`` NEWTON: Potential Energy Stored in Buoyant System

 Potential Energy Stored in Buoyant System ``` Name: Christian Status: other Grade: other Location: OK Country: USA Date: Fall 2013 ``` Question: My question has to do with the energy of buoyancy. If I am submerged 60 feet underwater and fill a balloon with 1 cubic meter of air, then release it to the surface, how much energy does the buoyant force produce in joules? As it rises, and the air expands and displaces more water, will it produce more energy from 30 feet to the surface than it did from 60 feet to 30 feet? Replies: Hi Christian, Buoyancy is not about energy, but about a force balance. If the an object is submerged in a fluid such as water, and it weighs more than the volume of fluid it displaces, then it will sink. If it weighs less, it will rise. If it is the same, the object is said to be neutrally buoyant. As an example, submarines control their ballast tanks to take on weight in the form of water or release water weight in order to rise, sink or cruise underwater. But if something like an air-filled balloon is released underwater, the balloon will do work on the surrounding water as it rises. Work is defined as force applied through a distance. So the balloon being released to rise to the surface does represent a release of energy, and it can be calculated if you know the drag force on the balloon as a function of depth underwater. If you know this, you can integrate this force with respect to depth and calculate a net work done. Now what you ask on top of that is a really good question. You are correct that at 60 foot depth the balloon will be smaller due to the surrounding water pressure. A smaller balloon will move through the water with less drag. On the other hand, if you are only at 30 foot depth that is less distance to travel, yet the balloon will probably create greater drag just due to expanded size. So which would expend more work? You would have to do the integration to work that out. You would have to know the diameter of the balloon as a function of the net pressure on the balloon, and you could probably assume that the drag equation works here with a drag coefficient equal to that of a sphere as an approximation. Regards, John C Strong Make sure your units are consistent. At 60ft the balloon is buoyed up by the mass of the fluid (being displaced) --- in your case the mass of the volume of 60 feet of water (Be careful to keep the various units consistent) As the 1 cubic air rises to the surface, it expands, so its volume increases. This causes the bubble of air to increase in its volume, which in turn, causes the bubble to expand. This, in turn increases the mass of the displaced water. I will leave you to work out the details, but the key issue is the mass of water being displaced at any given position. You can also attack the problem as an ?Ideal Gas? problem: PV=nRT , n,R, and T are fixed so P is proportional to 1/V. I will leave the number crunching to you, but be careful the units are consistent. Vince Calder Click here to return to the Physics Archives

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