buoyancy again

slider201

Does anyone have any real scientific/experimental data to prove whether Archimedes' principle should be taken literally?

Let me explain: As I remember, Archimedes' principle goes something like upthrust is equal to the weight of the displaced fluid. Archimedes was saying that the force down was being overcome by the force up, which is proportional to the weight of the displaced fluid. Most take the easy route and calculate buoyancy from the standpoint of the displaced fluid.

I can calculate the buoyancy of an object from the standpoint of the displaced fluid, from the standpoint of the object, and from pressure-area calculations of down force versus up force. All three methods give the exact same result - IF the object is suspended in the surrounding fluid.

If the pressure-area method is the real correct scientific method to calculate buoyancy and if the lower surface area is isolated from the surrounding fluid, then what is buoyancy? What is correct? Do we take Archimedes' principle literally or ...?

Real example: running steel casing in a vertical oil well:
14644 feet of 88.2 lbs/ft casing with end surface areas of 25.3 sq in.
the pipe is otherwise smooth
The weight of pipe in air is 1,291,601 lbs, and its buoyant weight in 13.9 ppg drilling mud is 1,017,505 lbs. If the bottom surface area is somehow blocked from the surrounding fluid, what is the pipe's weight?

russ_watters

I don't know what you mean by "isolating the lower surface area from the surrounding fluid", but the answer to the basic question is yes, Archimedes principle works.

Doc Al

The real deal is always: What net force does the surrounding fluid exert on an object? As long as certain conditions are met--for example, the object is suspended in the surrounding fluid--then Archimedes's principle holds. For example, imagine a cube of metal under water. If it's completely surrounded by water, then the water pressure creates an upward force equal to the weight of the displaced water. But if the bottom of the block makes a water-tight seal with the bottom of the container--then the "buoyant" force becomes downward; Archimedes's doesn't apply. But in all cases what you want is the actual net force on the object; when it applies, Archimedes's principle is the easy--and correct--way to go.

slider201

Good answer. But there is the other school that claims that Archimedes' principle is absolute law - buoyancy is buoyancy - always there no matter what. I won't tell you which camp I am in. Are there papers, experiments, etc., that prove one or the other?

russ_watters

Every scientific theory has a domain of applicability. As Doc says, as long as you are in its domain, it holds.

slider201

I guess I need to explain my motive for this thread. In the oil well drilling industry, there are engineers very adamantly in one camp or the other. Camp #1 thinks the Archimedes' principle is absolute - buoyancy is always there no matter what. Camp #2 thinks more in terms of pressure-area and that an object is not buoyant if the bottom facing surface area(s) are not subject to the surrounding fluid's pressure. In the oil & gas industry, we have massive loads (tens of thousands of tons) where the load bearing gear is engineered to fit-for-purpose situations. Buoyancy plays a big factor. If buoyancy is always there per camp #1, lesser rated load bearing gear can be many $ cheaper. If we design for camp #1 but camp #2 is really true, then the load bearing gear may fail, costing many $. Also, if we design for camp #2 but camp #1 is really true, then we have wasted $ on over rated load bearing equipment.

Obviously I don't know any of your backgrounds, but I thought I would give your forum a shot.

arildno

If there is a side of an object not exposed to the incessant pounding of the fluid's molecules (i.e the physical mechanism producing pressure), then, trivially, there won't be any fluid pressure force acting upon that side.

Doc Al

There are no "schools" on this issue, at least not in physics. Buoyant force is found by adding up the force on each surface due to the surrounding fluid--period! That's where buoyant force comes from. In many cases, you can get the answer fast using Archimedes's principle--but only where it applies.

This is physics 101. If anyone disagrees, the burden of proof is on them to derive Archimedes's principle from scratch in the case of interest.
Camp #2 is right; Camp #1 is wrong. Why not set things up and actually measure the the buoyant force in the cases of interest?

slider201

DOC AL, thank you for your answers and time. I appreciate it.

It has been a few years since I had Physics 101, and I am currently about 150miles offshore, so I can't go to the library. It would be helpful to have a reference - textual from a book, the internet, etc. As you probably well know, old engineers know what they know and what they know is right no matter what. In other words, a preconception is hard to fight without a good reference.

In my initial example of the casing at 30000', the bottom surface area of the casing actually was inadvertently isolated from the drilling mud column and we did see the full non-buoyed weight. Our landing string approached/exceeded plastic yield. I could not convince camp #2 of this. I am in camp #1.

To experiment with loads at the water and well depths we have here would cost millions of $. Our rig rate runs about $25000/hr.

Thanks again for your help.

Doc Al

Right.
Do you have your camps mixed up? You had defined them:

"Camp #1 thinks the Archimedes' principle is absolute - buoyancy is always there no matter what. Camp #2 thinks more in terms of pressure-area and that an object is not buoyant if the bottom facing surface area(s) are not subject to the surrounding fluid's pressure."

I hope you are in camp #2!

arildno

Again, it must be stressed:
There are NO camps on this!
Consider the equation of motion when the fluid is at rest:
[tex]0=-\frac{1}{\rho}\nabla{p}-g\vec{k}[/tex]
where [itex]\rho[/itex] is the fluid density, p is the fluid pressure, g the acceleration of gravity, and [itex]\vec{k}[/itex] is a unit vector anti-parallell to the direction of gravity.
Thus, we get the rewriting for the pressure gradient:
[tex]\nabla{p}=-\rho{g}\vec{k}[/tex]

Now, suppose we have an object O wholly submerged in the water, and yet a part of its surface S not in contact with the fluid. We call this part of the surface M, and the part of S in contact with the fluid S/M. The volume of O is V

The fluid pressure force on O is now given by the integral:
[tex]\vec{F}=-\int_{S/M}p\vec{n}dS[/tex]
We now define a pseudo-pressure function p* on V and S which satisfies p*(S/M)=p(S/M), and [itex]\nabla{p}*=\nabla{p}=-\rho{g}\vec{k}[/itex]
Thus, we have:
[tex]\vec{F}=-\int_{S/M}p\vec{n}dS=-\int_{S/M}p*\vec{n}dS-\int_{M}p*\vec{n}dS+\int_{M}p*\vec{n}dS=-\int_{S}p*\vec{n}dS+\int_{M}p*\vec{n}dS=-\int_{S}\nabla{p}*dV+\int_{M}p*\vec{n}dS=\rho{g}V\vec{k}+\int_{M}p*\vec {n}dS[/tex]
Note that the first term is the standard buoyancy term, whereas the second one goes to zero when M goes to zero.
Thus, we have agreement with Archimedes' principle in the limit

This formula is nice in order to see how Archimedes' principle appears in the limit; in order to to evaluate it, it is, however, trivial to show that the net pressure force acting upon O must equal the weight of the fluid column resting upon O.

Q_Goest

hi slider,
What Doc Al and arildo are pointing out, which I'd fully agree with, is that the force on an object in water or any fluid (in a gravitational field) is the sum of all the forces on every portion of the object. In other words, one needs to integrate the force (which is a vector quanity) over the object's entire area which is exposed to the fluid. One can derive Archimedes' principal from that integration. So if there's a long, straight verticle pipe in 30,000 feet of water, and the lowest end of the pipe is somehow isolated from water pressure, then I'd have to assume the forces on the outside, which are only in the horizontal direction, do not provide any verticle force whatsoever.

I have to wonder though, what does this pipe actually look like? Making a simplistic assumption as to how this pipe looks is probably where reality takes over and errors in the assumptions will come into play. You say:
I'm having a lot of problems visualizing what exactly you have here and what the issue is. If you could post a drawing on this, that might help clear up any miss-communication issues.

Also, you may want to try your question on Engineering Tips forums. There's a lot of engineers there that have considerable experience with what you're dealing with, and may already understand the details of how these drilling rigs and pipes are designed. I'd suggest one of these two forums:
http://www.eng-tips.com/threadminder.cfm?pid=378
http://www.eng-tips.com/threadminder.cfm?pid=470

cesiumfrog

It's not so easy to "isolate one end from the pressure". From every day experience (doing so is the principle of a suction cap) such things are very rare: You can't just press a block of wood against the bottom of a swimming pool, and expect it to stick. Perhaps an equivalent way of thinking (rather than explicitly integrating pressures) is to take the Archimedian bouyancy force and subtract whatever suction force (plus also static friction force) you think mud can apply when you pull on a drill that's jammed into it. ('Course, another total in both cases if you're still pushing the drill in.. as Q says, hard to understand your specifics yet.)

slider201

Oops. You are right. I have my own camps backwards.

I know there shouldn't be. In academia, you are probably right. In industry, people believe what they believe.

My issue was what evidence I could show to prove that buoyancy is more complicated than simply the weight of fluid displaced. I think you all have given that to me.

You are correct. It's not easy but it can and will happen by accident.

Bramle

Slider.....

I am in the same field... oilwell casing, and the desk is strewn with the stuff.

Archimedes is right if the immersed body is wetted on all of the immersed surfaces.

The integration theory is always right.... always, always!!!!.... but man it can take time.

You are stabbing the end of your casing into a seal-bore receptacle. If, for example, your casing were dry inside, and had no connection upsets (flush), there would be NO fluid upthrust.... NONE!!!! The outside surface would be wetted, but the verticality would forbid the possibility of upthrust.

Bramle

If your pipe has no upsets, and is truly flush, and vertical, you could stab that pipe into the seal bore, evacuate the inside and fill the outside with liquid mercury and still see no change to the force at surface necessary to support it. The fluid cannot get an upward lift on that pipe no matter what the external fluid density is.

I can prove that one to you, mathematically.... it's a gift.

If there are external upsets, or couplings, there is a bit of lift, and the lift will increase as the external density increases, but it's not much in the scheme of things, and it's not there if the pipe is flush.

No pipe will be truly vertical though... they all wobble a bit.... but it won't change the result too much if it's close to vertical, and stabbed into a seal bore, and dry inside the bore.

Do not underestimate that Archimedian stuff though.... it's absolutely correct when applied to Archimedian systems.... it's just that when he got in the bath, he did not stick his feet into the plug-hole and seal off his ankles.

Bramle

It must be remembered that when he got into the bath that day, Archimedes did not stick his feet down the plug-hole and put a seal around his ankles.

If he had, he would NOT have experienced an upthrust equal to the weight of fluid displaced. There would have been upthrusts, yes, but the sum total of them would not have been Archimedian.

It is not intuitive but, for the case of your FLUSH vertical casing, you could fill volume external to your casing with liquid mercury, and as long as the pressure on the end of your pipe is isolated from the amplified mercury pressure, you will see NO change to the pipe tension at surface.

If your casing has upset connections, or uses couplings, there will be a change, but it is not a powerful effect as the mercury (or whatever is in there) can't get much lift on them.