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 Jan 3rd 2015, 09:55 AM #21 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,333 If the 1000 gallons of air is at 1 atmosphere, then adding another 2 gallons initially at 1 atmosphere won't take much pressure at all - about 1.002 atmospheres. Water (and many other fluids) are usually considered to be incompressible - though in reality they can be compressed, the pressure required is quite high. That's why hydraulic systems such as the brake system on your car use fluid as the medium for conveying pressure from the brake master cylinder the brake pistons at each wheel. Gasses on the other hand generally obey the Ideal Gas Law: PV=nRT where P = pressure, V = volume, n = amount of gas in the vessel (typically provided in moles, but could be some other unit), T = temp in Kelvins, and R is a the Ideal Gas Constant. If we assume that the temperature and volume remain constant, then the relationship between pressure and amount of gas is (P1/P2) = (n1/n2). Hence to add 10% more gas to a vessel of fixed size requires 10% more pressure.
 Jan 5th 2015, 01:23 AM #22 Junior Member   Join Date: May 2014 Posts: 20 Hi Chip, Another question. (they really do have something in common).... Assume that you have (1) 4 foot sphere that weighs 2000 pounds and has a volume of 33.52 cu/ft. (If it were neutrally buoyant, the sphere should weigh 2092 lbs.) So if your sphere only weighs 2000 lbs. it should be slightly buoyant and stay on the surface, although most of it would be below the water line. The first thing I would like to know is if that depiction is correct. If it is, what I would like to know is if you pulled that sphere down to a depth of 80 feet, is it still buoyant at that depth, and if it is, how long would it take to rise 80 feet to the surface? The formula would be great. As always, thank you very much for all of your help. Kenji.
 Jan 5th 2015, 06:44 AM #23 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,333 Yes, if you pull the sphere downward under the water it will try to rise to the surface. The buoyancy force on the sphere is equal to the weight of water displaced, so from your data that is 2092 pound-force. With force of gravity acting downward at 2000 pound-force, the net force acting on the sphere when it is submerged is 92 lbf upward. The calculation of how long it takes to rise to the surface is a bit complicated, as we must take into account the drag forces of the water acting to slow its rate of rise. A common model that is often used is that the drag force is equal to: F_drag = (1/2) rho C_d A v^2 where rho = density of the water, C_d is the coefficient of drag (which is dependent on the geometry, but for a sphere is assumed to be about 0.4), and A is the cross-section area of the sphere. We can calculate the terminal velocity of the sphere by setting the drag force equal to the net buoyancy force: (1/2) rho C_d A v_terminal^2 = 92 lbf Rearrange: v= sqrt(2 x 92 lbf/(rho C_d A)) Do the math and you find v_terminal = 4.3 ft/s. Since the sphere starts with v = 0 and takes a bit of time to get up to its terminal speed the time to rise to the surface is a bit greater than 80 ft/(4.3 ft/s) = 18.6 s. A more detailed calculation shows the expected time to reach the surface is about 20.7 seconds. I won't go into that detailed calculation here - a reasonably good approximation is the that it takes 8 seconds to reach terminal velocity, during which time it rises about 25 ft, so the remaining 55 ft takes another 55/4.3 = 12.8s, for a total of 20.8 s. A word of warning: the value of C_d is a bit of a SWAG, and may actually be as low as 0.1 to as great as 0.5, depending on how smooth the sphere is. So the calculation of rise time may vary from as low as 13 seconds to as great as 23 seconds. Last edited by ChipB; Jan 5th 2015 at 10:48 AM.
 Jan 5th 2015, 05:39 PM #24 Junior Member   Join Date: Oct 2014 Posts: 2 Compression vs Decompression was not really a factor I was considering. If a journey was 10 days of Compression at the start then near FTL speed for a year of travel then 30 days of Decompression at the other end, Vs a slower means of traveling the same distance. I guess the Velocity vs Gravity idea of using a forced movement threw a manipulated ocean to off set or reverse a high external velocity/gravity environment. (Making reversing Earths Gravity in a High Velocity Submarine an interesting challenge.)
 Apr 8th 2017, 07:55 PM #25 Junior Member   Join Date: May 2014 Posts: 20 Hi Chip, I have been away for awhile, but am back with a new quandary if you have a moment.... Premise: If you take a straw and put it in water and then hold your thumb over the top of the straw and lift the straw out of the water all the water will stay in the straw until you remove your thumb. Consider something on a much larger scale….Let’s say your straw now has a diameter of 6 feet and has a height of 100’. The top of the “straw” has a gate valve and there is a gate valve at the 90 foot level as well. The bottom of the “straw” is open and is sitting in a swimming pool a foot or two below the surface. The “straw” is filled with water. The top valve is closed and the 90 foot valve is open. My first question is does size matter? Will that 100 foot column of water stay in the “straw” until that top valve is opened just like the smaller straw held the water until you removed your thumb? If the answer is yes, can you please explain why. If yes to part one, part 2: Now close the valve at the 90 foot level. In the chamber between 90 and 100 feet there is a deflated balloon at the 95 foot level with a stem connected to the outside of the chamber. Now open the top valve and then blow up the balloon which will be a 4 foot sphere (V=33.51 cu/ft). Naturally 33.51 cu/ft of water will spill out of the top of the “straw”. Once this is done, close the top valve and then open the 90 foot valve. The equilibrium in the “straw” should be the same as it was without the balloon. My question is what happens inside the “straw” when you let out all of the air in the balloon? Does all of the air leave the balloon? Is some kind of vacuum created in the “straw”? Does 33.51 cu/ft of water get sucked into the “straw” from the pool 100 feet below to replace the volume of air from the balloon?
Apr 9th 2017, 05:26 AM   #26
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 Originally Posted by tonk Consider something on a much larger scale….Let’s say your straw now has a diameter of 6 feet and has a height of 100’. The top of the “straw” has a gate valve and there is a gate valve at the 90 foot level as well. The bottom of the “straw” is open and is sitting in a swimming pool a foot or two below the surface. The “straw” is filled with water. The top valve is closed and the 90 foot valve is open. My first question is does size matter? Will that 100 foot column of water stay in the “straw” until that top valve is opened just like the smaller straw held the water until you removed your thumb? If the answer is yes, can you please explain why.
The answer is no, because no straw can work the way you suggest if it's longer than about 32 feet - the reason is that what keeps the water in the straw is air pressure acting in the surface of the swimming pool, which is about equivalent to the pressure of 32 feet of water. So, for the remainder of this post I'm going to change your numbers so that tghe concept will work. Let's assume that the straw is 30 feet long, with a gate valve at 20 feet, and the balloon will be at the 25 foot level.

In step 1 you have a straw with the 30-foot gate closed, so you have a column of water 30 feet tall.

In step 2 you close the valve at the 20-foot level, open the top valve, and inflate the balloon at the 25-foot level so that it displaces 2 column feet of water. You now have water from 0 to 20 feet, then a closed valve, then water from 20 to 25 feet, then 2 feet of balloon, then water from 27-30 feet.

 Originally Posted by tonk My question is what happens inside the “straw” when you let out all of the air in the balloon? Does all of the air leave the balloon? Is some kind of vacuum created in the “straw”? Does 33.51 cu/ft of water get sucked into the “straw” from the pool 100 feet below to replace the volume of air from the balloon?
When you let air out of the balloon the weight of the water above the balloon forces the air out, so you end up with a column of water from 0 to 20 feet, then a closed valve, then water from 20 to 25 feet, then a thin layer of balloon, then water from 25 to 28 feet, and finally air from 28 feet to the top (which is still open).

It could get interesting if you consider what happens if you close the top valve before letting air out of the balloon. Under ideal conditions, if the balloon has insignificant tension in its skin, then when you open the neck of the balloon to let air out nothing happens - it remains in the condition of step 2 above. It might be clearer if we replace the balloon with an air bag (i.e. a container that does not rely on the "stretchiness" of its skin).

 Apr 9th 2017, 10:34 AM #27 Junior Member   Join Date: May 2014 Posts: 20 Hi Chip, Thank you for the reply. A couple of things. 1. On the new dimensions for the straw (D=6' H=30'). If you had a pressure gauge and swam inside the column of water, I assume the pressure at the entrance/base of the column would be 32PSI and if you swam to the top it would be close to zero? 2. If you had another swimming pool at 30 feet in elevation with the same 30 foot "straw" apparatus coming out of it, what would the pressures be at the base and the top of the column? Is there a formula I could use to play around with different heights of the swimming pool and "straw"? 3. As to your point "It could get interesting if you consider what happens if you close the top valve before letting air out of the balloon. Under ideal conditions, if the balloon has insignificant tension in its skin, then when you open the neck of the balloon to let air out nothing happens - it remains in the condition of step 2 above. It might be clearer if we replace the balloon with an air bag (i.e. a container that does not rely on the "stretchiness" of its skin)." I think you missed the part about closing the top valve prior to deflating the balloon in my post. If I understand correctly, if you open the valve to deflate the balloon (which has insignificant tension) the balloon will remain the same size with the air inside. If you wished to fully deflate the 4' balloon/sphere, how much force would it take in terms of PSI? (I'm guessing 32PSI) if that is correct, 32 PSI of what, the surface area of the sphere? The size of the opening to the sphere? And if you were able to generate that force I am guessing you would than be sucking two column feet of water into the straw rather than creating some kind of a vacuum.
Apr 10th 2017, 04:40 AM   #28
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 Originally Posted by tonk Hi Chip, Thank you for the reply. A couple of things. 1. On the new dimensions for the straw (D=6' H=30'). If you had a pressure gauge and swam inside the column of water, I assume the pressure at the entrance/base of the column would be 32PSI and if you swam to the top it would be close to zero?
Correct.

 Originally Posted by tonk 2. If you had another swimming pool at 30 feet in elevation with the same 30 foot "straw" apparatus coming out of it, what would the pressures be at the base and the top of the column? Is there a formula I could use to play around with different heights of the swimming pool and "straw"?
Same as above - simply raising the entire apparatus 30 feet makes no significant difference.

 Originally Posted by tonk 3. ... I think you missed the part about closing the top valve prior to deflating the balloon in my post.
Yes,I did miss that .

 Originally Posted by tonk If I understand correctly, if you open the valve to deflate the balloon (which has insignificant tension) the balloon will remain the same size with the air inside. If you wished to fully deflate the 4' balloon/sphere, how much force would it take in terms of PSI? (I'm guessing 32PSI)
Correct.

 Originally Posted by tonk if that is correct, 32 PSI of what, the surface area of the sphere? The size of the opening to the sphere?
32 PSI pressing downward on the balloon.

 Originally Posted by tonk And if you were able to generate that force I am guessing you would than be sucking two column feet of water into the straw rather than creating some kind of a vacuum.
No, the water level would drop by 2 feet, leaving a vacuum at the top of the column.

I should point out that a 6-foot diameter "straw" wont work, at least not with water. A straw larger than a 1/2 inch or so doesn't work, because you need surface tension to keep the surface of the water at the bottom together. If the diameter of the opening is larger than that and air can work its way in, destroying the vacuum.

 May 1st 2017, 02:03 AM #29 Junior Member   Join Date: May 2014 Posts: 20 Hi Chip, If you have a 100 foot column of water the pressure on the valve at the bottom holding it in would be 43.3 psi. If you had a valve at let's say the 50 foot level and you closed that valve, would the psi on the bottom valve still be 43.3 psi until you relieved pressure in the chamber or would it instantly be cut in half as soon as you closed the 50 foot valve?
May 1st 2017, 12:31 PM   #30
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 Originally Posted by tonk Hi Chip, If you have a 100 foot column of water the pressure on the valve at the bottom holding it in would be 43.3 psi. If you had a valve at let's say the 50 foot level and you closed that valve, would the psi on the bottom valve still be 43.3 psi until you relieved pressure in the chamber or would it instantly be cut in half as soon as you closed the 50 foot valve?
With the valve closed the pressure of the water inside the vessel wouldn't change - it would still be 43 psi at the bottom and half that at the 50-foot level.

The obvious next question is: what happens if you remove the water above the 50 foot level, while keeping the valve closed? Again, there should be no change of pressure in the enclosed bottom 50 feet, but only if the valve is very stiff and doesn't flex. Water is almost incompressible, which very small changes in volume results in large changes in pressure. So if the material of the valve flexes or stretches due to the the stresses created by the pressure inside the vessel the result will be reduced pressure in the vessel.

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