- #1

Boyles

- 5

- 0

- 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

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