AboutKevin Johnson Expertise I will try to answer any question in physics, but I do not provide homework solutions or project ideas. I have some teaching experience at university level but I am also skilled at explaining things in simple terms.
My specialty areas are granular materials, computational physics, particle physics, quantum physics, econophysics and general physics.
Questions in GERMAN are welcome, too.
Experience I have graduated in physics, specialized in theoretical particle physics and quantum field theory, worked in the area of econophysics and am currently working on my PhD in granular materials and computational physics. I have some tutoring and teaching experience at University level.
Education/Credentials German Diploma in Physics (equivalent to M.Sc)
I have not opportunity to study phisics in my country so I ask you a very dumb question... but I really want to know the answer.. Can you help ?
A Diver need air in botton of the sea..
There is a balloon filled with X liters of air on a boat. This balloon is connected by a hose to another ballon of same size, but the other balloon is empty and in botton of the ocean at a depht of Y meters.
The diver friend wants to transfer the air of the ballon on the top to fill the ballon on the botton.
For that he must choose a steel plate and place it over the balloon against the boat floor, if the weight is right he can transfer all air through the hose and the diver can survive...
How can he calculate how much should be the weight of the steel plate to empty the ballon on the top ?
I really want to find this out.. Can you help ?
Tks for reading.
ANSWER: Hello Joao,
Since the water pressure increases with increasing depth, the pressure difference between the surface and the diver's depth must be overcome for air to be transferred. The water pressure increses with aprrox. 15 psi (pounds per square inch) every 10 yards of depth.
This pressure must be created by your balloon-and-steel-plate pump.
Assuming that the balloon is not just "flattened", but is somehow kept at a constant base area A, the steel plate must have a weight W to create a pressure P:
W = P * A
(convert the area to square inches, use psi as pressure unit and the weight will come out in pounds.)
In case you need to convert units, you might find google's built-in calculator function helpful.
Just enter for example "1 square meter in square inches" at www.google.com and it will convert the units for you ;-)
I hope I could help you a bit.
Peace,
Kevin
---------- FOLLOW-UP ----------
QUESTION: Dear Master Kevin,
Thanks a lot for your great explanation (the google calculator was a must)!!!
I think I am dumb once I still have a doubt... Pls feel free not to answer if you think I am too stupid.
I got confused by the area concept... A baloon would have a volume in liters of air and once it is a sphere is also not so easy to calculate the m2 of the disc with the weight that should be put on top of it..
Also it raises me another question... if the area weight is enough to move 1 liter of air, will it move 100 or 1000 if the suface of the pressure area is same but the ballon shape is different... Imagine a long cilinder baloon standing up against the floor with a much lesser surface area in top but more volume once is a long cylinder ? I mean would a cilinder baloon means the pressure aplied would be lesser ?
To finish my stupidity can you give practical example ?
Lets say the ballon (you can choose the best format) in botton is at 100m depth, with 1000 liters air volume. How much weight should be the steel plate considering the format of the ballon choosen ..
I think I asked too much sorry !
Pls do not answer if you think I am not worthy of it ... and thank you a lot if you could at least stand reading untill here !!!
Best for you...Joao
Answer Hello again Joao,
Sorry for the late answer, I was away for the weekend.
For the balloon under water you do not need to take any form factors into account, since the force of the water pressures is equal from all directions on the balloon and thus it assumes a more of less spherical shape.
But the balloon on the boat has the weight force of the steel plate acting on it from above and the resistance from the floor acting on it from below. From the sides there is only the force of the tension of the balloon's shell acting on it. It depends on this tension how much the balloon deforms when it is squeezed. To make the calculations easier you could idealize the problem by assuming the balloon were a cylinder with rigid walls. The base area of this cylinder should be chosen as approximately the maximum cross-section of the squeezed balloon.
You might notice that the larger the base area of the (idealized) balloon (or cylinder), the higher the force that is needed to overcome a given pressure difference. So confining the balloon in a cylindrical casing with a small base area is probably a good idea.
I am sorry I cannot give you any practical examples, since engineering is way outside my area of expertise, unfortunately.
