Pressure underwater and equivalent weight

[Katpar]

Hello,

I am trying to calculate the equivalent amount of weight that a submarine would be under at 1000m below the sea. Apologies if this isn't that well explained!

At 1000m - 101 atmospheres - which is 14.5 psi per atmosphere - so at 1000m it would be 1464.5 psi.

Do I then multiple 1464.5 by the entire surface area of the submarine or just the top part that is facing the surface?

I am aiming to try and get a figure that will enable me to say 'at 1000m deep the submarine is supporting the equivalent weight of X jumbo jets / cars etc'

Any help would be so much appreciated!

 

[Bystander]

Apologies if this isn't that well explained!

The "journalistic/pop-sci" approach isn't that well defined. Your body's surface area is 2500 -3000 square inches and under a load of (omigosh!) 30,000- 40,000 pounds just here at the Earth's surface --- and pearl divers triple and quadruple that load.
Fifteen hundred pounds per square inch is fifteen hundred pounds per square inch --- it is not a concept that requires translation.

 

[Chestermiller]

Hello,

I am trying to calculate the equivalent amount of weight that a submarine would be under at 1000m below the sea. Apologies if this isn't that well explained!

At 1000m - 101 atmospheres - which is 14.5 psi per atmosphere - so at 1000m it would be 1464.5 psi.

Do I then multiple 1464.5 by the entire surface area of the submarine or just the top part that is facing the surface?

Neither. Have you ever heard of Archimedes Principle?

 

[Katpar]

yes I have read a little - but I am not sure how to apply to what I am trying to do. It's quite confusing!

 

[sophiecentaur]

yes I have read a little - but I am not sure how to apply to what I am trying to do. It's quite confusing!

If you want to know the 'upthrust' then all you need to know is the density of the water, g and the volume i.e the Weight of water displaced.
The difficulty would be in knowing the actual volume of the submarine and the actual volume of the compressed water. Of course, if the submarine is not going up or down, the upthrust is equal to its weight.

 

[Katpar]

Hello,

Neither. Have you ever heard of Archimedes Principle?

If you are at 1000m doesn't the pressure mean that you are negatively buoyant?

 

[Chestermiller]

yes I have read a little - but I am not sure how to apply to what I am trying to do. It's quite confusing!

Please tell us your understanding of Archimedes Principle so that we can help you understand it better and apply it.

Actually, the pressure at 1000 m has very little to do with what you want to determine.

Chet

 

[russ_watters]

I suppose people might be interested in how much water is sitting on top of a sub, so you could multiply by the horizontal cross sectional area...

 

[nasu]

From the OP it looks like he is not interested in the buoyancy but rather the forces acting to "crush" the body of the submarine.
But it's not completely clear what he is after.

 

[sophiecentaur]

From the OP it looks like he is not interested in the buoyancy but rather the forces acting to "crush" the body of the submarine.
But it's not completely clear what he is after.

In which case, it should be pointed out to him that it isn't the 'total' of the forces acting that counts but the pressure on the weakest part of the hull that will determine whether the sub will be crushed or not.

 

[DaveC426913]

In which case, it should be pointed out to him that it isn't the 'total' of the forces acting that counts but the pressure on the weakest part of the hull that will determine whether the sub will be crushed or not.

While factually true, it does not get the OP what he is looking for.

 

[DaveC426913]

I am aiming to try and get a figure that will enable me to say 'at 1000m deep the submarine is supporting the equivalent weight of X jumbo jets / cars etc'

Avoid the phrase "supporting", since that suggests the pressure is from above. It is actually from all sides.

Do I then multiple 1464.5 by the entire surface area of the submarine or just the top part that is facing the surface?

You could say "resisting a crushing force of" - and yes, you'd multiply the surface area of the sub by 1464.

So, a 100ft sub with a 20ft diameter will be resisting (2x10⁶)*1464 pounds, or about 3200 kilotons. That's about 10,000 Jumbo jets.

 

[sophiecentaur]

While factually true, it does not get the OP what he is looking for.

Should we really be answering such questions as if they are meaningful? The jumbo jet thing is for popular journalists and not appropriate for PF. There is a difference between intensive and extensive variables.

 

[nasu]

If you have the sub on the ground and put a jumbo jet on top, there will be the weight of the jet pushing down and a normal force pushing up, with a magnitude equal to the weight of the jumbo jet. Would it be meaningful to say that the sub supports (or resists) the weight of 2 jumbo jets?

The result of multiplying pressure by area gives a meaningful force only of the surface is flat. Otherwise the changing direction of force should be taken into account.
But anyway, the crushing effect depends on pressure (stress) and not on net force.

It will be more accurate to compare the pressure on the sub with the pressure applied by x jumbo jets stacked on in top of the other, on the landing gear of the bottom one. Will be a cool picture to imagine.

 

[oara27]

Pascal's princple states that the pressure exerted by a fluid (water) is equal over the entire surface of the walls containing the fluid (in this case the sub hull keeping the water out). Since pressure = force/area, it's correct to multiply the pressure times the total area of the sub, to get the total force acting on the sub from outside. Obviously, the crushing effect of this force applies to all sides, so it would make more sense to calculate how much force is supported by a significant surface element, say a square foot, which would be 144 sq.in.*1464.5 lb/sq.in = 210,888 lb or 105.4 long tons. This is the weight each square foot of the sub needs to be able to support.105.4 tons is about the weight of a mid-size airliner, or 50 passenger cars. Compare this with the force usually supported by building floors of about 100 lb. per sq. foot. .

 

[Chestermiller]

Pascal's princple states that the pressure exerted by a fluid (water) is equal over the entire surface of the walls containing the fluid (in this case the sub hull keeping the water out). Since pressure = force/area, it's correct to multiply the pressure times the total area of the sub, to get the total force acting on the sub from outside.

This is not correct. At each location on the hull, the pressure acts perpendicular to the surface, so one must integrate the pressure over the surface area vectorially to get the total force acting on the sub hull. Also, Pascal's principle states that the pressure at a given location acts equally in all directions, not that it is equal over the entire surface of the walls containing the fluid. Some parts of the hull are at greater depth than others, so the pressure acting on the parts of the surface at greater depth is higher.

Chet

 

[sophiecentaur]

"one must integrate the pressure over the surface area vectorially to get the total force acting on the sub hull."
But what on Earth does that mean and how is it relevant to anything? It's just an arbitrary summing of values.

 

[nasu]

It will mean the buoyant force, won't it?. Calculated the hard way.

 

[Chestermiller]

It will mean the buoyant force, won't it?. Calculated the hard way.

Yes, exactly.

I should add that I didn't think that any of my post was relevant to the OP's question, unless he was interested in the buoyant force. The primary reason for my response in post #16 was just to correct the misinformation that the responder in post #15 had dispersed.

As far as issues associated with deformation and crushing of submarine hulls under external pressure are concerned, we have the computational modelling tools to handle such problems. But certainly, a simplistic analysis (such as in some of the posts in this thread) is not adequate. Crushing of a hull, in particular, is a complicated buckling problem that takes into account the stress-strain behavior of the hull.

Chet

 

[DaveC426913]

Should we really be answering such questions as if they are meaningful? The jumbo jet thing is for popular journalists and not appropriate for PF.

I confess, after doing the back-of-napkin calcs, and coming up with the answer of 10,000 jumbo jets, I must concur.

Suggesting a sub is withstanding the weight of 10,000 jumbo jets is going to do more harm than good. Saying every square inch must withstand the weight-equivalent of a VW Beetle is more meaningful.

 

[sophiecentaur]

It will mean the buoyant force, won't it?. Calculated the hard way.

oh yes. The force will be about the weight of the submarine because it will have been designed to be about neutral buoyancy. Simple Archimedes, in fact. The massive pressures and the force you get with simple scalar addition of forces are pretty meaningless, though. That has been my point.

 

[oara27]

This is not correct. At each location on the hull, the pressure acts perpendicular to the surface, so one must integrate the pressure over the surface area vectorially to get the total force acting on the sub hull. Also, Pascal's principle states that the pressure at a given location acts equally in all directions, not that it is equal over the entire surface of the walls containing the fluid. Some parts of the hull are at greater depth than others, so the pressure acting on the parts of the surface at greater depth is higher.

Chet

The original posting seems to imply that the sub is (vertically) stationary, so there is no acceleration and therefore the buoyant force is null. This means that the pressure is equal on all sides, although Pascal's principle may not be relevant to the problem.

However, the pressure resisted by the hull is not zero. I believe the question is about the force that this pressure represents on the hull structure, as I explained in my first answer.

 

[jbriggs444]

The original posting seems to imply that the sub is (vertically) stationary, so there is no acceleration and therefore the buoyant force is null. This means that the pressure is equal on all sides, although Pascal's principle may not be relevant to the problem.

If there is no acceleration, the buoyant force must be equal to the sub's weight. That is distinctly non-zero. That means that the pressure must not be equal on all sides.

 

[Chestermiller]

The original posting seems to imply that the sub is (vertically) stationary, so there is no acceleration and therefore the buoyant force is null. This means that the pressure is equal on all sides, although Pascal's principle may not be relevant to the problem.

However, the pressure resisted by the hull is not zero. I believe the question is about the force that this pressure represents on the hull structure, as I explained in my first answer.

In my judgement, based on over 50 year of practical experience in fluid mechanics, saying that the pressure on the hull times the surface area of the hull is something relevant to the structural mechanics of a submarine is both incorrect and misleading.

Chet

 

[Monsterboy]

I am trying to calculate the equivalent amount of weight that a submarine would be under at 1000m below the sea.

The OP wants to calculate the weight of the water above the submarine + weight of the submarine,which is equal to the total weight the submarine is resisting right ? This looks like you are not interested in the pressure just in the weight.By 'weight' does it refer to just the downward force ? If so first we need to calculate the volume of the water directly above the submarine times the density ( taking into account the changes in density with depth? Is it correct Chet? How can this be done ? Some kind of integration ? ?) multiplied by g which gives the weight of the water then just add it to the weight of the submarine ,Right ?