not close? Is there a lawsuit possible? Do you think the docking mechanism was damaged by the fire? Are you responsible for changing the design or doing repairs? These questions relate to how detailed calculations need to be and what sources you may need to consult.

Do you have drawings of the docking portions of the chamber and the bell? You can use the Attach files button to include pictures in your post.
 
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  • #3
Thank you for replying.

Sorry my question(s) are not clear enough.

No I am not wanting to know why the clamp did not close as it is reported (see link to report) the clamp ("docking mechanism") did not close due to fire distortion. Although I believe it must have closed to allow the divers to transfer.

That is my question in a nutshell; was the internal force on the mating flange too much for the Bell weight alone to overcome? No a lawsuit is not possible as it was in the 70s and it was sorted then. No I was not involved in changing design/repairs after the incident it is a purely theoretical question. If you ignore the actual real life incident the question basically is what is the force acting on the flange mating face and how is it calculated?
 
  • #4
Sorry there are no drawings. It would be a 800 diameter trunk, likely 1000 mm long. The bell would be approx spherical about 2000 mm diameter, and the part of the system the trunk is bolted on to (various names mostly Entry Lock -EL or TUP) would be similar in dimension to the bell, and the EL is attached to the rest of the system but hopefully with the doors closed. The trunk is at atmospheric pressure, the bell and entry lock both have 800 mm doors sealed with 5 bar of pressure. The bell "mates" onto the trunk flange mating face , the clamp should close and then the Trubk can slowly be pressurised to the same depth as EL and Bell, the doors can open and the divers can transfer from Bell to EL. I hope this explains it?
 
  • #5
If I understand your question, the mating flange is the correct 'target' for your analysis. I get numbers comparable to yours. From the linked story, it sounds as if the divers were briefly subjected to reduced pressure - they would have been able to re-pressurize after getting into the EL. They apparently weren't 'bent' long enough to do any serious damage. Lucky.
 
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  • #6
Very interesting tale; a great good fortune that no one was injured. I am struck by the amount of colorful descriptive jargon in the article...hadn't heard much of it before.
I believe I understand your question: it has to do with how the force at the flange is related to the internal pressure, in particular do you use the flat area of the flange opening or instead the internal area of the bell itself. The short answer is either one gives the same answer, but you need be aware that pressure always pushes outward and you are only interested in the force upward. For the plane of the flange interface, all of the pressure is directed upward and so the calculation is simpler (and I believe you did it correctly). To do the calculation on the walls you would need to include the angle of each part of the wall relative to the vertical...the area is bigger but the angles reduce the upward effect of the pressure and the result is exactly the same (there are theorems that say this!).
Hope that helps. Saturation diving is clearly not for the faint of heart.
 
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  • #7
Thank you Dullard, do you think that the bell would sit on the flange, as gas pressure was added was added to the trunk? I think I worked out it would at 5 msw (0.5 barg). So I can't imagine that the leak would be overcome enough by pressurising to overcome the leak (constantly adding enough to overcome what is leaving and get to 5 bar pressure to allow the bell and EL doors to open at equal pressure). So I suspect the clamps did close (although there may have been a leak as the o-ring may have burned away and perhaps there may not have been a replacement).

Thank you hutchphd. If for some reason the trunk had held pressure (lets say it was actually 5 msw, and the force I worked out was equal to or slightly less than the weight and there was an o-ring in place), Because there is now total equalised pressure (imagine a totally hollow barbell subject to *internal pressure and the weak spot is the *joined band in the middle of the handle where it was badly welded together); how do I work out the force acting on that weak spot welded band?

*edited to be clearer
 
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  • #8
If I understand the geometry the total longitudinal (down the handle) force will be the (pressure) times (the "handle" interior cross section Area) . This will create tensile stress on the weld. Of course welds are brittle and so that stress may not distribute evenly around the weld circumference so the rest of the estimate is way beyond my experience but there will be stress concentrations to worry about.
There will also be some tensile stress circumferentially in the tube but that is unlikely to be an issue I think.
Please be aware that I am a pretty good physicist but a very bad welder.
 
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  • #9
Boyles said:
...
That is my question in a nutshell; was the internal force on the mating flange too much for the Bell weight alone to overcome? ... the question basically is what is the force acting on the flange mating face and how is it calculated?
If the unbolted mating surfaces could be considered flat flanges, then, the weight of the bell needs to be converted to pressure over the area of the annular mating surfaces (with or without a gasket) for the proper calculation.

If that pressure was higher than the 5 bars existing outside the chamber, then little or no leak could have happened.

Please, see:
http://www.mechanicalengineeringsite.com/gaskets-basics-types-working-principle-calculation-bolt-load/

https://www.pveng.com/wp-content/uploads/2016/06/Loads_On_Flanges.pdf

The weight of the bell may have been the only force required to keep the flange sealed against the in-out pressure differential.
 
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  • #10
The submerged weight required to seat the bell must be greater than the total area of the opening times the pressure differential. Not the area of some annular ring.
The size of the ring determine s stress on welds or bolts and gaskets, but that is not the question.
I believe your post is not correct.
 
  • #11
Lnewqban said:
If the unbolted mating surfaces could be considered flat flanges, then, the weight of the bell needs to be converted to pressure over the area of the annular mating surfaces (with or without a gasket) for the proper calculation.
The submerged weight required to seat the bell must be greater than the total area of the opening times the pressure differential. Not the area of some annular ring.
The size of the ring determine s stress on welds or bolts and gaskets, but that is not the question.
I believe your post is not correct.
 
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  • #12
From the provided link,
With the bell now out of the water, the through water comms inoperable in air signs were held up to the viewport of the bell, telling the divers inside not to open the bottom hatch. They apparently did not see this and opened the hatch and dropped down into the TUP and accessed the saturation chamber.
Since the hatch opens inwards, if there was pressure differential, the divers might not have been able to lift and open it.

Boyles said:
Summary:: trying to find the answer to a query on the force acting on a leak at a bell trunk.

believe the depth was 50 msw (5 barg), and it was a top mating bell (bell door is on bottom and sits on top of system). Apparently the weight (unknown but likely circa 6 t) of the bell
Could the actual depth and pressure have been less?
Also, if the umbilical was severed, could some of the internal pressure have escaped from the bell before valves were sealed and then be less than depth pressure?

Is the 6t the weight of the bell - the added cage and equipment would add to the weight if not removed before mating, maybe another 10t ( ? just a guess ) additional - though still not up to the the calculated weight.

Interesting forensic on a past event, with sketchy information.👍
Makes one wonder how investigators do really finally piece together a report on an incident.
 
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  • #13
Thank you. Yes a lot of unknowns with the original story. The witness is a photographer and recalling something that he witnessed 35 years ago, so undoubtedly some of the facts may be wrong or not known.

I believe the clamps holding the mating flanges together must have been closed. Nowadays the clamp would be secured closed by a safety interlock, perhaps a similar device was what was distorted by the heat.

But it is really the theory of the forces I was wondering about.

Is it initially the surface area of the bell closed door the force (weight) the pressure is acting upon?
If equalised (equal pressure in the trunk, bell and system) what is the force acting on now, the internal area of the bell sphere? When equalised does the bell or system doors being closed affect this force?
 

Related to Internal forces acting on a dive system

1. What are internal forces?

Internal forces are forces that act within a system, such as a dive system, and are caused by interactions between the components of the system. These forces can include tension, compression, shear, and bending.

2. How do internal forces affect a dive system?

Internal forces can affect a dive system in several ways. They can determine the strength and stability of the system, as well as its ability to withstand external forces. Internal forces can also cause deformation or failure of the system if they exceed its structural limits.

3. What is the difference between internal and external forces?

External forces act on a system from outside, while internal forces act within the system. External forces are typically caused by the environment or other external factors, while internal forces are a result of interactions between components within the system.

4. How can internal forces be calculated in a dive system?

Internal forces can be calculated using principles of mechanics, such as Newton's laws of motion and the equations of statics. These calculations can help determine the magnitude and direction of internal forces within a dive system.

5. How can internal forces be minimized in a dive system?

Internal forces can be minimized by designing the system to distribute loads evenly and efficiently. This can be achieved through proper material selection, structural design, and regular maintenance to ensure the system is in good condition. Additionally, following safety guidelines and using proper techniques while diving can also help minimize internal forces on the system.

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